Quantum Field Theory: Lecture Notes

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Quantum ﬁeld theory: Lecture Notes

Rodolfo Alexander Diaz Sanchez Universidad Nacional de Colombia Departamento de F´ısica Bogot´a, Colombia

August 23, 2015 Contents

1 Relativistic quantum mechanics 7 1.1 Surveyonquantummechanics ...... 7 1.1.1 Vector subspaces generated by eigenvalues ...... 8 1.2 Symmetriesinquantummechanics ...... 9 1.3 Irreducible inequivalent representations of groups ...... 12 1.4 ConnectedLiegroups ...... 14 1.5 Lorentztransformations ...... 17 1.6 The inhomogeneous Lorentz Group (or Poincarégroup) ...... 19 1.6.1 Four-vectorsandtensors...... 20 1.7 SomesubgroupsofthePoincarégroup ...... 22 1.7.1 Proper orthochronous Lorentz group ...... 23 1.7.2 Discrete transformations in the Lorentz group ...... 24 1.7.3 Infinitesimal transformations within the proper orthochronus Lorentz group ...... 25 1.8 QuantumLorentzTransformations ...... 25 1.8.1 Four-vector and tensor operators ...... 26 1.8.2 Infinitesimal quantum Lorentz transformations ...... 27 1.8.3 Lorentz transformations of the generators ...... 28 1.8.4 Lie algebra of the Poincarégenerators ...... 29 1.8.5 Physical interpretation of Poincare’s generators ...... 30 1.9 One-particlestates ...... 32 1.9.1 One-particle states under pure translations ...... 32 1.9.2 One-particle states under homogeneous Lorentz transformations ...... 32 1.9.3 Physicallittlegroups...... 35 1.9.4 Normalization of one-particle states ...... 36 1.10 One-particle states with non-null mass ...... 38 1.10.1 Wigner rotation and standard boost ...... 40 1.11 One-particle states with null mass ...... 44 1.11.1 Determination of the little group ...... 44 1.11.2 Lie algebra of the little group ISO (2) ...... 48 1.11.3 Massless states in terms of eigenvalues of the generators of ISO (2)...... 50 1.11.4 Lorentz transformations of massless states ...... 51 1.12 Space inversion and time-reversal ...... 53 1.13 Parity and time-reversal for one-particle states with M > 0...... 56 1.13.1 Parity for M > 0...... 56 1.13.2 Time reversal for M > 0...... 58 1.13.3 Parity for null mass particles ...... 59 1.13.4 Time-reversal for null mass particles ...... 62 1.14 Action of T 2 andKramer’sdegeneracy...... 63

2 CONTENTS 3

2 Scattering theory 65 2.1 Construction of “in” and “out” states ...... 65 2.2 The S matrix ...... 70 − 2.3 Symmetries of the S matrix ...... 75 − 2.3.1 Lorentzinvariance ...... 75 2.3.2 Internalsymmetries ...... 81 2.3.3 Parity ...... 83 2.3.4 Time-reversal...... 86 2.3.5 PTsymmetry...... 89 2.3.6 Charge-conjugationC,CPandCPT ...... 90 2.4 Ratesandcross-sections ...... 91 2.4.1 One-particle initial states ...... 94 2.4.2 Two-particles initial states ...... 95 2.4.3 Multi-particle initial states ...... 95 2.4.4 Lorentz transformations of rates and cross-sections ...... 95 2.5 Physical interpretation of the Dirac’s phase space factor δ4 (p p ) dβ ...... 98 β − α 2.5.1 The case of Nβ =2...... 98 2.5.2 The case with Nβ =3andDalitzplots...... 101 2.6 Perturbationtheory ...... 102 2.6.1 Distorted-wave Born approximation ...... 107 2.7 Implicationsofunitarity ...... 108 2.7.1 Generalized optical theorem and CPT invariance ...... 111 2.7.2 Unitarity condition and Boltzmann H-theorem ...... 112

3 The cluster decomposition principle 115 3.1 Physicalstates ...... 115 3.1.1 Interchange of identical particles ...... 116 3.1.2 Interchange of non-identical particles ...... 117 3.1.3 Normalization of multi-particle states ...... 117 3.2 Creation and annihilation operators ...... 118 3.2.1 Commutation and anti-commutation relations of a (q) and a† (q) ...... 119 3.3 Arbitrary operators in terms of creation and annihilationoperators ...... 120 3.4 Transformation properties of the creation and annihilation operators ...... 121 3.5 Cluster decomposition principle and connected amplitudes...... 122 3.5.1 Someexamplesofpartitions...... 124 3.6 Structureoftheinteraction ...... 127 3.6.1 Asimpleexample ...... 129 3.6.2 Connected and disconnected parts of the interaction ...... 131 3.6.3 Some examples of the diagrammatic properties ...... 133 3.6.4 Implications of the theorem ...... 136

4 Relativistic quantum field theory 137 4.1 Freefields...... 137 4.2 Lorentz transformations for massive fields ...... 139 4.2.1 Translations...... 142 4.2.2 Boosts...... 144 4.2.3 Rotations ...... 144 4.3 Implementation of the cluster decomposition principle ...... 145 4.4 Lorentz invariance of the S matrix...... 146 − 4 CONTENTS

4.5 Internal symmetries and antiparticles ...... 147 4.6 Lorentz irreducible ﬁelds and Klein-Gordon equation ...... 149

5 Causal scalar fields for massive particles 151 5.1 Scalar fields without internal symmetries ...... 151 5.2 Scalar fields with internal symmetries ...... 155 5.3 Scalar fields and discrete symmetries ...... 158

6 Causal vector fields for massive particles 162 6.1 Vector fields without internal symmetries ...... 162 6.2 Spinzerovectorfields ...... 165 6.3 Spinonevectorfields...... 166 6.4 Spin one vector fields with internal symmetries ...... 174 6.4.1 Field equations for spin one particles ...... 175 6.5 Inversion symmetries for spin-one fields ...... 176

7 Causal Dirac fields for massive particles 178 7.1 Spinor representations of the Lorentz group ...... 178 7.2 Some additional properties of the Dirac matrices ...... 182 7.3 The chiral representation for the Dirac matrices ...... 185 7.4 CausalDiracfields ...... 190 7.5 Dirac coefficients and parity conservation ...... 193 7.6 Charge-conjugation properties of Dirac fields ...... 201 7.7 Time-reversal properties of Dirac fields ...... 204 7.8 Majoranafermionsandfields ...... 207 7.9 Scalar interaction densities from Dirac fields ...... 207 7.10TheCPTtheorem ...... 209

8 Massless particle ﬁelds 211

9 The Feynman rules 223 9.1 Generalframework ...... 223 9.2 Rules for the calculation of the S matrix ...... 226 − 9.3 Diagrammatic rules for the S matrix ...... 227 − 9.4 Calculation of the S matrixfromthefactorsanddiagrams ...... 229 − 9.5 Afermion-bosontheory ...... 234 9.5.1 Fermion-boson scattering ...... 234 9.5.2 Fermion fermion scattering ...... 238 9.5.3 Boson-bosonscattering ...... 238 9.6 Aboson-bosontheory ...... 238 9.7 Calculation of the propagator ...... 240 9.7.1 Other deﬁnitions of the propagator ...... 248 9.8 Feynman rules as integrations over momenta ...... 249 9.9 Examples of application for the Feynman rules with integration over four-momenta variables . . . 252 9.9.1 Fermion-boson scattering ...... 252 9.9.2 Fermion-fermion scattering ...... 255 9.9.3 Boson-bosonscattering ...... 255 9.10 Examples of Feynman rules as integrations over momenta ...... 258 9.10.1 Fermion-boson scattering ...... 259 9.10.2 Fermion-fermion and Boson-boson scattering ...... 260 CONTENTS 5

9.11 Topological structure of the lines ...... 260 9.12 Off-shell and on-shell four-momenta ...... 262 9.12.1 The r thderivativetheorem ...... 264 − 10 Canonical quantization 268 10.1 Canonicalvariables...... 268 10.1.1 Canonical variables for scalar fields ...... 269 10.1.2 Canonical variables for vector fields ...... 271 10.1.3 Canonical variables for Dirac fields ...... 274 10.2 Functional derivatives for canonical variables ...... 275 10.3 FreeHamiltonians ...... 277 10.3.1 Free Hamiltonian for scalar fields ...... 277 10.4 Interacting Hamiltonians ...... 280 10.5 TheLagrangianformalism...... 281 10.6 From Lagrangian to Hamiltonian formalism ...... 283 10.6.1 Setting the Hamiltonian for the use of perturbation theory ...... 286 10.7 Gauges of the Lagrangian formalism ...... 289 10.8 Globalsymmetries ...... 290 10.9 Conserved quantities in quantum field theories ...... 293 10.9.1 Space-time translation symmetry ...... 294 10.9.2 Conserved currents and Lagrangian densities for space-time symmetries ...... 295 10.9.3 Additional symmetry principles ...... 296 10.9.4 Conserved current for a two scalars Lagrangian ...... 298 10.10Lorentzinvariance ...... 300 10.10.1 Currents and time-independent operators ...... 300 10.10.2 Generators and Lie algebra between the homogeneous and inhomogeneous generators . . . 303 10.10.3 Invariance of the S matrix ...... 305 − 10.10.4 Lie algebra within the homogeneous generators ...... 305 10.11The transition to the interaction picture ...... 306 10.11.1 Scalar field with derivative coupling ...... 306 10.11.2 Spin one massive vector field ...... 307 10.11.3 Dirac Fields of spin 1/2 ...... 314 10.12Constraints and Dirac Brackets ...... 316

11 Quantum electrodynamics 323 11.1 Gaugeinvariance ...... 323 11.1.1 Currents and their coupling with Aµ ...... 325 11.1.2 Action for the photons (radiation) ...... 326 11.1.3 General overview of gauge invariance ...... 327 11.2 Constraints and gauge conditions ...... 328 11.3 Quantization in Coulomb Gauge ...... 332 11.3.1 Canonical quantization of the constrained variables...... 332 11.3.2 Quantization with the solenoidal part of Π~ ...... 335 11.3.3 Constructing the Hamiltonian ...... 337 11.4 Formulation of QED in the interaction picture ...... 339 11.5 Thepropagatorofthephoton ...... 342 11.6 Feynman rules in spinor electrodynamics ...... 343 11.6.1 DrawingtheFeynmandiagrams ...... 344 11.6.2 Factors associated with vertices ...... 344 6 CONTENTS

11.6.3 Factors associated with external lines ...... 344 11.6.4 Factors associated with internal lines ...... 345 11.6.5 Construction of the S matrixprocess ...... 346 − 11.7 General features of the Feynman rules for spinor QED ...... 346 11.7.1 Photon polarization ...... 347 11.7.2 Electron and positron polarization ...... 349 11.8 Example of application: Feynman diagrams for electron-photon (Compton) scattering ...... 350 11.9 Calculation of the cross-section for Compton scattering...... 352 11.9.1 Feynman amplitude for Compton scattering ...... 353 11.9.2 Feynman amplitude for the case of linear polarization ...... 356 11.9.3 Diﬀerential cross-section for Compton scattering ...... 358 11.9.4 Diﬀerential cross-section in the laboratory frame ...... 360 11.10Traces of Dirac gamma matrices ...... 361 11.11Some properties of “slash” momenta ...... 365

12 Path integral approach for bosons in quantum field theory 366 12.1 The general path-integral formula for bosonic operators ...... 367 12.1.1 Probability amplitude for infinitesimal time-intervals...... 369 12.1.2 Probability amplitude for finite time intervals ...... 369 12.1.3 Calculation of matrix elements of operators through the path-integral formalism ...... 371 12.2 Path formalism for the S matrix...... 373 − Chapter 1

Relativistic quantum mechanics

1.1 Survey on quantum mechanics

A Hilbert space , is a complete vector space with inner product. Given two vectors ψ , and φ in such a space, E | i | i there is a complex number φ ψ that satisfies the following axioms h | i φ ψ = ψ φ ∗ h | i h | i φ αψ + βψ = α φ ψ + β φ ψ h | 1 2i h | 1i h | 2i αφ + βφ ψ = α∗ φ ψ + β∗ φ ψ h 1 2 | i h 1 | i h 2 | i ψ ψ 0 ; and ψ ψ = 0 ψ = 0 (1.1) h | i ≥ h | i ⇔ | i where we define the norm of a vector ψ as | i ψ 2 ψ ψ ; ψ 2 = 0 ψ = 0 k| ik ≡ h | i k| ik ⇔ | i which is positive definite i.e. it is positive for any non-zero vector and zero for the null vector. A vector is normalized if its norm is equal to unity. Physical states in quantum mechanics are described by normalized vectors (or kets) on the Hilbert space . Physical observables in quantum mechanics are eigenvalues of hermitian E operators with a complete spectrum (that is, the eigenvectors of each one of these operators form a basis on the Hilbert space). We recall that the adjoint A† of a linear operator A, is another linear operator on a Hilbert space that satisfies the condition Aφ ψ = φ A†ψ ; ψ , φ h | i h ∀ | i | i∈E E further, it also happens that two linearly dependent (normalized) vectors ψ describe the same physical state. | i This fact induces the following definition

Definition 1.1 Given a normalized state ψ , A ray induced by a ket ψ is the set of all normalized vectors | i∈E | i that are linearly dependent with ψ | i iθ ψ e ψ : θ [0, 2π) R| i ≡ | i ∈ n o Two vectors belonging to the same ray describe the same physical states. On the other hand, two vectors belonging to different rays are linearly independent so that they represent different physical states. Therefore, if we think on each ray as a single object, we could say that a given physical state is represented by a single ray and that a given ray represents a unique physical state. This one-to-one relation between rays and physical states justify the introduction of such a concept. For a given observable A, a given ray posseses a unique eigenvalue α if the vectors of are eigenvectors of R R A A ψ = α ψ ψ | i | i ∀ | i ∈ R 7 8 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS the eigenvalues of an hermitian operator are real (this fact is necessary to interpret such eigenvalues as physical observables), and eigenvectors associated with different eigenvalues are orthogonal. When the spectrum of a given observable is degenerate, a given eigenvalue αn could be associated with several linearly independent eigenvectors as follows

A ψm = α ψm ; m = 1,...,g | n i n | n i n where g is the degree of degeneracy. Therefore, there will be g rays m associated with a given eigenvalue α . n n Rn n If a system is in a state described by the ray , and we measure an observable A, the probability of ﬁnding R the eigenvalue αk of A is given by

gk P (α )= ψm ψ 2 ; ψ and ψm m k |h k | i| | i ∈ R | k i ∈ Rk m=1 X since observables (hermitian complete linear operators on ), have eigenvectors that form a basis of the sum of E E these probabilities is equal to the unity

gn gn gn P (α ) = ψm ψ 2 = ψ ψm ψm ψ = ψ ψm ψm ψ = ψ I ψ k |h k | i| h | k i h k | i h | | k i h k | | i h | | i m=1 m=1 " m=1 # Xk Xk X Xk X Xk X P (αk) = 1 Xk thus, the feature for the observables of having a complete spectrum, is essential to keep the conservation of probability.

1.1.1 Vector subspaces generated by eigenvalues Let , ,..., be a set of subspaces of a given vector space . We say that is the direct sum of such a {E1 E2 Eq} E E set of subspaces and denote it as = . . . E E1 ⊕E2 ⊕ ⊕Eq if any given arbitrary vector x is expresible in a unique way in the form ∈E x = x + x + . . . + x such that x 1 2 q k ∈Ek in words, given an arbitrary x , it can be expressed by a sum of q vectors x , where each x belongs to a ∈E − { k} k subspace in the set. In addition, there is one and only one vector x belonging to , that can be part of this Ek k Ek decomposition. We say that x is the projection of x into the subspace . k Ek Let A be a linear operator of a vector space into itself. We say that a vector subspace is invariant under E Ek the action of A if for any x , we have that Ax = x . In words, a subspace is invariant under A if by ∈Ek ′ ∈Ek Ek restricting the domain of A to , the resultant range is also contained in . Ek Ek Let A be an observable (hermitian complete linear operator on ). For simplicity, we shall assume that its E spectrum is discrete. Its eigenvalues and eigenvectors are given by

A ψm = a ψm ; m = 1, ..., g | n i n | n i n with gn the degeneracy of the eigenvalue an. By taking all linearly independent vectors of the form

ψ1 , ψ2 , ψgn n n | n i and forming all possible linear combinations (including the null linear combination) we obtain the set of all eigenvectors of A, with eigenvalue a (plus the null vector). This set forms a subspace of , called the vector n E 1.2. SYMMETRIES IN QUANTUM MECHANICS 9 subspace induced by the eigenvalue a of A, and is denoted by . The dimensionality of such a subspace is n Ean clearly the degree gn of degeneracy of an. On the other hand, by virtue of the completeness of A, the set of all its linearly independent eigenvectors forms a basis of . Consequently, given the complete set ψm an arbitrary E {| n i} vector x , can be written in a unique way, as a linear combination of the form ∈E gn ∞ x = β ψm (1.2) n,m | n i Xn=1 mX=1 g1 g2 gn x = β ψm + β ψm + . . . + β ψm + . . . 1,m | 1 i 2,m | 2 i n,m | n i m=1 m=1 m=1 X X X gk x = x + x + . . . + x + . . . ; x β ψm (1.3) 1 2 n k ≡ k,m | k i mX=1 several observations are in order at this step: (1) a given vector x as defined in (1.3) belongs to . (2) Since k Eak the complete set of scalars that define the linear combination in (1.2) is unique (for a given order of the basis), each vector x defined in (1.3) is also unique for a given x. In conclusion, the Hilbert space , can be decomposed k E in a direct sum of subspaces generated by the eigenvalues a of a given observable A { n} = ...... (1.4) E Ea1 ⊕Ea2 ⊕ ⊕Ean ⊕ where the dimension of each is the degree of degeneracy of the eigenvalue a associated. In particular, if the Ean n eigenvalue is non-degenerate, the associated subspace is one-dimensional. Further, it is quite obvious that each subspace is invariant under the observable A. Ean We should keep in mind that the decomposition (1.4) depends on the observable chosen. By choicing another observable B, we should take its eigenvalues b and construct the associated subspaces , in order to construct m Ebm the decomposition of induced by B. E 1.2 Symmetries in quantum mechanics

A (passive) symmetry transformation is a change in the point of view that does not change the results of possible experiments. If an observer O sees the system in a state described for the ray , an equivalent observer O could R ′ see the same physical system in another state described by the ray . Nevertheless, both observers must ﬁnd the R′ same physics, for example they should ﬁnd the same probabilities

P ( , α )= P ′, α R k R k indeed this is only a necessary condition. Additional condi tions are necessary for the transformation to be a symmetry transformation1. We shall establish without proof a theorem owe to Wigner concerning with the characterization of possible structures for symmetry operators

Theorem 1.1 The symmetry representation theorem: Any symmetry transformation ′can be characterized R → R by an operator U on the Hilbert space, such that if ψ then U ψ with U being either linear unitary | i ∈ R | i ∈ R′ U (αφ + βψ) = αU (φ)+ βU (ψ) ; φ , ψ , α, β C ∀ | i | i∈E ∀ ∈ Uφ Uψ = φ ψ ; φ , ψ h | i h | i ∀ | i | i∈E or antilinear antiunitary

U (αφ + βψ) = α∗U (φ)+ β∗U (ψ) ; φ , ψ , α, β C ∀ | i | i∈E ∀ ∈ Uφ Uψ = φ ψ ∗ ; φ , ψ h | i h | i ∀ | i | i∈E 1In this context, the rays R and R′ are associated with the same physical state because both are seen from diﬀerent observers. If they are seen by the same observer and R= 6 R′, they must be associated with diﬀerent physical states. 10 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS

The adjoint of a linear operator is deﬁned as

Uφ ψ = φ U †ψ ; φ , ψ (1.5) h | i h ∀ | i | i∈E E

We shall see that the relation (1.5) is not consistent for antilinear operators. To prove it, let us consider an arbitrary complex linear combination of the form

φ = α1φ1 + α2φ2 (1.6) substituting (1.6) on the LHS of Eq. (1.5), using the antilinearity of U and the axioms (1.1) we have

U (α φ + α φ ) ψ = α∗Uφ + α∗Uφ ψ h 1 1 2 2 | i h 1 1 2 2 | i U (α φ + α φ ) ψ = α Uφ ψ + α Uφ ψ (1.7) h 1 1 2 2 | i 1 h 1 | i 2 h 2 | i on the other hand, using Eq. (1.5), the antilinearity of U and the axioms (1.1), we can write the same expression as

U (α φ + α φ ) ψ = α φ + α φ U †ψ = α∗ φ U †ψ + α∗ φ U †ψ h 1 1 2 2 | i h 1 1 2 2 1 h 1 2 h 2 E E E U (α1φ1 + α2φ2) ψ = α1∗ Uφ1 ψ + α2∗ Uφ2 ψ (1.8) h | i h | i h | i equating equations (1.7, 1.8) we obtain that αi∗ = αi which cannot be hold by arbitrary complex values of α1, α2. In other words2, the condition (1.5) cannot be satisﬁed by an antilinear operator because Eq. (1.7) says that the left-hand-side (LHS) of Eq. (1.5) is linear in φ, while Eq. (1.8) says that the same expression is antilinear in φ. Therefore, we shall deﬁne the adjoint of an antilinear operator as

Uφ ψ ∗ φ U †ψ = ψ Uφ ; φ , ψ (1.9) h | i ≡ h h | i ∀ | i | i∈E E with this deﬁnition, the conditions for both unitarity or antiunitarity take the form

1 U † = U − (1.10) the identity is a trivial symmetry transformation which is linear and unitary. Many symmetries in Physics are continuous in the sense that we can connect the associated operator U with the identity by means of a continuous change in some parameters. This is the case in rotations, translations and Lorentz transformations. In that case, the requirement of continuity demands for the symmetry to be represented by a unitary linear transformation. To see it, we observe that we cannot pass continuously from a linear unitary operator (the identity) to an antilinear antiunitary operator, such a transition requires at least one discrete transformation. Symmetries represented by antilinear antiunitary operators involve a reversal in the direction of time’s flow. If a symmetry transformation is infinitesimally closed to the identity, it can be represented by a linear unitary operator written as U = I + iεT with ε a real infinitesimal quantity. For U to be unitary and linear T must be linear hermitian. Most of observables in quantum mechanics such as the angular momentum, momentum, Hamiltonian etc, arise from symmetry transformations in this way. The set of symmetry transformations satisfy the axioms of a group. If T is a transformation that takes i Rn into , we see that (a) the identity is a symmetry transformation, (b) composition of symmetry transformations Rn′ yields another symmetry transformation, (c) transformations are associative, (d) each symmetry transformation has an inverse that is also a symmetry transformation

2We should take into account that the Hilbert space of quantum mechanics is a complex (rather than real) vector space. 1.2. SYMMETRIES IN QUANTUM MECHANICS 11

The unitary or antiunitary operators U (T ) associated with these symmetry transformations carry the group { i } properties of the set T . However, the operators U (T ) act on vectors of the Hilbert space instead of rays. If { i} T takes into , then when U (T ) acts on a vector ψ , it yields a vector U (T ) ψ . Further 1 Rn Rn′ 1 | ni ∈ Rn 1 | ni ∈ Rn′ if T takes into then U (T ) acting on U (T ) ψ must yield a vector in the ray . On the other hand 2 Rn′ Rn′′ 2 1 | ni Rn′′ T T also takes into . Consequently U (T T ) ψ is also in the ray , so these vectors can only diﬀer by 2 1 Rn Rn′′ 2 1 | ni Rn′′ a phase φn (T2, T1) U (T ) U (T ) ψ = eiφn(T2,T1)U (T T ) ψ (1.11) 2 1 | ni 2 1 | ni we shall see that with one important exception, the linearity or antilinearity of U (T ) tells us that these phases are independent of the state ψ . Consider two linearly independent states ψ and ψ , applying Eq. (1.11) | ni | Ai | Bi to the states ψ and ψ as well as to the state ψ ψ + ψ , we have | Ai | Bi | ABi ≡ | Ai | Bi U (T ) U (T ) ψ = eiφAB (T2,T1)U (T T ) ψ 2 1 | ABi 2 1 | ABi eiφAB (T2,T1)U (T T ) [ ψ + ψ ] = U (T ) U (T ) [ ψ + ψ ] (1.12) 2 1 | Ai | Bi 2 1 | Ai | Bi eiφAB (T2,T1)U (T T ) ψ + eiφAB (T2,T1)U (T T ) ψ = U (T ) U (T ) ψ + U (T ) U (T ) ψ 2 1 | Ai 2 1 | Bi 2 1 | Ai 2 1 | Bi eiφAB (T2,T1)U (T T ) ψ + eiφAB (T2,T1)U (T T ) ψ = eiφA U (T T ) ψ + eiφB U (T T ) ψ (1.13) 2 1 | Ai 2 1 | Bi 2 1 | Ai 2 1 | Bi any linear unitary or antilinear antiunitary operator has an inverse (its adjoint) which is also linear unitary or 1 antilinear antiunitary respectively. Multiplying (1.13) by U − (T2T1) we ﬁnd

1 iφAB (T2,T1) iφAB (T2,T1) 1 iφA U − (T T ) e U (T T ) ψ + e U (T T ) ψ = U − (T T ) e U (T T ) ψ 2 1 2 1 | Ai 2 1 | Bi 2 1 2 1 | Ai n o n +eiφB U (T T ) ψ 2 1 | Bi o 1 when the operator U − (T2T1) jumps over the complex numbers, the latter become their complex conjugates when the operator is antilinear thus

iφAB (T2,T1) iφAB (T2,T1) iφA iφB e± ψ + e± ψ = e± ψ + e± ψ (1.14) | Ai | Bi | Ai | Bi where the minus sign in the phases occurs when the operator is antilinear. Since ψ and ψ are linearly | Ai | Bi independent, we equate coeﬃcients in (1.14) to obtain

eiφAB = eiφA = eiφB hence the phases are independent of the states and Eq. (1.11) can be written as an equation of operators

iφ(T2,T1) U (T2) U (T1)= e U (T2T1) (1.15) in the case φ = 0, U (T ) provides a representation of the group of symmetry transformations. For non-null phases we obtain a projective representation or a representation up to a phase. The structure of the Lie group cannot tell us whether physical state vectors furnish an ordinary or a projective representation, but it can tell us whether the group has any intrinsically projective representation. The exception for the preceding argument has to do with the fact that it may not be possible to prepare the system in a state represented by ψ + ψ . For instance, it is widely believed that we cannot prepare a system | Ai | Bi in a superposition of two states whose total angular momentum are integer and half-integer respectively. In that case there is a superselection rule between diﬀerent classes of states and the phases φ (T2, T1) could depend on which of these classes of states are acting the operators U (T2) U (T1) and U (T2T1). It could be shown that any symmetry group with projective representations can be enlarged in such a way that its representations can all be deﬁned as ordinary i.e. with φ = 0, without changing the physical contents. 12 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS

1.3 Irreducible inequivalent representations of groups

For the sake of simplicity, we shall restrict our discussion to ﬁnite-dimensional vector spaces but most of our results are applied to inﬁnite-dimensional vector spaces. Let G = T be a group. Each element of the group T can be mapped into a linear operator U (T ) of a { i} n n vector space V onto itself, in such a way that

U (TkTm)= U (Tk) U (Tm) (1.16) relation (1.16) guarantees that the group structure of G is preserved in the mapping. The set of linear operators U (T ) is called a representation of the group G, in the vector space V . If the mapping T U (T ) is one-to-one, { i } i → i we say that the representation is faithful and that there is an isomorphism between G and U (T ) . In that case, { i } both sets are totally identical as groups. However, in some cases, the mapping T U (T ) is not one-to-one, in i → i that case the representation is degenerate and we say that the mapping is a homomorphism. In that case, some information about G is not carried by U (T ) . { i } Let U (T ) be a representation of a group G in a vector space V . If we take an arbitrary non-singular { i } operator Q of V onto itself, it is obvious that the set of operators W (T ) QU (T ) Q 1 also forms a { i } ≡ i − representation. But equally obvious is the fact that this representation does not contain any new information. Conversely, suppose that we have two representations U (Ti) and W (Ti) of the same group in the same { } { 1 } vector space V , if it exists an operator Q such that W (Ti) = QU (Ti) Q− , for all W (Ti) and for all U (Ti) of each representation, then we say that U (T ) and W (T ) are equivalent representations, and we take them { i } { i } essentially as a single representation. However, it happens in some cases that there is not a non-singular operator Q that connects the two representations U (T ) and W (T ) (defined on the same vector space) in the way described above. In that case we { i } { i } say that we have two (or more) inequivalent representations of G in the vector space V . Let U (T ) be a representation of a group G in a vector space V . Suppose that exists a proper subspace V of { i } k V that is invariant under all linear operators in the set U (T ) . It means that we can restrict to V the domain of { i } k each U (Ti), because the range of each U (Ti) under such a restriction will be contained in Vk. As a consequence, we can form a representation U (T ) of G in the vector space V V . We say that the representation U (T ) { i } k ⊂ { i } of G in V , is reducible because such a representation can be restricted to a proper subspace of V . Even more, suppose that V can be decomposed in non-null vector subspaces V such that { p} V = V V . . . V (1.17) 1 ⊕ 2 ⊕ ⊕ m and that each subspace Vp is invariant under all operators U (Ti). In that case we say that the representation defined in V , is fully reducible into representations defined on each proper subspace Vk. Of course, it could happen that a given subspace Vp could be further reduced in smaller non-null vector subspaces invariant under all U (Ti), so that the representation on Vp is in turn reducible. The idea is to find a decomposition like (1.17) such that none of the subspaces Vp can be decomposed into smaller non-null subspaces in which we can form representations of U (T ) . In that case, we say that our representation is irreducible and that each V is a { i } p minimal invariant subspace under U (T ) . In addition, for most of the cases of interest, the subspaces of the { i } decomposition (1.17) are orthogonal each other, i.e. for any given vector x V and any given vector x V we i ∈ i k ∈ k have that x x =0 if k = i. We say that (1.17) is an orthogonal decomposition and we denote it as V V . h k | ii 6 i ⊥ k We shall assume from now on that we are dealing with orthogonal decompositions unless otherwise indicated. When we have a representation U (T ) of G in V , we can form the matrix representations of each U (T ) by { i } i taking any orthonormal basis V v (1.18) → {| ai} of V (we shall call it the “original” basis). Now, suppose that U (T ) is reducible in V , and that such a { i } reduction can be carried out as in Eq. (1.17). In that case, it is more convenient to choose the basis in the following way: we take an orthonormal basis of the subspace V of dimension d , that is a set of vectors w 1 1 {| 1,r1 i} 1.3. IRREDUCIBLE INEQUIVALENT REPRESENTATIONS OF GROUPS 13

where r1 runs over d1 linearly independent vectors within V1. We proceed in the same way with V2 and so on, then we form a basis for the whole space V as follows

w ; i = 1, 2,...,m ; r = 1, 2,...,d (1.19) {| i,ri i i i} it is easy to see that in this basis ordered as

w , w ,..., w , w , w ,..., w ,..., w , w ,..., w (1.20) {| 1,1i | 1,2i | 1,d1 i | 2,1i | 2,2i | 2,d2 i | m,1i | m,2i | m,dm i} the matrix representatives of each U (Tp) in V are all block-diagonal. To see it, we observe that each Vi is invariant under all U (T ). Hence, w V implies that U (T ) w V and taking into account that w V and p | i,ri i∈ i p | i,ri i∈ i | k,rk i∈ k that V V for i = k, we have i ⊥ k 6 w U (T ) w =0 If i = k h k,rk | p | i,ri i 6 Therefore, the matrix representation of each U (Tp) in the ordered basis (1.20) does not connect two vectors associated with diﬀerent subspaces Vi and Vk. Thus, we form submatrices associated with each Vk, with zeros in the other entries. Since the basis given by (1.20) simpliﬁes considerably the texture of the matrix representation of the operators U (T ), we call it the canonical basis associated with the representation U (T ) in V . p { p } Let us illustrate these facts with an example. Assume that U (T ) is a representation of a group G on a { i } seven dimensional vector space V , and that V can be decomposed in three minimal orthogonal invariant subspaces under U (T ) , as { i } V = V V V 1 ⊕ 2 ⊕ 3 where V1 is 2-dimensional, V2 is 3-dimensional and V3 is 2-dimensional, let us take an orthonormal basis on each subspace as follows

V w , w ; V w , w , w ; V w , w 1 → {| 1,1i | 1,2i} 2 → {| 2,1i | 2,2i | 2,3i} 3 → {| 3,1i | 3,2i} so that we shall use the following orthonormal ordered basis in V

w , w , w , w , w , w , w {| 1,1i | 1,2i | 2,1i | 2,2i | 2,3i | 3,1i | 3,2i} under this ordered basis, the matrix representative of each U (Ti) will have the following texture

00000 × × 00000  × ×  0 0 0 0 A2 2 (U (Ti)) 0 0 × × × × D (U (Ti))  0 0 0 0  = 0 B3 3 (U (Ti)) 0 (1.21) →  × × ×   ×   0 0 0 0  0 0C2 2 (U (Ti))  × × ×  ×  00000     × ×   00000     × ×  D (U (Ti)) = A2 2 (U (Ti)) B3 3 (U (Ti)) C2 2 (U (Ti)) (1.22) × ⊕ × ⊕ × where the “ ” symbol denotes elements that could be non-null. The matrices A2 2, B3 3, C2 2 are the matrix × × × × representatives of each U (Ti) in the subspaces V1,V2,V3 respectively. So the block-diagonal texture of Eq. (1.21), shows that the representation in V can be expressed as a direct sum of representations in V1,V2,V3. Conversely, if we have two (or more) representations in spaces V1,V2, it is clear that we can form a new representation by taking the direct sum of them which is a representation in V V . However, it is equally clear that the new representation 1 ⊕ 2 formed that way does not contain any new information with respect to the component representations. Notwithstanding, we should keep in mind that even if U (T ) in V is reducible, by choicing an arbitrary basis { i } such as (1.18), the matrices will not have the texture (1.21). To exhibit such a texture, an apropriate ordered basis such as the canonical basis given by (1.20) must be chosen. Therefore, changing from our “original” basis to a canonical basis in which the reduction is apparent, is one of the main challenges of group representation theory. 14 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS

The previous discussion, shows us that in characterizing representations of a given group, we should take over two types of redundancies: (a) Given two representations in the same vector space, we consider them different only if they are inequivalent. Equivalent representations are consider as a single one. (b) Given a representation on a vector space V , we should reduce it (if possible) in order to find the irreducible representations. The direct sum of these irreducible representations has no more information with respect to the irreducible ones. Thus, only irreducible representations are considered. Therefore, in characterizing the representations of a given group, we intend to classify all (or as many as possible) irreducible inequivalent representations. The technics and criteria for this classification are out of the scope of the present treatment. By now and for future purposes, we only mention a couple of Lemmas that are crucial in the theory of irreducible representations of groups.

Lemma 1 (Schur’s lemma 1) Let U (G) and U ′ (G) be two irreducible representations of a group G in V and V ′ respectively. Let A be a linear transformation from V to V which satisﬁes AU (g) = U (g) A for all g G. It ′ ′ ∈ follows that either (i) A = 0, or (ii) A is an isomorphism from V ′ onto V (i.e. V and V ′ are isomorphic) and U (G) is equivalent to U ′ (G).

Lemma 2 (Schur’s lemma 2) Let U (G) be an irreducible representation of a group G on the ﬁnite-dimensional vector space V . Let A be an arbitrary operator in V . If A commutes with all the operators in the representation, that is if AU (g)= U (g) A, g G, then A must be a multiple of the identity operator. ∀ ∈ By now we shall only discuss an important consequence of Schur’s lemma 2

Theorem 1.2 All irreducible representations of any abelian group must be 1-dimensional.

Proof: Let U (G) be an irreducible representation of an abelian group G. Let p be a ﬁxed element of the group. Now, U (p) U (g)= U (g) U (p) g G, because of the abelianity of the group. Hence U (p) is an operator ∀ ∈ that commutes with all the U (g)′ s, we conclude from Schur’s lemma 2 that U (p) = λpE. Since p is arbitrary, the representation U (g) is equivalent to the set of operators λ E . But this representation is reducible in { } { p } contradiction with our hypothesis, unless E is the identity in one dimension. Therefore, U (G) is equivalent to the representation p λ C for all p G. QED. → p ∈ ∈ 1.4 Connected Lie groups

These are groups of transformations T (θ) that are described by a ﬁnite set of continuous parameters

θ θ1,θ2, ..., θr { }≡ in such a way that each element of the group is continuosly con nected with the identity by a path within the group. The group multiplication rule takes the form

1 2 r 1 2 r T θ¯ T (θ)= T f θ,θ¯ ; f θ,θ¯ f θ,θ¯ , f θ,θ¯ ,...,f θ,θ¯ = θ′ ,θ′ ,...,θ′ (1.23) ≡ where f θ,θ¯ is a set of r functions such that for a given function f a θ,θ¯ an a given couple of sets θ,θ¯ we have f a θ,θ¯ = θ a with θ another set of r parameters. According with (1.23), the set of r functions ′ ′ − f θ,θ¯ provides the law of combination for the two sets of parameters θ and θ¯ which in turn provides the law of combination of the group. By convention, it is customary to choose θa = 0 as the coordinates of the identity, in that case we have

T (θ) = T (0) T (θ)= T (f (0,θ)) ; T (θ)= T (θ) T (0) = T (f (θ, 0)) ⇒ T (θ) = T (f (0,θ)) = T (f (θ, 0)) 1.4. CONNECTED LIE GROUPS 15 consequently f a (θ, 0) = f a (0,θ)= θa (1.24) since these transformations are continuously connected with the identity, they must be represented on the Physical Hilbert space by unitary (rather than antiunitary) operators U (T (θ)). Operators U (T (θ)) can be represented by a power series, at least in a ﬁnite neighborhood of the identity

1 U (T (θ)) = I + iθaT + θbθcT + θ3 (1.25) a 2 bc O where Ta and Tbc = Tcb are operators independent of the θs, with Ta hermitian. Suppose that U (T (θ)) provides an ordinary (non-projective) representation of this group of transformations, thus

U T θ¯ U (T (θ)) = U T θ¯ T (θ) and using (1.23) we ﬁnd U T θ¯ U (T (θ)) = U T f θ,θ¯ (1.26) by expanding condition (1.26) in powers of θ and θ¯, we shall obtain a condition. The expansion of f a θ,θ¯ around the identity (i.e. around θ¯ = θ = 0) up to second order gives ∂f a θ,θ¯ ∂f a θ,θ¯ 1 ∂2f a θ,θ¯ f a θ,θ¯ = f a (0, 0) + θb + θ¯b + θ¯bθc ∂θb ∂θ¯b 2 ∂θ¯b∂θc θ=θ¯=0 θ=θ¯=0 θ=θ¯=0 1 ∂2f a θ,θ¯ 1 ∂2f a θ,θ¯ + θbθc + θ¯bθ¯c + (3) b c ¯b ¯c 2 ∂θ ∂θ 2 ∂θ ∂θ O θ=θ¯=0 θ=θ¯=0

∂f a (0,θ) ∂f a θ,¯ 0 1 ∂2f a θ,θ¯ f a θ,θ¯ = f a (0, 0) + θb + θ¯b + θ¯bθc b ¯b ¯b c ∂θ θ=0 ∂θ 2 ∂θ ∂θ θ¯=0 θ=θ¯=0 1 ∂2f a θ,θ¯ 1 ∂2f a θ,θ¯ + θbθc + θ¯bθ¯c + (3) b c ¯b ¯c 2 ∂θ ∂θ 2 ∂θ ∂θ O θ=θ¯=0 θ=θ¯=0

where (3) denotes terms of third order i.e. proportional to θ¯3,θ3 ,θθ¯2,θ2θ¯. From Eq. (1.24), such an expansion O becomes

f a θ,θ¯ = 0+ δa θb + δa θ¯b + f a θ¯bθc + ga θbθc + ha θ¯bθ¯c + (3) b b bc bc bc O f a θ,θ¯ = θa + θ¯a + f a θ¯bθc + ga θbθc + ha θ¯bθ¯c + (3) (1.27) bc bc bc O 2 a ¯ 2 a ¯ 2 a ¯ a 1 ∂ f θ,θ a 1 ∂ f θ,θ a 1 ∂ f θ,θ f bc ; g bc ; h bc ≡ 2 ∂θ¯b∂θc ≡ 2 ∂θb∂θc ≡ 2 ∂θ¯b∂θ¯c θ=θ¯=0 θ=θ¯=0 θ=θ¯=0

setting θ¯ = 0 in (1.27) we obtain f a (0,θ)= θa + ga θbθc + (3) bc O a thus, in order to be consistent with the condition (1.24) for arbitrary values of θ, we require that g bc = 0. a Similarly, setting θ = 0 we observe that we require that h bc = 0. In other words, the second order terms of the type θbθc or θ¯bθ¯c would violate the condition (1.24), but second order terms of the type θ¯bθc are in principle allowed. Therefore, Eq. (1.27) becomes

1 ∂2f a θ,θ¯ f a θ,θ¯ = θa + θ¯a + f a θ¯bθc + (3) ; f a (1.28) bc bc ¯b c O ≡ 2 ∂θ ∂θ θ=θ¯=0

16 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS

From Eqs. (1.25, 1.28) the RHS of Eq. (1.26) becomes 1 U T f θ,θ¯ = I + if a θ,θ¯ T + f b θ,θ¯ f c θ,θ¯ T + . . . a 2 bc 1 U T f θ,θ¯ = I + i θa + θ¯a + f a θ¯bθc + . . . T + θb + θ¯b + . . . θc + θ¯c + . . . T + . . . bc a 2 bc h i 1 U T f θ,θ¯ = I + i θa + θ¯a + f a θ¯bθc T + θb + θ¯b θc + θ¯c T + (3) (1.29) bc a 2 bc O h i and substituting (1.25) on the LHS of Eq. (1.26) we have

1 1 U T θ¯ U (T (θ)) = I + iθ¯aT + θ¯bθ¯cT I + iθdT + θeθf T + (3) (1.30) a 2 bc d 2 ef O Substituting (1.29) and (1.30) in Eq. (1.26) yields

1 1 I + iθ¯aT + θ¯bθ¯cT I + iθdT + θeθf T + (3) = I + i θa + θ¯a + f a θ¯bθc T a 2 bc d 2 ef O bc a 1 h i + θb + θ¯b θc + θ¯c T + (3) (1.31) 2 bc O 1 1 I + iθdT + θeθf T + iθ¯aT θ¯aθdT T + θ¯bθ¯cT + (3) = I + iθaT + iθ¯aT + if a θ¯bθcT d 2 ef a − a d 2 bc O a a bc a 1 1 1 + θbθcT + θbθ¯cT + θ¯bθcT 2 bc 2 bc 2 bc 1 + θ¯bθ¯cT + (3) 2 bc O

1 1 I + i θa + θ¯a T + θbθc + θ¯bθ¯c T θ¯bθcT T = I + i θa + θ¯a T + θbθc + θ¯bθ¯c T a 2 bc − b c a 2 bc h i + [if a T + T ] θ¯bθc +h (3)i (1.32) bc a bc O 2 where we have used the fact that indices of sum are dummy and that Tbc = Tcb. The terms of order 1, θ, θ,θ¯ and θ¯2 match automatically in Eq. (1.32). However, by matching coeﬃcients of θθ¯ in such an equation we ﬁnd a non-trivial condition

T T = [if a T + T ] − b c bc a bc T = T T if a T (1.33) bc − b c − bc a a a therefore given the structure of the group (1.26), i.e. the functions f θ,θ¯ , we have its quadratic coeﬃcient f bc as can be seen in (1.28). From them, the second order terms in U (T (θ)) [Eq. (1.25)], can be calculated from the generators Ta that appear in the ﬁrst-order terms. Moreover there is a consistency condition: the operator Tbc must be symmetric in b and c since it is the second derivative of U (T (θ)) with respect to θb and θc, as can be seen from Eq. (1.25). Consequently, Eq. (1.33) demands that

T = T = T T if a T (1.34) bc cb − c b − cb a substracting (1.34) from (1.33) we obtain

0 = T T if a T + T T + if a T − b c − bc a c b cb a T T T T = i [ f a + f a ] T b c − c b − bc cb a 1.5. LORENTZ TRANSFORMATIONS 17 and we obtain finally [T , T ]= iCa T ; Ca f a + f a (1.35) b c bc a bc ≡− bc cb the set of commutation relations in (1.35) defines a Lie algebra. It can also be proved that condition (1.35) is the only condition required to ensure that the process can be continued. In other words, the whole power series in (1.25) for U (T (θ)) can be calculated from an infinite sequence of relations like (1.33), as long as we know the first order terms, the generators Ta. It does not necessarily mean that the operators U (T (θ)) are uniquely determined for all θa if we know the generators T a, but it does mean that the operators U (T (θ)) are uniquely determined in at least a finite neighborhood of the coordinates θa = 0 associated with the identity, such that Eq. (1.26) is satisfied if θ, θ¯ and f θ, θ¯ are in this neighborhood. In some cases, it happens that the function f θ,θ¯ satisfies the condition (at least for some subset of the coordinates θa) f a θ,θ¯ = f a θ, θ¯ = θ¯a + θa (1.36) as it is the case in space-time translations or for rotations about a given fixed axis. In that case the coefficients a f bc in Eq. (1.28) vanish and so do the structure constants in (1.35). Hence, all generators commute

[Tb, Tc] = 0 from (1.36) the laws of combination (1.23, 1.26) for the group representation yield

U T θ¯ U (T (θ)) = U T f θ,θ¯ = U T θ¯ + θ = U T θ + θ¯ ¯ ¯ = U T f θ, θ = U (T (θ)) U T θ U T θ¯ U (T (θ)) = U (T (θ)) U T θ¯ (1.37) ⇒ hence the elements of the group commute each other. Consequently, when the function f (θ) is given by (1.36), the connected Lie group becomes abelian. In that case, we can calculate U (T (θ)) for all θa and not only a neighborhood of the identity. From Eqs. (1.26, 1.36), we ﬁnd

U (T (θ2)) U (T (θ1)) = U (T (θ1 + θ2))

U (T (θN )) ...U (T (θ2)) U (T (θ1)) = U (T (θ1 + θ2 + . . . + θN )) by defining θ θ/N, we see that for any positive integer N we have i ≡ θ N U (T (θ)) = U T (1.38) N setting N , the angle θ/N becomes infinitesimal. Thus, we can use expansion (1.25) for U (T (θ/N)) keeping →∞ only the terms at first-order in θ. In this way we get

N i a U (T (θ)) = lim 1+ θ Ta N N →∞ a U (T (θ)) = exp [iTaθ ] (1.39)

1.5 Lorentz transformations

Special relativity establishes the existence of certain special reference frames called inertial frames which are in constant relative motion among them. Special relativity is based on two basic postulates: (a) The laws of nature are the same in all inertial reference frames. (b) The speed of light in vacuum, measured in any inertial reference frame is the same regardless of the motion of the light source relative to that reference frame. The second postulate represents a significant deviation with respect to Galilean and Newtonian mechanics. It leads in 18 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS turn to different transformations connecting coordinate systems in different inertial frames. We denote as xµ the coordinates in one inertial frame S, with x1,x2,x3 cartesian space coordinates while x0 = t is a time coordinate, the speed of light will be settled equal to unity. We shall use latin indices such as i, j, k when running over the three space components, and greek indices such as µ, ν, ρ when running over the four space-time indices. Quantities for which we obtain the same value in any inertial frame are called Lorentz invariants. A well-known invariant in special relativity is the quantity

2 2 2 2 2 2 2 (dτ)2 dx1 + dx2 + dx3 (dt)2 = dx1 + dx2 + dx3 dx0 (1.40) ≡ − − where the invariance of this “proper time” is related with the invariance ofc, the speed of light in vacuum. This Lorentz invariant can be written as 100 0 dx1 2 2 1 2 3 0 010 0 dx µ ν (dτ) dx dx dx dx     = dx gµν dx ≡ 001 0 dx3  0 0 0 1   dx0   −    100 0     010 0 gµν =   (1.41) 001 0  0 0 0 1   −    where gµν deﬁned in (1.41) is called the metric tensor. We use from now on a convention of sum over repeated µ ν upper and lower indices. By using this invariant, we can relate the coordinates x of S, with the coordinates x′ in any other inertial frame S′, in the following way

µ ν µ ν gµν dx′ dx′ = gµν dx dx (1.42) this equation leads to ∂x µ ∂x ν ∂xµ ∂xν g ′ ′ = g = g δµ δν µν ∂xρ ∂xσ µν ∂xρ ∂xσ µν ρ σ ∂x µ ∂x ν g ′ ′ = g (1.43) µν ∂xρ ∂xσ ρσ it is also easy to arrive to (1.42) from (1.43). Hence Eqs. (1.43) and (1.42) are equivalent. A light wave travelling at unit speed satisﬁes dx = 1 g dxµdxν = (dx)2 (dt)2 = 0 dt ⇒ µν −

µ µ and the same holds for S′. Any coordinate transformation x x′ that satisﬁes Eq. (1.43) is linear → µ µ ν µ x′ = Λ νx + a (1.44)

µ µ 3 with a arbitrary constants, and Λ ν a constant matrix . Restricting for a while to a homogeneous transformation i.e. with aµ = 0, the Jacobian of such a transformation gives µ µ µ ∂x′ ν µ ∂x′ x′ = x ; Λ ∂xν ν ≡ ∂xν µ further, from (1.43) we see that Λ ν satisﬁes the condition

µ ν gµν Λ ρΛ σ = gρσ (1.45)

3 µ ′ The matrix Λ ν depends on the velocity of S with respect to S. However, since both frames are inertial, such a velocity is constant µ and so Λ ν is. 1.6. THE INHOMOGENEOUS LORENTZ GROUP (OR POINCARE´ GROUP) 19 it is convenient for some purposes to write the Lorentz transformation condition (1.45) in a different way. It is µν easy to check that the matrix gµν in (1.41) coincides with its inverse (that we denote as g ). Multiplying Eq. στ κ (1.45) by g Λ τ we find µ ν στ κ στ κ µ ν κ στ στ κ τ κ gµν Λ ρΛ σ (g Λ τ ) = gρσ (g Λ τ ) gµν Λ ρ (Λ σΛ τ g ) = (gρσg ) Λ τ = δρ Λ τ µ ν κ στ κ κ µ ⇒ νκ µ (gµν Λ ρ) (Λ σΛ τ g ) = Λ ρ = δµ Λ ρ = gµν g Λ ρ µ ν κ στ µ νκ (gµν Λ ρ) (Λ σΛ τ g ) = gµν Λ ρg the relativity principle demands for the Lorentz transformations to have an inverse. Defining M g Λµ , and ρν ≡ µν ρ multiplying with the inverse of this matrix we find4

ν κ στ νκ 1 αρ ν κ στ 1 αρ νκ Mρν (Λ σΛ τ g ) = Mρν g M − Mρν (Λ σΛ τ g )= M − Mρν g α ν κ στ α νκ ⇒ α κ στ ακ δ ν (Λ σΛ τ g ) = δ νg Λ σΛ τ g = g ν κ στ νκ ⇒ Λ σΛ τ g = g (1.46) the condition (1.45) or equivalently (1.46) is usually called the generalized orthogonality condition in the Minkowski metric space.

1.6 The inhomogeneous Lorentz Group (or Poincar´egroup)

The set of all Lorentz (inhomogeneous) transformations (Λ, a) form a group. If we perform a Lorentz transfor- { } mation (1.44) and then a second Lorentz transformation x µ x µ, then the resultant transformation xµ x µ ′ → ′′ → ′′ is described by

µ µ ρ µ µ ρ ν ρ µ x′′ = Λ¯ ρx′ +a ¯ = Λ¯ ρ (Λ νx + a ) +a ¯ µ µ ρ ν µ ρ µ x′′ = Λ¯ ρΛ ν x + Λ¯ ρa +a ¯ (1.47) The bar in Λ¯ is used to distinguish one Lorentz transformation from the other, and same fora ¯, with respect to a. We should show that the effect is the same as a Lorentz transformation xµ x µ. In other words, we have to → ′′ verify that (1.47) defines a Lorentz transformation. To do this, let us define µ Λ¯ µ Λρ Z ν ≡ ρ ν we should check that µ obeys the relation (1.45). For this, we take into account that both Λ and Λ¯ must obey Z ν such a relation g µ ν = g Λ¯ µ Λβ Λ¯ ν Λγ = g Λ¯ µ Λ¯ ν Λβ Λγ µν Z ρZ σ µν β ρ γ σ µν β γ ρ σ = g Λβ Λγ =g βγ ρ σ ρσ ⇒ g µ ν = g µν Z ρZ σ ρσ Hence µ defines a Lorentz transformation. Equation (1.47) shows us how is the composition rule for the Z ν transformations T (Λ, a) induced on physical states T Λ¯, a¯ T (Λ, a)= T ΛΛ¯ , Λ¯a +a ¯ (1.48) each Λ admits an inverse. To find it, we start from the definitio n of inverse and use again the condition (1.45) to write

1 ρ ν ρ ρσ ρσ µ ν Λ− νΛ β = δ β = g gσβ = g gµν Λ σΛ β 1 ρ ν ρσ µ ν Λ− νΛ β = (gνµg Λ σ) Λ β

4 µ µ The matrix Aνρ ≡ gµν Λ ρ = gνµΛ ρ = (gΛ)νρ is non-singular since both g and Λ are non-singular. Therefore its transpose M ≡ A is also invertible. e 20 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS

Thus, condition (1.45) says that the inverse of Λ takes the form

1 ρ ρ ρσ µ Λ− Λ = g g Λ (1.49) ν ≡ ν νµ σ The reader can also check that (det Λ)2 = 1 (1.50) from the composition law (1.48) it is easy to see what is the identity for the transformations T (Λ, a)

T (1, 0) T (Λ, a) = T (1Λ, 1a + 0)= T (Λ, a) ; T (Λ, a) T (1, 0)= T (Λ1, Λ0 + a)= T (Λ, a) I = T (1, 0) (1.51) ⇒ where 1 is the identity matrix 4 4, and 0 the null 4-components vector. The inverse of a given T (Λ, a) can also × be obtained from

1 1 1 1 1 T Λ− , Λ− a T (Λ, a) = T Λ− Λ, Λ− a Λ− a = T (1, 0) − 1 1 1 − 1 T (Λ, a) T Λ− , Λ− a = T ΛΛ− , Λ Λ− a +a = T (1, 0) −1 1 −1 T − (Λ, a) = T Λ− , Λ− a (1.52) − 1.6.1 Four-vectors and tensors The Lorentz invariant (1.40) suggests to deﬁne a Lorentz invariant norm for the 4-components vectors xµ as follows

2 0 2 1 2 2 2 3 2 x (x,x)= x + x + x + x = x′,x′ (1.53) k k ≡ − 0 2 i strictly speaking this is a pseudonorm because it is not positive-deﬁnite since in some cases x >x xi. Taking into account (1.41) this relation can be written as µ ν (x,x)= x gµν x (1.54) it is easy to see the invariance of this norm under a homogeneous Lorentz transformation, by using condition (1.45) we ﬁnd

µ ν µ α ν β µ ν α β α β x′,x′ = x′ gµν x′ = (Λ αx ) gµν Λ βx = (gµν Λ αΛ β) x x = gαβx x µ ν α β x′,x′ = x′ gµν x′ = x gαβx = (x,x) (1.55) a more convenient way to write this norm is the following

(x,x) = xµg xν xµx ; x g xν (1.56) µν ≡ µ µ ≡ µν xµ = x1,x2,x3,x0 ; x = (x ,x ,x ,x )= x1,x2,x3, x0 (1.57) µ 1 2 3 0 − we deﬁne a four-vector as any arrangement of four components that under a homogeneous Lorentz transformation, changes under the same prescription of xµ. That is V µ is a four-vector if under a homogeneous Lorentz transformation we have µ µ α V ′ = Λ αV (1.58) µ for any “contravariant” four-vector V ′ we can deﬁne a “covariant” four-vector as in Eq. (1.56) V g V ν µ ≡ µν multiplying by the inverse of gµν we obtain the inverse relation

αµ αµ ν α ν α g Vµ = g gµν V = δ νV = V α αµ ⇒ V = g Vµ 1.6. THE INHOMOGENEOUS LORENTZ GROUP (OR POINCARE´ GROUP) 21 we can deﬁne the inner product between two (contravariant) four vectors V µ,W µ as

(V,W ) V µg W ν = V µW ≡ µν µ with the same procedure used to prove the Lorentz invariance of (x,x), we see that (V,W ) is also Lorentz invariant. The summation of upper and lowered indices is called a contraction. The Lorentz transformation of a covariant four-vector gives

ν ν α ν αβ βα ν Vµ′ = gµν V ′ = gµν Λ αV = gµν Λ αg Vβ = gµν g Λ α Vβ where we have used the symmetrical nature of gαβ. Applying Eq. (1.49) we obtain

β 1 β Vµ′ = Λµ Vβ = Λ− µVβ (1.59) so covariant four-vectors transform with the inverse transformation with respect to contravariant four-vectors. It justiﬁes the names covariant and contravariant. Two adjacent four-vectors transform as

µ ν µ α ν β µ ν α β V ′ W ′ = (Λ αV ) Λ βW = Λ αΛ βV W (1.60) an arrangement of numbers characterized by two indices of the form T µν is called a second-rank Lorentz tensor if under a homogeneous Lorentz transformation, it changes in a way similar to the two adjacent four-vectors in Eq. (1.60), that is µν µ ν αβ T ′ = Λ αΛ βT (1.61) taking into account the expression (1.49) for the inverse of Λ, we can write this transformation as

µν µ αβ ν µ αβ 1 ν T ′ = Λ αT Λ β = Λ αT Λ− β (1.62) in matrix from it yields 1 Tcont′ = ΛTcontΛ− which is a similarity transformation of Tcont under Λ. We can deﬁne in an analogous way second-rank covariant tensors Tµν based on covariant four-vectors. By using Eq. (1.59) we have

1 α 1 β 1 α 1 β Vµ′Wν′ = Λ− µVα Λ− νVβ = Λ− µ Λ− νVαVβ 1 αh 1 β i T′ = Λ− Λ− T (1.63) ⇒ µν µ ν αβ which can also be written as

1 α 1 β α 1 β Tµν′ = Λ− µTαβ Λ− ν = Λµ Tαβ Λ− ν (1.64) 1 Tcov′ = ΛTcovΛ− (1.65) from Eqs. (1.62, 1.64), it is easy to show that the contraction of two second rank tensors (one contravariant and other covariant) is a Lorentz invariant (also known as a Lorentz scalar or a zero-rank Lorentz tensor)

µν µ αβ 1 ν γ 1 δ µ γ 1 ν 1 δ αβ T ′ Hµν′ = Λ αT Λ− β Λµ Hγδ Λ− ν = Λ αΛµ Λ− β Λ− ν T Hγδ h 1 µ γ ih1 ν δ αβ i = Λ− α Λµ Λ− β Λν T Hγδ γ δ αβ h αβ i µν = δα δβ T Hγδ =T Hαβ = T Hµν 22 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS thus we can denote this contraction with a quantity without Lorentz indices, and it is equal in any inertial reference frame µν µν T Hµν = T ′ Hµν′ = C

µα in a similar way, we can show that the contraction of a second-rank tensor T with a four-vector Vα, gives a four-vector W µ µα µ T Vα = W (1.66) the previous developments justify the convention that indices may be lowered or raised by contraction with gµν or gµν . For instance T g T µ ; T µ gµσT (1.67) σρ ≡ σµ ρ ρ ≡ σρ a very important four-vector in special relativity is the four-momentum

pµ = p1,p2,p3,p0 = p,p0 (p, E) , p = (p, E) (1.68) ≡ µ − hence p is the three momentum and the energy is the zeroth component. Taking into account the fundamental relation p2 + m2 = E2 (1.69) the pseudo-norm of the four momentum gives

2 p2 = pµg pν = pµp = p0 + p2 = E2 + p2 µν µ − − p2 = m2 (1.70) ⇒ − note that four-momenta with positive pseudonorm leads to m2 > 0, i.e. to a non-physical mass. Thus, physical − states are related with four-momenta with non-positive pseudonorm. It is important to keep in mind that if the metric tensor is chosen as η = (1, 1, 1, 1), we obtain p2 = m2 and the positive values of p2 are the physical µν − − − ones. Hence, it is extremely important to know the conventions used in each case.

1.7 Some subgroups of the Poincar´egroup

The whole group of transformations T (Λ, a) is called the inhomogeneous Lorentz group, or the Poincar´e group. It has several important subgroups. First the set of transformations with a = 0, T (Λ, 0) clearly forms a { } subgroup. To see it we observe ﬁrst that the identity T (1, 0) of T (Λ, a) is also contained in T (Λ, 0) . Further { } { } Eq. (1.48) gives us the composition law, on this subset

T Λ¯, 0 T (Λ, 0) = T ΛΛ¯ , Λ¯0 + 0 = T ΛΛ¯ , 0 and the composition law is closed within the subset T (Λ, 0) . Finally, the inverse of any given element in { } T (Λ, 0) , also belongs to such a subset as can be seen from (1.52) { } 1 1 1 1 T − (Λ, 0) = T Λ− , Λ− 0 = T Λ− , 0 − this subgroup of the inhomogeneous Lorentz group is called the homogeneous Lorentz group. When we work on the homogeneous Lorentz group, we usually simplify the notation and write

T (Λ, 0) T (Λ) ; I T (1) ≡ ≡ 1 1 T Λ¯ T (Λ) = T ΛΛ¯ ; T − (Λ, 0) = T Λ− 1.7. SOME SUBGROUPS OF THE POINCARE´ GROUP 23 further, we note that Eq. (1.50) gives two possibilities (a) detΛ = +1 (b) detΛ = 1. Those transformations − with det Λ = +1 obviously form a subgroup of either the homogeneous or inhomogeneous Lorentz groups5. On the other hand, by taking the 00 components of Eq. (1.45) we have

µ ν gµν Λ 0Λ 0 = g00 (1.71) expanding the LHS explicitly and taking into account that gµν is diagonal, we obtain

µ ν µ µ 0 0 i i gµν Λ 0Λ 0 = gµµΛ 0Λ 0 = g00Λ 0Λ 0 + giiΛ 0Λ 0 2 g Λµ Λν = Λi Λi Λ0 (1.72) µν 0 0 0 0 − 0 where we are using sum over repeated (upper) indices i. Substituting (1.72) into (1.71) and using g = 1, we 00 − find 2 2 Λi Λi Λ0 = 1 Λ0 =1+Λi Λi 0 0 − 0 − ⇒ 0 0 0 0 2 0 0 similarly from Eq. (1.46) we can find that Λ 0 =1+Λ iΛ i. We obtain finally

0 2 i i 0 0 Λ 0 =1+Λ 0Λ 0 =1+Λ iΛ i (1.73) since the matrices Λµ are real, we have Λ0 Λ 0 0. Hence, we see from (1.73) that Λ0 1. Consequently, we ν i i ≥ 0 ≥ have that either Λ0 1 or Λ0 1. 0 ≥ 0 ≤−

1.7.1 Proper orthochronous Lorentz group 0 µ Those transformations with Λ 0 1 form a subgroup. To show it, we assume by hypothesis that both Λ ν and µ ≥ Λ¯ ν satisfy the conditions Λ0 1, and Λ¯ 0 1 (1.74) 0 ≥ 0 ≥ we have 0 0 µ 0 0 0 i ΛΛ¯ 0 = Λ¯ µΛ 0 = Λ¯ 0Λ 0 + Λ¯ iΛ 0 (1.75) if Λ¯ 0 Λi 0, Eqs. (1.74, 1.75) yield immediately that ΛΛ¯ 0 1. Now we should examine the case in which i 0 ≥ 0 ≥ Λ¯ 0 Λi < 0. First we prove the inequality i 0 a + b + ab a (a + 2) b (b + 2) 0 ; if a 0 and b 0 (1.76) − ≥ ≥ ≥ we prove it as follows p p

(a b)2 0 (a + b + ab)2 a (a + 2) b (b + 2) 0 − ≥ ⇒ − ≥ ⇒ (a + b + ab)2 a (a + 2) b (b + 2) ≥ if a, b are non-negative, we can take positive square roots on both sides and preserve the order relation in the inequality, thus a + b + ab a (a + 2) b (b + 2) ; if a 0 and b 0 ≥ ≥ ≥ then we obtain Eq. (1.76). p p Deﬁning the three vectors

v Λ1 , Λ2 , Λ3 ; ¯v Λ¯ 0 , Λ¯0 , Λ¯ 0 (1.77) ≡ 0 0 0 ≡ 1 2 3 i i 0 0 v = √v v = Λ 0Λ 0 ; ¯v = √¯v ¯v = Λ kΛ k (1.78) k k · k k · 5It is also clear that the set of Lorentz transformationsp with detΛ = −1, does not formp a subgroup of the Lorentz group. For instance, it does not contain the identity. 24 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS

Eq. (1.73) shows that the lengths of these two three-vectors are given by

v = (Λ0 )2 1 ; ¯v = Λ¯ 0 2 1 (1.79) k k 0 − k k 0 − q q and using the inequality ¯v v ¯v v (1.80) | · | ≤ k k k k substituting (1.77) and (1.79) in (1.80) and taking into account that Λ¯ 0 Λi 0, we have i 0 ≤ Λ¯ 0 , Λ¯ 0 , Λ¯ 0 Λ1 , Λ2 , Λ3 (Λ0 )2 1 Λ¯ 0 2 1 1 2 3 · 0 0 0 ≤ 0 − 0 − q q Λ¯ 0 Λi (Λ0 )2 1 Λ¯ 0 2 1 i 0 ≤ 0 − 0 − q q Λ¯ 0 Λi (Λ0 )2 1 Λ¯ 0 2 1 − i 0 ≥ − 0 − 0 − q q Λ¯ 0 Λi (Λ0 )2 1 Λ¯ 0 2 1 (1.81) i 0 ≥ − 0 − 0 − q q substituting (1.81) in (1.75), we ﬁnd

0 0 0 0 i ΛΛ¯ 0 = Λ¯ 0Λ 0 + Λ¯ iΛ 0

ΛΛ¯ 0 Λ¯ 0 Λ0 (Λ0 )2 1 Λ¯ 0 2 1 (1.82) 0 ≥ 0 0 − 0 − 0 − q q since Λ¯ 0 1 and Λ0 1, we can express them as 0 ≥ 0 ≥ Λ¯ 0 =1+ a , Λ0 =1+ b ; a 0 , b 0 (1.83) 0 0 ≥ ≥ using (1.83) on the RHS of (1.82) we have

Λ¯ 0 Λ0 (Λ0 )2 1 Λ¯ 0 2 1 = (1+ a)(1+ b) (1 + a)2 1 (1 + b)2 1 0 0 − 0 − 0 − − − − q q q q Λ¯ 0 Λ0 (Λ0 )2 1 Λ¯ 0 2 1 = 1+ a + b + ab a (a + 2) b (b + 2) 0 0 − 0 − 0 − − q q and using Eq. (1.76) we ﬁnd p p Λ¯ 0 Λ0 (Λ0 )2 1 Λ¯ 0 2 1 1 (1.84) 0 0 − 0 − 0 − ≥ q q and combining Eqs. (1.82, 1.84) we ﬁnd ΛΛ¯ 0 1, and the subset of Lorentz transformations with Λ0 1 0 ≥ 0 ≥ is closed under composition. It is left to the reader to show that the inverse of an element in this subset, also belongs to the subset [homework!!(2)]. Hence such a subset forms a subgroup. The subgroup with detΛ = +1 and Λ0 +1, is known as the proper orthochronus Lorentz group 0 ≥ (either homogeneous or inhomogeneous). It is not possible by a continuous change of parameters to jump from detΛ = +1 to detΛ = 1 (or vice versa), neither from Λ0 +1 to Λ0 1. Consequently, any Lorentz − 0 ≥ 0 ≤ − transformation that can be obtained through a continuous change of parameters from the identity, must have the 0 same sign of det Λ and Λ 0 as the identity. Therefore, those elements connected continuously with the identity must belong to the proper orthochronus Lorentz group.

1.7.2 Discrete transformations in the Lorentz group Any Lorentz transformation is either proper orthochronus, or can be written as an element of the proper orthochronus subgroup with one of the discrete transformations , or where is the parity or space inversion P T PT P operator, which is diagonal with elements

0 = 1 , i = 1 (1.85) P 0 P i − 1.8. QUANTUM LORENTZ TRANSFORMATIONS 25 while is the time-reversal matrix, also diagonal with elements T 0 = 1 , i = 1 (1.86) T 0 − T i consequently, the study of the whole Lorentz group reduces to the study of the (connected) proper orthochronus Lorentz group, along with the study of the discrete space inversion and time reversal operators. By now, we shall study the proper orthochronus Lorentz group, which is a connected Lie group.

1.7.3 Inﬁnitesimal transformations within the proper orthochronus Lorentz group Much of the information about connected Lie groups can be extracted by examining the behavior of the group elements in a neighborhood of the identity. For the inhomogeneous Lorentz group, the identity is characterized by the transformations µ µ µ Λ ν = δ ν , a = 0 (1.87) therefore a neighborhood of the identity can be written as

µ µ µ µ µ Λ ν = δ ν + ω ν , a = ε (1.88)

µ µ where ω ν and ε are inﬁnitesimals. The Lorentz condition (1.45) taken over the inﬁnitesimal transformations (1.88) gives

µ µ ν ν µ ν µ ν µ ν µ ν gρσ = gµν (δ ρ + ω ρ) (δ σ + ω σ)= gµν δ ρδ σ + gµν δ ρω σ + gµν ω ρδ σ + gµν ω ρω σ = g + g ων + g ωµ + ω2 ρσ ρν σ σµ ρ O g = g + ω + ω + ω2 ρσ ρσ ρσ σρ O therefore, keeping terms up to ﬁrst order, the Lorentz condition lead to

ω = ω (1.89) ρσ − σρ hence ω is a second-rank antisymmetric tensor, whose degrees of freedom are given by N (N 1) /2. For N = 4 ρσ − (dimensions) we obtain 6 independent components. By adding the 4 degrees of freedom associated with the components of εµ, we obtain a total of 10 parameters for an inhomogeneous Lorentz transformation6.

1.8 Quantum Lorentz Transformations

By now we shall restrict ourselves to proper orthochronus Lorentz transformations for which all transformations are continuous. As we have already seen, in the framework of quantum mechanics the elements T (Λ, a) of the Lorentz group are operators applied on rays. However, in practice we transform kets of the Hilbert space and not rays. Thus we should construct a representation U (Λ, a) consisting of sets of operators acting on the Hilbert space. From the Wigner’s representation theorem and the connected nature of the proper orthochronus Lorentz group, we see that the transformations T (Λ, a) induce a linear unitary (rather than an antilinear antiunitary) transformation on vectors of the Hilbert space of states

Ψ U (Λ, a) Ψ | i→ | i in order to carry the group structure, the operators U (Λ, a) must respect the composition law (1.48)

U Λ¯, a¯ U (Λ, a)= U ΛΛ¯ , Λ¯a +a ¯ (1.90)

6In a more physical point of view, we require three parameters for a boost (the components of a velocity), three for a rotation (e.g. the Euler angles), and four to perform a space-time translation. Hence, six parameters belong to the homogeneous Lorentz transformations (boosts and rotations), and four parameters belong to the inhomogeneous part. 26 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS strictly speaking we could also form projective representations or representations “up to a phase” as shown in Eq. (1.15). In general, it is necessary to enlarge the Lorentz group to avoid the appearance of phase factors on the RHS of Eq. (1.90). The inverse of U (Λ, a) is indicated by Eq. (1.52)

1 1 1 U − (Λ, a)= U Λ− , Λ− a (1.91) − Since U (1, 0) carries any ray into itself, it must be proportional to the unit operator. By an apropriate choice of phase, it can be settled as equal to the identity.

1.8.1 Four-vector and tensor operators In quantum Mechanics, observables are eigenvalues of a complete hermitian operator. For example the three components of the classical linear momentum is replaced by a corresponding set of three-Hermitian operators

p (p ,p ,p ) P (P , P , P ) ≡ 1 2 3 → ≡ 1 2 3 for the energy we have the Hamiltonian. On the other hand, we have seen that the four components consisting of the three momentum components plus the energy forms a four-vector in special relativity. Thus, the corresponding arrangement in quantum mechanics

pµ (p ,p ,p , E) P µ (P , P , P ,H) ≡ 1 2 3 → ≡ 1 2 3 could be taken as the prototype of a “four-vector operator”. Thus, we shall study the transformation of P µ under a quantum Lorentz transformation in order to deﬁne other four-vector operators as the ones that under a Lorentz transformation change in the way prescribed by P µ. We then start with four-momentum eigenstates

P µ p = pµ p (1.92) | i | i Then we apply a quantum homogeneous Lorentz transformation U (Λ) on both sides of (1.92)

U (Λ) P µ p = pµU (Λ) p (1.93) | i | i it is important to emphasize that U (Λ) acts on vectors of the Hilbert space and not on four-vectors pµ of the Minkowski space. Owing to it, U (Λ) passes by the eigenvalue pµ. Inserting an identity in Equation (1.93) we have µ 1 µ µ 1 µ U (Λ) P U − (Λ) U (Λ) p = p U (Λ) p U (Λ) P U − (Λ) U (Λ) p = p U (Λ) p | i | i ⇒ { | i} | i which can be rewritten as µ µ 1 U ′ (Λ) p′ = p p′ ; p′ U (Λ) p ,U ′ (Λ) U (Λ) P U − (Λ) (1.94) ≡ | i ≡ it is convenient to introduce the eigenvalues p µ associated with the transformed state p . Obviously, the trans- ′ | ′i formation pµ p µ must be carried out with the transformation Λ in the Minkowski space associated with the → ′ quantum Lorentz transformation U (Λ). Therefore

ν ν µ µ 1 µ ν µ ν p′ = Λ p p = Λ− p′ = Λ p′ (1.95) µ ⇒ ν ν substituting (1.95) in (1.94) we ﬁnd

µ ν µ ν U ′ (Λ) p′ = Λν p′ p′ = Λν P p′ since this is valid for p arbitrary we obtain | ′i µ ν U ′ (Λ) = Λν P µ 1 µ ν ⇒ U (Λ) P U − (Λ) = Λν P this induces the following deﬁnition 1.8. QUANTUM LORENTZ TRANSFORMATIONS 27

Definition 1.2 (four-vector operators) An arrangement of four component operators µ 1, 2, 3, 0 is O ≡ O O O O called a (contravariant) four-vector operator if under a quantum homogeneous Lorentz transformation, it changes under the formula µ 1 µ ν U (Λ) U − (Λ) = Λ (1.96) O ν O By superposing four-vector operators we can define second-rank (or higher order) Lorentz tensors, in a similar way as we did in section 1.6.1. For instance, a contravariant second-rank Lorentz tensor transform under a homogeneous Lorentz transformations as µν µ ν αβ T ′ = Λα Λβ T (1.97) If we compare Eq. (1.58) that defines a contravariant four-vector in Minkowski space with Eq. (1.96) that defines a contravariant four-vector operator on a Hilbert space, we realize that the RHS of these equations differ µ 1 µ since Λν = Λ− ν. Something similar can be seen by comparing (1.61) with (1.97) for second-rank contravariant tensors.

1.8.2 Infinitesimal quantum Lorentz transformations We shall use the formalism developed in Sec. 1.4 about connected Lie groups. In such a section, we saw that much of the properties of the group structure can be developed from expansions of the elements of the group up to first order in the parameters. For an infinitesimal Lorentz transformation (1.88) the corresponding operator U (Λ, a) acting on the Hilbert space, is given by

U (Λ, a)= U (1+ ω, ε) (1.98)

7 and must be equal to the identity plus terms linear in ωρσ and ερ, that we shall parameterize as in Eq. (1.25)

1 U (1+ ω, ε)=1+ iω J ρσ iε P ρ + ω2, ε2, ωε (1.99) 2 ρσ − ρ O where J ρσ, P ρ are operators independent of the parameters ω and ε. For U (1 + ω, ε) to be unitary, the operators J ρσ, P ρ must be hermitian ρσ ρσ ρ ρ J † = J ; P † = P (1.100)

ρσ 8 since ωρσ is antisymmetric, we can take the operators J to be antisymmetric as well

J ρσ = J σρ (1.101) − 1 on the other hand, the expansion of U − (1 + ω, ε) up to ﬁrst order is given by

1 1 ρσ ρ 2 2 U − (1 + ω, ε) = 1 iω J + iε P + ω , ε , ωε (1.102) − 2 ρσ ρ O it can be seen by multiplying Eqs. (1.99, 1.102), and observing that we obtain the identity plus terms of second order in ω and/or ε. In addition each parameter is accompanied by an associated generator ω J ρσ and ρσ ↔ ερ P ρ. We have ten independent parameters and so ten independent generators (owing to the antisymmetry ↔ ρσ of ωρσ and J ).

7 a Comparing Eqs. (1.25, 1.99) we see that we have chosen our parameters in the form θ → (ωρσ, −ερ). 8As a matter of example, for a given couple of numbers say 2, 3; what we have to ﬁx is the quantity 1 ω ω i ω J 23 + ω J 32 = 23 i J 23 − J 32 ≡ 23 iJ (23) 2 23 32 2 2 so that only J (23) is ﬁxed. Thus we can choose J 23 ≡ J (23)/2, which leads to J 23 = −J 32. 28 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS

1.8.3 Lorentz transformations of the generators We now examine the Lorentz transformation properties of the generators J ρσ, P ρ of the continuous Poincar´e group. To do it, we examine the transformation properties of U (1 + ω, ε) induced by another new (and in general ﬁnite) transformation U (Λ, a) 1 U (Λ, a) U (1 + ω, ε) U − (Λ, a) (1.103) of course, (Λ, a) are totally independent of (ω, ε). From Eqs. (1.90, 1.91) we can write this product as

1 1 1 U (Λ, a) U (1 + ω, ε) U − (Λ, a) = [U (Λ, a) U (1 + ω, ε)] U Λ− , Λ− a 1− 1 = [U (Λ(1 + ω) , Λε + a)] U Λ− , Λ− a 1 −1 = U Λ(1+ ω) Λ− , Λ(1+ω) Λ− a + Λε + a 1 − 1 = U 1 + ΛωΛ− , a ΛωΛ− a + Λε + a 1 1 − − 1 U (Λ, a) U (1 + ω, ε) U − (Λ, a) = U 1 + ΛωΛ− , Λε ΛωΛ− a − then we have 1 U (Λ, a) U (1 + ω, ε) U − (Λ, a)= U (1 +ω, ¯ ε¯) (1.104) 1 1 ω¯ ΛωΛ− , ε¯ Λε ΛωΛ− a (1.105) ≡ ≡ − on the other hand, by using the expansion (1.99) on the LHS of Eq. (1.104) we ﬁnd

1 1 ρσ ρ 1 U (Λ, a) U (1+ ω, ε) U − (Λ, a) = U (Λ, a) 1+ iω J iε P U − (Λ, a) 2 ρσ − ρ 1 1 ρσ ρ 1 U (Λ, a) U (1+ ω, ε) U − (Λ, a) = 1+ U (Λ, a) iω J iε P U − (Λ, a) (1.106) 2 ρσ − ρ now using the expansion (1.99) on the RHS of Eq. (1.104) we obtain 1 U (1 +ω, ¯ ε¯)=1+ iω¯ J µν iε¯ P µ (1.107) 2 µν − µ equating Eqs. (1.106, 1.107) and using the deﬁnitions (1.105) we ﬁnd

1 ρσ ρ 1 1 1 µν 1 µ U (Λ, a) ω J ε P U − (Λ, a)= ΛωΛ− J Λε ΛωΛ− a P (1.108) 2 ρσ − ρ 2 µν − − µ now, equating coeﬃcients of ωρσ on both sides of Eq. (1.108), we have

1 ρσ 1 1 1 µν 1 µ U (Λ, a) ω J U − (Λ, a) = ΛωΛ− J + ΛωΛ− a P 2 ρσ 2 µν µ 1 1 µν 1 1 ν 1 1 µ = ΛωΛ− J + ΛωΛ− a P + ΛωΛ− a P 2 µν 2 ν 2 µ ρσ 1 1 µν 1 ν 1 µ ωρσU (Λ, a) J U − (Λ, a)= ΛωΛ− µν J + ΛωΛ− a ν P + ΛωΛ− a µ P (1.109) manipulating the RHS of Eq. (1.109), and using Eq. (1.49) as well as the antisymmetry ofωρσ, we have 1 µν 1 ν 1 µ K ΛωΛ− J + ΛωΛ− a P + ΛωΛ− a P ≡ µν ν µ 1 µν 1 µ ν 1 ν µ = ΛωΛ− µν J + ΛωΛ− νµ a P + ΛωΛ− µν a P ρ 1 σ h µν σ i 1 ρh µ ν iρ 1 σ ν µ K = Λµ ωρσ Λ− ν J + Λν ωσρ Λ− µa P + Λµ ωρσ Λ− ν a P ρ σ µν σ ρ µ ν ρ σ ν µ = [Λµ ωρσΛν ] J + [Λν ωσρΛµ a ]P + [Λµ ωρσΛν a ] P = ω Λ ρΛ σJ µν [Λ σω Λ ρaµ] P ν + ω Λ ρΛ σaνP µ ρσ µ ν − ν ρσ µ ρσ µ ν K = ω Λ ρΛ σ [J µν aµP ν + aνP µ] (1.110) ρσ µ ν − 1.8. QUANTUM LORENTZ TRANSFORMATIONS 29

substituting (1.110) in (1.109), and taking into account that ωρσ is infinitesimal but otherwise arbitrary, we find ρσ 1 ρ σ µν µ ν ν µ U (Λ, a) J U − (Λ, a) = Λ Λ [J a P + a P ] µ ν − now, equating coefficients of ερ on both sides of Eq. (1.108), we have ρ 1 µ ρ µ ερU (Λ, a) P U − (Λ, a) = (Λε)µ P = (Λµ ερ) P ρ 1 ρ µ U (Λ, a) P U − (Λ, a) = Λµ P putting them together we have ρσ 1 ρ σ µν µ ν ν µ U (Λ, a) J U − (Λ, a) = Λµ Λν [J a P + a P ] (1.111) ρ 1 ρ µ − U (Λ, a) P U − (Λ, a) = Λµ P (1.112) For homogeneous Lorentz transformations (aµ = 0), Eqs. (1.111, 1.112) give ρσ 1 ρ σ µν ρ 1 ρ µ U (Λ) J U − (Λ) = Λµ Λν J ; U (Λ) P U − (Λ) = Λµ P (1.113) comparing with Eqs. (1.96, 1.97), the transformations (1.113) say that J µν is a second-rank tensor operator and µ µ ρ ρ P a four-vector operator. On the other hand, by using pure translations (ω ν = 0, Λµ = δµ ), Eqs. (1.111, 1.112) give ρσ 1 ρ σ µν µ ν ν µ ρσ ρ σ σ ρ U (1, a) J U − (1, a) = δµ δν [J a P + a P ]= J a P + a P ρ 1 ρ − − U (1, a) P U − (1, a) = P hence P µ is translation-invariant but J ρσ is not. In particular, by applying Eqs. (1.111, 1.112) to the space-space components of J ρσ (i.e. components of the type J ij), we obtain the usual change of angular momentum under a change of the origin relative to which we calculate such an angular momentum [homework!!(3)].

1.8.4 Lie algebra of the Poincarégenerators Now we apply transformations (1.111, 1.112) to the case in which U (Λ, a) is infinitesimal by itself, so that µ µ µ µ µ Λ ν = δ ν + ω ν , a = ε (1.114) where we use the notation (ω, ε), to emphasize that this infinitesimal parameters are totally independent of the ones previously used. By applying the expansions (1.99,e 1.102) on thee LHS of Eq. (1.111) and keeping terms up to first order in (ω, ε) we havee e

ρσ ρσ 1 1 αβ α ρσ 1 γδ γ J ′ U (1 + ω, ε) J U − (1 + ω, ε)= 1+ iω J iε P J 1 iω J + iε P ≡ e e 2 αβ − α − 2 γδ γ 1 1 = J ρσ +e ieω J αβJ ρσ iεe PeαJ ρσ 1 e iω J γδ +eiε P γ e e 2 αβ − α − 2 γδ γ 1 1 = J ρσ iωe J ρσJ γδ + iε eJ ρσP γ + iω J αβeJ ρσ iε eP αJ ρσ + ω2, ε2, ω ε − 2 γδ γ 2 αβ − α O · 1 1 = J ρσ + iω J αβJ ρσ iω J ρσJ αβ + iε J ρσP α iε P αJ ρσ + ω2, ε2, ω ε 2 eαβ − e2 αβ e α − e α O e e e ·e ρσ ρσ 1 αβ ρσ α ρσ 2 2 J ′ = J + iω J , J iε [P , J ]+ ω , ε , ω ε (1.115) 2 eαβ −e α eO e· e e e e h i Further, using (1.114) on the RHS of Eq. (1.111), and keeping terms up to ﬁrst order in (ω, ε) we have e e e e e e ρσ ρ ρ σ σ µν µ ν ν µ J ′ = (δµ + ωµ ) (δν + ων ) [J ε P + ε P ] − e e = (δ ρ + ω ρ) δ σ [J µν εµP ν + ενP µ] + (δ ρ + ω ρ) ω σJ µν + ω2, ε2, ω ε µ µ ν − µ µ ν O · = (δ ρ + ωe ρ) [J µσ eεµP σ + εσPeµ]+ ω σeJ ρν + ω2, ε2, ω ε µ µ − ν O · ρσ ρ µσe µ σ σ eµ ρe µσ σ ρν e e2 2 e e e e J ′ = δµ [J ε P + ε P ]+ ωµ J + ων J + ω , ε , ω ε (1.116) e− e e e Oe e e e· e e e e e e e e 30 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS

µν and using the symmetry of g and the antisymmetry of ωµν we ﬁnd

ρσ ρσ ρ σ σ ρ αρ µσ ασ ρν 2 2 J ′ = J ε P + ε P + ω g J + ω g J + ω , ε , ω ε − µα να O · = J ρσ gραε P σ + gσαε P ρ ω gραJ µσ ω gσαJ ρν + ω2, ε2, ω ε − α α − αµ − αν O · ρσ ρσ e σα eρ ραe σ ραe βσ σα ρβ e e e2 e2 J ′ = J + εα (g P g P ) ωαβ g J + g J + ω , ε , ω ε (1.117) e − e − e e O e e e· e Alternatively, from (1.116) wee can also write e e e e e ρσ ρσ ρ σ σ ρ νρ µσ σ ρµ 2 2 J ′ = J ε P + ε P + ω g J + ω J + ω , ε , ω ε − µν µ O · = J ρσ + ε (gσαP ρ gραP σ)+ ω gνρJ µσ + ω gνσJ ρµ + ω2, ε2, ω ε α − µν µν O · ρσ ρσ e σα eρ ραe σ νρe µσ νσ ρµe e e e2 2 J ′ = J + εα (g P g P )+ ωµν (g J + g J )+ ω , ε , ω ε (1.118) e − e e O e e e· e equating Eqs. (1.115, 1.117) we have e e e e e e 1 iω [J µν , J ρσ]+ iε [J ρσ, P µ]= ε (gσµP ρ gρµP σ) ω (gρµJ νσ + gσµJ ρν ) 2 µν µ µ − − µν since (ω, ε) are infinitesimale but otherwisee arbitrarye we equate coefficients ofe ω, and ε on both sides of this equation 1 e e i [J µν , J ρσ] = (gρµJ νσ + gσµJ ρν )e e (1.119) 2 − i [J ρσ, P µ] = (gσµP ρ gρµP σ) (1.120) − on the other hand, equating Eqs. (1.115, 1.118), we obtain the same condition with ε, but equating the coefficients of ωµν we find 1 i [J µν, J ρσ] = (gνρJ µσ + gνσJ ρµ)e (1.121) e 2 Equations (1.119) and (1.121) are essentially identical. We can put such equations in a more symmetrical form by adding them, from which we have

i [J µν , J ρσ]= (gρµJ νσ + gσµJ ρν) + (gνρJ µσ + gνσJ ρµ) − performing a similar procedure from Eq. (1.112) we reproduce the result (1.120) and obtain the additional condition [P µ, P ρ] = 0 (1.122) collecting all equations (1.121, 1.120, 1.122) and taking into account that gµν is symmetric, we obtain

i [J µν , J ρσ] = gνρJ µσ gµρJ νσ gσµJ ρν + gσν J ρµ (1.123) − − i [P µ, J ρσ] = gµρP σ gµσP ρ (1.124) − [P µ, P ρ] = 0 (1.125)

1.8.5 Physical interpretation of Poincare’s generators In some senses the physical interpretation of the operators J µν and P µ is easier in a three-dimensional notation. Conserved quantities in quantum mechanics are related with operators that commute with the Hamiltonian or energy operator H = P 0. We deﬁne then the momentum three-vector

P P 1, P 2, P 3 (1.126) ≡ and the angular momentum three-vector

J J 23, J 31, J 12 J , J , J J km = ε J (1.127) ≡ ≡ { 1 2 3} ⇔ kmn n 1.8. QUANTUM LORENTZ TRANSFORMATIONS 31 the energy operator (Hamiltonian) H = P 0 (1.128) and the remaining generators form what is called the “boost” three-vector

K J 01, J 02, J 03 K , K , K J 0i = K (1.129) ≡ ≡ { 1 2 3} ⇔ i So we have all ten degrees of freedom (the remaining components of J µν are not independent because of the antisymmetry of it). In a three-dimensional notation, the commutation relations (1.123, 1.124, 1.125) can be written as [homework!!(4)]

[Ji, Jj ] = iεijkJk (1.130)

[Ji, Kj] = iεijkKk (1.131) [K , K ] = iε J (1.132) i j − ijk k [Ji, Pj ] = iεijkPk (1.133) [K , P ] = iHδ (1.134) i j − ij [Ji,H] = [Pi,H] = [H,H] = 0 (1.135) [K ,H] = iP (1.136) i − i [Pi, Pj ] = 0 (1.137) from relations (1.135) we can observe that operators P, J commute with H, and so they are conserved. However Eq. (1.136) shows that K is not conserved. Consequently, we shall not use the eigenvalues of K to label physical states. The commutation relations (1.130) forms the well-known Lie algebra of angular momentum operators. Moreover commutation relations (1.133, 1.137), coincides with the ones expected between a linear momentum operator and an orbital angular momentum operator. Commutation relations (1.130), show that the set of generators Ji forms a closed algebra, such an algebra in turn generates a subgroup [the subgroup of three-dimensional rotations SO (3)]. In the same way, commutation relations (1.137) deﬁne a closed algebra of the Pi generators, since these generators commute with each other, the subgroup generated (the subgroup of space translations), is abelian. However, we usually work with the subgroup of space-time translations instead of the group of space translations. Further, commutation relations (1.132) show that boosts generators do not form a closed algebra and do not generate a subgroup. Indeed, Eq. (1.132) shows the well-known feature that two succesive boosts might generate a rotation. For Λ = 1, we obtain pure space-time translations. The set T (1, a) of all pure translations forms an abelian subgroup of the inhomogeneous Lorentz group with a group multiplication rule given by Eq. (1.48)

T (1, a¯) T (1, a)= T (1, a¯ + a) (1.138) considering that this subgroup is characterized by only four parameters aµ, the multiplication rule is given by a relation of the type (1.36) as it corresponds to an additive abelian subgroup. Hence we can use Eq. (1.39) valid for rules of multiplication of the type (1.36). Thus, any ﬁnite translation is represented on the physics Hilbert space by U (1, a) = exp ( iP µa ) (1.139) − µ in the same way, a rotation Rθ by an angle θ around the direction θ, is represented on the physical Hilbert space | | by U (Rθ, 0) = exp (iJ θ) (1.140) · to obtain (1.140) the argument is the same as to obtain (1.139), if we take into account that rotations around a ﬁxed axis commute each other. 32 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS

1.9 One-particle states

We shall classify one-particle states according with their transformation properties under inhomogeneous Lorentz transformations. Equation (1.125), shows that the components of the four-momentum commute with each other. Therefore, they admit a complete set of common eigenstates that we denote as p,σ , where σ denotes the remaining degrees | i of freedom. Then our starting point will be momentum eigenstates

P µ p,σ = pµ p,σ (1.141) | i | i in the most general states (for instance those describing several unbound particles), the remaining labels could be either discrete or continuous and even both. Nevertheless, we shall take as part of the definition of a one-particle state, that the label σ is purely discrete, and we shall restrict ourselves to study that case. In particular, specific bound states of two or more particles, such as the lowest state of the hydrogen atom, is to be considered as a one-particle state though this is not an elementary particle. By now we shall not distinguish between composite or elementary particles. On the other hand, all eigenstates of P with a given eigenvalue p (plus the null vector) form a subpace of 0 Ep0 the Hilbert space (see section 1.1.1). The dimensionality of such a subspace is equal to the degree of degeneracy of the eigenvalue p , that is the number of linearly independent vectors of the form p ,σ for a fixed p . 0 | 0 i 0 1.9.1 One-particle states under pure translations We shall start the characterization of states of the type p,σ , under inhomogeneous Lorentz transformations. | i From Eqs. (1.141, 1.139) we see the action of pure translations over those states

µ ipa U (1, a) p,σ = exp [ ia P ] p,σ = e− p,σ (1.142) | i − µ | i | i therefore, p,σ is an eigenstate of pure translation operators U (1, a). In addition, we observe that is the | i Ep subspace induced by each eigenvalue e ipa. Even more, the one-dimensional subspaces are invariant9 (and − Ep,σ of course minimal) under the representation U (1, a) of the subgroup of pure translations. Since any given { } minimal subspace invariant under U (1, a) is one-dimensional, we say that the vectors p,σ are singlets with { } | i respect to U (1, a) . This is consistent with the fact that all irreducible representations of any abelian group are { } one-dimensional (see theorem 1.2 page 1.2).

1.9.2 One-particle states under homogeneous Lorentz transformations As for quantum homogeneous Lorentz transformations U (Λ, 0) U (Λ), equation (1.112) shows that they trans- ≡ form a state p,σ in another momentum eigenstate but with eigenvalue Λp | i µ 1 µ 1 µ P U (Λ) p,σ = U (Λ) U − (Λ) P U (Λ) p,σ = U (Λ) U − (Λ) P U (Λ) p,σ | i | i | i 1 µ 1 1 1 µ ρ = U (Λ) U Λ− P U − Λ− p,σ = U (Λ) Λ− P p,σ | i ρ | i µ ρ = Λ ρp [U (Λ) p,σ ] h i | i P µ [U (Λ) p,σ ] = (Λp)µ [U (Λ) p,σ ] | i | i hence U (Λ) p,σ must be an eigenstate of P with eigenvalue Λp. From the previous facts, we could say that | i U (Λ) p,σ belongs to the subspace . Hence it is a linear combination of a basis within this subspace | i EΛp

U (Λ) p,σ = C ′ (Λ,p) Λp,σ′ (1.143) | i σ ,σ σ′ X 9 The subspace Ep,σ consists of all vectors of the form α |p,σi, with (p,σ) fixed and with α running over all complex scalars. 1.9. ONE-PARTICLE STATES 33 equation (1.143) shows that, by using the basis p,σ we can find a matrix representation of U (Λ), i.e. a {| i} matrix representation of the Lorentz group in the Hilbert space . The idea now is to characterize the irreducible E representations and also the minimal invariant subspaces of under U (Λ, a) . Therefore, we should look for the E { } apropriate canonical basis in which the reduction is apparent. It is natural to associate states of a specific particle type with the components of a irreducible representation of the inhomogeneous Lorentz group. It could happen that different particle species may correspond to isomorphic representations10. First we note that a subspace of the type is not invariant under U (Λ) , since a given element p,σ is Ep { } | i∈Ep mapped through U (Λ) into an element that belongs to another subspace . Notice however that this mapping EΛp preserves the norm (or pseudonorm) of the four-vector p i.e. the quantity p2 g pµpν. In addition, if p2 0 ≡ µν ≤ then the sign of p0 is also preserved [homework!!(5)]. It is then convenient for each value of p2 and (for p2 0) ≤ each sign of p0 to choose a standard four-momentum kµ. Hence we write any pµ of this class11 as

µ µ ν p = L ν (p) k (1.144)

µ where L ν is a Lorentz transformation that connects two four-vectors within the same class. It is clear that such a transformation must depend on p, but also implicitly on the chosen standard vector k. Equation (1.143) shows that a quantum operator U (Λ) takes p into another element of the same class. Then we can define the states p,σ of momentum p, as a transformation from the reference point k,σ such that | i | i p,σ N (p) U (L (p)) k,σ (1.145) | i≡ | i where N (p) is a constant of normalization that we shall choose later. By comparing Eqs. (1.144, 1.145) we see that if we associate pµ of the Minkowski space with p,σ of the Hilbert space, and the standard vector kµ is | i associated with k,σ , it is logical that the transformation L (p) on the Minkowski space that connects k with p | i should have an associated operator U (L (p)) that connects k,σ with p,σ , except for a possible normalization | i | i constant. Note that the defined operator U (L (p)) only transforms the momenta degrees of freedom but not the other degrees of freedom σ. Indeed, equation (1.145), says how the σ labels are related for different momenta. Applying a homogeneous Lorentz transformation U (Λ) on Eq. (1.145) we find

U (Λ) p,σ N (p) U (Λ) U (L (p)) k,σ = N (p) U (ΛL (p)) k,σ | i ≡ 1| i | i = N (p) U (L (Λp)) U − (L (Λp)) U (ΛL (p)) k,σ 1 | i = N (p) U (L (Λp)) U L− (Λp) U (ΛL (p)) k,σ 1 | i U (Λ) p,σ = N (p) U (L (Λp)) U L− (Λp) ΛL (p) k,σ (1.146) | i | i 1 observe that the Lorentz transformation L− (Λp) ΛL (p) transforms k into itself as can be seen by applying Eq. (1.144)

1 1 1 L− (Λp) ΛL (p) k = L− (Λp) Λ L (p) k = L− (Λp) Λp 1 = L− (Λp) (Λp)= k so it belongs to the set of all Lorentz transformations that leave kµ invariant

µ ν µ W νk = k (1.147)

10Sometimes it is even convenient to define particle types as irreducible representations of a group that contains the proper orthochronus Lorentz group as a proper subgroup. For example, we shall see later that for massless particles with space inversion symmetry, it is usual to associate a given irreducible representation of the proper orthochronus inhomogeneous Lorentz group including space inversion, with a single-particle type. 11We are forming a partition of the set of all four-vectors {p} (that is, a disjoint collection of subsets whose union equals the set). 2 b 2 2 For p > 0, a subset Sp2 (or class) of this collection consist of all vectors with a fixed (positive) norm p . For p ≤ 0, we define a+ 2 0 a− partitions of the form Sp2 consisting of all vectors with a fixed (negative) norm p and p > 0. Finally a class Sp2 consists of all vectors with a fixed (negative) norm p2 and p0 < 0. Of course, to obtain the set of all {p} we must run over all values of p2 within each subset. 34 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS this set forms a group [homework!!(6)] known as the little group. For any W belonging to the Little group i.e. satisfying (1.147), equation (1.143) becomes

U (W ) k,σ = D ′ (W ) k,σ′ (1.148) | i σ σ σ′ X

We have said that subspaces of the type are not invariant under homogeneous Lorentz transformations. Ep Equation (1.148), shows that by restricting to the little group W (in the Minkowski space), we are generating invariant subspaces under the representation U (W ) of the little group in the Hilbert space. Ek It can be seen that the coeﬃcients D (W ) provides a representation of the little group, we see it as follows

D ′ WW k,σ′ = U WW k,σ = U W U (W ) k,σ σ σ | i | i σ′ X

= U W Dσ′′σ (W ) k,σ′′ = Dσ′′σ (W ) U W k,σ′′ σ′′ σ′′ X X

= Dσ′′σ (W ) Dσ′σ′′ W k,σ′ = Dσ′σ′′ W Dσ′′σ (W ) k,σ′ σ′′ " σ′ # σ′ (" σ′′ # ) X X X X ′ Dσ σ WW k,σ′ = D W D (W ) σ′σ k,σ′ σ′ σ′ X X and resorting to the linear independence of states k,σ we ﬁnally obtain | ′i

′ Dσ σ WW = D W D (W ) σ′σ we have already seen that the transformation

1 W (Λ,p) L− (Λp) Λ L (p) (1.149) ≡ belongs to the little group. Substituting (1.149) in Eq. (1.146), and using (1.148) we have

U (Λ) p,σ = N (p) U (L (Λp)) [U (W (Λ,p)) k,σ ]= N (p) U (L (Λp)) D ′ (W (Λ,p)) k,σ′ | i | i σ σ " σ′ # X

U (Λ) p,σ = N (p) D ′ (W (Λ,p)) U (L (Λp)) k,σ′ | i σ σ " σ′ # X and from the deﬁnition (1.145) we obtain

Λp,σ′ U (Λ) p,σ = N (p) D ′ (W (Λ,p)) U (L (Λp)) k,σ′ = N (p) D ′ (W (Λ,p)) | i | i σ σ σ σ N (Λp) σ′ σ′ X X N (p) U (Λ) p,σ = D ′ (W (Λ,p)) Λp,σ′ (1.150) | i N (Λp) σ σ σ′ X comparing Eq. (1.150) with Eq. (1.143) we observe that apart from the problem of the normalization, we have replaced the problem of determining the coeﬃcients Cσ,σ′ in the transformation rule (1.143), to the (reduced) problem of determining the coeﬃcients Dσ,σ′ which are the matrix elements associated with the representation of the little group [see Eq. (1.148)]. This approach of deriving representations of a group (the homogeneous Lorentz group in our case) from the representations of a little group, is called the induced representation method. 1.9. ONE-PARTICLE STATES 35

Type of pµ Standard kµ Little group (a) p2 = M 2 < 0, p0 > 0 (0, 0, 0, M) SO (3) − (b) p2 = M 2 < 0, p0 < 0 (0, 0, 0, M) SO (3) − − (c) p2 = 0, p0 > 0 (0, 0, κ, κ) ISO (2) (d) p2 = 0, p0 < 0 (0, 0, κ, κ) ISO (2) − (e) p2 = N 2 > 0 (0, 0, N, 0) SO (2, 1) (f) pµ = 0 (0, 0, 0, 0) SO (3, 1) Table 1.1: This table displays the six types of classes of four-momenta. We also display a convenient choice for the standard four-vector kµ, and the most general subgroup of the proper orthochronus Lorentz group that leave kµ invariant (little group). The quantities M,N, κ, are all positive.

1.9.3 Physical little groups

12 a+ It is clear that the partition we have defined consists of six types of classes : (a) classes of the type Sp2 in which 2 2 0 a 2 2 0 a+ p = M < 0, and p > 0. (b) classes of the type Sp2− in which p = M < 0, and p < 0. (c) classes S0 with 2 − 0 a 2 0 − b 2 2 p = 0, and p > 0. (d) classes S0− with p = 0, and p < 0. (e) Classes Sp2 with p = N > 0, finally (f) the 0 µ class S0 with p = 0. The numbers M and N are positive. Now, it is well-known that the case p2 > 0 lead to a non-physical mass [see Eq. (1.70) and discussion below], and p2 0 with p0 < 0 is a non-physical configuration. Therefore, only the cases (a), (c), (f) leads to an ≤ interpretation in terms of physical states (as far as we know). For each class we want to establish a suitable standard vector kµ and the associated little groups. For the class (a) with p2 = M 2 < 0, and p0 > 0 a natural standard vector is kµ (0, 0, 0, M). The little − ≡ group is the set of all homogeneous proper orthochronus Lorentz transformations that leave the vector (0, 0, 0, M) invariant. This kµ describes a particle with non-null mass at rest. It is a fact that a proper orthochronus Lorentz transformation can always be expressed as a composition of a boost followed by a rotation in the three space coordinates. From physical grounds it is pretty obvious that a transformation containing a boost does not leave this four-vector invariant, since in that case we are changing to another reference frame with relative velocity with respect to the first, in this new reference frame the particle is not at rest any more. By contrast, a rotation of the three space coordinates keeps the particle at rest and does not alter the rest mass or the rest energy of the particle13. Consequently, the little group will be SO (3) which is the group of all continuous rotations in three dimensions. For the case (c) with p2 = 0, p0 > 0, a good choice is kµ = (0, 0, κ, κ) that describes a particle of null rest mass, µ which travels to the speed of light along the axis x3, the vector k is invariant under rotations around the axis x3 (two dimensional rotations). We shall see later that its little group is ISO (2) which is the group of euclidean geometry consisting of rotations and translations in two dimensions. The case (f) is trivial, the condition pµ = 0 indicates the vacuum (no particles at all) which is left invariant by any U (Λ), hence the little group is the whole proper orthochronus Lorentz group in 3 + 1 dimensions (three space + 1 time coordinates). We symbolize it as SO (3, 1). Table 1.1, shows the six types of classes with the apropriate standard vectors and little groups. The little groups SO (2, 1) and SO (3, 1) for p2 > 0 and pµ = 0 have no non-trivial finite dimensional unitary representations, so if we had states with a given momentum pµ with p2 > 0 or pµ = 0 that transform non-trivially under the little group, there would have to be an infinite number of them. However, in what follows we only consider classes of the type (a) and (c) (with p2 0 and pµ = 0), which are the only ones that posseses a ≤ 6 non-trivial Physical interpretation.

12 b The number of classes is inﬁnite. For instance, we have as much classes of the type Sp2 , as the number of diﬀerent positive values of p2. The number of classes is then continuous. 13The invariance of the value of p2 and the sign of p0 are valid for arbitrary proper orthochronus Lorentz transformations. Therefore, they are valid for the little group in particular. 36 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS

1.9.4 Normalization of one-particle states Now we should take over the problem of the normalization of the states. A convenient starting point is to choose the states with standard momentum kµ to be orthonormal in the extended sense, that is14

3 k′,σ′ k,σ = δ k′ k δ ′ (1.151) i − σ σ from condition (1.151), the representation of the little group in Eqs. (1.148, 1.150) must be unitary 1 D† (W )= D− (W ) now we wonder about the scalar products between states of arbitrary momenta. We can calculate that inner product by using the unitarity of U (Λ) in Eqs. (1.145, 1.150)

1 p′,σ′ p,σ = N (p) p′,σ′ U (L (p)) k,σ = N (p) k,σ U † (L (p)) p′,σ′ ∗ = N (p) k,σ U − (L (p)) p′,σ′ ∗ | i | i h | h | 1 = N (p) k,σ U L− (p) p′,σ′ ∗ h | ∗ N (p′) 1 1 = N (p) k,σ D ′′ ′ W L− (p) ,p′ L− (p) p′,σ′′ h | N (L 1 (p) p ) σ σ ( " − ′ σ′′ #) X N ∗ (p′) 1 = N (p) D∗′′ ′ W L− (p) ,p′ k,σ k′,σ′′ ∗ N (L 1 (p) p ) σ σ h | ∗ − ′ σ′′ X N (p) N ∗ (p′) 1 1 p′,σ′ p,σ = D∗′′ ′ W L− (p) ,p′ k′,σ′′ k,σ ; k′ L− (p) p′ | i N (k ) σ σ i ≡ ∗ ′ σ′′ X and then using the orthonormalization condition (1.151) we ﬁnd

N (p) N ∗ (p′) 1 3 p′,σ′ p,σ = D∗′′ ′ W L− (p) ,p′ δ k′ k δ ′′ | i N (k ) σ σ − σ σ ∗ ′ σ′′ X N (p) N ∗ (p′) 1 3 1 p′,σ′ p,σ = Dσσ∗ ′ W L− (p) ,p′ δ k′ k ; k′ L− (p) p′ (1.152) | i N ∗ (k′) − ≡ 1 finally by the definition (1.145), we have N (k′) = 1. Further, since also k = L− (p) p, then k = k′ if and only 3 3 if p = p . Therefore, the delta function δ k k′ is proportional to δ (p p ). As a consequence, the inner ′ − − ′ product (1.152) is non-null only for p = p . Now, for p = p , the little group transformation W L 1 (p) ,p here ′ ′ − ′ is trivial. To see it we use definition (1.149) to write 1 W (Λ,p) L− (Λp) Λ L (p) 1 ≡ 1 1 1⇒ 1 1 1 W L− (p) ,p = L− L− (p) p L− (p) L (p)= L− L− (p) p = L− (k)

1 but deﬁnition (1.144) with p = k, shows that L (k) must be the identity and so L− (k), therefore

1 W L− (p) ,p = 1 from these facts, the scalar product in (1.152) becomes

2 3 1 p′,σ′ p,σ = N (p) δ ′ δ k′ k ; k′ L− (p) p′ (1.153) | i | | σσ − ≡ nevertheless, it is desirable to write the inner product in terms ofδ (p p ). Thus, we should ﬁnd the constant of − ′ proportionality that relates δ3 (k k) with δ3 (p p). Note that the Lorentz-invariant integral of an arbitrary ′ − ′ − 14We are writing hk′,σ′| k,σi as proportional to δ (k − k′) instead of δ (k − k′). It is because given the three spatial components, the constraint k2 = −M 2, along with the condition k0 > 0 (or k0 < 0), provides us the zeroth temporal component. Hence k = k′ is equivalent to k = k′. 1.9. ONE-PARTICLE STATES 37 scalar (Lorentz-invariant) function f (p) over four-momenta with p2 = M 2 0 and p0 > 0, which corresponds − ≥ to cases (a) and (c) in table 1.1, can be written as

I = d4p δ p2 + M 2 θ p0 f (p) Z 2 I = d3p dp0 δ p0 p2 + M 2 θ p0 f p,p0 (1.154) − Z where the delta function ensures the satisfaction of the condition p2 = M 2, and the step function guarantees − that only the terms with p0 > 0 contribute in the integral. We can rewrite the integral by using the property 1 δ x2 e2 = [δ (x + e)+ δ (x e)] (1.155) − 2 e − | | hence performing the integral over p0 in (1.154) by using (1.155), we obtain

δ p0 p2 + M 2 + δ p0 + p2 + M 2 I = d3p dp0 − θ p0 f p,p0 h p 2 2 p i Z 2 p + M p since θ p0 forbids negative values of p0, the term δ p0 + p2 + M 2 does not contribute, hence p δ p0 p2 + M 2 I = d3p dp0 − θ p0 f p,p0 p2 2 Z 2 p + M f p, pp2 + M 2 I = d3p (1.156) p2 2 Z 2 p + M since we have taken the positive square root for p0, the step function is not necessary anymore. When integrating f p,p0 on the “mass shell” p2 + M 2 = 0, the integral (1.156) shows that the invariant volume element is

d3p (1.157) p2 + M 2 the delta function is deﬁned as p

3 3 3 2 2 3 d p′ F (p)= F p′ δ p p′ d p′ = F p′ p′ + M δ p′ p (1.158) − − p 2 2 Z Z ′ + M hp i hence combining Eqs. (1.158, 1.157), we see that the invariant delta function is p

2 2 3 0 3 p + M δ p′ p = p δ p′ p (1.159) ′ − − thus the form of the “new” delta functionp deﬁned in (1.159), must be preserved under a Lorentz transformation. As a consequence, since p′ and p are related with k′ and k, by the same Lorentz transformation L (p), we have

0 3 0 3 p δ p′ p = k δ k′ k (1.160) − − which is the relation between δ3 (p p) and δ3(k k) that we were looking for. Substituting (1.160) in (1.153), ′ − ′ − the inner product becomes 0 2 p 3 p′,σ′ p,σ = N (p) δ ′ δ p′ p | i | | σσ k0 − 38 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS sometimes the normalization is chosen such that N (p) = 1. However, in that case we should keep track of the factor p0/k0 in scalar products. The most usual convention (which is the one we adopt here) is that

k0 N (p)= (1.161) s p0 with this choice, the ﬁnal form of the inner product yields 3 p′,σ′ p,σ = δ ′ δ p′ p (1.162) | i σσ − we shall study now the cases (a) and (c) of table 1.1, that is the case of a particle with non-null mass M > 0, and the case of particles with zero mass p2 = 0.

1.10 One-particle states with non-null mass

When p2 < 0, table 1.1 shows that the little group is SO (3). For this group, all irreducible representations are finite-dimensional and can be settled as unitary15. Further, a unitary representation can be broken up into a direct sum of irreducible unitary representations. The irreducible representations characterized by the matrices (j) Dσ′σ (R) have dimensionality 2j + 1, with j = 0, 1/2, 1, 3/2, 2,.... Infinitesimal rotations can be written in the form (1.88) R = δ +Θ ;Θ = Θ ik ik ik ik − ki the antisymmetry comes from Eq. (1.89), and the fact that SO (3) is a subgroup of SO (3, 1). Considered as a subgroup of SO (3, 1), we can write the infinitesimal rotation as in Eq. (1.99), but taking into account that for pure rotations, the only generators that appear in that expansion are the angular momentum operators defined in Eq. (1.127) 1 1 1 1 U (1+Θ, 0) = 1+ iΘ J ik =1+ iΘ J 23 + iΘ J 31 + iΘ J 12 2 ik 2 23 2 31 2 12 1 1 1 U (1+Θ, 0) = 1+ iΘ J + iΘ J + iΘ J (1.163) 2 1 1 2 2 2 2 3 3 J J 23, J 31, J 12 J , J , J ; Θ Θ , Θ , Θ Θ , Θ , Θ ≡ ≡ { 1 2 3} ≡ { 23 31 12} ≡ { 1 2 3} hence, the matrix representative associated with the (j) representation reads − (j) i (j) i (j) D ′ (1+Θ) = δσ′,σ + Θik Jik = δσ′,σ + Θp Jp (1.164) σ ,σ 2 σ′σ 2 σ′σ the canonical basis of 2j + 1 orthonormal vectors for the (j) representation is denoted by j, σ ; σ = j, j 1, j 2,..., j {| i − − − } and the matrix representations of the generators in the canonical basis are well-known from the theory of angular momenta (j) 1 J1 = j (j + 1) σ (σ + 1) δσ′,σ+1 + j (j + 1) σ (σ 1) δσ′,σ 1 (1.165) σ′σ 2 − − − − (j) 1hp p i J2 = j (j + 1) σ (σ + 1) δσ′,σ+1 j (j + 1) σ (σ 1) δσ′,σ 1 (1.166) σ′σ 2i − − − − − (j) h(jp) (j) (j) (j) p i J J23 iJ31 = J1 iJ2 = δσ′,σ 1 (j σ) (j σ + 1) (1.167) ± σ′σ ≡ ± σ′σ ± σ′σ ± ∓ ± (j) (j) p J12 = J3 = σ δσ′,σ (1.168) σ′σ σ′,σ σ = j, j 1, j 2,..., j − − − 15It means that, if a given irreducible representation of SO (3) is non-unitary, it could become unitary through a similarity transformation. 1.10. ONE-PARTICLE STATES WITH NON-NULL MASS 39

For future purposes we shall construct a rotation R (p) that takes the three-axis into the direction of the three- vector p. The determination of a direction requires two angles, in spherical coordinates we use the azimuthal φ and polar θ angles (both are measured with respect to the three-axis).b The direction of a unitary vector determined by these angles is specified as p p = sin θ cos φ u + sin θ sin φ u + cos θ u (1.169) ≡ p 1 2 3 | | it is clear on geometrical groundsb that the rotation R (p) can be carried out as follows: (a) a rotation R2 (θ) around X2 by the angle θ [it takes (0, 0, 1) into (sin θ, 0, cos θ)], and then (b) a rotation R3 (φ) around X3 by an angle φ [it takes the vector (sin θ, 0, cos θ) into the direction definedb by (1.169)]. Let us do this process explicitly (for simplicity, we shall only use the three-space coordinates), the first rotation R2 (θ) clearly gives cos θ 0 sin θ 0 sin θ R (θ) u = 0 1 0 0 = 0 2 3       sin θ 0 cos θ 1 cos θ −       the second rotation R3 (φ) is described by

sin θ cos φ sin φ 0 sin θ cos φ sin θ − R (φ) 0 = sin φ cos φ 0 0 = sin θ sin φ = p 3         cos θ 0 0 1 cos θ cos θ         b the complete process is then R (p) u3 = R3 (φ) R2 (θ) u3 = p (1.170) the complete matrix of rotation yields b b cos φ sin φ 0 cos θ 0 sin θ − R (p) = R (φ) R (θ)= sin φ cos φ 0 0 1 0 3 2     0 0 1 sin θ 0 cos θ − b cos θ cos φ sin φ cos φ sin θ    − R (p) = cos θ sin φ cos φ sin θ sin φ (1.171)   sin θ 0 cos θ − b   it is easy to check that its inverse corresponds to the transpose as it must be for three-dimensional rotations

cos θ cos φ cos θ sin φ sin θ 1 − R− (p)= R (p)= sin φ cos φ 0 (1.172)  −  cos φ sin θ sin θ sin φ cos θ b e b   it is convenient to write R (p) in terms of the cartesian components pi of the unitary vector p. According with Eq. (1.169) such components yield b b b p1 = sin θ cos φ , p2 = sin θ sin φ , p3 = cos θ (1.173) from Eq. (1.173) the matrix in (1.171) becomes b b b

pb3pb1 pb2 sin θ sin θ p1 b b − b R (p)= p3p2 p1 p (1.174)  sin θ sin θ 2  sin θ 0 pb3 − b  b  using (1.173) again, we have b p2 + p2 = sin2 θ = p2 p2 = 1 p2 1 2 − 3 − 3 b b b b b 40 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS since 0 θ π then sin θ 0, so that sin θ = √sin2 θ. Therefore ≤ ≤ ≥ sin θ = 1 p2 (1.175) − 3 substituting (1.175) in (1.174) we have q b pb1pb3 pb2 p1 √1 pb2 √1 pb2 − 3 − − 3 pb2pb3 pb1 R (p)=  p2  (1.176) √1 pb2 √1 pb2 − 3 − 3 b  1 p2 0 p   − − 3 3  b  b  Finally, taking into account that the transformationp is embedded in the four-dimensional Minkowski space Eqs. (1.171, 1.172, 1.176) must be written as b b

pb1pb3 pb2 p1 0 cos θ cos φ sin φ cos φ sin θ 0 √1 pb2 √1 pb2 − 3 − − 3 − pb2pb3 pb1 cos θ sin φ cos φ sin θ sin φ 0  p2 0  R (p) =   = √1 pb2 √1 pb2 (1.177) sin θ 0 cos θ 0 − 3 − 3 b −  1 p2 0 p 0   0 0 01   − − 3 3     0 0b 01  b    p  b b b b  p1p3 b p2p3 b 2  b2 b2 1 p3 0 cos θ cos φ cos θ sin φ sin θ 0 √1 p3 √1 p3 − − − −b b− sin φ cos φ 0 0 p2 p1 1  b2 b2 p 0 0  R− (p) =  −  = −√1 p3 √1 p3 (1.178) cos φ sin θ sin θ sin φ cos θ 0 − − b  p p p 0   0 0 01   1 2 3     0 0 01  b       The rotation R (p) is carried out at the Minkowski four-dimensionalb space, thb ough it actsb non-trivially only within the three-space coordinates. According with Eq. (1.170), the corresponding representation on the Hilbert space becomes b U (R (p)) = U (R3 (φ) R2 (θ)) = U (R3 (φ)) U (R2 (θ)) which in terms of the generators is written as b U (R (p)) = exp ( iφJ ) exp ( iθJ ) ; 0 θ π, 0 φ< 2π (1.179) − 3 − 2 ≤ ≤ ≤ there is an important difference between the rotation itself and their representations: in representations associated with integer values of (j) we obtainb the same element by shifting θ or φ by 2π; by contrast, in representations associated with half-integer values of (j), we obtain the element with opposite sign by shifting θ or φ by 2π. Thus the only representations that can describe geometrical objects are integer representations, while half-integer representations can only describe intrinsic variables16. Representations in quantum mechanics such as Eq. (1.179), could be associated either to integer or half-integer representations. Since the rotation R (p) described by (1.177) has the role of taking the three-axis into the direction (1.169), it is clear that we could have added an initial rotation around the three-axis without affecting that result. Thus, any other choice on R (p) in theb Minkowski space would differ from this one at most in an initial rotation around the three-axis. For the quantum operator U (R (p)) in Eq. (1.179), such a difference leads just to a mere redefinition of the phase of theb one-particle states. b 1.10.1 Wigner rotation and standard boost For a particle of mass M > 0, and spin j, Eq. (1.150) combined with normalization (1.161), becomes17

0 (Λp) (j) U (Λ) p,σ = D ′ (W (Λ,p)) Λp,σ′ (1.180) | i s p0 σ σ σ′ X 16 It is because of these reasons that the spherical harmonics Yl,m (θ,φ) only admit integer values of l and m. 17By now, we say “spin” referring to the label (j) associated with a given irreducible representation of SO (3). 1.10. ONE-PARTICLE STATES WITH NON-NULL MASS 41 now we shall calculate the little-group element (Wigner rotation) W (Λ,p) deﬁned by Eq. (1.149)

1 W (Λ,p)= L− (Λp) Λ L (p) (1.181) to calculate this Wigner rotation, we should choose a “standard boost” L (p) that carries the “standard” four momentum kµ = (0, 0, 0, M) to pµ = p1,p2,p3,p0 [as required by Eq. (1.144)]. Indeed, it is easier to construct the inverse transformation 1 k = L− (p) p in Eq. (1.177) we have characterized the three-dimensional rotation R (p) that takes the three-axis into the 1 direction of the unitary vector p. Thus, R− (p) applied on p preserves the norm of the three-vector and put it along with the three-axis, while the time-component remains invariant b b b1 0 R− (p) p = 0, 0, p ,p | | next we construct a boost B 1 ( p ) that eliminates the third component of the new four-vector. It is clear that the − | | b first and second components do not require modification. Since the pseudonorm must be preserved and the final vector has no space components, the time-component must be the rest mass M of the particle (otherwise it would not be a Lorentz transformation). It is clearly carried out by a pure boost B 1 ( p ) along the three-component − | | to a new reference frame in which the particle is at rest, so we find

0 10 0 0 0 0 0 01 0 0 0 1 0 0 0 0 B− ( p )   =  p0 p    =  p p 0  =   | | p 0 0 | | p M p |M| p 0 M M | |− 2 | 0|  p −p0  | 0|  2 (p0)   p   0 0 | |   p   p +   M     − M M     M M           −    then we have already arrived to kµ by using the sequence

1 1 k = B− ( p ) R− (p) p (1.182) | | Moreover, since kµ contains a null three-vector, it is clear that we can add an arbitrary three-dimensional rotation, b and we shall not alter kµ. For reasons to be understood later, we shall add the rotation R (p) to the previous sequence of transformations. Then we ﬁnd

1 1 1 b k = L− (p) p R (p) B− ( p ) R− (p) p (1.183) ≡ | | on the other hand, we write B 1 as a function of p because p0 is not independent of p . Such a fact is more − | | b b | | apparent if we deﬁne

2 2 0 2 2 2 p + M p 2 p + M M p γ = γ 1= 2 2 = | | (1.184) ≡ p M M ⇒ − r M − M M p so that B 1 ( p ) can be written as a function of the single parameter γ − | | 10 0 0

1 01 0 0 B− ( p )=   | | 0 0 γ γ2 1 − −  2   0 0 γ 1 p γ   − −  and B ( p ) is easily obtain p | | 10 0 0 01 0 0 B ( p )=   (1.185) | | 0 0 γ γ2 1 −  2   0 0 γ 1 p γ   −  p 42 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS from Eq. (1.183) we ﬁnally obtain the standard boost L (p)

1 L (p)= R (p) B ( p ) R− (p) (1.186) | | whose explicit form is obtained by picking up Eqs. (1.177,b 1.178, 1.185)b

pb1pb3 pb2 p1 0 √1 pb2 √1 pb2 − 3 − − 3 pb2pb3 pb1  p2 0  L (p) = √1 pb2 √1 pb2 − 3 − 3 b ×  1 p2 0 p 0   − − 3 3   0 0b 01   p    pb pb pb pb b b 1 3 2 3 1 p2 0 10 0 0 b2 b2 3 √1 p3 √1 p3 − − −b b− 01 0 0 p2 p1 0 0  b2 b2 p    −√1 p3 √1 p3 × 0 0 γ γ2 1 − − b −  p1 p2 p3 0   2     0 0 γ 1 p γ   0 0 01   −    p  b b b  b b b b b b b p1p3 p2 p1p3 p2p3 1 p2 0 b2 b2 p1 0 b2 b2 3 √1 p3 −√1 p3 √1 p3 √1 p3 − − b −b b − −b b− p2p3 p1 p 0 p2 p1 p 0 0  b2 b2 2   b2 b2  L (p)= √1 p √1 p −√1 p3 √1 p3 − 3 − 3 b − − b 2 2  1 p 0 p 0   γp1 γp2 γp3 γ 1   − − 3 3   −   0 0b 01   p γ2 1 p γ2 1 p γ2 1 γ   p   1 2 3 p     − − −  b b p b p b p b (pb2+pb2pb2+γpb2 γpb2pb2) b b b 2 1 3 1− 1 3 2 1 pb2 p1p2 (γ 1) p1p3 (γ 1) p1 γ 1 − 3 − − − (pb2+pb2pb2+γpb2 γpb2pb2)  1 2 3 2− 2 3 p 2  L (p)= p1p2 (γ 1) 1 pb2 p2p3 (γ 1) p2 γ 1 (1.187) − b b − 3 b b − b −  2 2   p1p3 (γ 1) p2p3 (γ 1) 1 + (γ 1) p3 p3pγ 1   2 − 2 − −2 −   bp1b γ 1 p2 γ 1 bp3b γ 1 b γ   − − − p   b b b b b b  p1 2 p p the ﬁrst two diagonal elementsbL 1 (p) and L 2 (p) canb be simpliﬁed asb follows

1 1 L1 (p) = 1 p2 − p2 + p2p2 + γp2 γp2p2 = 1 p2 − p2 + p2p2 + γp2 1 p2 1 − 3 2 1 3 1 − 1 3 − 3 2 1 3 1 − 3 2 1 2 2 2 2 2 2 2 = 1 p3− p2 + p1 p1 + p1p3 + γp1 1 p3 (1.188) − b b b b − b b b − b b b b b b 2 1 2 2 2 2 2 2 2 = 1 p3− 1 p3 p1 1 p3 + γp1 1 p3 = 1 p1 + γp1 1 − 2b b− b − b −b b b −b − L 1 (p) = 1+ p1 (γ 1) (1.189) b − b b b b b b b similarly b L2 (p)=1+ p2 (γ 1) (1.190) 2 2 − substituting (1.189) and (1.190) in (1.187) the standard bobost L (p) becomes

2 2 1 + (γ 1) p1 (γ 1) p1p2 (γ 1) p1p3 p1 γ 1 − − 2 − 2 −  (γ 1) p1p2 1 + (γ 1) p2 (γ 1) p2p3 p2pγ 1  L (p)= − − − 2 2 − (1.191) (γ 1) p1pb3 (γ 1) pb2pb3 1 + (γ b1)bp3 pb3pγ 1  − 2 − 2 −2 −   p1 γ b b1 p2 γ 1b p3 γ b b1 b γ   − − − p   p b b p b b p b b  b b b 1.10. ONE-PARTICLE STATES WITH NON-NULL MASS 43

µ µ µ as a matter of consistency, it can be checked explicitly that L ν (p) applied on k = (0, 0, 0, M) yields p = p1,p2,p3,p0 .

2 2 (γ 1)p (γ 1)p p (γ 1)p p p1√γ 1 2 1 1 2 1 3 Mp1√γ 1 1+ − 2 − 2 − 2 − − p p p p p 2 | |2 0 (γ 1)p p (γ 1)p (γ 1)p p p2√γ 1 | | 2  1 2 2 2 3  Mp2√γ 1 − 2 1+ − 2 − 2 − 0   p p p p p − L (p) k = 2 | |2   = (γ 1)p p (γ 1)p p (γ 1)p p3√γ 1 | | 2  1 3 2 3 3  0 Mp3√γ 1  − 2 − 2 1+ − 2 −    p p p p  p −   2 2 2 | |   M     p1√γ 1 p2√γ 1 p3√γ 1    | | − − −  γM   p p p γ       | | | | | |   1    p1 p 2 p2 p L (p) k =   =  3  = p p3 p  p0   p0          where we have used Eqs. (1.184). In summary, the standard boost L (p) that carries the “standard” four momentum kµ = (0, 0, 0, M) to pµ = p1,p2,p3,p0 is given by Eq. (1.191) which can be expressed in the form

Li (p) = δ + (γ 1) p p k ik − i k i 0 2 0 L 0 (p) = L i (p)= pi γ 1 ; L 0 = γ b b − p p2 + M 2 p0 pi pi/ p ,b γ = (1.192) ≡ | | ≡ p M M It is important that when Λµ is an arbitrary three-dimensional rotation , the Wigner rotation W ( ,p) is the ν b R R same as for all p. In other words, if Λ belongs to the little group, then the element W (Λ,p) of the little group R induced by Λ must coincide with Λ itself18. We see it by noticing that the boost (1.192) can be expressed as in Eq. (1.186) 1 L (p)= R (p) B ( p ) R− (p) (1.193) | | where R (p) is the rotation (1.177) that takes the three-axis into the direction of p, and b b 10 0 0 b 01 0 0 B ( p )=   (1.194) | | 0 0 γ γ2 1 −  2   0 0 γ 1 p γ   −  is a pure boost along the three-axis. Then for an arbitraryp rotation we have from (1.181, 1.193) that R 1 1 1 1 W ( ,p) = L− ( p) L (p)= R ( p) B ( p ) R− ( p) − R (p) B ( p ) R− (p) R R R1 1 R |R | R 1 R | | W ( ,p) = R ( p) B− ( p ) R− ( p) R (p) B ( p ) R− (p) (1.195) R R | | R bR | | b b b but the rotation R 1 ( p) R (p) takes the three-axis (i.e. the unit vector u ) into the direction p, and then − R Rb b b b 3 into the direction p, and then back to the three-axis. Therefore, the whole operation can only change the unit R vectors u1 and u2. Henceb the wholeb operation is a rotation by some angle θ around the three-axis (orb around the unit vector u3) b cos θ sin θ 0 0 1 sin θ cos θ 0 0 R− ( p) R (p)= R (θ)  −  (1.196) R R ≡ 0 010  0 001    b b   18In order to obtain such a property we require the structure (1.193), which explains the convenience of the (a priori unnecessary) introduction of R (p) in the step from Eq. (1.182) to Eq. (1.183).

b 44 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS since matrices (1.194, 1.196) act non-trivially on diﬀerent subspaces of the Minkowski space, they commute with each other [R (θ) ,B ( p )] = 0 (1.197) | | from Eqs. (1.196, 1.197), the Wigner rotation (1.195) becomes

1 1 1 1 1 W ( ,p) = R ( p) B− ( p ) R− ( p) R (p) B ( p ) R− (p)= R ( p) B− ( p ) R (θ) B ( p ) R− (p) R R 1 | | R R 1 | | 1 R | | | | = R ( p) B− ( p ) B( p ) R (θ) R− (p)= R ( p) R (θ) R− (p) Rb | | | | b b R b b b and using (1.196) again,b we ﬁnally obtain b b b 1 1 1 W ( ,p) = R ( p) R (θ) R− (p)= R ( p) R− ( p) R (p) R− (p) R R R R R W ( ,p) = R R b b b b b b then states of a moving massive particle (and by extension multi-particle states), transform in the same way under rotations as in non-relativistic quantum mechanics. Consequently, all tools developed for rotations in non- relativistic quantum mechanics, such as the spherical harmonics, Clebsch-Gordan coeﬃcients etc. can be utilized in relativistic quantum mechanics.

1.11 One-particle states with null mass

1.11.1 Determination of the little group We will ﬁrst characterize the little group associated to case (c) in table 1.1. We shall choose the standard four- vector as kµ = (0, 0, 1, 1) (1.198) µ and we should ﬁnd the set of Lorentz transformations W ν, that satisfy

µ ν µ W νk = k since W is unitary and W k = k, we have the following properties for an arbitrary four vector tµ

Wt Wt = t t ; Wt k = Wt W k = t k (1.199) h | i h | i h | i h | i h | i applying (1.199) to a time-like four-vector of the form tµ (0, 0, 0, 1), we obtain ≡ (Wt)µ (Wt) = tµt = 1 (1.200) µ µ − (Wt)µ k = tµk = 1 (1.201) µ µ − by writing (Wt)µ = (α,β,ζ,η) we can rewrite Eq. (1.201) as

(α,β,ζ,η) (0, 0, 1, 1) = 1 ζ η = 1 η =1+ ζ − − ⇒ − − ⇒ consequently, any four-vector that satisﬁes the condition (1.201) can be parameterized as

(Wt)µ = (α, β, ζ, 1+ ζ) (1.202) in addition, the ﬁrst condition (1.200) combined with (1.202) yields

(Wt)µ (Wt) = α2 + β2 + ζ2 (1 + ζ)2 = 1 µ − − α2 + β2 ζ = (1.203) 2 1.11. ONE-PARTICLE STATES WITH NULL MASS 45 therefore, α, β are the independent parameters. However, we shall use the ζ parameter to simplify some expressions. µ ν The eﬀect of W ν on t is the same as that of the following Lorentz transformation

1 0 α α − µ 0 1 β β S ν (α, β)  −  (1.204) ≡ α β 1 ζ ζ −  α β ζ 1+ ζ   −    it can be checked out explicitly

1 0 α α 0 α − µ ν 0 1 β β 0 β µ ν S νt =  −    =   = W ν t α β 1 ζ ζ 0 ζ −  α β ζ 1+ ζ   1   1+ ζ   −            where we have used equation (1.202). It does not mean that W = S (α, β), but it means that

1 1 1 S− (α, β) W t = S− (α, β) [Wt]= S− (α, β) [S (α, β) t]= t 1 hence, S− (α, β) W is a Lorentz transformation that leaves the time-like four-vector (0, 0, 0, 1) invariant. There- 1 19 fore, S− (α, β) W is a pure rotation . It could also be seen that

1 0 α α 0 0 − µ ν 0 1 β β 0 0 µ S νk =  −    =   = k α β 1 ζ ζ 1 1 −  α β ζ 1+ ζ   1   1   −           

µ µ µ 20 hence S ν like W ν, leaves the reference light-like vector k invariant . Since S and W belong to the little group 1 1 µ µ then S− (α, β) W also does. Further since S− (α, β) W leaves t = (0, 0, 0, 1) and k = (0, 0, 1, 1) invariant, we 1 conclude that S− (α, β) W leaves the three-axis invariant. Thus, such an operator must be a rotation around the three-axis

cos θ sin θ 0 0 1 µ sin θ cos θ 0 0 S− (α, β) W = R (θ) ; R ν (θ)  −  (1.205) ≡ 0 010  0 001      from Eq. (1.205) we obtain

W (θ, α, β)= S (α, β) R (θ) (1.206)

19Since a Lorentz transformation leaves the pseudonorm invariant, and this transformation leaves the zeroth component invariant, the three-vector norm remains invariant. 20 µ In that sense, S ν is something like a “standard element” of the little group. 46 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS since W is arbitrary, we conclude that the most general element W of the little group can be parameterized as in Eq. (1.206). To characterize this little group, we ﬁrst establish the multiplication rule for S (α, β)

1 0 α¯ α¯ 1 0 α α − − 0 1 β¯ β¯ 0 1 β β S α,¯ β¯ S (α, β) =  −   −  α¯ β¯ 1 ζ¯ ζ¯ α β 1 ζ ζ − −  α¯ β¯ ζ¯ 1+ ζ¯   α β ζ 1+ ζ   −   −   1 0  α α¯ α+α ¯ − − 0 1 β β¯ β + β¯ =  − −  α +α ¯ β + β¯ 1 αα¯ ββ¯ ζ ζ¯ ζ¯ + αα¯ + ββ¯ + ζ − − − −  α +α ¯ β + β¯ ζ¯ αα¯ ββ¯ ζ 1+ ζ¯ + αα¯ + ββ¯ + ζ   − − − −   1 0 (α +α ¯) α +α ¯  − 0 1 β + β¯ β + β¯ S α,¯ β¯ S (α, β) =  −  (1.207) α +α ¯ β + β¯ 1 αα¯ + ββ¯ + ζ¯ + ζ αα¯ + ββ¯ + ζ¯ + ζ −  α +α ¯ β + β¯ αα¯ + ββ¯ + ζ¯ + ζ 1+ αα¯ + ββ¯ + ζ¯ + ζ   −    and applying Eq. (1.203), we see that

α¯2 + β¯2 α2 + β2 αα¯ + ββ¯ + ζ¯ + ζ = αα¯ + ββ¯ + + 2 2 2 α¯2 + α2 + 2¯αα + β¯2 + β2 + 2ββ¯ (¯α + α)2 + β¯ + β = = 2 2 αα¯ + ββ¯ + ζ¯ + ζ = ζ α¯ + α, β¯ + β (1.208) substituting (1.208) in (1.207) we ﬁnd

1 0 (α +α ¯) α +α ¯ − 0 1 β + β¯ β + β¯ S α,¯ β¯ S (α, β)  −  ≡ α +α ¯ β + β¯ 1 ζ α¯ + α, β¯ + β ζ α¯ + α, β¯ + β −  α +α ¯ β + β¯ ζ α¯ + α, β¯ + β 1+ ζ α¯ + α, β¯ + β   −  ¯  ¯  S α,¯ β S (α, β) = S α¯ + α, β + β (1.209) the rotations R (θ) are obviously abelian since they are around the three-axis only

R θ¯ R (θ)= R θ¯ + θ (1.210) and it is easy to check explicitly that

1 1 S− (α, β)= S ( α, β) ; R− (θ)= R ( θ) (1.211) − − − by reasons to be understood later, we call the little group deﬁned in (1.206) as ISO (2). From Eqs. (1.209, 1.210, 1.211) we see that the sets S (α, β) and R (θ) form subgroups of the little group ISO (2). Besides, the { } { } subgroup formed by S (α, β) [or equivalently by W (α,β,θ = 0) ], is invariant in ISO (2). It means that for { } { } any given element g ISO (2) and any given element h S (α, β) we have ∈ ∈

1 ghg− S (α, β) ; g ISO (2) ; h S (α, β) (1.212) ∈ ∀ ∈ ∀ ∈ 1.11. ONE-PARTICLE STATES WITH NULL MASS 47 to see it, we observe ﬁrst that

cos θ sin θ 0 0 1 0 α α − 1 sin θ cos θ 0 0 0 1 β β R (θ) S (α, β) R− (θ) =  −   −  0 010 α β 1 ζ (α, β) ζ (α, β) × −  0 001   α β ζ (α, β) 1+ ζ (α, β)     −   cos θ sin θ 0 0   − sin θ cos θ 0 0   × 0 0 10  0 0 01     1 0 α cos θ β sin θ α cos θ + β sin θ − − 1 0 1 α sin θ β cos θ α sin θ + β cos θ R (θ) S (α, β) R− (θ) =  − −  α cos θ + β sin θ α sin θ + β cos θ 1 ζ (α, β) ζ (α, β) − −  α cos θ + β sin θ α sin θ + β cos θ ζ (α, β) ζ (α, β) + 1   − −    by deﬁning α α cos θ + β sin θ ; β α sin θ + β cos θ (1.213) ≡ ≡− we obtain e e 1 0 α α − 1 0 1 β β R (θ) S (α, β) R− (θ)=  −  (1.214) α β 1 ζ (eα, β) ζ (α,e β)  −   α β ζ (α,e β) ζ (α, βe) + 1   −   e e  and e e α2 + β2 (α cos θ + β sin θ)2 + ( α sin θ + β cos θ)2 ζ α, β = = − 2 2 αe2 cose2 θ + sin2 θ + β2 cos2 θ + sin2 θ e e = 2 α2 + β2 ζ α, β = = ζ (α, β) (1.215) 2 substituting (1.213, 1.215)e ine (1.214) we ﬁnd

1 0 α α − 0 1 β β 1  −  R (θ) S (α, β) R− (θ) = α β 1 ζ eα, β ζ α,e β = S α, β  − e e     α β ζ α, β 1+ζ α,β  e  e e − e e e e  e 1   R (θ) S (α, β) R− (θ) = S (α cos θ + β sin θ, α sin θ + β cos θ) (1.216) e e e e− e e using Eqs. (1.206, 1.216) we have

1 1 W θ, α,¯ β¯ S (α, β) W − θ, α,¯ β¯ = S α,¯ β¯ R (θ) S (α, β) S α,¯ β¯ R (θ) − ¯ 1 1 ¯ = S α,¯ β R (θ)S (α, β) R − (θ) S− α,¯β 1 1 W θ, α,¯ β¯ S (α, β) W − θ, α,¯ β¯ = S α,¯ β¯S α, β S− α,¯ β¯ = S α¯ + α α,¯ β¯ + β β¯ − − 1 W θ, α,¯ β¯ S (α, β) W − θ, α,¯ β¯ = S α, β ; α α cos θ + βsin θ , β α sin θ + β cos θ (1.217) e ≡e ≡−e e e e e e 48 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS since the RHS of Eq. (1.217) clearly belongs to the subgroup S (α, β) , we observe by comparing with Eq. { } (1.212), that S (α, β) is invariant in the little group ISO (2). The multiplication rule for the whole little group { } ISO (2) is obtained by combining Eqs. (1.206, 1.209, 1.210, 1.216)

1 W θ,¯ α,¯ β¯ W (θ, α, β) = S α,¯ β¯ R θ¯ [S (α, β) R (θ)] = S α,¯ β¯ R θ¯ S (α, β) R− θ¯ R θ¯ R (θ) = S α,¯ β¯ S α cos θ¯ + β sin θ,¯ α sin θ¯ + β cos θ¯ R θ¯ + θ − = S α¯ + α cos θ¯ + β sin θ,¯ α sin θ¯ + β cos θ¯ + β¯ R θ¯ + θ − = W θ¯ + θ, α¯ + α cos θ¯ + β sin θ,¯ α sin θ¯ + β cos θ¯ + β¯ − the composition law is then given by

W θ,¯ α,¯ β¯ W (θ, α, β)= W θ¯ + θ, α, β ; α α¯ + α cos θ¯ + β sin θ¯ , β = α sin θ¯ + β cos θ¯ + β¯ (1.218) ≡ − the multiplication rule (1.218) is the oneb associatedb b with the group of euclideanb geometry consisting of translations in two dimensions (by a vector α, β ) and rotations in two dimensions by an angle θ. However, it is important to | i clarify that we call this little group ISO (2), because it is isomorphic with the euclidean group of rotations and translations in two dimensions, but our little group is a proper subgroup of the homogeneous Lorentz group, hence translations in the Minkowski space (or in the Hilbert space of quantum mechanics) are not contained in this little group but only rotations and boosts. In other words, the transformations contained in our little group (understood as transformations in the Minkowski or Hilbert space) are not rotations and translations in two dimensions, they are just isomorphic with the group of bidimensional translations and rotations in the euclidean space. When a group does not have invariant abelian groups it is called semi-simple, showing that it has certain simple properties. We have just seen that the little group ISO (2) is not semi-simple [according with Eq. (1.217), it contains the translations in two dimensions S (α, β) as an abelian invariant subgroup]. In the same way the inhomogeneous Lorentz group is not semi-simple either.

1.11.2 Lie algebra of the little group ISO (2)

To obtain more information about the way in which ISO (2) is embedded in SO (3, 1), we shall study the Lie algebra of ISO (2), and its relation with the Lie algebra of SO (3, 1). If θ, α, β become infinitesimal parameters we obtain an infinitesimal transformation of the little group. Taking Eqs. (1.204, 1.205) with infinitesimal parameters up to first order in θ, α and/or β, we have

1 0 α α 1 θ 0 0 1 θ α α − − 0 1 β β θ 1 0 0 θ 1 β β W (θ, α, β) = S (α, β) R (θ)=  −   −  =  − −  α β 1 0 0 010 α θβ β + θα 1 0 −  α β 0 1   0 001   α θβ β + θα 0 1       −  1 000  0 θ α  α    − 0 100 θ 0 β β W (θ, α, β) =   +  − −  + α2, β2, αβ (1.219) 0 010 α β 0 0 O  0 001   α β 0 0          Parameterizing the transformation (1.219) as in Eq. (1.88) page 25, we ﬁnd

0 θ α α − µ µ µ µ θ 0 β β W (θ, α, β) ν = δ ν + ω ν ; ω ν =  − −  α β 0 0  α β 0 0      1.11. ONE-PARTICLE STATES WITH NULL MASS 49

and the totally covariant transformation ωµν becomes

100 0 0 θ α α − α 010 0 θ 0 β β ωµν = gαµω ν =    − −  001 0 α β 0 0  0 0 0 1   α β 0 0   −    0 θ α α    − θ 0 β β ωµν =  − −  (1.220) α β 0 0  α β 0 0   − −    observe that ωµν is antisymmetric, in consistence with Eq. (1.89), page 25, and the fact that W (θ, α, β) is a µ particular case of a Lorentz transformation (notice however that ω ν is not antisymmetric). Substituting Eq. (1.220) in Eq. (1.99) page 27, we obtain the generators J ρσ (which have also be taken as antisymmetric) that contributes in the expansion of the associated operator U (W (θ, α, β)) in the Hilbert space 03 (for θ, α, β inﬁnitesimal). Equation (1.220) shows that ω03 = ω30 = 0. Hence, the associated generators J and J 30 do not contribute in such an expansion21 1 U (W (θ, α, β)) = 1+ iω J ρσ + ω2 2 ρσ O 1 1 1 = 1+ i ω J 12 + ω J21 + i ω J 13 + ω J 31 + i ω J 10 + ω J 01 2 12 21 2 13 31 2 10 01 1 1 + i ω J 23 + ω J 32 + i ω J 20 + ω J 02 2 23 32 2 20 02 ρσ ρσ since both ωρσ and J are antisymmetric, the products ωρσJ are symmetric. Using this fact and the explicit form of the elements ωµν given by Eq. (1.220) we ﬁnd

12 13 10 23 20 U (W (θ, α, β)) = 1+ iω12J + iω13J + iω10J + iω23J + iω20J = 1+ iθJ 12 iαJ 13 + iαJ 10 iβJ 23 + iβJ 20 − − U (W (θ, α, β)) = 1+ iα J 10 J 13 + iβ J 20 J 23 + iθJ 12 − − U (W (θ, α, β)) = 1+ iαA + iβB + iθJ 12 ; A J 10 J 13 , B J 20 J 23 ≡ − ≡ − using Eqs. (1.127, 1.129) page 30, we obtain

U (W (θ, α, β)) = 1+ iαA + iβB + iθJ3 (1.221) A J 31 J 01 = J K ; B J 02 J 23 = K J (1.222) ≡ − 2 − 1 ≡− − − 2 − 1 the Lie algebra of ISO (2), i.e. between the generators A, B and J3 can be obtained from the commutation relations (1.130, 1.131, 1.132) page 31. For instance, using Eqs. (1.130, 1.131) we ﬁnd

[J , A] = [J , J K ] = [J , J ] [J , K ]= iJ iK = iB 3 3 2 − 1 3 2 − 3 1 − 1 − 2 proceeding in the same way with the other generators, the Lie algebra of ISO (2) (written as a subgroup of the Lorentz group) becomes [J , A]= iB , [J ,B]= iA , [A, B] = 0 (1.223) 3 3 − these commutation relations can also be obtained from Eqs. (1.209, 1.210, 1.216) [homework!!(7)]. In the framework of the euclidean space in two dimensions, J3 is the generator of rotations and A, B are generators of translations (owing to it, A and B commute each other). However, deﬁnitions (1.222) show that in the framework

21Of course, the generators J µµ are null even in the general case because of the antisymmetry. 50 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS of the Minkowski space (or in the Hilbert space in quantum mechanics), A and B are generators that combine rotations with boosts. For future purposes we shall find the transformation properties of the generators A and B under rotations U [R (θ)] within the little group. To do it, we shall use an infinitesimal transformation U [S (α, β)], such that α and β are infinitesimals, but θ is arbitrary. From Eq. (1.216) we find

1 1 U [R (θ)] U [S (α, β)] U − [R (θ)] = U R (θ) S (α, β) R− (θ) = U [S (α cos θ + β sin θ, α sin θ + β cos θ)] − 1 U [R (θ)] U [S (α, β)] U − [R (θ)] = U S α, β ; α α cos θ + β sin θ , β α sin θ + β cos θ(1.224) ≡ ≡− h i now using the expansion (1.221) we can write e e e e 1 1 U [R (θ)] U [S (α, β)] U − [R (θ)] = U [R (θ)] 1+ iαA + iβB U − [R (θ)] 1 { 1 } 1 U [R (θ)] U [S (α, β)] U − [R (θ)] = 1+ iαU [R (θ)] AU − [R (θ)] + iβU [R (θ)] BU − [R (θ)] (1.225) using again the expansion (1.221) on the RHS of Eq. (1.224) we have

1 U [R (θ)] U [S (α, β)] U − [R (θ)] = U S α, β =1+ iαA + iβB = 1+h i (α cosiθ + β sin θ) A + i ( α sin θ + β cos θ) B 1 e e e e− U [R (θ)] U [S (α, β)] U − [R (θ)] = 1+ iα (A cos θ B sin θ)+ iβ (A sin θ + B cos θ) (1.226) − equating Eqs. (1.225, 1.226) yields

1 1 α U [R (θ)] AU − [R (θ)] + β U [R (θ)] BU − [R (θ)] = α (A cos θ B sin θ)+ β (A sin θ + B cos θ) − since α and β are inﬁnitesimal but otherwise arbitrary, we can equate the coeﬃcients of them to obtain

1 U [R (θ)] AU − [R (θ)] = A cos θ B sin θ (1.227) 1 − U [R (θ)] BU − [R (θ)] = A sin θ + B cos θ (1.228)

1.11.3 Massless states in terms of eigenvalues of the generators of ISO (2) Since A and B are hermitian commuting operators, they admit a common basis of eigenvectors

A k, a, b, γ = a k, a, b, γ , B k, a, b, γ = b k, a, b, γ (1.229) | i | i | i | i where γ denotes any additional quantum numbers required to deﬁne physical states. Since k, a, b, γ forms a {| i} basis, the physical states are superpositions of such vectors

k,σ = c k, a, b, γ (1.230) | i abγ | i a,b,γX However, we shall see that if we ﬁnd a set of non-zero eigenvalues of A, B they would belong to a continuous spectrum. From Eqs. (1.229) and utilizing (1.227) we obtain

1 1 1 AU − [R (θ)] k, a, b, γ = U − [R (θ)] U [R (θ)] AU − [R (θ)] k, a, b, γ | i 1 | i = U − [R (θ)] A cos θ B sin θ k, a, b, γ 1 { − } | i = U − [R (θ)] cos θ A k, a, b, γ sin θ B k, a, b, γ 1 { | i− | i} = U − [R (θ)] cos θ a k, a, b, γ sin θ b k, a, b, γ 1 { | 1 i− | i} A U − [R (θ)] k, a, b, γ = a cos θ b sin θ U − [R (θ)] k, a, b, γ | i { − } | i 1.11. ONE-PARTICLE STATES WITH NULL MASS 51 a similar procedure can be done from Eqs. (1.229) and (1.228), so that we obtain A k, a, b, γ θ = (a cos θ b sin θ) k, a, b, γ θ (1.231) | iθ − | iθ θ 1 B k, a, b, γ = (a sin θ + b cos θ) k, a, b, γ ; k, a, b, γ U − [R (θ)] k, a, b, γ (1.232) | i | i | i ≡ | i note that θ is a continuous parameter, hence Eqs. (1.231, 1.232) show that if a and/or b are diﬀerent from zero, the spectra of A and B are continuous. Such a continuous degree of freedom like θ is not observed in massless particles22. Consequently, to avoid this additional continuous degree of freedom, we should demand that the only possible eigenvectors involved in the superposition (1.230) that forms physical states k,σ , are eigenvectors with | i eigenvalues a = b = 0. k,σ = c k, 0, 0, γ c k, γ (1.233) | i 00γ | i≡ γ | i γ γ X X applying operators A or B on the physical states (1.233) we ﬁnd A k,σ = B k,σ = 0 (1.234) | i | i hence particle states cannot be characterized by eigenvalues of A or B. These states are then distinguished by their eigenvalues associated with the remaining generator J3 of ISO (2) J k,σ = σ k,σ (1.235) 3 | i | i since the momentum k [the three-momentum associated with the “standard” four-momentum (1.198)] is in the three-direction u3, we have that σ gives the component of angular momentum in the direction of motion, called the helicity.

1.11.4 Lorentz transformations of massless states From the previous properties, we are able to ﬁnd the Lorentz transformations of general states of masless particles. By using the general arguments described in Sec. 1.4 [see Eq. (1.39), Page 17], we obtain that Eqs. (1.221) can be generalized to the case of ﬁnite values of the parameters α, β and θ, since the associated subgroups are abelian U (S (α, β)) = exp(iαA) exp (iβB) = exp (iαA + iβB) (1.236)

U (R (θ)) = exp (iJ3θ) (1.237) where we have taken into account that A and B commute. According with Eqs. (1.206, 1.234), an arbitrary element W of the little group acts on a massless state k,σ as | i U (W ) k,σ = U (S (α, β)) U (R (θ)) k,σ = exp (iαA + iβB) exp (iJ θ) k,σ | i | i 3 | i = exp (iσθ) exp (iαA + iβB) k,σ | i U (W ) k,σ = exp (iσθ) k,σ | i | i hence the representation Dσσ′ (W ) of the little group in Eq. (1.148) yields

Dσ′σ (W ) = exp (iθσ) δσ′σ (1.238) from this we ﬁnd the general Lorentz transformation rule (i.e. under U (Λ)), for a general massless particle state p,σ of arbitrary helicity, by using Eqs. (1.150, 1.161) | i N (p) k0/p0 U (Λ) p,σ = Dσ′σ (W (Λ,p)) Λp,σ′ = exp [iσθ (Λ,p)] δσ′σ Λp,σ′ | i N (Λp) 0 0 σ′ (Λpk) / (Λp) σ′ X X q (Λp)0 U (Λ) p,σ = exp [iσθ (Λ,p)] Λp,σ (1.239) | i s p0 | i

22 All particles observed so far, are such that the only continuous degrees of freedom are related with the orbital Hilbert space Er. Thus the continuous degrees of freedom are determined by the commuting operators {R1, R2, R3} or by the commuting observables {K1, K2, K3}. The remaining degrees of freedom (such as spin or similar variables) are always discrete. 52 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS from Eqs. (1.149) and (1.206) we see that θ (Λ,p) is deﬁned by

1 W (Λ,p) L− (Λp) Λ L (p) S (α (Λ,p) , β (Λ,p)) R (θ (Λ,p)) (1.240) ≡ ≡ we shall see later that electromagnetic gauge invariance arise from the part of the little group parameterized by α and β. Note that we have not restricted the spectrum of σ. Indeed, by contrast to the case of massive particles, the spectrum of σ for massless particles is not restricted by the algebra. To see it, we observe that for massive particles it is the Lie algebra of angular momentum (with all its three generators) associated with the little group SO (3), that predicts the fact that σ is only integer or half-integer. The litte group ISO (2) of massless particles only contain one generator of angular momentum (J3 by convention) that is not enough to restrict the spectrum of σ. Despite σ is not restricted on algebraic grounds, there are topological considerations that restrict the allowed values of σ to be integer of half-integer, like in the case of massive particles [see section 2.7 of Ref. [1]]. In order to calculate the little group element (1.240) for a given Λ and p, we proceed similarly as in the case of massive particles: we have to fix a convention for the standard Lorentz transformation that takes us from the stardard four-momentum kµ (0, 0, κ, κ) to pµ. We can choose such a stardard Lorentz transformation to have ≡ the form L (p)= R (p) B ( p /κ) (1.241) | | where B (u) is a pure boost along the u3 direction b 10 0 0 01 0 0 B (u)=  (u2+1) (u2 1)  (1.242) − 0 0 2u 2u  (u2 1) (u2+1)   0 0 −   2u 2u    and R (p) is a pure rotation [see Eq. (1.177), page 40] that carries the three-axis into the direction of the momentum determined by the unitary vector p. Note that in passing from kµ to pµ, it is enough to determine the changeb from k to p because of the conservation of the pseudonorm. Hence, the boost in (1.241) accomplishes the role of changing the norm of the three-vector from k to p , while the rotation changes the direction to its b | | | | final one. The arguments to build up the standard boost (1.241) for massless particle states, are similar to the ones we followed to obtain the standard boost (1.193) for massive particles. We can see from Eq. (1.239) that helicity is Lorentz-invariant; a massless particle of a given helicity looks the same (aside from its momentum) in all inertial reference frames23. Therefore, based on SO (3, 1) transformations, we could a priori think of massless particles of different helicities as different species of particles (in the same way as particles with different electric charge are considered different species). Nevertheless, this is not the case when the discrete Lorentz transformations are included. We shall see in the next section that particles of opposite helicity are related by parity or space inversion. For electromagnetic and gravitational forces (that obey space inversion symmetry), the massless particles of opposite helicity are called the same. The massless particles of helicity 1, ± associated with electromagnetic phenomena are both called photons. In the same way, the (hypothetical) massless particles of helicity 2, that are supposed to mediate the gravitational interactions are both called gravitons. ± On the other hand, the almost massless particles of helicity 1/2 that are emitted in nuclear beta decay24 ± have no interaction (apart from gravitation which is negligible for many purposes) that respect the symmetry of space inversion (weak interactions do not respect space inversion symmetry). Therefore, these particles receive different names: neutrinos for helicity 1/2, and antineutrinos for helicity +1/2. − 23Observe that for a massless particle, a positive helicity cannot become a negative one (or vice versa). For such a projection to change its sign, it is necessary that the momentum changes its sense. However, for a massless particle (which travels at the speed of light) there is no any boost to another reference frame that can reverse the direction of motion i.e. the momentum of the particle. 24For a long time, it was supposed that such particles were massless (neutrinos and antineutrinos). Nowadays we know these particles are massive (though with masses much smaller that electron masses) so their helicities are not Lorentz-invariants anymore. 1.12. SPACE INVERSION AND TIME-REVERSAL 53

Though helicity of massless particles is Lorentz-invariant, Eq. (1.239) shows that the state itself is not. For instance, owing to the helicity-dependent phase exp (iσθ) in Eq. (1.239), a state formed by a linear superposition of one-particle states with opposite helicities will be changed by a Lorentz transformation into a diﬀerent superposition. As an example, a general one-photon state of four-momentum p might be written as

2 2 p; α = α+ p, +1 + α p, 1 ; α+ + α = 1 | i | i − | − i | | | −| if α+ = α and both are non-zero, we are in the case of elliptic polarization. Circular polarization occurs when | | 6 | −| one of the coefficients vanishes, in the opposite extreme we have the linear polarization when α+ = α . The | | | −| overall phase of α+ and α has no physical significance, and for linear polarization they could be adjusted such − that α = α+∗ (1.243) − but the relative phase is still important. For linear polarization with the choice (1.243), the phase of α+ can be identified as the angle between the plane of polarization and a given fixed reference direction perpendicular to p. µ Equation (1.239) shows that under a Lorentz transformation Λ ν , this angle rotates by an amount θ (Λ,p).

1.12 Space inversion and time-reversal

We have seen that the complete Lorentz group also contains some improper (discrete) transformations that leave the pseudonorm of four-vectors invariant. Any non-continuous homogeneous Lorentz transformation can be written as the product of a proper orthochronus transformation (with detΛ = +1, and Λ0 1) with a discrete operation 0 ≥ that can be either (space inversion), (time reversal) or . P T PT 10 00 100 0 − µ 0 1 0 0 µ 010 0 ν =  −  ; ν =   (1.244) P 0 0 1 0 T 001 0 −  0 0 01   0 0 0 1     −      We should study how to extend these discrete operations to the representations U (Λ, a) of the inhomogeneous Lorentz group in the Hilbert space. We shall ﬁrst mount what we could call a scenario with conservation of time reversal and parity. In this scenario, the fundamental multiplication rule (1.90) of the Poincar´egroup

U Λ¯, a¯ U (Λ, a)= U ΛΛ¯ , Λ¯a +a ¯ (1.245) would be valid even if Λ and/or Λ¯ involves a discrete transformation, and there are discrete operators P and T on the Hilbert space associated with the discrete operators and deﬁned on the Minkowski space P T P U ( , 0) ; T U ( , 0) (1.246) ≡ P ≡ T to induce these discrete operators on the Hilbert space, we note that the parameters (Λ, a) which correspond to an operator and a four-vector on the Minkowski space respectively, transform under the discrete symmetries as follows 1 1 Λ P Λ − , Λ T Λ − ; a P a , a T a → P P →T T → P →T therefore, it is natural to deﬁne the transformation of a continuous operator U (Λ, a) on the Hilbert space through the discrete operators P and T , as follows

1 1 1 1 PU (Λ, a) P − = U Λ − , a ; TU (Λ, a) T − = U Λ − , a (1.247) P P P T T T for any proper orthochronus transformation Λ and for any translation a. However, a series of experiments have shown that weak interactions such as those produced in nuclear beta decay, violates the conservation of these discrete symmetries. In 1956-57 it was shown that parity is violated in 54 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS decays involving weak interactions. In 1964, indirect evidence appeared of violation of the time reversal symmetry. Therefore, the present scenario will be only approximately true, and should be corrected when parity-violating theories (theories involving the nuclear weak interaction) are treated. By now, we shall use a generic operator to indicate either or . An inﬁnitesimal transformation associated W P T with Eqs. (1.247) is given by

1 1 WU (1 + ω, ε) W − = U (1 + ω) − , ε 1 W W1 W WU (1 + ω, ε) W − = U 1+ ω − , ε ; W = P, T and , (1.248) W W W W ≡ P T where we have taken into account that there is no inﬁnitesima l parameter associated with . Applying the W expansion (1.99) page 27, on both sides of Eq. (1.248) we ﬁnd

1 ρσ ρ 1 1 1 µν µ W 1+ iω J iε P W − = 1+ i ω − J i ( ε) P 2 ρσ − ρ 2 W W µν − W µ 1 ρσ 1 ρ 1 1 ρ 1 σ µν ρ µ ω W [iJ ] W − ε W [iP ] W − = i ω − J i ε P (1.249) 2 ρσ − ρ 2 Wµ ρσ W ν − Wµ ρ now we should take into account that the fact that prevented us to consider antilinear antiunitary operators in representations of SO (3, 1) (according to Wigner’s representation theorem) was the continuous connection of all its elements with the identity. However, discrete symmetries have no these properties and the door is open to consider the possibility (and even the necessity) of antilinear antiunitary operators in the representations of the extended Lorentz group. Therefore, we shall not take the “i” factor out of our W operators by now. Equating coeﬃcients of ωρσ and ερ in Eq. (1.249) we obtain

ρσ 1 ρ σ µν W [iJ ] W − = i µ ν J (1.250) ρ 1 W ρWµ W [iP ] W − = i P ; W = P, T and , (1.251) Wµ W ≡ P T where we have used the fact that 1 σ = σ for both and as can be seen from Eq. (1.244) [and using W− ν Wν P T a definition like (1.49)]. Equations (1.250, 1.251) are quite similar to Eqs. (1.111, 1.112) page 29, except from the fact that we have not cancelled the “i” factor on both sides of Eqs. (1.250, 1.251), because of the possibility of having antilinear antiunitary operators. Setting ρ = 0 and = in Eq. (1.251) we get W P 0 1 0 µ 0 0 P iP P − = i µ P = i 0 P 1 P P P(iH) P − = iH where H is the energy operator. If P were antiunitary and antilinear then it would anticommute with i. Therefore, we would obtain P HP 1 = H. From which we would find − − 1 H Ψ = E Ψ P HP − P Ψ = EP Ψ HP Ψ = EP Ψ | i | i ⇒ | i | i ⇒ − | i | i H P Ψ = E P Ψ ⇒ { | i} − { | i} Then, for any eigenstate Ψ of the Hamiltonian with energy E > 0, there would be another state P Ψ of energy | i | i E < 0. There are no states of negative energy (energy less than that of the vacuum)25. Hence, it is necessary − for the operator P to be linear and unitary. Now, setting ρ = 0 and = in Eq. (1.251) we get W T 1 0 µ 0 0 T (iH) T − = i µ P = i 0 P 1 T T T (iH) T − = iH − 25This also implies that the spectrum is not bounded from below, hence states would decay indefinitely toward lower and lower energies. Stability would be impossible. 1.12. SPACE INVERSION AND TIME-REVERSAL 55 if we suppose that T is linear and unitary, we can cancel the “i” factor on both sides of this equation. Hence, T HT 1 = H, leading again to the conclusion that for any state Ψ with energy E > 0, there would be a state − − | i T Ψ of energy E < 0. To avoid this, we must conclude that the time reversal operator T is antilinear and | i − antiunitary. Therefore, Eqs. (1.250, 1.251) for each discrete symmetry become

ρσ 1 ρ σ µν ρ 1 ρ µ P J P − = µ ν J ; P P P − = µ P (1.252) ρσ 1 P Pρ σ µν ρ 1 P ρ µ T J T − = J ; T P T − = P (1.253) −Tµ Tν −Tµ moreover, Eqs. (1.252, 1.253) can be rewritten in terms of the generators in three-dimensional notation given by kl Eqs. (1.126, 1.127, 1.128, 1.129). For example, Eqs. (1.127, 1.85) says that Ji = J for some three-component indices k, l and that i = 1. Hence, we can use J ρσ = J kl in Eq. (1.252) and obtain Pi − 1 kl 1 k l µν k l kl P J P − = P J P − = J = J = ( 1) ( 1) J = J i Pµ Pν Pk Pl − − i i there is no sum over k, l. Similarly we have

1 kl 1 k l µν k l kl T J T − = T J T − = J = J = (+1) (+1) J = J i −Tµ Tν −Tk Tl − i − i proceeding similarly with the other generators we obtain

1 1 1 P JP − = +J ; P KP − = K ; P PP − = P (1.254) 1 1 − 1 − T JT − = J ; T KT − =+K ; T PT − = P (1.255) 1 − 1 − P HP − = T HT − = H (1.256) it is sensible that P must reverse sign under parity. Further, it is reasonable that parity preserves the sign of J, at least for the orbital angular momentum since both the position and the linear momentum change signs, thus the cross product must be invariant. On the other hand time reversal reverses J, once again it is sensible, since under time reversal and observer would see the “ﬁlm backwards”, hence the observer see that the spinning goes on the opposite direction26. Observe for instance that the ﬁrst of Eqs. (1.255) is consistent with the angular momentum commutation relations

1 1 1 1 1 1 1 T [J , J ] T − = T J J T − T J J T − = T J T − T J T − T J T − T J T − i j i j − j i i j − j i = ( Ji) ( Jj) ( Jj) ( Ji) 1 − − − − − T [Ji, Jj] T − = [Ji, Jj ] on the other hand 1 1 T (iε J ) T − = iε T J T − = iε J ijk k − ijk k ijk k showing that 1 1 [J , J ]= iε J T [J , J ] T − = T (iε J ) T − i j ijk k ⇒ i j ijk k it worths emphasizing that the antilinearity of T (i.e. the fact that T anticommutes with i) was essential in obtaining this consistency. In general, it can be checked that Eqs. (1.254, 1.255, 1.256) are consistent with all the commutation relations (1.130-1.137) [homework!!(8)]. Now we examine these discrete symmetries for massive particles and particles with null mass

26We could think (classically) in time reversal as taking t → −t, and keeping positions unaltered r → r. Since p = mdr/dt under time reversal we obtain p′ = mdr/d (−t)= −p. Therefore J′ = r′ × p′ = −J. 56 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS

1.13 Parity and time-reversal for one-particle states with M > 0.

1.13.1 Parity for M > 0 In Sec. 1.10, we took the reference standard vector kµ = (0, 0, 0, M), the little group was SO (3) with generators µ J1, J2, J3. This k corresponds to a particle at rest with energy M (only self-energy). Therefore, the associated eigenvectors k,σ in the Hilbert space are eigenvectors of P,H and J with eigenvalues p = 0, p0 = E = M and | i 3 σ respectively. From Eqs. (1.254) and (1.256) we see that P anticommutes with P, and it commutes with J and H, thus

P [P k,σ ] = P [P k,σ ] = 0 | i − | i H [P k,σ ] = P [H k,σ ]= M [P k,σ ] | i | i | i J [P k,σ ] = P [J k,σ ]= σ [P k,σ ] 3 | i 3 | i | i consequently the state P k,σ has the same quantum numbers 0,M,σ as k,σ . Therefore27 except for degeneracies | i | i (arising from additional quantum numbers), both states can only diﬀer by a phase

P k,σ = η k,σ ; η = 1 (1.257) | i σ | i | σ| we shall see that the phase is independent of σ. To see it, we shall apply Eqs. (1.148, 1.163, 1.164) for an inﬁnitesimal transformation of the little group SO (3)

U (W ) k,σ = D ′ (W ) k,σ′ | i σ σ σ′ X (j) (j) U (1+Θ , 0) k,σ = D ′ (1+Θ ) k,σ′ p | i σ σ p σ′ X 1 i (j) 1+ iΘpJp k,σ = δσ′σ + Θp Jp k,σ′ 2 | i 2 σ′σ σ′ X (j) Jp k,σ = Jp k,σ′ | i σ′σ σ′ X applying this for p = 1, 2 and using Eqs. (1.167) we ﬁnd

(j) (j) J1 k,σ = J1 k,σ′ ; iJ2 k,σ = iJ2 k,σ′ | i σ′σ ± | i ± σ′σ σ′ σ′ X X (j) (j) (J1 iJ2) k,σ = J1 iJ2 k,σ′ = δσ′,σ 1 (j σ) (j σ + 1) k,σ′ ± | i ± σ′σ ± ∓ ± σ′ σ′ X X h p i (J iJ ) k,σ = (j σ) (j σ + 1) k,σ 1 (1.258) 1 ± 2 | i ∓ ± | ± i p where j is the particle’s spin. Applying P on the LHS of Eq. (1.258), we ﬁnd

P (J iJ ) k,σ = (J iJ ) P k,σ = (J iJ ) η k,σ = η (j σ) (j σ + 1) k,σ 1 (1.259) 1 ± 2 | i 1 ± 2 | i 1 ± 2 σ | i σ ∓ ± | ± i p and applying P on the RHS of Eq. (1.258), we ﬁnd

(j σ) (j σ + 1)P k,σ 1 = (j σ) (j σ + 1) ησ 1 k,σ 1 (1.260) ∓ ± | ± i ∓ ± ± | ± i 27Observe that the factp that P does not modify the quantum numbersp 0, M (i.e. the eigenvalues of the four-momentum operator) of the state |k,σi, is related with the invariance under P of the associated classical four-momentum k ≡ (0, M). 1.13. PARITY AND TIME-REVERSAL FOR ONE-PARTICLE STATES WITH M > 0. 57 and equating Eqs. (1.259, 1.260) we find ησ = ησ 1 ± then ησ is independent of σ, so we can rewrite Eq. (1.257) as P k,σ = η k,σ ; η = 1 (1.261) | i | i | | the phase η is called the intrinsic parity. It depends only on the species of particle on which P is acting. Further, in Sec. 1.10, we saw that to pass from kµ to an arbitrary four-momentum pµ, we use the standard boost L (p) defined by Eq. (1.192). Now, to obtain states p,σ in the Hilbert space with arbitrary momentum p, | i we apply the associated unitary operator U (L (p)) according with definition (1.145)

k0 p,σ = N (p) U (L (p)) k,σ = U (L (p)) k,σ | i | i s p0 | i M p,σ = U (L (p)) k,σ (1.262) | i s p0 | i By using the matrix representation of , we obtain p = p,p0 . Now, we shall examine the transformation P P − properties of the standard boost L (p) given by Eqs. (1.191, 1.192) through the parity transformation. It is expected that such a transformation (p) 1 be related with L ( p). To see this relation, we first evaluate PL P− P L ( p). To do this, we first observe that according with Eq. (1.192) we find P γ ( p)= γ p,p0 = γ p,p0 P − now, using the other Eqs. (1.192), we find

Li ( p) = Li p,p0 = δ + [γ ( p) 1] ( p ) ( p )= δ + [γ (p) 1] p p = Li (p) j P j − ij P − − i − j ij − i j j L0 ( p) = γ ( p)= γ (p)= L0 ( p) 0 P P 0 P Li ( p) = L0 ( p)= p γ2 ( p) 1= pb γ2 (bp) 1= Li (p)= bLb0 (p) 0 P i P − i P − − i − − 0 − i in summary we have p p b b Li ( p) = Li (p) ; L0 ( p)= L0 (p) j P j 0 P 0 Li ( p) = L0 ( p)= Li (p)= L0 (p) (1.263) 0 P i P − 0 − i combining (1.191) with (1.263) the explicit form of Lµ ( p) yields ν P 2 2 1 + (γ 1) p1 (γ 1) p1p2 (γ 1) p1p3 p1 γ 1 − − 2 − − 2 −  (γ 1) p1p2 1 + (γ 1) p2 (γ 1) p2p3 p2pγ 1  L ( p)= − − − 2 − 2 − (1.264) P (γ 1) p1pb3 (γ 1) pb2pb3 1 + (γ b1)bp3 pb3pγ 1  − 2 − 2 −2 − −   p1 γ b b1 p2 γ b1 p3 γ b b1 b γ   − − − − − − p   b b b b b b  On the other hand, we can evaluatep the transformationp (p) p 1 explicitly by taking Eqs. (1.191, 1.244) b b PL bP− 2 2 10 00 1 + (γ 1) p1 (γ 1) p1p2 (γ 1) p1p3 p1 γ 1 − − − 2 − 2 − 1 0 1 0 0 (γ 1) p1p2 1 + (γ 1) p2 (γ 1) p2p3 p2 γ 1 (p) − =  −   − − − p −  PL P 0 0 1 0 (γ 1) p pb (γ 1) pb pb 1 + (γ b1)bp2 pb γ2 1 × − 1 3 2 3 3 3p  0 0 01   − 2 − 2 −2 −     p1 γ b b1 p2 γ 1b p3 γ b b1 b pγ     − − −  10 00  p b b p b b p b b  − 0 1 0 0 b b b  −  × 0 0 1 0 −  0 0 01      58 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS after multiplying these matrices we obtain the matrix in (1.264). Thus we ﬁnd

1 0 2 2 L (p) − = L ( p) ; p = p,p = p, p + M (1.265) P P P P − − p applying P on Eq. (1.262) and using Eqs. (1.247, 1.265), we have

P p,σ = η p,σ (1.266) | i |P i 1.13.2 Time reversal for M > 0 Equations (1.255, 1.256), show that the eﬀect of T on the zero three-momentum state of reference k,σ yields | i P (T k,σ )=0 ; H (T k,σ )= M (T k,σ ) ; J (T k,σ )= σ (T k,σ ) | i | i | i 3 | i − | i therefore, the quantum numbers 0, M of P and H remain invariant while the quantum number σ of J3 changes sign T k,σ = ζ k, σ (1.267) | i σ | − i where ζσ is a phase factor that could depend on σ. Applying T to (1.258) and recalling that T anticommutes with J and i, we have

T (J iJ ) k,σ = (j σ) (j σ + 1) T k,σ 1 1 ± 2 | i ∓ ± | ± i ( J1 iJ2) T k,σ = p(j σ) (j σ + 1) ζσ 1 k, σ 1 − ± | i ∓ ± ± | − ∓ i (J1 iJ2) ζσ k, σ = p(j σ) (j σ + 1) ζσ 1 k, σ 1 (1.268) − ∓ | − i ∓ ± ± | − ∓ i and using again Eq. (1.258) in the LHS of Eq. (1.268),p it becomes

(J iJ ) ζ k, σ = [j ( σ)] [j ( σ) + 1] ζ k, σ 1 − 1 ∓ 2 σ | − i − ± − ∓ − σ | − ∓ i (J iJ ) ζ k, σ = [j σ] [j σ + 1] ζ k, σ 1 (1.269) − 1 ∓ 2 σ | − i −p ∓ ± σ | − ∓ i and equating Eqs. (1.268, 1.269) we ﬁnd p ζσ = ζσ 1 (1.270) − ± we write the solution in the form j σ ζ = ( 1) − ζ (1.271) σ − where ζ is another phase that depends only on the species of particle. Combining Eqs. (1.267, 1.271), we obtain

j σ T k,σ = ( 1) − ζ k, σ (1.272) | i − | − i but unlike the intrinsic parity η, the time-reversal phase ζ has no physical signiﬁcance because we can redeﬁne the one-particle states in such a way that this phase is removed

i θ iθ k,σ k,σ ′ = e 2 k,σ ; ζ e | i → | i | i ≡ 1.13. PARITY AND TIME-REVERSAL FOR ONE-PARTICLE STATES WITH M > 0. 59 the new state under time-reversal gives

i θ i θ j σ i θ iθ j σ i θ i θ i θ T k,σ ′ = T e 2 k,σ = e− 2 T k,σ = ( 1) − e− 2 e k, σ = ( 1) − e− 2 e 2 e 2 k, σ | i | i | i − | − i − | − i j σ T k,σ ′ = ( 1) − k, σ ′ n o | i − | − i nevertheless, we shall continue using the phase in Eq. (1.272) in order to be free of choosing diﬀerent phases. But we should keep in mind that such a phase is physically irrelevant. Once again, in order to study arbitrary states p,σ , we use the standard Lorentz boost deﬁned by Eq. (1.192). | i From Eqs. (1.191, 1.244) we can check explicitly that

1 2 2 L (p) − = L ( p) ; p = p = p, p + M (1.273) T T P P T − µ p in words, changing the sign of each element of L ν with an odd number of time-indices is the same as changing the signs of elements with an odd number of space-indices. Applying P on Eq. (1.262) and using Eqs. (1.247, 1.273), we have

j σ T p,σ = ( 1) − ζ p, σ (1.274) | i − |P − i It worths observing that time-reversal understood as “putting the ﬁlm backwards” clearly reverse the sense of the three-momentum and left the energy invariant. Consequently, it transforms the four-momentum in the same way as space-inversion. By contrast, we observe that when applying time-reversal to the four-vector of position xµ, the three space-components remain invariant while the time-coordinate obviously reverse sign. Consequently, time-reversal applied on the position four-vector gives minus the space-inversion applied on such a four-vector. Therefore, the matrix representation of given by Eq. (1.244) works well on position four-vectors but not on T four-momenta. In turn it has to do with the fact that a matrix representation works well in all circumstances only if the associated operator is linear. However, time-reversal is antilinear so that the matrix representation (1.244) is not always adequate. From the previous discussion the statement p = p given in Eq. (1.273) must P T be interpreted in terms of the transformation of the operators but not in terms of the matrix representations (1.244). On the other hand, we also note that if time-reversal acted on the four-momentum according with the representation of in Eq. (1.244), the sign of p0 would be reversed leading to a non-physical state. T Finally, we observe that parity is a linear operator, so that the matrix representation (1.244) for , works well P for both the four-momentum and the position four-vector.

1.13.3 Parity for null mass particles In section 1.11 we took as the reference four-momentum kµ = (0, 0, κ, κ), which describes a particle of null mass traveling at the speed of light along the three-axis. Thus, the associated quantum state k,σ is eigenvector of P µ µ | i with eigenvalue k = (0, 0, κ, κ), and it is also eigenvector of J3 with eigenvalue σ (helicity). In the Minkowski space the parity operator gives a state with four-momentum P kµ = (0, 0, κ, κ) P − 60 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS thus the associated quantum state must be associated with opposite three-momentum. It can also be seen by taking into account that k,σ is eigenvector of P with eigenvalue k = (0, 0, κ), as follows | i P [P k,σ ]= P [P k,σ ]= P [k k,σ ]= k [P k,σ ] | i − | i − | i − | i as for J3 we see that J [P k,σ ]= P [J k,σ ]= σ [P k,σ ] 3 | i 3 | i | i hence P does not change the eigenvalue of J3. However, since J3 is the component of angular momentum along u (i.e. the original direction of motion), and P k,σ has the opposite direction of motion k, we see that the 3 | i − component of spin along the direction of motion of P k,σ has opposite sign with respect to k,σ (precisely | i | i because σ did not change its value). Thus, parity inverts the direction of motion and helicity of the state. This shows that the existence of a space inversion symmetry requires that any species of massless particle with non-zero helicity must be accompanied with another of opposite helicity. Since P does not leave the standard momentum invariant, it is convenient to use an operator that provides such an invariance. Hence we shall consider an operator28 1 U R2− P (1.275) 1 µ µ where R− is a rotation that takes k to k (i.e. that reverses the transformation of P over k,σ ). Consequently, 2 P | i R is a rotation that takes k to k. Of course, the simplest transformation that takes (0, 0, κ, κ) to (0, 0, κ, κ) 2 P − is a rotation of π around the two-axis29 U (R2) = exp (iπJ2) (1.276) 1 Since U R2− reverses the sign of J3, we have

1 U R− P k,σ = η k, σ (1.277) 2 | i σ | − i where ησ is a phase factor that could depend on σ. Now we characterize the action of P on an arbitrary one-particle state p,σ . We do it by applying P on Eq. | i (1.145) with L (p) given by (1.241) and using Eq. (1.277) to obtain

κ p 1 1 1 P p,σ = N (p) PU (L (p)) k,σ = PU R (p) B | | U R− P − U R− P k,σ | i | i p0 κ 2 2 | i r κ p 1 1 = PU R (p) B | | P − U (Rb) U R− P k,σ p0 κ 2 2 | i r κ p 1 = U R (pb) B | | − U (R ) η k, σ p0 P κ P 2 σ | − i r κ p 1 P p,σ = U R (pb) B | | − R η k, σ (1.278) | i p0 P κ P 2 σ | − i r 1 Further, it can be checked that theb Lorentz “boost” (1.242) commutes with R− (and so with its inverse 2 P 1R )30, and that commutes with the rotation R (p) of Eq. (1.177) that takes the three-axis into the direction P− 2 P 28Since we are considering the extended Poincar´egroup (including discrete operations), we could think about operator (1.275), as an extension of the Little group that includes discrete symmetriesb that leave kµ invariant. We shall also consider an operator of the −1 form U R2 T as another extension of the little group associated with the discrete time-reversal symmetry. 29This is a convention since it could also be around the one-axis. An additional original rotation around the three-axis could be considered, but it is not necessary. 30If [A, B] = 0, then A−1 also commutes with B:

A−1, B = A−1B − BA−1 = A−1BA A−1 − A−1 ABA−1 = BA−1 − A−1B A−1, B = − A−1, B = 0 1.13. PARITY AND TIME-REVERSAL FOR ONE-PARTICLE STATES WITH M > 0. 61 of p. Therefore we have

κ p P p,σ = η U R (p) R B | | k, σ (1.280) | i σ p0 2 κ | − i r it worths emphasizing that R (p) R is a rotation thatb takes the three-axis into the direction of p, however 2 − U (R (p) R ) is quite diﬀerent from U (R ( p)), as can be seen by applying Eq. (1.179) 2 − b b U (R ( p)) = exp[ i (φ π) J ] exp [ i (π θ) J ] (1.281) b − b − ± 3 − − 2 the azimuthal angle is chosen as φ+π if 0 φ<π or as φ π if π φ< 2π. It is in order that the azimuthal angle b ≤ − ≤ remains within the interval [0, 2π). On the other hand, the two-component of p is given by p u = p sin θ sin φ, · 2 since 0 θ π we have sin θ 0 so that p u > 0 if 0 <φ<π, and p u < 0 if π<φ< 2π. Then we could say ≤ ≤ ≥ · 2 · 2 alternatively that the azimuthal angle in (1.281) is chosen as φ + π or φ π according whether the two-component − of p is positive or negative respectively31. From Eq. (1.281) we obtain

1 1 U − (R ( p)) U (R (p) R ) = U − (R ( p)) U (R (p)) U (R ) − 2 − 2 = exp [i (π θ) J ] exp [i (φ π) J ] exp [ iφJ ] exp [ iθJ ] exp (iπJ ) { − 2 ± 3 } { − 3 − 2 } 2 1 b b b b U − (R ( p)) U (R (p) R ) = exp [i (π θ) J ] exp [ iπJ ] exp [i (π θ) J ] − 2 − 2 ± 3 − 2 or equivalentlyb b U (R (p) R )= U (R ( p)) exp [i (π θ) J ] exp [ iπJ ] exp [i (π θ) J ] 2 − − 2 ± 3 − 2 but a rotation of π around the three–axis reverses the sign of J2. Therefore ± b b U (R (p) R ) = U (R ( p)) exp [ iπJ ] exp [i (π θ) ( J )] exp [i (π θ) J ] 2 − ± 3 − − 2 − 2 U (R (p) R2) = U (R ( p)) exp [ iπJ3] (1.282) b −b ± Note that despite U (R (p) R ) does not coincide with U (R ( p)) as anticipated, they diﬀer by an initial rotation b 2 b − around the three-component, and the associated initial rotation around the three-axis in the Minkowski space, clearly does not alter theb action of R (p) R2 over the three-axis.b From the form of the standard boost (1.241), we can obtain L ( p) as follows P b p p L (p)= R (p) B | | L ( p)= R ( p) B | | ; p = p,p0 (1.283) κ ⇒ P − κ P − using (1.282, 1.283), we have b b

p p p U (R (p) R ) U B | | = U (R ( p)) exp [ iπJ ] U B | | = U (R ( p)) U B | | exp [ iπJ ] 2 κ − ± 3 κ − κ ± 3 (1.284) b b b 31If the two-component of p is zero, we have φ = 0 (positive one-component) or φ = π (negative one-component). For φ = 0 (positive one-component) we choose φ + π, and for φ = π (negative one-component) we choose φ − π. 62 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS and substituting (1.284) in equation (1.280) we have κ p P p,σ = η U (R ( p)) U B | | exp [ iπJ ] k, σ | i σ p0 − κ ± 3 | − i r κ p = η U R ( bp) B | | exp [ iπσ] k, σ σ p0 − κ ∓ | − i r κ P p,σ = η exp [ iπσ] b U (L ( p)) k, σ | i σ ∓ p0 P | − i r and according with (1.145) page 33, we ﬁnally obtain

P p,σ = η exp [ iπσ] p, σ (1.285) | i σ ∓ |P − i where the phase is πσ or +πσ if the two-component of p is positive or negative respectively. This produces a − change in sign only if σ is half-integer. Such a change in sign in the operation of parity for masless particles with half-integer spin is due to the convention adopted in Eq. (1.179) for the rotation used to deﬁne massless particles of arbitrary momentum. Such kind of discontinuities are unavoidable owing to the fact that SO (3) is not simply connected32.

1.13.4 Time-reversal for null mass particles We already saw that the reference four-momentum is kµ = (0, 0, κ, κ), and that the quantum state k,σ is µ µ | i eigenvector of P , J3 with eigenvalues k = (0, 0, κ, κ), and σ respectively. We also see that in the Minkowski space, the time reversal operator gives a state with four-momentum T kµ = kµ = (0, 0, κ, κ)= p,p0 T P − − Further, Eq. (1.255) shows that T anticommutes with P and J. Therefore, it reverses the sign of p and σ.

P [T k,σ ]= T P k,σ = k [T k,σ ] ; J [T k,σ ]= T J k,σ = σ [T k,σ ] | i − | i − | i 3 | i − 3 | i − | i hence T does not change the helicity J k, because it reverses the sign of both quantities. Consequently, time- · reversal says nothing about whether massless particles of one helicity σ are accompanied with others of helicity σ. Like in the case of P , time-reversalb does not leave the stardard four-momentum invariant, from which it is − µ 1 more convenient to work with a related operator that does leave k invariant. We see that U R2− T does this role where R is the rotation deﬁned in (1.276), and takes k into k. This commutes with J , so 2 P 3 1 U R− T k,σ = ζ k,σ (1.286) 2 | i σ | i where ζσ is in principle a helicity-dependent phase. Once again, we shall characterize the action of T on an arbitrary one-particle state p,σ . We do it by applying | i T on the state (1.145) and using Eq. (1.286) we ﬁnd

1 Since R− commutes with the boost B ( p /κ), and commutes with the rotation R (p), we obtain 2 T | | T

T p,σ = ζ exp [ iπσ] p,σ (1.288) | i σ ± |P i again the positive or negative sign appears according to whether the two-component of p is positive or negative respectively. If the two-component of p is null, the positive or negative sign appears according to whether the one-component of p is positive or negative respectively.

1.14 Action of T 2 and Kramer’s degeneracy

Using (1.274) for massive one-particle states, and taking into account that T is antilinear antiunitary, we have

2 j σ j σ j σ j+σ 2 T p,σ = T [T p,σ ]= T ζ ( 1) − p, σ = ( 1) − ζ∗ [T p, σ ] = ( 1) − ζ∗ ζ ( 1) p,σ | i | i − |P − i − |P − i − − P T 2 p,σ = ( 1)2j p,σ h i h (1.289) i | i − | i we shall see that for massless particles the result is the same. For massless particles, if the two-component of p is positive (negative), then the two-component of p is negative (positive). Therefore, since T p,σ = P | i ζ exp [ iπσ] p,σ , we have σ ± |P i T p,σ = ζ exp [ iπσ] 2p,σ = ζ exp [ iπσ] p,σ (1.290) |P i σ ∓ P σ ∓ | i

From Eqs. (1.288, 1.290), we obtain

2 T p,σ = T ζ exp [ iπσ] p,σ = ζ∗ exp [ iπσ] T p,σ = ζ∗ exp [ iπσ] ζ exp [ iπσ] p,σ | i { σ ± |P i} σ ∓ |P i σ ∓ σ ∓ | i T 2 p,σ = exp [ 2iπσ] p,σ | i ∓ | i as long as σ is integer or half-integer, we can rewrite it as

2 2 σ T p,σ = ( 1) | | p,σ (1.291) | i − | i we usually deﬁne the “spin” of a massless particle as the absolute value of the helicity, from which Eq. (1.289) is equivalent to (1.291). 64 CHAPTER 1. RELATIVISTIC QUANTUM MECHANICS

Suppose now that T 2 acts on a state associated with a system of non-interacting particles, either massive or 2j 2 σ massless. Such an action yields a factor ( 1) or ( 1) | | for each particle. Hence if the system consists of an − − odd number of particles of half-integer spin or helicity (perhaps with some additional particles of integer spin or helicity), we obtain an overall phase given by

T 2 Ψ = Ψ (1.292) | i − | i if we now “turn on” several interactions, all of which respect time-reversal invariance, this result will be preserved even if rotational invariance is not respect. As an example, we could add an arbitrary static gravitational and/or electric field. Now suppose that Ψ is an eigenstate of the Hamiltonian. Since T commutes with the Hamiltonian, T Ψ is | i | i also an eigenstate of H with the same eigenvalue (same energy). It is natural to ask whether T Ψ is the same | i state or a different one. In the latter case we would have a degeneracy (two or more stationary states with the same energy). Assuming that T Ψ is the same state as Ψ , both of them can only differ at most in a phase | i | i T Ψ = ζ Ψ | i | i in that case 2 2 T Ψ = T [ζ Ψ ]= ζ∗T Ψ = ζ∗ζ Ψ = ζ Ψ = Ψ | i | i | i | i | | | i | i in contradiction with Eq. (1.292). Therefore, any energy eigenstate Ψ that satisfies Eq. (1.292) must be | i degenerate with (at least) another eigenstate of the same energy. This is known as “Kramer’s degeneracy”. This conclusion is trivial if the system is in a rotationally invariant environment, because the total angular momentum j of such an state would have to be half-integer, leading to 2j +1=2, 4, 6,...degenerate states. The surprising result is that even if the rotational invariance is broken (for instance by introducing external fields such as an electrostatic field) a two-fold degeneracy persists as long as the fields introduced are time-reversal invariant. As a particular case, if any particle had an electric or gravitational dipole moment then the degeneracy among its 2j + 1 spin states would be entirely removed in a static electric or gravitational field. It implies that such dipole moments are forbidden by time-reversal invariance. Chapter 2

Scattering theory

We have constructed so far one-particle states in the framework of relativistic quantum mechanics. Nevertheless, in nature and experiments we have sets of interacting particles. By now, we shall study predictions concerning experiments in which a set of non-interacting particles (that are very far from each other) approach each other from macroscopically large distances, and interact (or collide) in a microscopically small region, after which the products of the interaction fly off in such a way that they become far from each other again. We use to say that we prepare the system of non-interacting particles in a region far from the collision region (the incoming or simply the “in” region) at a time t , they approach to the “collision” or “interaction” region at a time t 0, and → −∞ ∼ after the products travel out again they arrive to the outcoming region (the “out” region) at a time t , in → ∞ which the particles are non-interacting again. The in and out regions corresponding to t and t → −∞ → ∞ respectively, are called asymptotic regions. In the asymptotic regions particles are effectively non-interacting, hence they can be described as direct products of the one-particle states that we have constructed in section 1.9. In such experiments, we detect particles and infer probability distributions or “cross sections”. Our aim in this chapter is to develop a formalism that permits to calculate such probabilities and cross-sections.

2.1 Construction of “in” and “out” states

A state that describes several non-interacting particles can be considered as a state that transforms under the inhomogeneous Lorentz group, as a direct product of one-particle states. We shall label the one-particle states with the four-momenta pµ, the third component of spin (or the helicity for massless particles) σ and an additional discrete label n that take into account that we could be dealing with several species of particles. The index n could mean several indices that speciﬁes the particle type by indicating its mass, charge, spin, etc. The general transformation rule under the inhomogeneous Lorentz transformations for a one-particle state of a massive particle is obtained by using the transformation rule (1.180) page 40, valid for homogeneous Lorentz transformations and adding the inhomogeneous part (space-time translations) given by (1.142) page 32

0 µ (Λp) (j) U (Λ, a) p,σ = exp [ iaµ (Λp) ] D ′ (W (Λ,p)) Λp,σ′ (2.1) | i − s p0 σ σ σ′ X and for massless particles, we combine Eq. (1.239) page 51, with Eq. (1.142) page 32 to ﬁnd

0 µ (Λp) U (Λ, a) p,σ = exp [ iaµ (Λp) ] exp [iσθ (Λ,p)] Λp,σ (2.2) | i − s p0 | i where W (Λ,p) is the Wigner rotation deﬁned in Eq. (1.181) page 41, associated with the little group of massive particles SO (3) or the one associated with masless particles ISO(2). The general transformation rule for a set of

65 66 CHAPTER 2. SCATTERING THEORY non-interacting particles is then obtained by making direct products of the transformation rule (2.1) or (2.2) for massive and massless particles respectively

0 0 µ µ (Λp1) (Λp2) U (Λ, a) p1,σ1,n1; p2,σ2,n2; . . . = exp iaµ [(Λp1) + (Λp2) + . . .] · · · | i {− } s p0p0 1 2 · · · Fσ′ σ (W (Λ,p1)) Fσ′ σ (W (Λ,p2)) Λp1,σ′ ,n1; Λp2,σ′ ,n2; . . . × 1 1 2 2 · · · 1 2 σ′ σ′ X1 2··· (2.3) where (ji) 2 Dσ′ σ (W (Λ,pi)) if p < 0 F ′ (W (Λ,pi)) = i i (2.4) σiσi 2 ( exp [iσiθ (Λ,pi)] if p = 0 (j) where Dσ′σ (W (Λ,p)) are the unitary matrices associated with the (j) irreducible representation of SO (3) with dimension 2j + 1, and the θ (Λ,p) angle is the one deﬁned in Eq. (1.240) page 52. The states are normalized according with Eq. (1.162) page 38

3 3 p′ ,σ′ ,n′ ; p′ ,σ′ ,n′ ; . . . p1,σ1,n1; p2,σ2,n2; . . . = δ p′ p1 δσ′ σ δn′ n δ p′ p2 δσ′ σ δn′ n 1 1 1 2 2 2 | i 1 − 1 1 1 1 2 − 2 2 2 2 · · · permutations (2.5) ± the term “ permutations” is added to take into account of the possibility that it is some permutation of the ± particle types n1′ ,n2′ ,...that are of the same species as the particle types n1,n2,...; we shall see later that its sign is 1 if this permutation includes an odd permutation of half-integer spin particles, and it is +1 otherwise. − We shall often abbreviate the notation by using a single symbol (a greek letter) for the whole collection of quantum numbers α (p ,σ ,n ; p ,σ ,n ; . . .) (2.6) ≡ 1 1 1 2 2 2 then we have the following assigments

α p ,σ ,n ; p ,σ ,n ; . . . ; dα d3p d3p (2.7) | i ≡ | 1 1 1 2 2 2 i ···≡ · · · 1 2 · · · n ,σ n ,σ Z X1 1 X2 2 Z in summing and integrating over states it is understood that we include only conﬁgurations that do not diﬀer simply by the exchange of identical particles. In this notation, orthonormalization (2.5) and completeness for states normalized as in (2.5) are written as

3 α′ α = δ α′ α (2.8) | i − Ψ = dα α α Ψ (2.9) | i | i h | i Z we should keep in mind that the transformation rule (2.3) is only valid for non-interacting particles. Setting µ µ µ Λ ν = δ ν and a = (0, 0, 0,τ), that is, using a pure time-translation U (Λ, a) = exp (iHτ), Eq. (2.3) requires among other conditions, that α be an energy eigenstate | i H α = E α (2.10) | i α | i with energy equal to the sum of one-particle energies

0 0 Eα = p1 + p2 + . . . (2.11) without interaction terms. That is, no terms involving two or more particles at a time. 2.1. CONSTRUCTION OF “IN” AND “OUT” STATES 67

In scattering processes, equation (2.3) applies to both asymptotic states that we call “in” states denoted as α+ (at t ) and “out” states denoted as α (at t + ), the asymptotic states α+ and α will | i → −∞ | −i → ∞ | i | −i be found to contain the particles described by the label α if observations are made at t or t + → −∞ → ∞ respectively. To keep manifest Lorentz invariance, it is more convenient to use the Heisenberg picture, in which state- vectors do not change in time, so that a state-vector Ψ describes the whole space time history of a system of | i particles. In this picture, operators carry all time-dependence. Therefore, we do not say that α are the limits | ±i at t of a time dependent state vector α (t) . → ∓∞ | i By defining a state we are implicitly choicing an inertial reference frame. Different observers see equivalent state-vectors but not the same state-vector. Suppose that a standard observer sets its clock so that t = 0 is a O given time during the collision process, and another observer (at rest with respect to ) sets its clock so that O′ O t = 0 is at a time t = τ. The two observers time-coordinates are related by t = t τ. If sees the system in a ′ ′ − O state Ψ , will see the system in a state | i O′ U (1, τ) Ψ = exp [ iHτ] Ψ − | i − | i hence, the appearance of the state long before or after the collision (asymptotic states in whatever the basis used by ) is found by applying a time-translation operator exp [ iHτ] with τ or τ + , respectively. If O − → −∞ → ∞ the state is an energy eigenstate, it cannot be localized in time (because of the time-energy uncertainty principle, the characteristic time of evolution of the system is infinite). In that case, the operator exp [ iHτ] simply yields − an irrelevant phase exp [ iE τ]. However, the situation is different when the state Ψ consists of a wave-packet − α | i of energy eigenstates Ψ = dα g (α) α (2.12) | i | i Z we shall assume that the amplitude g (α) is non-zero and varies smoothly over some finite range ∆E of energies. The “in” and “out” states are defined such that the superposition

exp [ iHτ] dα g (α) α± = dα g (α) exp [ iE τ] α± − − α Z Z

1 1 has the form of a corresponding superposition of free particle states for τ << (∆E)− or τ >> (∆E)− , − respectively. To construct this “in” and “out” states, suppose we can divide the time-translation generator (Hamiltonian) into two terms, a free-particle Hamiltonian H0 and an interaction V

H = H0 + V in such a way that H has eigenstates α (0) with the same appearance as the eigenstates α of the complete 0 | i | ±i Hamiltonian

(0) (0) H0 α = Eα α (2.13)

(0) (0)E E α′ α = δ α ′ α (2.14) − D E where we have assumed that H0 has the same spectrum as the full Hamiltonian H. It demands that the masses appearing in H0 be the physical masses that are actually measured, which are not necessarily the bare mass terms appearing in H. If there is any difference, it must be included in V and not in H0. Further, any relevant bound states in the spectrum of H should be introduced into H0 as if they were elementary particles. For instance, in “rearrangement collisions” in which some bound states appear in the initial state but not the final state or vice-versa, one must use a different split of H into H0 and V in the initial and final states. In non-relativistic problems it is usual to include the binding potential in H0. 68 CHAPTER 2. SCATTERING THEORY

The “in” and “out” states can now be deﬁned as eigenstates of H, not H0

H α± = Eα α± (2.15) since the form or appearance of α(0) is the same as the one of eigenstates α , the “in” and “out” states satisfy | ±i the condition (0) dα g (α) exp [ iE τ] α± dα g (α) exp [ iE τ] α (2.16) − α → − α Z Z E for τ and τ + respectively. Equation (2.16) can be rewritten as → −∞ → ∞ (0) lim exp [ iHτ] dα g (α) α± lim exp [ iH0τ] dα g (α) α (2.17) τ − → τ − →∓∞ Z →∓∞ Z E we can write Eq. (2.17) as

(0) lim dα g (α) α± lim exp [+iHτ] exp [ iH0τ] dα g (α) α (2.18) τ → τ − →∓∞ Z →∓∞ Z E now, apart from being smooth, g (α) is arbitrary from which Eq. (2.18) leads to a formula for the “in” and “out” states (0) α± = Ω ( ) α ; Ω (τ) exp[+iHτ] exp [ iH τ] (2.19) ∓∞ ≡ − 0 E nevertheless, we should take into account that Ω ( ) in Eq. (2.19) gives meaningful results only when acting ∓∞ on a smooth superposition of energy states. As a consequence of deﬁnition (2.16), the “in” and “out” states are normalized like the free-particle states. We can see it by observing that the LHS of Eq. (2.16) is obtained by applying the unitary operator exp [ iHτ] − to a time independent state (since we are in the Heisenberg picture)

dα g (α) exp [ iE τ] α± = exp [ iHτ] dα g (α) α± − α − Z Z therefore, its norm is time-independent, in particular equals the norm of its limit τ , which is the norm of → ±∞ the RHS of Eq. (2.16). The equality of the square of the norms of the wave-packets on both sides of (2.16) can be written as

(0) (0) dα dβ exp [ i (E E ) τ] g (a) g∗ (β) β± α± = dα dβ exp [ i (E E ) τ] g (a) g∗ (β) β α − α − β − α − β Z Z D E since this equality holds for all smooth functions g (α), the scalar products must be equal, thus

(0) (0) β± α± = β α = δ (β α) − D E for future purposes we shall write an explicit formal soluti on of the energy eigenvalue equation (2.15) that satisﬁes the conditions (2.16). To do it, we rewrite Eq. (2.15) as

(H0 + V ) α± = Eα α± (Eα H0) α± = V α ± − the operator E H contains at least one null eigenvalue [since any given E is part of the spectrum of H α − 0 α 0 according with Eqs. (2.13, 2.15)], therefore it is not invertible. Such an operator annihilates not only the free- particle state α(0) but also the continuum of other free particle states β(0) of the same energy. In order to permit the invertibility of the operator we shall shift it by a quantity iε, where ε is a positive inﬁnitesimal ± number1 (E H iε) α± = V α± (2.20) α − 0 ± 1 The shift is made in the imaginary part to ensure that Eα ± iε is not part of the spectrum of H0 (which must be real).

2.1. CONSTRUCTION OF “IN” AND “OUT” STATES 69 we shall write a temptative solution of Eq. (2.20), as the solution of the homogeneous equation (with V 0) → which is given by α(0) , plus a particular solution obtained by taking into account that the operator E H iε α − 0 ± is invertible (0) 1 α± = α + (E H iε)− V α± (2.21) α − 0 ± E using the completeness of the free-particle states β(0) this solution becomes

(0) (0) (0) 1 α± = α + dβ β β (E H iε)− V α± α − 0 ± E Z ED (0) (0) (0) 1 = α + dβ β β (E H iε)− V α ± α − 0 ± E Z EnD o (0) (0) (0) 1 α± = α + dβ β β (Eα Eβ iε)− V α± − ± E Z EnD o where we have used the fact that H0 is assumed to have the same spectrum as the full Hamiltonian H. We then have ﬁnally T (0) βα± (0) (0) α± = α + dβ β ; Tβα± β V α± (2.22) (Eα Eβ iε) ≡ E Z − ± E D

Expresions (2.21, 2.22) are known as the Lippmann-Schwinger equations. Now we should show that equations (2.22) with +iε or iε in the denominator, satisﬁes the condition (2.16) for an “in” or an “out” state, respectively. − To show it, let us consider the wave-packets

iEαt g± (t) dα e− g (α) α± (2.23) ≡ Z (0) iEαt (0) g (t) dα e− g (α) α (2.24) ≡ E Z E we want to show that g+ (t) and g (t) approach g(0) (t) when t and t + , respectively. | i | − i → −∞ → ∞ Substituting (2.22) in (2.23) we obtain

T ± iEαt (0) βα (0) g± (t) dα e− g (α) α + dβ β ≡ ( (Eα Eβ iε) ) Z E Z − ± E iEαt e− g (α) T ± iEαt (0) βα (0) g± (t) = dα e− g (α) α + dα dβ β (Eα Eβ iε) Z E Z Z − ± E by using Eq. (2.24) and interchanging the order of integration we ﬁnd

e iEαtg (α) T (0) (0) − βα± g± (t) = g (t) + dβ β dα (E E iε) Z Z α β E E − ± (0) (0) g± (t) = g (t) + dβ β ± (2.25) Iβ Z EiEαt E e− g (α) T βα± ± dα (2.26) Iβ ≡ (E E iε) Z α − β ± we examine ﬁrst the case for t . By making the complex extension of E , the exponential in Eq. (2.26) for → −∞ α t yields → −∞ iEαt i(ReEα+iImEα) t i(ReEα) t (ImEα) t e− = e | | = e | |e− | | (2.27) the integration in (2.26) is over all quantum numbers, in particular over the energy. As for the integration with respect to energy, we can close the contour of integration for the energy variable in the upper half complex 70 CHAPTER 2. SCATTERING THEORY plane with a large semi-circle. It is clear that at any point of the semi-circle (except over the real axis) we have ImEα > 0. Therefore, it is clear from (2.27) that the contribution from this semi-circle is killed by the factor exp [ iE t], which is exponentially small for t and ImE > 0. The integral is then given by a sum over − α → −∞ α the singularities of the integral in the upper half plane.

The functions g (α) and Tβα± could in general have singularities at values of Eα with finite positive imaginary parts, but their contribution is exponentially attenuated for t . For this fact to be satisfied, t must be → −∞ − much greater than both the time uncertainty in the wave-packet g (α) and the duration of the collision, which respectively govern the location of the singularities of g (α) and Tβα± in the complex Eα plane. This leaves the 1 singularity in (Eα Eβ iε)− . In this factor, the singularity in the upper-half plane is at ImEα = ε, since + − ± ε 0 then ImEα 0 and the exponential in Eq. (2.27) does not necessarily damp this contribution. However → | |→ + + such a singularity in the upper half-plane appears for − but not for . We conclude then that vanishes for Iβ Iβ Iβ t . → −∞ In a similar way, for t + we must close the contour of integration in the lower half-plane, and we see that → ∞ + (0) β− vanishes in this limit. This reasoning along with Eq. (2.25) shows that g (t) approaches g (t) when I (0) | i t , and that g− (t) approaches g (t) when t + , showing that the condition (2.16) is satisfied. → −∞ | i → 1∞ For future purposes, we shall represent (Eα Eβ iε)− in a more convenient way as − ± 1 1 (E iε) (E iε) E ε = ∓ = ∓ = iπ E iε E iε (E iε) E2 + ε2 E2 + ε2 ∓ π (E2 + ε2) ± ± ∓ Hence we write

1 Pε (E iε)− = iπδ (E) (2.28) ± E ∓ ε P E ε ε ; δ (E) (2.29) E ≡ E2 + ε2 ε ≡ π (E2 + ε2) the function P/E in (2.29) approaches 1/E for E >> ε, and vanishes for E 0, so for ε 0 is behaves like the | | → → “principal value function” P/E, that permits to integrate 1/E times any smooth function of E, by excluding an inﬁnitesimal interval around E = 0. On the other hand, the function δ (E) in (2.29) is of order ε for E >> ε, and ε | | gives unity when integrated over all E. Indeed ε/ π x2 + ε2 is one of the well-known functions that approaches the Dirac delta function when ε approaches zero from the positive side. From this discussion, we can drop the ε label in Eq. (2.28) to write 1 P (E iε)− = iπδ (E) ± E ∓

2.2 The S matrix − In scattering experiments, the initial state (the “in” state α+ ) is usually prepared to have a deﬁnite particle | i content (deﬁned by the set of quantum numbers α) at t , and then it is measured how the state looks like → −∞ at t (the “out” state β ) with particle content β. The probability amplitude for the transition α β is → ∞ | −i → thus the scalar product + Sβα = β− α (2.30) this array of complex amplitudes is called the S matrix. If there were no interactions, the “in” and “out” states − would be the same so that S = δ (α β) for non-interaction (2.31) βα − The rate for a reaction α β is proportional to → S δ (α β) 2 (2.32) | αβ − − | 2.2. THE S MATRIX 71 −

It is important to take into account that “in” and “out” states belong to the same Hilbert space. They only diﬀer in the way they are labelled: by their appearance at t or at t + . Any “in” state can be → −∞ → ∞ expanded as a superposition of “out” states by using the completeness of the latter

+ + α = dβ β− β− α = dβ β− Sβα (2.33) Z Z thus the coeﬃcients of the expansion are the S matrix elements (2.30). Thus, the role of this matrix is to connect − two complete sets of orthonormal states, hence it must be unitary. This fact is more apparent by applying the orthonormality and completeness of both “in” and “out” states

+ + + + δ (γ α) = γ α = dβ γ β− β− α = dβ Sβγ∗ Sβα = dβ S† Sβα − γβ Z Z Z + + δ (γ α) = γ− α− = dβ γ− β β α− = dβ SγβSαβ∗ = dβ Sγβ S† − βα Z Z Z so we have obtained

dβ S† Sβα = δ (γ α) S†S = 1 (2.34) γβ − ⇔ Z dβ Sγβ S† = δ (γ α) SS† = 1 (2.35) βα − ⇔ Z it worths emphasizing that the conditions S†S = 1 and SS† = 1 are not equivalent for matrices in inﬁnite dimensions. In many cases, instead of dealing with the S matrix, it is convenient to work with an operator S, deﬁned − such as its matrix elements between free-particle states coincide with the corresponding elements of the S matrix − β(0) S α(0) S (2.36) ≡ βα D E since the formula (2.19) connects “in” and “out” states with states of free particles (but with the same particle content), it provides a formula for the S operator − + (0) (0) S = β− α = β Ω† (+ ) Ω ( ) α βα ∞ −∞ (0) nD (0) on Eo S = β Ω† (+ ) Ω ( ) α (2.37) βα ∞ −∞ D E observe that Ω (t) does not depend on the particle content. Comparing Eqs. (2.36, 2.37) and using the second of Eqs. (2.19) we have

S = Ω† (+ ) Ω ( ) U ( , ) (2.38) ∞ −∞ ≡ −∞ ∞ U (τ,τ ) Ω† (τ) Ω (τ ) = exp[+iH τ] exp [ iH (τ τ )] exp [ iH τ ] (2.39) 0 ≡ 0 0 − − 0 − 0 0 we shall derive an alternative formula for the S matrix by returning to Eqs. (2.25, 2.26), for the “in” state − g+ (t) , and then taking t + | i → ∞ e iEαtg (α) T + + (0) (0) + + − βα g (t) = g (t) + dβ β β ; β dα (2.40) I I ≡ (Eα Eβ + iε) E Z E Z −

We now close the contour of integration for E in the lower half-plane, running from E = to E = + , α α −∞ α ∞ and then back to E = through a large semi-circle in the lower half-plane. Once again, the singularities in α −∞ T + and g (α) are damped by the exponential at t + in the lower half-plane, so they give no contribution. βα → ∞ 72 CHAPTER 2. SCATTERING THEORY

1 But we now pick up a contribution coming from the factor (E E + iε)− . With our choice of circulation, α − β the singularity is circled in a clockwise sense. We have a pole of ﬁrst order and by the method of residues, the contribution to the integral over Eα is given by

iEαt + e− g (α) T βα iEαt + 2πi (Eα Eβ + iε) = 2πi e− g (α) Tβα − − (Eα Eβ + iε) − Eα=Eβ iε − Eα Eβ= iε − − − i.e. the value of the integrand at E = E iε times a factor 2πi. Therefore, in the limit ε 0+ and t + , α β − − → → ∞ the integral + over α in (2.40) has the asymptotic behavior Iβ

+ iE t + 2πie− β dα δ (E E ) g (α) T (2.41) Iβ →− α − β βα Z hence for t + , and using (2.24, 2.41), equation (2.40) gives → ∞ + iE t (0) iE t (0) + g (t + ) = dβ e− β g (β) β 2πi dβ e− β β dα δ (E E ) g (α) T → ∞ − α − β βα Z E Z E Z + (0) iE t + g (t + ) = dβ β e− β g (β) 2πi dα δ (E E ) g (α) T (2.42) → ∞ − α − β βα Z E Z and expanding (2.23) in a complete set of out states, we have

+ iEαt + iEαt + g (t) dα e− g (α) α = dα e− g (α) dβ β− β− α ≡ Z Z Z + iEαt g (t) = dα e− g (α) dβ β− Sβα Z Z and since S contains a factor δ (E E ), we can write βα β − α

+ iEβ t g (t) = dβ β− e− dα g (α) Sβα Z Z and using the property (2.16) that deﬁnes the “out” states, the asymptotic behavior of g+ (t) for t + yields | i → ∞ + (0) iE t g (t + ) = dβ β e− β dα g (α) S (2.43) → ∞ βα Z E Z comparing Eqs. (2.42, 2.43), and using the linear independe nce of β(0) , we have dα g (α) S = g (β) 2πi dα δ (E E ) g (α) T + βα − α − β βα Z Z dα g (α) S = dα g (α) δ (β α) 2πi δ (E E ) T + βα − − α − β βα Z Z h i + using the arbitariness of g (α) and Tβα we obtain

S = δ (β α) 2πi δ (E E ) T + (2.44) βα − − α − β βα For a weak interaction V , equation (2.44) suggests a simple aproximation for the S matrix: for small V we can − + neglect the diﬀerence between the “in” states and the free particles states in the deﬁnition of Tβα given in Eq. (2.22). In that case Eq. (2.44) becomes

S δ (β α) 2πi δ (E E ) β(0) V α(0) (2.45) βα ≃ − − α − β D E

2.2. THE S MATRIX 73 − equation (2.45) is known as the Born approximation. We shall study later how to calculate higher-order terms. On the other hand, the Lippmann-Schwinger equations (2.21) for the “in” and “out” states can provide an alternative proof of the orthonormality of these states, the unitarity of the S matrix, and the validity of Eq. − (2.44) without dealing with limits at t . → ∓∞ First by applying the Lippmann-Schwinger equations (2.21) either on the LHS or the RHS of the matrix element β V α we ﬁnd h ±| | ±i (0) 1 (0) 1 β± V α± = β + β± V (E H iε)− V α± = β V α± + β± V (E H iε)− V α± β − 0 ∓ β − 0 ∓ nD (0) 1o D (0) 1 β± V α± = β± V α + (E H iε)− V α ± = β± V α + β± V (E H iε)− V α± α − 0 ± α − 0 ± n E o E and equating both expressions, we have

(0) 1 (0) 1 β± V α + β± V (E H iε)− V α± = β V α± + β± V (E H iε)− V α± α − 0 ± β − 0 ∓ E 1 D 1 T ±∗ + β± V (Eα H0 iε)− V α± = T ± + β ± V (E β H0 iε)− V α± αβ − ± βα − ∓ 1 1 T ±∗ T ± = β± V (E H iε)− V α± β ± V (E H iε)− V α ± (2.46) αβ − βα β − 0 ∓ − α − 0 ± inserting an identity with free-states γ(0) on the RHS of (2.46) we have

1 (0) (0) D β± V ( E H iε)− dγ γ γ V α± αβ ≡ β − 0 ∓ Z ED 1 (0) (0) β ± V (E H iε)− dγ γ γ V α± − α − 0 ± Z ED 1 (0) (0) = dγ β± V (E H iε)− γ γ V α± β − 0 ∓ Z n Eo D 1 (0) (0) dγ β ± V (Eα H0 iε)− γ γ V α± − − ± Z n EoD 1 (0) (0) = dγ (Eβ Eγ iε)− β± V γ γ V α ± − ∓ Z ED 1 (0) (0) dγ (Eα Eγ iε)− β ± V γ γ V α± − − ± Z ED 1 1 Dαβ = dγ (Eβ Eγ iε)− T ±∗T ± dγ (Eα E γ iε)− T ±∗T ± (2.47) − ∓ γβ γα − − ± γβ γα Z Z and equating Eqs. (2.46, 2.47) we obtain

1 1 T ±∗ T ± = dγ (E E iε)− (E E iε)− T ±∗T ± (2.48) αβ − βα β − γ ∓ − α − γ ± γβ γα Z h i By dividing Eq. (2.48) by E E 2iε we get α − β ±

Tαβ±∗ Tβα± = dγF T ±∗T ± (2.49) E E 2iε − E E 2iε αβγ γβ γα α − β ± α − β ± Z 1 1 F αβγ ≡ (E E iε) (E E 2iε) − (E E iε) (E E 2iε) β − γ ∓ α − β ± α − γ ± α − β ± as for Fαβγ we have (E E iε) (E E iε) (E E 2iε) F = α − γ ± − β − γ ∓ = α − β ± αβγ (E E iε) (E E 2iε) (E E iε) (E E iε) (E E 2iε) (E E iε) α − γ ± α − β ± β − γ ∓ α − γ ± α − β ± β − γ ∓ 1 F = (2.50) αβγ (E E iε) (E E iε) α − γ ± β − γ ∓ 74 CHAPTER 2. SCATTERING THEORY substituting (2.50) in (2.49) we ﬁnd

T ±∗ T ± T ±∗Tγα± αβ βα = dγ γβ E E 2iε − E E 2iε (E E iε) (E E iε) α − β ± α − β ± Z α − γ ± β − γ ∓ T ±∗ T ± T ±∗ T ± αβ + βα = dγ γβ γα E E 2iε E E 2iε − (E E iε) (E E iε) β − α ∓ α − β ± Z β − γ ∓ α − γ ± T ± ∗ T ± T ± ∗ T ± αβ + βα = dγ γβ γα E E 2iε E E 2iε − E E iε (E E iε) β − α ± ! α − β ± Z β − γ ± ! α − γ ± the factors ε in the denominators on the integrals on the RHS can be replaced with 2ε, since the only important thing is that these are positive inﬁnitesimals. Hence

T ± ∗ T ± T ± ∗ T ± αβ + βα = dγ γβ γα (2.51) E E iε E E iε − E E iε (E E iε) β − α ± ! α − β ± Z β − γ ± ! α − γ ±

1 + By applying Eq. (2.51), we shall show now that the quantity δ (β α) + (E E iε)− T deﬁnes a unitary − α − β ± βα matrix.

+ + Tβα Tηα∗ Kβα = δ (β α) + ; K† = Kηα∗ = δ (η α)+ (2.52) − E E iε αη − E E iε α − β ± α − η ∓ + + Tβα Tηα∗ Kβα K† = δ (β α) + δ (η α)+ αη " − Eα Eβ iε#" − Eα Eη iε# − ± − ∓ + + T ∗ T = δ (β α) δ (η α)+ δ (β α) ηα + βα δ (η α) − − − E E iε E E iε − α − η ∓ α − β ± + + T T ∗ + βα ηα E E iε E E iε α − β ± α − η ∓ hence the matrix product gives

+ Tηα∗ dα Kβα K† = dα δ (β α) δ (η α)+ δ (β α) αη " − − − Eα Eη iε Z Z − ∓ + + + T T T ∗ + βα δ (η α)+ βα ηα E E iε − E E iε E E iε α − β ± α − β ± α − η ∓ # + + + + Tηβ∗ Tβη Tβα Tηα∗ dα Kβα K† = δ (β η)+ + + dα αη − Eβ Eη iε Eη Eβ iε Eα Eβ iε Eα Eη iε Z − ∓ − ± Z − ± − ∓ dγ Kβγ K† = δ (β η)+ Fβη (2.53) γη − Z T + ∗ T + T + ∗ T + F ηβ + βη + dγ ηγ βγ βη ≡ E E iε E E iε E E iε E E iε β − η ± ! η − β ± Z γ − η ± ! γ − β ± and using Eq. (2.51) we see that Fβη = 0. Therefore, Eqs. (2.52, 2.53) become

+ Tβα Kβα = δ (β α) + ; dγ Kβγ K† = δ (β η) (2.54) − Eα Eβ iε γη − − ± Z 2.3. SYMMETRIES OF THE S MATRIX 75 − on the other hand Eq. (2.22) can be rewritten as

T β(0) T (0) βα± βα± (0) α± = α + dβ = dβ δ (β α) + β E E iε − E E iε Z α − β ± Z " α − β ± # E E (0) α± = dβ β Kβα (2.55) Z E

since β(0) is an orthonormal basis and Eqs. (2.52, 2.55) show that β(0) is connected with α through a | ±i unitary operation, we conclude that α is also an orthonormal basis. | ±i The unitarity of the S matrix is proved similarly by multiplying (2.48) with δ (E E ). − β − α

2.3 Symmetries of the S matrix − We shall study what is meant by the invariance of the S matrix under various symmetries. Further, we shall − study the conditions on the Hamiltonian, to ensure such invariance properties.

2.3.1 Lorentz invariance

For any proper orthochronus Lorentz transformation in the Minkowski space x Λx + a, we cna deﬁne a unitary → operator on the Hilbert space U (Λ, a) by specifying that it acts as in Eq. (2.3) on either the “in” or the “out” states. We say that a theory is Lorentz invariant when the same operator U (Λ, a) acts as in (2.3) on both “in” and “out” states. Since the operator U (Λ, a) is unitary, we may write

+ + Sβα = β− α = U (Λ, a) β− U (Λ, a) α (2.56) and combining Eqs. (2.3, 2.56) we obtain the Lorentz invariance (indeed covariance) property of the S matrix, µ µ − for arbitrary Lorentz transformations Λ ν and four-translations a . We start rewriting Eq. (2.3) as

0 0 µ ν µ (Λp1) (Λp2) U (Λ, a) p1,σ1,n1; p2,σ2,n2; . . . = exp iaµΛ ν [ p1 p . . .] · · · | i { − − 2 − } s p0p0 1 2 · · · F (W (Λ,p )) F (W (Λ,p )) Λp , σ¯ ,n ; Λp , σ¯ ,n (2.57); . . . × σ¯1σ1 1 σ¯2σ2 2 ···| 1 1 1 2 2 2 i σ¯1σ¯2 X··· and the adjoint of this equation is

0 0 µ ν µ (Λp1′ ) (Λp2′ ) p1′ ,σ1′ ,n1′ ; p2′ ,σ2′ ,n2′ ; . . . U † (Λ, a) = exp iaµΛ ν p1′ + p2′ + . . . 0 0 · · · s p1′ p2′ · · · Fσ¯′ σ′ W Λ,p′ Fσ¯′ σ′ W Λ,p′ Λp′ , σ¯′ ,n′ ; Λp′ , σ¯′ ,n(2.58)′ ; . . . × 1 1 1 2 2 2 · · · 1 1 1 2 2 2 σ¯′ σ¯′ X1 2···

Now if we assume that the “in” and “out” states are given by

+ α = p ,σ ,n ; p ,σ ,n ; . . . ; β− = p′ ,σ′ ,n′ ; p′ ,σ′ ,n′ ; . . . (2.59) | 1 1 1 2 2 2 i 1 1 1 2 2 2

76 CHAPTER 2. SCATTERING THEORY then we can combine Eqs. (2.56, 2.59) with Eqs. (2.57, 2.58) to obtain

µ ν µ ν ν Sp′ ,σ′ ,n′ ;p′ ,σ′ ,n′ ;...;p ,σ ,n ;p ,σ ,n ;... = exp iaµΛ ν p′ + p′ + . . . p p . . . 1 1 1 2 2 2 1 1 1 2 2 2 1 2 − 1 − 2 − 0 0 0 0 (Λp1) (Λp2) (Λp1′ ) (Λp2′ ) 0 0 · · · 0 0 · · · ×s p p p′ p′ 1 2 · · · 1 2 · · · F (W (Λ,p )) F (W (Λ,p )) × σ¯1σ1 1 σ¯2σ2 2 · · · σ¯1σ¯2 X··· Fσ¯′ σ′ W Λ,p′ Fσ¯′ σ′ W Λ,p′ × 1 1 1 2 2 2 · · · σ¯′ σ¯′ X1 2··· SΛp′ ,σ¯′ ,n′ ;Λp′ ,σ¯′ ,n′ ;...;Λp ,σ¯ ,n ;Λp ,σ¯ ,n ;... (2.60) × 1 1 1 2 2 2 1 1 1 2 2 2 where the factors Fσ¯iσi (W (Λ,pi)) are defined by Eq. (2.4). In particular, since the LHS of Eq. (2.60) is independent of a , the RHS must also be. Consequently, the S matrix vanishes unless the four-momentum is µ − conserved. We can then parameterize the part of the S matrix that represents actual interactions among the − particles in the form S δ (β α)= 2πiM δ4 (p p ) (2.61) βα − − − βα β − α where we have used a structure similar to (2.44) but including the conservation of the four-momentum (not only the conservation of the energy). We shall see later that Mβα contains additional delta factors. Indeed, equation (2.60) is not a theorem but a definition of what we mean by the Lorentz invariance of the S matrix. We shall see that only certain chosen Hamiltonians lead to the existence a unitary operator that acts − as in (2.3) [or equivalently as in (2.57)] on both “in” and “out” states. Hence, one of our main tasks is to find the conditions on the Hamiltonian that ensures the Lorentz invariance of the S matrix. To find them, it is useful to − work with the S operator − (0) (0) Sβα = β S α D E the free particle states defined in Section 1.9 provides a rep resentation of the inhomogeneous Lorentz group, such that we can always define a unitary operator U0 (Λ, a) that induces the transformation (2.3) [or equivalently the transformation (2.57)] on these states

0 0 (0) µ ν ν (Λp1) (Λp2) U0 (Λ, a) p1,σ1,n1; p2,σ2,n2; . . . = exp iaµΛ ν [p1 + p2 + . . .] 0 0 · · · {− } s p1p2 E · · · (0) Fσ′ σ (W (Λ,p1)) Fσ′ σ (W (Λ,p2)) Λp1,σ′ ,n1; Λp2,σ′ ,n2; .(2.62) . . × 1 1 2 2 · · · 1 2 σ′ σ′ 1 2 E X···

Equation (2.62) will hold if the unitary operator U (Λ, a) commutes with the S operator 0 − 1 U0 (Λ, a)− SU0 (Λ, a)= S (2.63) as in section 1.7.3, the condition (2.63) can be expressed in terms of infinitesimal transformations, leading to commutation relations with the generators. We shall denote the generators of these infinitesimal transformations as a momentum P0, angular momentum J0, and a boost generator K0, that along with H0 generates the infinitesimal inhomogeneous Lorentz transformations, when acting on free-particle states. Hence Eq. (2.60) is equivalent to say that the S matrix is unaffected by these transformations [which is more clear in the equivalent equation (2.56)]. − In turn, it is equivalent to the fact that the S operator commutes with these generators −

[H0,S] = [P0,S] = [J0,S] = [K0,S] = 0 (2.64) 2.3. SYMMETRIES OF THE S MATRIX 77 −

(0) Since the operators H0, P0, J0, K0 generate inﬁnitesimal inhomogeneous Lorentz transformations on states α , they obey the commutation relations (1.130-1.137) page 31

i j k J0, J0 = iεijkJ0 (2.65)

h i ji k J0, K0 = iεijkK0 (2.66) h i Ki, Kj = iε J k (2.67) 0 0 − ijk 0 h i ji k J0, P0 = iεijkP0 (2.68) h i Ki , P j = iH δ (2.69) 0 0 − 0 ij h i i i i j J0,H0 = P0,H0 = P0, P0 = 0 (2.70) Ki ,H = iP i h i (2.71) 0 0 − 0 in the same way, we can deﬁne a set of “exact generators” P, J, K,H (H is of course the full Hamiltonian), such that they generate the transformations (2.3) [or equivalently (2.57)] on, say the “in” states. It is not obvious however that the same operators generate the same transformations on the “out” states. The group structure says that these “exact generators” satisfy the same commutation relations (1.130-1.137)

i j k J , J = iεijkJ (2.72) i j k J , K = iεijkK (2.73) Ki, Kj = iε J k (2.74) − ijk i j k J , P = iεijkP (2.75) Ki, P j = iHδ (2.76) − ij i i i j J ,H = P ,H = P , P = 0 (2.77) Ki,H = iP i (2.78) − In almost all known ﬁeld theories, the eﬀect of interactions is to add a potential term V to the Hamiltonian, but leaving the momentum and angular momentum unchanged

H = H0 + V , P = P0 , J = J0 (2.79) the only known exceptions are theories with topologically twisted ﬁelds, for instance theories with magnetic monoples, where the angular momentum of the states depends on the interactions. Eq. (2.79) implies that the commutation relations (2.65, 2.68) become the commutation relations (2.72, 2.75). If in addition the interaction commutes with the free-particle momentum and angular momentum i.e.

[V, P0] = [V, J0] = 0 (2.80) and using (2.70) we obtain

i i i J ,H = J0,H0 + V = J0,H0 = 0 i i i P ,H = P0,H0 + V = P0,H0 = 0 then Equation (2.70) also becomes Eq. (2.77). On the other hand, the Lippmann-Schwinger equation (2.21) or equivalently Eq. (2.19) show that the operators that generate translations and rotations on the “in” and “out” states are simply P0 and J0. To see it, we observe that P0 is the generator of translations for the free-particle 78 CHAPTER 2. SCATTERING THEORY

(0) states α and P0 commute with both H0 and H. Let U (1, a) be a pure translation operator on free-particle states (hence its generators are P ). Using the previous facts and multiplying Eq. (2.19) by U (1, a) yields 0

(0) (0) ( U (1, a) α± = U (1, a) Ω ( ) α = U (1, a) exp (+iHτ) exp ( iH τ) α = exp(+iHτ) exp ( iH τ) U (1, a) α ∓∞ − 0 − 0 E E = exp (+iHτ) exp ( iH τ) exp [ ia (pµ + pµ + . . .)] α(0) − 0 − µ 1 2 E = exp [ ia (pµ + pµ + . . .)] exp (+iHτ) exp ( iH τ ) α(0) − µ 1 2 − 0 µ µ n Eo U (1, a) α± = exp [ iaµ (p1 + p2 + . . .)] α± − and we conclude that U (1, a) is also a pure translation operator on “in” and “out” states. A similar argument shows that J0 is the generator of rotations for “in” and “out” states. From the fact that P0 and J0 commute with H0 and H, we also see that P0 and J0 commute with the operator U (t,t0) deﬁned in Eq. (2.39). Therefore, P0 and J commute with the S operator U ( , ) deﬁned by Eq. (2.38). Finally, since there are energy-conservation 0 − −∞ ∞ delta functions in both terms of (2.44), we see that the S operator commutes with H . − 0 We still have to show that K commutes with the S operator. Unlike the case of P and J, we cannot set the 0 − boost generator K as equal to the free-particle counterpart K0, because Eqs. (2.69) and (2.76) would lead to i i i i H = i K , P = i K0, P0 = H0 and we obtain H = H0, in contradiction with the fact that the interaction V is added to the Hamiltonian. Then, when we add a “correction” V to the time-translation generator H0, we must also add a “correction” W to the boost generator K0 K = K0 + W (2.81) and we shall concentrate on the commutation relation (2.78). The LHS of Eq. (2.78) yields

i i i i i K ,H = K0 + W ,H = K0,H + W ,H = Ki ,H + Ki ,V + W i,H = iP i + Ki,V + W i,H 0 0 0 − 0 0 Ki,H = iP i + Ki,V + W i,H (2.82) − 0 where we have used (2.71). Substituting (2.82) in (2.78) we have

iP + Ki,V + W i,H = iP i Ki ,V + W i,H − 0 − ⇒ 0 so we obtain the condition [K ,V ]= [W,H] (2.83) 0 − The condition (2.83) is empty by itself, since for any V we could deﬁne W to satisfy such a condition: the matrix representations of operators on both sides of Eq. (2.83) must coincide in any basis. In particular, by using the basis of H eigenstates α and β equation (2.83) becomes − | i | i α [K ,V ] β = α [W,H] β (2.84) h | 0 | i − h | | i and the RHS of Eq. (2.84) yields

α [W,H] β = α [WH HW] β = α [WE E W] β = (E E ) α W β − h | | i − h | − | i − h | β − α | i α − β h | | i such that Eq. (2.84) becomes α [K ,V ] β = (E E ) α W β h | 0 | i α − β h | | i consequently, the condition (2.83) is satisﬁed for any V by deﬁning the matrix elements of W between H eigenstates − α and β as | i | i α [K ,V ] β W = α W β = h | 0 | i (2.85) αβ h | | i − E E α − β 2.3. SYMMETRIES OF THE S MATRIX 79 −

We should observe that to guarantee the Lorentz invariance of a theory is not enough to show the existence of a set of exact generators satisfying the Lie algebra (2.72-2.78), but also that these operators should act the same way on the “in” and the “out” states. Therefore, it is not enough to find an operator K that satisfies Eq. (2.83), we also require the remaining Lorentz invariance condition that K commutes with the S operator [see Eq. (2.64)]. 0 − To show the conditions to obtain [K0,S] = 0, we shall consider the commutator of K0 with the operator U (t,t0) defined by Eq. (2.38). We shall apply now the following result

Theorem 2.1 Suppose we have two operators A and B such that B commutes with their commutator, that is

[B,C]=0 ; C [A, B] (2.86) ≡ if F (B) is a function of the operator B then we have

[A, F (B)] = [A, B] F ′ (B) (2.87) where F ′ (B) is the derivative of F (B) “with respect to B” deﬁned as

∞ n ∞ n 1 F (B)= f B F ′ (B) nf B − (2.88) n ⇒ ≡ n n=0 n=0 X X in particular, it is easy to show that [exp (αB)]′ = α exp (αB).

We start evaluating the commutator [K0, exp (iH0t)] with A = K and B = H Eq. (2.71) gives C = [K ,H ]= iP , and since P commutes with H , we can apply 0 0 0 0 − 0 0 0 Eq. (2.87) to obtain

[K , exp (iH t)] = [K ,H ] it exp (iH t) = iP it exp (iH t) 0 0 0 0 { 0 } − 0 { 0 } [K0, exp (iH0t)] = tP0 exp (iH0t) (2.89) further if Eq. (2.80) is satisﬁed, then P0 = P commutes with H and from Eqs. (2.87, 2.78) we obtain

[K, exp (iHt)] = [K,H] it exp (iHt) = tP exp (iHt) { } [K, exp (iHt)] = tP0 exp (iHt) (2.90) and using Eqs. (2.89, 2.90) along with (2.38) we have

[K ,U (τ,τ )] = [K , exp +iH τ exp iH (τ τ ) exp iH τ ] 0 0 0 { 0 } {− − 0 } {− 0 0} = exp +iH τ [K , exp iH (τ τ ) exp iH τ ] { 0 } 0 {− − 0 } {− 0 0} + [K , exp +iH τ ]exp iH (τ τ ) exp iH τ 0 { 0 } {− − 0 } {− 0 0} [K ,U (τ,τ )] = exp +iH τ exp iH (τ τ ) [K , exp iH τ ] 0 0 { 0 } {− − 0 } 0 {− 0 0} +exp +iH τ [K , exp iH (τ τ ) ]exp iH τ { 0 } 0 {− − 0 } {− 0 0} +τP exp (iH τ)exp iH (τ τ ) exp iH τ 0 0 {− − 0 } {− 0 0}

[K ,U (τ,τ )] = exp +iH τ exp iH (τ τ ) [ τ P exp ( iH τ )] 0 0 { 0 } {− − 0 } − 0 0 − 0 0 +exp +iH τ [K , exp iH (τ τ ) ]exp iH τ { 0 } 0 {− − 0 } {− 0 0} +τP exp (iH τ)exp iH (τ τ ) exp iH τ 0 0 {− − 0 } {− 0 0} 80 CHAPTER 2. SCATTERING THEORY

[K ,U (τ,τ )] = (τ τ ) P exp +iH τ exp iH (τ τ ) exp ( iH τ ) 0 0 − 0 0 { 0 } {− − 0 } − 0 0 +exp +iH τ [K W, exp iH (τ τ ) ]exp iH τ { 0 } − {− − 0 } {− 0 0} [K ,U (τ,τ )] = (τ τ ) P U (τ,τ )+exp +iH τ [K, exp iH (τ τ ) ]exp iH τ 0 0 − 0 0 0 { 0 } {− − 0 } {− 0 0} exp +iH τ [W, exp iH (τ τ ) ]exp iH τ − { 0 } {− − 0 } {− 0 0}

[K ,U (τ,τ )] = (τ τ ) P U (τ,τ )+exp +iH τ (τ τ ) P exp iH (τ τ ) exp iH τ 0 0 − 0 0 0 { 0 } {− − 0 0 {− − 0 }} {− 0 0} exp +iH τ [W, exp iH (τ τ ) ]exp iH τ − { 0 } {− − 0 } {− 0 0} [K ,U (τ,τ )] = (τ τ ) P U (τ,τ ) (τ τ ) P U (τ,τ ) 0 0 − 0 0 0 − − 0 0 0 exp +iH τ [W, exp iH (τ τ ) ]exp iH τ − { 0 } {− − 0 } {− 0 0} [K ,U (τ,τ )] = exp +iH τ [exp iH (τ τ ) , W]exp iH τ 0 0 { 0 } {− − 0 } {− 0 0} [K ,U (τ,τ )] = exp +iH τ exp iH (τ τ ) W exp iH τ 0 0 { 0 } {− − 0 } {− 0 0} exp +iH τ W exp iH (τ τ ) exp iH τ − { 0 } {− − 0 } {− 0 0} inserting an identity we ﬁnd

[K ,U (τ,τ )] = exp [+iH τ] exp [ iH (τ τ )] exp [ iH τ ] exp [iH τ ] W exp [ iH τ ] 0 0 0 − − 0 { − 0 0 0 0 } − 0 0 exp [+iH τ] W exp [ iH τ] exp [iH τ] exp [ iH (τ τ )] exp [ iH τ ] − 0 { − 0 0 } − − 0 − 0 0 [K ,U (τ,τ )] = U (τ,τ ) exp [iH τ ] W exp [ iH τ ] exp [+iH τ] W exp [ iH τ] U (τ,τ ) 0 0 0 { 0 0 − 0 0 } − { 0 − 0 } 0 and we obtain ﬁnally

[K ,U (τ,τ )] = U (τ,τ ) W (τ ) W (τ) U (τ,τ ) (2.91) 0 0 0 0 − 0 W (t) exp [iH t] W exp [ iH t] (2.92) ≡ 0 − 0 now taking the limits τ and τ + we have → −∞ 0 → ∞ [K ,U ( , + )] = U ( , + ) W (+ ) W ( ) U ( , + ) 0 −∞ ∞ −∞ ∞ ∞ − −∞ −∞ ∞ [K ,S] = S W (+ ) W ( ) S (2.93) 0 ∞ − −∞ then for [K0,S] to vanish, a suﬃcient condition is that

W (t ) 0 (2.94) → ±∞ → It occurs if the matrix elements of W between H eigenstates are suﬃciently smooth functions of energy, then 0− matrix elements of W (t) between smooth superpositions of energy eigenstates vanish for t . → ±∞ In conclusion, the condition (2.83) along with the condition W (t ) 0, are suﬃcient to obtain → ±∞ →

[K0,S] = 0 in particular the condition W (t ) 0 is satisfied if the matrix elements of W between H eigenstates are → ±∞ → 0− sufficiently smooth functions of energy. If we add the condition (2.80) we have a set of sufficient conditions for the Lorentz invariance of the S matrix [i.e. to satisfy Eqs. (2.64)]. − Note that the smoothness condition on W is a natural one, since it is like the condition on the matrix elements of V that is necessary to make V (t) 0 when t . The latter is required for the very idea of the S matrix, → → ±∞ − since it is the condition to obtain asymptotic free states. On the other hand, by using τ = 0 and τ = in Eq. (2.91) we obtain 0 ∓∞ [K ,U (0, )] = U (0, ) W ( ) W (0) U (0, ) 0 ∓∞ ∓∞ ∓∞ − ∓∞ [K ,U (0, )] = W (0) U (0, ) (2.95) 0 ∓∞ − ∓∞ 2.3. SYMMETRIES OF THE S MATRIX 81 − from definitions (2.19, 2.39, 2.92) we have

U (0,τ ) = exp [iHτ ] exp [ iH τ ]=Ω(τ ) 0 0 − 0 0 0 W (0) = W so that equation (2.95) becomes

[K , Ω ( )] = W Ω ( ) K Ω ( ) Ω ( ) K + W Ω ( ) = 0 0 ∓∞ − ∓∞ ⇒ 0 ∓∞ − ∓∞ 0 ∓∞ (K + W) Ω ( ) Ω ( ) K = 0 0 ∓∞ − ∓∞ 0 ﬁnally KΩ ( )=Ω( ) K (2.96) ∓∞ ∓∞ 0 let us recall that according with Eq. (2.19), Ω ( ) is the operator that converts free particle states α(0) into ∓∞ the corresponding “in” or “out” states α . | ±i Further, by applying Eqs. (2.79, 2.80) we see that P0 and J0 commutes with both H and H0. Therefore P0 and J commutes with Ω ( ), but P = P and J = J so this commutativity could be expressed as 0 ∓∞ 0 0 PΩ ( )=Ω( ) P ; JΩ ( )=Ω( ) J ∓∞ ∓∞ 0 ∓∞ ∓∞ 0 ﬁnally, since all α(0) and α are eigenstates of H and H respectively with the same eigenvalue E , we obtain | ±i 0 α HΩ ( )=Ω( ) H ∓∞ ∓∞ 0 in summary we can write

GΩ ( ) = Ω( ) G0 ; G K, P, J,H 1 ∓∞ ∓∞ ≡ Ω− ( ) GΩ ( ) = G ; G K, P, J,H (2.97) ∓∞ ∓∞ 0 ≡ showing that with our assumptions “in” and “out” states transform under inhomogeneous Lorentz transformations just like the free-particle states. In addition, since (2.97) are similarity transformations, the exact generators K, P, J,H satisfy the same commutation relations as K0, P0, J0,H0. Observe that in deriving these results we have used the commutation relations (2.72, 2.75, 2.77, 2.78), while the remaining ones (2.73, 2.74, 2.76) are automatically obtained.

2.3.2 Internal symmetries Up to now we have only consider space-time symmetries (continuous or discrete). Notwithstanding, there are other symmetries like the symmetry in nuclear physics under interchange of neutrons and protons, or the “charge conjugation” symmetry between particles and antiparticles. Such kind of symmetries have nothing directly to do with Lorentz invariance, and also appear the same in all inertial frames. We shall denote a symmetry transformation of this type as T , and acting on the Hilbert space of Physical states as a unitary operator U (T ), which induces linear transformations on the indices ni that label the particle species.

U (T ) p ,σ ,n ; p ,σ ,n ; . . . = (T ) (T ) p ,σ , n¯ ; p ,σ , n¯ ; . . . (2.98) | 1 1 1 2 2 2 i Dn¯1,n1 Dn¯2,n2 ···| 1 1 1 2 2 2 i n¯ ,n¯ ,... 1X2 if T¯ is some other transformation, the operators U (T ) and U T¯ must satisfy the group multiplication rule

U T¯ U (T )= U T¯ T (2.99) where T¯ T is the transformation obtained by first performing the trans formation T and then the transformation T¯. By applying U T¯ on Eq. (2.98) we see that the matrices (T ) must satisfy the same rule D T¯ (T )= T¯ T (2.100) D D D 82 CHAPTER 2. SCATTERING THEORY on the other hand, by taking the scalar product of the states obtained by applying U (T ) on two different “in” states (or two different “out” states), along with the normalization condition (2.5) page 66, we see that (T ) D must be unitary 1 † (T )= − (T ) (2.101) D D Finally, by taking the scalar product of the states obtained by applying U (T ) on one “out” state and one “in” state shows that commutes with the S matrix, in the sense that D −

∗¯ ′ ′ (T ) ∗¯ ′ ′ (T ) ¯ (T ) ¯ (T ) S ′ ′ ¯ ′ ′ ′ ¯ ′ ¯ ¯ N1,n1 N2,n2 N1,n1 N2,n2 p1,σ1,N1;p2,σ2,N2 ,p1,σ1,N1;p2,σ2,N2; ′ ′ D D ···D D · · · ··· ··· N¯1,N¯2 N¯ ,N¯ X··· 1X2··· ′ ′ ′ ′ ′ ′ = Sp ,σ ,n ;p ,σ ,n ,p1,σ1,n1;p2,σ2,n2; (2.102) 1 1 1 2 2 2··· ··· once again this is what we mean by a theory to be invariant under the internal symmetry T , since to derive (2.102) we still have to show that the same unitary operator U (T ) will induce the transformation (2.98) on both “in” and “out” states. This will be the case if there is an “unperturbed” transformation operator U0 (T ) that induces these transformations on free-particle states

U (T ) p ,σ ,n ; p ,σ ,n ; . . . = ¯ (T ) ¯ (T ) p ,σ , N¯ ; p ,σ , N¯ ; . . . 0 | 1 1 1 2 2 2 i DN1,n1 DN2,n2 · · · 1 1 1 2 2 2 N¯ ,N¯ ,... 1X2 and that commutes with both the free-particle and interaction parts of the Hamiltonian

1 U0− (T ) H0 U0 (T ) = H0 1 U0− (T ) VU0 (T ) = V by using the Lippmann-Schwinger equations (2.22) or Eqs. (2.19), we obtain that U0 (T ) will induce the transformations (2.98) on “in” and “out” states as well as free-particle states. Therefore, we can derive Eq. (2.98) taking U (T ) as U0 (T ). A particularly important case is that of a one-parameter Lie group in which T is a function of one parameter θ and T θ¯ T (θ)= T θ¯ + θ as we saw in Eq. (1.39) page 17, in that case the corresponding operators on the Hilbert space must have the form

U (T (θ)) = exp (iQθ) (2.103) with Q a Hermitian operator. The corresponding matrices (T ) take the form D (T (θ)) = δ exp (iq θ) (2.104) Dmn mn n where qn are a set of real numbers that depend on the species of particles. In this case, Eq. (2.102) says that the q′s are conserved, because Sβα vanishes unless

qm1 + qm2 + . . . = qn1 + qn2 + . . . (2.105) perhaps the most classical example of this kind of conservation law is that of conservation of electric charge. Further, all known processes conserve baryon number (the number of baryons, such as protons, neutrons, and hyperons, minus the number of their antiparticles), and most of processes seem to conserve the lepton number2 (the number of leptons, such as electrons, muons, taos, and neutrinos minus the number of their antiparticles). However, it is generally believed that this conservation laws are only very good aproximations. There are other conservation laws that are deﬁnitely only approximate such as the conservation of strangeness and isotopic spin.

2The recent observations of neutrino oscillations seem to violate the lepton number. 2.3. SYMMETRIES OF THE S MATRIX 83 −

2.3.3 Parity As customary, our starting point is the transformation of single particle states under the symmetry involved. They are given by Eq. (1.266) for massive particles and by (1.285) for massless particles.

P p,σ = η p,σ massive particles | i |P i P p,σ = η exp [ iπσ] p, σ null mass particles | i σ ∓ |P − i As long as the symmetry under the transformation x x is valid, there must exist a unitary operator P under →− which both “in” and “out” states transform as a direct product of single particle states

P p ,σ ,n ; p ,σ ,n ; . . .± = η η p , ε σ ,n ; p , ε σ ,n ; . . .± (2.106) 1 1 1 2 2 2 a1 a2 · · · P 1 1 1 1 P 2 2 2 2 where we have deﬁned 2 2 2 2 ηni if p = M < 0 +1 if p = M < 0 ηai = −2 ; εi = −2 ησ exp [ iπσn ] if p = 0 1 if p = 0 ni ∓ i − where is the operator on the Minkowski space that reverses the space components of pµ and η is the intrinsic P ai parity. By denoting the “in” and “out” states as

+ α = p ,σ ,n ; p ,σ ,n ; . . . ; β− = p′ ,σ′ ,n′ ; p′ ,σ′ ,n′ ; . . . | 1 1 1 2 2 2 i 1 1 1 2 2 2 The parity conservation condition for the S matrix reads −

′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ Sp σ n ;p σ n ; ,p1σ1n1;p2σ2n2; = ηa∗′ ηa∗′ ηa1 ηa2 S p ,ε σ ,n ; p ,ε σ ,n ; , p1,ε1σ1,n1; p2,ε2σ2,n2; (2.107) 1 1 1 2 2 2 ··· ··· 1 2 · · · ···× P 1 1 1 1 P 2 2 2 2 ··· P P ··· as in the case of internal symmetries, an operator P satisfying Eq. (2.106) will actually exists if the operator P0 deﬁned to act this way on free-particle states commute with V and with H0

1 1 P0H0P0− = H0 ; P0V P0− = V

We shall consider the case of massive particles from now on. The phases ηn could be inferred from either dynamical models or experiments. However, neither can provide a unique determination of the η′s. It is owe to the fact that we can always redeﬁne P by combining it with any conserved symmetry operator. As an example, If B and L are conserved (recall that we believe these symmetries are only good approximations), and if P is also conserved, the following operator is conserved as well

P ′ P exp (iαB + iβL + iγQ) (2.108) ≡ where B, L, Q are baryon number, lepton number and electric charge respectively, and α, β, γ are arbitrary real phases. In other words, if B and L are conserved, then P ′ is conserved whenever P is conserved, and any of them can be called the parity. The neutron, proton and electron have diﬀerent combinations of values of B,L,Q and by apropriate choices of the phases α, β, γ we can deﬁne the intrinsic parities of all three particles to be +1. Protons have Bp = 1, Lp = 0, Qp = e, neutrons have Bn = 1, Ln = 0, Qn = 0, and electrons have Be = 0, L = 1, Q = e. We can settle the intrinsic parity of all of them to be equal by imposing that the total phase e e − of the exponential in (2.108) be equal for all three particles, that is

αBk + βLk + γQk = N0 with k = p,n,e with N0 being an apropriate constant. These equations ﬁxed the phases α, β, γ. Consequently, Once we have settle the intrinsic parity of p,n,e to be equal (say to the value +1), the intrinsic parity of other particles are not arbitrary anymore. For example, the intrinsic parity of charged pion (which can be emitted in the transition n p + π ) is no longer arbitrary. In addition, the intrinsic parity of any particle like the neutral pion π0 which → − 84 CHAPTER 2. SCATTERING THEORY

carries no conserved quantum numbers (such as B,L,Q) is always meaningful, though in that case P ′ and P in Eq. (2.108) coincide. Note in particular that the combination of parity with other symmetries, decreases the number of diﬀerent intrinsic parities assigned to the particles. The space inversion P has the group multiplication P 2 = 1, hence its eigenvalues are 1. Hence, it is natural ± to ask whether the intrinsic parities must always have the values 1. The fact is that sometimes the parity ± operator that is conserved is not the space inversion operator, and could diﬀer from the latter by a certain phase transformation. Regardless P 2 = 1 or not, the operator P 2 behaves just like an internal symmetry transformation:

2 2 2 2 2 2 2 P p ,σ ,n ; p ,σ ,n ; . . .± = η η p , ε σ ,n ; p , ε σ ,n ; . . .± 1 1 1 2 2 2 a1 a2 · · · P 1 1 1 1 P 2 2 2 2 P 2 p ,σ ,n ; p ,σ ,n ; . . . = η2 η2 p ,σ ,n ; p ,σ ,n ; . . . 1 1 1 2 2 2 ± a1 a2 1 1 1 2 2 2 ± · · ·

If this internal symmetry is part of a continuous symmetry gr oup of phase transformations, such as the group of multiplication by the phases exp(iαB + iβL + iγQ) with arbitrary values of α, β, γ, then its inverse square root 3 IP must also be a member of this group 2 2 IP P = 1 it is also quite obvious that [IP , P ] = 0 for instance if P 2 = exp [iαB + . . .], then we take I = exp 1 iαB + . . . . As a consequence, we can deﬁne a P − 2 new parity operator 2 P ′ PI P ′ = 1 ≡ P ⇒ and P ′ is conserved with the same extent as P . Hence, there is no reason why we should not call this the parity operator. In that case, since P 2 = 1, then the intrinsic parities can only take the values 1. ′ ± However, we could have a theory in which there is some discrete internal symmetry which is not a member of any continuous symmetry group of phase transformations. In that case, it is not necessarily possible to deﬁne parity in such a way that intrinsic parities can only take the values 1. ± For instance, it is a consequence of angular-momentum conservation that the total number F of all particles of half-integer spin can only change by even numbers4, so the internal symmetry operator ( 1)F is conserved. On − the other hand, all known half-integer spin particles have odd values of the sum B + L. Let us take as examples of half-integer spin particles the proton, neutron and electron, we have

Bp + Lp = Bn + Ln =1+0=1 ; Be + Le =0+1=1 and B + L is odd for all of them. Combining these two facts we observe that if F is odd since each Bi + Li is odd the sum of Bi + Li over all F particles is also odd. Similarly, if F is even the total value of B + L is also even. As a consequence, as far as we know we have ( 1)B+L = ( 1)F − − if this is true, the discrete symmetry ( 1)F is part of a continuous symmetry group of internal symmetries, − consisting of the operators exp [iα (B + L)] with arbitrary real α. It has an inverse square root exp [ iα (B + L) /2]. − In this case, if P 2 = ( 1)F then P can be redeﬁned such that all intrinsic parities are given by 1 as discussed − ± above. Let us take as examples of half-integer spin particles the proton, neutron and electron, we have

Bp + Lp = Bn + Ln =1+0=1 ; Be + Le =0+1=1 and B + L is odd for all of them.

3If P is part of the group hence P 2 also is. Further, any element of the group must have an inverse. In particular P 2 has an inverse 2 2 that we denote as IP . Now since P is in general diﬀerent from the identity, it is followed that P is not necessarily its own inverse. Hence IP is not necessarily P . 4If an odd number of half-integer spin particles appear or dissapear the total change of spin is half-integer, and it cannot be compensated by an opposite change in the orbital angular momentum because such a momentum only takes integer values. 2.3. SYMMETRIES OF THE S MATRIX 85 −

However, if we discovered a particle of half-integer spin, and even value of B + L, then it would be possible to have P 2 = ( 1)F without being able to redefine the parity operator to have eigenvalues 1. However, in this − ± case we have P 4 = 1, so all particles would have intrinsic parities either 1, or i. This would be the case of ± ± the so-called Majorana neutrinos with j = 1/2 and B + L = 0, in which we expect that intrinsic parities take the values i. ± From Eq. (2.107) we see that if the product of intrinsic parities in the final state is equal to the product of intrinsic parities in the initial state, the S matrix must be even overall in the three-momenta. If the product of − intrinsic parities in the final state is minus the product of intrinsic parities in the initial state, the S matrix must − be odd overall in the three-momenta. For instance, in 1951 it was observed that a pion can be absorbed by a deuteron (a deuteron is a bound state of a proton and neutron with even orbital angular-momentum, chiefly l = 0) from the l = 0 ground state of the π−d atom, the reaction reads π− + d n + n (2.109) → the initial state have total angular momentum j = 1, where the pion has spin 0 and the deuteron has spin 1. Therefore, the final state must have orbital angular momentum l = 1 and total neutron spin s = 1. It is because the other possibilities

l = 0, s = 1 l = 1, s = 0 l = 2 , s = 1 which are allowed by angular momentum conservation (compatible with the addition rules of angular momenta)5, are forbidden by the requirement that the final state is antisymmetric in the two neutrons (two identical fermions). The allowed final state has l = 1 which corresponds to the vector representation of SO (3) for the orbital variables. Thus, the matrix element is odd under reversal of the direction of all three momenta. Consequently, the S matrix is odd overall in the three-momenta. In turn, it implies that the product of intrinsic parities in − the final state is minus the product of intrinsic parities in the initial state. We then conclude that the intrinsic parities of the particles in the reaction (2.109) must be related as

2 η η − = η d π − n on the other hand, we have seen that we are able to choose the neutron and proton to have the same intrinsic 2 parity. Further, since a deuteron is a bound state of a proton and a neutron we have ηd = ηn obtaining

η − = 1 π − hence, the negative pion is a speudoscalar particle. Similar analyses show that π+ and π0 has also negative parity, as expected from the isospin invariance of the three particles.

Since pions have negative intrinsic parity, a spin zero particle that decays into three pions must have intrinsic parity η3 = ( 1)3 = 1. To see it, we observe that in the Lorentz frame in which the decaying particle is at rest π − − we have l = 0. Consequently, l = s = j = 0 and we are in the scalar representation of SO (3). Thus, rotational invariance only allows the matrix elements to depend on scalars or pseudoscalars. As for the scalar quantities they can only be formed by scalar products of the pion three-momenta with each other, and all of them are even under reversal of all momenta. Pseudoscalars can be formed by products of the type

p (p p ) i · j × k 5For instance, l = 2 and s = 1 gives three possible values of total angular momentum j = 3, 2, 1. Conservation of angular momentum leads us to j = 1. Similarly the (allowed) case gives l = s = 1, which yields j = 2, 1, 0, hence we take j = 1 because of the conservation of angular momentum. 86 CHAPTER 2. SCATTERING THEORY

Note that, since we are in the proper frame in which p + p + p = 0 the triple scalar product p (p p ) 1 2 3 1 · 2 × 3 formed from the three pion momenta vanishes

p = (p + p ) p (p p ) = (p + p ) (p p )= p (p p )+ p (p p )= 0 + 0 = 0 1 − 2 3 ⇒ 1 · 2 × 3 2 3 · 3 × 2 2 · 3 × 2 3 · 3 × 2 and similarly with the other triple products. Therefore, the S matrix does not depend on pseudoscalar quantities. − We then conclude that the S matrix must be even overall in the three-momenta. Consequently, the product of the 3 − intrinsic parities ηπ in the final states must equal the product of intrinsic parities η of the initial states (consisting in this case of a single particle). 2 For the same reason, a spin zero particle that decays into two pions must have intrinsic parity ηπ = +1. Among the strange particles discovered in the late 1940’s there were two particles of zero spin (inferred from the angular distribution of their decay products): The τ particle was identified by its decay into three pions, and − hence was assigned to it the intrinsic parity 1. On the other hand, the θ particle was identified by its decays − − into two pions and was given a parity +1. After more detailed studies the τ and θ particle seemed increasingly to have identical masses and lifetimes. After many attempts of solution to the so-called τ θ puzzle, Lee and Yang − suggested in 1956 that both are the same particle (currently called K±) and that parity is not conserved in the weak interactions involved in these decays. We shall see later that the rate for a physical process α β (with α = β) is proportional to S 2, where the → 6 | βα| proportionality factors are invariant under reversal of all three-momenta. Hence, as long as the states α and β contains a definite number of particles of each type, the phase factors in Eq. (2.107) clearly do not play any role in the quantity S 2. Consequently, equation (2.107) implies that the rate α β is invariant under the reversal | βα| → of direction of all three-momenta. We have seen that for the decay of a K meson into two or three pions, it is a trivial consequence of the rotational invariance, but it is a non-trivial restriction on rates associated with more complicated processes. As an example, following the suggestions of Lee and Yang concerning parity violation in weak interactions, Wu and her collaborators measured the angular distribution of the electron in the final state of the decay

60 60 Co Ni + e− + ν (2.110) → the momentum of the antineutrino and nickel nucleus were not measured. The experiment focused on the angular distribution of the electrons, ﬁnding that they are preferably emitted in a direction opposite to that of the spin of the decaying nucleus. Recalling that parity reverses all three-momenta but not the spin, we see that by applying parity to this process we would obtain that the parity transformed state is one in which electrons are preferably emitted in the direction of the spin of the decaying nucleus. Therefore the parity transformed state is a physically diﬀerent one (and not observed in the laboratory), showing that the interaction (weak interaction) that governs this decay violates parity conservation. A similar result was found for the decay of a positive muon µ+ which is polarized in its production through the process π+ µ+ + ν with the subsequent decay of µ+ e+νν. → → π+ νµ+ νe+νν (2.111) → → The latter experimental observations show that for the processes (2.110, 2.111), Eq. (2.107) is untenable and parity is not conserved in the weak interactions responsible for these decays. We shall see however that parity is conserved in strong and electromagnetic interactions, so it has still an important role in Physics.

2.3.4 Time-reversal We saw that the time-reversal operator T transforms one-particle states according with Eqs. (1.274) and (1.288) for massive and masless particles respectively

j σ 2 2 T p,σ = ( 1) − ζ p, σ for p = M < 0 | i − |P − i − T p,σ = ζ exp [ iπσ] p,σ for p2 = 0 | i σ ± |P i 2.3. SYMMETRIES OF THE S MATRIX 87 −

And the multi-particle state transforms as customary, as the direct product of one-particle states, except that under time-reversal transformation in which we “watch the ﬁlm in the opposite sense”, we expect “in” and “out” states to be interchanged

T p ,σ ,n ; p ,σ ,n ; . . .± = ζ ζ p , ε σ ,n ; p , ε σ ,n ; . . .∓ (2.112) 1 1 1 2 2 2 a1 a2 · · · P 1 − 1 1 1 P 2 − 2 2 2 where ji σi 2 2 2 2 ζni ( 1) − if p = M < 0 +1 if p = M < 0 ζai = − −2 ; εi = −2 (2.113) ζσ exp [ iπσn ] if p = 0 1 if p = 0 ni ± i − we abbreviate the assumption (2.112) in the form

T α± = α∓ (2.114) T where indicates a reversal of the three-momenta, multiplication of the spins σ by ε and multiplication of T ni − i the state by the phase factors ζ deﬁned in Eq. (2.113). Since T is antiunitary, we have σni + + β− α = T β− T α ∗ + + β− α = T α T β− therefore, time-reversal invariance of the S matrix is expressed by − Sβ,α = S α, β (2.115) T T or more explicitly

′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ Sp σ n ;p σ n ; ,p1σ1n1;p2σ2n2; = ζa ζa ζa∗ ζa∗ S p1, ε1σ1,n1; p2, ε2σ2,n2; ; p , ε σ ,n ; p , ε σ ,n 1 1 1 2 2 2 ··· ··· 1 2 · · · 1 2 ···× P − P − ··· P 1 − 1 1 1 P 2 − 2 2 2··· where we notice once again that besides the reversal of momenta and the transformation of spins, the role of the initial and ﬁnal states is interchanged. The S matrix will satisfy this transformation rule if the operator T deﬁned on free-particle states − 0 T α(0) α(0) (2.116) 0 ≡ T E E commutes with the free-particle Hamiltonian [which is auto matic, see Eq. (1.256)] and also with the interaction

1 1 T0− H0T0 = H0 ; T0− V T0 = V (2.117) in that case we can take T = T0 and use either (2.19) or (2.21) to show that time-reversal transformation acts as stated in Eq. (2.114). For instance, applying T on the Lippmann-Schwinger equation (2.21) and using Eqs. (2.116, 2.117) we obtain

(0) 1 T α± = T α + T (E H iε)− V α± α − 0 ± (0) E 1 T α± = α + (E H iε)− V T α± T α − 0 ∓ E observe that the sign of iε is reversed because of the antilinearity of T . We can rewrite both sides as ± (0) 1 α∓ = α + (E H iε)− V α∓ T T α − 0 ∓ T E 6 which is clearly the Lippmann-Schwinger equation for the state α∓ , thus justifying Eq. (2.114) . Similarly |T i applying T on the factor Ω (τ) deﬁned in Eq. (2.19) and taking into account the antilinearity of T and the fact that T commutes with H0 and H we ﬁnd T Ω (τ)= T exp (+iHτ) exp ( iH τ) = exp ( iHτ) exp (iH τ) T = Ω ( τ) T { − 0 } { − 0 } − 6In the free-particle states we have not “incident” or “outgoing” states. Thus, there is nothing like that to interchange under time-reversal. 88 CHAPTER 2. SCATTERING THEORY and applying T on both sides of Eq. (2.19)

(0) (0) T α± = T Ω ( ) α = Ω ( ) T α ∓∞ ±∞ (0)E E α∓ = Ω ( ) α T ±∞ T E which is the equation (2.19) for the state α , again leading to Eq. (2.114). |T ∓i In contrast to the case of parity conservation, time-reversal invariance condition (2.115) does not in general imply that the rate for the process α β is the same as for the process α β due to the fact that time- → T → T reversal also interchanges the “in” and “out” states. However, something like this is true when the S matrix can − be expressed as (0) (1) Sβα = Sβα + Sβα (2.118) where matrix elements S(1) are much smaller that the ones of S(0) except for some particular process of interest for which the matrix element of S(0) vanishes. Sometimes S(0) is simply the unit operator. Calculating the unitarity condition up to ﬁrst order in S(1) we have

(0) (1) † (0) (1) (0) (0) (0) (1) (1) (0) (1)2 1= S†S = S + S S + S = S †S + S †S + S †S + S O h i h i (0) (0) (1) and using the zeroth-order relation S †S = 1 we have to ﬁrst order in S

(0) (1) (1) (0) (0) (1) (1) (0) 1 = 1+ S †S + S †S S †S + S †S = 0 (1) (0) (1) (0) ⇒ S = S S †S (2.119) − which yields a reality condition for S(1). If S(1) as well as S(0) satisﬁes the time-reversal condition (2.115), the matrix multiplication in (2.119) can be put in the form

(1) (0) (1) (0) (0) (1) (0) S = dγ dγ′ S ′ S ′ †S = dγ dγ′ S ′ S ′∗S βα − βγ γ γ γα − βγ γγ γα Z Z Z Z (1) (0) (1) (0) S = dγ dγ′ S ′ S ∗′ S (2.120) βα − βγ γ , γ γα Z Z T T and recalling that S(0) does not contribute to the particular process α β by hypothesis, we see that Eq. (2.120) → represents the S matrix for such a process. Since S(0) is unitary, the rates for the processes α β and α β − → T →T are then the same if summed over sets and of final and initial states that are complete with respect to S(0). I F In this context, completeness of with respect to S(0) means that if S(0) is non-zero, and either α or α are in I α′α ′ , then both states are in . Hence, there is no transition from a state of the set to a state outside of such a I I I set. Completeness for is defined in the same way. When restricted to the subsets of states and , equation F I F (2.120) becomes (1) (0) (1) (0) Sβα = dγ dγ′ Sβγ′ S γ∗′, γ Sγα − T T ZF ZI In the simplest case, we have complete sets and consisting of just one state each; that is both the initial (0) I F 2iδα i2iδβ and final states are eigenvectors of S with eigenvalues e and e respectively, where δα and δβ are both real since S(0) is unitary and so its spectrum lies on the unit complex circle. In this case Eq. (2.120) simplifies to

(1) 2iδβ (1) 2iδα 2iδβ (1) 2iδα Sβα = dγ dγ′ e δ β γ′ S γ∗′, γ e δ (α γ) = e S β,∗ αe − − T T − − T T Z Z n o n o (1) 2i(δα+δβ ) (1) Sβα = e S β,∗ α (2.121) − T T where δα and δβ are called phase shifts. In this scenario, it is clear that the absolute value of the matrix for the process α β is the same as the one for the process α β. For instance, if all particles in both the initial → T → T 2.3. SYMMETRIES OF THE S MATRIX 89 − and final states are massive, and satified the conditions above and time-reversal invariance, the differential rate for the specific process remains unchanged if we reverse both the momenta and the spin three-component σ of all particles. A good example is the nuclear beta decay N N + e +ν ¯, with S(0) the S matrix produced by the strong → ′ − − nuclear and electromagnetic interactions alone, and S(1) the correction to the S matrix generated by the weak − interactions. The process α β is the same as the one for the process α β as long as we neglect the weak → T →T Coulomb interaction between the electron and the nucleus N ′ in the final state. For this beta decay, the initial and final states are eigenstates of the strong interaction S matrix with δ = δ = 0. Under the assumption of − α β time-reversal invariance, the differential rate for a beta decay is unaltered if we reverse both the momenta and the third componente σ of the spin of all particles7. This prediction was not contradicted by the experiment in 1957 that led to the violation of parity. Time-reversal invariance is compatible with the observation that electrons from the decay 60 60 Co Ni + e− + ν → are emmitted preferentially in a direction opposite to that of the spin of Co60. To see it, we observe that by reversing all three momenta and spins, the transformed state is such that electrons are preferentially emmitted in the opposite direction of the spin of Co60. We shall see later that indirect evidence against time-reversal invariance was discovered in 1964. However, such a symmetry remains being a good aproximation in weak, strong and electromagnetic interactions. It is sometimes possible to use a basis of states for which α = α and β = β. In that case8, Eq. (2.121) T T simplifies even more (1) 2i δ +δ (1) S = e ( α β )S ∗ (2.122) βα − β,α (1) which just says that iSβα has the phase δα + δβ mod π. This statement is known as Watson’s theorem. The phases in Eqs. (2.121, 2.122) can be measured in processes with interference between different final states. An example is the decay of the spin 1/2 hyperon Λ into a nucleon and a pion. The final state can only have orbital angular momentum l =0 or l = 1, and the angular distribution of the pion relative to the Λ spin involves the interference between these states. Therefore, according with Watson’s theorem this interference depends on the difference δ δ of their phase shifts. s − p

2.3.5 PT symmetry

The 1957 experiment did not rule out the conservation of time-reversal, but showed that P T was not conserved. If conserved this operator must be antiunitary for the same reasons as T . Hence, in processes like beta decay P T conservation would lead to relations similar to (2.121)

(1) 2i(δα+δβ) (1) Sαβ = e S ∗β, α (2.123) − PT PT for massive particles, it is clear that reverses the sign of σ (spin three-component), but not the three momenta. PT In the beta decay 60 60 Co Ni + e− +ν ¯ → 60 neglecting the Coulomb interaction between Ni and e− in the ﬁnal state, it would lead to no preference for the electron to be emitted in the same or opposite direction to the Co60 spin, in contradiction with the observations.

7All particles involved in this interaction are massive (including the neutrino). 8Imagine for instance two identical particles of three component of spin zero, and opposite momenta in the initial state, and a final state with similar features (though with different species of particles for the initial and final states). 90 CHAPTER 2. SCATTERING THEORY

2.3.6 Charge-conjugation C, CP and CPT Charge conjugation is a special case of internal symmetry that interchanges particles and antiparticles. It implies the existence of a unitary operator C, with an eﬀect on multi-particle states described by

c c C p σ n ; p σ n ; . . .± = ξ ξ p σ n ; p σ n ; . . .± (2.124) 1 1 1 2 2 2 n1 n2 · · · 1 1 1 2 2 2 c where n is the antiparticle of the particle of the type n, where ξn is a phase related with this transformation. If this is the rule of transformation for both the “in” and “out” states, the S matrix will satisfy the condition −

′ ′ ′ ′ ′ ′ ′ ′ c′ ′ ′ c′ c c Sp σ n ,p σ n , ;p1σ1n1,p2σ2n2, = ξn∗′ ξn∗′ ξn1 ξn2 Sp σ n ,p σ n , ;p1σ1n ,p2σ2n , (2.125) 1 1 1 2 2 2 ··· ··· 1 2 · · · · · · 1 1 1 2 2 2 ··· 1 2 ··· as in any other internal symmetry, condition (2.125) is satisﬁed if the operator C0 that acts as in (2.124) on free-particle states, commutes with H0 and V

1 1 C− H0C = H0 ; C− VC = V and in that case we take C = C0. The phase ξn is called a charge-conjugation parity. Like in the case of intrinsic parities ηn, the ξn are in general not uniquely defined, because for any C operator defined to satisfy Eq. (2.124) we can define another such operator with different ξn by multiplying C by a phase transformation such as exp [iαB + iβL + iγQ]. The only particles whose charge conjugation parities can be measured individually, are those completely neutral like the photon or the neutral pion (that is particles carrying no conserved quantum numbers), and they coincide with their own antiparticles. In reactions involving only completely neutral particles, Eq. (2.125) says that the product of initial charge- conjugation parities must be the same as the product of final states charge-conjugation parities. As an example, we shall see later that quantum electrodynamics requires that photon has charge-conjugation parity ξ = 1, so γ − the observation of the process π0 2γ → 0 requires that η 0 = +1. In turn, it implies that the process π 3γ should be forbidden as effectively occurs π → as far as we know. For γ and π0 we have real charge-conjugation parities either +1 or 1. As in the case − of intrinsic parity, the charge-conjugation parity can always be defined as 1, if all internal charge-conjugate ± phase transformations are part of continuous groups of phase transformations. In that case we can redefine C by multiplying by the inverse square root of the internal symmetry C2

2 2 IC C =1 ; [IC ,C] = 0 and the new operator C CI , satisfies the condition C 2 = 1. ′ ≡ C ′ For general reactions, the satisfaction of condition (2.125) requires that the rate for a process equals the rate for the same process replacing particles by their antiparticles. Once again this was not contradicted directly by the results of the 1957 experiment, since the analogous experiment with the corresponding antiparticles was not carried out. However, these experiments showed that C is not conserved in the theory of weak interactions as proposed by Lee and Yang to take into account on parity non-conservation. Indeed, we shall see later that the observed violation of P T imply a violation of C conservation in any field theory of weak interactions. At the present state of the art, we know that C and P are not conserved in the weak interactions responsible of processes such as the beta decay, and the decay of a pion and muon. Notwithstanding, C and P are conserved in the strong and electromagnetic interactions. Further, the violation of parity and charge-conjugation symmetries, keep the door open for the conservation of the CP symmetry. It had important implications on the properties of neutral K mesons. In 1954 Gellmann and Pais pointed out that since K0 does not coincide with its antiparticle (K0 carries a non-zero strangeness 2.4. RATES AND CROSS-SECTIONS 91 which is an approximate symmetry) the particles with definite decay rates would be not K0 or K0 but the linear combinations K0 K0 ± this was originally explained in terms of C conservation but once the C violation was stated, it was explained in the framework of CP conservation. If we arbitrarily define the phases in the CP operator and in the K0 and K0 states as CP K0 = K0 ; CP K0 = K0 E E we can define self-charge-conjugate one-particle states

1 1 K0 K0 + K0 ; K0 K0 K0 1 ≡ √2 2 ≡ √2 − h Ei h Ei which have CP eigenvalues +1 and 1 respectively. The fastest decay mode of these particles is into two pions but − CP conservation would allow it for K0 but not for K0. The neutral K mesons have spin zero, so the two-pion 1 2 − final state has l = 0 so that P = +1. On the other hand, C = +1 for 2 neutral pions because the pion has + C = +1 (in addition, C = +1 for a state π π− with l = 0, because C interchanges the two pions). 0 − Therefore, the K2 state is expected to decay in slower modes, into three pions or a pion, muon or electron and neutrino, thus enlarging its mean lifetime. However, Fitch and Cronin in 1964 found that the long-lived neutral K0 meson, has a small probability for decaying into two pions. They conclude that CP is slightly violated in 2 − weak interactions, but seems to be more nearly conserved than C or P individually. We shall see later that although neither C or CP are exactly conserved, we expect that CP T is exactly conserved in all interactions at least in any quantum field theory. It is CP T the operator that provides the correspondence between particles and antiparticles, and the fact that CP T commutes with the Hamiltonian tells us that stable particles and antiparticles have exactly the same mass. Further, since CP T is antiunitary, it relates the S matrix for an arbitrary process to the S matrix for the − − inverse process with all spin three-components inverted and particles replaced by antiparticles (but three-momenta are unchanged). In cases in which the S matrix can be divided as in Eq. (2.118) into a weak term S(1) that produces a given − reaction and a strong term S(0) that acts in the initial and final states, we can use an argument similar to the one followed in the T conservation scenario to show that the rate of any process is equal to the rate of the same process with particles replaced by antiparticles and spin three-components reversed, as long as we sum over sets of initial and final states that are complete with respect to S(0). In particular, despite the partial rates for decay (0) of the particle into a pair of final states β1 and β2 with S = 0 may differ from the partial rates for the decay β1β2 6 of the antiparticle into the corresponding final states β and β , we shall see that the total decay rate CPT 1 CPT 2 of any particle is exactly equal to the total decay rate of its antiparticle. From the discussion above, we can understand that the 1957 experiment shows strong C, P and P T violation but not CP violation. It is because such an experiment is consistent with T conservation and any quantum field theory must preserve CP T . Thus such an experiment is consistent with CP conservation. Similarly, the observation in 1964 of small CP violation in the weak interactions9 along with the CP T conservation assumption, leads to an indirect evidence of a tiny violation of time-reversal symmetry. This has been verified by a more detailed studies of the K0 K0 system. − 2.4 Rates and cross-sections

The S matrix with elements S provides the amplitude of probability associated with the process α β. − βα → However, we still have to relate such an amplitude with observables measured by experimentalists. Observables are in some way related with the probability density given by S 2. On the other hand, Eq. (2.61) shows that | βα| 9More recently, evidence of CP violation has been found in a B − B system, where B is another neutral meson. 92 CHAPTER 2. SCATTERING THEORY

S contains a factor δ4 (p p ) that ensures the conservation of three-momentum and energy. Hence, we should βα β − α manage properly the factor δ4 (p p ) 2 in calculations. β − α From a fundamental point of view, we should manage with wave packets that represent particles localized far from each other before the collision (which is the way the experiments are prepared), and then characterizing the time evolution of these superpositions of multi-particle states. Our derivation will be however quite practical. We consider that our whole system of physical particles is enclosed in a macroscopic volume V . This box could be taken as a cube, but with points on opposite sides identiﬁed such that the single-valuedness of the spatial wave-function requires the momenta to be quantized10 2π p = (n ,n ,n ) (2.126) L 1 2 3

3 where ni are integers, and L = V . In that case, all three dimensional delta functions are redeﬁned by an integration over a bounded volume (instead of the whole space) of the same integrand that deﬁnes the original delta function 3 1 3 i(p p′) x δV p′ p d x e − · (2.127) − ≡ (2π)3 ZV therefore, the new “bounded delta function” becomes discrete. We observe that if p′ = p, the RHS of Eq. (2.127) becomes the (bounded) macroscopic volume of the box, hence

3 1 3 i(p p′) x V δV p′ p d x e − · = δp,p′ (2.128) − ≡ (2π)3 (2π)3 ZV where δp,p′ is an ordinary (discrete!) Kronecker delta. Moreover, substituting the Dirac delta functions in the normalization condition (2.5) by the “bounded Dirac delta functions” in (2.128), we obtain that the states involved in such a condition have scalar products in a box that are not only sums of products of Kronecker deltas, but also N contains a factor V/ (2π)3 for each particle in the state. Thus we also have a factor V/ (2π)3 , where N is the number of particlesh in thei state. Calculations of transition probabilities should useh states ofi unit norm. We shall introduce states that are normalized approximately for our box

3 Nα Box (2π) α v α (2.129) ≡ u" V # | i u t whose norm is given by Box Box β α = δβα where δ is a product of Kronecker deltas, one for each three-momentu m, spin, and species label, plus terms with βα particles permuted as in the normalization condition (2.5). Thus, the S matrix associated with states (2.129) − normalized in the box yields

Nα Nβ + V V Box(+) Box( ) Sβα = α β− = 3 3 α β − s(2π) s(2π) D E (Nα+Nβ) 2 + V Box Sβα = α β− = 3 Sβα (2.130) (2π)

10It is a prominent feature on quantum mechanics that bounded systems posseses a discrete spectrum of momenta. Such a feature depends on the particle-wave nature of physical systems, so we can extrapolate it to the case of relativistic quantum mechanics. For macroscopic volumes a huge quantity of allowed discrete values is expected, and only in the limit in which the box becomes unbounded in all its dimensions, we obtain a totally continuous spectrum of momenta. 2.4. RATES AND CROSS-SECTIONS 93

Box where Sβα is calculated from states (2.129). If we left the particles in the box forever, then every possible transition would occur again and again. A meaningful transition probability occurs when we also put our system in a “time box”. That is, we assume that the interaction is turned on for a time T . It is conventional to define that the particles arrive at the region of collision at t 0. Thus, we define the “box of time” in which the interaction occurs as [ T/2, T/2]. As a ∼ − consequence, the energy conservation delta function is replaced with11 1 T/2 δT (Eα Eβ)= exp [i (Eα Eβ) t] dt (2.131) − 2π T/2 − Z− If the multi-particle system is in the state α before the interaction is turned on, the probability of finding it in the state β after the interaction is turned off, reads

3 (Nα+Nβ ) Box 2 (2π) 2 P (α β)= Sβα = Sβα (2.132) → " V # | | where we have used (2.130). It is the probability of transition into one specific box state β. The number of 3 one-particle box states within a momentum space volume d p is the number of triplets of integers n1,n2,n3 for which the momentum (2.126) lies within the volume d3p centered at p. On the other hand, Eq. (2.128) says that V/ (2π)3 is the density of such states in the three-momentum space. Therefore, the number of one-particle box states within a momentum space volume d3p is given by d3p V (2π)3 we shall define the final state interval dβ as a product of d3p for each final particle, hence the total number of states in such a range is given by V Nβ d = dβ (2.133) Nβ 3 (2π) and the total probability for the system initially in the α state, to lie within a range dβ of final states is obtained | i by combining Eqs. (2.132, 2.133) and gives

(Nα+Nβ) (2π)3 V Nβ dP (α β) = P (α β) d = S 2 dβ → → Nβ V | βα| 3 " # (2π) 3 Nα (2π) 2 dP (α β) = P (α β) d β = Sβα dβ (2.134) → → N " V # | | In this section, we shall restrict our attention to final states β that are different from the initial states α, but that also satisfy the condition that no subset of the particles in the state β (other than the whole system itself) has precisely the same four-momentum as some corresponding subset of the particles in the state α. For such states, we can define a delta-free matrix element Mβα as follows S 2iπδ3 (p p ) δ (E E ) M (2.135) βα ≡− V β − α T β − α βα the introduction of the box permits to interpret the squares of the delta functions in S 2 for β = α. Since δ3 (p p ) | βα| 6 V β − α is zero for p = p we can write this square as α 6 β 2 δ3 (p p ) = δ3 (p p ) δ3 (0) V β − α V β − α V [δ (E E )]2 = δ (E E ) δ (0) T β − α T β − α T 11Observe that (2.128) along with (2.131) can be interpreted as the “bounded Dirac delta function” in the four-momentum, defined in a bounded space-time box. 94 CHAPTER 2. SCATTERING THEORY

3 and δV (0), δT (0) can be obtained by using p = p′ and Eα = Eβ in Eqs. (2.128, 2.131). We obtain ﬁnally

3 2 3 3 3 V δV (pβ pα) = δV (pβ pα) δV (0) = δV (pβ pα) (2.136) − − − (2π)3 T [δ (E E )]2 = δ (E E ) δ (0) = δ (E E ) (2.137) T β − α T β − α T T β − α 2π from Eqs. (2.135, 2.136, 2.137) the diﬀerential of transition probability (2.134) becomes

3 Nα 3 Nα (2π) 2 (2π) 3 2 dP (α β) = Sβα dβ = 2iπδV (pβ pα) δT (Eβ Eα) Mβα dβ → " V # | | " V # − − −

3 Nα 2 (2π) 3 2 2 2 = (2π) δV (pβ pα) [δT (Eβ Eα)] Mβα dβ " V # − − | | Nα (2π)3 V T dP (α β) = (2π)2 δ3 (p p ) δ (E E ) M 2 dβ → V V β − α 3 T β − α 2π | βα| " # (2π) and we obtain ﬁnally

Nα 1 (2π)3 − T dP (α β) = (2π)2 M 2 δ3 (p p ) δ (E E ) dβ (2.138) → V 2π | βα| V β − α T α − β " # and letting V and T to be very large, the delta function product becomes an ordinary four-dimensional delta function δ4 (p p ). In this limit the transition probability is proportional to the time T during which the β − α interaction is acting, with a coefficient that can be interpreted as a differential transition rate (differential of probability per unit time)

dP (α β) 3Nα 2 1 Nα 2 4 dΓ (α β) → = (2π) − V − M δ (p p ) dβ (2.139) → ≡ T | βα| β − α where the structure (2.135) of Sαβ in this limit becomes S 2πiδ4 (p p ) M (2.140) βα ≡− β − α βα this is the master formula to connect the S matrix elements with experimental measurements. There are two − cases of particular importance

2.4.1 One-particle initial states

When the initial state consists of a one-particle state then Nα = 1, and Eq. (2.139) becomes independent of the volume V . It gives the transition rate for a single-particle state α to decay into a general multi-particle state β dΓ (α β) = 2π M 2 δ4 (p p ) dβ (2.141) → | βα| β − α however, we should take into account that many particles are unstable so that they decay spontaneously (that is, they decay even in the absence of interaction). For a given (very big) sample of unstable particles, the number of particles as a function of time is approximately given by a model like

Γt N = N0e− where N0 is the initial number of particles. From this model we can deﬁne a characteristic time of decay (mean lifetime of the particle) as the time in which the number of particles in the sample has decayed by a factor of 1/e. Thus it is clear that the mean lifetime of the particle described by the state α is given by

1 τα =Γα− 2.4. RATES AND CROSS-SECTIONS 95

and Γα is called the spontaneous decay width of the particle. In this case, we are characterizing not the spontaneous decay but the decay induced by an interaction. Consequently, Eq. (2.141), is only valid if we can neglect the spontaneous decay rate. In other words, such an equation only makes sense if the time T during which the interaction acts is much less than the mean lifetime τα of the particle α. However, it leads to the problem that this condition could prevents us to pass to the limit T in δ (E E ). There is an unremovable width →∞ T α − β 1 1 ∆E & ≈ T τα

1 in this delta function, such that Eq. (2.141) is only useful if the total decay rate τα− is much less than any of the characteristic energies of the process.

2.4.2 Two-particles initial states

When Nα = 2, Eq. (2.139) is proportional to 1/V . In other words, it is proportional to the density of either particle at the position of the other one. In experiments it is usually reported the transition rate per flux (instead of the transition rate per density), also known as the cross section. The flux of either particle at the position of the other one is defined as the product of the density 1/V and the relative velocity uα u Φ = α (2.142) α V where we define (by now) uα as the velocity of one particle if the other is at rest. Therefore, the differential cross-section is given by

4 1 uα 2 4 dΓ (α β) (2π) Mβα δ (pβ pα) dβ dσ (α β) → = uα V | | − → ≡ Φα uα/V 4 dΓ (α β) (2π) 2 4 dσ (α β) → = Mβα δ (pβ pα) dβ (2.143) → ≡ Φα uα | | −

2.4.3 Multi-particle initial states The cases N = 1, 2 are the most important, but transition rates with N 3 are also observable, and some of α α ≥ them are important in branches such as Particle Physics, Chemistry, Astrophysics etc. As an example, in the main reactions that release energy from the sun, two protons and an electron turn into a deuteron and a neutrino. We shall see later some applications of the master transition rate formula (2.139), to the case of arbitrary number of initial particle states.

2.4.4 Lorentz transformations of rates and cross-sections The Lorentz transformation rule (2.60) for the S matrix is complicated by the momentum-dependent matrices − associated with each particle’s spin. We can avoid such a complication by squaring the absolute value of (2.60) after factoring out the Lorentz-invariant delta function in Eq. (2.140), and then sum over all spins. The unitarity of the matrices deﬁned in Eq. (2.4), page 66, shows that apart from the energy factors in (2.60), the sum is Lorentz-invariant. It means that the quantity

M 2 E E R (2.144) | βα| ≡ βα α spinsX Yβ Y is a scalar function of the four-momenta of the particles in states α and β. By E we mean the product of all β single-particle energies p0 = p2 + m2 for the particles in the state α. Q p 96 CHAPTER 2. SCATTERING THEORY

We can then write the spin-summed single-particle decay rate (2.141) in the form

2 spins Mβα E E | | β α dΓ (α β) = 2π M 2 δ4 (p p ) dβ = 2πδ4 (p p ) dβ → | βα| β − α β − α P E QE Q spins spins X X β α Q Q taking into account that there is a single particle in the state α we have R dΓ (α β) = 2πδ4 (p p ) dβ βα → β − α E E spins α X β 2πE 1R δ4 (p p )Qdβ dΓ (α β) = α− βα β − α → E spins X β Q the factor dβ/ E can be recognized as the product of the Lorentz-invariant momentum-space volume elements β deﬁned in (1.157)Q 12. Since this product is Lorentz invariant, R and δ4 (p p ) also are. The only non-invariant βα β − α factor is 1/Eα, where Eα is the energy of the single-particle initial state. We have found that the decay rate has the same Lorentz transformation property of 1/Eα. This is a manifestation of the timedilation property of special relativity, the faster the particle the slower it decays. In the same way, after summing the spins, the cross-section (2.143) yields

2 4 spins Mβα E E (2π) | | β α dσ (α β)= δ4 (p p ) dβ → u β − α P E QE Q spins α X β α Q Q but in this case there are two particles in the initial state α, therefore

2 4 spins Mβα E E (2π) | | β α dσ (α β) = δ4 (p p ) dβ → u β − α P E E QE Q spins α 1 2 X β 4 Q (2π) R δ4 (p p ) dβ dσ (α β) = βα α − β → u E E E spins α 1 2 X β Q where E1 and E2 are the energies of the two particles in the initial state α. It is conventional to deﬁne the cross-section (when summed over spins) to be a Lorentz invariant function of the four-momenta. Further, the factors dβ R ; ; δ4 (p p ) βα E α − β β Q are already Lorentz invariant. Therefore, we must deﬁne the relative velocity uα such that uαE1E2 is a scalar (Lorentz invariant) function of four-momenta. We have said that in the Lorentz frame in which one of the particles (say particle 1) is at rest, uα is the velocity of the other particle. From the previous facts, the quantity uα is uniquely determined in an arbitrary Lorentz frame

2 2 2 (p1 p2) m1m2 uα = · − (2.145) q E1E2 12Note that we are identifying dβ with d3p. This is another way to say that the orbital variables are the only ones that can be continuous. 2.4. RATES AND CROSS-SECTIONS 97

where p1, p2 and m1, m2 are the four-momenta and proper mass of the two particles in the state α. From Eq. (2.145) it is obvious that E1E2uα is a Lorentz invariant function of the four-momenta (recall that the contraction of two four-vectors is a scalar). Moreover, when the particle 1 is at rest, we have

µ µ p1 = (p1, E1) = (0,m1) ; p2 = (p2, E2) p p = pµp = (p , E ) (p , E )= m E (2.146) 1 · 2 1 2µ 1 1 · 2 − 2 − 1 2 so that in this frame, equation (2.145) gives

2 2 2 2 2 2 2 2 (p1 p2) m1m2 ( m1E2) m1m2 m E m u = · − = − − = 1 2 − 2 α q q E1E2 E1E2 pm1E2 2 2 E2 m2 uα = − (2.147) p E2 further, we also have p2 = (p , E ) (p , E )= p 2 E2 2 2 2 · 2 − 2 | 2| − 2 and p2 = m2, then 2 − 2 p 2 E2 = m2 | 2| − 2 − 2 p 2 = E2 m2 (2.148) | 2| 2 − 2 substituting (2.148) in (2.147) we obtain p2 uα = | | E2 which is precisely the velocity of particle 2, as required. Similarly, uα becomes the velocity of particle 1 in the reference frame in which the particle 2 is at rest13. It is also useful to see the form of uα in the “center of mass” frame, in which the total three-momentum is null. Hence, in such a frame we have

p = (p, E ) , p = ( p, E ) (2.149) 1 1 2 − 2 p p = p2 E E (2.150) 1 · 2 − − 1 2 on the other hand, we also have

2 2 2 2 2 2 E1 = p + m1 ; E2 = p + m2 (2.151) m2 = E2 p2 ; m2 = E2 p2 (2.152) 1 1 − 2 2 − using (2.150) and (2.152) we ﬁnd

2 (p p )2 m2m2 = p2 E E E2 p2 E2 p2 1 · 2 − 1 2 − − 1 2 − 1 − 2 − = p4 + E2E2 + 2E E p2 E2E2 + E2p2 + p2E2 p4 1 2 1 2 − 12 1 2 − (p p )2 m2m2 = 2E E p2 + E2p2 + p2E2 = p2 2E E + E2 + E2 1 · 2 − 1 2 1 2 1 2 1 2 1 2 (p p )2 m2m2 = p2 (E + E )2 (2.153) 1 · 2 − 1 2 1 2 and applying (2.153) in Eq. (2.145), we obtain p (E + E ) p p u = | | 1 2 = 1 2 (2.154) α E E E − E 1 2 1 2

13 It is important to say that uα has nothing to do with the four-velocity usually deﬁned in special relativity. 98 CHAPTER 2. SCATTERING THEORY

as could be expected from a relative velocity. We should say however that in this frame uα does not correspond to a physical velocity. In particular, if we apply Eq. (2.154) to two ultrarelativistic particles we have14 p p u = + 2 α E E ≈ 1 2

such that uα = 2 which is greater than c (recall that c = 1, so that velocities are dimensionless in natural units).

2.5 Physical interpretation of the Dirac’s phase space factor δ4 (p p ) dβ β − α The phase-space factor δ4 (p p ) dβ appears in the general formula (2.139) and of course in the particular β − α cases (2.141, 2.143) for decay rates and cross-sections. We shall study the case in which we are in the center of mass reference frame, where the total three-momentum of the initial state vanishes

pα = 0

In particular if Nα = 1, it corresponds to the reference frame in which the initial particle is at rest. Returning to the general case, if the thrre momenta of the ﬁnal states are denoted by p1′ , p2′ , p3′ ,..., the phase-space factor then becomes

4 3 0 0 0 3 3 3 δ (pβ pα) dβ = δ p1′ + p2′ + . . . 0 δ p1′ + p2′ + . . . p d p1′ d p2′ d p3′ 4 − 3 − − 3 3 3 · · · δ (p p ) dβ = δ p′ + p′ + . . . δ E′ + E′ + . . . E d p′ d p′ d p′ (2.155) β − α 1 2 1 2 − 1 2 3 · · · where E E is the total energy of the initial state. We can perform the integration over each p , (say p ) just α ≡ k′ 1′ by dropping the momentum delta function

4 3 2 δ (p p ) dβ δ E′ + E′ + . . . E d p′ d p′ (2.156) β − α → 1 2 − 2 3 · · · with the undestanding that whenever p1′ appears (as in E1′ ), it must be replaced with

p′ = p′ p′ . . . (2.157) 1 − 2 − 3 − similarly, we can use the remaining delta function to eliminate any one of the remaining integrals.

2.5.1 The case of Nβ =2 Let us now study the special case in which there are two particles in the ﬁnal state β. In that case Eq. (2.155) yields 4 3 3 δ (p p ) dβ δ E′ + E′ E d p′ d p′ β − α → 1 2 − 1 2 after integrating over p1′ we can write the delta function as in (2.156) 4 3 δ (p p ) dβ δ E′ + E′ E d p′ (2.158) β − α → 1 2 − 2 as long as we take into account that condition (2.157) must be satisﬁed, that is

p′ = p′ (2.159) 2 − 1 from Eqs. (2.151) we can write the phase-space form factor (2.158) as

4 2 2 2 2 2 δ (p p ) dβ δ p + m + p + m E p′ d p′ dΩ (2.160) β − α → 1′ 1′ 1′ 2′ − 1 1 q q 14Recall that an ultrarelativistic particle is the one in which the kinetic energy is much greater than the self-energy of the particle. 2 2 2 2 2 Therefore, in such a limit we have E ≈ p . In our case, if both particles are ultrarelativistic we have E1 ≈ E2 ≈ p . 2.5. PHYSICAL INTERPRETATION OF THE DIRAC’S PHASE SPACE FACTOR δ4 (P P ) Dβ 99 β − α where dΩ = sin θ dθ dφ is the solid angle diﬀerential for p1′ . Equation (2.160) can be simpliﬁed by using the property δ (x x ) δ (f (x)) = − 0 (2.161) f (x ) | ′ 0 | where f (x) is an arbitrary real function with a single simple zero at x = x0. From Eq. (2.160) our function within the Dirac’s delta function is given by

2 2 2 2 f p′ p + m + p + m E 1 ≡ | 1′ | 1′ | 1′ | 2′ − q q it is convenient to redeﬁne 2 2 2 f (x) x + m + x + m E ; x = p′ ≡ 1′ 2′ − 1 q q the argument y in which f (x) vanishes, i.e. f (y) = 0, yields

2 2 y + m1′ + y + m2′ = E q q we can obtain y by squaring two times as follows

2 2 2 2 2 y + m1′ + y + m2′ + 2 y + m1′ y + m2′ = E

q 2 2 2 2 2 2y + m′ +m′ E = 2 y + m y + m 1 2 − 1′ 2′ 2 2 2 2 q 2 2 2y + m′ + m′ E = 4 y+ m′ y + m′ 1 2 − 1 2 simplifying these equations we ﬁnd

2 2 2 2 2 2 2y + m1′ + m2′ E 4 y + m1′ y + m2′ = 0 4 2 2 2 2 − 2 − 2 2 4 4 E 2E m2′ 2E m1′ 4Ey 2m1′ m2′ +m1′ + m2′ = 0 2 4 −4 2 − 2 4 − 2 2− 2 2 2 2 4E y + m′ + m′ + 2m′ m′ + E 2E m′ 2E m′ 4m′ m′ = 0 − 1 2 1 2 − 1 − 2 − 1 2 2 2 2 2 2 2 2 4E y + E m′ m′ 4m′ m′ = 0 − − 1 − 2 − 1 2 from which the root gives 2 2 2 2 2 2 2 E m1′ m2′ 4m1′ m2′ y = p1′ = − − 2 − 4E since p′ is positive there is a simple zero at p′ = √y k′ | 1| | 1| ≡

2 2 2 2 2 2 E m1′ m2′ 4m1′ m2′ k′ = − − − q 2E in addition

E2 m 2 m 2 2 4m 2m 2 2 2 2 2 2 2 1′ 2′ 1′ 2′ 4E m1′ E′ = p + m = k + m = − − − + 1 1′ 1′ ′ 1′ s 4E2 4E2 q q E2 m 2 m 2 2 4m 2m 2 + 4E2m 2 E2 m 2 + m 2 2 − 1′ − 2′ − 1′ 2′ 1′ − 2′ 1′ E1′ = = q 2E q 2E 2 2 2 E m2′ + m1′ E1′ = − 2E 100 CHAPTER 2. SCATTERING THEORY

and similarly for E2′ . Now the derivative evaluated at the simple zero gives

d 2 2 2 2 p1′ p1′ f ′ k′ = p1′ + m1 + p1′ + m2 E = | | + | | d p′ | | | | − ′ ′  2 2 2 2  1 p1 =k | | q q p1′ + m1 p1′ + m2 ′ ′ | | | | | | p1 =k q q | | k k = ′ + ′ 2 2 2 2 k′ + m1 k′ + m2 k′ k′ k′ (E1′ + E2′ ) k′E f ′ k′ = p + = p = E1′ E2′ E1′ E2′ E1′ E2′ In summary, the argument E + E E of the delta function in (2.160) has a unique zero at p = k , where 1′ 2′ − | 1′ | ′ 2 2 2 2 2 E m1′ m2′ 4m1′ m2′ k′ = − − − (2.162) q 2E 2 2 2 2 2 E m2′ + m1′ E′ = k + m = − (2.163) 1 ′ 1′ 2E q 2 2 2 2 2 E m1′ + m2′ E′ = k + m = − (2.164) 2 ′ 2′ 2E q with derivative

d 2 2 2 2 k′ k′ k′E f ′ k′ = p1′ + m1 + p1′ + m2 E = + = (2.165) d p1′ | | | | − p′ =k′ E1′ E2′ E1′ E2′ | | 1 q q | | Therefore applying property (2.161) to our speciﬁc case Eq. (2.160) yields

4 2 2 2 2 2 δ ( p1′ k′) 2 δ (p p ) dβ δ p + m + p + m E p′ d p′ dΩ= | |− p′ d p′ dΩ β − α → 1′ 1′ 1′ 2′ − 1 1 f (k ) 1 1 ′ ′ q q | | 4 E1′ E2′ δ ( p1′ k′) 2 δ (pβ pα) dβ | |− k′ d p1′ dΩ (2.166) − → k′E

Therefore, after performing the integration over d p , we drop the delta function and the diﬀerential d p | 1′ | | 1′ | and replace p by k in the remaining terms on the RHS of Eq. (2.166) | 1′ | ′ k E E δ4 (p p ) dβ ′ 1′ 2′ dΩ (2.167) β − α → E and understanding that k′, E1′ and E2′ are given everywhere by (2.162, 2.163, 2.164). For the particular case Nα = 1, the diﬀerential decay rate (2.141) that describes the decay of one particle at rest (zero three-momentum) and energy E into two particles is

dΓ (α β) 2πk E E → = ′ 1′ 2′ M 2 (2.168) dΩ E | βα| and the diﬀerential cross-section for a two-body scattering process 1 2 1 2 is given by Eq. (2.143) and yields → ′ ′ 4 dσ (α β) (2π) k′E1′ E2′ 2 → = Mβα dΩ Euα | | and using Eq. (2.154) for the relative velocity we ﬁnd

dσ (α β) (2π)4 k E E E E → = ′ 1′ 2′ 1 2 M 2 dΩ E p (E + E ) | βα| | 1| 1 2 2.5. PHYSICAL INTERPRETATION OF THE DIRAC’S PHASE SPACE FACTOR δ4 (P P ) Dβ 101 β − α hence, we obtain ﬁnally

4 4 dσ (α β) (2π) k′E1′ E2′ 2 (2π) k′E1′ E2′ E1E2 2 → = Mβα = 2 Mβα dΩ Euα | | E k | | k p = p . (2.169) ≡ | 1| | 2| recall that all these formulas have been obtained in the CM reference frame.

2.5.2 The case with Nβ =3 and Dalitz plots

There are still many cases of interest in the case of Nβ = 3 as it is the case of decyas of a single intitial particle into three bodies. For the case with Nβ = 3 equation (2.156) yields

4 3 3 2 2 2 2 2 2 δ (p p ) dβ d p′ d p′ δ (p + p ) + m + p + m + p + m E (2.170) β − α → 2 3 × 2′ 3′ 1′ 2′ 2′ 3′ 3′ − q q q the momentum-space volume in spherical coordinates is given by

3 3 2 2 d p2′ d p3′ = p2′ d p2′ p3′ d p3′ dΩ3 dφ23 d cos θ23

where dΩ3 is the differential element of solid angle for p3′ and θ23, φ23 are the polar and azimuthal angles of p2′ reltive to the p3′ direction. The orientation of the plane spanned by p2′ and p3′ is specified by φ23 and the direction of p3′ , and the remaining angle θ23 is fixed by the energy conservation condition

p 2 + 2 p p cos θ + p 2 + m 2 + p 2 + m 2 + p 2 + m 2 = E 2′ | 2′ | | 3′ | 23 3′ 1′ 2′ 2′ 3′ 3′ q q q the derivative of the argument of the delta function with respect to cos θ23 is

∂E p p 1′ = | 2′ | | 3′ | ∂ cos θ23 E1′ so the integral over cos θ23 can be done just by dropping the delta function and dividing by this derivative

4 δ (p p ) dβ p′ d p′ p′ d p′ E′ dΩ dφ β − α → 2 2 3 3 1 3 23 replacing mometa with energies , we obtain ﬁnally

4 δ (p p ) dβ E′ E′ E′ dE′ dE′ dΩ dφ (2.171) β − α → 1 2 3 2 3 3 23 now we recall that the expression (2.144) obtained by summing M 2 over spins and multiplying with the product | βα| of energies. Such an expression is a scalar function of four-momenta. If we approximate this scalar as a constant, Eq. (2.171) tells us that for a given initial state, the distribution of events plotted in the E2′ , E3′ plane is uniform. Any deviation from this uniform distribution of events is a useful clue to the dynamics of the decay process, including possible centrifugal barriers or resonant intermediate states. This are known as Dalitz plots, since it was used by Dalitz in 1953 to analize the decay

+ + + K π π π− → 102 CHAPTER 2. SCATTERING THEORY

2.6 Perturbation theory

One of the most useful technique to calculate the S matrix is perturbation theory. It consists of obtaining an − expansion in powers of the interaction term V when the Hamiltonian is given by H = H0 + V . We start with the expressions (2.44) and (2.22) of the S matrix − S = δ (β α) 2πiδ (E E ) T + ; T + = β(0) V α+ (2.172) βα − − β − α βα βα D where α+ satisfies the Lippmann-Schwinger equations (2.22) | i T + α+ = α(0) + dγ γα γ(0) (2.173) (Eα Eγ + iε) E Z − E + from Eq. (2.172) it is clear that all non-trivial contents of the S matrix is concentrated on the factor T . To − βα find an approximate value of such a factor, we apply β(0) V on both sides of Eq. (2.173), and we obtain T + β(0) V α+ = β(0) V α(0) + dγ γα β(0) V γ(0) (Eα Eγ + iε) D D E Z − D E T + V + γα βγ (0) (0) Tβα = Vβα + dγ ; Vβα β V α (2.174) (Eα Eγ + iε) ≡ Z − D E + which is an integral equation for T . The perturbation series is obtained by iterating Eq. (2.174). The first iteration gives V V V T + = V + dγ βγ1 T + = V + dγ βγ1 V + dγ γ1γ2 T + βα βα 1 (E E + iε) γ1α βα 1 (E E + iε) γ1α 2 (E E + iε) γ2α Z α − γ1 Z α − γ1 Z α − γ2 V V V V T + = V + dγ βγ1 γ1α + dγ dγ βγ1 γ1γ2 T + (2.175) βα βα 1 (E E + iε) 1 2 (E E + iε) (E E + iε) γ2α Z α − γ1 Z Z α − γ1 α − γ2 the second iteration yields V V T + = V + dγ βγ1 γ1α βα βα 1 (E E + iε) Z α − γ1 + V V T Vγ γ + dγ dγ βγ1 γ1γ2 V + dγ γ3α 2 3 1 2 (E E + iε) (E E + iε) γ2α 3 (E E + iε) Z Z α − γ1 α − γ2 " Z α − γ3 # V V V V V T + = V + dγ βγ1 γ1α + dγ dγ βγ1 γ1γ2 γ2α βα βα 1 (E E + iε) 1 2 (E E + iε) (E E + iε) Z α − γ1 Z Z α − γ1 α − γ2 V V V + dγ dγ dγ βγ1 γ1γ2 γ2γ3 T + (2.176) 1 2 3 (E E + iε) (E E + iε) (E E + iε) γ3α Z Z Z α − γ1 α − γ2 α − γ3 Eqs. (2.175, 2.176), show that the RHS depends in turn on T +. However, it can also be observed that each term (from left to right) increases the power of V and the last term on the RHS of these equations (which is the only one that contains T +) is of higher order (in powers of V ) than the other ones15. Thus if V is sufficiently small, higher order terms are smaller. Then, these expressions are calculable if we drop the latter term on the RHS on either of these equations. If we drop the last term on the RHS of Eq. (2.175), we are calculating at second order in V , while if we drop the last term in Eq. (2.176) we are calculating at third order in V . The expansion at third order in V obtained from (2.176) reads V V V V V T + = V + dγ βγ1 γ1α + dγ dγ βγ1 γ1γ2 γ2α + . . . (2.177) βα βα 1 (E E + iε) 1 2 (E E + iε) (E E + iε) Z α − γ1 Z Z α − γ1 α − γ2 15We should take into account that T + depends in turn on V according with the definition (2.172), thus the integral that contains T + is of order V n+1 in the potential, where n is the number of integrals in that term. 2.6. PERTURBATION THEORY 103

The method of calculation based on Eqs. (2.175, 2.176) is called old-fashioned perturbation theory. The presence of the energy in the denominators obscure the underlying Lorentz invariance of the S matrix. Notwith- − standing, it is still useful in clarifying the way that singularities of the S matrix arise from several intermediate − states. We shall work mostly on a rewritten version of Eqs. (2.175, 2.176) known as time-dependent perturbation theory, in which Lorentz invariance is more apparent, though it somewhat obscures the contribution of individual intermediate states. The simplest way to derive the time-ordered perturbation expansion is by starting with the S operator deﬁned − by Eqs. (2.38, 2.39)

S = U ( , ) ; U (τ,τ ) exp [iH τ] exp [ iH (τ τ )] exp [ iH τ ] (2.178) −∞ ∞ 0 ≡ 0 − − 0 − 0 0

Diﬀerentiating this formula for U (τ,τ0) with respect to τ gives a diﬀerential equation for U (τ,τ0)

d d U (τ,τ ) = exp [iH τ] exp [ iH (τ τ )] exp [ iH τ ] dτ 0 dτ { 0 − − 0 − 0 0 } = exp [iH τ] (iH ) exp [ iH (τ τ )] exp [ iH τ ] { 0 } 0 { − − 0 − 0 0 } + exp [iH τ] ( iH) exp [ iH (τ τ )] exp [ iH τ ] { 0 } − { − − 0 − 0 0 } d U (τ,τ ) = i exp [iH τ] (H H) exp [ iH (τ τ )] exp [ iH τ ] dτ 0 { 0 } 0 − { − − 0 − 0 0 } and inserting an identity operator we ﬁnd

d U (τ,τ ) = i exp [iH τ] V exp [ iH τ] exp [+iH τ] exp [ iH (τ τ )] exp [ iH τ ] dτ 0 − { 0 } { − 0 0 } { − − 0 − 0 0 } d U (τ,τ ) = i exp [iH τ] V exp [ iH τ] U (τ,τ ) dτ 0 − { 0 − 0 } 0 so we obtain ﬁnally

d i U (τ,τ ) = V (τ) U (τ,τ ) (2.179) dτ 0 0 V (t) exp [iH τ] V exp [ iH τ] (2.180) ≡ 0 − 0 observe that even if V is time-independent the operator V (t) is time-dependent. Note that V (t) is the operator associated with V through the interaction picture (not the Heisenberg picture)16. Equation (2.179) along with the initial condition U (τ0,τ0) = 1 (2.181) are satisﬁed by the following solution of the integral equation

τ U (τ,τ ) = 1 i dt V (t) U (t,τ ) (2.182) 0 − 0 Zτ0 once again, we obtain an expansion of U (τ,τ0) in powers of V by iteration of Eq. (2.182). The ﬁrst iteration gives

τ τ t1 U (τ,τ ) = 1 i dt V (t ) U (t ,τ ) = 1 i dt V (t ) 1 i dt V (t ) U (t ,τ ) 0 − 1 1 1 0 − 1 1 − 2 2 2 0 Zτ0 Zτ0 Zτ0 τ τ t1 U (τ,τ ) = 1 i dt V (t ) + ( i)2 dt dt V (t ) V (t ) U (t ,τ ) 0 − 1 1 − 1 2 1 2 2 0 Zτ0 Zτ0 Zτ0 16Remember that in the Heisenberg picture the operators are transformed in the same way as in equation (2.180), but with the whole Hamiltonian H instead of H0. 104 CHAPTER 2. SCATTERING THEORY further iterations are given by

τ τ t1 t2 U (τ,τ ) = 1 i dt V (t ) + ( i)2 dt dt V (t ) V (t ) 1 i dt V (t ) U (t ,τ ) 0 − 1 1 − 1 2 1 2 − 3 3 3 0 Zτ0 Zτ0 Zτ0 Zτ0 τ τ t1 U (τ,τ ) = 1 i dt V (t ) + ( i)2 dt dt V (t ) V (t ) 0 − 1 1 − 1 2 1 2 Zτ0 Zτ0 Zτ0 τ t1 t2 + ( i)3 dt dt dt V (t ) V (t ) V (t ) U (t ,τ ) − 1 2 3 1 2 3 3 0 Zτ0 Zτ0 Zτ0 τ τ t1 U (τ,τ ) = 1 i dt V (t ) + ( i)2 dt dt V (t ) V (t ) 0 − 1 1 − 1 2 1 2 Zτ0 Zτ0 Zτ0 τ t1 t2 t3 + ( i)3 dt dt dt V (t ) V (t ) V (t ) 1 i dt V (t ) U (t ,τ ) − 1 2 3 1 2 3 − 4 4 4 0 Zτ0 Zτ0 Zτ0 Zτ0 the expansion yields

τ τ t1 U (τ,τ ) = 1 i dt V (t ) + ( i)2 dt dt V (t ) V (t ) 0 − 1 1 − 1 2 1 2 Zτ0 Zτ0 Zτ0 τ t1 t2 + ( i)3 dt dt dt V (t ) V (t ) V (t ) − 1 2 3 1 2 3 Zτ0 Zτ0 Zτ0 τ t1 t2 t3 + ( i)4 dt dt dt dt V (t ) V (t ) V (t ) V (t ) U (t ,τ ) − 1 2 3 4 1 2 3 4 4 0 Zτ0 Zτ0 Zτ0 Zτ0 once again for V (t) suﬃciently small, we can neglect the term containing U (t,τ0) on the RHS, because it is the term of highest order. We then have

τ τ t1 U (τ,τ ) = 1 i dt V (t ) + ( i)2 dt dt V (t ) V (t ) 0 − 1 1 − 1 2 1 2 Zτ0 Zτ0 Zτ0 τ t1 t2 + ( i)3 dt dt dt V (t ) V (t ) V (t )+ . . . (2.183) − 1 2 3 1 2 3 Zτ0 Zτ0 Zτ0 from Eq. (2.178) we see that we obtain the S operator expansion by taking τ = and τ = in Eq. (2.183) − 0 −∞ ∞ t1 S = 1 i ∞ dt V (t ) + ( i)2 ∞ dt dt V (t ) V (t ) − 1 1 − 1 2 1 2 Z−∞ Z−∞ Z−∞ t1 t2 + ( i)3 ∞ dt dt dt V (t ) V (t ) V (t )+ . . . (2.184) − 1 2 3 1 2 3 Z−∞ Z−∞ Z−∞ this equation can also be derived directly from the old-fashioned perturbation expansion (2.177), by using the Fourier representation of the energy factors in such an equation

1 ∞ (E E + iε)− = i dτ exp [i (E E ) τ] (2.185) α − γ − α − γ Z0 ετ with the understanding that the integrals must be evaluated by inserting a convergence factor e− in the integrand with ε 0+. → We can rewrite Eq. (2.184) in a way useful to carry out calculations that are manifestly Lorentz invariant. We deﬁne the time-ordered product of any product of operators that depend on time as follows

T (t1) (t2) . . . (tn 1) (tn) = (tp1 ) (tp2 ) . . . tpn−1 (tpn ) (2.186) {O O O − O } O O O O tpn < tpn−1 <...

T V (t) = V (t) { } T V (t ) V (t ) = θ (t t ) V (t ) V (t )+ θ (t t ) V (t ) V (t ) { 1 2 } 1 − 2 1 2 2 − 1 2 1 where θ (τ) is the step function. The time-ordered product of n operators is a sum over all n! permutations of such products, and each one gives the same integral over all times t t t . Therefore, Eq. (2.184) can be written as 1 2 · · · n n ∞ ( i) ∞ S =1+ − dt dt dt T V (t ) V (t ) V (t ) (2.188) n! 1 2 · · · n { 1 2 · · · n } nX=1 Z−∞ which is sometimes called the Dyson series. If all V (ti)’s commute with each other, this sum becomes

S = exp i ∞ dt V (t) − Z−∞ However, this is not usually the case. Sometimes Eq. (2.188) even does not converge, and is at most an asymptotic expansion in some coupling factors that appear in V . Nevertheless, Eq. (2.188) can be written as

S = T exp i ∞ dt V (t) − Z−∞ where T indicates that the expression must be evaluated by time-ordering each term in the series expansion of the exponential. Now, we shall find some sufficient conditions (satisfied by a large class of theories) for which the S matrix − is manifestly Lorentz invariant. Recalling that the S matrix elements are the elements of the S operator in − − the basis of free particles α(0) , β(0) etc. We want to find the conditions for which the S operator commute − with the operator U (Λ, a) that produces Lorentz transformations on the free-particle states. This is equivalent 0 to find the conditions for which the S operator commute with the generators H , P , J and K . To satisfy that − 0 0 0 0 requirement we shall try by using the hypothesis that V (t) is an integral over three-space

V (t)= d3x (x,t)= d3x (x) (2.189) H H Z Z where (x) is a scalar in the sense that H 1 U (Λ, a) (x) U − (Λ, a)= (Λx + a) (2.190) 0 H 0 H it can be checked that (x) has a time-dependence consistent with Eq. (2.180). To see it, we write an inﬁnitesimal H pure translation operator in the Lorentz group Eq. (1.99), page 27

U (1, ε) = 1 iε P ρ 0 − ρ 0 and we specify it even more as a pure time-translation

U 1, ε0 = 1 iε P 0 =1+ iH ε0 (2.191) 0 − 0 0 0 applying (2.191) in Eq. (2.190) and equating the coeﬃcients of ε0, we obtain that the time dependence of (x) H is consistent with Eq. (2.180) [homework!!(9)]. 106 CHAPTER 2. SCATTERING THEORY

By using the hypothesis (2.189), the S operator (2.188) may be written as a sum of four-dimensional integrals − n ∞ ( i) ∞ S = 1+ − dt dt dt T d3x (x ) d3x (x ) d3x (x ) n! 1 2 · · · n 1 H 1 2 H 2 · · · n H n n=1 Z−∞ Z Z Z X n ∞ ( i) ∞ S = 1+ − d4x d4x d4x T (x ) (x ) (x ) (2.192) n! 1 2 · · · n {H 1 H 2 ···H n } n=1 X Z−∞ in Eq. (2.192) everything is manifestly Lorentz invariant except for the time-ordering of the operator product. Now, recall that for two given events x1 and x2, the time ordering of two causally connected events [time-like events i.e. (x x )2 < 0] is a Lorentz invariant, but it is not a Lorentz invariant for causally disconnected events 1 − 2 [space-like events i.e. (x x )2 > 0]. Therefore, it is clear that a suﬃcient condition for Eq. (2.192) to be Lorentz 1 − 2 invariant, is that (x) commute at space-like or light-like separations17 H 2 (x) , x′ =0 for x x′ 0 (2.193) H H − ≥ On the other hand, we shall use the results of section 2.3.1, to prove with a non-perturbative argument that an interaction of the type (2.189) satisfying Eqs. (2.190) and (2.193) does lead to an S matrix with the appropriate − Lorentz transformation properties. For an inﬁnitesimal boost, Eq. (2.190) yields [homework!!(10), reproduce equations (2.194), and (2.195)] ∂ i [K , (x,t)] = t (x,t)+ x (x,t) (2.194) − 0 H ∇H ∂tH hence, integrating over x and setting t = 0, we obtain

[K ,V ] = K , d3x (x, 0) = [H , W] (2.195) 0 0 H 0 Z W d3x x (x, 0) (2.196) ≡ − H Z If (as it is usually the case) the matrix elements of (x, 0) between eigenstates of H are smooth functions of the H 0 energy eigenvalues, then the same is true for V , as is necessary for the validity of the scattering theory, and also true for W, which is necessary in the proof of Lorentz invariance [This smooth behavior leads to Eq. (2.94), page 80]. From Eq. (2.195) we see that the other condition for Lorentz invariance, given by commutation relation (2.83) page 78, is also valid if and only if [H0, W] = [H, W] = [H0 + V, W] or equivalently 0 = [W,V ]= d3x d3y x [ (x, 0) , (y, 0)] (2.197) H H Z Z It is clear that the “causality condition” (2.193) leads to (2.197). However, Eq. (2.197) provides a less restrictive sufficient condition for the Lorentz invariance of the S matrix. − Since the conditions we have found are sufficient but not necessary, it is clear that the class of theories that satisfy these conditions are not the only ones that are Lorentz invariant. However, the most general Lorentz invariant theories are not very different. There is always a commutation condition quite similar to (2.193) that should be satisfied. Note that the “causality condition” (2.193), has no counterpart in non-relativistic theories. It owes to the fact that time-ordering is always Galilean invariant, it is obvious since Galilean transformations keep the time coordinate invariant. Consequently, condition (2.193) is the one that makes the combination of Lorentz invariance and quantum mechanics very restrictive.

17We prefer to include the light-like condition because we shall see later that Lorentz invariance can be disturbed by troublesome singularities at x = x′. 2.6. PERTURBATION THEORY 107

2.6.1 Distorted-wave Born approximation The methods described above are useful as long as the interaction term V is sufficiently small. A modified method of approximation known as the distorted-wave Born approximation, is useful when the interaction contains two terms V = Vs + VW (2.198) such that V is weak but V is strong. We can define α as the “in” and “out” states when we consider V as W s | s±i s the whole interaction. In that case, we can write the Lippmann-Schwinger equation (2.21) associated with the strong interaction.

(0) 1 α± = α + (E H iε)− V α± (2.199) s α − 0 ± s s (0)E 1 α± = α + α± V (E H iε )− (2.200) s s s α − 0 ∓ D 18 from (2.198, 2.200) the second of Eqs. (2.22) associated wit h the complete interaction yields

+ (0) + 1 + T = β V α = β− β− V (E H + iε)− (V + V ) α βα s − s s β − 0 s W D + h 1 i + = β− V α + β− V V (E H + iε)− (V + V ) α s W s s − s β − 0 s W + h + 1 i + = β− V α + β− V α (E H + iε)− (V + V ) α s W s s − β − 0 s W + + n + 1 + o T = β− V α + β− V α (E H + iε)− V α βα s W s s − β − 0 n o and using the Lippmann-Schwinger equations associated with the whole interaction V , we obtain

+ + (0) Tβα = βs− VW α + βs− Vs α (2.201) E + The second term on the RHS of Eq. (2.201) is the term Tβα when it is associated with the strong interaction Vs alone

s+ (0) + (0) T β V α = β− V α (2.202) βα ≡ s s s s D E To prove Eq. (2.202) we just follow the same procedure that le d to Eq. (2.201) but dropping VW everywhere.

s+ (0) + 1 + T = β V α = β− β− V (E H + iε)− V α βα s s s − s s β − 0 s s D h 1 + i = β− V V (E H + iε)− V α s s − s β − 0 s s h + 1i + = β− V α (E H + iε)− V α s s s − β − 0 s s + n (0) o Tβα = βs− Vs α E Equation (2.201) is most useful when the second term on the RHS vanishes. That is, when the process α β → cannot be produced by the strong interaction alone. For such processes the matrix element (2.202) vanishes. Consequently, Eq. (2.201) becomes + + Tβα = βs− VW α (2.203) up to now the equation is exact, at least under our assumption s. Nevertheless, this equation is useful when VW can be considered so weak to neglect its eﬀect on the state α+ in Eq. (2.203). In that case, we can replace the | i 18Of course, there is no distinction between the free states α(0) in the Lippmann-Schwinger equations associated with the strong E interaction alone, with respect to the free states in the Lippmann-Schwinger equations associated with the whole interaction.

108 CHAPTER 2. SCATTERING THEORY state α+ (associated with the whole interaction) by the state α+ (due to the strong interaction alone), in Eq. | i | s i (2.203) + + T β− V α (2.204) βα ≃ s W s this is valid to ﬁrst order in V but to all orders in V . W s This approximation is used in many scenarios of Physics. A good example is the nuclear beta or gamma decay for which the S matrix element is calculated by using Eq. (2.204) with V being the strong nuclear interaction − s and V is either the weak nuclear interaction or the electromagnetic interaction, and with β and α+ the ﬁnal W | s−i | s i and initial nuclear states. In particular, in nuclear beta decay we require a weak nuclear force to turn neutrons into protons, even though we cannot ignore the presence of the strong nuclear force. Thus, the process cannot be produced by the strong interaction alone, and the second term on the RHS of Eq. (2.201) vanishes, as required in the distorted-wave Born approximation.

2.7 Implications of unitarity

Equation (2.61) shows that the non-trivial part of the S matrix is contained in the amplitude M . We shall see − βα that the unitarity condition of the S matrix imposes a useful condition relating the amplitude M for forward − αα scattering in an arbitrary multiparticle state α, to the total rate for all reactions in that state. We start with the parameterization (2.61) of the S matrix for a general process α β − → S = δ (β α) 2πiδ4 (p p ) M (2.205) βα − − β − α βα the unitarity condition is 1= S†S (2.206) writing the unitarity condition (2.206) explicitly and using the parameterization (2.205) we ﬁnd

δ (γ α) = dβ S† S = dβ S∗ S − γβ βα βγ βα Z Z 4 4 = dβ δ (β γ) + 2πiδ (p p ) M ∗ δ (β α) 2πiδ (p p ) M − β − γ βγ − − β − α βα Z δ (γ α) = dβ δ (β γ) δ (β α) 2πi dβ δ (β γ) δ4 (p p ) M − − − − − β − α βα Z Z 4 2 4 4 +2πi dβ δ (β α) δ (p p ) M ∗ + 4π dβ δ (p p ) δ (p p ) M ∗ M − β − γ βγ β − γ β − α βγ βα Z 4 4 Z δ (γ α) = δ (γ α) 2πi δ (p p ) M + 2πi δ (p p ) M ∗ − − − γ − α γα α − γ αγ 2 4 4 +4π dβ δ (p p ) δ (p p ) M ∗ M β − γ β − α βγ βα Z cancelling δ (γ α) we obtain −

4 4 2 4 4 2πi δ (p p ) M + 2πi δ (p p ) M ∗ + 4π dβ δ (p p ) δ (p p ) M ∗ M = 0 − γ − α γα α − γ αγ β − γ β − α βγ βα Z and taking into account that

δ4 (p p ) δ4 (p p )= δ4 (p p ) δ4 (p p ) β − γ β − α γ − α β − α we have

4 4 2 4 4 2πi δ (p p ) M + 2πi δ (p p ) M ∗ + 4π δ (p p ) dβ δ (p p ) M ∗ M = 0 (2.207) − γ − α γα α − γ αγ γ − α β − α βγ βα Z 2.7. IMPLICATIONS OF UNITARITY 109 we can factor 2πδ4 (p p ) in (2.207). Thus, for p = p we obtain19 γ − α α γ

4 iM + iM ∗ + 2π dβ δ (p p ) M ∗ M = 0 (2.208) − γα αγ β − α βγ βα Z if we further assume that α = γ, the ﬁrst two terms on the LHS of Eq. (2.208) become

iM + iM ∗ = iM + ( iM )∗ = 2Re [ iM ] − αα αα − αα − αα − αα = 2Re [ i (ReM + iImM )] = 2Re [ iReM + ImM ] − αα αα − αα αα iM + iM ∗ = 2ImM − αα αα αα therefore, Eq. (2.208) for α = γ yields

4 2ImM + 2π dβ δ (p p ) M ∗ M = 0 αα β − α βα βα Z ImM = π dβ δ4 (p p ) M 2 (2.209) αα − β − α | βα| Z which is the most useful form of Eq. (2.208). We shall relate ImMαα with the total rate for all reactions produced by an initial state α in a volume V . To do it we integrate Eq. (2.139) over all ﬁnal states β

dΓ (α β) 3Nα 2 1 Nα 2 4 Γ dβ → = dβ (2π) − V − M δ (p p ) α ≡ dβ | βα| β − α Z Z h i 3Nα 2 1 Nα 4 2 Γ = (2π) − V − dβ δ (p p ) M (2.210) α β − α | βα| Z substituting Eq. (2.209) in Eq. (2.210) we ﬁnd

1 3Nα 2 1 Nα Γ = (2π) − V − ImM (2.211) α −π αα Let us examine the particular case in which α is a two-particle state. In that case, using Eq. (2.143), we see that the total cross-section in the state α is given by

dσ (α β) (2π)4 σ dβ → = M 2 δ4 (p p ) dβ (2.212) α ≡ dβ u | βα| β − α Z α Z once again substituting Eq. (2.209) in Eq. (2.212) we obtain

1 (2π)4 σα = ImMαα −π uα 16π3 σα = ImMαα (2.213) − uα where uα is the relative velocity given by (2.145). This is usually expressed in terms of a scattering amplitude f (α β) coming from the cross-section in the → center-of-mass reference frame. The diﬀerential cross-section for two-body scattering in the center-of-mass frame is given by Eq. (2.169)

4 dσ (α β) (2π) k′E1′ E2′ E1E2 2 → = M ; k p = p ; k′ p′ = p′ . (2.214) dΩ kE2 | βα| ≡ | 1| | 2| ≡ 1 2

19 Of course, for pα 6= pγ Eq. (2.207) becomes trivial. 110 CHAPTER 2. SCATTERING THEORY so we deﬁne the scattering amplitude as

2 4π k′E1′ E2′ E1E2 f (α β) Mβα (2.215) → ≡− E r k so that the differential cross-section is written as dσ (α β) → = f (α β) 2 (2.216) dΩ | → | it is clear that the phase of f (α β) is conventional, and is usually motivated by the wave mechanical interpre- → tation of f as the coefficient of the outgoing wave in the solution of the time-independent Schrödinger equation. In particular for elastic two-body scattering (so that k = k′ and Ei′ = Ei for i = 1, 2), we have

2 4π kE1E2E1E2 f (α β) = Mβα → − E r k 4π2E E f (α β) = 1 2 M (2.217) → − E βα now taking the imaginary part on Eq. (2.217), setting α = β, and using Eq. (2.213) we have

4π2E E 4π2E E σ u Im f (α α) = 1 2 Im M = 1 2 α α → − E αα E 16π3 E E σ Im f (α α) = 1 2 α u → E 4π α and using the expression (2.154) for the relative velocity uα in the center-of-mass frame, we have E E σ k (E + E ) 1 σ Im f (α α) = 1 2 α 1 2 = α kE → E 4π E1E2 E 4π k Im f (α α) = σ (2.218) → 4π α This form of the unitarity condition (2.213) for elastic scattering of two-body states α is known as the optical theorem. We can obtain information about the pattern of scattering at high energy from the optical theorem. The scattering amplitude can be expected to be a smooth function of angles. Hence, there should be some solid angle ∆Ω within which f 2 has nearly the same value (say within a factor of 2) as in the forward direction (θ = 0). In | | that case, the total cross-section is bounded by 1 1 σ f 2 dΩ f (α α) 2 ∆Ω Im f (α α) 2 ∆Ω α ≥ | | ≥ 2 | → | ≥ 2 | → | Z combining these inequalities with Eq. (2.218), we ﬁnd an upper bound on ∆Ω

1 1 k2 32π2 σ Im f (α α) 2 ∆Ω = σ2 ∆Ω ∆Ω α ≥ 2 | → | 2 16π2 α ⇒ k2σ ≥ α 32π2 ∆Ω 2 (2.219) ≤ k σα total cross-section is expected to approach constants or grow slowly at high energies. Consequently, Eq. (2.219) shows that the solid angle around the forward direction within which the diﬀerential cross-section is roughly constant shrinks at least as fast as k 2 for k . This increasingly narrow peak in the forward direction at high − →∞ energies is called a diﬀraction peak. 2.7. IMPLICATIONS OF UNITARITY 111

2.7.1 Generalized optical theorem and CPT invariance Returning to the general case of reactions involving an arbitrary number of particles, we shall relate the total interaction rates of particles and antiparticles by combining Eq. (2.209) with CPT invariance. Since CPT is antiunitary, it does not imply a simple relation between the process α β, and the corresponding → process obtained by changing particles by their antiparticles. Due to the role of the time-reversal (that leads to the antiunitarity of CP T ) it shall provide a relation between a process and the inverse process involving antiparticles. By using the same arguments that led to Eq. (2.115) page 87 for time-reversal invariance, we can see that CPT invariance requires the S matrix to satisfy the condition −

Sβ,α = S α, β (2.220) CPT CPT we recall that implies that we must reverse all spin x components, change all particles by their correspond- CPT 3− ing antiparticles, and multiply the matrix element by several phase factors for the particles in the initial state and by their complex conjugates for the particles in the ﬁnal state (all three-momenta are left invariant). Since CP T invariance requires that the mass of each particle is the same as the mass of the corresponding antiparticle, the relation (2.220) must also be hold by the amplitude or coeﬃcient of δ4 (p p ) in S shown in Eq. (2.205) α − β βα

Mβ,α = M α, β (2.221) CPT CPT in the particular case in which initial and ﬁnal states are the same (α = β), all phases cancel since their module is the unity. Hence, when α = β Eq. (2.221) becomes

c c c c Mp1σ1n1;p2σ2n2; ,p1σ1n1;p2σ2n2; = Mp1( σ1)n ;p2( σ2)n ; ,p1( σ1)n ;p2( σ2)n ; (2.222) ··· ··· − 1 − 2 ··· − 1 − 2 ··· so that the amplitude Mα,α must coincide with the amplitude associated with states of the corresponding antiparticles (characterized by the quantum numbers nc) with opposite spin x components20. Consequently, the i 3− generalized optical theorem (2.209) or equivalently (2.211), says that the total reaction rate from an initial state consisting of some set of particles is the same as for an initial state consisting with the corresponding antiparticles with spins reversed.

c c Γp1σ1n1;p2σ2n2; =Γp1( σ1)n ;p2( σ2)n ; (2.223) ··· − 1 − 2 ··· In the particular case of one-particle states, the decay rate of any particle equals the decay rate of the antiparticle with reversed spin. On the other hand, rotational invariance does not allowed particle decay rates to depend on the spin x component of the decaying particle. Hence, as a special case of the general result (2.223) we see 3− that unstable particles and their corresponding antiparticles have the same lifetimes. Equation (2.209) was obtained from the unitarity condition S†S = 1. By using the unitarity condition SS† = 1 very similar arguments lead to the relation (recall that both unitarity conditions are not equivalent in an inﬁnite dimensional Hilbert space) ImM = π dβ δ4 (p p ) M 2 (2.224) αα − β − α | αβ| Z such that Eqs. (2.209, 2.224) lead to a reciprocity relation

dβ δ4 (p p ) M 2 = dβ δ4 (p p ) M 2 (2.225) β − α | βα| β − α | αβ| Z Z it is important to say that this reciprocity relation is far-from trivial since in the general case there is not any simple relation between Mβα and Mαβ. By using Eq. (2.139), we can rewrite equation (2.225) in the following

20Note that setting α = β simpliﬁes the relation (2.221) because the processes α → β and β → α are the same, and also because the phase factors are all cancelled. 112 CHAPTER 2. SCATTERING THEORY way

dΓ (α β) V Nα 1 dΓ (β α) V Nβ 1 dβ → − = dβ → − dβ 3Nα 2 dα 3Nβ 2 Z (2π) − Z (2π) − V 1 dΓ (α β) V Nα V 1 dΓ (β α) V Nβ − dβ → = − dβ → 2 3Nα 2 3N (2π)− dβ (2π)− dα β Z (2π) Z (2π) obtaining ﬁnally dΓ (α β) dΓ (β α) V Nα dβ cα → = dβ cβ → ; cα (2.226) dβ dα ≡ (2π)3 Z Z 2.7.2 Unitarity condition and Boltzmann H-theorem

Equation (2.226) can be used to derive the “Boltzmann H-theorem” which is crucial in kinetic theory. Let Pα dα be the probability of finding the system in a volume dα of the space of multiparticle states α . We start by | i calculating the rate of decrease in Pα due to transitions to all other states. In other words, the probability per unit time that flows outside the volume of multi-particle states. The probability per unit time for a transition of a fixed α state into a set of states within the volume dβ is given by

dΓ (α β) dβ → dβ and integrating over β we obtain the probability per unit time to obtain the transition from a fixed α into any final state β. However, we have not certainty that the initial state is precisely α, hence we have to multiply this expression by Pα. We then finally obtain

dP dΓ (α β) α = P dβ → dt α dβ in Z now we calcualte the rate of increase of Pα due to transitions from all other states into the state α. In other words, the probability per unit time that ﬂows inside the volume of multiparticle states. It can be derived with a similar argument as dP dΓ (β α) α = dβ P → dt β dα out Z therefore, the rate of change of Pα reads

dP dP dP dΓ (β α) dΓ (α β) α = α α = dβ P → P dβ → (2.227) dt dt − dt β dα − α dβ in out Z Z ﬁrst of all we integrate over α states on both sides

d dΓ (β α) dΓ (α β) P dα = dα dβ P → dα P dβ → (2.228) dt α β dα − α dβ Z Z Z Z Z interchanging the labelling of the integration variables in the integral of the second term in Eq. (2.228), we obtain that d dΓ (β α) dΓ (β α) P dα = dα dβ P → dβ P dα → dt α β dα − β dα Z Z Z Z Z d dΓ (β α) dΓ (β α) P dα = dα dβ P → dα dβ P → dt α β dα − β dα Z Z Z Z Z 2.7. IMPLICATIONS OF UNITARITY 113 such that d P dα = 0 dt α Z leads to the conservation of probability. On the other hand, we can calculate the rate of change of the entropy dS d P d P dP P d P dα P ln α = dα P ln α = dα α ln α + P ln α dt ≡ −dt α c − dt α c − dt c α dt c Z α Z α Z α α dP P c d P dP P dP = dα α ln α + P α α = dα α ln α + α − dt c α P dt c − dt c dt Z α α α Z α dS P dP = dα ln α + 1 α (2.229) dt − c dt Z α Substituting (2.227) into Eq. (2.229) we ﬁnd

dS P dΓ (β α) dΓ (α β) = dα dβ ln α + 1 P → P → (2.230) dt − c β dα − α dβ Z Z α dS P dΓ (β α) P dΓ (α β) = dα dβ ln α + 1 P → + dα dβ ln α + 1 P → (2.231) dt − c β dα c α dβ Z Z α Z Z α interchanging the labelling of the integration variables in the second term, we write Eq. (2.231) as follows

dS P dΓ (β α) P dΓ (β α) = dα dβ ln α + 1 P → + dβ dα ln β + 1 P → dt − c β dα c β dα Z Z α Z Z β P dΓ (β α) dΓ (β α) = dα dβ ln α P → dα dβ P → − c β dα − β dα Z Z α Z Z P dΓ (β α) dΓ (β α) + dα dβ ln β P → + dα dβ P → c β dα β dα Z Z β Z Z P P dΓ (β α) = dα dβ ln β ln α P → c − c β dα Z Z β α dS P c dΓ (β α) = dα dβ P ln β α → (2.232) dt β P c dα Z Z α β now for any two positive quantities x and y, the following inequality holds y y ln y x (2.233) x ≥ − setting x = Pαcβ and y = Pβcα the inequality (2.233) becomes

P c P c P c P c ln β α P c P c P ln β α P α β β α P c ≥ β α − α β ⇒ β P c ≥ β − c α β α β α P c P P P ln β α β α c (2.234) β P c ≥ c − c β α β β α substituting (2.234) into (2.232), we see that the rate of change of entropy is bounded by

dS P P dΓ (β α) dα dβ β α c → dt ≥ c − c β dα Z Z β α dS P dΓ (β α) P dΓ (β α) dα dβ β c → dα dβ α c → dt ≥ c β dα − c β dα Z Z β Z Z α 114 CHAPTER 2. SCATTERING THEORY and interchanging variables of integration in the second term we have

dS P dΓ (β α) P dΓ (α β) dα dβ β c → dβ dα β c → dt ≥ c β dα − c α dβ Z Z β Z Z β dS P dΓ (β α) dΓ (α β) dα dβ β c → c → dt ≥ c β dα − α dβ Z Z β dS P dΓ (β α) dΓ (α β) dβ β dα c → c → dt ≥ c β dα − α dβ Z β Z Now, the unitarity relation (2.226) (with α and β interchanged) says that the integral over α on the RHS of this inequality vanishes. Therefore, we conclude that the entropy never decreases

dS d P dα P ln α 0 dt ≡−dt α c ≥ Z α which is the so-called “Boltzmann H-theorem”. In textbooks of statistical mechanics, such a theorem is deduced using the Born approximation, for which M 2 is symmetric in α and β so that | βα| dΓ (α β) dΓ (β α) c → = c → α dβ β dα or assuming time-reversal invariance, which would imply that M 2 is unchanged by the interchange of α and | βα| β, combined with the reversal of all momenta and spins. In that sense the present derivation is more general since we have only used the unitarity result (2.226), to derive the “Boltzmann H-theorem”. By contrast the Born approximation and time-reversal invariance are not exact. We can also observe that the conditions to obtain a steady entropy also depends only on the unitarity relation (2.226) and not of the Born approximation or time-reversal invariance. To see it we observe that the entropy becomes constant when Pα becomes a function only of conserved quantities such as the total energy and charge, times the factor cα. In this case the conservation laws require dΓ (β α) P P → =0 or α = β dα cα cβ therefore we can replace cβ Pβ Pα → cα in the ﬁrst term of Eq. (2.227). Now using the unitarity relation (2.226) shows that in this case, the probability Pα is time-independent. Chapter 3

The cluster decomposition principle

Now we shall take up the problem of constructing the Hamiltonian operator in a suitable way. Such an operator can be defined through its matrix elements between states of arbitrary numbers of particles. We shall see that another more suitable way to construct the Hamiltonian is by expressing it as a function of operators that create and destroy single particles. Historically, these operators appeared first after quantizing the electromagnetic field as well as other fields. However, there is another motivation aside of the quantization of a classical field or the question whether particles are created or destroyed. We shall see that if we express the Hamiltonian as a sum of products or creation and annihilation operators with suitable non-singular coefficients, the S matrix will satisfy − the cluster decomposition principle, which says that distant experiments produce uncorrelated results. It is for this reason that the formalism of creation and annihilation operators is widely used in quantum statistical mechanics, even if the number of particles is fixed. In relativistic quantum theories the cluster decomposition principle leads inevitably to quantum field theory. Many attempts have been done to construct a relativistic quantum theory that would not be a local field theory. Though it is possible to construct theories that are not field theories and yet lead to a Lorentz-invariant S matrix for two-particle scattering, such theories have always problems for systems − of more than two particles: Either the three-particle S matrix is not Lorentz invariant, or else violates the cluster − decomposition principle. We first construct the basis of states with arbitrary number of bosons and fermions, then define the creation and annihilation operators, and show how we can construct Hamiltonians based on those operators, that yield S matrices that satisfy the cluster decomposition principle. − 3.1 Physical states

The Hilbert space whose vectors describe the physical states must contain states describing arbitrary numbers of particles. If is the Hilbert space whose vectors describe physical one-particle states, then a Hilbert space Ei describing N-particles is given by (N) = . . . E E1 ⊗ ⊗EN since we have to describe physical states of 0, 1, 2, 3,... particles, the total Hilbert space must be

= (1) (2) . . . (N) . . . E E ⊕E ⊕ ⊕E ⊕ We shall consider physical states of 0, 1, 2, 3,... free particles. We could consider either free-particles states, or “in” or “out” states. For deﬁniteness we shall use free-particle states

p σ n , p σ n ,... | 1 1 1 2 2 2 i but all our results will be valid for “in” or “out” states as well. As customary, σ denotes spin x components or 3− helicities for massless particles, and ni denotes particle species.

115 116 CHAPTER 3. THE CLUSTER DECOMPOSITION PRINCIPLE

3.1.1 Interchange of identical particles There is an essential point that has not been considered yet: the symmetrization postulate of quantum mechanics. According with it, a set of identical particles (that is particles that possess the same quantum number n that identiﬁes the species) is described by physical states that are either symmetric or antisymmetric under the interchange of two particles within the set of identical particles. When the set of identical particles is described by a symmetric (antisymmetric) state we call the particles bosons (fermions). As far as we know, all particles are either bosons or fermions. We describe this symmetry property of a set of identical bosons or fermions as

p σ n, p σ n,..., p σ n,..., p σ n,... = ε p σ n, p σ n,..., p σ n,..., p σ n,... (3.1) | 1 1 2 2 i i j j i n | 1 1 2 2 j j i i i +1 if the particles are bosons ε (3.2) n ≡ 1 if the particles are fermions − these two cases are called the boson and fermion “statistics”. We shall see later that Bose and Fermi statistics are only possible for particles with integer or half-integer spins respectively, but we shall not need this information by now. We shall set up suitable normalization conditions for multi-boson or multi-fermion states. We ﬁrst notice that if two particles with momenta and spins pi,σi and pj,σj belong to identical species n, the two state vectors

p σ n, p σ n,..., p σ n,..., p σ n,... ; p σ n, p σ n,..., p σ n,..., p σ n,... (3.3) | 1 1 2 2 i i j j i | 1 1 2 2 j j i i i represent the same physical state. Otherwise, the particles would be distinguished by their order in the labelling of the state-vector, and the ﬁrst listed would not be identical to the second. Since both states in (3.3) are physically indistinguishable, they must belong to the same ray, so that

p σ n, p σ n,..., p σ n,..., p σ n,... = α p σ n, p σ n,..., p σ n,..., p σ n,... (3.4) | 1 1 2 2 i i j j i n | 1 1 2 2 j j i i i where αn is a complex number lying in the unitary complex circle. We could consider it, as part of the deﬁnition of what we mean by identical particles. Now the gist of the reasoning is to decide of what variables could αn depends. If αn only depends on the species n, then interchanging the two particles in (3.4) again, we obtain

p σ n, p σ n,..., p σ n,..., p σ n,... = α2 p σ n, p σ n,..., p σ n,..., p σ n,... | 1 1 2 2 i i j j i n | 1 1 2 2 i i j j i 2 such that αn = 1, yielding Eq. (3.2) as the only two possibilities for the phase. Suppose now that αn could also depend on the numbers and species of the other particles in the state. It would lead to the unpleasant conclusion that the symmetry of the state-vectors under interchange of particles in a given place may depend on the presence of particles in any other place in the universe. These are the possibilities that we discard from the cluster decomposition principle. Another possibility is to assume that αn could depend on the spins of the two particles. In that case, the (j) spin-dependent αn phase factors would provide a one-dimensional representation of the rotation group in three- dimensional space SO (3) (or in SU (2)). However, the only one-dimensional irreducible inequivalent representation of SO (3), is the trivial one (consisting of the identity alone), there are not one-dimensional representations 1 consisting of various phase factors. Thus, αn must be independent of the spin . Moreover, let us examine the scenario in which αn could depend on the three-momenta of the two particles involved in the interchange. In this scenario, Lorentz-invariance would require that αn would only depend on the scalars µ µ µ p1 p1µ, p2 p2µ, p3 p3µ 1Note that for the group of rotations in two dimensions SO (2), there are non-trivial representations in one dimension consisting of various phases. This fact opens the window for the possibility of having phases that depend on the spin in models constructed on a two-dimensional space. 3.1. PHYSICAL STATES 117 which are symmetric under the interchange of the particles 1 and 2. Therefore, such dependence would not change 2 the argument, from which αn = 1. There is another possibility in which the states p σ n, p σ n,... could carry a phase factor that depends on | 1 1 2 2 i the path through the momentum space by which the momenta of the particles are brought to the values p1, p2 etc. In this case, the interchange of two particles twice might change the state by a phase factor for which α2 = 1. It n 6 could be shown that this is a real possibility in two-dimensional space, but not for three or more spatial dimensions.

3.1.2 Interchange of non-identical particles There is not any essential symmetry property under the interchange of particles of different species. However, we could agree to choose certain symmetry patterns under these interchanges that will show to be convenient for future purposes. For instance we could agree to label the state-vector by listing all photon momenta and helicities first then all electron momenta and spin x components, and so on by sweeping the elementary particle species. Alternatively, 3− we could allow the particle labels to appear in any order, and define the state-vectors with particle labels in an arbitrary order as equal to the state-vectors with particle labels in some standard order times phase factors, whose dependence on the interchange of particles of different species can be anything we like. For example, suppose that we have an electron e− a muon µ− and a photon γ, we can describe this multiparticle state by listing the quantum numbers of the electron followed by the ones of the muon and then the ones of the photon ψ p σ n ,p σ n ,p σ n | i ≡ | e e e µ µ µ γ γ γi but we can also list them (say) starting with the muon numbers, then the electron numbers and finally the photon numbers ψ′ p σ n ,p σ n ,p σ n ≡ | µ µ µ e e e γ γ γi and since the quantum numbers of each particle are the saem as in the ket ψ , it is clear that both kets must | i describe the same physical state. Therefore both must belong to the same ray

ψ = α ψ′ | i n However, since particles are not identical, the phase do not have to follow the symmetrization postulate pattern.

We can accomodate any pattern to define the phases when non-identical particles are interchanged. It is important to say that there are symmetries like the isospin invariance that relate particles of different species. From this fact, it would be convenient to choose a convention to generalize the symmetry pattern in (3.1, 3.2). We shall do it as follows: the state-vector will be chosen to be symmetric under the interchange of any bosons with each other, or any bosons with any fermions, and antisymmetric with respect to interchange of any two fermions with each other. This pattern is chosen regardless whether the particles interchanged are of the same species or not. Of course, this pattern contains the symmetrization postulate as a special case. Note that this reasoning also shows that the symmetry or antisymmetry of the state-vector under interchange of particles of the same species but different helicities (or different x components of the spin), is purely conventional. 3− It is because we could agree from the beginning to list first the momenta of photons of helicity +1, then the momenta of all photons with helicity 1, then the momenta of all electrons of σ = +1/2, then all three- momenta − of electrons with σ = 1/2 and so on. We shall choose the convention that the state-vector is symmetric or − antisymmetric under interchange of identical bosons or fermions of different helicities or spin u components, in 3− order to facilitate the use of rotational invariance.

3.1.3 Normalization of multi-particle states The normalization of the multi-particle states must be deﬁned in consistency with the previous symmetry conventions. We shall use a label q to denote all quantum numbers of a single particle q p,σ,n ≡ 118 CHAPTER 3. THE CLUSTER DECOMPOSITION PRINCIPLE thus, N particle states are denoted as − q ,q ,...,q | 1 2 N i the vacuum state with N = 0 is denoted as 0 . For N = 0, 1 the symmetry of interchange is not relevant, hence | i we normalize these states as

0 0 = 1 ; q′ q = δ q′ q h | i 3 | i − δ q′ q δ p′ p δ ′ δ ′ − ≡ − σ σn n On the other hand, for N = 2 the vectors q ,q and q ,q describe the same physical state, thus the normal- | 1′ 2′ i | 2′ 1′ i ization is written as

q′ ,q′ q ,q = δ q′ q δ q′ q + ε δ q′ q δ q′ q (3.5) 1 2 | 1 2i 1 − 1 2 − 2 χ 2 − 1 1 − 2 1 if both particles are fermions εχ − ≡ +1 otherwise More generally N! N q′ ,q′ ,...,q′ q ,q ,...,q = δ δP δ q q′ (3.6) 1 2 M | 1 2 N i NM i − Pi P=1 i=1 X Y where δ indicates that both states are orthogonal if M = N i.e. for states of diﬀerent number of particles. The NM 6 sum is over all N! permutations of the integers 1,...,N. Further

1 if P involves an odd permutation of fermions δP − ≡ +1 otherwise thus δP = 1 if the permutation implies an odd number of fermion interchanges and +1 otherwise. It is easy − to check that the orthonormalization condition (3.6) fullﬁlls all symmetry and antisymmetry requirements under interchange of the qi and also under the interchange of the qj′ .

3.2 Creation and annihilation operators

We shall define the creation and annihilation operators by their effect on normalized multi-particle states. We denote the creation operator as a† (q) (i.e. as the adjoint of certain operator a (q)) because of the analogy with creation operators in the quantum harmonic oscillator. We define the creation operator a (p,σ,n) a (q), as † ≡ † the operator that adds a particle with quantum numbers q at the front of the list of particles in the state2

a† (q) q q q qq q q (3.7) | 1 2 · · · N i ≡ | 1 2 · · · N i In particular, the N particle state can be obtained by applying N creation operators on the vacuum state −

a† (q1) a† (q2) . . . a† (qN 1) a† (qN ) 0 = q1q2 qN 1qN (3.8) − | i | · · · − i on the other hand, the adjoint of the creation operator is denoted as a (q). We shall show that a (q) removes a particle from any state on which it acts and then it is called the annihilation operator. We shall show it, for the case in which the particles qq q q are either all bosons or all fermions. We start by calculating the scalar 1 2 · · · N product of a (q) q q q with an arbitrary state q q q . Using Eq. (3.7) it yields | 1 2 · · · N i | 1′ 2′ · · · M′ i

q′ q′ q′ a (q) q q q a† (q) q′ q′ q′ q q q 1 2 · · · M | 1 2 · · · N i ≡ 1 2 · · · M 1 2 · · · N i D q1′ q2′ qM′ a (q) q1q2 qN = qq1′ q2′ qM′ q1q2 qN (3.9) · · · | · · · i · · · · · · i 2For the sake of phase conventions it is important to deﬁne in what place of the list the new particle is added.

3.2. CREATION AND ANNIHILATION OPERATORS 119 now we write the scalar product on the RHS of Eq. (3.9) according with Eq. (3.6).

N! M+1 q′ q′ q′ a (q) q q q = qq′ q′ q′ q q q = δ δP δ q q′ (3.10) 1 2 · · · M | 1 2 · · · N i 1 2 · · · M 1 2 · · · N i N,M+1 i − Pi P=1 i=1 X Y In equation (3.10), we shall separate the sum over all permut ations P in a sum over the integer r that is permuted into the ﬁrst place Pr = 1, and a sum over mappings P from the remaining integers

r¯ 1, 2, . . . , r 1, r + 1,...,N ≡ − into 1, 2,...,N 1. Moreover, since particles qq q q are either all bosons or all fermions, the sign factor is − 1 2 · · · N r 1 δP = εχ− δP (3.11) with ε equal to +1 for bosons and ( 1) for fermions [homework!!(11) ﬁll the details to arrive to Eq. (3.12)]. By χ − separating the permutations as described above, and using (3.11), Eq. (3.10) becomes

N M r 1 q′ q′ q′ a (q) q q q = δ ε − δ δ (q q ) δ q′ qP 1 2 · · · M | 1 2 · · · N i N,M+1 χ P − r i − i r=1 i=1 X XP Y N M r 1 q′ q′ q′ a (q) q q q = δ ε − δ (q q ) δ δ q′ qP 1 2 · · · M | 1 2 · · · N i N,M+1 χ − r P i − i r=1 P i=1 X X Y and using Eq. (3.6) again, we ﬁnd

N r 1 q1′ q2′ qM′ a (q) q1q2 qN = δN,M+1 εχ− δ (q qr) q1′ q2′ qM′ q1q2 qr 1qr+1 qN · · · | · · · i − · · · · · · − · · · i r=1 X ﬁnally, since the state q q q is arbitrary, we obtain | 1′ 2′ · · · M′ i N r+1 a (q) q1q2 qN = εχ δ (q qr) q1q2 qr 1qr+1 qN (3.12) | · · · i − | · · · − · · · i r=1 X r+1 r 1 where we have used tha fact that εχ = εχ− . Equation (3.12) shows that the operator a (q) removes a particle from any state in which it acts, as stated. As a special case of Eq. (3.12), we observe that for both bosons and fermions, a (q) annihilates the vacuum

a (q) 0 = 0 0 a† (q) = 0 (3.13) | i ⇔ h |

3.2.1 Commutation and anti-commutation relations of a (q) and a† (q)

Applying a (q′) on Eq. (3.7) and using Eq. (3.12) we have

a q′ a† (q) q q q = a q′ qq q q | 1 2 · · · N i | 1 2 · · · N i N r+2 a q′ a† (q) q1q2 qN = δ q′ q q1q2 qN + εχ δ q′ qr qq1q2 qr 1qr+1 qN (3.14) | · · · i − | · · · i − | · · · − · · · i r=1 X where the sign in the second term is εr+2 because q is in the (r + 1) th place in qq q q . χ r − | 1 2 · · · N i On the other hand, applying a† (q) in Eq. (3.12) we ﬁnd

N r+1 a† (q) a q′ q1 qN = εχ δ q′ qr qq1q2 qr 1qr+1 qN (3.15) | · · · i − | · · · − · · · i r=1 X 120 CHAPTER 3. THE CLUSTER DECOMPOSITION PRINCIPLE substracting or adding Eqs. (3.14, 3.15) we obtain

a q′ a† (q) ε a† (q) a q′ q q = δ q′ q q q (3.16) − χ | 1 · · · N i − | 1 · · · N i n o now Eq. (3.16) holds for arbitrary states q q containing either only bosons or only fermions (but could be | 1 · · · N i easily extended for states containing both bosons and fermions). Thus we obtain

a q′ a† (q) ε a† (q) a q′ = δ q′ q (3.17) − χ − moreover, Eq. (3.17) leads to a† q′ a† (q) ε a† (q) a† q′ = 0 (3.18) − χ and taking the adjoint of (3.18) we also have

a (q) a q′ ε a q′ a (q) = 0 (3.19) − χ recall that in Eqs. (3.17, 3.18, 3.19) the top and bottom sign s are for bosons and fermions respectively. Moreover, according with the phase conventions discussed in section 3.1.2, the creation and/or annihilation operators for particles of two different species commute if either particle is a boson, and anticommute if both are fermions. In many textbooks this discussion is given in the opposite order. We could start with the commutation of anticommutation relations (3.17, 3.18, 3.19) derived from the canonical quantization of some given filed theory. Then multi-particle states are defined by constructing from the vacuum as in Eq. (3.8), and their scalar product Eq. (3.6) is derived from the commutation or anticommutation relations. The latter order of reasoning is indeed closer to the historical development. However, we have followed this order to show the necessity of the field quantization later.

3.3 Arbitrary operators in terms of creation and annihilation operators

Let us deﬁne an operator in the form of a sum of products of creation and annihilation operators as follows O

∞ ∞ = dq′ dq′ dq dq a† q′ a† q′ a (q ) a (q ) C q′ q′ ; q q (3.20) O 1 · · · N 1 · · · M × 1 · · · N M · · · 1 × NM 1 · · · N 1 · · · M N=0 M=0 Z X X now we shall show that any linear operator can be expressed in the form prescribed by equation (3.20) by choicing suitable values of the coefficients CNM . To do it, we shall prove that the CNM coefficients can be chosen such that the matrix elements of this expression acquire any desired values. The proof will use mathematical induction. First we start by proving that 0 0 can take any desired value by the apropriate choice of the C h | O | i NM coefficients. By using Eq. (3.20) we obtain

∞ ∞ 0 0 = dq′ dq′ dq dq C q′ q′ ; q q 0 a† q′ a† q′ a (q ) a (q ) 0 h | O | i 1 · · · N 1 · · · M × NM 1 · · · N 1 · · · M ×h | 1 · · · N M · · · 1 | i N=0 M=0 Z X X however Eqs. (3.13) say that only the terms with N = M = 0 contribute in this expansion

0 0 = C h | O | i 00 thus to give the matrix element 0 0 any desired value, we only have to ﬁx the value of C regardless the h | O | i 00 values of the other CNM coeﬃcients. Now suppose that all matrix elements of between N and M particle states with N

∞ ∞ q′ q′ q q = dq′ dq′ dq dq C q′ q′ ; q q 1 · · · L O | 1 · · · Ki 1 · · · N 1 · · · M × NM 1 · · · N 1 · · · M N=0 M=0 Z X X q′ q′ a† q′ a† q′ a (q ) a (q ) q q × 1 · · · L 1 · · · N M · · · 1 | 1 · · · Ki

q′ q′ q q = L!K!C q′ q′ ; q q 1 · · · L O | 1 · · · Ki LK 1 · · · L 1 · · · K +terms involving C NM with N < L,M K or N L, M

F q q = [f (q )+ . . . + f (q )] q q (3.21) | 1 · · · N i 1 N | 1 · · · N i such an operator can be written as in Eq. (3.20) using only the term with N = M = 1

F = dq a† (q) a (q) f (q) (3.22) Z A very important particular case, is the free-particle Hamiltonian

2 2 H0 = dq a† (q) a (q) E (q) ; E (q)= p + mn (3.23) Z p where E (q) is clearly the single-particle energy3.

3.4 Transformation properties of the creation and annihilation operators

We should characterize the transformation properties of the creation and annihilation operators under the various symmetries we have already considered. We shall start with the inhomogeneous proper orthochronus Lorentz transformations for massive particles. We recall that the N particle states (of non-null mass) have the transformation property (2.57) under such a group − 0 0 (Λp1) (Λp2) U0 (Λ, a) p1σ1n1, p2σ2n2,... = exp [ i (Λp1) a] exp [ i (Λp2) a] · · · | i − · − · · · · s p0p0 1 2 · · · D(j1) (W (Λ,p )) D(j2) (W (Λ,p )) × σ¯1σ1 1 σ¯2σ2 2 · · · σ¯1σ¯2 X··· p σ¯ n , p σ¯ n ,... (3.24) | 1Λ 1 1 2Λ 2 2 i 3Recall that for interacting multi-particle systems, energy cannot be assigned uniquely to each particle, but only to the whole system. Thus, the energy operator (Hamiltonian) is not additive anymore. 122 CHAPTER 3. THE CLUSTER DECOMPOSITION PRINCIPLE where p is the three-vector part of the four-vector Λp, and D(j) is the irreducible unitary spin j representation Λ σσ¯ − of SO (3). Finally, W (Λ,p) is the particular rotation

1 W (Λ,p) L− (Λp) Λ L (p) ≡ where L (p) is the standard “boost” that takes a particle of mass m = 0, from rest to four momentum pµ. It 6 is obvious that m and j depends on the species label n. We insist that this is for m = 0, the case of massless 6 particles will be studied later. On the other hand, these states can be constructed from the vacuum as in Eq. (3.8)

p σ n , p σ n ,... = a† (p σ n ) a† (p σ n ) 0 | 1 1 1 2 2 2 i 1 1 1 2 2 2 ···| i where 0 is the vacuum state that we assume Lorentz-invariant | i U (Λ, a) 0 = 0 (3.25) 0 | i | i In order that the state (3.8) transforms properly [taking into account the transformation properties (3.24, 3.25)], it is necessary and suﬃcient that the creation operator have the transformation rule

0 1 (Λp) (j) U (Λ, a) a† (pσn) U − (Λ, a) = exp [ i (Λp) a] D (W (Λ,p)) a† (p σ¯ n) (3.26) 0 0 − · p0 × σσ¯ Λ s σ¯ X Similarly, the discrete operators C, P and T , that produce charge-conjugation, space inversion, and time-reversal on free-particle states, transform the creation operators as

1 c C a† (pσn) C− = ξna† (pσn ) (3.27) 1 P a† (pσn) P − = ηna† ( p σ n) (3.28) 1 −j σ T a† (pσn) T − = ζ ( 1) − a† ( p σ n) (3.29) n − − − as mentioned in Sec. 3.1 the whole formalism was constructed on free-particle states but can be extrapolated to “in” or “out” states. We can then introduce operators ain and aout deﬁned in the same way by their eﬀects on the “in” and “out” states. These operators satisfy the same Lorentz transformation rule described by Eq. (3.26), but with the true Lorentz transformation operator U (Λ, a) instead of the free-particle operator U0 (Λ, a).

3.5 Cluster decomposition principle and connected amplitudes

It is a fundamental principle in Physics and indeed in all sciences, that phenomena that are suﬃciently separated in space are not correlated. Otherwise, any result of any experiment would depend on all other experiments in the earth and even worse, on all other experiments and/or phenomena that occur in the earth and in the whole universe. If this principle (known as the cluster decomposition principle) were not true, we were unable to make any predictions on any experiment without knowing everything about the whole universe. As usual the greek letters α, β will denote a collection of particles including for each particle a speciﬁcation of its momentum, spin and species. We shall also denote α1 + α2 + . . . + α to the state formed by combining N all particles in the states α1, α2,...,α . With this notation, we shall establish how could we express the cluster N decomposition principle in S matrix theory. − Suppose we have a set of multi-particle processes α1 β1, α2 β2,...,α β and that each process is → → N → N very distant from each other. The fact that this set of processes produces uncorrelated results implies that the S matrix element for the overall process factorizes as follows − S S S S (3.30) β1+β2+...+βN , α1+α2+...+αN → β1α1 β2α2 · · · βN αN 3.5. CLUSTER DECOMPOSITION PRINCIPLE AND CONNECTED AMPLITUDES 123

Thus, Eq. (3.30) holds if for all i = j all of the particles in states α and β are at a large spatial distance from all 6 i i of the particles in states α and β . The factorization of the S matrix elements leads in turn to the factorization j j − of the corresponding transition probabilities, associated with uncorrelated experimental results. It says that the probability of ocurrence for two (or more) independent or uncorrelated events, is the product of each probability as it must be. We shall rewrite Eq. (3.30) in a more transparent way by using a combinatoric trick. We define the connected part of the S matrix SC by the formula − βα S = ε SC SC (3.31) βα χ β1α1 β2α2 · · · partitionsX where the sum is over all different ways of partitioning the particles in the state α into clusters α1, α2,..., and similarly a sum over all ways of partitioning the particles in the state β into clusters β1, β2,...,. We do not count as different those partitions that merely arrange particles within a given cluster or permute whole clusters. The factor ε is +1 or ( 1) according to whether the rearragements α α α and β β β involve χ − → 1 2 · · · → 1 2 · · · altogether an even or an odd number of fermion interchanges, respectively. We shall justify later the use of the term connected. Note that (3.31) is a recursive definition. For each α and β, the sum on the RHS of Eq. (3.31) consists of a C C term Sβα plus a sum Σ′ over products of two or more S matrix elements, with a total number of particles in − 4 each of the states αj and βj that is less than the number of particles in the states α and β . In other words, we can separate (3.31) in a connected term in which α and β are consider as single clusters, plus terms in which α and β are separated into two or more clusters

C C C Sβα = Sβα + ′εχSβ1α1 Sβ2α2 partitionsX C Suppose that Sβα in this sum have been chosen such that Eq. (3.31) is satisfied for states β, α containing together fewer than say N particles. Therefore, regardless the values found in this way for the S matrix elements appearing C − in the sum Σ′, we can always choose the remaining term Sβα in such a way that Eq. (3.31) is also satisfied by states α, β containing a total of N particles. Thus Eq. (3.31) has no information by itself, it is simply a definition of SC . This argument works only if we consider that each αj and βj form non-empty sets. We must define the C connected vacuum-vacuum element S0,0 to be zero. Equation (3.31) cannot be used to define the vacuum-vacuum S matrix S , which in the absence of time-varying external fields is simply defined to be the unity S = 1. − 0,0 0,0 The following definition is useful to obtain the numbers of partitions in clusters for a given couple of multi- particle states α and β. Definition 3.1 A partition λ (n) λ , λ , . . . , λ of the positive integer n is a sequence of positive integers λ , ≡ { 1 2 r} i arranged in descending order, whose sum is n, that is λ λ and r λ = n. i ≥ i+1 i=1 i It is clear that the number of partitions of a state α with N particlesP into clusters, is the number of partitions of the positive integer N (some permutations within a given partition must be included as we shall see later). The partitions of the first four positive integers are given by λ (1) = 1 ; λ(1) (2) = 2 ; λ(2) (2) = 1, 1 (3.32) { } { } { } λ(1) (3) = 3 ; λ(2) (3) = 2, 1 ; λ(3) (3) = 1, 1, 1 (3.33) { } { } { } λ(1) (4) = 4 ; λ(2) (4) = 3, 1 ; λ(3) (4) = 2, 1, 1 ; λ(4) (4) = 2, 1, 1 (3.34) { } { } { } { } 4 It is clear that if α1, α2,...,αp defines a partition of the state α with p ≥ 2, the number of particles Nk in each subset αk must satisfy the condition N1 + N2 + . . . + Np = N where N is the number of particles in the state α. Thus Nk < N for each αk. Similarly occurs for the state β and any partition of it, in two or more subsets. 124 CHAPTER 3. THE CLUSTER DECOMPOSITION PRINCIPLE

3.5.1 Some examples of partitions

Thus, the simplest case arises when both α and β are one-particle states with quantum numbers q and q′ respectively. In that case, the only term on the RHS of Eq. (3.31) is SC itself from which the connected S matrix βα − becomes C S ′ S ′ = δ q′ q (3.35) q q ≡ q q − apart from possible degeneracies, the proportionality of S ′ withδ (q q) comes from the conservation laws. q q ′ − The absence of a proportionality factor in Eq. (3.35) comes from a suitable choice of relative phase between “in” and “out” states. Here we are assuming that single-particle states are stable, such that there are no transitions between single-particle states and any others, e.g. the vacuum. The next simple case corresponds to transitions between two-particle states. In that case, Eq. (3.31) becomes

C C C C C S ′ ′ = S ′ ′ + S ′ S ′ + εχS ′ S ′ q1q2,q1q2 q1q2,q1q2 q1,q1 q2,q2 q1,q2 q2,q1 C ′ ′ Sq q ,q1q2 = Sq′ q′ ,q q + δ q1′ q1 δ q2′ q2 + εχδ q1′ q2 δ q2′ q1 (3.36) 1 2 1 2 1 2 − − − − where ε is ( 1) if both particles in the interchange q q are fermions, and +1 otherwise. Observe that we χ − 1 ↔ 2 have used Eq. (3.35). We recognize that the two delta function terms in (3.36) just add up to the norm (3.5). Therefore, in this case SC is just (S 1) . However, the general case is more complicated. βα − βα For transitions between three-particle states, equation (3.33) shows that we can divide α and β in the following three types of clusters (1) α and β as single clusters, (2) two clusters, (α1, β1) with one-particle and (α2, β2) with two-particles, (3) Three-clusters, each one with a single particle. Thus, Eq. (3.31) gives

C Sq′ q′ q′ ,q q q = S ′ ′ ′ 1 2 3 1 2 3 q1q2q3,q1q2q3 C C +Sq′ ,q Sq′ q′ ,q q permutations 1 1 2 3 2 3 ± C C C +Sq′ ,q Sq′ ,q Sq′ ,q permutations 1 1 2 2 3 3 ± using Eq. (3.35) we have

C S ′ ′ ′ = S ′ ′ ′ q1q2q3,q1q2q3 q1q2q3,q1q2q3 C +δ q1′ q1 Sq′ q′ ,q q permutations − 2 3 2 3 ± +δ q′ q1 δ q′ q2 δ q′ q3 permutations (3.37) 1 − 2 − 3 − ± where the connected S matrices for two-particle states are given by Eq. (3.36). Taking all permutations into − account the total number of terms in Eq. (3.37) is given by

1+9+6=16

C C as an example, the nine permutations associated with S ′ S ′ ′ in Eq. (3.37) are given by q1q1 q2q3,q2q3

C C C C C C S ′ S ′ ′ , S ′ S ′ ′ , S ′ S ′ ′ q1q1 q2q3,q2q3 q1q2 q2q3,q1q3 q1q3 q2q3,q1q2 C C C C C C S ′ S ′ ′ , S ′ S ′ ′ , S ′ S ′ ′ q2q1 q1q3,q2q3 q2q2 q1q3,q1q3 q2q3 q1q3,q1q2 C C C C C C S ′ S ′ ′ , S ′ S ′ ′ , S ′ S ′ ′ q3q1 q1q2,q2q3 q3q2 q1q2,q1q3 q3q3 q1q2,q1q2 In addition, we observe that we are not taking as diﬀerent permutations within a given cluster or permutations of complete clusters. For instance, we could express that the process 1 1′ is uncorrelated with the process C C → 23 2′3′ by means of Sq′ q Sq′ q′ ,q q . However, it is clearly equivalent to say (for instance) that 1 1′ is → 1 1 2 3 2 3 → C C uncorrelated with the process 32 2′3′, which is expressed by Sq′ q Sq′ q′ ,q q (that implies a permutation within → 1 1 2 3 3 2 3.5. CLUSTER DECOMPOSITION PRINCIPLE AND CONNECTED AMPLITUDES 125

a given cluster). It is also equivalent to say that the process 23 2′3′ is uncorrelated with the process 1 1′, C C → → which is expressed by S ′ ′ S ′ (that implies the permutation of complete clusters). q2q3,q3q2 q1q1 On the other hand, the transitions between four-particle states allow the following partitions (1) The four- particles as a whole (2) Two clusters, each one with two-particles (3) two clusters the ﬁrst with a single particle and the second with three-particles. (4) Three-clusters two of them with a single particle and the third one with two-particles. (5) Four clusters each one with a single-particle. Therefore, the S matrix reads − C Sq′ q′ q′ q′ ,q q q q = S ′ ′ ′ ′ 1 2 3 4 1 2 3 4 q1q2q3q4,q1q2q3q4 C C +Sq′ q′ ,q q Sq′ q′ ,q q permutations 1 2 1 2 3 4 3 4 ± C +δ q1′ q1 Sq′ q′ q′ ,q q q permutations − 2 3 4 2 3 4 ± C +δ q1′ q1 δ q2′ q2 Sq′ q′ ,q q permutations − − 3 4 3 4 ± +δ q′ q1 δ q′ q2 δ q′ q3 δ q′ q4 permutations (3.38) 1 − 2 − 3 − 4 − ± and including all permutations the total number of terms in Eq. (3.38) yields

1 + 18 + 16 + 72 + 24 = 131

If we had not assumed that one-particle states are stable, there would be more terms in Eqs. (3.37, 3.38). C C The process described above show that the definition of Sβα is recursive since we used Eq. (3.35) to define Sβα C for two-particle states, then use this definition in Eq. (3.37) when we define Sβα for three-particle states, then use C the previous definitions in Eq. (3.38) to obtain the definition of Sβα for four-particle states, and so on. The gist of the definition of the connected part of the S matrix is that the cluster decomposition principle is C − equivalent to demand that Sβα must vanish when any one or more of the particles in the states β and/or α are far away in space from the others. We shall show the statement above by supposing that the states β and α are grouped into clusters β1, β2,... and α1, α2,.... We also assume that all particles in the set αi + βi are far from all particles in the set αj + βj for C each j = i. On the other hand, suppose that Sβ′α′ vanishes if any particles in β′ or α′ are far from the others. 6 C This in turn implies that Sβ′α′ vanishes if any particles in these states are in different clusters, so we have

C (k) C C S εχS S βkαk → βk1αk1 βk2αk2 · · · X so that the deﬁnition (3.31) gives

S (1)ε SC SC (2)ε SC SC βα → χ β11α11 β12α12 ···× χ β21α21 β22α22 ···×··· X X (j) where Σ is a sum over all diﬀerent ways of partitioning the clusters βj and αj into subclusters βj1, βj2,... and αj1, αj2,.... But referred to Eq. (3.31) this is the desired factorization property (3.30), that is the manifestation of the cluster decomposition principle in the S matrix theory. − As a matter of example, let us take a four-particle reaction 1234 1′2′3′4′. Let the set of particles 1, 2, 1′, 2′ C → be very far from the set 3, 4, 3′, 4′. In that case, if Sβα vanishes when any particles in β and/or α are far from the others, the only terms in Eq. (3.38) that survive, are given by (we shorten the notation q i) i → C C S ′ ′ ′ ′ S ′ ′ S ′ ′ 1 2 3 4 ,1234 → 1 2 ,12 3 4 ,34 C + (δ ′ δ ′ δ ′ δ ′ ) S ′ ′ 1 1 2 2 ± 1 2 2 1 3 4 ,34 C + (δ ′ δ ′ δ ′ δ ′ ) S ′ ′ 3 3 4 4 ± 3 4 4 3 1 2 ,12 + (δ ′ δ ′ δ ′ δ ′ ) (δ ′ δ ′ δ ′ δ ′ ) (3.39) 1 1 2 2 ± 1 2 2 1 3 3 4 4 ± 3 4 4 3 126 CHAPTER 3. THE CLUSTER DECOMPOSITION PRINCIPLE as a matter of example the terms

C C Sq′ q′ q′ q′ ,q q q q ; δ q1′ q1 Sq′ q′ q′ ,q q q permutations (3.40) 1 2 3 4 1 2 3 4 − 2 3 4 2 3 4 ± must vanish when the set of particles 1, 2, 1′, 2′ is very far from the set 3, 4, 3′, 4′, because in the ﬁrst term all particles are in the same cluster (hence they are not very far from each other) while in the second three of the four particles are in the same cluster. Comparing (3.39) with (3.36) we observe that it is just the required factorization condition (3.30) S ′ ′ ′ ′ S ′ ′ S ′ ′ 1 2 3 4 ,1234 → 1 2 ,12 3 4 ,34 It is clear that the cluster decomposition principle is formulated in spatial coordinates. The meaning of “far” in such a principle refers to long distances in the coordinate space. We have in turn reestablished this principle as C the condition that Sβα vanishes if any particles in the states β or α are far from any others. In that sense, it would be convenient to rexpress this in momentum space. The coordinate space matrix elements are deﬁned as a Fourier transform

′ ′ ′ ′ C 3 3 3 3 C ip1 x1 ip2 x2 ip1 x1 ip2 x2 Sx′ x′ ...,x x ... d p1′ d p2′ d p1 d p2 Sp′ p′ ...,p p ...e · e · e− · e− · (3.41) 1 2 1 2 ≡ · · · · · · 1 2 1 2 · · · · · · Z where we have temporarily dropped spin and species labels. By now we know the implications of the cluster C C decomposition principle in S ′ ′ , but S ′ ′ is easier to construct and to connect with experiments x1x2...,x1x2... p1p2...,p1p2... (experiments usually measure momenta but not position). Thus we want to know what conditions demands the C cluster decomposition principle for S ′ ′ p1p2...,p1p2... C Let us start by assuming that S ′ ′ is well-behaved (Lebesgue-integrable). In that case, the Riemann- p1p2...,p1p2... Lebesgue theorem says that the integral (3.41) would vanish when any combination of spatial coordinates goes to infinity. This is a very strong requirement, and as we shall see, much stronger than necessary to satisfy the cluster decomposition principle. Translational invariance says that the connected S matrix like the S matrix − − itself, can only depend on differences of coordinate vectors. Hence such a matrix should not change when all C xi and xj′ vary together, as long as their differences remain constant. This requires that the elements of S in a momentum basis must (like those of S) be proportional to a three-dimensional delta function that ensures the momentum conservation5, as well as the energy-conservation delta function required by scattering theory. Thus, we can rewrite

C 3 ′ ′ Sp′ p′ ...,p p ... = δ p1′ + p2′ + p1 p2 δ E1′ + E2′ + E1 E2 Cp p ,p1p2 (3.42) 1 2 1 2 ···− − −··· ···− − −··· 1 2··· ··· On the other hand, the presence of the delta function (which has singular behavior) spoils the Lebesgue-integrability C of S ′ ′ . However, this is not a problem, since the cluster decomposition principle only requires that p1p2...,p1p2... (3.41) vanishes when the differences among some of the x and/or x coordinates become large. However, if C i i′ itself in Eq. (3.42) contained additional delta functions of linear combinations of three-momenta, this principle would not be satisfied. For example, suppose that C contains a delta function that says that the sum of the p and p for some subset of the particles vanished. In that case Eq. (3.41) would not change if all of the i′ − j xi′ and xj for the particles in that subset moved together (with constant differences) away from all the other xk′ and xl, in contradiction with the cluster decomposition principle. Roughly speaking, the cluster decomposition principle says that the connected part of the S matrix unlike the S matrix itself, contains only one momentum − − conservation delta function. In summary, we could say that the coefficient C ′ ′ in Eq. (3.42) should be smooth when it is p1p2 ,p1p2 considered as a function of its momentum labels. The simples··· t··· smoothness requirement would be to demand that such a coefficient be analytic at p1′ = p2′ = = p1 = p2 = 0. Such condition would guarantee that C · · · · · · S ′ ′ vanishes exponentially fast when any of the x and x is very distant from any of the other x and x . x x ,x1x2 ′ ′ 1 2··· ··· 5Recall that translational invariance leads to conservation of the momentum of the whole system. 3.6. STRUCTURE OF THE INTERACTION 127

Nevertheless an exponential decrease of SC is not essential in the cluster decomposition principle. In fact, not all theories satisfy this requirement of analiticity. For instance, in theories with massless particles, SC can have poles C at certain values of the momenta. In some cases, after fourier transforming, such poles give terms in S ′ ′ x1x2 ,x1x2 that decay only as negative powers of coordinate diﬀerences. Indeed it is not necessary to rule out this behavior··· ··· to satisfy the cluster decomposition principle. The smoothness condition on SC could allow various poles and branch-cuts at certain values of p and p′. However, strong singularities such as Dirac delta functions should be discarded.

3.6 Structure of the interaction

The next task is to construct a Hamiltonian that yields an S matrix that satisfy the cluster decomposition − principle. It is in this point that the creation and annihilation operators become important. The answer comes from the following theorem: The S matrix satisﬁes the cluster decomposition principle if − the Hamiltonian can be expressed as in Eq. (3.20)

∞ ∞ H = dq′ dq′ dq dq a† q′ a† q′ a (q ) a (q ) h q′ q′ ,q q (3.43) 1 · · · N 1 · · · M × 1 · · · N M · · · 1 × NM 1 · · · N 1 · · · M N=0 M=0 Z X X in which the coeﬃcients hNM contain just a single three-dimensional momentum-conservation delta function, that is

3 h p′ σ′ n′ , p′ σ′ n′ ; p σ n , p σ n = δ p′ + . . . + p′ p . . . p NM 1 1 1 · · · N N N 1 1 1 · · · M M M 1 N − 1 − − M hNM p1′ σ1′ n1′ , pN′ σN′ nN′ ; p1σ1n 1, pM σM n(3.44)M × · · · · · · where hNM does not contain delta function factors. It worthse emphasizing that the expression (3.43) is not enough to guarantee that the S matrix satisfies the cluster decomposition principle (CDP). Indeed, we have − seen ine section 3.3 [see Eq. (3.20)], that any operator can be written according with the structure followed in Eq. (3.43). It is essential the additional condition (3.44) that shows that the coefficients in the expansion (3.43) contains only a global delta function. To prove the theorem, we shall use the perturbation theory in its time-dependent form. An important advantage of the time-dependent perturbation theory (with respect to the old-fashioned perturbation theory) is that the combinatorics underlying the CDP is more apparent. For instance, if E = E1 + . . . + En is the sum of one-particle iEt 1 6 energies, then e is a product of functions of the individual energies while [E E + iε]− is not . − − α In the framework of time-dependent perturbation theory, the S operator is given by Eq. (2.188) − n ∞ ( i) ∞ S = − dt dt dt T V (t ) V (t ) V (t ) n! 1 2 · · · n { 1 2 · · · n } nX=0 Z−∞ where we adopt the convention that for n = 0, the time-ordered product is defined as the identity operator. From the definition (2.36) of the S operator, the S matrix yields − − n ∞ ( i) ∞ S : β(0) S α(0) = − dt dt dt β(0) T V (t ) V (t ) V (t ) α(0) (3.45) βα ≡ n! 1 2 · · · n { 1 2 · · · n } D E n=0 Z−∞ D E X where V (t) is defined as in Eq. (2.180)

V (t) exp [iH t] V exp [ iH t] ≡ 0 − 0 6Recall that old-fashioned perturbation theory is based on iterations coming from expression (2.174), in which the energy appears −1 in the factor [E − Eα + iε] . On the other hand, we can pass from the old-fashioned to the time-dependent perturbation theory by means of Eq. (2.185) in which the RHS depends on exponentials of the energy. 128 CHAPTER 3. THE CLUSTER DECOMPOSITION PRINCIPLE

where H0 is the free-particle Hamiltonian and V is the interaction. It is convenient to express the free-particle states by showing explicitly their particle content and the state of each particle

α(0) qαqα qα ; β(0) qβqβ qβ (3.46) ≡ | 1 2 · · · P i ≡ 1 2 · · · Q E E E (0) (0) On the other hand, according with Eq. (3.8), the free-partic le states α and β can be expressed as a product of creation operators acting on the vacuum 0 . | i

(0) α α α α α α α q q q = a† (q ) a† (q ) . . . a† (q ) 0 (3.47) ≡ | 1 2 · · · P i 1 2 P | i (0)E β β β β β β β q q q = a† q a† q . . . a† q 0 ≡ 1 2 · · · Q 1 2 Q | i E E further V (t) as any other operator, can be written as a sum of products of creation and annihilation operators

∞ ∞ k k k k k k V (tk) = dq1′ dqN¯′ dq1 dqM¯ a† q1′ a† qN¯′ ¯ ¯ · · · · · · × · · · NX=0 MX=0 Z k k k k k k a q a q C ¯ ¯ q ′ q ′; q q ; t (3.48) × M¯ · · · 1 × NM 1 · · · N¯ 1 · · · M¯ we should emphasize that the coefficients CN¯M¯ must be explicit functions of time. Thus the inner product in the integrand of Eq. (3.45) can be written by using Eqs. (3.47, 3.48). The complete expression is very complicated but the crucial point is that each term in the sum (3.45) could be written as a sum of vacuum expectation values of products of creation and annihilation operators. Now, by using the commutation and anticommutation relations (3.17) we can move each annihilation operator to the right passing the creation operators. Each time that we move an annihilation operator to the right passing a creation operator we obtain two terms, as shown in Eq. (3.17) in the form a q′ a† (q)= ε a† (q) a q′ + δ q′ q (3.49) χ − moving other creation operator past the annihilation operator in the first term generates yet more terms. However, Eq. (3.13) shows that any annihilation operator that moves all the way to the right and acts on the vacuum 0 | i gives zero, so at the end of the process we obtain only delta functions. In conclusion, the vacuum expectation value (VEV) of a product of creation and annihilation operators can be written as a sum of different terms, each term equal to the product of delta functions and signs from the commutators or anticommutators. In turn, it ± means that each term in Eq. (3.45) can be expressed by a sum of terms, each term equal to a product of delta functions and signs from the commutators or anticommutators and whatever factors are contributed by V (t), ± integrated over all the times and integrated and sum over the momenta, spins and species in the arguments of the delta functions. Each of the terms generated in this way can be symbolized by a diagram. The diagram will be constructed according with the following algorithm

1. We start by drawing n points called vertices, one for each V (t) operator.

2. For each delta function produced when an annihilation operator in one of these V (t) operators moves past a creation operator in the initial state α(0) , we draw a line coming into a diagram from below that ends at the corresponding vertex.

3. For each delta function produced when an annihilation operator in the adjoint of the ﬁnal state β(0) moves past a creation operator in one of the V (t), draw a line from the corresponding vertex upwards out of the

diagram.

4. For each delta function produced when an annihilation operator in one V (t) moves past a creation in another V (t), draw a line between the corresponding vertices. 3.6. STRUCTURE OF THE INTERACTION 129

5. For each delta function produced when an annihilation operator in the adjoint of the ﬁnal state β(0) moves past a creation operator in the initial state α(0) , draw a line from bottom to top, right through the

diagrams.

Each of the delta functions associated with one of these lines enforces the equality of the momentum arguments of the pair of creation and annihilation operators represented by the line. In addition, there is at least one delta function contributed by each of the vertices, which enforces the conservation of the total three-momentum at the vertex.

3.6.1 A simple example As a matter of example, let us assume that both the initial and ﬁnal states consists of two-particle states. Hence, equation (3.47) gives

(0) α α α α (0) β β β β α q q = a† (q ) a† (q ) 0 ; β q q = a† q a† q 0 (3.50) ≡ | 1 2 i 1 2 | i ≡ 1 2 1 2 | i E E E so that by using equations (3.50) the inner product in (3.45) can be written as

(n) (0) (0) β β α α D β T V (t ) V (t ) V (t ) α = 0 a q a q T V (t ) V (t ) V (t ) a† (q ) a† (q ) 0 βα ≡ { 1 2 · · · n } h | 2 1 { 1 2 · · · n } 1 2 | i D E (2) let us examine the term Dβα . For simplicity we assume that t1 >t2

(2) (0) (0) β β α α D β T V (t ) V (t ) α = 0 a q a q V (t ) V (t ) a† (q ) a† (q ) 0 βα ≡ { 1 2 } h | 2 1 1 2 1 2 | i D E in turn each V (tk) can be expanded as in Eq. (3.48), then we have

∞ ∞ 1 1 1 1 1 1 V (t1) V (t2) = dq ′ dq ¯′ dq dq ¯ a† q ′ a† q ¯′ 1 · · · N1 1 · · · M1 × 1 · · · N1 N¯1=0 M¯ 1=0 Z X X 1 1 1 1 1 1 a q ¯ a q C ¯ ¯ q ′ q ¯′ ; q q ¯ ; t M1 · · · 1 × N1M1 1 · · · N1 1 · · · M1 ∞ ∞ 2 2 2 2 2 2 dq ′ dq ¯′ dq dq ¯ a† q ′ a† q ¯′ × 1 · · · N2 1 · · · M2 × 1 · · · N2 N¯2=0 M¯ 2=0 Z X X 2 2 2 2 2 2 a q ¯ a q C ¯ ¯ q ′ q ¯′ ; q q ¯ ; t (3.51) M2 · · · 1 × N2M2 1 · · · N2 1 · · · M2 (2) therefore the inner product Dβα becomes

(2) ∞ ∞ ∞ ∞ 1 1 1 1 2 2 2 2 D = dq ′ dq ¯′ dq dq ¯ dq ′ dq ¯′ dq dq ¯ βα 1 · · · N1 1 · · · M1 1 · · · N2 1 · · · M2 ¯ ¯ ¯ ¯ NX1=0 MX1=0 NX2=0 MX2=0 Z β β 1 1 1 1 0 a q a q a† q ′ a† q ¯′ a q ¯ a q h | 2 1 1 · · · N1 M1 · · · 1 × 2 h2 2 2 α α i a† q ′ a† q ¯′ a q ¯ a q a† (q ) a† (q ) 0 1 · · · N2 M2 · · · 1 1 2 | i h i for instance the term with N¯1 = M¯ 1 = N¯2 = M¯ 2 = 2 reads

1 1 1 1 2 2 2 2 dq1′dq2′ dq1dq2 dq1′dq2′ dq1dq2 Z β β 1 1 1 1 2 2 2 2 α α 0 a q a q a† q ′ a† q ′ a q a q a† q ′ a† q ′ a q a q a† (q ) a† (q ) 0 × h | 2 1 1 2 2 1 1 2 2 1 1 2 | i h i h i 130 CHAPTER 3. THE CLUSTER DECOMPOSITION PRINCIPLE

or for N¯1 = M¯ 1 = N¯2 = M¯ 2 = 1 we have

1 1 2 2 β β 1 1 2 2 α α dq ′ dq dq ′ dq 0 a q a q a† q ′ a q a† q ′ a q a† (q ) a† (q ) 0 (3.52) 1 1 1 1 h | 2 1 1 1 1 1 1 2 | i Z h i h i let us concentrate on the vacuum expectation value (VEV) in the term (3.52)

β β 1 1 2 2 α α 0 A 0 0 a q a q a† q ′ a q a† q ′ a q a† (q ) a† (q ) 0 (3.53) h | | i ≡ h | 2 1 1 1 1 1 1 2 | i 2 we shall pass the term a q1 to the right by succesive use of the commutation or anticommutation relations β β 1 1 2 2 α α A a q a q a† q ′ a q a† q ′ a q a† (q ) a† (q ) ≡ 2 1 1 1 1 1 1 2 β β 1 1 2 h α i 2 α 2 α = a q a q a† q1′ a q1 a† q1′ εqα,q2 a† (q1 ) a q1 + δ q1 q1 a† (q2 ) 2 1 1 1 − β β 1 1 h 2 α 2 α i = ε α 2 a q a q a† q′ a q a† q ′ a† (q ) a q a† (q ) q1 ,q1 2 1 1 1 1 1 1 2 α 2 β β 1 1 2h α i +δ q q a q a q a† q′ a q a† q ′ a†(q ) 1 − 1 2 1 1 1 1 2 β β 1 1 h 2 α i α 2 α 2 = εqα,q2 a q a q a† q1′ a q1 a† q1′ a† (q1 ) εqα,q2 a† (q2 ) a q1 + δ q2 q1 1 1 2 1 2 1 − α 2 β β 1 2 h 1 2 1 α i +δ q1 q1 a q a q a† q1′ εq2′,q1 a† q1′ a q1 + δ q1′ q1 a† (q2 ) − 2 1 1 1 − h i β β 1 1 2 α α 2 A = ε α 2 ε α 2 a q a q a† q ′ a q a† q ′ a† (q ) a† (q ) a q q1 ,q1 q2 ,q1 2 1 1 1 1 1 2 1 α 2 β β 1 1 2 α +εqα,q2 δ q2 q1 a q a q a† q1′ a q1 a† q1′ a† (q1 ) 1 1 − 2 1 α 2 β β 1 2 1 α +δ q1 q1 εq2′,q1 a q a q a† q1′ a† q1′ a q1 a† (q2 ) − 1 1 2 1 α 2 2 1 β β 1 α +δ q q δ q ′ q a q a q a† q ′ a† (q ) (3.54) 1 − 1 1 − 1 2 1 1 2 however, from Eq. (3.53) we recall that we are interested in the VEV of the operator A. Thus, the ﬁrst line of Eq. (3.54) does not contribute because there is an annihilation operator on the right-side. Then we can write

α 2 β β 1 1 2 α A εqα,q2 δ q2 q1 a q a q a† q1′ a q1 a† q1′ a† (q1 ) → 1 1 − 2 1 α 2 β β 1h 2 i1 α +δ q1 q1 εq2′,q1 a q a q a† q1′ a† q1′ a q1 a† (q2 ) − 1 1 2 1 α 2 2 1 β β 1 h α i +δ q q δ q ′ q a q a q a† q ′ a† (q ) 1 − 1 1 − 1 2 1 1 2 h i α 2 β β 1 2 1 2 1 α A εqα,q2 δ q2 q1 a q a q a† q1′ εq2′,q1 a† q1′ a q1 + δ q1′ q1 a† (q1 ) → 1 1 − 2 1 1 1 − α 2 β β 1h 2 α 1 i α 1 +δ q1 q1 εq2′,q1 a q a q a† q1′ a† q1′ εqα,q1 a† (q2 ) a q1 + δ q2 q1 − 1 1 2 1 2 1 − α 2 2 1 β 1 hβ β 1 α i +δ q1 q1 δ q1′ q1 a q2 εq1′,qβ a† q1′ a q1 + δ q1 q1′ a† (q2 ) − − 1 1 − once again we drop the terms with an annihilation h operator on the right. Thus i

α 2 2 1 β β 1 α A εqα,q2 δ q2 q1 δ q1′ q1 a q a q a† q1′ a† (q1 ) → 1 1 − − 2 1 α 2 α 1 β β 1 2 +δ q1 q1 δ q2 q1 εq2′,q1 a q a q a† q1′ a† q1′ − − 1 1 2 1 α 2 2 1 β 1 β α +εq1′,qβ δ q1 q1 δ q1′ q1 a q2 a† q1′ a q1 a† (q2 ) 1 1 − − α 2 2 1 β 1 β α +δ q q δ q ′ q δ q q ′ a q a† (q ) 1 − 1 1 − 1 1 − 1 2 2 3.6. STRUCTURE OF THE INTERACTION 131 continuing the process, the “eﬀective operator” that contributes to the VEV is a sum of products of delta functions. With our notation, it is easy to track the origin of a given delta function. To show it, we remember what momenta corresponds to the initial state, ﬁnal state and interactions V (t1) and V (t2). They can be tracked from Eqs. (3.50, 3.51)

α(0) qαqα ; β(0) qβqβ ≡ | 1 2 i ≡ 1 2 E 1 1 1 1 E E V (t1) q1′,q2′ ; q1,q 2 → 2 2 2 2 V (t ) q ′,q ′ ; q ,q 2 → 1 2 1 2 Therefore, δ qα q2 is a delta function produced when an annihilation operator in V (t ) moves past a creation 2 − 1 2 operator in the initial state α. The delta factor δ qβ q1 is produced when an annihilation operator in the 1 − 1′ adjoint of the ﬁnal state β moves past a creation operator in V (t1). Assuming that the time-ordered operator gives V (t ) V (t ), the factor δ q2 q1 is produced when an annihilation operator in V (t ) moves past a creation 1 2 1′ − 1 1 operator in V (t ). 2

3.6.2 Connected and disconnected parts of the interaction A given diagram generated with the previous algorithm may be connected (i.e. every point connected to every other by a set of lines), and if not connected, it breaks up into a number of connected pieces. The V (t) operator associated with a vertex in one connected component eﬀectively commutes with the V (t) associated with any vertex in any other connected component, because for this diagram we are not including any terms in which an annihilation operator in one vertex destroys a particle that is produced by a creation operator in the other vertex, if we did, then the two vertices would be in the same connected component. Therefore, the matrix element in Eq. (3.45) can be expressed as a sum over products of contributions, one from each connected component

ν (0) (0) (0) (0) β T V (t1) V (t2) V (tn) α = εχ βj T V (tj1) V tj,nj αj (3.55) { · · · } · · · C D E clusterings j=1 D E X Y where the sum is over all ways of splitting up the incoming and outgoing particles and V (t) operators into ν clusters (including a sum over ν from 1 to n) with n operators V (t ) V t and the subsets of initial j j1 · · · j,nj particles α and ﬁnal particles β all in the j th cluster. It is clear that the product over the ν clusters on the j j − RHS of Eq. (3.55) must include all the n vertices on the LHS of that equation. It means that

n = n1 + . . . + nν and that α is the union of all particles in the subsets α1, α2,...,αν , and similarly for the final states. It may happen that some clusters in (3.55) have no vertices so that nj = 0. For these factors, we must take the matrix element factor in Eq. (3.55) to vanish unless βj and αj are both one-particle states, because the only connected diagrams without vertices are the ones of a single line running through the diagram from bottom to top. In that case, the matrix element is just the delta function δ (α β ). j − j The subscript C in Eq. (3.55) means that we exclude contributions associated with disconnected diagrams, i.e. contributions in which any V (t) operator or any initial or final particle is not connected to every other by a sequence of particle creations and annihilations. Now we substitute Eq. (3.55) into Eq. (3.45). We observe that each time variable is integrated from −∞ to , so it makes no difference which of the t s are sorted out into each cluster. Consequently, the sum over ∞ i′ clusterings yields a factor n! n !n ! n ! 1 2 · · · ν 132 CHAPTER 3. THE CLUSTER DECOMPOSITION PRINCIPLE

which is the number of ways of sorting out n vertices into ν clusters, each one containing n1,n2,...vertices:

I ∞ dt dt dt β(0) T V (t ) V (t ) V (t ) α(0) βα ≡ 1 2 · · · n { 1 2 · · · n } Z−∞ D E ν n! ∞ (0) (0) Iβα = εχ dtj1 dtjnj βj T V (tj1) V tj,nj αj (3.56) n1!n2! nν! · · · · · · C part n1 nν j=1 ··· · · · Z−∞ D E X X Y with n = n1 + . . . + nν (3.57) where the first sum is over all ways of partitioning the particles in the initial and final states into clusters α α 1 · · · ν and β β including a sum over the number ν of clusters. 1 · · · ν As a matter of example, let us assume that we have an initial state and a final state each one with four particles, and let us take the number of vertices as n = 4:

α(0) = qαqαqαqα ; β(0) = qβqβqβqβ V (t ) k = 1, 2, 3, 4 | 1 2 3 4 i 1 2 3 4 k → E E E (0) (0) and let us examine the case in which we have two partitions αj and βj with j = 1, 2 consisting of E E (0) α α (0) β β ﬁrst partition α = q q , β = q q ; 12 1′2′ 1 | 1 2 i 1 1 2 → (0)E α α (0)E β βE second partition α = q q , β = q q ; 34 3′4′ 2 | 3 4 i 2 3 4 → E E E in this case we have two clusters ν = 2 in which n1 = n2 = 2. Hence we have n! 4! = = 6 n1!n2! 2!2! the six number of ways we can sort out these 4 vertices into these two clusters is given by

V (t1) V (t2) V (t3) V (t4) V (t1) V (t3) V (t2) V (t4) V (t1) V (t4) V (t2) V (t3) V (t2) V (t3) V (t1) V (t4) V (t2) V (t4) V (t1) V (t3) V (t3) V (t4) V (t1) V (t2)

On the other hand, when Iβα of Eq. (3.56) is substituted in Eq. (3.45) the n! term is cancel by the 1/n! term in such an equation. Besides, owing to the constraint (3.57) the factor ( i)n in Eq. (3.45) can be expressed as − ( i)n = ( i)n1 ( i)n2 ( i)nν − − − · · · − from these facts instead of summing over n and then summing over n1,n2,...,nν constrained by (3.57), we can just sum independently over each n1,n2,...,nν. It yields

ν nj ∞ ( i) ∞ (0) (0) Sβα = εχ − dtj1 dtjnj βj T V (tj1) V tj,nj αj (3.58) nj! · · · · · · C partitions j=1 nj =0 Z−∞ D E X Y X C and comparing Eq. (3.58) with the deﬁnition (3.31) of the connected matrix elements Sβα, we see that such matrix elements are given by the factors in the product of Eq. (3.58)

n C ∞ ( i) ∞ (0) (0) Sβα = − dt1 dtn β T V (t1) V (tn) α n! · · · { · · · } C n=0 Z−∞ D E X

3.6. STRUCTURE OF THE INTERACTION 133

C C in conclusion the matrix elements Sβα are calculated with a simple prescription: Sβα is the sum of all contributions to the S matrix that are connected, it means that we drop all terms in which any initial or final state or any − operator V (t) is not connected to all the others by a sequence of particle creations and annihilations. It explains the adjective “connected” for SC . We have seen that momentum is conserved at each vertex and along every line. Hence, the connected parts C 3 of the S matrix individually conserve momentum. Therefore, Sβα contains a factor δ (pβ pα). We shall prove C − − that Sβα contains no other delta functions, such that the CDP holds. Now we start with the hypothesis that the coefficients hNM in the expansion (3.43) of the Hamiltonian in terms of creation and annihilation operators are proportional to a single three-dimensional delta function that provides momenta conservation. This is automatically true for the free-particle Hamiltonian H0 and using our hypothesis, it will be true separately for the interaction V . In our graphical interpretation of the matrix elements, it means that each vertex contributes one three-dimensional delta function. The other delta functions in matrix elements Vγδ just keep unchanged the momentum of any particle that is not created or annihilated at the corresponding vertex. Most of these delta functions just fix the momentum of intermediate particles. The only momenta that are left unfixed by the delta functions are the ones that circulate in loops of internal lines. Observe that any line that if cut leaves the diagram disconnected carries a momentum that is fixed by momentum conservation as some linear combination of the momenta of the lines coming into or going out of the diagram. If the diagram contains L lines that can all be cut at the same time without the diagram becoming disconnected, we say that it has L independent loops, and there are L momenta that are not fixed by the neither the delta functions nor the momentum conservation. With V vertices, I internal lines, and L loops there are V delta functions: I L delta functions fix internal − momenta, leaving V I + L delta functions relating the momenta of incoming or outgoing particles. From a − well-known topological identity, we know that for every graph consisting of C connected pieces, the number of vertices V , internal lines I, and loops L, we have the relation

V I + L = C (3.59) − C in the particular case of a connected matrix element like Sβα, that arises from graphs with C = 1, we have a single three-dimensional delta function δ3 (p p ), as we wanted to prove. β − α We can see the identity (3.59) by a heuristical argument as follows: A graph with a single vertex has VC = 1, L = 0, C = 1. If we add V 1 vertices with just enough internal lines to keep the graph connected, we have C − I = V 1, L = 0 and C = 1. Any additional internal lines attached (without new vertices) to the same C C − C connected graph produce an equal number of loops. Hence, I = V + L 1 and C = 1. If a disconnected graph C C C − consists of C connected parts, the sums of IC ,VC ,LC in each connected part will then satisfy the equation

I = V + L C C C C − XC XC XC where the sum is over each connected part. It leads to Eq. (3.59). In the above argument it was not important that the time variables were integrated from to + . Conse- −∞ ∞ quently, we can use the same argument to show that if the coefficients hNM in the Hamiltonian contain just single delta functions, U (t,t0) can also be decomposed into connected parts, each one containing a single momentum- conservation delta function factor. In addition, the connected part of the S matrix also contains an energy- C − conservation delta function, we shall see later that Sβα contains only a single energy-conservation delta fucntion factor δ (E E ), while U (t,t ) does not contain any energy conservation delta function at all. β − α 0 3.6.3 Some examples of the diagrammatic properties In Fig. 3.1a,b we show how the lines entering into a vertex are showing a “flux” of momentum that must be conserved. We could choose the convention of assuming that a momentum flux entering is positive and an 134 CHAPTER 3. THE CLUSTER DECOMPOSITION PRINCIPLE

Figure 3.1: (a) Vertex with three incoming momenta. (b) Vertex with one incoming momentum and two outgoing momenta. (c) A disconnected diagram consisting of two connected pieces.

Figure 3.2: (a) A diagram with one loop. (b) The previous diagram could be drawn with the same momentum in the incoming and in the outgoing external lines. The momenta k within the loop is totally indetermined. (c) A diagram with two loops. outgoing ﬂux of momentum is negative. Thus in Fig. 3.1a the conservation of momentum in the vertex gives

p1 + p2 + p3 = 0 while in the diagram of Fig. 3.1b, the conservation of momentum is expressed by

p p p = 0 1 − 2 − 3 in these ﬁgures the line with momentum p1 comes from a delta function produced when an annihilation operator in one of the V (tk) operators moves past a creation operator in the initial state α. The line with p2 (and also the line with p3), comes from a delta function produced when an annihilation operator in the adjoint of the ﬁnal state β moves past a creation operator in one of the V (tk). In the diagram of Fig. 3.1c, the line carrying momentum q comes from a delta function produced when an annihilation operator in one V (tk) moves past a creation operator in another V (tm). Moreover, the line carrying momentum p5 comes from a delta function produced when an annihilation operator in the adjoint state β moves past a creation operator in the initial state α. In Fig. 3.1c, the whole diagram is disconnected, and consists of two connected pieces. Figure 3.2a, shows a diagram that contains a loop, which is formed by the two internal lines with momenta p1 + k and k. Observe that we can cut one of these internal lines, and the diagram remains connected. Since we have two vertices, we have two delta functions associated with each vertex, they are given by

δ (p (p + k)+ k) ; δ ((p + k) k p ) 1 − 1 1 − − 2 we can observe that the ﬁrst delta is trivial, and says nothing about the momentum k. The another delta function simply says that p1 = p2, and says nothing about k. Therefore, in this diagram containing one loop, the momentum k of one of the internal lines in the loop remains totally indetermined. It is easy to generalize the fact 3.6. STRUCTURE OF THE INTERACTION 135

Figure 3.3: (a) A diagram with two loops, there is one momenta within each loop (k1 and k2) that are totally independent. (b) The previous diagram could be drawn with the same momentum for each internal line that is not part of a loop. that there are L totally indetermined momenta for a diagram with L loops. In practice, it is usual in this diagram to put p1 in both external lines in order to simplify calculations, it is shown in Fig. 3.2b. Moreover, in Fig. 3.2b, the diagram has V = 2, I = 2, L = 1 then applying Eq. (3.59) we have

V I + L = 2 2+1=1 − − showing that C = 1, i.e. that it is a connected diagram. In Fig. 3.2 we have V = 5, L = 2, I = 6. In particular, we observe that four internal lines are part of the loops while two internal lines do not form a loop. By using Eq. (3.59), we clearly have

V I + L = 5 6+2=1 − − so that C = 1 showing that it is a connected diagram. As a last example, in the diagram of Fig. 3.3a, we have V = 6, I = 7, L = 2. We have six delta functions owing to the six vertices, they are

δ (p + p q ) , δ (q (q + k )+ k ) , δ ((q + k ) k q ) 1 2 − 1 1 − 1 1 1 1 1 − 1 − 2 δ (q + k (q + k )) , δ ( q k + (q + k )) ; δ (q p p ) (3.60) 2 2 − 2 2 − 3 − 2 2 2 3 − 3 − 4 from the V = 6 delta functions, I L = 5 of them ﬁx internal momenta since L = 2 momenta are totally − indetermined (k1 and k2 in this case). Thus, we have 5 delta functions that determine internal momenta. Finally, V (I L) = 1 delta function is related with external momenta. Then, we only have one global delta function − − that contains non-trivial information. Eﬀectively, the six delta functions (3.60) says that

q1 = p1 + p2 , q1 = q2 , q2 = q3 , q3 = p3 + p4 or equivalently p1 + p2 = q1 = q2 = q3 = p3 + p4 136 CHAPTER 3. THE CLUSTER DECOMPOSITION PRINCIPLE thus all six deltas are equivalent to a single delta function δ (p + p p p ) 1 2 − 3 − 4 Owing to this fact, these kind of diagrams are usually written only in terms of the independent momenta as shown in Fig. 3.3b.

3.6.4 Implications of the theorem

The fact that hNM in Eq. (3.43) should have only one three-dimensional momentum conservation delta function factor is far from trivial and has deep implications. For example, let us assume that V has non-vanishing matrix elements between two-particle states. Thus, Eq. (3.43) applied for the interaction

∞ ∞ V = dq′ dq′ dq dq a† q′ a† q′ a (q ) a (q ) v q′ q′ ,q q 1 · · · N 1 · · · M × 1 · · · N M · · · 1 × NM 1 · · · N 1 · · · M N=0 M=0 Z X X must contain a term with N = M = 2, i.e. a term with coefficient v2,2 (p1′ p2′ , p1p2) associated with two creation and two annihilation operators. Therefore, such a coefficient is associated with two initial paritcle states and two final particle states v p′ p′ , p p = V ′ ′ (3.61) 2,2 1 2 1 2 p1p2,p1p2 where we have dropped the spin and species labels. But then the matrix element of the interaction between three-particle states is7 3 Vp′ p′ p′ ,p p p = v3,3 p′ p′ p′ , p1p2p3 + v2,2 p′ p′ , p1p2 δ p′ p3 permutations (3.62) 1 2 3 1 2 3 1 2 3 1 2 3 − ± For instance, we could try to do a relativistic quantum theory that is not a field theory, by choosing v2,2 in such a way that the two-body S matrix is Lorentz invariant, and adjusting the rest of the Hamiltonian so that − there is no scattering in states containing three or more particles. We would then have to take v3,3 to cancel the other terms in Eq. (3.62) 3 v p′ p′ p′ , p p p = v p′ p′ , p p δ p′ p permutations (3.63) 3,3 1 2 3 1 2 3 − 2,2 1 2 1 2 3 − 3 ∓ nevertheless, recalling that v (p p , p p ) has a factor δ3 (p + p p p ), Eq. (3.63) means that each term 2,2 1′ 2′ 1 2 1′ 2′ − 1 − 2 in v3,3 contains two delta function factors, violating the CDP. Therefore, in a theory that satisfies the CDP, the existence of scattering processes involving two particles makes processes involving three or more particles inevitable. To solve problems involving three-bodies in quantum theories that satisfy the CDP, the term v3,3 in Eq. (3.62) has no particular problems, but the extra delta function in the other terms make the Lippmann-Schwinger 1 equation difficult to solve directly. The problem is that these delta functions makes the kernel [E E + iε]− V α − β βα of this equation not square-integrable, even after factorizing out an overall momentum conservation delta function. Hence, it cannot be approximated by a finite matrix, even of very large rank. Therefore, to solve problems with three or more particles, we should replace the Lippmann-Schwinger equation with one that has a connected RHS. Such equations have been developed and solved recursively but in a non-relativistic regime, while they have not been succesful in relativistic theories. The proof of the main theorem of this section is heavily relied on perturbation theory. However, it has been shown that the reformulated Lippmann-Schwinger equations with non-perturbative dynamics are consistent with C C the requirement that U (t,t0) (and so S ) should also contain only one momentum-conservation delta fucntion factor, as required by the CDP, as long as the Hamiltonian satisfies the condition that the coefficient functions hNM each contain a single momentum-conservation delta function. 7In Eq. (3.62) we are omitting the possibility of having three clusters each of one-particle states, since they are in the free-particle ′ ′ hamiltonian H0 and not in the interaction term V . Similarly, in Eq. (3.61) we do not include in the interaction term Vp1p2,p1p2 the case of two clusters each consisting of one-particle states, since such a term would be in the free-Hamiltonina H0 and not in the interaction term V . Chapter 4

Relativistic quantum ﬁeld theory

In this chapter we shall show the necessity of the introduction of ﬁelds to unite succesfully the special relativity with quantum mechanics while maintaining the Cluster Decomposition Principle (CDP). Further we shall discuss some arising properties of the development of a relativistic quantum ﬁeld theory: the connection between spin and statistics, the existence of antiparticles, and several relationships between the particles and antiparticles being the most remarkable the so-called CPT theorem.

4.1 Free ﬁelds

We saw in section 2.6 that the S matrix is Lorentz invariant if the interaction can be written as − V (t)= d3x (x,t) (4.1) H Z where (x,t) is a scalar in the sense that H 1 U (Λ, a) (x) U − (Λ, a)= (Λx + a) (4.2) 0 H 0 H and satisfying the conditions 2 (x) , x′ =0 for x x′ 0 (4.3) H H − ≥ we shall see later that the possibilities are more general than (4.3) but they are not very different from it. It is also interesting to ask whether Λ in Eq. (4.2) must be restricted to proper orthochronus Lorentz transformations, or can also include discrete space inversions. In order to satisfy the CDP we shall express (x) in terms of H creation and annihilation operators. Nevertheless, in doing this we shall confront the following problem: equation (3.26) shows that under Lorentz transformations each creation or annihilation operator is multiplied by a matrix that depends on the momentum carried by the operator. We should then find the way to couple such operators together to form a scalar1. The solution is to build (x) based on fields. For this, we build annihilation fields + H ψk (x) and creation fields ψk− (x) as follows:

ψ+ (x) = d3p u (x; p,σ,n) a (p,σ,n) d3q u (x; q) a (q) (4.4) k k ≡ k σn X Z Z 3 3 ψ− (x) = d p v (x; p,σ,n) a† (p,σ,n) d q v (x; q) a† (q) (4.5) k k ≡ k σn X Z Z in other words, we avoid the problem of momentum dependence of the Lorentz transformation of annihilation and creation operators, by integrating out all those momenta (and also the spin and species degrees of freedom). In

1It is clear that condition (4.2) for H (x) to be a scalar, requires that the Lorentz transformation for H (x) be independent of any momenta.

137 138 CHAPTER 4. RELATIVISTIC QUANTUM FIELD THEORY

addition, the coeﬃcients uk (x; p,σ,n) and vk (x; p,σ,n) are chosen such that under Lorentz transformations each creation and annihilation ﬁeld is multiplied by a position-independent matrix2:

+ 1 1 + U0 (Λ, a) ψk (x) U0− (Λ, a) = Dkk¯ Λ− ψk¯ (Λx + a) (4.6) k¯ X 1 1 U0 (Λ, a) ψk− (x) U0− (Λ, a) = Dkk¯ Λ− ψk¯− (Λx + a) (4.7) ¯ Xk In principle, we could have different transformation matrices D± for the annihilation and creation fields, but we will see that it is always possible to choose the fields such that both matrices are equal. By applying a second (inhomogeneous) Lorentz transformation Λ¯, a¯, the total Lorentz transformation becomes

U0 Λ¯, a¯ U0 (Λ, a)= U0 ΛΛ¯ , Λ¯a +a ¯ (4.8) for the Lorentz transformation written as in the LHS of Eq. (4.8), we ﬁnd from Eq. (4.6) that

+ + 1 ψ (x) U Λ¯, a¯ U (Λ, a) ψ (x) U Λ¯, a¯ U (Λ, a) − k ≡ 0 0 k 0 0 ¯ + 1 1 ¯ = U0 Λ, a¯ U0 (Λ, a)ψk (x) U0− (Λ , a) U0− Λ, a¯ (4.9) e 1 + 1 = U Λ¯, a¯ D ¯ Λ− ψ (Λx + a) U − Λ¯, a¯ 0  kk k¯  0 k¯ X  1 ¯ +  1 ¯ = Dkk¯ Λ− U0 Λ, a¯ ψk¯ (Λx + a) U0− Λ, a¯ (4.10) k¯ X h i 1 ¯ 1 + ¯ = Dkk¯ Λ− Dkm¯ Λ− ψm Λ (Λx + a) +a ¯ k¯ " m # X X 1 1 + = D ¯ Λ− D¯ Λ¯ − ψ Λ¯ (Λx + a) +a ¯ (4.11)  kk km  m m ¯ X Xk +  1 ¯ 1 + ¯ ψk (x) = D Λ− D Λ− km ψm Λ (Λx + a) +a ¯ (4.12) m X e and for the Lorentz transformation written as in the RHS of Eq. (4.8), we ﬁnd from Eq. (4.6) that

+ + 1 1 + ¯ ¯ ¯ ¯ ¯ − ¯ ¯ ψk (x) U0 ΛΛ, Λa +a ¯ ψk (x) U0− ΛΛ, Λa +a ¯ = Dkk¯ ΛΛ ψk¯ ΛΛ x + Λa +a ¯ ≡ ¯ Xk e+ 1 ¯ 1 + ¯ ψk (x) = Dkk¯ Λ− Λ− ψk¯ Λ (Λx + a) +a ¯ (4.13) k¯ X e thus, equating equations (4.12, 4.13), we ﬁnd that

1 1 1 D Λ− D Λ¯ − = D ΛΛ¯ − (4.14) and deﬁning Λ Λ 1 and Λ Λ¯ 1 we obtain 1 ≡ − 2 ≡ −

D (Λ1) D (Λ2)= D (Λ1Λ2) (4.15) so that D matrices provide a representation of the homogeneous Lorentz group. There are many representations − µ µ starting with the trivial representation D (Λ) = 1, the vector representation D (Λ) ν = Λ ν, and a host of tensor

2In other words, we demand that the Lorentz transformation of the creation and annihilation fields, be homogeneous in space-time. 4.2. LORENTZ TRANSFORMATIONS FOR MASSIVE FIELDS 139 and spinor representations. The representations mentioned above are irreducible. Notwithstanding, we do not require now that the D matrix representation be irreducible. In general, in certain basis (the canonical basis) − it is a set of block-diagonal matrices, with an arbitrary array of irreducible representations in the blocks. The index k here includes a label that runs over the types of particles described and the irreducible representations in the different blocks, and also another that runs over the components of the individual irreducible representations. We shall separate these fields later into irreducible fields each one describing a single particle species (and its antiparticle), and transforming irreducibly under the Lorentz group. Once we learn the way to construct fields satisfying the Lorentz transformation rules (4.6) and (4.7), we can construct the interaction density as

+ + ′ ′ (x)= gk k , k1 kM ψ−′ (x) ψ−′ (x) ψ (x) ψ (x) (4.16) 1 N k1 kN k1 kM H ′ ′ ··· ··· · · · · · · MN k k k1 kM X 1X··· N X··· where Eq. (4.16) is the analogous of expansion (3.43) but based on creation and annihilation fields instead of creation and annihilation operators. The integrations over momenta, spins and species of Eq. (3.43) do not appear explicitly in Eq. (4.16) because we have already carried out such an integration in defining the creation and annihilation fields as can be seen in Eqs. (4.4). The interaction density (4.16) will be a scalar in the sense of Eq. (4.2) if the constant coefficients g ′ ′ k1 kN , k1 kM are chosen to be Lorentz covariant, in the sense that for all Λ : ··· ···

1 1 1 1 g¯′ ¯′ ¯ ¯ = D ′ ¯′ Λ− D ′ ¯′ Λ− D ¯ Λ− D ¯ Λ− g ′ ′ (4.17) k1 kN , k1 kM k1k1 kN kN k1k1 kM kM k1 kN , k1 kM ··· ··· ′ ′ · · · · · · ··· ··· k k k1 kM 1X··· N X··· in which we have not included derivatives because we regard the derivatives of components of these fields as simply additional sorts of field components. The task of finding coefficients g ′ ′ that satisfy Eq. (4.17) k1 kN , k1 kM is similar to the task of obtaining the Clebsch-Gordan coefficients to couple together··· ··· several representations of SO (3) to form rotational scalars. Another important task is to set up the interaction density so that it satisfies (4.3). We shall later combine creation and annihilation operators in such a way that this density commutes with itself at space-like and light-like separations.

4.2 Lorentz transformations for massive ﬁelds

By now we shall study the case of massive particles. In order to obtain the coeﬃcient functions uk (x; p,σ,n) and vk (x; p,σ,n), we observe that Eq. (3.26), gives the transformation rules for the creation operators.

0 1 (Λp) (jn) U (Λ, b) a† (p,σ,n) U − (Λ, b) = exp [ i (Λp) b] D (W (Λ,p)) a† (p σ¯ n) 0 0 − · p0 × σσ¯ Λ s σ¯ X where jn is the spin of particles of species n, and pΛ is the three-vector part of Λp. Using the unitarity of the (jn) rotation matrices Dσσ¯ this equation becomes

0 1 (Λp) (jn) 1 U (Λ, b) a† (p,σ, n) U − (Λ, b) = exp [ i (Λp) b] D ∗ W − (Λ,p) a† (p σ¯ n) (4.18) 0 0 − · p0 × σσ¯ Λ s σ¯ X and taking the adjoint of (4.18) we obtain the transformation rule of the annihilation operators

0 1 (Λp) (jn) 1 U (Λ, b) a (p,σ, n) U − (Λ, b) = exp [i (Λp) b] D W − (Λ,p) a (p σ¯ n) (4.19) 0 0 · p0 × σσ¯ Λ s σ¯ X 140 CHAPTER 4. RELATIVISTIC QUANTUM FIELD THEORY

We also saw in Eq. (1.157), page 37 that the volume element d3p/p0 is Lorentz invariant. Consequently, we have

d3p d3 (Λp) d3 (Λp) = ; d3p = p0 (4.20) p0 (Λp)0 (Λp)0 replacing d3p by the expression (4.20) in Eqs. (4.4), we ﬁnd

3 + 0 d (Λp) ψk (x)= p 0 uk (x; p,σ,n) a (p,σ,n) (4.21) σn (Λp) X Z + from Eqs. (4.21, 4.19) the Lorentz transformation of ψk (x) gives

3 + + 1 0 d (Λp) 1 ψ (x) U0 (Λ, b) ψ (x) U − (Λ, b)= p uk (x; p,σ,n) U0 (Λ, b) a (p,σ,n) U − (Λ, b) k ≡ k 0 (Λp)0 0 σn Z X e 3 0 0 d (Λp) (Λp) (jn) 1 = p uk (x; p,σ,n) exp [i (Λp) b] Dσσ¯ W − (Λ,p) a (pΛ σ¯ n) (Λp)0  · s p0 ×  σn Z σ¯ X X   obtaining ﬁnally

+ 1 3 U (Λ, b) ψ (x) U − (Λ, b) = d (Λp) u (x; p,σ,n) exp [i (Λp) b] 0 k 0 k · σσn¯ X Z 0 (jn) 1 p Dσσ¯ W − (Λ,p) 0 a (pΛ, σ,n¯ ) (4.22) × s(Λp) ) a similar exercise for Eq. (4.5) yields

1 3 U (Λ, b) ψ− (x) U − (Λ, b) = d (Λp) v (x; p,σ,n) exp [ i (Λp) b] 0 k 0 k − · σσn¯ X Z 0 (jn) 1 p Dσσ¯ ∗ W − (Λ,p) 0 a† (pΛ, σ,n¯ ) (4.23) × s(Λp) ) By comparing Eq. (4.6) with Eq. (4.22), we have

1 + 3 D ¯ Λ− ψ (Λx + b) = d (Λp) u (x; p,σ,n) exp [i (Λp) b] kk k¯ k · k¯ σσn¯ Z X X 0 (jn) 1 p Dσσ¯ W − (Λ,p) 0 a (pΛ, σ,n¯ ) (4.24) × s(Λp) ) and substituting (4.4) in (4.24) we have

1 3 3 D ¯ Λ− d p u¯ (Λx + b; p ,σ,n) a (p ,σ,n) = d (Λp) u (x; p,σ,n) exp [i (Λp) b] kk k Λ Λ k · k¯ " σn Z # σσn¯ Z X X X 0 (jn) 1 p Dσσ¯ W − (Λ,p) 0 a (pΛ, σ,n¯ ) × s(Λp) ) (4.25) 4.2. LORENTZ TRANSFORMATIONS FOR MASSIVE FIELDS 141 and using Eq. (4.20) on the LHS of Eq. (4.25) we have

3 1 0 d (Λp) D ¯ Λ− p u¯ (Λx + b; pΛ,σ,n) a (pΛ,σ,n) kk (Λp)0 k k¯ " σn Z # X X 0 3 (jn) 1 p = d (Λp) uk (x; p,σ,n) exp [i (Λp) b] Dσσ¯ W − (Λ,p) a (pΛ, σ,n¯ ) (4.26) · × (Λp)0 σσn¯ (Z s ) X reorganizing the series and integrals in both sides of Eq. (4.26) we have

0 0 p 3 p 1 d (Λp) a (pΛ, σ,n¯ ) D ¯ Λ− u¯ (Λx + b; pΛ, σ,n¯ ) (Λp)0  (Λp)0 kk k  σn¯ Z s s k¯ X  X  0   p 3 (jn) 1 = d (Λp) a (pΛ, σ,n¯ ) uk (x; p,σ,n) exp [i (Λp) b] Dσσ¯ W − (Λ,p) (4.27) (Λp)0 · σn¯ Z s ( σ ) X X since this relation must hold for arbitrary x, Λ and b, the terms in the brackets on both sides of Eq. (4.27) must be equal, hence

0 p 1 (jn) 1 D ¯ Λ− u¯ (Λx + b; pΛ, σ,n¯ )= uk (x; p,σ,n) exp [i (Λp) b] Dσσ¯ W − (Λ,p) (Λp)0 kk k · s k¯ σ¯ X X + ???therefore, we can see that in order that the field ψk (x) satisfies the Lorentz transformation rule (4.6), it is necessary and sufficient that

0 1 p (jn) 1 D ¯ Λ− u¯ (Λx + b; p ,σ,n)= D W − (Λ,p) exp [+i (Λp) b] u (x; p, σ,n¯ ) (4.28) kk k Λ 0 σσ¯ · k ¯ s(Λp) σ¯ Xk X

Similarly, by comparing Eqs. (4.7, 4.23), the necessary and suﬃcient condition for the ﬁeld ψk− (x) to satisfy the Lorentz transformation rule (4.7), yields

0 1 p (jn) 1 D ¯ Λ− v¯ (Λx + b; p ,σ,n)= D ∗ W − (Λ,p) exp [ i (Λp) b] v (x; p, σ,n¯ ) (4.29) kk k Λ 0 σσ¯ − · k ¯ s(Λp) σ¯ Xk X 1 1 1 we can put Eq. (4.28) in a slightly diﬀerent form by using the fact that Dkk¯ Λ− = Dk−k¯ (Λ) and Dσσ¯ W − = 1 Dσ−σ¯ (W ). Multiplying Eq. (4.28) by Dk′k (Λ) and summing over k we ﬁnd 0 1 p D ′ (Λ) D ¯ Λ− u¯ (Λx + b; p ,σ,n) = D ′ (Λ) k k kk k Λ k k 0 × k ¯ k s(Λp) X Xk X (jn) 1 D W − (Λ,p) exp [+i (Λp) b] u (x; p, σ,n¯ ) σσ¯ · k σ¯ X

0 p (jn) 1 δ ′¯ u¯ (Λx + b; pΛ,σ,n) = Dk′k (Λ) Dσσ¯ W − (Λ,p) exp [+i (Λp) b] uk (x; p, σ,n¯ ) k k k (Λp)0 · k¯ k s σ¯ X X X (jn) 1 uk′ (Λx + b; pΛ,σ,n) = Dσσ¯ W − (Λ,p) σ¯ X p0 Dk′k (Λ) 0 exp [+i (Λp) b] uk (x; p, σ,n¯ ) (4.30) × s(Λp) · Xk 142 CHAPTER 4. RELATIVISTIC QUANTUM FIELD THEORY

(jn) similarly, when multiplying Eq. (4.30) by Dσ′σ (W (Λ,p)) and summing over σ we have

(jn) (jn) (jn) 1 Dσ′σ (W (Λ,p)) uk¯ (Λx + b; pΛ,σ,n) = Dσ′σ (W (Λ,p)) Dσσ¯ W − (Λ,p) σ σ σ¯ X X X p0 Dkk¯ (Λ) 0 exp[+i (Λp) b] uk (x; p, σ,n¯ ) × s(Λp) · Xk

0 (jn) p D ′ (W (Λ,p)) u¯ (Λx + b; p ,σ,n) = δ ′ D¯ (Λ) exp [+i (Λp) b] u (x; p, σ,n¯ ) σ σ k Λ σ σ¯ kk 0 · k σ σ¯ s(Λp) X X Xk 0 (jn) p D ′ (W (Λ,p)) u¯ (Λx + b; p ,σ,n) = D¯ (Λ) exp [+i (Λp) b] u x; p,σ′,n σ σ k Λ kk 0 · k σ k s(Λp) X X renaming σ σ¯ and σ σ we obtain → ′ → 0 (jn) p D (W (Λ,p)) u¯ (Λx + b; p , σ,n¯ )= D¯ (Λ) exp [+i (Λp) b] u (x; p,σ,n) σσ¯ k Λ kk 0 · k σ¯ s(Λp) X Xk ???therefore, in a slightly diﬀerent form, Eq. (4.28) becomes

0 (jn) p u¯ (Λx + b; p , σ,n¯ ) D (W (Λ,p)) = D¯ (Λ) exp [i (Λp) b] u (x; p,σ,n) (4.31) k Λ σσ¯ 0 kk · k σ¯ s(Λp) X Xk with a similar procedure Eq. (4.29), becomes

0 (jn) p v¯ (Λx + b; p , σ,n¯ ) D ∗ (W (Λ,p)) = D¯ (Λ) exp [ i (Λp) b] v (x; p,σ,n) (4.32) k Λ σσ¯ 0 kk − · k σ¯ s(Λp) X Xk [homework!!(12) arrive to Eqs. (4.28), (4.32) with a correct procedure]The advantage of Eqs. (4.31, 4.32) with respect to Eqs. (4.28, 4.29) is that the former are in terms of the Lorentz transformations instead of their inverses. In summary Eqs. (4.31, 4.32), are the fundamental requirements that allow to calculate the coefficient functions uk and vk, in terms of finite number of free parameters. We shall use Eqs. (4.31) and (4.32) in three steps, considering in turn the three different types of proper orthochronus Lorentz transformations.

4.2.1 Translations We shall start by studying Eqs. (4.31, 4.32) for the case of U (1, b) corresponding with pure translations. By setting Λ = 1 and b arbitrary in Eq. (4.31) we ﬁnd

0 (jn) p u¯ (x + b; p, σ,n¯ ) D (W (1,p)) = D¯ (1) exp [ip b] u (x; p,σ,n) (4.33) k σσ¯ p0 kk · k σ¯ s X Xk from Eq. (1.181), page 41 we see that

1 1 W (Λ,p) = L− (Λp) Λ L (p) W (1,p)= L− (p) 1 L (p) ⇒ W (1,p) = 1 (4.34) 4.2. LORENTZ TRANSFORMATIONS FOR MASSIVE FIELDS 143

Then Eq. (4.33) becomes u¯ (x + b; p,σ,n) = exp [ip b] u¯ (x; p,σ,n) (4.35) k · k observe that the traslation in x of the coefficient u is carried out by a factor exp [ip b] which resembles the k · translation operator with generator p and parameter b. Now, we define the “standard” coefficient uk (p,σ,n) in the form 3/2 u (x = 0; p,σ,n) (2π)− u (p,σ,n) (4.36) k ≡ k 3/2 where the factor (2π)− is defined for convenience. From this definition, we obtain by setting x = 0 in Eq. (4.35) that 3/2 u (b; p,σ,n)=(2π)− exp [ip b] u (p,σ,n) k · k the procedure for vk (x; p,σ,n) from Eq. (4.32) is similar. Since b is arbitrary, the coefficients uk (x; p,σ,n) and vk (x; p,σ,n) acquire the form

3/2 ip x uk (x; p,σ,n) = (2π)− e · uk (p,σ,n) (4.37) 3/2 ip x vk (x; p,σ,n) = (2π)− e− · vk (p,σ,n) (4.38)

From which we have obtained the x dependence of the coeﬃcients u and v . Substituting (4.37, 4.38) in Eqs. − k k (4.4, 4.5) we see that the annihilation and creation ﬁelds are their Fourier transforms

+ 3/2 3 ip x ψk (x) = (2π)− d p uk (p,σ,n) e · a (p,σ,n) (4.39) σ,n X Z 3/2 3 ip x ψk− (x) = (2π)− d p vk (p,σ,n) e− · a† (p,σ,n) (4.40) σ,n X Z 3/2 At this step we understand the convenience of deﬁning the factor (2π)− in Eq. (4.36). On the other hand, equations (4.37, 4.38) are valid for arbitrary values of p. Consequently, if it is valid for p it is also valid for pΛ, where pΛ is the three-vector associated with Λp. In other words, Eqs. (4.37, 4.38) must hold for an arbitrary Lorentz transformation. Thus, taking Eq. (4.31)

0 (jn) p u¯ (Λx + b; p , σ,n¯ ) D (W (Λ,p)) = D¯ (Λ) exp [i (Λp) b] u (x; p,σ,n) (4.41) k Λ σσ¯ 0 kk · k σ¯ s(Λp) X Xk and substituting (4.37) in (4.41) we obtain

0 i(Λp) (Λx+b) (jn) p i[(Λp) b] ip x e · uk¯ (pΛ, σ,n¯ ) Dσσ¯ (W (Λ,p)) = 0 Dkk¯ (Λ) e · e · uk (p,σ,n) σ¯ s(Λp) X Xk 0 i(Λp) (Λx+b) (jn) p i[(Λp) b+p x] e · uk¯ (pΛ, σ,n¯ ) Dσσ¯ (W (Λ,p)) = 0 Dkk¯ (Λ) e · · uk (p,σ,n) (4.42) σ¯ s(Λp) X Xk now, since the dot product of two four-vectors is Lorentz invariant, we have

(Λp) b + p x = (Λp) b + (Λp) (Λx) = (Λp) (Λx + b) · · · · · therefore, the phases on both sides of Eq. (4.42) are equal and we obtain

0 (jn) p uk¯ (pΛ, σ,n¯ ) Dσσ¯ (W (Λ,p)) = 0 Dkk¯ (Λ) uk (p,σ,n) σ¯ s(Λp) X Xk 144 CHAPTER 4. RELATIVISTIC QUANTUM FIELD THEORY

a similar result is obtained for the coeﬃcients vk (p,σ,n) by substituting (4.38) in Eq. (4.32). In conclusion, from Eqs. (4.37, 4.38) we see that Eqs. (4.31, 4.32) are satisﬁed if and only if

0 (jn) p uk¯ (pΛ, σ,n¯ ) Dσσ¯ (W (Λ,p)) = 0 Dkk¯ (Λ) uk (p,σ,n) (4.43) σ¯ s(Λp) X Xk 0 (jn) p ∗ vk¯ (pΛ, σ,n¯ ) Dσσ¯ (W (Λ,p)) = 0 Dkk¯ (Λ) vk (p,σ,n) (4.44) σ¯ s(Λp) X Xk Equations (4.43, 4.44) must be valid for arbitrary homogeneous Lorentz transformations. Note that with respect to Eqs. (4.31, 4.32), equations (4.43, 4.44) are for the standard coeﬃcients deﬁned in (4.36), so that the x dependence − has dissapeared.

4.2.2 Boosts Now, let us take p = 0 in Eqs. (4.43) and (4.44) and use Λ as the standard boost L (q) that takes a particle of mass m from rest to a given four-momentum qµ. For p = p, p0 = (0,m), it is clear that L (p) = 1, and from deﬁnition (1.181), page 41 we have 1 1 1 W (Λ,p) L− (Λp) Λ L (p)= L− (q) L (q) 1 = L− (q) L (q) = 1 ≡ for this special case, it is also clear that pΛ = q. Using all these facts in Eq. (4.43) we ﬁnd

(jn) m u¯ (q, σ,n¯ ) D (1) = D¯ (L (q)) u (0,σ,n) k σσ¯ q0 kk k σ¯ X r Xk m u¯ (q, σ,n¯ ) δ = D¯ (L (q)) u (0,σ,n) k σσ¯ q0 kk k σ¯ X r Xk and the same replacements can be done in Eq. (4.44). Therefore, in this special case, Eqs. (4.43) and (4.44) yield

m u¯ (q,σ,n) = D¯ (L (q)) u (0,σ,n) (4.45) k q0 kk k r Xk m v¯ (q,σ,n) = D¯ (L (q)) v (0,σ,n) (4.46) k q0 kk k r Xk hence, if we know the quantities uk¯ (0,σ,n) and vk¯ (0,σ,n) for zero momentum, we can obtain the corresponding quantities uk¯ (p,σ,n) and vk¯ (p,σ,n) for arbitrary momentum p, for a given representation D (Λ) of the homogeneous Lorentz group.

4.2.3 Rotations Now we take again p = 0, but with Λ being a rotation R. Since rotations are Lorentz transformations that preserve the norm of the three-momentum, we have pΛ = 0. Here it is clear that

p = Λp Rp = (0,m) ≡ In addition, since the standard boost L (p) takes a particle from rest to its ﬁnal state also at rest, we have L (p) = 1. Then we have

1 1 W (Λ,p) W (R,p) L− (Rp) R L (p)= L− (p) R L (p) ≡ ≡ W (Λ,p) = R 4.3. IMPLEMENTATION OF THE CLUSTER DECOMPOSITION PRINCIPLE 145

From these considerations Eq. (4.43), yields

(jn) m u¯ (0, σ,n¯ ) D (R)= D¯ (R) u (0,σ,n) k σσ¯ m kk k σ¯ r k X X and a similar procedure can be carried out for Eq. (4.44). In summary, Eqs. (4.43) and (4.44) give

(jn) uk¯ (0, σ,n¯ ) Dσσ¯ (R) = Dkk¯ (R) uk (0,σ,n) (4.47) σ¯ X Xk (jn) vk¯ (0, σ,n¯ ) Dσσ¯ ∗ (R) = Dkk¯ (R) vk (0,σ,n) (4.48) σ¯ X Xk further, by applying Eqs. (4.47, 4.48) to inﬁnitesimal rotations, such relations can be written in terms of the generators of the rotation group (which are independent of the speciﬁc rotation R i.e. of the parameters)

(jn) u¯ (0, σ,n¯ ) J = ¯ u (0,σ,n) (4.49) k σσ¯ Jkk k σ¯ X Xk (jn) v¯ (0, σ,n¯ ) J ∗ = ¯ v (0,σ,n) (4.50) k σσ¯ − Jkk k σ¯ X Xk where J(j) and are the angular-momentum matrices in the representations D(j) (R) and D (R), respectively. It J is clear that any representation D (Λ) of the homogeneous Lorentz group, provides a representation of the three- dimensional rotation group when Λ is restricted to rotations R. Equations (4.49) and (4.50) say that if the fields ψ± (x) described particles of a given spin j, then the representation D (R) must contain among its irreducible (j) components the spin -j representation D (R), where the coefficients uk (0,σ,n) and vk (0,σ,n) describes how the spin j representation of the rotation group is embedded in D (R). We shall see later that each irreducible − representation of the proper orthochronus Lorentz group contains any given irreducible representation of the + rotation group at least once. Therefore, if the fields ψk (x) and ψk− (x) transform irreducibly (so that Dkk¯ (R) corresponds to an irreducible representation) then they are unique up to overall scale. More generally, the number of parameters in the annihilation or creation fields (including their overall scales) is equal to the number of irreducible representations in the field. It can be shown that coefficient functions uk (p, σ,n¯ ) and vk (p, σ,n¯ ) given by Eqs. (4.45) and (4.46), with uk (0,σ,n) and vk (0,σ,n) satisfying Eqs. (4.47) and (4.48), automatically satisfy the more general requirements (4.43) and (4.44).

4.3 Implementation of the cluster decomposition principle

Substituting Eqs. (4.39, 4.40) in Eq. (4.16) and integrating over x, the interaction potential becomes [homework!!(13), prove Eqs. (4.51, 4.52, 4.53)]

3 3 3 3 V = d p′ d p′ d p d p 1 · · · N 1 · · · M NM Z σ′ σ′ σ1 σM n′ n′ n1 nM X 1X··· N X··· 1X··· N X··· a† p′ σ′ n′ a† p′ σ′ n′ a (p σ n ) a (p σ n ) × 1 1 1 · · · N N N M M M · · · 1 1 1 p′ σ′ n′ p′ σ′ n′ , p σ n p σ n (4.51) ×VNM 1 1 1 · · · N N N 1 1 1 · · · M M M where the coeﬃcient functions are given by

3 p′ σ′ n′ p′ σ′ n′ , p σ n p σ n = δ p′ + . . . p . . . (4.52) VNM 1 1 1 · · · N N N 1 1 1 · · · M M M 1 − 1 − × p′ σ′ n′ p′ σ′ n′ , p σ n p σ n VNM 1 1 1 · · · N N N 1 1 1 · · · M M M e 146 CHAPTER 4. RELATIVISTIC QUANTUM FIELD THEORY with 3 3N/2 3M/2 p′ σ′ n′ p′ σ′ n′ , p σ n p σ n = (2π) − − VNM 1 1 1 · · · N N N 1 1 1 · · · M M M g ′ ′ v ′ p′ σ′ n′ v ′ p′ σ′ n′ u (p1σ1n1) u (pM σM nM ) (4.53) ek1 kN ,k1 kM k1 1 1 1 kN N N N k1 kM × ′ ′ ··· ··· · · · · · · k k k1 kM 1X··· N X··· the form of the interaction guarantees that the Cluster Decomposition Principle (CDP) is satisfied by the S matrix, since has a single delta function factor, with a coefficient that at least for a finite number − VNM VNM of field types, has at most branch point singularities at zero particle momenta. We could say conversely that any operator can be written as in Eq. (4.51) and the CDP demands that the coefficiente may be written as in VNM Eq. (4.52), i.e. as a product of a single momentum-conservation delta function times a smooth coefficient function . Any sufficiently smooth function (but not one that contains a delta function) can be expressed as in Eq. VNM (4.53). For general functions the indices k and k′ can have infinite range. However, we shall restrict k and k′ to a finitee range because of the principle of renormalizability to be discussed later. We could summarize our results by saying that the CDP along with Lorentz invariance makes it natural that the interaction density should be constructed out of the annihilation and creation operators.

4.4 Lorentz invariance of the S matrix − By combining annihilation and creation operators in arbitrary polynomials (4.16), where the coupling coefficients ′ ′ gk k ,k1 kM are subjected only to the invariance condition (4.17), and a suitable reality condition (for (x) to 1··· N ··· H be hermitian), we construct a scalar density that satisfies the CDP. Now, for the Lorentz invariance of the S matrix, it is also necessary that the interaction density satisfies the − commutation relation (4.3). 2 (x) , x′ =0 for x x′ 0 (4.54) H H − ≥ To check the conditions to obtain (4.54) we start by calculating the commutation and anti-commutation relations between the creation and annihilation fields

+ I ψ (x) , ψ¯− (y) ∓ ≡ k k h i∓ where [. . .] and [. . .]+ denotes commutation and anticommutation respectively. We then use Eqs. (4.39, 4.40) and the commutation− and anti-comutation relations between creation and annihilation operators (3.17)

3/2 3 ip x 3/2 3 ip′ y I = (2π)− d p uk (p,σ,n) e · a (p,σ,n) , (2π)− d p′ vk¯ p′, σ,¯ n¯ e− · a† p′, σ,¯ n¯ ∓ " σ,n Z σ,¯ n¯ Z # X X ∓ 3 3 ip x 3 ip′ y = (2π)− d p uk (p,σ,n) e · d p′ vk¯ p′, σ,¯ n¯ e− · a (p,σ,n) , a† p′, σ,¯ n¯ σ,n σ,¯ n¯ Z Z ∓ X X h i 3 3 ip x 3 ip′ y I = (2π)− d p uk (p,σ,n) e · d p′ vk¯ p′, σ,¯ n¯ e− · δσσ¯ δnn¯δ p p′ ∓ − σ,n σ,¯ n¯ Z Z X X Therefore, the commutation or anti-commutation relations between annihilation and creation ﬁelds yield

+ 1 3 ip (x y) ψk (x) , ψk¯− (y) = 3 d p uk (p,σ, n) vk¯ (p,σ, n) e · − (4.55) (2π) σ,n h i∓ X Z where the sign indicates a commutator or anticommutator if the particles destroyed and created by the com- + ∓ ponents ψk and ψk¯− are bosons or fermions respectively. We see from (4.55) that the commutation relation (4.54) is not satisfied automatically by arbitrary functions of the annihilation and creation fields, because in general the integral in (4.55) does not vanish even if (x y) is space-like. We cannot avoid this problem by making the − 4.5. INTERNAL SYMMETRIES AND ANTIPARTICLES 147 interaction density out of annihilation or creation fields alone, since in that case the interaction would not be hermitian. The only way to avoid such a difficulty is by making linear combinations of annihilation and creation fields + ψ (x) κ ψ (x)+ λ ψ− (x) (4.56) k ≡ k k k k where the constants κk and λk and any other arbitrary constants in the fields, are adjusted such that 2 [ψk (x) , ψk′ (y)] = ψk (x) , ψk†′ (y) = 0 if (x y) 0 (4.57) ∓ − ≥ h i∓ note that by including explicit constants in Eq. (4.56) we are still leaving free the overall scale of the annihilation and creation fields to choose it at our convenience. We shall see later how to choose the constants in the linear combination (4.56) for several irreducibly transforming fields. The Hamiltonian density (x) will satisfy the H commutation condition (4.54) if it is constructed out of such fields and their adjoints, with an even number of any field components that destroy and create fermions. The condition (4.57) is frequently described as a causality condition, because if x y is space-like, no signal can − reach y from x or vice versa. Therefore, the measurement of ψk at a space-time point x should not interfere with a measurement of ψk′ or ψk†′ at a point y. Such considerations of causality makes sense for the electromagnetic field, since each of its components can be measured at a given space-time point. However, the fields we shall be dealing with (such as the Dirac field of the electron) seem not to be measurable in any sense. Hence, we shall better take Eq. (4.57) as a necessary condition for Lorentz invariance of the S matrix but without any association with − measurability or causality.

4.5 Internal symmetries and antiparticles

We then construct fields (4.56) that satisfy Eq. (4.57). It could happen that particles that are destroyed and created by these fields carry non-zero values of one or more conserved quantum numbers. We shall take as an example the conservation of the electric charge. If particles of species n carry a value q (n) for the electric charge, we can define the charge operator Q, as3 Q q ,...,q = q (n )+ q (n )+ . . . + q (n ) q ,...,q (4.58) | 1 N i { 1 2 N } | 1 N i Q q ,...,q = Q q ,...,q (4.59) | 1 N i | 1 N i where we have defined Q as the total charge of the state q ,...,q . That is | 1 N i Q q (n )+ q (n )+ . . . + q (n ) (4.60) ≡ 1 2 N since q (nk) is the charge of the species nk, and using the action of a (qr) on an arbitrary multi-particle state [see Eq. (3.12), Page 119] we obtain N r+1 Q a (q) q1,...,qN = Q ( 1) δ (q qr) q1 ...qr 1qr+1 ...qN | i ± − | − i Xr=1 N r+1 = ( 1) δ (q qr) [q (n1)+ . . . + q (nr 1)+ q (nr+1)+ . . . + q (nN )] ± − − r=1 X q1 ...qr 1qr+1 ...qN × | − i N r+1 = ( 1) δ (q qr) [Q q (nr)] q1 ...qr 1qr+1 ...qN ± − − | − i Xr=1 N r+1 Q a (q) q1,...,qN = [Q q (n)] ( 1) δ (q qr) q1 ...qr 1qr+1 ...qN | i − ± − | − i r=1 X 3 Do not confuse the notation qr for the state of the particle qr ≡ pr,σr, nr, with the notation q (nk) that gives the charge of the nk species. 148 CHAPTER 4. RELATIVISTIC QUANTUM FIELD THEORY where we have taken into account that because of the delta function, the term Q q (n ) only contributes if q = q − r r so that n = nr, where n denotes the species of particle destroyed by a (q). We obtain finally Q a (q) q ,...,q = [Q q (n)] a (q) q ,...,q (4.61) | 1 N i − | 1 N i On the other hand,

a (q) Q q ,...,q = a (q) Q q ,...,q | 1 N i | 1 N i a (q) Q q ,...,q = Q a (q) q ,...,q (4.62) | 1 N i | 1 N i substracting Eqs. (4.61, 4.62) we obtain

Q a (q) a (q) Q q ,...,q = q (n) a (q) q ,...,q { − } | 1 N i − | 1 N i since the state q ,...,q is arbitrary we obtain | 1 N i [Q, a (p,σ,n)] = q (n) a (p,σ,n) − similarly by using the deﬁnition of the creation operator Eq. (3.7), page 118, we have

Q a† (q) q ,...,q = Q q,q ,...,q = [Q + q (n)] q,q ,...,q = [Q + q (n)] a† (q) q ,...,q | 1 N i | 1 N i | 1 N i | 1 N i a† (q) Q q ,...,q = Q a† (q) q ,...,q | 1 N i | 1 N i so that Q, a† (p,σ,n) =+q (n) a† (p,σ,n) In conclusion, if particles of speciesh n carry a valuei q (n) for the electric charge, the commutator of the charge operator with the creation and annihilation ﬁelds yield4

[Q, a (p,σ,n)] = q (n) a (p,σ,n) (4.63) − Q, a† (p,σ,n) = +q (n) a† (p,σ, n) (4.64) h i Note that for Eqs. (4.63, 4.64) to be fullfilled, we only require that the conserved quantum number be additive i.e. we only require the property (4.58) for a “generalized conserved charge”. Now, let us Q denotes the charge operator or any other symmetry operator. Since it must be a constant of motion, it should commute with the Hamiltonian, and so with the interaction density (x). In order that (x) H H should commute with the charge operator Q (or some other symmetry generator) it is necessary that it be formed with fields that have simple commutation relations with the charge operator [similar to the relation given by Eq. (4.63]: [Q, ψ (x)] = q ψ (x) (4.65) k − k k since in that case, we can achieve that (x) commute with Q, by constructing it as a sum of products of fields H ψ ψ and adjoints ψm† ψm† such that k1 k2 · · · 1 2 · · · q + q + q q = 0 (4.66) k1 k2 ···− m1 − m2 −··· Now, according with (4.16), (x) is written as a sum of products of the creation and annihilation fields. Thus, H we should examine the commutation relations of creation and annihilation fields with the charge operator, and check for the conditions to satisfy Eq. (4.65). To do it, we use Eqs. (4.39, 4.63) to obtain

+ 3/2 3 ip x 3/2 3 ip x Q, ψ (x) = (2π)− d p u (p,σ,n) e · [Q, a (p,σ,n)] = (2π)− d p u (p,σ,n) e · [ q (n) a (p,σ,n)] k k k − σ,n Z σ,n Z X X 4It is worth emphasizing that Eqs. (4.63, 4.64) are only commutator relations, and have no anti-commutator counter-parts. They are valid regardless the species of particles that carry the conserved quantum number are bosons or fermions. 4.6. LORENTZ IRREDUCIBLE FIELDS AND KLEIN-GORDON EQUATION 149 and the satisfaction of Eq. (4.65) requires that we are able to factorize q (n) out of the sum such that

+ 3/2 3 ip x Q, ψ (x) = q (2π)− d p u (p,σ,n) e · [ a (p,σ,n)] = q ψ (x) (4.67) k − k k − k k σ,n Z X + from which it is clear that Eq. (4.65) is satisfied for a given component of the annihilation field ψk (x) if and only if all particle species n that are destroyed by the field carry the same charge q (n)= qk. Thus, the sum over species in Eq. (4.67) is only over species of the same charge. Similarly, Eq. (4.65), is satisfied by one particular component ψk− (x) of the creation field if and only if all particle speciesn ¯ that are created by the field, carry the charge q (¯n) = q . We then conclude that in order that such a theory conserve quantum numbers like electric − k charge, there must be a doubling of particle species carrying non-zero values of the conserved quantum numbers: if a particular component of the annihilation field destroys a particle of species n, then the same component of the creation field must create particles of a speciesn ¯, called the antiparticles of the particles of species n, that have opposite values of all conserved quantum numbers. These arguments then, predicts the existence of antiparticles.

4.6 Lorentz irreducible ﬁelds and Klein-Gordon equation

We have said that the Lorentz representations D (Λ) in Eq. (4.6, 4.7) do not have to be irreducible. By choicing an apropriate basis (canonical basis) the matrix representation D (Λ) could acquire a block diagonal form, so that fields that belong to different blocks cannot transform into each other under Lorentz transformations. On the other hand, the Lorentz transformations do not have effect on the particle species. Consequently, instead of using one big field, including many irreducible components and many particle species, we shall restrict our attention to fields that destroy only a single type of particle (since in this case the label n is fixed, we shall omit it) and create only the corresponding antiparticle, and that also transform irreducibly under the Lorentz group (as mentioned above we could include or not include space inversion). It is understood that in general we shall have to consider many different such fields, some perhaps formed by the derivatives of other fields. The next task is to finish the determination of the coefficient functions uk (p,σ) and vk (p,σ), fix the values of the constants κ and λ of Eqs. (4.56) and deduce relations between the properties of particles and antiparticles for fields that belong to the simplest irreducible representations of the Lorentz group: the scalar, the vector, and Dirac spinor representations. Let us then concentrate on all components of a field of definite mass m. For our purposes we first calculate µ ip x the quantities ∂ ∂µe± · . For them, we obtain

α β β µ ip x µν ipαx µν ip x α µν ip x α ∂ ∂ e± · = g ∂ ∂ e± = g ∂ ip e± β ∂ x = g ∂ ip e± β δ µ µ ν µ ± α ν µ ± α ν α µν ip xβ h α β µνi ip x h βα ip x i = ip δ g ∂ e± β = p p δ δ g e± · = p p g e± · ± α ν µ − α β ν µ − α β µ ip x 2 ip x 2 ip x ∂ ∂ e± · = p e± · = mhe± · i µ − now, from Eq. (4.39) and assuming that the sum is only over species of the same mass, we have

µ + 3/2 3 µ ip x ∂µ∂ ψk (x) = (2π)− d p uk (p,σ,n) ∂µ∂ e · a (p,σ,n) σ,n Z X 2 3/2 3 ip x = m (2π)− d p uk (p,σ,n) e · a (p,σ,n) σ,n X Z µ + 2 + ∂µ∂ ψk (x) = m ψk (x) (4.68) similarly, from Eq. (4.40) we obtain

µ 3/2 3 µ ip x ∂µ∂ ψk− (x) = (2π)− d p vk (p,σ,n) ∂µ∂ e− · a† (p,σ,n) σ,n X Z µ 2 ∂µ∂ ψk− (x) = m ψk− (x) (4.69) 150 CHAPTER 4. RELATIVISTIC QUANTUM FIELD THEORY

and combining Eqs. (4.68, 4.69) with (4.56), we can obtain a ﬁeld equation for ψk (x)

µ µ + µ + µ ∂µ∂ ψk (x) ∂µ∂ κkψk (x)+ λkψk− (x) = κk∂µ∂ ψk (x)+ λk∂µ∂ ψk− (x) ≡ 2 + = m κkψk (x)+ λkψk− (x) µ 2 ∂µ∂ ψk (x) = m ψk (x) (4.70) rewriting the second-order diﬀerential operator as

∂ ∂µ (4.71) µ ≡ we can rewrite the ﬁeld equation (4.70) in the form

m2 ψ (x) = 0 (4.72) − k Expression (4.72), is called the Klein-Gordon equation. This will be one of our most important field equations to work with. We recall that the validity of the equation requires that the expansion of the fields ψk± (x) in Eqs. (4.39, 4.40) be over particle species of the same mass. For example, Klein Gordon equation is valid if the sum is over a particle and its corresponding antiparticle. Of course, such an equation is also valid if the expansion of the fields is over a single particle species. In conclusion, the combination of Eqs. (4.39, 4.40) with Eq. (4.56) shows that all components of a field of definite mass m satisfy the Klein-Gordon equation (4.72). Some fields satisfy other field equations, depending on whether or not there are more field components than independent particle states. The most usual approach is to start with the field equation (e.g. the Klein-Gordon equation) or from the Lagrangian density that generates it5, and uses them to derive the expansion of the fields in terms of one-particle annihilation and creation operators. In the present approach, the starting point are the particles, and we derive the fields according with the dictates of CDP and Lorentz invariance, in this way the field equations arise as a byproduct of this construction. We have already proved in Sec. 3.6, that the condition that guarantees that a theory satisfies the CDP, is that the interaction can be expressed as a sum of products of creation and annihilation operators, with all creation operators to the left of all annihilation operators, and with coefficients that contain only a single momentum conservation delta function. Owing to it, we should write the interaction in the “normal ordered form”

3 V = d x : ψ (x) , ψ† (x) : F Z the colons indicate that the enclosed expression is to be rewritten so that all creation operators stand to the left of all annihilation operators. In doing this, we should ignore non-vanishing commutators or anticommutators, but we should include minus signs for permutations of fermionic operators. Moreover, by using the commutation or anticommutation relations of the fields, any such normal ordered function of the fields can be written as a sum of ordinary products of the fields with complex number coefficients. Rewriting : : in this way we can see that F if it is constructed out of fields that satisfy Eq. (4.57), with even numbers of any fermionic field components, we shall obtain that : ψ (x) , ψ (x) : will commute with : ψ (y) , ψ (y) : when x y is space-like, despite the F † F † − normal ordering.

5We can see the plausibility of the Klein-Gordon equation by a principle of correspondence for the relativistic free-particle relation 2 2 2 µ 2 E − p − m = 0 or equivalently −pµp − m = 0. Resorting to the principle of correspondence pµ → −i∂µ = −i (∇, −∂t) we have µ µ pµp ψk (x) → −∂µ∂ ψk (x) = ψk (x), and we obtain the Klein-Gordon equation. Note that the same principle of correspondence along with the non-relativistic relation E = p2/2m + V (x,t), lead us to the Schr¨odinger equation. Chapter 5

Causal scalar ﬁelds for massive particles

We shall consider first the so-called scalar representation of the homogeneous Lorentz group in which D (Λ) = 1. Since this irreducible representation is one-dimensional, we consider one-component annihilation and creation + fields φ (x) and φ− (x) that transform according with the scalar representation of the Lorentz group. When we restrict us to rotations, it is the scalar representation of the SO (3) group, for which all matrix representatives of rotations and/or generators collapse to the identity matrix in one dimension. The label of the identity (or scalar) representation of SO (3) corresponds to j = 0 (zero spin particles), from which the labelsσ,σ ¯ in Eqs. (4.49) and (4.50) can only take a single value. Consequently, a scalar field can only describe particles of zero spin.

5.1 Scalar ﬁelds without internal symmetries

Let us assume by now that the field describes only a single species of particle, with no distinct anti-particle, the quantities uk (0,σ,n) and vk (0,σ,n) can be written just as the numbers u (0) and v (0) (we omit the species label n, the field component label k, and the spin label σ because each of these labels takes only one value). Thus, the annihilation and creation fields (4.39) and (4.40) are written as

+ 1 3 ip x 1 3 ip x φ (x)= d p u (p) e · a (p) ; φ− (x)= d p v (p) e− · a† (p) (5.1) 3 3 (2π) Z (2π) Z q q now since we are in the identity representation Dkk¯ (L (p)) = δkk¯ , and Eqs. (4.45, 4.46) become m m u (p)= u (0) ; v (p)= v (0) (5.2) p0 p0 r r so that the ﬁelds read

+ 1 3 m ip x 1 3 m ip x φ (x)= d p u (0) e · a (p) ; φ− (x)= d p v (0) e− · a† (p) 3 p0 3 p0 (2π) Z r (2π) Z r

q + q now, the simplest way to settle φ (x) and φ− (x) as the adjoint of each other, is by choosing u (0) and v (0) to be real and equal. Therefore, it is conventional to adjust the overall scales of creation and annihilation operators such that these constants acquire the values 1 u (0) = v (0) = (5.3) √2m then, combining Eqs. (5.2) with conventions (5.3) we ﬁnd 1 u (p)= v (p)= (5.4) 2p0

151 p 152 CHAPTER 5. CAUSAL SCALAR FIELDS FOR MASSIVE PARTICLES

Hence, in the scalar case without internal symmetries, the annihilation and creation ﬁelds (5.1) are written as

+ 1 3 1 ip x φ (x) = d p a (p) e · (5.5) 3 0 (2π) Z 2p

q 1 3 p1 ip x + φ− (x) = d p a† (p) e− · = φ † (x) (5.6) 3 0 (2π) Z 2p q p A Hamiltonian density (x) that is formed as a polynomial in φ+ (x) and φ (x), automatically satisfy the H − requirement (4.16) to transform as a scalar. We should also ensure that it also satisﬁes the condition of the Lorentz-invariance for the S matrix, expressed by the fact that (x) must commute with (y) at space-like − H H separations x y. This condition would be satisﬁed automatically if (x) were a polynomial in φ+ (x) alone, − H because all annihilation operators commute (anticommute) if the particles are bosons (fermions). Then we have1

φ+ (x) , φ+ (y) = 0 (5.7) ∓ regardless the distance (x y) is space-like or not. We denote [. . .] to mean the commutator and [. . .] to mean − + the anti-commutator. However, if the Hamiltonian density (x) were− a polynomial in φ+ (x) alone, it would be + H + not hermitian. To be hermitian, (x) must involve φ (x) but also its adjoint φ † (x)= φ− (x). The problem then + H is that φ (x) does not commute nor anti-commute with φ− (y) for general space-like separations. To characterize + the commutation or anti-commutation algebra between φ (x) and φ− (y) we utilize the expressions (5.5, 5.6) for such scalar ﬁelds

+ 1 3 1 ip x 3 1 ip′ y φ (x) , φ− (y) = d p a (p) e · , d p′ a† p′ e− · (2π)3 0 0 ∓ "Z 2p Z 2p′ # ∓ ip x ip′ y 1 3 p3 e · e− · p = d p d p′ a (p) , a† p′ (2π)3 2p0 2p 0 Z Z ′ ∓ h i and using the commutation and anti-commutation relationsp forp bosons and fermions respectively, Eqs. (3.17), we obtain 3 3 + d p d p′ ip x ip′ y 3 φ (x) , φ− (y) = e · e− · δ p p′ (2π)3 (2p0 2p 0) − ∓ Z ′ 3 · + 1 d p ip (x y) φ (x) , φ− (y) = p e · − (2π)3 2p0 ∓ Z hence this commutator or anti-commutator can be expressed in terms of a single integral

+ φ (x) , φ− (y) = ∆+ (x y) (5.8) ∓ − 3 1 d p ip x ∆+ (x) e · (5.9) ≡ (2π)3 2p0 Z 3 0 since d p/p is the Lorentz invariant volume [see Eq. (1.157), page 37)], the integral ∆+ (x) is manifestly Lorentz- invariant. Consequently, for space-like x, it can depend only on the invariant square x2 > 0, and can be evaluated in any convenient reference frame. We can evaluate ∆+ (x) for space-like x by choosing the coordinate system such that2 x0 = 0, x = √x2 p x = p0x0 + p x = p x (5.10) k k ⇒ · − · · 1 ′ Equation (5.7) comes from the fact that [a (p) ,a (p )]∓ = 0, combined with equation (5.5) 2It is important to recall that the choice (5.10) is only possible for space-like events. This choice means that in this reference frame, the event (0, 0) is simultaneous with the event x,x0 with x 6= 0. Hence it is only possible for causal disconnected (space-like) events. 5.1. SCALAR FIELDS WITHOUT INTERNAL SYMMETRIES 153 from which Eq. (5.9) becomes 3 1 d p ip x ∆+ (x)= e · (5.11) (2π)3 p2 2 Z 2 + m for a while we shall denote p = p , so by now p stands for thep magnitude of the three-vector (instead of denoting | | the four-vector). By using spherical coordinates in the three-momentum space we have d3p = p2 dp sin θ dθ dφ. By convention we choose3

x = x u = √x2 u so that p x = p x cos θ = p√x2 cos θ k k 3 3 · k k k k picking up all these facts, Eq. (5.11) becomes

2 2 π 2π 1 p dp sin θ dθ dφ ip√x2 cos θ 1 ∞ p dp ip√x2 cos θ ∆+ (x) = e = e sin θ dθ dφ (2π)3 2 2 (2π)3 2 2 Z 2 p + m Z0 2 p + m Z0 Z0 2 π 2π ∞ pp dp ia cos θ p 2 ∆+ (x) = I ; I e sin θ dθ , a p√x (5.12) (2π)3 2 2 ≡ ≡ Z0 2 p + m Z0 let us ﬁrst evaluate the integralp I

π ia cos θ π ia ia ia cos θ ie i ia ia 2 e e− 2sin a I e sin θ dθ = = e− e = − = ≡ a a − a 2i a Z0 0

√ 2 2sin a 2sin p x I = = (5.13) a p√x2 substituting (5.13) in Eq. (5.12), we ﬁnd

2 sin p√x2 2π ∞ p dp ∆+ (x)= (2π)3 2 2 √ 2 Z0 p + m p x and changing the variable of integration to p p dp u ; du = ≡ m m it yields

2 2 sin mu√x2 3 sin mu√x2 1 ∞ m u m du 1 ∞ m u du ∆+ (x) = = (2π)2 √m2u2 + m2 √ 2 4π2 m√u2 + 1 √ 2 Z0 mu x Z0 m x m ∞ u du 2 ∆+ (x) = sin m√x u (5.14) 2 4π2√x2 0 √u + 1 Z which can be written in terms of a Hankel function

m 2 ∆+ (x)= K1 m√x (5.15) 4π2√x2 we emphasize again that expression (5.15) is only valid for x2 > 0, i.e. for space-like x. This function is not zero. 2 µ + However, we observe from (5.15) that for x > 0, ∆+ (x) is an even function in x . Instead of using only φ (x), we shall try to construct (x) from a linear combination of φ+ (x) and φ+ (x)= φ (x) H † − + φ (x) κφ (x)+ λφ− (x) (5.16) ≡ 3Since x is ﬁxed (and so x), during the integration, choosing x along the three-axis is part of the deﬁnition of the reference frame. 154 CHAPTER 5. CAUSAL SCALAR FIELDS FOR MASSIVE PARTICLES from (5.16) and using (5.7) and its adjoint we have

+ + φ (x) , φ† (y) = κφ (x)+ λφ− (x) , κ∗φ− (y)+ λ∗φ (y) ∓ ∓ h i + + + = κφ (x) , κ∗φ− (y)+ λ∗φ (y) + λφ−(x) , κ∗φ− (y)+ λ∗φ (y) ∓ ∓ + + + + = κφ (x) , κ∗φ− (y) + κφ (x) , λ∗φ (y) + λφ− (x) , κ∗φ− (y) + λφ− (x) , λ∗φ (y) + ∓ + + ∓ ∓ + ∓ = κκ∗ φ (x) , φ− (y) +κλ∗ φ (x) , φ (y) +λκ∗ φ− (x) , φ− (y) +λλ∗ φ− (x) , φ (y) ∓ ∓ ∓ ∓ 2 + 2 + φ (x) , φ† (y) = κ φ (x) , φ− (y) + λ φ− (x) , φ (y) | | | | ∓ ∓ ∓ h i and using Eqs. (5.8), we have for x y being space-like, that − 2 + 2 + 2 2 φ (x) , φ† (y) = κ φ (x) , φ− (y) λ φ (y) , φ− (x) = κ ∆ (x y) λ ∆ (y x) | | ∓ | | | | + − ∓ | | + − ∓ ∓ ∓ h i 2 2 φ (x) , φ† (y) = κ λ ∆ (x y) (5.17) | | ∓ | | + − h i∓ where we have used the fact that the commutator (anticommutator) is antisymmetric (symmetric) under the interchange of the ﬁelds. We also used the fact that ∆+ (x) is an even function of x. Similarly

+ + [φ (x) , φ (y)] = κλ φ (x) , φ− (y) + φ− (x) , φ (y) = κλ (1 1) ∆+ (x y) (5.18) ∓ ∓ ∓ ∓ − note that for the case of anticommutators, in which the “+” sign applies, expressions (5.17, 5.18) never vanishes. Consequently, the particle cannot be a fermion. Thus, those expressions vanish if and only if the particle is a boson and κ and λ are equal in magnitude κ = λ (5.19) | | | | Let us redefine the phases of the states such that a (p) eiαa (p), then a (p) e iαa (p). According with → † → − † Eqs. (5.5, 5.6) this redefinition leads to φ+ (x) eiαφ+ (x) and φ (x) e iαφ (x). Finally, for φ (x) to be → − → − − invariant in Eq. (5.16) we should redefine

iα iα κ e− κ and λ e λ → → ﬁnally, by taking 1 κ α arg ≡ 2 λ we can make the phases of λ and κ to coincide. Therefore, under this convention κ = λ. Now, absorbing the overall factor κ = λ in Eq. (5.16) we obtain

+ φ (x) φ (x)+ φ− (x) ≡ from which it is clear that φ (x) is self-adjoint

+ + φ (x) φ (x)+ φ † (x)= φ† (x) (5.20) ≡ and using (5.5) and (5.6) we obtain its fourier expansion

+ + 1 3 1 ip x ip x φ (x) φ (x)+ φ † (x)= d p a (p) e · + a† (p) e− · (5.21) ≡ 3 2p0 (2π) Z h i thus, the interaction density (x) will commuteq with (y) forp space-like separations x y, if it is constructed as H H − a normal-ordered polynomial in the self-adjoint scalar ﬁeld φ (x) deﬁned by (5.20). 5.2. SCALAR FIELDS WITH INTERNAL SYMMETRIES 155

Despite the choice of relative phase in the two terms in Eq. (5.20) is conventional, it is also a convention that once a given convention is taken, it will be applied wherever a scalar field for this particle appears in the interaction Hamiltonian density. For example, suppose that besides the field (5.20) the interaction density contains another scalar field φ (x) for the same particle

iα + iα + e φ (x)= e φ (x)+ e− φ † (x) where α is an arbitrary phase. This φ likee φ, would be causal in the sense that φ (x) commutes with φ (y) when x y is space-like. However, φ (x) would not commute with φ (y) for space-like separations of x and y. Therefore, − both ﬁelds cannot appear in the samee theory. e e e 5.2 Scalar ﬁelds with internal symmetries

In the previous treatment, we assumed that we have only one species of particles with no distinct antiparticles. In other words, we have supposed that the particle described is totally neutral, i.e. that all intrinsic quantum numbers of it are null. Nevertheless, if the particles that are destroyed and created by φ (x) carry some quantum number like electric charge, we see that (x) conserve this number if and only if each term in (x) contains H H equal numbers of operators a (p) and a† (p). For instance, if a given term contains two operators of the type a† (p), then it would create two particles and so two units of charge, with the consequent increasing of the electric charge of the multi-particle state. Therefore it is necessary that the same term contains two operators of the type a (p) that destroy two particles and hence two units of charge, in order to obtain the same net charge at the end of the process in the multi-particle state. Nevertheless, it is impossible that each term in the interaction density (x), contains the same number of H+ + operators a (p) and a† (p), if (x) is constructed as a polynomial in φ (x)= φ (x)+φ † (x). To see it, we observe H + that according with Eqs. (5.5, 5.6) the ﬁeld φ (x) contains an operator a (p), while the ﬁeld φ− (x) contains an operator a† (p). Let us take for example a polynomial of second degree in φ (x), it would be of the form

2 + + 2 αφ + βφ = α φ (x)+ φ− (x) + β φ (x)+ φ− (x) 2 + + 2 2 + + αφ + βφ = α φ (x)+ φ− (x) + β φ (x) + β φ− (x) + β φ (x) φ− (x)+ φ− (x) φ (x) indeed only the last two terms on the RHS of this equation cont ain the same number of operators a (p) and a† (p). We can see the problem from another point of view. For (x) to commute with the charge operator Q (or any H other symmetry operator) it is necessary that (x) be constructed out of ﬁelds that have simple commutation H + relations with Q such as Eq. (4.65). This is true for φ (x) and its adjoint φ− (x) as can be seen from Equations (5.5, 4.63)

3 ip x 3 + 1 d p ip x e · d p Q, φ (x) = Q, a (p) e · = [Q, a (p)] 3 0 3 0 − (2π) " Z 2p # (2π) Z 2p − − q eip x d3pp q p = q · a (p)= qφ+ (x) − 3 0 − (2π) Z 2p q p + and similarly for φ− (x)= φ † (x), we then obtain

Q, φ+ (x) = qφ+ (x) − − + + Q, φ † (x) = Q, φ− (x) =+qφ † (x) h i− + + however, the self-adjoint ﬁeld φ (x)= φ (x)+ φ † (x) deﬁned in Eq. (5.20), does not satisfy a relation of the type (4.65). 156 CHAPTER 5. CAUSAL SCALAR FIELDS FOR MASSIVE PARTICLES

We deal with this problem by assuming that there are two spinless bosons in which one of them is the “charge conjugate” of the other. We shall see that the two bosons should be of equal mass and opposite generalized charges. We denote φ+ (x) and φc+ (x) the annihilation ﬁelds for these two particles, where the label “c” denotes “charge conjugate”. Then we have

Q, φ+ (x) = qφ+ (x) ; Q, φc+ (x) = qcφc+ (x) (5.22) − − − − of course the ﬁelds φ+ (x) and φc+ (x) are expanded as in Eq. (5.5) in terms of annihilation operators a (p) and ac (p) respectively. Now, we deﬁne the linear combination

+ c+ φ (x)= κφ (x)+ λφ † (x) (5.23) from Eqs. (5.22) the commutation relation of the operator Q with the ﬁeld φ (x) deﬁned by Eq. (5.23), yields

+ c+ + c+ [Q, φ (x)] = Q, κφ (x)+ λφ † (x) = κ Q, φ (x) + λ Q, φ † (x) − − − − h + c c+ i h i [Q, φ (x)] = κqφ (x)+ λq φ † (x) (5.24) − − now, to ﬁnd a relation of the form (4.65) for φ (x), we require to reconstruct φ (x) at the RHS of Eq. (5.24). To do it we demand that qc = q, from which we obtain a relation of the form (4.65) −

+ c+ [Q, φ (x)] = q κφ (x)+ λφ † (x) − − [Q, φ (x)] = qφh(x) i − − we conclude that for the ﬁeld φ (x) deﬁned by Eq. (5.23), we obtain a relation of the form (4.65) [which is the same commutation relation with Q of the operator φ+ (x) alone], if we demand that qc = q. Therefore, Eqs. − (5.22) become

Q, φ+ (x) = qφ+ (x) (5.25) − c+ − c+ Q, φ (x) = +qφ (x) (5.26) − We should keep in mind that a particle that carries no conserved quantum numbers may or may not be its own antiparticle, with a (p)= ac (p). By now we are assuming that the two spinless bosons have opposite charges (or any other conserved quantum number), so that a (p) = ac (p). 6 On the other hand, since a (p) and ac (p) destroy different particles they must commute or anticommute. In a similar way, the annihilation fields φ+ (x) and φc+ (x) destroy different particles. A similar argument follows for the creation operators and fields. Thus, we have4

c c a (p) , a p′ = a† (p) , a † p′ = 0 ∓ + c+ h c i∓ φ (x) , φ (y) = φ− (x) , φ −(y) = 0 ∓ ∓ 4We can also see it in the following way

′ ′ ′ ′ ′ ′ a (p, σ, n) , a p ,σ , n ∓ = δ p − p δσσ δnn (5.27) and ac (p′,σ′, n′) ≡ a (p′,σ′, nc). If the particle does not coincide with the antiparticle then n 6= nc such that (5.27) is clearly zero. 5.2. SCALAR FIELDS WITH INTERNAL SYMMETRIES 157 from which the commutator or anticommutator of φ (x) with its adjoint at a space-like separation is given by

+ c+ + c+ φ (x) , φ† (y) = κφ (x)+ λφ † (x) , κ∗φ † (y)+ λ∗φ (y) ∓ ∓ h i h + + c+ c+ i + c+ = κφ (x) , κ∗φ † (y)+ λ∗φ (y) + λφ † (x) , κ∗φ † (y)+ λ∗φ (y) ∓ ∓ h 2 + + +i h c+ i = κ φ (x) , φ † (y) + κλ∗ φ (x) , φ (y) | | ∓ ∓ h c+ + i 2 c+ c+ +λκ∗ φ † (x) , φ † (y) + λ φ † (x) , φ (y) | | ∓ ∓ 2 h+ + i 2 c+h c+ i φ (x) , φ† (y) = κ φ (x) , φ † (y) λ φ (y) , φ † (x) (5.28) | | ∓ | | h i∓ h i∓ h i∓ + + c+ c+ further it is clear that Eq. (5.8) could be applied for both pairs φ (x) , φ † (y) and φ (y) , φ † (x), since their c c associated annihilation and creation operators a (p) , a† (p) and a (p) , a † (p) have the same commutation or anticommutation algebra. Thus for space-like separations x y we obtain − 2 2 c φ (x) , φ† (y) = κ ∆ (x y) λ ∆ (x y) | | + − ∓ | | + − ∓ where we have used the even propertyh of thisi function for space-like separations. Now, for this commutator or anticommutator to be null for all space-like separations x y, we require to factorize the functions ∆ and ∆c . − + + Then we demand that ∆ (x y)=∆c (x y) + − + − on the other hand, expression (5.11) says such a condition is satisﬁed for all space-like separations x y, if and − only if both functions are associated with the same mass. In other words, the causal condition leads us to the condition that the particle and antiparticle have the same mass. Then, combining Eqs. (5.8) and (5.28) we have

2 2 φ (x) , φ† (y) = κ λ ∆ (x y) (5.29) | | ∓ | | + − h i∓ while φ (x) and φ (y) automatically commute or anticommute with each other for all x and y (regardless its + c+ separation is space-like or not) because φ and φ † destroy and create different particles. Once again, Eq. (5.29) shows that Fermi statistics is ruled out here, because φ (x) cannot anticommute with φ† (y) at space-like separations unless κ = λ = 0, so that the fields (5.23) simply vanish. We conclude that a spinless particle must be a boson5. Now restricting to Bose statistics, we see from Eq. (5.29), that in order that a complex φ (x) should commute with φ (y) at space-like separation, it is necessary and sufficient that κ 2 = λ 2, as well as for the particle and † | | | | antiparticle to have the same mass. As before, we can redefine the relative phase of states of these two particles, and give κ and λ the same phase such that κ = λ. Once again, we can eliminate the resultant common factor by redefinition of the field φ (x) in (5.23) to obtain + c+ φ (x)= φ (x)+ φ † (x) (5.30) + c+ and expanding φ (x)+ φ † (x) as in (5.5) we obtain 3 d p ipx c ipx φ (x)= a (p) e + a † (p) e− (5.31) (2π)3/2 (2p0)1/2 Z h i which is the essentially unique causal scalar field [notice however that this φ (x) is not hermitian anymore]. Expression (5.31) is valid both for purely neutral spinless particles6 [in that case we take ac (p)= a (p)], and for particles with distinct antiparticles for which ac (p) = a (p). 6 5We recall here that in the symmetrization postulate we have associated bosons with symmetric physical states and fermions with antisymmetric physical states. We had not done any association between bosons and fermions with the spin of the particles. 6By purely neutral we mean that the particle carries no conserved quantum numbers. 158 CHAPTER 5. CAUSAL SCALAR FIELDS FOR MASSIVE PARTICLES

We note for future purposes, that the commutator of the complex ﬁeld φ (x) with its adjoint is

+ c+ + c+ φ (x) , φ† (y) = φ (x)+ φ † (x) , φ † (y)+ φ (y) − − h i h + + c+ c+ i = φ (x) , φ † (y) + φ † (x) , φ (y) − − h + + i h c+ c+ i = φ (x) , φ † (y) φ (y) , φ † (x) − h i− h i− φ (x) , φ† (y) = ∆ (x y) ∆ (y x) + − − + − h i− where we have used Eq. (5.8). The commutator φ (x) , φ† (y) then becomes

φ (x) , φ† (y) = ∆ (x y) − − h i 3 d p ip (x y) ip (x y) ∆ (x y) ∆+ (x y) ∆+ (y x)= e · − e− · − (5.32) − ≡ − − − 2p0 (2π)3 − Z h i

2 this is a general relation that is for any kind of separation between x and y. We already saw that for x > 0, ∆+ (x) is an even function of x. Hence, the function ∆ (x y) deﬁned by Eq. (5.32) is null for space-like separations of − x and y.

5.3 Scalar ﬁelds and discrete symmetries

We shall consider now the effect of the inversion symmetries on the field φ (x) defined in (5.31). To do this, we take the results obtained in section 3.4, in which we studied the transformation properties of the creation operators a† (p,σ,n) under C, P and T . Equation (3.28), page 122 gives us the effect of the space-inversion operator on the creation operators, it is clear that the effect on the annihilation operator is obtained by simply taking the adjoint of (3.28) and that the c c operators a † (p) and a (p) must obey identical rules of transformation, then

1 P a (p) P − = η∗a ( p) (5.33) c 1 c c − P a † (p) P − = η a † ( p) (5.34) − where η and ηc are the intrinsic parities of the particle and antiparticle respectively. We can then apply the result (5.33) to the annihilation ﬁeld (5.5) to ﬁnd

+ 1 1 3 1 1 ip x P φ (x) P − = d p P a (p) P − e · 3 0 (2π) Z 2p q 1 3 p1 i(p x p0x0) = d p η∗a ( p) e · − 3 0 − (2π) Z 2p

q p 0 0 η∗ 3 1 i( p ( x) p x ) = d p a ( p) e − · − − 3 0 − (2π) Z 2p q p 5.3. SCALAR FIELDS AND DISCRETE SYMMETRIES 159 and changing the variable of integration from p to p we have7 − + 1 η∗ 3 1 i[p ( x) p0x0] η∗ 3 1 i[(p,p0) ( x,x0)] P φ (x) P − = d p a (p) e · − − = d p a (p) e · − 3 0 3 0 (2π) Z 2p (2π) Z 2p

+ 1 q η∗ 3 p1 i[p ( x)] q 0 p P φ (x) P − = d p a (p) e · P ; x x,x 3 0 P ≡ − (2π) Z 2p + 1 + 0 P φ (x) P − = ηq∗φ ( x) ; px x,x P P ≡ − similarly, applying the result (5.34) to the charge-conjug ate of the creation ﬁeld (5.6) we ﬁnd

c+ 1 1 3 1 c 1 ip x P φ † (x) P − = d p P a † (p) P − e− · 3 2p0 (2π) Z h i q 1 3 p1 c c i[p x p0x0] = d p η a † ( p) e− · − 3 2p0 − (2π) Z h i q p and changing the variable of integration from p to p we have − c c c+ 1 η 3 a † (p) i[ p x p0x0] c c+ P φ † (x) P − = d p e− − · − = η φ † ( x) 3 0 P (2π) Z 2p q p + c+ in summary, the ﬁelds φ (x) and φ † (x) transform under space inversion according to the following prescription

+ 1 + P φ (x) P − = η∗φ ( x) (5.35) c+ 1 c c+ P 0 P φ † (x) P − = η φ † ( x) ; x = x,x (5.36) P P − It is worth emphasizing that when space-inversion is applied to the scalar ﬁeld given by

+ c+ φ (x)= φ (x)+ φ † (x) (5.37) we obtain a diﬀerent ﬁeld φP (x)

1 + c+ 1 + 1 c+ 1 φ (x) P φ (x) P − = P φ (x)+ φ † (x) P − = P φ (x) P − + P φ † (x) P − P ≡ 1 h + c c+ i φ (x) P φ (x) P − = η∗φ ( x)+ η φ † ( x) (5.38) P ≡ P P

Both ﬁelds φ (x) and φP (x) are separately causal. However, in general φ and φP† do not commute at space-like separations. In that case, the ﬁelds φ and φP† cannot appear in the same interaction. So the only way to preserve Lorentz invariance, parity conservation and the hermiticity of the interaction is by demanding that φP (x) be proportional to φ ( x). In turn, it means that we should be able to factorize the phases on the RHS of Eq. (5.38) P and hence that c η = η∗ (5.39) consequently, the intrinsic parity ηηc = η 2 of a state that contains a spinless particle and its antiparticle is even. | | Combining equations (5.38, 5.39) we then have

1 + c+ φ (x) P φ (x) P − = η∗ φ ( x)+ φ † ( x) P ≡ P P h i 7In the change of variable p →−p, the differential of volume changes the sign but at the same time the three intervals of integration (in cartesian coordinates) are also inverted. By inverting again the intervals of integration we reverse the sign again. Thus, the net result is that the differential of volume does not change the sign. 160 CHAPTER 5. CAUSAL SCALAR FIELDS FOR MASSIVE PARTICLES obtaining finally, 1 P φ (x) P − = η∗φ ( x) (5.40) P the results are also valid when the spinless particle coincides with its antiparticle, since in this case η = ηc, which combined with Eq. (5.39), implies that the intrinsic parity of such a particle must be real: η = 1. ± Now we deal with time-reversal. From Eq. (3.29), page 122, we have

1 T a (p) T − = ζ∗a ( p) c 1 c c − T a † (p) T − = ζ a † ( p) − and taking into account the antilinearity of T , we have

3 3 1 + 1 d p 1 ip x 1 d p T a (p) T − ip x T φ (x) T − = T a (p) e · T − = e− ·  3 2p0  3 2p0 Z (2π)  Z (2π) q p q p ζ∗ 3 1 i[ p x+p0x0] =  d p a ( p) e − · 3 0 − (2π) Z 2p q p with the change of variable p p, we obtain →− + 1 ζ∗ 3 1 i[p x+p0x0] ζ∗ 3 1 i[(p,p0) (x, x0)] T φ (x) T − = d p a (p) e · = d p a (p) e · − 3 0 3 0 (2π) Z 2p (2π) Z 2p

q ζ∗ 3 p1 i[p ( x)] q p = d p a (p) e · −P 3 0 (2π) Z 2p + 1 + T φ (x) T − = ζq∗φ ( x) p −P c+ and similarly for φ † (x). Thus we ﬁnd

+ 1 + T φ (x) T − = ζ∗φ ( x) (5.41) c+ 1 c c+ −P T φ † (x) T − = ζ φ † ( x) (5.42) −P applying time-reversal to the field φ (x) of Eq. (5.31), we obtain a new field φ (x) T φ (x) T 1. For this field to T ≡ − be simply related with the field φ at the time-reversed point x, we must have −P c ζ = ζ∗ (5.43) obtaining 1 T φ (x) T − = ζ∗φ ( x) (5.44) −P Charge-conjugation can be managed similarly. By using the results of Sec. 3.4, Eq. (3.27), page 122, we have

1 c Ca (p) C− = ξ∗a (p) (5.45) c 1 c Ca † (p) C− = ξ a† (p) (5.46) where ξ and ξc are the phases associated with the operation of charge conjugation on one-particle states. Using Eqs. (5.45, 5.46) in the ﬁeld expansion (5.5) and the charge-conjugate of (5.6) we ﬁnd

+ 1 c+ Cφ (x) C− = ξ∗φ (x) (5.47) c+ 1 c + Cφ † (x) C− = ξ φ † (x) (5.48) 5.3. SCALAR FIELDS AND DISCRETE SYMMETRIES 161

again the charge-conjugation operation applied on the ﬁeld φ (x) of Eq. (5.31) gives another ﬁeld φC (x) 1 ≡ Cφ (x) C− . In order that φC (x) commutes with φ† (x) at space-like separations, φC (x) must be proportional to φ† (x), for which it is necessary that c ξ = ξ∗ (5.49) just as for the ordinary parity, the intrinsic charge-conjugation parity ξξc of a state consisting of a spinless particle and its antiparticle is even. We just have

1 Cφ (x) C− = ξ∗φ† (x) (5.50) again, these results also apply to the case in which the particle is its own antiparticle, so ξc = ξ. In that case, the charge-conjugation parity like the ordinary parity must be real ξ = 1. ± Finally, we point out that the results shown in this section are valid only if the discrete symmetry involved is a good symmetry of the system. Chapter 6

Causal vector ﬁelds for massive particles

The next simplest case arise when we deal with ﬁelds that transform as a four-vector, which is associated with the so-called vector representation of the homogeneous Lorentz group. This is the four-dimensional representation with which we constructed originally the Lorentz transformations in the Minkowski space, that is

D (Λ)µ Λµ (6.1) ν ≡ ν 0 It is important to say that there are massive particles W ± and Z that at low energies are described by vector ﬁelds, these particles are responsible for what we call the nuclear weak interaction. Further, one possible approach to quantum electrodynamics (QED) is to describe the photon in terms of a massive vector ﬁeld in the limit of very small mass.

6.1 Vector ﬁelds without internal symmetries

By now, we shall assume that only one type of particle is described by this field, hence we drop the label n of species. We shall consider later the possibility that the field describes both a particle and a distinct antiparticle. The components of the annihilation and creation fields are obtained simply by taking (4.39, 4.40) by dropping the label n. Further, we shall substitute the label k (component label) for a label µ (more usual for a four-vector at the Minkowski space)

+µ 3/2 3 µ ip x φ (x) = (2π)− d p u (p,σ) a (p,σ) e · (6.2) σ X Z µ 3/2 3 µ ip x φ− (x) = (2π)− d p v (p,σ) a† (p,σ) e− · (6.3) σ X Z In addition, the coeﬃcient functions uµ (p,σ) and vµ (p,σ) for arbitrary momentum can be obtained from those of zero momentum from Eqs. (4.45) and (4.46)

m m uµ (p,σ)= D (L (p))µ uν (0,σ) ; vµ (p,σ)= D (L (p))µ vν (0,σ) (6.4) p0 ν p0 ν ν ν r X r X and using the convention of sum over repeated lower-upper indices as well as Eq. (6.1) we ﬁnd

m uµ (p,σ) = L (p)µ uν (0,σ) (6.5) p0 ν r m vµ (p,σ) = L (p)µ vν (0,σ) (6.6) p0 ν r 162 6.1. VECTOR FIELDS WITHOUT INTERNAL SYMMETRIES 163 in turn, the coeﬃcient functions at zero momentum are subject to the conditions (4.49) and (4.50)

uµ (0, σ¯) J(j) = µ uν (0,σ) (6.7) σσ¯ J ν σ¯ X µ (j) µ ν v (0, σ¯) J ∗ = v (0,σ) (6.8) σσ¯ −J ν σ¯ X the rotation generators µ in the four-vector representation (in cartesian coordinates) are given by [homework] J ν ( )µ = ( )0 = 0 (6.9) Jk 0 Jk ν ( )i = iε ; i, j, k = 1, 2, 3 (6.10) Jk j − ijk it is clear from Eqs. (6.9) that the rotation generators act only on the three-dimensional space-coordinates. In particular, we can calculate 2 J 2 2 2 2 = 1 + 2 + 3 Ji J J J ⇒ 2 = ( )i ( )ν + ( )i ( )ν + ( )i ( )ν J j J1 ν J1 j J2 ν J2 j J3 ν J3 j = ( )i ( )m + ( )i ( )m + ( )i ( )m J1 m J1 j J2 m J2 j J3 m J3 j = ε ε ε ε ε ε − im1 mj1 − im2 mj2 − im3 mj3 = ε ε ε ε ε ε ε ε ε ε ε ε − i21 2j1 − i31 3j1 − i12 1j2 − i32 3j2 − i13 1j3 − i23 2j3 = δ δ ε ε δ δ ε ε δ δ ε ε − i3 j3 321 231 − i2 j2 231 321 − i3 j3 312 132 δ δ ε ε δ δ ε ε δ δ ε ε − i1 j1 132 312 − i2 j2 213 123 − i1 j1 123 213 the product of Levi-Civit´aterms are of the form ε ε which are equal to ( 1) because both terms have opposite ijk jik − sign (they diﬀer each other by an interchange), then we have

i 2 = δ δ + δ δ + δ δ + δ δ + δ δ + δ δ J j i3 j3 i2 j2 i3 j3 i1 j1 i2 j2 i1 j1 i m i = 2 [δi1δj1 + δi2δj2 + δi3δj3] = 2δ mδ j = 2δ j the other terms are easier to calculate

0 2 = ( )0 ( )ν + ( )0 ( )ν + ( )0 ( )ν = 0 J µ J1 ν J1 µ J2 ν J2 µ J3 ν J3 µ we then obtain

0 µ 2 = 2 = 0 (6.11) J µ J 0 2 i i j = 2δ j (6.12) J alternatively, we can obtain Eq. (6.12) by observing that 2 is a casimir of the group SO (3), in the three- J dimensional coordinate space associated with the j = 1 irreducible representation1. Thus, according with the Schur’s lemma, it must be proportional to the identity within the three-dimensional coordinate space, and specif- ically of the form j (j + 1) I = 2I. Setting µ = 0 in Eq. (6.7) and using (6.9) we obtain

u0 (0, σ¯) J(j) = 0 uν (0,σ) = 0 (6.13) σσ¯ J ν σ¯ X 1Note however that J 2 is not a Casimir of the Lorentz group, because it does not commute with all generators of such a group. Hence, J 2 is not in general proportional to the identity at the Minkowski space. 164 CHAPTER 6. CAUSAL VECTOR FIELDS FOR MASSIVE PARTICLES

2 it is more convenient to obtain an expression in terms of J(j) , because it is a Casimir operator of SO (3) and (j) thus proportional to the identity. To do this, we multiply Eq. (6.13) by Jσβ and sum over σ on both sides of this equation, then we have 2 0 (j) (j) 0 (j) u (0, σ¯) Jσσ¯ Jσβ = 0 u (0, σ¯) J = 0 (6.14) ⇒ σβ¯ σ σ¯ σ¯ X X X now setting µ = i in Eq. (6.7) and using (6.9) we have

ui (0, σ¯) J(j) = i uν (0,σ) σσ¯ J ν σ¯ X ui (0, σ¯) J(j) = i um (0,σ) σσ¯ J m σ¯ X (j) as before, we multiply by Jσβ and sum over σ on both sides of this equation, then we use equation (6.7) once again, to obtain ui (0, σ¯) J(j)J(j) = i um (0,σ) J(j) σσ¯ σβ J m σβ σ σ¯ σ X X 2 X i (j) i m ν u (0, σ¯) J = m νu (0, β) σβ¯ J J σ¯ X and using Eqs. (6.12) we find 2 i (j) i m n 2 i n i n u (0, σ¯) J = m nu (0, β)= nu (0, β) = 2δ nu (0, β) (6.15) σβ¯ J J J σ¯ X thus, picking up Eqs. (6.14, 6.15) we obtain finally 2 u0 (0, σ¯) J(j) = 0 (6.16) σσ¯ σ¯ X 2 ui (0, σ¯) J(j) = 2ui (0,σ) (6.17) σσ¯ σ¯ X an analogous procedure can be done for the vµ coefficient from Eq. (6.8) to obtain 2 0 (j) v (0, σ¯) J ∗ = 0 (6.18) σσ¯ σ¯ X 2 i (j) i v (0, σ¯) J ∗ = 2v (0,σ) (6.19) σσ¯ σ¯ X (j) 2 now we recall that J σσ¯ = j (j + 1) δσσ¯ then Eqs. (6.16-6.19) become j (j + 1) u0 (0,σ) = 0 (6.20) j (j + 1) ui (0,σ) = 2ui (0,σ) (6.21) j (j + 1) v0 (0,σ) = 0 (6.22) j (j + 1) vi (0,σ) = 2vi (0,σ) (6.23) a non-trivial solution requires that u0 (0,σ) and/or ui (0,σ) be non-null, and same for v0 (0,σ) and vi (0,σ). Equations (6.20, 6.22) show that u0 (0,σ) and v0 (0,σ) can be non-null only if j = 0. In that case, Eqs. (6.21, 6.23) say that ui (0,σ) and vi (0,σ) must be zero. On the other hand, we see that Eqs. (6.21, 6.23) are consistent for non-null values of ui (0,σ) and vi (0,σ) only if j (j + 1) = 2 or equivalently j = 1. In that case Eqs. (6.20, 6.22) show that u0 (0,σ) and v0 (0,σ) must be null. In conclusion consistent solutions for the coefficients at p = 0 can only be obtained in two cases 6.2. SPIN ZERO VECTOR FIELDS 165

1. If j = 0, the representation is one-dimensional and hence σ acquires a single value, therefore we omit the label σ. In this case, ui (0) = vi (0) = 0, while u0 (0) and v0 (0) could be non-zero.

2. If j = 1, the representation is three-dimensional and σ acquires three-diﬀerent values. In this case, u0 (0) = v0 (0) = 0, while the ui (0) and vi (0) could be non-zero.

In other words, we have two possibilities for the spin of the particle described by the vector ﬁeld: j = 0 or j = 1. Let us now examine both possibilities in detail.

6.2 Spin zero vector ﬁelds

We shall study ﬁrst the case of j = 0, in which

ui (0) = vi (0) = 0

By an appropriate normalization of the non-null components of uµ (0) and vµ (0), we can take the non-vanishing components of these coeﬃcients to have the values (recall that σ takes a single value so we drop it)

m m u0 (0) i ; v0 (0) = i (6.24) ≡ 2 − 2 r r 0 0 where we choose u (0) = v ∗ (0) for the annihilation and creation ﬁelds to be the adjoint of each other, the convenience of introducing the “i” factor in the choice of phase (6.24) will be clear later. The coeﬃcient uµ (p) for p = 0 can be obtained from Eq. (6.5) so that 6 m m m uµ (p) = L (p)µ uν (0)= L (p)µ i δν p0 ν p0 ν 2 0 r r r 1 uµ (p) = im L (p)µ (6.25) 2p0 0 r similarly we obtain 1 vµ (p)= im L (p)µ (6.26) − 2p0 0 r we recall that L (p) is the “standard boost” that carries the “standard” four momentum kµ = (0, 0, 0,m) for a massive particle to the four-momentum pµ. Such a boost is given by Eq. (1.192) page 43

Li (p) = δ + (γ 1) p p k ik − i k i 0 2 0 L 0 (p) = L i (p)= pi γ 1 ; L 0 = γ b b − p p2 + m2 pi pi/ p ,b γ (6.27) ≡ | | ≡ p m then we have b p p p2 + m2 p p2 pi Li (p) = i γ2 1= i 1= i = 0 p − p m2 − p m2 m | | | |r | |r p 2 p2 + m2 (p0) p0 L0 = γ = = = 0 q p m m m in the last step we have used the fact that p0 > 0. We finally obtain pµ Lµ = (6.28) 0 m 166 CHAPTER 6. CAUSAL VECTOR FIELDS FOR MASSIVE PARTICLES substituting (6.28) in (6.25) we find 1 uµ (p)= ipµ (6.29) 2p0 r similarly, from Eq. (6.26) we obtain 1 vµ (p)= ipµ (6.30) − 2p0 r substituting (6.29, 6.30) in the expressions (6.2, 6.3) for the annihilation and creation fields (in which there is no sum over σ) we find

+µ 3/2 3 µ 1 ip x 3/2 3 1 µ ip x φ (x) = (2π)− d p (ip ) a (p) e · = (2π)− d p a (p) ∂ e · (6.31) 2p0 2p0 Z r Z r +µ µ 3/2 3 1 ip x φ (x) = ∂ (2π)− d p a (p) e · (6.32) 2p0 Z r Similarly we obtain2

µ µ 3/2 3 1 ip x φ− (x)= ∂ (2π)− d p a† (p) e− · 2p0 Z r Taking into account Eqs. (5.5, 5.6) we obtain

+µ µ + µ µ φ (x)= ∂ φ (x) ; φ− (x)= ∂ φ− (x) (6.33) where φ± (x) are the scalar annihilation and creation fields that we obtained for a spinless particle. So the vector annihilation and creation fields here are nothing but the derivatives of the scalar annihilation and creation fields for a spinless particle. It is immediate that the causal vector field for a spinless particle is also simply the derivative of the causal scalar field µ +µ µ µ φ (x)= φ (x)+ φ− (x)= ∂ φ (x) (6.34)

6.3 Spin one vector ﬁelds

We have already seen that for the case j = 1, we have [see Eqs. (6.20, 6.22)]

u0 (0,σ)= v0 (0,σ) = 0 (6.35)

Before continuing, it is important to take into account that in Eqs. (6.7, 6.8) the matrix representations of the generators on the LHS correspond to the irreducible representation j in the canonical basis j, σ , while the | i RHS provides the matrix representations of the generators in the cartesian basis described by Eqs. (6.9, 6.10). Therefore, for our present work we need besides the matrix representations of the generators in the cartesian basis Eqs. (6.9, 6.10), the matrix representations for the generators associated with the irreducible representation j = 1 in the canonical basis. Ordering the canonical basis as

j = 1,σ = 1 , j = 1,σ = 0 , j = 1,σ = 1 | i | i | − i

2At this step we understand the introduction of the “i” factor in the choice of phase (6.24). 6.3. SPIN ONE VECTOR FIELDS 167 the matrix representation of the generators are given by 0 1 0 0 i 0 1 1 − (J )(j=1) = 1 0 1 ; (J )(j=1) = i 0 i 1 √   2 √  −  2 0 1 0 2 0 i 0 1 0 0  1 0 0  (j=1) (J )(j=1) = 0 0 0 ; J2 = 2 0 1 0 3     0 0 1 0 0 1 −  0 √2 0  0 0 0 (j=1) (j=1) (J+) = 0 0 √2 ; (J ) = √2 0 0 (6.36)   −   0 0 0 0 √2 0     By setting σ = 0 in Eq. (6.7), and using Eq. (6.9) and (6.35), we obtain

uµ (0, σ¯) (J )(j) = ( )µ uν (0,σ) k σσ¯ Jk ν σ¯ X uµ (0, σ¯) (J )(j) = ( )µ uν (0, 0) k σ¯0 Jk ν σ¯ X ( )i un (0, 0) = ui (0, σ¯) (J )(j) Jk n k σ¯0 σ¯ X using (6.10), and taking into account thatσ ¯ = 1, 0, 1; we have − n i (j=1) i (j=1) i (j=1) iεink u (0, 0) = u (0, +1) (Jk)1,0 + u (0, 0) (Jk)0,0 + u (0, 1) (Jk) 1,0 (6.37) − − − it is convenient to redeﬁneσ ¯ as a label of matrix elements, hence 1, 0, 1 1, 2, 3 − → iε un (0, 0) = ui (0, +1) (J )(j=1) + ui (0, 0) (J )(j=1) + ui (0, 1) (J )(j=1) (6.38) ikn k 1,2 k 2,2 − k 3,2 using (6.36) and setting k = 3, equation (6.38) becomes

n i (j=1) i (j=1) i (j=1) iεi3n u (0, 0) = u (0, +1) (J3)1,2 + u (0, 0) (J3)2,2 + u (0, 1) (J3)3,2 n − iεi3n u (0, 0) = 0 (6.39) for i = 1, 2 in (6.39) we ﬁnd iε u2 (0, 0) = 0 ; iε u1 (0, 0) = 0 132 231 ⇒ u1 (0, 0) = u2 (0, 0) = 0 (6.40) for i = 3 Eq. (6.39) gives no information. In addition for n = 3 in Eq. (6.39) we have no information either. Therefore, u3 (0, 0) is arbitrary. On the other hand, Eqs. (6.35, 6.40) say that uµ (0, 0) = 0 for µ = 3. Ordering 6 the components as 1, 2, 3, 0; we shall normalize the ﬁelds so that the vector uµ (0, 0) takes the value 0 1 0 uµ (0, 0) =   (6.41) √2m 1  0      now setting k = 1 in Eq. (6.38)

iε un (0, 0) = ui (0, +1) (J )(j=1) + ui (0, 0) (J )(j=1) + ui (0, 1) (J )(j=1) i1n 1 1,2 1 2,2 − 1 3,2 i i n u (0, +1) + u (0, 1) iεi1n u (0, 0) = − (6.42) √2 168 CHAPTER 6. CAUSAL VECTOR FIELDS FOR MASSIVE PARTICLES taking i = 1 in (6.42) we have

0= u1 (0, +1) + u1 (0, 1) (6.43) − with i = 2 in (6.42) and taking into account the normalization (6.41), we ﬁnd

i√2ε u3 (0, 0) = u2 (0, +1) + u2 (0, 1) 213 − 1 i√2 = u2 (0, +1) + u2 (0, 1) − √2m − i = u2 (0, +1) + u2 (0, 1) (6.44) −√m − setting i = 3 in (6.42), and using (6.40) we obtain

3 3 2 u (0, +1) + u (0, 1) iε312 u (0, 0) = − √2 0 = u3 (0, +1) + u3 (0, 1) (6.45) −

Finally, by setting k = 2 in Eq. (6.38), and using (6.36) we ﬁnd

iε un (0, 0) = ui (0, +1) (J )(j=1) + ui (0, 0) (J )(j=1) + ui (0, 1) (J )(j=1) i2n 2 1,2 2 2,2 − 2 3,2 i i n i u (0, +1) + i u (0, 1) iεi2n u (0, 0) = − − √2 √2ε un (0, 0) = ui (0, 1) ui (0, +1) (6.46) i2n − − and for i = 1 in Eq. (6.46) and using the normalization (6.41), we get

√2ε u3 (0, 0) = u1 (0, 1) u1 (0, +1) 123 − − 1 √2 = u1 (0, 1) u1 (0, +1) √2m − − 1 = u1 (0, 1) u1 (0, +1) (6.47) √m − − with i = 2 in Eq. (6.46)

0= u2 (0, 1) u2 (0, +1) (6.48) − − for i = 3 in Eq. (6.46)

√2ε u1 (0, 0) = u3 (0, 1) u3 (0, +1) 321 − − 0 = u3 (0, 1) u3 (0, +1) (6.49) − − 6.3. SPIN ONE VECTOR FIELDS 169 picking up all relations obtained so far [Eqs. (6.35, 6.41, 6.43, 6.44, 6.45, 6.47, 6.48, 6.49)] we write

u0 (0,σ) = 0 (6.50) 0 1 0 uµ (0, 0) =   (6.51) √2m 1  0    u1 (0, +1) + u1 (0, 1) = 0   (6.52) − i u2 (0, +1) + u2 (0, 1) = (6.53) − −√m u3 (0, +1) + u3 (0, 1) = 0 (6.54) − 1 u1 (0, 1) u1 (0, +1) = (6.55) − − √m u2 (0, 1) u2 (0, +1) = 0 (6.56) − − u3 (0, 1) u3 (0, +1) = 0 (6.57) − − where we have ordered the components as 1, 2, 3, 0. These relations are enough to determine completely the vectors uµ (0, 0) , uµ (0, +1) and uµ (0, 1), so we have obtained all vectors3 of the form uµ (0,σ). For example, − by substracting equations (6.52, 6.55) and substracting equations (6.53, 6.56) we obtain

1 i 2u1 (0, +1) = ; 2u2 (0, +1) = (6.58) −√m −√m further by substracting equations (6.54, 6.57) and taking Eqs. (6.50) we ﬁnd

2u3 (0, +1) = 0 ; u0 (0, +1) = 0 (6.59) from Eqs. (6.58, 6.59) we have the complete vector uµ (0, +1). Ordering the components as 1, 2, 3, 0; it becomes

1 1 1 +i uµ (0, +1) =   −2 √m 0  0      by adding the equations that we substract previously we obtain the complete vector uµ (0, 1). Ordering the − components as 1, 2, 3, 0; we obtain 1 1 1 i uµ (0, 1) =  −  − 2 √m 0  0      A totally similar exercise can be done with the coeﬃcients vk from Eqs. (6.8) to obtain all vectors of the form vµ (0,σ). In summary, by setting σ = 0 in Eqs. (6.7, 6.8) and examining the behavior of J3 we ﬁnd

0 i 3 3 u (0, 0) = 0 ; u (0, 0) = u (0, 0) δ kuk (6.60) 0 i 3 3 v (0, 0) = 0 ; v (0, 0) = v (0, 0) δ kuk (6.61)

3Notice however that we can determine such vectors only after ﬁxing the value of u3 (0, 0). 170 CHAPTER 6. CAUSAL VECTOR FIELDS FOR MASSIVE PARTICLES so that the vectors uµ (0, 0) and vµ (0, 0) for σ = 0 are in the three-direction. A suitable normalization of the ﬁelds, permits us to take these vectors as

0 1 0 uµ (0, 0) = vµ (0, 0) =   (6.62) √2m 1  0      where the four components are ordered as 1, 2, 3, 0. Equations (6.7, 6.8) [or equivalently Eqs. (6.50)-(6.57)] also provide the values of uµ (0, 1)4. We then obtain ± 1 1 1 +i uµ (0, +1) = vµ (0, 1) =   − − −√2 √2m 0  0    1  1 1 i uµ (0, 1) = vµ (0, +1) =  −  (6.63) − − √2 √2m 0  0      and applying Eq. (6.5) we obtain

2m 1 uµ (p,σ) = L (p)µ uν (0,σ)= L (p)µ √2m uν (0,σ) 2p0 ν 2p0 ν r r h i 1 uµ (p,σ) = L (p)µ eν (0,σ) ; eν (0,σ) √2m uν (0,σ) 2p0 ν ≡ r h i a similar exercise can be done from Eq. (6.6) for vk, we ﬁnally obtain

µ µ 0 1/2 µ u (p,σ) = v ∗ (p,σ)= 2p − e (p,σ) (6.64) eµ (p,σ) Lµ (p) eν (0,σ) ; eν (0,σ) √2m uν (0,σ) (6.65) ≡ ν ≡ from the deﬁnition (6.65)5 and expressions (6.62, 6.63) we obtain the explicit form of eµ (0,σ)

0 1 1 0 1 +i 1 i eµ (0, 0) =   , eµ (0, +1) =   , eµ (0, 1) =  −  (6.66) 1 −√2 0 − √2 0  0   0   0              substituting (6.64) into (6.2, 6.3) we obtain the annihilation and creation ﬁelds

3 +µ µ 1 d p µ ip x φ (x)= φ− † (x)= e (p,σ) a (p,σ) e · (6.67) 3 0 (2π) σ 2p X Z q p 4 (1) (1) Alternatively, the other components can be found by calculating the effect of the raising and lowering operators J1 ± J2 on u and v. 5Note that definition (6.65) simply establishes that the vector eµ at an arbitrary three-momentum p is connected to eµ at zero three-momentum, by means of the standard boost that passes from zero three-momentum to the three-momentum p. 6.3. SPIN ONE VECTOR FIELDS 171 it is clear that the fields φ+µ (x) and φ+ν (y) commute or anticommute for all x and y. However, φ+µ (x) and ν φ− (y) do not. Their commutator for bosons or anticommutator for fermions yields

3 3 +µ ν 1 d p µ ip x d p′ ν ip′ y φ (x) , φ− (y) = e (p,σ) a (p,σ) e · , e ∗ p′, σ¯ a† p′, σ¯ e− · (2π)3 0 0 ∓ " σ Z 2p σ¯ Z 2p′ # X X ∓ 3 3 1 dpp µ ip x d p′ ν p ip′ y = e (p,σ) e · e ∗ p′, σ¯ e− · a (p,σ) , a† p′, σ¯ (2π)3 2p0 2p 0 σ Z σ¯ Z ′ h i∓ X 3 X 3 1 pd p µ ip x pd p′ ν ip′ y = e (p,σ) e · e ∗ p′, σ¯ e− · δ p p′ δσσ¯ (2π)3 0 0 − σ Z 2p σ¯ Z 2p′ X 3 X +µ ν 1 d ppip (x y) µ ν p φ (x) , φ− (y) = e · − e (p,σ) e ∗ (p,σ) (2π)3 2p0 ∓ Z σ X we write this commutation or anticommutation relation as

3 +µ ν d p ip (x y) µν φ (x) , φ− (y) = e · − Π (p) (6.68) (2π)3 2p0 ∓ Z µν µ ν Π (p) e (p,σ) e ∗ (p,σ) (6.69) ≡ σ X based on Eq. (6.66) we can obtain Πµν (0). For example we have

3 12 1 2 1 2 1 2 1 2 Π (0) = e (0,σ) e ∗ (0,σ)= e (0, 0) e ∗ (0, 0) + e (0, +1) e ∗ (0, +1) + e (0, 1) e ∗ (0, 1) − − σ=1 X 0 ∗ 12 0 1 1 i Π (0) = √2 √2  √2  − − i √  − 2  1 1 ∗  1 1 ∗ Π12 (0) = 0 0+ i + i = 0 · −√2 · −√2 √2 · −√2 with that procedure for each component we obtain

100 0 010 0 Πµν (0)=   (6.70) 001 0  000 0      applied on an arbitrary four vector, we have

1 00 0 x1 x1 0 10 0 x2 x2 Πµν (0) xν =     =   0 01 0 x3 x3  0 00 0   x0   0        Πµν (0) xν = (x, 0)      we can see that Πµν (0) is the projection matrix on the space orthogonal to the time direction. By using Eqs. (6.65, 6.66, 6.69) we can obtain Πµν (p) 172 CHAPTER 6. CAUSAL VECTOR FIELDS FOR MASSIVE PARTICLES

µν µ ν µ ρ ν β ∗ Π (p) e (p,σ) e ∗ (p,σ) [L (p) e (0,σ)] L (p) e (0,σ) ≡ ≡ ρ β σ σ X X h i µ ρ ν β µ ρ ν β = [L ρ (p) e (0, +1)] L β (p) e ∗ (0, +1) + [L ρ (p) e (0, 0)] L β (p) e ∗ (0, 0)

µ ρ h ν β i h i + [L (p) e (0, 1)] L (p) e ∗ (0, 1) ρ − β − µ 1 h µ 2 i ν 1 ν 2 = L 1 (p) e (0, +1) + L 2 (p) e (0, +1) L 1 (p) e ∗ (0, +1) + L 2 (p) e ∗ (0, +1) µ 3 ν 3 + L 3 (p) e (0, 0) L 3 (p) e ∗ (0, 0) µ 1 µ 2 ν 1 ν 2 + L (p) e (0, 1) + L (p) e (0, 1) L (p) e ∗ (0, 1) + L (p) e ∗ (0, 1) 1 − 2 − 1 − 2 − 1 i 1 i = Lµ (p) Lµ (p) Lν (p)+ Lν (p) + Lµ (p) Lν (p) −√ 1 − √ 2 −√ 1 √ 2 3 3 2 2 2 2 1 µ i µ 1 ν i ν + L 1 (p) L 2 (p) L 1 (p)+ L 2 (p) √2 − √2 √2 √2

1 Πµν (p) = [Lµ (p)+ iLµ (p)] [ Lν (p)+ iLν (p)] + Lµ (p) Lν (p) −2 1 2 − 1 2 3 3 1 + [Lµ (p) iLµ (p)] [Lν (p)+ iLν (p)] 2 1 − 2 1 2 1 i i 1 = Lµ (p) Lν (p) Lµ (p) Lν (p)+ Lµ (p) Lν (p)+ Lµ (p) Lν (p)+ Lµ (p) Lν (p) 2 1 1 − 2 1 2 2 2 1 2 2 2 3 3 1 i i 1 + Lµ (p) Lν (p)+ Lµ (p) Lν (p) Lµ (p) Lν (p)+ Lµ (p) Lν (p) 2 1 1 2 1 2 − 2 2 1 2 2 2 µν µ ν µ ν µ ν Π (p) = L 1 (p) L 1 (p)+ L 2 (p) L 2 (p)+ L 3 (p) L 3 (p) (6.71) and applying (6.27) in Eq. (6.71) we have

ij i j i j i j Π (p) = L 1 (p) L 1 (p)+ L 2 (p) L 2 (p)+ L 3 (p) L 3 (p) = [δ + (γ 1) p p ] [δ + (γ 1) p p ] + [δ + (γ 1) p p ] [δ + (γ 1) p p ] i1 − i 1 j1 − j 1 i2 − i 2 j2 − j 2 + [δ + (γ 1) p p ] [δ + (γ 1) p p ] i3 − i 3 j3 − j 3 = δ δ + (γ 1)b bδ p p + (γ 1)bp bp δ + (γ 1)2 p pbpb2 b b i1 j1 − i1 j 1 − i 1 j1 − i j 1 +δ δ + (γ 1)b δb p p + (γ 1) bp pb δ + (γ 1)2 p p p2 i2 j2 − i2 j 2 − i 2 j2 − i j 2 b b b b b2 b b 2 +δi3δj3 + (γ 1) δi3pjp3 + (γ 1) pip3δj3 + (γ 1) pipjp3 − b b − b b − b b b Πij (p) = δi δk + (γ 1)bδkbp p + (γ b1)b p p δk + (γ b1)b2 pb p p pk k j − i j k − i k j − i j k = δi + (γ 1) p p + (γ 1) p p + (γ 1)2 p p p pk j − j i − i j − i j k ij b b kb b ij pbipbj b b = g + (γ 1) pipj 2 + (γ 1) pkp = g + (γ 1) [2+(γ 1)] − b b − b b b b b−b p2 − h i k where we have used the fact that pk is normalizedb b so that bpkbp = 1. p p p p Πij (p) = gij + (γ 1) i j (γ +1) = gij + i j γ2 1 b − p2 b b p2 − p p p2 + m2 p p p2 = gij + i j 1 = gij + i j p2 m2 − p2 m2 p p Πij (p) = gij + i j (6.72) m2 6.3. SPIN ONE VECTOR FIELDS 173

From Eqs. (6.28, ,6.27, 6.71) we also have

i0 i 0 i 0 i 0 Π (p) = L 1 (p) L 1 (p)+ L 2 (p) L 2 (p)+ L 3 (p) L 3 (p) p p p = [δ + (γ 1) p p ] 1 + [δ + (γ 1) p p ] 2 + [δ + (γ 1) p p ] 3 i1 − i 1 m i2 − i 2 m i3 − i 3 m p p p (γ 1) = δ 1 + δ 2 + δ 3 + − p [p p + p p + p p ] i1 m i2 mb b i3 m m i b1b1 2 2 3 3 b b b b b b p (γ 1) p (γ 1) p (γ 1) p Πi0 (p) = δi k + − p p pk = i + − p p pk p = i + − i p k m m i k m m i k | | m m p | | | | pi (γ 1) γ p0 = + − pi =b b pi = pi b b b m m m m2 pip0 Πi0 (p) = Π0i (p)= (6.73) m2

ﬁnally, from Eqs. (6.28, 6.71) we have

00 0 0 0 0 0 0 Π (p) = L 1 (p) L 1 (p)+ L 2 (p) L 2 (p)+ L 3 (p) L 3 (p) p p p p p p p2 p2 + m2 p0p0 = 1 1 + 2 2 + 3 3 = = 1= 1+ m m m m m m m2 m2 − − m2 p0p0 Π00 (p) = g00 + (6.74) m2 picking up Eqs. (6.72, 6.73, 6.74) we obtain

pµpν Πµν (p)= gµν + (6.75) m2 In addition Eqs. (6.65, 6.75) shows that Πµν (p) is the projection matrix on the space orthogonal to the four- vector pµ

pµp pν m2pν Πµν (p) p = gµν p + µ = pν µ µ m2 − m2 µν Π (p) pµ = 0 in a similar way we can show that the four-vectors eµ (p,σ) are orthogonal to pµ. To see it, we use Eqs. (6.65, 6.66). Using these equations for σ = 0 we obtain

µ µ ν e (p,σ) pµ L ν (p) e (0,σ) pµ µ ≡ µ ν µ 3 µ 0 i e (p, 0) pµ L ν (p) e (0, 0) pµ = L 3 (p) e (0, 0) pµ = L 3 (p) pµ = L 3 (p) p0 + L 3 (p) pi ≡ p p = 3 p + δi + (γ 1) pip p = 3 p0 + p + (γ 1) pip p m 0 3 − 3 i − m 3 − i 3 0 0 p3 0 p i p3 0 p = p + p3 + 1b bp pi p p3 = p + p3 + b b1 p3 − m m − | | − m m − eµ (p, 0) p = 0 µ b b b and we can proceed similarly for σ = 1. Then we obtain ± µ e (p,σ) pµ = 0 (6.76) 174 CHAPTER 6. CAUSAL VECTOR FIELDS FOR MASSIVE PARTICLES

On the other hand, the commutator or anticommutator (6.68) can be written in terms of the function ∆+ deﬁned in Eq. (5.9), as

3 3 µ ν +µ ν d p ip (x y) µν d p ip (x y) µν p p φ (x) , φ− (y) = e · − Π (p)= e · − g + (2π)3 2p0 (2π)3 2p0 m2 ∓ Z Z 3 3 µν d p ip (x y) 1 d p µ ν ip (x y) = g e · − + p p e · − (2π)3 2p0 m2 (2π)3 2p0 Z Z 3 3 µν d p ip (x y) 1 d p µ ν ip (x y) = g e · − ∂ ∂ e · − (2π)3 2p0 − m2 (2π)3 2p0 Z Z µ ν +µ ν µν ∂ ∂ φ (x) , φ− (y) = g ∆ (x y) (6.77) − m2 + − ∓ by now, what really matters is that this expression does not vanish for x y space-like and is even in x y. Hence, − − we can repeat the reasoning of section 5 to construct a causal ﬁeld: we start by forming a linear combination of annihilation and creation ﬁelds µ +µ µ v (x) κφ (x)+ λφ− (x) (6.78) ≡ and for space-like separations of x and y, we obtain

µ ν µ ν µν ∂ ∂ [v (x) , v (y)] = κλ [1 1] g 2 ∆+ (x y) ∓ ∓ − m − µ ν µ ν 2 2 µν ∂ ∂ v (x) , v † (y) κ λ g ∆ (x y) ≡ | | ∓ | | − m2 + − h i∓ once again, for both commutators or anticommutators to vanish at space-like separations x y, it is necessary − and sufficient that the spin one particles be bosons and that κ = λ . By a suitable choice of the phase of the | | | | one-particle states we can settle κ = λ, and drop the common factor κ by redefining the overall normalization of the field. With those facts, the causal vector field (6.78) for a massive particle of spin one yields

µ +µ +µ v (x)= φ (x)+ φ † (x) (6.79) note that vµ (x) is real µ µ v (x)= v † (x) (6.80)

6.4 Spin one vector ﬁelds with internal symmetries

The previous framework is suitable for totally neutral spin one particles. If the particles described carry a non-zero value of some conserved quantum number Q, we cannot construct an interaction that conserves Q out of such a ﬁeld. Hence, we must suppose that there is another boson of the same mass and spin, which carries an opposite value of Q. We then construct the causal ﬁeld in the form

µ +µ c+µ v (x)= φ (x)+ φ † (x) (6.81) which can be expanded as

3 µ 1 d p µ ip x µ c ip x v (x)= e (p,σ) a (p,σ) e · + e ∗ (p,σ) a † (p,σ) e− · (6.82) 3 0 (2π) σ 2p X Z h i q p where the superscript c indicates operators that create the antiparticle that is charge-conjugate to the particle annihilated by φ+µ (x). This is a causal field but not real (hermitian) anymore. Once again, we can use Eq. (6.82) 6.4. SPIN ONE VECTOR FIELDS WITH INTERNAL SYMMETRIES 175 for the case of a purely neutral spin one particle that it is its own antiparticle. In that case we should take simply ac (p)= a (p) . In either case, the commutator of a vector field with its adjoint gives µ ν µ ν µν ∂ ∂ v (x) , v † (y) = g ∆ (x y) (6.83) − m2 − h i where ∆(x y) is the function defined by Eq. (5.32). − 6.4.1 Field equations for spin one particles The real and complex fields we have constructed obey some interesting field equations. For example, since pµ in the exponential of Eq. (6.82) represents a physical four-momentum, we have p2 = m2. Thus, v (x) are the − µ components of a field with definite mass. Therefore, according with the discussion given at section 4.6, the field vµ (x) must satisfy the Klein-Gordon equation (4.72) page 150 m2 vµ (x) = 0 (6.84) − as it also happens for scalar fields. On the other hand, we have already seen that [see Eq. (6.76)] µ e (p,σ) pµ = 0 (6.85) µ therefore by applying ∂µ on both sides of (6.82) and using (6.85), we see that the spin one field v (x) obeys another field equation µ ∂µv (x) = 0 (6.86) Note that when v (x) corresponds to the four-vector potential A (x) of electrodynamics, and taking m 0, the µ µ → Klein Gordon equation (6.84) becomes the wave equation, while equation (6.86) becomes the Lorentz gauge. In other words, in the limit of small mass, Eqs. (6.84, 6.86) become the equations for the potential four-vector of electrodynamics in the Lorentz gauge. Notwithstading we do not obtain the electrodynamics by simply taking the limit of m going to zero. We can see that by taking the rate of production of a spin one particle through an interaction density = J vµ where J H µ µ is an arbitrary four-momentum current6. Squaring the matrix elements and summing over the three-component of spins we find a rate proportional to 3 µ 2 µν J e (p,σ)∗ = J J ∗ Π (p) |h µi | h µi h νi σ=1 X where p is the three-momentum of the emitted spin-one particle, and J is the matrix element of the current h µi (say at x = 0) between the initial and final states of all other particles. From Eq. (6.75) we see that Πµν (p) contains a term of the form pµpν/m2 that clearly blows up when m 0. The only way to avoid the problem is → by assuming that J pµ vanishes, which in the coordinate space is equivalent to h µi µ ∂µJ = 0 which is the continuity equation that leads to the conservation of the current J µ or more precisely, the conservation of the generalized charge Q ρ dV = J 0 dV ≡ Z Z We can see the need for the conservation of the current by counting degrees of freedom: A massive spin one particle has three spin states of “helicities” σ = +1, 0, 1. By contrast, any massless particle like the photon have − only two helicity states 1. Thus, the current conservation ensures that the zero helicity states of the spin-one ± particle (of very small mass) are not emitted in the limit of zero mass.

6 µ This will be the form of any interaction density that depends linearly on the ﬁeld v (x). Of course, Jµ must be a four-vector for µ H to be Lorentz invariant. Further for v (x) hermitian, Jµ (x) must be hermitian as well. 176 CHAPTER 6. CAUSAL VECTOR FIELDS FOR MASSIVE PARTICLES

6.5 Inversion symmetries for spin-one ﬁelds

The procedure is quite similar to the case of scalar fields. To evaluate the effect of space inversion, we need a formula that connects eµ ( p,σ) with eµ ( p,σ). To do it, we shall use Eq. (1.265) page (58) − − Lµ ( p)= µ Lρ (p) δ (6.87) ν − P ρ δ P ν as well as Eq. (6.65) and the fact that e0 (0,σ)=0, we find

eµ ( p,σ) Lµ ( p) eρ (0,σ)= Lµ ( p) em (0,σ)= Lµ ( p) m em (0,σ) − ≡ ρ − m − − m − P m = Lµ ( p) α eα (0,σ)= Lµ ( p) η eα (0,σ)= ( µ Lρ (p) ν ) η eα (0,σ) − α − P α − η − P α − P ρ ν P η P α = µ Lρ (p) ( ν η ) eα (0,σ)= µ Lρ (p) δν eα (0,σ)= µ Lρ (p) eν (0,σ) −P ρ ν P ηP α −P ρ ν α −P ρ ν eµ ( p,σ) = µ eρ (p,σ) − −P ρ Then we obtain the desired formula eµ ( p,σ)= µ eρ (p,σ) (6.88) − −P ρ on the other hand, to to evaluate the time-reversal effects, we need to relate eµ ( p, σ) with eµ (p,σ). To do ∗ − − it, we use the identity 1+σ µ µ ( 1) e ∗ (0, σ)= e (0,σ) (6.89) − − − which can be checked explicitly from Eqs. (6.66). For example for σ = 1, equations (6.66) yield − 1 1+( 1) µ µ 1 i µ ( 1) − e ∗ (0, +1) = e ∗ (0, +1) =  −  = e (0, 1) − −√2 0 − −  0      and similarly for σ = 0, +1. Combining Eqs. (6.88, 6.89) and using the definition (6.65) we find

1+σ µ 1+σ µ ρ 1+σ µ ρ ( 1) e ∗ ( p, σ) = ( 1) [L ( p) e (0, σ)]∗ = ( 1) L ( p) e ∗ (0, σ) − − − − ρ − − − ρ − − = Lµ ( p) eρ (0,σ)= eµ ( p,σ)= µ eρ (p,σ) − ρ − − − P ρ once again we obtain the relation desired

1+σ µ µ ν ( 1) e ∗ ( p, σ)= e (p,σ) (6.90) − − − P ν from these results, along with the transformation properties of creation and annihilation operators under space- inversion [Eq. (3.28) page 122] and time-reversal [see Eq. (3.29)] we obtain the space-time inversion transformation properties of the annihilation and creation ﬁelds (6.67). For example, for space inversion the annihilation ﬁeld (6.67) transforms as

3 +µ 1 1 d p µ 1 ip x P φ (x) P − = e (p,σ) P a (p,σ) P − e · 3 0 (2π) σ Z 2p X q p3 1 d p µ i(p x p0x0) = e (p,σ) [η∗a ( p,σ)] e · − 3 0 − (2π) σ 2p X Z q p 6.5. INVERSION SYMMETRIES FOR SPIN-ONE FIELDS 177 changing p p, and using (6.88) we obtain →− 3 +µ 1 η∗ d p µ i( p x p0x0) P φ (x) P − = e ( p,σ) a (p,σ) e − · − 3 0 − (2π) σ 2p X Z q p3 η∗ d p µ ρ i[(p,p0) ( x,x0)] = [ ρe (p,σ)] a (p,σ) e · − 3 0 −P (2π) σ 2p X Z q p 3 µ 1 d p ρ i[p x] = η∗ ρ e (p,σ) a (p,σ) e ·P − P  3 0  (2π) σ 2p  X Z  q p then we obtain finally   +µ 1 µ +ρ P φ (x) P − = η∗ φ ( x) (6.91) − P ρ P As before, for the causal fields to be transformed into other fields with which they commute at space-like separations, it is necessary that the intrinsic space inversion, time-reversal and charge-conjugation phases for spin-one particles and their antiparticles be related by

c η = η∗ (6.92) c ξ = ξ∗ (6.93)

ζ = ζ∗ (6.94) and all phases must be real if the spin-one particle is its own antiparticle. Under the phase conditions (6.92, 6.93, 6.94) the causal vector ﬁeld (6.82) has the following properties

µ 1 µ ν P v (x) P − = η∗ νv ( x) (6.95) µ 1 − Pµ P Cv (x) C− = ξ∗v † (x) (6.96) µ 1 − µ ν T v (x) T − = ζ∗ v ( x) (6.97) P ν −P In particular, Eq. (6.95) says that a vector field that transforms as a polar vector, with no extra phases or signs multiplying the matrix µ , describes a spin-one particle with intrinsic parity η = 1. Effectively, if the three P ν − spatial components of vµ (x) define a polar three-vector field we have

i 1 i P v (x) P − = v ( x) (6.98) − P while equation (6.95) yields

i 1 i ν i i i P v (x) P − = η∗ v ( x)= η∗ v ( x)= η∗v ( x) (6.99) − P ν P − P i P P equating Eqs. (6.98, 6.99) we obtain η = 1. We recall once again that the relations developed in this section are − valid only if the inversion symmetries involved, are good symmetries of the physical system. Chapter 7

Causal Dirac ﬁelds for massive particles

We have taken the point of view that the structure and properties of any quantum field theory are given by the irreducible representation of the Homogeneous Lorentz group under which it transforms. Following that principle, we should mention that from the mathematical point of view, there are two broad classes of representations of the rotation and Lorentz groups (more precisely, of their covering groups) that are generically called tensor and spinor representations. We shall describe briefly the difference between the spinor and tensor representations of SO (3) [or more precisely, of its covering group SU (2)]. We can use (for instance) the Euler angles to define an element of SO (3). On one hand, for the representations with j integer (tensor representations), we can use the same Euler angles and obtain a one-to-one representation with the required properties of periodicity. On the other hand, for spinor representations (with j being a half-odd integer) we have for one of the Euler angles that U (ψ + 2π) = U (ψ) − which has no the periodicity required on geometrical grounds. Owing to it these representations are discarded when we deal with space variables, this is the reason to rule out half-odd integer values of the orbital angular momentum. We can recover the one-to-one character of the representation if we define an “extended manifold” in which 0 ψ 4π, in that way we arrive to the group SU (2) which is the universal covering group of SO (3). ≤ ≤ A similar reasoning shows that the spinor representation of the Lorentz group cannot be used to describe purely orbital systems. However, it was already known from non-relativistic quantum mechanics that the spinor representation of SO (3) was adequate to describe the electronic spin. Therefore, it is reasonable to use the spinor representation of the Lorentz group to describe relativistic electrons. Historically, the spinor representation was first introduced in Physics by Dirac in his theory of relativistic electrons.

7.1 Spinor representations of the Lorentz group

In this section, we shall treat the spinor representation (that we call the Dirac formalism) from the mathematical point of view. We do it because according with our approach, we start with the representation to construct the ﬁelds and then predict what kind of particles can be described with this kind of ﬁelds. Then we shall start by forming a representation U (Λ)

U (Λ) U Λ¯ = U ΛΛ¯ in a similar way in which we characterized the unitary repres entations U (Λ). That is by starting with inﬁnitesimal transformations1 as in Eqs. (1.88, 1.89), page 25

µ µ µ Λ ν = δ ν + ω ν (7.1) ω = ω (7.2) µν − νµ 1We omit the generators of translations because we shall deal only with the homogeneous Lorentz group.

178 7.1. SPINOR REPRESENTATIONS OF THE LORENTZ GROUP 179 so that the infinitesimal transformations can be written in terms of antisymmetric generators µν as in Eqs. (1.99, J 1.101), page 27 i D (1+ ω)=1+ ω µν ; µν = νµ (7.3) 2 µνJ J −J the set of matrices µν satisfy the commutation relations (1.123), page 30 J i [ µν , ρσ]= gνρ µσ gµρ νσ gσµ ρν + gσν ρµ (7.4) J J J − J − J J in order to find the set of matrices µν, we first construct matrices γµ that satisfies the anticommutation J relations γµ, γν = 2gµν (7.5) { } and we define by now i µν = [γµ, γν ] (7.6) J −4 which has the antisymmetry required for µν . J It is convenient to define the commutator between two γ matrices. By using Eq. (7.5) we can show that − γµγν + γν γµ = 2gµν γµγν γνγµ + 2γν γµ = 2gµν − [γµ, γν] = 2(gµν γνγµ) (7.7) − then from Eqs. (7.6, 7.7) we have

i i i i [ µν , γρ] = [γµ, γν ] , γρ = [γνγµ γµγν, γρ]= [γνγµ, γρ] [γµγν, γρ] J −4 4 − 4 − 4 i i i i = γν [γµ, γρ]+ [γν, γρ] γµ γµ [γν , γρ] [γµ, γρ] γν 4 4 − 4 − 4 i i i i = γν (gµρ γργµ)+ (gνρ γργν ) γµ γµ (gνρ γργν) (gµρ γργµ) γν 2 − 2 − − 2 − − 2 − i [ µν , γρ] = [γνgµρ + gνργµ γµgνρ gµργν ] J 2 − − i + [ γνγργµ γργνγµ + γµγργν + γργµγν] 2 − − and using (7.5) we have

i [ µν, γρ] = [ (γνγρ + γργν ) γµ + (γµγρ + γργµ) γν] J 2 − i = [ γν, γρ γµ + γµ, γρ γν ] 2 − { } { } = i [ gνργµ + gµργν] − obtaining ﬁnally, [ µν, γρ]= iγµgνρ + iγνgµρ (7.8) J − from Eq. (7.8) we can verify in turn that Eq. (7.6) satisﬁes the required commutation relations (7.4). We do it as follows 180 CHAPTER 7. CAUSAL DIRAC FIELDS FOR MASSIVE PARTICLES

i 1 1 i [ µν , ρσ] = i µν , [γρ, γσ] = [ µν , [γρ, γσ]] = [ µν , γργσ γσγρ] J J J −4 4 J 4 J − 1 1 = [ µν, γργσ] [ µν, γσγρ] 4 J − 4 J 1 1 1 1 = γρ [ µν , γσ]+ [ µν, γρ] γσ γσ [ µν , γρ] [ µν , γσ] γρ 4 J 4 J − 4 J − 4 J 1 1 i [ µν , ρσ] = γρ ( iγµgνσ + iγν gµσ)+ ( iγµgνρ + iγν gµρ) γσ J J 4 − 4 − 1 1 γσ ( iγµgνρ + iγν gµρ) ( iγµgνσ + iγνgµσ) γρ −4 − − 4 − i i i i i [ µν, ρσ] = γργµgνσ + γµgνσγρ + γργνgµσ γνgµσγρ J J −4 4 4 − 4 i i i i γµgνργσ + γσγµgνρ + γν gµργσ γσγν gµρ −4 4 4 − 4 i i i [ µν, ρσ] = ( γργµ + γµγρ) gνσ + (γργν γνγρ) gµσ J J 4 − 4 − i i + (γσγµ γµγσ) gνρ + (γνγσ γσγν) gµρ 4 − 4 − i i i i = [γµ, γρ] gσν + [γρ, γν] gσµ + [γσ, γµ] gνρ + [γν, γσ] gµρ 4 4 4 4 = µρgσν ρνgσµ σµgνρ νσgµρ −J −J −J −J i [ µν, ρσ] = gσν ρµ gσµ ρν + gνρ µσ gµρ νσ J J J − J J − J so that µν are valid representations of the generators of the homogeneous Lorentz group. J It can be proved that the matrices γµ are irreducible, in the sense that there is not any proper subspace that is left invariant under all these matrices2. If these matrices were reducible, we would be able to choose some smaller sets of ﬁelds components, which would transform as in Eqs. (7.3) and (7.6) with an irreducible set of γµ′ s. A set of matrices that satisfy the relations (7.5) or the Euclidean analog with the Kronecker delta instead of gµν , is called a Cliﬀord algebra. From the mathematical point of view, it can be shown that the most general irreducible representation of the Lorentz group (more precisely of its covering group) is either a tensor or a spinor representation transforming as in Eqs. (7.3) and (7.6) or a direct product of a spinor and a tensor. From Eq. (7.3), and using (7.2, 7.8) we can see how γρ transforms under a homogeneous Lorentz transformation

ρ 1 i µν ρ i αβ ρ i µν ρ i αβ D (Λ) γ D− (Λ) = 1+ ω γ 1 ω = γ + ω γ 1 ω 2 µνJ − 2 αβJ 2 µν J − 2 αβJ i i = γρ + ω µν γρ ω γρ αβ + ω2 2 µνJ − 2 αβ J O i i = γρ + ω µν γρ ω γρ µν + ω2 2 µνJ − 2 µν J O i i = γρ + ω [ µν, γρ]+ ω2 = γρ + ω ( iγµgνρ + iγν gµρ)+ ω2 2 µν J O 2 µν − O 1 1 1 1 = γρ + ω gνργµ ω gµργν = δ ργσ + ω ργµ + ω gµργν 2 µν − 2 µν σ 2 µ 2 νµ 1 1 = δ ργσ + ω ργµ + ω ργν = δ ργσ + ω ργσ σ 2 µ 2 ν σ σ ρ 1 ρ ρ σ D (Λ) γ D− (Λ) = (δσ + ωσ ) γ 2We should emphasize however, that it does not mean that the group representation is irreducible, since the γ−matrices are not the generators of the group [recall that the generators are the operators J µν deﬁned in Eq. (7.6)]. Indeed, we shall see later that the representation we are constructing is reducible. 7.1. SPINOR REPRESENTATIONS OF THE LORENTZ GROUP 181 which according with Eq. (7.1) gives ρ 1 ρ σ D (Λ) γ D− (Λ) = Λσ γ (7.9) and comparing with Eq. (1.96) page 27, we can see that γρ transform as a four-vector operator3. Similarly, the unit matrix is obviously a scalar 1 D (Λ) 1D− (Λ) = 1 (7.10) from the deﬁnition (7.6) of the generators and the rule of transformation (7.9) we have

′ρσ ρσ 1 i ρ σ 1 i ρ σ σ ρ 1 D (Λ) D− (Λ) = D (Λ) [γ , γ ] D− (Λ) = D (Λ) [γ γ γ γ ] D− (Λ) J ≡ J −4 −4 − i ρ σ 1 σ ρ 1 = D (Λ) γ γ D− (Λ) D (Λ) γ γ D− (Λ) −4 − i ρ 1 σ 1 σ 1 ρ 1 = D (Λ) γ D− (Λ) D (Λ) γ D− (Λ) D (Λ) γ D− (Λ) D (Λ) γ D− (Λ) −4 − i i = [Λ ργµ] [Λ σγν ] [Λ σγν ] [Λ ργµ] = Λ ρΛ σ γµγν γνγµ −4 { µ ν − ν µ } −4 µ ν { − } i = Λ ρΛ σ [γµ, γν ] = Λ ρΛ σ µν −4 µ ν µ ν J hence, under a homogeneous proper orthochronus Lorentz transformation, the generators ρσ transform as J ρσ 1 ρ σ µν D (Λ) D− (Λ) = Λ Λ (7.11) J µ ν J then Eqs. (7.3, 7.11) say that µν is an antisymmetric tensor. Other totally antisymmetric tensors can be J constructed from γµ as follows:

ρστ γ[ργσγτ] (7.12) A ≡ ρστη γ[ργσγτ γη] (7.13) P ≡ where the brackets indicate a sum over all permutations of the indices within the brackets, with a minus sign for odd permutations. For example, Eq. (7.12) can be written explicitly as [homework write Eq. (7.13) explicitly]

ρστ γργσγτ γργτ γσ γσγργτ + γτ γργσ + γσγτ γρ γτ γσγρ (7.14) A ≡ − − − by using repeatedly Eq. (7.5) we can express any product of γ matrices as a sum of antisymmetrized products of γ′s times a product of metric tensors. Therefore, the totally antisymmetric tensors form a complete basis for the set of all matrices that can be constructed from the Dirac matrices. By deﬁning the matrix β iγ0 (7.15) ≡ a similarity transformation of the Dirac matrices through the β matrix yields

i 1 0 i 0 1 0 i 0 1 0 i i 0 0 1 βγ β− = iγ γ iγ − = γ γ γ − = γ , γ γ γ γ − − 1 1 = g0i γiγ0 γ0 − = γiγ0 γ0 − = γi − − − 0 1 0 0 0 1 0 0 0 1 0 βγ β− = iγ γ iγ− = γ γ γ −= γ then we obtain i 1 i 0 1 0 βγ β− = γ ; βγ β− =+γ (7.16) − that we can rewrite as µ 1 0 0 βγ β− = γ, γ γ, γ (7.17) − ≡ P 3We recall that it means that the array γi,γ0 has the same property of transformation as the four-vector operator Pi,H , where each γµ represents a single operator at the Minkowski space (see section 1.8.1 page 26). 182 CHAPTER 7. CAUSAL DIRAC FIELDS FOR MASSIVE PARTICLES thus β can be considered as a “parity” transformation for the “four-vector operator” γµ. Note that here we are ordering the label as µ = 0, 1, 2,...(this order is more convenient because the dimensionality is arbitrary so far). We can say alternatively that β anticommutes with γi and commutes with γ0. Consequently, the same similarity transformation applied to any product of γ matrices gives a minus or plus sign according whether the product of γ′s contain even or odd number of γ′s with space-like indices, respectively. For instance,

0 1 0 2 1 0 1 0 2 0 1 3 2 1 0 1 3 2 βγ γ γ γ β− = γ γ γ γ ; βγ γ γ γ β− = γ γ γ γ − in particular we have ij 1 ij i0 1 i0 00 β β− = ; β β− = ; = 0 (7.18) J J J −J J all the properties we shall develop in this section are valid in any number of space-time dimensions and for any “metric” gµν . However, in four spacetime dimensions we have the particular feature that no totally antisymmetric tensor can have more than four indices [homework!!(14a)], so that the complete sequence of tensors is given by

1, γρ, ρσ, ρστ , ρστη (7.19) J A P Each of these tensors transform diﬀerently under Lorentz and/or parity transformations so that they are all linearly independent. Another way to check such a linear independence is the following: we deﬁne the scalar product of two matrices by the trace of the product that is

µ µ ν (B, C)= T r [B∗C]= (B∗C) µ = B∗ νC µ (7.20) µ X it can be shown that the rule (7.20) satisfies the axioms of a scalar product, and that the matrices (7.19) form an orthogonal set under this scalar product[homework!!(15)]4. Now an orthogonal set is linearly independent unless some of its vectors are null. However, none of the matrices (7.19) vanish, since each component of each of these tensors is proportional to a product of different γ matrices, and in turn such a product has a square equal to − plus or minus the product of the associated squares, and then equal to 1. ± Now we count the number of linearly independent components on each of the matrices contained in the set (7.19). The identity has only one independent component, γρ has four, ρσ has six, ρστ has four, while ρστη J A P has only one independent component [homework!!(14b) explain this counting in detail]. So we have a total of 16 independent components. Thus an arbitrary 4 4 matrix can be written as a complex linear combination of the × 16 linearly independent matrices given by (7.19). According with the Clifford algebra (7.5) we have

i 2 0 2 γ = 14 4 ; γ = 14 4 (7.21) × − × and recalling the deﬁnition of β iγ0 we obtain that ≡ 2 β = 14 4 (7.22) × 1 Then we shall not distinguish between β and β− from now on. Hence, all similarity transformations through the “parity operator” β, will be written as β (. . .) β.

7.2 Some additional properties of the Dirac matrices

It is convenient to write the totally antisymmetric tensors (7.12, 7.13) in a more suitable way. It is well-known that in a n dimensional space there is only one linearly independent totally antisymmetric tensor of n indices. − 4It is easier to show that by using a speciﬁc representation (for instance, the one that we shall develop later). The results obtained in a given representation are valid for any other representation, because the trace is invariant under a similarity transformation. 7.2. SOME ADDITIONAL PROPERTIES OF THE DIRAC MATRICES 183

Consequently, the totally antisymmetric matrix (7.13) must be proportional to the pseudotensor ερστη, deﬁned as a totally antisymmetric quantity with ε0123 = +1. Setting ρ, σ, τ, η equal to 0, 1, 2, 3 respectively, we ﬁnd

0123 = γ0γ1γ2γ3 permut (7.23) P ± for an arbitrary ρστη, we have that ρστη is diﬀerent from zero only if all symbols ρστη are diﬀerent. Thus, for P a non-null value of ρστη , the set of symbols ρστη is a permutation P of 0, 1, 2, 3. When writing ρστη it is P ρστη P clear that we obtain exactly the same 4! terms as in Eq. (7.23)

ρστη = γργσγτ γη permut (7.24) P ± however all 4! terms in (7.24) will have opposite sign with respect to (7.23) if the permutation Pρστη is odd (Pρστη is the permutation from the sequence 0, 1, 2, 3 to the sequence ρ, σ, τ, η). It owes to do with the fact that for ρστη P we assign by deﬁnition the positive sign to the sequence γργσγτ γη, while in Eq. (7.24) this sequence will be assigned the negative sign if the permutation from 0, 1, 2, 3 to ρ, σ, τ, η is odd, then we obtain

ρστη = ερστη 0123 = ερστη γ0γ1γ2γ3 permut (7.25) P P ± we now take into account that γµ anticommutes with γν as long as µ = ν. In that case an odd number of jumps 6 simply gives a minus sign (as long as we jump over a diﬀerent matrix). For instance we have

0123 = γ0γ1γ2γ3 γ0γ1γ3γ2 γ0γ2γ1γ3 + γ0γ3γ1γ2 other permut P − − ± = γ0γ1γ2γ3 + γ0γ1γ2γ3 + γ0γ1γ2γ3 γ0γ1γ3γ2 other permut − ± = γ0γ1γ2γ3 + γ0γ1γ2γ3 + γ0γ1γ2γ3 + γ0γ1γ2γ3 other permut ± we see that all terms are identical, therefore

0123 = 4! γ0γ1γ2γ3 (7.26) P substituting (7.26) in (7.25) we ﬁnd

ρστη = 4!ερστη γ0γ1γ2γ3 = 4!iερστη iγ0γ1γ2γ3 P − then we obtain ﬁnally

ρστη = 4!iερστη γ (7.27) P 5 γ iγ0γ1γ2γ3 (7.28) 5 ≡ − in the same way ρστ must be proportional to ερστη contracted with some matrix . To show it, let us evaluate A Mη the antisymmetric tensor ρστ of Eq. (7.14), ρ, σ, τ = 0, 1, 2 A 012 γ0γ1γ2 γ0γ2γ1 γ1γ0γ2 + γ2γ0γ1 + γ1γ2γ0 γ2γ1γ0 A ≡ − − − = γ0γ1γ2 + γ0γ1γ2 + γ0γ1γ2 + γ0γ1γ2 + γ0γ1γ2 + γ0γ1γ2 012 = 3!γ0γ1γ2 = 3!ε0123γ0γ1γ2 (7.29) A now deﬁning γ g γν (7.30) µ ≡ µν and taking into account the Cliﬀord algebra (7.5) we have

i 2 ii i i i γ = g = 1 ; γ γi = γ γ gii = 1 No sum over i (7.31) 184 CHAPTER 7. CAUSAL DIRAC FIELDS FOR MASSIVE PARTICLES using (7.31) in Eq. (7.29) we have

012 = 3!ε0123γ0γ1γ2 γ3γ = 3!iε0123 iγ0γ1γ2γ3 γ A 3 − 3 012 = 3!iε0123γ γ A 5 3 and we can do a similar process for all possible sets of indices ρ, σ, τ in Eq. (7.14). It can be checked that by setting ρ, σ, τ equal to 0,1,2 or 0,1,3 or 0,2,3 or 1,2,3 we ﬁnd

ρστ = 3!iερστη γ γ (7.32) A 5 η µ it is easy to show that the square of γ5 is the unit matrix and that γ5 anticommutes with all γ

γ2 = 1 , γ , γµ = 0 (7.33) 5 { 5 } it can be shown as follows

γ , γ0 = i γ0γ1γ2γ3 γ0 + γ0 γ0γ1γ2γ3 = i γ0γ0γ1γ2γ3 + γ0γ0γ1γ2γ3 = 0 5 − − − γ , γ1 = i γ0γ1γ2γ3 γ1 + γ1 γ0γ1γ2γ3 = i γ0γ1γ1γ2γ3 γ0γ1γ1γ2γ3 = 0 5 − − − Proceeding similarly with the other γ matrices we obtain γ , γµ = 0. Moreover − { 5 } γ2 = γ0γ1γ2γ3 γ0γ1γ2γ3 = γ0γ1γ2γ3γ0γ1γ2γ3 = γ0γ0γ1γ2γ3γ1γ2γ3 5 − − 2 2 2 2 = γ0γ0γ1γ1γ2γ3γ2γ3 = γ0γ0γ1γ1γ2γ2γ3γ3 = γ0 γ1 γ2 γ3 − − γ0, γ0 γ1, γ1 γ2, γ2 γ3, γ3 = = g00g11g22g33 = 1 − 2 2 2 2 − µ since γ5 anticommutes with γ , and using the deﬁnition of β in Eq. (7.15), we see that

1 0 0 0 0 0 2 βγ β− = βγ β = iγ γ iγ = γ γ γ = γ γ = γ 5 5 5 − 5 5 − 5 µ and the commutator between γ and γ5 becomes

µ µ µ µ µ [γ , γ5] = γ γ5 γ5γ = γ γ5 + γ γ5 µ µ − [γ , γ5] = 2γ γ5 (7.34) we can obtain the commutator between the generators ρσ and γ by taking into account Eqs. (7.33, 7.34) and J 5 the deﬁnition (7.6)

4i [ µν , γ ] = [[γµ, γν ] , γ ] = [γµγν , γ ] [γνγµ, γ ]= γµ [γν, γ ] + [γµ, γ ] γν γν [γµ, γ ] [γν, γ ] γµ J 5 5 5 − 5 5 5 − 5 − 5 = 2γµγνγ + 2γµγ γν 2γν γµγ 2γν γ γµ = 2γµγν γ 2γµγν γ 2γν γµγ + 2γν γµγ 5 5 − 5 − 5 5 − 5 − 5 5 4i [ µν , γ ] = 0 J 5

Therefore, the matrix γ5 is a pseudoscalar in the sense that

ρσ [ , γ5] = 0 (7.35) J 1 βγ β− = βγ β = γ (7.36) 5 5 − 5

Note that γ5 is a Casimir of the homogeneous proper orthochronus Lorentz group since it commutes with all the generators. However it is not in general proportional to the identity at the Minkowski space because this representation is not irreducible in such a space. We should remember that Schur’s lemmas are only valid within minimal vector spaces associated with irreducible representations. 7.3. THE CHIRAL REPRESENTATION FOR THE DIRAC MATRICES 185

From Eqs. (7.27, 7.32) we can see that the 16 independent 4 4 matrices deﬁned in Eq. (7.19) can be × reorganized in matrices with more suitable properties as follows

1 : scalar

γ5 : pseudoscalar γρ : vector ρσ : antisymmetric tensor J γ5γη : axial vector (7.37) the notation γ5 has to do with the fact that the anticommutation relations (7.33) along with (7.5) show that the set 0 1 2 3 γ , γ , γ , γ , γ5 provide a Cliﬀord algebra in ﬁve space-time dimensions.

7.3 The chiral representation for the Dirac matrices

We shall choose an explicit set of 4 4 γ matrices as follows × −

0 02 2 12 2 02 2 σ γ i × × ; γ i × ; σ (σ1,σ2,σ3) (7.38) ≡− 12 2 02 2 ≡− σ 02 2 ≡ × × − × where σi are the usual Pauli matrices

0 1 0 i 1 0 σ1 = , σ2 = − , σ3 = (7.39) 1 0 i 0 0 1 − let us recall some basic properties of the Pauli matrices

[σ ,σ ] = 2iε σ ; σ ,σ = 2δ i j ijk k { i j} ij σi† = σi ; Trσi = 0 σiσj = iσk with i, j, k a cyclic permutation of 1, 2, 3 (7.40) it can be shown that any other irreducible set of γ matrices that satisfy the Clifford algebra (7.5), are related with − this through a similarity transformation. Thus a given set of γ matrices define a unique irreducible inequivalent − representation of the Clifford algebra. However, several representations (though equivalent) of the γ matrices are − used. By returning to our specific representation, we first observe from (7.38) that the γ0 matrix is anti-hermitian while the γi matrices are hermitian

0 02 2 12 2 0 02 2 σ† 02 2 σ γ † i × × = γ ; γ† i × − = i × − = γ (7.41) ≡ 12 2 02 2 − ≡ σ† 02 2 σ 02 2 × × × × we can write it in a shorten notation as

µ µµ µ γ † = g γ no sum over µ (7.42) we should keep in mind that the hermitian or anti-hermitian property of any matrix is preserved under a similarity transformation carried out by a unitary matrix. 186 CHAPTER 7. CAUSAL DIRAC FIELDS FOR MASSIVE PARTICLES

We shall now calculate the product of two gamma matrices. For instance from Eq. (7.38) we can obtain the product between two Dirac matrices with space-like components

i j 02 2 σi 02 2 σj σiσj 02 2 γ γ = × × = − × − σi 02 2 σj 02 2 − 02 2 σiσj − × − × × − i j σiσj 02 2 γ γ = × 02 2 σiσj × and we also have

0 i 02 2 12 2 02 2 σi σi 02 2 γ γ = × × × = − × − 12 2 02 2 σi 02 2 − 02 2 σi × × − × × i 0 02 2 σi 02 2 12 2 σi 02 2 γ γ = × × × = × − σi 02 2 12 2 02 2 − 02 2 σi − × × × × − 0 0 02 2 12 2 02 2 12 2 12 2 02 2 γ γ = × × × × = × × − 12 2 02 2 12 2 02 2 − 02 2 12 2 × × × × × × picking up all these results we have

i j σiσj 02 2 0 i i 0 σi 02 2 0 0 γ γ = × ; γ γ = γ γ = × ; γ γ = 14 4 (7.43) 02 2 σiσj − 02 2 σi − × × × − from the properties of the Pauli matrices (7.40) we can verify that the representation (7.38) reproduces the Cliﬀord algebra (7.5) as it must be

i j σi,σj 02 2 δij 02 2 ij ij γ , γ = { } × = 2 × = 2δ4 4 = 2g 02 2 σi,σj 02 2 δij × × { } × γi, γ0 = γ0γi + γiγ0 = γ0γi γ0γi =0=2gi0 0 0 0 0 − 00 γ , γ = 2γ γ = 2 14 4 = 2g 14 4 − · × · × on the other hand, we can calculate the Lorentz group generators (7.6). From Eq. (7.38), and using Eqs. (7.40, 7.43), they are given by

ij i i j i [σi,σj] 02 2 = γ , γ = × J −4 −4 02 2 [σi,σj] × ij i 2iεijkσk 02 2 1 σk 02 2 = × = εijk × J −4 02 2 2iεijkσk 2 02 2 σk × × i0 i i 0 i 0 i i 0 i 0 i 0 i i 0 i i σi 02 2 = γ , γ = γ γ γ γ = γ γ + γ γ = γ γ = × J −4 4 − 4 2 2 02 2 σi × − 00 i 0 0 = γ , γ = 04 4 J −4 × then we obtain for the generators in this representation

ij 1 σk 02 2 i0 i σi 02 2 00 = εijk × ; = × ; = 04 4 (7.44) J 2 02 2 σk J 2 02 2 σi J × × × − note that the generators (7.44) are block-diagonal. Therefore, the Dirac matrices give a reducible representation of the proper orthochronus Lorentz group. The four-dimensional representation is thus decomposed as the direct sum of two irreducible two-dimensional representations with ij = iε k0. J ± ijkJ 7.3. THE CHIRAL REPRESENTATION FOR THE DIRAC MATRICES 187

For the speciﬁc representation given by Eq. (7.38) the γ5 matrix becomes

γ iγ0γ1γ2γ3 5 ≡ − 5 02 2 12 2 02 2 σ1 02 2 σ2 02 2 σ3 = ( i) × × × × × − 12 2 02 2 σ1 02 2 σ2 02 2 σ3 02 2 × × − × − × − × σ1 02 2 σ2σ3 02 2 σ1σ2σ3 02 2 = i − × − × = i × − 02 2 σ1 02 2 σ2σ3 − 02 2 σ1σ2σ3 × × − × − and the product of Pauli matrices can be obtained by explicit calculation from (7.39) or by using the properties (7.40)5 i σ σ σ = (σ σ ) σ = iσ σ = σ ,σ = i (7.45) 1 2 3 1 2 3 3 3 2 { 3 3} such that γ5 in this representation becomes

12 2 02 2 γ5 = × × (7.46) 02 2 12 2 × − × this representation has the advantage of reducing ρσ and γ to block-diagonal form (this is the so-called chiral J 5 representation). Note that both 2 2 submatrices are proportional to the identity. It is because in the basis × chosen the representation is decomposed in two bidimensional irreducible representations. Since ρσ and γ have J 5 the same block-diagonal form, equation (7.35) shows that the 2 2 submatrices that represent γ are Casimirs in × 5 each bidimensional subspace. Since these Casimirs are associated with irreducible representations they must be proportional to the identity within each minimal invariant bidimensional subspace. From the physical point of view, the chiral representation is suitable to describe particles in the ultra-relativistic limit v c. However, in the non-relativistic limit v << c, it is more convenient to choose a representation in →0 6 which γ is diagonal instead of γ5, which is the case in the Dirac representation . The representation of the homogeneous Lorentz group that we have constructed is not unitary, since the generators ρσ are not all represented by hermitian matrices. In the speciﬁc representation given by Eqs. (7.38) J ij are hermitian while i0 are anti-hermitian. It can be seen from the fact that γ0 is anti-hermitian while the Jk J γ ′s are hermitian [see Eq. (7.42)]

iµ i i µ i i µ µ i i µ i i µ † = + γ , γ † = γ γ † γ γ † = γ †γ † γ †γ † J 4 4 − 4 − i h i i h i = gµµγµγi gµµγiγµ = gµµ γi, γµ 4 − − 4 iµ µµ iµ † = g (7.47) J J It is also convenient to deﬁne the matrix β iγ0 of Eq. (7.15), such that the reality conditions are manifestly ≡ Lorentz -invariant. In the representation (7.38) β has the form

0 1 β = (7.48) 1 0 1 recalling that β = β− , all similarity transformations through the “parity operator” β can be written as β (. . .) β. 0 k Using again the fact that γ is anti-hermitian while the γ ′s are hermitian, we see that

0 0 0 0 0 0 0 0 0 βγ †β = βγ β = iγ γ iγ = γ γ γ = γ − − − k k 0 k 0 0 k 0 0 0 k k βγ †β = βγ β = iγ γ iγ = γ γ γ = γ γ γ = γ − − 5The advantage of using the properties (7.40) instead of the explicit forms (7.39), is that we can be sure that the result (7.45) is independent of the representation chosen. 6There is still another widely used representation: the so-called Majorana representation. 188 CHAPTER 7. CAUSAL DIRAC FIELDS FOR MASSIVE PARTICLES that can be written as µ µ βγ †β = γ (7.49) − and equation (7.49) in turn leads to

ρσ i ρ σ σ ρ † i σ ρ ρ σ β †β = β (γ γ γ γ ) β = β γ †γ † γ †γ † β J −4 − 4 − i σ ρ ρ σ i σ ρ ρ σ = βγ †β βγ †β βγ †β βγ †β = [( γ ) ( γ ) ( γ ) ( γ )] 4 − 4 − − − − − i h i i = [γσγρ γργσ]= [γρ, γσ]= ρσ 4 − −4 J so we also obtain the identity ρσ ρσ β †β = (7.50) J J it is also important to characterize the transformation of the matrices D (Λ) under the parity transformation taking into account that such matrices are not unitary. To do it, we use the expression (7.3) page 179 for inﬁnitesimal Lorentz transformations, as well as Eq. (7.50)

i µν † i µν i µν βD† (Λ) β = β 1+ ω β = β 1 ω † β = 1 ω β †β 2 µν J − 2 µνJ − 2 µν J i µν 1 βD† (Λ) β = 1 ω = D− (Λ) − 2 µνJ and since any finite transformation can be written as a product of infinitesimal transformations, we can extend this result for any finite proper orthochronus homogeneous Lorentz transformation. In conclusion, though the matrices D (Λ) are not unitary, they satisfy the “pseudounitarity” condition

1 βD (Λ)† β = D (Λ)− (7.51) moreover, it is easy to check that γ5 is hermitian and anticommutes with β

γ = γ† , γ , β = 0 (7.52) 5 5 { 5 } from which we obtain βγ†β = γ (7.53) 5 − 5 combining equations (7.33, 7.49, 7.53) we also obtain

µ µ µ µ µ µ β (γ γ )† β = βγ †γ†β = βγ †β βγ†β = ( γ ) ( γ )= γ γ = γ γ 5 5 5 − − 5 5 − 5 so we have µ µ β (γ γ )† β = γ γ (7.54) 5 − 5 The Dirac and related matrices obey some symmetry properties. From Eqs. (7.38) and (7.39) and the fact that σ2 is antisymmetric while σ1 and σ3 are symmetric, we see that γµ is symmetric for µ = 0, 2 and antisymmetric for µ = 1, 3.

0 02 2 12 2 0 2 02 2 σ2 02 2 σ2 2 γ i × × = γ ; γ i × − = i × = γ ≡ − 12 2 02 2 ≡− σ2 02 2 − σ2 02 2 × × × − × k 02 2 σk 02 2 σk k e γe = i × − = i ×e − = γ ; k = 1, 3 − σk 02 2 − σk 02 2 e − × × 0 0 2 2e 1 1 2 2 eγ = γ ; γ = γ ; γ = γ ; γ = γ (7.55) e − − e e e e 7.3. THE CHIRAL REPRESENTATION FOR THE DIRAC MATRICES 189

Note that the property that is independent of the basis is the hermiticity of the Pauli matrices. In the particular representation described by equation (7.39), σ2 is antisymmetric because its non-null elements are purely imaginary, while σ1 and σ3 are symmetric because they are real. Consequently, equation (7.55) can be guaranteed only for the representation given by Eqs. (7.38) and (7.39). This can be summarized as

µ µ 1 2 σ2 0 γ = Cγ C− ; C γ β = i (7.56) − ≡ − 0 σ2 − to show that, it is convenient to expresse Eqs. (7.38) in a more condensed notation as

0 σµ γµ ; σµ σk, 1 = (σ , 1) ; no sum over µ (7.57) ≡ gµµσµ 0 ≡ k − from the properties (7.40) of the Pauli matrices and the deﬁnition (7.57) of σµ, we obtain

2 k 2 µ 2 σ = σk =1 ; (σ ) = 1 (7.58) from which we can easily check that 1 C− = C (7.59) − using the deﬁnition of C Eq. (7.56) as well as Eqs. (7.57, 7.59), we have

µ µ µ 1 σ2 0 0 σ σ2 0 P Cγ C− = CγµC = i i µµ µ i ≡ − − 0 σ2 − g σ 0 − 0 σ2 − − − σ 0 0 σµσ = i 2 − 2 0 σ gµµσ σ 0 − 2 − µ 2 µ µ µ 0 σ2σ σ2 0 Z P = i µµ µ i µµ µ (7.60) − g σ2σ σ2 0 ≡− g Z 0 − − now we can evaluate the products of Pauli matrices by using properties (7.40) and (7.58). If µ = 0, 2; we ﬁnd

Z0 = σ σ0σ = σ σ = 1 ; Z2 = σ σ2σ = σ σ σ = σ 2 2 2 2 − 2 2 − 2 2 2 − 2 and if µ = k = 1, 3; we obtain

σ σkσ = σ σ σ = σ σ σ = σ ; k = 1, 3 − 2 2 − 2 k 2 k 2 2 k so that we have Z0 = 1 ,Z2 = σ ,Z1 = σ ,Z3 = σ (7.61) − 2 1 3 substituting Eqs. (7.61) in (7.60) for µ = 0, 1, 2, 3; and taking into account Eqs. (7.55), we obtain Pµ = γµ showing the validity of Eq. (7.56). Moreover, from Eq. (7.56) we can obtain the transposes of the matrices in the basis (7.37). For instance e

µν i ν µ µ ν i ν 1 µ 1 µ 1 ν 1 = [γ γ γ γ ]= Cγ C− Cγ C− Cγ C− Cγ C− J −4 − −4 − − − − − i ν µ 1 µ ν 1 i ν µ µν 1 νµ 1 µν 1 e = Cγ γ C− Cγ γ C− = C [γ γ γ γ ] C− = C C− = C C− −4 e e e e− −4 − J − J therefore, from Eq. (7.56) we derive the following results

µν µν 1 = C C− (7.62) J − J 1 γ5 = +Cγ5C− (7.63) e ^µ µ 1 (γ5γ ) = +Cγ5γ C− (7.64) e 190 CHAPTER 7. CAUSAL DIRAC FIELDS FOR MASSIVE PARTICLES the signs in Eqs. (7.62, 7.63, 7.64) will be important when we consider the charge-conjugation properties of currents formed with these matrices. The results obtained for adjoints and transposes can be combined to obtain the properties of complex conjugate of the Dirac and related matrices. For example, by combining Eqs. (7.49, 7.56) and taking into account that β is real in the chiral representation [see Eq. (7.48)] we obtain

µ 1 µ µ µ ∗ µ βCγ C− β = βγ β = β∗ (γ ∗)∗ β∗ = βγ †β = γ ∗ − − − h i and with an analogous procedure we obtaine the conjugatee of the matrices in the basis (7.37) µ µ 1 γ ∗ = βCγ C− β (7.65) µν µν 1 ∗ = βC C− β (7.66) J − J 1 γ5∗ = βCγ5C− β (7.67) µ − µ 1 (γ γ )∗ = βCγ γ C− β (7.68) 5 − 5 for future purposes we also observe that

C, β = Cβ + βC γ2β β + β γ2β = γ2 γ2β2 = γ2 γ2 = 0 { } ≡ − − [C, γ ] = Cγ γ C iγ2γ0 γ γ iγ2γ0 = iγ2γ0 γ iγ2γ0 γ = 0 5 5 − 5 ≡ 5 − 5 5 − 5 µ where we have used the fact that γ5 anticommutes with γ . We then obtain

C, β = [C, γ ] = 0 (7.69) { } 5 7.4 Causal Dirac ﬁelds

We shall assume since the beginning that the particle does not coincide with the associated antiparticle, and assume that we run over a single species of particle (so we have two species n andn ¯ corresponding to the particle and its antiparticle respectively). As customary, we shall construct particle annihilation and antiparticle creation ﬁelds, that in this case transform under the Lorentz group according with the Dirac (or spinor) representation. By using Eqs. (4.39, 4.40) we write

+ 3/2 3 ip x ψk (x) = (2π)− d p uk (p,σ) e · a (p,σ) (7.70) σ X Z c 3/2 3 ip x c ψk− (x) = (2π)− d p vk (p,σ) e− · a † (p,σ) (7.71) σ X Z where the particle species has been omitted7. We calculate the coefficient functions u (p,σ) and v (p,σ) as usual: we start with Eqs. (4.49, 4.50) to find uk and vk for zero momentum, then we apply Eqs. (4.45, 4.46) to find them for arbitrary momenta, we should use D ¯ (Λ) in this case as the 4 4 Dirac representation of the homogeneous kk × Lorentz group constructed from the generators ρσ of Eq. (7.3). J The zero momentum conditions (4.49) become

(j) u¯ (0, σ¯) J = ¯ u (0,σ) k σσ¯ Jkk k σ¯ X Xk uµ (0, σ¯) J(j) = µ uν (0,σ)= µ u0 (0,σ)+ µ ui (0,σ) σσ¯ J ν J 0 J i σ¯ X 7 It worths pointing out that the value of the coefficients uk (p,σ) and vk (p,σ) only depends on the irreducible representation of the Lorentz group to which the fields are associated. They do not depend on the specific species of particles. Therefore, since particles and antiparticles transform under the same irreducible representation, their coefficients are the same. Thus the only difference between c− − † c† ψk (x) and ψk (x) is the creation operators a and a as can be seen in Eq. (7.71). 7.4. CAUSAL DIRAC FIELDS 191 and using the explicit form of the cartesian generators of SO (3), Eqs. (6.9, 6.10), page 163 we have

uµ (0, σ¯) (J )(j) = ( )µ ui (0,σ) k σσ¯ Jk i σ¯ X for µ = 0,n respectively we have

u0 (0, σ¯) (J )(j) = ( )0 ui (0,σ) ; un (0, σ¯) (J )(j) = ( )n ui (0,σ) k σσ¯ Jk i k σσ¯ Jk i σ¯ σ¯ X X u0 (0, σ¯) (J )(j) = 0 ; un (0, σ¯) (J )(j) = iε ui (0,σ) k σσ¯ k σσ¯ − nik σ¯ σ¯ X X ??? —————————- ————————- ———————– Something similar can be done for the zero momentum condition (4.50). On the other hand, the four dimensional representation we are dealing with, is reducible. Thus, it is convenient to replace the four component index µ by a couple of indices: one 2-valued index m that labels the rows and columns of the submatrices in Eqs. (7.44) and a second index that takes the values , and that labels the rows and columns of the supermatrix in Eqs. ± (7.44). With this notation, we ﬁnally obtain [homework!!(16)].

(j) 1 um¯ (0, σ¯) Jσσ¯ = σmm¯ um (0,σ) (7.72) ± 2 ± σ¯ m X X (j) 1 vm¯ (0, σ¯) Jσσ¯ ∗ = σmm¯ um (0,σ) (7.73) − ± 2 ± σ¯ m X X we can rewrite these equations by regarding um (0,σ) and vm (0,σ) as the m,σ elements of matrices U and ± ± ± V , that is ±

(U+)mσ um+ (0,σ) , (U )mσ um (0,σ) ≡ − ≡ − (V+)mσ vm+ (0,σ) , (V )mσ vm (0,σ) (7.74) ≡ − ≡ − By now, such matrices (of dimension m σ) are rectangular but not neccesarily square, because m takes two values × but we do not know yet how many values takes the quantum number σ [the number of values of σ is 2j +1 but we have not determined j so far]. In the matrix notation described by (7.74), equations (7.72, 7.73) are written as

(j) 1 (U )m¯ σ¯ Jσσ¯ = σmm¯ (U )mσ (0,σ) ± 2 ± σ¯ m X X (j) 1 (V )m¯ σ¯ Jσσ¯ ∗ = σmm¯ (U )mσ (0,σ) − ± 2 ± σ¯ m X X which can be ﬁnally written as 1 U J(j) = σU (7.75) ± 2 ± (j) 1 V J ∗ = σV (7.76) − ± 2 ± now, the (2j + 1) dimensional matrices J(j) and J(j) provide irreducible representations of the Lie algebra of − − ∗ rotation8, the same occurs for the 2 2 matrices σ/2. Consequently, according with Schur’s Lemma 1 [see Lemma × 8If we have a given matrix representation D (G) of a group, it is straightforward to see that the conjugate matrices D∗ (G) also give a representation on the same vector space. The natural question is whether the conjugate representation is equivalent to the original representation or not. In the case of SO (3), all conjugate representations are equivalent to the original ones. 192 CHAPTER 7. CAUSAL DIRAC FIELDS FOR MASSIVE PARTICLES

1, page 14] the matrices U and V either vanish (which is not a case in which we are interested) or is square ± ± and non-singular. Therefore, m takes the same number of values as σ (since U and V are matrices of dimension ± ± m σ). Consequently, since m takes two values we have that 2j + 1 = 2, and the Dirac ﬁeld can only describe × particles of spin j = 1/2. Each matrix U and V in Eqs. (7.75, 7.76) is a 2 2 square matrix. Further the matrices ± ± × J(1/2), J(1/2) and σ/2 must describe the same irreducible representation because there is only one irreducible − ∗ inequivalent representation of SO (3) in a given vector space. In other words, the matrices J(1/2) and J(1/2) must − ∗ be the same as σ/2 up to a similarity transformation. Indeed, in the standard (or canonical) representation Eqs. (1.165-1.168) of the rotation generators, we have J(1/2) = σ/2. From which we have

1 1 1 1 (1/2) σ σ σ = σ σ σ = σ = σ∗ = J ∗ ; k = 1, 3 2 2 k 2 −2 2 2 k −2 k −2 k − k 1 1 1 (1/2) σ σ σ = σ = σ∗ = J ∗ ; k = 2 2 2 2 2 2 2 −2 2 − 2

Thus in the canonical representation described by Eqs. (1.165-1.168), we have

(1/2) σ (1/2) 1 J = and J ∗ = σ σσ . (7.77) 2 − 2 2 2 combining Eqs. (7.75, 7.76) with Eq. (7.77), it follows that

1 1 1 U J(1/2) = σU U σ = σU [U , σ] = 0 ± 2 ± ⇒ ± 2 2 ± ⇒ ± (1/2) 1 1 1 1 2 1 V J ∗ = σV V σ2σσ2 = σV V σ2σσ2 = σV σ2 − ± 2 ± ⇒ 2 ± 2 ± ⇒ 2 ± 2 ± 1 1 V σ2σ = σV σ2 [V σ2, σ] = 0 ⇒ 2 ± 2 ± ⇒ ± therefore

[U , σ] = [V σ2, σ] = 0 (7.78) ± ± 9 so that according with Schur’s lemma 2 [see 2, page 14] U and V σ2 must be proportional to the unit matrix , ± ± hence

U = c I2 2 ; V σ2 = id I2 2 V = id σ2 ± ± × ± − ± × ⇒ ± − ± (U )σ = c δmσ ; (V )mσ = id (σ2)σ ± ± ± − ± and recalling deﬁnitions (7.74) we ﬁnd

um, (0,σ)= c δmσ ; vm, (0,σ)= id (σ2)mσ (7.79) ± ± ± − ± as a matrix index it is convenient to redeﬁne σ 1/2, 1/2 1, 2. From which the ﬁrst of Eqs. (7.79) can be ≡ − → written explicitly as

1 1 u1,+ 0, 2 c+δ1,1 c+ u1,+ 0, 2 c+δ1,2 0 1 − 1 u2,+ 0, 2 c+δ2,1 0 u2,+ 0, 2 c+δ2,2 c+  1  =   =   ;  − 1  =   =   u1, 0, 2 c δ1,1 c u1, 0, 2 c δ1,2 0 − 1 − − − − 1 −  u2, 0,   c δ2,1   0   u2, 0,   c δ2,2   c   − 2   −     − − 2   −   −              9 Of course, the proportionality with the unit matrix is for each two-component matrix U+ and U−,. because the four component matrix is reducible and the Schur’s lemmas are only valid for irreducible representations. Owing to it, we have two diﬀerent coeﬃcients for the + and − parts in Eq. (7.79). Similar argument follows for V±. 7.5. DIRAC COEFFICIENTS AND PARITY CONSERVATION 193

By using the explicit form of σ2 in Eq. (7.39), the second of Eqs. (7.79) yields 1 id (σ ) v1,+ 0, + 2 + 2 1,1 0 0 1 − v2,+ 0, + 2 id+ (σ2)2,1 id+i d+  1  =  −  =  −  =   v1, 0, + 2 id (σ2)1,1 0 0 − 1 − −  v2, 0, + 2   id (σ2)2,1   id i   d   −   −   − −   −     −      1 id (σ ) v1,+ 0, 2 + 2 1,2 id+ ( i) d+ − 1 − − − − v2,+ 0, 2 id+ (σ2)2,2 0 0  − 1  =  −  =   =   v1, 0, 2 id (σ2)1,2 id ( i) d − − 1 − − − − − − −  v2, 0, 2   id (σ2)2,2   0   0   − −   −         −      thus the coeﬃcients at zero momentum yield

c+ 0 1 0 1 c+ u 0, =   ; u 0, =   (7.80) 2 c −2 0 −  0   c     −      0 d+ 1 d+ 1 0 v 0, =   ; v 0, =   (7.81) 2 0 −2 − d −  d   0       −    from Eqs. (4.45, 4.46), the spinors at arbitrary momentum read m u (p,σ) = D (L (p)) u (0,σ) (7.82) p0 r m v (p,σ) = D (L (p)) v (0,σ) (7.83) p0 r 7.5 Dirac coeﬃcients and parity conservation

In general the constants c and d in Eqs. (7.80, 7.81) are rather arbitrary. We could for instance choose c+ and ± ± d+ (or d and c ) to be zero, such that the Dirac ﬁeld would have only two non-vanishing components. The only − − way to say something else about the relative values of c or the d on physical grounds, is by considering the ± ± parity conservation scenario. Equations (5.33, 5.34) page 158, tell us how the particle annihilation and antiparticle creation operators transform under space inversion

1 P a (p,σ) P − = η∗ a ( p,σ) (7.84) c 1 c c − P a † (p,σ) P − = η a † ( p,σ) (7.85) − substituting (7.84, 7.85) in the ﬁeld expansions (7.70) we have

+ 1 η∗ 3 ip ( x) P ψk (x) P − = d p uk ( p,σ) e · P a (p,σ) (7.86) 3 − (2π) σ X Z c c 1 q η 3 ip ( x) c P ψk− (x) P − = d p vk ( p,σ) e− · P a † (p,σ) (7.87) 3 − (2π) σ X Z q then we have to characterize the coeﬃcients u ( p,σ) and v ( p,σ). We can do it by observing from Eqs. (6.27) k − k − page 165 that Li ( p)= Li (p) ; Li ( p)= L0 ( p)= L0 (p) ; L0 ( p)= L0 (p) k − k 0 − i − − i 0 − 0 194 CHAPTER 7. CAUSAL DIRAC FIELDS FOR MASSIVE PARTICLES from this and Eq. (7.18), we ﬁnd βD (L (p)) β = D (L ( p)) (7.88) − combining Eqs. (7.82, 7.88) we have

m m u ( p,σ)= D (L ( p)) u (0,σ)= βD (L (p)) β u (0,σ) − p0 − p0 r r a similar procedure can be done for v ( p,σ) from Eq. (7.83). We then obtain − m u ( p,σ) = β D (L (p)) β u (0,σ) (7.89) − p0 r m v ( p,σ) = β D (L (p)) β v (0,σ) (7.90) − p0 r substituting (7.89, 7.90) in (7.86, 7.87) we ﬁnd

+ 1 η∗ m 3 ip ( x) P ψk (x) P − = β d p [D (L (p)) β u (0,σ)] e · P a (p,σ) (7.91) 3 p0 (2π) σ r X Z q c c 1 η m 3 ip ( x) c P ψk− (x) P − = β d p [D (L (p)) β v (0,σ)] e− · P a † (p,σ) (7.92) 3 p0 (2π) σ r X Z q once again, in order to preserve causality, the parity operator should transform the annihilation and creation ﬁelds at the point x into something proportional to these ﬁelds evaluated at x. To satisfy that condition, it is P necessary that βu (0,σ) and βv (0,σ) be proportional to u (0,σ) and v (0,σ) respectively, then

βu (0,σ)= buu (0,σ) ; βv (0,σ)= bvv (0,σ) (7.93) where Eq. (7.93) says that u (0,σ) is eigenvector of β with eigenvalue bu, and v (0,σ) is eigenvector of β with eigenvalue b . Since β2 = 1, its eigenvalues are 1, hence b and b are sign factors b2 = b2 = 1. In this case, by ν ± u v u v substituting (7.93) in (7.91, 7.92) we have

+ 1 1 m 3 ip ( x) P ψk (x) P − = η∗buβ d p [D (L (p)) u (0,σ)] e · P a (p,σ) 3 p0 (2π) σ r X Z q c 1 c 1 m 3 ip ( x) c P ψk− (x) P − = η bvβ d p [D (L (p)) v (0,σ)] e− · P a † (p,σ) 3 p0 (2π) σ r X Z q the ﬁrst of this equation can be written as

+P + 1 1 3 m ip ( x) ψk (x) P ψk (x) P − = η∗buβ d p D (L (p)) u (0,σ) e · P a (p,σ) ≡ 3 p0 (2π) σ X Z r 1 q3 ip ( x) + = η∗buβ d p u (p,σ) e · P a (p,σ)= η∗buβψk ( x) 3 P (2π) σ X Z q c 1 where we have used Eqs. (7.70, 7.82). A similar procedure can be done for P ψk− (x) P − . Consequently, the ﬁelds have the following properties under space inversion

+ 1 + P ψ (x) P − = η∗buβψ ( x) (7.94) c 1 c c P P ψ − (x) P − = η b βψ − ( x) (7.95) v P 7.5. DIRAC COEFFICIENTS AND PARITY CONSERVATION 195 now we shall obtain information about the coeﬃcients c and d in Eqs. (7.80, 7.81). Let us do the exercise ± ± explicitly for u (0, 1/2). Equation (7.93) says that u (0, 1/2) must be an eigenvector of β iγ0 with eigenvalue b ≡ u 1 1 iγ0u 0, = b u 0, 2 u 2 By using Eqs. (7.38, 7.80), we obtain the explicit form of this equation

001 0 c+ c+ 000 1 0 0     = bu   (7.96) 100 0 c c − −  010 0   0   0              as in any equation for eigenvectors, we can define one of the components of the eigenvector arbitrarily. Let us set c 1/√2. With this assignment, Eq. (7.96) becomes + ≡ 0 01 0 1/√2 1/√2 0 00 1 0 0     = bu   1 00 0 c c − −  0 10 0   0   0           c  bu  − √2 0 0  1  =   √2 buc    −   0   0      which leads to two equations bu 1 c = ; buc = (7.97) − √2 − √2 2 substituting the first of these equations into the second we obtain the already known condition bu = 1. Hence √ 1 using the convention c+ = 1/ 2 and the first of equations (7.97) the vector u 0, 2 yields 1/√2 1 0 u 0, =   2 b /√2 u  0      the same procedure can be carried out for the other zero momentum vectors u (0, 1/2) and v (0, 1/2). − ± In summary, we can adjust the overall scale of the fields for the coefficient functions at zero momentum to have the specific form

1 0 1 1 0 1 1 1 u 0, =   , u 0, =   (7.98) 2 √2 bu −2 √2 0  0   b     u   0   1 1 1 1 1 1 0 v 0, =   , v 0, =   (7.99) 2 √ 0 −2 −√ bv 2 2  b   0   v        in order to build up a causal ﬁeld we make a linear combination of the annihilation and creation ﬁelds

+ c ψ (x)= κψ (x)+ λψ − (x) (7.100) 196 CHAPTER 7. CAUSAL DIRAC FIELDS FOR MASSIVE PARTICLES that commutes or anticommutes with itself and with its adjoint at space-like separations. To ﬁx the values of κ † and λ, we shall impose causality to the commutation or anticommutation relation ψk (x) , ψk¯ (y) . First from ∓ expansions (7.70, 7.71) the ﬁelds ψ (x) and ψ† (y) become h i

+ c ψ (x) = κψ (x)+ λψ − (x)

3/2 3 ip x ip x c ψk (x) = (2π)− d p κ uk (p,σ) e · a (p,σ)+ λvk (p,σ) e− · a † (p,σ) σ X Z h i 3/2 3 ip′ y ip′ y c † − ψk¯ (y) = (2π) d p′ κ∗ uk∗¯ p′,σ′ e− · a† p′,σ′ + λ∗vk¯∗ p′,σ′ e · a p′,σ′ σ′ Z X h i such that its commutation or anticommutation relations give.

1 3 3 ip x I ψk (x) , ψ¯† (y) = d p d p′ κ uk (p,σ) e · a (p,σ) ≡ k (2π)3 ∓ σ Z σ′ Z h i X X ip x c ip′ y ip′ y c +λvk (p,σ) e− · a † (p,σ) , κ∗uk∗¯ p′,σ′ e− · a† p′,σ′ + λ∗vk¯∗ p′,σ′ e · a p′,σ′ ∓ i 1 ′ I 3 3 2 ip x ip y = 3 d p d p′ κ uk (p,σ) uk∗¯ p′,σ′ e · e− · a (p,σ) , a† p′,σ′ (2π) × | | ∓ σ Z σ′ Z X X h i 2 ip x ip′ y c c + λ vk (p,σ) vk¯∗ p′,σ′ e− · e · a † (p,σ) , a p′,σ′ | | ∓ h i 1 3 3 2 ip x ip′ y I = d p d p′ κ uk (p,σ) u∗¯ p′,σ′ e · e− · δ p p′ δσσ′ (2π)3 × | | k − σ Z σ′ Z X X n 2 ip x ip′ y c c λ v (p,σ) v∗ p′,σ′ e− · e · a p′,σ′ , a † (p,σ) ∓ | | k k¯ h io

1 3 2 ip x ip y 2 ip x ip y I = d p κ u (p,σ) u∗ (p,σ) e · e− · λ v (p,σ) v∗ (p,σ) e− · e · 3 | | k k¯ ∓ | | k k¯ (2π) σ X Z n o 1 3 2 ip (x y) 2 ip (x y) I = d p κ u (p,σ) u∗ (p,σ) e · − λ v (p,σ) v∗ (p,σ) e− · − 3 | | k k¯ ∓ | | k k¯ (2π) ( " σ # " σ # ) Z X X

† In summary, the commutation or anticommutation relation ψk (x) , ψk¯ (y) gives h i∓

1 3 2 ip (x y) 2 ip (x y) ψk (x) , ψ¯† (y) = d p κ N ¯ (p) e · − λ M ¯ (p) e− · − (7.101) k (2π)3 | | kk ∓ | | kk h i∓ Z h i N ¯ (p) u (p,σ) u∗ (p,σ) (7.102) kk ≡ k k¯ σ X M ¯ (p) v (p,σ) v∗ (p,σ) (7.103) kk ≡ k k¯ σ X

In order to find N (p) we obtain first N (0) and use an apropriate boost transformation to find N (p), and similarly 7.5. DIRAC COEFFICIENTS AND PARITY CONSERVATION 197

for M (p). For example, the matrix Nkk¯ (0) for p = 0, can be evaluated explicitly from Eqs. (7.98)

N ¯ (0) u (0,σ) u∗ (0,σ) kk ≡ k k¯ σ X 1 1 1 1 1 1 1 N = u 0, u∗ 0, + u 0, u∗ 0, = + 0 0= 11 1 2 1 2 1 −2 1 −2 √ √ · 2 2 2 1 1 1 1 1 1 N = u 0, u∗ 0, + u 0, u∗ 0, = 0 + 0 = 0 12 1 2 2 2 1 −2 2 −2 √ · · √ 2 2 1 1 1 1 1 bu∗ bu N13 = u1 0, u∗ 0, + u1 0, u∗ 0, = + 0 0= 2 3 2 −2 3 −2 √2 · √2 · 2 1 1 1 1 1 bu∗ N = u 0, u∗ 0, + u 0, u∗ 0, = 0 + 0 = 0 10 1 2 0 2 1 −2 0 −2 √ · · √ 2 2 proceeding the same for the other components we obtain explicitly

1 1 2 0 2 bu 0 1 1 0 2 0 2 bu N (0) =  1 1  2 bu 0 2 0  0 1 b 0 1   2 u 2   10 00  0 01 0 1 01 00 0 00 1 1 N (0) =   + bu   = [1 + buβ] 2 00 10 1 00 0 2    00 01   0 10 0            we can do the same with the vectors vk (0,σ) to obtain M (0). In conclusion, by using either the eigenvalue conditions (7.93) or the expressions (7.98, 7.99) we ﬁnd for the coeﬃcients Nkk¯ and Mkk¯ at zero momentum that

1+ b β 1+ b β N (0)= u ; M (0)= v (7.104) 2 2 Now, to obtain N (p) at arbitrary momentum, we apply (7.82), so we have

m m ∗ N ¯ (p) u (p,σ) u∗ (p,σ)= D (L (p)) u (0,σ) D¯ (L (p)) u (0,σ) kk ≡ k k¯ p0 km m p0 kn n σ σ "r m #"r n # Xm X X X = D (L (p)) D¯ (L (p)) u (0,σ) u∗ (0,σ) p0 km kn m n m n σ m X X X m = D (L (p)) D¯ (L (p)) N (0)= D (L (p)) N (0) D ¯ (L (p)) p0 km kn mn p0 km mn nk m n m n mX X X X = D (L (p))[1 + b β] D∗ (L (p)) e 2p0 km u mn nk¯ m n m X X N ¯ (p) = D (L (p))[1 + b β] De† (L (p)) kk 2p0 km u mn nk¯ m n X X and similarly for M (p). Therefore, from Eqs. (7.82, 7.83) we obtain those coeﬃcients at arbitrary momentum m N (p) = D (L (p))[1 + b β] D† (L (p)) (7.105) 2p0 u m M (p) = D (L (p))[1 + b β] D† (L (p)) (7.106) 2p0 v 198 CHAPTER 7. CAUSAL DIRAC FIELDS FOR MASSIVE PARTICLES

The pseudounitarity condition (7.51) applied to Λ = L (p) gives

1 βD (L (p))† β = D (L (p))− (7.107) that can be written as

D (L (p)) βD (L (p))† β = 1 2 D (L (p)) βD (L (p))† β = β since β2 = 1 we get D (L (p)) β D† (L (p)) = β alternatively Eq. (7.107) also yields

1 2 2 1 βD (L (p))† β = D (L (p))− β D (L (p))† β = βD (L (p))− β 1 ⇒ 1 D (L (p))† = βD (L (p))− β D (L (p)) D (L (p))† = D (L (p)) βD (L (p))− β ⇒ therefore the pseudounitarity condition can be written in either of these ways

D (L (p)) β D† (L (p)) = β (7.108) 1 D (L (p)) D† (L (p)) = D (L (p)) β D− (L (p)) β (7.109) and using the fact that β = iγ0 and the “four-vector” character of γµ Eq. (7.9) we ﬁnd

1 0 1 0 µ 1 0 µ β′ D (L (p)) β D− (L (p)) = iD (L (p)) γ D− (L (p)) = iL (p) γ = i L− (p) γ ≡ µ µ = i [L ( p)]0 γµ = i [L ( p)]0 γ0 + i [L ( p)]0 γi − µ − 0 − i p0 (p0)2 p (p0)2 m2 = i γ0 + i p 1 γi = i 0 γ0 i p − γi m − is m2 −  − m −  is m2  p  pb  b  = i 0 γ0 i i p γi − m − m | | b where we have used Eqs. (1.49, 6.27), we ﬁnally obtain

µ 1 0 µ pµγ D (L (p)) β D− (L (p)) = iL (p) γ = i (7.110) µ − m substituting (7.108), (7.109) and (7.110) in Eq. (7.105), the factor N (p) becomes m m m N (p) = D (L (p)) [1 + b β] D† (L (p)) = D (L (p)) D† (L (p)) + b D (L (p)) βD† (L (p)) 2p0 u 2p0 2p0 u µ m 1 m pµγ m = D (L (p)) β D− (L (p)) β + b β = i β + b β 2p0 2p0 u − 2p0 2p0 u something similar can be done for M (p) in Eq. (7.106) we then obtain

1 N (p) = [ ipµγ + b m] β (7.111) 2p0 − µ u 1 M (p) = [ ipµγ + b m] β (7.112) 2p0 − µ v 7.5. DIRAC COEFFICIENTS AND PARITY CONSERVATION 199 substituting (7.111, 7.112) in Eq. (7.101) the commutator or anticommutator of the ﬁelds become

1 3 2 1 µ ip (x y) I ψk (x) , ψ¯† (y) = d p κ [ iγµp + bum] β ¯ e · − ≡ k (2π)3 | | 2p0 − km mk h i∓ Z 2 1 µ ip (x y) λ [ iγ p + b m] β ¯ e− · − ∓ | | 2p0 − µ v km mk 1 3 2 1 µ ip (x y) 2 1 µ ip (x y) = 3 d p κ 0 [ γµ∂ + bum] β e · − λ 0 [ γµ∂ + bvm] β e− · − (2π) | | 2p − ∓ | | 2p − ¯ Z kk 3 3 2 µ 1 d p ip (x y) 2 µ 1 d p ip (y x) = κ [ γµ∂ + bum] β e · − λ [ γµ∂ + bvm] β e · − | | − (2π)3 2p0 ∓ | | − (2π)3 2p0 Z kk¯ Z so we obtain

2 µ 2 µ ψk (x) , ψ¯† (y) = κ [ γ ∂µ + bum] β ∆+ (x y) λ [ γ ∂µ + bvm] β ∆+ (y x) (7.113) k | | − − ∓ | | − − kk¯ h i∓ n o where ∆+ is the function deﬁned in Eq. (5.9), page 152

3 1 d p ip x ∆+ (x) e · (7.114) ≡ (2π)3 2p0 Z we saw in section 5 that ∆ (x y) is an even function of x y for space-like separations between x and y. It + − − implies that its ﬁrst derivatives are odd functions of x y. Then the commutation or anticommutation relations − give

2 µ 2 µ ψk (x) , ψ¯† (y) = κ [ γ ∂µ + bum] β ∆+ (x y) λ [γ ∂µ + bvm] β ∆+ (x y) k | | − − ∓ | | − kk¯ ∓ h i n 2 2 µ 2 2 o ψ (x) , ψ† (y) = κ λ γ ∂ β ∆ (x y)+ κ b λ b mβ ∆ (x y) − | | ∓ | | µ + − | | u ∓ | | v + − h i∓ n o h i Therefore, in order that both the derivative and non-derivative terms in the commutator or anticommutator vanish at space-like separations, it is necessary and suﬃcient that

κ 2 = λ 2 (7.115) | | ∓ | | and κ 2 b = λ 2 b (7.116) | | u ± | | v it is clear that (7.115) discards the possibility of a minus sign which corresponds to the case of commutators. Consequently, the particles described by Dirac fields must be fermions10. Combining Eqs. (7.115, 7.116) we see that it is also necessary that κ 2 = λ 2 and b = b . As in the case of scalars we can redefine the phases | | | | u − v such that κ/λ be real positive, in that case we have κ = λ, and absorbing the overall phase of the field ψ we obtain finally κ = λ = 1

In addition, it is possible to replace ψ by γ5ψ, which changes the sign of both bu and bv, then we can choose

b = b = +1 (7.117) u − v from Eqs. (7.70, 7.71) along with (7.100), the Dirac ﬁeld becomes

1 3 ip x ip x c ψk (x)= d p uk (p,σ) e · a (p,σ)+ vk (p,σ) e− · a † (p,σ) (7.118) 3 (2π) σ X Z h i q 10Further, we already saw that j = 1/2. So we have predicted once more that fermions are half-odd integer spin particles. 200 CHAPTER 7. CAUSAL DIRAC FIELDS FOR MASSIVE PARTICLES where the coeﬃcients at zero momentum can be obtained by combining (7.98, 7.99, 7.117) and they are given by 1 0 1 1 0 1 1 1 u 0, =   , u 0, =   (7.119) 2 √2 1 −2 √2 0  0   1       0  1 − 1 1 1 1 1 0 v 0, =   , v 0, =   (7.120) 2 √2 0 −2 √2 1  1   0   −        the spin sums are 1 N (p) = [ ipµγ + m] β (7.121) 2p0 − µ 1 M (p) = [ ipµγ m] β (7.122) 2p0 − µ − so that the anticommutator in Eq. (7.101), yields

µ ψk (x) , ψ¯† (y) = [ γ ∂µ + m] β ¯ ∆ (x y) (7.123) k + { − }kk − h i where ∆(x y) is defined in Eq. (5.32) page 158. We now come back to the requirement that for a parity − conserving theory, under space inversion the field ψ (x) must transform into something proportional to ψ ( x). P For this to be possible the phases in Eqs. (7.94, 7.95) must be equal. Therefore, the intrinsic parities of particles and their antiparticles are related by c η = η∗ − then we obtain that the intrinsic parity ηηc = η 2 of a state consisting of a spin 1/2 particle and its antiparticle − | | is odd. Now, equations (7.94, 7.95) provide the transformation of the field ψ (x) under space inversion

1 P ψ (x) P − = η∗βψ ( x) P Applying u (p,σ) on both sides of Eq. (7.110) and using Eq. (7.82) we ﬁnd µ 1 pµγ D (L (p)) β D− (L (p)) u (p,σ) = i u (p,σ) − m m p γµ D (L (p)) β u (0,σ) = i µ u (p,σ) p0 − m r and using Eqs. (7.93, 7.117) and (7.82) we obtain m p γµ b D (L (p)) u (0,σ) = i µ u (p,σ) u p0 − m r p γµ u (p,σ) = i µ u (p,σ) − m a similar procedure can be done for v (p,σ), and we get p γµ v (p,σ)= i µ v (p,σ) − − m therefore, we see that u (p,σ) and v (p,σ) are eigenvectors of ipµγ /m with eigenvalues +1 and 1 respectively. − µ − We can rewrite these equations as

(ipµγ + m) u (p,σ) = 0 , ( ipµγ + m) v (p,σ) = 0 (7.124) µ − µ 7.6. CHARGE-CONJUGATION PROPERTIES OF DIRAC FIELDS 201

µ applying the operator (γ ∂µ + m) on the ﬁeld ψ (x) of Eq. (7.118), and using Eqs. (7.124), we obtain

µ 3/2 3 µ ip x µ ip x c (γ ∂µ + m) ψ (x) = (2π)− d p (γ ∂µ + m) e · u (p,σ) a (p,σ)+ (γ ∂µ + m) e− · v (p,σ) a † (p,σ) σ Z X n o 3/2 3 ip x µ ip x µ c = (2π)− d p e · [(iγ p + m) u (p,σ)] a (p,σ)+ e− · [( iγ p + m) v (p,σ)] a † (p,σ) µ − µ σ Z n = 0 X

Therefore, from Eqs. (7.124) we find that the field (7.118) satisfies the following differential equation

µ (γ ∂µ + m) ψ (x) = 0 (7.125) which is the so-called Dirac equation for a free particle of spin 1/2. In this approach the free-particle Dirac equation is the Lorentz invariant way in which we have put together the two irreducible representations of the proper orthochronus Lorentz group in order to form a ﬁeld with a simple transformation rule under space inversion.

7.6 Charge-conjugation properties of Dirac ﬁelds

In order to characterize the charge conjugation and time-reversal properties of the Dirac field, we require expressions for the complex conjugates of u and v. According with Eqs. (7.119, 7.120) they are real for zero momentum. To obtain them at arbitrary momentum we should multiply these coefficients with the complex matrix D (L (p)). Hence we require first an expression for D∗ (L (p)). To obtain it, we start with Eq. (7.66) for a general real value of ωµν, to get

1 µν ∗ 1 µν 1 µν 1 1 ∗ = i ω = iω ∗ = iω − = − ; βC Z 2 J µν −2 µν J 2 µν KJ K KZK K≡ 2 1 1 2 1 n n 1 ∗ = − − = − ∗ = − Z KZK KZK KZ K ⇒ Z KZ K hence the same transformation (7.66) can be applied to a power series of i µν ω /2 as long as it is convergent. J µν In particular, we have 1 µν ∗ 1 µν 1 exp i ω = βC exp i ω C− β (7.126) 2 J µν 2 J µν the LHS of this equation gives an arbitrary element D (Λ) of the spinor representation of the proper orthochronus homogeneous Lorentz group. Therefore, we have in particular

1 1 D∗ (L (p)) = βCD (L (p)) C− β = (βC) D (L (p)) (βC)− (7.127) then for an arbitrary p, the coeﬃcient u∗ (p,σ) yields

m ∗ m u∗ (p,σ) = D (L (p)) u (0,σ) = D∗ (L (p)) u (0,σ) p0 p0 r r m 1 u∗ (p,σ) = βCD (L (p)) C− βu (0,σ) (7.128) p0 r On the other hand, by using Eqs. (7.93, 7.117) as well as Eqs. (7.56, 7.59), we have

1 1 σ2 0 C− βu (0,σ)= b C− u (0,σ)= Cu (0,σ)= i u (0,σ) u − 0 σ − 2 202 CHAPTER 7. CAUSAL DIRAC FIELDS FOR MASSIVE PARTICLES evaluating at σ = 1/2, we obtain 0 i 0 0 1 0 − 1 1 i i 0 0 0 0 1 1 1 C− βu 0, =     =   = v 0, 2 √2 0 0 0 i 1 −√2 0 − 2  0 0 i 0   0   1   −     −        and similarly for u (0, 1/2) and also for v (0, 1/2). Then we have the properties − ± 1 1 C− βu (0,σ)= v (0,σ) ; C− βv (0,σ)= u (0,σ) (7.129) − − substituting (7.129) in (7.128) we obtain m u∗ (p,σ)= βCD (L (p)) v (0,σ)= βCv (p,σ) − p0 − r and we can do the same for v∗ (p,m). We ﬁnally obtain

u∗ (p,σ)= βCv (p,σ) ; v∗ (p,σ)= βCu (p,σ) (7.130) − − therefore, the adjoint of the ﬁeld (7.118), gives

1 3 T ip x T ip x c ψ† (x) = d p u∗ (p,σ) e− · a† (p,σ)+ v∗ (p,σ) e · a (p,σ) 3 (2π) σ X Z h i q 1 3 T ip x T ip x c ψ∗ (x) = d p [βCv (p,σ)] e− · a† (p,σ) [βCu (p,σ)] e · a (p,σ) 3 − − (2π) σ X Z h i e q 1 3 ip x ip x c ψ∗ (x) = d p βCv (p,σ) e− · a† (p,σ) βCu (p,σ) e · a (p,σ) 3 − − (2π) σ X Z h i q 1 3 ip x ip x c ψ∗ (x) = βC d p v (p,σ) e− · a† (p,σ)+ u (p,σ) e · a (p,σ) (7.131) − 3 (2π) σ X Z h i q As always, for the field to transform under charge-conjugation into another field ψC (x) with which it commutes at space-like separations, we require that the charge-conjugation parities of the particle and antiparticle be related by c ξ = ξ∗ (7.132) Now,using Eqs. (3.27, 7.132), the charge-conjugation of the field (7.118) becomes

1 1 3 ip x 1 ip x c 1 ψC (x) Cψ (x) C− = d p u (p,σ) e · Ca (p,σ) C− + v (p,σ) e− · Ca † (p,σ) C− ≡ 3 (2π) σ X Z h i 1 q3 ip x c ip x c ψC (x) = d p u (p,σ) e · ξ∗a (p,σ)+ v (p,σ) e− · ξ a† (p,σ) 3 (2π) σ X Z h i q ξ∗ 3 ip x c ip x ψC (x) = d p u (p,σ) e · a (p,σ)+ v (p,σ) e− · a† (p,σ) (7.133) 3 (2π) σ X Z h i q and comparing Eqs. (7.131, 7.133) we have

ξξ∗ 3 ip x ip x c ψ∗ (x) = βC d p v (p,σ) e− · a† (p,σ)+ u (p,σ) e · a (p,σ) − 3 (2π) σ X Z h i 2 1 ψ∗ (x) = ξβCqψ (x) ξ∗βψ∗ (x)= ξ∗ξβ Cψ (x) ξ∗βψ∗ (x)= Cψ (x)= C− ψ (x) − C ⇒ − C ⇒ − C C ξ∗Cβψ∗ (x) = ψC (x) 7.6. CHARGE-CONJUGATION PROPERTIES OF DIRAC FIELDS 203 and from Eq. (7.69) page 190, we obtain ﬁnally

1 Cψ (x) C− = ξ∗Cβψ∗ (x)= ξ∗βCψ∗ (x) (7.134) − where we write ψ∗ (x) instead of ψ† (x) on the RHS of this equation to emphasize that it is a column vector, not a row. For a system consisting of a particle and its antiparticle, there is an important diﬀerence between fermions and bosons concerning the intrinsic charge-conjugation phase. Such a state can be written by applying a creation operator for the particle and a creation operator for the antiparticle on the vacuum state 0 as follows | i

3 3 c Φ = d p d p′ χ p, σ, p′,σ′ a† (p,σ) a † p′,σ′ 0 (7.135) | i | i σ,σ′ Z Z X Now, we shall assume that the vacuum is invariant under charge conjugation

C 0 = 0 | i | i Therefore, under charge conjugation this state transforms as

3 3 c C Φ = d p d p′ χ p, σ, p′,σ′ C a† (p,σ) a † p′,σ′ 0 | i | i σ,σ′ Z Z X 3 3 1 c 1 = d p d p′ χ p, σ, p′,σ′ C a† (p,σ) C− C a † p′,σ′ C− C 0 | i σ,σ′ Z Z X h ih i c 3 3 c C Φ = ξξ d p d p′ χ p, σ, p′,σ′ a † (p,σ) a† p′,σ′ 0 | i | i σ,σ′ Z Z X interchanging the variables of integration and summation, and using Eq. (7.132) we have

3 3 c C Φ = ξξ∗ d p′ d p χ p, σ, p′,σ′ a † (p,σ) a† p′,σ′ 0 | i | i σ′,σ Z Z X 3 3 c = d p′ d p χ p, σ, p′,σ′ a † (p,σ) a† p′,σ′ 0 | i σ′,σ Z Z X and utilizing the anticommutation relations for the creation operators we obtain

3 3 c C Φ = d p′ d p χ p, σ, p′,σ′ a† p′,σ′ a † (p,σ) 0 | i − | i σ′,σ Z Z X and interchanging the dummy indices σ σ and p p ↔ ′ ↔ ′

3 3 c C Φ = d p d p′ χ p′,σ′, p,σ a† (p,σ) a † p′,σ′ 0 (7.136) | i − | i σ,σ′ Z Z X Now, if the wave function χ of the state is even or odd under the interchange of the momenta and spins of the particle and antiparticle, we have χ p′,σ′, p,σ = χ p, σ, p′,σ′ (7.137) ± in that case substituting (7.137) in (7.136) and comparing with (7.135) we obtain

C Φ = Φ | i ∓ | i 204 CHAPTER 7. CAUSAL DIRAC FIELDS FOR MASSIVE PARTICLES from which we conclude that the charge-conjugation parity of a state consisting of a particle described by a Dirac ﬁeld and its antiparticle is odd, in the sense that if the wave function χ of the state is even or odd under the interchange of the momenta and spins of the particle and antiparticle, then the charge-conjugation operator applied on such a state gives a sign ( 1) or (+1) respectively. − A classical example is the positronium, which is the bound state of an electron and a positron. The two lowest states of positronium are a pair of nearly degenerate states with total spin S = 0 and S = 1, called para-positronium and ortho-positronium respectively. The wave function of these two states is even under the interchange of momenta and odd or even respecively under the interchange of spin three-components. Hence the para- and ortho-positronium have C = +1 and ( 1) respectively. it leads to very diﬀerent decay pattern of them. − The para-positronium decays rapidly into a pair of photons each of which has C = 1. The ortho-positronium − can only decay much more slowly into three or more photons. Another example is given by the ρ0 and ω0 mesons. They are produced as resonances coming from high-energy electron-positron annihilation, with one photon as an intermediate state, showing that they must have C = 1, − which is consistent with the interpretation of these mesons as a pair of quark anti quark bound states with L = 0 and S = 1.

7.7 Time-reversal properties of Dirac ﬁelds

We start again from the transformation properties of the particle annihilation and antiparticle creation operators give by Eqs. (3.29) page 122, but with j = 1/2

1 1 σ T a (p,σ) T − = ζ∗ ( 1) 2 − a ( p, σ) − − − c 1 c 1 σ c T a † (p,σ) T − = ζ ( 1) 2 − a † ( p, σ) (7.138) − − − and recalling the antilinearity of T , the ﬁeld (7.118) transforms under time-reversal as

1 1 3 ip x ip x c 1 T ψk (x) T − = d p T uk (p,σ) e · a (p,σ)+ vk (p,σ) e− · a † (p,σ) T − 3 (2π) σ X Z h i q 1 3 ip x 1 ip x c 1 = d p uk∗ (p,σ) e− · T a (p,σ) T − + vk∗ (p,σ) e · T a † (p,σ) T − 3 (2π) σ X Z h i q and applying (7.138) we get

1 3/2 3 1 σ ip x c ip x c T ψ (x) T − = (2π)− d p ( 1) 2 − ζ∗u∗ (p,σ) e− · a ( p, σ)+ ζ v∗ (p,σ) e · a † ( p, σ) k − k − − k − − σ X Z h i by redeﬁning the variables of summation and integration as p and σ we ﬁnd − −

1 3/2 3 1 +σ i( p,p0) (x,x0) c i( p,p0) (x,x0) c T ψ (x) T − = (2π)− d p ( 1) 2 ζ∗u∗ ( p, σ) e− − · a (p,σ) +ζ v∗ ( p, σ) e − · a † k − k − − k − − σ X Z h 3/2 3 1 +σ i(p,p0) ( x,x0) c i(p,p0) ( x,x0) c = (2π)− d p ( 1) 2 ζ∗u∗ ( p, σ) e− · − a (p,σ) +ζ v∗ ( p, σ) e · − a † − k − − k − − σ X Z h 1 3/2 3 1 +σ ip ( x) T ψ (x) T − = (2π)− d p ( 1) 2 ζ∗u∗ ( p, σ) e− · P a (p,σ) k − k − − σ X Z h c ip ( x) c +ζ v∗ ( p, σ) e · P a † (p,σ) k − − i 7.7. TIME-REVERSAL PROPERTIES OF DIRAC FIELDS 205 so that we need formulas for u ( p, σ) and v ( p, σ) in terms of u (p,σ) and v (p,σ). To do it, we use k∗ − − k∗ − − k k Eqs. (7.82, 7.83)

m m u ( p, σ) = D (L ( p)) u (0, σ) ; v ( p, σ)= D (L ( p)) v (0, σ) (7.140) − − p0 − − − − p0 − − r r m m u∗ ( p, σ) = D∗ (L ( p)) u (0, σ) ; v∗ ( p, σ)= D∗ (L ( p)) v (0, σ) (7.141) − − p0 − − − − p0 − − r r where we have recalled that u (0,σ) and v (0,σ) are real. Then we need expressions for D (L ( p)) and also for ∗ − u (0, σ) , v (0, σ) in terms of u (0,σ) , v (0,σ). − − We start by obtaining D (L ( p)) in terms of D (L (p)). For this, we take into account that L (p) is a pure ∗ − boost, then 1 1 1 L− (p)= L ( p) D (L ( p)) = D L− (p) = D− (L (p)) − ⇒ − so that D (L ( p)) is the representation of a pure boost. Hence according wit h Eqs. (1.129, 7.3), in an inﬁnitesimal ± transformation of the type D (L ( p)) only the generators = 0i appear in the expansion of D (L ( p)) − Ki J ± 1 D (L ( p)) = 1 i ω 0i + ω i0 = 1 iω i0 (7.142) ± ± 2 0iJ i0J ± i0J we shall also use equations (7.59, 7.69, 7.18, 7.35, 7.127)

1 i0 µν C− = C , C, β = [C, γ5]= β, = [γ5, ] = 0 − { } 1 J J D∗ (L (p)) = βCD (L (p)) C− β from the previous properties we obtain

1 1 1 i0 1 D∗ (L ( p)) = βCD (L ( p)) C− β = βCD− (L (p)) C− β = βC 1 iω C− β − − − i0J = βC 1 iω i0 Cβ = Cβ 1 iω i0 βC − − i0J − − i0J i0 i0 = Cββ 1+ iωi0 C = C 1+ iωi0 C − 1 J − J D∗ (L ( p)) = CD (L (p)) C− − alternatively, it can be written as

i0 2 i0 i0 D∗ (L ( p)) = C 1+ iωi0 C = C (γ5) 1+ iωi0 C = Cγ5 1+ iωi0 γ5C − − J 1 − J − J D∗ (L ( p)) = γ CD (L (p)) C− γ − 5 5 or equivalently as

1 1 D∗ (L ( p)) = γ CD (L (p)) C− γ = γ β βCD (L (p)) C− β βγ = γ βD∗ (L (p)) βγ − 5 5 5 5 5 5 from which the matrix representation D (L ( p)) can be written in several equivalent forms ∗ − 1 1 D∗ (L ( p)) = CD (L (p)) C− = γ CD (L (p)) C− γ = γ βD∗ (L (p)) βγ (7.143) − 5 5 5 5 The next step is to characterize u (0, σ) , v (0, σ) in terms of u (0,σ) , v (0,σ). For this we can combine − − Eqs. (7.56, 7.59, 7.93, 7.117) to obtain

1 2 2 γ5C− u (0, σ) = γ5Cu (0, σ)= γ5γ βu (0, σ)= buγ5γ u (0, σ) 1 − − 2 − − − − − γ C− u (0, σ) = γ γ u (0, σ) 5 − − 5 − 206 CHAPTER 7. CAUSAL DIRAC FIELDS FOR MASSIVE PARTICLES and using the explicit expressions (7.46, 7.38, 7.39, 7.119) we see that

0 2 1 1 2 1 γ5γ 1 γ5C− u 0, = γ5γ u 0, =   −2 − −2 − √ 0 2  1    10 0 0 0 0 0 i 0 1 − i 01 0 0 0 0 i 0 1 1 0 =       =   √2 0 0 1 0 0 i 0 0 0 √2 1 −  0 0 0 1   i 0 0 0   1   0   −   −       10 0 0   0 0 0 i   1   0 − 1 1 i 01 0 0 0 0 i 0 0 1 1 γ5C− u 0, + =       =   2 √2 0 0 1 0 0 i 0 0 1 −√2 0 −  0 0 0 1   i 0 0 0   0   1   −   −              then we ﬁnd 1 1 1 γ C− u 0, = u 0, 5 ∓2 ± ±2 which can also be written as 1 1 σ γ C− u (0, σ) = ( 1) 2 − u (0,σ) 5 − − proceeding similarly for v (0, σ) we ﬁnd −

1 1 σ γ C− u (0, σ) = ( 1) 2 − u (0,σ) (7.144) 5 − − 1 1 σ γ C− v (0, σ) = ( 1) 2 − v (0,σ) (7.145) 5 − − Now, substituting Eqs. (7.143, 7.144) in Eq. (7.141) we can obtain the coeﬃcients u ( p, σ) , v ( p, σ) in ∗ − − ∗ − − terms of u (p,σ) , v (p,σ) as required

1 +σ 1 +σ m 1 +σ m 1 ( 1) 2 u∗ ( p, σ) = ( 1) 2 D∗ (L ( p)) u (0, σ) = ( 1) 2 γ CD (L (p)) C− γ u (0, σ) − − − − p0 − − − p0 5 5 − r r 1 +σ m 1 = ( 1) 2 γ CD (L (p)) γ C− u (0, σ) − p0 5 5 − r 1 +σ m 1 σ m = ( 1) 2 γ CD (L (p)) ( 1) 2 − u (0,σ) = γ C D (L (p)) u (0,σ) − p0 5 − − 5 p0 r n o r = γ C u (p,σ) − 5 and similarly for v ( p, σ). From which we obtain the desired relations ??? ∗ − −

1 +σ ( 1) 2 u∗ ( p, σ) = γ Cu (p,σ) (7.146) − − − − 5 1 +σ ( 1) 2 v∗ ( p, σ) = γ Cv (p,σ) (7.147) − − − − 5 again, in order for time-reversal to take the Dirac field into a field proportional to itself evaluated at the time reversed point (with which it would anticommute at space-like separations) it is necessary that the time-reversal phases be related by c ζ = ζ∗ (7.148) 7.8. MAJORANA FERMIONS AND FIELDS 207 in that case, by substituting (7.146, 7.147) and (7.148) in (7.139) we find

1 3/2 3 1 +σ ip ( x) ip ( x) c T ψ (x) T − = ζ∗ (2π)− d p ( 1) 2 u∗ ( p, σ) e− · P a (p,σ)+ v∗ ( p, σ) e · P a † (p,σ) k − k − − k − − σ X Z h i 1 3/2 3 ip ( x) ip ( x) c T ψ (x) T − = ζ∗ (2π)− d p γ Cu (p,σ) e− · P a (p,σ) γ Cv (p,σ) e · P a † (p,σ) k − 5 − 5 σ X Z h i 1 3/2 3 ip ( x) ip ( x) c T ψ (x) T − = ζ∗ (2π)− γ C d p u (p,σ) e · −P a (p,σ)+ v (p,σ) e− · −P a † (p,σ) k − 5 σ X Z h i and comparing with (7.118) we have ﬁnally

1 T ψ (x) T − = ζ∗γ Cψ ( x) (7.149) − 5 −P

7.8 Majorana fermions and ﬁelds

We have been distinguishing particles and antiparticles. However, it not not ruled out the scenario in which they are identical. Spin 1/2 particles that coincide with their antiparticles are called Majorana fermions. Using the same reasoning that led to (7.134), the Dirac ﬁeld of a Majorana particle must satisfy the reality condition

ψ (x)= βCψ∗ (x) for Majorana fermions − In addition, the intrinsic space-inversion parity of a Majorana particle must be imaginary, η = i, while charge ± conjugation parity must be real ξ = 1. ±

7.9 Scalar interaction densities from Dirac ﬁelds

The Lorentz transformation of the ﬁeld (7.118), can be obtained from the Lorentz transformation of the creation ﬁeld Eq. (3.26) page 122, applied on a homogeneous Lorentz transformation

0 1 (Λp) (j) U (Λ) a† (pσn) U − (Λ) = D (W (Λ,p)) a† (p σ¯ n) (7.150) 0 0 p0 σσ¯ Λ s σ¯ X The ﬁeld transforms as 1 1 U [Λ] ψ (x) U − [Λ] = D− (Λ) ψ (Λx) (7.151) and its adjoint transforms as

1 1 † U [Λ] ψ (x) U − [Λ] † = U − [Λ] † ψ† (x) U † [Λ] = βU † [Λ] β ψ† (x) U † [Λ]

1n o 1 = β†U [Λ] β† ψ† (x) βU − [Λ] β = βU [Λ] βψ† (x) βU − [Λ] β

n o 1 = βU [Λ] βψ† (x) β βU [Λ] − ???? { } { } n o For future purposes, it is important to find out how to construct interaction densities out of Dirac fields and their adjoints. Since the Dirac representation is not unitary, the bilinear form ψ† (x) ψ (x) is not a scalar. To solve that problem it is convenient to define another kind of “adjoint”

ψ¯ (x) ψ†β (7.152) ≡ 208 CHAPTER 7. CAUSAL DIRAC FIELDS FOR MASSIVE PARTICLES from the pseudounitarity condition (7.51) we could show that the bilinear forms ψ¯ (x) Mψ (x) have the following transformation properties

¯ 1 ¯ 1 U0 (Λ) ψ (x) Mψ (x) U0− (Λ) = ψ (Λx) D (Λ) M D− (Λ) ψ (Λx) (7.153) and under space inversion 1 P ψ¯ (x) Mψ (x) P − = ψ¯ ( x) βMβψ ( x) (7.154) P P by comparing Eqs. (7.153, 7.154) we see that both transformations have a similar structure by making the assignments U (Λ) P , Λ and D (Λ) β, emphasizing once more the role of β as a parity operation for 0 ↔ ↔ P ↔ Dirac ﬁelds. Taking M as the basis matrices I, γµ, µν , γ , γ γµ we obtain bilinears that transform as J 5 5 ψ¯ (x) Iψ (x) scalar ψ¯ (x) γµψ (x) four-vector µ ψ¯ (x) γ5γ ψ (x) axial or pseudo four-vector

ψ¯ (x) γ5ψ (x) pseudoscalar ψ¯ (x) µν ψ (x) second rank tensor J we recall that the terms axial or pseudo mean that they have opposite properties of transformation under space- inversion with respect to ordinary vectors and scalars. A pseudoscalar has negative parity. Further, the space and time components of an axial vector have positive and negative parity respectively. These results apply also when the two fermion fields in the bilinear refer to different particle species (and not particle and antiparticle), except that in the latter case a space inversion also gives a ratio of the intrinsic parities. By the law of transformation of these matrices we can find any bilinear form ψ¯ (x) Mψ (x) because such matrices form a basis. For future purposes, it is important to characterize the charge-conjugation properties of these bilinears. From Eqs. (7.134, 7.59, 7.56) as well as Eqs. (7.62-7.64) we have

1 ^ ^ 1 C ψ¯ (x) Mψ (x) C− = (βCψ)βM (βCψ∗)= (βCψ ) M Cψ = ψ¯ (x) C− MCψ = ψ¯ (x) Mψ (x) − ∗ ± where the sign in the last expression is + for the matrices I, γ γfµ, γ and ( ) for thef matrices γµ and µν . Note 5 5 − J that the first minus in this sequence comes from the Fermi statistics, and we ignore a c number anticommutator. − Consequently, if charge-conjugation is preserved, a boson field that interacts with the current must have C = +1 for scalars, pseudoscalars or axial vectors, and C = 1 for polar vectors or antisymmetric tensors. For example, − we can see in this way that the neutral pion π0 which couples with pseudoscalar or axial vector nucleon currents, must have C = +1, while the photon that couples with polar vectors has C = 1. − The currents of the form ψ¯ (x) Mψ (x) are very important in modelling the interactions. For example, the original Fermi theory of beta decay contain a couple of polar vector currents (or an interaction density) of the form µ ψ¯pγ ψnψ¯eγµψν which was parity conserving. It was discovered later that the most general non-derivative Lorentz-invariant and parity-conserving beta decay interactions has the form of a linear combination of products of two currents like this but with γµ replaced by any one of the five covariant types of 4 4 matrices I, γ , µν, γ γµ or γ . We are × 5 J 5 5 assuming that the space-inversion operator is defined so that the proton, electron and neutron all have intrinsic parity +1. If we consider the neutrino as massless (which is only an approximation) its parity can also be defined as +1, if necessary the neutrino field can be replaced by γ5ψν . When it was realized that weak interactions are non-parity conserved, the list of non-derivative interactions added ten terms proportional to

ψ¯ Mψ ψ¯ Mψ and ψ¯ Mψ ψ¯ Mγ ψ ; M I, γµ, µν, γ γµ, γ p n e ν p n e 5 ν ≡ J 5 5 7.10. THE CPT THEOREM 209

7.10 The CPT theorem

We have seen that the combination of quantum mechanics with special relativity leads to the existence of antiparticles. It is necessary that each particle has an antiparticle, though of course it is possible in some cases that a given particle be its own antiparticle. The CPT theorem relates properties of particles and antiparticles in the following way

Theorem 7.1 (CPT theorem) for an appropiate choice of inversion phases under C, P and T , the product CP T of all inversions is conserved.

To prove it, we begin by characterizing the action of CP T on scalar, vector and Dirac ﬁelds. They can be obtained by composition of the inversions studied individually for each ﬁeld [see sections 5.3, 6.5, and 7.5-7.7]. The results are

1 [CP T ] φ (x) [CP T ]− = ζ∗ξ∗η∗φ† ( x) (7.155) 1 − [CP T ] φ (x) [CP T ]− = ζ∗ξ∗η∗φ† ( x) (7.156) µ − µ − 1 [CP T ] ψ (x) [CP T ]− = ζ∗ξ∗η∗γ ψ∗ ( x) (7.157) − 5 − the phases ζ,ξ and η depend on the species of particle described by each field. We can choose the phases so that fo all particles we have ζξη = 1 (7.158) taking it into account, a tensor φµ1µ2 µn must transform as a superposition of fields of the form φµi as follows ··· 1 1 1 [CP T ] φ (x) φ (x) ...φ (x) [CP T ]− = [CP T ] φ (x) [CP T ]− [CP T ] φ (x) [CP T ]− . . . { µ1 µ2 µn } µ1 µ2 × n on o 1 . . . [CP T ] φ (x) [CP T ]− = φ† ( x) φ† ( x) . . . φ† ( x) × µn − µ1 − − µ2 − − µn − n o n 1 on n o n o [CP T ] φ (x) φ (x) ...φ (x) [CP T ]− = ( 1) φ† ( x) φ† ( x) ...φ† ( x) { µ1 µ2 µn } − µ1 − µ2 − µn − therefore, any tensor φµ1µ2 µn formed from any set of scalar and vector fields and their derivatives transforms ··· into 1 n [CP T ] φµ1µ2 µn (x) [CP T ]− = ( 1) φµ† 1µ2 µn ( x) (7.159) ··· − ··· − it is important to take into account that any complex number appearing in these tensor is transformed into its complex conjugate since CP T is an antiunitary operator. It can be checked that the same rule is applied to bilinear forms of Dirac fields. Using Eq. (7.157) such bilinears have the following rule of transformation

1 [CP T ] ψ¯ (x) Mψ (x) [CP T ]− = ψ ( x) γ βM ∗γ ψ∗ ( x)= ψ¯ ( x) γ M γ ψ ( x) † (7.160) 1 2 1 − 5 5 2 − 1 − 5 5 2 − note that the minus sign coming from thee anticommutation of β and γ5 is cancelled by the anticommutation of fermionic operators. If the bilinear is a tensor of rank n, we have that M is a product of n modulo 2 Dirac matrices. Consequently γ Mγ = ( 1)n M 5 5 − so that the bilinear satisfies relation (7.159). On the other hand, equation (4.16) page 139, says that a Hermitian scalar interaction density (x) must H have the same number of creation fields and annihilation fields. Therefore, a Hermitian scalar interaction density (x) must be formed from tensors with an even total number of space-time indices (i.e. tensors of even rank), so H that 1 [CP T ] [ (x)] [CP T ]− = ( x) (7.161) H H − 210 CHAPTER 7. CAUSAL DIRAC FIELDS FOR MASSIVE PARTICLES

AB it is easy to see that the same is true for Hermitian scalar constructed from the fields ψab (x) belonging to one or more of the general irreducible representations of the homogeneous Lorentz group.From the effects of inversions on these fields we find AB 1 2B AB [CP T ] ψ (x) [CP T ]− = ( 1) ψ † ( x) ab − ab − For the Dirac field, the factor ( 1)2B is supplied by the matrix γ in Eq. (7.157). Since the scalar interaction − 5 density (x) is constructed out of products of the form ψA1B1 (x) ψA2B2 (x) . . .. In order to couple the fields in H a1b1 a2b2 this way it is necessary that both A1 + A2 + . . .and B1 + B2 + . . .be integers. Therefore

( 1)2B1+2B2+... = 1 − so that a hermitian scalar automatically satisﬁes Eq. (7.161). From Eq. (7.161) it is straightforward that CP T commutes with the interaction V give by V d3x (x, 0) ≡ H Z hence 1 [CP T ] V [CP T ]− = V and in any theory CP T commutes with the free-hamiltonian H0. Consequently, CP T commutes with the total Hamiltonian H so that it is a constant of motion. Therefore, the operator CP T that has been deﬁned here by its action on free-particle operators, acts on “in” and “out” states in the way described in sections 2.3.2-2.3.6. The physical consequences of CPT theorem has been discussed in sections 2.3.6 and 2.7.1. Chapter 8

Massless particle ﬁelds

We have constructed fields associated with massive particles so far. For scalar and Dirac fields, it is not difficult to construct zero mass states through the zero mass limit. However, we have already discussed some difficulties in taking the zero mass limit for vector fields of spin one. It owes to do with the fact that at least one of the polarization vectors blows up in this limit. More generally, it can be shown that the creation and annihilation operators for physical masssless particles of spin j 1 cannot be used to construct all of the irreducible (A, B) ≥ fields that can be constructued for finite mass. This particular limitation of field types lead us naturally to the introduction of gauge invariance. We shall construct a general free field for a massless particle by means of a linear combination of the annihilation c operators a (p,σ) for particles of momentum p and helicity σ, and the associated creation operators a † (p,σ) for the antiparticles. We shall deal with only one species of particle so that we drop the label n. In addition, we shall introduce at once the linear combination of creation and annihilation fields with coefficients κ, λ that we shall adjust to preserve causality. Under all those considerations we find

3/2 3 ip x c ip x ψk (x)=(2π)− d p κ a (p,σ) uk (p,σ) e · + λa † (p,σ) vk (p,σ) e− · (8.1) σ Z X h i for massless particles we have p0 = p . The rule of transformation of creation operators under homogeneous | | Lorentz transformations is given generically for equations 3.26, page 122

0 1 (Λp) (j) U (Λ) a† (p,σ) U − (Λ) = D (W (Λ,p)) a† (p σ¯) (8.2) 0 0 p0 × σσ¯ Λ s σ¯ X (j) where Dσσ¯ (W (Λ,p)) is an irreducible representation of the little group characterized by the elements W (Λ,p). We recall that the little group associated with the massless particles is ISO (2) and its representations are determined by Eqs. (1.238) page 51 Dσ′σ (W ) = exp (iθσ) δσ′σ (8.3) Combining equations (8.2, 8.3) we obtain the transformation rule of creation operators for massless particles under homogeneous Lorentz transformations

0 1 (Λp) U (Λ) a† (p,σ) U − (Λ) = exp [iσθ (p, Λ)] a† (pΛ,σ) (8.4) s p0 0 c 1 (Λp) c U (Λ) a † (p,σ) U − (Λ) = exp [iσθ (p, Λ)] a † (pΛ,σ) (8.5) s p0 and for the annihilation operator we have

0 1 (Λp) U (Λ) a (p,σ) U − (Λ) = exp [ iσθ (p, Λ)] a (pΛ,σ) (8.6) s p0 −

211 212 CHAPTER 8. MASSLESS PARTICLE FIELDS where p Λp, and the angle θ (p, Λ) is deﬁned by Eq. (1.240) page 52. If we want the ﬁeld to transform Λ ≡ according with a given representation D (Λ) of the homogeneous Lorentz group, we have [see Eq. (7.151)]

1 1 U (Λ) ψk (x) U − (Λ) = Dkk¯ Λ− ψk¯ (Λx) (8.7) ¯ Xk By using four-translations, we obtained equations (4.43, 4.44) that are in terms of irreducible representations of the little group elements W (Λ,p). Therefore, such equations are valid for either the massive or massless case, by replacing the apropriate little group in each case. Hence, substituting the little group representation (8.3) associated with the massless case in equations (4.43, 4.44), we see that the coeﬃcients u and v have to satisfy the conditions

p0 uk¯ (pΛ,σ) exp [iσθ (p, Λ)] = 0 Dkk¯ (Λ) uk (p,σ) (8.8) s(Λp) Xk p0 vk¯ (pΛ,σ) exp [ iσθ (p, Λ)] = 0 Dkk¯ (Λ) vk (p,σ) (8.9) − s(Λp) Xk By a procedure similar to the one follows in section 4.2.2, we can start from the standard four-momentum (0, 0,k,k), and apply a standard boost (p) that takes such a four momentum to an arbitrary physical massless L four-momentum (p, p ). The result is given by | | k uk¯ (p,σ) = | | Dkk¯ ( (p)) uk (k,σ) (8.10) s p0 L Xk k vk¯ (p,σ) = | | Dkk¯ ( (p)) vk (k,σ) (8.11) s p0 L Xk with k (0, 0, k) being the standard three-momentum. These equations are the massless analogous of Eqs. (4.45, ≡ 4.46) for massive particles. Note that here we have started from a standard non-zero three-momentum k, instead of a zero three-momentum as we did in section 4.2.2. It owes to the fact that we cannot have physical massless particles at zero three-momentum, since they travel at the speed of light in the vacuum. We can also see it by observing that four-vectors of massless particles are of the form (p, p ) which for p = 0 would indicate a zero | | four-momentum i.e. no particle at all. Now in section 4.2.3 we obtained a relation between coefficients at zero three-momentum for massive particles, by using a transformation of the little group (i.e. rotations for the massive case). The analogous procedure for massless particles permits to obtain a relation between coefficients at the standard three-momentum k, by using transformations associated with the little group ISO (2). Therefore, the role of Eqs. (4.47, 4.48) [or equivalently equations (4.49, 4.50)] for the coefficients at zero momentum for massive particles, is taken by the following relations for the coefficient functions evaluated at the standard momentum k

uk¯ (k,σ) exp [iσθ (k,W )] = Dkk¯ (W ) uk (k,σ) (8.12) Xk v¯ (k,σ) exp [ iσθ (k,W )] = D¯ (W ) v (k,σ) (8.13) k − kk k Xk where W µ is an arbitrary element of the little group associated with the four-momentum k = (k, k ), i.e. a ν | | transformation that leaves such a standard four-momentum invariant. We can extract the content of Eqs. (8.12, 8.13) by considering separately the two transformations that provides the little group as shown in Eqs. (1.206) page 45

W (θ, α, β)= S (α, β) R (θ) (8.14) 213

First, we consider the rotation R (θ) around the three-axis given by Eq. (1.205) page 45 cos θ sin θ 0 0 µ sin θ cos θ 0 0 R ν (θ)=  −  (8.15) 0 010  0 001      from this rotation Eqs. (8.12, 8.13) yield

iσθ uk¯ (k,σ) e = Dkk¯ (R (θ)) uk (k,σ) (8.16) Xk iσθ vk¯ (k,σ) e− = Dkk¯ (R (θ)) vk (k,σ) (8.17) Xk in addition, by using the other transformation S (α, β) (which is a combination of a rotation and a boost) in the x y plane, given by Eq. (1.204) page 45 − 1 0 α α − 2 2 µ 0 1 β β α + β S ν (α, β)=  −  ; ζ (8.18) α β 1 ζ ζ ≡ 2 −  α β ζ 1+ ζ   −    equations (8.12, 8.13) yield

uk¯ (k,σ) = Dkk¯ (S (α, β)) uk (k,σ) (8.19) Xk vk¯ (k,σ) = Dkk¯ (S (α, β)) vk (k,σ) (8.20) Xk in summary, Eqs. (8.16, 8.17) and (8.19, 8.20) are the ones that determine the coefficient functions at the standard momentum k. Then, Eqs. (8.10, 8.11) give us the corresponding coefficients at arbitrary momenta. Note that the equations for v are just the complex conjugates of the equations for u. With a suitable choice of phases of κ and λ it is possible to settle the coefficient functions such that

vk (p,σ)= uk∗ (p,σ) (8.21) the problem is that we cannot obtain a coefficient uk that satisfies Eq. (8.19) for general representations of the homogeneous Lorentz group, even for those representations for which we were able to construct fields for particles of a given helicity in the case of massive particles. We shall see the inconsistency by trying to construct the field for a massless particle of helicity 1 from the ± four-vector representation. Such a representation is characterized by µ µ D ν (Λ) = Λ ν (8.22) since we shall be in the four-vector representation, it is convenient to change the index notation from k, k,....¯ to the usual Minkowski notation µ,ν,.... As in the case of massive particles Eqs. (6.64, 6.65) page 170, it is customary to write the coefficient function uµ in terms of a “polarization vector” eµ (p,σ)

0 1/2 1/2 u (p,σ) 2p − e (p,σ) = (2 p )− e (p,σ) (8.23) µ ≡ µ | | µ using (8.22, 8.23), equation (8.10) yields

ν µ k µ ν k µ e (k,σ) u (p,σ) = | |D ν ( (p)) u (k,σ)= | | ν (p) s p0 L s p L 2 k | | | | µ e (p,σ) 1 µ ν p = ν (p) e (k,σ) 2 p s2 p L | | | | p 214 CHAPTER 8. MASSLESS PARTICLE FIELDS so that eµ (p,σ)= (p)µ eν (k,σ) (8.24) L ν on the other hand, from Eq. (8.16) and using Eqs. (8.22, 8.23) we have

ν µ iσθ µ ν µ e (k,σ) u (k,σ) e = D ν (R (θ)) u (k,σ)= R (θ) ν 2 k µ ν | | e (k,σ) iσθ µ e (k,σ) e = R (θ) ν p 2 k 2 k | | | | and we can proceed the samep with Eq. (8.19). Hence, inp the four-vector representation Eqs. (8.16, 8.19) become

µ iσθ µ ν e (k,σ) e = R (θ) ν e (k,σ) (8.25) µ µ ν e (k,σ) = S (α, β) ν e (k,σ) (8.26) setting σ = +1, and using µ = 1, 2, 3, 0 in Eq. (8.25) and using (8.15) we get

µ iθ µ ν e (k, +1) e = R (θ) νe (k, +1) 1 iθ 1 1 1 2 1 3 1 0 e (k, +1) e = R (θ) 1e (k, +1) + R (θ) 2e (k, +1) + R (θ) 3e (k, +1) + R (θ) 0e (k, +1) e1 (k, +1) eiθ = cos θ e1 (k, +1) + sin θ e2 (k, +1) 2 iθ 2 1 2 2 1 2 e (k, +1) e = R (θ) 1e (k, +1) + R (θ) 2e (k, +1) = sin θ e (k, +1) + cos θ e (k, +1) 3 iθ 3 3 3 0 − iθ 0 0 0 e (k, +1) e = R (θ) 3e (k, +1) = e (k, +1) ; e (k, +1) e = R (θ) 0e (k, +1) = e (k, +1) in summary we obtain

e1 (k, +1) eiθ = cos θ e1 (k, +1) + sin θ e2 (k, +1) e2 (k, +1) eiθ = sin θ e1 (k, +1) + cos θ e2 (k, +1) − e3 (k, +1) eiθ = e3 (k, +1) ; e0 (k, +1) eiθ = e0 (k, +1) the only way that the last two equations can be satisﬁed for all θ is by setting

e3 (k, +1) = e0 (k, +1) = 0 and the ﬁrst two equations for e1 (k, +1) and e2 (k, +1) can be rewritten as

cos θ eiθ e1 (k, +1) + e2 (k, +1) sin θ = 0 (8.27) − e1 (k, +1) sin θ + e2 (k, +1) cos θ eiθ = 0 (8.28) − − then we should ﬁnd a non-trivial solution for this homogeneous system of two equations with two variables. Setting e1 (k, +1) = 1, we obtain from equation (8.28)

sin θ + cos θ eiθ e2 (k, +1) = 0 cos θ eiθ e2 (k, +1) = sin θ − − ⇒ − using the identity1 cos θ eiθ = i sin θ, we find − − ie2 (k, +1) sin θ = sin θ ie2 (k, +1) = 1 e2 (k, +1) = i − ⇒ − ⇒ 1The identity follows from eiθ + e−iθ − 2eiθ −eiθ + e−iθ eiθ − e−iθ cos θ − eiθ = = = −i = −i sin θ 2 2 2i 215 and replacing e1 (k, +1) = 1 and e2 (k, +1) = i in Eq. (8.27) we see that it is a consistent solution. Then we have e1 (k, +1) = 1, e2 (k, +1) = i, e3 (k, +1) = e0 (k, +1) = 0. From Eq. (8.25) it is clear that we can multiply this vector by a constant and it remains being a valid solution. Thus we normalize it to obtain finally 1 eµ (k, +1) = (1, +i, 0, 0) √2 we can do the same for eµ (k, 1). We conclude that equation (8.25) requires that (up to a constant that can be − absorbed in the coefficients κ and λ) the polarization vector be 1 eµ (k, 1) = (1, i, 0, 0) (8.29) ± √2 ± on the other hand, substituting (8.18) and (8.29) in Eq. (8.26) we obtain the condition

e (k, 1) = S (α, β) e (k, 1) ± ± 1 1 0 α α 1 − i 0 1 β β i  ±  =  −   ±  0 α β 1 ζ ζ 0 −  0   α β ζ 1+ ζ   0     −     1   1    i i  ±  =  ±  0 α iβ ±  0   α iβ     ±      In summary, Eq. (8.25) leads to a solution of the form (8.29) for the polarization vectors, but then Eq. (8.26) also requires that α iβ = 0 which cannot be true for arbitrary real values of α and β (recall that the parameters α,β,θ ± of our little group must be all real). Consequently, we cannot satisfy the requirement (8.19) or (8.12). Instead, we can obtain the transformation of eν (k, 1) through a complete transformation W (θ, α, β) of the Little group ± Dµ (W (θ, α, β)) eν (k, 1) = Sµ (α, β) Rλ (θ) eν (k, 1) ν ± λ ν ±

1 0 α α cos θ sin θ 0 0 1 − 1 0 1 β β sin θ cos θ 0 0 i =  −   −   ±  √2 α β 1 ζ ζ 0 010 0 −  α β ζ 1+ ζ   0 001   0   −       1 0 α α   cos θ i sin θ    cos θ i sin θ − ± ± 1 0 1 β β i cos θ sin θ 1 i cos θ sin θ =  −   ± −  =  ± −  √2 α β 1 ζ ζ 0 √2 α (cos θ i sin θ) β (sin θ i cos θ) − ± − ∓  α β ζ 1+ ζ   0   α (cos θ i sin θ) β (sin θ i cos θ)   −     ± − ∓       

iθ iθ e± e± 1 iθ iθ 1 i (cos θ i sin θ) 1 ie± e± i =  iθ± ±  =  iθ± iθ  =  ±  √2 αe± iβ (cos θ i sin θ) √2 αe± iβe± √2 α iβ ± ± ± ±  αe iθ iβ (cos θ i sin θ)   αe iθ iβe iθ   α iβ   ± ± ±   ± ± ±   ±   1 0  1   0  iθ iθ e± i e± 0 iθ 1 i iθ (α iβ) 0 =  ±  + (α iβ)   = e±  ±  + e± ±   √2 0 √2 ± 1 √2 0 √2 k k | |  0   1   0  | |  k         | |          216 CHAPTER 8. MASSLESS PARTICLE FIELDS then we obtain ﬁnally

µ ν µ λ ν µ (α iβ) µ D ν (W (θ, α, β)) e (k, 1) = S λ (α, β) R ν (θ) e (k, 1) = exp ( iθ) e (k, 1) + ± k (8.30) ± ± ± ± √2 k | | hence we conclude that we cannot construct a four-vector field from the annihilation and creation operators for a particle of null mass and helicity 1. ± Let us ignore this difficulty for now. From Eqs. (8.24, 8.29) we are able to define a polarization vector at arbitrary momentum. Moreover taking into account Eqs. (8.21, 8.23) the field (8.1) can be written in terms of the polarization vectors as follows

3/2 3 ip x c ip x aµ (x) = (2π)− d p κ a (p,σ) uµ (p,σ) e · + λa † (p,σ) uµ∗ (p,σ) e− · σ= 1 Z X± h i 3/2 3 0 1/2 ip x ip x c aµ (x) = (2π)− d p 2p − κeµ (p,σ) e · a (p,σ)+ λeµ∗ (p,σ) e− · a † (p,σ) (8.31) Z σ= 1 X± h i we shall see later the utility of this field in physical theories. Since the field (8.31) is defined on a single species of particle plus its antiparticle, it is clear that such a field satisfies the Klein-Gordon equation [see discussion below Eq. (4.72) page 150] aµ (x) = 0 (8.32) which for null mass becomes the wave equation. Other properties of the field follow from the properties of the polarization vector. On the other hand, we saw in section 1.11.4, that the Lorentz transformation (p) that takes a massless L particle momentum from k to p can be decomposed as a “boost” ( p / k ) along the X axis that takes the B | | | | 3− particle from energy k to energy p ; followed by a rotation R (p) that takes the X direction to the p direction | | | | 3− [see Eqs. (1.241, 1.242) page 52, and Eq. (1.177), page 40]. b 10 0 0 p 01 0 0 (p) = R (p) B | | ; (u)=  (u2+1) (u2 1)  (8.33) L k B 0 0 − 2u 2u | |  (u2 1) (u2+1)   −  b  0 0 2u 2u   pb1pb3  pb2 p1 0 cos θ cos φ sin φ cos φ sin θ 0 √1 pb2 √1 pb2 − 3 − − 3 − pb2pb3 pb1 cos θ sin φ cos φ sin θ sin φ 0  p2 0  R (p) =   = √1 pb2 √1 pb2 (8.34) sin θ 0 cos θ 0 − 3 − 3 b −  1 p2 0 p 0   0 0 01   − − 3 3     0 0b 01  b    p   b b  On the other hand, from Eq. (8.29) it is clear that eν (k, 1) is a purely spatial vector with only x and x ± 1 2 components. Therefore, eν (k, 1) is left invariant under the transformation of a boost along the X axis. It can ± 3− be checked explicitly from Eqs. (8.33, 8.29)

10 0 0 1 1 01 0 0 i i (u) e (k, 1) =  (u2+1) (u2 1)  = = e (k, 1) −  ±   ±  B ± 0 0 2u 2u 0 0 ±  (u2 1) (u2+1)   −   0   0   0 0 2u 2u            Consequently, equation (8.24) gives

e (p, 1) = (p) e (k, 1) = R (p) ( p / k ) e (k, 1) = R (p) e (k, 1) µ ± L µ ν ± B | | | | ± ± e (p, 1) = R (p) νe (k, 1) (8.35) ± ± b b b 217 we also see from Eq. (8.29) that

e0 (k, 1) = 0 kµe (k, 1) = kie (k, 1) = k e (k, 1) = 0 ± ⇒ µ ± i ± | | 3 ± hence e0 (k, 1)=0 and k e (k, 1) = k e (k, 1) = 0 ± · ± · ± we can see that the same is true for an arbitrary p. To show it, we substitute Eqs. (8.29, 8.34) in Eq. (8.35) to have

µ µ ν e (p, 1) = R (p) νe (k, 1) ± b b ± b b b p1p3 p2 p 0 p p3 i p2 b2 b2 1 1 b2 b2 √1 p3 −√1 p3 1 √1 p3 ∓ √1 p3 b b −b b − b− −b p2p3 p1 i p1 p3  b2 b2 p2 0   i b2 + p2 b2  = √1 p3 √1 p3  ±  = ± √1 p3 √1 p3 − − b 0 b − −  1 p2 0 p 0   1 p2   − − 3 3   0   − − 3   0 0b 01     0 b   p     p   b p p ip b   b  1 3 ∓ 2 1 ip + p p p p eµ (p, 1) = 1 2 3 ; p i = i (8.36) 2  ± 2  i ± 1 p b b1 pb3 ≡ p E − 3 − − | |  b 0 b b  p   b b  b  therefore e0 (p, 1) = 0 and then ± 2 µ i i p1 (p1p3 ip2)+ p2 ( ip1 + p2p3) p3 1 p3 pµe (p, 1) = pie (p, 1) = p pie (p, 1) = p ∓ ± − − ± ± | | ± | | 1 p2 − 3 2 b b b b b b b b b b 2 2 2 p 1 p p + p + p b1 p3 3 p = p 1 2 3 − = p | | − = 0 b | | 2 | | 2 1 p3 1 p3 b b −b b b − b now since eν (k, 1) = e ν (k, 1),p and R (p) is real, we alsop obtain ± ∗ ∓ b b ν ν e (p, 1) = e∗ (p, 1) (8.37) b ± ∓ then we also obtain p e (p, 1) = 0. Summarizing we have · ∗ ± 0 µ e (p, 1) = 0 ; p e (p, 1) = p e (p, 1) = p e∗ (p, 1) = 0 (8.38) ± µ ± · ± · ± from which we conclude that the polarization vector is purely spatial and orthogonal to the direction of propagation. Setting µ = 0 in (8.31) and using e0 (p, 1) = 0, we see that a0 (x) = 0. Applying the operator ∂ on both sides ± µ of Eq. (8.31) we have

µ 3/2 3 0 1/2 ip x µ ip x µ c ∂µa (x) = (2π)− d p 2p − κ ∂µe · e (p,σ) a (p,σ)+ λ ∂µe− · e ∗ (p,σ) a † (p,σ) Z σ= 1 X± n o µ 3/2 3 0 1/2 µ ip x ip x µ c ∂ a (x) = (2π)− d p 2p − iκp e (p,σ) e · a (p,σ) iλe− · p e ∗ (p,σ) a † (p,σ) µ µ − µ Z σ= 1 X± n o µ 0 i and using (8.38) we obtain ∂µa (x) = 0, but a (x) = 0, so that ∂ia (x) = 0. We summarize it as follows

a0 (x)=0 ; a (x)=0 ; a (x) a (x) , a0 (x) (8.39) ∇ · µ ≡ in quantum electrodynamics, these are the conditions that are satisﬁed by the vacuum potential four-vector in the Coulomb or radiation gauge. 218 CHAPTER 8. MASSLESS PARTICLE FIELDS

Since a0 (x) is null in all reference frames, it is clear that aµ (x) is not a Lorentz four-vector. Instead, Eq. (8.30) show that for an arbitrary momentum p and an arbitrary homogeneous Lorentz transformation Λ, in place of Eq. (8.8), the polarization vector must hold the relation

µ µ ν µ e (pΛ, 1) exp [ iθ (p, Λ)] = D ν (Λ) e (p, 1) + p Ω (p, Λ) ± ± ± ± then, for a general homogenous Lorentz transformation we have

1 ν U (Λ) aµ (x) U − (Λ) = Λ µaν (Λx)+ ∂µΩ (x, Λ) (8.40) where Ω(x, Λ) is constructed from a linear combination of annihilation and creation operators. However, its explicit form is not relevant for our present discussion. A field like aµ (x) can be part in a Lorentz invariant 2 Physical theory if the couplings of aµ (x) besides being formally Lorentz invariant , is also invariant under the µ gauge transformation aµ aµ + ∂µΩ. It is carried out by demanding the couplings of aµ to have the form aµj µ → µ where j is a four-vector current that satisfies the continuity equation ∂µj = 0. Despite there is no ordinary four-vector field associated with massless particles of helicities 1, we can construct ± an antisymmetric tensor field associated with such particles. Now we recall that

µ ν µ ν D ρ (Λ) D σ (Λ) = Λ ρΛ σ (8.41) provides the tensor representation of the homogenous Lorentz group coming from the tensor product of two vector representations. We shall apply the representation (8.41) on an antisymmetric tensor ρσ deﬁned by Bk ρσ = kρeσ (k, 1) kσeρ (k, 1) Bk ± − ± from Eq. (8.30) and using the invariance of the standard four-momentum kµ under the little group, we obtain

Dµ (W (θ, α, β)) Dν (W (θ, α, β)) [kρeσ (k, 1) kσeρ (k, 1)] ρ σ ± − ± = Dµ (W (θ, α, β)) Dν (W (θ, α, β)) kρeσ (k, 1) Dµ (W (θ, α, β)) Dν (W (θ, α, β)) kσeρ (k, 1) ρ σ ± − ρ σ ± = [Dµ (W (θ, α, β)) kρ] [Dν (W (θ, α, β)) eσ (k, 1)] [Dν (W (θ, α, β)) kσ] [Dµ (W (θ, α, β)) eρ (k, 1)] ρ σ ± − σ ρ ±

= kµ [Dν (W (θ, α, β)) eσ (k, 1)] kν [Dµ (W (θ, α, β)) eρ (k, 1)] σ ± − ρ ± iθ µ ν (α iβ) ν iθ ν µ (α iβ) µ = e± k e (k, 1) + ± k e± k e (k, 1) + ± k ± √2 k − ± √2 k | | | | then we obtain ﬁnally

µ ν ρ σ σ ρ iθ µ ν ν µ D ρ (W (θ, α, β)) D σ (W (θ, α, β)) [k e (k, 1) k e (k, 1)] = e± [k e (k, 1) k e (k, 1)](8.42) µ ±ν − ± ρσ iθ µν ± − ± D (W (θ, α, β)) D (W (θ, α, β)) = e± (8.43) ρ σ Bk Bk Hence, Eqs. (8.42, 8.43) show that under an appropriate choice of normalization, the coeﬃcient function that satisﬁes Eq. (8.8) for the antisymmetric tensor representation of the homogeneous Lorentz group is

µν 3/2 0 3/2 µν u (p, 1) = i (2π)− 2p − ± Bp µν 3/2 0 3/2 µ ν ν µ u (p, 1) = i (2π)− 2p − [p e (p, 1) p e (p, 1)] ± ± − ± with eµ (p, 1) given by Eq. (8.35). From this along with Eq. (8.31) we obtain the general antisymmetric tensor ± ﬁeld for massless particles of helicity 1, as follows ± f (x)= ∂ a ∂ a (8.44) µν µ ν − ν µ 2 µ µ ν It means that aµ must be invariant under formal Lorentz transformations under which a → Λ ν a . 219

it worths pointing out that this is a tensor despite aµ is not a four-vector. It owes to the fact that the extra term in Eq. (8.40) that prevents aµ to be a four-vector is cancelled in Eq. (8.44). For future purpose, by using the total antisymmetry of ερσµν we evaluate the quantities ερσµν ∂ ∂ a = ερσµν ∂ ∂ a = ερµσν ∂ ∂ a = ερσµν ∂ ∂ a σ µ ν µ σ ν − µ σ ν − σ µ ν ερσµν ∂ ∂ a = ερσµν ∂ ∂ a = ερνµσ∂ ∂ a = ερσµν ∂ ∂ a σ ν µ ν σ µ − ν σ µ − σ ν µ where in the last step of each line we have used the fact that all indices are dummy. Hence such quantities are null ρσµν ρσµν ε ∂σ∂µaν = ε ∂σ∂νaµ = 0 (8.45) In addition, from Eqs. (8.44, 8.32) and (8.39) we see that µν µ ν ν µ µ ν ν µ ν ν µ ∂µf (x) = ∂µ (∂ a ∂ a )= ∂µ (∂ a ∂ a )= a ∂ (∂µa ) µν − − − ∂µf (x) = 0 and using (8.45) we obtain ερσµν ∂ f = ερσµν ∂ (∂ a ∂ a )= ερσµν ∂ ∂ a ερσµν ∂ ∂ a = 0 σ µν σ µ ν − ν µ σ µ ν − σ ν µ hence from the field equations (8.44, 8.32, 8.39) for the vector field aµ (x), and the antisymmetric nature of ερσµν , we derive the following field equations for the tensor field f µν µν ρσµν ∂µf = 0 ; ε ∂σfµν = 0 (8.46) which are identical in form to the vacuum Maxwell equations. Now, it is important to calculate the commutation relations for the tensor fields. To do it, we require to sum µ ν over helicities of the bilinears e e ∗. ij i j i j T e (k,σ) e ∗ (k,σ)= e (k,σ) e (k, σ) k ≡ − σ= 1 σ= 1 X± X± where we have used ei (k, 1) = ei (k, 1). With the assignment σ σ it becomes ∗ ± ∓ →− ij i j j i T e (k, σ) e (k,σ)= e (k,σ) e ∗ (k,σ) k ≡ − σ= 1 σ= 1 X± X± ij ji i j T = T e (k,σ) e ∗ (k,σ) k k ≡ σ= 1 X± . Using Eq. (8.29) and taking into account that k1 = k2 = 0 and k3 = k , we obtain | | 1 1 1 1 1 1 1 1 k k e (k,σ) e ∗ (k,σ) = + =1= δ √ √ √ √ 11 − 2 σ= 1 2 2 2 2 k X± | | 1 2 1 2 1 1 k k e (k,σ) e ∗ (k,σ) = i i =0= δ √ − √ 12 − 2 σ= 1 2 2 k X± | | 1 3 1 3 k k e (k,σ) e ∗ (k,σ) = 0= δ 13 − 2 σ= 1 k X± | | 2 2 2 2 1 1 k k e (k,σ) e ∗ (k,σ) = i ( i)+ ( i) i =1= δ 2 − 2 − 22 − 2 σ= 1 k X± | | 2 3 2 3 k k e (k,σ) e ∗ (k,σ) = 0= δ 13 − 2 σ= 1 k X± | | 3 3 3 3 k k k k e (k,σ) e ∗ (k,σ) = 0= δ | | | | = δ 33 − 2 33 − 2 σ= 1 k k X± | | | | 220 CHAPTER 8. MASSLESS PARTICLE FIELDS so that we obtain i j ij i j k k T e (k,σ) e ∗ (k,σ)= δ (8.47) k ≡ ij − 2 σ= 1 k X± | | further, using (8.35) and R (p)i = e0 (k, 1) = 0, we find 0 ± ij i j i m j n T e (p,σ) e ∗ (p,σ)= R (p) e (k,σ) R (p) e ∗ (k,σ) p ≡ b m n σ= 1 σ= 1 ± ± h ih i X X m n i j m n b i b j k k = R (p) mR (p) n e (k,σ) e ∗ (k,σ)= R (p) mR (p) n δmn − k 2 σ= 1 X± | | i m j n b b R (p) mk R (p) nk b b i j = R (p) nR (p) n − h kih2 i b | | b i m j n b R (pb) mk R (p) nk ij Tp = δij − h kih2 i b | | b µ ν where we have used the orthogonality condition for rotations in the last step. Let us do explicitly R (p) νk by using (8.34) pb1pb3 pb2 p1 0 b √1 pb2 √1 pb2 0 k p1 − 3 − − 3 pb2pb3 pb1 | |  p2 0  0 k p2 1 pb2 1 pb2 R (p) k = R (p)= √ 3 √ 3 b   =  | |  − 2 − k k pb3  1 p3 0 p3 0  | | | |  − − b   k   kb  b b  0 0 01   | |   | |   p     b  then we have  b b  i j ij i j k p k p i j T e (p,σ) e ∗ (p,σ)= δ | | | | = δ p p p ≡ ij − 2 ij − σ= 1 k X± b| | b thus, we obtain finally b b i j ij i j p p T e (p,σ) e ∗ (p,σ)= δ (8.48) p ≡ ij − 2 σ= 1 p X± | | now we are able to calculate the commutation relations between the tensor fields. First, from the expression (8.31) for aµ the explicit form of the tensor fµν gives

f (x) = ∂ a ∂ a µν µ ν − ν µ 3/2 3 0 1/2 ip x = (2π)− d p 2p − κ a (p,σ) [e (p,σ) ∂ e (p,σ) ∂ ] e · ν µ − µ ν Z σ= 1 X± c ip x +λ a † (p,σ) e∗ (p,σ) ∂ e∗ (p,σ) ∂ e− · ν µ − µ ν o and deﬁning ∂ ∂ ∂ ; ∂ µ ≡ ∂xµ µ ≡ ∂yµ the commutator between two of those tensors yields

3 3 0 1/2 3 0 1/2 ip x f (x) ,f † (y) = (2π)− d p 2p − d p′ 2p′ − κ a (p,σ) e (p,σ) ∂ e (p,σ) ∂ e · µν ρσ { ν µ − µ ν } Z Z σ= 1 σ′= 1 h i X± X± c ip x ip +λa† (p,σ) e∗ (p,σ) ∂ e∗ (p,σ) ∂ e− · , κ∗ a† p′,σ′ e∗ p′,σ′ ∂ e∗ p′,σ′ ∂ e− ν µ − µ ν ν µ − µ ν c ip′ y +λ∗a p′,σ′ e p′,σ′ ∂ e p′,σ ′ ∂ e · ν µ − µ ν i 221

3 3 0 1/2 3 0 1/2 fµν (x) ,fρσ† (y) = (2π)− d p 2p − d p′ 2p′ − Z Z σ= 1 σ′= 1 h i X± X± 2 ip x ip′ y κ e (p,σ) ∂ e (p,σ) ∂ e∗ p′,σ′ ∂ e∗ p′,σ′ ∂ e · e− · a (p,σ) , a† p′,σ′ | | { ν µ − µ ν} ν µ − µ ν n 2 ip x ip′ yh c c i + λ e∗ (p,σ) ∂ e∗ (p,σ) ∂ e p′,σ′ ∂ e p′,σ′ ∂ e− · e · a† (p,σ) , a p′,σ ′ | | ν µ − µ ν ν µ − µ ν h

3 3 0 1 fµν (x) ,fρσ† (y) = (2π)− d p 2p − Z σ= 1 h i X± 2 ip (x y) κ e (p,σ) ∂ e (p,σ) ∂ e∗ (p,σ) ∂ e∗ (p,σ) ∂ e · − | | { ν µ − µ ν} ν µ − µ ν n 2 ip (x y) λ e∗ (p,σ) ∂ e∗ (p,σ) ∂ e (p,σ) ∂ e (p,σ) ∂ e− · − − | | ν µ − µ ν ν µ − µ ν o but ∂ e ip (x y) = ∂ e ip (x y), therefore µ ± · − − µ ± · −

3 3 0 1 fµν (x) ,fρσ† (y) = (2π)− d p 2p − Z σ= 1 h i X± 2 ip (x y) κ e (p,σ) ∂ e (p,σ) ∂ e∗ (p,σ) ∂ e∗ (p,σ) ∂ e · − − | | { ν µ − µ ν } ν µ − µ ν n 2 ip (x y) + λ e∗ (p,σ) ∂ e∗ (p,σ) ∂ e (p,σ) ∂ e (p,σ) ∂ e− · − | | ν µ − µ ν { ν µ − µ ν} o

3 3 0 1 f (x) ,f † (y) = (2π)− d p 2p − e (p,σ) ∂ e (p,σ) ∂ e∗ (p,σ) ∂ e∗ (p,σ) ∂ µν ρσ { ν µ − µ ν } ν µ − µ ν Z σ= 1 h i X± 2 ip (x y) 2 ip (x y) κ e · − + λ e− · − − | | | | n o

3 3 0 1 f (x) ,f † (y) = (2π)− d p 2p − e (p,σ) e∗ (p,σ) ∂ ∂ e (p,σ) e∗ (p,σ) ∂ ∂ µν ρσ ν ν µ µ − ν µ µ ν Z σ= 1 h i X± e (p,σ) e∗ (p,σ) ∂ ∂ + e (p,σ) e∗ (p,σ) ∂ ∂ − µ ν ν µ µ µ ν ν × 2 ip (x y) 2 ip (x y) κ e · − + λ e− · − × − | | | | n o

3 3 0 1 f (x) ,f † (y) = (2π)− d p 2p − e (p,σ) e∗ (p,σ) ∂ ∂ + e (p,σ) e∗ (p,σ) ∂ ∂ µν ρσ − ν ν µ µ ν µ µ ν Z ( σ= 1 σ= 1 h i X± X±

+ e (p,σ) e∗ (p,σ) ∂ ∂ e (p,σ) e∗ (p,σ) ∂ ∂ µ ν ν µ − µ µ ν ν × σ= 1 σ= 1 ) X± X± 2 ip (x y) 2 ip (x y) κ e · − λ e− · − × | | − | | n o obtaining ﬁnally ???

3 f (x) ,f † (y) = (2π)− [ g ∂ ∂ + g ∂ ∂ + g ∂ ∂ g ∂ ∂ ] µν ρσ − µρ ν σ νρ µ σ µσ ν ρ − νσ µ ρ h i 3 1 2 i p(x y) 2 i p(x y) d p κ e · − λ e− · − × 2p0 | | − | | Z h i 222 CHAPTER 8. MASSLESS PARTICLE FIELDS which clearly vanishes for x0 = y0 if and only if

κ 2 = λ 2 (8.49) | | | | in that case since fµν is a tensor, the commutator also vanishes for all space-like separations. On the other hand, Eq. (8.49) also implies that the commutators of the aµ vanish at equal times, and it can be shown that it is enough to provide the Lorentz invariance of the S matrix. − Once again, the relative phase between the creation and annihilation operators can be defined so that κ = λ. Moreover, if the particles are their own antiparticles the fields will be hermitian. This is the case of the photon. A priori we could think in using only fields of the type fµν instead of fields of the type aµ (x). After all fµν is a Lorentz tensor while aµ is not. However, and interaction density constructed only from fµν and its derivatives will have matrix elements that vanish very quickly (faster than in theories that uses the aµ field) for small energy and momentum of massless particles, those interactions at large distances would fall off faster than the usual inverse square law. It owes to the presence of derivatives in the definition (8.44) of fµν. Though it is in principle possible, gauge invariant theories that use vector fields for massless spin one particles represent a more general class of theories, that in particular include the interactions realized in nature. Similar features appear when apply the theory to the (hypothetical) gravitons, which are massless particles of helicity 2. From the creation and annihilation operators of those particles, we can construct a tensor R ± µνρσ with the algebraic properties of the Riemann-Christoffel curvature, that is they are antisymmetric within the pairs µ,ν and ρ, σ and symmetric between the pairs. However, to obtain an interaction with the inverse square law characteristic of gravitational interactions, we have to introduce a field hµν that transforms as a symmetric tensor upon to gauge transformations [analogous to Eq. (8.40)] associated with the general relativity with general coordinate transformations. As in the case of electromagnetic gauge invariance, we achieve long-range interactions by requiring general µν µν covariance that is satisfied by coupling the field to a conserved “tensor” current θ satisfying ∂µθ = 0. Roughly speaking, the tensor structure of the formalism is duplicated with respect to electrodynamics. Chapter 9

The Feynman rules

The requirement that the S matrix satisﬁes Lorentz invariance and the Cluster Decomposition Principle (CDP) − has led us naturally to the construction of the Hamiltonian density based on covariant free ﬁelds. This construction has the advantage of satisfying automatically the Lorentz invariance and the CDP properties of the S matrix − in each order of the interaction density, whatever the form of the perturbation theory chosen. However, in the approach called “old-fashioned perturbation theory” described in Sec. 2.6, the satisfaction of such conditions is not manifest at each stage of the calculation. In the present chapter we shall describe the perturbative approach developed by Feynman, Tomonaga and Schwinger in the formed described by Dyson (1949). Such an approach has the advantage that Lorentz invariance and clustering conditions are apparent at each step of the calculations.

9.1 General framework

The starting point is obtained by combining the Dyson series (2.192) for the S matrix, with Eq. (3.8) for − free-particle states

′ ′ ′ ′ ′ ′ Sp σ n ;p σ n ; ,p1σ1n1;p2σ2n2; 1 1 1 2 2 2 ··· ··· N ∞ ( i) 4 4 = − d x d x 0 a p′ σ′ n′ a p′ σ′ n′ T (x ) (x ) a† (p σ n ) a† (p σ n ) 0 N! 1 · · · N h |··· 2 2 2 1 1 1 {H 1 ···H N } 1 1 1 2 2 2 ···| i N=0 Z X (9.1) we recall that p,σ,n denote particle momenta, spin and species, 0 denotes the free-particle vacuum state, the | i primes denote labels for particles in the ﬁnal state, while non-prime denotes particles in the initial state. Further, a and a are annihilation and creation operators, T means time-ordering such that the interaction densities (x) † H are put in an order in which the arguments x0 decrease from left to right. On the other hand, according with Eq. (4.16), page 139, the Hamiltonian density can be written as a polynomial in the ﬁelds and their adjoints1

(x)= g (x) (9.2) H iHi Xi where (x) is a product of definite numbers of fields and field adjoints of each type. Moreover, Eqs. (4.39, 4.40) Hi page 143 along with Eq. (4.56) page 147, describe the field (and the field adjoint) associated with a particle species n and its associated antiparticle nc, that transforms under a given representation of the homogeneous Lorentz

1 + − In Eq. (4.16) Pág. 139, the Hamiltonian density is written in terms of the creation and annihilation fields ψk (x) and ψk (x). However, to show manifest Lorentz covariance and the conservation of internal symmetries, such creation and annihilation fields should be written in terms of fields and field adjoints of the type given by Eq. (5.23) Pág. 156.

223 224 CHAPTER 9. THE FEYNMAN RULES group (with or without space inversions)

3/2 3 ip x c ip x ψk (x) = (2π)− d p uk (p,σ,n) a (p,σ,n) e · + vk (p,σ,n) a† (p,σ, n ) e− · (9.3) σ X Z h i 3/2 3 ip x c ip x ψk† (x) = (2π)− d p uk∗ (p,σ,n) a† (p,σ, n) e− · + vk∗ (p,σ, n) a (p,σ, n ) e · (9.4) σ X Z h i the exponential exp ( ip x) is calculated by taking ± · 0 2 2 p = p + mn in chapters 5, 6, 7, we saw that the coefficients uk andpvk depend on the Lorentz transformation properties of the field and the spin of the particle that it describes. For example Eq. (5.4) page 151 show that for the scalar field 1/2 with energy E the coefficient uk is given by (2E)− . We also recall that the index k, on the field should denote the particle type and the representation of the Lorentz group under which the field transforms, and also includes a running index over the components in this representation. We shall not consider separately the interactions involving derivatives of fields, they will just be considered as other fields of the type (9.3) with different coefficients uk and vk. We shall call by convention certain species as “particles” such as electrons, protons or neutrons, and their corresponding conjugates will be called “antiparticles” such as positrons, antiprotons and antineutrons. The field operators that destroy particles and create antiparticles are called by convention simply as fields, while their corresponding adjoints which destroy antiparticles and create particles are called field adjoints. As we have seen, there are some particle species that coincide with their antiparticles. In that case, the field adjoints are proportional to the fields. Now, with a procedure similar to the one described in section 3.6.1 page 129, we proceed to move all annihilation operators to the right in Eq. (9.1). It can be done by repeated use of the commutation or anticommutation relations (3.17, 3.18, 3.19) page 120

3 a (p,σ,n) a† p′σ′n′ = εa† p′σ′n′ a (p,σ,n)+ δ p′ p δ ′ δ ′ (9.5) − σ σ n n a (p,σ,n) a p′σ′n′ = εa p′σ′n′ a (p,σ,n) (9.6) a† (p,σ,n) a† p′σ′n′ = εa† p′σ′n′ a† (p,σ,n) (9.7) +1 if n and/or n are bosons ε = ′ (9.8) 1 if n,n are both fermions − ′ also, as discussed in section 3.6.1, when an annihilation operator appears on the extreme right (or a creation operator appears on the extreme left) its contribution associated with Eq. (9.1) vanishes since according with Eq. (3.13) these operators annihilate the vacuum state

a (p,σ,n) 0 = 0 0 a† (p,σ,n) = 0 (9.9) | i ⇔ h | and the remaining contributions to Eq. (9.1) are the ones associated with Dirac functions terms on the RHS of Eq. (9.5), with every creation and annihilation operator in the initial or ﬁnal states or in the interaction density paired in this way with some other annihilation or creation operator. Therefore, each pairing of the form a (p,σ,n) a† (p′σ′n′) is ﬁnally replaced by the corresponding delta factors given by Eq. (9.5) 3 a (p,σ,n) a† p′σ′n′ δ p′ p δ ′ δ ′ → − σ σ n n or equivalently according with Eq. (9.5) 3 a (p,σ,n) a† p′σ′n′ δ p′ p δ ′ δ ′ = a (p,σ,n) a† p′σ′n′ εa† p′σ′n′ a (p,σ,n) → − σ σ n n − a (p,σ,n) a† p′σ′n′ a(p,σ,n), a† p′σ′n′ → ∓ h i 9.1. GENERAL FRAMEWORK 225

so that the pairing a (p,σ,n) a† (p′σ′n′) should be associated with its corresponding commutator or anticommutator. Note that in the Dyson series (9.1) several kind of pairings of operators can occur, the sequence of operators in such a series has the order

O a p′ σ′ n′ a p′ σ′ n′ T (x ) (x ) a† (p σ n ) a† (p σ n ) (9.10) ≡··· 2 2 2 1 1 1 {H 1 ···H N } 1 1 1 2 2 2 · · · for instance, an annihilation operator a (p′σ′n′) of a ﬁnal particle could be paired with a ﬁeld ψk (x) (contained in a given interaction (x)), as discussed above, such a pairing should be associated with its commutator or Hi anticommutator

3/2 3 ip x c ip x a p′σ′n′ , ψk (x) = (2π)− d p a p′σ′n′ , uk (p,σ, n) a (p,σ,n) e · + vk (p,σ,n) a† (p,σ, n ) e− · ∓ σ Z X nh io 3/2 3 c ip x = (2π)− d p a p′σ′n′ , a† (p,σ, n ) vk (p,σ,n) e− · σ Z X h i 3/2 3 ip x a p′σ′n′ , ψ (x) = (2π)− δ ′ δ ′ c d p v (p,σ,n) δ p p′ e− · k σσ n n k − ∓ σ Z X a p′σ′n′ , ψk (x) = 0 ∓ this commutator or anticommutator vanishes since n′ is a particle so that δn′nc = 0. For the pairing of a (p′σ′n′) with a ﬁeld adjoint we have

3/2 3 ip x c ip x a p′σ′n′ , ψk† (x) = (2π)− d p a p′σ′n′ , uk∗ (p,σ, n) a† (p,σ, n) e− · + vk∗ (p,σ, n) a (p,σ, n ) e · ∓ σ Z h i X nh io 3/2 3 ip x = (2π)− d p a p′σ′n′ , a† (p,σ, n) uk∗ (p,σ, n) e− · σ Z X h i 3/2 3 ip x a p′σ′n′ , ψ† (x) = (2π)− δ ′ δ ′ d p u∗ (p,σ,n) δ p p′ e− · k σσ n n k − ∓ σ Z h i X 3/2 ip′ x a p′σ′n′ , ψk† (x) = (2π)− uk∗ p′,σ′,n′ e− · ∓ h i note that the term [a (p′σ′n′) , ψk (x)] would be nonzero if the particle coincides with its own antiparticle. However, in that case the field must be∓ proportional to the field adjoint so that we would be able to keep only the commutator or anticommutator a (p′σ′n′) , ψk† (x) . c ∓ In addition, an annihilation operatorh a (p′σ′n ′) ofi a final antiparticle could be paired with a field ψk (x) or

ﬁeld adjoint ψk† (x). With the same reasoning as before we obtain

c a p′σ′n ′ , ψk† (x) = 0 ∓ h c i 3/2 ip′ x ap′σ′n ′ , ψk (x) = (2π)− e− · vk p′σ′n′ ∓ It can also occurs the pairing of a creation operator a† (pσn) of an initial particle with a field ψk (x) or a field adjoint ψ† (x) in an interaction (x). From Eq. (9.10) we see that this pairing occurs with the field to the left k Hi of the creation operator

ψk† (x) , a† (pσn) = 0 ∓ h i 3/2 ip x ψk (x) , a† (pσn) = (2π)− e · uk (pσn) h i∓ 226 CHAPTER 9. THE FEYNMAN RULES

c the pairing of a creation operator a† (pσn ) of an initial antiparticle with a ﬁeld ψk (x) or a ﬁeld adjoint ψk† (x) yields

c 3/2 ip x ψk† (x) , a† (pσn ) = (2π)− e · vk∗ (pσn) ∓ h c i ψk (x) , a† (pσn ) = 0 h i∓ pairing of a ﬁnal particle (or antiparticle) with an initial particle (or antiparticle) yields2

3 a p′σ′n′ , a† (pσn) = δ p′ p δ ′ δ ′ − σ σ n n ∓ h i it is clear that such a paring vanishes when it occurs between a particle and an antiparticle. Finally the paring between a field (or field adjoint) in an interaction term (x) with a field (or field adjoint) Hi in another interaction term (x). In this case, it is not convenient to express the pairing in terms of fields or field Hj adjoints. It is because when expanding a Hamiltonian density the creation and annihilation fields are in a very specific order (normal order) as can be seen in Eq. (4.16), page 139. By contrast, when replacing the creation and annihilation fields by the covariant fields and field adjoints, it is not clear the order in which the latter appear. Thus, it is more convenient to express this pairing in terms of creation and annihilation fields. From Eq. (4.4) page 137, we then see that + + + + ψk (x) , ψm (y) = ψk † (x) , ψm† (y) ∓ h i∓ for the non-trivial pairings we should preserve causality then we write

+ + θ (x y) ψ (x) , ψ † (y) θ (y x) ψ−† (y) , ψ− (x) i∆ (x,y) (9.11) − k m ± − m k ≡ − km ∓ ∓ h i h i +1 if x0 >y0 θ (x y) (9.12) − ≡ 0 if x0

1. Pairing of a final particle with quantum numbers p ,σ ,n with a field adjoint ψ† (x) in (x) provides a ′ ′ ′ k Hi factor 3/2 ip′ x a p′σ′n′ , ψk† (x) = (2π)− e− · uk∗ p′σ′n′ (9.13) ∓ h i 2. Pairing of a final antiparticle with quantum numbers p ,σ ,nc with a field ψ (x) in (x) provides a factor ′ ′ ′ k Hi c 3/2 ip′ x a p′σ′n ′ , ψk (x) = (2π)− e− · vk p′σ′n′ (9.14) ∓ 3. Pairing of an initial particle with quantum numbers p,σ,n with a field ψ (x) in (x) provides a factor k Hi 3/2 ip x ψk (x) , a† (pσn) = (2π)− e · uk (pσn) (9.15) h i∓ 2In Eq. (9.10), we see that such a pairing occurs either (a) when N = 0, i.e. when there are no vertices of interaction or (b) when the corresponding annihilation operator has jumped over all interaction operators. 9.3. DIAGRAMMATIC RULES FOR THE S MATRIX 227 −

c 4. Pairing of an initial antiparticle state with quantum numbers p,σ,n with a ﬁeld adjoint ψ† (x) in (x) k Hi gives a factor c 3/2 ip x ψk† (x) , a† (pσn ) = (2π)− e · vk∗ (pσn) (9.16) h i∓ 5. Pairing of a ﬁnal particle (or antiparticle) with numbers p′,σ′,n′ with an initial particle (or antiparticle) with numbers p,σ,n yields a factor

3 a p′σ′n′ , a† (pσn) = δ p′ p δ ′ δ ′ (9.17) − σ σ n n ∓ h i

6. Pairing of a ﬁeld ψ (x) in (x) with a ﬁeld adjoint ψm† (x) in (y) provides a factor k Hi Hj

+ + θ (x y) ψ (x) , ψ † (y) θ (y x) ψ−† (y) , ψ− (x) i∆ (x,y) (9.18) − k m ± − m k ≡ − km ∓ ∓ h i h i +1 if x0 >y0 θ (x y) (9.19) − ≡ 0 if x0

+ 3/2 3 ip x ψk (x) = (2π)− d p uk (p,σ,n) e · a (p,σ,n) (9.20) σ Z X 3/2 3 ip x c ψm− (x) = (2π)− d p vm (p,σ,n) e− · a† (p,σ,n ) (9.21) σ Z X the step function in (9.18) arise from the time ordering in (9.1).

It worths saying that if the interaction density (x) is written in the normal ordered form as in Eq. (4.16), H there will be no pairing of ﬁelds and ﬁeld adjoints in the same interaction (x), so there are no terms of the form Hi

ψ† (x) , ψ (x) ∆ (x x) k m ∼ − h i∓ Otherwise, some kind of regularization is required to give a meaning to ∆km (0). Moreover, we can encounter a pairing between an annihilation field ψ+ (x) in (x) with a creation field ψ+ (y) in (y) only if (x) was H † H H initially to the left of (y) in Eq. (9.1) or equivalently if x0 >y0. Similarly, we have a pairing of an annihilation H field ψ (y) in (y) with a creation field ψ (x) in (x) only if (y) was initially to the left of (x) in (9.1) −† H − H H H or equivalently, if y0 > x0. The appearance of in the second term of Eq. (9.18) will be clarified later. The ± quantity (9.18) is called a propagator. From Eq. (9.1), it is clear that the S matrix is obtained by multiplying all those factors, along with additional − numerical factors that we shall see later, then integrating over x x , summing over all pairings, and then over 1 · · · N the numbers of interactions of each type.

9.3 Diagrammatic rules for the S matrix − It is convenient to represent each factor described in section 9.2 to calculate the S matrix, through suitable − diagrams (see Fig. 9.1). To do it, we construct diagrams consisting of points called vertices, where each vertex represents one of the (x), and lines that represent the pairing of a creation with an annihilation operator. All Hi this with the following algorithm 228 CHAPTER 9. THE FEYNMAN RULES

Figure 9.1: For each pairing of operators that arise in the coordinate-space evaluation of the S matrix, there is a − factor and a diagram associated. The lines represent the diagrams and the expressions on the right represent the factor associated.

1. The pairing of a ﬁnal particle with a ﬁeld adjoint in one of the (x), will be represented by a line running H from the vertex representing that (x), upwards out of the diagram, carrying an arrow pointed upwards H [see Fig. 9.1 (1)].

2. The pairing of a ﬁnal antiparticle with a ﬁeld in one of the (x), will be represented by a line running from H the vertex that represents this (x) upwards out of the diagram, and carrying an arrow pointed downwards H [see Fig. 9.1 (2)]. Arrows are omitted for particles that coincide with their antiparticles (sometimes arrows pointing in both directions are used for these kind of particles).

3. The pairing of an initial particle with a ﬁeld in one of the (x) will be represented by a line running into H the diagram from below, ending in the vertex that represents this (x), and carrying an arrow pointing H upwards [see Fig. 9.1 (3)].

4. The pairing of an initial antiparticle with a ﬁeld in one of the (x) is represented by a line running into the H diagram from below, ending in the vertex that represents this (x), carrying an arrow pointing downwards H [see Fig. 9.1 (4)].

5. The pairing of a ﬁnal particle or antiparticle with an initial particle or antiparticle will be represented by a line running clear through the diagram from bottom to top, not touching any vertex, with an arrow pointing upwards for particles and downwards for antiparticles [see Fig. 9.1 (5)].

6. The pairing of a field in (x) with a field adjoint in (y) will be represented by a line joining the vertices H H associated with (x) and (y), with an arrow pointing from y to x [see Fig. 9.1 (6)]. H H Observe that arrows always point in the direction in which a particle is moving and in the opposite direction of motion of an antiparticle. Thus it is reasonable not to put any arrow (or put arrows in both senses), when the particle coincides with its own antiparticle. Note that the arrow direction in rule 6 is consistent with this convention since a field adjoint in (y) can either create a particle destroyed by a field in (x), or destroy an Hj Hi antiparticle created by a field in (x). Further, since every field or field adjoint in (x) must be paired with Hi Hi 9.4. CALCULATION OF THE S MATRIX FROM THE FACTORS AND DIAGRAMS 229 − something, the total number of lines at a vertex of type i, associated with a term (x) in Eq. (9.2), is equal to Hi the total number of field or field adjoint factors in (x). Of these lines, the number with arrows pointed into the Hi vertex or out of it, equals the number of fields or field adjoints respectively in the associated interaction term. For example, each diagram in Fig. 9.4, page 232, contains two vertices, each one associated with an interaction term of the type given by Eq. (9.24) which contains two fields and one field adjoint. Each vertex in each diagram contains three lines equal to the total number of fields (two) or field adjoint factors (one) in each (x). Hi 9.4 Calculation of the S matrix from the factors and diagrams − From the rules previously established we can calculate the contribution to the S matrix for a given process, of a − given order N in each of the interaction terms (x) in Eq. (9.2), as follows i Hi

1. We start by drawing all Feynman diagrams that contain Ni vertices of each type i, and containing a line coming into the diagrams from below for each particle or antiparticle in the initial state, and a line going upward out of the diagram for each particle or antiparticle in the final state. In addition, the diagram should contain internal lines connecting one vertex to another, in a number so that we give each vertex the proper number of lines attached to it. The lines carry arrows pointing upwards or downwards as we described in section 9.3. Each vertex is labelled with an interaction type i and a space coordinate xµ. Each line (internal or external) is labelled at the end where it runs into a vertex with a field (or a field adjoint) type k associated

with a field ψk (x) or a field adjoint ψk† (x) which creates or destroys the particle or antiparticle at the given vertex. Finally, each external line that enters or leave the diagram are labelled with the quantum numbers p,σ,n or p′,σ′,n′ associated with the initial or final particle (or antiparticle) respectively [See Fig. 9.1, page 228].

2. For each vertex of type i, we must include a factor ( i) [coming from the factor ( i)N in Eq. (9.1)] and a − − factor g [that deﬁnes the coupling constant in the product of ﬁelds (9.2) in (x)]. For each line running i Hi upwards out of the diagram, include a factor (9.13) or (9.14) for the arrow pointing up or down respectively. For each line running from below into the diagram, include a factor (9.15) or (9.16) for the arrow pointing up or down respectively. For each line running straight through the diagram include a factor (9.17). Finally, for each internal line connecting two vertices include a factor (9.18).

3. We then integrate the product of these factors over the coordinates x1,x2,... of each vertex.

4. We then sum the contribution of each Feynman diagram. The complete perturbation series for the S matrix − is obtained by adding up all the contributions of each order in each interaction type, up to the order we want to calculate.

Note that we have not included a factor 1/N!, that appears in Eq. (9.1). It is because the time-ordered product in (9.1) is a sum over the N! permutations of x x x ., where each permutation gives the same contribution 1 2 · · · N to the final result. In terms of diagrams it means that a given Feynman diagram with N vertices is on a set of N! identical diagrams, that differ from each other only by permutations of the labels of the vertices. Hence, by adding up all these N! identical diagrams we cancel the 1/N! in Eq. (9.1). In summary, we do not include more than one diagram that differ from the others only by the labelling of the vertices. However, there are some exceptions to the rules described above. In some cases there are additional combinatoric factors or signs that must be included in the contribution of a given Feynman diagram

Suppose that an interaction (x) contains among other fields and field adjoints, M factors of the same • Hi field. Now suppose that each of these fields is paired with a field adjoint in a different interaction (different for each one), or in the initial or final state. Thus, the first of these field adjoints can be paired with any of the M identical fields in (x); the second with any of the remaining M 1 identical fields, the third with the Hi − 230 CHAPTER 9. THE FEYNMAN RULES

remaining M 2 identical fields, and so on. It gives an extra factor M!. It is customary to compensate it by − defining coupling constants g so that an explicit factor 1/M! appears in any (x) containing M identical i Hi fields or field adjoints. For instance, the interaction of Mth order in a scalar single field φ (x) would be − written gφM /M!, because of the presence of the M identical fields φ (x). Further, it is customary to display a 1/M! factor when the interaction involves a sum of M factors of fields from the same symmetry multiplet, or when for any reason the coupling coefficient is totally symmetric or antisymmetric under permutations of M boson or fermion fields. We have described in the previous item the situation in which each of the M identical fields in (x), is • Hi paired with a field adjoint in a different interaction. When this is not the case, the cancellation of the M! is not complete. Let us take the opposite scenario with a Feyman diagram in which the M identical fields in one interaction (x) are paired with M corresponding field adjoints in a single other interaction (y). Hi Hj In this case, we find only M! different pairings cancelling only one of the two factors of 1/M! in the two different interactions. This is due to the fact that it makes no difference which of the field adjoints is called the first, second, third, and so on. In this case, we would require to insert an extra factor M! “by hand” into the contribution of that given Feynman diagram. For instance, Fig. 9.2 shows a diagram in which three identical fields in (x) are paired with three corresponding field adjoints in a single other interaction Hi (y). Thus an assignment of the form g g /3!, would give a contribution of the form g g / (3!)2 due to Hj i → i i j the presence of two vertices. However, the multiplicity is only 3! (the 3! permutations of the internal lines in the diagram).

Figure 9.2: This particular diagram requires an extra combinatoric factor in the S matrix. For an interaction − including say, three factors of some field (plus other fields) it is customary to include a factor 1/3! in the interaction Hamiltonian, in order to cancel factors coming from sums over ways of pairing these fields with their adjoints in other interactions. However, in this diagram there are two of those factors of 1/3! and only 3! different pairings. Hence, we are left with an extra factor of 1/3!.

Another important scenario arises when some of the permutations of the vertices have no eﬀect on the Feynman diagram. In that case other combinatoric factors appear because the cancellation of the factor 1/N! in the series (9.1) is not complete when relabelling the vertices does not give a new diagram. An example is given by Fig. 9.3 which is a Feynman diagram of Nth order in with the shape of a ring with N corners. It is clear that any H cyclic change of the labels of the vertices give the same diagram. Thus, there are only (N 1)! diﬀerent diagrams − since a permutation of labels that moves each label to the next vertex around the ring provides the same diagram. Therefore, such a diagram is accompanied by the factor (N 1)! 1 − = (9.22) N! N Physically, this situation is usual in the calculation of vacuum-to-vacuum S matrix elements in a theory with a − quadratic interaction = ψ† M ψ (9.23) H k km m 9.4. CALCULATION OF THE S MATRIX FROM THE FACTORS AND DIAGRAMS 231 −

Figure 9.3: This is an eighth-order graph describing a vacuum-to-vacuum amplitude with particles interacting with only one external field represented by wiggly lines. For a given set of labels for the vertices a cyclic change of such labels leads to the same diagram, applying other cyclic relabelling provides the same diagram again and so on until we perform seven succesive cyclic permutations (the eighth cycle would lead to the same labelling as the beginning). Therefore, there are 7! of these eight-order diagrams differing only by relabelling the vertices, but we do not count as different those labellings that simply rotates the ring. Consequently, the 1/8! factor appearing in the Dyson series (9.1) is not completely cancelled, leaving us with an extra factor 1/8. where M could depend on external fields.

In theories with fermion fields, the use of Eqs. (9.5), (9.6) and (9.7) to move annihilation operators to the • right and creation operators to the left, introduces minus signs into the contributions of several pairings. In particular, we get a minus sign when the permutation of the operators in Eq. (9.1) required to put all paired operators in the “appropriate ordered” (with annihilation operators just to the left of the paired creation operators) involves an odd number of interchanges of fermion operators. We see that by taking into account that to calculate the contribution of a given pairing, we first permute all operators in (9.1) in such a way that each annihilation operator is just to the left of the creation operator with which it is paired, ignoring all commmutators and anticommutators of unpaired operators, and replace each product of paired operators with their commutators or anticommutators. Therefore, there will arise a minus sign in the relative sign of the two terms in Eq. (9.18) for the fermion propagator. Given the permutation that puts the annihilation part ψ+ (x) of a field in (x) just to the left of the creation part ψ+ (y) of a field adjoint in (y), the H † H associated permutation that puts the annihilation part ψ−† (y) of the field adjoint just to the left of the creation part ψ− (x) of the field, involves an additional interchange of fermion operators, giving a minus sign in the second term of Eq. (9.18) for fermions.

Some additional minus signs can appear in the contribution of whole Feynman diagrams. To see it, let us • take a theory in which the sole interaction of fermions has the form

(x)= g ψ† (x) ψ (x) φ (x) (9.24) H kmn k m n kmnX 232 CHAPTER 9. THE FEYNMAN RULES

Figure 9.4: The connected second-order diagrams for fermion-fermion scattering for a type of interaction described by (9.24). The ﬁelds and ﬁeld adjoints are associated with the vertices y (left) and x (right). Straight lines represent fermions and dotted lines represent neutral bosons. There is a relative minus sign when adding the contributions of both diagrams, which arises from an extra interchange of fermion operators in the pairings associated with the second diagram.

where gkmn are general coupling constants, ψk (x) are a set of complex fermion ﬁelds, and φn (x) are a set of real bosonic (not necessarily scalar) ﬁelds. Let us study the process of fermion-fermion scattering 12 1 2 , → ′ ′ up to second order in . The fermion operators in the second order term of Eq. (9.1) appear in the order3 H a 2′ a 1′ ψ† (x) ψ (x) ψ† (y) ψ (y) a† (1) a† (2) (9.25) there are two connecting diagrams associated with the two pairings4

a 2′ ψ† (x) a 1′ ψ† (y) ψ (y) a† (1) ψ (x) a† (2) (9.26) h ih ih ih i a 1′ ψ† (x) a 2′ ψ† (y) ψ (y) a† (1) ψ (x) a† (2) (9.27) h ih ih ih i these two pairings are displayed in Fig. 9.4. In order to go from (9.25) to (9.26) we require an even permutation of fermionic operators. For example, by moving ψ (x) past three operators to the right and then move a (1′) past one operator to the right. Consequently, there is no any minus sign in going from (9.25) to (9.26) i.e. in the contribution of the pairing (9.26). On the other hand, the only difference between the pairings (9.26) and (9.27) is the interchange of two fermionic operators a (1′) and a (2′). In turn, this relative minus sign is what we require to be compatible with Fermi statistics, since it makes the scattering amplitude antisymmetric under the interchange of particles 1′ and 2′ (or 1 and 2). On the other hand, we recall that the overall sign of the S matrix is irrelevant in calculating transition rates (such a global sign − depends on sign conventions for the initial and final states). Therefore, what really matters is the relative sign between the pairings (9.26) and (9.27) instead of the sign of each pairing. Nevertheless, not all sign factors are so simply related to the antisymmetry of the final or initial states, • even in the lowest order of perturbation theory. To see it, we take as an example the fermion-antifermion scattering 1 2c 1 2c , to second order in the same interaction (9.24). The fermionic operators in the → ′ ′ second-order term in Eq. (9.1) appear in the order c c a 2 ′ a 1′ ψ† (x) ψ (x) ψ† (y) ψ (y) a† (1) a† (2 ) (9.28) and again we have two Feynman diagrams associated with the pairings

c c a 2 ′ ψ (x) a 1′ ψ† (x) ψ (y) a† (1) ψ† (y) a† (2 ) (9.29)

c h ih ih c i a 2 ′ ψ (x) a 1′ ψ† (y) ψ (y) a† (1) ψ† (x) a† (2 ) (9.30) h ih ih i 3Note that the order of the fields and field adjoints here are well defined because of the form in which we proposed our Hamiltonian density Eq. (9.24). Otherwise, the order that is well defined within a Hamiltonian density is the one concerning the creation and annihilation fields as we can see in the expansion (4.16), page 139. 4Note for instance that pairings of the form [a (2′) ψ (x)] or of the form ψ† (x) a† (2) gives no contribution as discussed in section 9.1, page 223. 9.4. CALCULATION OF THE S MATRIX FROM THE FACTORS AND DIAGRAMS 233 −

Figure 9.5: The connected second-order diagrams for fermion-antifermion scattering for a type of interaction described by (9.24). Straight lines pointing upward (downward) represent fermions (antifermions), while dotted lines represent neutral bosons. There is a relative minus sign when adding the contributions of both diagrams, which arises from an extra interchange of fermion operators in the pairings associated with the second diagram.

as displayed by Fig. 9.5, where Fig. 9.5(a) corresponds to the pairings 9.30, and Fig. 9.5(b) corresponds to the pairings in (9.29). In order to go from (9.28) to Eq. (9.29) we require an even permutation of fermionic operators. For example, we can move ψ (x) past two operators to the left and move ψ† (y) past two operators to the right. Hence, there is not extra minus sign in the contribution of the pairing (9.29). But to go from (9.28) to (9.30) requires an odd permutation [since in passing from (9.29) to (9.30) we require the interchange of ψ (x) and ψ† (y)]. Hence, both pairings (9.29) and (9.30) must have opposite signs. This sign has an indirect relation with the Fermi statistics5.

Since the same field can destroy a particle and create an antiparticle, there is a “crossing symmetry” between • processes in which initial particles or antiparticles are exchanged with final antiparticles and particles. For instance, the amplitudes for the process 1 2c 1 2c are related with the amplitudes of the “crossed process” → ′ ′ 1 2 1 2; the two pairings (9.29) and (9.30) are associated with the two diagrams for this process, that ′ → ′ differ by an interchange of 1 and 2′ (or 1′ and 2). Hence, the antisymmetry of the scattering amplitude under the interchange of initial (or final) particles requires a minus sign in the relative contribution of these two pairings. Nevertheless the “crossing symmetry” in general requires an analytic continuation in kinematic variables, making it very difficult to use in practice.

When we consider higher order contributions, additional signs appear. For example, in a theory of the • type described by (9.24), the fermion lines form either chains of lines that pass through the diagram with arbitrary numbers of interactions with boson ﬁelds as displayed by Fig. 9.6, or fermionic loops as shown in Fig. 9.7. Let us see the eﬀect of adding a fermionic loop with M corners to the Feynman diagram for any process. The diagram is associated with the following pairing of fermionic operators

ψ (x ) ψ¯ (x ) ψ (x ) ψ¯ (x ) ψ (x ) ψ¯ (x ) (9.31) 1 2 2 3 · · · M 1 that is a given vertx (say x ) starts a sequence x x . . . x x that ends into itself. But in the 1 1 → 2 → → M → 1 Dyson series Eq. (9.1) these operators appear in the order

ψ¯ (x ) ψ (x ) ψ¯ (x ) ψ (x ) ψ¯ (x ) ψ (x ) (9.32) 1 1 2 2 · · · M M 5Note however that the initial and final states in both diagrams of Fig. 9.5 are in the same “position”. We cannot see a difference by checking only the external lines. The difference has to do with the form in which the internal bosonic line connects the initial and final states. 234 CHAPTER 9. THE FEYNMAN RULES

Figure 9.6: The connected second-order diagrams for boson-fermion scattering for a type of interaction described by (9.24). Once again, straight lines represent fermions and dashed lines represent neutral bosons.

and going from (9.32) to (9.31) requires an odd number of permutations of fermionic operators (for instance, we can move ψ¯ (x ) to the right past 2M 1 operators). Therefore, the contribution of each such fermionic 1 − loop is accompanied by a minus sign.

The Feynman rules described above provides the full S matrix, including processes in which several clusters of − particles are in space-time regions widely separated from each other. According with the discussion in chapter 3, we can exclude the contributions of such disconnected clusters by taking only the connected Feynman diagrams. It excludes in particular lines passing clean through the diagram without interacting, that are associated with the factors (9.17). We shall illustrate the use of the Feynman rules by calculating the low-order contributions to the S matrix − for particle scattering using two diﬀerent theories

9.5 A fermion-boson theory

We shall consider ﬁrst the theory described by Eq. (9.24) involving interactions of fermions and self-charge conjugate bosons. Each vertex in this theory contains three lines, two fermion lines and one boson line. Based on the Feynman rules we shall construct the S matrix associated with fermion-boson and fermion-fermion scattering. −

9.5.1 Fermion-boson scattering The lowest order connected diagrams for fermion boson scattering are the ones shown in Fig. 9.6. Using the rules described in Fig. 9.1 and Secs. 9.2, 9.3, 9.4, we obtain the associated S matrix element. We show in Fig. 9.8 − all labels associated with external lines and vertices. Since all are particles, the fermion arrows point upwards (it is not usual to paint arrows for scalars). Now following the rules illustrated in Fig. 9.1 we begin by adding the factors associated with each line and vertex. We start with diagram 9.8(a).

1. For the initial (fermion) particle p1σ1n1 with a line pointing upwards, with index n for its ﬁeld component , and ending into the vertex (x,m) we put a term

ip1 x e · un (p1,σ1,n1) (2π)3/2

2. For the initial (boson) particle p2σ2n2 with a line pointing upwards, with index k for its ﬁeld component, 9.5. A FERMION-BOSON THEORY 235

Figure 9.7: The lowest order connected diagram for boson-boson scattering with an interaction density of the type (9.24). These fermion loop diagrams give an extra minus sign coming from permutations of the paired fermion ﬁelds.

and ending into the vertex (x,m) we put a term

ip2 x e · uk (p2,σ2,n2) (2π)3/2

3. Now for the vertex (x,m) we add a factor ( ig ) − mnk where m is the index of the vertex, n is the index component of the external incoming fermion and k the index of component of the ﬁeld associated with the external incoming boson. Note that the order of the labels is consistent with the one given in Eq. (9.24) in which the last index correspond to the index of the boson.

4. For the internal line connecting the vertices (x,m) and (y,m′) we put a propagator

i∆ ′ (y x) − m m −

5. Now for the vertex (y,m′) we put a factor ig ′ ′ ′ − n m k

6. For the ﬁnal (fermion) state p1′ σ1′ n1′ with a line starting at the vertex (y,m′) going upwards, with component index n′ we put a factor ′ ip1 y e− · un∗ ′ (p1′ ,σ1′ ,n1′ ) (2π)3/2 236 CHAPTER 9. THE FEYNMAN RULES

Figure 9.8: Labels to apply the Feynman rules for the lowest order diagrams associated with the fermion-boson scattering in Fig. 9.6.

7. For the ﬁnal (boson) state p2′ σ2′ n2′ with a line starting at the vertex (y,m′) going upwards, with component index k′ we put a factor ′ ip2 y e− · uk∗′ (p2′ ,σ2′ ,n2′ ) (2π)3/2 8. We put all this stuﬀ together

′ ′ ip2 y ip1 y e− · uk∗′ (p2′ ,σ2′ ,n2′ ) e− · un∗ ′ (p1′ ,σ1′ ,n1′ ) ( ign′m′k′ ) [ i∆m′m (y x)] " (2π)3/2 #" (2π)3/2 # − − −

ip2 x ip1 x e · uk (p2,σ2,n2) e · un (p1,σ1,n1) ( igmnk) × − " (2π)3/2 #" (2π)3/2 #

9. We then integrate the product of these factors over the coordinates x and y of each vertex and sum over the components of the ﬁelds

′ ′ ip2 y ip1 y 4 4 e− · uk∗′ (p2′ ,σ2′ ,n2′ ) e− · un∗ ′ (p1′ ,σ1′ ,n1′ ) D d x d y ( ig ′ ′ ′ ) [ i∆ ′ (y x)] 1 3/2 3/2 n m k m m ≡ ′ ′ ′ " (2π) #" (2π) # − − − kXn m knmX Z Z ip2 x ip1 x e · uk (p2,σ2,n2) e · un (p1,σ1,n1) ( igmnk) × − " (2π)3/2 #" (2π)3/2 #

6 2 4 4 D1 (2π)− ( i) gn′m′k′ gmnkun∗ ′ p1′ ,σ1′ ,n1′ un (p1,σ1,n1) d x d y [ i∆m′m (y x)] ≡ ′ ′ ′ − − − kXn m knmX Z Z ′ ′ ip y ip1 x ip y ip2 x e− 1· e · e− 2· u∗′ p′ ,σ′ ,n′ e · u (p ,σ ,n ) (9.33) × k 2 2 2 k 2 2 2 n o A similar procedure should be done for the second diagram to obtain its associated contribution D2. For the diagram 2 [see Fig. 9.8(b)], the initial particles p1σ1n1 (fermion) and p2σ2n2(boson) with ﬁeld components n and k respectively and connected to the vertices (x,m) and (y,m′) respectively, yield a contribution

ip2 y ip1 x e · uk (p2,σ2,n2) e · un (p1,σ1,n1) " (2π)3/2 #" (2π)3/2 # 9.5. A FERMION-BOSON THEORY 237

the vertices (x,m) and (y,m′) and the propagator (internal line) yield

( ig ′ ) , ( ig ′ ′ ) , [ i∆ ′ (y x)] − mnk − n m k − m m − and the external particles p1′ σ1′ n1′ (fermion) and p2′ σ2′ n2′ (boson) with ﬁeld components n′ and k′ respectively and connected to the vertices (y,m′) and (x,m) respectively, yield a contribution

′ ′ ip1 y ip2 x e− · un∗ ′ (p1′ ,σ1′ ,n1′ ) e− · uk∗′ (p2′ ,σ2′ ,n2′ ) " (2π)3/2 #" (2π)3/2 # making the product

ip2 y ip1 x e · uk (p2,σ2,n2) e · un (p1,σ1,n1) ( igmnk′ ) ( ign′m′k) " (2π)3/2 #" (2π)3/2 # − − ′ ′ ip1 y ip2 x e− · un∗ ′ (p1′ ,σ1′ ,n1′ ) e− · uk∗′ (p2′ ,σ2′ ,n2′ ) [ i∆m′m (y x)] − − " (2π)3/2 #" (2π)3/2 # integrating over the vertices and summing over ﬁeld components

ip2 y ip1 x 4 4 e · uk (p2,σ2,n2) e · un (p1,σ1,n1) D d x d y ( ig ′ ) ( ig ′ ′ ) 2 ≡ 3/2 3/2 − mnk − n m k k′n′m′ knm Z Z " (2π) #" (2π) # X X ′ ′ ip1 y ip2 x e− · un∗ ′ (p1′ ,σ1′ ,n1′ ) e− · uk∗′ (p2′ ,σ2′ ,n2′ ) [ i∆m′m (y x)] − − " (2π)3/2 #" (2π)3/2 #

6 2 D (2π)− ( i) g ′ ′ g ′ u∗ ′ p′ ,σ′ ,n′ u (p ,σ ,n ) 2 ≡ − n m k mnk n 1 1 1 n 1 1 1 k′n′m′ knm X X ′ ′ 4 4 ip y ip1 x ip x ip2 y d x d y [ i∆ ′ (y x)] e− 1· e · e− 2· u∗′ p′ ,σ′ ,n′ e · u (p ,σ ,n ) − m m − k 2 2 2 k 2 2 2 Z Z n o we can interchange k k within the sum since they are dummy indices ↔ ′ 6 2 4 4 D2 (2π)− ( i) gn′m′k′ gmnkun∗ ′ p1′ ,σ1′ ,n1′ un (p1,σ1,n1) d x d y [ i∆m′m (y x)] ≡ ′ ′ ′ − − − kXn m knmX Z Z ′ ′ ip y ip1 x ip x ip2 y e− 1· e · e− 2· u∗ p′ ,σ′ ,n′ e · u ′ (p ,σ ,n ) (9.34) × k 2 2 2 k 2 2 2 n o Then, the S matrix element associated with fermion-boson scattering at lowest order, is the sum D + D (in − 1 2 this case there is no a minus sign coming from odd interchanges of fermion operators). Thus adding Eqs. (9.33) and (9.34) we obtain

6 2 Sp′ σ′ n′ p′ σ′ n′ ;p σ n p σ n = (2π)− ( i) gn′m′k′ gmnku∗ ′ p′ ,σ′ ,n′ un (p1,σ1,n1) 1 1 1 2 2 2 1 1 1 2 2 2 − n 1 1 1 k′n′m′ knm X X ′ 4 4 ip y ip1 x d x d y i∆ ′ (y x) e− 1· e · × − m m − Z ′ Z h i ip y ip2 x e− 2· u∗′ p′ ,σ′ ,n′ e · u (p ,σ ,n ) × k 2 2 2 k 2 2 2 n ′ ip2 x ip2 y +e− · uk∗ p2′ ,σ2′ ,n2′ e · uk′ (p2,σ2,n2) (9.35) o where according with Fig. 9.8, the labels 1 and 2 are used for fermions and bosons respectively. 238 CHAPTER 9. THE FEYNMAN RULES

9.5.2 Fermion fermion scattering For fermion-fermion scattering there are also two second-order diagrams displayed in Fig. 9.4. The associated S matrix yields (Homework!! B1) − 6 2 S ′ ′ ′ ′ ′ ′ = (2π)− ( i) g ′ ′ g ′ p1σ1n1p2σ2n2;p1σ1n1p2σ2n2 m mk n nk ′ ′ ′ − kXn m knmX u∗ ′ p′ ,σ′ ,n′ u∗ ′ p′ ,σ′ ,n′ u (p ,σ ,n ) u (p ,σ ,n ) × m 2 2 2 n 1 1 1 m 2 2 2 n 1 1 1 ′ ′ 4 4 ip x ip y ip2 x ip1 y dx d y e− 2·e− 1· e · e · [ i∆ ′ (x y) ] × − k k − Z Z 1′ 2′ (9.36) − ⇔ where the last term indicates the interchange of the two fermions 1′ and 2′ (or the fermions 1 and 2) in the previous terms accompanied by a minus sign.

9.5.3 Boson-boson scattering There are no second-order graphs for boson-boson scattering in this theory. It is because second order graphs for a boson-boson scattering requires vertices of three lines with two bosons and one fermion. But Eq. (9.24) shows that only vertices with two fermions and one boson appears in this theory. The lowest order terms for this process are of fourth order, such as the ones displayed in Fig. 9.7. Note that each vertex in Fig. 9.7, contains three lines with two fermions and one boson as demanded by the theory in Eq. (9.24).

9.6 A boson-boson theory

Figure 9.9: The connected second-order diagrams for boson-boson scattering with an interaction of the type (9.37).

In the theory described by the interaction (9.24), the three fields are all different6. Thus, it is useful to consider an example with a trilinear interaction involving three identical fields or at least entering into the interaction in a symmetrical way. In that case, more combinatoric factors are introduced as discussed in Sec. 9.4. We shall consider an interaction density that is a sum of terms that are thrilinear in a set of real bosonic fields φk (x) 1 (x)= g φ (x) φ (x) φ (x) (9.37) H 3! kmn k m n kmnX where gkmn are coupling constants that are real and totally symmetric. Note that we have included a correction factor 1/3! to the coupling constant according with the discussion in section 9.4. We shall characterize a scattering

6The fermions fields are a field and an adjoint field. They are different even if they describe the same species of fermions. 9.6. A BOSON-BOSON THEORY 239 process 1 2 1 2 , up to second order in this interaction. Each of the two vertices must have two of the four → ′ ′ external lines attached to it. A priori, the other possibility is that one of the external lines is attached to one of the vertices while the three remaining external lines are attached to the other vertex. Nevertheless, in the latter case we would not have remaining lines to connect the two vertices, so that it would be a disconnected diagram i.e. a disconnected contribution. Coming back to our connected graph, the additional line required at each vertex should be used to connect the two vertices between them. For the vertex attach to line 1, the same vertex could be attach to either line 2 or line 1′ or line 2′. Therefore, three graphs of that type arise, as can be seen in Fig. 9.9. From the Feynman rules, the contribution of these three diagrams to the S matrix reads (homework!! B2) − 2 6 4 4 S S ′ ′ ′ ′ ′ ′ = ( i) (2π)− g ′ ′′ g ′ ′′ d x d y [ i∆ ′′ ′′ (x,y)] p1σ1n1p2σ2n2;p1σ1n1p2σ2n2 nn n mm m n m ≡ − ′ ′′ ′ ′′ − nn nXmm m Z Z ′ ′ ip x ip x ip1 y ip2 y u∗ p′ ,σ′ ,n′ e− 1· u∗ ′ p′ ,σ′ ,n′ e− 2· u (p ,σ ,n ) e · u ′ (p ,σ ,n ) e · × n 1 1 1 n 2 2 2 m 1 1 1 m 2 2 2 ′ ′ h ip1 x ip1 x ip2 y ip2 y +un∗ ′ p1′ ,σ1′ ,n1′ e− · un (p1,σ1,n1) e · um∗ ′ p2′ ,σ2′ ,n2′ e− · um (p2,σ2,n2) e · ′ ′ ip2 x ip1 x ip1 y ip2 y +un∗ ′ p2′ ,σ2′ ,n2′ e− · un (p1,σ1,n1) e · um∗ ′ p1′ ,σ1′ ,n1′ e− · um (p2,σ2,n2) e · i (9.38) we can simplify this model by assuming that the boson particles in this theory are spinless (scalars) of a single species. In that case all indices of field components dissapear, and the interaction (9.37) becomes φ3 (x) (x)= g (9.39) H 3! substituting (9.39) in (9.38) the S matrix becomes − 3 6 2 4 4 S Sp′ p′ ;p p = ( i) (2π)− g d x d y [∆F (x,y)] ≡ 1 2 1 2 − ′ Z ′ Z ip x ip x ip1 y ip2 y u∗ p′ e− 1· u∗ p′ e− 2· u (p ) e · u (p ) e · × 1 2 1 2 ′ ′ h ip1 x ip1 x ip2 y ip2 y +u∗ p1′ e− · u (p1) e · u∗ p2′ e− · u (p2) e · ′ ′ ip2 x ip1 x ip1 y ip2 y +u∗ p2′ e− · u (p1) e · u∗ p1′ e− · u (p2) e · (9.40) i where we have taken into account that σ only can takes one value and there is only one species of particle. Hence the S matrix element only depend on the momenta. Now, for a scalar neutral field, we have [see Eq. (5.4) Page − 151] 1 u (p)= v (p)= (9.41) √2E substituting (9.41) in (9.40) we find

′ ′ ip x ip x ip1 y ip2 y 3 6 2 4 4 e− 1· e− 2· e · e · S Sp′ p′ ;p p = ( i) (2π)− g d x d y [∆F (x,y)] ≡ 1 2 1 2 − 2E 2E √2E √2E Z Z " 1′ 2′ 1 2 ip′ x ip x ip′ y ip y ip′ x ip x ip′ y ip y e 1 e 1 e 2 e 2 e 2 e 1 e 1 pe 2 p + − · · − · · + − · · − · · 2E1′ √2E1 2E2′ √2E2 2E2′ √2E1 2E1′ √2E2 # and we obtain ﬁnallyp p p p 2 ig 4 4 Sp′ p′ ;p p = d x d y ∆F (x y) 1 2 1 2 (2π)6 16E E E E − 1′ 2′ 1 2 Z Z ′ ′ ′ ′ ′ ′ i(p +p ) x i(p1+p2) y i(p1 p ) x i(p2 p ) y i(p1 p ) x i(p2 p ) y e− p1 2 · e · + e − 1 · e − 2 · + e − 2 · e − 1 · × h i 240 CHAPTER 9. THE FEYNMAN RULES

Figure 9.10: Connected third-order diagrams for boson-boson scattering would have at least one vertex attached to four lines, but such kind of vertices are not included in the theory described by (9.37). where ∆ (x y) is the scalar ﬁeld propagator that we shall calculate soon. There are no terms of third order F − or of any odd order in (x). To see it, we observe that third order connected diagrams would have at least one H vertex connected to four lines (see Fig. 9.10), but the theory described by (9.37) only contains vertices attached to three lines. Diagrams of fourth order are similar to the one shown in Fig. 9.7, page 235, except that all lines involved are boson type lines.

9.7 Calculation of the propagator

As we explained in Sec. 9.2, in the pairing of a ﬁeld ψk (x) with a ﬁeld adjoint ψm† (y), a factor of the form (9.18) that is called the propagator arises. Using Eqs. (9.20) and (9.21) and the commutation and anticommutation relations for annihilation and creation operators we have

+ + 1 3 ip x 3 ip′ y ψ (x) , ψm† (y) = d p uk (p,σ,n) e · a (p,σ,n) , d p′ um∗ p′,σ′,n′ e− · a† p′,σ′,n′ k (2π)3 ∓ "Z σ Z σ′ # h i X X ∓ 1 3 3 ip′ y ip x = d p d p′ uk (p,σ,n) um∗ p′,σ′,n′ e− · e · a (p,σ,n) , a† p′,σ′,n′ (2π)3 Z Z σ′ σ X X h i 1 3 3 ip′ y ip x = d p d p′ uk (p,σ,n) um∗ p′,σ′,n′ e− · e · δ p p′ δσσ′ δnn′ (2π)3 − Z Z σ′ σ X X + + 1 3 ip (x y) ψk (x) , ψm† (y) = 3 d p uk (p,σ,n) um∗ (p,σ,n) e · − (9.42) (2π) σ h i∓ Z X similarly 1 3 ip (y x) ψm−† (y) , ψk− (x) = 3 d p vm∗ (p,σ,n) vk (p,σ,n) e · − (9.43) (2π) σ h i∓ Z X substituting (9.42) and (9.43) in Eq. (9.18), we obtain

3 3 ip (x y) i∆ (x,y) = θ (x y) (2π)− d p u (pσn) u∗ (pσn) e · − − km − k m σ Z X 3 3 ip (y x) θ (y x) (2π)− d p v∗ (pσn) v (pσn) e · − (9.44) ± − m k σ Z X when we derived the commutation and anticommutation relations for fields and field adjoints, we also found expressions for the sum over spins of bilinear forms of the coefficients. For vector fields they were given by Eqs. 9.7. CALCULATION OF THE PROPAGATOR 241

(6.64, 6.69) and (6.75) page 173, while for Dirac ﬁelds they were given by Eqs. (7.102, 7.103) and Eqs. (7.121, 7.122) page 200. Such bilinear forms posses the following general structure7

2 2 Pkm p, p + mn uk (pσn) u∗ (pσn) = (9.45) m 2 2 σ 2 pp+ mn X P p p, p2 + m2 km − − n vk (pσn) v∗ (pσn) = (9.46) m ± 2 2 σ 2 p p+ mn X where P (p, λ) is a polinomial in p and λ. In Eqs. (9.44, 9.45, 9.46)p the signs refer to bosonic and fermionic km ± fields respectively. If ψ (x) φ (x) and ψ (y) φ (y) are scalar fields for a particle of spin zero, Eqs. (5.4) k → m → page 151, shows that 1 1 uk (pσn) u∗ (pσn)= u (p) u∗ (p)= = (9.47) m 2p0 2 2 σ 2 p + mn X comparing Eqs. (9.45, 9.47) the polynomial P (p, λ) for scalar fields becomesp P p,p0 = P (p) = 1 (9.48) if ψ (x) ψµ (x) and ψ (y) ψν (y) are Dirac fields for a particle of spin 1/2, Eqs. (7.102, 7.103) and Eqs. k → m → (7.121, 7.122) page 200, can be combined to obtain

1 µ N ¯ (p) u (p,σ) u∗ (p,σ)= [( ip γ + m) β] (9.49) kk ≡ k k¯ 2p0 − µ kk¯ σ X 1 µ M ¯ (p) v (p,σ) v∗ (p,σ)= [( ip γ m) β] (9.50) kk ≡ k k¯ 2p0 − µ − kk¯ σ X comparing Eqs. (9.45, 9.46) with Eqs. (9.49, 9.50) we find that for Dirac fields of spin 1/2, the polynomial reads P (p)=[( iγ pα + m) β] (9.51) µν − α µν µ ν note that the matrix β appears here because we are considering the pairing of ψ (x) with ψ † (y) instead of a pairing of ψµ (x) with ψ¯ν (y) ψν (y) β. ≡ † Finally, if ψk (x) and ψm (y) are vector fields Vµ (x) and Vν (y) for a particle of spin one, the combination of Eqs. (6.64, 6.69) and (6.75) lead to µ ν µν µ ν 0 1/2 µ 0 ν µν p p Π (p) e (p,σ) e ∗ (p,σ)= 2p u (p,σ) 2p u ∗ (p,σ) = g + ≡ m2 σ σ X X h µ ν i µ ν 1 µν p p u (p,σ) u ∗ (p,σ)= g + (9.52) ⇒ 2p0 m2 σ X once again comparison of Eqs. (9.45) and (9.52) shows that the polynomial for spin one vector fields is given by p p P (p)= g + µ ν (9.53) µν µν m2 Now, substituting Eqs. (9.45) and (9.46) in Eq. (9.44) we have

2 2 Pkm p, p + mn 3 3 ip (x y) i∆km (x,y) = θ (x y) (2π)− d p e · − − − p2p0 Z 2 2 Pkm p, p + mn 3 3 ip (y x) +θ (y x) (2π)− d p − − e · − (9.54) − 2pp0 Z 7For scalar ﬁelds we can obtain similar expressions from Eqs. (5.4) page 151. 242 CHAPTER 9. THE FEYNMAN RULES now taking into account that n ∂ ip x n ip x i e · = p e · − ∂x we can replace a polynomial in p by

0 ip (x y) ∂ ip (x y) P p,p e · − = P i e · − (9.55) km km − ∂x substituting (9.55) in (9.54) the propagator becomes

∂ ip (x y) ∂ ip (y x) θ (x y) 3 Pkm i ∂x e · − θ (y x) 3 Pkm i ∂x e · − i∆km (x,y) = − d p − + − d p − − (2π)3 2p0 (2π)3 2p0 Z Z 3 ip (x y) 3 ip (y x) ∂ d p e · − ∂ d p e · − i∆km (x,y) = θ (x y) Pkm i + θ (y x) Pkm i − − − ∂x 2p0 (2π)3 − − ∂x 2p0 (2π)3 Z Z then the propagator takes the form

∂ i∆ (x,y) = θ (x y) P i ∆ (x y) − km − km − ∂x + − ∂ +θ (y x) P i ∆ (y x) (9.56) − km − ∂x + − where ∆+ (x) is the function deﬁned in Eq. (5.9) page 152

3 1 d p ip x 0 2 2 ∆+ (x) e · ; p + p + m (9.57) ≡ (2π)3 2p0 ≡ Z p Now we have to extend the definition of the polynomial P (p). The definition of P (p) given by Eqs. (9.45) and (9.46) are only valid for four-momenta “on the mass shell”. That is, when p0 = p2 + m2. We require such an ± extension because as we shall see later, some internal lines will carry momenta that are off the mass shell. p Since any power of the form p0 2n or p0 2n+1 can be written as p2 + m2 n or p0 p2 + m2 n; we see that any polynomial function of such four-momentum can be taken as linear in p0. Consequently, we can define a generalized polynomial P (L) (p) as follows

P (L) (p) = P (p) for p0 = p2 + m2 (9.58) (L) (0) 0 (1) µ P (q) = P (q)+ q P (q)p for general q (9.59) with P (0,1) being polynomials that depend only on q. Now, using the properties

∂ ∂ θ x0 y0 = θ y0 x0 = δ x0 y0 ∂x0 − −∂x0 − − we are able to move the derivative operators to the left of the θ functions in Eq. (9.56) to obtain ———————————————— ————————————————open ????

∂ ∂ P (L) i = P (0) ( i ) iP (1) ( i ) km − ∂x − ∇ − − ∇ ∂x0 9.7. CALCULATION OF THE PROPAGATOR 243

∂ E = P (L) i ∆ (x y)+ δ x0 y0 P (1) ( i )[∆ (x y) ∆ (y x)] km − ∂x F − − km − ∇ + − − + − ∂ = P (0) ( i ) iP (1) ( i ) iθ (x y) ∆ (x y)+ iθ (y x) ∆ (y x) + δ x0 y0 P (1) ( i ) [∆ (x − ∇ − − ∇ ∂x0 { − + − − + − } − km − ∇ + ∂ ∂ = P (0) ( i ) iP (1) ( i ) iθ (x y) ∆ (x y) + P (0) ( i ) iP (1) ( i ) iθ (y x) ∆ (y x − ∇ − − ∇ ∂x0 { − + − } − ∇ − − ∇ ∂x0 { − + − +δ x0 y0 P (1) ( i )[∆ (x y) ∆ (y x)] − km − ∇ + − − + − ∂ = P (0) ( i ) iθ (x y) ∆ (x y) iP (1) ( i ) iθ (x y) ∆ (x y) + P (0) ( i ) iθ (y x) ∆ ( − ∇ { − + − }− − ∇ ∂x0 { − + − } − ∇ { − + h i ∂ h i iP (1) ( i ) iθ (y x) ∆ (y x) + δ x0 y0 P (1) ( i ) [∆ (x y) ∆ (y x)] − − ∇ ∂x0 { − + − } − km − ∇ + − − + − ∂θ (x y) ∂∆ (x y) E = P (0) ( i ) iθ (x y) ∆ (x y) iP (1) ( i ) i − ∆ (x y)+ iθ (x y) + − − ∇ { − + − }− − ∇ ∂x0 + − − ∂x0 h i ∂θ (y x) ∂∆ (y x) + P (0) ( i ) iθ (y x) ∆ (y x) iP (1) ( i ) i − ∆ (y x)+ iθ (y x) + − − ∇ { − + − }− − ∇ ∂x0 + − − ∂x0 h i +δ x0 y0 P (1) ( i )[∆ (x y) ∆ (y x)] − km − ∇ + − − + − ∂∆ (x y) E = P (0) ( i ) iθ (x y) ∆ (x y) iP (1) ( i ) iδ x0 y0 ∆ (x y)+ iθ (x y) + − − ∇ { − + − }− − ∇ − + − − ∂x0 h i ∂∆ (y x) + P (0) ( i ) iθ (y x) ∆ (y x) iP (1) ( i ) iδ x0 y0 ∆ (y x)+ iθ (y x) + − − ∇ { − + − }− − ∇ − − + − − ∂x0 h i +δ x0 y0 P (1) ( i )[∆ (x y) ∆ (y x)] − km − ∇ + − − + − ∂∆ (x E = P (0) ( i ) iθ (x y) ∆ (x y) iP (1) ( i ) iδ x0 y0 ∆ (x y) iP (1) ( i ) iθ (x y) + − ∇ { − + − }− − ∇ − + − − − ∇ − ∂x0 h i ∂∆ + P (0) ( i ) iθ (y x) ∆ (y x) iP (1) ( i ) iδ x0 y0 ∆ (y x) iP (1) ( i ) iθ (y x) + − ∇ { − + − }− − ∇ − − + − − − ∇ − h i +δ x0 y0 P (1) ( i )∆ (x y) δ x0 y0 P (1) ( i )∆ (y x) − km − ∇ + − − − km − ∇ + − ∂∆ (x y) E = P (0) ( i ) iθ (x y) ∆ (x y) + δ x0 y0 P (1) ( i ) ∆ (x y) iP (1) ( i ) iθ (x y) + − − ∇ { − + − } − − ∇ + − − − ∇ − ∂x0 h i ∂∆ (y + P (0) ( i ) iθ (y x) ∆ (y x) δ x0 y0 P (1) ( i ) ∆ (y x) iP (1) ( i ) iθ (y x) + − − ∇ { − + − }− − − ∇ + − − − ∇ − ∂x0 h i +δ x0 y0 P (1) ( i )∆ (x y) δ x0 y0 P (1) ( i )∆ (y x) − km − ∇ + − − − km − ∇ + − ∂∆ (x y) E = P (0) ( i ) iθ (x y) ∆ (x y) iP (1) ( i ) iθ (x y) + − − ∇ { − + − }− − ∇ − ∂x0 h i ∂∆ (y x) + P (0) ( i ) iθ (y x) ∆ (y x) iP (1) ( i ) iθ (y x) + − − ∇ { − + − }− − ∇ − ∂x0 h i +δ x0 y0 P (1) ( i )∆ (x y)+ δ x0 y0 P (1) ( i ) ∆ (x y) − km − ∇ + − − − ∇ + − 0 0 (1) 0 0 (1) δ x y P ( i )∆+ (y x) δ x y P ( i ) ∆+ (y x) − − km − ∇ − − − − ∇ − 244 CHAPTER 9. THE FEYNMAN RULES

???? ————————————————– ————————————————–close

∂ ∆ (x,y)= P (L) i ∆ (x y)+ δ x0 y0 P (1) ( i ) [∆ (x y) ∆ (y x)] (9.60) km km − ∂x F − − km − ∇ + − − + − where ∆F is deﬁned by i∆ (x) θ (x) ∆ (x)+ θ ( x) ∆ ( x) (9.61) − F ≡ + − + −

0 and it is called the Feynman propagator. Now, we observe that for x = 0, the function ∆+ (x) is even with respect to x. We see it by noting that the change x x in Eq. (9.57) can be compensated by a change p p →− →− in the integration variable. Thus, we can omit the second term in Eq. (9.60) and write

∂ ∆ (x,y)= P (L) i ∆ (x y) (9.62) km km − ∂x F −

It is convenient to express the Feynman propagator as a Fourier integral. To do it, we write the step function in its Fourier representation

1 ∞ exp ( ist) θ (t)= − ds (9.63) −2πi s + iε Z−∞ the validity of Eq. (9.63) can be shown as follows: if t> 0 the numerator yields

i(Re s+iIm s)t it Re s t Im s exp ( ist)= e− = e− e − we can close the contour of integration with a large clockwise semi-circle in the lower half-plane (such that Ims< 0 and the numerator does not diverge), from which the integral has a contribution of 2πi from the pole at s = iε. − − Now, if t < 0 we can close the contour by a large counter-clockwise semi-circle in the upper half-plane, within which the integrand is analytic, so that the integral vanishes. Now to express the Feynman propagator as a Fourier integral, we combine Eq. (9.63) with Eq. (9.61) as well as the Fourier integral (9.57) for ∆+ (x). Doing it we obtain

i∆ (x) θ (x) ∆ (x)+ θ ( x) ∆ ( x) − F ≡ + − + − 0 3 0 3 1 ∞ exp isx 1 d p ip x 1 ∞ exp isx 1 d p ip x = − ds e · + ds e− · −2πi s + iε (2π)3 2p0 2πi s + iε (2π)3 2p0 "Z−∞ # Z "Z−∞ # Z

By redeﬁning the variables q p and q0 = p0 + s, in the ﬁrst term of Eq. (9.61) we have8 ≡ ———————————– ———————————–

8Note that by assuming that the spatial components of q coincides with the spatial components of p, while the temporal components do not coincide, means that q is oﬀ the mass shell. 9.7. CALCULATION OF THE PROPAGATOR 245

0 3 1 ∞ exp isx 1 d p θ (x) ∆+ (x) = − ds exp [ip x] −2πi s + iε (2π)3 2p0 · "Z−∞ # Z 0 0 0 3 1 ∞ exp i q p x 0 1 d p 0 0 = −0 0− dq 3 0 exp ip x ip x −2πi " (q p )+ iε # (2π) 2p · − Z−∞ − Z 0 0 exp i q p2 + m2 x 3 1 ∞ − − 0 1 d p 2 2 0 = dq 3 exp ip x i p + m x −2πi  hq0 p2p+ m2 + iε i  "(2π) 2 p2 + m2 · − # Z−∞ − Z h p i   p exp i qp0 q2 + m2 x0 3 1 ∞ − − 0 1 d q 2 2 0 = dq 3 exp iq x i q + m x −2πi  hq0 q2p+ m2 + iε i  "(2π) 2 q2 + m2 · − # Z−∞ − Z h p i   p p3 exp iq0x0 + i q2 + m2x0 exp iq x i q2 + m2x0 1 1 d q ∞ 0 − · − = 3 dq −2πi 2 (2π) q2 + m2 h p q0 q2i+ m2h + iε p i Z Z−∞ −   p 0 0 p 1 3 ∞ 0 exp iq x iq x 1 θ (x) ∆+ (x) = d q dq · − 3 2 2 −2πi  2 (2π) q + m q0 q2 + m2 + iε Z Z−∞ −  p  we can calculate θ ( x) ∆ ( x) with a similar procedure, so thatp the Feynman propagator becomes − + − —————————- —————————

0 0 1 3 ∞ 0 exp iq x iq x i∆F (x) = d q dq · − − −2πi 2 (2π)3 q2 + m2 Z Z−∞ 1 p 1 + (9.64) ×  q0 q2 + m2 + iε q0 q2 + m2 + iε  − − −   the denominators in parenthesis give p p 1 1 D + ≡ q0 q2 + m2 + iε q0 q2 + m2 + iε − − − p 1 p 1 = q0 q2 + m2 + iε − q0 + q2 + m2 iε − − h i h i p q0 + q2 + m2 iεp q0 q2 + m2 + iε = − − − h p2 i h p i (q0)2 q2 + m2 iε q0 q2 + m2 + iε q0 + q2 + m2 + O (ε2) − − − p2 q2 + m2 2iε p 2 qp2 + m2 2iε = − = − 2 2 (q0) (q2p+ m2) + 2iε q2 + m2 (q0) q2p m2 + 2iε q2 + m2 − − − h 2 q2 + m2p i h p i D = − (9.65) 2 q2 (q0) + mp2 2iε q2 + m2 − − h i replacing (9.65) in (9.64) we have p 246 CHAPTER 9. THE FEYNMAN RULES

0 0 2 2 1 3 ∞ 0 exp iq x iq x 2 q + m i∆F (x) = d q dq · − − (9.66) 3 2 2 2 − −2πi 2 (2π) q + m  q2 (q0) + mp2 2iε q2 + m2  Z Z−∞  − −  p 0 0 h p i 1 3 ∞ 0 exp iq x iq x  1  i∆F (x) = d q dq · − (9.67) 3  2  − 2πi (2π) q2 (q0) + m2 2iε q2 + m2 Z Z−∞  − −  0 0 h p i i 3 ∞ 0 exp i q x q x  1  i∆F (x) = − d q dq · − (9.68) 3  2  − 2π (2π) q2 (q0) + m2 iε Z Z−∞  − −  h i Note that in the denominator we have replaced 2ε q2 + m2 with ε, because the only relevant thing about that quantity is that it is a positive inﬁnitesimal. In a four-dimensional notation ∆ (x) can be written as p F

1 4 exp (iq x) 2 2 0 2 ∆F (x)= d q · ; q q q (9.69) (2π)4 q2 + m2 iε ≡ − Z − Moreover, it shows that ∆F is a Green’s function for the Klein-Gordon diﬀerential operator, to see it we apply the Klein gordon operator on ∆F (x)

2 2 2 2 1 4 m exp (iq x) 1 4 q m exp (iq x) m ∆F (x) = d q − · = d q − − · − (2π)4 q2 + m2 iε (2π)4 q2 + m2 iε Z − Z − 2 1 4 m ∆F (x) = d q exp (iq x) − −(2π)4 · Z m2 ∆ (x)= δ4 (x) (9.70) − F − with boundary conditions determined by the inﬁnitesimal quantity iε in the denominator, since Eq. (9.61) − shows that ∆ (x) for x0 + involves only positive frequency terms exp ix0 p2 + m2 , while for x0 F → ∞ − → −∞ involves only negative terms exp +ix0 p2 + m2 . h p i By substituting Eq. (9.69) inh Eq. (9.62)p the propagatori becomes

(L) ∂ (L) ∂ 1 4 exp [iq (x y)] ∆km (x,y) = P i ∆F (x y)= P i d q · − km − ∂x − km − ∂x (2π)4 q2 + m2 iε Z − (L) 1 P i ∂ exp [iq (x y)] = d4q km − ∂x · − (2π)4 q2 + m2 iε Z − (L) iq (x y) 1 4 Pkm (q) e · − ∆km (x,y)= d q (9.71) (2π)4 q2 + m2 iε Z − The polynomial P (p) is Lorentz-covariant when p is “on the mass shell”. That is, when p2 = m2. Nevertheless, − in Eq. (9.71) we are integrating over all qµ, such that p is “oﬀ the mass shell”. On the other hand, Eq. (9.59) (L) µ 0 shows that the polynomial Pkm (q) for a general q is linear in q , such a condition clearly does not respect Lorentz covariance unless the polynomial is also linear in each spatial component qk. Hence we could alternatively deﬁne the extension of P (p) to general momentum qµ, such that P (q) is Lorentz-covariant for general qµ as follows

Pkm (Λq)= Dkk′ (Λ) Dmm∗ ′ (Λ) Pk′m′ (q) where Λ is a general Lorentz transformation on the Minkowski space, while D (Λ) is the associated representation of the Lorentz group. 9.7. CALCULATION OF THE PROPAGATOR 247

In the case of scalar, Dirac and vector ﬁelds the covariant extensions of the polynomial are obtained by substituting pµ by a general four-momentum qµ in Eqs. (9.48), (9.51) and (9.53). In the case of scalar and Dirac ﬁelds they are already linear in q0 such that P (L) (q) and P (q) are identical

(L) Pkm (q)= Pkm (q) for scalar and Dirac fields but for a vector field of a particle of spin one, we see that the 00 components of the covariant polynomial q q P (q)= g + µ ν µν µν m2 are quadratic in q0. Thus, both polynomials are different

0 0 2 2 2 0 0 2 2 2 qµqν δ δ q q m q q δ δ q + q + m P (L) (q) = g + − µ ν 0 − − = g + µ ν + µ ν − 0 µν µν m2 µν m2 m2 2 2 0 0 (L) q + m δµδν Pµν (q) = Pµν (q)+ 2 for vector bosons of spin one (9.72) m the additional term is ﬁxed by imposing two conditions (a) the cancellation of the quadratic term in P00 (q) and (b) it must vanish when qµ is on the mass shell. Now, substituting (9.72) in Eq. (9.71) yields the propagator of a vector ﬁeld of spin one

iq (x y) 4 0 0 1 4 Pµν (q) e · − δ (x y) δµ δν ∆µν (x,y)= d q + − (9.73) (2π)4 q2 + m2 iε m2 Z − where the first term is manifestly covariant. As for the second term, it is not covariant, but it is local, so that it can be cancelled by a local non-covariant term added to the Hamiltonian density. If a vector field Vµ (x) interacts with other fields by means of a coupling of the form V (x) J µ (x) in (x), the second term in Eq. (9.73) produces µ H an effective interaction described by

0 0 1 µ ν δµ δν i eff (x)= [ iJ (x)] [ iJ (x)] i 2 (9.74) − H 2 − − "− m # where the factors i are those factors that appear for each vertex and propagator. The factor 1/2 appears − because there are two ways to pair other fields with (x), that differ in the interchange of J µ (x) and J ν (x). Heff Consequently, the effect of the non-covariant second term in Eq. (9.73) can be cancelled by adding to (x) the H term 1 2 (x)= (x)= J 0 (x) (9.75) HNC −Heff 2m2 further, the singularity that appears at equal-time commutators of vector fields at zero separation requires to use a wider class of interactions than those of the scalar density. It is in that way that we can obtain a totally Lorentz invariant S matrix. − It worths pointing out that the previous phenomenon not only occurs for spins j 1. For example, in the ≥ case of vector fields associated with j = 0 (see section 6.2, page 165) which consists of the derivative of a scalar field ∂µφ (x), its pairing with a scalar φ† (y) generates the polynomial P (p) on the mass shell given by

Pλ (p)= ipλ (9.76) while the pairing of ∂λφ (x) with ∂ηφ† (y) generates the polynomial

Pλ,η (p)= pλpη (9.77) 248 CHAPTER 9. THE FEYNMAN RULES

Once again, the polynomial associated to the off shell four-momentum qµ is obtained by replacing qµ for pµ in Eqs. (9.76) and (9.77). Further, Eq. (9.76) shows that Pλ (q) is already linear in q0, from which Pλ (q) and (L) Pλ (q) are identical. By contrast, they differ in Eq. (9.77) such that P (L) (q) = q q q2 q2 m2 δ0 δ0 λ,η λ η − 0 − − λ η 2 2 0 0 = Pλ,η (q)+ q + m δλ δη from which the propagator becomes iq x 1 4 qλqη e · 0 0 4 ∆λ,η (x,y)= d q + δ δη δ (x y) (2π)4 q2 + m2 iε λ − Z − and we can also cancel the effect of the non-covariant second term by adding to the interaction another non- covariant term given by 1 2 (x)= J 0 (x) HNC 2 with J µ (x) being the current that couples to ∂ φ (x) in the covariant part of (x). µ H It is in general always possible to cancel the effects of non-covariant parts of the propagator of a massive (L) particle by adding non-covariant local terms to the Hamiltonian density. It is because the numerator Pkm (q) in µ the propagator must equal the covariant polynomial Pkm (q) when q is on the mass shell, so that the difference between P (L) (q) and P (q) must contain a factor q2 + m2. Such a factor cancels the denominator q2 + m2 iε km km − in the contribution of this difference to Eq. (9.71), hence Eq. (9.71) is always equal to a covariant term plus a term proportional to δ4 (x y) or its derivatives. The effect of the latter term can be cancelled by adding to − the interaction density a term which is quadratic in the currents to which the paired fields couple (or in their derivatives). From now on, we shall assume that such a term has been properly included, and therefore we shall use the covariant polynomial Pkm (q) in the propagator (9.71). Hence we shall omit the label “(L)” in that polynomial. At this step, we have simply added the non-covariant term to recover the covariance of the theory, by this addition has not been justified satisfactorily. We shall see that in the canonical formalism the non-covariant term in the Hamiltonian density required to cancel non-covariant terms in the propagator, appears automatically. Indeed, it is one of the main motivations for the introduction of the canonical quantization.

9.7.1 Other deﬁnitions of the propagator There are other deﬁnitions of the propagator in the literature, equivalent to Eq. (9.44). By taking the vacuum expectation value (VEV) of Eq. (9.18) page 227, we have

+ + i∆km (x,y)= θ (x y) ψk (x) , ψm† (y) θ (y x) ψm−† (y) , ψk− (x) (9.78) − − 0 ± − 0 h i∓ h i∓ where ABC . . . 0 ABC . . . 0 h i0 ≡ h | | i + is the vacuum expectation value (VEV) of the product of operators ABC . . .. The ﬁelds ψk (x) and ψm−† (y) annihilate the vacuum. Consequently, only one term in each commutator or anticommutator contributes to the propagator in Eq. (9.78)

+ + i∆km (x,y)= θ (x y) ψk (x) , ψm† (y) θ (y x) ψm−† (y) , ψk− (x) (9.79) − − 0 ± − 0 + D E D +E in addition, ψ−† and ψ would annihilate the vacuum state on the right, while ψ− and ψ † would annihilate + the vacuum state on the left. Consequently, ψ and ψ− could be substituted everywhere in Eq. (9.79) with the complete ﬁeld ψ ψ+ + ψ , we then obtain ≡ −

i∆km (x,y)= θ (x y) ψk (x) , ψm† (y) θ (y x) ψm† (y) , ψk (x) − − 0 ± − 0 D E D E 9.8. FEYNMAN RULES AS INTEGRATIONS OVER MOMENTA 249 which can be written in the form

i∆km (x,y)= T ψk (x) ψm† (y) − 0 D n oE where T means a time-ordered product extended to all fields, with a minus sign for any odd permutation of fermionic operators. Notice that this definition of a generalized time-ordered product is not consistent with our definition in Eq. (2.188) page 105 of the time-ordered product of Hamiltonian densities, because the Hamiltonian density only contains even numbers of fermionic field factors.

9.8 Feynman rules as integrations over momenta

Figure 9.11: Feynman diagram in which we only indicate the ﬂux of momenta and their associated exponential. There is another exponential coming from the internal line, owing to the Fourier representation of the propagator. The four-momentum q of the internal line ﬂows from the vertex y to the vertex x.

The Feynman rules for a diagram of N th order have been described by means of integrals over N space-time − coordinates where the integrands are in turn products of space-time dependent factors. Nevertheless, for many reasons integrations over momenta are more advantageous. On one hand, experiments in particle physics usually measure momenta but not positions or times. On the other hand, momenta can be related through kinematics coming from the principle of conservation of four-momentum, and the relation p2 = m2 is very useful when we − are dealing with on-shell particles. When establishing those Feynman rules we found that for a final particle or antiparticle line with momentum p µ leaving a vertex with space-time coordinate xµ, we obtain a factor proportional to exp ( ip x). We also ′ − ′ · found that for an initial particle line with momentum pµ entering a vertex with a space-time coordinate xµ the associated factor becomes exp (+ip x). Finally, we saw that the factor associated with an internal line running · from y to x can be written as a Fourier integral over off-shell four momenta qµ with an integrand proportional to exp [iq (x y)]. We could consider that the momentum qµ is flowing along the internal line in the direction of · − 250 CHAPTER 9. THE FEYNMAN RULES the arrow i.e. from y to x. For the diagram in Fig. 9.11, we show the exponentials associated with each external and internal lines. Since the four-momentum must be conserved at each vertex we have

p1 + p2 = q = p1′ + p2′ and the product P of all exponentials coming from external and internal lines yield

′ ′ ′ ′ ip x ip x iq(x y) ip1y ip2y i(p +p )x iqx iqy i(p1+p2)y P e− 1 e− 2 e − e e = e− 1 2 e e− e ≡ iqx iqx iqy iqy P = e− e e− e = 1 therefore, all exponential cancel each other and should not be included when we express the propagator explicitly as a Fourier expansion. Besides, in order to account for conservation of four-momentum, the integral over each vertex’s space-time position provides a factor

4 4 (2π) δ p + q p′ q′ (9.80) − − X X X X where p′ and p denote the total four-momentum of all the final or initial particles leaving or entering the vertex, while q and q denote the total four-momentum of all the internal lines with arrows leaving or entering P P′ the vertex respectively. Then instead of integrals over spacetime coordinates xµ, we have to do integrals over the P P Fourier variables qµ (in general off shell) one for each internal line (momenta associated with external lines are on the mass shell, so they are fixed). From the discussion above, it is convenient to contextualize the Feynman rules to adapt them for calculations of contributions to the S matrix by means of integrals over momenta (see Fig. 9.12). − 1. The Feynman diagrams of a given order are the ones described in Sec. 9.3. However, we shall not label each vertex with spacetime coordinates. Instead, each internal line is labelled with an off-mass-shell four- momentum, that by convention flows in the direction of the arrow. For neutral particle lines without arrows the momentum flows in either direction.

2. For a given vertex of type i, we put a factor

4 4 i (2π) g δ p + q p′ q′ (9.81) − i − − X X X X with the conventions deﬁned below equation (9.80). Note that the Dirac delta guarantees the conservation of momentum at each vertex in the diagram.

(a) For each external line running upwards out of the diagram we include a factor

1 uk∗ p′,σ′,n′ for arrows pointing upwards (particle) (2π)3/2 1 vk p′,σ′,n′ for arrows pointing downwards (antiparticle) (2π)3/2 (b) For each external line running from below into the diagram the associated factor is

1 uk (p,σ,n) for arrows pointing upwards (particle) (2π)3/2 1 vk∗ (p,σ,n) for arrows pointing downwards (antiparticle) (2π)3/2 9.8. FEYNMAN RULES AS INTEGRATIONS OVER MOMENTA 251

Figure 9.12: Feynman representations of pairings of operators in the momentum-space evaluation of the S matrix. − On the right we have the factors that must be included in the momentum space integrand of the S matrix for each − line of the Feynman diagram.

(c) For each internal line running with edges labelled as k and m, with the arrow flowing from m to k, µ iq x with a momentum label q , we include as a factor the coefficient associated with e · in the integral for i∆ (x) [see Eq. (9.71) page 246]9: − km 4 i (2π)− Pkm (q) − q2 + m2 iε km − 2 where mkm denotes the mass of the particle in the internal line defined between the vertices m and k. We recall that the factors u (p,σ, n) and v (p,σ, n) and the polynomial P (q) are given by 1 u (q)= v (q)= ; P (q) = 1 for scalars and pseudoscalars 2q0

For Dirac spinors of mass M andp four-momentum q, the u (q,σ, n) and v (q,σ, n) factors are the ones described in Sec. 7.4, and the polynomial is the matrix P (q) = ( iγ qµ + M) β. − µ 3. Once all these factors are obtained, we integrate the product of all of them over the four-momenta carried by internal lines, then sum over all ﬁeld indices k,m etc.

4. Add up the result obtained from each diagram.

9keep in mind that the exponential has been cancelled with exponentials of other lines. 252 CHAPTER 9. THE FEYNMAN RULES

5. As discussed in section 9.4, additional combinatoric factors and fermionic signs are required.

According to these rules we have a four-momentum integration variable for each internal line. However, the Dirac factors associated with the vertices described in Eq. (9.81) eliminates many integrals. Energy and momentum are separately conserved for each connected part of a Feynman diagram. Thus, there are C remaining delta functions in a graph with C connected parts (C = 1 for each connected diagram). Therefore in a diagram with I internal lines and V vertices, we shall have I (V C) independent four-momenta (i.e. not ﬁxed by the − − delta functions), and as we discussed in section 3.6.2 [see Eq. (3.59) page 133], this is precisely the number of independent loops L L = I V + C (9.82) − which is deﬁned as the maximum number of internal lines that can be cut without disconnecting the diagram, since any such (and only such) internal lines are associated with an independent four-momentum. From this discussion the independent momenta characterizes the momenta that circulate in each loop. We call a tree graph as a diagram without loops, for such diagrams after integrating the delta functions there are no momentum-space integrals.

9.9 Examples of application for the Feynman rules with integration over four-momenta variables

We take as a ﬁrst example, a theory with an interaction of the type (9.24)

(x)= g ψ† (x) ψ (x) φ (x) (9.83) H kmn k m n kmnX As before, we shall calculate the S matrix elements associated with fermion-boson, fermion-fermion, and boson- − boson scattering. But we shall express such matrix elements as integrals over momenta.

9.9.1 Fermion-boson scattering The diagrams for fermion boson scattering in the theory (9.83) are the ones in Fig. 9.8, Page 236. Let us start with the diagram in Fig. 9.8(a)

1. For the two initial particles (p1,σ1,n1) and (p2,σ2,n2) (fermion and boson respectively) we associate factors u (p ,σ ,n ) u (p ,σ ,n ) n 1 1 1 and k 2 2 2 (2π)3/2 (2π)3/2

2. For the vertex (x,m) we add a factor

i (2π)4 g δ4 (p + p q) − mnk 1 2 −

3. For the internal line with the arrow ﬂowing from m to m′, we add the coeﬃcient of the propagator

4 i (2π)− Pm′m (q) 2 2 − q + m ′ iε m m −

where mm′m refers to the mass of the particle in the internal line between vertices m and m′.

4. For the vertex (y,m′) we add a factor

4 4 i (2π) g ′ ′ ′ δ (q p ′ p ′ ) − n m k − 1 − 2 9.9. EXAMPLES OF APPLICATION FOR THE FEYNMAN RULES WITH INTEGRATION OVER FOUR-MOMENTA

5. For the ﬁnal fermion and boson states (p1′ ,σ1′ ,n1′ ) and (p2′ ,σ2′ ,n2′ ) we associate factors

u ′ (p ,σ ,n ) u ′ (p ,σ ,n ) n∗ 1′ 1′ 1′ and k∗ 2′ 2′ 2′ (2π)3/2 (2π)3/2

6. Let us put all this stuﬀ together

4 u (p ,σ ,n ) u (p ,σ ,n ) i (2π)− P ′ (q) P(a) n 1 1 1 k 2 2 2 i (2π)4 g δ4 (p + p q) m m 3/2 3/2 mnk 1 2 2 2 ≡ " (2π) #" (2π) # − − "− q + mm′m iε # h i − 4 4 un∗ ′ (p1′ ,σ1′ ,n1′ ) uk∗′ (p2′ ,σ2′ ,n2′ ) i (2π) gn′m′k′ δ (q p1′ p2′ ) × − − − " (2π)3/2 #" (2π)3/2 # h i 7. This product is integrated over the momenta of internal lines, and sum over all ﬁeld indices (knm) and (k′n′m′). In this case we have only one internal line carrying a momentum q ﬂowing upwards in the diagram in Fig. 9.8(a). We then obtain

(a) 4 (a) 2 8 SF B d q P = ( i) (2π) gn′m′k′ gmnkun∗ ′ p1′ ,σ1′ ,n1′ un (p1,σ1,n1) ≡ ′ ′ ′ ′ ′ ′ − Z k nXm knm k nXm knm 4 4 i (2π)− Pm′m (q) 6 4 4 ′ ′ d q 2 2 (2π)− uk∗′ p2′ ,σ2′ ,n2′ uk (p2,σ2,n2) δ (p1 + p2 q) δ (q p1 p2 ) × − q + m ′ iε − − − Z " m m # − 8. Now we add the contribution of diagram 9.8(b) (there are no extra combinatoric factors for these diagrams). From which we ﬁnally obtain the S matrix (9.35) page 237 for fermion-boson scattering with the momentum- − space rules

(a) (b) SF B S + S = Sp′ σ′ n′ , p′ σ′ n′ ; p σ n , p σ n ≡ F B F B 1 1 1 2 2 2 1 1 1 2 2 2 2 8 4 4 Pm′m (q) S ′ ′ ′ F B = ( i) (2π) gn m k gmnkun∗ ′ p1′ σ1′ n1′ un (p1σ1n1) d q i (2π)− 2 2 ′ ′ ′ − − q + mm′m iε k nXm knm Z − 6 4 4 (2π)− uk∗′ p2′ σ2′ n2′ uk (p2σ2n2) δ (p1 + p2 q) δ (q p1′ p2′ ) × 4 4− − − +u∗ p′ σ′ n′ u ′ (p σ n ) δ (p p ′ + q) δ (p p ′ q) k 2 2 2 k 2 2 2 2 − 1 1 − 2 − where the labels 1 and 2 are denoting fermions and bosons respectively. After integrating over the oﬀ-mass- shell momentum q, the Dirac delta functions demand that

q = p′ + p′ = p + p = p′ p = p′ p 1 2 1 2 1 − 2 2 − 1 and we are left with a single Dirac function that provides the global conservation of four-momentum (because this is a connected diagram). In addition, we have no integrals over oﬀ-shell momenta (owing to the absence of loops, since this is a tree graph without independent momenta).

2 4 SF B Sp′ σ′ n′ , p′ σ′ n′ ; p σ n , p σ n = i (2π)− δ p1 + p2 p′ p′ ≡ 1 1 1 2 2 2 1 1 1 2 2 2 − 1 − 2 g ′ ′ ′ g u∗ ′ p′ σ′ n′ u (p σ n ) × n m k mnk n 1 1 1 n 1 1 1 k′n′m′knm X Pm′m (p1 + p2) u∗′ p′ σ′ n′ u (p σ n ) × 2 2 k 2 2 2 k 2 2 2 "(p1 + p2) + mm′m iε − Pm′m (p2′ p1) + − u∗ p′ σ′ n′ u ′ (p σ n ) (9.84) 2 2 k 2 2 2 k 2 2 2 (p2′ p1) + mm′m iε # − − 254 CHAPTER 9. THE FEYNMAN RULES

It is convenient to deﬁne a more compact notation as follows, we can deﬁne the fermion-boson coupling matrix as [Γ ] g (9.85) k nm ≡ nmk in this matrix notation, the matrix element (9.84) for fermion-boson scattering could be rewritten in the form ————————————– ————————————-

2 4 SF B Sp′ σ′ n′ , p′ σ′ n′ ; p σ n , p σ n = i (2π)− δ p1 + p2 p′ p′ ≡ 1 1 1 2 2 2 1 1 1 2 2 2 − 1 − 2 Pm′m (p1 + p2) [Γ ′ ] ′ ′ [Γ ] u∗ ′ p′ σ′ n′ u (p σ n ) u∗′ p′ σ′ n′ u (p σ n ) × k n m k mn n 1 1 1 n 1 1 1 2 2 k 2 2 2 k 2 2 2 (k′n′m′knm (p1 + p2) + mm′m iε X − Pm′m (p2′ p1) + [Γk′ ]n′m′ [Γk]mn un∗ ′ p1′ σ1′ n1′ un (p1σ1n1) 2 − uk∗ p2′ σ2′ n2′ uk′ (p2σ2n2) ′ 2 k′n′m′knm (p2 p1) + mm′m iε ) X − −

2 4 SF B Sp′ σ′ n′ , p′ σ′ n′ ; p σ n , p σ n = i (2π)− δ p1 + p2 p′ p′ ≡ 1 1 1 2 2 2 1 1 1 2 2 2 − 1 − 2 Pm′m (p1 + p2) u∗ ′ p′ σ′ n′ [Γ ′ ] ′ ′ [Γ ] u (p σ n ) u∗′ p′ σ′ n′ u (p σ n ) × n 1 1 1 k n m 2 2 k mn n 1 1 1 k 2 2 2 k 2 2 2 (k′n′m′knm (p1 + p2) + mm′m iε X − Pm′m (p2′ p1) ′ ′ + un∗ ′ p1′ σ1′ n1′ [Γk ]n′m′ 2 − [Γk]mn un (p1σ1n1) uk∗ p2′ σ2′ n2′ uk (p2σ2n2) ′ 2 k′n′m′knm (p2 p1) + mm′m iε ) X − − we have the following matrix multiplications

Pm′m (p1 + p2) M ′ u∗ ′ p′ σ′ n′ [Γ ′ ] ′ ′ [Γ ] u (p σ n ) 1k k ≡ n 1 1 1 k n m 2 2 k mn n 1 1 1 n′m′nm (p1 + p2) + mm′m iε X − P (p1 + p2) = u† p′ σ′ n′ Γ ′ Γ u (p1σ1n1) 1 1 1 k 2 2 k (p1 + p2) + M iε − Pm′m (p2′ p1) M ′ ′ 2k k un∗ ′ p1′ σ1′ n1′ [Γk ]n′m′ 2 − [Γk]mn un (p1σ1n1) ≡ ′ 2 n′m′nm (p2 p1) + mm′m iε X − − P (p2′ p1) = u† p′ σ′ n′ Γ ′ − Γ u (p1σ1n1) 1 1 1 k 2 2 k (p2′ p1) + M iε − − ————————————- ————————————-

2 4 SF B Sp′ σ′ n′ , p′ σ′ n′ ; p σ n , p σ n = i (2π)− δ p1 + p2 p′ p′ ≡ 1 1 1 2 2 2 1 1 1 2 2 2 − 1 − 2 P (p1 + p2) u† p1′ σ1′ n1′ Γk′ Γk u (p1σ1n1) u∗′ p2′ σ2′ n2′ uk (p2σ2n2) × (p + p )2 + M 2 iε k k′k 1 2 X − P (p2′ p1) + u† p′ σ′ n′ Γ ′ − Γ u (p1σ1n1) u∗ p′ σ′ n′ u ′ (p2σ2n2) (9.86) 1 1 1 k 2 2 k k 2 2 2 k " (p1 p2′ ) + M iε #) − − where M 2 is the diagonal mass matrix of the fermions in the internal lines associated with each propagator in Eqs. (9.86). 9.9. EXAMPLES OF APPLICATION FOR THE FEYNMAN RULES WITH INTEGRATION OVER FOUR-MOMENTA

9.9.2 Fermion-fermion scattering Similarly, in the same theory the S matrix element for fermion-fermion scattering given by Eq. (9.36) page 238, − and Fig. 9.4, page 232, yields (Homework!! B3)

2 4 SFF Sp′ σ′ n′ , p′ σ′ n′ ; p σ n , p σ n = i (2π)− δ p1 + p2 p′ p′ ≡ 1 1 1 2 2 2 1 1 1 2 2 2 − 1 − 2 Pk′k (p1′ p1) g ′ ′ g ′ − m mk n nk 2 2 × ′ ′ ′ ′ (p1 p1) + mk′k iε k nXm knm − − u∗ ′ p′ σ′ n′ u∗ ′ p′ σ′ n′ u (p σ n ) u (p σ n ) 1′ 2′ (9.87) × m 2 2 2 n 1 1 1 m 2 2 2 n 1 1 1 − ⇔ in the matrix notation deﬁned by Eq. (9.85), the matrix element (9.87) for fermion-fermion scattering could be rewritten in the form (Homework!! B4)

2 4 SFF Sp′ σ′ n′ , p′ σ′ n′ ; p σ n , p σ n = i (2π)− δ p1 + p2 p′ p′ ≡ 1 1 1 2 2 2 1 1 1 2 2 2 − 1 − 2 Pk′k (p1′ p1) − u† p′ σ′ n′ Γ ′ u (p σ n ) u† p′ σ′ n′ Γ u (p σ n ) 2 2 2 2 2 k 2 2 2 1 1 1 k 1 1 1 × ′ k′k (p1 p1) + mk′k iε X − − h ih i 1′ 2′ (9.88) − ⇔ where m2 is the diagonal mass matrix of the bosons in the internal lines in Eq. (9.88) page 255, and Fig. 9.4, page 232.

9.9.3 Boson-boson scattering

Figure 9.13: Feynman diagram (box diagram) with four vertices and four internal lines. Since this diagram has a loop. we have one independent oﬀ-shell momentum q over which we shall integrate.

The rule to write the S matrix contributions in matrix notation, is that we write coefficient functions, coupling − matrices and propagators in an ordered ruled by following fermion lines backwards from the ordered determined 256 CHAPTER 9. THE FEYNMAN RULES by the arrows10. For example, in the matrix notation, the S matrix for boson-boson scattering in the same − theory is given by the sum of one loop diagrams, as shown in Fig. 9.7 page 235. We show in Fig. 9.13, the flow of momenta in that loop diagram. Note that with the labelling of momenta in each internal line of Fig. 9.13, the conservation of four-momentum is guaranteed in each vertex. We see it by using the labelling of vertices in Fig. 9.7, and using positive sign for momenta flowing into the vertex, and negative sign for momenta flowing out of the vertex, we see that

V q + p′ p q + p′ + p = 0 k1 → 1 − 1 − 1 1 Vk′ q + p′ p′ q = 0 1 → 1 − 1 − Vk′ q q p2′ p2′ = 0 2 → − − − V q p′ + p q + p′ p = (p + p ) p′ + p′ = q q = 0 k2 → − 2 2 − 1 − 1 1 2 − 1 2 − Thus, we shall not introduce the Dirac delta for each vertex but only the global Dirac delta δ (p + p p p ) associated 1 2 − 1′ − 2′ with the global conservation of four-momentum. In that case the S matrix element can be written as follows − —————————————- —————————————-

1. Initial particles (p1σ1n1) and (p2σ2n2) (both are bosons) provide factors u (p σ n ) u (p σ n ) k1 1 1 1 ; k2 2 2 2 (2π)3/2 (2π)3/2

2. Final particles (p1′ σ1′ n1′ ) and (p2′ σ2′ n2′ ) (both are bosons) provide factors

uk∗′ (p1′ σ1′ n1′ ) uk∗′ (p2′ σ2′ n2′ ) 1 ; 2 (2π)3/2 (2π)3/2

3. The vertex k1 gives (not Dirac delta is introduced) i (2π)4 g − n1n4k1

4. The propagator from k1 to k1′ yields 4 i (2π)− P ′ (q + p ) k1k1 1′ − 2 2 (q + p1′ ) + mk′ k iε 1 1 −

5. Vertex k1′ yields 4 i (2π) gn n k′ − 2 1 1

6. Propagator from k1′ to k2′ gives 4 i (2π)− P ′ ′ (q) k2k1 2 2 − q + mk′ k′ iε 2 1 −

7. Vertex k2′ yields 4 i (2π) gn n k′ − 3 2 2

8. Propagator from k2′ to k2 gives 4 i (2π)− Pk k′ (q p′ ) 2 2 − 2 − 2 2 (q p2′ ) + mk k′ iε − 2 2 − 10The reader can check such a rule by contrasting Eq. (9.86) with Fig. 9.8, Page 236, as well as Eq. (9.88) with Fig. 9.4, page 232. 9.9. EXAMPLES OF APPLICATION FOR THE FEYNMAN RULES WITH INTEGRATION OVER FOUR-MOMENTA

9. Vertex k2 gives i (2π)4 g − n4n3k2

10. Propagator from k2 to k1 yields 4 i (2π)− P (q + p p ) k1k2 1′ − 1 2 2 −(q + p′ p1) + m iε 1 − k1k2 − 11. Let us multiply all these factors, put a minus sign associated with the fermionic loops, and the global Dirac delta function

u∗′ (p′ σ′ n′ ) u∗′ (p′ σ′ n′ ) uk1 (p1σ1n1) uk2 (p2σ2n2) k1 1 1 1 k2 2 2 2 P δ p1 + p2 p1′ p2′ ≡ − − − " (2π)3/2 (2π)3/2 #" (2π)3/2 (2π)3/2 # 4 4 i (2π)− P ′ (q + p ) i (2π)− P ′ ′ (q) 4 k1k1 1′ 4 k2k1 i (2π) gn n k i (2π) g ′ 1 4 1 2 2 n2n1k1 2 2 − −  − − q + m ′ ′ iε (q + p1′ ) + mk′ k iε " k k # h i 1 1 − h i 2 1 −  4  4 i (2π)− P ′ (q p ) 4 k2k2 2′ 4 i (2π)− Pk1k2 (q + p1′ p1) i (2π) gn n k′ − i (2π) gn n k − − 3 2 2 − 2 2  − 4 3 2 − 2 2 (q p2′ ) + mk k′ iε " (q + p1′ p1) + mk k iε# h i − 2 2 − h i − 1 2 −  

δ (p1 + p2 p1′ p2′ ) P − − − uk∗′ p1′ σ1′ n1′ uk∗′ p2′ σ2′ n2′ [uk1 (p1σ1n1) uk2 (p2σ2n2)] ≡ (2π)6 1 2 h i P ′ ′ (q) P ′ (q + p ) k2k1 k1k1 1′ g ′ g ′ n3n2k2 2 2 n2n1k1 2 2 q + m ′ ′ iε   " k k # (q + p1′ ) + mk′ k iε h i 2 1 − h i 1 1 −   ′ P (q + p p ) Pk2k (q p2′ ) [g ] k1k2 1′ − 1 [g ] 2 − n1n4k1 2 2 n4n3k2  2 2  "(q + p1′ p1) + mk k iε# (q p2′ ) + mk k′ iε − 1 2 − − 2 2 −   we can replace the couple of indices for vertices by the corresponding index of the internal line thus

k′ k′ n ; k′ k n ; k k n ; k k′ n 2 1 → 2 1 1 → 1 1 2 → 4 2 2 → 3 using this, and applying the matrix notation we have

δ (p1 + p2 p1′ p2′ ) P − − − uk∗′ p1′ σ1′ n1′ uk∗′ p2′ σ2′ n2′ [uk1 (p1σ1n1) uk2 (p2σ2n2)] ≡ (2π)6 1 2 h i Pn2 (q) Pn1 (q + p1′ ) Γk′ Γk′ 2 n n 2 2 1 n n 2 2 3 2 q + mn2 iε 2 1 "(q + p′ ) + mn iε# h i − h i 1 1 −

Pn4 (q + p1′ p1) Pn3 (q p2′ ) [Γk ] − [Γk ] − 1 n1n4 2 2 2 n4n3 2 2 "(q + p′ p1) + mn iε# "(q p′ ) + mn iε# 1 − 4 − − 2 3 −

δ (p1 + p2 p1′ p2′ ) P − − − uk∗′ p1′ σ1′ n1′ uk∗′ p2′ σ2′ n2′ [uk1 (p1σ1n1) uk2 (p2σ2n2)] ≡ (2π)6 1 2 h i P (q) P (q + p1′ ) Γk′ Γk′ 2 q2 + M 2 iε 1 2 2 n3n2 " (q + p1′ ) + M iε# − − n2n1 P (q + p p ) P (q p ) Γ 1′ − 1 Γ − 2′ k1 2 2 k2 2 2 " (q + p1′ p1) + M iε# " (q p2′ ) + M iε# − − n1n4 − − n4n3 258 CHAPTER 9. THE FEYNMAN RULES

note that the chain of matrix multiplications (in the two last lines) starts and ends in the same index (n3 in this case), thus when summing over the indices n1,n2,n3,n4 what we obtain is the trace of such product of matrices

12. Now we integrate over the oﬀ-shell momentum q (after integrating over all four momenta associated with each internal line, we are left with only one integral and one global delta because of the Dirac delta functions over each vertex). We also sum over ﬁeld indices (n1,n2,n3,n4) and (k1, k2, k1′ , k2′ )

S(1) = d4q P k k k′ k′ n1n2n3n4 Z 1 X2 1 2 X

the ﬁrst set of sums gives the trace of the product of matrices as mentioned above. Thus we ﬁnally obtain

——————————————- ——————————————–

6 4 SBB Sp′ σ′ n′ , p′ σ′ n′ ; p σ n , p σ n = (2π)− δ p1 + p2 p′ p′ ≡ 1 1 1 2 2 2 1 1 1 2 2 2 − − 1 − 2

uk∗′ p1′ ,σ1′ ,n1′ uk∗′ p2′ ,σ2′ ,n2′ uk1 (p1,σ1,n1) uk2 (p2,σ2,n2) × 1 2 k k k′ k′ 1 X2 1 2 4 P (q) P (q + p1′ ) d q T r Γ ′ Γ ′ k2 2 2 k1 2 2 × ( q + M iε (q + p′ ) + M iε Z − 1 − P (q + p p ) P (q p ) Γ 1′ − 1 Γ − 2′ k1 2 2 k2 2 2 × (q + p′ p1) + M iε (q p′ ) + M iε) 1 − − − 2 − + . . . (9.89) the ellipsis in the last line indicates terms obtained by permuting bosons 1′, 2′, and 2. The minus sign at the beginning of the RHS is the extra minus sign associated with the fermionic loops. After eliminating all delta functions we are left with one momentum-space integral, as it must be for a one loop diagram. By comparing (9.89) with Fig. 9.13 the reader can check once again that the rule to write the S matrix − contributions in matrix notation, is that we write coeﬃcient functions, coupling matrices and propagators in an ordered ruled by following lines backwards from the ordered determined by the arrows.

9.10 Examples of Feynman rules as integrations over momenta

Consider a theory involving a Dirac spinor field ψ (x) of mass M and a pseudoscalar field φ (x) of mass m, with an interaction given by igφψγ¯ ψ (9.90) − 5 where the factor i is included for the interaction to be hermitian with a real coupling constant g. Recalling that − µ the polynomial P (q) is the unity for scalars and [ iγµq + M] β for the Dirac spinors, and that the coefficient 1/2 − functions are u = (2E)− for a scalar of energy E while for the normalized Dirac spinors u is the one shown in Sec. 7.4. With this input, equations (9.86), (9.88) and (9.89) give the lowest-order connected S matrix elements − for fermion-boson scattering, fermion-fermion scattering and boson-boson scattering 9.10. EXAMPLES OF FEYNMAN RULES AS INTEGRATIONS OVER MOMENTA 259

9.10.1 Fermion-boson scattering Let us start with Eq. (9.86)

2 4 SF B Sp′ σ′ n′ , p′ σ′ n′ ; p σ n , p σ n = i (2π)− δ p1 + p2 p′ p′ ≡ 1 1 1 2 2 2 1 1 1 2 2 2 − 1 − 2 P (p1 + p2) u† p1′ σ1′ n1′ Γk′ Γk u (p1σ1n1) u∗′ p2′ σ2′ n2′ uk (p2σ2n2) × (p + p )2 + M 2 iε k k′k 1 2 X − P (p2′ p1) + u† p′ σ′ n′ Γk′ − Γk u (p1σ1n1) u∗ p′ σ′ n′ uk′ (p2σ2n2) (9.91) 1 1 1 2 2 k 2 2 2 " (p1 p2′ ) + M iε #) − − by taking into account that in theory (9.90) there is only one type of fermion and one type of boson and that there is only one value of σ for scalar bosons, we write

2 4 SF B Sp′ σ′ , p′ ; p σ , p = i (2π)− δ p1 + p2 p′ p′ ≡ 1 1 2 1 1 2 − 1 − 2 P (p1 + p2) u† p′ σ′ Γ Γ u (p1σ1) u∗ p′ u (p2) × 1 1 2 2 2 (p1 + p2) + M iε − P (p2′ p1) + u† p′ σ′ Γ − Γ u (p1σ1) u∗ p′ u (p2) (9.92) 1 1 2 2 2 " (p1 p2′ ) + M iε #) − − we start by inserting the polynomials for fermions, the coeﬃcient functions for scalars and the fermion-fermion- boson coupling according with Eq. (9.90)

µ 1 P (q) = [ iγµq + M] β ; u (p)= ; Γ= igγ5 (9.93) − √2E − inserting (9.93) in (9.92) we have

2 4 SF B Sp′ σ′ , p′ ; p σ , p = i (2π)− δ p1 + p2 p′ p′ ≡ 1 1 2 1 1 2 − 1 − 2 µ [ iγµ (p1 + p2) + M] β 1 1 u† p′ σ′ [ igγ5] − [ igγ5] u (p1σ1) 1 1 2 2 × (" − (p1 + p2) + M iε − 2E′ √2E2 # − 2 µ [ iγµ (p2′ p1) + M] β p1 1 + u† p′ σ′ [ igγ5] − − [ igγ5] u (p1σ1) (9.94) 1 1 2 2 " − (p1 p2′ ) + M iε − 2E2′ √2E2 #) − − p 2 2 2 4 1 1 S S ′ ′ ′ = i ( i) (2π)− g δ p + p p′ p′ F B p1σ1, p2; p1σ1, p2 1 2 1 2 ≡ − − − 2E2′ √2E2 µ [ iγµ (p1 + p2) + M] β p u† p1′ σ1′ γ5 − γ5 u (p1σ1) × (p + p )2 + M 2 iε 1 2 − µ [ iγµ (p2′ p1) + M] β + u† p′ σ′ γ5 − − γ5 u (p1σ1) (9.95) 1 1 2 2 " (p1 p2′ ) + M iε #) − − ???

2 2 1 4 S S ′ ′ ′ = i (2π)− g δ p + p p′ p′ F B p1σ1, p2; p1σ1, p2 1 2 1 2 ≡ − 4E2′ E2 − − µ [ iγµ (p1 + p2)p+ M] u† p1′ σ1′ β γ5 − γ5 u (p1σ1) × (p + p )2 + M 2 iε 1 2 − µ [ iγµ (p2′ p1) + M] + u† p′ σ′ β γ5 − − γ5 u (p1σ1) (9.96) 1 1 2 2 " (p1 p2′ ) + M iε #) − − 260 CHAPTER 9. THE FEYNMAN RULES

2 2 1/2 4 S S ′ ′ ′ = i (2π)− g 4E′ E − δ p + p p′ p′ F B p1σ1p2;p1σ1p2 2 2 1 2 1 2 ≡ − µ − − iγµ (p1 + p2) + M u¯ p1′ σ1′ γ5 − γ5u (p1σ1) × (p + p )2 + M 2 iε 1 2 − µ iγµ (p1 p2′ ) + M +u ¯ p′ σ′ γ5 − − γ5u (p1σ1) 1 1 2 2 (p1 p2′ ) + M iε ) − − 9.10.2 Fermion-fermion and Boson-boson scattering For fermion-fermion and boson-boson scattering we can use Eq. (9.88) page 255 and Eq. (9.89) page 258, to obtain the S matrix elements for these processes in the framework of the theory (9.90). We obtain −

2 2 4 SFF Sp′ σ′ p′ σ′ ;p σ p σ = i (2π)− g δ p1 + p2 p′ p′ ≡ 1 1 2 2 1 1 2 2 − − 1 − 2 u¯ p′ σ′ γ u (p σ ) u¯ p′ σ′ γ u (p σ ) × 2 2 5 2 2 1 1 5 1 1 1 1′ 2′ 2 2 ×(p1′ p1) + m iε − ⇔ − −

6 4 1/2 4 SBB Sp′ p′ ;p p = (2π)− g 16E1E2E′ E′ − δ p1 + p2 p′ p′ ≡ 1 2 1 2 − 1 2 − 1 − 2 µ µ 4 iγµq + M iγµ (q+ p1′ ) + M d q T r γ5 − γ5 − 2 2 2 2 × ( q + M iε (q + p′ ) + M iε Z − 1 − µ µ iγµ (q + p1′ p1) + M iγµ (q p2′ ) + M γ5 − − γ5 − − + . . . 2 2 2 2 × (q + p′ p1) + M iε (q p′ ) + M iε) 1 − − − 2 − once again the ellipsis mean a sum over permutations of particles 2, 1′, 2′. The factors u† have been replaced by u¯ by using the deﬁnition (7.152) page 207.

9.11 Topological structure of the lines

The topological structure of the diagrams suggests a kind of conservation law of lines. It is useful if we consider the internal and external lines as being created at vertices and destroyed in pairs at the centers of internal lines or when external lines leave the diagram. Note that it is not related with the direction of the arrows carried by the lines. Let I and E be the number of internal and external lines respectively, where Vi denotes the numbers of vertices of various types labelled by i, and ni are the number of lines attached to each vertex. Equating the number of lines that are created and destroyed we obtain

2I + E = niVi (9.97) Xi this also holds separately for ﬁelds of each type. Let us take an example, assume that we have a theory with four types of interactions described as follows

Type 1 fermion fermion boson vertex(trilinear) ; n = 3 ≡ − − 1 Type 2 boson boson boson vertex (trilinear); n = 3 ≡ − − 2 Type 3 boson boson fermion vertex(trilinear); n = 3 ≡ − − 3 Type 4 four bosons vertex ; n = 4 ≡ 4 9.11. TOPOLOGICAL STRUCTURE OF THE LINES 261

Figure 9.14: Two connected diagrams for a theory with four types of interactions. (a) Scattering from fermion- fermion to fermion-boson. (b) Scattering from boson-boson to fermion-boson.

In Fig. 9.14 (a), we have a connected diagram of the scattering from fermion-fermion to fermion-boson. This diagram as a whole cannot describe a physical process11, but it could be a piece of a greater diagram. For this diagram we have

E = 4 ; I = 7 ; V1 = 4 (vertices a, d, e, f) ;

V2 = 1 (vertex b) ; V3 = 1(vertex c) ; V4 = 0 thus Eq. (9.97) gives in this case 2 7+4=3 4 + 3 1 + 3 1 + 4 0 · · · · · In Fig. 9.14 (b), we have a connected diagram of the scattering from boson-boson to fermion-boson. For the same reasons as before, this diagram cannot account on a physical process, but can be part of a greater diagram. For this diagram we have

E = 4 ; I = 6 ; V1 = 3 (vertices c, d, e) ;

V2 = 0 ; V3 = 1 (vertex b) ; V4 = 1 (vertex a) thus Eq. (9.97) gives in this case 2 6+4=3 3 + 3 0 + 3 1 + 4 1 · · · · · 11The initial state (fermion-fermion) has an integer spin, while the final process (fermion- boson) has a half-odd integer spin. Therefore, the difference between final and initial spins is a half-odd integer, and we cannot balance such a difference with an orbital angular momentum (since orbital angular momenta are always integer). Thus, the process is forbidden by conservation of total angular momentum. 262 CHAPTER 9. THE FEYNMAN RULES

In the special case in which all interactions involve the same number of ﬁelds, we have ni = n so that Eq. (9.97) becomes 2I + E = nV (9.98) where V is the total number of all vertices. In this case we can combine Eqs. (9.82) and (9.98) to eliminate I, and for the case of a connected diagram (C = 1) we have

L = I V + C = I V + 1 I = L + V 1 − − ⇒ − nV = 2I + E = 2 (L + V 1) + E ⇒ − (n 2) V = 2L + E 2 ⇒ − − 2L + E 2 V = − (9.99) n 2 − For example, if we have a trilinear interaction, the diagrams for a scattering process of two particles into two particles (E = 4) with L = 0, 1, 2,... we have V = 2, 4, 6,.... In general, the expansion in powers of the coupling constants is an expansion in increasing numbers of loops.

9.12 Oﬀ-shell and on-shell four-momenta

Figure 9.15: (a) A one-loop diagram. (b) The previous diagram could be constructed by connecting two pairs of external lines of tree graph diagrams.

In the Feynman diagrams for the S matrix, the external lines are “on the mass shell” (or simply on-shell), − i.e. the four-momentum associated with each external line obeys the restriction p2 = m2 for a particle of mass − m. It is however important to consider also external lines in which the associated energies are not related with the three-momenta (like the case of internal lines) i.e. diagrams “off the mass shell” (or simply off shell). This off shell diagrams are usually part of a larger Feynman diagram. For example, a loop appearing as an insertion in some internal line of a diagram could be seen as a Feynman diagram with two external lines, in which both are off the mass shell. Figure 9.15, shows a one loop diagram that can be formed by connecting two diagrams in which the external lines that form the loop should be off-shell. In the path integral approach, it is usual to derive first the Feynman rules with all external lines off-shell and then construct the S matrix elements by applying the on-shell limit. − Once the contribution of a diagram off the mass shell is calculated, the associated contribution to the S − matrix can be calculated by taking the on shell restriction and considering the four-momentum flowing along the line into the diagram with p0 = p2 + m2 for particles in the initial state and with p0 = p2 + m2 for particles − in the final state inserting the apropriate external line factors: p p u v k ; k∗ for initial particles or antiparticles respectively (2π)3 (2π)3 q u q v k∗ ; k for final particles or antiparticles respectively (2π)3 (2π)3 q q 9.12. OFF-SHELL AND ON-SHELL FOUR-MOMENTA 263

Feynman diagrams off the mass shell are a particular case of some types of diagrams that take into account the effects of various possible external fields. Let us consider some additional terms in the Hamiltonian involving some external fields ε (x), such that the interaction V (t) in the Dyson series (2.188) for the S matrix is replaced by a − V (t)= V (t)+ d3x ε (x) (x,t) (9.100) ε a Oa a X Z where the “currents” (t) have the time-dependence typical of operators in the interaction picture Oa (t) = exp (iH t) (0) exp ( iH t) (9.101) Oa 0 Oa − 0 but otherwise such operators are arbitrary. The S matrix for a given transition α β becomes a functional − → S [ε] in terms of the c number function ε (t). We require an extension of the usual Feynman rules to obtain βα − a such a functional. Besides the usual vertices obtained from V (t) we must include some extra vertices:

1. If the current (x) is a product of n field factors, each vertex with position label x, must be attached Oa a Oa to na lines of the corresponding types. So its contribution to the position-space Feynman rules, is equal to iε (x) times numerical factors that appear in (x). (Note that ε (x) is acting as though it were a − a Oa a constant coupling). 2. The r th variational derivative of S [ε] with respect to ε (x ) , ε (x ) ,...,ε (x ) at ε = 0 is given − βα a1 1 a2 2 ar r by position space diagrams with r additional vertices, to which internal lines na1 ,na2 ,... are attached respectively (i.e. the number n of fields associated with each current (x,t)), and no external lines ak Oak (since we are evaluating at ε = 0 i.e. eliminating the external fields εa (x)). These vertices have position labels x,y,... over which we do not integrate. Each of these vertices provides a contribution of i times − numerical factors that appear in the corresponding current . To see it, we observe that the non-vanishing Oa terms of S [ε] when we take the r th variational derivatives are precisely the terms that contain the βα − r external fields εa1 (x1) , εa2 (x2) ,...,εar (xr) (and nothing else). On one hand, the variational derivative − 12 with respect to εak (xk) vanishes if such a field is not contained in the term . On the other hand, if there are some extra fields in the term the derivative vanishes when evaluating at ε = 0.

——————————————————- An important particular case appears when these currents are all single ﬁeld factors, it can be written as

3 Vε (t)= V (t)+ d x εk (x,t) ψk (x,t) Xk Z the r th variational derivative of S [ε] with respect to ε (x ) , ε (x ) ,...,ε (x ) at ε = 0 is represented by − βα k1 1 k2 2 kr r space diagrams with r additional vertices carrying space-time labels x1,x2,...,xr and to each of then is attached a single internal particle line of type k1, k2,...,kr We can figure out them as off shell external lines, but with the 3/2 ip x difference that their contribution to the matrix element is not a coefficient of the form (2π)− uk (p,σ) e · or 3/2 ip x 13 of the form (2π)− u (p,σ) e but a propagator but also a factor ( i) from the vertex at the end of the k∗ − · − line. The result is then a momentum space Feynman diagram with particles in states α and β on the mass shell with the addition of r external lines of type k1, k2,...,kr which carry momenta p1,p2,...,pr from the variational derivative r δ Sβα [ε] δε (x ) δε (x ) ...δε (x ) k1 1 k2 2 kr r ε=0 by removing the propagators on each off-shell line and taking the adequate Fourier transform , and multiplying with the proper coefficient functions u , u etc, and a factor ( i)r . k k∗ − 12 Thus in the Dyson series (2.188), page 105, only survives terms with r−vertices Vε1 (t1) Vε2 (t2) · · · Vεr (tr) which contains precisely the set of external fields εa1 (x1) εa2 (x2) ...εar (xr), through the vertex correction (9.100). 13 It has to do with the fact that the external lines εa (x) of these diagrams will be part of another diagrams in which such external lines become internal lines, as illustrated in Fig. 9.15, page 262. 264 CHAPTER 9. THE FEYNMAN RULES

9.12.1 The r th derivative theorem − There is a simple relation between the sum of contributions coming from all diagrams associated with perturbation theory for any oﬀ shell amplitude and a matrix element, between eigenstates of the full Hamiltonian, of a time- ordered product of corresponding operators in the Heisenberg picture. The theorem that provides such a relations, states that to all orders of perturbation theory

r δ Sβα [ε] r + δ S = ( i) β− T (x ) (x ) α (9.102) r ≡ δε (x ) δε (x ) ...δε (x ) . . . − {Oa1 1 ···Oar r } a1 1 a2 2 ar r ε=0 where ¯ (x), are the counterparts of (x) in the Heisenberg picture Oa Oa 1 ¯ (x,t) = exp (iHt) (x, 0) exp ( iHt)=Ω(t) (x,t)Ω− (t) (9.103) Oa Oa − Oa iHt iH0t Ω (t) e e− (9.104) ≡ we recall that β+ and β are “in” and “out” eigenstates of the full hamiltonian H. | i | −i We prove it as follows. From Eqs. (2.36, 2.188) we have

N (0) (0) (0) ∞ ( i) ∞ (0) Sβα [ε] = β S [ε] α = β − dτ1 dτ2 dτN T Vε1 (τ1) Vε2 (τ2) VεN (τN ) α " N! · · · { · · · }# D E D NX=0 Z−∞ E N ∞ ( i) ∞ S [ε] = − dτ dτ dτ β(0) T V (τ ) V (τ ) V (τ ) α(0) βα N! 1 2 · · · N { ε1 1 ε2 2 · · · εN N } N=0 Z−∞ D E X as discusssed above, when we apply the r th derivative only terms with r operators of the form (x ,t ) − − Oak k k survive in the Dyson series

N P V (τ ) [V (τ ) V (τ ) V (τ )] = V (τ )+ d3y ε (y ) (y ) V (τ )+ d3y ε (y ) ≡ εI I ≡ ε1 1 ε2 2 · · · εN N  1 b1 1 Ob1 1   2 b2 2 IY=1 Xb1 Z Xb2 Z    V (τ )+ d3y ε (y ) (y ) · · ·  N bN N ObN N  XbN Z  r  δ P r = ( i) V (τ1) V (τN ) a1 (x1) a2 (x2) ar (xr) δεa1 (x1) ...δεar (xr) − · · · O O ···O Therefore, from the Dyson series Eq. (2.188) page 105 we observe that the left-hand side of Eq. (9.102) reads

r N+r δ Sβα [ε] ∞ ( i) ∞ δrS = − dτ1dτ2 dτN ≡ δεa1 (x1) ...δεar (xr) ε=0 N! · · · NX=0 Z−∞ β(0) T V (τ ) V (τ ) (x ) (x ) α(0) (9.105) × { 1 · · · N Oa1 1 ···Oar r } D E let us assume that x0 x0 . . . x 0 such that the currents (x ) are not reordered by the time ordering 1 ≥ 2 ≥ ≥ r Oak k operator. However, the original interaction V (τ ) V (τ ) are not necessarily ordered neither they are in order 1 · · · N with respect to the currents. Hence, a given subset of the N vertices are on left of (x ), another subset is on Oa1 1 the left of (x ) and so on. Of course there could also be a subset of vertices on the right of (x ). Oa2 2 Oar r We shall denote as τ τ all τ s that are greater than x0; as τ τ all τ s between x0 and x0 and so 01 · · · 0N0 ′ 1 11 · · · 1N1 ′ 1 2 on. We ﬁnally denote as τ τ all τ s that are less than x0. With this notation, the time ordering yields r1 · · · rNr ′ r T T V (τ ) V (τ ) (x ) (x ) = T V (τ ) V (τ ) (x ) T V (τ ) V (τ ) (x ) ≡ { 1 · · · N Oa1 1 ···Oar r } { 01 · · · 0N0 } Oa1 1 { 11 · · · 1N1 } Oa2 2 T V (τr 1,1) V τr 1,Nr−1 ar (xr) T V (τr1) V (τrNr ) ···×··· − · · · − O { · · · } 9.12. OFF-SHELL AND ON-SHELL FOUR-MOMENTA 265 note that we have r + 1 “cases” to ﬁll between the r currents (x ) (x ) in the time ordering process. − Oa1 1 ···Oar r That is 1 (x ) 2 (x ) . . . r (x ) r + 1 {· · · ···}Oa1 1 {··· ···}Oa2 2 {··· ···}Oar r {··· ···} with

1 = T V (τ ) V (τ ) ; 2 = T V (τ ) V (τ ) {··· ···} { 01 · · · 0N0 } {··· ···} { 11 · · · 1N1 } r = T V (τr 1,1) V τr 1,Nr−1 ; r + 1 = T V (τr1) V (τrNr ) {··· ···} − · · · − {··· ···} { · · · } it is clear that all N vertices V(τi) must be sort out into the r+1 cases, and that the order in each case is not relevant (because of the presence of the time ordering operator for each subset). Therefore this is a typical combinatory and the possible ways of sorting the N vertices into the r + 1 subsets each one containing N0,N1,...,Nr vertices is given by N! with N + N + . . . + N = N N !N ! N ! 0 1 r 0 1 · · · r for each subset we can reformulate the limits of integration in time according with its position with respect to the currents. For instance, for the first subset, we had by definition that τ τ are greater than x0 then we can 01 · · · 0N0 1 replace ∞ ∞ dτ01 dτ0N0 dτ01 dτ0N0 · · · → x0 · · · Z−∞ Z 1 we also had that τ τ are between x0 and x0 thus 11 · · · 1N1 1 2 x0 ∞ 1 dτ11 dτ1N1 dτ11 dτ1N1 · · · → x0 · · · Z−∞ Z 2 similarly x0 ∞ r−1 dτr 1,1 dτr 1,Nr−1 dτr 1,1 dτr 1,Nr−1 − · · · − → x0 − · · · − Z−∞ Z r finally, τ τ are less than x0 so r1 · · · rNr r 0 xr ∞ dτ dτ dτ dτ r1 · · · rNr → 11 · · · 1N1 Z−∞ Z−∞ with all these considerations, Eq. (9.105) becomes

r N δ Sβα [ε] r ∞ ( i) N! δN,N0+N1+ +Nr δrS = ( i) − ··· ≡ δεa1 (x1) ...δεar (xr) ε=0 − N! N0!N1! Nr! N=0 N0N1 Nr X X··· · · · x0 x0 x0 ∞ 1 r−1 r dτ01 dτ0N0 dτ11 dτ1N1 dτr 1,1 dτr 1,Nr−1 dτr1 dτrNr × x0 · · · x0 · · · · · · x0 − · · · − · · · Z 1 Z 2 Z r Z−∞ β T V (τ ) V (τ ) (x ) T V (τ ) V (τ ) (x ) × h | { 01 · · · 0N0 } Oa1 1 { 11 · · · 1N1 } Oa2 2 · · · T V (τr 1,1) V τr 1,Nr−1 ar (xr) T V (τr1) V (τrNr ) α ×··· − · · · − O { · · · } | i where the factor N! δN,N0+N1+ +Nr ··· N !N ! N ! 0 1 · · · r is the combinatoric factor which provides the number of ways of sorting N τ ′s into r + 1 subsets, where each subset contains N0,N1,...,Nr of these τ ′s, with the constraint N0 + N1 + . . . + Nr = N. Then we sum over N. Now, since in the process of summing over N such a number takes any non-negative integer, we observe that summing over N and then summing over N0,N1,...,Nr with the constraint N0 + . . . + Nr = N, is equivalent to 266 CHAPTER 9. THE FEYNMAN RULES

sum over N0,N1,...,Nr independently, that is with each Nk taking all non-negative integer values (i.e. omitting the constraint). Further we can decompose ( i)N as ( i)N0 ( i)Nr . From this process we obtain − − · · · − r N0 N1 Nr δ Sβα [ε] r ( i) ( i) ( i) δrS = ( i) β − − · · · − ≡ δεa1 (x1) ...δεar (xr) ε=0 − h | · · · N0!N1! Nr! XN0 XN1 XNr · · · x0 x0 x0 ∞ 1 r−1 r dτ01 dτ0N0 dτ11 dτ1N1 dτr 1,1 dτr 1,Nr−1 dτr1 dτrNr × x0 · · · x0 · · · · · · x0 − · · · − · · · Z 1 Z 2 Z r Z−∞ T V (τ ) V (τ ) (x ) T V (τ ) V (τ ) (x ) × { 01 · · · 0N0 } Oa1 1 { 11 · · · 1N1 } Oa2 2 · · · T V (τr 1,1) V τr 1,Nr−1 ar (xr) T V (τr1) V (τrNr ) α ×··· − · · · − O { · · · } | i separating each sum and respecting the time ordering we ﬁnd

r N0 δ Sβα [ε] r ( i) ∞ δrS = ( i) β − dτ01 dτ0N0 T V (τ01) V (τ0N0 ) a1 (x1) ≡ δε (x ) ...δε (x ) − h |  N ! 0 · · · { · · · } O a1 1 ar r ε=0 0 x1 XN0 Z 0   N1 x ( i) 1 − dτ11 dτ1N1 T V (τ11) V (τ1N1 ) a2 (x2)  N ! 0 · · · { · · · } O × 1 x2 XN1 Z  N 0  ( i) r−1 xr−1 − dτr 1,1 dτr 1,Nr−1 T V (τr 1,1) V τr 1,Nr−1 ×···  Nr 1! x0 − · · · − − · · · −  Nr−1 − Z r X  Nr 0  ( i) xr ar (xr) − dτr1 dτrNr T V (τr1) V (τrNr ) α O " Nr! · · · { · · · }# | i XNr Z−∞ defining N ′ ∞ ( i) t U t′,t − dτ dτ T V (τ ) V (τ ) (9.106) ≡ N! 1 · · · N { 1 · · · N } N=0 Zt X we can write the r th derivative as − δrS [ε] δ S βα = ( i)r β U ,x0 (x ) U x0,x0 r ≡ δε (x ) ...δε (x ) − h | ∞ 1 Oa1 1 1 2 a1 1 ar r ε=0 0 0 0 0 0 a2 (x2) U x2,x3 U xr 1,xr ar (xr) U xr, α (9.107) O · · · − O −∞ | i Note that the operator (9.106) has a structure similar to the Dyson series Eq. (2.188) page 105 (except that U (t′,t) has finite limits of integration). The operator U (t′,t) obeys the differential equation d U t′,t = iV t′ U t′,t ; U (t,t) = 1 dt′ − whose solution reads

1 U t′,t = exp iH t′ exp iH t′ t exp ( iH t)=Ω− t′ Ω (t) (9.108) 0 − − − 0 where Ω is deﬁned by Eq. (9.104). Substituting Eq. (9.108) in Eq. (9.107) we obtain ————————————–

r δ Sβα [ε] r 1 0 1 0 0 δ S = ( i) β Ω− ( )Ω x (x ) Ω− x Ω x r ≡ δε (x ) ...δε (x ) − h | ∞ 1 Oa1 1 1 2 a1 1 ar r ε=0 1 0 0 1 0 0 1 0 a2 (x2) Ω− x2 Ω x3 Ω− xr 1 Ω xr ar (xr) Ω− xr Ω ( ) α (9.109) O · · · − O −∞ | i 9.12. OFF-SHELL AND ON-SHELL FOUR-MOMENTA 267 taking into account that Ω 1 ( )=Ω ( ) and using Eq. (9.103) we have − ∞ † ∞ r δ Sβα [ε] r 0 1 0 δ S = ( i) Ω ( ) β Ω x (x )Ω− x r ≡ δε (x ) ...δε (x ) − h ∞ | 1 Oa1 1 1 a1 1 ar r ε=0 0 1 0 0 1 0 0 1 0 Ω x2 a2 (x2)Ω− x2 Ω x3 Ω− xr 1 Ω xr ar (xr)Ω− xr Ω ( ) α (9.110) O · · · − O | −∞ i δrS [ε] δ S βα = ( i)r Ω ( ) β ¯ (x ) ¯ (x ) ¯ (x ) Ω ( ) α (9.111) r ≡ δε (x ) ...δε (x ) − h ∞ | Oa1 1 Oa2 2 Oar r | −∞ i a1 1 ar r ε=0 On the other hand we saw in 2.1 that [in the sense of Eq. (2.16) page 68] the “in” and “out” states are related with the free states in the form given by Eq. (2.19) page 68

(0) β± = Ω ( ) β (9.112) ∓∞ E hence Eq. (9.111) becomes r δ Sβα [ε] r + δ S = ( i) β− ¯ (x ) ¯ (x ) ¯ (x ) α (9.113) r ≡ δε (x ) ...δε (x ) − Oa1 1 Oa2 2 Oar r a1 1 ar r ε=0

————————————– Recalling that we have assumed the condition x0 x0 . . . x0, we could replace the product of operators 1 ≥ 2 ≥ ≥ r on the right-hand side of Eq. (9.113) with a time-ordered product of operators

r δ S [ε] r + δ S = ( i) β− T (x ) (x ) α (9.114) r ≡ δε (x ) ...δε (x ) − {Oa1 1 ···Oar r } a1 1 ar r ε=0

Both sides of Eq. (9.114) are completely symmetric (or completely antisymmetric for fermions) in the a′s and 0 0 0 x′s. Consequently, this expression is satisﬁed regardless the order of the times x1,x2,...,xr. Equation (9.114) coincides with the result (9.102) we were looking for. Chapter 10

Canonical quantization

The canonical quantization of postulated Lagrangians has been the first historical approach for quantum field theories, and it is also the initial approach in most books of quantum field theory. This starting point has the advantage that most of the known quantum fields theories can be easily formulated in a Lagrangian form. Moreover, a classical theory with a Lorentz invariant Lagrangian density, leads when canonically quantized to a Lorentz invariant quantum theory. The canonical formalism will lead to quantum mechanical operators that obey the commutation relations of the Poincaréalgebra, leading in turn to a Lorentz-invariant S matrix. − The preservation of the Lorentz invariance after canonical quantization should not be taken for granted and it is and outstanding property. On one hand, some symmetries could be broken after a process of quantization (phenomenon known as anomaly), and on the other hand we saw in section 9.7 that in theories with derivative couplings or spins j 1, that it does not suffice to construct the interaction Hamiltonian as the integral over ≥ space of a scalar interaction density. It is necessary to add a non-scalar Hamiltonian density to compensate non-invariant terms in the propagators. The canonical formalism with a scalar Lagrangian density provides the additional required terms for such a cancellation. Further, in the case of non-abelian gauge theories is particularly difficult to guess the form of the extra terms without starting with a Lorentz-invariant and gauge-invariant Lagrangian density.

10.1 Canonical variables

Our present developments will lead to commutation rules and equations of motions proper for a Hamiltonian version of the canonical formalism. This is the Hamiltonian formalism we require to calculate the S matrix − regardless we do it with a canonical or path integral formalism. However, it is not in general easy to find Hamiltonians that provides a Lorentz-invariant S matrix. Thus, the starting point will be a Lagrangian version − of the canonical formalism in order to derive satisfactory Hamiltonians. To do it, we should identify properly the canonical fields and their associated conjugates in various field theories. In this way we shall learn how to separate the free-field terms in the Lagrangian, to finally check that physically realistic theories are possible in the canonical formalism. We shall start by proving that the free fields for scalar, vector and Dirac fields provide a system of quantum op- n erators q (x,t) and canonical conjugates pn (x,t) that satisfy the appropriate (equal-time) canonical commutation and anticommutation relations

n 3 n [q (x,t) ,pn¯ (y,t)] = iδ (x y) δn¯ (10.1) n n¯ ∓ − q (x,t) ,q (y,t) = 0; [pn (x,t) ,pn¯ (y,t)] = 0 (10.2) ∓ ∓ the subscripts indicates commutators if either of the particles that are created and destroyed by the two ∓ operators are bosons, and anticommutators if both particles are fermions.

268 10.1. CANONICAL VARIABLES 269

10.1.1 Canonical variables for scalar fields Let us start with a real scalar field φ (x) that describes a self–charge-conjugate particle of zero spin. By combining Eqs. (5.8, 5.9) page 152 with Eq. (5.20) page 154, we obtain the commutation relation for such a field + + + + + [φ (x) , φ (y)] = φ (x)+ φ− (x) , φ (y)+ φ− (y) = φ (x) , φ (y) + φ (x) , φ− (y) − + − − − + φ− (x) , φ (y) + φ− (x) , φ−(y) − − + + = φ (x) , φ− (y) φ (y) , φ− (x) =∆+ (x y) ∆+ (y x) − − − − − − 3 1 d p ip (x y) ip (x y) [φ (x) , φ (y)] = 3 0 e · − e− · − − (2π) 2p − Z h i µ let us recall the reader that ∆+ (x) is even in x only for space-like separations of x and y. Hence for time-like separations this commutator does not vanish in general. Hence the self-charge-conjugate scalar field obeys the commutation relation

[φ (x) , φ (y)] = ∆ (x y) (10.3) − − 3 1 d k ik x ikx 0 2 2 ∆ (x) e · e− ; k k + m (10.4) ≡ (2π)3 2k0 − ≡ Z h i p the function ∆ (x) and its time derivative can be written as 3 1 d k i(k x k0t) i(k x k0t) ∆ (x,t) e · − e− · − ≡ (2π)3 2k0 − Z 3 h i 3 1 d k 0 i(k x k0t) 0 i(k x k0t) i d k i(k x k0t) i(k x k0t) ∆˙ (x,t) = ik e · − ik e− · − = e · − + e− · − (2π)3 2k0 − − −(2π)3 2 Z h i Z h i i 3 i(k x k0t) ∆˙ (x,t) = d ke · − −(2π)3 Z where we have denoted with a dot, the derivative with respect to the time x0 = t. Such functions evaluated at t = 0 yield 3 3 1 d k ik x ik x 2i d k ∆ (x, 0) e · e− · = sin k x ≡ (2π)3 2k0 − (2π)3 2√k2 + m2 · Z h i Z i 3 ik x ∆˙ (x, 0) = d k e · −(2π)3 Z the integral of ∆ (x, 0) vanishes since the integrand is odd in k. On the other hand, the integral of ∆˙ (x, 0) is the Fourier representation of Dirac’s delta. Hence, we have seen that ∆ (x, 0) = 0, ∆˙ (x, 0) = iδ3 (x) (10.5) − if we think that φ (x,t) is an appropriate generalized coordinate, a good prospect for the canonical conjugate momenta is φ˙ (x,t). Let us examine our hypothesis by calculating the commutator of φ (x,t) and φ˙ (x,t). First, by recalling the explicit form of the creation and annihilation ﬁelds Eqs. (5.5, 5.6) page 152 we can obtain φ˙+ (x) 0 0 and φ˙− (y) at equal times i.e. x = y = t, so that the time derivative has the same meaning regardless the point x or y in which we evaluate it 3 3 + d p 1 i(p x p0t) d p′ 1 i(p′ y p′0t) φ (x,t) = a (p) e · − ; φ− (y,t)= a† p′ e− · − 3 0 3 0 Z (2π) 2p Z (2π) 2p′ q 3 p 0 q 3 p 0 + d p ip i(p x p0t) d p′ ip′ i(p′ y p′0t) φ˙ (x,t) = − a (p) e · − ; φ˙− (y,t)= a† p′ e− · − (10.6) 3 0 3 0 Z (2π) 2p Z (2π) 2p′ q p q p 270 CHAPTER 10. CANONICAL QUANTIZATION

0 + 1 3 1 ip x 3 ip′ i(p′ y p′0t) φ (x) , φ˙− (y) = d p a (p) e · , d p′ a† p′ e− · − (2π)3 2p0 2p 0 − "Z Z ′ # h i ∓ ip x ip′ y 1 3 p3 e · e− · 0 p = d p d p′ ip′ a (p) , a† p′ (2π)3 2p0 2p 0 Z Z ′ ∓ 3 3 h i 3 0 d p d p′ ip x ip′ y 3 1 0 d p ip (x y) = ip′ p ep· e− · δ p p′ = ip e · − (2π)3 (2p0 2p 0) − (2π)3 2p0 Z · ′ Z 3 + i d ppip (x y) φ (x) , φ˙− (y) = e · − (2π)3 2 h i− Z since we are evaluating at equal times then x y = x y. Thus − − 3 + i d p ip (x y) i 3 φ (x,t) , φ˙− (y,t) = e · − = δ (x y) (2π)3 2 2 − h i− Z in a similar way we can obtain

+ i 3 + + φ˙ (x,t) , φ− (y,t) = δ (x y) ; φ˙ (x) , φ (y) = φ˙− (x) , φ− (y) = 0 −2 − h i− h i− h i− now we are ready to obtain the commutation relation between φ (x) and φ˙ (y)

+ + + + φ (x,t) , φ˙ (y,t) = φ (x,t)+ φ− (x,t) , φ˙ (y,t)+ φ˙− (y,t) = φ (x,t) , φ˙− (y,t) + φ− (x,t) , φ˙ (y,t) h i− h i− h i− h i− + + i 3 i 3 = φ (x,t) , φ˙− (y,t) φ˙ (y,t) , φ− (x,t) = δ (x y)+ δ (y x) − 2 − 2 − h i− h i− and we ﬁnally obtain φ (x,t) , φ˙ (y,t) = ∆˙ (x y, 0) = iδ3 (x y) − − − h i− and combining Eqs. (10.5) and (10.3) for equal time events we ﬁnd

[φ (x,t) , φ (y,t)] =∆ x y,x0 y0 = ∆ (x y, 0) = 0 − − − − Then, we can see that the ﬁeld and its time derivative φ˙ satisfy the equal-time commutation relations:

φ (x,t) , φ˙ (y,t) = iδ3 (x y) − h i− [φ (x,t) , φ (y,t)] = φ˙ (x,t) , φ˙ (y,t) = 0 − h i− Consequently, we can deﬁne them as the canonical variables

q (x,t) φ (x,t) , p (x,t) φ˙ (x,t) (10.7) ≡ ≡ that satisfy the canonical commutation relations (10.1, 10.2). In the case of a complex scalar ﬁeld of a particle of spin zero (in which the particle is diﬀerent from the antiparticle) we have the commutation relations given by Eqs. (5.32) page 158

φ (x) , φ† (y) = ∆ (x y) ; [φ (x) , φ (y)] = 0 (10.8) − − h i− 10.1. CANONICAL VARIABLES 271 comparing Eq. (10.8) with Eq. (10.3) it is quite natural to define φ (x) as the generalized coordinate and use φ˙† (y) as a trial field for the conjugate canonical momentum. With a procedure similar to the one carried out for the slef-charge conjugate scalars, we can show that such a definition is consistent. Therefore, we can define the free-particle canonical variables as the complex operators

q (x,t) φ (x,t) ; p (x,t) φ˙† (x,t) (10.9) ≡ ≡ Equivalently by deﬁning 1 φ (φ1 + iφ2) ; with φ† = φ1 and φ† = φ2 ≡ √2 1 2 we can deﬁne canonical variables as follows

k q (x,t)= φk (x,t) ; pk (x,t)= φ˙k (x,t) (10.10) and they satisfy the commutation relations (10.1, 10.2).

10.1.2 Canonical variables for vector ﬁelds Once again let us start with real vector ﬁelds of a particle of spin one that coincides with its antiparticle. The commutation relations can be obtained by combining Eqs. (6.77, 6.79) page 174

∂µ∂ν [vµ (x) , vν (y)] = gµν ∆ (x y) (10.11) − m2 − − where we shall use the notation vµ instead of V µ because the latter will be used for the ﬁelds in the Heisenberg picture. In this case we can take the canonical variables as follows

∂v0 (x,t) qi (x,t)= vi (x,t) ; p (x,t) =v ˙i (x,t)+ ; i = 1, 2, 3 (10.12) i ∂xi It can be checked that Eqs. (10.12) satisfy the commutation relations (10.1, 10.2). Let us see it —————————————————————— ——————————————————————— From Eqs. (10.11, 10.12) and taking into account Eq. (10.5) we have

∂k∂n qk (x,t) ,qn (y,t) = vk (x,t) , vn (y,t) = gkn ∆ (x y, 0) = 0 − m2 − h i h i− i now for the commutation relations between q (x,t) and pi (y,t) we have

3 µ +µ +µ 1 d p µ ip x ip0t v (x,t) = φ (x,t)+ φ † (x,t)= e (p,σ) a (p,σ) e · − 3 0 (2π) σ 2p X Z h µ ip xq+ip0t p +e ∗ (p,σ) a† (p,σ) e− · (10.13) i 3 k +k +k i d p 0 k ip x ip0t v˙ (x,t) = φ˙ (x,t)+ φ˙ † (x,t)= p e (p,σ) a (p,σ) e · − 3 0 − (2π) σ 2p X Z h 0 k ipqx+ip0t p +p e ∗ (p,σ) a† (p,σ) e− · (10.14) 0 3 i ∂v (x,t) i d p k 0 ip x ip0t k 0 ip x+ip0t = p e (p,σ) a (p,σ) e · − p e ∗ (p,σ) a† (p,σ) e− · (10.15) ∂xk 3 0 − (2π) σ 2p X Z h i q p 272 CHAPTER 10. CANONICAL QUANTIZATION

3 i d q k 0 0 k iq y iq0t pk (y,t) = q e (q, σ¯) q e (q, σ¯) a (q, σ¯) e · − 3 0 − (2π) σ¯ 2q X Z nh i q 0 k p k 0 iq y+iq0t + q e ∗ (q, σ¯) q e ∗ (q, σ¯) a† (q, σ¯) e− · − h i o 3 k k +k +k 1 d p k ip x ip0t q (x,t) = v (x,t)= φ (x,t)+ φ † (x,t)= e (p,σ) a (p,σ) e · − 3 0 (2π) σ 2p X Z h k ip x+ip0t q p +e ∗ (p,σ) a† (p,σ) e− · i 3 3 n i d p d q n 0 k k 0 [q (x,t) ,pk (y,t)] = e (p,σ) q e ∗ (q, σ¯) q e ∗ (q, σ¯) 3 0 0 − (2π) 2p 2q σ σ¯ Z Z X X h i ip x ip0t iq y+iq0t e · − pe− · p a (p,σ) , a† (q, σ¯) × 3 h3 i i d p d q n k 0 0 k + e ∗ (p,σ) q e (q, σ¯) q e (q, σ¯) 3 0 0 − (2π) 2p 2q σ σ¯ Z Z X X h i ip x+ip0t iq y iq0t e− · pe · − pa† (p,σ) , a (q, σ¯) × h i 3 3 n i d p d q n 0 k k 0 [q (x,t) ,pk (y,t)] = e (p,σ) q e ∗ (q, σ¯) q e ∗ (q, σ¯) 3 0 0 − (2π) 2p 2q σ σ¯ Z Z X X h i ip x ip0t iq y+iq0t e · − pe− · p δ (p q) δσσ¯ × 3 3 − i d p d q n k 0 0 k e ∗ (p,σ) q e (q, σ¯) q e (q, σ¯) − 3 0 0 − (2π) 2p 2q σ σ¯ Z Z X X h i ip x+ip0t iq y iq0t e− · pe · − pδ (p q) δ × − σσ¯

3 n i d p n 0 k k 0 ip x ip0t ip y+ip0t [q (x,t) ,pk (y,t)] = e (p,σ) p e ∗ (p,σ) p e ∗ (p,σ) e · − e− · (2π)3 2p0 − Z σ h i 3 X i d p n k 0 0 k ip x+ip0t ip y ip0t e ∗ (p,σ) p e (p,σ) p e (p,σ) e− · e · − − 3 2p0 − (2π) σ Z X h i

3 n i d p n 0 k k 0 ip (x y) [q (x,t) ,pk (y,t)] = e (p,σ) p e ∗ (p,σ) p e ∗ (p,σ) e · − (2π)3 2p0 − Z σ h i 3 X i d p n k 0 0 k ip (x y) e ∗ (p,σ) p e (p,σ) p e (p,σ) e− · − − 3 2p0 − (2π) σ Z X h i changing p p in the ﬁrst integral, we have →− 3 i d p n 0 k k 0 ip (x y) I e ( p,σ) p e ∗ ( p,σ)+ p e ∗ ( p,σ) e− · − 1 ≡ 3 2p0 − − − (2π) σ Z X h i using the relations (6.88) page 176 we have that

eµ ( p,σ) = µ eρ (p,σ) − −P ρ en ( p,σ) = n eρ (p,σ)= n en (p,σ)= en (p,σ) − −P ρ −P n e0 ( p,σ) = 0 eρ (p,σ)= 0 e0 (p,σ)= e0 (p,σ) − −P ρ −P 0 − 10.1. CANONICAL VARIABLES 273

so that I1 becomes

3 i d p n 0 k k 0 ip (x y) I = e (p,σ) p e ∗ (p,σ) p e ∗ (p,σ) e− · − 1 3 2p0 − (2π) σ Z X h i then we obtain 3 n i d p n 0 k k 0 ip (x y) [q (x,t) ,pk (y,t)] = e (p,σ) p e ∗ (p,σ) p e ∗ (p,σ) e− · − (2π)3 2p0 − Z σ h i 3 X i d p n k 0 0 k ip (x y) e ∗ (p,σ) p e (p,σ) p e (p,σ) e− · − − 3 2p0 − (2π) σ Z X h i

3 n i d p 0 n k n k [q (x,t) ,p (y,t)] = p e (p,σ) e ∗ (p,σ)+ e ∗ (p,σ) e (p,σ) k 3 2p0 (2π) ( σ Z X h i k n 0 n 0 ip (x y) p e (p,σ) e ∗ (p,σ)+ e ∗ (p,σ) e (p,σ) e− · − − σ ) X

3 n i d p ip (x y) [q (x,t) ,pk (y,t)] = B e− · − (10.16) (2π)3 2p0 Z 0 n k n k k n 0 n 0 B p e (p,σ) e ∗ (p,σ)+ e ∗ (p,σ) e (p,σ) p e (p,σ) e ∗ (p,σ)+ e ∗ (p,σ) e (10.17)(p,σ) ≡ − σ σ X h i X let us evaluate the term B. By using the deﬁnition (6.69) page 6.69, and Eq. (6.75) page 173 the term in brackets yield

0 n k n k k n 0 n 0 B p e (p,σ) e ∗ (p,σ)+ e ∗ (p,σ) e (p,σ) p e (p,σ) e ∗ (p,σ)+ e ∗ (p,σ) e (p,σ) ≡ − σ σ X h i X = p0 Πnk (p) + Πkn (p) + pk Πn0 (p) + Π0n (p) = 2p0Πnk (p) + 2pkΠn0 (p) h pnpk i pnp0 pnpk pnp0 = 2p0 δnk + 2pk gn0 + = 2p0δnk + 2p0 2pk m2 − m2 m2 − m2 B = 2p0δnk (10.18) substituting (10.18) in (10.16) we ﬁnd

3 n i d p 0 nk ip (x y) [q (x,t) ,pk (y,t)] = 2p δ e− · − (2π)3 2p0 Z nk 1 3 ip (x y) = iδ d pe− · − (2π)3 Z [qn (x,t) ,p (y,t)] = iδnδ (x y) k k − which is the expected commutation relation for two canonical variables. It can also be checked that

[pn (x,t) ,pk (y,t)]=0 from which Equations (10.12) deﬁnes consistent canonical variables. ———————————————————————– ———————————————————————— 274 CHAPTER 10. CANONICAL QUANTIZATION

The Klein Gordon equation (6.84), the ﬁeld equation (6.86) page 175, along with Eq. (10.12) permits to write v0 in terms of the other variables as

p v0 = ∇ · (10.19) m2 therefore, v0 is not independent, and it not seen as one of the canonical variables. The extension to complex vector ﬁelds (in which particles are distinct to antiparticles), is similar to the case of complex scalar ﬁelds ——————————————————- Let us obtain (10.19). Separating Eqs. (6.84, 6.86) in its time and coordinate derivatives we obtain

i 0 ∂iv (x)+ ∂0v (x) = 0 (10.20) i u 0 µ 2 µ ∂ ∂iv + ∂ ∂0v = m v (10.21) with µ = 0 in Eq. (10.21) and using Eq. (10.20) we have

m2v0 = ∂i∂ v0 + ∂0 ∂ v0 = ∂i∂ v0 ∂0 ∂ vi = ∂ ∂iv0 ∂0vi i 0 i − i i − ∂v0 (x,t) ∂vi (x,t) ∂v0 (x,t) = ∂i ∂ v0 + ∂vi =∂i + = ∂i +v ˙i (x,t) i 0 ∂xi ∂x0 ∂xi 2 0 i m v = ∂ pi (x,t) which reproduces Eq. (10.19).

10.1.3 Canonical variables for Dirac ﬁelds

For the Dirac ﬁeld of a non-Majorana particle of spin 1/2 (that is with particle diﬀerent from antiparticle), we saw in Eq. (7.123) page 200 that the anticommutator yields

µ ψn (x) , ψn†¯ (y) = [( γ ∂µ + m) β] ∆ (x y) + − n,n¯ − h i and

[ψn (x) , ψn¯ (y)]+ = 0

Note that the anticommutator of ψn and ψm† does not vanish at equal times, so that we cannot take them as independent canonical variables (that is we cannot take all of them as q′s). A consistent possibility is given by

n q (x) ψ (x) ; p (x) iψ† (x) (10.22) ≡ n n ≡ n we can see that Eqs. (10.22) satisfy the anticommutation relations (10.1, 10.2). For instance, we have

n µ 0 0 [q (x,t) ,pk (y,t)]+ = ψn (x,t) , iψ† (y,t) = [( γ ∂µ + m) β] ∆ x y,x y = 0 k + − nk − − [qn (x,t) ,p (y,t)] =h [ γµ∂ β] ∆ (x yi,t = 0)+ mβ ∆ x y,x0 y0 = 0 k + − µ nk − nk − − 10.2. FUNCTIONAL DERIVATIVES FOR CANONICAL VARIABLES 275 but according to Eq. (10.5) we have ∆ (x y, 0) = 0 so the second term on the right-hand-side vanishes. Let us − deﬁne z x y. Hence ≡ − 3 n µ 1 d k ik z ik z [q (x,t) ,pk (y,t)]+ = [ γ ∂µβ] e · e− · = − nk (2π)3 2k0 − z0=0 Z 3 h i µ 1 d k ik z ik z = [ γ β]nk 3 0 ikµe · + ikµe− · − (2π) 2k 0 Z z =0 3 h i µ 1 d k ik z ik z = i [ γ β]nk 3 0 kµ e · + e− · − (2π) 2k 0 Z z =0 3 h i n 0 1 d k ik z ik z [q (x,t) ,pk (y,t)] = i γ β k0 e · + e− · + − nk (2π)3 2k0 Z 3 h i i 1 d k ik z ik z +i γ β ki e · + e− · − nk (2π)3 2k0 Z h i the second integral is odd in k so it vanishes. Further

iγ0β = β2 = 1 − − − thus

3 n 1 d k ik (x y) ik (x y) [q (x,t) ,pk (y,t)] = δnk e · − + e− · − + − (2π)3 2 Z h i 1 1 = δ δ (x y)+ δ (y x) − nk 2 − 2 − [qn (x,t) ,p (y,t)] = δnδ (x y) k + − k −

10.2 Functional derivatives for canonical variables

Let us f (x,y) be a function of two sets of variables x and y . We shall denote as F [f (y)] a functional that { } { } depends on the values of f (x,y) for all x at fixed value of y (thus x appears usually integrated). We define a bosonic functional as a functional for which each term contains only even number of fermionic fields (i.e. so that the total spin of the system associated is integer). Let us assume a system of canonical variables i.e. that satisfy the commutation or anticommutation relations (10.1), (10.2). For a set of canonical variables we can define a quantum mechanical functional derivative as follows: for an arbitrary bosonic functional F [q (t) ,p (t)] at a given time t, we define

δF [q (t) ,p (t)] i [p (x,t) , F [q (t) ,p (t)]] (10.23) δqn (x,t) ≡ n δF [q (t) ,p (t)] i [F [q (t) ,p (t)] ,qn (x,t)] (10.24) δpn (x,t) ≡ to understand the sense of such a deﬁnition we observe that if F [q (t) ,p (t)] were written in such a way that all q′s are to the left of all p′s, equations (10.23) and (10.24) become the left and right -derivatives respectively with k respect to q and pk. As a matter of example, let us assume a functional of the form

F q1 (t) ,q2 (t) ,p (t) ,p (t) d3y aq1 (y,t) p (y,t)+ bq1 (y,t) q2 (y,t) p (y,t) p (y,t) 1 2 ≡ 1 1 2 Z 276 CHAPTER 10. CANONICAL QUANTIZATION

i i this functional is written such that all q are on the left of all pi the left-derivative with respect to q gives 1 2 1 1 δF q (t) ,q (t) ,p1 (t) ,p2 (t) δq (y,t) δq (y,t) = a d3y p (y,t)+ b d3y q2 (y,t) p (y,t) p (y,t) δq1 (x,t) δq1 (x,t) 1 δq1 (x,t) 1 2 Z Z = a d3y δ (y x) p (y, t)+ b d3y δ (y x) q2 (y,t) p (y,t) p (y,t) − 1 − 1 2 1 2 Z Z δF q (t) ,q (t) ,p1 (t) ,p2 (t) 2 1 = ap1 (x, t)+ bq (x,t) p1 (x,t) p2 (x,t) (10.25) δq (x,t) on the other hand we have

3 1 1 2 i [p1 (x,t) , F [q (t) ,p (t)]] = i p1 (x,t) , d y aq (y,t) p1 (y,t)+ bq (y,t) q (y,t) p1 (y,t) p2 (y,t) Z 3 1 = i d y p1 (x,t) , aq (y,t) p1 (y,t) Z 3 1 2 +i d y p1 (x,t) , bq (y,t) q (y,t) p1 (y,t) p2 (y,t) Z 3 1 = ia d y p1 (x,t) ,q (y,t) p1 (y,t) Z 3 1 2 +ib d y p1 (x,t) ,q (y,t) q (y,t) p1 (y,t) p2 (y,t) Z i [p (x,t) , F [q (t) ,p (t)]] = ia d3y ( i) δ (x y) p (y,t) 1 − − 1 Z +ib d3y ( i) δ (x y) q2 (y,t) p (y,t) p (y,t) − − 1 2 Z 2 i [p1 (x,t) , F [q (t) ,p (t)]] = ap1 (x, t)+ bq (x,t) p1 (x,t) p2 (x,t) (10.26) comparing Eqs. (10.25, 10.26) we have

1 2 δF q (t) ,q (t) ,p1 (t) ,p2 (t) 1 = i [p1 (x,t) , F [q (t) ,p (t)]] δq (x,t) and we can obtain similar relations for the remaining variables. Thus deﬁnitions (10.23) and (10.24) are now clearly motivated. For an arbitrary c number variation of δq and δp yields − δF [q (t) ,p (t)] δF [q (t) ,p (t)] δF [q (t) ,p (t)] = d3x δqn (x,t) + δp (x,t) δqn (x,t) δp (x,t) n n n Z X k k k where q and p are bosonic or fermionic and δq , δpk are assumed to commute or anticommute with all fermionic operators respectively, and to commute with all bosonic operators. More general functionals require some extra signs and equal time commutators in their deﬁnitions (10.23) and (10.24). In particular, H0 is the generator of time-translations on free-particle states in the following sense qn (x,t) = exp (iH t) qn (x, 0) exp ( iH t) (10.27) 0 − 0 p (x,t) = exp (iH t) p (x, 0) exp ( iH t) (10.28) n 0 n − 0 taking the time derivative of Eq. (10.27) we have

n n n q˙ (x,t) = iH0 exp (iH0t) q (x, 0) exp ( iH0t) i exp (iH0t) q (x, 0) exp ( iH0t) H0 {n n − }− { − } = iH0q (x,t) iq (x,t) H0 n n − q˙ (x,t) = i [H0,q (x,t)] 10.3. FREE HAMILTONIANS 277

and appealing to the deﬁnition (10.23) and seen H0 as a functional of the canonical variables we ﬁnd

n n δH0 q˙ (x,t)= i [H0,q (x,t)] = δpn (x,t) and a similar relation can be obtained for pn. Therefore, the free particle operators has the time-dependence given by

n n δH0 q˙ (x,t) = i [H0,q (x,t)] = (10.29) δpn (x,t) δH p˙ (x,t) = i [p (x,t) ,H ]= 0 (10.30) n − n 0 −δqn (x,t) which has the form of Hamilton’s equations similar to the ones in classical mechanics, in terms of either the Poisson brackets (replaced by commutators) or Hamiltonian derivatives (replaced by hamiltonian functinal derivatives).

10.3 Free Hamiltonians

An energy term has the form 2 2 a† (k,σ,n) a (k,σ,n) k + mn (10.31) 2 2 the operator a† (k,σ,n) a (k,σ,n) extracts the number of particlesp in the state given by k,σ,n; and k + mn is the energy of each of these (free) states. Therefore, when applied to a multiparticle state, a term of the form p (10.31) provides the energy of all particles in the state k,σ,n. By summing over all possible states (of course, integrating on the continuum), provides the total energy of the multiparticle state. Thus, the free-Hamiltonian can be written as 3 2 2 H0 = d k a† (k,σ,n) a (k,σ,n) k + mn (10.32) n,σ X Z p Such a Hamiltonian can be written in terms of the canonical variables (q,p) at time t.

10.3.1 Free Hamiltonian for scalar fields For example, the free Hamiltonian (10.32) for a real scalar field φ (x) can be written (up to a constant term) as the functional 1 1 1 H = d3x p2 + ( q)2 + m2q2 (10.33) 0 2 2 ∇ 2 Z we can see it by applying Eq. (10.7) and the Fourier expansion of the real scalar field φ (x) Eq. (5.21), then equation (10.33) gives 1 1 1 H = d3x φ˙ (x,t)2 + (∂ φ (x,t)) ∂iφ (x,t) + m2φ2 (x,t) (10.34) 0 2 2 i 2 Z from the Fourier expansion (5.21) we have

1 3 1 i(p x p0t) i(p x p0t) φ (x,t) d p a (p) e · − + a† (p) e− · − ≡ 3 2p0 (2π) Z h i q p 0 i 3 p ip x ip x φ˙ (x,t) = d p a (p) e · + a† (p) e− · 3 2p0 − (2π) Z h i q p i i 3 p ip x ip x ∂iφ (x) = d p a (p) e · a† (p) e− · 3 2p0 − (2π) Z h i q p 278 CHAPTER 10. CANONICAL QUANTIZATION

so that H0 in Eq. (10.34) gives

2 0 i 1 3 3 k ik x ik x H0 = d x d k a (k) e · + a† (k) e− · (2π)3 2 √ 0 − Z Z 2k 0 h i 3 q iq x iq x d q a (q) e · + a† (q) e− · × 0 − Z 2q 2 h i i i 1p 3 3 k ik x ik x + d x d k a (k) e · a† (k) e− · (2π)3 2 √2k0 − Z Z h i 3 qi iq x iq x d q a (q) e · a† (q) e− · × ( 2q0 − ) Z h i 2 1 mp 3 3 1 ik x ik x + d x d k a (k) e · + a† (k) e− · 3 2 √2k0 (2π) Z Z h i q 3 1 iq x iq x d q a (q) e · + a† (q) e− · × 2q0 Z h i p

0 0 1 1 3 3 3 k q i(k+q) x i(k q) x H0 = d x d k d q a (k) a (q) e · a (k) a† (q) e − · −(2π)3 2 √2k0 2q0 − Z Z Z h i(k q) x i(k+q) x a† (k) a (q) e− − · + a† (k) a†p(q) e− · − i i 1 1 3 3 3 k qi i(k+q) x i(k q) x d x d k d q a (k) a (q) e · a (k) a† (q) e − · −(2π)3 2 √2k0 2q0 − Z Z Z h i(k q) x i(k+q) x a† (k) a (q) e− − · + a† (k) a† (pq) e− · − 2 i 1 m 3 3 3 1 i(k+q) x i(k q) x + d x d k d q a (k) a (q) e · + a (k) a† (q) e − · (2π)3 2 √2k0 2q0 Z Z Z h i(k q) x i(k+q) x +a† (k) a (q) e− − · + a† (k) a† (q)pe− · i integrating over x we obtain Dirac deltas

1 3 i(k+q) x 1 3 i(k+q) x i(k0+q0)t i(k0+q0)t 1 3 i(k+q) x i(k0+q0)t d x e · = d x e · e− = e− d x e · = e− δ (k + q) (2π)3 (2π)3 (2π)3 Z Z Z 1 3 i(k+q) x i(k0+q0)t d x e− · = e δ (k + q) (2π)3 Z 1 3 i(k q) x i(k0 q0)t 1 3 i(k q) x i(k0 q0)t d x e − · = e− − δ (k q) ; d x e− − · = e − δ (k q) (2π)3 − (2π)3 − Z Z now since all our four-vectors are on-shell we see that if k = q then k0 = q0, therefore k k k k

1 3 i(k+q) x 2iq0t d x e± · = e∓ δ (k + q) (2π)3 Z 1 3 i(k q) x d x e± − · = δ (k q) (2π)3 − Z 10.3. FREE HAMILTONIANS 279 then the Hamiltonian becomes 0 0 1 3 3 k q 2iq0t H0 = d k d q a (k) a (q) e− δ (k + q) a (k) a† (q) δ (k q) −2 √2k0 2q0 − − Z Z h 2iq0t a† (k) a (q) δ (k qp)+ a† (k) a† (q) e δ (k + q) − − i i 1 3 3 k qi 2iq0t d k d q a (k) a (q) e− δ (k + q) a (k) a† (q) δ (k q) −2 √2k0 2q0 − − Z Z h 2iq0t a† (k) a (q) δ (k qp)+ a† (k) a† (q) e δ (k + q) − − 2 i m 3 3 1 2iq0t + d k d q a (k) a (q) e− δ (k + q)+ a (k) a† (q) δ (k q) 2 √2k0 2q0 − Z Z h 2iq0t +a† (k) a (q) δ (k q)+pa† (k) a† (q) e δ (k + q) − i 0 0 1 3 q q 2iq0t 2iq0t H = d q a ( q) a (q) e− a (q) a† (q) a† (q) a (q)+ a† ( q) a† (q) e 0 −2 2q0 − − − − Z i h i 1 3 q qi 2iq0t 2iq0t d q a ( q) a (q) e− a (q) a† (q) a† (q) a (q)+ a† ( q) a† (q) e −2 2q0 − − − − Z2 h i m 3 1 2iq0t 2iq0t + d q a ( q) a (q) e− + a (q) a† (q)+ a† (q) a (q)+ a† ( q) a† (q) e 2 2q0 − − Z h i

0 2 2 2 0 2 2 2 1 3 2iq0t q q m 2iq0t q q m H = d q a ( q) a (q) e− + + a† ( q) a† (q) e + 0 2 − − 2q0 − 2q0 2q0 − − 2q0 − 2q0 2q0 Z ( " # " # 2 2 q0 q2 m2 q0 q2 m2 +a (q) a† (q) 0 + 0 + 0 + a† (q) a (q) 0 + 0 + 0 "2q 2q 2q # "2q 2q 2q #)

0 2 2 2 1 3 2iq0t 2iq0t q q m H0 = d q a ( q) a (q) e− + a† ( q) a† (q) e 0 0 + 0 2 ( − − "− 2q − 2q 2q # Z h i 2 q0 q2 m2 + a (q) a† (q)+ a† (q) a (q) 0 + 0 + 0 " 2q 2q 2q #) h i

1 3 0 H = d q a (q) a† (q)+ a† (q) a (q) q 0 2 Z h i therefore from (10.33) we obtain the Hamiltonian

1 3 0 3 0 1 3 H0 = d k k a (k) , a† (k) = d k k a† (k) a (k)+ δ (k k) (10.35) 2 + 2 − Z h i Z which differs from (10.32) only by a constant divergent term. Such terms only affect the zero of energy and are not physically observable in the absence of gravity or in the case in which we change the boundary conditions for the fields. For instance, quantizing in the space between parallel plates instead of infinite space, leads to divergent significant terms. Equation (10.35) could be used to test whether a free-field Lagrangian is valid for the given theory. Any valid free-field Lagrangian must lead to Eq. (10.35) up to a constant. For instance, returning to the case of the scalar field φ (x), we could start by finding a free-field Lagrangian that leads to (10.35) for spinless particles or 280 CHAPTER 10. CANONICAL QUANTIZATION equivalently leads to the free-Hamiltonian (10.32) (up to an unphysical constant). We could check it by starting with the free-field Lagrangian and applying the Legendre transformation

L [q (t) , q˙ (t)] = d3x p (x,t)q ˙n (x,t) H (10.36) 0 n − 0 n X Z n n where pn must be replaced by its expression in terms of q andq ˙ (and in some cases some auxiliary ﬁelds). In particular from the Hamiltonian (10.33) and the Legendre transformation (10.36) we can derive the free-ﬁeld scalar Lagrangian

1 1 1 L = d3x pq˙ p2 ( q)2 m2q2 0 − 2 − 2 ∇ − 2 Z and from the canonical relations (10.7) this free Lagrangian could be written in terms of the field and its derivatives 1 1 1 L = d3x φ˙φ˙ φ˙2 (∂ φ) ∂iφ m2φ2 0 − 2 − 2 i − 2 Z 1 1 1 = d3x φ˙φ˙ (∂ φ) ∂iφ m2φ2 2 − 2 i − 2 Z 1 1 1 = d3x (∂ φ) (∂ φ) (∂ φ) ∂iφ m2φ2 2 0 0 − 2 i − 2 Z 1 1 1 = d3x (∂ φ) ∂0φ (∂ φ) ∂iφ m2φ2 −2 0 − 2 i − 2 Z In summary, from the Hamiltonian (10.33) and from the canonical relations (10.7) we can derive the free-field Lagrangian given by 1 1 1 L = d3x pq˙ p2 ( q)2 m2q2 (10.37) 0 − 2 − 2 ∇ − 2 Z 1 1 L = d3x ∂ φ∂µφ m2φ2 (10.38) 0 −2 µ − 2 Z in terms of the scalar field φ (x) and its derivatives. Whatever interaction we add to the theory, this free-field Lagrangian must be taken as the zeroth-order term in a perturbative approach.

10.4 Interacting Hamiltonians

We have formulated only free-field theories in canonical form so far. To do the same for interacting fields as well, we introduce canonical variables in the Heisenberg picture defined by

Qn (x,t) exp (iHt) qn (x, 0) exp ( iHt) (10.39) ≡ − P (x,t) = exp (iHt) p (x, 0) exp ( iHt) (10.40) n n − where H is the full Hamiltonian. Note that we are “turning on” the interaction at time t = 0, so that Q′s and P ′s coincide with q′s and p′s at t = 0. This is a similarity transformation that commutes with H and also preserves all products of canonical variables, for instance

(QP )′ exp (iHt) (QP ) exp ( iHt) = [exp (iHt) Q exp ( iHt)] [exp (iHt) P exp ( iHt)] ≡ − − − (QP )′ = Q′P ′ form this preservation of products we can see that

1 A′ BAB− F A′ = F ′ (A) , A′,B′ = [A, B]′ (10.41) ≡ ⇒ 10.5. THE LAGRANGIAN FORMALISM 281

From these facts, the full Hamiltonian is the same functional of the Heisenberg picture operators in terms of the canonical variables (q,p)

iHt iHt H [Q, P ]= e H [q,p] e− = H [q,p] Further, the similarity transformations (10.39, 10.40) preserve the commutators or anticommutators with respect to (q,p), so that they are also canonical variables

n 3 n [Q (x,t) , Pn¯ (y,t)] = iδ (x y) δn¯ (10.42) n n¯ ∓ − Q (x,t) ,Q (y,t) = [Pn (x,t) , Pn¯ (y,t)] = 0 (10.43) ∓ ∓ with the same procedure that led from Eqs. (10.27, 10.28) to Eqs. (10.29, 10.30) we can start from Eqs. (10.39, 10.40) to obtain the time-dependence of the (interacting) canonical variables

δH Q˙ n (x,t) = i [H,Qn (x,t)] = (10.44) δPn (x,t) δH P˙ (x,t) = i [P (x,t) ,H]= (10.45) n − n −δQn (x,t)

As a matter of example, an interacting Hamiltonian for a real scalar ﬁeld can be constructed by using the free- particle term (10.33) plus the integral of a scalar interaction density . Hence, in terms of the Heisenberg picture H the full Hamiltonian reads 1 1 1 H = d3x P 2 + ( Q)2 + m2Q2 + (Q) (10.46) 2 2 ∇ 2 H Z in this example, the canonical conjugate to Q has the same expression as for free ﬁelds

P = Q˙ (10.47) but we shall see later that in the general case, the relation between canonical conjugates Pn (x) and the ﬁeld variables and their time-derivatives is diﬀerent from the case of free particle operators. Such relation can be inferred from Eqs. (10.44) and (10.45).

10.5 The Lagrangian formalism

Now we are challenged to choose appropriate Hamiltonians for realistic theories. A simple form to ensure Lorentz invariance and other symmetries is by choosing a suitable Lagrangian and use it to derive the Hamiltonian. In general, we can derive Lagrangians from Hamiltonians and vice versa. Equation (10.36) is the clue for a derivation in either direction. In general it is more feasible to explore physically realistic theories by listing possible Lagrangians instead of Hamiltonians. The Lagrangian is a functional of a set of fields Ψk (x,t) and their time derivatives Ψ˙ (x,t). The conjugate fields are defined in a way analogous with classical mechanics as the variational derivatives

δL Ψ (t) , Ψ˙ (t) Πk (x,t) (10.48) ≡ hδΨ˙ k (x,t) i where we are using upper case greek letters to indicate that these are interacting instead of free ﬁelds. As in classical mechanics, the ﬁeld equations can be generated in terms of a Hamilton’s variational principle. Thus we start with the action I [Ψ] ∞ dt L Ψ (t) , Ψ˙ (t) (10.49) ≡ Z−∞ h i 282 CHAPTER 10. CANONICAL QUANTIZATION

Under an arbitrary variation of Ψ (x) the change in I [Ψ] yields δL δL δI [Ψ] = ∞ dt d3x δΨk (x)+ δΨ˙ k (x) δΨk (x) δΨ˙ k (x) Z−∞ Z assuming that δΨk (x) vanishes for t , we can integrate by parts and obtain → ±∞ 4 δL d δL k 4 δL d k δI [Ψ] = d x δΨ (x)= d x Πk (x,t) δΨ (x) δΨk (x) − dt δΨ˙ k (x) δΨk (x) − dt Z Z where we have used the deﬁnition (10.48) of conjugate canonical momenta. Now, imposing the stationarity of the action with respect to all variations δΨk (x) that vanish at t , we see that the necessary and suﬃcient → ±∞ condition is that the conjugate canonical momenta satisfy the following equations of motion

δL Ψ (t) , Ψ˙ (t) Π˙ (x,t)= (10.50) k hδΨk (x,t) i Now, since the action is the generator of the equations of motion, a natural choice to obtain a Lorentz invariant theory is to make I [Ψ] a functional Lorentz scalar. Further, since I [Ψ] is a time-integral of L Ψ (t) , Ψ˙ (t) , we could expect that L should be a space-integral of an ordinary scalar function of Ψ (x) and ∂Ψ (xh) /∂xµ knowni as the Lagrangian density . L L Ψ (t) , Ψ˙ (t) = d3x Ψ (x,t) , Ψ (x,t) , Ψ˙ (x,t) (10.51) L ∇ h i Z such that the action can be written in a manifestly invariant way in terms of the scalar density L ∂Ψ (x) I [Ψ] = d4x Ψ (x) , (10.52) L ∂xµ Z All field theories of elementary particles have Lagrangians of this form. As in any (classic or quantum) field theory, it is convenient to express the equations of motion in terms of local quantities (generalized densities) instead of global ones, (e.g. Lagrangian densities instead of Lagrangians). Thus, we intend to write equation of motion (10.50) in terms of the Lagrangian density . To do it, we start by L making the variation of Ψm (x) by an amount δΨm (x) and integrating by parts, we find a variation in L given by ∂ ∂ ∂ δL = d3x L δΨm + L δΨm + L δΨ˙ m ∂Ψm ∂ ( Ψm)∇ ∂Ψ˙ m Z ∇ ∂ ∂ ∂ = d3x L L δΨm + L δΨ˙ m ∂Ψm − ∇ · ∂ ( Ψm) ∂Ψ˙ m Z ∇ (homework!! B5 make the integration by parts) so that

δL ∂ ∂ δΨm (y,t) ∂ δΨ˙ m (y,t) = d3y L L + L δΨk (x,t) ∂Ψm (y,t) − ∇ · ∂ ( Ψm (y,t)) δΨk (x,t) ∂Ψ˙ m (y,t) δΨk (x,t) Z " ∇ # δL ∂ ∂ ∂ ∂ = d3y L L δ δ (x y)= L L δΨk (x,t) ∂Ψm (y,t) − ∇ · ∂ ( Ψm (y,t)) mk − ∂Ψk (x,t) − ∇ · ∂ ( Ψk (x,t)) Z ∇ ∇ a similar procedure can be done for δL and we obtain δΨ˙ k δL ∂ ∂ = L L (10.53) δΨk (x,t) ∂Ψk (x,t) − ∇ · ∂ ( Ψk (x,t)) ∇ δL ∂ = L (10.54) δΨ˙ k (x,t) ∂Ψ˙ k (x,t) 10.6. FROM LAGRANGIAN TO HAMILTONIAN FORMALISM 283 using deﬁnition (10.48) and Eq. (10.54) we have

˙ δL Ψ (t) , Ψ (t) ∂ Πk (x,t) = L ≡ hδΨ˙ k (x,t) i ∂Ψ˙ k (x,t) ⇒ ∂ ∂ ∂ ∂ Π˙ (x,t) = L = L (10.55) k ∂x0 ˙ k ∂x0 ∂Ψk (x,t) /∂x0 ∂Ψ (x,t) on the other hand, by using the ﬁeld equations (10.50) along with Eq. (10.53), we ﬁnd

˙ δL Ψ (t) , Ψ (t) ∂ ∂ Π˙ (x,t) = = L L k hδΨk (x,t) i ∂Ψk (x,t) − ∇ · ∂ ( Ψk (x,t)) ∇ ∂ ∂ ∂ Π˙ (x,t) = L L (10.56) k ∂Ψk (x,t) − ∂xi ∂ (∂Ψk (x,t) /∂xi) equating Eqs. (10.55, 10.56) we obtain

∂ ∂ ∂ ∂ ∂ L = L L ∂x0 ∂Ψk (x,t) /∂x0 ∂Ψk (x,t) − ∂xi ∂ (∂Ψk (x,t) /∂xi) Hence, the ﬁeld equations (10.50), become

∂ ∂ ∂ L = L (10.57) ∂xµ ∂ (∂Ψk/∂xµ) ∂Ψk which are the well known Euler-Lagrange equations. If is a scalar these equations are Lorenz-invariant. L Another important condition for the action (besides being a Lorentz scalar) is that we require it to be real. This owes to the fact that we want as many fields equations as the number of fields. By splitting the complex fields into their real and imaginary parts we can figure out I as being a functional of N real fields. If I were complex with independent real and imaginary parts, we could settle the stationary condition for the real and stationary parts leading to a set of 2N Euler-Lagrange equations of motion for N fields, which overdetermines the problem in general. We shall see later that the reality condition for the action also guarantees that we obtain Hermitian generators associated with several symmetry transformations.

10.6 From Lagrangian to Hamiltonian formalism

We have already said that the Lagrangian formalism is easier to construct realistic Lorentz invariant (and other symmetries invariant) theories. However, to calculate the S matrix we should calculate the interaction Hamil- − tonian. Such interaction Hamiltonian (as in the case of the free Hamiltonian) is connected with the interacting Lagrangian through a Legendre transformation

H = d3x Π (x,t) Ψ˙ k (x,t) L Ψ (t) , Ψ˙ (t) (10.58) k − Xk Z h i

k k Equation (10.48) does not in general allow to express Ψ˙ (x) uniquely in terms of Ψ (x) and Πk. However, it can be seen that Eq. (10.58) has null variational derivatives with respect to Ψ˙ (x) for any Ψ˙ (x) that satisﬁes Eq. 284 CHAPTER 10. CANONICAL QUANTIZATION

(10.48)

δH δΨ˙ m (x,t) δΠ (y,t) = d3y Π (y,t) + d3y m Ψ˙ m (x,t) ˙ k m ˙ k ˙ k δΨ (x) m δΨ (x) m δΨ (x) X Z X Z ˙ δL Ψ (t) , Ψ (t) δL δΨm (y,t) d3y − h ˙ k i − δΨm (y,t) ˙ k δΨ (x,t) m δΨ (x,t) Z X δL Ψ (t) , Ψ˙ (t) = d3y Π (y,t) δ δ (x y) m mk − − h ˙ k i m δΨ (x,t) X Z δH = Πk (x,t) Πk (x,t) = 0 δΨ˙ k (x) −

k Hence, the Hamiltonian (10.58) is only a functional of Ψ (x) and Πk. Then, its variational derivatives with respect to these two sets of variables read

˙ m δH 3 δΨ (y, t) δL k = d y Πm (y,t) k k δΨ (x,t) Π δΨ (x,t) − δΨ (x,t) Ψ˙ Z m Π X δL δΨ˙ m (y, t) d3y − δΨ˙ m (y,t) δΨk (x,t) Z m Ψ Π X

˙ m δH ˙ k 3 δΨ (y,t) = Ψ (x,t)+ d y Πm (y,t) k δΠk (x,t) Ψ δΠ (x,t) Z m Ψ X δL δΨ˙ m (y,t) d3y − δΨ˙ m (y,t) δΠk (x,t) Z m Ψ Ψ X the subscripts denote the variables that are kept ﬁxed in the variational derivatives. From the deﬁnition (10.48) of Πk, such derivatives become δH δL = δΨk (x,t) − δΨk (x,t) Π Ψ˙ and δH = Ψ˙ k (x,t) (10.59) δΠ (x,t) k Ψ and the equations of motion (10.50) are equivalent to

δH = Π˙ (x,t) (10.60) δΨk (x,t) − k Π

so these are the equations of motion (10.50) but in terms of the Hamiltonian. We could a priori identify the k k ﬁeld variables Ψ (x) and their conjugates Πk with the canonical variables Q and Pk, and impose the same canonical commutation relations (10.42, 10.43) on them, such that Eqs. (10.59) and (10.60) are the same as the Hamiltonian equations of motion (10.44) and (10.45). This is not the case in the most general context as we shall see. Nevertheless, this association is correct for the simple case of the scalar ﬁeld Φ with non-derivative coupling. For this, we consider the Lagrangian density 1 1 = ∂ Φ∂µΦ m2Φ2 (Φ) (10.61) L −2 µ − 2 −H 10.6. FROM LAGRANGIAN TO HAMILTONIAN FORMALISM 285

It worths remarking that we are not including a free constant factor in the term (1/2) ∂ Φ∂µΦ because such a − µ constant (if positive) can be absorbed in the normalization of Φ, and a negative constant would lead to an spectrum not bounded from below. Note that the Lagrangian density (10.61) is obtained by adding a real function (Φ) −H of Φ to the free-field Lagrangian density for the scalar free-field (10.38) (of course replacing the free-field φ (x) by the interacting field Φ (x)). From the Lagrangian density (10.61) we obtain

∂ ∂ ∂ 1 α 1 2 2 1 ∂ αβ µ kL µ = ∂µ (∂αΦ) (∂ Φ) m Φ (Φ) = ∂µ g (∂αΦ) (∂βΦ) ∂x ∂ (∂Ψ /∂x ) ∂ (∂µΦ) −2 − 2 −H −2 ∂ (∂µΦ) h i 1 ∂ (∂ Φ) ∂ (∂ Φ) 1 = ∂ gαβ α (∂ Φ) + gαβ (∂ Φ) β = ∂ gαβδ ∂ Φ+ gαβδ ∂ Φ −2 µ ∂ (∂ Φ) β α ∂ (∂ Φ) −2 µ αµ β βµ α µ µ ∂ ∂ 1 h i L = ∂ 2gµβ ∂ Φ = ∂ ∂µΦ (10.62) ∂xµ ∂ (∂Ψk/∂xµ) −2 µ β − µ ∂ ∂ h1 i 1 ∂ (Φ) L = ∂ Φ∂µΦ m2Φ2 (Φ) = m2Φ H (10.63) ∂Ψk ∂Φ −2 µ − 2 −H − − ∂Φ taking into account the Euler-Lagrange equation (10.57) we equate Eqs. (10.62, 10.63), so that the ﬁeld equation becomes ∂ (Φ) m2 Φ= H (10.64) − ∂Φ The canonical conjugate variable associated with Φ can be calculated from the Lagrangian density (10.61) as

∂ ∂ 1 1 1 Π = L = ∂ Φ∂0Φ ∂ Φ∂iΦ m2Φ2 (Φ) ˙ ˙ −2 0 − 2 i − 2 −H ∂Φ ∂Φ ∂ 1 1 ∂ = ∂ Φ ∂ Φ = Φ˙ 2 ˙ 2 0 0 2 ˙ ∂Φ ∂Φ ∂ Π= L = Φ˙ (10.65) ∂Φ˙ which is the same association given by Eq. (10.47) if we identify Φ and Π with the canonical (interacting) variables Q and P . From the Legendre transformation Eq. (10.58) the full Hamiltonian yields

1 1 1 H = d3x ΠΦ˙ = d3x Π2 + ( Φ)2 + m2Φ2 + (Φ) (10.66) −L 2 2 ∇ 2 H Z Z which is the Hamiltonian (10.46). In order to interpret the Hamiltonian as an energy is must be bounded from below. The positivity of the first two terms in Eq. (10.66) shows that the sign postulated in the first term of Eq. (10.61) was correct. We also require the condition that (1/2) m2Φ2 + (Φ) must be bounded from below as a H function of Φ. As discusssed below, this example basically validates the Lagrangian (10.61) as a possible theory for scalar fields (as well as the association of Φ, Π with the canonical variables Q, P ). There are however field variables (such as the time component of a vector field or the hermitian conjugate of a Dirac field) that are not canonical field variables Qn neither have canonical conjugates. Nevertheless, Lorentz invariance requires that they must be present in the Lagrangians for the vector and Dirac fields respectively. This type of variables have the feature that they appear in the Lagrangian but not their time-derivatives. The field variables Ψk whose time-derivatives do not appear in the Lagrangian will be denoted as Cr. The remaining independent field variables are the canonical variables Qn. The canonical conjugates of Qn are given by

δL Q (t) , Q˙ (t) ,C (t) Pn (x,t)= (10.67) h δQ˙ n (x,t) i 286 CHAPTER 10. CANONICAL QUANTIZATION

n The pairs Q , Pk satisfy the canonical commutation relations (10.42)-(10.43). But as we already said, there are no canonical conjugates associated with Cr. Since δL/δC˙ r = 0, the Hamiltonian (10.58) reads

H = d3x P Q˙ n L Q (t) , Q˙ (t) ,C (t) n − n X Z h i r k but it is not useful until we express the variables C and Q˙ in terms entirely of Q′s and P ′s. For the variables Cr the left-hand side of the equations of motion (10.50) vanishes

δL Q (t) , Q˙ (t) ,C (t) 0= (10.68) h δCr (x,t) i hence the equations of motion associated with Cr involve only ﬁelds and their ﬁrst time-derivatives. We shall treat here simple cases in which Eqs. (10.67) and (10.68) can be solved directly to express the Cr and Q˙ k in terms of Q′s and P ′s. In gauge theories we can solve it by either choosing a particular gauge or by using more modern covariant methods.

10.6.1 Setting the Hamiltonian for the use of perturbation theory

n Once we construct a Hamiltonian as a functional in the Heisenberg picture in terms of canonical variables Q , Pn, we should pass to the interaction picture in order to use perturbation theory. Since the Hamiltonian is time- 1 n independent we can express it in terms of Q and Pn at t = 0, which coincides with the asssociated operators n pn and q in the interaction picture at t = 0. Then we can express the Hamiltonian in terms of the q′s and p′s in the interaction picture, and split it into two parts, one corresponding to a free-Hamiltonian H0 and another associated with an interaction V . Finally, we can use the time-dependent equations (10.29) and (10.30) as well as the commutation or anticommutation relations (10.1, 10.2) in order to express the q′s and p′s in V (t) as linear combinations of annihilation and creation operators. By now we shall provide a simple example of this procedure: the scalar ﬁeld with Hamiltonian (10.66). We start by splitting it into a free-particle Hamiltonian plus an interaction

H = H0 + V (10.69) 1 1 1 H = d3x Π2 + ( Φ)2 + m2Φ2 (10.70) 0 2 2 ∇ 2 Z V = d3x (Φ) (10.71) H Z where Φ and Π are evaluated at the same time t. H is independent of t despite H0 and V usually depend on t. As described above, the next step is to pass to the interaction picture. Thus, by taking t = 0 in Eqs. (10.70) and (10.71) we can replace Φ, Π with the interaction picture variables φ, π because according with Eqs. (10.39) and (10.40) they coincide at t = 0. Now, to calculate the interaction V (t) in the interaction picture we apply the similarity transformation given by Eq. (2.180) page 103

V (t) = exp (iH t) V (t = 0) exp ( iH t)= d3x exp (iH t) (φ (x, 0)) exp ( iH t) 0 − 0 0 H − 0 Z = d3x (exp (iH t) φ (x, 0) exp ( iH t)) = d3x (φ (x,t)) H 0 − 0 H Z Z 1Recall that in the Heisenberg picture we “absorb” all time dependence through a similarity transformation with the (complete) time-evolution operator. 10.6. FROM LAGRANGIAN TO HAMILTONIAN FORMALISM 287 where in the last step we use the fact that the similarity transformations preserve the product [see Eq. (10.41)]. Thus we obtain V (t) = exp (iH t) V exp ( iH t)= d3x (φ (x,t)) (10.72) 0 − 0 H Z the same transformation must be applied to H0, but it is clearly kept unchanged

1 1 1 H = exp (iH t) H exp ( iH t)= d3x π2 (x,t)+ ( φ (x,t))2 + m2φ2 (10.73) 0 0 0 − 0 2 2 ∇ 2 Z We can relate the variables π and φ˙ by substituting (10.73) in Eq. (10.29)

δH φ˙ (x,t)= 0 = π (x,t) (10.74) δπ (x,t)

Note that this is the same relation as (10.65) but this is not a general feature. The equation of motion for φ is given by combining Eqs. (10.73) and (10.30)

δH π˙ (x,t)= 0 = 2φ (x,t) m2φ (x,t) (10.75) −δφ (x,t) ∇ − from Eq. (10.74) we have

π (x,t) = ∂ φ (x,t) 0 ⇒ π˙ (x,t) = ∂ ∂ φ (x,t)= ∂ ∂0φ (x,t) (10.76) 0 0 − 0 and comparing Eqs. (10.75, 10.76) we obtain

∂ ∂0φ (x,t)= ∂ ∂iφ (x,t) m2φ (x,t) − 0 i − therefore, by combining Eqs. (10.74) and (10.75) we ﬁnd the ﬁeld equation

m2 φ (x) = 0 − whose general real solution is given by [see procedure in section 4.6, page 149]

3 3/2 d p ip x ip x φ (x) = (2π)− e · a (p)+ e− · a† (p) (10.77) 2p0 Z h i p with p0 = p2 + m2, and a (p) are some unknown operator function of p to be determined. The canonical conjugate is given by Eq. (10.74) p

0 3/2 3 p ip x ip x π (x)= φ˙ (x)= i (2π)− d p e · a (p) e− · a† (p) (10.78) − r 2 − Z h i then we adjust the unknown operators a (p) and a† (q) in order to obtain the desired commutation relations

[φ (x,t) ,π (y,t)] = iδ3 (x y) (10.79) − − [φ (x,t) , φ (y,t)] = [π (x,t) ,π (y,t)] = 0 (10.80) − − ———————————————— ————————————————– 288 CHAPTER 10. CANONICAL QUANTIZATION

by imposing the ﬁrst of conditions (10.80) on the expansions (10.77) and (10.78), we obtain

3 3 3/2 d p ip x ip x 3/2 d q iq y iq y 0 = [φ (x,t) , φ (y,t)] = (2π)− e · a (p)+ e− · a† (p) , (2π)− e · a (q)+ e− · a† (q) − " 2p0 2q0 # Z n o Z n o 3 3 3 d p d q ip x p ip x iq y iq y p 0 = (2π)− e · a (p)+ e− · a† (p) , e · a (q)+ e− · a† (q) 0 0 Z 2p Z 2q 3 3 hn o n oi 3 pd p pd q ip x iq y ip x iq y 0 = (2π)− e · e · [a (p) , a (q)]+ e · e− · a (p) , a† (q) 2p0 2q0 Z Z n h i ip x iq y ip x iq y +e− · e p· a† (p) p, a (q) + e− · e− · a† (p) , a† (q) Iφφ + Iφφ + Ihφφ + Iφφ i h io ≡ 1 2 3 4 interchanging the dummy variables q p in Iφφ we have ↔ 2 3 3 φφ 3 d p d q ip x iq y I (2π)− e · e− · a (p) , a† (q) 2 ≡ 0 0 Z 2p Z 2q 3 3 h i 3 d q d p iq x ip y = (2π)− p p e · e− · a (q) , a† (p) 0 0 Z 2q Z 2p 3 3 h i 3 d p d q iq x ip y = (2π)− p p e · e− · a† (p) , a (q) − 2p0 2q0 Z Z h i 3 p 3 p φφ φφ 3 d p d q ip x iq y iq x ip y I + I = (2π)− e− · e · e · e− · a† (p) , a (q) 2 3 0 0 − Z 2p Z 2q h i applying the second of conditions (10.80)p on thep expansions (10.77) and (10.78), we have

0 0 3/2 3 p ip x ip x 3/2 3 q iq y 0 = [π (x,t) ,π (y,t)] = i (2π)− d p e · a (p) e− · a† (p) , i (2π)− d q e · a (q) e− − "− r 2 − − r 2 − Z n o Z n 0 0 3 3 p 3 q ip x ip x iq y iq y = (2π)− d p d q e · a (p) e− · a† (p) , e · a (q) e− · a† (q) − 2 2 − − Z r Z r 0 0 hn o n oi 3 3 p 3 q ip x iq y ip x iq y 0 = (2π)− d p d q e− · e · a† (p) , a (q) e− · e− · a† (p) , a† (q) 2 2 − Z r Z r ip x iq y ipnx iq y h i h i e · e · [a (p) , a (q)] + e · e− · a (p) , a† (q) − h io by imposing condition (10.79) on the expansions (10.77) and (10.78), we obtain

3 0 3 3/2 d p ip x ip x 3/2 3 q iq y iδ (x y) = [φ (x,t) ,π (y,t)] = (2π)− e · a (p)+ e− · a† (p) , i (2π)− d q e · a (q) − − " 2p0 − r 2 − Z n o Z n 3 0 3 d p 3 q pip x ip x iq y iq y = i (2π)− d q e · a (p)+ e− · a† (p) , e · a (q) e− · a† (q) − 2p0 r 2 − Z Z hn o n oi 3 0 3 3 pd p 3 q ip x iq y ip x iq y iδ (x y) = i (2π)− d q e · e · [a (p) , a (q)] e · e− · a (p) , a† (q) − − 2p0 r 2 − Z Z n h i ip x iq y ip x iq y +e− · e · pa† (p) , a (q) e− · e− · a† (p) , a† (q) − h i h io ???? 10.7. GAUGES OF THE LAGRANGIAN FORMALISM 289

———————————————- ———————————————–

Therefore, to satisfy Eqs. (10.79, 10.80) the unknown operators must fulfill the commutation relations (Home- work!! B6) 3 a (p) , a† (q) = δ (p q) ; [a (p) , a (q)] = 0 (10.81) − h i we have already seen that expansion (10.73) leads to the usual free Hamiltonian Eq. (3.23) page 121, up to an unphysical constant. It is important to emphasize that this should not be considered as an alternative derivation of Eqs. (10.77), and (10.81) but rather as a proof of validation of the first two terms in Eq. (10.61) as the correct free-particle Lagrangian for a real scalar field. The way is paved to use perturbation theory to calculate the S matrix, by taking Eq. (10.72) as V (t), with the scalar field φ (x) given by Eq. (10.77) where a (p) and a (p) − † obeys the commutation relations (10.81).

10.7 Gauges of the Lagrangian formalism

Since the Hamilton’s stationary principle usually requires an integration by parts and we usually assume that the fields vanish at infinity, Lagrangian densities that differ only by total derivatives ∂ µ do not contribute to the µF action leading to the same field equations. It is also clear that a space derivative term in the Lagrangian ∇ · F density does not contribute to the Lagrangian so that it does not affect the quantum field theory defined by such a Lagrangian. It is important however to emphasize that it is not true anymore when the fields do not vanish in the boundaries (it happens in some theories in which we define boundaries not at infinity). Furthermore, a time derivative ∂ 0 in the Lagrangian density does not affect the quantum structure of the theory, though such a fact 0F is less apparent. We shall see it by considering the effect of adding a more general term to the Lagrangian in the form

3 ∆L (t)= d x Dn,x [Q (t)] Q˙ (x,t) (10.82) Z where D is an arbitrary functional of the values of Q at a given time that in general depend on n and x. According with Eq. (10.48) the formula for the canonical conjugate momenta P (t) change as a functional of Q (t) and Q˙ (t) as δ∆L (t) ∆Pn (x,t)= = Dn,x [Q (t)] (10.83) δQ˙ n (x,t) But taking into account Eqs. (10.82, 10.83) we see that there is no change in the Hamiltonian expressed as a functional of Q (t) and Q˙ (t) [so that the change in variables only appears through the canonical momenta Pn]

∆H Q (t) , Q˙ (t) = d3x ∆P (x,t) Q˙ n (x,t) ∆L (t) n − h i Z 3 n 3 = d x D x [Q (t)] Q˙ (x,t) d x D x [Q (t)] Q˙ (x,t) = 0 n, − n, Z Z

∆H Q (t) , Q˙ (t) = d3x ∆P (x,t) Q˙ n (x,t) ∆L (t) = 0 (10.84) n − Z h i n Therefore, the Hamiltonian written as a functional of the old canonical variables Q and Pn do not change either. We should notice however that the Hamiltonian is not the same functional of the new canonical variables Qn and n Pn +∆Pn, and in a theory described by the new Lagrangian density +∆ it is the new canonical set Q and n L L Pn +∆Pn instead of the old one Q and Pn, that satisﬁes the canonical commutation relations. The commutators n k n of Q with Q and of Q with Pk obey the usual commutation relations, but now the commutators of Pn with Pk are given by 290 CHAPTER 10. CANONICAL QUANTIZATION

[P (x,t) , P (y,t)] = [P (x,t)+∆P (x,t) , P (y,t)+∆P (y,t)] [∆P (x,t) , P (y,t)+∆P (y,t)] n m n n m m − n m m [P (x,t)+∆P (x,t) , ∆P (y,t)]+[∆P (x,t) , ∆P (y,t)] − n n m n m δD [Q (t)] δD [Q (t)] [P (x,t) , P (y,t)] = i n,x + i m,y (10.85) n m − δQm (y,t) δQn (x,t) which is in general non-null. Nevertheless, if the additional term in the Lagrangian is a total time-derivative dG δG [Q (t)] ∆L = = d3x Q˙ n (x,t) (10.86) dt δQn (x,t) Z we see that D in Eq. (10.82) acquires a particular form

δG [Q (t)] D [Q]= (10.87) n,x δQn (x,t)

n and in that case the commutator (10.85) vanishes, from which the variables Q and Pk satisfies the usual commutation relations. Thus we have shown that a change of the form (10.82) in the Lagrangian does not change n the form of the Hamiltonian as a functional of Q and Pk, and taking into account that the commutation relations of those variables do not change, we conclude that the addition to the Lagrangian of the term (10.82) does lead to the same quantum field theory. Thus, different Lagrangian densities obtained from each other by partial integration are equivalent in both quantum and classical field theories.

10.8 Global symmetries

Like in classical mechanics, Lagrangian formalism is very suitable in quantum mechanics to implement symmetry principles. It has to do with the fact that the dynamical equations of motion has the form of a variational principle in the Lagrangian formalism i.e. the Hamilton’s variational principle. We shall start with an inﬁnitesimal transformation of the ﬁelds

Ψk (x) Ψk (x)+ iε k (x) (10.88) → F that keeps the action (10.49) invariant

δI [Ψ] δI [Ψ] 0= δI = d4x δΨk (x)= iε d4x k (x) (10.89) δΨk (x) δΨk (x)F Z Z when ε is constant we call them as global symmetries. The factor k (x) depends in general on the fields and their F derivatives at x. Equation (10.89) is automatically satisfied for all infinitesimal variations of the fields if those fields obey the dynamical equations. Thus we define an infinitesimal symmetry transformation as one that leaves the action invariant even when the dynamical equations are not satisfied. Let us now consider a local transformation, that is, one in which ε is a function of x

Ψk (x) Ψk (x)+ iε (x) k (x) (10.90) → F in this case the variation of the action does not vanish, but it acquires the form ∂ε (x) δI = d4x J µ (x) (10.91) − ∂xµ Z for it to vanish when ε becomes constant. Here J µ is a functional of the fields and its first-derivatives. If we assume that the fields in I [Ψ] satisfy the field equations we obtain that I [Ψ] is stationary with respect to arbitrary 10.8. GLOBAL SYMMETRIES 291

field variations that vanish at large space-time distances, including local variations of the type (10.90), hence in that case the variation (10.91) should vanish. Integrating by parts we find that J µ must satisfy a conservation law (Homework!! B7) ∂J µ (x) 0= (10.92) ∂xµ it is straightforward that dF 0= , with F d3x J 0 (x) (10.93) dt ≡ Z so that there is one conserved current J µ and one constant of the motion F for each independent infinitesimal symmetry transformation. This leads to the fact that continuous symmetries imply conservation laws2, statement usually called Noether’s theorem. For some symmetry transformations the Lagrangian is invariant which is a condition stronger than the invariance of the action. Good examples are the translations and rotations in space and also isospin transformations as well as other internal symmetry transformations (but it is not the case for general Lorentz transformations). For the stronger condition of invariance of the Lagrangian we correspondingly obtain a stronger form of the conservation law, i.e. we can obtain an explicit formula for the conserved quantities F . To see it, let us consider a field variation of the type (10.90) with ε (x) depending on t but not on x. Under this restricted transformation, the variation of the action becomes

δL Ψ (t) , Ψ˙ (t) δL Ψ (t) , Ψ˙ (t) δI = δ dt L Ψ (t) , Ψ˙ (t) = dt d3x δΨk (x,t)+ δΨ˙ k (x,t)  hδΨk (x,t) i hδΨ˙ k (x,t) i  Z h i Z Z   ˙ ˙ δL Ψ (t) , Ψ (t)  δL Ψ (t) , Ψ (t) d  δI = i dt d3x ε (t) k (x,t)+ ε (t) k (x,t)  hδΨk (x,t) i F hδΨ˙ k (x,t) i dt F  Z Z  h i ˙ δL Ψ (t) , Ψ (t)  δI = i dt d3x ε (t) k (x,t)  hδΨk (x,t) i F Z Z  ˙ ˙ δL Ψ (t) , Ψ (t) δL Ψ (t) , Ψ (t) d + ε˙ (t) k (x,t)+ ε (t) k (x,t) (10.94) hδΨ˙ k (x,t) i F hδΨ˙ k (x,t) i dtF   and the condition of invariance of the Lagrangian under this transformation when ε becomes constant gives

˙ ˙ δL Ψ (t) , Ψ (t) δL Ψ (t) , Ψ (t) d 0= d3x k (x,t)+ k (x,t) (10.95)  hδΨk (x,t) i F hδΨ˙ k (x,t) i dtF  Z     assuming that the transformations k (x,t) are independent we obtain F ˙ ˙ δL Ψ (t) , Ψ (t) δL Ψ (t) , Ψ (t) d k (x,t)= k (x,t) (10.96) hδΨk (x,t) i F − hδΨ˙ k (x,t) i dtF

2The condition of continuity is essential because we are using inﬁnitesimal transformations. However, in practice we often ﬁnd conservation laws associated with discrete symmetries. Nevertheless, the latter case is not guaranteed by Noether’s theorem. 292 CHAPTER 10. CANONICAL QUANTIZATION substituting (10.96) in Eq. (10.94) we obtain

˙ δL Ψ (t) , Ψ (t) d δI = i dt d3x ε (t) k (x,t) − hδΨ˙ k (x,t) i dtF Z Z  ˙ ˙ δL Ψ (t) , Ψ (t) δL Ψ (t) , Ψ (t) d + ε˙ (t) k (x,t)+ ε (t) k (x,t) hδΨ˙ k (x,t) i F hδΨ˙ k (x,t) i dtF   so regardless the ﬁeld equations are satisﬁed or not, the variation of the action gives 

δL Ψ (t) , Ψ˙ (t) δI = i dt d3x ε˙ (t) k (x,t) (10.97)  hδΨ˙ k (x,t) i F  Z Z   Now let us rewrite Eq. (10.91) recalling thatε depends on t but not on x 

∂ε (t) ∂ε (t) δI = dt d3x J 0 (x) + J i (x) = dt ε˙ (t) d3x J 0 (x) − ∂x0 ∂xi − Z Z Z Z δI = dt ε˙ (t) F (10.98) − Z and comparing Eqs. (10.97) and (10.98) we obtain

δL Ψ (t) , Ψ˙ (t) F = i d3x k (x,t) (10.99) − hδΨ˙ k (x,t) iF Z so we have obtained an explicit expression for the conserved generalized charge F . As a matter of consistency, by taking the symmetry condition (10.95) it could be checked that F is time-independent for any ﬁelds that satisfy the dynamical equations (10.50). An even stronger condition that posseses some symmetry transformations (e.g. isospin symmetry) leave the action, Lagrangian, and Lagrangian density invariant. In that case we can additionally obtain an explicit formula for the current J µ (x). The action (10.52) (written in terms of the Lagrangian density), has a variation under the transformation (10.90) with the local inﬁnitesimal parameter ε (x) of the form

4 ∂ (Ψ (x) , ∂µΨ (x)) k ∂ (Ψ (x) , ∂µΨ (x)) k δI [Ψ] = i d x L k δΨ (x)+ L k δ ∂µΨ (x) ∂Ψ (x) ∂ (∂µΨ (x)) Z 4 ∂ (Ψ (x) , ∂µΨ (x)) k ∂ (Ψ (x) , ∂µΨ (x)) k δI [Ψ] = i d x L k (x) ε (x)+ L k ∂µ (x) ε (x) ∂Ψ (x) F ∂ (∂µΨ (x)) F Z h i the invariance of the Lagrangian density when ε becomes constant, provides the condition

∂ (Ψ (x) , ∂µΨ (x)) k ∂ (Ψ (x) , ∂µΨ (x)) k 0= L k (x)+ L k ∂µ (x) (10.100) ∂Ψ (x) F ∂ (∂µΨ (x)) F so for arbitrary ﬁelds, the variation of the action gives

∂ (Ψ (x) , ∂ Ψ (x)) δI [Ψ] = i d4x L µ k (x) ∂ ε (x) ∂ (∂ Ψk (x)) F µ Z µ 10.9. CONSERVED QUANTITIES IN QUANTUM FIELD THEORIES 293 and comparing with (10.91) we see that ∂ J µ = i L k (10.101) − ∂ (∂Ψk/∂xµ)F As a matter of consistency we can check that the symmetry condition (10.100) leads to the continuity equation µ ∂µJ = 0 when the ﬁelds obey the Euler-Lagrange equation (10.57) [homework!! B8]. It can also be checked that the time component of the current (10.101) integrated over spatial coordinates has the value calculated above Eq. (10.99) in agreement with Eq. (10.93) [homework!! B9].

10.9 Conserved quantities in quantum ﬁeld theories

The developments done so far are equally valid for classical and quantum ﬁeld theories. The quantum properties of the conserved quantities F are more apparent when they come from symmetries of the Lagrangian (not necessarily of the Lagrangian density) that transform the canonical ﬁelds Qn (x,t) (i.e. the ones of the Ψk whose time- derivatives appear in the Lagrangian) into x dependent functionals of themselves at the same time. Those types − of transformations yield

n (x,t)= n [Q (t) ; x] (10.102) F F we shall see later that infinitesimal translations, rotations as well as infinitesimal internal symmetry transformations have the form of Eqs. (10.88) and (10.102), with n being a linear functional of the Q s, though we shall not F ′ require the linearity here. For those symmetries in quantum mechanics besides being conserved, such operators are generators of the symmetry. To see it, we first observe that when Ψk is a canonical variable Qk it has an associated canonical momentum k k r Pk given by the functional derivative δL/δΨ˙ , while when Ψ is an auxiliary field C , such a functional derivative is null. From these facts Eq. (10.99) can be written as

F = i d3x P (x,t) n (x,t)= i d3x P (x,t) n [Q (t) ; x] (10.103) − n F − n F Z Z To calculate the commutator (not the anticommutator) of F with the canonical ﬁeld Qk (x,t) at an arbitrary time t, we can use Eq. (10.93) to evaluate F as a functional of the Q′s and P ′s (at the time t), and then use the equal-time canonical commutation relations (10.42) and (10.43) and obtain [homework!! Eq. (10.104) (with commutator) should be true also for fermionic variables Q, P ]

n 3 m n 3 m n [F,Q (x,t)] = i d y Pm (y,t) [Q (t) ; y] ,Q (x,t) = i d y [Pm (y,t) [Q (t) ; y] ,Q (x,t)] − − F − F − Z − Z = i d3y P (y,t) [ m [Q (t) ; y] ,Qn (x,t)] i d3y [P (y,t) ,Qn (x,t)] m [Q (t) ; y] − m F − m F Z − Z − − = i d3y iδ3 (x y) δ m [Q (t) ; y] − − − nm F Z − [F,Qn (x,t)] = n (x,t) (10.104) − −F where we have assumed that for Qn being bosonic or fermionic the variation n is bosonic or fermionic respectively, F from which F is bosonic. Nevertheless, in the case of some symmetries known as supersymmetries, for which F is fermionic and (10.104) is an anticommutator if Qn is fermionic as well. It is in the sense of Eq. (10.104) that F can be considered the generator of the transformation with (10.102) and (10.103). The canonical commutation rules also lead to [homework!! B10 do it for fermionic operators Q, P ]

3 m 3 m [F, Pn (x,t)] = i d y Pm (y,t) [Q (t) ; y] , Pn (x,t) = i d y Pm (y,t) [ [Q (t) ; y] , Pn (x,t)] − − F − F − Z − Z = i d3y [P (y,t) m [Q (t) ; y] , P (x,t)] − m F n Z − 294 CHAPTER 10. CANONICAL QUANTIZATION and using the deﬁnition (10.23) page 275 we ﬁnally obtain

δ m (Q (t) ; y) [F, P (x,t)] = d3y P (y,t) F (10.105) n m δQn (x,t) − Z k n when F is a linear functional, equation (10.105) shows that Pn transforms contragradiently with respect to Q , n it justiﬁes the notation of indices followed so far for Q and Pn.

10.9.1 Space-time translation symmetry We shall ﬁrst examine the space-time translation symmetry

Ψk (x) Ψk (x + ε)=Ψk (x)+ εµ∂ Ψk (x) (10.106) → µ equation (10.106) has the structure of Eq. (10.88) with four independent parameters εµ with four associated transformation functions

k = i∂ Ψk (10.107) Fµ − µ so that we have four independent conserved currents which are conveniently grouped into the energy-momentum µ tensor T ν µ ∂µT ν = 0 where we can derive four “generalized charges” which are global conserved quantities written as spatial integrals of the time components of each of the four “translation currents” denoted by Pν.

d P = d3x T 0 ; P = 0 (10.108) ν ν dt ν Z

In order to avoid confusion with the canonical variables Pn (x,t) we have used greek letters for the subscript of the four generalized charges.

Spatial translations The Lagrangian L (t) is invariant under spatial translations (there is an integration over all spatial components to deﬁne L Ψ (t) , Ψ˙ (t) ), so that the spatial components Pk of Eq. (10.108) acquires an explicit form according with Eq. (10.99)h or equivalentlyi Eq. (10.103). Hence, combining Eqs. (10.103) and (10.107) the spatial components of Pν acquires the form (we use for a moment the notation Pk to avoid confusion with the canonical variable Pk)

P = i d3x P (x,t) n [Q (t) ; x]= i d3x P (x,t) [ i∂ n] k − n Fk − n − k⊖ Z Z

P d3x P (x,t) Qn (x,t) (10.109) ≡− n ∇ Z from the canonical equal-time comutation relations (10.42) and (10.43) we can ﬁnd the commutator of P with the canonical variables

n 3 m n 3 m n [P,Q (x,t)] = d y Pm (y,t) Q (y,t) , Q (x,t) = d y [ Pm (y,t) Q (y,t) , Q (x,t)] − − ∇ − ∇ − Z − Z = d3y [ P (y,t) , Qn (x,t)] Qm (y,t)= i d3y δ3 (x y) δ Qm (y,t)= i Qn (x,t) − m ∇ − mn ∇ ∇ Z − Z 10.9. CONSERVED QUANTITIES IN QUANTUM FIELD THEORIES 295

3 m 3 m [P, Pn (x,t)] = d y Pm (y,t) Q (y,t) , Pn (x,t) = d yPm (y,t) [Pn (x,t) , Q (y,t)] − − ∇ ∇ − Z − Z = i d3yP (y,t) δ3 (x y) δ P (x,t) m − nm∇ n Z

[P,Qn (x,t)] = i Qn (x,t) (10.110) − ∇ [P, Pn (x,t)] = i Pn (x,t) (10.111) − ∇ from Eqs. (10.110, 10.111), it can be shown that for a function of the canonical variables Qk, P that does not ℑ k also depend on x, we have [P, (x)] = i (x) (10.112) ℑ ∇ℑ from these results we conclude that P is the generator of spatial translations.

Time translations In contrast to the space translations, time translations do not leave the Lagrangian L (t) invariant. But we already know that its generator is the Hamiltonian P 0 H. And we also know that the Hamiltonian satisﬁes the ≡ commutation relation

[H, (x,t)] = i ˙ (x,t) ℑ − ℑ for any function (x,t) of Heisenberg picture operators. ℑ 10.9.2 Conserved currents and Lagrangian densities for space-time symmetries By assuming that the Lagrangian is a space integral of a Lagrangian density, we can also obtain an explicit formula for the energy-momentum tensor T µ . Nevertheless, the Lagrangian density is not invariant under space-time ν L translations, so we cannot apply Eq. (10.101). Hence, we shall go back to examine the change in the action under a space-time dependent translation of the form

Ψk (x) Ψk (x + ε (x))=Ψk (x)+ εµ (x) ∂ Ψk (x) (10.113) → µ the variation of the action under transformation (10.113) reads

4 ∂ µ k ∂ µ k δI [Ψ] = d x Lk ε ∂µΨ + L k ∂ν ε ∂µΨ (10.114) ∂Ψ ∂ (∂ν Ψ ) Z h i from the Euler-Lagrange equations (10.57) we see that the terms proportional to ε add up to εµ∂ , so that the µL variation (10.114) becomes

4 ∂ µ k ∂ µ k ∂ k µ δI [Ψ] = d x Lk ε ∂µΨ + L k ε ∂ν∂µΨ + L k ∂µΨ (∂νε ) ∂Ψ ∂ (∂ν Ψ ) ∂ (∂ν Ψ ) Z 4 µ ∂ k ∂ k ∂ k µ = d x ε Lk ∂µΨ + L k ∂ν∂µΨ + L k ∂µΨ (∂νε ) ∂Ψ ∂ (∂νΨ ) ∂ (∂ν Ψ ) Z 4 µ ∂ k ∂ k k ∂ ∂ k µ = d x ε Lk ∂µΨ + ∂ν L k ∂µΨ ∂µΨ ∂ν L k + L k ∂µΨ (∂ν ε ) ∂Ψ ∂ (∂ν Ψ ) − ∂ (∂νΨ ) ∂ (∂νΨ ) Z 4 µ ∂ ∂ k µ ∂ k ∂ k µ = d x ε Lk ∂ν L k ∂µΨ + ε ∂ν L k ∂µΨ + L k ∂µΨ (∂νε ) ∂Ψ − ∂ (∂νΨ ) ∂ (∂ν Ψ ) ∂ (∂ν Ψ ) Z 4 µ ∂ k ∂ k µ = d x ε ∂ν L k ∂µΨ + L k ∂µΨ (∂νε ) ∂ (∂νΨ ) ∂ (∂ν Ψ ) Z 296 CHAPTER 10. CANONICAL QUANTIZATION

∂ ∂ δI [Ψ] = d4x L εµ + L ∂ Ψk∂ εµ ∂xµ ∂ (∂ Ψk) µ ν Z ν and integrating by parts we are led to the form of Eq. (10.91) (Homework!! B11)

δI [Ψ] = d4x T ν ∂ εµ − µ ν Z with the associated currents ν ν ∂ k T µ = δµ L k ∂µΨ (10.115) L− ∂ (∂νΨ ) by consistency we can see that the spatial components of Eq. (10.108) coincide with the formula (10.109) for P, while for µ = 0 Eq. (10.108) gives the well known expression for the Hamiltonian

H P 0 = P = d3x P Q˙ n (10.116) ≡ − 0 n −L " n # Z X 10.9.3 Additional symmetry principles In some theories there are other symmetry principles that provides the invariance of the action under a set of coordinate independent transformations of the canonical ﬁelds

Qn (x) Qn (x)+ iεa (t )n Qm (x) (10.117) → a m along with a set of transformations of the auxiliary ﬁelds Cr

Cr (x) Cr (x)+ iεa (τ )r Cs (x) (10.118) → a s where ta and τa denote Hermitian matrices associated with representations of the Lie algebra of the symmetry group, with sum over repeated group indices a, b, . . .. For example, we shall see that for electrodynamics we have n a symmetry of this kind with a single diagonal matrix t m, with the charges associated with each ﬁeld in the diagonal. µ For these kind of symmetries we can ﬁnd a set of conserved currents Ja

µ ∂µJa = 0 (10.119) where the time components are the densities of a set of time-independent “generalized charge operators” (quantized charges) 3 0 Ta = d x Ja (10.120) Z As before, when the Lagrangian as well as the action is invariant under the transformation (10.117) then Eq. (10.99) provides an explicit expression for the generalized charged operator

T = i d3x P (x,t) (t )n Qm (x,t) (10.121) a − n a m Z in this case the equal-time commutation relations give

n n m [Ta,Q (x)] = (ta) mQ (x) (10.122) − m [Ta, Pn (x)] = + (ta) nPm (x) (10.123) 10.9. CONSERVED QUANTITIES IN QUANTUM FIELD THEORIES 297

n where ta is diagonal. These relations show that Q and Pn lower and raise the value of Ta by an amount equal to n (ta) n. From the previous results, we can calculate the commutator between the generalized charge operators

3 n m 3 k p [Ta, Tb] = i d x Pn (x,t) (ta) mQ (x,t) , i d y Pk (y,t) (tb) pQ (y,t) − − − Z Z − [T , T ] = d3x d3y (t )n (t )k [P (x,t) Qm (x,t) , P (y,t) Qp (y,t)] (10.124) a b − a m b p n k − Z Z − let us evaluate the commutators

m p C [Pn (x,t) Q (x,t) , Pk (y,t) Q (y,t)] ≡ m p − p m = Pn (x,t) [Q (x,t) , Pk (y,t) Q (y,t)] + [Pn (x,t) , Pk (y,t) Q (y,t)] Q (x,t) m p − m − p = Pn (x,t) Pk (y,t) [Q (x,t) , Q (y,t)] + Pn (x,t) [Q (x,t) , Pk (y,t)] Q (y,t) p m − p − m +Pk (y,t) [Pn (x,t) , Q (y,t)] Q (x,t) + [Pn (x,t) , Pk (y,t)] Q (y,t) Q (x,t) − − C = iδ3 (x y) δ P (x,t) Qp (y,t) iδ3 (x y) δ P (y,t) Qm (x,t) (10.125) − mk n − − np k substituting (10.125) in (10.124) we obtain

[T , T ] = i d3x d3y (t )n (t )k δ3 (x y) [δ P (x,t) Qp (y,t) δ P (y,t) Qm (x,t)] a b − a m b p − mk n − np k − Z Z = i d3x (t )n (t )m P (x,t) Qp (x,t) + (t )n (t )k P (x,t) Qm (x,t) − a m b p n a m b n k Z n o 3 m n k n k m [Ta, Tb] = i d x Pm (ta) n (tb) kQ + Pn (tb) k (ta) mQ (10.126) − − Z h i if we deﬁne P, Q as matrix arrangements (column vectors) for Pm,Qm, and ta the matrix associated with the transformation (10.117) we can write

3 [Ta, Tb] = i d x P tatb Q + P tbta Q − − Z h i 3 [Ta, Tb] = i d x eP [ta, tb] Qe − − Z n o c and hence, if the matrices ta form a Lie algebra with structuree constants fab

c [ta,tb] = ifab tc (10.127) −

3 c 3 c [Ta, Tb] = i d x P [ta, tb] Q = fab d x P tc Q = fab Tc − − Z n o Z where we have used (10.121) in the last step.e Hence, the quantum operatorse Ta form the same Lie algebra as the matrices ta c [Ta, Tb] = ifab Tc (10.128) − which shows that the operators (10.121) are properly normalized to be considered as generators of the symmetry group. Once again when there is a Lagrangian density that is invariant under the transformations (10.117) and (10.118) we can obtain an expression for the current J µ associated with the global symmetry. From Eq. (10.101) and using Eqs. (10.117) and (10.118) we have

∂ ∂ J µ i L (t )n Qm i L (τ )r Cs (10.129) a ≡− ∂ (∂Qn/∂xµ) a m − ∂ (∂Cr/∂xµ) a s 298 CHAPTER 10. CANONICAL QUANTIZATION note that the construction of the current requires the auxiliary fields Cr as well as their transformation properties (10.118), despite Ta in Eq. (10.121) only depends on the fields Q and their canonical conjugates P . The explicit expression (10.129) for the current lead to other useful commutation relations. In particular, when the Lagrangian density does not contain time-derivatives of the auxiliary fields, we obtain from (10.129)

0 ∂ n m ∂ n m J i L (ta) mQ = i L (ta) mQ a ≡ − ∂ (∂Qn/∂x0) − ∂Q˙ n J 0 = iP (t )n Qm a − n a m then we can derive the equal-time commutators of general ﬁelds not only with the symmetry operators Ta, but 0 also with the densities Ja

J 0 (x,t) ,Qn (y,t) = δ3 (x y) (t )n Qm (x,t) (10.130) a − − a m J 0 (x,t) , P (y,t) = δ3 (x y) (t )n P (x,t) (10.131) a m − a m n Moreover, if the auxiliary ﬁelds are constructed such that they are local functions of Q′s and P ′s, such that they transform according with a representation of the Lie symmetry algebra with generators τa, we also have commutation relations involving auxiliary ﬁelds of the form

J 0 (x,t) , Cr (y,t) = δ3 (x y) (τ )r Cs (x,t) (10.132) a − − a s It is usual to shorten the set of commutation relations (10.130) and (10.132) as follows

J 0 (x,t) , Ψk (y,t) = δ3 (x y) (t )k Ψm (x,t) a − − a m h i Commutation relations of the type given by (10.130)-(10.132) are used to derive some relations for matrix elements related with the current J µ known as Ward identities.

10.9.4 Conserved current for a two scalars Lagrangian

As a matter of example, let us take the case of two real ﬁelds with the same mass, with a Lagrangian density given by

1 1 1 1 = ∂ Φ ∂µΦ m2Φ2 ∂ Φ ∂µΦ m2Φ2 Φ2 +Φ2 (10.133) L −2 µ 1 1 − 2 1 − 2 µ 2 2 − 2 2 −H 1 2 this Lagrangian density is invariant under a transformation of the type (10.117) given by3

Φ′ Φ δΦ = εΦ ;Φ′ Φ δΦ = εΦ (10.134) 1 − 1 ≡ 1 − 2 2 − 2 ≡ 2 1 3This transformation correspond to an inﬁnitesimal rotation in two-dimensions as we can see from

cos ε − sin ε Φ Φ′ 1 = 1 sin ε cos ε Φ2 Φ´ 2 at ﬁrst order cos ε ≈ 1, sin ε ≈ ε, so that

Φ′ 1 −ε Φ Φ − εΦ 1 = 1 = 1 2 Φ´ 2 ε 1 Φ2 Φ2 + εΦ1 10.9. CONSERVED QUANTITIES IN QUANTUM FIELD THEORIES 299 we see it as follows

1 µ 1 µ 1 2 2 1 2 2 2 2 ′ = ∂ Φ′ ∂ Φ′ ∂ Φ′ ∂ Φ′ m Φ′ m Φ′ Φ′ +Φ′ L −2 µ 1 1 − 2 µ 2 2 − 2 1 − 2 2 −H 1 2 1 1 1 = ∂ (Φ εΦ ) ∂µ (Φ εΦ ) ∂ (Φ + εΦ ) ∂µ (Φ + εΦ ) m2 (Φ εΦ )2 −2 µ 1 − 2 1 − 2 − 2 µ 2 1 2 1 − 2 1 − 2 1 m2 (Φ + εΦ )2 (Φ εΦ )2 + (Φ + εΦ )2 −2 2 1 −H 1 − 2 2 1 1 µ 1 µ 1 µ 2 1 µ ′ = ∂ Φ ∂ Φ + ε ∂ Φ ∂ Φ + ε ∂ Φ ∂ Φ ε ∂ Φ ∂ Φ L −2 µ 1 1 2 µ 1 2 2 µ 2 1 − 2 µ 2 2 1 1 1 1 ∂ Φ ∂µΦ ε ∂ Φ ∂µΦ ε ∂ Φ ∂µΦ ε2 ∂ Φ ∂µΦ −2 µ 2 2 − 2 µ 2 1 − 2 µ 1 2 − 2 µ 1 1 1 1 1 1 m2Φ2 + εm2Φ Φ ε2 m2Φ2 m2Φ2 εm2Φ Φ ε2 m2Φ2 −2 1 1 2 − 2 2 − 2 2 − 2 1 − 2 1 ε2Φ2 2εΦ Φ +Φ2 + ε2Φ2 + 2εΦ Φ +Φ2 −H 2 − 1 2 1 1 1 2 2 1 µ 1 µ 2 1 µ 2 1 µ ′ = ∂ Φ ∂ Φ ∂ Φ ∂ Φ ε ∂ Φ ∂ Φ ε ∂ Φ ∂ Φ L −2 µ 1 1 − 2 µ 2 2 − 2 µ 2 2 − 2 µ 1 1 1 1 1 1 m2Φ2 m2Φ2 ε2 m2Φ2 ε2 m2Φ2 Φ2 +Φ2 + ε2 Φ2 +Φ2 −2 1 − 2 2 − 2 2 − 2 1 −H 1 2 1 2 2 ′ = + ε L L O comparing Eq. (10.134) with Eq. (10.117) we have

Φ′ Φ εΦ ;Φ′ =Φ + εΦ 1 ≡ 1 − 2 2 2 1 Q1 (x) Q1 (x)+ iεa (t )1 Qm (x)= Q1 (x) εQ2 → a m − Q1 (x)+ iεt1 Qm (x)= Q1 (x) εQ2 ⇒ m − Q1 (x)+ iεt1 Q1 (x)+ iεt1 Q2 (x)= Q1 (x) εQ2 ⇒ 1 2 − it1 = 0 ; it1 = 1 ⇒ 1 2 − similarly

2 2 m 2 1 Q (x)+ iεt mQ (x) = Q (x)+ εQ 2 2 iεt 1 = 1 ; iεt 2 = 0 note that despite we are transforming two ﬁelds, we only have one independent inﬁnitesimal parameter ε and so only one matrix of the form ta. The matrix ta yields

t1 = t2 = 0 ; t1 = i ; t2 = i (10.135) 1 2 2 1 − in this case there is only one current that we obtain by replacing (10.135) in (10.129)

∂ ∂ ∂ J µ i L (t )n Qm i L (τ )r Cs = i L tn Qm a ≡ − ∂ (∂Qn/∂xµ) a m − ∂ (∂Cr/∂xµ) a s − ∂ (∂Qn/∂xµ) m ∂ ∂ = i L t1 Q1 i L t1 Q2 − ∂ (∂Q1/∂xµ) 1 − ∂ (∂Q1/∂xµ) 2 ∂ ∂ i L t2 Q1 i L t2 Q2 − ∂ (∂Q2/∂xµ) 1 − ∂ (∂Q2/∂xµ) 2

∂ 1 ∂ 2 ∂ ∂ = i L t 2Φ2 i L t 1Φ1 = L Φ2 L Φ1 − ∂ (∂µΦ1) − ∂ (∂µΦ2) ∂ (∂µΦ1) − ∂ (∂µΦ2) 300 CHAPTER 10. CANONICAL QUANTIZATION and using the Lagrangian density (10.133) we ﬁnally obtain

1 µ 1 µ µ ∂ 2 ∂µΦ1∂ Φ1 ∂ 2 ∂µΦ2∂ Φ2 µ µ J = − Φ2 − Φ1 = Φ2∂ Φ1 +Φ1∂ Φ2 ∂ (∂µΦ1) − ∂ (∂µΦ2) − in summary Lagrangian (10.133) posseses a symmetry transformation (10.134) that lead to the following conserved current J µ =Φ ∂µΦ Φ ∂µΦ 2 1 − 1 2

10.10 Lorentz invariance

We shall see that the Lorentz invariance of the Lagrangian density implies the Lorentz invariance of the S matrix. − We proceed in three steps (a) We generate the currents and charges generated by Lorentz symmetry. (b) We construct the generators of the Lorentz group by using the previous currents and (c) We prove that such generators commute with the S operator. − 10.10.1 Currents and time-independent operators We start as usual with inﬁnitesimal Lorentz transformations

Λµ = δµ + ωµ ; ω = ω ν ν ν µν − νµ From our previous analysis the invariance of the action under these transformations lead to a set of conserved currents that we denote as ρµν M ∂ ρµν = 0 ; ρµν = ρνµ ρM M −M i.e. one current for each independent component of ωµν (six independent components owing to the antisymmetry of ωµν). We have generalized charges in the form of time-independent tensors of the form

d J µν d3x 0µν ; J µν = 0 ≡ M dt Z The tensors J µν will become the generators of the homogeneous Lorentz group. It is not immediate to guess the explicit expression for the tensor ρµν , because Lorentz transformations act on M the coordinates so that they do not leave the Lagrangian density invariant. Notwithstanding, translation invariance permits to formulate Lorentz invariance as a symmetry of the Lagrangian density under certain transformations on the fields and field derivatives alone (not including coordinate transformations). The transformations on the fields of the form (10.117) acquires the matrix form given by [note that the indices µν are making the role of the index a in Eq. (10.117)] i δΨk = ωµν ( )k Ψm (10.136) 2 Jµν m where the matrices satisfy the Lie algebra of the Lorentz homogeneous group Jµν [ , ]= i g i g i g + i g Jµν Jρσ Jρν µσ − Jσν µρ − Jµσ νρ Jµρ νσ for instance, in the case of a scalar field φ we have δφ = 0 then = 0, while for a covariant vector field we have Jµν λ δVκ = ωκ Vλ (10.137) so the tensors are given by ( ) λ = ig δ λ + ig δ λ (10.138) Jρσ κ − ρκ σ σκ ρ 10.10. LORENTZ INVARIANCE 301

we can check the validity of the tensor (10.138) by calculating δVk using the covariant form of Eq. (10.136) and using the tensor (10.138) i i δV = ωµν ( ) mV = ωµν [ ig δ m + ig δ m] V κ 2 Jµν κ m 2 − µκ ν νκ µ m 1 1 = ωµν [g V g V ]= [g ωµν V g ωµν V ] 2 µκ ν − νκ µ 2 κµ ν − κν µ 1 1 = [g ωµνV + g ωνµV ]= [ω ν V + ω µV ] 2 κµ ν κν µ 2 κ ν κ µ µ δVκ = ωκ Vµ hence the tensor (10.138) generates the correct covariant variation (10.137) of the vector ﬁeld Vκ. Now, the derivative of a ﬁeld that transforms as (10.136) has the same structure of transformation but adding an extra vector index 1 δ (∂ Ψ )= iωµν ( ) m∂ Ψ + ω λ∂ Ψ (10.139) ρ k 2 Jµν k ρ m ρ λ k Since we assume that the Lagrangian density is invariant under the combined transformations (10.136) and (10.139) we have ∂ k ∂ k 0= Lk δΨ + L k δ ∂ρΨ (10.140) ∂Ψ ∂ (∂ρΨ ) substituting (10.136) and (10.139) in (10.140) we have

∂ i µν k m ∂ i µν k m ∂ λ k 0= Lk ω ( µν) mΨ + L k ω ( µν) m∂ρΨ + L k ωρ ∂λΨ (10.141) ∂Ψ 2 J ∂ (∂ρΨ ) 2 J ∂ (∂ρΨ ) µν λ the first two terms are written in terms of ω but the latter is written in terms of ωρ . It is convenient to write the third term in terms of ωµν in order to factorize it. To do it, we rewrite the third term on the right-hand side of Eq. (10.141) as follows 1 1 1 ω λ∂ Ψk = ω ν∂ Ψk + ω µ∂ Ψk = g ωµν∂ Ψk + g ωνµ∂ Ψk ρ λ 2 ρ ν 2 ρ µ 2 ρµ ν ρν µ 1 ω λ∂ Ψk = ωµν (g ∂ g ∂ )Ψk (10.142) ρ λ 2 ρµ ν − ρν µ Replacing (10.142) in (10.141) we can factorize ωµν as we desired i ∂ i ∂ 1 ∂ 0= ωµν L ( )k Ψm + L ( )k ∂ Ψm + L (g ∂ g ∂ )Ψk 2 ∂Ψk Jµν m 2 ∂ (∂ Ψk) Jµν m ρ 2 ∂ (∂ Ψk) ρµ ν − ρν µ ρ ρ ωµν is infinitesimal but otherwise arbitrary, setting the coefficients of ωµν to zero we have

i ∂ k m i ∂ k m 1 ∂ k 0= Lk ( µν) mΨ + L k ( µν) m∂ρΨ + L k (gρµ∂ν gρν∂µ)Ψ (10.143) 2 ∂Ψ J 2 ∂ (∂ρΨ ) J 2 ∂ (∂ρΨ ) − By applying the Euler-Lagrange equations (10.57) and the expression (10.115) for the energy-momentum tensor Tµν equation (10.143) can be rewritten as

i ∂ k m i ∂ k m 1 ∂ k k 0 = ∂ρ L k ( µν) mΨ + L k ( µν) m∂ρΨ + L k gρµ∂νΨ gρν∂µΨ 2 ∂ (∂ρΨ ) J 2 ∂ (∂ρΨ ) J 2 ∂ (∂ρΨ ) − i ∂ 1 ∂ ∂ = ∂ L ( )k Ψm + g L ∂ Ψk g L ∂ Ψk 2 ρ ∂ (∂ Ψk) Jµν m 2 ρµ ∂ (∂ Ψk) ν − ρν ∂ (∂ Ψk) µ ρ ρ ρ i ∂ 1 = ∂ L ( )k Ψm + g (δρ T ρ ) g δρ T ρ ρ 2 ∂ (∂ Ψk) Jµν m 2 µρ ν L− ν − νρ µL− µ ρ i ∂ 1 0 = ∂ L ( )k Ψm + [(g T ) (g T )] ρ 2 ∂ (∂ Ψk) Jµν m 2 µν L− µν − νµL− νµ ρ 302 CHAPTER 10. CANONICAL QUANTIZATION

i ∂ 1 0= ∂ L ( )k Ψm (T T ) (10.144) ρ 2 ∂ (∂ Ψk) Jµν m − 2 µν − νµ ρ which suggest to redeﬁne a new energy-momentum tensor, called the Belinfante tensor i Θµν T µν ∂ Zρµν (10.145) ≡ − 2 ρ ρµν ∂ µν k m ∂ ρν k m ∂ ρµ k m Z L k ( ) mΨ L k ( ) mΨ L k ( ) mΨ (10.146) ≡ ∂ (∂ρΨ ) J − ∂ (∂µΨ ) J − ∂ (∂ν Ψ ) J because of the antisymmetry of ρν, the term Zρµν is clearly antisymmetric in the indices µ and ρ. J µρν ∂ ρν k m ∂ µν k m ∂ µρ k m Z = L k ( ) mΨ L k ( ) mΨ L k ( ) mΨ ∂ (∂µΨ ) J − ∂ (∂ρΨ ) J − ∂ (∂νΨ ) J

∂ µν k m ∂ ρν k m ∂ ρµ k m = L k ( ) mΨ + L k ( ) mΨ + L k ( ) mΨ −∂ (∂ρΨ ) J ∂ (∂µΨ ) J ∂ (∂ν Ψ ) J Zµρν = Zρµν − therefore we have ρµν ∂µ∂ρZ = 0 ρµν µν because ∂µ∂ρ is symmetric in µ, ρ while Z is antisymmetric in those indices. Hence, Θ obeys the same conservation law as T µν i ∂ Θµν ∂ T µν ∂ ∂ Zρµν = ∂ T µν = 0 µ ≡ µ − 2 µ ρ µ µν ∂µΘ = 0 (10.147) Owing to the antisymmetry in µ, ρ we have Z00ν = 0. Hence when we set µ = 0 in Eq. (10.145), the other index runs over space components only i i Θ0ν T 0ν ∂ Zρ0ν = T 0ν ∂ Zj0ν ≡ − 2 ρ − 2 j therefore by using the divergence theorem and taking into account that all ﬁelds vanish at inﬁnity, the derivative term in the Belinfante tensor disappear when we integrate over all space i i Θ0ν d3x T 0ν d3x ∂ Zj0ν d3x = T 0ν d3x Zj0ν dS ≡ − 2 j − 2 j Z Z Z Z Z Θ0ν d3x = T 0ν d3x = P ν (10.148) Z Z from Eqs. (10.115) and (10.148) the time component of P ν yields

P Θ d3x = T d3x = g T µ d3x = g T 0 d3x = 0 ≡ 00 00 0µ 0 00 0 Z Z Z Z 0 3 0 ∂ k 3 3 ∂ ˙ k = T 0 d x = δ0 L k ∂0Ψ d x = d x L Ψ − − L−∂ (∂0Ψ ) − L − ∂Ψ˙ k Z Z Z P = d3x ΠΨ˙ k = H 0 − L − Z we conclude that P 0 H ≡ from which we can regard Θµν as the energy-momentum tensor just as well as T µν. 10.10. LORENTZ INVARIANCE 303

Now we shall show that the Belinfante tensor Θµν is symmetric (in contrast to the tensor T µν). To prove it, we use the deﬁnition (10.145) to evaluate

i ∂ ∂ I Θµν Θνµ = T µν T νµ ∂ L ( µν )k Ψm L ( νµ)k Ψm ≡ − − − 2 ρ ∂ (∂ Ψk) J m − ∂ (∂ Ψk) J m ρ ρ ∂ ∂ ∂ ∂ L ( ρν )k Ψm + L ( ρµ)k Ψm L ( ρµ)k Ψm + L ( ρν )k Ψm −∂ (∂ Ψk) J m ∂ (∂ Ψk) J m − ∂ (∂ Ψk) J m ∂ (∂ Ψk) J m µ ν ν µ the last four terms cancel directly. Now from the antisymmetry of µν and using Eq. (10.144) we obtain J i ∂ I Θµν Θνµ = T µν T νµ ∂ 2 L ( µν )k Ψm = 0 ≡ − − − 2 ρ ∂ (∂ Ψk) J m ρ Consequently, we can see from Eq. (10.144) that the Belinfante tensor is symmetric

Θµν =Θνµ (10.149) in contrast to the tensor T µν. Indeed in gravitational theories is Θµν and not T µν the appropriate energy- momentum tensor. Because of the symmetry of the Belinfante tensor we can obtain another conserved tensor density as λµν xµΘλν xνΘλµ (10.150) M ≡ − it is immediate to see that such a tensor is conserved i.e. leads to a continuity equation by using (10.149, 10.147

∂ λµν = ∂ xµΘλν xν Θλµ = δµΘλν + xµ∂ Θλν δν Θλµ xν∂ Θλµ λM λ − λ λ − λ − λ = δµΘλν δν Θλµ =Θµν Θνµ = 0 λ − λ − ∂ λµν = 0 (10.151) λM So Lorentz invariance leads to another set of conserve charges (time-independent operator)

J µν = 0µν d3x = d3x xµΘ0ν xνΘ0µ (10.152) M − Z Z 10.10.2 Generators and Lie algebra between the homogeneous and inhomogeneous generators Recalling the assignments of generators for the Lorentz group in section 1.8.5, Eqs. (1.126)-(1.129), page 30

P P 1, P 2, P 3 (10.153) ≡ J J 23, J 31, J 12 J , J , J J km = ε J (10.154) ≡ ≡ { 1 2 3} ⇔ kmn n H = P 0 (10.155) K J 01, J 02, J 03 K , K , K J 0i = K (10.156) ≡ ≡ { 1 2 3} ⇔ i we shall show that by assigning the time-independent operators P ν , J µν of Eqs. (10.148, 10.152) as generators we recover the Lie algebra of the inhomogeneous Lorentz group through definitions (10.153)-(10.156). Since the rotation operator it not time-dependent neither has explicit time-dependence it is a constant of motion so that it commutes with the Hamiltonian 1 [H, J]=0 ; J = ε J ij k 2 ijk 304 CHAPTER 10. CANONICAL QUANTIZATION applying the commutation identity (10.112) to the tensor Θ0ν we have 1 1 [P , J ] = ε P , J mk = ε P , d3x xmΘ0k xkΘ0m j i 2 imk j 2 imk j − Z 1 h i i = ε d3x P , xmΘ0k xkΘ0m = ε d3x ∂ xmΘ0k xkΘ0m 2 imk j − 2 imk j − Z Z i h ∂ i ∂ = ε d3x δmΘ0k + xm Θ0k + δkΘ0m xk Θ0m 2 imk j ∂xj j − ∂xj Z i i = ε d3x δmΘ0k + ε d3x δkΘ0m 2 imk j 2 imk j Z Z i ∂ ∂ + ε d3x xm Θ0k xk Θ0m 2 imk ∂xj − ∂xj Z by applying the antisymmetry of εijk we see that the first term cancel each other i i i i A ε d3x δmΘ0k + ε d3x δkΘ0m = ε d3x Θ0k + ε d3x Θ0m ≡ 2 imk j 2 imk j 2 ijk 2 imj Z Z Z Z i i = ε d3x Θ0k ε d3x Θ0m = 0 2 ijk − 2 ijm Z Z then we have i ∂ ∂ [P , J ]= ε d3x xm Θ0k xk Θ0m = iε d3x Θ0k j i 2 imk ∂xj − ∂xj − ijk Z Z where we have integrated by parts in the last step. We finally obtain the correct Lie algebra Eq. (1.133) page 31 [P , J ]= iε P j i − ijk k By contrast, the “boost” generator K J k0 is time-independent but has an explicit time-coordinate k ≡ K J k0 = d3x xkΘ00 x0 Θ0k k ≡ − Z h i that can be rewritten as K d3x xkΘ00 x0 d3x Θ0k k ≡ − Z Z K = tP + d3x x Θ00 (x,t) (10.157) − Z K is a constant so that 0 = K˙ = P + i [H, K] − [H, K] = iP (10.158) − so that we obtain the proper Lie algebra Eq. (1.136). Using again the identity (10.112) we have

[P , K ] = P , tP + d3x xk Θ00 (x,t) = d3x P ,xk Θ00 (x,t) j k j − k j Z Z h i 3 k 00 3 k 00 3 k 00 = i d x ∂j x Θ (x,t) = i d x δj Θ (x,t)+ i d x x ∂jΘ (x,t) Z Z Z h∂ i = i d3x xk Θ00 = iδ d3x Θ00 ∂xj − jk Z Z where we integrate by parts in the last step. So that [P , K ]= iδ H j k − jk 10.10. LORENTZ INVARIANCE 305

10.10.3 Invariance of the S matrix − For most of realistic Lagrangian densities, the operator (10.157) is smooth in the sense explained in Sec. 2.3.1 [see discussion below Eq. (2.94) page 80], that is the interaction terms

iH0t 3 00 iH0t e d x x Θ (x, 0) e− (10.159) Z vanish at t . i.e. the matrix elements between states that are smooth superpositions of energy eigenstates → ±∞ vanish in the limit t . Indeed, the interaction term in Eq. (10.159) must vanish if we want to deﬁne → ±∞ proper “in” and “out” states as well as the S matrix. With the same arguments of section 2.3.1, we can derive − the S matrix Lorentz invariance by using such a smoothness assumption along with the commutation relation − (10.158).

10.10.4 Lie algebra within the homogeneous generators In addition the same arguments were used in section 2.3.1, to show that the remaining commutation relations of J µν between themselves, acquire the apropriate form. To prove it, we should write the operators J ij in terms of fields. We start by using equations (10.145, 10.146) and (10.115) to express the Belinfante tensor Θ0j in terms of fields i i i i Θ0j = T 0j ∂ Zρ0j = T 0 gµj ∂ Zρ0j = T 0 gjj ∂ Zρ0j = T 0 ∂ Zρ0j − 2 ρ µ − 2 ρ j − 2 ρ j − 2 ρ ∂ i ∂ k = δ0 L ∂ Ψk ∂ L 0j Ψm j L− ∂ (∂ Ψk) j − 2 ρ ∂ (∂ Ψk) J m 0 ρ ∂ k ∂ k L ρj Ψm L ρ0 Ψm −∂ (∂ Ψk) J m − ∂ (∂ Ψk) J m 0 j 0j ∂ k i ∂ 0j k m Θ = L ∂jΨ ∂ρ L k mΨ − ˙ k  − 2 ∂ (∂ρΨ ) J ∂ Ψ  ∂ k ∂ k L ρj Ψm L ρ0 Ψm −∂ (∂ Ψk) J m − ∂ (∂ Ψk) J m 0 j then such a tensor is replaced in Eq. (10.152) to obtain J ij in terms of fields

J ij = d3x xiΘ0j xjΘ0i =??? − Z the rotation generators in terms of ﬁelds take the form

ij 3 ∂ i k j k ij k m J = d x L x ∂jΨ + x ∂iΨ i mΨ (10.160) ∂Ψ˙ k − − J Z h i Now, by definition there are no present time-derivatives of the auxiliary fields in the Lagrangian density so that ∂ /∂C˙ r = 0, and the rotation generators do not mix the canonical and auxiliary fields. From those facts we see L that in Eq. (10.160) auxiliary fields are absent. Thus, we can rewrite (10.160) in terms of canonical variables only

n J ij = d3x P xi∂ Qn + xj∂ Qn i ij Qm (10.161) n − j i − J m Z h i From Eq. (10.161) we can also verify the correctness of the commutation relations for rotation generators. For instance, from the canonical commutation relations and Eq. (10.161) we obtain

ij n n ij n m J ,Q (x) = i ( xi∂j + xj∂i) Q (x) mQ (x) (10.162) − − − − J ij ij m J , Pn (x) = i ( xi∂j + xj∂i) Pn (x)+ nPm (x) (10.163) − − J 306 CHAPTER 10. CANONICAL QUANTIZATION from the results (10.162) and (10.163) we can obtain the commutation relations between J ij by themselves and ij with other generators (since all generators are written as functions of Q′s and P ′s). For example, J commutes n ij with H and with PnQ˙ so that it commutes with the Lagrangian L. Thus, the commutator of J with the auxiliary fields (when they exist) must be consistent with the rotational invariance of L. In the absence of auxiliary fields we can apply the same reasoning to the “boosts” generators to prove that the generators P µ and J µν obey the commutation relations of the Poincarégroup. Notwithstanding, when there are auxiliary fields the “boost” matrices mix auxiliary fields with canonical ones (e.g. the components V i and V 0 of a vector field), since boosts mix naturally time-components with spatial components. Hence, in the presence of auxiliary fields the commutation relations of the J i0 between themselves and with other generators should be checked for the specific case. However, it is not necessary for the proof of Lorentz-invariance of the S matrix obtained in section 2.3.1. −

10.11 The transition to the interaction picture

We have already seen in Sec. 10.6.1, that we have to go to the interaction picture to have a theory ready for the use of perturbation theory, and we saw how to construct the structure of the Lagrangian including the free and interaction parts. We shall solve some additional more complicated examples here.

10.11.1 Scalar ﬁeld with derivative coupling

We shall start again with a neutral scalar ﬁeld, but with derivative coupling. Let us assume the Lagrangian density

1 1 = ∂ Φ∂µΦ m2Φ2 J µ∂ Φ (Φ) L −2 µ − 2 − µ −H with J µ being either a c number external current (not to be confused with the conserved currents worked pre- − viously) or a functional of several fields different from Φ (in the latter case we should add terms involving these other fields to the Lagrangian density). The canonical conjugate of Φ yields

∂ ∂ 1 1 1 Π = L = ∂ Φ∂0Φ ∂ Φ∂iΦ m2Φ2 J 0∂ Φ J i∂ Φ (Φ) ˙ ˙ −2 0 − 2 i − 2 − 0 − i −H ∂Φ ∂Φ ∂ 1 = Φ˙ 2 J 0Φ˙ ˙ 2 − ∂Φ Π = Φ˙ J 0 (10.164) −

The Hamiltonian is obtained from the Legendre transformation and using (10.164) to obtain the Hamiltonian in terms of Φ and Π

1 1 H = d3x ΠΦ˙ = d3x Π Π+ J 0 + ∂ Φ∂µΦ+ m2Φ2 + J µ∂ Φ+ (Φ) −L 2 µ 2 µ H Z Z h i 1 1 1 = d3x Π Π+ J 0 Φ˙ 2 + ( Φ)2 + m2Φ2 + J 0Φ+˙ J Φ+ (Φ) − 2 2 ∇ 2 ·∇ H Z 1 2 1 1 H = d3x Π Π+ J 0 Π+ J 0 + ( Φ)2 + m2Φ2 + J 0 Π+ J 0 + J Φ+ (Φ) (10.165) − 2 2 ∇ 2 ·∇ H Z 10.11. THE TRANSITION TO THE INTERACTION PICTURE 307 now, Π and J 0 commute, because J 0 contains other ﬁelds diﬀerent from Φ. From this fact the Hamiltonian can be reorganized it as

1 1 2 1 H = d3x Π2 + ΠJ 0 Π2 J 0 ΠJ 0 + ( Φ)2 − 2 − 2 − 2 ∇ Z 1 2 + m2Φ2 + J 0Π+ J 0 + J Φ+ (Φ) 2 ·∇ H 1 1 2 H = d3x Π2 + ΠJ 0 + J 0 2 2 Z 1 1 + ( Φ)2 + m2Φ2 + J Φ+ (Φ) 2 ∇ 2 ·∇ H This Hamiltonian can be separated into the free and interacting parts as follows

H = H0 + V 1 1 1 H = d3x Π2 + ( Φ)2 + m2Φ2 0 2 2 ∇ 2 Z 1 2 V = d3x ΠJ 0 + J Φ+ J 0 + (Φ) ·∇ 2 H Z as we saw in Sec. 10.6.1, we can go to the interaction picture by simply replacing the set Π, Φ with the set π, φ (which corresponds to evaluate the Hamiltonian at t = 0). Of course, we should do the same replacement in the ﬁelds contained in J µ though we do not indicate it explicitly.

1 1 1 H = d3x π2 (x,t)+ [ φ (x,t)]2 + m2φ2 (x,t) (10.166) 0 2 2 ∇ 2 Z 1 2 V (t) = d3x π (x,t) J 0 (x,t)+ J (x,t) φ (x,t)+ J 0 (x,t) + (φ (x,t)) (10.167) ·∇ 2 H Z The free Hamiltonian H0 coincides with the Hamiltonian (10.73) of Sec. 10.6.1, thus it leads to Eqs. (10.74)- (10.81). Indeed, regardless the form of the total Hamiltonian we should take (10.166) as the free part of it, because it is this Hamiltonian the one that leads to the appropriate expansion (10.77) of the scalar ﬁeld in terms of creation an annihilation operators that obey the commutation relations (10.81). Next, we should substitute π in the interaction Hamiltonian with its value φ˙ in the interaction picture (not with its value φ˙ J 0 in the Heisenberg − picture), so Eq. (10.167) becomes

1 2 V (t)= d3x ∂ φ (x,t) J 0 (x,t)+ J i (x,t) ∂ φ (x,t)+ J 0 (x,t) + (φ (x,t)) 0 i 2 H Z 1 2 V (t)= d3x J µ (x,t) ∂ φ (x,t)+ J 0 (x,t) + (φ (x,t)) (10.168) µ 2 H Z note that the non-invariant term in Eq. (10.168) is precisely the one required to recover Lorentz invariance in the propagator of ∂φ in section 9.7.

10.11.2 Spin one massive vector ﬁeld

We shall start the canonical quantization of the vector field Vµ for a particle of spin one, by writing a quite general Lagrangian density 1 1 1 = α∂ V ∂µV ν β∂ V ∂ν V µ m2V V µ + J V µ (10.169) L −2 µ ν − 2 µ ν − 2 µ µ 308 CHAPTER 10. CANONICAL QUANTIZATION where α,β, m2 are arbitrary constants so far. As before, J is either a c number external current, or an operator µ − that depends on other fields. Once again, in the latter case we should add to the Lagrangian density, terms involving the other fields. By taking the Euler-Lagrange equation for the vector field ∂ ∂ ∂µ L = L ∂ (∂µVν) ∂Vν and applying it to the Lagrangian density (10.169) the left-hand side of the Euler-Lagrange equation yields ∂ ∂ 1 1 A ∂ L = ∂ α∂ V ∂αV β β∂ V ∂βV α 1 ≡ µ ∂ (∂ V ) µ ∂ (∂ V ) −2 α β − 2 α β µ ν µ ν ∂ 1 1 = ∂ α (∂ V ) gαδgβω (∂ V ) β (∂ V ) gαδgβω (∂ V ) µ ∂ (∂ V ) −2 α β δ ω − 2 α β ω δ µ ν ∂ 1 1 = gαδgβω∂ α (∂ V ) (∂ V ) β (∂ V ) (∂ V ) µ ∂ (∂ V ) −2 α β δ ω − 2 α β ω δ µ ν 1 1 = gαδgβω∂ αδαµδβν (∂ V ) α (∂ V ) δµδδνω µ −2 δ ω − 2 α β 1 1 βδαµδβν (∂ V ) β (∂ V ) δµωδνδ −2 ω δ − 2 α β 1 1 A = ∂ αgµδ gνω (∂ V ) α (∂ V ) gαµgβν 1 µ −2 δ ω − 2 α β 1 1 βgµδgνω (∂ V ) β (∂ V ) gαν gβµ −2 ω δ − 2 α β 1 1 1 1 = ∂ α (∂µV ν ) α (∂µV ν) β (∂ν V µ) β (∂ν V µ) µ −2 − 2 − 2 − 2 A = α∂ ∂µV ν β (∂ν ∂ V µ) 1 − µ − µ ∂ ν ν µ A1 ∂µ L = αV β (∂ ∂µV ) (10.170) ≡ ∂ (∂µVν) − − the right-hand side of the Euler-Lagrange equation gives ∂ ∂ 1 ∂ 1 A L = m2V V µ + J µV = m2gµαV V + J µV 2 ≡ ∂V ∂V −2 µ µ ∂V −2 µ α µ ν ν ν 1 1 = m2gµα (δ V + δ V )+ δ J µ = m2 (gναV + gµν V )+ J ν −2 µν α αν µ µν −2 α µ ∂ 2 ν ν A2 L = m V + J (10.171) ≡ ∂Vν − equating the left and right sides of the Euler-Lagrange equation (10.170) and (10.171) we find αV ν β (∂ν∂ V µ)= m2V ν + J ν − − µ − In summary, the Euler-Lagrange equations (10.57) applied to the vector field Vν with the Lagrangian density (10.169) yields αV ν β∂ν (∂ V µ)+ m2V ν = J ν (10.172) − − µ and taking the divergence we have α∂ V ν β∂ ∂ν (∂ V µ)+ m2∂ V ν = ∂ J ν − ν − ν µ ν ν α∂ V ν β (∂ V µ)+ m2∂ V ν = ∂ J ν − ν − µ ν ν 10.11. THE TRANSITION TO THE INTERACTION PICTURE 309

(α + β) ∂ V λ + m2∂ V λ = ∂ J λ (10.173) − λ λ λ m2 ∂ J λ ∂ V λ = λ (10.174) − (α + β) λ −(α + β)

λ 2 which provides the equation for an ordinary scalar field (∂λV is clearly scalar) with mass m / (α + β) and source λ ∂λJ / (α + β). We shall describe a particle of spin one (not of spin zero). Thus, we can avoid the appearance of an independently propagating scalar field term ∂ V λ by taking α = β. In that case we cannot divide by (α + β) λ − and according with Eq. (10.173) we have 2 λ λ m ∂λV = ∂λJ therefore, when α = β, the term ∂ V λ can be written in terms of an external current or other fields, as ∂ J λ/m2. − λ λ The Lagrangian density (10.169) becomes

1 1 1 = α∂ V ∂µV ν + α∂ V ∂νV µ m2V V µ + J V µ (10.175) L −2 µ ν 2 µ ν − 2 µ µ

2 so that the scalar α can be absorbed in the normalization of the vector field Vµ (recall that m by now is just a coefficient and Jµ can also be normalized), then we can define

α = β = 1 (10.176) − so that the Lagrangian density becomes

1 1 1 = (∂ V ) ∂µV ν + (∂ V ) ∂ν V µ m2V V µ + J V µ (10.177) L −2 µ ν 2 µ ν − 2 µ µ and taking into account that indices µ and ν are dummy

1 1 = [2 (∂ V ) ∂µV ν 2 (∂ V ) (∂µV ν)] m2V V µ + J V µ L −4 µ ν − ν µ − 2 µ µ 1 1 = [(∂ V ) ∂µV ν (∂ V ) (∂µV ν ) (∂ V ) ∂µV ν + (∂ V ) ∂µV ν] m2V V µ + J V µ −4 µ ν − ν µ − ν µ µ ν − 2 µ µ 1 1 = [(∂ V ) ∂µV ν (∂ V ) (∂ν V µ) (∂ V ) ∂µV ν + (∂ V ) ∂ν V µ] m2V V µ + J V µ −4 µ ν − µ ν − ν µ ν µ − 2 µ µ 1 1 = (∂ V ∂ V ) (∂µV ν ∂ν V µ) m2V V µ + J V µ L −4 µ ν − ν µ − − 2 µ µ so the Lagrangian density yields

1 1 = F F µν m2V V µ + J V µ (10.178) L −4 µν − 2 µ µ F ∂ V ∂ V (10.179) µν ≡ µ ν − ν µ it is clear that F µν is an antisymmetric tensor. Under the assumption (10.176) the Euler-Lagrange equations become

∂ ∂µV ν + (∂ν ∂ V µ)+ m2V ν = J ν − µ µ ∂ [(∂νV µ) ∂µV ν ]+ m2V ν = J ν µ − ∂ F µν + m2V ν = J ν (10.180) − µ 310 CHAPTER 10. CANONICAL QUANTIZATION

µν The derivative of the tensors F and Fµν with respect to V˙µ yields

∂Fαβ ∂ = (∂αVβ ∂βVα)= δ0αδµβ δ0βδµα ∂V˙µ ∂ (∂0Vµ) − − αβ ∂F ∂ α β β α αδ βγ ∂ = ∂ V ∂ V = g g (∂δVγ ∂γ Vδ) ∂V˙µ ∂ (∂0Vµ) − ∂ (∂0Vµ) − αβ ∂F αδ βγ = g g (δ0δδµγ δ0γ δµδ) ∂V˙µ − ∂F αβ = gα0gβµ gαµgβ0 ∂V˙µ − thus, the derivative of the Lagrangian density (10.178) with respect to V˙µ gives

∂ 1 ∂ 1 ∂F αβ 1 ∂F L = F F αβ = F αβ F αβ ˙ ˙ αβ αβ ˙ ˙ ∂Vµ −4 ∂Vµ −4 ∂Vµ − 4 ∂Vµ ! 1 1 = F gα0gβµ gαµgβ0 (δ δ δ δ ) F αβ −4 αβ − − 4 0α µβ − 0β µα ∂ 1 1 L = F 0µ F µ0 F 0µ F µ0 ∂V˙µ −4 − − 4 − and taking into account the antisymmetry of F µν we ﬁnally obtain

∂ L = F 0µ (10.181) ∂V˙µ − which is non-zero for spatial components µ i. Thus, V i are canonical ﬁelds, and their conjugates are given by ≡

i ∂ 0i i0 i 0 0 i 0 i Π = L = F = F = ∂ V ∂ V = ∂iV + ∂0V ∂V˙ i − −

i i0 i 0 Π = F = V˙ + ∂iV (10.182) However, because of the antisymmetry, F 00 = 0. Consequently, V˙ 0 does not appear in the Lagrangian density, so that V 0 is an auxiliary field. The fact that ∂ /∂V˙ 0 = 0, implies that the field equation for V 0 does not contain L second time-derivatives, from which it can be used as a constraint that eliminates the field variable. For ν = 0, the Euler-Lagrange equation (10.180) gives

∂ F µ0 + m2V 0 = J 0 − µ ∂ F 00 ∂ F i0 = m2V 0 + J 0 − 0 − i − so we have ﬁnally ∂ F i0 = m2V 0 J 0 (10.183) i − combining Eqs. (10.183) and Eq. (10.182) we ﬁnd

1 1 V 0 = ∂ F i0 + J 0 = ∂ Πi + J 0 m2 i m2 i 1 V 0 = Π+~ J 0 (10.184) m2 ∇ · 10.11. THE TRANSITION TO THE INTERACTION PICTURE 311 showing again the fact that V 0 is not an independent variable. Now we calculate the Hamiltonian. For this, we ﬁrst observe that Eqs. (10.182) and (10.184) permits to express V˙ in terms of Π~ and J 0 1 V˙ = Π~ V 0 = Π~ Π+~ J 0 (10.185) −∇ − m2 ∇ ∇ · Now, we intend to evaluate the Hamiltonian. We start by reexpressing the Lagrangian density (10.178) in terms of Πi and V i (canonical variables) and eliminating V 0. We do this by using Eqs. (10.184, 10.182) as followsevaluating the quantity ——————————–

1 1 = F F µν m2V V µ + J V µ L −4 µν − 2 µ µ 1 1 1 1 = F F 0ν F F iν m2V V 0 m2V V i + J V 0 + J V i −4 0ν − 4 iν − 2 0 − 2 i 0 i 1 1 1 1 2 1 = F F 0i F F i0 F F ij + m2 V 0 m2V2 + J V 0 + J V −4 0i − 4 i0 − 4 ij 2 − 2 0 · 1 1 1 2 1 J0 = Π Πi F F ij + Π+~ J 0 m2V2 + Π+~ J 0 + J V −2 i − 4 ij 2 ∇ · − 2 m2 ∇ · · 1 1 1 2 1 J 0 = Π~ 2 F F ij + Π+~ J 0 m2V2 Π+~ J 0 + J V L −2 − 4 ij 2 ∇ · − 2 − m2 ∇ · · we can show that 1 1 F F ij = ( V)2 −4 ij −2 ∇× and the Lagrangian density yields

1 1 1 2 1 J 0 = Π~ 2 ( V)2 + Π+~ J 0 m2V2 Π+~ J 0 + J V (10.186) L −2 − 2 ∇× 2 ∇ · − 2 − m2 ∇ · · ——————————– using Eqs. (10.185) and (10.186), the Hamiltonian becomes

1 H = d3x Π~ V˙ = d3x Π~ Π~ Π+~ J 0 −L − m2 ∇ ∇ · −L Z Z 1 1 1 = d3x Π~ 2 + Π~ Π+~ J 0 Π~ 2 + ( V)2 m2 ∇ · ∇ · − 2 2 ∇× Z 1 1 2 1 + m2V2 Π+~ J 0 J V+ J 0 Π+~ J 0 2 − 2m2 ∇ · − · m2 ∇ · then we split it into the free and interacting part

H = H0 + V then we pass to the interaction picture by replacing the quantities V, Π~ (in the Heisenberg picture) with the corresponding interaction picture variables v and ~π (the same replacements should be done for the ﬁelds contained in J µ).

1 1 1 1 H = d3x ~π2 + ( ~π)2 + ( v)2 + m2v2 (10.187) 0 2 2m2 ∇ · 2 ∇× 2 Z 1 1 2 V = d3x J v + J 0 ~π + J 0 (10.188) − · m2 ∇ · 2m2 Z 312 CHAPTER 10. CANONICAL QUANTIZATION and the relation between ~π and v is given by

δH (v, ~π) 1 ˙v = 0 = ~π ( ~π) (10.189) δ~π − m2 ∇ ∇ · while the field equation yields δH (v, ~π) ~π· = 0 = 2v ( v) m2v (10.190) − δv ∇ −∇ ∇ · − as for V 0, since it is not an independent field variable, it is not related with an interaction picture field v0 through a similarity transformation. It is convenient to define the following quantity

~π v0 = ∇ · (10.191) m2 then combining Eqs. (10.189) and (10.191) we can express ~π in the form

~π ~π = ˙v + ∇ · ∇ m2 ~π = ˙v + v0 (10.192) ∇ Thus, we see that by deﬁning v0 as in Eq. (10.191) we obtain a relation (10.192) between ~π and ˙v similar to the associated relation in the Heisenberg picture Eq. (10.182). Taking divergence in Eq. (10.192) and using Eq. (10.191) we have

~π = ˙v + 2v0 (10.193) ∇ · ∇ · ∇ m2v0 = ˙v + 2v0 (10.194) ∇ · ∇ on the other hand, deriving (10.192) with respect to time

~π· = v·· + v˙0 ∇ and inserting it in Eq. (10.190) we obtain

v·· + v˙0 = 2v ( v) m2v ∇ ∇ −∇ ∇ · − therefore, the ﬁeld equations become

2v0 + ˙v m2v0 = 0 (10.195) ∇ ∇ · − 2v ( v) v·· v˙0 m2v = 0 (10.196) ∇ −∇ ∇ · − −∇ − that can be combined to be written in the covariant form. To do it, we rewrite Eq. (10.195) as

2v0 + ˙v m2v0 = 0 ∇ ∇ · − ∂i∂ v0 + ∂ ∂ vi m2v0 = 0 i i 0 − ∂i∂ v0 + ∂0∂ v0 ∂0∂ v0 ∂0∂ vi m2v0 = 0 i 0 − 0 − i − ∂ν ∂ v0 ∂0∂ vν m2v0 = 0 ν − ν − v0 ∂0∂ vν m2v0 = 0 (10.197) − ν − 10.11. THE TRANSITION TO THE INTERACTION PICTURE 313 and the k th component of Eq. (10.196) is rewritten as − .. 2vk vk ∂kv˙0 ∂k ( v) m2vk = 0 ∇ − − − ∇ · − ∂ ∂ivk ∂ ∂ vk ∂k∂ v0 ∂k∂ vi m2vk = 0 i − 0 0 − 0 − i − ∂ ∂ivk + ∂ ∂0vk ∂ ∂kv0 ∂ ∂kvi m2vk = 0 i 0 − 0 − i − ∂ ∂νvk ∂ ∂kvν m2vk = 0 ν − ν − vk ∂k∂ vν m2vk = 0 (10.198) − ν − thus Eqs. (10.197) and (10.198) can be written in a single covariant equation as

vµ ∂µ∂ vν m2vµ = 0 (10.199) − ν − taking the divergence we have

∂ vµ ∂ ∂µ∂ vν m2∂ vµ = 0 µ − µ ν − µ ∂ vµ ∂ vν m2∂ vµ = 0 µ − ν − µ ∂ vµ ∂ vµ m2∂ vµ = 0 µ − µ − µ which leads to µ ∂µv = 0 (10.200) from which the second term in Eq. (10.199) vanishes. Thus such an equation becomes

m2 vµ = 0 (10.201) − a real vector ﬁeld that obeys Eqs. (10.200) and (10.201) can be expanded as a Fourier transform in the form

3 µ 3/2 d p µ ip x µ ip x v (x)=(2π)− e (p,σ) a (p,σ) e · + e ∗ (p,σ) a† (p,σ) e− · (10.202) 0 σ 2p X Z n o p with p0 = p2 + m2; the terms eµ (p,σ) with σ = 1, 0, 1 are three independent vectors with the constraint [see − Eq. (6.76) page 173] p µ pµe (p,σ) = 0 (10.203) and normalized in the form [see Eqs. (6.69, 6.75) page 171]

µ ν µ ν µν p p e (p,σ) e ∗ (p,σ)= g + (10.204) m2 σ X and a (p,σ) , a† (p,σ) are operator coeﬃcients to be determined. From Eqs. (10.192), (10.202) and (10.204) we can check that v and ~π follow the correct commutation relations

vi (x,t) , πj (y,t) = iδ δ3 (x y) ij − i j i j v (x,t) , v (y,t) = π (x,t) , π (y,t) = 0 as long as the operators a (p,σ) , a† (p,σ) satisfy the commutation relations

3 a (p,σ) , a† q,σ′ = δ ′ δ (p q) σσ − h i a (p,σ) , aq,σ′ = a† (p,σ) , a† q,σ′ = 0. h i 314 CHAPTER 10. CANONICAL QUANTIZATION since we already knew that the vector ﬁeld for a neutral spin one particle must have the form (10.202), the derivation of our present results permits to verify that Eq. (10.187) provides a correct form for the free Hamiltonian for a massive particle of spin one. It could also be checked that the Hamiltonian (10.187) can be written up to a constant term in the standard form for a free-particle energy, i.e. as (homework!! B12)

3 0 H0 = d p p a† (p,σ) a (p,σ) σ X Z Further, applying Eq. (10.191) in Eq. (10.167) gives us the interaction Hamiltonian in the interaction picture

1 2 V (t)= d3x J vµ + J 0 (10.205) − µ 2m2 Z the non-invariant term in Eq. (10.205) is the one we needed to cancel the non-invariant term coming from the propagator of the vector ﬁeld [see Eqs. (9.73), (9.74), and (9.75) on page 247, and discussion around them].

10.11.3 Dirac Fields of spin 1/2 For a Dirac ﬁeld of spin 1/2, we shall start with the trial Lagrangian density

= Ψ¯ (γµ∂ + m)Ψ Ψ, Ψ¯ L − µ −H where Ψ, Ψ¯ is a real function of Ψ and Ψ.¯ The Lagrangian is not real but the action is because H µ µ µ µ A Ψ¯ γ ∂ Ψ Ψ¯ γ ∂ Ψ † = Ψ¯ γ ∂ Ψ (∂ Ψ)† (γ )† Ψ¯ † ≡ µ − µ µ − µ µ µ † µ µ = Ψ¯ γ ∂ Ψ (∂ Ψ)† [ βγ β] Ψ†β = Ψ¯ γ ∂ Ψ+ ∂ Ψ†βγ β (βΨ) µ − µ − µ µ µ µ 2 = Ψ¯ γ ∂µΨ+ ∂µ Ψ†β γ β Ψ µ µ µ = Ψ¯ γ ∂µΨ+ ∂µΨ¯ γ Ψ= ∂µ Ψ¯ γ Ψ therefore the field equations obtained from the stationary condition for the action with respect to Ψ¯ are the adjoints of the equations obtained applying the stationary condition with respect to Ψ. Thus both sets of equations are not independent as we require to avoid too many field equations that overdetermine the problem. The canonical conjugate of Ψ yields ∂ Π= L = Ψ¯ γ0 (10.206) ∂Ψ˙ − we should not consider Ψ¯ as a field like Ψ but as proportional to the canonnical conjugate momentum of Ψ. The Hamiltonian is given by

H = d3x ΠΨ˙ = d3x Πγ0 [γ + m]Ψ+ −L ·∇ H Z Z h i as customary we should split it in a free and interacting Hamiltonian

H = H0 + V (10.207) H = d3x Πγ0 [γ + m] Ψ ; V = d3x (10.208) 0 ·∇ H Z Z as we know, the next step is to pass to the interaction picture. We observe that Eq. (10.206) does not involve the time, thus the similarity transformation (10.39), (10.40) gives

π = ψγ¯ 0 − 10.11. THE TRANSITION TO THE INTERACTION PICTURE 315

in the same way, H0 and V (t) can be calculated by substituting Ψ, Π by ψ, π in Eqs. (10.208). From this we obtain the equation of motion ∂H ψ˙ = 0 = γ0 (γ + m) ψ δπ ·∇ or equivalently µ (γ ∂µ + m) ψ = 0 (10.209) as explained above, the other equation of motion

δH π˙ = 0 − δψ provides just the adjoint of Eq. (10.209). A solution of Eq. (10.209) can be expanded as a Fourier transform

1 ψ (x) = d3p u (p,σ) eip x a (p,σ)+ v (p,σ) e ip xb (p,σ) (10.210) 3/2 ·· − · † (2π) σ Z X n o p0 p2 + m2 ≡ p where a (p,σ) and b (p,σ) are operator coefficients, u p, 1 are two independent solutions of † ± 2 µ (iγ pµ + m) u (p,σ) = 0 (10.211) similarly v p, 1 are two independent solutions of ± 2 ( iγµp + m) v (p,σ) = 0 (10.212) − µ The matrix iγµp has eigenvalues m. Consequently, the bilinears uu¯ and vv¯ must be proportional to µ ± the projection matrices P P ( iγµp + m) (iγµp + m) u (p,σ)u ¯ (p,σ) ∝ − µ ; v (p,σ)v ¯ (p,σ) ∝ µ 2m 2m σ σ X X thus the proportionality factor can be fit up to a sign by absorbing it into the definition of u (p,σ) and v (p,σ). Then we can fit the sign by a positivity condition

T r u (p,σ)u ¯ (p,σ) β = u† (p,σ) u (p,σ) β > 0 σ σ X X and similarly for v (p,σ). We then will normalize the spinors as

( iγµp + m) u (p,σ)u ¯ (p,σ) = − µ 2p0 σ X (iγµp + m) v (p,σ)v ¯ (p,σ) = µ − 2p0 σ X

Once again we can determine the properties of the undetermined operators a (p,σ) and b† (p,σ) by imposing the canonical conditions to the ﬁelds ψ and canonical conjugates π. So in order to obtain the canonical anticommutation relations

ψ (x,t) , ψ¯ (y,t) = [ψ (x,t) ,π (y,t)] γ0 = i γ0 δ3 (x y) α β + α γ + γβ αβ − [ψα (x,t) , ψβ (y,t)]+ = 0 316 CHAPTER 10. CANONICAL QUANTIZATION

we must ﬁt the anticommutation relations for a (p,σ) , a† (p,σ) and b (p,σ) , b† (p,σ) as follows

3 a (p,σ) , a† q,σ′ = b (p,σ) , b† q,σ′ = δ (q p) δσσ′ + + − h i h i a (p,σ) , a q,σ′ + = b (p,σ) , b q,σ′ + = 0

a (p,σ) , b q,σ′ = a (p,σ) , b† q,σ′ = 0 + + h i and their adjoints. These developments are in agreement with the results obtained in chapter 7. Hence, it conﬁrms the validity of H0 in Eq. (10.208) as a free Hamiltonian for Dirac ﬁelds of spin 1/2. In terms of the operators a (p,σ) and b (p,σ) the Hamiltonian reads (homework!! B13)

3 0 H = d p p a† (p,σ) a (p,σ) b (p,σ) b† (p,σ) (10.213) 0 − σ X Z h i once again we can rewrite it as a more standard (normal ordered) Hamiltonian plus an inﬁnite c number −

3 0 H0 = d p p a† (p,σ) a (p,σ)+ b† (p,σ) b (p,σ) (10.214) σ X Z h i the c number divergent term in Eq. (10.214) only matters in graviational theories. For our case like in the case − of the scalar theory, this is just a shift of the zero of energy. When we drop out the c number, H is a positive − 0 operator as it was the case for bosons.

10.12 Constraints and Dirac Brackets

Since the starting point is usually the Lagrangian, we should confront the task of passing to the Hamiltonian, which implies to pass to the system of canonical variables Q, P . This task is particularly difficult when we have constraints. We shall use the analysis of Dirac to manage this problem. As a matter of example we shall use the vector field theory described by Lagrangian (10.178), and we shall use the same analysis in more complex (and realistic) problems later. Primary constraints appear in two cases (a) when we impose a constraint to the system (this is the case when we choose a particular gauge in any gauge field theory), or (b) when the constraint comes from the structure of the Lagragian itself. For this second case we shall take as an example the Lagrangian (10.178) that describes a µ real vector field V of spin one, coupled to a current Jµ 1 1 = F F µν m2V V µ + J V µ (10.215) L −4 µν − 2 µ µ F ∂ V ∂ V (10.216) µν ≡ µ ν − ν µ Let us try to work with the four components on V µ on the same foot. We then define an extension of the conjugates (10.182) ∂ Πµ L = F 0µ ≡ ∂ (∂0Vµ) − because of the antisymmetry of the F µν tensor we arrive to the primary constraint

Π0 = 0 (10.217)

In general we get a primary constraint when the equations δL Πk = k δ (∂0Ψ ) 10.12. CONSTRAINTS AND DIRAC BRACKETS 317

k k cannot be solved to obtain all the ∂0Ψ (at least locally) in terms of Πk and Ψ . The necessary and sufficient condition for it, is that the matrix δ2L k m δ (∂0Ψ ) δ (∂0Ψ ) be singular i.e. its determinant must be zero. Lagrangians of this kind are called irregular. On the other hand, we have secondary constraints which come from the condition that the primary constraints be consistent with the equations of motion. In the case of the massive vector field, it is the Euler-Lagrange equation (10.183) for V 0 ∂ Π = m2V 0 J 0 (10.218) i i − these are all constraints we shall encounter in the present example. However, in other theories further constraints could appear by requiring consistency of the secondary constraints will field equations, and so on. Nevertheless, the distinction between primary, secondary and other constraints will not be relevant in our present approach. Let us assume a Lagrangian L Ψ, Ψ˙ that depends on a set of variables Ψa (t) and time-derivatives Ψ˙ a (t). The Lagrangians of a Field theoryh are ai special case in which the index a runs over all pairs of k and x. For all these variables we can define canonical conjugate momenta as

∂L Πa ≡ ∂Ψ˙ a The set of variables Π and Ψ are not in general independent but can be connected by several constraints equations, both primary or secondary (Hamilton’s approach requires that the canonical variables be independent so that we should get rid of the constrained variables). We deﬁne the Poisson bracket as

∂A ∂B ∂B ∂A [A, B]P a a ≡ ∂Ψ ∂Πa − ∂Ψ ∂Πa

a where we ignore the constraints in calculating the derivatives with respect to Ψ and Πa (the calculation of a partial derivative implies to ignore constraints, since we are moving each single variable without moving the others). It is clear that a a [Ψ , Πb]P = δb once again we emphasize that in a ﬁeld theory this delta has kronecker deltas for discrete indices and a Dirac delta for positions. In evaluating the Poisson bracket, all ﬁelds are evaluated at the same time, thus we omit the time argument. The Poisson brackets possess the same algebraic properties of commutator including the Jacobi identity (Homework!! B14)

[A, B] = [B, A] (10.219) P − P [A,BC]P = B [A, C]P + [A, B]P C (10.220)

[A, [B,C]P ]P + [B, [C, A]P ]P + [C, [A, B]P ]P = 0 (10.221) for example property (10.220) can be veriﬁed as follows

∂A ∂ (BC) ∂ (BC) ∂A ∂A ∂C ∂B ∂C ∂B ∂A [A,BC] = = B + C B + C P ∂Ψa ∂Π − ∂Ψa ∂Π ∂Ψa ∂Π ∂Π − ∂Ψa ∂Ψa ∂Π a a a a a ∂A ∂C ∂A ∂B ∂C ∂A ∂B ∂A = B + C B + C ∂Ψa ∂Π ∂Ψa ∂Π − ∂Ψa ∂Π ∂Ψa ∂Π a a a a ∂A ∂C ∂C ∂A ∂A ∂B ∂B ∂A = B + C ∂Ψa ∂Π − ∂Ψa ∂Π ∂Ψa ∂Π − ∂Ψa ∂Π a a a a [A,BC]P = B [A, C]P + [A, B]P C 318 CHAPTER 10. CANONICAL QUANTIZATION as usual we deﬁne a function of operators as the corresponding series expansion of the ordinary function

∞ F ( ) f n n O ≡ n O n=0 X and its derivative with respect to as O dF ( ) ∞ n 1 O nf − d ≡ nO O nX=0 we can deﬁne functions and derivatives of two operators f (A, B) accordingly. If we have two operators and O1 that commutes with their commutators that is O2 [ , [ , ]] = [ , [ , ]] = 0 O1 O1 O2 O2 O1 O2 we ﬁnd the property

dF ( ) dG ( ) [ , F ( )] = [ , ] O2 ; [G ( ) , ] = [ , ] O1 O1 O2 O1 O2 d O1 O2 O1 O2 d O2 O1 a very important particular case is the following

[ , ]= αI O1 O2 which is clearly the case when we have canonical variables. In that case we have

dF ( ) dG ( ) [ , F ( )] = α O2 ; [G ( ) , ]= α O1 O1 O2 d O1 O2 d O2 O1 ????? ——————————- From the previous results ???? if we were able to adopt the usual commutation relations

a a a b [Ψ , Πb]= δb ; Ψ , Ψ = [Πa, Πb] = 0 h i the commutator of any two functions of Ψ′s and Π′s would satisfy

[A, B]= i [A, B]P but the constraints do not permit it in general. We can generically express the constraints in the form

χN (Ψ, Π) = 0 (10.222) where χN is a set of functions of the canonical variables. Since we are including both primary and secondary constraints, the set (10.222) of all constraints is necessarily consistent with the equations of motion ˙ A = [A, H]P so that

χ˙ N = [χN ,H]P from which

[χN ,H]P = 0 (10.223) 10.12. CONSTRAINTS AND DIRAC BRACKETS 319

when we have the constraint χN = 0. A constraint is of first class if its poisson bracket with all the other constraints vanishes when (after calculating the Poisson bracket) we impose the constraint. We shall see an example in quantum electrodynamics where the first class constraint comes from gauge invariance which is a symmetry of the action. In general the set of first class constraints χN = 0 is always associated with a group of symmetries, from which any quantity A undergoes the infinitesimal transformation δ A ε [χ , A] N ≡ N N P XN recalling that in field theory the index N has a space-time coordinate, these theories contain local transformations. From Eq. (10.223) it can be seen that this transformation leaves the Hamiltonian invariant, and for first class constraints it respects all other constraints as well. First class constraints can be dropped out by an appropriate gauge choice. When all first class constraints are dropped out by choicing the adequate gauge, the remaining constraints χN = 0 satisfy the condition that any non-trivial linear combination of the Poisson brackets between each other

uN [χN ,χM ]P XN does not vanish. Therefore, the matrix associated with the Poisson brackets of the remaining constraints must have non-null determinant

det C = 0 ; C [χ ,χ ] 6 NM ≡ N M P such a matrix is then non-singular. These types of constraints are second class constraints. It is clear that CNM is antisymmetric. Besides, an antisymmetric matrix of odd dimensionality has null determinant4. Consequently, the number of second class constraints must be even. We have seen in Eqs. (10.217) and (10.218) that for the case of the massive vector ﬁeld the constraints are given by χ1x = χ2x = 0 (10.224) with 2 0 0 χ x = Π (x) ; χ x = ∂ Π (x) m V (x)+ J (x) 1 0 2 i i − the Poisson brackets of these constraints read

2 0 0 C x y = [χ x,χ y] = Π (x) , ∂ Π (y) m V (y)+ J (y) 1 ,2 1 2 P 0 i i − P 2 0 0 2 0 0 ∂Π0 (x) ∂ ∂iΠ i (y) m V (y)+ J (y) ∂ ∂iΠi (y) m V (y)+ J (y) ∂Π0 (x) = µ − − µ ∂V ∂Πµ − ∂V ∂Πµ (y) 2 0 0 0 ∂ ∂iΠi (y) m V (y)+ J (y) ∂Π0 (x) 2 ∂V (y) ∂Π0 (x) = − µ = m 0 − ∂V ∂Πµ (y) ∂V (y) ∂Π0 (y) 2 3 C x y = C y x = [χ x,χ y] = m δ (x y) (10.225) 1 ,2 − 2 ,1 1 2 P − C1x,1y = C2x,2y = 0 (10.226) 0 m2δ3 (x y) C = − (10.227) m2δ3 (x y) 0 − − 4For an antisymmetric matrix we have

det A = det −A =(−1)n det A det A = (−1)n dete A e which for n even becomes a triviality and for n odd gives det A = − det A, so that the determinat becomes zero. 320 CHAPTER 10. CANONICAL QUANTIZATION

It is clear that this “matrix” is non-singular5. Then, the constraints (10.224) are of second class. Dirac suggested that when all constraints are of second class, the commutation relations satisfy the property

[A, B]= i [A, B]D (10.228)

6 where [A, B]D is a generalization of the Poisson bracket, known as the Dirac bracket

1 NM [A, B] [A, B] [A, χ ] C− [χ ,B] (10.229) D ≡ P − N P M P We recall again that the indices N and M include the position in space, taking values of the form 1, x and 2, x in the example of the massive vector ﬁeld. The Dirac brackets satisfy the same basic algebraic properties of commutators and Poisson brackets, that is (homework!! B15)

[A, B] = [B, A] D − D [A,BC]D = [A, B]D C + B [A, C]D

[A, [B,C]P ]P + [B, [C, A]P ]P + [C, [A, B]P ]P = 0 now setting A = χN in the deﬁnition (10.229) we have

1 QM [χ ,B] [χ ,B] [χ ,χ ] C− [χ ,B] N D ≡ N P − N Q P M P NQ 1 QM = [χN ,B] C C− [χM ,B] P − P = [χ ,B] δNM [χ ,B] = 0 N P − M P so we obtain the additional relations

[χN ,B]D = 0 (10.230) property (10.230) makes the commutation relations (10.228) consistent with the constraints χN = 0. It can also be shown that the Dirac brackets are left invariant when we substitute χN with any functions χN′ for which the equations χN′ = 0 and χN = 0 generate the same submanifold of phase space. But none of the previous properties prove Dirac’s conjecture (10.228). The conjecture is illuminated by a theorem proved by Maskawa and Nakajima. They proved that for any set a of canonical variables Ψ and Πa governed by second class constraints, it can always be constructed by a canonical transformation7 two sets of variables Qn, r and their corresponding canonical conjugates P , , such that the Q n Pr constraints give r = = 0 Q Pr n 8 while Q and Pn are unconstrained canonical variables . Now, using these coordinates to calculate the Poisson brackets, and redeﬁning the constraint functions as

χ = r ; χ = 1r Q 2r Pr we obtain

5 Strictly speaking, this matrix has continuous entries. We see it by recalling that CNM corresponds to indices N and M in which N contains a discrete index and a position index. In Eq. (10.227), we see that the non-diagonal submatrices are proportional to the identity in the continuum. 6Note that the deﬁnition (10.229) requires that the constraints be of second class for C−1 to exist. 7A canonical transformation keeps the poisson bracket invariant as well as the structure of Hamilton’s equations. 8In other words exists a canonical transformation for which a given subset of the canonical variables absorbs all the constraints when all cosntraints are of second class. 10.12. CONSTRAINTS AND DIRAC BRACKETS 321

C = [ r, ] = δr ; C = C 1r,2s Q Ps P s 2s,1r − 1r,2s C = [ r, s] = 0 ; C = [ , ] = 0 1r,1s Q Q P 2r,2s Pr Ps P 0 δ (x y) 1 0 δ (x y) C = − ; C− = − − = C δ (x y) 0 δ (x y) 0 − − − − r where we have taken into account that δs implies a kronecker delta and a Dirac delta in the position. For arbitrary functions A and B of the complete set of canonical variables Qn, r and P , we obtain Q n Pr ∂A ∂ r ∂ r ∂A ∂A ∂ r ∂ r ∂A [A, χ ] = [A, r] = Q Q + Q Q 1r P Q P ∂Qn ∂P − ∂Qn ∂P ∂ s ∂ − ∂ s ∂ n n Q Ps Q Ps ∂ r ∂A ∂A ∂A = Q = δr = −∂ s ∂ − s ∂ −∂ Q Ps Ps Pr 9 and similarly for [A, χ2r]P then for arbitrary functions A and B of the canonical variables we have ∂A ∂A [A, χ ] = ; [A, χ ] = (10.231) 1r P −∂ 2r P ∂ r Pr Q This C matrix is non-singular since C 1 = C, as it must be for second class constraints. Hence, the Dirac − − − brackets (10.229) give here

1 NM [A, B] [A, B] [A, χ ] C− [χ ,B] D ≡ P − N P M P 1 1r,1s 1 1r,2s = [A, B] [A, χ1r] C− [χ1s,B] [A, χ1r] C− [χ2s,B] P − P P − P P 1 2s,1r 1 2r,2s [A, χ2s] C− [χ1r,B] [A, χ2r] C− [χ2s,B] − P P − P P [A, B] = [A, B] + [A, χ ] δr [χ ,B] [A, χ ] δr [χ ,B] D P 1rP s 2s P − 2s P s 1r P [A, B] = [A, B] + [A, χ ] [χ ,B] [A, χ ] [χ ,B] (10.232) D P 1r P 2r P − 2r P 1r P and applying Eqs. (10.231) to Eq. (10.232) we have

∂A ∂B ∂A ∂B [A, B]D = [A, B]P + r r ∂ r ∂ − ∂ ∂ r ∂AP ∂BQ ∂BQ ∂AP [A, B] = [A, B] + (10.233) D P − ∂ r ∂ ∂ r ∂ Q Pr Q Pr taking into account that the Poisson bracket must be taken with respect to the whole system of canonical variables Qn, r and P , ; we ﬁnd Q n Pr ∂A ∂B ∂B ∂A ∂A ∂B ∂B ∂A ∂A ∂B ∂B ∂A [A, B] = + + D ∂Qn ∂P − ∂Qn ∂P ∂ r ∂ − ∂ r ∂ − ∂ r ∂ ∂ r ∂ n n Q Pr Q Pr Q Pr Q Pr ∂A ∂B ∂B ∂A [A, B]D = n n (10.234) ∂Q ∂Pn − ∂Q ∂Pn In words we can say that the Dirac bracket coincides with the Poisson bracket calculated in terms of the reduced n set of unconstrained canonical variables Q , Pn. It shows the utility of the Dirac brackets, since the Dirac bracket of two arbitrary functions A, B can be evaluated as a Poisson bracket and only with the independent degrees of n freedom Q , Pn.

9In this point we emphasize again that by taking partial derivatives we are ignoring the constraints even for the set Q, P. 322 CHAPTER 10. CANONICAL QUANTIZATION

Now, if we assume that the unconstrained canonical variables satisfy the canonical commutation relations, we see that the commutators of general operators A, B are given by Eq. (10.228) in terms of the Dirac brackets. It is important however, to say that it is not clear yet whether we should adopt canonical commutation relations n for the unconstrained variables Q and Pn constructed from theMaskawa-Nakajima canonical transformation. Indeed the test of such canonical commutation relations is their consistency with the free-field commutation relations derived for scalar, vector and Dirac fields (see chapters 5, 6, 7, 8). Nevertheless, in order to apply this n test it is necessary toknow what the canonical variables Q and Pn are. We shall restrict to show it for some classes of theories. We shall also see that for these classes of theories the Hamiltonian expressed in terms of the unconstrained Ψ′s and P ′s can also be written in terms of the constrained variables. Let us go back to our special case of massive vector field, and quantize it in terms of Dirac brackets. For this 0 i example it is easy to write the constrained variables V and Π0 in terms of the unconstrained variables V and Πi, since they are given by Eqs. (10.217) and (10.218). 1 Π = 0 ; V 0 = J 0 + ∂ Π (10.235) 0 m2 i i

From Eqs. (10.225) and (10.226) we see that the matrix CNM has an inverse given by

1 1x,2y 1 2y,1x 1 3 C− = C− = δ (x y) − −m2 − 1 1x,1y 1 2x,2y C− = C− = 0

From which the Dirac prescription gives the following equal-time commutators i [A, B]= i [A, B] + d3z [A, Π (z)] ∂ Π (z) m2V 0 (z) J 0 (z) , B A B P m2 0 P i i − − − ↔ Z By deﬁnition we have

[V µ (x) , Π (y)] = δ3 (x y) δµ ; [V µ (x) ,V ν (y)] = [Π (x) , Π (y)] = 0 ν P − ν P µ ν P so that i V i (x) ,V j (y) = V 0 (x) ,V 0 (y) = 0 ; V i (x) ,V 0 (y) = ∂ δ3 (x y) −m2 i − i i 3 0 µ V (x) , Πj (y) = iδ jδ (x y) ; V (x) , Πj (y) = [V (x) , Π0 (y)]=0 µ ν − [Π (x) , Π (y)] = 0 which are the commutation relations that we obtain by assuming that the unconstrained variables obey the canonical commutation relations

V i (x) , Π (y) = iδi δ3 (x y) j j − i j V (x) ,V (y) = [Πi (x) , Πj (y)]=0

0 and using the constraints (10.235) to evaluate the commutators involving the constrained variables Π0 and V . Chapter 11

Quantum electrodynamics

The usual starting point to construct quantum electrodynamics is to start with Maxwell’s equations and quantize them. In the present approach we shall instead start showing the necessity of a gauge invariance principle coming from the diﬃculties that arise when trying to quantize massless particles with spin. From the gauge invariance principle we shall infer many of the properties of the quantum theory of electromagnetism, in particular to deduce the existence of a vector potential to describe massless particles of spin one.

11.1 Gauge invariance

In trying to construct a four-vector covariant field as a linear combination of creation and annihilation fields for helicity 1, we saw in chapter 8, that it was not possible to build up a truly covariant four-vector of this kind, ± but a four-vector field aµ (x) that transforms as a four-vector up to a gauge transformation [see Eq. (8.40) page 218] 1 ν U0 (Λ) aµ (x) U0− (Λ) = Λµ aν (Λx)+ ∂µΩ (x, Λ) (11.1) it is because in trying to adjust the coefficients of the expansion, we had to use the generators of the little group ISO (2), whose elements can be written as

W (θ, α, β)= S (α, β) R (θ) then we adjust the coeﬃcients by using only the subgroup R (θ), and after trying to do the same with the subgroup S (α, β) we obtain a contradiction. However, it was possible to construct a truly covariant antisymmetric second-rank tensor free-ﬁeld of the form

f (x)= ∂ a (x) ∂ a (x) µν µ ν − ν µ The impossibility of building a true four-vector permits to understand the presence of singularities at m =0 in the Feynman propagator of a massive vector ﬁeld of spin one

qµqν 4 4 iq (x y) gµν + m2 ∆ (x,y)=(2π)− d q e · − µν q2 + m2 iε Z − which forbids us to obtain the propagator of massless vector particles of helicity 1 through the limit m 0. ± → From now on we shall use capitol letters because we shall be dealing with interacting ﬁelds. We can avoid the problems above by using only F (x)= ∂ A (x) ∂ A (x) (11.2) µν µ ν − ν µ because it is a truly covariant second-rank tensor. Nevertheless, this is not the most general possibility neither is realized in nature. Thus, instead of getting rid of Aµ (x) in the action, we shall demand for the part of the

323 324 CHAPTER 11. QUANTUM ELECTRODYNAMICS

action IM that describes the matter and the radiation-matter interaction to be invariant under the general gauge transformation A (x) A (x)+ ∂ ε (x) (11.3) µ → µ µ at least when the matter fields satisfy the field equations. In that way the extra term in Eq, (11.1) has no physical consequences. The change in the matter action under the transformation (11.3) reads δI δI = d4x M ∂ ε (x) (11.4) M δA (x) µ Z µ ————————-open —————————- ???? We first observe that under the condition of null fields at infinity, the four dimensional integrals of any four- µ µ divergence ∂µF of any function F involving those fields must vanish. In particular

δI δI δI 0 = d4x ∂ M ε (x) = d4x ε (x) ∂ M + d4x M ∂ ε (x) µ δA (x) µ δA (x) δA (x) µ Z µ Z µ Z µ δI 0 = d4x ε (x) ∂ M + δI (11.5) µ δA (x) M Z µ where we have used Eq. (11.4). Demanding Lorentz invariance we have δIM = 0 and appealing to the arbitrariness of ε (x) the integrand in Eq. (11.5) must vanish ???? —————————- ——————————–close Thus, Lorentz invariance of IM demands δIM ∂µ = 0 (11.6) δAµ (x)

Let us take ﬁrst the case in which IM depends only on Fµν (x) and its derivatives, together with matter ﬁelds. In that case it can be shown that

δIM δIM = 2∂ν (11.7) δAµ (x) δFµν (x) by applying a divergence on both sides we obtain

δIM δIM ∂µ = 2∂µ∂ν = 0 δAµ (x) δFµν (x) the latter step comes from the symmetric nature of the tensor derivative ∂µ∂ν and the antisymmetric nature of Fµν . We conclude that the Lorentz invariance condition (11.6) is obviously satisfied if IM depends only on Fµν (x) and its derivatives, together with matter fields. Nevertheless, if IM depends on Aµ (x) itself, equation (11.6) becomes a non-trivial constraint on the theory. ———————open ———————– ???? Now we shall prove Eq. (11.7). If IM depends only on Fµν (x) and its derivatives, together with matter fields, we take into account that matter fields do not contribute to the variation since they are independent of Aµ, therefore δI δF δI δ (∂ A ∂ A ) δI M = αν = α ν − ν α δAµ δAµ δFαν δAµ δFαν δI δ (∂ A ) δI δ (∂ A ) δI M = α ν ν α (11.8) δAµ δAµ δFαν − δAµ δFαν 11.1. GAUGE INVARIANCE 325 now the variation under a Lorentz transformation of the derivative of a field is given by Eqs. (10.136) and (10.139) pages 300 and 301, that can be combined as

λ δ (∂αAν )= ∂αδAν + ωα ∂λAν

δλ δ (∂αAν )= ∂αδAν + g ωαδ∂λAν (11.9) Applying Eq. (11.9) in Eq. (11.8) gives

δI δA ∂ A δI δA ∂ A δI M = ∂ ν + gδλω λ ν ∂ α + gδλω λ α δA α δA αδ δA δF − ν δA νδ δA δF µ µ µ αν µ µ αν δI δA δA δI ∂ A ∂ A δI M = ∂ ν ∂ α + gδλω λ ν gδλω λ α (11.10) δA α δA − ν δA δF αδ δA − νδ δA δF µ µ µ αν µ µ αν since Aµ and their derivatives are consider independent (as in the Euler-Lagrange equations), only the ﬁrst term in parenthesis contributes in Eq. (11.10)

δIM δAν δI δAα δI = ∂α ∂ν δAµ δAµ δFαν − δAµ δFαν δI δI δI δI = δµν ∂α δαµ∂ν = ∂α ∂ν δFαν − δFαν δFαµ − δFµν δIM δI δI δI = ∂α + ∂ν = 2∂ν δAµ δFαµ δFνµ δFνµ which proves Eq. (11.7). ————— —————-close

11.1.1 Currents and their coupling with Aµ

We wonder now about a theory that provides a conserved current to be coupled to the ﬁeld Aµ (x). In section 10.8, we saw that inﬁnitesimal internal symmetries of the action lead to the existence of conserved currents. In particular assuming that the transformation

k k δΨ (x)= iε (x) qkΨ (x) nosumover k (11.11) leaves the matter action invariant for constant ε, the change in the matter action for arbitrary inﬁnitesimal values ε (x) must have the form [see Eqs. (10.90) and (10.91) page 290]

δI = d4x J µ (x) ∂ ε (x) (11.12) M − µ Z if the matter ﬁelds satisfy the ﬁeld equations, the matter action is stationary with respect to arbitrary variations k of the Ψ , thus in that case δIM in Eq. (11.12) must vanish, therefore

µ ∂µJ = 0 (11.13) further, if I is the integral of a function of Ψk and ∂ Ψk, the conserved current becomes [see section 10.8] M LM µ

µ ∂ M k J = i L k qkΨ (11.14) − ∂ (∂µΨ ) Xk 326 CHAPTER 11. QUANTUM ELECTRODYNAMICS

Where Ψk runs over all independent fields different from Aµ. The use of capitol Ψ indicates that they are fields in the Heisenberg picture, whose time-dependence includes the effects of interactions. This generates the transformations (11.11) i.e. [see Eqs. (4.65), page 148]

Q, Ψk (x) = q Ψk (x) − k h i where Q is the generalized conserved charge operator

Q = d3x J 0 Z µ Thus, we can construct a Lorentz-invariant theory by coupling Aµ with the conserved current J , in the sense µ that δIM /δAµ (x) is taken as proportional to J (x) as can be seen by comparing Eqs. (11.6) and (11.13). Any constant of proportionality can be absorbed in the overall scale of the charges qk, hence we can settle the constant of proportionality equal to unity δI M = J µ (x) (11.15) δAµ (x) the conservation of the electric charge only permits to adjust the values of all charges in terms of the value of a given one. The reference charge is usually taken to be the electron charge e. Equation (11.15) is the one that − provides a definite meaning to the value of e. It is important to take into account that Eq. (11.15) ony fixes the definition of e after we have carried out the normalization of the field Aµ (x). We can express the condition (11.15) in the form of a principle of invariance: The matter action is invariant under the joint transformations

δAµ (x) = ∂µε (x) (11.16)

δΨk (x) = iε (x) qkΨk (x) (11.17) this kind of symmetry is called a local symmetry because of the local nature of ε (x). It is also called a gauge invariance of the second kind. When ε is constant we call them global symmetries or gauge invariance of the ﬁrst kind. We have many examples of exact local symmetries while global symmetries appear only as accidental symmetries coming from other principles.

11.1.2 Action for the photons (radiation) As for the action of the photons (radiation) themselves, we can take it as the one for massive vector fields with m = 0 1 I = d4x F F µν (11.18) γ −4 µν Z which coincides with the action of classical electrodynamics. However, we justify it by observing that (up to a constant) it is the unique gauge invariant functional quadratic in Fµν , without higher derivatives. We shall see later that it leads to a consistent quantum field theory. Any terms in the action with higher derivatives and/or of higher order in Fµν can be translated into the so called matter action. It can be shown from Eq. (11.18) that δIγ µν = ∂µF (11.19) δAν From Eqs. (11.15) and (11.19) the field equation for electrodynamics gives

δ µν ν 0= [Iγ + IM ]= ∂µF + J (11.20) δAν 11.1. GAUGE INVARIANCE 327

ν which are the well-known inhomogeneous Maxwell equations with source J . From the deﬁnition (11.2) of Fµν , other homogeneous Maxwell equations arise

0 = (∂ ∂ A ∂ ∂ A ) + (∂ ∂ A ∂ ∂ A ) + (∂ ∂ A ∂ ∂ A ) µ ν ε − ν µ ε ε µ ν − µ ε ν ν ε µ − ε ν µ 0 = (∂ ∂ A ∂ ∂ A ) + (∂ ∂ A ∂ ∂ A ) + (∂ ∂ A ∂ ∂ A ) µ ν ε − µ ε ν ε µ ν − ε ν µ ν ε µ − ν µ ε 0 = ∂ (∂ A ∂ A )+ ∂ (∂ A ∂ A )+ ∂ (∂ A ∂ A ) µ ν ε − ε ν ε µ ν − ν µ ν ε µ − µ ε

0= ∂µFνε + ∂εFµν + ∂ν Fεµ (11.21) which is obtained from cyclic permutation of the symbols µ,ν,ε.

11.1.3 General overview of gauge invariance We have started with the existence of massless spin one particles and we arrived to the invariance of the matter action under the combined local gauge transformations described by Eqs. (11.16) and (11.17). However, in the standard literature it is usual to start with a global internal symmetry

δΨk (x)= iε qkψk (x) (11.22) and wonder about the conditions to extend it to a local symmetry

δΨk (x)= iε (x) qkΨk (x) (11.23)

For the case of a Lagrangian density that depends only on the fields Ψk (x) but not on their derivatives, L the local or global nature of ε is irrelevant. Notwithstanding, most of realistic Lagrangian densities depend on the fields and their derivatives. In the latter case we should work out the problem that the derivatives of fields transform differently under global or local gauges (i.e. transform differently from fields themselves)

k k k δ∂µΨ (x)= iε (x) qk∂µΨ (x)+ iqkΨ (x) ∂µε (x) (11.24) we can cancel the second term that “spoils” gauge invariance by introducing a vector ﬁeld Aµ (x) with transformation rule δAµ (x)= ∂µε (x) (11.25) k and demand that the Lagrangian density depend on ∂µΨ and Aµ only through the combination

D Ψk ∂ Ψk iq A Ψk (11.26) µ ≡ µ − k µ which transforms like the ﬁelds Ψk themselves as we wanted

δD Ψk (x) = δ ∂ Ψk iq A Ψk = δ ∂ Ψk iq δ A Ψk µ µ − k µ µ − k µ = δ ∂ Ψk iq (δA )Ψk iq A δΨk µ − k µ − k µ and applying Eqs. (11.23), (11.24) and (11.25) we have

k k k k k k δDµΨ (x) = iε (x) qk∂µΨ (x)+ iqkΨ (x) ∂µε (x) iqk [∂µε (x)] Ψ + qkAµε (x) q Ψ (x) k k k − = iε (x) qk∂µΨ (x)+ qkAµε (x) q Ψ (x) δD Ψk (x) = iε (x) q ∂ Ψk (x) iA qkΨk (x) µ k µ − µ h i δD Ψk (x) iε (x) q D Ψk (x) (11.27) µ ≡ k µ 328 CHAPTER 11. QUANTUM ELECTRODYNAMICS

A matter Lagrangian (Ψ, DΨ) that only depends on Ψk and D Ψk will be invariant under the transformations LM µ (11.23) and (11.25), with ε (x) an arbitrary function, if it is invariant when ε is constant. With a Lagrangian density of this form we obtain

δIM ∂ M k ∂ M k = L k iqkΨ = i L k qkΨ δAµ ∂DµΨ − − ∂ (∂µΨ ) Xk Xk and from Eq. (11.14) we have δI M = J µ (x) δAµ which coincides with Eq. (11.15). Note that we could include F and its derivatives in . With this approach, µν LM the masslessness of the particles described by Aµ is a consequence of the gauge invariance (instead of being an assumption), since a term in the Lagragian density of the form 1 m2A Aµ Lm ≡−2 µ would violate gauge invariance. From an eﬀective point of view, the global symmetry (11.22) becomes a local symmetry (11.23), by replacing the ordinary derivative ∂µ by the covariant derivative Dµ

∂ D ∂ iq A (11.28) µ → µ ≡ µ − k µ and imposing the rule of transformation (11.25) on the gauge ﬁeld Aµ.

11.2 Constraints and gauge conditions

n We shall adopt a notation in which Q and Pn denotes canonical matter fields and their canonical conjugates, while Ai, Πi are the canonical electromagnetic fields and their canonical conjugates. Thus, we could start by defining the canonical conjugates of the electromagnetic vector potential field as ∂ Πµ L ≡ ∂ (∂0Aµ) imposing canonical quantization conditions we would have

[A (x,t) , Πν (y,t)] = iδν δ3 (x y) µ µ − ν but we cannot do it, because Aµ and Π posses several constraints. We see it by observing that for F F µν = µν Lγ − 4 with a procedure similar to the one that led to Eq. (10.181) page 310, we obtain ∂ Lγ = F 0µ ∂ (∂0Aµ) − which vanishes for µ = 0, owing to the antisymmetry of F µν . As for the matter Lagrangian , if it involves only LM Ψk and D Ψk, the prescription (11.26) says that does not depend on any derivatives of any Aν . Furthermore, µ LM even if depends also on F we have that ∂ /∂ (∂ A ) is also antisymmetric in µ and ν, so that it vanishes LM µν LM ν µ for µ = ν = 0. Thus we have ∂ ( + ) Π0 Lγ LM = 0 ≡ ∂ (∂0A0) 11.2. CONSTRAINTS AND GAUGE CONDITIONS 329

From the discussion above, we ﬁnd a ﬁrst constraint coming from the fact that the Lagrangian density + L≡Lγ LM does not depend on the time-derivative of A0, consequently

Π0 (x) = 0 (11.29) this is a primary constraint, since it comes directly from the structure of the Lagrangian. We also have a secondary constraint in this theory, that comes from the field equation (Euler.Lagrange equation) for the quantity fixed by the primary constraint. With a procedure similar to the one that led to Eq. (10.218) page 10.218, we obtain i ∂ ∂ 0 ∂iΠ = ∂i L = L = J (11.30) − ∂Fi0 −∂A0 − once again the time-derivative term does not appear becuase F00 = 0. Despite the matter Lagrangian density 0 0 n could depend on A , the charge density J only depends on the canonical matter fields and momenta Q , Pn. We can see it by using Eq. (11.14) page 325 with µ = 0 ∂ J 0 = i L q Ψn = i P q Qn (11.31) − ∂ (∂ Ψn) n − n n n 0 n X X therefore, Eq. (11.30) is a functional relation between canonical variables. Both Eqs. (11.29) and (11.30) are not consistent with the usual quantization relations [A (x,t) , Πν (y,t)] = iδν δ3 (x y) ; [Qn (x,t) , Πν (y,t)]=[P (x,t) , Πν (y,t)] = 0 (11.32) µ µ − n for instance by using µ = ν = 0 in the first set of relations (11.32) we obtain A (x,t) , Π0 (y,t) = iδ3 (x y) 0 − but constraint (11.29) says that Π0 = 0 so that this commutation relation must vanish. On the other hand, by applying ∂ν in the second of relations (11.32) and using the constraint (11.30) we have [Qn (x,t) , ∂ Πν (y,t)]=0 Qn (x,t) , ∂ Πi (y,t) = 0 Qn (x,t) , J 0 (y,t) = 0 ν ⇒ i ⇒ − and using (11.31) we obtain

n m i qm [Q (x,t) , Pm (y,t) Q (y,t)] = 0 m X n m i qm [Q (x,t) , Pm (y,t)] Q (y,t) = 0 m X the left-hand side of this equation is not zero in general. In the case of massive vector ﬁelds we can solve that problem in two equivalent forms (a) By using the Dirac i brackets or (b) Treating only A and Πi as canonical variables solving the analog of Eq. (11.30) in order to obtain A0 in terms of such variables. However, in our present context we cannot use Dirac brackets. We see it by observing that our constraint functions χ are

0 0 χ1 = Π ; χ2 = ∂iΠi + J (11.33) for massive vector ﬁelds the second constraint is ∂ Π + J 0 m2A0 [see Eq. (10.218) page 10.218]. But constraints i i − (11.33) have vanishing Poisson brackets

0 0 0 0 ∂Π ∂ ∂iΠi + J ∂ ∂iΠi + J ∂Π [χ ,χ ] = Π0, ∂ Π + J 0 = + 1x 2y P i i P ∂Aa ∂Πa ∂Aa ∂Πa 0 [χ1x,χ2y]P = 0 (since Π = 0) 330 CHAPTER 11. QUANTUM ELECTRODYNAMICS consequently, constraints (11.29) and (11.30) are of first class. On the other hand, we cannot either eliminate A0 as a dynamical variable by writing it in terms of the other variables. Note that Eq. (11.30) is just an initial condition (not an equation for A0 at all times). Nevertheless, if (11.30) is valid at a given time, it is valid for all times, because the field equations for the other fields Ai yield ∂ ∂ ∂ ∂ ∂ L J 0 = ∂ ∂ L ∂ J 0 = ∂ ∂ L ∂ J 0 0 i ∂F − − 0 i ∂F − 0 − i 0 ∂F − 0 i0 0i 0i ∂ 0 = ∂i∂j L ∂iJi ∂0J ∂Fji − − ∂ ∂ ∂ ∂ L J 0 = ∂ ∂ L ∂ J µ (11.34) 0 i ∂F − i j ∂F − µ i0 ji the first term on the right-hand side vanishes because of the symmetry of ∂i∂j and the antisymmetry of Fji, while the second term vanishes because of the current coservation condition. Then Eq. (11.34) becomes

∂ ∂ ∂ L J 0 = 0 0 i ∂F − i0 note that we have three field equations for four components of Aµ. It is because we have a local gauge symmetry that prevents us to predict the values of the fields at all times from their values and rates of change at a given time. In other words, we do not have a unique time evolution for the field Aµ (x,t) even knowing the initial conditions for Aµ and its derivatives. Instead, for a given solution Aµ (x,t) of the three field equations, by choosing ε (x,t) so that its first and second derivatives vanish at t = 0, we can obtain another solution of the form

Aµ (x,t)+ ∂µε (x,t) (11.35) each one with the same value and time-derivative at t = 0. Indeed there are infinite solutions of the type (11.35) with the same initial conditions. Of course each solution of the form (11.35) differs from each other [and from Aµ (x,t)] for later times. Owing to the partial arbitrariness of Aµ (x,t) we cannot impose the quantization canonical conditions to that field directly (or for finite mass to Ai). The most common strategies to work it out are (a) The Lorentz-invariant method of BRST-quantization (very useful in the quantization of Yang-Mills theories) and (b) By making profit of the gauge invariance of the theory to choose an apropriate gauge. We shall follow the second procedure here, that is we shall quantize in a particular gauge. Then we should do a particular transformation of the form

A (x) A (x)+ ∂ λ (x);Ψ (x) exp [iq λ (x)] Ψ (x) µ → µ µ k → k k in order to impose a condition on Aµ (x) that permits to apply the methods of canonical quantization. In principle we have an inﬁnite set of choices. However, the most common gauges are the following

µ Lorentz (or Landau) gauge : ∂µA = 0 Coulomb Gauge : A = 0 ∇ · Temporal Gauge : A0 = 0 Axial Gauge : A3 = 0 Unitarity Gauge : Φ real

In the latter gauge, Φ is a complex scalar ﬁeld with q = 0. The unitarity gauge is used when the gauge symmetry 6 is spontaneously broken by a non-zero vacuum expectation value of Φ. For the canonical quantization the most convenient gauges are the axial and the Coulomb gauges. We shall choose the Coulomb gauge because the rotational invariance is more manifest in such a gauge. 11.2. CONSTRAINTS AND GAUGE CONDITIONS 331

We ﬁrst should be sure that this gauge is possible. To see it, we observe that if Aµ (x,t) does not already i satisfy the condition ∂iA = 0, we can make a gauge transformation µ µ µ A′ (x,t) A (x,t)+ ∂ λ (x,t) ≡ in particular for the spatial coordiante it yields

i i i A′ (x,t) A (x,t)+ ∂ λ (x,t) ≡ by applying divergence on both sides and demanding the condition A = 0 we have ∇ · ′ i i i 0 = ∂ A′ (x,t) ∂ A (x,t)+ ∂ ∂ λ (x,t) i ≡ i i 0 = A (x,t)+ 2λ (x,t) ∇ · ∇ hence, to satisfy the Coulomb gauge the function λ (x,t) must be chosen such that it satisﬁes the equation

2λ (x,t)= A (11.36) ∇ −∇ · that is we basically have to solve a Poisson equation with source A. It is the existence of a solution for this −∇ · differential equation that guarantees the possibility of choosing such a gauge. Indeed, we only have to mind about the existence of the solution, but we do not need to solve equation (11.36) explicitly. We just impose the condition1 A = 0 (11.37) ∇ · and perform a quantization that respect that constraint. For simplicity we shall limit ourselves to theories in which the matter Lagrangian density could depend LM on matter fields and their time-derivatives and on Aµ but not on derivatives of Aµ. For instance, this is the case when depends only on Ψ and D Ψ with D defined by Eq. (11.28). Quantum Electrodynamics has a LM µ µ Lagrangian density that fulfills this condition. From this assumption, the only term that could depend on Fµν is the kinematic term 1 F F µν Lγ ≡−4 µν and the constraint equation (11.30) gives

i 0i 0 ∂iΠ = ∂iF = J ∂ F i0 = J 0 (11.38) − i and applying the Coulomb condition (11.37) on the right-hand side of this equation we have

∂ F i0 = ∂ ∂iA0 ∂0Ai = 2A0 ∂0 ∂ Ai − i − i − −∇ − i ∂ F i0 = 2A0 (11.39) − i −∇ equating Eqs. (11.38) and (11.39) we have 2A0 = J 0 (11.40) ∇ − Equation (11.40) is the well-known Poisson equation with source J 0, which can be solved as J 0 (y,t) A0 (x,t)= d3y (11.41) 4π x y Z | − | where the remaining three degrees of freedom Ai are subject to the gauge condition (11.37). Note that Eq. (11.40) can conversely show that J 0 cannot depend exclusively on matter ﬁelds. 0 n As we already said, the charge density J only depends on the matter variables Q and Pn as we see in Eq. (11.31). Hence, Eq. (11.41) provides an explicit solution for the auxiliary ﬁeld A0.

1We forget the prime notation A′, since we just impose the Coulomb condition since the beginning. 332 CHAPTER 11. QUANTUM ELECTRODYNAMICS

11.3 Quantization in Coulomb Gauge

11.3.1 Canonical quantization of the constrained variables We shall omit the time argument, since all functions and operators are evaluated at the same time. Despite we have been able to eliminate Π0 and A0 through Eqs. (11.29) and (11.41), we still have two more constraints on i the set of variables A and Πi that do not permit to apply the canonical quantization yet. They are the Coulomb gauge condition i χ x ∂ A (x) = 0 (11.42) 1 ≡ i and the secondary constraint (11.30) that provides the constraint

i 0 χ x ∂ Π (x)+ J (x) = 0 (11.43) 2 ≡ i these constraints are not compatible with the quantization conditions

[A (x) , Π (y)] = iδ δ3 (x y) (11.44) i j ij − since by operating on the right-hand side with either ∂/∂xi or ∂/∂yj does not give zero. We shall see that the constraints (11.42) and (11.43) are of second class, from which we can follow the prescription described in section 10.12 for the commutation relations2. We shall see it by observing that the constraint functions have Poisson brackets that forms a non-singular “matrix” CNM . In this context, the deﬁnition of the Poisson bracket for two arbitrary functionals U and V is3 δU δV δV δU [U, V ] d3x (11.45) P ≡ δAi (x) δΠ (x) − δAi (x) δΠ (x) Z i i ——————————– —————————–open ???? By applying the deﬁnition (11.45) of the Poisson brackets and the constraints (11.42) and (11.43) we have

3 δχ1x δχ2y δχ2y δχ1x C1x y [χ x,χ y] = d z ,2 ≡ 1 2 P δAi (z) δΠ (z) − δAi (z) δΠ (z) Z i i i i 0 i 0 i 3 δ ∂iA (x) δ ∂iΠ (y)+ J (y) δ ∂iΠ (y)+ J (y) δ ∂iA (x) C1x y = d z ,2 δAi (z) δΠ (z) − δAi (z) δΠ (z) Z " i i # 0 n 0 Since according with Eq. (11.31), J only depends on the matter canonical variables Q and Pn, then J does not contribute to the Poisson brackets, thus

p k k p 3 δ [∂pA (x)] δ ∂kΠ (y) δ ∂kΠ (y) δ [∂pA (x)] C1x y = d z ,2 δAi (z) δΠ (z) − δAi (z) δΠ (z) Z " i i # p k 3 δ [∂pA (x)] δ ∂kΠ (y) C1x y = d z ,2 δAi (z) δΠ (z) Z i by using Eq. (11.9) and assuming that the derivatives ∂µAν are independent of the ﬁelds Aµ we have

i i δ ∂iA (x) δ A (x) j = ∂i j δA (z) δA (z) 2Recall that the constraints (11.33) were of ﬁrst class, so that we were not able to use the formalism of Dirac brackets there. 3What really matters is that this Poisson brackets obey the algebraic rules (10.219, 10.220, 10.221) page 317. So that we can apply the results obtained in section 10.12. 11.3. QUANTIZATION IN COULOMB GAUGE 333 and similarly for the canonical ﬁeld momenta

i i δ ∂iΠ (y) δ Π (y) j = ∂i j δA (z) δA (z) therefore we have [Homework!! C1 prove equation 11.46)]

p k 3 δA (x) δΠ (y) 3 3 C1x y = d z ∂ ∂ = d z [δ ∂ δ (x z)] δ ∂ δ (y z) ,2 p δAi (z) k δΠ (z) pi p − ki k − Z i Z 3 3 3 3 C1x y = d z ∂ δ (x z) ∂ δ (y z) = ∂ ∂ δ (x y) (11.46) ,2 i − i − − i i − Z 2 3 C1x y = δ (x y) (11.47) ,2 −∇ −

————————————— close We can do a similar procedure for the other Poisson brackets to obtain

2 3 C1x y = C y x [χ x,χ y] = δ (x y) ,2 − 2 ,1 ≡ 1 2 P −∇ − C x y [χ x,χ y] = 0 ; C x y [χ x,χ y] = 0 1 ,1 ≡ 1 1 P 2 ,2 ≡ 2 2 P 0 2δ3 (x y) C = −∇ − (11.48) 2δ3 (x y) 0 ∇ − The “matrix” CNM is non-singular, so the constraints (11.42) and (11.43) are of second class as we anticipated. On the other hand, the ﬁeld variables Ai can be written in terms of independent canonical variables. For instance, we can take 1 2 Q x A (x) , Q x A (x) 1 ≡ 2 ≡ as independent variables while A3 (x) is given by the solution of Eq. (11.42) that makes A compatible with the Coulomb gauge. x3 A3 (x)= ds ∂ A1 x1,x2,s + ∂ A2 x1,x2,s − 1 2 Z i in the same way, we can write Πi and A in terms of the canonical (independent) conjugates P1x, P2x and Q1x, Q2x by using Eq. (11.43). As we said in section 10.12, it can be shown that if the independent variables Q1x, Q2x and P1x, P2x satisfy the usual commutation relations, the commutators of the constrained variables and their canonical conjugates are given by the Dirac brackets according with Eqs. (10.228) and (10.229). This prescription is convenient since we do not have to write explicit expressions of the dependent variables in terms of independent ones. In order to calculate the Dirac brackets we start by writing the inverse of the C matrix. For this we ﬁrst − write the quantity 2δ3 (x y) in its Fourier representation ∇ − 3 3 2 3 2 d p ip (x y) d p 2 ip (x y) δ (x y) = e− · − = e− · − ∇ − ∇ (2π)3 (2π)3 ∇ Z Z 3 2 3 d p 2 ip (x y) δ (x y) = p e− · − ∇ − − (2π)3 Z from this it can be shown that the inverse of such a quantity is given by

3 ip (x y) 1 d p e · − 2δ3 (x y) − = (11.49) ∇ − − (2π)3 p2 Z 334 CHAPTER 11. QUANTUM ELECTRODYNAMICS now the matrix (11.48) has the texture 0 a 0 2δ3 (x y) C = − = −∇ − a 0 2δ3 (x y) 0 ∇ − whose inverse has the form 1 1 2 3 − 0 a 0 δ (x y) C = 1 = 1 ∇ − (11.50) 0 2δ3 (x y) − 0 − a − ∇ − ! from Eqs. (11.49) and (11.50) the inverse of CNM reads

3 ik (x y) 1 1 d k e · − 1 C− = C− = = 1x,2y − 2y,1x − (2π)3 k2 −4π x y Z | − | 1 1 C− 1x,1y = C− 2x,2y = 0 (11.51) i And the non-zero Poisson brackets of the A and Πi variables with the constraint functions yield

i ∂ 3 A (x) ,χ y = δ (x y) (11.52) 2 P −∂xi − ∂ 3 [Π (x) , χ y] = δ (x y) (11.53) i 1 P ∂xi − now, the commutator can be evaluated by taking equations (10.228) and (10.229) for the Dirac brackets. Let us i evaluate the commutator of A (x) with Πj (y). From Eqs. (10.228) and (10.229) we have

i i i i 1 NM A (x) , Π (y) = i A (x) , Π (y) = i A (x) , Π (y) i A (x) ,χ C− [χ , Π (y)] j j D j P − N P M j P i i 1 1z,1w = i A (x) , Π (y) i A (x) ,χ z C − [χ w, Π (y)] j P − 1 P 1 j P i 1 1z,2w i 1 2w,1z i A (x) ,χ z C− [χ w, Π (y)] i A (x) ,χ w C− [χ z, Π (y)] − 1 P 2 j P − 2 P 1 j P i 1 2z,2w i A (x) ,χ z C− [χ w, Π (y)] − 2 P 2 j P but Eqs. (11.52) and (11.53) give us the only non-zero Poisson brackets, then we have

i i i 1 2w,1z A (x) , Π (y) = i A (x) , Π (y) i A (x) ,χ w C− [χ z, Π (y)] j j P − 2 P 1 j P i i i 1 2w,1z A (x) , Πj (y) = i A (x) , Πj (y)P + i A (x) ,χ2wP C− [Πj (y) ,χ1z]P ∂ 1 ∂ Ai (x) , Π (y) = i Ai (x) , Π (y) + i d3w d3z δ3(x w) δ3 (y z) j j P −∂xi − 4π x y ∂yj − Z | − | in the last step we take into account that we sum over repeated indices, that in the case of the indices z and w implies integration. Further, taking into account that

Ai (x) , Π (y) = δi δ3 (x y) j P j − we ﬁnally obtain [Homework!! C2] ∂2 1 Ai (x) , Π (y) = iδi δ3 (x y)+ i j j − ∂xj∂xi 4π x y | − | Therefore, from Eqs. (10.228) and (10.229), the equal-time commutators read ∂2 1 Ai (x) , Π (y) = iδi δ3 (x y)+ i (11.54) j j − ∂xj∂xi 4π x y | − | i j A (x) , A (y) = [Πi (x) , Πj (y)]=0 (11.55) 11.3. QUANTIZATION IN COULOMB GAUGE 335 which are consistent with the Coulomb gauge conditions (11.42) and (11.43), as it must be from the general properties of the Dirac Bracket. Comparing (11.54) with (11.44), we see that the constraints prevent us for using the usual quantization rules for the constrained variables. Indeed, we modify the quantization conditions for the constrained variables, precisely to mantain the usual quantization rules on the independent variables.

11.3.2 Quantization with the solenoidal part of Π~ We then wonder about the meaning of Π~ in electrodynamics. For the class of theories in which only the kinematic term 1 d3x F F µν −4 µν Z in the Lagrangian depend on A˙ , by varying the Lagrangian with respect to A˙ without considering the constraint given by the Coulomb gauge (11.42), yields

δL 1 3 δ µν 1 3 δFµν (y) µν Πj = = d y [Fµν (y) F (y)] = d y F (y) δA˙ j (x) −4 δA˙j (x) −2 δA˙j (x) Z Z 1 3 δ [∂µAν (y) ∂νAµ (y)] µν 1 3 δ [∂µAν (y)] δ [∂ν Aµ (y)] µν = d x − F (y)= d x j j F (y) −2 δA˙j (x) −2 δ [∂0A (x)] − δ [∂0A (x)] Z Z 1 1 = d3x δ δ δ3 (y x) δ δ δ3 (y x) F µν (y)= [δ δ δ δ ] F µν (x) −2 µ0 νj − − ν0 µj − −2 µ0 νj − ν0 µj Z 1 1 = [δ δ F µν (x)+ δ δ F νµ (x)] = F 0j (x)+ F 0j (x) = F j0 (x) −2 µ0 νj ν0 µj −2 = ∂jA0 (x) ∂0Aj (x) = ∂ A0 (x)+ ∂ Aj (x) − j 0 δL j ∂ 0 Πj = = A˙ (x)+ A (x) (11.56) δA˙j (x) ∂xj However, when the Coulomb condition is taken into account, variational derivatives with respect to A˙ are not well-deﬁned. If the variation of L under δA˙ in A˙ is given by

δL = d3x δA˙ (11.57) P · Z let us assume an arbitrary scalar function (x) with the only condition that it vanishes at inﬁnity. Therefore, we F can construct a null integral via divergence theorem (in three-dimensions)

0 = d3x ∂ i (x) δA˙ = d3x ∂ i (x) δA˙ + i (x) ∂ δA˙ i F i iF i F i i Z Z h i n o 0 = d3x (x) δA˙ + (x) δA˙ ∇F · F ∇ · Z n o 0 = d3x (x) δA˙ ∇F · Z n o but taking into account that δA˙ = 0, we then have ∇ · 0= d3x (x) δA˙ (11.58) ∇F · Z n o the we can add a zero on both sides of Eq. (11.57) through Eq. (11.58) to obtain

δL = d3x [ + (x)] δA˙ P ∇F · Z 336 CHAPTER 11. QUANTUM ELECTRODYNAMICS for an arbitrary scalar function (x) with the only condition that it vanishes at inﬁnity. Thus, after applying the F Coulomb gauge condition, by examining the Lagrangian we only can say that Π=~ A˙ (x)+ A0 (x)+ (x) (11.59) ∇ ∇F where the scalar (x) is arbitrary. By applying condition (11.43) we can remove the ambiguity. Such a condition F along with Eq. (11.40) (valid only in the Coulomb gauge) requires Π=~ J 0 = 2A0 (11.60) ∇ · − ∇ now, applying the divergence on Eq. (11.59) and taking into account Eq. (11.60) and that A˙ = 0, we have ∇ · Π=~ 2A0 (x)+ 2 (x)= 2A0 ∇ · ∇ ∇ F ∇ it can be seen that Eq. (11.56) provides a correct (though perhaps not unique) expression for Πi. The commutation relations (11.54, 11.55) are relatively simple but we should confront the fact that Π~ does n not commute with matter ﬁelds and their canonical conjugates Q , Pn. Thus, if F is any functional of matter variables, its Dirac bracket with A is null (Homework!! C3), but its Dirac bracket with Π~ yields

~ ~ 1 NM ~ F, Π (z) = F, Π (z) [F,χN ]P C− χM , Π (z) (11.61) D P − P from the deﬁnition of Poissonh bracketsi inh our presenti context Eq. (11.45) pageh 332 iti is clear that F, Π~ (z) = 0 P h i since F does not depend on Π~ and A (it only depends on matter ﬁelds). With these considerations Eq. (11.61) becomes ~ 1 1x,1y ~ 1 1x,2y ~ F, Π (z) = [F,χ1x]P C− χ1y, Π (z) [F,χ1x]P C− χ2y, Π (z) D − P − P h i 12y,1x h ~ i 1 2x,2y h ~ i [F,χ2y] C− χ1x, Π (z) [F,χ2x]P C− χ2y, Π (z) − P P − P h i h i but recalling that the only non-zero Poisson brackets between Π~ and χN are the ones in Eqs. (11.53) we have ~ 1 1x,1y ~ 1 2y,1x ~ F, Π (z) = [F,χ1x]P C− χ1y, Π (z) [F,χ2y] C− χ1x, Π (z) D − P − P P h i h 1 i h i but from Eq. (11.51) we see that the diagonal terms of C− are null. Hence 1 2y,1x F, Π~ (z) = [F,χ2y] C− χ1x, Π~ (z) D − P P Besides, we should take into accounth thati there is sum over repeated indicesh 2y andi 1x. But we should recall that these indices contain a continuous part and a discrete part, thus we shall also have integrals in x and y. From 1 these facts and using expression (11.51) of C− , we have

3 3 1 F, Π~ (z) = d x d y [F,χ2y] χ1x, Π~ (z) D − P 4π x y P h i Z | − | h i and interchanging the dummy indices x and y we obtain ~ 3 3 1 ~ F, Π (z) = d x d y [F,χ2x]P χ1y, Π (z) D − 4π x y P Z | − | h i ∂ h 1i ∂ = d3x d3y F, Πi (x)+ J 0 (x) Ak (y) , Π~ (z) − ∂xi 4π x y ∂yk Z P | − | P 1 = d3x d3y F, J 0 (x) δ3 (y z) − P 4π x y ∇ − Z | − | = d3y F, A0 (y) δ3 (y z) − P ∇ − Z = F, A0 (z) = F, A0 (z) ∇ P ∇ D 11.3. QUANTIZATION IN COULOMB GAUGE 337 in order to obtain a “redeﬁned” canonical momentum whose Dirac bracket with F vanishes, it is natural to deﬁne the solenoidal part of Π~ as Π~ Π~ A0 (11.62) ⊥ ≡ −∇ combining (11.56) con (11.62) we see that

0 j 0 0 j Π~ j (x) Π~ j (x) A (x) = A˙ (x)+ ∂jA (x) ∂jA (x)= A˙ (x) ⊥ ≡ − ∇ j − Π~ (x) Π~ (x) A0 (x)=A˙ (x) (11.63) ⊥ ≡ −∇ note one additional advantage of the use of Π~ (x): Its relation with its canonical coordinate is simpler [compare Eq. (11.63) with Eq. (11.56)]. We shall write⊥ the Hamiltonian in terms of A and Π~ (instead of A and Π)~ because it is more convenient for the pass to the interaction picture. From this deﬁnition⊥ it is clear that

F, Π~ (z) = 0 (11.64) ⊥ D h i Π~ (x) , A0 (y) = Π~ (x) , Π~ (y) Π~ (y) = 0 (11.65) ⊥ ∇ ⊥ − ⊥ h 0 0 i h i ∂iA (x) , ∂jA (y) = 0 (11.66) from Eqs. (11.65) and (11.66) it can be verified that Π~ (x) obeys the same commutation relations as Π~ (x). Further by applying Eq. (11.60) we find that Π~ obeys a⊥ constraint ⊥ Π~ = Π~ A0 = Π~ 2A0 = 2A0 2A0 ∇ · ⊥ ∇ · −∇ ∇ · −∇ ∇ −∇ Π~ = 0 (11.67) ∇ · ⊥ since Π~ is divergenceless, it justifies the name solenoidal part of Π.~ ———————————————-⊥ ——————————————– ————————————————

11.3.3 Constructing the Hamiltonian In order to construct the Hamiltonian we can use the customary relation between the Hamiltonian and the Lagrangian using the constrained variables A and Π~ , without ﬁrst having to express explicitly the Hamiltonian n ⊥ in terms of the unconstrained variables Q and Pn. For electrodynamics we obtain

3 i n 3 0 i n H = d x Π~ iA˙ + PnQ˙ = d x Π~ i + A A˙ + PnQ˙ −L ⊥ ∇ −L Z h i Z h i with a procedure similar to the one that lead to Eq. (11.58) by using A = 0, and taking (x) A0 we can ∇ · F → show that the term with A0 vanishes. Hence ∇

3 i n H = d x Π~ iA˙ + PnQ˙ (11.68) ⊥ −L Z n h i as we said before Q and Pn will denote the matter canonical ﬁelds and their canonical conjugates respectively. As a matter of example let us consider a Lagrangian density of the form 1 = F F µν + J Aµ + (11.69) L −4 µν µ Lmatter where the current Jµ does not depend on Aµ, and matter is the Lagrangian density that involves any other ﬁelds L µ that appear in Jµ aside from their electromagnetic interactions [given explicitly by the term JµA in Eq. (11.69)]. 338 CHAPTER 11. QUANTUM ELECTRODYNAMICS

The electrodynamics of particles of spin 1/2 has the form of the Lagragian density (11.69). By substituting A˙ everywhere with Π~ 2 [according with Eq. (11.63)], the Hamiltonian (11.68) becomes ⊥

3 i n 1 µν µ H = d x Π~ iΠ~ + PnQ˙ + Fµν F JµA matter ⊥ ⊥ 4 − −L Z 3 i 1 µν µ 3 n H = d x Π~ iΠ~ + Fµν F JµA + d x PnQ˙ matter ⊥ ⊥ 4 − −L Z Z h i 3 2 1 0i i0 ij 0 i 3 n H = d x Π~ + F0iF + Fi0F + FijF J0A JiA + d x PnQ˙ matter ⊥ 4 − − −L Z Z h i 3 2 1 0i ij 0 0 3 n H = d x Π~ + 2F0iF + FijF + J A J A + d x PnQ˙ matter ⊥ 4 − · −L Z Z h i 3 2 1 0i 1 2 0 0 3 n H = d x Π~ + F0iF + ( A) + J A J A + d x PnQ˙ matter (11.70) ⊥ 2 2 ∇× − · −L Z Z h i let us develop the term

1 1 1 F F 0i = (∂ A ∂ A ) ∂0Ai ∂iA0 = (∂ A ∂ A ) (∂ A ∂ A ) 2 0i 2 0 i − i 0 − −2 0 i − i 0 0 i − i 0 1 1 = (∂ A ) (∂ A ) + (∂ A ) (∂ A ) (∂ A ) (∂ A ) −2 0 i 0 i 0 i i 0 − 2 i 0 i 0 1 1 2 1 2 0 1 0 2 = Π~ i Π~ i + Π~ i (∂iA0) ( A0) = Π~ Π~ A A −2 ⊥ ⊥ ⊥ − 2 ∇ −2 ⊥ − ⊥ · ∇ − 2 ∇ 2 1 0i 1 0 F0iF = Π~ + A (11.71) 2 −2 ⊥ ∇ substituting (11.71) in (11.70) the Hamiltonian becomes

2 3 2 1 2 1 0 0 0 H = d x Π~ + ( A) Π~ + A J A + J A + HM ⊥ 2 ∇× − 2 ⊥ ∇ − · Z with HM being the Hamiltonian for the matter ﬁelds, without their electromagnetic interactions

H d3x P Q˙ n M ≡ n −Lmatter Z Applying the expression for A0 in Eq. (11.41) we have

3 1 2 1 2 1 0 0 H = d x Π~ + ( A) J A + J A + HM (11.72) 2 ⊥ 2 ∇× − · 2 Z from Eq. (11.41), it can be seen that the term (1/2) J 0A0 is the Coulomb energy

1 1 J 0 (x) J 0 (y) V = d3x J 0A0 = d3x d3y (11.73) Coul 2 2 4π x y Z Z Z | − | By applying the commutation relations (11.54, 11.55) it can be checked that any operator function F of A and Π~ obeys the ﬁeld equation (Homework!! C4) iF˙ = [F,H] as expected. 11.4. FORMULATION OF QED IN THE INTERACTION PICTURE 339

11.4 Formulation of QED in the interaction picture

As customary, we split the Hamiltonian (11.72) into a free term H0 and an interacting term V

H = H0 + V (11.74)

3 1 2 1 2 H0 = d x Π~ + ( A) + Hmatter,0 (11.75) 2 ⊥ 2 ∇× Z V = d3x J A + V + V (11.76) − · Coul matter Z where Hmatter,0 is the free-particle term for matter and Vmatter its corresponding interacting term. Finally, VCoul is the Coulomb interaction term (11.73). Recalling that the total Hamiltonian H in Eq. (11.74) is time-independent, we can evaluate H0 and V in Eqs. (11.75) and (11.76) at any time. In particular we can evaluate those terms at t = 0. As usual, the transition to the interaction picture is carried out through a similarity transformation

V (t) = exp (iH0t) V A, Π~ , Q, P exp ( iH0t) ⊥ t=0 − = V [a (t) , π (t)h, q (t) , p (t)] i

Where we are omitting the subscript on π (x). Here P means the canonical conjugates to the matter ﬁelds ⊥ Q. If (x) is an operator in the interaction picture we denote as ¯ (x), its counterpart in the Heisenberg picture. O O In particular, (x,t) in the interaction picture can be expressed by its value ¯ (x, 0) at t = 0 in the Heisenberg O O picture as (x,t) = exp [iH t] ¯ (x, 0) exp [ iH t] O 0 O − 0 deriving with respect to time on both sides

˙ (x,t) = iH exp [iH t] ¯ (x, 0) exp [ iH t] + exp [iH t] ¯ (x, 0) exp [ iH t] ( iH ) O 0 0 O − 0 0 O − 0 − 0 ˙ (x,t) = iH (x,t) i (x,t) H O 0 {O }− O 0 then its equation of motion yields i ˙ (x,t) = [ (x,t) ,H ] (11.77) O O 0 of course, a similarity transformation does not change the equal-time commutation relations, so they are the same as in the Heisenberg picture

∂2 1 ai (x,t) , πj (y,t) = i δ δ3 (x y)+ (11.78) ij − ∂xi∂xj 4π x y | − | ai (x,t) , aj (y,t) = πi (x,t) ,πj (y,t) = 0 (11.79) and similarly for the matter fields and their canonical conjugates. For the same reason, the constraints (11.37) and (11.67) preserve their form (recall that ~π means ~π in this context) ⊥ a = 0 ; ~π = 0 (11.80) ∇ · ∇ · now, we can find the relation between ~π and ˙a, by evaluating ˙a through Eq. (11.77) and using the commutation 340 CHAPTER 11. QUANTUM ELECTRODYNAMICS relations (11.78) and (11.79) we have 1 1 ia˙ (x,t) = [a (x,t) ,H ]= a (x,t) , d3y ~π2 (y)+ ( a (y))2 + H i i 0 i 2 2 ∇× matter,0 Z 3 1 2 1 2 1 3 2 = ai (x,t) , d y ~π (y)+ ( a (y)) = d y ai (x,t) , ~π (y) 2 ⊥ 2 ∇× 2 Z Z 1 = d3y [a (x,t) , π (y)] π (y)+ π (y) [a (x,t) , π (y)] 2 { i j j j i j } Z ∂2 1 = d3y i δ δ3 (x y)+ π (y) ij − ∂xi∂xj 4π x y j Z | − | and we obtain ∂2 1 ia˙ (x,t) = [a (x,t) ,H ]= i d3y δ δ3 (x y)+ π (y,t) i i 0 ij − ∂xi∂xj 4π x y j Z | − | we can replace ∂/∂xj by ∂/∂yj, and integrate by parts then use the second of Eqs. (11.80) to obtain [Homework!! − C5] ˙a = ~π (11.81) like in the Heisenberg picture [see Eq. (11.63)]. Similarly, the field equation is given by 1 1 iπ˙ (x,t) = [π (x,t) ,H ]= π (x,t) , d3y ~π2 (y)+ ( a (y))2 + H i i 0 i 2 2 ∇× matter,0 Z 1 = d3y π (x,t) , ( a (y))2 2 i ∇× Z h i ∂2 1 iπ˙ (x,t) = [π (x,t) ,H ]= i d3y δ δ3 (x y)+ ( a (y,t)) (11.82) i i 0 − ij − ∂xi∂xj 4π x y ∇×∇× j Z | − | from Eq. (11.82) and using the first of Eqs. (11.80) and Eq. (11.81) we arrive at the usual wave equation

a = 0 (11.83) recalling that in the Heisenberg picture A0 is not an independent variable but a functional (11.41) of the matter ﬁelds and their canonical conjugates that vanishes in the limit of zero charges, we do not introduce a corresponding degree of freedom in the interaction picture. Instead, for convenience we deﬁne

a0 = 0 (11.84)

The most general real solution of the ﬁrst of Eqs. (11.80), along with Eqs. (11.83) and (11.84) gives

3 µ 1 d p ip x µ ip x µ a (x)= e · e (p,σ) a (p,σ)+ e− · e ∗ (p,σ) a† (p,σ) (11.85) 3 0 √2π 2p σ Z X h i with p0 = p . The coefficients eµ (pp,σ) are two independent degrees of freedom that satisfy the relations k k p e (p,σ)=0 ; e0 (p,σ) = 0 (11.86) · we call these coefficients polarization vectors, name that we shall justify later. Again, a (p,σ) are a pair of operator coefficients, where σ is a two-valued index. By an appropriate normalization of a (p,σ) we can normalize the polarization vectors eµ (p,σ) so that the completeness relation gives

i j pipj e (p,σ) e (p,σ)∗ = δ (11.87) ij − 2 σ p X k k 11.4. FORMULATION OF QED IN THE INTERACTION PICTURE 341 in particular we can choose the polarization vectors that we obtained in chapter 8

1/√2 i/√2 eµ (p, 1) = R (p)  ±  (11.88) ± 0  0    b   where R (p) is the standard rotation that takes the unitary vector u3 into the direction p. From Eqs. (11.87) and (11.81) we obtain that the commutation relations (11.78) and (11.79) are fulﬁlled if and only if the operator coeﬃcientsb in Eq. (11.85) hold the conditions [Homework!! C6] b

3 a (p,σ) , a† q,σ′ = δ (p q) δ ′ (11.89) − σσ h i a (p,σ) , a† q,σ′ = 0 (11.90) h i as we have said several times, this is not an alternative derivation of Eqs. (11.89) and (11.90) but a validation of Eq. (11.75) as the correct free Hamiltonian for massless particles of helicity 1. In a similar way we can apply ± Eqs. (11.81), and (11.85) in Eq. (11.75) to calculate the free-photon Hamiltonian (Homework!! C7)

3 1 0 H0 = d p p a (p,σ) , a† (p,σ) 2 + σ Z X h i 3 0 1 3 H = d p p a† (p,σ) a (p,σ)+ δ (p p) (11.91) 0 2 − σ Z X which has the expected form of the free-Hamiltonian in terms of creation and annihilation operators [see Eq. (10.32) page 277], plus an irrelevant divergent term. Thus, Eq. (11.91) completes the validation of Eq. (11.75) as a correct expression for the Free Hamiltonian. The interaction term in the Heisenberg picture is given by Eq. (11.76), the corresponding interaction term in the interaction picture reads

V (t)= d3x j (x,t) aµ (x,t)+ V (t)+ V (t) (11.92) − µ Coul matter Z in Eq. (11.92) we write a jµ instead of a j because a0 = 0. With respect to the current in the Heisenberg picture µ · Jµ, the associated current jµ in the interaction picture reads

j (x,t) exp [iH t] J (x,t) exp [ iH t] µ ≡ 0 µ − 0 and VCoul (t) is the associated Coulomb term

V (t) = exp [iH t] V exp [ iH t] Coul 0 Coul − 0 1 j0 (x,t) j0 (x,t) V (t) = d3x d3y (11.93) Coul 2 4π x y Z k − k

ﬁnally, (in the interaction picture) Vmatter (t) is the non-electromagnetic part of the matter ﬁeld interaction

V (t) = exp [iH t] V exp [ H t] matter 0 matter − 0 342 CHAPTER 11. QUANTUM ELECTRODYNAMICS

11.5 The propagator of the photon

According with the Feynman rules discusssed in chapter 9, an internal photon line in a Feynman diagram provides a factor in the S matrix associated with the process, with the propagator − i∆ (x y) 0 T a (x) , a (y) 0 (11.94) − µν − ≡ h | { µ ν } | i with T denoting the time-ordered product. Substituting the expansion (11.85) in (11.94) the propagator becomes

3 d p ip (x y) ip (y x) i∆µν (x y)= Pµν (p) e · − θ (x y)+ e · − θ (y x) (11.95) − − (2π)3 (2 p ) − − Z k k h i with

0 P (p) e (p,σ) e (p,σ)∗ ; p = p µν ≡ µ ν k k σ= 1 X± From Eqs. (11.87) and (11.86) we have

pipj Pij (p)= δij ; P0i (p)= Pi0 (p)= P00 (p) = 0 (11.96) − p 2 k k We showed in Chapter 9, that the theta function in Eq. (11.95) can be written in terms of integrals over an independent time component q0 belonging to an oﬀ-shell four-momentum qµ. From this, equation (11.95) can be expressed by an integral over a four-momentum but with the oﬀ-mass-shell condition i.e. q0 is independent of q

4 4 Pµν (q) iq (x y) ∆ (x y)=(2π)− d q e · − µν − q2 iε Z − we are considering an internal photon line carrying a four-momentum q running between the vertices in which the photon is created and destroyed by ﬁelds aµ and aν . According with the Feynman rules in momentum space, the contribution of such an internal line gives i P (q) − µν (2π)4 q2 iε − It is convenient to reexpress Eq. (11.96) as follows

0 0 2 q qµnν + q qνnµ qµqν + q nµnν Pµν (q) = gµν + − (11.97) q 2 k k 2 nµ (0, 0, 0, 1) ; q2 = q2 q0 with q0 arbitrary (11.98) ≡ − thus nµ is a constant time-like vector. Since q0 is arbitrary we shall determine it in Eq. (11.97) from four- momentum conservation. That is, it will be taken as the difference of the matter p0’s flowing in and out of the vertex in which the photon is created. Thus, the terms proportional to qµ and/or qν do not contribute to the S matrix. It owes to the fact that the factors qµ or qν act like derivatives ∂µ and ∂ν , while the photon fields aµ − µ ν µ and aν are coupled to currents j and j that satisfy the conservation condition ∂µj = 0. On the other hand, the 2 2 4 term proportional to nµnν has a factor q that is cancelled by the factor q in the denominator of the propagator , giving a term that coincides with the one that would be generated by the term in the action

4 1 4 4 0 0 i d q iq (x y) i d x d y i j (x) i j (y) − e · − − 2 − − (2π)4 q 2 Z Z Z k k 2 4 q 1 1 2 A term of the form q2−iε = 1−i ε = 1−iε′ shows that we can cancel q completely, after all what really matters for ε it that it q2 tends to zero from the positive side. 11.6. FEYNMAN RULES IN SPINOR ELECTRODYNAMICS 343 the integration over q0 provides a delta function in time. Thus, this is equivalent to a correction for V (t) given by

1 j0 (x,t) j0 (y,t) d3x d3y −2 4π x y Z Z k − k which is the precise term to cancel the Coulomb interaction Eq. (11.93). In summary, the covariant quantity

eff 4 4 gµν iq (x y) ∆ (x y) = (2π)− d q e · − µν − q2 iε Z − can be used as an eﬀective photon propagator. That is, the Coulomb term does not appear henceforth. We can see that the apparent violation of Lorentz invariance in the (instantaneous) Coulomb interaction is cancelled by another apparent violation of the Lorentz invariance coming from the fact that the ﬁelds aµ are not truly four-vectors, leading to a non-covariant propagator. After this cancellation is carried out, we are left with a term of the form i g − µν (2π)4 q2 iε − as the contribution of an internal photon line in the momentum space Feynman rules. The Coulomb interaction term is then absent in the rule.

11.6 Feynman rules in spinor electrodynamics

We shall study the electrodynamics of a single species of particles of spin 1/2, charge q = e and mass m. We − shall make the treatment for electrons but it works equally well for muons, taos and other similar particles. We shall start with the simplest gauge and Lorentz invariant Lagrangian density for this theory. Such a theory contains a kinetic term for the photons and a (locally gauge invariant) Dirac’s Lagrangian density. Since the Dirac’s Lagrangian density is locally gauge invariant, it will contain a covariant derivative Dµ. Therefore, the Dirac’s Lagrangian density contains the matter radiation coupling plus the matter part of the Lagrangian density

1 = F F µν Ψ¯ (γµD + m)Ψ L −4 µν − µ 1 = F F µν Ψ¯ (γµ [∂ + ieA ]+ m) Ψ (11.99) L −4 µν − µ µ The four-vector that describes the electric current is given by

∂ J µ = L = ieΨ¯ γµΨ (11.100) ∂Aµ − and the interaction term (11.92) in the interaction picture yields

3 µ V (t)= ie d x ψ¯ (x,t) γ ψ (x,t) aµ (x,t)+ VCoul (t) (11.101) Z 5 and we have no a Vmatter term . As already discussed, the Coulomb term cancels with the non-covariant part of the photon propagator (where both terms are local in time i.e. instantaneous). We shall then deﬁne the Feynman rules in the momentum space to calculate the connected part of the S matrix − in this spinor electrodynamics.

5That is this Lagrangian does not contain terms of interaction with only matter ﬁelds (the purely matter terms contains the matter propagator and the mass term for fermions). All interaction terms are of the matter-radiation type. 344 CHAPTER 11. QUANTUM ELECTRODYNAMICS

11.6.1 Drawing the Feynman diagrams We start by drawing all connected Feynman diagrams for a given number of vertices.

1. There are two types of lines: electron lines carrying arrows and drawed as continuous lines, and photon lines that do not carry arrows and represented by waving lines6. Lines are joined at vertices

2. Each vertex contains three lines: one incoming electron line, one outcoming electron line and one photon line.

3. For each initial particle we have an external line coming from below into the diagram.

4. For each ﬁnal particle we have an external line going upwards out of the diagram.

5. Electrons in the diagram correspond to lines with arrows pointing upwards (regardless whether they are internal lines or external lines into or out of the diagram).

6. Positrons correspond to lines with arrows pointing downwards (regardless whether they are internal lines or external lines into or out of the diagram).

7. There are as many internal lines as the ones to attach each vertex to the required number of lines (three lines for each vertex).

8. Each internal line will be labeled by an oﬀ-mass-shell four-momentum ﬂowing in a given sense on the line (conventionally in the same direction of the arrow for electron lines)7

9. Each external line is labelled with the momentum and third component of spin or helicity for an electron or a photon respectively, in the initial and ﬁnal states.

11.6.2 Factors associated with vertices For a given vertex we shall have three labels (one for each attached line): (a) one four-component Dirac index α at the electron line with arrow coming into the vertex (b) A four-component Dirac index β at the electron line with arrow going out of the vertex, and (c) a space-time vertex µ associated with the photon line. Then from these three labels α,β,µ; we construct an associated vertex factor given by

4 µ 4 (2π) e (γ ) δ k k′ + q (11.102) βα − with k and k′ denoting the electron four-momenta entering and leaving the vertex respectively, and q denotes the photon four-momentum entering the vertex (if the photon four-momentum is leaving the vertex, we put a minus).

11.6.3 Factors associated with external lines 1. We label a given external line with the three-momentum p and third component of spin or helicity σ (for electrons and photons respectively) of the particle in the initial or ﬁnal state.

2. For an electron line in the ﬁnal state going out of a vertex with a Dirac label β on this line, we associate a factor u¯ (p,σ) β (11.103) (2π)3/2

6In general, it is conventional to draw scalar bosons as dashed lines, fermions as continuous lines, and vector bosons as waving lines. 7The arrow mentioned before (upward for electrons and downward for positrons), defines what we call the “fermionic flux”. The arrows of the fermionic flux do not necessarily coincide with the arrows of the momentum flux. It is convenient (but not mandatory) to make them coincide. 11.6. FEYNMAN RULES IN SPINOR ELECTRODYNAMICS 345

observe that we have extracted a matrix β from the interaction (11.102). It is because of this, thatu ¯ andv ¯ will appear in the Feynman rules instead of u† and v†. The four-component spinors u (p,σ) and v (p,σ) are the ones dicussed in section 7.4. 3. For a positron line in the ﬁnal state coming into a vertex with a Dirac label α on this line, we associate a factor vα (p,σ) (2π)3/2 4. For an electron line in the initial state coming into a vertex with a Dirac label α on this line, the associated factor will be uα (p,σ) (2π)3/2 5. For a positron line in the initial state going out of the vertex with a Dirac label β on this line, the factor associated yields v¯β (p,σ) (2π)3/2 6. For a photon line in the ﬁnal state attached to a vertex with space-time index µ on this line, the factor is

eµ∗ (p,σ) (2π)3/2 2p0 where eµ (p,σ) are the polarization vectors described inp section 11.4. 7. For a photon line in the initial state attached to a vertex with space-time label µ on this line, the factor reads eµ (p,σ) (2π)3/2 2p0 If we call u (p,σ) and v (p,σ) as spinors, andu ¯ (p,σ) andp v ¯ (p,σ) as adjoint spinors, we could say that we use spinors when the arrow of the electron line comes into a vertex and we use adjoint spinors when the electron arrow goes out of the vertex. Further we use u (p,σ) oru ¯ (p,σ) for electrons (particles i.e. upward lines) and v (p,σ) orv ¯ (p,σ) for positrons (antiparticles i.e. downward lines). Note also that the “fermionic ﬂux” never changes its sense so that it forms a “fermionic current”. For instance, the two fermion arrows in a given vertex never goes both out (or both come into) the vertex, always one of them is entering and the other one is leaving the vertex.

11.6.4 Factors associated with internal lines

From now on we shall use a “Dirac slash” notation for a given four vector kµ deﬁned by k γ kµ (11.104) 6 ≡ µ 1. For each internal electron line labelled by a four momentum k and running from a vertex with a Dirac label β to another vertex with Dirac label α, the factor is given by i [ i k + m] − − 6 αβ (2π)4 k2 + m2 iε − 2. For each internal photon line labelled by a four-momentum q running between two vertices with spacetime labels µ and ν the factor gives i g − µν (2π)4 q2 iε − 346 CHAPTER 11. QUANTUM ELECTRODYNAMICS

11.6.5 Construction of the S matrix process − To construct the S matrix assocaited with the process we then proceed as follows − 1. Integrate the product of all the previous factors over the four-momenta associated with internal lines (since they are oﬀ the mass shell), and sum over all Dirac and spacetime indices (we usually used the convention of sum over repeated indices).

2. Add up the results above for each Feynman diagram.

3. Some additional combinatoric factors and fermionic signs might appear as discussed in section 9.4.

11.7 General features of the Feynman rules for spinor QED

As the number of internal lines and vertices increases, there is a corresponding increase in the diﬃculty of evalauting the diagram. Therefore, it is useful to have some idea of the numerical order of magnitude of the contribution of the diagram. We shall estimate these numerical factors by including factors e coming from the electronic charge associated with the vertices and also factors 2 and π that comes from vertices, propagators, and momentum space integrals. We have already seen some diagrammatical identities concerning the number of vertices V , internal lines I, external lines E and loops L. They are given by [see Eqs. (9.82) and (9.98)) pages 252, 262]

L = I V +1 ; 2I + E = 3V (11.105) − solving for I on both sides we have

3V E L + V 1 = − − 2 ⇒ V = 2L + E 2 (11.106) − 4 4 we have a factor e (2π) from each vertex, a factor (2π)− from each internal line, and a four–dimensional momentum space integral for each loop. The element of volume in the four-dimensional euclidean space with a radius parameter k is π2k2dk2. Consequently, each loop gives a factor π2. From these facts and using Eqs. (11.105, 11.106), the diagram contains a numerical factor F given by

I I 4 L V 1 L V 1 (2π) F = e (2π)4 π2 = [e]V (2π)4 4 4 16π2 (2π) (2π) " # h i h i V I+L L L 2L+E 2 4 − 1 2 L E 2 4 1 = [e] − (2π) = e e − (2π) 16π2 16π2 h i 2 L 4 E 2 e F = (2π) e − (11.107) 16π2 the number E of external lines is the same for each diagram associated with the same physical process. Then from Eq. (11.107), we ﬁnd that the expansion parameter that gives us an idea of the suppression of the Feynman diagrams for each additional loop reads

2 e α 4 = = 5.81 10− 16π2 4π × this number is small enough to ensure a good perturbative behavior of the Feynman diagrams. 11.7. GENERAL FEATURES OF THE FEYNMAN RULES FOR SPINOR QED 347

11.7.1 Photon polarization We should take into account the spin states in which photons and electrons are typically found in experiments. In most of experiments the electrons and photons have no well-deﬁned values of the z component of spin neither the − helicity. In the case of photons, they are usually found in a state of transverse or elliptical polarization (instead of helicity states). For photons states with well-deﬁned helicity the polarization vectors are given by Eq. (11.88)

1/√2 i/√2 eµ (p, 1) = R (p)  ±  (11.108) ± 0  0    b   the most general photon state is a linear superposition of helicity states8

α+Ψp,+1 + α Ψp, 1 (11.109) − − if the helicity states are normalized, the normalization of this superposition yields

2 2 α+ + α = 1 (11.110) | | | −| If we want to calculate the S matrix element for absorbing or emitting a photon in the state (11.109) we should − replace the vector polarization e (p, 1) (that describes a helicity eigenstate) with µ ± µ µ eµ (p)= α+e (p, +1) + α e (p, 1) (11.111) − − in the Feynman rules. The polarization vectors with well deﬁned helicity satisfy the normalization condition

µ eµ∗ p, λ′ e (p, λ)= δλλ′ (11.112) so that if the normalization condition (11.110) is fulfilled , the polarization vector (11.111) also satisfies a normalization condition of the type µ eµ∗ (p) e (p) = 1 we have two extreme cases of polarizations of the type (11.111) (a) when either α+ =0 or α = 0 corresponding − to circular polarization, (b) the case α+ = α = 1/√2, corresponding to linear polarization. | | | −| For linear polarization, by properly choosing the overall phase of the state (11.109), we can settle the coefficients α+ and α to be complex conjugates − 1 α = exp ( iφ) (11.113) ± √2 ∓ and substituting (11.113, 11.108) in (11.111) the polarization vector yields

iφ iφ e− /2 e /2 iφ iφ µ µ +ie− /2 ie /2 eµ (p)= α+e (p, +1) + α e (p, 1) = R (p)   +  −  − − 0 0    0   0      b       cosφ  sin φ eµ (p)= R (p)   (11.114) 0  0    b   8Of course we are still assuming it as a three-momentum eigenstate. 348 CHAPTER 11. QUANTUM ELECTRODYNAMICS then we can use the polarization vector (11.114) in the Feynman rules. So that φ is the azimuthal angle of the photon polarization in the plane perpendicular to p [recall that eµ (p,σ) is perpendicular to p, according with Eq. (11.86) page 340]. The photon polarization vector (11.114) is real, and it is only possible for linear polarization. The most general state correspond to elliptical polarization in the case in which both α+ and α are non-zero − and they are non-equal. In a more general experimental context, the initial photons could be prepared in a statistical mixture of helicity (r) states. In the most general case an initial photon could have any number of possible polarization vectors eµ (p) 9 each one with probability Pr. The rate of abosortion of such a photon in a given process will have the form

2 (r) µ (r) µ ∗ (r) ν Γ = Pr eµ (p) M = Pr eµ (p) M eν (p) M r r X X µ ν (r) (r) Γ = M ∗M Preµ ∗ (p) eν (p) r X

2 (r) µ µ ν Γ = Pr eµ (p) M = M ∗M ρνµ (11.115) r X (r) (r) ρνµ Pre (p) e ∗ ( p) (11.116) ≡ ν µ r X where ρ is the density matrix. Let us recall that ρ is a Hermitian positive matrix of unit trace (that account on the fact that r Pr = 1). Further, from Eqs. (11.86) the density matrix satisﬁes

P ν µ ρν0 = ρ0µ =0 and ρνµp = ρνµp = 0 (11.117) the matrix density in its canonical form, can be expressed as

ρνµ = λseν (p; s) eµ∗ (p; s) s=1,2 X where eµ (p; s) are the two orthonormal eigenvectors of ρ, and λs denote the associated eigenvalues, with

µ e0 (p; s)= eµ (p; s) p = 0 owing to the positivity of the matrix and the conservation of probability (unit trace), the associated eigenvalues λs must satisfy λ 0 ; λ = 1 s ≥ s s=1,2 X therefore, the rate for the photon absorption process can be reexpressed by

Γ= λ e (p; s) M ν 2 s | ν | sX=1,2 so that the existence of a statistical mixture of initial photon states is equivalent to the superposition of two orthonormal polarizations eν (p; s) with probabilities λs. In many experiments the photons and electrons are not prepared in well deﬁned polarization states, neither is measured the ﬁnal polarization states of such particles.

9Note that the expression (11.115) for the rate of absorption is a sum over probabilities and not a sum over amplitudes, indicating that we are considering a statistical mixture of helicity states instead of a linear superposition of quantum helicity states. 11.7. GENERAL FEATURES OF THE FEYNMAN RULES FOR SPINOR QED 349

If we have no information about the initial photon polarization, we consider that both helicity states have the 1 same probability λ1 = λ2 = 2 . Using that assumption and Eq. (11.87), the density matrix (and so the absorption rate) which is an average over both possible initial polarizations, becomes 1 1 ρ = e (p; s) e∗ (p; s)= (δ p p ) ij 2 i j 2 ij − i j s=1,2 X where we have used the fact that the unitary transformation that takesb b the density matrix to its canonical form, leaves the identity invariant as well as the factor pipj since this rotation is in the helicity space but not in the momentum space. Consequently, this result is independent of the particular pair of polarization vectors ei (p; s) chosen to make the average. Thus, for unpolarizedb b photons the absorption rate can be averaged over any pair of orthonormal polarization vectors. Further if we also ignore the ﬁnal polarization state of the photon, we can calculate the rate by summing over any pair of orthonormal ﬁnal photon polarization vectors.

11.7.2 Electron and positron polarization A similar result follows for electrons and positrons. If they are not prepared in well deﬁned states of spin, the rate is calculated by averaging over any two orthonormal initial spin states10, such as those with z component of spin − equal to σ = 1/2. If we have no information about the ﬁnal spin states of the electrons and positrons, we sum ± the rate over any two orthonormal spin states, such as the ones with z component of spin given by σ = 1/2. − ± Those sums are obtained by using Eqs. (7.102, 7.103) as well as Eqs. (7.121, 7.122) i p + m u (p,σ)u ¯ (p,σ) = − 6 (11.118) α β 2p0 σ αβ X i p m v (p,σ)v ¯ (p,σ) = − 6 − (11.119) α β 2p0 σ αβ X where p0 = p2 + m2. As a matter of example, if we have an electron in the state p,σ and a positron in the | i state p ,σ in the initial state, the S matrix of the process will have the form | ′ ′ip − v¯ (p,σ) u (p,σ) α Mαβ β if the electron and positron spin states are not determined, the rate is proportional to

1 1 2 Γ v¯ p′,σ′ u (p,σ) ∼ 2 · 2 α Mαβ β σ′,σ X where each 1/2 factor comes from the average of electron and positron initial spin states.

1 2 1 Γ v¯ p′,σ′ u (p,σ) = v¯ p′,σ′ u (p,σ) † v¯ p′,σ′ u (p,σ) ∼ 4 M 4 M M σ′,σ σ′,σ X X 1 = u† (p,σ) † v¯† p′,σ′ v¯ p′,σ′ u (p,σ) 4 M M σ′,σ X h i 1 † = u† (p,σ) † v† p′,σ′ β v¯ p′,σ′ u (p,σ) 4 M M σ′,σ X h i 1 = u† (p,σ) † β v p′,σ′ v¯ p′,σ′ u (p,σ) 4 M M σ′,σ X n o 1 = u† (p,σ) ββ † β v p′,σ′ v¯ p′,σ′ u (p,σ) 4 M M σ′,σ X n o 10This is another way to say that if we have no information about spin states, any direction can be equally well chosen to be the direction of the axis of quantization. 350 CHAPTER 11. QUANTUM ELECTRODYNAMICS where we have inserted a factor β2 = 1, in the last step

1 Γ u¯ (p,σ) β † β v p′,σ′ v¯ p′,σ′ u (p,σ) ∼ 4 M M σ′,σ X n o 1 = u¯ (p,σ) β † β v p′,σ′ v¯ p′,σ′ u (p,σ) 4 µ µβMβα αρ ρ γ Mγδ δ σ′,σ X n o 1 = u (p,σ)u ¯ (p,σ) v p′,σ′ v¯ p′,σ′ β † β 4 δ µ ρ γ µβMβα αρ Mγδ ( σ ) ( σ′ ) X X n o and using Eqs. (11.118) and (11.119), we have

1 i p + m i p′ m † Γ − 6 0 − 6 0− βµβ βα βαρ γδ ∼ 4 2p δµ 2p′ ργ M M n o 1 i p′ m i p + m = β † β − 6 − − 6 4 µβMβα αρ 2p 0 Mγδ 2p0 ( ′ ργ δµ)

1 2 1 i p′ m i p + m Γ v¯ p′,σ′ u (p,σ) = T r β †β − 6 − − 6 (11.120) ∼ 4 α Mαβ β 4 M 2p 0 M 2p0 σ′,σ ′ X

11.8 Example of application: Feynman diagrams for electron-photon (Comp- ton) scattering

Figure 11.1: At the tree-level, there are two diagrams that contribute to the electron-photon (Compton) scattering.

At the tree level, there are two Feynman diagrams for electron-photon scattering shown in Fig. 11.1. Let us write the Feynman rules for the diagram in Fig. 11.1(a). 11.8. EXAMPLE OF APPLICATION: FEYNMAN DIAGRAMS FOR ELECTRON-PHOTON (COMPTON) SCATTERING

According with the rules explained above, for the initial electron line coming into a vertex we have a Dirac label α on this line, with momentum p and third component of spin σ: Thus, this external line gives a factor

uα (p,σ) (2π)3/2 the initial photon state has label µ on this line, momentum k and helicity eµ. This external line provides a factor

eµ (k) (2π)3/2 √2k0 now for the vertex 1, we have the Dirac indices α, β for the electron lines coming into and going out of the vertex respectively, the photon index is µ. The momenta p and k are entering, while q is leaving the vertex. Hence, the vertex factor yields (2π)4 e (γµ) δ4 (p + k q) βα − the internal electron line is labelled by a momentum q running from a vertex with Dirac label β to a vertex with Dirac label α′. The factor of the internal line becomes

i [ i q + m] ′ − − 6 α β (2π)4 q2 + m2 iε − now for the vertex 2, we have Dirac indices α′, β′ for the electron lines coming into and going out of the vertex respectively, the photon index is ν. The momenta p′ and k′ are leaving, while q is entering the vertex. Hence, the vertex factor yields 4 ν 4 (2π) e (γ ) ′ ′ δ p′ k′ + q β α − − the ﬁnal state of electron goes out of the vertex with Dirac label β′ on this line. Further, its momentum is p′ and its third component of spin is σ′, so the factor gives

u¯β′ (p′,σ′) (2π)3/2 the ﬁnal photon line state has space-time index ν on this line with momentum k′ and helicity eν′ thus the factor gives eν′∗ (k′) 3/2 0 (2π) √2k′ putting all these factors together we ﬁnd

eν′∗ (k′) u¯β′ (p′,σ′) 4 ν 4 i [ i q + m]α′β P (2π) e (γ )β′α′ δ p′ k′ + q − − 6 ≡ (2π)3/2 √2k 0 (2π)3/2 − − (2π)4 q2 + m2 iε ′ ! ! − h i 4 µ 4 eµ (k) uα (p,σ) (2π) e (γ )βα δ (p + k q) × − (2π)3/2 √2k0 ! (2π)3/2 ! h i the amplitude associated with this diagram is obtained by summing over Dirac and photon indices α, β, α′, β′,µ,ν and integrating over momenta of internal lines which is this case is only q.

4 eν′∗ (k′) u¯β′ (p′,σ′) 4 ν 4 a = d q (2π) e (γ ) ′ ′ δ p′ k′ + q M 3/2 √ 0 3/2 β α − − α,β α′,β′ µ,ν Z (2π) 2k′ ! (2π) ! X X X h i i [ i q + m] ′ e (k) u (p,σ) − 6 α β 4 µ 4 µ α − 4 2 2 (2π) e (γ )βα δ (p + k q) × (2π) q + m iε − (2π)3/2 √2k0 ! (2π)3/2 ! − h i 352 CHAPTER 11. QUANTUM ELECTRODYNAMICS using convention of sum over repeated indices and reordering terms we have

u¯ ′ (p ,σ ) e (k ) u (p,σ) e (k) i i q + m = β ′ ′ ν′∗ ′ α µ d4q a 3/2 3/2 3/2 3/2 − 4 2− 6 2 M (2π) (2π) √2k 0 (2π) (2π) √2k0 (2π) q + m iε ′ ′ Z − α β 4 ν 4 4 µ 4 e (2π) (γ ) ′ ′ δ q p′ k′ e (2π) (γ ) δ (q p k) (11.121) × β α − − βα − − adding the contributionh of the second diagram we obtainih the tree-level contributioni of the S matrix element for − Compton scattering.

u¯ ′ (p ,σ ) e (k ) u (p,σ) e (k) i i q + m S (p,σ; k, e) p ,σ ; k , e = β ′ ′ ν′∗ ′ α µ d4q ′ ′ ′ ′ 3/2 3/2 3/2 3/2 − 4 2− 6 2 → √ 0 √ 0 (2π) q + m iε ′ (2π) (2π) 2k′ (2π) (2π) 2k Z − α β 4 ν 4 4 µ 4 e (2π) (γ ) ′ ′ δ q p′ k′ e (2π) (γ ) δ (q p k) × β α − − βα − − nh 4 µ 4 ih 4 ν 4 i + e (2π) (γ ) ′ ′ δ q + k p′ e (2π) (γ ) δ q + k′ p (11.122) β α − βα − h ih io 11.9 Calculation of the cross-section for Compton scattering

We have shown the Feynman diagrams for electron-photon (Compton) scattering to lowest order (tree-level) in e. Integrating over the internal line momentum in Eq. (11.122) we replace q p + k for the ﬁrst diagram and → q p k for the second → − ′

2 4 ie2 (2π) 1 S = − u¯β′ p′,σ′ e′∗ k′ uα (p,σ) eµ (k) 4 h 4i √ 0√ 0 ν (2π)3/2 (2π) 2k′ 2k h i i ( p+ k)+ m ν 4 µ − 6 6 (γ ) ′ ′ δ p + k p′ k′ (γ ) × 2 2 β α − − βα ((p + k) + m iεα′β − h i i ( p k′)+ m µ 4 ν + − 6 − 6 (γ ) ′ ′ δ p k′ + k p′ (γ ) 2 2 β α − − βα (p k′) + m iεα′β ) − − h i factorizing the delta and simplifying factors 2 4 ie δ (p + k p′ k′) S = − − − u¯β′ p′,σ′ 2 0 0 (2π) √2k′ √2k i ( p+ k)+ m ν µ e′∗ k′ eµ (k) − 6 6 (γ ) ′ ′ (γ ) × ν 2 2 β α βα ( (p + k) + m iεα′β − i ( p k′)+ m µ ν + e′∗ k′ eµ (k) − 6 − 6 (γ ) ′ ′ (γ ) uα (p,σ) ν 2 2 β α βα (p k′) + m iεα′β ) − − reordering terms 2 4 ie δ (p + k p′ k′) S = − − − u¯β′ p′,σ′ 2 0 0 (2π) √2k′ √2k ν i ( p+ k)+ m µ e′∗γ ′ ′ − 6 6 (eµγ ) × ν β α 2 2 βα ( (p + k) + m iεα′β − µ i ( p k′)+ m ν + (eµγ ) ′ ′ − 6 − 6 e′∗γ uα (p,σ) β α 2 2 ν βα (p k′) + m iεα′β ) − − 11.9. CALCULATION OF THE CROSS-SECTION FOR COMPTON SCATTERING 353 using the notation (11.104) and deﬁning µ e∗ e∗ γ 6 ≡ µ where we emphasize that e is not ( e)∗. With this notation we ﬁnd 6 ∗ 6 2 4 ie δ (p + k p′ k′) S = − − − u¯β′ p′,σ′ 2 0 0 (2π) √2k′ √2k i ( p+ k)+ m e′∗ − 6 6 ( e) × 6 β′α′ 2 2 6 βα ( (p + k) + m iεα′β − i ( p k′)+ m + ( e) ′ ′ − 6 − 6 e′∗ uα (p,σ) 6 β α 2 2 6 βα (p k′) + m iεα′β ) − − and rewriting the result in matrix notation gives

2 4 ie δ (p′ + k′ p k) i ( p+ k)+ m S = − − − u¯ p′,σ′ e′∗ − 6 26 e 2 √ 0 0 6 (p + k) + m2 6 (2π) 2 k k′ i (h p k′)+i m + e − 6 − 6 e′∗ u (p,σ) (11.123) 6 (p k )2 + m2 6 − ′ where we have dropped the factor iε since the denominators here do not acquire a singularity. Since external lines refer to on-shell particles and phtons are massless, we have

2 2 2 2 p = m ; k = k′ = 0 − with m being the electron mass. Thus, the denominators become

(p + k)2 + m2 = p2 + k2 + 2p k + m2 · = m2 +0+2p k + m2 = 2p k − · · and similarly for the other denominator, hence

(p + k)2 + m2 = 2p k (11.124) · 2 2 p k′ + m = 2p k′ (11.125) − − · The Compton S matrix element becomes − 2 4 e i ( p+ k)+ m S = 2πiδ p′ + k′ p k u¯ p′,σ′ e′∗ − 6 6 e − − − 3 √ 0 0 6 2p k 6 (2π) 2 k k′ · i ( p k′)+ m h i e − 6 − 6 e′∗ u (p,σ) (11.126) − 6 2p k 6 · ′ 11.9.1 Feynman amplitude for Compton scattering The Feynman amplitude M deﬁned in Eq. (2.61) page 76 is given by

4 S = 2πiδ p′ + k′ p k M (11.127) − − − comparing (11.127) with (11.126) the Feynman amplitude becomes

2 e i ( p+ k)+ m i ( p k′)+ m M = 3 u¯ p′,σ′ e′∗ − 6 6 e e − 6 − 6 e′∗ u (p,σ) (11.128) 4 (2π) √k0k 0 6 p k 6 − 6 p k′ 6 ′ · · 354 CHAPTER 11. QUANTUM ELECTRODYNAMICS

To calculate the square of this amplitude we additionally sum over σ and σ′ because we shall later use an average over initial states and sum over ﬁnal states. Then we shall utilize the normalization

( i p + m) u (p,σ)u ¯ (p,σ)= − 6 αβ α β 2p0 σ X Further, for an arbitrary 4 4 matrix , with a procedure similar to the one used to obtain Eq. (11.120) we have × M 2 u¯ p′,σ′ u (p,σ) = u¯ p′,σ′ u (p,σ) u¯ (p,σ) β † β u p′,σ′ M M M σ,σ′ σ,σ′ X X h i

= βαuα (p,σ)u ¯γ (p,σ) β †β uδ p′,σ′ u¯β p′,σ′ M M γδ σ,σ′ X 2 i p + m i p′ + m u¯ p′,σ′ u (p,σ) = T r − 6 β †β − 6 (11.129) M M 2p0 M 2p 0 σ,σ′ ′ X

Combining Eqs. (11.128) and (11.129) the square of the amplitude (sum over spin states) yields

4 2 2 e i ( p+ k)+ m i ( p k′)+ m M = u¯ p′,σ′ e′∗ − 6 6 e e − 6 − 6 e′∗ u (p,σ) | | 16 (2π)6 k0k 0 6 p k 6 − 6 p k 6 σ,σ′ ′ σ,σ′ ′ X X · · 4 e i ( p+ k)+ m i ( p k′)+ m i p + m = T r e′∗ − 6 6 e e − 6 − 6 e′∗ − 6 16 (2π)6 k0k 0 6 p k 6 − 6 p k 6 2p0 ′ · · ′ i ( p+ k)+ m i ( p k′)+ m † i p′ + m β e′∗ − 6 6 e e − 6 − 6 e′∗ β − 6 (11.130) 6 p k 6 − 6 p k 6 2p 0 · · ′ ′ )

and recalling the property (7.49) page 188 βγµ† β = γ , we have − µ µ µ µ µ β z†β = β (z γ )† β = β z ∗γ† β = z ∗βγ† β = z ∗γ 6 µ µ µ − µ β z†β = z∗ (11.131) 6 − 6 more generally for a product of terms with slash

β ( z z z . . . z )† β = β z† . . . z† z† z†β = β z† ββ...ββ z†ββ z†ββ z†β 6 1 6 2 6 3 6 n 6 n 6 3 6 2 6 1 6 n 6 3 6 2 6 1 = β z† β β...β β z†β β z†β β z†β 6 n 6 3 6 2 6 1 = ( zn∗) . . . ( z3∗) ( z2∗) ( z1∗) − 6 n − 6 − 6 − 6 β ( z z z . . . z )† β = ( 1) z∗ . . . z∗ z∗ z∗ (11.132) 6 1 6 2 6 3 6 n − 6 n 6 3 6 2 6 1 from the property (11.132) and taking into account that p,p′,k,k′ are all real, the ﬁrst three products of the last line of Eq. (11.130) becomes

i ( p+ k)+ m i ( p k′)+ m † β e′∗ − 6 6 e e − 6 − 6 e′∗ β F ≡ 6 p k 6 − 6 p k 6 · · ′ i e ( p+ k) e + m e e i e ( p k ) e + m e e † = β − 6 ′∗ 6 6 6 6 ′∗ 6 − 6 6 − 6 ′ 6 ′∗ 6 6 ′∗ β p k − p k · · ′ i [ e ( p+ k) e]† + m [ e e]† i [ e ( p k ) e ]† + m [ e e ]† = β 6 ′∗ 6 6 6 6 ′∗ 6 6 6 − 6 ′ 6 ′∗ 6 6 ′∗ β p k − p k " · ! · ′ !# 11.9. CALCULATION OF THE CROSS-SECTION FOR COMPTON SCATTERING 355

iβ [ e ( p+ k) e]† β + mβ [ e e]† β iβ [ e ( p k ) e ]† β + mβ [ e e ]† β = 6 ′∗ 6 6 6 6 ′∗ 6 6 6 − 6 ′ 6 ′∗ 6 6 ′∗ F p k − p k · ! · ′ ! i ( 1)3 [ e ( p+ k) e ]+ m ( 1)2 [ e e ] i ( 1)3 [ e ( p k ) e ]+ m ( 1)2 [ e e ] = − 6 ∗ 6 6 6 ′ − 6 ∗ 6 ′ − 6 ′ 6 − 6 ′ 6 ∗ − 6 ′ 6 ∗ p k − p k · ! · ′ ! i ( p+ k)+ m i ( p k′)+ m = e∗ − 6 6 e′ e′ − 6 − 6 e∗ 6 p k 6 − 6 p k 6 · · ′ therefore we have

i ( p+ k)+ m i ( p k′)+ m † β e′∗ − 6 6 e e − 6 − 6 e′∗ β F ≡ 6 p k 6 − 6 p k 6 · · ′ i ( p+ k)+ m i ( p k′)+ m = e∗ − 6 6 e′ e′ − 6 − 6 e∗ (11.133) 6 p k 6 − 6 p k 6 · · ′ substituting (11.133) in Eq. (11.130) we have

e4 M 2 = 6 0 0 ′ | | 64 (2π) ωω′p p′ Xσ,σ [ i ( p+ k)+ m] [ i ( p k′)+ m] T r e′∗ − 6 6 e e − 6 − 6 e′∗ ( i p + m) × 6 p k 6 − 6 p k 6 − 6 · · ′ [ i ( p+ k)+ m] [ i ( p k′)+ m] e∗ − 6 6 e′ e′ − 6 − 6 e∗ i p′ + m (11.134) × 6 p k 6 − 6 p k 6 − 6 · · ′ we shall use the following “gauge”

e p = e∗ p = e′ p = e′∗ p = 0 (11.135) · · · · 0 0 such as for instance the Coulomb gauge in the laboratory frame, in which e = e′ = 0 and p = 0. Using these facts as well as Eqs. (11.200) and (11.201) we obtain

[ i p + m] e [ i p + m] = e [i p + m] [ i p + m] − 6 6 − 6 6 6 − 6 = e p2 + m2 = e p2 + m2 = 0 6 6 6 [ i p + m] e [ i p + m] = 0 − 6 6 − 6 and similarly for e , e and e because all of them are orthogonal to p 6 ′∗ 6 ′ 6 ∗

[ i p + m] e [ i p + m] = [ i p + m] e′ [ i p + m] = 0 − 6 6 − 6 − 6 6 − 6 [ i p + m] e∗ [ i p + m] = [ i p + m] e′∗ [ i p + m] = 0 (11.136) − 6 6 − 6 − 6 6 − 6 From Eq. (11.136), many products in Eq. (11.134) vanish and we can express Eq. (11.134) in a simpler form

4 2 e e′∗ k e e k′ e′∗ e∗ k e′ e′ k′ e∗ M = T r 6 6 6 + 6 6 6 ( i p + m) 6 6 6 + 6 6 6 i p′ + m | | −64 (2π)6 ωω p0p 0 p k p k − 6 p k p k − 6 σσ′ ′ ′ · · ′ · · ′ X (11.137) The trace of an odd number of Dirac matrices vanishes. As a consequence, terms linear in m vanish since they contain seven Dirac matrices. We also see that terms independent of m contains eight Dirac matrices while terms 356 CHAPTER 11. QUANTUM ELECTRODYNAMICS proportional to m2 contain six Dirac matrices. Hence, Eq. (11.137) splits into terms of zeroth and second order in m. [Homework!! C8, goes from Eq. (11.137) to Eq. (11.138)]

e4 T T T M 2 = 1 + 2 + 3 6 0 0 2 | | 64 (2π) ωω p p (p k) (p k) (p k′) (p k) (p k′) σσ′ ′ ′ · · · · · X 2 2 2 2 T4 m t1 m t2 m t3 m t4 + 2 2 2 (11.138) (p k ) − (p k) − (p k) (p k′) − (p k) (p k′) − (p k ) · ′ · · · · · · ′ with the deﬁnitions

T = T r e′∗ k e p e∗ k e′ p′ (11.139) 1 6 6 6 6 6 6 6 6 T = T r e′∗ k e p e′ k′ e∗ p′ (11.140) 1 6 6 6 6 6 6 6 6 T = T r e k′ e′∗ p e∗ k e′ p′ (11.141) 3 6 6 6 6 6 6 6 6 T = T r e k′ e′∗ p e′ k′ e∗ p′ (11.142) 4 6 6 6 6 6 6 6 6 t = T r e′∗ k e e∗ k e′ (11.143) 1 6 6 6 6 6 6 t = T r e′∗ k e e′ k′ e∗ (11.144) 2 6 6 6 6 6 6 t = T r e k′ e′∗ e∗ k e′ (11.145) 3 6 6 6 6 6 6 t = T r e k′ e′∗ e′ k′ e∗ (11.146) 4 6 6 6 6 6 6 we shall see later how to calculate the traces of the type T r ( a a a . . .) as a sum of products of scalar products 6 1 6 2 6 3 of the four-vectors a1, a2, a3 . . .. Traces of products of 6 or 8 gamma matrices (as is the case of ti and Tk) contain 15 and 105 terms in the sum respectively. However, in this case many scalar products vanish reducing the number of terms considerably. We see it by taking into account Eq. (11.135) as well as the relations (coming form the masslessness of photons) k k = k′ k′ = 0 (11.147) · · moreover, the normalization condition e e∗ = e′ e′∗ = 1 (11.148) · · also simpliﬁes the calculation.

11.9.2 Feynman amplitude for the case of linear polarization

µ µ We make an additional simpliﬁcation by assuming linear polarization. Therefore, e and e′ are real. Hence, we can omit the asterisk in Eqs. (11.139)-(11.146). For instance

T = T r e′ k e p e k e′ p′ (11.149) 1 6 6 6 6 6 6 6 6 we shall see that by virtue of some orthonormal relations the trace (11.149) will be reduced to the trace of only µ µ four slash four-vectors. We start by using e pµ = 0 with e eµ = 1, along with properties (11.200, 11.201) we obtain e p e = p e e = pe2 = p 6 6 6 − 6 6 6 − 6 − 6 and T1 becomes T = T r e′ k p k e′ p′ 1 − 6 6 6 6 6 6 µ since the photons are massless we have k kµ = 0, and using (11.199) then

k p k = k ( p k)= k [2 (p k) k p]= k k p + 2 k (p k)= k2 p + 2 k (p k) 6 6 6 6 6 6 6 · − 6 6 − 6 6 6 6 · − 6 6 · k p k = 2 k (p k) 6 6 6 6 · 11.9. CALCULATION OF THE CROSS-SECTION FOR COMPTON SCATTERING 357 from which

T = 2 (p k) T r e′ k e′ p′ 1 − · 6 6 6 6 Now using the identity (11.202) T1 yields

T r a b c d = 4[(a b) (c d) (a c) (b d) + (a d) (b c)] {6 6 6 6 } · · − · · · ·

T = 8 (p k) e′ k e′ p′ e′ e′ k p′ + e′ p′ k e′ 1 − · · · − · · · · T = 8 (p k) 2 e′ k e′ p′ k p′ (11.150) 1 − · · · − · where we have used (11.148) with e′ real. We shall make the following substitutions coming from the conservation of four-momentum and the orthogonal conditions (11.135) page 355

e′ p′ = e′ p + k k′ · · − e′ p′ = e′ k (11.151) · · now the conservation of four-momentum yields

p + k = p′ + k′ p k′ = p′ k 2 2⇒ − − ⇒ p k′ = p′ k 2 − 2 2 − 2 p 2p k′ + k′ = p′ 2k p′ + k − 2· −2 · m 2p k′ = m 2k p′ − − · − − · obtaining ﬁnally

p k′ = k p′ (11.152) · · substituting (11.151) and (11.152) in (11.150) we ﬁnd

T = 8 (p k) 2 e′ k e′ k p k′ 1 − · · · − · 2 T1 = 16 (p k) e′ k + 8 (p k) p k′ (11.153) − · · · · with a similar procedure we have [Homework!! C9, obtain expressions (11.154)-(11.157)]

2 2 2 2 T = T = 8 e k′ (p k) + 16 e e′ p k′ (p k) + 8 e e′ k k′ m 2 3 − · · · · · · · 2 2 8 e e′ m k e′ k′ e + 8 e′ k p k′ − · · · · · 2 4 (k p) + 4 k k′ p p′ 4 k p′ p k′ (11.154) − · · · − · · 2 T4 = 16 p k′ e k′ + 8 (p k) p k′ (11.155) · · · · t1 = t4 = 0 (11.156) 2 t = t = 8 e e′ k e′ k′ e + 8 k k′ e e′ 4 k k′ (11.157) 2 3 − · · · · · − · substituting all these terms in Eq. (11.138) we obtain [Homework!! C10, obtain Eq. (11.158)]

4 2 2 e 8 (k k′) 2 M = · + 32 e e′ (11.158) | | 64 (2π)6 ωω p0p 0 (k p) (k p) · σσ′ ′ ′ " · ′ · # X 358 CHAPTER 11. QUANTUM ELECTRODYNAMICS

11.9.3 Diﬀerential cross-section for Compton scattering The diﬀerential cross-section for processes with two particles in the initial state, is given by Eq. (2.143) page 95

4 1 2 4 3 3 dσ = (2π) u− M δ p′ + k′ p k d p′ d k′ (11.159) | | − − let us recall the expression (2.145) page 96 for the relative velocity u

2 2 2 (p1 p2) m1m2 u = · − q E1E2 where particles 1 and 2 refer to initial states. For our case, since the photons are massless, the initial relative velocity reads p k u = | · | (11.160) p0k0 For high-energy photon-electron dispersion (i.e. X ray or gamma ray regime), it is reasonable to assume that the electron can be taken at rest with respect to the laboratory. Thus we shall take a reference frame in which the electron is initially at rest

p = 0, p0 = m (11.161) ⇒ p k = p0k = p0k0 (11.162) · 0 − and the velocity u in Eq. (11.160) yields u = 1 (11.163) we denote the energy of the initial photon as ω, by using (11.162) such an energy yields

p0k0 p k ω = k0 = k = = · | | p0 − m and similarly for the ﬁnal photon, thus p k ω = k0 = k = · (11.164) | | − m 0 p k′ ω′ = k′ = k′ = · (11.165) − m taking into account that we are in the frame in which p = 0 , the Dirac’s delta function in (11.159) becomes

4 3 0 0 0 0 δ p′ + k′ p k = δ p′ + k′ p k δ p′ + k′ p k − − 3 − − 0 0 0− 0− = δ p′ + k′ k δ p′ + k′ p k − − − after integrating d3p we settle p = k k , and we are left only with the energy part of the delta function ′ ′ − ′ 0 0 0 0 2 2 δ p′ + k′ p k = δ p + m + ω′ m ω − − ′ − − 0 0 0 0 p 2 2 δ p′ + k′ p k = δ (k k ) + m + ω′ m ω (11.166) − − − ′ − − q from which ω′ (the ﬁnal photon energy) must satisfy

2 2 (k k ) + m = m + ω ω′ − ′ − 2 2 q 2 k + k 2 k k cos θ + m = ω′ + m + ω ′ − | | | ′| − 0 2 0 2 0 0 2 (kp) + (k ) 2k k cos θ + m = ω′ + m + ω ′ − ′ − q 11.9. CALCULATION OF THE CROSS-SECTION FOR COMPTON SCATTERING 359 so that 2 2 2 ω 2ωω cos θ + ω + m = ω + m ω′ − ′ ′ − with θ being the angle between k andp k′, i.e. the angle that measures the deﬂection of the photon. Squaring both sides and cancelling terms yields

2 2 2 2 2 2 ω 2ωω′ cos θ + ω′ + m = ω + m + ω′ + 2mω 2ωω′ 2mω′ − − − 2ωω′ cos θ = 2mω 2ωω′ 2mω′ − − − (m + ω ω cos θ) ω′ = mω − mω ω′ = ω (θ) (11.167) m + ω (1 cos θ) ≡ c − by using Eq. (11.167) and the property (2.161) page 99, the energy delta function (11.166) can be rewritten as

0 0 0 0 2 2 δ p′ + k′ p k = δ (k k ) + m + ω′ m ω − − − ′ − − q 2 2 2 = δ ω 2ωω cos θ + ω + m + ω′ m ω − ′ ′ − − 0 0 0 0 p δ (ω′ ωc (θ)) δ p′ + k′ p k = − − − ∂ 2 2 2 ′ √ω 2ωω′ cos θ + ω′ + m + ω′ ∂ω − δ (ωh ω (θ)) i = ′ − c (ω′ ω cos θ) 1+ −p′0 0 0 0 0 0 p ′ ω′ δ p′ + k′ p k = δ ω′ ω (θ) (11.168) − − mω − c

3 from the previous facts, after integrating Eq. (11.159) over d p′ we are left in that equation with the energy delta function (11.168) and taking into account the constraint p = k k′ (coming from the three-momentum part of ′ − the delta) and using Eq. (11.163), the equation (11.159) for the diﬀerential cross section reads

4 1 2 4 3 3 dσ = (2π) u− M δ p′ + k′ p k d p′ d k′ | | − − 4 2 0 0 0 0 3 = (2π) M δ p′ + k′ p k d k′ | | − − 0 4 p′ ω′ 2 3 dσ = (2π) M δ ω′ ω (θ) d k′ with p′ = k k′ (11.169) mω | | − c −

3 and the diﬀerential d k′ can be written is spherical coordinates as

3 2 d k′ = ω′ dω′ dΩ (11.170) where dΩ is the solid angle within which the final photon is scattered. The delta function in (11.169) permits to eliminate the differential dω′ in Eq. (11.170) obtaining a differential cross-section

0 4 p′ ω′ 2 2 dσ = (2π) M δ ω′ ω (θ) ω′ dω′ dΩ mω | | − c p 0ω 3 dσ = (2π)4 M 2 ′ ′ dΩ (11.171) | | mω with 0 p′ = k k′ ; p′ = m + ω ω′ and ω′ = ω (θ) (11.172) − − c and ωc (θ) is given by Eq. (11.167). 360 CHAPTER 11. QUANTUM ELECTRODYNAMICS

As we already discussed, it is not usual to measure the spin z component of the initial or ﬁnal electron. Then − we sum over σ′ and average over σ. Hence, 1 dσ¯ (p,σ; k, e) p′,σ′; k′, e′ dσ (p,σ; k, e) p′,σ′; k′, e′ (11.173) → ≡ 2 → σ,σ′ X It worths noticing that expression (11.167) can be written in the form

1 1 1 cos θ = − (11.174) ω′ − ω m so that there is an increase in the wavelength. Equation (11.174) is the usual form of the Compton formula in te scattering of X rays by electrons. − 11.9.4 Diﬀerential cross-section in the laboratory frame All previous expressions are covariant, i.e. valid in any inertial Lorentz reference frame. Since results are usually measured and analized in the laboratory, it is convenient to especialize these results to the Laboratory reference frame. In such a frame we have

k k′ = (k,ω) k′,ω′ = k k′ ωω′ = k k′ cos θ ωω′ · · · − | | − (cos θ 1) 1 1 k k′ = ωω′ cosθ ωω ′ = mωω′ − = mωω′ · − m ω − ω ′ where we have used Eq. (11.174). From this fact and picking up equations (11.164) and (11.165) we obtain

1 1 k k′ = ωω′ (cos θ 1) = mωω′ · − ω − ω ′ p k = mω ; p k′ = mω′ (11.175) · − · − substituting equations (11.158) and (11.175) in Eq. (11.171) page 359, we obtain the laboratory frame cross-section given by

4 0 3 1 (2π) 2 p′ ω′ dσ¯ dσ (p,σ; k, e) p′,σ′; k′, e′ = M dΩ ≡ 2 → 2 | | mω σ,σ′ σ,σ′ X X 4 4 2 0 3 (2π) e 8 (k k′) 2 p′ ω′ = · + 32 e e′ dΩ 2 64 (2π)6 ωω p0p 0 (k p) (k p) · mω ( ′ ′ " · ′ · #) 4 4 1 1 2 0 3 (2π) e 8 mωω′ ω ω′ 2 p′ ω′ dσ¯ = − + 32 e e′ dΩ 2 64 (2π)6 ωω p0p 0 ( mω) ( mω ) · mω ( ′ ′ " − − ′ #) And we ﬁnally obtain [Homework!! C11]

4 2 1 e ω′ dΩ ω ω′ 2 dσ (p,σ; k, e) p′,σ′; k′, e′ = + 2 + 4 e e′ (11.176) 2 → 64π2m2ω2 ω ω − · σσ′ ′ X This formula was derived by O. Klein and Y. Nishina in 1929 by using old-fashioned perturbation theory. Assuming that the initial photon is not prepared in a given polarization eigenstate, we must average over the two orthonormal vectors e. It yields 1 1 e e = δ k k 2 i j 2 ij − i j e X b b 11.10. TRACES OF DIRAC GAMMA MATRICES 361 so that the diﬀerential cross-section becomes [Homework!! C12]

4 2 1 e ω′ dΩ ω ω′ 2 dσ (p,σ; k, e) p′,σ′; k′, e′ = + 2 k e′ (11.177) 4 → 64π2m2ω2 ω ω − · e σ,σ′ ′ X X b it occurs that the scattered photon is prererentially polarized in a direction perpendicular to the initial and also the ﬁnal photon direction. So it is polarized mainly perpendicular to the plane in which the scattering of the photon occurs. Further, if the ﬁnal photon polarization is not measured in the experiment, we must sum (11.177) over both polarization states, such that

ei′ ej′ = δij ki′ kj′ ′ − Xe then we obtain [Homework!! C13] b b

4 2 1 e ω′ dΩ ω ω′ 2 dσ (p,σ; k, e) p′,σ′; k′, e′ = + 1 + cos θ (11.178) 4 → 32π2m2ω2 ω ω − e,e′ σ,σ′ ′ X X with θ being the angle between k and k′ (scattered angle of the photon). In the non-relativistic limit ω << m, Eq. (11.178) becomes b b 4 1 e dΩ 2 dσ (p,σ; k, e) p′,σ′; k′, e′ = 1 + cos θ (11.179) 4 → 32π2m2 e,e′ σ,σ′ X X by integrating over the solid angle we have

π 2π 16π 1 + cos2 θ dΩ= 1 + cos2 θ sin θ dθ dϕ = 3 Z Z0 Z0 so that the total cross-section for ω <

4 2 e 8π 2 e 15 σ = = r ; r 2.818 10− m (11.180) T 6πm2 3 0 0 ≡ 4πm ≃ × where r0 is known as the classical electron radius. Equation (11.180) is called the Thompson cross-section. Equations (11.179) and (11.180) were originally derived using classical mechanics and electrodynamics, calculating the reemission of light by a non-relativistic point charge in a plane-wave electromagnetic ﬁeld.

11.10 Traces of Dirac gamma matrices

In calculating many observables related with Dirac ﬁelds, we encounter traces of products of Dirac gamma matrices. We start by proving that for an even number of gamma matrices the trace is given by

T r γ γ γ = 4 δ g ′ (11.181) { µ1 µ2 · · · µ2N } P paired µ s pairingsX Y where the sum is over all ways of pairing the indices µ1,µ2,...,µ2N . We can see a pairing as a permutation of the integers 1, 2,..., 2N into some order P1, P2,...,P2N , in which we pair as folows

(µP1 ,µP2 ) , (µP3 ,µP4 ) , , µP2N−1 ,µP2N 362 CHAPTER 11. QUANTUM ELECTRODYNAMICS

It is clear that permuting complete pairs or permuting the two µ′s into a given pair gives the same pairing. Therefore, the number of pairings is

(2N)! = = (2N 1) (2N 3) 1 (2N 1)!! (11.182) NN N! 2N − − · · · ≡ − · In order to avoid summing over equivalent pairings we can demand

P1 < P2, P3 < P4, P2N 1 < P2N (11.183) − and P < P < P < (11.184) 1 3 5 · · · the requirement (11.183) order the elements within a pair, avoiding duplication of pairs. The requirement (11.184) avoids replications due to permutations of complete sets of pairs. With this convention the factor δ = 1 P ± according whether the pairing involves an even or odd permutation of indices. Thus, the product in Eq. (11.181) is over all N pairs, and the nth pair contributes with a factor g . µP2n−1 µP2N As an example, for N = 1 there is only one possible pairing

(µ ,µ ) (11.185) { 1 2 } which coincides with the result of formula (11.182)

= (2 1 1)!! = 1!! = 1 N1 × − for N = 2 in Eq. (11.182), the number of pairings is

= (2 2 1)!! = 3!! = 3 N2 × − that could be listed as

(µ ,µ ) , (µ ,µ ) ; (µ ,µ ) , (µ ,µ ) (µ ,µ ) , (µ ,µ ) (11.186) { 1 2 3 4 } { 1 3 2 4 } { 1 4 2 3 } note that each pairing in (11.186) has been chosen to satisfy conditions (11.183) and (11.184). For N = 3 the number of pairings yields

= (2 3 1)!! = 5!! = 5 3 1 = 15 N3 × − × × and the list of all 15 pairings that follows the rules (11.183) and (11.184) is the following

(µ ,µ ) , (µ ,µ ) , (µ ,µ ) ; (µ ,µ ) , (µ ,µ ) , (µ ,µ ) ; (µ ,µ ) , (µ ,µ ) , (µ ,µ ) { 1 2 3 4 5 6 } { 1 2 3 5 4 6 } { 1 2 3 6 4 5 } (µ ,µ ) , (µ ,µ ) , (µ ,µ ) ; (µ ,µ ) , (µ ,µ ) , (µ ,µ ) ; (µ ,µ ) , (µ ,µ ) , (µ ,µ ) { 1 3 2 4 5 6 } { 1 3 2 5 4 6 } { 1 3 2 6 4 5 } (µ ,µ ) , (µ ,µ ) , (µ ,µ ) ; (µ ,µ ) , (µ ,µ ) , (µ ,µ ) ; (µ ,µ ) , (µ ,µ ) , (µ ,µ ) { 1 4 2 3 5 6 } { 1 4 2 5 3 6 } { 1 4 2 6 3 5 } (µ ,µ ) , (µ ,µ ) , (µ ,µ ) ; (µ ,µ ) , (µ ,µ ) , (µ ,µ ) ; (µ ,µ ) , (µ ,µ ) , (µ ,µ ) { 1 5 2 3 4 6 } { 1 5 2 4 3 6 } { 1 5 2 6 3 4 } (µ ,µ ) , (µ ,µ ) , (µ ,µ ) ; (µ ,µ ) , (µ ,µ ) , (µ ,µ ) ; (µ ,µ ) , (µ ,µ ) , (µ ,µ ) { 1 6 2 3 4 5 } { 1 6 2 4 3 5 } { 1 6 2 5 3 4 } the parity of each term is dictated by the permutation from the standard order to the order induced by the pairing for instance for the pairing (µ ,µ ) , (µ ,µ ) , (µ ,µ ) P9 ≡ { 1 4 2 6 3 5 } the associated permutation is 123456 142635 11.10. TRACES OF DIRAC GAMMA MATRICES 363 which is a even permutation. And writing

µ,ν,ρ,σ,κ,η µ ,µ ,µ ,µ ,µ ,µ ↔ 1 2 3 4 5 6 we ﬁnd

T r γ γ = 4g (11.187) { µ ν} µν T r γ γ γ γ = 4 [g g g g + g g ] (11.188) { µ ν ρ σ} µν ρσ − µρ νσ µσ νρ T r γ γ γ γ γ γ = 4 [g g g g g g + g g g g g g + g g g { µ ν ρ σ κ η} µν ρσ κη − µν ρκ ση µν ρη σκ − µρ νσ κη µρ νκ ση g g g + g g g g g g + g g g g g g − µρ νη σκ µσ νρ κη − µσ νκ ρη µσ νη ρκ − µκ νρ ση +g g g g g g + g g g g g g + g g g ] (11.189) µκ νσ ρη − µκ νη ρσ µη νρ σκ − µη νσ ρκ µη νκ ρσ On the other hand, the trace of an odd product of Dirac gamma matrices is zero

T r γ γ γ = 0 1 2 · · · µ2N+1 We can prove Eq. (11.181) by using mathematical induction. First of all, by using the Cliﬀord algebra of gamma matrices and the cyclic invariance of traces we have11

T r γµγν = T r 2gµν I4 4 γνγµ = 2gµν T r I4 4 T r γνγµ { } { × − } { × }− { } T r γ γ = 8g T r γ γ 2T r γ γ = 8g { µ ν} µν − { µ ν } ⇒ { µ ν} µν T r γ γ = 4g (11.190) ⇒ { µ ν} µν as we already discussed for N = 1 (which is the case here) we have only one pairing of the form (11.185) so that Eq. (11.190) coincides with (11.181) for N = 1. Now that we have proved that formula (11.181) works for N = 1, we assume that it is true for N M 1. ≤ − Under such a hypothesis we ﬁnd

T r γ γ γ = 2g g T r γ γ T r γ γ γ γ { µ1 µ2 · · · µ2M } µ1 µ2 { µ3 · · · µ2M }− { µ2 µ1 µ3 · · · µ2M } T r γ γ γ = 2g g T r γ γ 2g T r γ γ γ { µ1 µ2 · · · µ2M } µ1 µ2 { µ3 · · · µ2M }− µ1µ3 { µ2 µ4 · · · µ2N } +T r γ γ γ γ γ { µ2 µ3 µ1 µ4 · · · µ2M } T r γ γ γ = 2g g T r γ γ 2g T r γ γ γ { µ1 µ2 · · · µ2M } µ1 µ2 { µ3 · · · µ2M }− µ1µ3 { µ2 µ4 · · · µ2M } +2g g T r γ γ γ γ µ1 µ4 { µ2 µ3 µ5 · · · µ2M }−··· +2g g T r γ γ T r γ γ γ µ1 µ2M µ2 · · · µ2M−1 − { µ2 · · · µ2M µ1 } Now, recalling that the trace is cyclic invariant, the last term substracted in the last expression coincides with the left hand side. Consequently

2T r γ γ γ = 2g g T r γ γ 2g T r γ γ γ { µ1 µ2 · · · µ2M } µ1 µ2 { µ3 · · · µ2M }− µ1µ3 { µ2 µ4 · · · µ2M } +2g g T r γ γ γ γ + 2g g T r γ γ µ1 µ4 { µ2 µ3 µ5 · · · µ2M }−··· µ1 µ2M µ2 · · · µ2M−1 so that

T r γ γ γ = g g T r γ γ g T r γ γ γ { µ1 µ2 · · · µ2M } µ1 µ2 { µ3 · · · µ2M }− µ1µ3 { µ2 µ4 · · · µ2M } +g g T r γ γ γ γ + g g T r γ γ (11.191) µ1 µ4 { µ2 µ3 µ5 · · · µ2M }−··· µ1 µ2M µ2 · · · µ2M−1 now if we assume that Eq. (11.181) provides the correct trace for any product of 2N 2 gamma matrices, we − conclude that Eq. (11.191) shows that Eq. (11.181) provides the correct trace for any product of 2N gamma matrices.

11 We should take into account that gµν is NOT a matrix in this context, since µ and ν are ﬁxed indices. So gµν is a number and as a matrix is gµν times the identity 4×4. 364 CHAPTER 11. QUANTUM ELECTRODYNAMICS

We can see that the trace of an odd number of gamma matrices vanishes by recalling the property

1 γ = γ γ (γ )− − µ 5 µ 5 So for an od number 2N + 1 of gamma matrices we have

1 1 1 1 1 1 γ γ γ γ γ (γ )− = γ γ (γ )− γ γ (γ )− γ (γ )− γ γ (γ )− γ γ (γ )− 5 µ1 µ2 · · · µ2N µ2N+1 5 5 µ1 5 5 µ2 5 5 · · · 5 5 µ2N 5 5 µ2N+1 5 1 1 1 1 = γ γ (γ )− γ γ (γ )− γ γ (γ )− γ γ (γ )− 5 µ1 5 5 µ2 5 · · · 5 µ2N 5 5 µ2N+1 5 =h ( 1)2N+1 γ γih γ γ i h ih i − µ1 µ2 · · · µ2N µ2N+1 1 γ γ γ γ γ (γ )− = γ γ γ γ (11.192) 5 µ1 µ2 · · · µ2N µ2N+1 5 − µ1 µ2 · · · µ2N µ2N+1 taking traces on both sides of Eq (11.192) and using the fact that traces are invariant under a similarity transformation, we obtain

1 T r γ γ γ γ γ (γ )− = T r γ γ γ γ 5 µ1 µ2 · · · µ2N µ2N+1 5 − µ1 µ2 · · · µ2N µ2N+1 n T r γ γ γ γ o = T r γ γ γ γ µ1 µ2 · · · µ2N µ2N+1 − µ1 µ2 · · · µ2N µ2N+1 T r γ γ γ γ = 0 (11.193) ⇒ µ1 µ2 · · · µ2N µ2N+1 We also encounter traces of the form T r γ γ γ γ (11.194) { 5 µ1 µ2 · · · µn } taking into account that γ iγ γ γ γ (11.195) 5 ≡ 0 1 2 3 if n is odd in Eq. (11.194), then substituting Eq. (11.195) into Eq. (11.194) we obtain the trace of n + 4 gamma matrices which is also odd. Thus the trace (11.194) vanish for odd n

T r γ γ γ γ = 0 5 µ1 µ2 · · · µ2N+1 It also vanishes for n = 0 and n = 2 T r γ = T r γ γ γ = 0 { 5} { 5 µ ν} we can show that by observing that from expression (11.195) we cannot pair the indices of

T r γ γ γ γ or T r γ γ γ γ γ γ (11.196) { 0 1 2 3} { 0 1 2 3 µ ν} in such a way that the spacetime indices in each pair are equal. Thus all possible pairings of

0, 1, 2, 3 or 0, 1, 2, 3,µ,ν

contain at least one non-diagonal gµiµj = 0 in the product of g′s. We can see it explicitly by applying Eqs. (11.188) and (11.189) to obtain the traces (11.196). By contrast, for n = 4 we can pair the indices in

T r γ γ γ γ γ = iT r γ γ γ γ γ γ γ γ (11.197) { 5 µ ν ρ σ} { 0 1 2 3 µ ν ρ σ} in such a way that the spacetime indices in each pair are all equal so that all g′s in the product are diagonal. However, it is possible only if the set µ, ν, ρ, σ is some permutation of the set 0, 1, 2, 3. Now, since gamma matrices with diﬀerent indices anticommute, we see that the trace (11.197) must be odd under permutations of µ, ν, ρ, σ. Thus by deﬁning T T r γ γ γ γ γ µνρσ ≡ { 5 µ ν ρ σ} 11.11. SOME PROPERTIES OF “SLASH” MOMENTA 365

This is a totally antisymmetric structure of four indices in which each index can take four values. Therefore, the trace (11.197) must be proportional to the totally antisymmetric tensor εµνρσ. We can determine the constant of proportionality by setting µ, ν, ρ, σ 0, 1, 2, 3 recalling that → ε = ε0123 = 1 0123 − − then we obtain T r γ γ γ γ γ = 4iε (11.198) { 5 µ ν ρ σ} µνρσ using the same procedure to obtain Eq. (11.181) we can obtain the trace of products of γ5 with six, eight or more gamma matrices.

11.11 Some properties of “slash” momenta

From the Cliﬀrod algebra of the Dirac matrices we have

a b = (a γµ) (b γν )= a b γµγν = a b (2gµν γν γµ) 6 6 µ ν µ ν µ ν − a b = (2a gµν b b γνa γµ) 6 6 µ ν − ν µ a b = 2a b b a (11.199) 6 6 · − 6 6 in particular if the four-momenta a and b are orthogonal we obtain a b = b a if a b = 0 (11.200) 6 6 − 6 6 · another interesting particular case appears from Eq. (11.199) by setting a = b a2 = 2a2 a2 a2 = a2 (11.201) 6 − 6 ⇒ 6 of course, in the scalars that appear in these identities we should multiply by a 4 4 unit matrix. × We shall also encounter traces of products of slash momenta. We can calculate them by using the traces of products of Dirac gamma matrices. For instance, using Eq. (11.193) we have

T r a a a = T r (aµ1 γ ) (aµ2 γ ) aµ2N+1 γ {6 1 6 2 · · · 6 2N+1} µ1 µ2 · · · µ2N+1 = aµ1 aµ2 aµ2N+1 T r γ γ γ · · · µ1 µ2 · · · µ2N+1 T r a a a = 0 {6 1 6 2 · · · 6 2N+1} Using Eq. (11.187) we have T r a b = T r aµγ aνγ = aµaν T r γ γ = 4aµaνg {6 6 } { µ ν} { µ ν} µν T r a b = 4a b {6 6 } · this result can also be obtained from Eq. (11.199) and the cyclic invariance of the trace

T r a b = T r 2 (a b) 14 4 b a = 2 (a b) T r 14 4 T r b a {6 6 } { · × − 6 6 } · { × }− {6 6 } T r a b = 8 (a b) T r a b 2T r a b = 8 (a b) {6 6 } · − {6 6 }⇒ {6 6 } · T r a b = 4 (a b) {6 6 } · now, from Eq. (11.188) we ﬁnd T r a b c d = aµbνcρdσT r γ γ γ γ = 4aµbνcρdσ [g g g g + g g ] {6 6 6 6 } { µ ν ρ σ} µν ρσ − µρ νσ µσ νρ = 4 [(aµg bν ) (cρg dσ) (aµg cρ) (bνg dσ) + (aµg dσ) (bνg cρ)] µν ρσ − µρ νσ µσ νρ T r a b c d = 4 [(a b) (c d) (a c) (b d) + (a d) (b c)] (11.202) {6 6 6 6 } · · − · · · · Chapter 12

Path integral approach for bosons in quantum ﬁeld theory

By applying the canonical quantization methods we were able to derive the Feynman rules for many theories. However, in the case of scalar fields with derivative couplings or the vector fields, we saw that the interaction Hamiltonian contains a non-covariant term that is cancelled by a non-covariant term in the propagator1. In quantum electrodynamics, the non-covariant term in the interacting Hamiltonian (the Coulomb energy) is not spatially local but it is local in time. Forgetting the cancellation process we can manage to use only the covariant terms to obtain the Feynman rules. However, this process of cancellation of non-covariant terms could be very difficult to manage in non-abelian theories and in general relativity. Consequently, it is preferable a method that uses the Lagrangian directly to derive the Feynman rules in a manifestly covariant form. The path integral approach is a good alternative to work directly with a Lagrangian rather than a Hamiltonian. Indeed their are useful in non-abelian theories even when they exhibit spontaneous symmetry breaking, as is the case of the standard model of weak and electromagnetic interactions. Moreover, path integral methods permit to account on the contributions to the S matrix with an essential singularity at zero coupling constant that cannot − be discovered at any givne order in perturbation theory. The reason to start with a canonical formalism lies in the fact that in such a formalism the S matrix is − clearly unitary. On the other hand, the path-integral formalism provides manifestly Lorentz invariance in the diagrammatic rules but the only way to show that the path-integral approach gives a unitary S matrix is by − reconstructing the canonical formalism in which unitarity is apparent. Hence, in the canonical formalism unitarity is apparent while Lorentz invariance is obscure and the opposite occurs with the path-integral formalism. The advantage of deriving the path-integral formalism from the canonical approach (as we shall do) is that we are certain of having the same S matrix in both formalisms. Consequently, we can guarantee for the S matrix both − − the Lorentz-invariance and unitarity. A second reason to start with the canonical formalism has to do with the fact that for some important theories the simplest version of the path-integral approach (without starting with a canonical formalism) in which propagators and interaction vertices are taken directly from the Lagragian are wrong. This is the case with the non-linear σ model − 1 = g φ ∂ φk (∂µφm) L −2 km µ in which the Feynman ruels derived directly from the Lagrangian density would lead to a non-unitary and wrong S matrix, that even depends on the way in which we define the scalar field. By deriving the path-integral − approach from a canonical formalism we are able to see the additional sorts of vertices required to correct the simplest version of the Feynman path-integral formulation.

1The covariant term in the interaction Hamiltonian equals the negative of the interaction term in the Lagrangian.

366 12.1. THE GENERAL PATH-INTEGRAL FORMULA FOR BOSONIC OPERATORS 367

12.1 The general path-integral formula for bosonic operators

Let us assume a general quantum mechanical system with Hermitian bosonic operator “coordiantes” Qa and conjugate momenta Pb that satisfy canonical (bosonic) commutation (and NOT anticommutation) relations

[Qa, Pb] = iδab (12.1)

[Qa,Qb] = [Pa, Pb] = 0 (12.2) we are assuming that any ﬁrst class constraints are eliminated by choosing an apropriate gauge, and that the remaining second class constraints are solved by expressing the constrained variables in terms of the unconstrained variables Qa and Pa. Thus, we are dealing with independent canonical variables only. The index a, is condensing continuous (position) indices as well as discrete ones (discrete Lorentz and species m indices)

Q Qx Q (x) (12.3) a ≡ ,m ≡ m P Px P (x) (12.4) a ≡ ,m ≡ m in the same way the kronecker delta in Eq. (12.1) means

3 δ δx y δ (x y) δ (12.5) ab ≡ ,m ; ,n ≡ − mn the operators defined so far are in the Schrödinger picture taken at a fixed time (e.g. t = 0). Since the set of Q s commute, we can find a basis of simultaneous eigenstates q with eigenvalues q a′ | i a Q q = q q (12.6) a | i a | i note that the lower case notation qa denotes eigenvalues instead of operators in the interaction picture. Such a notation is not misleading since we shall not use the interaction picture by now. The eigenvectors of such a basis can be taken as orthonormal (in the extended sense)

q′ q = δ q′ q δ q′ q | i a − a ≡ − a Y with the corresponding completeness relation

1= dq q q a | i h | a Z Y in the same way we can ﬁnd an orthonormal basis p of eigenvectors common to all P s: | i a′ P p = p p (12.7) a | i a | i p′ p = δ p′ p δ p′ p (12.8) | i a − a ≡ − a Y 1 = dp p p (12.9) a | i h | a Z Y Now from Eq. (12.1) it can be shown that P acts as a derivative operator i∂/∂q on wave functions on the b − b q basis. Then the wave function p in the basis q is given by − | i {| i} 1 q p = exp (iq p ) (12.10) h | i √ a a a 2π Y 368 CHAPTER 12. PATH INTEGRAL APPROACH FOR BOSONS IN QUANTUM FIELD THEORY

1 where the factor √ is obtained from the normalization condition (12.8). On the other hand, Eq. (12.10) can a 2π also be seen as theQ scalar product between the two complete orthonormal sets q and p . {| i} {| i} In the Heisenberg picture, the canonical operators acquires time-dependence

Q (t) exp (iHt) Q exp ( iHt) (12.11) a ≡ a − P (t) exp (iHt) P exp ( iHt) (12.12) a ≡ a − where H is the total Hamiltonian. The canonical operators in the Heisenberg picture have eigenstates q; t and | i p; t | i Q (t) q; t = q q; t a | i a | i P (t) p; t = p p; t a | i a | i it is easy to obtain the basis q; t and p; t in terms of q and p | i | i | i | i Q (t) q; t = q q; t exp ( iHt) Q (t) exp (iHt) exp ( iHt) q; t = q exp ( iHt) q; t a | i a | i ⇒ − a − | i a − | i Q [exp ( iHt) q; t ] = q [exp ( iHt) q; t ] a − | i a − | i and comparing with (12.6) we obtain q = exp ( iHt) q; t | i − | i and similarly for momenta eigenstates then we have

q; t = exp (iHt) q ; p; t = exp (iHt) p (12.13) | i | i | i | i it worhts emphasizing that q; t is the eigenstate of Q (t) with eigenvalue q and it is NOT the time-evolution | i a a of the state q for a time t. Owing to it, its time dependence is given by a factor exp (iHt) (i.e. the inverse | i of the time-evolution operator) instead that exp ( iHt). Once again, these states satisfy the completeness and − orthonormality conditions

Q (t) q; t = q q; t ; P (t) p; t = p p; t a | i a | i a | i a | i q′; t q; t = δ q′ q δ q′ q ; p′; t p; t δ p′ p | i a − a ≡ − | i≡ − a Y dq q; t q; t = dp p; t p; t = 1 a | i h | a | i h | a a Z Y Z Y as well as the product 1 q; t p; t = exp (iq p ) (12.14) h | i √ a a a 2π Y If after measuring at time t, our system becomes prepared at the state q; t , the probability amplitude of ﬁnding | i it at the state q ; t when we measure at time t , is given by the scalar product | ′ ′i ′

q′; t′ q; t (12.15) A≡ | i we shall focus on calculating this probability amplitude. 12.1. THE GENERAL PATH-INTEGRAL FORMULA FOR BOSONIC OPERATORS 369

12.1.1 Probability amplitude for inﬁnitesimal time-intervals The amplitude (12.15) is easier to calculate when the time interval involved is inﬁnitesimal. From Eq. (12.13) we have

q′; τ + dτ q; τ = q′; τ exp ( iH dτ) q; τ (12.16) | i − | i Originally the Hamiltonian is expressed as a function H ( Q, P ), but taking into account that the Hamiltonian iHt commutes with itself (and so with e± ), and that Eqs. (12.11) and (12.12) are similarity transformations, the Hamiltonian can equally be written as the same function H (Q (t) , P (t))

iHt iHt H H (Q, P )= e H (Q, P ) e− = H (Q (t) , P (t)) ≡ The Hamiltonian can be expressed in several forms through the commutation relations (12.1) and (12.2) by moving the operators Q and P to obtain a given order. We shall settle it in a “standard form” or “normal order” with all Q′s to the left of all P ′s. As a matter of example, if a term of the form PaQbPc appears in the Hamiltonian, by using Eq. (12.1), we have P Q P = Q P P iδ P a b c b a c − ab c by setting this normal order the Qa (t)′ s in the Hamiltonian of Eq. (12.16) could be replaced by the eigenvalue qa′ of Q (t) associated with the bra. Notice that such a replacement is possible only because the operator exp [ iH dτ] a − is linear in H for inﬁnitesimal dτ. Now to deal with the P (t) operators (on the right) it is convenient to insert an identity involving their eigenvectors p; τ | i

q′; τ + dτ q; τ = dp q′; τ exp [ iH (Q (τ) , P (τ)) dτ] p; τ p; τ q; τ | i a − | i h | i Z a Y

= dp q′; τ 1 iH (Q (τ) , P (τ)) dτ p; τ p; τ q; τ a { − } | i h | i Z a Y

= dp q′; τ p; τ p; τ q; τ i dp q′; τ H (Q (τ) , P (τ)) dτ p; τ p; τ q; τ a i h | i− a { } | i h | i Z a Z a Y Y taking into account the order chosen for the Q, P operators in the Hamiltonian, we can replace H (Q (τ) , P (τ)) by H (q′,p). In addition, by using the relation (12.14) we obtain

dpa q′; τ + dτ q; τ = exp iH q′,p dτ + i q′ q p (12.17) | i 2π − a − a a Z a " a # Y X with p integrated over the interval ( , ). a −∞ ∞ 12.1.2 Probability amplitude for ﬁnite time intervals We shall calculate q ; t q; t , for t

q′; t′ q; t = dq1 dqN q′; t′ qN ; τN qN ; τN qN 1; τN 1 qN 1; τN 1 q1; τ1 q1; τ1 q; t | i · · · i h | − − i h − − |···| i h | i Z

(12.19) 370 CHAPTER 12. PATH INTEGRAL APPROACH FOR BOSONS IN QUANTUM FIELD THEORY by deﬁning q q ; q = q′ 0 ≡ N+1 along with deﬁnitions (12.18) for τN+1 and τ0. We can rewrite (12.19) as

q′; t′ q; t = dq1 dqN qN+1; τN+1 qN ; τN qN ; τN qN 1; τN 1 qN 1; τN 1 q1; τ1 q1; τ1 q0; τ0 | i · · · h | i h | − − i h − − |···| i h | i Z (12.20)

We have inner products of the type

q ; τ q ; τ ; k = 0, 1, 2,...,N h k+1 k+1| k ki if N is large enough, the time intervals become small enough to apply the identity (12.17) valid for inﬁnitesimal time intervals

dp q ; τ q ; τ a exp iH (q ,p ) ∆τ + i (q q ) p (12.21) h k+1 k+1 | k ki≈ 2π − k+1 k k+1,a − k,a k,a a " a # Z Y X substituting (12.21) in (12.20) we ﬁnd

N N dpk,a q′; t′ q; t dq | i ≈ k,a 2π " a #" a # Z kY=1 Y kY=0 Y N exp i (q q ) p H (q ,p ) ∆τ (12.22) × k+1,a − k,a k,a − k+1 k ( " a #) Xk=0 X where we have deﬁned q (τ ) q ; p (τ ) p (12.23) a k ≡ k,a a k ≡ k,a now we take the limit in which N so that ∆τ dτ 0. In that case the aproximation (12.22) becomes → ∞ ≡ → exact and we obtain N N (q q ) p H (q ,p ) dτ ≡ k+1,a − k,a k,a − k+1 k ( a ) Xk=0 X N = q˙ (τ ) p (τ ) H (q (τ ) ,p (τ )) dτ + (dτ)2 a k a k − k k O ( a ) Xk=0 X t′ N q˙ (τ) p (τ) H (q (τ) ,p (τ)) dτ (12.24) → a a − t ( a ) Z X additionally we can deﬁne integrals over the functions q (τ) and p (τ) as follows

dpb (τ) dpk,b dqa (τ) lim dqk,a (12.25) 2π ···≡ dτ 0 2π · · · τ,a → Z Y Yτ,b Z Yk,a Yk,b therefore, Eq. (12.22) becomes a constrained path integral

dpb (τ) q′; t′ q; t = dqa (τ) qa(t)=qa | i ′ ′ 2π Zqa(t )=qa τ,a τ,b Y Y t′ exp i dτ q˙ (τ) p (τ) H (q (τ) ,p (τ)) (12.26) × a a − ( t " a #) Z X 12.1. THE GENERAL PATH-INTEGRAL FORMULA FOR BOSONIC OPERATORS 371

we call it a path integral because we are integrating over all paths that take q (τ) from q at τ = t to q′ at τ = t′, and also over all p (τ). We shall see that by writing the matrix elements as path integrals, we can calculate them easily by expanding in powers of the coupling constants in H.

12.1.3 Calculation of matrix elements of operators through the path-integral formalism In addition to amplitudes of probabilities we can obtain matrix elements between states q ; t and q; t of time- h ′ ′| | i ordered products of general operators of the form (P (τ) ,Q (τ)). In this case it is more convenient to deﬁne O these operators with all P ′s moved to the left of all Q′s. Thus, by inserting such an operator in Eq. (12.17) we obtain

q′; τ + dτ (P (τ) ,Q (τ)) q; τ M ≡ O | i = dp q′; τ exp [ iH (Q (τ) , P (τ)) dτ] p; τ p; τ (P (τ) ,Q (τ)) q; τ a − | i h | O | i Z a Y

Owing to the order chosen for the operators Q, P in the Hamiltonian and in the operator , we can make the O replacements H (Q (τ) , P (τ)) H q′,p and (P (τ) ,Q (τ)) (p,q) → O → O within the inner products. Then we can use the relation (12.14). The procedure is similar to the one that led to Eq. (12.26)

dpa q′; τ + dτ (P (τ) ,Q (τ)) q; τ = exp iH q′,p dτ + i q′ q p (p,q) (12.27) O | i 2π − a − a a O Z a " a # Y X

Now, to calculate the matrix element of a product of time ordered operators

(P (τ ) ,Q (τ )) (P (τ ) ,Q (τ )) ; t >t > OA1 A1 A1 OA2 A2 A2 · · · A1 A2 · · · we should insert these operators between the apropriate states in Eq. (12.19) [or Eq. (12.20)], and then utilize Eq. (12.27). For example, if t lies between τ and τ , we should insert (P (τ ) ,Q (τ )) between A1 k k+1 OA1 A1 A1 q ; τ and q ; τ . We observe that in Eq. (12.19) each sucessive sum over states is at a later time, and it is h k+1 k+1| | k ki possible because of the assumption that t >t > . With the same procedure as before, we ﬁnd the general A1 A2 · · · path-integral formula given by

q′; t′ A1 (P (τA1 ) ,Q (τA1 )) A2 (P (τA2 ) ,Q (τA2 )) q; t MO ≡ O O ···| i dpb (τ) = dqa (τ) A1 (p (τA1 ) ,q (τA1 )) A2 (p (τA2 ) ,q (τA2 )) qa(t)= qa 2π O O · · · ′ ′ τ,a Zqa(t )=qa Y Yτ,b t′ exp i dτ q˙ (τ) p (τ) H (q (τ) ,p (τ)) (12.28) × a a − ( t " a #) Z X in which we are assuming the time ordering given by

t′ >t >t > >t A1 A2 · · · nevertheless, there is nothing on the right-hand side of Eq. (12.28) that refers to the order of time-arguments. Therefore, if we have a path integral of the form of the right-hand side of Eq. (12.28) but with t , t in A1 A2 · · · arbitrary order (except that they are all between t and t′ with t

T q′; t′ T A1 (P (τA1 ) ,Q (τA1 )) A2 (P (τA2 ) ,Q (τA2 )) q; t M O ≡ {O O ···}| i dpb (τ) = dqa (τ) A1 (p (τA1 ) ,q (τA1 )) A2 (p (τA2 ) ,q (τA2 )) qa(t)= qa 2π O O · · · ′ ′ τ,a Zqa(t )=qa Y Yτ,b t′ exp i dτ q˙ (τ) p (τ) H (q (τ) ,p (τ)) (12.29) × a a − ( t " a #) Z X where as customary, T represents time-ordering of the operators. Note that the c number functions q (τ) and p (τ) in Eq. (12.29) are unconstrained variables of integration − a a (each one is swept independently) and in particular they are not constrained to obey the classical equations of motion ∂H (q (τ) ,p (τ)) ∂H (q (τ) ,p (τ)) q˙a (τ) =0;p ˙a (τ)+ = 0 (12.30) − ∂pa (τ) ∂qa (τ) owing to it, the Hamiltonian H (q (τ) ,p (τ)) in Eq. (12.29) is not constant in τ. Notwithstanding, in some sense path integrals respect those equations of motion. Let us assume that one of the functions in Eq. (12.29) e.g. (P (τ ) ,Q (τ )), is the left hand side of either of Eqs. (12.30). It can be seen that for t

t′ I [q,p] dτ q˙ (τ) p (τ) H (q (τ) ,p (τ)) ≡ a a − t ( a ) Z X whenever tA1 does not approach t nor t′, the integration over the variables qa (tA1 ) and pa (tA2 ) are not constrained, and if they have well behavior in convergence, the integral of these variational derivatives must vanish. Consequently, the path integral given by (12.29) vanishes if (p,q) is taken to be the left.hand side of either of OA1 the equations of motion (12.30).

The rule above is valid only if the integration variables qa (tAi ) , pa (tAi ) are independent of any other pair of variables q (t ) , p (t ) that appears in any other function different from in Eq. (12.29). Then, the a Ak a Ak OAk OAi rule follows only if we forbid t to approach t with k = i, and also different from t and t . If t approaches Ai Ak 6 ′ Ai a given t , the path integral will contain a non-zero term proportional to δ (t t ) or its derivatives. These Ak Ai − Ak are the same delta function that we encounter in the operator formalism coming from time derivatives of step functions that are implicit in the definition of time-ordered products. In order to evaluate the path integrals (12.26) or (12.29) it suffices to know the classical Hamiltonian as a c number function H (q,p). In formulating a path integral approach a natural question is how to choose the − Hamiltonian after the quantization H (Q, P ). Certainly, there are several Hamiltonians that differs only in the order in which the Q′s and P ′s are put after quantization. According to our developments the apropriate choice seems to be the one with all Q′s on the left and all P ′s on the right. However, such a choice depends on the interpretation given to the measure dqa (τ) dpb (τ) that appears in the path integrals (12.26) or (12.29).Y TheY prescription of all Q′s on the left and all P ′s on the right is correct only if the measured is interpreted according with Eqs. (12.23), (12.24) and (12.25). With other 12.2. PATH FORMALISM FOR THE S MATRIX 373 − measures we would have other prescription for the ordering of the operators. Nevertheless, different prescriptions for ordering the operators in the Hamiltonian lead to different choices of the constants that appear as coefficients in the various terms of the Hamiltonian. Since such coefficients are considered arbitrary parameters (to be adjusted by phenomenology) any prescription is physically equivalent to any other. The path integral form (12.29) is not adequate for numerical calculations or to prove theorems. In that case it is more advantageous to use a path integral method in which we calculate amplitudes in the Euclidean space with t substituted by ix , so that the exponential in Eq. (12.29) becomes a negative real quantity. In the Minkowski − 4 space, we have oscillating paths generating rapid oscillations of the integrand from one path to another. By using the Euclidean space, the oscillating paths are exponentially supressed. Indeed we could start directly from a path integral formulation in the Euclidean space or pass from one formulation to the other. By now, we use the Minkowski space since our main goal is to calculate Feynman amplitudes with a perturbation approach.

12.2 Path formalism for the S matrix − The results so far are valid in the framework of ordinary quantum mechanics. A more suitbale notation for quantum ﬁeld theory is obtained by setting the index a to run over points x in space and over a spin and species index m so that we replace Q (t) Q (x,t) ; P (t) P (x,t) a → m a → m further, we also rewrite

H (q (t) ,p (t)) H [q (t) ,p (t)] ; (p (t) ,q (t)) H [p (t) ,q (t)] → O → to emphasize that the Hamiltonian and operators are functionals of qm (x,t) and pm (x,t) at a given time t. From these facts, we rewrite Eq. (;;;) as

T q′; t′ T A1 [P (tA1 ) ,Q (tA1 )] A2 [P (tA2 ) ,Q (tA2 )] q; t M O ≡ {O O ···}| i dpm (x,τ) = dqm (x,τ) A1 [p (tA1 ) ,q (tA1 )] A2 [p (tA2 ) ,q (tA2 )] qm(x ,t)=qm(x) 2π O O · · · ′ ′ τ,x,m τ,x,m Zqm(x,t )=qm(x) Y Y t′ exp i dτ d3x q˙ (x,τ) p (x,t) H [q (τ) ,p (τ)] (12.31) × m m − ( t " m #) Z Z X Bibliography

[1] Steven Weinberg, The Quantum Theory of Fields. Vol. I: Foundations. Cambridge University Press (1995).

[2] Steven Weinberg, The Quantum Theory of Fields. Vol. II: Modern Applications. Cambridge University Press (1996).

[3] Steven Weinberg, The Quantum Theory of Fields. Vol. III: Supersymmetry. Cambridge University Press (2000).

[4] Lewis H. Ryder, Quantum Field Theory (Second Ed.). Cambridge University Press (1996).

[5] Michio Kaku, Quantum Field Theory, a modern introduction. Oxford University Press (1993).

[6] John Collins, Renormalization. Cambridge Monographs on Mathematical Physics. Cambridge University Press (1984).

[7] Claude Itzykson, Jean-Bernard Zuber, Quantum Field Theory. McGraw-Hill International Editions, Physics Series (1980).

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