Quantum Field Theory I

Physics 523 & 524: Quantum Field Theory I

Kevin Cahill Department of Physics and Astronomy University of New Mexico, Albuquerque, NM 87131

Contents

1 Quantum fields and special relativity page 1 1.1 States 1 1.2 Creation operators 1 1.3 How fields transform 3 1.4 Translations 7 1.5 Boosts 8 1.6 Rotations 9 1.7 Spin-zero fields 9 1.8 Conserved charges 12 1.9 Parity, charge conjugation, and time reversal 13 1.10 Vector fields 15 1.11 Vector field for spin-zero particles 18 1.12 Vector field for spin-one particles 18 1.13 Lorentz group 23 1.14 Gamma matrices and Clifford algebras 25 1.15 Dirac’s gamma matrices 27 1.16 Dirac fields 28 1.17 Expansion of massive and massless Dirac fields 41 1.18 Spinors and angular momentum 44 1.19 Charge conjugation 49 1.20 Parity 51 2 Feynman diagrams 53 2.1 Time-dependent perturbation theory 53 2.2 Dyson’s expansion of the S matrix 54 2.3 The Feynman propagator for scalar fields 56 2.4 Application to a cubic scalar field theory 60 2.5 Feynman’s propagator for fields with spin 62 ii Contents 2.6 Feynman’s propagator for spin-one-half fields 63 2.7 Application to a theory of fermions and bosons 66 2.8 Feynman propagator for spin-one fields 71 2.9 The Feynman rules 73 2.10 Fermion-antifermion scattering 74 3 Action 76 3.1 Lagrangians and hamiltonians 76 3.2 Canonical variables 76 3.3 Principle of stationary action in field theory 78 3.4 Symmetries and conserved quantities in field theory 79 3.5 Poisson Brackets 84 3.6 Dirac Brackets 85 3.7 Dirac Brackets and QED 86 4 Quantum Electrodynamics 93 4.1 Global U( 1) Symmetry 93 4.2 Abelian gauge invariance 94 4.3 Coulomb-gauge quantization 96 4.4 QED in the interaction picture 97 4.5 Photon propagator 98 4.6 Feynman’s rules for QED 100 4.7 Electron-positron scattering 101 4.8 Trace identities 104 4.9 Electron-positron to muon-antimuon 105 4.10 Electron-muon scattering 111 4.11 One-loop QED 112 4.12 Magnetic moment of the electron 124 4.13 Charge form factor of electron 132 5 Nonabelian gauge theory 134 5.1 Yang and Mills invent local nonabelian symmetry 134 5.2 SU(3) 135 6 Standard model 138 6.1 Quantum chromodynamics 138 6.2 Abelian Higgs mechanism 138 6.3 Higgs’s mechanism 140 6.4 Quark and lepton interactions 146 6.5 Quark and Lepton Masses 148 6.6 CKM matrix 153 6.7 Lepton Masses 156 6.8 Before and after symmetry breaking 157 Contents iii 6.9 The Seesaw Mechanism 161 6.10 Neutrino Oscillations 163 7 Path integrals 165 7.1 Path integrals and Richard Feynman 165 7.2 Gaussian integrals and Trotter’s formula 165 7.3 Path integrals in quantum mechanics 166 7.4 Path integrals for quadratic actions 170 7.5 Path integrals in statistical mechanics 175 7.6 Boltzmann path integrals for quadratic actions 180 7.7 Mean values of time-ordered products 183 7.8 Quantum field theory 185 7.9 Finite-temperature field theory 189 7.10 Perturbation theory 192 7.11 Application to quantum electrodynamics 196 7.12 Fermionic path integrals 200 7.13 Application to nonabelian gauge theories 208 7.14 Faddeev-Popov trick 208 7.15 Ghosts 211 7.16 Integrating over the momenta 212 8 Landau theory of phase transitions 214 8.1 First- and second-order phase transitions 214 9 Effective field theories and gravity 217 9.1 Effective field theories 217 9.2 Is gravity an effective field theory? 218 9.3 Quantization of Fields in Curved Space 221 9.4 Accelerated coordinate systems 228 9.5 Scalar field in an accelerating frame 229 9.6 Maximally symmetric spaces 233 9.7 Conformal algebra 236 9.8 Conformal algebra in flat space 237 9.8.1 Angles and analytic functions 241 9.9 Maxwell’s action is conformally invariant for d = 4 247 9.10 Massless scalar field theory is conformally invariant if d = 2 248 9.11 Christoffel symbols as nonabelian gauge fields 249 9.12 Spin connection 254 10 Field Theory on a Lattice 256 10.1 Scalar Fields 256 10.2 Finite-temperature field theory 257 iv Contents 10.3 The Propagator 261 10.4 Pure Gauge Theory 261 10.5 Pure Gauge Theory on a Lattice 264 References 266 Index 267 1 Quantum fields and special relativity

1.1 States A Lorentz transformation Λ is implemented by a unitary operator U(Λ) which replaces the state |p, σi of a massive particle of momentum p and spin σ along the z-axis by the state s 0 (Λp) X (j) U(Λ)|p, σi = D (W (Λ, p)) |Λp, s0i (1.1) p0 s0σ s0 where W (Λ, p) is a Wigner rotation

W (Λ, p) = L−1(Λp)ΛL(p) (1.2) and L(p) is a standard Lorentz transformation that takes (m,~0) to p.

1.2 Creation operators The vacuum is invariant under Lorentz transformations and translations

U(Λ, a)|0i = |0i. (1.3)

A creation operator a†(p, σ) makes the state |p, σi from the vacuum state |0i |pσi = a†(p, σ)|0i. (1.4)

The creation and annihilation operators obey either the commutation relation

† 0 0 † 0 0 † 0 0 (3) 0 [a(p, s), a (p , s )]− = a(p, s) a (p , s ) − a (p , s ) a(p, s) = δss0 δ (p − p ) (1.5) 2 Quantum ﬁelds and special relativity or the anticommutation relation

† 0 0 † 0 0 † 0 0 (3) 0 [a(p, s), a (p , s )]+ = a(p, s) a (p , s ) + a (p , s ) a(p, s) = δss0 δ (p − p ). (1.6) The two kinds of relations are written together as

† 0 0 † 0 0 † 0 0 (3) 0 [a(p, s), a (p , s )]∓ = a(p, s) a (p , s ) ∓ a (p , s ) a(p, s) = δss0 δ (p − p ). (1.7) A bracket [A, B] with no signed subscript is interpreted as a commutator. Equations (1.1 & 1.4) give

s 0 (Λp) X (j) U(Λ)a†(p, σ)|0i = D (W (Λ, p)) a†(Λp, s0)|0i. (1.8) p0 s0σ s0

And (1.3) gives

s 0 (Λp) X (j) U(Λ)a†(p, σ)U −1(Λ)|0i = D (W (Λ, p)) a†(Λp, s0)|0i. (1.9) p0 s0σ s0

SW in chapter 4 concludes that

s 0 (Λp) X (j) U(Λ)a†(p, σ)U −1(Λ) = D (W (Λ, p)) a†(Λp, s0). (1.10) p0 s0σ s0

If U(Λ, b) follows Λ by a translation by b, then

s 0 (Λp) X (j) U(Λ, b)a†(p, σ)U −1(Λ, b) = e−i(Λp)·a D (W (Λ, p)) a†(Λp, s0) p0 s0σ s0 s 0 (Λp) X †(j) = e−i(Λp)·a D (W −1(Λ, p)) a†(Λp, s0) p0 s0σ s0 s 0 (Λp) X ∗(j) = e−i(Λp)·a D (W −1(Λ, p)) a†(Λp, s0) p0 σs0 s0 (1.11) 1.3 How ﬁelds transform 3 The adjoint of this equation is s 0 (Λp) X ∗(j) U(Λ, b)a(p, σ)U −1(Λ, b) = ei(Λp)·a D (W (Λ, p)) a(Λp, s0) p0 s0σ s0 s 0 (Λp) X †(j) = ei(Λp)·a D (W (Λ, p)) a(Λp, s0) p0 σs0 s0 s 0 (Λp) X (j) = ei(Λp)·a D (W −1(Λ, p)) a(Λp, s0). p0 σs0 s0 (1.12) These equations (1.11 & 1.12) are (5.1.11 & 5.1.12) of SW.

1.3 How fields transform The “positive frequency” part of a field is a linear combination of annihilation operators Z + X 3 ψ` (x) = d p u`(x; p, σ) a(p, σ). (1.13) σ The “negative frequency” part of a field is a linear combination of creation operators of the antiparticles Z − X 3 † ψ` (x) = d p v`(x; p, σ) b (p, σ). (1.14) σ To have the fields (1.13 & 1.14) transform properly under Poincarétrans- formations X U(Λ, a)ψ+(x)U −1(Λ, a) = D (Λ−1)ψ+(Λx + a) ` ``¯ `¯ `¯ Z X −1 X 3 = D``¯(Λ ) d p u`¯(Λx + a; p, σ) a(p, σ) `¯ σ X U(Λ, a)ψ−(x)U −1(Λ, a) = D (Λ−1)ψ−(Λx + a) ` ``¯ `¯ `¯ Z X −1 X 3 † = D``¯(Λ ) d p v`¯(Λx + a; p, σ) b (p, σ) `¯ σ (1.15) the spinors u`(x; p, σ) and v`(x; p, σ) must obey certain rules which we’ll now determine. 4 Quantum fields and special relativity First (1.12 & 1.13) give

Z + −1 X 3 −1 U(Λ, a)ψ` (x)U (Λ, a) = U(Λ, a) d p u`(x; p, σ) a(p, σ)U (Λ, a) σ Z X 3 −1 = d p u`(x; p, σ) U(Λ, a)a(p, σ)U (Λ, a) σ (1.16) s Z 0 X (Λp) X (j) = d3p u (x; p, σ) ei(Λp)·a D (W −1(Λ, p)) a(Λp, s0). ` p0 σs0 σ s0

Now we use the identity

d3p d3(Λp) = (1.17) p0 (Λp)0 to turn (1.16) into

Z + −1 X 3 i(Λp)·a U(Λ, a)ψ` (x)U (Λ, a) = d (Λp) u`(x; p, σ) e σ s 0 p X (j) × D (W −1(Λ, p)) a(Λp, s0). (1.18) (Λp)0 σs0 s0

Similarly (1.11, 1.14, & 1.17) give

Z − −1 X 3 † −1 U(Λ, a)ψ` (x)U (Λ, a) = U(Λ, a) d p v`(x; p, σ) b (p, σ)U (Λ, a) σ Z X 3 † −1 = d p v`(x; p, σ) U(Λ, a)b (p, σ)U (Λ, a) (1.19) σ s Z 0 X (Λp) X ∗(j) = d3p v (x; p, σ) e−i(Λp)·a D (W −1(Λ, p)) b†(Λp, s0) ` p0 σs0 σ s0 s Z 0 X p X ∗(j) = d3(Λp) v (x; p, σ) e−i(Λp)·a D (W −1(Λ, p)) b†(Λp, s0). ` (Λp)0 σs0 σ s0

So to get the ﬁelds to transform as in (1.15), equations (1.18 & 1.19) say 1.3 How ﬁelds transform 5 that we need

X X X Z D (Λ−1)ψ+(Λx + a) = D (Λ−1) d3p u (Λx + a; p, σ) a(p, σ) ``¯ `¯ ``¯ `¯ `¯ `¯ σ Z X −1 X 3 = D``¯(Λ ) d (Λp) u`¯(Λx + a;Λp, σ) a(Λp, σ) `¯ σ Z X 3 i(Λp)·a = d (Λp) u`(x; p, σ) e (1.20) σ s 0 p X (j) × D (W −1(Λ, p)) a(Λp, s0) (Λp)0 σs0 s0 Z X 3 0 i(Λp)·a = d (Λp) u`(x; p, s ) e s0 s 0 p X (j) × D (W −1(Λ, p)) a(Λp, σ) (Λp)0 s0σ σ and

X X X Z D (Λ−1)ψ−(Λx + a) = D (Λ−1) d3p v (Λx + a; p, σ) b†(p, σ) ``¯ `¯ ``¯ `¯ `¯ `¯ σ Z X −1 X 3 † = D``¯(Λ ) d (Λp) v`¯(Λx + a;Λp, σ) b (Λp, σ) `¯ σ Z X 3 −i(Λp)·a = d (Λp) v`(x; p, σ) e (1.21) σ s 0 p X ∗(j) × D (W −1(Λ, p)) b†(Λp, s0) (Λp)0 σs0 s0 Z X 3 0 −i(Λp)·a = d (Λp) v`(x; p, s ) e s0 s 0 p X ∗(j) × D (W −1(Λ, p)) b†(Λp, σ). (Λp)0 s0σ σ

Equating coefficients of the red annihilation and blue creation operators, we find that the fields will transform properly if the spinors u and v satisfy the 6 Quantum fields and special relativity rules

s 0 X −1 p X (j) −1 0 i(Λp)·a D ¯(Λ ) u¯(Λx + a;Λp, σ) = D (W (Λ, p))u (x; p, s ) e `` ` (Λp)0 s0σ ` `¯ s0 (1.22) s 0 X −1 p X ∗(j) −1 0 −i(Λp)·a D ¯(Λ )v¯(Λx + a;Λp, σ) = D (W (Λ, p))v (x; p, s )e `` ` (Λp)0 s0σ ` `¯ s0 (1.23)

which diﬀer from SW’s by an interchange of the subscripts σ, s0 on the rotation matrices D(j). (I think SW has a typo there.) If we multiply both sides of these equations (1.22 & 1.23) by the two kinds of D matrices, then we get ﬁrst

X −1 D`0`(Λ)D``¯(Λ ) u`¯(Λx + a;Λp, σ) = u`0 (Λx + a;Λp, σ) `,`¯ s 0 p X (j) −1 0 i(Λp)·a = D (W (Λ, p))D 0 (Λ)u (x; p, s ) e (1.24) (Λp)0 s0σ ` ` ` s0,` X −1 D`0`(Λ)D``¯(Λ )v`¯(Λx + a;Λp, σ) = v`0 (Λx + a;Λp, σ) `,`¯ s 0 p X ∗(j) −1 0 −i(Λp)·a = D (W (Λ, p))D 0 (Λ)v (x; p, s )e (1.25) (Λp)0 s0σ ` ` ` s0,` 1.4 Translations 7 and then with W ≡ W (Λ, p)

X (j) Dσs¯ (W )u`0 (Λx + a;Λp, σ) σ s 0 p X (j) −1 (j) 0 i(Λp)·a = D (W )D (W )D 0 (Λ)u (x; p, s ) e (Λp)0 s0σ σs¯ ` ` ` s0,σ,` s 0 p X i(Λp)·a = D 0 (Λ)u (x; p, s¯) e (1.26) (Λp)0 ` ` ` ` X ∗(j) Dσs¯ (W )v`0 (Λx + a;Λp, σ) σ s 0 p X ∗(j) −1 ∗(j) 0 −i(Λp)·a = D (W )D (W )D 0 (Λ)v (x; p, s )e (Λp)0 s0σ σs¯ ` ` ` σ,s0,` s 0 p X −i(Λp)·a = D 0 (Λ)v (x; p, s¯)e (1.27) (Λp)0 ` ` ` ` which are equations (5.1.13 & 5.1.14) of SW: s 0 X (j) p X i(Λp)·a u¯(Λx + a;Λp, s¯)D (W (Λ, p)) = D¯ (Λ)u (x; p, σ) e ` sσ¯ (Λp)0 `` ` s¯ ` s 0 X ∗(j) p X −i(Λp)·a v¯(Λx + a;Λp, s¯)D (W (Λ, p)) = D¯ (Λ)v (x; p, σ)e . ` sσ¯ (Λp)0 `` ` s¯ ` (1.28) These are the equations that determine the spinors u and v up to a few arbitrary phases.

1.4 Translations When Λ = I, the D matrices are equal to unity, and these last equations (1.28) say that for x = 0

ip·a u`(a; p, σ) = u`(0; p, σ) e −ip·a (1.29) v`(a; p, σ) = v`(0; p, σ)e . Thus the spinors u and v depend upon spacetime by the usual phase e±ip·x

−3/2 ip·x u`(x; p, σ) = (2π) u`(p, σ) e (1.30) −3/2 −ip·x v`(x; p, σ) = (2π) v`(p, σ)e 8 Quantum fields and special relativity in which the 2π’s are conventional. The fields therefore are Fourier transforms: Z + −3/2 X 3 ip·x ψ` (x) = (2π) d p e u`(p, σ) a(p, σ) σ Z (1.31) − −3/2 X 3 −ip·x † ψ` (x) = (2π) d p e v`(p, σ) b (p, σ) σ and every field of mass m obeys the Klein-Gordon equation

2 2 2 2 (∇ − ∂0 − m ) ψ`(x) = (2 − m ) ψ`(x) = 0. (1.32)

Since exp[i(Λp · (Λx + a))] = exp(ip · x + iΛp · a), the conditions (1.28) simplify to s 0 X (j) p X u¯(Λp, s¯)D (W (Λ, p)) = D¯ (Λ)u (p, σ) ` sσ¯ (Λp)0 `` ` s¯ ` s (1.33) 0 X ∗(j) p X v¯(Λp, s¯)D (W (Λ, p)) = D¯ (Λ)v (p, σ) ` sσ¯ (Λp)0 `` ` s¯ ` for all Lorentz transformations Λ.

1.5 Boosts Set p = k = (m,~0) and Λ = L(q) where L(q)k = q. So L(p) = 1 and

W (Λ, p) ≡ L−1(Λp)ΛL(p) = L−1(q)L(q) = 1. (1.34)

Then the equations (1.33) are

rm X u¯(q, σ) = D¯ (L(q))u (~0, σ) ` q0 `` ` ` (1.35) rm X v¯(q, σ) = D¯ (L(q))v (~0, σ). ` q0 `` ` `

Thus a spinor at finite momentum is given by a representation D(Λ) of the Lorentz group (see the online notes of chapter 10 of my book for its finite-dimensional nonunitary representations) acting on the spinor at zero 3-momentum p = k = (m,~0). We need to find what these spinors are. 1.6 Rotations 9 1.6 Rotations Now set p = k = (m,~0) and Λ = R a rotation so that W = R. For rotations, the spinor conditions (1.33) are X ~ (j) X ~ u`¯(0, s¯)Dsσ¯ (R) = D``¯ (R)u`(0, σ) s¯ ` (1.36) X ~ ∗(j) X ~ v`¯(0, s¯)Dsσ¯ (R) = D``¯ (R)v`(0, σ). s¯ `

(j) The representations Dsσ¯ (R) of the rotation group are (2j + 1) × (2j + 1)- dimensional unitary matrices. For a rotation of angle θ about the θ~ = θ axis, they are the ones taught in courses on quantum mechanics (and discussed in the notes of chapter 10)

(j) h −iθ·J (j) i Dsσ¯ (θ) = e (1.37) sσ¯ where [Ja,Jb] = iabcJc. The representations D``¯ (R) of the rotation group are ﬁnite-dimensional unitary matrices. For a rotation of angle θ about the θ~ = θ axis, they are h −iθ·J i D``¯ (θ) = e (1.38) ``¯ in which [Ja, Jb] = iabcJ . For tiny rotations, the conditions (1.36) require (because of the complex conjugation of the antiparticle condition) that the spinors obey the rules X ~ (j) X ~ u`¯(0, s¯)(Ja )sσ¯ = (Ja)``¯ u`(0, σ) s¯ ` (1.39) X ~ ∗(j) X ~ v`¯(0, s¯)(−Ja)sσ¯ ) = (Ja)``¯ v`(0, σ) s¯ ` for a = 1, 2, 3.

1.7 Spin-zero ﬁelds Spin-zero ﬁelds have no spin or Lorentz indexes. So the boost conditions p 0 p 0 (1.210) merely require that u(q) = m/q √u(0) and v(q) = √m/q v(0). The conventional normalization is u(0) = 1/ 2m and v(0) = 1/ 2m. The spin-zero spinors then are

u(p) = (2p0)−1/2 and v(p) = (2p0)−1/2. (1.40)

For simpicity, let’s first consider a neutral scalar field so that b(p, s) = 10 Quantum fields and special relativity a(p, s). The definitions (1.13) and (1.14) of the positive-frequency and negative- frequency fields and their behavior (1.30) under translations then give us Z d3p φ+(x) = a(p) eip·x p(2π)32p0 (1.41) Z d3p φ−(x) = a†(p) e−ip·x. p(2π)32p0 Note that φ±(x)† = φ∓(x). (1.42) 0 Since [a(p), a(p )]± = 0, it follows that + + − − [φ (x), φ (y)]∓ = 0 and [φ (x), φ (y)]∓ = 0 (1.43) whatever the values of x and y as long as we use commutators for bosons and anticommutators for fermions. But the commutation relation † (3) [a(p, s), a (q, t)]∓ = δst δ (p − q) (1.44) makes the commutator Z 3 3 0 + − d pd p ip·x −ip0·y 3 0 [φ (x), φ (y)]∓ = e e δ (p − p ) (2π)3p2p02p00 (1.45) Z d3p = eip·(x−y) = ∆ (x − y) (2π)32p0 + nonzero even for (x − y)2 > 0 as we’ll now verify. For space-like x, the Lorentz-invariant function ∆+(x) can only depend 2 0 upon x > 0 since the time x and its sign are√ not Lorentz invariant. So we choose a Lorentz frame with x0 = 0 and |x| = x2. In this frame, Z 3 d p ip·x ∆+(x) = e (2π)32pp2 + m2 (1.46) Z p2dp d cos θ = eipx cos θ (2π)22pp2 + m2 where p = |p| and x = |x|. Now Z d cos θ eipx cos θ = eipx − e−ipx /(ipx) = 2 sin(px)/(px), (1.47) so the integral (1.46) is 1 Z ∞ sin(px) pdp ∆+(x) = 2 p (1.48) 4π x 0 p2 + m2 1.7 Spin-zero fields 11 with u ≡ p/m Z ∞ m sin(mxu) udu m 2 ∆+(x) = √ = K1(mx ) (1.49) 2 2 2 4π x 0 u + 1 4π x a Hankel function. To get a Lorentz-invariant, causal theory, we use the arbitrary parameters κ and λ setting φ(x) = κφ+(x) + λφ−(x) (1.50) Now the adjoint rule (1.42) and the commutation relations (1.45 and 1.45) give † + − ∗ − + [φ(x), φ (y)]∓ = [κφ (x) + λφ (x), κ φ (y) + λφ (y)]∓ 2 + − 2 − + = |κ| [φ (x), φ (y)]∓ + |λ| [φ (x), φ (y)]∓ 2 2 = |κ| ∆+(x − y) ∓ |λ| ∆+(y − x) + − + − (1.51) [φ(x), φ(y)]∓ = [κφ (x) + λφ (x), κφ (y) + λφ (y)]∓ + − − + = κλ [φ (x), φ (y)]∓ + [φ (x), φ (y)]∓

= κλ (∆+(x − y) ∓ ∆+(y − x)) .

2 But when (x − y) > 0, ∆+(x − y) = ∆+(y − x). Thus these conditions are † 2 2 [φ(x), φ (y)]∓ = |κ| ∓ |λ| ∆+(x − y) (1.52) [φ(x), φ(y)]∓ = κλ∆+(x − y)(1 ∓ 1). The first of these equations implies that we choose the minus sign and so that we use commutation relations and not anticommutation relations for spin-zero fields. This is the spin-statistics theorem for spin-zero fields. SW proves the theorem for arbitrary massive fields in section 5.7. We also must set |κ| = |λ|. (1.53) The second equation then is automatically satisfied. The common magnitude and the phases of κ and λ are arbitrary, so we choose κ = λ = 1. We then have φ(x) = φ+(x) + φ−(x) = φ+(x) + φ+†(x) = φ†(x). (1.54) Now the interaction density H(x) will commute with H(y) for (x − y)2 > 0, and we have a chance of having a Lorentz-invariant, causal theory. The field (1.54) Z d3p h i φ(x) = a(p) eip·x + a†(p) e−ip·x (1.55) p(2π)32p0 12 Quantum fields and special relativity obeys the Klein-Gordon equation

2 2 2 2 (∇ − ∂0 − m ) φ(x) ≡ (2 − m ) φ(x) = 0. (1.56)

1.8 Conserved charges If the field φ adds and deletes charged particles, an interaction H(x) that is a polynomial in φ will not commute with the charge operator Q because φ+ will lower the charge and φ− will raise it. The standard way to solve this problem is to start with two hermitian fields φ1 and φ2 of the same mass. One defines a complex scalar field as a complex linear combination of the two fields 1 φ(x) = √ (φ1(x) + iφ2(x)) 2 Z 3 d p 1 ip·x 1 † † −ip·x = √ (a1(p) + ia2(p)) e + √ a (p) + ia (p) e . p(2π)32p0 2 2 1 2 (1.57) Setting

1 † 1 † † a(p) = √ (a1(p) + ia2(p)) and b (p) = √ a (p) + ia (p) (1.58) 2 2 1 2 so that

1 † 1 † † b(p) = √ (a1(p) − ia2(p)) and a (p) = √ a (p) − ia (p) (1.59) 2 2 1 2 we have Z d3p h i φ(x) = a(p) eip·x + b†(p) e−ip·x (1.60) p(2π)32p0 and Z d3p h i φ†(x) = b(p) eip·x + a†(p) e−ip·x . (1.61) p(2π)32p0 Since the commutation relations of the real creation and annihilation operators are for i, j = 1, 2

† 0 3 0 0 † † 0 [ai(p), aj(p )] = δij δ (p − p ) and [ai(p), aj(p )] = 0 = [ai (p), aj(p )] (1.62) the commutation relations of the complex creation and annihilation operators are [a(p), a†(p0)] = δ3(p − p0) and [b(p), b†(p0)] = δ3(p − p0) (1.63) 1.9 Parity, charge conjugation, and time reversal 13 with all other commutators vanishing. Now φ(x) lowers the charge of a state by q if a† adds a particle of charge q and if b† adds a particle of charge −q. Similarly, φ†(x) raises the charge of a state by q [Q, φ(x)] = − qφ(x) and [Q, φ†(x)] = qφ†(x). (1.64) So an interaction with as many φ(x)’s as φ†(x)’s conserves charge.

1.9 Parity, charge conjugation, and time reversal If the unitary operator P represents parity on the creation operators

† −1 † † −1 † Pa1(p)P = η a1(−p) and Pa2(p)P = η a2(−p) (1.65) with the same phase η. Then

−1 ∗ −1 ∗ Pa1(p)P = η a1(−p) and Pa2(p)P = η a2(−p) (1.66) and so both Pa†(p)P−1 = ηa†(−p) and Pa(p)P−1 = η∗a(−p) (1.67) and Pb†(p)P−1 = ηb†(−p) and Pb(p)P−1 = η∗b(−p). (1.68) Thus if the field Z 3 d p h ip·x † −ip·xi φ1(x) = a1(p) e + a (p) e (1.69) p(2π)32p0 1 or φ2(x), or the complex field (1.60) is to go into a multiple of itself under parity, then we need η = η∗ so that η is real. Then the fields transform under parity as −1 ∗ 0 0 Pφ1(x)P = η φ1(x , −x) = ηφ1(x , −x) −1 ∗ 0 0 Pφ2(x)P = η φ(x , −x) = ηφ2(x , −x) (1.70) Pφ(x)P−1 = η∗φ(x0, −x) = ηφ(x0, −x). Since P2 = I, we must have η = ±1. SW allows for a more general phase by having parity act with the same phase on a and b†. Both schemes imply that the parity of a hermitian field is ±1 and that the state Z |abi = d3p f(p2) a†(p) b†(−p) |0i (1.71) has even or positive parity, P|abi = |abi. 14 Quantum fields and special relativity Charge conjugation works similarly. If the unitary operator C represents charge conjugation on the creation operators

† −1 † † −1 † Ca1(p)C = ξa1(p) and Ca2(p)C = − ξa2(p) (1.72) with the same phase ξ. Then

−1 ∗ −1 ∗ Ca1(p)C = ξ a1(p) and Ca2(p)C = − ξ a2(p) (1.73) √ √ and so since a = (a1 + ia2)/ 2 and b = (a1 − ia2)/ 2 Ca(p)C−1 = ξ∗ b(p) and Cb(p)C−1 = ξ∗a(p) (1.74) √ √ † † † † † † and since a = (a1 − ia2)/ 2 and b = (a1 + ia2)/ 2 Ca†(p)C−1 = ξ b†(p) and Cb†(p)C−1 = ξa†(p). (1.75)

Thus under charge conjugation, the ﬁeld (1.60) becomes Z d3p h i Cφ(x)C−1 = ξ∗ b(p) eip·x + ξ a†(p) e−ip·x (1.76) p(2π)32p0 and so if it is to go into a multiple of itself or of its adjoint under charge conjugation then we need ξ = ξ∗ so that ξ is real. We then get

Cφ(x)C−1 = ξ∗ φ†(x) = ξ φ†(x). (1.77)

Since C2 = I, we must have ξ = ±1. SW allows for a more general phase by having charge conjugation act with the same phase on a and b†. Both schemes imply that the charge-conjugation parity of a hermitian ﬁeld is ±1 and that the state Z |abi = d3p f(p2) a†(p) b†(p) |0i (1.78) has even or positive charge-conjugation parity, C|abi = |abi. The time-reversal operator T is antilinear and antiunitary. So if −1 ∗ −1 ∗ Ta1(p)T = ζ a1(−p) and Ta2(p)T = − ζ a2(−p) (1.79) † −1 † † −1 † Ta1(p)T = ζa1(−p) and Ta2(p)T = − ζa2(−p) then

−1 1 −1 1 −1 −1 Ta(p)T = T√ a1(p) + ia2(p) T = √ Ta1(p)T − iTa2(p)T 2 2 ∗ 1 ∗ = ζ √ a1(−p) + ia2(−p) = ζ a(−p) 2 (1.80) 1.10 Vector ﬁelds 15 and 1 1 Tb†(p)T−1 = T√ a†(p) + ia†(p)T−1 = √ Ta†(p)T−1 − iTa†(p)T−1 2 1 2 2 1 2 1 = ζ √ a†(−p) + ia†(−p) = ζ b†(−p) 2 1 2 (1.81) then one has

Z d3p h i Tφ(x)T−1 = T a(p) eip·x + b†(p) e−ip·x T−1 p(2π)32p0 Z d3p h i = Ta(p)T−1 e−ip·x + Tb†(p)T−1 eip·x (1.82) p(2π)32p0 Z d3p h i = ζ∗ a(−p) e−ip·x + ζ b†(−p) eip·x . p(2π)32p0

So if ζ is real, then after replacing −p by p, we get

Z 3 −1 ∗ d p h ip0x0+ip·x † −ip0x0+ip·xi Tφ(x)T = ζ p a(p) e + b (p) e (2π)32p0 (1.83) = ζ∗φ(−x0, x) = ζφ(−x0, x).

Since T2 = I, the phase ζ = ±1. SW lets ζ be complex but deﬁned only for complex scalar ﬁelds and not for their real and imaginary parts.

1.10 Vector ﬁelds Vector ﬁelds transform like the 4-vector xi of spacetime. So

`¯ D``¯ (Λ) = Λ ` (1.84) for `,¯ ` = 0, 1, 2, 3. Again we start with a hermitian ﬁeld labelled by i = 0, 1, 2, 3

X Z φ+i(x) = (2π)−3/2 d3p eip·xui(p, s) a(p, s) s (1.85) X Z φ−i(x) = (2π)−3/2 d3p e−ip·xvi(p, s) a†(p, s). s 16 Quantum ﬁelds and special relativity The boost conditions (1.210) say that

r m X ui(p, s) = L(p)i uk(~0, s) p0 k k (1.86) r m X vi(p, s) = L(p)i vk(~0, s). p0 k k

The rotation conditions (1.39) give

X i ~ (j) X i k ~ u (0, s¯)(Ja )ss¯ = (Ja) ku (0, s) s¯ k (1.87) X i ~ ∗(j) X i k ~ − v (0, s¯)(Ja )ss¯ = (Ja) kv (0, s). s¯ k

(j) The (2j + 1) × (2j + 1) matrices (Ja )ss¯ are the generators of the (2j + 1) × (2j + 1) representation of the rotation group. (See my online notes on group theory.) You learned that

3 3 j X h (j) 2i X X (j) (j) (Ja ) = (Ja )ss¯ (Ja )ss0 = j(j + 1)δss¯ 0 (1.88) ss¯ 0 a=1 a=1 s=−j and that

0 1 0 0 −i 0  1 0 0  1 1 J1 = √ 1 0 1 ,J2 = √ i 0 −i ,J3 = 0 0 0  2 0 1 0 2 0 i 0 0 0 −1 (1.89) in courses on quantum mechanics. i For k = 1, 2, 3, the three 4 × 4 matrices (Jk) j are the generators of rotations in the vector representation of the Lorentz group. Their nonzero components are

i (Jk) j = − iijk (1.90)

0 0 i for i, j, k = 1, 2, 3, while (Jk) 0 = 0, (Jk) j = 0, and (Jk) 0 = 0 for i, j, k = 1, 2, 3. So

2 i i (J ) j = 2δ j (1.91)

2 0 2 0 2 i with (J ) 0 = 0, (J ) j = 0, and (J ) 0 = 0 for i, j = 1, 2, 3. Apart from a factor of i, the Jk’s are the 4 × 4 matrices Ja = iRa of my online notes on 1.10 Vector ﬁelds 17 the Lorentz group

0 0 0 0  0 0 0 0 0 0 0 0 0 0 0 0  0 0 0 i 0 0 −i 0 J1 =   J2 =   J3 =   . 0 0 0 −i 0 0 0 0 0 i 0 0 0 0 i 0 0 −i 0 0 0 0 0 0 (1.92) 0 Since (Ja) k = 0 for a, k = 1, 2, 3, the spin conditions (1.87) give for i = 0 X 0 ~ (j) X 0 ~ ∗(j) u (0, s¯)(Ja )ss¯ = 0 and − v (0, s¯)(Ja )ss¯ = 0. (1.93) s¯ s¯

(j) Multiplying these equations from the right by (Ja )ss0 while summing over (j) 2 a = 1, 2, 3 and using the formula (1.88) [(J ) ]ss0 = j(j + 1) δss0 , we ﬁnd

j(j + 1) u0(~0, s) = 0 and j(j + 1) v0(~0, s) = 0. (1.94)

Thus u0(~0, σ) and v0(~0, σ) can be anything if the ﬁeld represents particles of spin j = 0, but u0(~0, σ) and v0(~0, σ) must both vanish if the ﬁeld represents particles of spin j > 0. Now we set i = 1, 2, 3 in the spin conditions (1.87) and again multiply (j) from the right by (Ja )ss0 while summing over a = 1, 2, 3 and using the formula (1.88) (J (j))2 = j(j + 1). The Lorentz rotation matrices generate a j = 1 representation of the group of rotations.

3 X i k i i (Ja) k (Ja) ` = j(j + 1) δ` = 2δ`. (1.95) ka=1 So the remaining conditions on the ﬁelds are

i ~ 0 X i ~ (j) (j) X i k ~ (j) j(j + 1) u (0, s ) = u (0, s¯)(Ja )ss¯ (Ja )ss0 = (Ja) ku (0, s)(Ja )ss0 ssa¯ ksa X i k ` ~ 0 X i ` ~ 0 i ~ 0 = (Ja) k (Ja) ` u (0, s ) = 2 δ` u (0, s ) = 2 u (0, s ) k`a k i ~ 0 X i ~ ∗(j) ∗(j) X i k ~ (j) j(j + 1) v (0, s ) = v (0, s¯)(Ja )ss¯ (Ja )ss0 = (Ja) kv (0, s)(Ja )ss0 ssa¯ ksa X i k ` ~ 0 X i ` ~ 0 i ~ 0 = (Ja) k (Ja) ` v (0, s ) = 2 δ` v (0, s ) = 2 v (0, s ). k`a k (1.96)

Thus if j = 0, then for i = 1, 2, 3 both ui(~0, s) and vi(~0, s) must vanish, while if j > 0, then since j(j + 1) = 2, the spin j must be unity, j = 1. 18 Quantum ﬁelds and special relativity 1.11 Vector ﬁeld for spin-zero particles The only nonvanishing components are constants taken conventionally as

u0(~0) = ipm/2 and v0(~0) = − ipm/2. (1.97)

At ﬁnite momentum the boost conditions (1.210) give them as p p uµ(~p) = ipµ/ 2p0 and vµ(~p) = − ipµ/ 2p0. (1.98)

The vector ﬁeld φµ(x) of a spin-zero particle is then the derivative of a scalar ﬁeld φ(x) Z d3p h i φµ(x) = ∂µφ(x) = ipµ a(p) eip·x − ipµ b†(p) e−ip·x (1.99) p(2π)32p0

1.12 Vector ﬁeld for spin-one particles We start with the s = 0 spinors ui(~0, 0) and vi(~0, 0) and note that since (j) (J3 )s¯0 = 0, the a = 3 rotation conditions (1.87) imply that i k ~ i ~ i k ~ i ~ (J3) k u (0, 0) = iR3 u (0, 0) = 0 and (J3) k v (0, 0) = iR3 v (0, 0) = 0. (1.100) Referring back to the explicit formulas for the generators of rotations and setting u, v = (0, x, y, z) we see that

0 0 0 0 0  0  0 0 0 −i 0 x −iy 0 J3 u =     =   =   (1.101) 0 i 0 0 y  x  0 0 0 0 0 z 0 0 and 0 0 0 0 0  0  0 0 0 −i 0 x −iy 0 J3 v =     =   =   . (1.102) 0 i 0 0 y  x  0 0 0 0 0 z 0 0 Thus only the 3-component z can be nonzero. The conventional choice is 0 µ µ 1 0 u (~0, 0) = v (~0, 0) = √   . (1.103) 2m 0 1

We now form the linear combinations of the rotation conditions (1.87) 1.12 Vector ﬁeld for spin-one particles 19

(1) (1) (1) that correspond to the raising and lowering matrices J± = J1 ± iJ2

0 1 0 0 0 0 (1) √ (1) √ J+ = 2 0 0 1 and J− = 2 1 0 0 . (1.104) 0 0 0 0 1 0

Their Lorentz counterparts are

0 0 0 0  (1) (1) (1) 0 0 0 ∓1 J = J ± iJ =   . (1.105) ± 1 2 0 0 0 −i 0 ±1 i 0

In these terms, the rotation conditions (1.87) for the j = 1 spinors ui(~0, s) are

X i ~ (1) X i k ~ u (0, s¯)(J± )ss¯ = (J±) ku (0, s). (1.106) s¯ k

But

∗(1) ∗(1) (1) (1) J1 ± iJ2 = J1 ∓ iJ2 = J∓. (1.107)

So the rotation conditions (1.87) for the j = 1 spinors vi(~0, s) are

X i ~ (1) X i k ~ − v (0, s¯)(J∓ )ss¯ = (J±) kv (0, s). (1.108) s¯ k

So for the plus sign and the choice s = 0, the condition (1.106) gives ui(~0, 1) as

0 0 0 0  0 X (1) √ 0 0 0 −1 1 0 ui(~0, s¯) J = 2 ui(~0, 1) = (J )i uk(~0, 0) =   √   +¯s0 + k 0 0 0 −i 0 s¯   2m   0 1 i 0 1 (1.109) or  0  i 1 −1 u (~0, 1) = √   . (1.110) 2 m −i 0 20 Quantum ﬁelds and special relativity Similarly, the minus sign and the choice s = 0 give for ui(~0, −1) 0 0 0 0  0 X (1) √ 0 0 0 1 1 0 ui(~0, s¯) J = 2 ui(~0, −1) = (J )i uk(~0, 0) =   √   −s¯0 − k 0 0 0 −i 0 s¯   2m   0 −1 i 0 1 (1.111) or  0  i 1  1  u (~0, −1) = √   . (1.112) 2 m −i 0

The rotation condition (1.108) for the j = 1 spinors vi(~0, s) with the minus sign and the choice s = 0 gives 0 0 0 0  0 X (1) √ 0 0 0 −1 1 0 − vi(~0, s¯)J = − 2 vi(~0, −1) = (J )i vk(~0, 0) =   √   −s¯0 + k 0 0 0 −i 0 s¯   2m   0 1 i 0 1 (1.113) or 0 i 1 1 v (~0, −1) = √   . (1.114) 2 m i 0 Similarly, the plus sign and the choice s = 0 give 0 0 0 0  0 X (1) √ 0 0 0 1 1 0 − vi(~0, s¯)J = − 2 vi(~0, 1) = (J )i vk(~0, 0) =   √   +¯s0 − k 0 0 0 −i 0 s¯   2m   0 −1 i 0 1 (1.115) or  0  i 1 −1 v (~0, 1) = √   . (1.116) 2 m  i  0 The boost conditions (1.210) now give for i, k = 0, 1, 2, 3

i ∗i p 0 i k ~ i p 0 u (~p,s) = v (~p,s) = m/p L k(~p) u (0, s) = e (~p,s)/ 2p (1.117) 1.12 Vector ﬁeld for spin-one particles 21 where i i k ~ e (~p,s) = L k(~p) e (0, s) (1.118) and 0 0  0  0 1 1 1  1  e(~0, 0) =   , e(~0, 1) = − √   , and e(~0, −1) = √   . 0 2 i 2 −i 1 0 0 (1.119) A single massive vector ﬁeld is then

1 X Z d3p φi(x) = φ+i(x)+φ−i(x) = ei(~p,s) a(~p,s)eip·x+e∗i(~p,s) a†(~p,s)e−ip·x. p 3 0 s=−1 (2π) 2p (1.120) The commutator/anticommutator of the positive and negative frequency parts of the ﬁeld is

Z 3 +i −k d p ip·(x−y) ik [φ (x), φ (y)]∓ = e Π (~p) (1.121) p(2π)32p0 where Π is a sum of outer products of 4-vectors

1 X Πik(~p) = ei(~p,s)e∗k (~p,s). (1.122) s=−1 At ~p = 0, the matrix Π is the unit matrix on the spatial coordinates

0 0 0 0 1 X 0 1 0 0 Π(~0) = ei(~0, s)e∗k (~0, s) =   . (1.123) 0 0 1 0 s=−1   0 0 0 1

So Π(~p) is

1 0 0 0 0 0 0 0 Π(~p) = LΠ(0)LT = LηLT + L   LT (1.124) 0 0 0 0 0 0 0 0 or Π(~p)ik = ηik + pipk/m2. (1.125) 22 Quantum ﬁelds and special relativity This equation lets us write the commutator (1.126) in terms of the Lorentz- invariant function ∆+(x − y) (1.45) as Z 3 +i −k ik i k 2 d p ip·(x−y) [φ (x), φ (y)]∓ = (η − ∂ ∂ /m ) p e (2π)32p0 (1.126) ik i k 2 = (η − ∂ ∂ /m ) ∆+(x − y). As for a scalar ﬁeld, we set vi(x) = κ φ+i(x) + λ φ−i(x) (1.127)

2 and find for (x − y) > 0 since ∆+(x − y) = ∆+(y − x) for x, y spacelike † 2 2 ik i k 2 [v(x), v (y)]∓ = |κ| ∓ |λ| (η − ∂ ∂ /m ) ∆+(x − y) (1.128) ik i k 2 [v(x), v(y)]∓ = (1 ∓ 1) κλ(η − ∂ ∂ /m ) ∆+(x − y). So we must choose the minus sign and set |κ| = |λ|. So then vi(x) = v+i(x) + v−i(x) = v+i(x) + v+i†(x) (1.129) is real. This is a second example of the spin-statistics theorem. If two such fields have the same mass, then we can combine them as we combined scalar fields i +i −i v (x) = v1 (x) + iv2 (x). (1.130) These fields obey the Klein-Gordon equation (2 − m2)vi(x) = 0. (1.131) And since both i i j k k ` p = L jk and e (~p) = L è (0) (1.132) it follows that p · e(~p) = k · e(0) = 0. (1.133) So the field vi also obeys the rule i ∂i v (x) = 0. (1.134) These equations (1.133) and 1.134 are like those of the electromagnetic field in Lorentz gauge. But one can’t get quantum electrodynamics as the i m → 0 limit of just any such theory. For the interaction H = Ji v would lead to a rate for v-boson production like ik JiJkΠ (~p) (1.135) which diverges as m → 0 because of the pipk/m2 term in Πik(~p). One can 1.13 Lorentz group 23

i avoid this divergence by requiring that ∂iJ = 0 which is current conservation. Under parity, charge conjugation, and time reversal, a vector ﬁeld transforms as a −1 ∗ a b Pv (x)P = − η P bv (Px) Cva(x)C−1 = ξ∗va†(x) (1.136) a −1 ∗ a b Tv (x)T = ζ P bv (−Px).

1.13 Lorentz group The Lorentz group O(3, 1) is the set of all linear transformations L that leave invariant the Minkowski inner product

xy ≡ x · y − x0y0 = xTηy (1.137) in which η is the diagonal matrix

−1 0 0 0  0 1 0 0 η =   . (1.138)  0 0 1 0 0 0 0 1

So L is in O(3, 1) if for all 4-vectors x and y

(Lx) T ηL y = xTLT η Ly = xTη y. (1.139)

Since x and y are arbitrary, this condition amounts to

LTη L = η. (1.140)

Taking the determinant of both sides and recalling that det AT = det A and that det(AB) = det A det B, we have

(det L)2 = 1. (1.141)

So det L = ±1, and every Lorentz transformation L has an inverse. Multi- plying (1.140) by η, we get

ηLTηL = η2 = I (1.142) which identifies L−1 as L−1 = ηLTη. (1.143) 24 Quantum fields and special relativity The subgroup of O(3, 1) with det L = 1 is the proper Lorentz group SO(3, 1). The subgroup of SO(3, 1) that leaves invariant the sign of the time component of timelike vectors is the proper orthochronous Lorentz group SO+(3, 1). To find the Lie algebra of SO+(3, 1), we take a Lorentz matrix L = I + ω that differs from the identity matrix I by a tiny matrix ω and require L to obey the condition (1.140) for membership in the Lorentz group

I + ωT η (I + ω) = η + ωTη + η ω + ωTω = η. (1.144)

Neglecting ωTω, we have ωTη = −η ω or since η2 = I

ωT = − η ω η. (1.145)

This equation implies that the matrix ωab is antisymmetric when both indexes are down

ωab = −ωba. (1.146)

e b ce To see why, we write it (1.145) as ω a = − ηab ω c η and the multiply both e b ce c sides by ηde so as to get ωda = ηde ω a = −ηab ω c η ηde = −ωac δd = −ωad. The key equation (1.145) also tells us that under transposition the time- time and space-space elements of ω change sign, while the time-space and spacetime elements do not. That is, the tiny matrix ω is for inﬁnitesimal θ and λ a linear combination

ω = θ · R + λ · B (1.147) of three antisymmetric space-space matrices

0 0 0 0  0 0 0 0 0 0 0 0 0 0 0 0  0 0 0 1 0 0 −1 0 R1 =   R2 =   R3 =   0 0 0 −1 0 0 0 0 0 1 0 0 0 0 1 0 0 −1 0 0 0 0 0 0 (1.148) and of three symmetric time-space matrices

0 1 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 B1 =   B2 =   B3 =   0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 (1.149) all of which satisfy condition (1.145). The three R` are 4 × 4 versions of the familiar rotation generators; the three B` generate Lorentz boosts. 1.14 Gamma matrices and Cliﬀord algebras 25 If we write L = I + ω as

L = I − iθ` iR` − iλ` iB` ≡ I − iθ`J` − iλ`K` (1.150) then the three matrices J` = iR` are imaginary and antisymmetric, and therefore hermitian. But the three matrices K` = iB` are imaginary and symmetric, and so are antihermitian. The 4×4 matrix L = exp(iθ`J`−iλ`K`) is not unitary because the Lorentz group is not compact.

1.14 Gamma matrices and Cliﬀord algebras In component notation, L = I + ω is

a a a L b = δ b + ω b, (1.151) cd T the matrix η is ηcd = η , and ω = − η ω η is

a T a a c da da a ω b = (ω )b = − (ηωη)b = − ηbc ω d η = − ωbd η = − ωb . (1.152) Lowering index a we get

a da d ωeb = ηea ω b = −ωbd η ηea = −ωbd δ e = − ωbe (1.153)

That is, ωab is antisymmetric

ωab = − ωba. (1.154) A representation of the Lorentz group is generated by matrices D(L) that represent matrices L close to the identity matrix by sums over a, b = 0, 1, 2, 3 i D(L) = 1 + ω J ab. (1.155) 2 ab The generators J ab must obey the commutation relations

i[J ab, J cd] = ηbc J ad − ηac J bd − ηda J cb + ηdb J ca. (1.156)

A remarkable representation of these commutation relations is provided by matrices γa that obey the anticommutation relations

{γa, γb} = 2 ηab. (1.157)

One sets i J ab = − [γa, γb] (1.158) 4 where η is the usual ﬂat-space metric (1.138). Any four 4 × 4 matrices that 26 Quantum ﬁelds and special relativity satisfy these anticommutation relations form a set of Dirac gamma matrices. They are not unique. Is S is any nonsingular 4×4 matrix, then the matrices

γ0a = S γa S−1 (1.159) also are a set of Dirac’s gamma matrices. Any set of matrices obeying the anticommutation relations (1.157) for any n × n diagonal matrix η with entries that are ±1 is called a Cliﬀord algebra. As a homework problem, show that

[J ab, γc] = − i γa ηbc + i γb ηac. (1.160)

One can use these commutation relations to derive the commutation relations (1.156) of the Lorentz group. The gamma matrices are a vectors in the sense that for L near the identity i i D(L) γc D−1(L) ≈ (I + ω J ab) γc (I − ω J ab) 2 ab 2 ab i = γc + ω [J ab, γc] 2 ab i = γc + ω (−i γa ηbc + i γb ηac) 2 ab 1 1 = γc + ω γa ηbc − ω γb ηac 2 ab 2 ab 1 1 = γc − ηcb ω γa − ηca ω γb (1.161) 2 ba 2 ab 1 1 = γc − ωc γa − ωc γb 2 a 2 b c c a = γ − ω a γ c c a = γ + ωa γ c c a = (δa + ωa ) γ c a = La γ c c in which we used (1.152) to write − ω a = ωa . The ﬁnite ω form is a −1 a c D(L) γ D (L) = Lc γ . (1.162) The unit matrix is a scalar

D(L) ID−1(L) = I. (1.163)

The generators of the Lorenz group form an antisymmetric tensor

ab −1 a b cd D(L) J D (L) = Lc Ld J . (1.164) 1.15 Dirac’s gamma matrices 27 Out of four gamma matrices, one can also make totally antisymmetric tensors of rank-3 and rank-4 Aabc ≡ γ[a γb γc] and Babcd ≡ γ[a γb γc γd] (1.165) where the brackets mean that one inserts appropriate minus signs so as to achieve total antisymmetry. Since there are only four γ matrices in four spacetime dimensions, any rank-5 totally antisymmetric tensor made from them must vanish, Cabcde = 0. Notation: The parity transformation is β = iγ0. (1.166) It ﬂips the spatial gamma matrices but not the temporal one β γi β−1 = − γi and β γ0 β−1 = γ0. (1.167) It ﬂips the generators of boosts but not those of rotations β J i0 β−1 = − J i0 and β J ik β−1 = J ik. (1.168)

1.15 Dirac’s gamma matrices Weinberg’s chosen set of Dirac matrices is 0 1 0 σi γ0 = − i = −γ0† and γi = −i = γi† (1.169) 1 0 − σi 0 in which the σ’s are Pauli’s 2 × 2 hermitian matrices 0 1 0 −i 1 0 σ = , σ = , and σ = , (1.170) 1 1 0 2 i 0 3 0 −1 which are the gamma matrices of 3-dimensional spacetime. With this choice of γ’s, the matrix β is 0 1 β = i γ0 = = β†. (1.171) 1 0 In spacetimes of five dimensions, the fifth gamma matrix γ4 which tradi- 5 tionally is called γ = γ5 is 1 0 γ5 = γ = −iγ0γ1γ2γ3 = . (1.172) 5 0 −1 It anticommutes with all four Dirac gammas and its square is unity, as it must if it is to be the fifth gamma in 5-space: {γa, γb} = 2ηab (1.173) 28 Quantum fields and special relativity for a, b = 0, 1, 2, 3, 4 with η44 = 1 and η40 = η04 = 0. With Weinberg’s choice of γ’s, the Lorentz boosts are

i i 0 σi 0 1 J i0 = − [γi, γ0] = − −i , −i 4 4 − σi 0 1 0 (1.174) i σi 0 −σi 0 i σi 0 = − = . 4 0 −σi 0 σi 2 0 −σi

The Lorentz rotation matrices are i i 0 σi 0 σk J ik = − [γi, γk] = − −i , −i 4 4 − σi 0 −σk 0 i −σiσk 0 −σkσi 0 = − 4 0 −σiσk 0 −σkσi (1.175) i k j −i [σ , σ ] 0 −i 2iikjσ 0 = i k = j 4 0 [σ , σ ] 4 0 2iikjσ 1 σj 0 = . 2 ikj 0 σj

The Dirac representation of the Lorentz group is reducible, as SW’s choice of gamma matrices makes apparent. The Dirac rotation matrices are

1 σi 0 J = . (1.176) i 2 0 σi

Some useful relations are

β γa† β = − γa, β J ab† β = J ab and β D(L)† β = D(L)−1 (1.177) as well as

† a † a β γ5 β = − γ5 and β(γ5γ ) β = − γ5γ . (1.178)

1.16 Dirac fields The positive- and negative-frequency parts of a Dirac field are Z + −3/2 X 3 ip·x ψ` (x) = (2π) d p u`(~p,s) e a(~p,s) s Z (1.179) − −3/2 X 3 −ip·x † ψ` (x) = (2π) d p v`(~p,s) e b (~p,s). s 1.16 Dirac fields 29 The rotation conditions (1.39) are

X ~ (j) X ~ u`¯(0, s¯)(Ji )ss¯ = (Ji)``¯ u`(0, s) s¯ ` (1.180) X ~ (j) ∗ X ~ v`¯(0, s¯)(−Ji )ss¯ ) = (Ji)``¯ v`(0, s). s¯ ` The Dirac rotation matrices (1.176) are

1 σi 0 J = (1.181) i 2 0 σi

¯ 1 so we set the four values `, ` = 1, 2, 3, 4 to ` = (m, ±) with m = ± 2 . And we + − consider u`(s) to be um(s) stacked upon um(s) and similarly take v`(s) to + − ± ± be vm±(s) above vm±(s) where um(s) and vm±(s) are, a priori, 2 × (2j + 1)- dimensional matrices with indexes m = ±1/2 and s = −j, . . . , j. That is,

  u+ (s)   v+ (s) u1(s) +1/2 v1(s) +1/2 + + u (s) u (s) v (s) v (s)  2  =  −1/2  and  2  =  −1/2  . (1.182) u (s) u− (s) v (s) v− (s)  3   +1/2   3   +1/2  − − u4v v4(s) u−1/2(s) v−1/2(s) We then have four equations

X + ~ (j) X 1 i + ~ um¯ (0, s¯)(Ji )ss¯ = 2 σmm¯ um(0, s) s¯ m X − ~ (j) X 1 i − ~ um¯ (0, s¯)(Ji )ss¯ = 2 σmm¯ um(0, s) s¯ m (1.183) X + ~ ∗(j) X 1 i + ~ vm¯ (0, s¯)(−Ji)ss¯ ) = 2 σmm¯ vm(0, s) s¯ m X − ~ ∗(j) X 1 i − ~ vm¯ (0, s¯)(−Ji)ss¯ ) = 2 σmm¯ vm(0, s). s¯ m SW deﬁnes the four 2 × (2j + 1) matrices

U + = u+ (~0, s) and U − = u− (~0, s) ms m ms m (1.184) + + ~ − − ~ Vms = vm(0, s) and Vms = vm(0, s). in terms of which the four Dirac rotation conditions (1.183) are

+ (j) 1 + − (j) 1 − U J = σi U and U J = σi U i 2 i 2 (1.185) + ∗(j) 1 + − ∗(j) 1 − V (−Ji ) = 2 σi V and V (−Ji ) = 2 σi V . 30 Quantum ﬁelds and special relativity Taking the complex conjugate of the second of these equations, we get

(j) +∗−1 1 i∗ +∗ +∗−1 1 i +∗ −J = V ( σ )V = V (− σ2 σ σ2)V i 2 2 (1.186) (j) −∗−1 1 i∗ −∗ −∗−1 1 i −∗ −Ji = V ( 2 σ )V = V (− 2 σ2 σ σ2)V or more simply

(j) +∗ −1 1 i +∗ J = (σ2V ) σ (σ2V ) i 2 (1.187) (j) −∗ −1 1 i −∗ Ji = (σ2V ) 2 σ (σ2V ). The 2 × 2 Pauli matrices ~σ and the (2j + 1) × (2j + 1) matrices J~(j) both generate irreducible representations of the rotation group. So by writing

+ (j) (j) 1 + (j) 1 1 + U Ji Jk = 2 σi U Jk = = 2 σi 2 σk U (1.188) and similar equations for U −,V +,V −, we see that

~(j) ~ U + D(j)(θ~) = U + e−iθ·J = e−iθ·~σ U + = D(1/2)(θ~) U + (1.189) ~(j) ~ U − D(j)(θ~) = U − e−iθ·J = e−iθ·~σ U − = D(1/2)(θ~) U − and similar equations for V ±.

+∗ (j) +∗ −iθ·J~(j) −iθ~·~σ +∗ (1/2) +∗ σ2V D (θ~) = σ2V e = e σ2V = D (θ~) σ2V −∗ (j) −∗ −iθ·J~(j) −iθ~·~σ −∗ (1/2) −∗ σ2V D (θ~) = σ2V e = e σ2V = D (θ~) σ2V . (1.190) Now recall Schur’s lemma (section 10.7 of PM):

Part 1: If D1 and D2 are inequivalent, irreducible representations of a group G, and if D1(g)A = AD2(g) for some matrix A and for all g ∈ G, then the matrix A must vanish, A = 0. Part 2: If for a finite-dimensional, irreducible representation D(g) of a group G, we have D(g)A = AD(g) for some matrix A and for all g ∈ G, then A = cI. That is, any matrix that commutes with every element of a finite-dimensional, irreducible representation must be a multiple of the identity matrix. Part 1 tells us that D(j)(θ~) and D(1/2)(θ~) must be equivalent. So j = 1/2 and 2j + 1 = 2. A Dirac field must represent particles of spin 1/2. Part 2 then says that the matrices U ± must be multiples of the 2 × 2 identity matrix + − U = c+ I and U = c− I (1.191) ±∗ and that the matrices σ2V must be multiples of the 2 × 2 identity matrix +∗ 0 −∗ 0 σ2V = d+ I and σ2V = d− I (1.192) 1.16 Dirac fields 31 or more simply

+ − V = − id+ σ2 and V = − id− σ2. (1.193) That is, 0 −1 0 −1 v+(~0, s) = d and v−(~0, s) = d . (1.194) m + 1 0 m − 1 0 Going back to ` = (m, ±) by using the index code (1.182)

  u+ (s)     v+ (s)   u1(s) +1/2 c+ δ1/2,s v1(s) +1/2 d+ δ−1/2,s + + u (s) u (s) c δ v (s) v (s) d δ  2  =  −1/2  =  + −1/2,s and  2  =  −1/2  =  + 1/2,s  . u (s) u− (s)  c δ  v (s)  v− (s)  d δ   3   +1/2   − 1/2,s   3   1/2   − −1/2,s − − u4(s) c− δ v4(s) d− δ u−1/2(s) −1/2,s v−1/2(s) +1/2,s (1.195) we have for the u’s

+ + u1/2(1/2) = c+ and u−1/2(1/2) = 0 (1.196) − − u1/2(1/2) = c− and u−1/2(1/2) = 0 (1.197) + + u1/2(−1/2) = 0 and u−1/2(−1/2) = c+ (1.198) − − u1/2(−1/2) = 0 and u−1/2(−1/2) = c− (1.199)

+ + v1/2(1/2) = 0 and v−1/2(1/2) = d+ (1.200) − − v1/2(1/2) = 0 and v−1/2(1/2) = d− (1.201) + + v1/2(−1/2) = − d+ and v−1/2(−1/2) = 0 (1.202) − − v1/2(−1/2) = − d− and v−1/2(−1/2) = 0 (1.203) So  u+ (1/2)     u+ (−1/2)    1/2 c+ 1/2 0 + + u (1/2) 0 u (−1/2) c u(~0, m = 1 ) =  −1/2  =   and u(~0, m = − 1 ) =  −1/2  =  + , 2  u− (1/2)  c  2  u− (−1/2)   0   1/2   −  1/2    − − 0 c− u−1/2(1/2) u−1/2(−1/2)  v+ (1/2)     v+ (−1/2)    1/2 0 1/2 d+ + + v (1/2) d v (−1/2) 0 v(~0, m = 1 ) =  −1/2  =  + and v(~0, m = − 1 ) =  −1/2  = −   . 2  v− (1/2)   0  2  v− (−1/2)  d   1/2     1/2   − − − d− 0 v−1/2(1/2) v−1/2(−1/2) (1.204) 32 Quantum ﬁelds and special relativity

To put more constraints on c± and d±, we recall that under parity −1 ∗ † −1 † Pa(~p,s)P = ηa a(−~p,s) and Pb (~p,s)P = ηb β (−~p,s) (1.205) and so Z + −1 −3/2 X 3 ip·x ∗ Pψ` (x)P = (2π) d p u`(~p,s) e ηa a(−~p,s) s Z −3/2 X 3 ip·Px ∗ = (2π) d p u`(−~p,s) e ηa a(~p,s) s Z (1.206) − −1 −3/2 X 3 −ip·x † Pψ` (x)P = (2π) d p v`(~p,s) e ηb b (−~p,s) s Z −3/2 X 3 −ip·Px † = (2π) d p v`(−~p,s) e ηb b (~p,s). s We recall the relations (1.177) β γa† β = − γa, β J ab† β = J ab, and β D(L)† β = D(L)−1 (1.207) and in particular, since J 0i† = − J 0i, the rule βJ 0iβ = J 0i† = −J 0i. (1.208) We also have the pseudounitarity relation β D†(L) β = D−1(L). (1.209) In general spinors at ﬁnite momentum are related to those at zero momentum by rm X u¯(q, s) = D¯ (L(q))u (~0, s) ` q0 `` ` ` (1.210) rm X v¯(q, s) = D¯ (L(q))v (~0, s) ` q0 `` ` ` which for Dirac spinors is r m u(p, s) = D(L(p)) u(~0, s) p0 (1.211) r m v(p, s) = D(L(p)) v(~0, s). p0 So now by using the boost rule (1.208) we have

p 0 p 0 −1 u`(−~p,s) = m/p D(L(−~p))u(0, s) = m/p D(L(~p)) u(0, s) (1.212) p = m/p0 β D(L(~p)) β u(0, s) (1.213) 1.16 Dirac ﬁelds 33 and

p 0 p 0 −1 v`(−~p,s) = m/p D(L(−~p))v(0, s) = m/p D(L(~p)) v(0, s) (1.214) p = m/p0 β D(L(~p)) β v(0, s). (1.215) So under parity Z + −1 −3/2 X 3 p 0 ip·Px ∗ Pψ (x)P = (2π) d p m/p β D(L(~p)) β u(0, s) e ηa a(~p,s) s Z − −1 −3/2 X 3 p 0 −ip·Px † Pψ (x)P = (2π) d p m/p β D(L(~p)) β v(0, s) e ηb b (~p,s). s (1.216)

± −1 ± So to have Pψ` (x)P ∝ ψ` (x), we need

β u(0, s) = bu u(0, s) and β v(0, s) = bv u(0, s). (1.217) We then get

+ −1 ∗ + − −1 − Pψ` (t, ~x)P = bu β ηa ψ` (t, −~x) and Pψ` (t, ~x)P = bv β ηb ψ` (t, −~x). (1.218) 2 2 2 Here since P = 1, these factors are just signs, bu = bv = 1. The eigenvalue equations (1.217) tell us that c− = bu c+ and that d− = bv d+. So rescaling the ﬁelds we get  1   0  ~ 1 1  0  ~ 1 1  1  u(0, m = 2 ) = √   and u(0, m = − 2 ) = √   , 2 bu 2  0  0 bu (1.219)  0   1  ~ 1 1  1  ~ 1 −1  0  v(0, m = 2 ) = √   and v(0, m = − 2 ) = √   . 2  0  2 bv bv 0 If the annihilation and creation operators a(p, s) and a†(p, s) obey the rule

† 0 0 3 0 [a(p, s), a (p , s )]∓ = δss0 δ (~p − ~p ) (1.220) and if the field is the sum of the positive- and negative-frequency parts (2.55) Z + −3/2 X 3 ip·x ψ` (x) = (2π) d p u`(~p,s) e a(~p,s) s Z (1.221) − −3/2 X 3 −ip·x † ψ` (x) = (2π) d p v`(~p,s) e b (~p,s) s 34 Quantum fields and special relativity with arbitrary coefficients κ and λ

+ − ψ`(x) = κψ` (x) + λψ` (x) (1.222) then

† + − ∗ +† ∗ −† [ψ`(x), ψ`0 (y)]∓ = [κψ` (x) + λψ` (x), κ ψ`0 (y) + λ ψ`0 (y)] Z 3 d p X h 2 ∗ ip·(x−y) = |κ| u (~p,s) u 0 (~p,s)e (2π)3 ` ` s 2 ∗ −ip·(x−y)i ∓|λ| v`(~p,s) v`0 (~p,s)e Z 3 d p X h 2 ip·(x−y) 2 −ip·(x−y)i = |κ| N 0 (p) e ∓ |λ| N 0 (p) e (2π)3 `` `` s (1.223) where

X ∗ N``0 (p) = u`(~p,s) u`0 (~p,s) s (1.224) X ∗ M``0 (p) = v`(~p,s) v`0 (~p,s). s

When ~p = 0, these matrices are

X ~ ∗ ~ N``0 (0) = u`(0, s) u`0 (0, s) s  1   0  1  0  1  1  N(0) =   1 0 bu 0 +   0 1 0 bu 2 bu 2  0  (1.225) 0 bu     1 0 bu 0 0 0 0 0 1  0 0 0 0 1 0 1 0 bu 1 + bu β =   +   = 2 bu 0 1 0 2 0 0 0 0  2 0 0 0 0 0 bu 0 1 1.16 Dirac ﬁelds 35 and X ~ ∗ ~ M``0 (0) = v`(0, s) v`0 (0, s) s  0   1  1  1  1  0  M(0) =   0 1 0 bv +   1 0 bv 0 2  0  2 bv (1.226) bv 0     0 0 0 0 1 0 bv 0 1 0 1 0 bv 1  0 0 0 0 1 + bv β =   +   = . 2 0 0 0 0  2 bv 0 1 0 2 0 bv 0 1 0 0 0 0 So then using the boost relations (1.211) we ﬁnd

X ∗ m X ∗ † N(~p) = u (~p,s) u 0 (~p,s) = D(L(p)) u (~0, s) u 0 (~0, s)D (L(p)) ` ` p0 ` ` s s m = D(L(p)) (1 + b β) D†(L(p)) 2p0 u X ∗ m X ∗ † M(~p) = v (~p,s) v 0 (~p,s) = D(L(p)) v (~0, s) v 0 (~0, s)D (L(p)) ` ` p0 ` ` s s m = D(L(p)) (1 + b β) D†(L(p)). 2p0 v (1.227) The pseudounitarity relation (1.209) β D†(L) β = D−1(L). (1.228) gives β D†(L) = D−1(L) β (1.229) which implies that D(L) β D†(L) = β. (1.230) The pseudounitarity relation also says that D†(L) = β D−1(L) β (1.231) so that D(L) D†(L) = D(L) β D−1(L) β. (1.232) Also since the gammas form a 4-vector (1.162)

a −1 a c D(L) γ D (L) = Lc γ (1.233) 36 Quantum ﬁelds and special relativity and since β = iγ0, we have

−1 0 −1 0 c c D(L(p)) β D (L(p)) = D(L(p)) iγ D (L(p)) = iLc (p) γ = −iL 0γc. (1.234) Now a a b a p = L b(p)k = L 0(p)m (1.235) so −1 c D(L(p)) β D (L(p)) = −i p γc/m (1.236) which implies that

† c D(L) D (L) = − i (p γc/m) β. (1.237) Thus m m N(~p) = D(L(p)) (1 + b β) D†(L(p)) = [(−i pcγ /m) β + b β] 2p0 u 2p0 c u 1 = [−i pcγ + b m] β (1.238) 2p0 c u and m m M(~p) = D(L(p)) (1 + b β) D†(L(p)) = [(−i pcγ /m) β + b β] 2p0 v 2p0 c v 1 = [−i pcγ + b m] β. (1.239) 2p0 c v We now put the spin sums (1.238) and (1.239) in the (anti)commutator (1.223) and get Z 3 † d p h 2 c ip·(x−y) [ψ (x), ψ (y)] = |κ| [( − i p γ + b m)β] 0 e ` `0 ∓ (2π)32p0 c u `` 2 c −ip·(x−y)i ∓ |λ| [( − i p γc + bv m)β]``0 e Z 3 2 c d p ip·(x−y) = |κ| [( − ∂ γ + b m)β] 0 e c u `` (2π)32p0 Z 3 2 c d p −ip·(x−y) ∓ |λ| [( − ∂ γ + b m)β] 0 e c v `` (2π)32p0 2 c = |κ| [( − ∂cγ + bu m)β]``0 ∆+(x − y) 2 c ∓ |λ| [( − ∂cγ + bv m)β]``0 ∆+(y − x). (1.240)

> Recall that for (x − y) 0, i.e. spacelike, ∆+(x − y) = ∆+(y − x). So its ﬁrst 1.16 Dirac ﬁelds 37 derivatives are odd. So for x − y spacelike

† 2 c [ψ`(x), ψ`0 (y)]∓ = |κ| [( − ∂cγ + bu m)β]``0 ∆+(x − y) 2 c ∓ |λ| [(∂ γ + b m)β] 0 ∆ (x − y) c v `` + (1.241) 2 2 c = |κ| ± |λ| [( − ∂cγ )β]``0 ∆+(x − y) 2 2 + |κ| bu ∓ |λ| bv m β``0 ∆+(x − y). To get the ﬁrst term to vanish, we need to choose the lower sign (that is, use anticommutators) and set |κ| = |λ|. To get the second term to be zero, we must set bu = − bv. We may adjust κ and and bu so that

κ = λ and bu = − bv = 1. (1.242) In particular, a spin-one-half ﬁeld must obey anticommutation relations

† † 2 [ψ`(x), ψ`0 (y)]+ ≡ {ψ`(x), ψ`0 (y)} = 0 for (x − y) > 0. (1.243) Finally then, the Dirac ﬁeld is

X Z d3p h i ψ (x) = u (~p,s) eip·x a(~p,s) + v (~p,s) e−ip·x b†(~p,s) . ` (2π)3/2 ` ` s (1.244) The zero-momentum spinors are 1 0

1 1 0 1 1 1 u(~0, m = ) = √   and u(~0, m = − ) = √   , 2 2 1 2 2 0 0 1 (1.245)  0  −1

1 1  1  1 1  0  v(~0, m = ) = √   and v(~0, m = − ) = √   . 2 2  0  2 2  1  −1 0 The spin sums are

X ∗ h 1 c i N(~p) = u`(~p,s)um(~p,s) = (−i p γc + m) β `m 2p0 `m s (1.246) X ∗ h 1 c i M(~p) = v`(~p,s)vm(~p,s) = (−i p γc − m) β . `m 2p0 `m s The Dirac anticommutator is

† † c [ψ`(x), ψ`0 (y)]+ ≡ {ψ`(x), ψ`0 (y)} = [( − ∂cγ + m)β]``0 ∆+(x − y). (1.247) 38 Quantum ﬁelds and special relativity Two standard abbreviations are 0 1 0 1 β ≡ iγ0 = and ψ ≡ ψ†β = iψ†γ0 = ψ∗ ψ∗ = ψ∗ ψ∗ . 1 0 ` r 1 0 r ` (1.248) A Majorana fermion is represented by a ﬁeld like

X Z d3p h i ψ (x) = u (~p,s) eip·x a(~p,s) + v (~p,s) e−ip·x a†(~p,s) . ` (2π)3/2 ` ` s (1.249)

−1 ∗ab ab −1 Since C = γ2β it follows that C = βγ2 and so that J = −βCJ C β = ab 2 2 ∗ab −βγ2βJ βγ2β. But βγ2β = − ββγ2 = −i γ0 γ2 = −γ2. So J = − ab γ2J γ2. Thus

∗ab ab ab ∗ −iωabJ −iωab(−γ2J γ2) iωabJ D (L) = e = e = γ2e γ2 = γ2 D(L) γ2. (1.250) Now with SW’s γ’s,

~ 1 ~ 1 ~ 1 ~ 1 γ2 u(0, ± 2 ) = v(0, ± 2 ) and γ2 v(0, ± 2 ) = u(0, ± 2 ). (1.251) Thus the hermitian conjugate of a Majorana ﬁeld is

X Z d3p h i ψ∗(x) = u∗(~p,s) e−ip·x a†(~p,s) + v∗(~p,s) eip·x a(~p,s) (2π)3/2 s X Z d3p h i = v∗(~p,s) eip·x a(~p,s) + u∗(~p,s) e−ip·x a†(~p,s) (2π)3/2 s X Z d3p h i = D∗(L(p)) v∗(~0, s) eip·x a(~p,s) + D∗(L(p)) u∗(~0, s) e−ip·x a†(~p,s) (2π)3/2 s X Z d3p h i = γ D(L(p))γ v(~0, s) eip·x a(~p,s) + D(L(p))γ u(~0, s) e−ip·x a†(~p,s) 2 (2π)3/2 2 2 s X Z d3p h i = γ D(L(p))u(~0, s) eip·x a(~p,s) + D(L(p))v(~0, s) e−ip·x a†(~p,s) 2 (2π)3/2 s X Z d3p h i = γ u(~p,s) eip·x a(~p,s) + v(~p,s) e−ip·x a†(~p,s) = γ ψ(x). 2 (2π)3/2 2 s (1.252)

This works because the spinors obey the Majorana conditions

u(p, s) = γ2 v∗(p, s) and v(p, s) = γ2 u∗(p, s). (1.253) 1.16 Dirac ﬁelds 39 The parity rules (1.254) now are

+ −1 ∗ + − −1 − Pψ` (t, ~x)P = β ηa ψ` (t, −~x) and Pψ` (t, ~x)P = − β ηb ψ` (t, −~x). (1.254) So to have a Dirac field survive a parity transformation, we need the phase of the particle to be minus the complex conjugate of the phase of the antiparticle ∗ ∗ ηa = − ηb or ηb = − ηa. (1.255) So the intrinsic parity of a particle-antiparticle state is odd. So negative- 0 parity bosons like π , ρ0, J/ψ can be interpreted as s-wave bound states of quark-antiquark pairs. Under parity a Dirac field goes as Pψ(t, ~x)P−1 = η∗ β ψ(t, −~x). (1.256) If a Dirac particle is the same as its antiparticle, then its intrinsic parity must be odd under complex conjugation, η = −η∗. So the intrinsic parity of a Majorana fermion must be imaginary η = ± i. (1.257) But this means that if we express a Dirac field ψ as a complex linear combination 1 ψ = √ (χ1 + iχ2) (1.258) 2 ∗ ∗ of two Majorana fields with intrinsic parities η1 = ±i and η2 = ±i, then under parity

−1 1 ∗ ∗ Pψ(t, ~x)P = √ η β φ1(t, −~x) + iη β φ2(t, −~x) (1.259) 2 1 2 ∗ ∗ so we need η1 = η2 to have

−1 1 ∗ ∗ ∗ Pψ(t, ~x)P = √ η β φ1(t, −~x) + iη β φ2(t, −~x) = η β ψ(t, −~x). 2 1 2 (1.260) But in that case the Dirac ﬁeld has intrinsic parity η = ±i. The equation (1.236) that shows how beta goes under D(L(p))

−1 c D(L(p)) β D (L(p)) = − i p γc/m (1.261) tells us that the spinors (1.211) r m r m u(p, s) = D(L(p)) u(~0, s) and v(p, s) = D(L(p)) v(~0, s) p0 p0 (1.262) 40 Quantum ﬁelds and special relativity

c are eigenstates of −i p γc/m with eigenvalues ±1 r m ( − i pcγ /m) u(p, s) = D(L(p)) β D−1(L(p)) D(L(p)) u(~0, s) c p0 r m r m = D(L(p)) β u(~0, s) = D(L(p)) u(~0, s) = u(p, s) p0 p0 r m ( − i pcγ /m) v(p, s) = D(L(p)) β D−1(L(p)) D(L(p)) v(~0, s) c p0 r m r m = D(L(p)) β v(~0, s) = − D(L(p)) v(~0, s) = − v(p, s). p0 p0 (1.263) So the spinors u and v obey Dirac’s equation in momentum space

c c (i p γc + m)u(p, s) = 0 and (−i p γc + m)v(p, s) = 0 (1.264) which implies that a Dirac ﬁeld obeys Dirac’s equation

X Z d3p h i (γa ∂ + m)ψ(x) = (γa ∂ + m) u(~p,s) eip·x a(~p,s) + v(~p,s) e−ip·x b†(~p,s) a a (2π)3/2 s X Z d3p h = (γa ∂ + m)u(~p,s) eip·x a(~p,s) (2π)3/2 a s a −ip·x † i + (γ ∂a + m)v(~p,s) e b (~p,s) X Z d3p h = (iγa p + m)u(~p,s) eip·x a(~p,s) (2π)3/2 a s a −ip·x † i + (−iγ pa + m)v(~p,s) e b (~p,s) = 0. (1.265)

The Majorana conditions (1.253) imply that

u∗(p, s) = − βCv(p, s) and v∗(p, s) = − βCu(p, s). (1.266)

So for a Dirac ﬁeld to survive charge conjugation, the particle-antiparticle phases must be related ∗ ξb = ξa. (1.267) Then under charge conjugation a Dirac ﬁeld goes as

C ψ(x) C−1 = − ξ∗ βC ψ∗(x). (1.268)

If a Dirac particle is the same as its antiparticle, then ξ must be real (and 1.17 Expansion of massive and massless Dirac ﬁelds 41 η imaginary), ξ = ±1, and must satisfy the reality condition ψ(x) = − β C ψ∗(x). (1.269) Suppose a particle and its antiparticle form a bound state X Z |Φi = d3p d3p0χ(p, s; p0, s0) a†(p, s) b†(p0s0) |0i. (1.270) ss0 Under charge conjugation Z X 3 3 0 0 0 † † 0 0 C |Φi = ξaξb d p d p χ(p, s; p , s ) b (p, s) a (p s ) |0i ss0 Z X 3 3 0 0 0 † 0 0 † = − ξaξb d p d p χ(p, s; p , s ) a (p s ) b (p, s) |0i ss0 (1.271) Z X 3 3 0 0 0 † † 0 0 = − ξaξb d p d p χ(p , s ; p, s) a (p, s) b (p , s ) |0i ss0 ∗ = − ξaξb |Φi = −ξaξa |Φi = −|Φi. The intrinsic charge-conjugation parity of a bound state of a particle and its antiparticle is odd.

1.17 Expansion of massive and massless Dirac ﬁelds A Dirac ﬁeld is a sum over s = ±1/2 of Z 3 X d p ip·x −ip·x † ψa(x) = ua(~p,s) e a(~p,s) + va(~p,s) e b (~p,s) (1.272) (2π)3/2 s in which m − ip/ m + ip/ u(p, s) = u(0, s) and v(p, s) = v(0, s) p2p0(p0 + m) p2p0(p0 + m) (1.273) and the zero-momentum spinors (1.245) are 1 0

1 1 0 1 1 1 u(0, s = ) = √   and u(0, s = − ) = √   , 2 2 1 2 2 0 0 1 (1.274)  0  −1

1 1  1  1 1  0  v(0, s = ) = √   and v(0, s = − ) = √   . 2 2  0  2 2  1  −1 0 42 Quantum ﬁelds and special relativity These static spinors and the matrix formulas (1.273) give explicit expressions for the spinors at arbitrary momentum p:

 0  m + p − p3 1 1  −p1 − ip2  u(p, 2 ) = p  0  (1.275) 2 p0(p0 + m) m + p + p3 p1 + ip2

  −p1 + ip2 0 1 1 m + p + p3 u(p, − 2 ) = p   (1.276) 2 p0(p0 + m)  p1 − ip2  0 m + p − p3

  −p1 + ip2 0 1 1  m + p + p3  v(p, 2 ) = p   (1.277) 2 p0(p0 + m)  −p1 + ip2  0 −m − p + p3

 0  −m − p + p3 1 1  p1 + ip2  v(p, − 2 ) = p  0  . (1.278) 2 p0(p0 + m)  m + p + p3  p1 + ip2

In the m → 0 limit with p = p0zˆ in the 3-direction, they are

0 0

0 1 0 0 1 1 u(p zˆ, ) =   and u(p zˆ, − ) =   (1.279) 2 1 2 0 0 0

0 0

0 1 1 0 1 0 v(p zˆ, ) =   and v(p zˆ, − ) =   . (1.280) 2 0 2 1 0 0

They tell us that the upper two components which are those of the left- 0 1 handed Dirac field represent particles of negative helicity u(p zˆ, − 2 ) and 0 1 antiparticles of positive helicity v(p zˆ, 2 ). They also say that the lower two components which are those of the right-handed field represent particles of 0 1 0 1 positive helicity u(p zˆ, 2 ) and antiparticles of negative helicity v(p zˆ, − 2 ). 1.17 Expansion of massive and massless Dirac fields 43 In the m → 0 limit, the spinors are for the particles     0 −p1 + ip2 0 1 1  0  1 1  p + p3  u(p, 2 ) = 0  0  and u(p, − 2 ) = 0   (1.281) 2p  p + p3  2p  0  p1 + ip2 0 and for the antiparticles     −p1 + ip2 0 0 1 1  p + p3  1 1  0  v(p, 2 ) = 0   and v(p, − 2 ) = 0  0  . (1.282) 2p  0  2p  p + p3  0 p1 + ip2 The massless Dirac field is 1 2 X Z d3p h i ψ (x) = u (p, s)eip·x a(p, s) + v (p, s) e−ip·x b†(p, s) . a (2π)3/2 a a 1 s=− 2 (1.283) The left-handed field is Z d3p h i ψ (x) = u (p, − 1 )eip·x a(p, − 1 ) + v (p, 1 ) e−ip·x b†(p, 1 ) à (2π)3/2 à 2 2 à 2 2 Z 3 d p h −p1 + ip2 ip·x 1 = 0 e a(p, − 2 ) (2π)3/22p0 p + p3 −p1 + ip2 −ip·x † 1 i + 0 e b (p, 2 ) . p + p3 (1.284) The right-handed field is Z d3p h i ψ (x) = u (p, 1 )eip·x a(p, 1 ) + v (p, − 1 ) e−ip·x b†(p, − 1 ) ra (2π)3/2 ra 2 2 ra 2 2 Z 3 0 d p h p + p3 ip·x 1 = e a(p, + 2 ) (2π)3/22p0 p1 + ip2 0 p + p3 −ip·x † 1 i + e b (p, − 2 ) . p1 + ip2 (1.285) The massless left-handed field and the massless right-handed field create and absorb different particles. These are different fields. That is why the unbroken gauge group SU(2)` can act on ψ` without affecting ψr. 44 Quantum fields and special relativity 1.18 Spinors and angular momentum Under a Lorentz transformation a Dirac field goes as −1 X −1 U(L)ψa(x)U (L) = Dab(L )ψb(Lx) (1.286) b 1 1 where D(L) is the ( 2 , 0)⊕(0, 2 ) representation of the Lorentz group. If L = R is an active rotation of |θ~| radians about the vector θ~, then the matrix is ! eiθ~·~σ/2 0 D(R−1) = . (1.287) 0 eiθ~·~σ/2

In particular, if θ~ = θzˆ, then eiθ/2 0 0 0  iθσ3/2 −iθ/2 −1 e 0  0 e 0 0  D(R ) = =   . (1.288) 0 eiθσ3/2  0 0 eiθ/2 0  0 0 0 e−iθ/2 So since the ﬁeld is X Z d3p ψ (x) = u (~p,s) eip·x a(~p,s) + v (~p,s) e−ip·x a†(~p,s), a (2π)3/2 a a c s (1.289) we need −1 −1 U(R)ψa(x)U (R) = D(R )abψb(Rx) X Z d3p h = u (~p,s)eip·xU(R)a(~p,s)U −1(R) (2π)3/2 a s −ip·x † −1 i + va(~p,s)e U(R)ac(~p,s)U (R) . (1.290) A state of momentum ~p, kind n, and spin s = ±1/2 in the z direction transforms under a right-handed rotation R(θ, z) of θ radians about the z axis as U(R(θ, z))|pz,ˆ s, ni = e−iθJz |pz,ˆ s, ni = e−iθs|pz,ˆ s, ni. (1.291) The state |pz,ˆ s, ni is made from the vacuum by the creation operator for a particle of momentum ~p, kind n, and spin s = ±1/2 in the z direction |pz,ˆ s, ni = a†(pz,ˆ s, n)|0i. (1.292) The creation operator transforms as U(R(z, θ))a†(pz,ˆ s, n)U(R(z, θ))−1 = e−iθsa†(pz,ˆ s, n). (1.293) 1.18 Spinors and angular momentum 45 The antiparticle creation operator goes as

† −1 X −iθ~·~σ/2 † 0 U(R)ac(~p,s)U (R) = e ac(R~p,s ), (1.294) ss0 s0 and so the particle annihilation operator goes as X ~ ∗ X ~ U(R)a(~p,s)U −1(R) = e−iθ·~σ/2 a(R~p,s0) = eiθ·~σ/2 a(R~p,s0). ss0 s0s s0 s0 (1.295) Our requirement (1.290) is −1 −1 U(R)ψa(x)U (R) = D(R )abψb(Rx) Z 3 X d p h ip·x X iθ~·~σ/2 0 = ua(~p,s)e e a(R~p,s ) (2π)3/2 s0s s s0 −ip·x X −iθ~·~σ/2 † 0 i + va(~p,s)e e ac(R~p,s ) . (1.296) ss0 s0 But iθ~·~σ/2 ! −1 e 0 D(R )abψb(Rx) = ψb(Rx) 0 eiθ~·~σ/2 ab iθ~·~σ/2 ! Z 3 e 0 X d p h ip·Rx = ub(~p,s)e a(~p,s) 0 eiθ~·~σ/2 (2π)3/2 ab s −ip·Rx † i + vb(~p,s)e ac(~p,s) . (1.297) Equivalently since Rp · Rx = p · x,

iθ~·~σ/2 ! −1 e 0 D(R )abψb(Rx) = ψb(Rx) 0 eiθ~·~σ/2 ab iθ~·~σ/2 ! Z 3 e 0 X d p h ip·x = ub(Rp,~ s)e a(R~p,s) 0 eiθ~·~σ/2 (2π)3/2 ab s ~ −ip·x † i + vb(Rp, s)e ac(R~p,s) . (1.298) Let’s focus on the zero-momentum terms. We need iθ~·~σ/2 ! X iθ~·~σ/2 ~ 0 e 0 ~ e ua(0, s ) = ~ ub(0, s) (1.299) ss0 0 eiθ·~σ/2 s0 ab and iθ~·~σ/2 ! X −iθ~·~σ/2 ~ 0 e 0 ~ e va(0, s ) = ~ vb(0, s). (1.300) s0s 0 eiθ·~σ/2 s0 ab 46 Quantum ﬁelds and special relativity The spinors 1 0

1 1 0 1 1 1 u(~0, ) = √   and u(~0, − ) = √   (1.301) 2 2 1 2 2 0 0 1 satisfy the condition (1.299) on the spinor ua(~0, s). Let’s look at the condition ~ ~ 1 (1.300) on va(0, s) for the simpler case of θ = θzˆ. We set s = 2 or equivalently s = + and also a = 1. Since the rotation is aboutz ˆ, −iθ~·~σ/2 −iθσ3/2 −iθs e = e = e δs0s. (1.302) s0s s0s So the condition (1.300) says that −iθ/2 ~ 1 iθ/2 ~ 1 e v1(0, 2 ) = e v1(0, 2 ). (1.303) ~ 1 So v1(0, 2 ) = 0. Now we set a = 2. Then the condition says that −iθ/2 ~ 1 −iθ/2 ~ 1 e v2(0, 2 ) = e v2(0, 2 ). (1.304) ~ 1 So v2(0, 2 ) can be anything. Now we set a = 3. Then the same condition (1.300) says that −iθ/2 ~ 1 iθ/2 ~ 1 e v3(0, 2 ) = e v3(0, 2 ) (1.305) ~ 1 which implies that v3(0, 2 ) = 0. Finally, we set a = 4. Now the same condition (1.300) says that −iθ/2 ~ 1 −iθ/2 ~ 1 e v4(0, 2 ) = e v4(0, 2 ). (1.306) ~ 1 ~ 1 So v4(0, 2 ) can be anything. That is, v(0, 2 ) must be of the form 0

1 α v(~0, ) =   (1.307) 2 0 β in which α and β are arbitrary. 1 Now we set s = − 2 or equivalently s = − and also a = 1. Then the condition (1.300) says that iθ/2 ~ 1 iθ/2 ~ 1 e v1(0, − 2 ) = e v1(0, − 2 ). (1.308) ~ 1 So v1(0, − 2 ) can be anything. But the condition (1.300) for a = 2 says iθ/2 ~ 1 −iθ/2 ~ 1 e v2(0, − 2 ) = e v2(0, − 2 ) (1.309) 1.18 Spinors and angular momentum 47 ~ 1 which means that v2(0, − 2 ) = 0. For a = 3, we get iθ/2 ~ 1 iθ/2 ~ 1 e v3(0, − 2 ) = e v3(0, − 2 ), (1.310) ~ 1 so v3(0, − 2 ) is unrestricted. Finally, for a = 4, we have iθ/2 ~ 1 −iθ/2 ~ 1 e v4(0, − 2 ) = e v4(0, − 2 ) (1.311) ~ 1 which makes v4(0, − 2 ) vanish. So γ

1 0 v(~0, − ) =   (1.312) 2 δ 0 in which γ and δ are arbitrary. To make the v’s suitably normalized and orthogonal to the u’s, one may set  0  −1

1 1  1  1 1  0  v(~0, ) = √   and v(~0, − ) = √   . (1.313) 2 2  0  2 2  1  −1 0

† 0 The state a (0, s, nc)|0i is (summed over a and s ) 1 Z Z d3xd3p d3x v¯ (0, s)ψ (x)|0i =v ¯ (0, s) v(~p,s0) e−ip·x a†(~p,s0, n )|0i (2π)3/2 a a a (2π)3 c Z 3 3 0 † 0 =v ¯a(0, s) d p δ (p)v(~p,s ) a (~p,s , nc)|0i

To determine its spin, we act on it with the operator e−iθJ3 that rotates states about the z axis by angle θ in a right-handed way. 1 X Z e−iθJ3 |0, s, n i = e−iθJ3 d3x v¯ (0, s)ψ (x)|0i (1.316) c (2π)3/2 a a a 1 X Z = d3x v¯ (0, s)e−iθJ3 ψ (x)eiθJ3 e−iθJ3 |0i. (2π)3/2 a a a 48 Quantum ﬁelds and special relativity Since the vacuum is invariant, this is

1 X Z e−iθJ3 |0, s, n i = d3x v¯ (0, s)e−iθJ3 ψ (x)eiθJ3 |0i c (2π)3/2 a a a 1 X Z = d3x v¯ (0, s)D(R−1) ψ (Rx)|0i. (1.317) (2π)3/2 a ab b a Since the jacobian of a rotation is unity, we have

1 X Z e−iθJ3 |0, s, n i = d3x v¯ (0, s)D(R−1) ψ (x)|0i (1.318) c (2π)3/2 a ab b a in which eiθ/2 0 0 0  −iθ/2 −1  0 e 0 0  D(R )ab =   . (1.319)  0 0 eiθ/2 0  0 0 0 e−iθ/2

So this is 1 X Z e−iθJ3 |0, s, n i = d3x v¯ (0, s)eiσ(a)θ/2ψ (x)|0i (1.320) c (2π)3/2 a a a in which σ(a) = 1 for a = 1 & 3, and σ(a) = −1 for a = 2 & 4. For s = ±1/2, the spinors v(0, s) are

 0  −1

1 1  1  1 1  0  v(~0, ) = √   and v(~0, − ) = √   . (1.321) 2 2  0  2 2  1  −1 0

So for s = 1/2, slots a = 2 & 4 are nonzero, and

1 X Z e−iθJ3 |0, 1/2, n i = e−iθ/2 d3x v¯ (0, s)ψ (x)|0i (1.322) c (2π)3/2 a a a which means that the state has spin 1/2 in the z direction. For s = −1/2, slots a = 1 & 3 are nonzero, and

1 X Z e−iθJ3 |0, 1/2, n i = eiθ/2 d3x v¯ (0, s)ψ (x)|0i (1.323) c (2π)3/2 a a a which means that the state has spin −1/2 in the z direction. 1.19 Charge conjugation 49 1.19 Charge conjugation Under charge conjugation, a massive Dirac ﬁeld

X Z d3p ψ(x) = u(~p,s) eip·x a(~p,s) + v(~p,s) e−ip·x a†(~p,s) (1.324) (2π)3/2 c s transforms as

−1 ∗ ∗ ∗ ∗ Cψ(x)C = −ξ βγ2βψ (x) = ξ γ2ψ (x) (1.325) in which ξ is the charge-conjugation parity of the particle annihilated by a (which is the complex conjugate of that of the particle annihilated by ac) and 0 1 0 σ2 0 1 0 σ2 βγ2β = (−i) = i = −γ2. (1.326) 1 0 −σ2 0 1 0 −σ2 0 Under charge conjugation, the creation operators for particles and their antiparticles, transform as

† −1 † † −1 † Ca (~p,s)C = ξac(~p,s) and Cac(~p,s)C = ξca (~p,s). (1.327) So

−1 ∗ −1 ∗ Ca(~p,s)C = ξ ac(~p,s) and Cac(~p,s)C = ξc a(~p,s). (1.328) So, suppressing explicit mention of the sum over spins, we have Z d3p Cψ(x)C−1 = u(~p,s) eip·x Ca(~p,s)C−1 + v(~p,s) e−ip·x Ca†(~p,s)C−1 (2π)3/2 c Z d3p = u(~p,s) eip·x ξ∗a (~p,s) + v(~p,s) e−ip·x ξ a†(~p,s)C−1. (2π)3/2 c c Under charge conjugation, the ﬁeld changes to −1 ∗ 0 σ2 ∗ ∗ ∗ Cψ(x)C = − iξ ψ (x) = ξ γ2ψ (x). (1.329) −σ2 0

∗ Explicitly and with ξc = ξ , we get

0 σ X Z d3p h Cψ(x)C−1 = − iξ∗ 2 u∗(~p,s) e−ip·x a†(~p,s) −σ 0 (2π)3/2 2 s ∗ ip·x i + v (~p,s) e ac(~p,s) (1.330) 50 Quantum ﬁelds and special relativity

∗ So leaving aside the factor −iξ , the spinors for the ac particles (aka, antiparticles) are 0 σ2 ∗ ub(~p,s) = v (~p,s) (1.331) −σ2 0 while the antispinors for the a particles are 0 σ2 ∗ va(~p,s) = u (~p,s). (1.332) −σ2 0 Let’s focus on the zero-momentum spinors. The zero-momentum spinors are

1 0

1 1 0 1 1 1 u(~0, ) = √   and u(~0, − ) = √   , 2 2 1 2 2 0 0 1 (1.333)  0  −1

1 1  1  1 1  0  v(~0, ) = √   and v(~0, − ) = √   . 2 2  0  2 2  1  −1 0

The spinor ub(0, +) is 0 σ2 ∗ ub(~0, +) = v (~0, +) (1.334) −σ2 0 or in all its components,

 0 0 0 −i  0  1  0 0 i 0  1  1  i 0 ub(~0, +) =   √   = √   = i u(~0, +) (1.335)  0 i 0 0  2  0  2 1 −i 0 0 0 −1 0 which is what we expect. Similarly, 0 σ2 ∗ ub(~0, −) = v (~0, −) (1.336) −σ2 0 or in all its components,

 0 0 0 −i −1 0  0 0 i 0  1  0  i 1 ub(~0, −) =   √   = √   = i u(~0, −) (1.337)  0 i 0 0  2  1  2 0 −i 0 0 0 0 1 1.20 Parity 51 which is what we expect. Similarly, the spinor va(0, +) is 0 σ2 ∗ va(~0, +) = u (~0, +) (1.338) −σ2 0 or in all its components,  0 0 0 −i 1  0   0 0 i 0  1 0 i  1  va(~0, +) =   √   = √   = i v(~0, +) (1.339)  0 i 0 0  2 1 2  0  −i 0 0 0 0 −1 which is what we expect. Finally, the spinor va(0, −) is 0 σ2 ∗ va(~0, −) = u (~0, −) (1.340) −σ2 0 or in all its components,  0 0 0 −i 0 −1  0 0 i 0  1 1 i  0  va(~0, −) =   √   = √   = i v(~0, −) (1.341)  0 i 0 0  2 0 2  1  −i 0 0 0 1 0 which is what we expect.

1.20 Parity Under parity the ﬁeld ψ(x) (1.324) changes to Pψ(x)P−1 = η∗βψ(P x) (1.342) where η is the parity of the particle annihilated by a, which is minus the complex conjugate of that of the particle annihilated by b. The annihilation and creation operators transform as † −1 † † −1 † Pa (~p,s)P = ηa (−~p,s) and Pac(~p,s)P = ηcac(−~p,s) (1.343) and −1 ∗ −1 ∗ Pa(~p,s)P = η a(−~p,s) and Pa(~p,s)P = ηc a(−~p,s). (1.344) So Z d3p Pψ(x)P−1 = u(~p,s) eip·x Pa(~p,s)P−1 + v(~p,s) e−ip·x Pa†(~p,s)P−1 (2π)3/2 c Z d3p = u(~p,s) eip·x η∗a(−~p,s) + v(~p,s) e−ip·x η a†(−~p,s) (2π)3/2 c c 52 Quantum ﬁelds and special relativity

∗ But ηc = −η , so Z d3p Pψ(x)P−1 = η∗ u(~p,s) eip·x a(−~p,s) − v(~p,s) e−ip·x a†(−~p,s). (2π)3/2 c Now Pψ(x)P−1 = η∗βψ(P x), that is, Z d3p Pψ(x)P−1 = η∗β u(~p,s) eiP p·x a(~p,s) + v(~p,s) e−iP p·x a†(~p,s) (2π)3/2 c Z d3p = η∗β u(−~p,s)eip·xa(−~p,s) + v(−~p,s)e−ip·xa†(−~p,s) (2π)3/2 c So we need βu(−~p,s) = u(~p,s) and βv(−~p,s) = − v(~p,s). (1.345) These rules are easy to check for ~p = 0 βu(~0, s) = u(~0, s) and βv(~0, s) = − v(~0, s) (1.346) which the spinors (1.333) satisfy. 2 Feynman diagrams

2.1 Time-dependent perturbation theory Most physics problems are insoluble, and we must approximate the un- known solution by numerical or analytic methods. The most common analytic method is called perturbation theory and is based on an assumption that something is small compared to something simple that we can analyze. In time-dependent perturbation theory, we write the hamiltonian H as the sum H = H0 + V of a simple hamiltonian H0 and a small, complicated part V which we give the time dependence induced by H0

V (t) = eiH0t/~ V e−iH0t/~. (2.1)

We alter the time dependence of states by the same simple exponential (and set ~ = 1) |ψ, ti = eiH0te−iHt|ψi (2.2) and ﬁnd as its time derivative d i |ψ, ti = eiH0t(H − H )e−iHt|ψi = eiH0t V e−iH0t eiH0te−iHt|ψi dt 0 (2.3) = V (t) |ψ, ti. We can formally solve this diﬀerential equation by time-ordering factors of V (t) in expansion of the exponential

−i R t V (t0) dt0 |ψ, ti = T e 0 |ψi n Z t 1 Z o (2.4) = 1 − i V (t0) dt0 − T V (t0)V (t00) dt0dt00 + ... |ψi 0 2! so that V ’s of later times occur to the left of V ’s of earier times, that is, if t > t0 then T [V (t)V (t0)] = V (t)V (t0) and so forth. 54 Feynman diagrams 2.2 Dyson’s expansion of the S matrix The time-evolution operator in the interaction picture is the time-ordered exponential of the integral over time of the interaction hamiltonian h R i T e−i V (t) dt . (2.5)

Time ordering has V (t>) to the left of V (t<) if the time t> is later than the time V (t<). The interaction hamiltonian is Z V (t) = H(t, x) d3x. (2.6)

The density of the interaction hamiltonian is (or is taken to be) a sum of terms X H(t, x) = giHi(t, x) (2.7) i

† each of which is a monomial in the ﬁelds ψ`(x) and their adjoints ψ` (x). A generic ﬁeld is an integral over momentum

X Z d3p h i ψ (x) = u (p, s, n) a(p, s, n) eip·x +v (p, s, n) b†(p, s, n) e−ip·x ` (2π)3/2 ` ` s (2.8) in which n labels the kind of ﬁeld. The elements of the S matrix are amplitudes for an initial state |p1, s1, n1; ... ; pk, sk, nki 0 0 0 0 0 0 to evolve into a ﬁnal state |p1, s1, n1; ... ; pk0 , sk0 , nk0 i. The case considered most often is when k = 2 incoming particles and k0 = 2 or 3 outgoing particles, but in high-energy collisions, k0 can be much larger than 2 or 3. An S-matrix amplitude is

0 0 0 0 0 0 h −i R H(x) d4xi S 0 0 0 = hp , s , n ; ... ; p 0 , s 0 , n 0 |T e |p , s , n ; ... ; p , s , n i p1,s1,n1;...;p1,s1,n1;... 1 1 1 k k k 1 1 1 k k k

= h0|a(pk0 , sk0 , nk0 ; ... ; a(p1, s1, n1) (2.9) ∞ X (−i)n Z h i × d4x . . . d4x T H(x ) ...H(x ) N! 1 N 1 N N=0 † † × a (p1, s1, n1; ... ; a (pk, sk, nk)|0i in which |0i is the vacuum state. It is the mean value in the vacuum state of an inﬁnite polynomial in creation and annihilation operators. To make sense of it, the ﬁrst step is to normally order the time-evolution operator by moving all annihilation operators to the right of all creation operators. We assume for now that H(x) itself is already normally ordered. 2.2 Dyson’s expansion of the S matrix 55 To do that we use these commutation relations with plus signs for bosons and minus signs for fermions:

† 0 0 0 † 0 0 0 3 0 a(p, s, n) a (p , s , n ) = ± a (p , s , n ) a(p, s, n) + δ (~p − ~p ) δss0 δnn0 a(p, s, n) a(p0, s0, n0) = ± a(p0, s0, n0) a(p, s, n) (2.10) a†(p, s, n) a†(p0, s0, n0) = ± a†(p0, s0, n0) a†(p, s, n).

Since a(p, s, n) |0i = 0 and h0| a†(p, s, n) = 0, (2.11) the S-matrix amplitude is just the left-over pieces proportional to products of delta functions multiplied by u’s and v’s and Fourier factors all divided by 2π’s. Each such piece arises from the pairing of one annihilation operator with one creation operator. Such pairs arise in six ways:

0 0 0 † 1. A particle p , s , n in the ﬁnal state can pair with a ﬁeld adjoint ψ` (x) in H(x) and yield the factor

0 0 0 † ∗ 0 0 0 −ip0·x −3/2 [a(p , s , n ), ψ` (x)]∓ = u` (p , s , n )e (2π) . (2.12) 0 0 0 2. An antiparticle p , s , n in the ﬁnal state can pair with a ﬁeld ψ`(x) in H(x) and yield the factor

0 0 0 0 0 0 −ip0·x −3/2 [b(p , s , n ), ψ`(x)]∓ = v`(p , s , n )e (2π) . (2.13)

3. A particle p, s, n in the initial state can pair with a ﬁeld ψ`(x) in H(x) and yield the factor

† ip·x −3/2 [ψ`(x), a (p, s, n)]∓ = u`(p, s, n)e (2π) . (2.14)

4. An antiparticle p, s, n in the initial state can pair with a ﬁeld adjoint † ψ` (x) in H(x) and yield the factor † † ∗ ip·x −3/2 [ψ` (x), b (p, s, n)]∓ = v` (p, s, n)e (2π) . (2.15) 5. A particle (or antiparticle) p0, s0, n0 in the ﬁnal state can pair with a particle (or antiparticle) p, s, n in the initial state and yield

0 0 0 † 3 0 [a(p , s , n ), a (p, s, n)]∓ = δ (~p − ~p ) δss0 δnn0 . (2.16)

+ − 6. A field ψ`(x) = ψ` (x) + ψ` (x) in H(x) can pair with a field adjoint in 0 † +† −† + +† H (y), ψm(y) = ψm (y) + ψm (y). But for ψ` (x) to cross ψm (y) the time 0 0 −† − 0 x must be later than y , and for ψm (y) to cross ψ` (x), the time y must 56 Feynman diagrams be later than x0. If for instance H(x) = ψ†(x)ψ(x)φ(x), then in Dyson’s expansion these terms would appear θ(x0−y0) ψ†(x)ψ(x)φ(x) ψ†(y)ψ(y)φ(y)+θ(y0−x0) ψ†(y)ψ(y)φ(y) ψ†(x)ψ(x)φ(x). (2.17) The resulting pairings are the propagator 0 0 + +† 0 0 −† − θ(x − y )[ψ` (x), ψm (y)]∓ ± θ(y − x )[ψm (y), ψ` (x)]∓ ≡ −i∆`m(x, y) (2.18) in which the ± signs will be explained later. One then integrates over N spacetimes and the implicit momenta. The result is defined by a set of rules and Feynman diagrams. In general, the 1/N! in Dyson’s expansion is cancelled by the N! ways of labelling the xi’s. For example, in the term 1 Z d4x d4x T H(x )H(x ) (2.19) 2! 1 2 1 2 one can have H(x1) absorb an incoming electron of momentum p and have 0 H(x2) absorb an incoming electron of momentum p or the reverse. But some processes require special combinatorics. So people often write g g H(x) = φ3(x) or H(x) = φ4(x) (2.20) 3! 4! to compensate for multiple possible pairings. But these factorials don’t al- ways cancel. Fermions introduce minus signs. The surest way to check the signs and factorials in each process until one has gained sufficient experience.

2.3 The Feynman propagator for scalar fields Adding ±i to the denominator of a pole term of an integral formula for a function f(x) can slightly shift the pole into the upper or lower half plane, causing the pole to contribute if a ghost contour goes around the upper half- plane or the lower half-plane. Such an i can impose a boundary condition on a Green’s function. The Feynman propagator ∆F (x) is a Green’s function for the Klein- Gordon differential operator (Weinberg, 1995, pp. 274–280) 2 4 (m − 2)∆F (x) = δ (x) (2.21) in which x = (x0, x) and ∂2 ∂2 2 = 4 − = 4 − (2.22) ∂t2 ∂(x0)2 2.3 The Feynman propagator for scalar fields 57 is the four-dimensional version of the laplacian 4 ≡ ∇ · ∇. Here δ4(x) is the four-dimensional Dirac delta function Z d4q Z d4q δ4(x) = exp[i(q · x − q0x0)] = eiqx (2.23) (2π)4 (2π)4 in which qx = q · x − q0x0 is the Lorentz-invariant inner product of the 4-vectors q and x. There are many Green’s functions that satisfy Eq.(2.21). Feynman’s propagator ∆F (x) Z d4q exp(iqx) Z d3q Z ∞ dq0 eiq·x−iq0x0 ∆F (x) = 4 2 2 = 3 2 2 . (2.24) (2π) q + m − i (2π) −∞ 2π q + m − i is the one that satisfies boundary conditions that will become evident when p 2 2 we analyze the effect of its i. The quantity Eq = q + m is the energy of a particle of mass m and momentum q in natural units with the speed of 0 light c = 1. Using this abbreviation and setting = /2Eq, we may write the denominator as

2 2 02 2 0 0 0 0 02 q + m − i = q · q − q + m − i = Eq − i − q Eq − i + q + (2.25) in which 02 is negligible. Dropping the prime on , we do the q0 integral Z ∞ 0 dq −iq0x0 1 I(q) = − e 0 0 . (2.26) −∞ 2π [q − (Eq − i)] [q − (−Eq + i)] As shown in Fig. 2.1, the integrand

−iq0x0 1 e 0 0 (2.27) [q − (Eq − i)] [q − (−Eq + i)]

0 has poles at Eq − i and at −Eq + i. When x > 0, we can add a ghost contour that goes clockwise around the lower half-plane and get

0 1 I(q) = ie−iEqx x0 > 0. (2.28) 2Eq When x0 < 0, our ghost contour goes counterclockwise around the upper half-plane, and we get

0 1 I(q) = ieiEqx x0 < 0. (2.29) 2Eq Using the step function θ(x) = (x + |x|)/2, we combine (2.28) and (2.29)

1 h 0 0 i − iI(q) = θ(x0) e−iEqx + θ(−x0) eiEqx . (2.30) 2Eq 58 Feynman diagrams

0 Figure 2.1 In equation (2.27), the function f(q ) has poles at ±(Eq − i), and the function exp(−iq0x0) is exponentially suppressed in the lower half plane if x0 > 0 and in the upper half plane if x0 < 0. So we can add a ghost contour (. . . ) in the LHP if x0 > 0 and in the UHP if x0 < 0.

In terms of the Lorentz-invariant function

Z 3 1 d q 0 ∆+(x) = 3 exp[i(q · x − Eqx )] (2.31) (2π) 2Eq and with a factor of −i, Feynman’s propagator (2.24) is

0 0 0 − i∆F (x) = θ(x ) ∆+(x) + θ(−x ) ∆+(x, −x ). (2.32) 2.3 The Feynman propagator for scalar ﬁelds 59

The integral (2.31) deﬁning ∆+(x) is insensitive to the sign of q, and so Z 3 1 d q 0 ∆+(−x) = 3 exp[i(−q · x + Eqx )] (2.33) (2π) 2Eq Z 3 1 d q 0 0 = 3 exp[i(q · x + Eqx )] = ∆+(x, −x ). (2π) 2Eq Thus we arrive at the Standard form of the Feynman propagator

0 0 − i∆F (x) = θ(x ) ∆+(x) + θ(−x ) ∆+(−x). (2.34) The annihilation operators a(q) and the creation operators a†(p) of a scalar field φ(x) satisfy the commutation relations [a(q), a†(p)] = δ3(q − p) and [a(q), a(p)] = [a†(q), a†(p)] = 0. (2.35) Thus the commutator of the positive-frequency part Z d3p φ+(x) = exp[i(p · x − p0x0)] a(p) (2.36) p(2π)32p0 of a scalar field φ = φ+ + φ− with its negative-frequency part Z d3q φ−(y) = exp[−i(q · y − q0y0)] a†(q) (2.37) p(2π)32q0 is the Lorentz-invariant function ∆+(x − y) Z d3p d3q [φ+(x), φ−(y)] = eipx−iqy [a(p), a†(q)] (2π)32pq0p0 Z d3p = eip(x−y) = ∆ (x − y) (2.38) (2π)32p0 + in which p(x − y) = p · (x − y) − p0(x0 − y0). 2 2 0 2 2 At points x that are space-like, that is, for which x = x −(x √) ≡ r > 0, 2 the Lorentz-invariant function ∆+(x) depends only upon r = + x and has the value (Weinberg, 1995, p. 202) m ∆ (x) = K (mr) (2.39) + 4π2r 1 in which the Hankel function K1 is π 1 z z 1 K (z) = − [J (iz) + iN (iz)] = + ln + γ − + ... (2.40) 1 2 1 1 z 2 2 2 where J1 is the first Bessel function, N1 is the first Neumann function, and γ = 0.57721 ... is the Euler-Mascheroni constant. 60 Feynman diagrams The Feynman propagator arises most simply as the mean value in the vacuum of the time-ordered product of the fields φ(x) and φ(y)

T{φ(x)φ(y)} ≡ θ(x0 − y0)φ(x)φ(y) + θ(y0 − x0)φ(y)φ(x). (2.41)

The operators a(p) and a†(p) respectively annihilate the vacuum ket a(p)|0i = 0 and bra h0|a†(p) = 0, and so by (2.36 & 2.37) do the positive- and negative- frequency parts of the ﬁeld φ+(z)|0i = 0 and h0|φ−(z) = 0. Thus the mean value in the vacuum of the time-ordered product is

h0|T {φ(x)φ(y)} |0i = h0|θ(x0 − y0)φ(x)φ(y) + θ(y0 − x0)φ(y)φ(x)|0i = h0|θ(x0 − y0)φ+(x)φ−(y) + θ(y0 − x0)φ+(y)φ−(x)|0i = h0|θ(x0 − y0)[φ+(x), φ−(y)] + θ(y0 − x0)[φ+(y), φ−(x)]|0i. (2.42)

But by (2.38), these commutators are ∆+(x − y) and ∆+(y − x). Thus the mean value in the vacuum of the time-ordered product of two real scalar ﬁelds

0 0 0 0 h0|T {φ(x)φ(y)} |0i = θ(x − y )∆+(x − y) + θ(y − x )∆+(y − x) Z d4q −i = −i∆ (x − y) = eiq·x (2.43) F (2π)4 q2 + m2 − i is the Feynman propagator (2.32) multiplied by −i.

2.4 Application to a cubic scalar field theory The action density 1 1 g L = − ∂ φ ∂νφ − m2φ2 − φ3, (2.44) 2 ν 2 3! describes a scalar field with cubic interactions. We let g Z V (t) = φ(x)3 d3x (2.45) 3! in which the field φ has the free-field time dependence Z d3p h i φ(x) = a(p)eip·x + a†(p)e−ip·x (2.46) p(2π)32p0 in which p · x = ~p · x − p0t and p0 = pp2 + m2. The amplitude for the scattering of bosons with momenta p and k to momenta p0 and k0 is to 2.4 Application to a cubic scalar field theory 61 lowest order in the coupling constant g 1 Z A = hp0, k0| − T V (t0)V (t00) dt0 dt00 |p, ki 2! (2.47) g2 Z = − hp0, k0| T φ3(x)φ3(y) d4xd4y |p, ki. 2!(3!)2

This process happens in three ways, so A = A1 + A2 + A3. The first way is for the incoming particles to be absorbed at the same vertex x or y and for the outgoing particles to be emitted at the other vertex y or x. We cancel the 2! by choosing the initial particles to be absorbed at y and the final particles to be emitted at x. Then using the expansion (2.46) of the field, we find 2 Z 3 00 3 00 3 000 3 000 g 0 0 d p d k d p d k † 000 † 000 A1 = − h0|a(k )a(p ) a (p )a (k ) 4 (2π)6p2p0002k0002p00002k0000 000 000 00 00 × e−i(p +k )·xT φ(x)φ(y)a(p00)a(k00)ei(p +k )·yd4xd4y a†(p)a†(k)|0i (2.48) after canceling two factors of 3 because each of the 3 fields φ3(x) and each of the 3 fields φ3(y) could be the one to remain in the time-ordered product T [φ(x)φ(y)]. The commutation relations [a(p), a†(k)] = δ3(p − k) now give Z 4 4 2 d xd y i(p+k)·y−i(p0+k0)·x A1 = − g e h0|T φ(x)φ(y) |0i (2π)6p2p002k002p02k0 (2.49) in which the mean value in the vacuum of the time-ordered product Z d4q exp(iqx) h0|T φ(x)φ(y)|0i = − i∆ (x − y) = −i (2.50) F (2π)4 q2 + m2 − i is Feynman’s propagator (2.24 & 2.43). Thus the amplitude A1 is Z 4 4 4 i(p+k)·y−i(p0+k0)·x+iq·(x−y) 2 d xd yd q e A1 = ig . (2.51) (2π)10p2p002k002p02k0 q2 + m2 − i Thanks to Dirac’s delta function, the integrals over x, y, and q are easy, and in the smooth → 0 limit A1 is Z 4 4 4 0 0 2 d q δ (p + k − q)δ (q − p − k ) A1 = ig (2π)2p2p002k002p02k0 q2 + m2 − i (2.52) δ4(p + k − p0 − k0) ig2 = . (2π)2p2p002k002p02k0 (p + k)2 + m2 62 Feynman diagrams The sum of the three amplitudes is

δ4(p + k − p0 − k0)h ig2 ig2 ig2 i A = + + 16π2pp00k00p0k0 (p + k)2 + m2 (p − k0)2 + m2 (p − p0)2 + m2 (2.53) in which the delta function conserves energy and momentum.

2.5 Feynman’s propagator for fields with spin The time-ordered product for fields with spin is defined so as to compensate for the minus sign that arises when a Fermi field is moved past a Fermi field. Thus the mean value in the vacuum of the time-ordered product † T{ψ`(x)ψm(y)} is

n † o 0 0 † 0 0 † h0|T ψ`(x)ψm(y) |0i = h0|θ(x − y )ψ`(x)ψm(y) ± θ(y − x )ψm(y)ψ`(x)|0i 0 0 + †− 0 0 †+ − = h0|θ(x − y )ψ` (x)ψm (y) ± θ(y − x )ψm (y)ψ` (x)|0i 0 0 + †− = h0|θ(x − y )[ψ` (x), ψm (y)]∓ 0 0 †+ − ± θ(y − x )[ψm (y), ψ` (x)]|0i∓. (2.54) in which the upper signs are used for bosons and the lower ones for fermions. The expansions

Z + −3/2 X 3 ip·x ψ` (x) = (2π) d p u`(~p,s) e a(~p,s) s Z (2.55) − −3/2 X 3 −ip·x † ψ` (x) = (2π) d p v`(~p,s) e b (~p,s) s give for the (anti)commutators

Z + †− h −3/2 X 3 ip·x [ψ` (x), ψm (y)]∓ = (2π) d p u`(~p,s) e a(~p,s), s Z −3/2 X 3 ∗ ~ −ik·y † ~ i (2π) d k um(k, r)e a (k, r) (2.56) ∓ r X Z d3p = u (~p,s) u∗ (~p,s) eip·(x−y) (2π)3 ` m s 2.6 Feynman’s propagator for spin-one-half ﬁelds 63 and

Z †+ − h −3/2 X 3 ∗ ip·y [ψm (y), ψ` (x)]∓ = (2π) d p vm(~p,s) e b(~p,s), s Z −3/2 X 3 ~ −ik·x † ~ i (2π) d k v`(k, r)e b (k, r) (2.57) ∓ r X Z d3p = v (~p,s) v∗ (~p,s) eip·(y−x). (2π)3 ` m s

Putting these expansions into the formula (2.54) for the time-ordered product, we get for its mean value in the vacuum

n o X Z d3p h0|T ψ (x)ψ† (y) |0i = θ(x0 − y0) u (~p,s) u∗ (~p,s) eip·(x−y) ` m (2π)3 ` m s X Z d3p ± θ(y0 − x0) v (~p,s) v∗ (~p,s) eip·(y−x). (2π)3 ` m s (2.58)

2.6 Feynman’s propagator for spin-one-half ﬁelds For ﬁelds of spin one half, the spin sums are

X ∗ h 1 c i N(~p) = u`(~p,s)um(~p,s) = (−i p γc + m) β `m 2p0 `m s (2.59) X ∗ h 1 c i M(~p) = v`(~p,s)vm(~p,s) = (−i p γc − m) β , `m 2p0 `m s so for spin-one-half ﬁelds Feynman’s propagator is

Z 3 n † o 0 0 d p h 1 c i ip·(x−y) h0|T ψ`(x)ψm(y) |0i = θ(x − y ) (−i pcγ + m) β e (2π)3 2p0 `m Z 3 0 0 d p h 1 c i ip·(y−x) − θ(y − x ) (−i pcγ − m) β e . (2π)3 2p0 `m (2.60) 64 Feynman diagrams

c c Using the derivative term −∂cγ to generate −ipcγ , we get Z 3 n † o 0 0 h c i d p ip·(x−y) h0|T ψ`(x)ψm(y) |0i = θ(x − y ) ( − ∂cγ + m) β e `m (2π)32p0 Z 3 0 0 h c i d p ip·(y−x) − θ(y − x ) (∂cγ − m) β e `m (2π)32p0 0 0 h c i = θ(x − y ) ( − ∂cγ + m) β ∆+(x − y) `m 0 0 h c i + θ(y − x ) ( − ∂cγ + m) β ∆+(y − x). `m (2.61) Since the derivative of the step function is a delta function 0 0 0 0 ∂0θ(x − y ) = δ(x − y ), (2.62) we can write Feynman’s propagator as

n † o h c i h0|T ψ`(x)ψm(y) |0i = (−∂cγ + m) β `m 0 0 0 0 × θ(x − y ) ∆+(x − y) + θ(y − x ) ∆+(y − x) 0 0 0 + ∂0γ βθ(x − y ) ∆+(x − y) 0 0 0 − ∂0γ βθ(y − x ) ∆+(y − x) (2.63) and so also as n † o h c i h0|T ψ`(x)ψm(y) |0i = (−∂cγ + m) β `m 0 0 0 0 × θ(x − y ) ∆+(x − y) + θ(y − x ) ∆+(y − x) 0 0 0 0 0 0 +γ βδ(x − y ) ∆+(x − y) − γ βδ(y − x ) ∆+(y − x) h c i = ( − ∂cγ + m) β `m 0 0 0 0 × θ(x − y ) ∆+(x − y) − θ(y − x ) ∆+(y − x) 0 0 0 +γ βδ(x − y ) ∆+(x − y) − ∆+(y − x) . (2.64)

But at equal times ∆+(x − y) = ∆+(y − x). So the ugly ﬁnal term vanishes, and Feynman’s propagator for spin-one-half ﬁelds is

n † o h c i h0|T ψ`(x)ψm(y) |0i = ( − ∂cγ + m) β `m 0 0 0 0 × θ(x − y )∆+(x − y) + θ(y − x )∆+(y − x) . (2.65) 2.6 Feynman’s propagator for spin-one-half ﬁelds 65 But this last piece is Feynman’s propagator for real scalar ﬁelds

0 0 0 0 h0|T {φ(x)φ(y)} |0i = θ(x − y )∆+(x − y) + θ(y − x )∆+(y − x) Z d4q −i = − i∆ (x − y) = eiq·(x−y). F (2π)4 q2 + m2 − i so Feynman’s propagator for spin-one-half ﬁelds is

n † o h0|T ψ`(x)ψm(y) |0i ≡ − i∆`m(x − y) Z 4 h c i d q −i iq·(x−y) = ( − ∂cγ + m) β e . `m (2π)4 q2 + m2 − i (2.66)

Letting the derivatives act on the exponential, we get

n † o h0|T ψ`(x)ψm(y) |0i ≡ − i∆`m(x − y) 4 a Z d q ( − iqaγ + m)β = − i `m eiq·(x−y). (2π)4 q2 + m2 − i (2.67)

Since a a 2 2 (−iqaγ + m)(iqaγ + m) = q + m , (2.68) people often write this as

n † o h0|T ψ`(x)ψm(y) |0i ≡ − i∆`m(x − y) 4 a Z d q ( − iqaγ + m)β = − i `m eiq·(x−y) (2π)4 q2 + m2 − i Z 4 d q h 1 i iq·(x−y) = − i 4 a β e . (2π) iqaγ + m − i `m (2.69)

a Feynman was a master of notation (and of everything else). He set p/ = paγ and wrote

n † o h0|T ψ`(x)ψm(y) |0i ≡ − i∆`m(x − y) Z d4q ( − i/q + m)β = − i `m eiq·(x−y) (2π)4 q2 + m2 − i (2.70) Z d4q 1 = − i β eiq·(x−y). (2π)4 i/q + m − i `m 66 Feynman diagrams In the common notation ψ = ψ†β, his propagator is

h0|T ψ`(x)ψm(y) |0i ≡ − i∆`m(x − y) β Z d4q ( − i/q + m) = − i `m eiq·(x−y) (2π)4 q2 + m2 − i (2.71) Z d4q 1 = − i eiq·(x−y). (2π)4 i/q + m − i `m

2.7 Application to a theory of fermions and bosons Let us consider ﬁrst the theory

† H(x) = g`m ψ` (x) ψm(x) φ(x) (2.72)

where g`m is a coupling constant, ψ(x) is the ﬁeld of a fermion f, and φ(x) = φ†(x) is a real boson b. Let’s compute the amplitude A for f + b → f 0 + b0. The lowest-order term is

1 Z h i A = − d4xd4y h0|a(p0, s0)b(k0) T H(x) H(y) b†(k)a†(p, s)|0i (2.73) 2! Z g`mg`0m0 4 4 0 0 0 h † † i = − d xd yh0|a(p , s )b(k )T ψ (x) ψ (x)φ(x)ψ (y) ψ 0 (y)φ(y) 2! ` m `0 m × b†(k)a†(p, s)|0i.

Here the operators b(k0) and b†(k) are the boson deletion and addition operators. Either the boson field φ(y) deletes the boson from the initial state and the boson field φ(x) deletes the boson from the final state or the boson field φ(x) deletes the boson from the initial state and the boson field φ(y) deletes the boson from the final state. These give the same result. So we cancel the 2! and let φ(y) delete the boson from the initial state and have 2.7 Application to a theory of fermions and bosons 67 φ(x) add the boson of the final state. We then get

Z 4 4 0 0 0 h † − † + i A = − g`mg`0m0 d xd yh0|a(p , s )b(k )T ψ` (x) ψm(x) φ (x)ψ`0 (y) ψm0 (y) φ (y) × b†(k)a†(p, s)|0i Z 4 4 0 0 0 h † = − g`mg`0m0 d xd y h0|a(p , s )b(k ) T ψ` (x) ψm(x) Z 3 00 d k † 00 −ik00·x † + i † † × b (k )e ψ 0 (y) ψm0 (y) φ (y) b (k)a (p, s)|0i p(2π)32k000 ` Z 4 4 0 0 h † = − g`mg`0m0 d xd y h0|a(p , s ) T ψ` (x) ψm(x) Z 3 00 d k 3 ~ 0 ~ 00 −ik00·x † + i † † × δ (k − k ) e ψ 0 (y) ψm0 (y) φ (y) b (k)a (p, s)|0i p(2π)32k000 ` (2.74) Z 4 4 0 0 h † 1 −ik0·x = − g`mg`0m0 d xd y h0|a(p , s ) T ψ (x) ψm(x) e ` p(2π)32k00 † + i † † × ψ`0 (y) ψm0 (y) φ (y) b (k)a (p, s)|0i.

We now let φ+(y) delete the boson from the initial state.

Z 2 4 4 0 0 h † 1 −ik0·x A = − g d xd y h0|a(p , s ) T ψ (x) ψm(x) e ` p(2π)32k00 Z 3 00 † d k 00 ik00·yi † † × ψ 0 (y) ψm0 (y) b(k )e b (k)a (p, s)|0i ` p(2π)32k000 Z 4 4 0 0 h † 1 −ik0·y = − g`mg`0m0 d xd y h0|a(p , s ) T ψ (x) ψm(x) e ` p(2π)32k00 Z 3 00 † d k 3 00 ik00·yi † × ψ 0 (y) ψm0 (y) δ (k − k)e a (p, s)|0i (2.75) ` p(2π)32k000 Z 4 4 0 0 h † 1 −ik0·x = − g`mg`0m0 d xd y h0|a(p , s ) T ψ (x) ψm(x) e ` p(2π)32k00

† 1 ik·yi † × ψ 0 (y) ψm0 (y) e a (p, s)|0i ` p(2π)32k0 Z 4 4 d xd y ik·y−ik0·x 0 0 h † † i † = − g`mg`0m0 e h0|a(p , s )T ψ (x)ψm(x)ψ 0 (y)ψm0 (y) a (p, s)|0i. p(2π)62k002k0 ` ` 68 Feynman diagrams Now there are two terms. In one the initial fermion is deleted at y and the ﬁnal fermion is added at x Z 4 4 d xd y ik·y−ik0·x 0 0 h +† † + i † A1 = − g`mg`0m0 e h0|a(p , s )T ψ (x)ψm(x)ψ 0 (y)ψ 0 (y) a (p, s)|0i p(2π)62k002k0 ` ` m Z 4 4 Z 3 00 d xd y ik·y−ik0·x 0 0 h d p −ip00·x † 00 00 ∗ 00 00 = − g`mg`0m0 e h0|a(p , s )T e a (p , s )u (p , s ) p(2π)62k002k0 p(2π)3 ` Z 3 000 † d p ip000·y 000 000 000 000 i † × ψm(x)ψ 0 (y) e a(p , s ) um0 (p , s ) a (p, s)|0i ` p(2π)3 (2.76) Z 4 4 Z 3 00 d xd y ik·y−ik0·x h d p −ip00·x ∗ 00 00 3 0 00 = − g`mg`0m0 e h0|T e u (p , s )δ (~p − ~p )δs0s00 p(2π)62k002k0 p(2π)3 ` Z 3 000 † d p ip000·y 000 000 3 000 i × ψm(x)ψ 0 (y) e um0 (p , s )δ (~p − ~p )δss000 |0i ` p(2π)3 Z 4 4 ∗ 0 0 d xd y ik·y−ik0·x −ip0·x ip·y h † i = − g`mg`0m0 u (p , s )um0 (p, s) e e e h0|T ψm(x)ψ 0 (y) |0i ` p(2π)122k002k0 ` Z 4 4 ∗ 0 0 d xd y i(k+p)·y−i(k0+p0)·x h † i = − g`mg`0m0 u (p , s )um0 (p, s) e h0| T ψm(x)ψ 0 (y) |0i. ` p(2π)122k002k0 `

In terms of SW’s deﬁnition (6.2.31) of the fermion propagator, A1 is

Z 4 4 ∗ 0 0 d xd y i(k+p)·y−i(k0+p0)·x A1 = −g`mg`0m0 u (p , s )um0 (p, s) e −i∆m`0 (x, y) . ` p(2π)122k002k0 (2.77) In the other term, the initial fermion is deleted at x and the ﬁnal fermion is added at y

Z 4 4 d xd y ik·y−ik0·x 0 0 h † + +† i † A2 = − g`mg`0m0 e h0|a(p , s )T ψ (x)ψm(x)ψ 0 (y)ψm0 (y) a (p, s)|0i p(2π)62k002k0 ` ` Z 4 4 d xd y ik·y−ik0·x 0 0 h † +† + i † = − g`mg`0m0 e h0|a(p , s )T ψ (x)ψ 0 (y)ψm0 (y)ψm(x) a (p, s)|0i p(2π)62k002k0 ` ` (2.78) Z 4 4 d xd y ik·y−ik0·x 0 0 h +† † + i † = g`mg`0m0 e h0|a(p , s )T ψ 0 (y)ψ (x)ψm0 (y)ψm(x) a (p, s)|0i p(2π)62k002k0 ` `

† +† +† † in which the minus sign arises from the transposition ψ` (x) ψ (y) → ψ (y) ψ` (x). + +† +† + The earlier transposition ψ (x) ψ (y) ψm0 (y) → ψ (y) ψm0 (y) ψ (x) pro- duced two minus signs or one plus sign. Inserting the expansions of ψ+†(y) 2.7 Application to a theory of fermions and bosons 69 and ψ+(x), we have Z 4 4 Z 3 00 d xd y ik·y−ik0·x 0 0 h d p ∗ 00 00 † 00 00 −ip00·y A2 = g`mg`0m0 e h0|a(p , s )T u 0 (p , s )a (p , s )e p(2π)62k002k0 p(2π)3 ` Z 3 000 † d p 000 000 000 000 ip000·xi † × ψ (x)ψm0 (y) um(p , s )a(p , s )e a (p, s)|0i ` p(2π)3 Z 4 4 ∗ 0 0 d xd y i(k−p0)·y−i(k0−p)·x h † i = g`mg`0m0 u 0 (p , s ) um(p, s) e h0|T ψ (x) ψm0 (y) |0i. ` p(2π)122k002k0 ` (2.79)

In terms of SW’s deﬁnition (6.2.31) of the fermion propagator, A2 is Z 4 4 ∗ 00 00 d xd y i(k−p0)·y−i(k0−p)·x h † i A2 = − g`mg`0m0 u 0 (p , s )um(p, s) e h0|T ψm0 (y)ψ (x) |0i ` p(2π)122k002k0 ` Z 4 4 ∗ 0 0 d xd y i(k−p0)·y−i(k0−p)·x = − g`mg`0m0 u 0 (p , s ) um(p, s) e − i∆m0`(y, x) . ` p(2π)122k002k0 (2.80) 0 0 The full amplitude for f + b → f + b is the sum A = A1 + A2 of the two amplitudes Z 4 4 ∗ 0 0 d xd y i(k+p)·y−i(k0+p0)·x A = − g`mg`0m0 u (p , s )um0 (p, s) e − i∆m`0 (x, y) ` p(2π)122k002k0 Z 4 4 ∗ 0 0 d xd y i(k−p0)·y−i(k0−p)·x − g`mg`0m0 u 0 (p , s ) um(p, s) e − i∆m0`(y, x) . ` p(2π)122k002k0 (2.81) 0 0 Interchanging x and y, m and m , and ` and ` in A1, we get Z 4 4 ∗ 0 0 d xd y i(k+p)·x−i(k0+p0)·y A = − g`0m0 g`mu 0 (p , s )um(p, s) e − i∆m0`(y, x) ` p(2π)122k002k0 Z 4 4 ∗ 0 0 d xd y i(k−p0)·y−i(k0−p)·x − g`mg`0m0 u 0 (p , s ) um(p, s) e − i∆m0`(y, x) . ` p(2π)122k002k0 (2.82) Combining terms, we get Z 4 4 ∗ 0 0 d xd y ip·x−ip0·y A = − g`0m0 g`mu 0 (p , s )um(p, s) e − i∆m0`(y, x) ` p(2π)122k002k0 0 0 × eik·x−ik ·y + eik·y−ik ·x (2.83) which agrees with SW’s (6.1.27) when his boson is restricted to a single scalar ﬁeld. 70 Feynman diagrams We now replace mean value in the vacuum of the time-ordered product by its value (2.70)

4 a n o Z d q ( − iqaγ + m)β h0|T ψ (x)ψ† (y) |0i ≡ − i∆ (x − y) = − i `m eiq·x. ` m `m (2π)4 q2 + m2 − i (2.84)

Replacing `, m, x, y by m0, `, y, x, we get

Z 4 a n † o d q ( − iqaγ + m)β m0` iq·(x−y) h0|T ψ 0 (y)ψ (x) |0i ≡ − i∆ 0 (y − x) = − i e . m ` m ` (2π)4 q2 + m2 − i (2.85)

We then ﬁnd

Z 4 4 ∗ 0 0 d xd y ip·x−ip0·y A = − g`0m0 g`mu`0 (p , s )um(p, s) √ e (2π)6 2k002k0 Z 4 0 0 d q ( − i/q + m)β 0 × eik·x−ik ·y + eik·y−ik ·x (−i) m ` eiq·(y−x). (2π)4 q2 + m2 − i (2.86)

In matrix notation, this is

4 4 Z d xd y 0 0 0 0 A = − √ ei(p+k)·x−i(p +k )·y + ei(k−p )·y+i(p−k )·x (2π)6 2k002k0 † 0 0 Z d4q u (p , s ) g ( − i/q + m)β g u(p, s) × (−i) eiq·(y−x). (2.87) (2π)4 q2 + m2 − i

The d4x and d4y integrations give

Z d4q A = i √ δ(q − k0 − p0)δ(p + k − q) + δ(q + k − p0)δ(p − k0 − q) (2π)2 2k002k0 † 0 0 u (p , s ) g ( − i/q + m)β g u(p, s) × (2.88) q2 + m2 − i which is " # iδ(p0 + k0 − p − k) −i(p/ + k/) + m −i(p/ + k/0) + m A = √ u†(p0, s0)g + βgu(p, s). 8π2 k00k0 (p + k)2 + m2 (p − k0)2 + m2 (2.89) 2.8 Feynman propagator for spin-one ﬁelds 71 2.8 Feynman propagator for spin-one ﬁelds The general form of the propagator is

X Z d3p h0|T ψ (x)ψ†(y)|0i = θ(x0 − y0) u (~p,s) u†(~p,s) eip·(x−y) a b (2π)3 a b s X Z d3p ± θ(y0 − x0) v (~p,s) v∗(~p,s) eip·(y−x). (2π)3 a b s (2.90)

We use the upper (+) sign for spin-one ﬁelds because they are bosons. For single real massive vector ﬁeld

1 X Z d3p h i ψa(x) = ea(~p,s)a(~p,s)eip·x + e∗a(~p,s)a†(~p,s)e−ip·x , p 3 0 s=−1 (2π) 2p (2.91) the mean value in the vacuum of its time-ordered product is

X Z d3p h0|T ψ (x)ψ†(y)|0i = θ(x0 − y0) e (~p,s) e∗(~p,s) eip·(x−y) a b (2π)32p0 a b s X Z d3p + θ(y0 − x0) e (~p,s) e∗(~p,s) eip·(y−x). (2π)32p0 a b s (2.92)

The spin-one spin sum is

X ∗ 2 ea(~p,s) eb (~p,s) = ηab + pa pb/m . (2.93) s

So the mean value (2.58) is

Z d3p p p h0|T ψ (x)ψ†(y)|0i = θ(x0 − y0) η + a b eip·(x−y) a b (2π)32p0 ab m2 Z d3p p p + θ(y0 − x0) η + a b eip·(y−x) (2π)32p0 ab m2 (2.94) 72 Feynman diagrams in which the momentum pa is physical (aka, on the mass shell) in that 0 p 2 2 2 2 2 p = ~p + m and p0 = ~p + m . In terms of derivatives, we have ∂ ∂ Z d3p h0|T ψ (x)ψ†(y)|0i = θ(x0 − y0) η − a b eip·(x−y) a b ab m2 (2π)32p0 ∂ ∂ Z d3p + θ(y0 − x0) η − a b eip·(y−x). ab m2 (2π)32p0 (2.95)

2 2 2 In this formula, we may interpret the double time derivative as ∂0 = ∇ −m . For spatial values of a and b, we can move the derivatives to the left of the step functions. And we can move the product ∂0 ∂i of one spatial and one time derivative by the argument we used for spin one-half. We ﬁnd

∂ ∂ h Z d3p h0|T ψ (x)ψ†(y)|0i = η − a b θ(x0 − y0) eip·(x−y) a b ab m2 (2π)32p0 Z d3p i + θ(y0 − x0) eip·(y−x) (2π)32p0 ∂ ∂ h i = η − a b θ(x0 − y0) ∆ (x − y) + θ(y0 − x0) ∆ (y − x) ab m2 + + ∂ ∂ = η − a b [−i∆ (x − y)] ab m2 F ∂ ∂ Z d4q eiq·(x−y) = − i η − a b (2.96) ab m2 (2π)4 q2 + m2 − i

2 2 2 in which ∂0 = ∇ − m . 2 2 2 We can relax the rule ∂0 = ∇ − m if we add an extra term to the propagator. The extra term is

i Z d4q eiq·(x−y) i Z d4q −~q2 − m2 + q2 ∇2 − m2 − ∂2 = 0 eiq·(x−y) m2 0 (2π)4 q2 + m2 − i m2 (2π)4 q2 + m2 − i i = − δ4(x − y). (2.97) m2 So the Feynman propagator for spin-one ﬁelds is

Z d4q (η + p p /m2) i h0|T ψ (x)ψ†(y)|0i = − i ab a b eiq·(x−y) − δ0 δ0 δ4(x − y). a b (2π)4 q2 + m2 − i m2 a b (2.98)

a Suppose the vector ﬁeld ψa(x) interacts with a current j (x) thru a term a ψa(x)j (x) in H(x). Then in Dyson’s expansion, the awkward second term 2.9 The Feynman rules 73 in the Feynman propagator (2.98) contributes the term 1 Z i i[j0(x)]2 − iH (x) = [−ija(x)][−ijb(y)] − δ0 δ0 δ4(x − y) d4y = . 2 2 m2 a b 2m2 (2.99) So if H(x) also contains the term [j0(x)]2 H (x) = , (2.100) C 2m2 then the eﬀect of the awkward second term in the Feynman propagator is cancelled. This actually happens in a natural way as SW explains in chapter 7.

2.9 The Feynman rules 1) Draw all diagrams of the order you are working at. Label each internal line with its (unphysical) 4-momentum considered to ﬂow in the direction of the arrow or in either direction if the particle is uncharged. 2) For each vertex of type i, include the factor

4 4 X 0 0 − i(2π) giδ p + q − p − q (2.101) which makes the sum of the 4-momenta p + q entering each vertex add up to sum of the 4-momenta p0 + q0 leaving each vertex. For each outgoing line include the factor u∗(~p0, s0, n0) v (~p0, s0, n0) ` or ` (2.102) (2π)3/2 (2π)3/2 ∗ 0 0 0 0 0 0 for arrows pointing out (u` (~p , s , n )) or pointing in (v`(~p , s , n )). For each incoming line include the factor u (~p,s, n) v∗(~p,s, n) ` or ` (2.103) (2π)3/2 (2π)3/2 ∗ 0 0 0 0 0 0 for arrows pointing in (u` (~p , s , n )) or pointing out (v`(~p , s , n )). For each internal line of a spin-zero particle carrying momentum qa include the factor −i 1 4 2 2 . (2.104) (2π) q + m` − i For each internal line of a spin-one-half particle with ends labelled by ` and m and carrying momentum qa include the factor −i ( − i/q + m)β `m 4 2 2 . (2.105) (2π) q + m` − i 74 Feynman diagrams For each internal line of a spin-one particle with ends labelled by ` and m and carrying momentum qa include the factor 2 −i η`m + q`qm/m 4 2 2 (2.106) (2π) q + m` − i and keep in mind the delta-function term in (2.98). 3). Integrate the product of all these factors over all the internal momenta and sum over ` and m, etc. 4) Add the results of all the Feynman diagrams.

2.10 Fermion-antifermion scattering We can watch Feynman’s rules emerge in fermion-antifermion scattering. We consider a fermion interacting with a neutral scalar boson (2.72) † H(x) = g`m ψ` (x) ψm(x) φ(x). (2.107) The initial state is |p, s; q, ti = a†(p, s)b†(q, t)|0i and the ﬁnal state is |p0, s0; q0, t0i = a†(p0, s0)b†(q0, t0)|0i. The lowest-order term is 1 Z h i A = − d4xd4y h0|b(q0, t0) a(p0, s0)T H(x) H(y) a†(p, s) b†(q, t)|0i 2! (2.108) Z g`mg`0m0 4 4 0 0 0 0 h † † i = − d xd yh0|b(q , t )a(p , s )T ψ (x) ψ (x)φ(x)ψ (y) ψ 0 (y)φ(y) 2! ` m `0 m × a†(p, s)b†(q, t)|0i in which the operators a and b delete fermions and antifermions. There is only one Feynman diagram (Fig. 2.2). We cancel the 2! by choosing to absorb the incoming fermion-antifermion pair at vertex y and to add the outgoing fermion-antifermion pair at vertex x. We are left with

g g 0 0 Z 0 0 h i A = − `m ` m d4xd4yh0|u∗(p0, s0) v (q0, t0) e−ix·(p +q ) T φ(x) φ(y) (2π)6 ` m ∗ iy·(p+q) × um0 (p, s) v`0 (q, t) e |0i. (2.109) Adding in the scalar propagator (2.50), we get

g g 0 0 Z 0 0 A = − `m ` m d4x d4y u∗(p0, s0) v∗ (q0, t0) e−ix·(p +q ) (2π)6 ` m Z 4 ik·(x−y) ∗ iy·(p+q) d k −ie × u 0 (p, s) v 0 (q, t) e . (2.110) m ` (2π)4 k2 + m2 − i 2.10 Fermion-antifermion scattering 75 Fermion-antifermion scattering

Figure 2.2 Fermion-antifermion scattering p+q → p0 +q0 (arbitrary colors).

The integration over y conserves 4-momentum at vertex y, and the integration over x conserves 4-momentum at vertex x. We then have

g g 0 0 Z A = i `m ` m d4k u∗(p0, s0) v (q0, t0) δ(k − p0 − q0) δ(p + q − k) (2π)2 ` m

∗ 1 × u 0 (p, s) v 0 (q, t) (2.111) m ` k2 + m2 − i or ∗ 0 0 0 0 ∗ g g 0 0 u (p , s ) v (q , t ) u 0 (p, s) v 0 (q, t) A = i δ(p + q − k0 − q0) `m ` m ` m m ` . (2π)2 (p + k)2 + m2 (2.112) 3 Action

3.1 Lagrangians and hamiltonians A transformation is a symmetry of a theory if the action is invariant or changes by a surface term. So we choose to work with actions that are symmetrical. The action is normally an integral over spacetime of an action density often called a lagrangian. Often the action density itself is invariant under the transformation of the symmetry. There are procedures, sometimes clumsy procedures, for computing the hamiltonian from the lagrangian. The hamiltonian often is not invariant under the transformation of the symmetry. So it’s very hard to ﬁnd a suitably symmetrical theory by starting with a hamiltonian. But once one has a hamiltonian, one can compute scattering amplitudes energies, and states with these energies.

3.2 Canonical variables In quantum mechanics, we use the equal-time commutation relations

[qi, pk] = iδik, [qi, qk] = 0, and [pi, pk] = 0. (3.1) iHt −iHt In general, the operators qi and qi(t) = e qie do not commute. Quan- tum field theory promotes these equal-time commutation relations to ones in which the indexes i and k denote different points of space n n [q (~x,t), pm(~y, t)]∓ = iδ(~x − ~y)δm, n m (3.2) [q (~x,t), q (~y, t)]∓ = 0, and [pn(~x,t), pm(~y, t)]∓ = 0. The commutator of a real scalar field Z d3p h i φ(x) = a(~p) eip·x + a†(~p) e−ip·x (3.3) p(2π)32p0 3.2 Canonical variables 77 is Z d3p [φ(x), φ(y)] = ∆(x − y) = eip·(x−y) − e−ip·(x−y) . (3.4) (2π)32p0

At equal times, one has

∂ 3 ∆(~x − ~y, 0) = 0, ∆(x − y)| 0 0 = − i δ (~x − ~y), and ∂x0 x =y (3.5) ∂2 ∆(x − y)| = 0. ∂x0∂y0 x0=y0

So a real ﬁeld φ and its time derivative φ˙ satisfy the equal-time commutation relations (3.60)

3 [φ(~x,t), φ˙(~y, t)]− = i δ (~x − ~y), (3.6) [φ(~x,t), φ(~y, t)]− = 0, and [φ˙(~x,t), φ˙(~y, t)]− = 0.

A complex scalar ﬁeld

1 φ(x) = √ φ1(x) + iφ2(x) (3.7) 2 obeys the commutation relations

† 1 [φ(x), φ (y)]− = [φ1(x) + iφ2(x), φ1(x) − iφ2(x)] 2 (3.8) 1 = ([φ (x), φ (x)] + [φ (x), φ (x)]) = ∆(x − y) 2 1 1 2 2 and 1 [φ(x), φ(y)]− = [φ1(x) + iφ2(x), φ1(x) + iφ2(x)] 2 (3.9) 1 = ([φ (x), φ (x)] − [φ (x), φ (x)]) = 0. 2 1 1 2 2 So the complex scalar ﬁelds φ(x) and φ˙†(x) obey the equal-time commutation relations (3.6)

† 3 † [φ(~x,t), φ˙ (~y, t)]− = i δ (~x − ~y) and [φ(~x,t), φ (~y, t)]− = 0 (3.10) [φ˙(~x,t), φ˙(~y, t)] = 0 and [φ˙†(~x,t), φ˙†(~y, t)] = 0 (3.11) [φ(~x,t), φ(~y, t)] = 0 and [φ†(~x,t), φ†(~y, t)] = 0. (3.12) 78 Action 3.3 Principle of stationary action in ﬁeld theory If φ(x) is a scalar ﬁeld, and L(φ) is its action density, then its action S[φ] is the integral over all of spacetime Z S[φ] = L(φ(x)) d4x. (3.13)

The principle of least (or stationary) action says that the field φ(x) that satisfies the classical equation of motion is the one for which the first-order change in the action due to any tiny variation δφ(x) in the field vanishes, δS[φ] = 0. And to keep things simple, we’ll assume that the action (or La- grange) density L(φ) is a function only of the field φ and its first derivatives a ∂aφ = ∂φ/∂x . The first-order change in the action then is Z ∂L ∂L 4 δS[φ] = δφ + δ(∂aφ) d x (3.14) ∂φ ∂(∂aφ) in which we sum over the repeated index a from 0 to 3. Now δ(∂aφ) = ∂a(φ + δφ) − ∂aφ = ∂aδφ. So we may integrate by parts and drop the surface terms because we set δφ = 0 on the surface at infinity Z ∂L ∂L 4 δS[φ] = δφ + ∂a(δφ) d x ∂φ ∂(∂aφ) Z (3.15) ∂L ∂L 4 = − ∂a δφ d x. ∂φ ∂(∂aφ) This first-order variation is zero, for arbitrary δφ only if the field φ(x) satisfies Lagrange’s equation

∂L ∂ ∂L ∂L ∂a ≡ a a = (3.16) ∂(∂aφ) ∂x ∂(∂φ/∂x ) ∂φ which is the classical equation of motion.

Example 3.1 (Theory of a scalar ﬁeld) The action density of a scalar ﬁeld φ of mass m is

1 ˙ 2 1 2 1 2 2 1 a 1 2 2 L = 2 (φ) − 2 (∇φ) − 2 m φ = − 2 ∂aφ ∂ φ − 2 m φ . (3.17) Lagrange’s equation (3.16) for this action density is

2 a 2 ∇ φ − φ¨ = ∂a ∂ φ = m φ (3.18) which is the Klein-Gordon equation (1.56). 3.4 Symmetries and conserved quantities in ﬁeld theory 79

In a theory of several ﬁelds φ1, . . . , φn with action density L(φk, ∂aφk), the ﬁelds obey n copies of Lagrange’s equation ∂ ∂L ∂L a = (3.19) ∂x ∂(∂aφk) ∂φk one for each k.

3.4 Symmetries and conserved quantities in ﬁeld theory

An action density L(φi, ∂aφi) that is invariant under a transformation of the a coordinates x or of the fields φi and their derivatives ∂aφi is a symmetry of the action density. Such a symmetry implies that something is conserved or time independent. Suppose that an action density L(φi, ∂aφi) is unchanged when the fields φi and their derivatives ∂aφi change by δφi and by δ(∂aφi) X ∂L ∂L 0 = δL = δφ + δ (∂ φ ). (3.20) ∂φ i ∂∂ φ a i i i a i For many transformations (but not for Lorentz transformations), we can set δ(∂aφi) = ∂a(δφi). In such cases by using Lagrange’s equations (3.19) to rewrite ∂L/∂φi, we find

X ∂L ∂L X ∂L 0 = ∂ δφ + ∂ δφ = ∂ δφ (3.21) a ∂∂ φ i ∂∂ φ a i a ∂∂ φ i i a i a i i a i which says that the current X ∂L J a = δφ (3.22) ∂∂ φ i i a i has zero divergence a ∂aJ = 0. (3.23)

Thus the time derivative of the volume integral of the charge density J 0 Z 0 3 QV = J d x (3.24) V is the ﬂux of current J~ entering through the boundary S of the volume V Z Z Z 0 3 k 3 k 2 Q˙ V = ∂0J d x = − ∂kJ d x = − J d Sk. (3.25) V V S 80 Action If no current enters V , then the charge Q inside V is conserved. When the volume V is the whole universe, the charge is the integral over all of space Z Z X ∂L Z X Q = J 0 d3x = δφ d3x = π δφ d3x (3.26) ˙ i i i i ∂φi i in which πi is the momentum conjugate to the ﬁeld φi ∂L πi = . (3.27) ∂φ˙i Example 3.2 (O(n) symmetry and its charge) Suppose the action density L is the sum of n copies of the quadratic action density (3.17)

n X 1 ˙ 2 1 2 1 2 2 1 a 1 2 2 L = 2 (φi) − 2 (∇φi) − 2 m φi = − 2 ∂aφ ∂ φ − 2 m φ , (3.28) i=1 and Aij is any constant antisymmetric matrix, Aij = − Aji. Then if the P ﬁelds change by δφi = j Aij φj, the change (3.20) in the action density n X 2 a δL = − m φiAijφj + ∂ φiAij∂aφj = 0 (3.29) i,j=1 vanishes. Thus the charge (3.26) associated with the matrix A Z Z X 3 X 3 QA = πi δφi d x = πi Aij φj d x (3.30) i i is conserved. There are n(n − 1)/2 antisymmetric n × n imaginary matrices; they generate the group O(n) of n × n orthogonal matrices.

An action density L(φi, ∂aφi) that is invariant under a spacetime transla- 0a a a a tion, x = x + δx , depends upon x only through the ﬁelds φi and their derivatives ∂aφi 2 ∂L X ∂L ∂φi ∂L ∂ φi = + . (3.31) ∂xa ∂φ ∂xa ∂∂ φ ∂xb∂xa i i b i

Using Lagrange’s equations (3.19) to rewrite ∂L/∂φi , we ﬁnd

2 ! X ∂L ∂L ∂ φi ∂L 0 = ∂b ∂aφi + b a − a ∂∂bφi ∂∂bφi ∂x ∂x ∂x i (3.32) " # X ∂L ∂φi 0 = ∂ − δbL b ∂∂ φ ∂xa a i b i 3.4 Symmetries and conserved quantities in ﬁeld theory 81 that the energy-momentum tensor

X ∂L ∂φi T b = − δbL (3.33) a ∂∂ φ ∂xa a i b i

b has zero divergence, ∂bT a = 0. Thus the time derivative of the 4-momentum PaV inside a volume V Z ! Z X ∂L ∂φi P = − δ0L d3x = T 0 d3x (3.34) aV ∂∂ φ ∂xa a a V i 0 i V is equal to the flux entering through V ’s boundary S Z Z Z 0 3 k 3 k 2 ∂0PaV = ∂0T a d x = − ∂kT a d x = − T a d Sk. (3.35) V V S The invariance of the action density L under spacetime translations implies the conservation of energy P0 and momentum P~ . The momentum πi(x) that is canonically conjugate to the field φi(x) is the derivative of the action density L with respect to the time derivative of the field ∂L πi = . (3.36) ∂φ˙i

If one can express the time derivatives φ˙i of the fields in terms of the fields φi and their momenta πi, then hamiltonian of the theory is the spatial integral of n ! 0 X ˙ H = P0 = T 0 = πi φi − L (3.37) i=1 in which φ˙i = φ˙i(φ, π). Example 3.3 (Hamiltonian of a scalar field) The hamiltonian density (3.37) of the theory (3.17) is

1 2 1 2 1 2 2 H = 2 π + 2 (∇φ) + 2 m φ . (3.38)

A Lorentz transformation (1.15) of a ﬁeld is like this

−1 X −1 U(Λ, a)ψ`(x)U (Λ, a) = D``¯(Λ )ψ`¯(Λx + a). (3.39) `¯ The action is invariant under Lorentz transformations. The action density 82 Action is constructed so as to be invariant when the ﬁelds transform this way:

0 X −1 ψ`(x) = D``¯(Λ )ψ`¯(x) (3.40) `¯ The action density is constructed so as to be invariant when the fields and their derivatives of the fields transform under infinitesimal Lorentz transformations as

` i ab ` m δψ (x) = ω (Jab) m ψ (x) 2 (3.41) i δ(∂ ψ`)(x) = ωab(J )` ∂ ψm(x) + ω j∂ ψ`(x). k 2 ab m k k j The invariance of the action density says that

∂L ` ∂L ` 0 = ` δψ + ` δ(∂k ψ ) (3.42) ∂ψ ∂∂kψ ∂L i ab ` m ∂L h i ab ` j `i = ` ω (Jab) m ψ + ` ω (Jab) m ∂k ψm + ωk ∂jψ . ∂ψ 2 ∂∂kψ 2 ab ba Remembering that ωab = −ωba and ω = −ω , we rewrite the last bit as 1 ω b∂ ψ = η ωab∂ = η ωab∂ ψ − η ωab∂ ψ . (3.43) k b ` ka b 2 ka b ` kb a ` We then have ∂L i 0 = ωab(J )` ψm ∂ψ` 2 ab m ∂L h i ab ` 1 ab ab i + ` ω (Jab) m ∂k ψm + ηka ω ∂b − ηkb ω ∂a ψ` (3.44) ∂∂kψ 2 2 or ∂L i 0 = (J )` ψm ∂ψ` 2 ab m ∂L h i ` 1 i + ` (Jab) m ∂k ψm + ηka ∂b − ηkb ∂a ψ` . (3.45) ∂∂kψ 2 2 The equations of motion now give ∂L i ` m 0 = ∂k ` (Jab) m ψ ∂∂kψ 2 ∂L h i ` 1 i + ` (Jab) m ∂k ψm + ηka ∂b − ηkb ∂a ψ` (3.46) ∂∂kψ 2 2 or ∂L i ` m ∂L 1 i 0 = ∂k ` (Jab) m ψ + ` ηka ∂b − ηkb ∂a ψ` ∂∂kψ 2 ∂∂kψ 2 3.4 Symmetries and conserved quantities in ﬁeld theory 83 or ∂L i ` m 1 ∂L ∂L 0 = ∂k ` (Jab) m ψ + a ` ∂b − b ` ∂aψ` . ∂∂kψ 2 2 ∂∂ ψ ∂∂ ψ We recall (3.33) the energy-momentum tensor

X ∂L ∂φi T b = − δbL (3.47) a ∂∂ φ ∂xa a i b i or ∂L ∂ψ` Tba = b a − ηbaL (3.48) ∂∂ ψ` ∂x b which has zero divergence, ∂bT a = 0. In terms of it, we have ∂L i ` m 1 0 = ∂k ` (Jab) m ψ − (Tab − Tba) . ∂∂kψ 2 2 So Belinfante deﬁned the symmetric energy-momentum tensor as

ab ab i h ∂L ab ` m Θ = T − ∂k ` (J ) m ψ 2 ∂∂kψ ∂L kb ` m ∂L ka ` mi − ` (J ) m ψ − ` (J ) m ψ . (3.49) ∂∂aψ ∂∂bψ The quantity in the square brackets is antisymmetric in a, k. So

ab ab i h ∂L ab ` m ∂aΘ = ∂aT − ∂a∂k ` (J ) m ψ 2 ∂∂kψ ∂L kb ` m ∂L ka ` mi (3.50) − ` (J ) m ψ − ` (J ) m ψ ∂∂aψ ∂∂bψ ab = ∂aT = 0. The Belinfante energy-momentum tensor is symmetric Θab = Θba (3.51) and so it is the one to use when gravity is involved. The quantity M abc = xbΘac − xcΘab (3.52) is a conserved current abc b ac c ab bc cb ∂aM = ∂a x Θ − x Θ = Θ − Θ = 0. (3.53) So the angular-momentum operators Z Z J bc = M 0bc d3x = xbΘ0c − xcΘ0b d3x (3.54) 84 Action are conserved.

3.5 Poisson Brackets

In a classical theory with coordinates qn and conjugate momenta ∂L pn = , (3.55) ∂q˙n the Poisson bracket is X ∂A ∂B ∂B ∂A [A, B] = − . (3.56) P ∂q ∂p ∂q ∂p n n n n n One of Dirac’s many insights was to say that in the quantum version of such a theory the equal-time commutation relations are

[A, B] = i~ [A, B]P . (3.57)

In ﬁeld theories, with coordinates φa = φn(x) and conjugate momenta ∂L(x) πa = πn(x) = , (3.58) ∂φ˙n(x) the Poisson bracket is X Z ∂A ∂B ∂B ∂A [A, B] = d3x − . P ∂φ (x) ∂π (x) ∂φ (x) ∂π (x) (3.59) n n n n n In the quantum theory, the equal-time commutator is

[A, B] = i~ [A, B]P . (3.60) When these equations and deﬁnitions (3.55–3.60) make sense, one sometimes can express the velocities φ˙n(x) in terms of the coordinates φn(x) and their conjugate momenta πn(x). In such cases, one can make a hamiltonian density

H = φ˙n(x)πn(x) − L(x) (3.61) in which L(x) is the action density. And the action principle gives us the equations of motion ∂L ∂L ∂i = . (3.62) ∂∂iφn ∂φn But things can go wrong. For instance, the Lagrange’s equations for the action densities L = q and L = φn lead to the inconsistent equation 0 = 1. (3.63) 3.6 Dirac Brackets 85 Similarly, So Nature doesn’t care about our ideas. The theory of a massive vector ﬁeld V 1 ik 1 i i L = − 4 FikF − 2 ViV − JiV (3.64) provides another example in which the momentum π0 is constrained to vanish ∂L(x) π0 = = 0 (3.65) ∂V˙0(x) and the equation of motion for the ﬁeld V 0 (3.62) ∂L ∂L ∂i 0 = 0 (3.66) ∂∂iV ∂V is i i0 2 − ∂iF 0 = ∂iF = −m V0 − J0 (3.67) 0i i0 or with πi = − F = F

i0 2 0 0 ∂iF = ∇ · π = m V + J . (3.68) This last equation constrains the ∇ · π to be m2V 0 + J 0 at each point x, t. To cope with such problems and thereby to extend the variety of quantum ﬁeld theories, Dirac invented a kind of bracket that is diﬀerent from his ha|bi bracket.

3.6 Dirac Brackets First, one makes a list of all the constraints

χ = 0 (3.69) that occur due to the deﬁnitions of the conjugate momenta and to the equations of motion. A constraint χf = 0 is ﬁrst class if its Poisson bracket with all the other constraints vanishes

[χf , χa]P = 0. (3.70) One satisﬁes these constraints by choosing a gauge. One then makes a matrix of the Poisson brackets of the second-class constraints

Cab = [χa, χb]P . (3.71) 86 Action In computing this matrix C, one does not apply the constraints (3.69) until after one has computed their Poisson brackets. Otherwise C would vanish. Dirac deﬁned his new bracket in terms of the second-class constraints as

−1 ab [A, B]D = [A, B]P − [A, χa]P (C ) [χb,B]P . (3.72) This new Dirac bracket inherits the algebraic properties of Poisson’s bracket

[A, B]D = − [B,A]D,

[A, BC]D = [A, B]D C + B [A, C]D, (3.73)

[A, [B,C]D]D + [B, [C,A]D]D + [C, [A, B]D]D = 0. Dirac then extended his rule that the commutator of two operators should be i~ times their Poisson bracket to his new rule that it should be i~ times their Dirac bracket

[A, B] = i~ [A, B]D. (3.74)

3.7 Dirac Brackets and QED The action density is 1 ik i L = − 4 FikF − JiA . (3.75) The usual rules give the momenta as ∂L πi = (3.76) ∂(∂0Ai) and the commutation relations as

k k 3 [Ai(x, t), π (y, t)] = iδi δ (x − y). (3.77) These equations contain the first-class constraints π0 = 0 (3.78) ∇ · π − J0 = 0. We first choose a gauge that satisfies these first-class constraints. One such gauge is the radiation or Coulomb gauge

∇ · A = 0. (3.79)

In this gauge, the ﬁrst-class constraints say that

i0 2 0 2 0 0 − ∂iF = −∇ A + ∇ · ∇˙ A = −∇ A = J . (3.80) 3.7 Dirac Brackets and QED 87 So we can solve for A0

Z J 0(y, t) A0(x, t) = d3y . (3.81) 4π|x − y|

Once we have eliminated A0 and π0 from the theory, we list the second- class constraints

χ1(x, t) = ∇ · A(x, t) = 0 0 χ2(x, t) = ∇ · π(x, t) + J (x, t) = 0. (3.82)

Their Poisson bracket is

X Z ∂∇ · A(x, t) ∂∇ · π(y, t) + J 0(y, t) [χ (x, t), χ (y, t)] = d3z 1 2 P ∂A (z) ∂π (z) i i i ∂∇ · π(y, t) + J 0(y, t) ∂∇ · A(x, t) − . (3.83) ∂Ai(z) ∂πi(z)

The second terms vanish as do terms with J 0. Since

∂∇ · A(x, t) 3 ∂∇ · π(y, t) 3 = ∂iδ (x − z) and = ∂iδ (y − z), (3.84) ∂Ai(z) ∂πi(z) the Poisson bracket is

2 3 [χ1(x, t), χ2(y, t)]P = − ∇ δ (x − y). (3.85)

The other Poisson brackets vanish, so

2 3 C1x,2y = − ∇ δ (x − y) = −C2y,1x

C1x,1y = 0 (3.86)

C2x,2y = 0.

The matrix C is nonsingular, det C 6= 0. Its inverse is

Z d3k eik·(x−y) 1 (C−1)1x,2y = − = − = − (C−1)2y,1x (3.87) (2π)3 k2 4π|x − y| in which the last minus sign is the one I missed in class. (Incidentally, there are typos in the b-ok.org versions of Weinberg’s QTFI which I put on the class website. Does anyone have a pdf of the 2005 paperback edition?) 88 Action To check the formula (3.87) for the inverse C−1, we do the integral

Z Z 3 −1 2z,1y 3 −1 1y,2z d z C1x,2z(C ) = − d z C1x,2z(C )

Z Z d3k eik·(y−z) = − d3z [−∇2δ3(x − z)][− ] (2π)3 k2 Z Z d3k eik·(y−z) = − d3z δ3(x − z)∇2 (2π)3 k2 Z Z d3k = d3z δ3(x − z) eik·(y−z) (2π)3 Z = d3z δ3(x − z)δ(y − z) = δ(x − y). (3.88)

i To evaluate the Dirac bracket [A (x, t), πk(y, t)]D we need to compute the Poisson brackets

Z i 0 i X 3 ∂A (x, t) ∂∇ · π(y, t) + J (y, t) [A (x, t), χ2y]P = d z ∂Ak(z, t) ∂πk(z, t) k ∂∇ · π(y, t) + J 0(y, t) ∂Ai(x, t) − ∂Ak(z, t) ∂πk(z, t) X Z ∂Ai(x, t) ∂∇ · π(y, t) = d3z (3.89) ∂Ak(z, t) ∂πk(z, t) k Z X 3 i 3 3 = d z δkδ (x − z) ∂kδ (y − z) k ∂ ∂ = δ3(x − y) = − δ3(x − y) ∂yi ∂xi 3.7 Dirac Brackets and QED 89 and Z X 3 ∂πi(x, t) ∂∇ · A(y, t) [πi(x, t), χ1y]P = d z Ak(z, t) ∂πk(z, t) k ∂∇ · A(y, t) ∂π (x, t) − i Ak(z, t) ∂πk(z, t) Z ∂∇ · A(y, t) ∂π (x, t) = − d3z i Ak(z, t) ∂πk(z, t) Z 3 ∂ = − d z δ(y − z) δki δ(x − z) (3.90) ∂yk Z ∂ = − d3z δ(y − z) δ(x − z) ∂yi ∂ = − δ(y − x) ∂yi ∂ ∂ = δ(y − x) = δ(x − y). ∂xi ∂xi i So ﬁnally we compute the Dirac bracket (3.72) of A (x) with πk(y) at equal times i i [A (x, t),πk(y, t)]D = [A (x, t), πk(y, t)]P (3.91) Z 3 3 i −1 2z,1w − d z d w [A (x, t), χ2(z, t)]P (C ) [χ1(w, t), πk(y, t)]P .

The ﬁrst Poisson bracket is Z i i X ∂A (x, t) ∂πk(y, t) ∂πk(y, t) ∂A (x, t) [Ai(x, t), π (y, t)] = d3z − k P ∂A (z) ∂π (z) ∂A (z) ∂π (z) j j j j j Z i X ∂A (x, t) ∂πk(y, t) = d3z (3.92) ∂A (z) ∂π (z) j j j Z X 3 3 3 3 = d z δijδ (x − z)δjkδ (y − z) = δik δ (x − y). j The matrix element (3.87) of C−1 is 1 (C−1)2z,1w = . (3.93) 4π|z − w| The Poisson brackets (3.89) and (3.90) are ∂ [Ai(x, t), χ ] = − δ3(x − y) 2y P ∂xi (3.94) ∂ [πi(x, t), χ1y]P = δ(x − y). ∂xi 90 Action

So the second term T2 in the Dirac bracket (3.91) is Z 1 T = − d3z d3w [Ai(x, t), χ (z, t)] [χ (w, t), π (y, t)] 2 2 P 4π|z − w| 1 k P Z 3 3 ∂ 3 1 ∂ = d z d w i δ (x − z) δ(w − y) ∂x 4π|z − w| ∂wk Z 3 3 3 ∂ 1 ∂ = d z d w δ (x − z) i δ(w − y) ∂z 4π|z − w| ∂wk Z ∂2 1 = d3z d3w δ3(x − z) δ(w − y) ∂zi∂zk 4π|z − w| ∂2 1 = . (3.95) ∂xi∂xk 4π|x − y| i So Dirac’s rule of quantization says that the commutator of A with πk is the transverse delta function i i [A (x, t), πk(y, t)] = i[A (x, t), πk(y, t)]D ∂2 1 = iδ δ3(x − y) + i (3.96) ik ∂xi∂xk 4π|x − y| and that i k [A (x, t),A (y, t)] = 0 and [πi(x, t), πk(y, t)] = 0. (3.97) Apart from matter terms, the hamiltonian for QED is Z 3 H = d x π⊥ · A˙ − L (3.98) in which π⊥ is the transverse part of π, that is, the part that has no divergence

∇ · π⊥ = 0. (3.99)

The longitudinal part πk of π does not contribute because it is the gradient of a scalar S which when dotted into a transverse vector vanishes as an integration by parts shows Z Z d3x ∇S · A˙ = − d3x S ∇ · A˙ = 0. (3.100)

So we can take the conjugate momentum π to be transverse

∇ · π⊥ = 0. (3.101) in the term π · A˙ The transverse part π⊥ is

∂L 0 π⊥ = − ∇A = A˙. (3.102) ∂A˙ 3.7 Dirac Brackets and QED 91 The hamiltonian for QED is then Z Z Z 3 3 2 3 2 H = d x π⊥ · A˙ − L = d x π⊥ − L = d x A˙ − L (3.103) in which L is the action density 1 ik i 1 ˙ 0 2 1 2 i L = − 4 FikF + JiA = 2 (A + ∇A ) − 2 (∇ × A) + JiA . (3.104) So H is Z 3 h 2 1 0 2 1 2 0 0i H = d x π⊥ − 2 (π⊥ + ∇A ) + 2 (∇ × A) − J · A + J A (3.105)

0 apart from terms due to the matter ﬁelds. The integral of π⊥ ·∇A vanishes so H is Z 3 h 1 2 1 2 0 0 1 0 2i H = d x 2 π⊥ + 2 (∇ × A) − J · A + J A − 2 (∇A ) . (3.106) By Gauss’s law and the Coulomb gauge condition ∇ · A = 0 i0 2 0 0 − ∂iF = −∇ A = J , (3.107) so the last term in H is Z Z Z 3 1 0 2 3 1 0 2 0 3 1 0 0 d x − 2 (∇A ) = d x 2 A ∇ A = d x − 2 A J . (3.108) The hamiltonian then is Z 3 h 1 2 1 2 1 0 0i H = d x 2 π⊥ + 2 (∇ × A) − J · A + 2 J A (3.109) in which the last term is the Coulomb energy Z 1 Z J 0(x, t)J 0(y, t) V = 1 d3x J 0A0 = d3x d3y . (3.110) C 2 2 4π|x − y| One may satisfy the commutation relations (3.96 & 3.97) by setting Z d3p X h i A(x) = eip·x e(p, s) a(p, s) + e−ip·x e∗(p, s) a†(p, s) p 3 0 (2π) 2p s (3.111) and using π⊥(x) = A˙ Z p 0 3 p d p X h ip·x −ip·x ∗ † i π⊥(x) = − i e e(p, s) a(p, s) − e e (p, s) a (p, s) p 3 (2π) 2 s (3.112) in both of which x = (x, t) and † 3 [a(p, s), a (q, t)] = δst δ (p − q) and [a(p, s), a(q, t)] = 0. (3.113) 92 Action One gets the transverse delta function because the two polarization 3-vectors e(p, s) are perpendicular to p and obey the sum rule

X pipk ei(p, s) ek∗(p, s) = δ − . (3.114) ik p2 s The 3-vectors e(p, s) can be thought of as the vector parts of 4-vectors with vanishing time components. 4 Quantum Electrodynamics

4.1 Global U( 1) Symmetry In most theories of charged ﬁelds, the action density is invariant when the charged ﬁelds change by a phase transformation

0 iθ` ψ`(x) = e ψ`(x). (4.1)

If the phase θ is independent of x, then the symmetry is called a global U(1) symmetry. Section 3.4 describes Noether’s theorem according to which a current

a X ∂L J = δψ` (4.2) ∂∂aψ` ` has zero divergence

a ∂aJ = 0. (4.3)

The charge Z Q = J 0 d3x (4.4) is conserved due to the global U(1) symmetry. Here for tiny θ`, δψ` = iθ`ψ`, and the charge density J 0 is

0 X ∂L X ∂L X J = δψ` = iθ`ψ` = π`iθ`ψ`. (4.5) ∂∂0ψ` ∂∂0ψ` ` ` `

One imagines that the angle is proportional to the charge of the ﬁeld ψ`, θ` = q`θ. 94 Quantum Electrodynamics 4.2 Abelian gauge invariance Quantum electrodynamics is a theory in which the action density is invariant when the charged ﬁelds change by a phase transformation that varies with the spacetime point x

0 iθ`(x) ψ`(x) = e ψ`(x). (4.6) Such a symmetry is called a local U(1) symmetry. A theory with a local U(1) symmetry also has a global U(1) symmetry and so it conserves charge. † Quantities like ψ` (x) ψ`(x) are intrinsically invariant under local U(1) symmetries because

† 0 0 † −iθ`(x) iθ`(x) † (ψ` (x)) ψ`(x) = ψ` (x)e e ψ(x) = ψ` (x) ψ(x). (4.7) Derivatives of ﬁelds present problems, however, because

iθ`(x) iθ`(x) iθ`(x) ∂a e ψ`(x) = e ∂a ψ`(x) + i(∂aθ(x)) ψ`(x) 6= e ∂a ψ`(x) (4.8) † a † so things like ψ` (x) ∂aψ`(x) and like (∂ ψ` (x)) ∂aψ`(x) are not invariant under local phase transformations. The trick is to introduce a ﬁeld Aa(x) that transforms so as to cancel the iθ awkward term e ` i(∂aθ) ψ` in the derivative (4.8). We want a new derivative Daψ` that transforms like ψ` .

0 iθ` (Daψ`) = e Daψ`. (4.9) So we set

Da = ∂a + iAa (4.10) and require that

0 0 0 iθ` iθ` (Daψ`) = (∂a + iAa)ψa = e Daψ` = e (∂a + iAa)ψ`. (4.11) That is, we insist that

0 0 0 iθ` iθ` (∂a + iAa)ψa = (∂a + iAa)e ψa = e (∂a + iAa)ψ`. (4.12) So we need 0 i∂aθ` + iAa = iAa (4.13) or 0 Aa(x) = Aa(x) − ∂aθ`. (4.14)

This is what we need except that the field Aa does not carry the index `. 4.2 Abelian gauge invariance 95 The solution to this problem is to define the symmetry transformation (4.6) so that the angle θ` is proportional to the charge of the field ψ`

0 iq`θ(x) ψ`(x) = e ψ`(x). (4.15) This deﬁnition has the advantage that the charge density (4.5) becomes

0 X J = iθ q` π`ψ` (4.16) ` which makes more sense than the old formula (4.5). More importantly, the deﬁnition (4.15) means that equations (4.9–4.22) change to

0 iq`θ (Daψ`) = e Daψ`. (4.17) So we make the covariant derivative

Daψ` = (∂a + iq`Aa)ψ` (4.18) depend upon the ﬁeld ψ` and require that

0 0 0 iq`θ iq`θ (Daψ`) = (∂a + iq`Aa)ψa = e Daψ` = e (∂a + iq`Aa)ψ`. (4.19) That is, we insist that

0 0 0 iq`θ iq`θ (∂a + iqÀa)ψa = (∂a + iqÀa)e ψa = e (∂a + iqÀa)ψ`. (4.20) So we need 0 iq`∂aθ + iqÀa = iqÀa (4.21) or 0 Aa(x) = Aa(x) − ∂aθ(x) (4.22) which is much better than the old rule (4.14). The twin rules

0 iq`θ(x) ψ`(x) = e ψ`(x) 0 (4.23) Aa(x) = Aa(x) − i∂aθ(x) constitute an abelian gauge transformation. Note that a field ψ` couples to the electromagnetic field Aa in a way qÀaψ` that is proportional to its charge q`. To insure that the right charge q` appears in the right place, we can introduce a charge operator q such that q ψ` = q`ψ` and redefine the abelian gauge transformation (4.23) as

0 iqθ(x) iq`θ(x) ψ`(x) = e ψ`(x) = e ψ`(x) 0 (4.24) Aa(x) = Aa(x) − ∂aθ(x). 96 Quantum Electrodynamics So 0 ∂bAa = ∂bAa − ∂b∂aθ (4.25) is not invariant, but the antisymmetric combination

0 0 ∂bAa − ∂aAb = ∂bAa − ∂b∂aθ − ∂aAb + ∂a∂bθ = ∂bAa − ∂aAb (4.26) is invariant. Maxwell introduced this combination

Fba = ∂bAa − ∂aAb. (4.27) Thus 1 L = − F F ba (4.28) 4 ba is a Lorentz-invariant, gauge-invariant action density for the electromagnetic ﬁeld.

4.3 Coulomb-gauge quantization The first step in the canonical quantization of a gauge theory is to pick a gauge. The most physical gauge for electrodynamics is the Coulomb gauge defined by the gauge condition 0 = ∇ · A.~ (4.29) If the action density (4.28) is modified by an interaction with a current J a 1 L = − F F ba + A J a (4.30) 4 ba a then the equation of motion is

ba a ∂bF = − J (4.31) while the homogeneous equations

0 = ∂aFbc + ∂bFca + ∂cFab (4.32) follow from the antisymmetry of Fab. Because of the antisymmetry of Fab, the derivative A˙ 0 does not appear in the action density (4.30). So for a = 0, the equation of motion (4.31) actually is a constraint

i0 i 0 ∂iF = − ∂iE = − J (4.33) known as Gauss’s law ∇ · E~ = ρ = J 0. (4.34) 4.4 QED in the interaction picture 97 This constraint together with the Coulomb gauge condition (4.29) lets us express A0 in terms of the charge density ρ = J 0. We ﬁnd

2 0 ∇ A0 − ∂0∇ · A~ = ∇ · E = J (4.35) or 2 0 ∇ A0 = J . (4.36) The solution is Z J 0(~y, t) A0(~x,t) = d3y. (4.37) 4π|~x − ~y|

The quantum ﬁelds in this gauge are the transverse parts of A~ and their conjugate momentum E~ .

4.4 QED in the interaction picture The Coulomb-gauge hamiltonian in the interaction picture is

H = H0 + V 1 Z H = E~ 2 + B~ 2 d3x + H 0 2 ψ,0 (4.38) Z 3 V = − J~ · A~ d x + VC + Vψ in which B = ∇ × A~ and both E~ and A~ are travsverse, that is ∇ · E~ = 0 and ∇ · A~ = 0 (4.39) and the Coulomb potential energy is 1 Z J 0(~x, 0)J 0(~y, 0) V = d3xd3y. (4.40) C 2 4π|~x − ~y| The electromagnetic ﬁeld is Z d3p X h i Ab(x) = eip·xeb(p, s)a(p, s) + e−ip·xeb∗(p, s)a†(p, s) . p 3 0 (2π) 2p s (4.41) The polarization vectors may be chosen to be  0  b 1  1  e (p, ±1) = R(ˆp)√   (4.42) 2 ±i 0 98 Quantum Electrodynamics where R(ˆp) rotatesz ˆ intop ˆ. Thus ~p · ~e(p, s) = 0 (4.43) and e0(p, s) = 0 (4.44) because A0 is a dependent variable (4.37). The commutation relations are † 0 0 3 0 0 0 [a(p, s), a (p , s )] = δss0 δ (~p − ~p ) and [a(p, s), a(p , s )] = 0. (4.45)

The term H0 in the hamiltonian (4.38) is Z X 1 H = p0[a(p, s), a†(p, s)] d3p 0 2 + s Z X 1 = p0 a†(p, s)a(p, s) + δ3(~p − ~p) d3p. (4.46) 2 s The interaction is h Z Z J 0(~x, 0)J 0(~y, 0) i V (t) = eiH0t − J~(~x, 0)·A~(~x, 0)d3x+ d3xd3y+V (0) e−iH0t 8π|~x − ~y| m (4.47) in which Vm(0) is the non-electromagnetic part of the matter interaction. Since A0 = 0, J~ · A~ = J · A.

4.5 Photon propagator The photon propagator is

− i∆ab(x − y) = h0|T [Aa(x),Ab(y)]|0i. (4.48) Inserting the formula for the electromagnetic ﬁeld (4.41), we get Z d3p p p −i∆ (x − y) = δ − i j ij (2π)32p0 ij ~p2 h i × eip·(x−y)θ(x0 − y0) + e−ip·(x−y)θ(y0 − x0) Z d4q −i q q = δ − i j eiq·(x−y) (4.49) (2π)4 q2 − i ij ~q2 with the understanding that this propagator vanishes when a or b is zero. If c 2 n = (1, 0, 0, 0) has only a time component, one can write the δij − qiqj/~q as q q q0(q n + q n ) − q q + q2n n δ − a b θ(ab) = η + a b b a a b a b (4.50) ab ~q2 ab ~q2 4.5 Photon propagator 99 in which q0 is arbitrary and may be chosen to be determined by conservation of energy. The terms qanb, qbna, and qaqb act like ∂anb, ∂bna, and ∂a∂b so they a b appear in the S-matrix as ∂aJ nb, and ∂bJ na, etc., which vanish because of current conservation. The remaining term q2n n −i −in n a b = a b (4.51) ~q2 q2 − i ~q2 gives in the S-matrix a term 1 Z −i Z d4q T = d4xd4y[−iJ 0(x)][−iJ 0(y)] eiq·(x−y) 2 (2π)4 ~q2 1 Z d3q δ(x0 − y0) = i d4xd4yJ 0(x)J 0(y) ei~q·(~x−~y) (4.52) 2 (2π)3 ~q2 1 Z J 0(~x,t)J 0(~y, t) = i d3x d3y dt (4.53) 2 4π|~x − ~y| which cancels the Coulomb term 1 Z J 0(~x,t)J 0(~y, t) T = −i d3x d3y dt . (4.54) C 2 4π|~x − ~y| The bottom line is that the eﬀective photon propagator is Z d4q −iη − i∆ (x − y) = ab eiq·(x−y) = −i∆ (y − x). (4.55) ab (2π)4 q2 − i ab Dirac’s action density is ¯ a Lψ = − ψ(γ ∂a + m)ψ (4.56) with ψ¯ = iψ†γ0. The conjugate momentum is ∂L π = = −ψγ¯ 0 = −iψ†γ0γ0 = iψ†. (4.57) ∂ψ˙ The hamiltonian is Z Z 3 † a 3 H = (πψ˙ − L)d x = [iψ ψ˙ + ψ¯(γ ∂a + m)ψ] d x Z (4.58) = ψ¯(~γ · ∇~ + m)ψ d3x.

The free Dirac ﬁeld aka the Dirac ﬁeld in the interaction picture is

Z d3p X h i ψ(x) = u(p, s)eipxa(p, s) + v(p, s)e−ipxb†(p, s) (4.59) (2π)3/2 s 100 Quantum Electrodynamics The addition and deletion operators obey the anticommutation relations

† 0 0 † 0 0 3 0 [a(p, s), a (p , s )]+ = [b(p, s), b (p , s )]+ = σss0 δ (~p − ~p ) (4.60) with the other anticommutators equal to zero. Putting ψ(x) into H gives Z X h i H = p0 a†(p, s)a(p, s) + b†(p, s)b(p, s) − δ3(~p − ~p) d3p. (4.61) s

4.6 Feynman’s rules for QED The action density is 1 L = − F F ab − ψ¯[γa(∂ + ieA ) + m]ψ. (4.62) 4 ab a a The electric current is ∂L J a = = −ieψγ¯ aψ. (4.63) ∂Aa The interaction is Z ¯ a 3 V (t) = ie ψ(~x,t)γ ψ(~x,t)Aa(~x,t) d x + VC (t) (4.64)

Draw all appropriate diagrams. Label each vertex with α = 1, 2, 3, or 4 on an electron line of momentum p entering the vertex and β = 1, .., 4 on an electron line of momentum p0 leaving the vertex, and an index a on the photon line of momentum q. The vertex carries the factor

4 a 4 0 (2π) eγβαδ (p − p + q). (4.65) An outgoing electron line gives u¯ (p, s) β . (4.66) (2π)3/2 An outgoing positron line gives v (p, s) α . (4.67) (2π)3/2 An incoming electron line gives u (p, s) α . (4.68) (2π)3/2 4.7 Electron-positron scattering 101 An incoming positron line gives

v¯ (p, s) β . (4.69) (2π)3/2

An outgoing photon gives

e∗(p, s) a . (4.70) p(2π)32p0

An incoming photon gives

e (p, s) a . (4.71) p(2π)32p0

An internal electron line of momentum p from vertex β to vertex α gives

−i (−ip/ + m)αβ . (4.72) (2π)4 p2 + m2 − i

An internal photon line of momentum q linking vertexes a and b gives

−i η ab . (4.73) (2π)4 q2 − i

Integrate over all momenta, sum over all indexes. Add up all terms and get the combinatorics and minus signs right.

4.7 Electron-positron scattering There are two Feynman diagrams for electron-positron scattering at order 2 2 e where α = e /(4π0~c) = 0.00729735256 ≈ 1/137 is the ﬁne-structure constant. The direct or s-channel diagram is represented in Fig. 4.1. 102 Quantum Electrodynamics S-channel electron-positron scattering

Figure 4.1 Direct or s-channel diagram for electron-positron scattering e− + e+ → e− + e+ via electron-positron annihilation into an electron- positron pair (arbitrary colors).

Feynman’s rules give for the annihilation diagram Z 4 4 a 4 4 b 4 0 0 As = d q (2π) eγβαδ (p + k − q)(2π) eγγδδ (−p − k + q) u¯ (p0, s0) v (k0, t0) u (p, s) v¯ (k, t) −i η × γ δ α β ab (2π)3/2 (2π)3/2 (2π)3/2 (2π)3/2 (2π)4 q2 − i 4 a 4 b 4 0 0 = (2π) eγβα(2π) eγγδδ (p + k − p − k ) u¯ (p0, s0) v (k0, t0) u (p, s) v¯ (k, t) −i η × γ δ α β ab (4.74) (2π)3/2 (2π)3/2 (2π)3/2 (2π)3/2 (2π)4 (p + k)2 a b 4 0 0 = eγβαeγγδδ (p + k − p − k ) −i η × u¯ (p0, s0)v (k0, t0)u (p, s)¯v (k, t) ab γ δ α β (2π)2 (p + k)2 2 a 0 0 b 0 0 4 0 0 = e v¯β(k, t)γβαuα(p, s)¯uγ(p , s )γγδvδ(k , t )δ (p + k − p − k ) −i η × ab (2π)2 (p + k)2 e2 δ4(p + k − p0 − k0) = − i v¯ (k, t)γa u (p, s)¯u (p0, s0)γ v (k0, t0). (2π)2 (p + k)2 β βα α γ aγδ δ 4.7 Electron-positron scattering 103 T-channel electron-positron scattering

Figure 4.2 Exchange or t-channel diagram for electron-positron scattering (arbitrary colors).

Suppressing indexes, we get

e2 vγ¯ au u¯0γ v0 A = − i δ4(p + k − p0 − k0) a . (4.75) s (2π)2 (p + k)2

The exchange diagram is

e2 u¯0γau vγ¯ v0 A = ± i δ4(p + k − p0 − k0) a , (4.76) t (2π)2 (p − p0)2 and is illustrated in Fig. 4.2 and we must check our minus signs. The initial state is |p, s; k, ti = a†(p, s)b†(k, t)|0i and the ﬁnal state is |p0, s0; k0, t0i = a†(p0, s0)b†(k0, t0)|0i. The direct diagram comes from

¯ a ¯ b T [ψ(x)γ ψ(x)Aa(x)ψ(y)γ ψ(y)Ab(y)] ¯ − a − ¯ + b + = (ψ) (x)γ ψ (x)T [Aa(x)Ab(y)](ψ) (y)γ ψ (y) (4.77) † 0 0 a † 0 0 b ∼ a (p s )γ b (k , t )T [Aa(x)Ab(y)]b(k, s)γ a(p, s), 104 Quantum Electrodynamics while the exchange diagram illustrated in Fig. 4.2comes from ¯ a ¯ b T [ψ(x)γ ψ(x)Aa(x)ψ(y)γ ψ(y)Ab(y)] ¯ + a − ¯ − b + = (ψ) (x)γ ψ (x)T [Aa(x)Ab(y)](ψ) (y)γ ψ (y) a † 0 0 † 0 0 b ∼ b(k, t)γ b (k , t )T [Aa(x)Ab(y)]a (p , s )γ a(p, s) (4.78) † 0 0 a † 0 0 b ∼ a (p , s )b(k, t)γ b (k , t )T [Aa(x)Ab(y)]γ a(p, s) † 0 0 a † 0 0 b ∼ − a (p , s )γ b (k , t )T [Aa(x)Ab(y)]b(k, t)γ a(p, s) which diﬀers by a minus sign. So the total amplitude is

A = As + At e2 vγ¯ au u¯0γ v0 = − i δ4(p + k − p0 − k0) a (2π)2 (p + k)2 h e2 u¯0γbu vγ¯ v0 i − − i δ4(p + k − p0 − k0) b (4.79) (2π)2 (p − p0)2 e2 vγ¯ au u¯0γ v0 u¯0γbu vγ¯ v0 i = − i δ4(p + k − p0 − k0) a − b . (2π)2 (p + k)2 (p − p0)2 The probability is the square |A|2. To compute such things, we need to know some trace identities.

4.8 Trace identities Since γ5 is a fourth spatial gamma matrix (1.173), its square is unity, γ5γ5 = 1, and it anticommutes with the ordinary gammas. So since the trace is cyclic,

Tr(γa) = Tr(γ5γ5γa) = −Tr(γ5γaγ5) = −Tr(γ5γ5γa) = −Tr(γa) (4.80) the trace of a single gamma matrix is zero. The trace of two gammas

Tr(γaγb) = Tr(2ηab − γbγa) (4.81) is Tr(γaγb) = ηabTr(1) = 4ηab. (4.82)

The trace of an odd number of gammas vanishes because

Tr(γa1 ··· γa2n+1 ) = Tr(γ5γ5γa1 ··· γa2n+1 ) = − Tr(γ5γa1 ··· γa2n+1 γ5) = − Tr(γ5γ5γa1 ··· γa2n+1 ) = −Tr(γa1 ··· γa2n+1 ) (4.83) 4.9 Electron-positron to muon-antimuon 105 which implies that Tr(γa1 ··· γa2n+1 ) = 0. (4.84) To ﬁnd the trace of four gammas, we use repeatedly the fundamental rule γaγb = 2ηab − γbγa. (4.85) We thus ﬁnd Tr(γaγbγcγd) = Tr[(2ηab − γbγa)γcγd] = Tr[2ηabγcγd − γbγaγcγd] = Tr[2ηabγcγd − γb(2ηac − γcγa)γd] = Tr[2ηabγcγd − 2ηacγbγd + γbγcγaγd] (4.86) = Tr[2ηabγcγd − 2ηacγbγd + 2ηadγbγc − γbγcγdγa] or Tr(γaγbγcγd) = Tr[ηabγcγd − ηacγbγd + ηadγbγc] = 4ηabηcd − ηacηbd + ηadηbc. (4.87) Reversing completely the order of the gammas changes nothing: Trγaγb ··· γyγz = Trγzγy ··· γbγa. (4.88)

4.9 Electron-positron to muon-antimuon

To avoid the extra complications that the sum of two amplitudes As + At entail, let’s consider the process e−(p) + e+(p0) → µ−(k) + µ+(k0) which requires only the direct amplitude e2 v¯(p0, s0)γau(p, s)u ¯(k, t)γ v(k0, t0) A = − i δ4(p + p0 − k − k0) a (2π)2 (p + p0)2 (4.89) = − 2πi δ4(p + p0 − k − k0) M where e2 v¯(p0, s0)γau(p, s)u ¯(k, t)γ v(k0, t0) M = a (4.90) (2π)3 (p + p0)2 in which I have relabeled the momenta so that k & k0 belong to the muons. The probability |A|2 includes the term a b ∗ vγ¯ u u¯µγavµ (¯vγ u u¯µγbvµ) (4.91) in which µ’s label muons. Sincev ¯ = iv†γ0 = v†β as well as β2 = (iγ0)2 = 1, and since (1.177) βγa†β = −γa, we get (¯vγau)∗ = (v†βγau)∗ = u†γa†βv = u†ββγa†βv = u†βγav =uγ ¯ av. (4.92) 106 Quantum Electrodynamics So the electron part of the term (4.91) is vγ¯ au (¯vγbu)∗ =vγ ¯ au uγ¯ bv. (4.93) The spin sums are

X ∗ h 1 c i u`(~p,s)um(~p,s) = (−i p γc + m) β 2p0 `m s (4.94) X ∗ h 1 c i v`(~p,s)vm(~p,s) = (−i p γc − m) β . 2p0 `m s Equivalently and more simply,

X h 1 c i u`(~p,s)¯um(~p,s) = (−i p γc + m) 2p0 `m s (4.95) X h 1 c i v`(~p,s)¯vm(~p,s) = (−i p γc − m) . 2p0 `m s So

X a b X 0 0 a b 0 0 vγ¯ u uγ¯ v = v¯j(p , s )γj`u`(p, s)u ¯m(p, s)γmnvn(p , s ) s,s0 s0,j,`,m,n

1 X 0 0 a b 0 0 = v¯j(p , s )γ (−ip/ + m)`mγ vn(p , s ) (4.96) 2p0 j` mn s0,j,`,m,n

1 X 0 0 0 0 a b = vn(p , s )¯vj(p , s )γ (−ip/ + m)`mγ 2p0 j` mn s0,j,`,m,n

1 X 0 a b = (−ip/ − m)njγ (−ip/ + m)`mγ 2p02p00 j` mn sj,`,m,n

1 0 a b = Tr[(−ip/ − me)γ (−ip/ + me)γ ]. 2p02p00 The muon part of the term (4.91) is

∗ uγ¯ av (¯uγbv) =uγ ¯ av vγ¯ bu =vγ ¯ bu uγ¯ av. (4.97) So the sum over all spins gives for the muons X 1 uγ¯ v vγ¯ u = Tr[(−ik/0 − m )γ (−ik/ + m )γ ]. (4.98) a b 2k02k00 µ b µ a t,t0 Using the formula (4.86) for the trace of 4 gammas, we evaluate the trace for the electrons in stages: 0 a b 0 a b 2 a b Tr[(−ip/ − me)γ (−ip/ + me)γ ] = − Tr p/γ pγ/ − meTr γ γ . (4.99) 4.9 Electron-positron to muon-antimuon 107 The 4-gamma term is

0 a b 0 c a d b Tr[(−ip/)γ (−ip/)γ ] = − Tr[pcγ γ pdγ γ ] 0 c a d b = − pcpdTr[γ γ γ γ ] 0 ca db cd ab cb ad = − 4pcpd η η − η η + η η = − 4p0apb − p0p ηab + p0bpc. (4.100)

The 2-gamma term is

2 a b 2 ab − meTr γ γ = − 4meη . (4.101)

So their sum is

0 a b 0a b 0b a ab 2 0 Tr[(−ip/ −me)γ (−ip/+me)γ ] = −4 p p +p p +η (me −p p) . (4.102)

The similar term for muons is

0 0 0 2 0 Tr[(−ik/ −mµ)γa(−ik/+mµ)γb] = −4 kakb +kbka +ηab(mµ −k k) . (4.103)

Their product is

0 a b 0 Tr[(−ip/ − me)γ (−ip/ + me)γ ] Tr[(−ik/ − mµ)γa(−ik/ + mµ)γb] (4.104) 0a b 0b a ab 2 0 0 0 2 0 = 16 p p + p p + η (me − p p) kakb + kbka + ηab(mµ − k k) h 0 0 0 0 2 0 2 0 i = 32 (p k )(pk) + (p k)(pk ) + mµ(pp ) + me(k k) .

So apart from the delta function and the (2π)’s, the squared amplitude summed over ﬁnal spins and averaged over initial spins is

4 0 0 0 0 2 0 2 0 1 X e (p k )(pk) + (p k)(pk ) + mµ(pp ) + me(k k) |M|2 = 4 2(2π)6p0p00k0k00 (p + p0)4 s,s0 (4.105) where (p + p0)4 = [(p + p0)2]2. In the rest frame of a collider, q2 = (p + p0)2 = 4E2 = 4p02. The cosine of the scattering angle θ is cos θ = ~p · ~k/(|~p||~k|). Since me ≈ mµ/200, I will neglect me in what follows. One then gets p · p0 = −2E2, p · k = E(|~k| cos θ − E), and p · k0 = p0 · k = −E(|~k| cos θ + E). 108 Quantum Electrodynamics Thus

4 2 2 2 2 2 2 1 X e E (E − k cos θ) + E (E + k cos θ) + 2mµE |M|2 = 4 2(2π)6E416E4 s,s0 e4 (E − k cos θ)2 + (E + k cos θ)2 + 2m2 = µ 32(2π)6E6 e4 2E2 + 2k2 cos2 θ + 2m2 e4 E2 + k2 cos2 θ + m2 = µ = µ 32(2π)6E6 16(2π)6E6 (4.106)

The squared amplitude is

1 X |A|2 = (2π)2|δ4(p + p0 − k − k0)|2 |M|2 (4.107) 4 ss0 VT 1 X = δ4(p + p0 − k − k0) |M|2. (4.108) (2π)2 4 ss0 since

Z d4x VT δ4(0) = = (4.109) (2π)4 (2π)4 in which V is the volume of the universe and T is its inﬁnite time. We also need to switch from delta-function normalization of states to unit normalization. The relation is that between a continuum delta function and a Kronecker delta

3 0 V δ (~p − ~p) = δ 0 . (4.110) (2π)3 pp

Since there are two particles in the initial and ﬁnal states, the probability is

(2π)3 P = 4|A|2N (4.111) V f

3 3 0 where Nf is the number of ﬁnal states. For a range of momenta d kd k , the number of ﬁnal states is

V N = 2d3kd3k0. (4.112) f (2π)3 4.9 Electron-positron to muon-antimuon 109 So the rate P/T is

(2π)3 |A|2 (2π)3 |A|2 V R = 4 N = 4 2d3kd3k0 V T f V T (2π)3 3 2 3 2 (2π) 4 |A| (2π) 2 |A| = N = d3kd3k0 (4.113) V T f V T (2π)4 1 X = δ4(p + p0 − k − k0) |M|2d3kd3k0 V 4 ss0

The ﬂux of incoming particles is u/V where u = |~v1 − ~v2 is the relative velocity, which with c = 1 for massless electrons is u = 2. So the diﬀerential cross-section is

1 1 X dσ = (2π)4δ4(p + p0 − k − k0) |M|2d3kd3k0. (4.114) 2 4 ss0

Integrating over d3k0 sets ~k0 = −~k, and so

1 1 X dσ = (2π)4δ(p0 + p00 − 2k0) |M|2d3k 2 4 0 ss (4.115) 1 q 1 X = (2π)4δ p0 + p00 − 2 k2 + m2 |M|2k2dkdΩ 2 µ 4 ss0

where k = |~k|. The derivative of the delta function is k2p0/k02, so

02 02 1 4 1 X 2 2 k 4 1 X 2 kk dσ = (2π) |M| k dΩ 0 = 2π |M| 0 dΩ 2 4 0 2kp 4 0 p ss ss (4.116) 1 X kk02 = 4π4 |M|2 dΩ. 4 p0 ss0 110 Quantum Electrodynamics So then

4 2 2 2 2 e E + k cos θ + m kk02 dσ = 4π4 µ dΩ 4(2π)6E6 p0 e4 E2 + k2 cos2 θ + m2 = µ k dΩ 4(2π)2E5 " # e4 k2 m2 = 1 + cos2 θ + µ k dΩ 4(2π)2E3 E2 E2 " # e4 E2 − m2 m2 = 1 + µ cos2 θ + µ k dΩ (4.117) 4(2π)2E3 E2 E2 s " ! # e4 m2 m2 m2 = 1 − µ 1 + µ + 1 − µ cos2 θ dΩ 4(2π)2E2 E2 E2 E2 s " ! # α2 m2 m2 m2 = 1 − µ 1 + µ + 1 − µ cos2 θ dΩ. 16E2 E2 E2 E2

The total cross-section is the inetgral over solid angle

s " ! ! # α2 m2 m2 m2 Z σ = 1 − µ 1 + µ 4π + 1 − µ cos2 θ dΩ 16E2 E2 E2 E2 s " ! ! # α2 m2 m2 m2 Z = 1 − µ 1 + µ 4π + 1 − µ cos2 θ 2πd cos θ 16E2 E2 E2 E2 s " ! !# πα2 m2 m2 1 m2 = 1 − µ 1 + µ + 1 − µ 4E2 E2 E2 3 E2 s ! πα2 m2 1 m2 = 1 − µ 1 + µ (4.118) 3E2 E2 2 E2

where E is the energy of each electron in the lab frame of the collider. The energy of the collider is Ecm = 2E. At very high energies, the x-section is σ = πα2/(3E2). 4.10 Electron-muon scattering 111 4.10 Electron-muon scattering − − − 0 − 0 For the process e (p1) + µ (p2) → e (p1) + µ (p2), the Feynman rules give 2 0 0 a 0 0 e 4 0 0 u¯(p1, s )γ u(p1, s)u ¯(p2, t )γau(p2, t) A = − i 2 δ (p1 + p2 − p1 − p2) 0 2 (2π) (p1 − p1) 4 0 0 = − 2πi δ (p1 + p2 − p1 − p2) M. (4.119)

2 0 2 The product of the traces over q = (p1 − p1) is 0 a b 0 Tr[(−ip/1 + me)γ (−ip/1 + me)γ ] Tr[(−ip/2 + mµ)γa(−ip/2 + mµ)γb] 0 2 . (p1 − p1) (4.120) For e+ − e− → µ+ + µ− we got

0 a b 0 Tr[(−ip/ − me)γ (−ip/ + me)γ ] Tr[(−ik/ − mµ)γa(−ik/ + mµ)γb] . (4.121) (p + p0)2

If in this amplitude for e+ − e− → µ+ + µ− we make these replacements

0 0 0 0 p → p1, p → −p1, k → p2, and k → −p2, (4.122) it becomes

Tr[(ip0 − m )γa(−ip0 + m )γb]Tr[(ip − m )γ (−ip0 + m )γ ]/(p + p0 )2 /1 e /1 e /2 µ a /2 µ b 1 2 Tr[(−ip0 + m )γa(−ip0 + m )γb] Tr[(−ip + m )γ (−ip0 + m )γ ] /1 e /1 e /2 µ a /2 µ b = 0 2 (p1 − p1) (4.123)

− − which is exactly the same as the amplitude (4.120) for e (p1) + µ (p2) → − 0 − 0 e (p1) + µ (p2) scattering. So apart from the delta function and the (2π)’s, the squared amplitude summed over ﬁnal spins and averaged over initial spins is the one for e+ − e− → µ+ + µ− but with the substitutions (4.122). − − − 0 − 0 That is, for e (p1) + µ (p2) → e (p1) + µ (p2) scattering, the summed amplitudes are

4 0 0 0 0 2 0 2 0 1 X 2 e (p1p2)(p1p2) + (p1p2)(p1p2) − mµ(p1p1) − me(p2p2) |M| = 6 0 00 0 00 0 4 . 4 2(2π) p p p p (p1 − p ) s,s0,t,t0 1 1 2 2 1 (4.124) The equality of the two amplitudes (4.120) and (4.121) under the corre- spondence (4.122) is an example of crossing symmetry. More generally, the scattering amplitude for a process in which an incoming particle of 4- momentum p that interacts with other particles is equal to the scattering 112 Quantum Electrodynamics amplitude for the process in which the antiparticle of the incoming particle leaves with 4-momentum −p after interacting with the same other particles

S(−p, nc; · · · | · · · ) = S(· · · |p, n; ··· ) (4.125)

0 0 in which type nc is labels the antiparticle of type n. Since −p + p = 0, this is not the amplitude for a physical process in which each particle has its 0 q 2 2 appropriate energy, pi = mi + ~pi . This symmetry caused huge excitement on the West Coast during the 1960s. 0 00 Let k = |~p1| = p1 be energy or equivalently the modulus (magnitude) of 0 the 3-momentum of the (massless) final electron. Let E = p2 be the energy of the initial muon. Then the differential x-section in the lab frame of the collider is dσ α2 h i = (E + k)2 + (E + k cos θ)2 − m2 (1 − cos θ) dΩ 2k2(E + k)2(1 − cos θ)2 µ (4.126) in which the cosine of the scattering angle of the electron is ~p · ~p cos θ = 1 1 . (4.127) |~p1||~p1| The total x-section σ is infinite.

4.11 One-loop QED The action density is 1 L = − F abF − ψ¯ [γ (∂a + ie Aa ) + m ] ψ (4.128) 4 B Bab B a B B B B in which the subscript B means that the values of these “bare” terms include infinite quantities that may be used to cancel the infinities that arise in perturbation theory. The actual physical terms are defined as −1/2 ψ ≡ Z2 ψB (4.129) a −1/2 a A ≡ Z3 AB (4.130) 1/2 e ≡ Z3 eB (4.131) m ≡ mB + δm (4.132)

a a so that in particular eA = eBAB. In these terms, the action density is

L = L0 + L1 + L2 (4.133) 4.11 One-loop QED 113 Vacuum polarization

Figure 4.3 One-loop Feynman diagram for vacuum polarization (arbitrary colors).

where L0 is the physical or naive free action density 1 L = − F abF − ψ¯ (γ ∂a + m) ψ, (4.134) 0 4 ab a

L1 is the interaction

a L1 = − ieAaψγ¯ ψ, (4.135) and L2 is a sum of inﬁnite “counterterms” 1 L = − (Z − 1)F abF − (Z − 1)ψ¯ (γ ∂a + m) ψ 2 4 3 ab 2 a a + Z2 δm ψψ¯ − ie(Z2 − 1)Aaψγ¯ ψ. (4.136)

Vacuum polarization is a one-loop correction to the photon propagator. Its Feynman diagram is in ﬁgure 4.3. Let’s use L1 to compute it. The amplitude S is

h R 4 i h R ¯ a 4 i S = hq0, s0|T ei L1 d x |q, si = hq, s|T ee Aa(x)ψ(x)γ ψ(x) d x |q, si. (4.137) 114 Quantum Electrodynamics The one-loop correction is due to the order-e2 term

1 Z Z S = hq0, s0|T e2 A (x)ψ¯(x)γaψ(x) d4x A (y)ψ¯(y)γbψ(y) d4y |q, si. 2 2 a b (4.138)

We choose to absorb the photon at y and cancel the factor of one-half. The electromagnetic ﬁeld is

Z 3 d p X h ip·y 0 0 −ip·y ∗ 0 † 0 i Ab(y) = e eb(p, s )a(p, s ) + e e (p, s )a (p, s ) . p 3 0 b (2π) 2p s0 (4.139) So S2 is

2 e Z 0 S = h0|T e−iq ·xe∗(q0, s0)ψ¯(x)γaψ(x) d4x 2 (2π)32q0 a Z iq·y ¯ b 4 × e eb(q, s)ψ(y)γ ψ(y) d y |0i. (4.140)

The term L1 is normally ordered (it should be surrounded by colons), so fermions emitted by ψ¯(y) and ψ(y) must be absorbed by ψ(x) and ψ¯(x). Thus S2 is

2 ∗ 0 0 e e (q , s )e (q, s) Z 0 S = a b e−iq ·xeiq·yh0|T [ψ¯ (x)γa ψ (y)]|0i 2 (2π)32q0 α αβ δ ¯ b 4 4 × h0|T [ψβ(x)ψγ(y)γγδ]|0i d x d y. (4.141) ¯ Moving ψδ(y) to the left of ψα(x) adds a minus sign, so

2 ∗ 0 0 e e (q , s )e (q, s) Z 0 S = − a b e−iq ·xeiq·yh0|T [ψ (y)ψ¯ (x)γa ]|0i 2 (2π)32q0 δ α αβ ¯ b 4 4 × h0|T [ψβ(x)ψγ(y)γγδ]|0i d x d y. (4.142)

This minus sign is an instance of Feynman’s rule that closed fermion loops make minus signs. In the common notation ψ = ψ†β, his propagator is h0|T ψ`(x)ψm(y) |0i ≡ − i∆`m(x − y) β 4 Z d q ( − i/q + m)`m = − i eiq·(x−y) (2π)4 q2 + m2 − i (4.143) Z d4q 1 = − i eiq·(x−y). (2π)4 i/q + m − i `m 4.11 One-loop QED 115 Putting all this together, we get

2 ∗ 0 0 4 e e (q , s )e (q, s) Z 0 Z d k ( − ik/ + m) S = a b e−iq ·xeiq·y δα eik·(y−x)γa 2 (2π)32q0 (2π)4 k2 + m2 − i αβ 4 Z d p ( − ip/ + m)βγ × eip·(x−y)γb d4x d4y. (4.144) (2π)4 p2 + m2 − i γδ

Dirac’s delta functions simplify this 16-dimensional integral to e2e∗(q0, s0)e (q, s)δ4(q0 − q) Z − i(p/ − /q) + m S = a b d4p δα 2 (2π)32q0 (p − q)2 + m2 − i ( − ip/ + m)βγ × γa γb . (4.145) αβ p2 + m2 − i γδ

The product of matrices is a trace

a b e2e∗(q0, s0)e (q, s)δ4(q0 − q) Z Tr − i(p/ − /q) + m γ ( − ip/ + m)γ S = a b d4p . 2 (2π)32q0 [(p − q)2 + m2 − i](p2 + m2 − i) (4.146)

Equivalently, since TrAB = TrBA, we have

b a e2e∗(q0, s0)e (q, s)δ4(q0 − q) Z Tr ( − ip/ + m)γ − i(p/ − /q) + m γ S = a b d4p . 2 (2π)32q0 (p2 + m2 − i)[(p − q)2 + m2 − i] (4.147) or interchanging a and b

a b e2e (q, s)e∗(q0, s0)δ4(q0 − q) Z Tr ( − ip/ + m)γ − i(p/ − /q) + m γ S = a b d4p 2 (2π)32q0 (p2 + m2 − i)[(p − q)2 + m2 − i] (4.148)

SW extracts from this the deﬁnition

a b −ie2 Z Tr ( − ip/ + m)γ − i(p/ − /q) + m γ Π∗ab(q) = d4p . (4.149) (2π)4 (p2 + m2 − i)[(p − q)2 + m2 − i]

We use the Feynman trick

1 Z 1 dx = 2 (4.150) AB 0 [(1 − x)A + xB] 116 Quantum Electrodynamics to write 1 Z 1 dx = 2 2 2 2 2 (p + m − i)[(p − q) + m − i] 0 (p2 + m2 − i)(1 − x)] + [(p − q)2 + m2 − i]x Z 1 dx = 2 0 p2 + m2 − i − 2p · q x + q2 x (4.151) Z 1 dx = . 2 0 (p − xq)2 + m2 − i + x(1 − x) q2 The next step is to make the integral finite somehow. One can use the counterterms, one can regard quantum field theory as merely a way to in- terpolate between values taken from experiment, or one can impose dimensional regularization as we’ll see momentarily. If we assume that we have made the integral finite one way or another, then we can shift p to p + x q. After this shift, the quantity (4.152) becomes a b −ie2 Z 1 Z Tr (−i(p/ + x/q) + m)γ − i(p/ − /q(1 − x)) + m γ Π∗ab(q) = dx d4p . 4 2 (2π) 0 p2 + m2 − i + x(1 − x) q2 (4.152) The trace identities of section 4.8 give a b Tr (−i(p/ + x/q) + m)γ − i(p/ − /q(1 − x)) + m γ h = 4 − (p + xq)a(p − (1 − x)q)b + (p + xq) · (p − (1 − x)q) ηab i − (p + xq)b(p − (1 − x)q)a + m2ηab . (4.153)

The variable p0 is being integrated along the real axis of the complex p0 plane from −∞ to +∞. We now imagine that we have made the p-integral suﬃciently ﬁnite that we can rotate the p0 contour by 90 degrees so that it runs up the imaginary axis in the complex p0 plane from p0 = −i∞ to p0 = i∞. This is a Wick rotation. We set p0 = ip4 and integrate over p4 = −ip0, which is real, from −∞ to ∞. We also set q0 = iq4 and q4 = −iq0. So now p · q = ~p · ~q − p0q0 = ~p · ~q + p4q4. (4.154) All the Lorentz signs are now positive and p0p0η00 = ip4ip4(−1) = p4p4 = p4p4δ44 (4.155) as well as q0q0η00 = iq4iq4(−1) = q4q4 = q4q4δ44. (4.156) 4.11 One-loop QED 117 The quantity q2 is real whether q0 is real or imaginary; it is positive when q0 < ~q2 whether or not q0 is imaginary. We now can drop i and express the quantity

Π∗ab(q) = Π∗ab(~q, q0) = Π∗ab(~q, iq4) (4.157) either as

−4ie2 Z 1 Z 1 Π∗ab(q) = dx d4p 4 2 (2π) 0 p2 + m2 − i + x(1 − x) q2 h × − (p + xq)a(p − (1 − x)q)b + (p + xq) · (p − (1 − x)q) ηab i − (p + xq)b(p − (1 − x)q)a + m2ηab (4.158)

4 0 4 1 2 3 4 or with p = −ip real and d pe = dp dp dp dp as

4e2 Z 1 Z 1 Π∗ab(q) = dx d4p 4 e 2 (2π) 0 p2 + m2 + x(1 − x) q2 h × − (p + xq)a(p − (1 − x)q)b + (p + xq) · (p − (1 − x)q) δab i − (p + xq)b(p − (1 − x)q)a + m2δab (4.159) in which all Lorentz signs are +1 and q4 = −iq0 is imaginary if q is real. The next step is to say how we regularize the integral over d4p. Since the 1970s, the method of choice has been dimensional regularization because it preserves Lorentz and gauge invariance. In dimensional regularization, the p integration is taken to be over an arbitrary number d of spacetime dimensions. The denominator of the integral (4.159) is a function of p2 which is taken to be the square of the radius of a sphere in d dimensions. Terms in the numerator that are odd powers of pa vanish. We set

p2 δab papp = . (4.160) d The area of a sphere of unit radius in d dimensions is

2πd/2 A = (4.161) d Γ(d/2) as we can verify by computing an integral of the gaussian exp(−x2) in d 118 Quantum Electrodynamics dimensions using both rectangular and spherical coordinates.

∞ d Z 2 Z 2 e−x ddx = e−x dx = πd/2 −∞ (4.162) Z ∞ d−1 −r2 1 = Ad r e dr = Ad Γ(d/2). 0 2 The integral (4.159) converges as long as d is complex or a fraction or anything but an even integer. Nonetheless, the divergencies in that integral reappear as powers of 1/(d−4) as d → 4. We’ll use the counterterms (4.136) in L2 to cancel them. p 2 d−1 The area of a sphere of radius k = p in d dimensions is Ad k . So using the formulas (4.160 and 4.161) of dimensional regularization, we can write the integral (4.159) as 4e2A Z 1 Z ∞ 1 Π∗ab(q) = d dx kd−1dk (4.163) 4 2 (2π) 0 0 k2 + m2 + x(1 − x) q2 h k2δab i × − 2 + 2qaqbx(1 − x) + [k2 − x(1 − x)q2]δab + m2δab . d The needed integrals are Z ∞ d−1 k 1 2 d/2−2 2 2 2 dk = 2 (c ) Γ(d/2) Γ(2 − d/2) (4.164) 0 (k + c ) Z ∞ d+1 k 1 2 d/2−1 2 2 2 dk = 2 (c ) Γ(1 + d/2) Γ(1 − d/2). (4.165) 0 (k + c ) Using these integrals, we get

2 Z 1 " ∗ab 2e Ad 2 ab 2 d/2−1 Π (q) = 4 dx 1 − δ (c ) Γ(1 + d/2) Γ(1 − d/2) (2π) 0 d (4.166) # h i + (2qaqb − q2δab)x(1 − x) + m2δab (c2)d/2−2Γ(d/2)Γ(2 − d/2) in which c2 = m2 + x(1 − x) q2. (4.167) The identity Γ(z + 1) = z Γ(z) (4.168) implies that Γ(2 − d/2) = (1 − d/2) Γ(1 − d/2), (4.169) 4.11 One-loop QED 119 and that Γ(1 + d/2) = (d/2) Γ(d/2), (4.170) so Γ(1 + d/2) Γ(1 − d/2) = (2/d − 1)−1Γ(d/2) Γ(2 − d/2). (4.171)

So we can combine terms in the integral (4.172)

2 Z 1 " ∗ab 2e Ad ab 2 Π (q) = 4 Γ(d/2)Γ(2 − d/2) dx − δ c (4.172) (2π) 0 # h i + (2qaqb − q2δab)x(1 − x) + m2δab (c2)d/2−2.

Using the value (4.167) of c2, we have

−δabc2 +(2qaqb −q2δab)x(1−x)+m2δab = 2(qaqb −q2δab)x(1−x). (4.173)

Thus we ﬁnd 4e2A Π∗ab(q) = d Γ(d/2)Γ(2 − d/2)(qaqb − q2δab) (2π)4 (4.174) Z 1 × dxx(1 − x)[m2 + x(1 − x) q2]d/2−2. 0 Current conservation and gauge invariance imply that

∗ab qa Π (q) = 0. (4.175) One of the advantages of dimensional regularization is that it preserves the vanishing of this quantity

∗ab a b 2 ab 2 b 2 b qa Π (q) ∝ qa(q q − q δ ) = q q − q q = 0. (4.176) This works because current conservation and the dimension of spacetime have nothing to do with each other: ∗ab The vanishing of qa Π (q) follows from the conservation

a a ∂aJ (x) = ∂a[ψ(x)γ ψ(x)] = 0 (4.177) of the current J a(x) = ψ(x)γaψ(x) (4.178) and from the neutrality of the current which is due to the fact that ψ lowers the charge while ψ¯ raises it. 120 Quantum Electrodynamics Apart from a constant f, the quantity Π∗ab(q) is by (4.142) Z Π∗ab(q) = f d4xd4yeiq(y−x)h0|T [J a(x)J b(y)]|0i. (4.179)

So Z ∗ab 4 4 iq(y−x) a b qaΠ (q) = f d xd y qa e h0|T [J (x)J (y)]|0i Z 4 4 iq(y−x) a b = f d xd y e (i∂a) h0|T [J (x)J (y)]|0i Z 4 4 iq(y−x) a b = f d xd y e h0|T [i∂aJ (x)J (y)]|0i (4.180) Z + f d4xd4y i δ(x0 − y0) eiq(y−x)h0|[J 0(x),J b(y)]|0i Z = f d4xd4y i δ(x0 − y0) eiq(y−x)h0|[J 0(x),J b(y)]|0i since the current is conserved. But the equal-time commutator of the charge J 0(x) with any operator is proportional to the charge of that operator because J 0(x) generates the U(1) rotation of the symmetry that conserves charge. So since currents are neutral, the commutator vanishes δ(x0 − y0) eiq(y−x)[J 0(x),J b(y)] = 0, (4.181) and we have ∗ab qaΠ (q) = 0. (4.182) The gamma function Γ(2 − d/2) in Π∗ab(q) (4.174) blows up as d → 4. So we bring in the contribution to Π∗ab(q) arising from the counterterm ab −(Z3 − 1)FabF . This quadratic form in derivatives of Aa and Ab adds to Π∗ab(q) the term

∗ab 2 ab a b Π (q)Z3 = − (Z3 − 1)(q δ − q q ) (4.183) in euclidian space. So to order e2,Π∗ab(q) is Π∗ab(q) = (q2δab − qaqb)π(q2) (4.184) where 2 Z 1 4e Ad 2 2 d/2−2 π(q) = − 4 Γ(d/2)Γ(2 − d/2) dxx(1 − x)[m + x(1 − x) q ] (2π) 0 − (Z3 − 1). (4.185)

So we can use Z3 to cancel the poles in the gamma functions. 4.11 One-loop QED 121 The quantity Π∗ab(q) is a correction to the photon propagator. And Π∗ab(0) is a correction to the mass of the photon which we want to remain at zero. So we set Π∗ab(0) = 0. (4.186) This means that set set 2 Z 1 4e Ad 2 d/2−2 Z3 = 1 − 4 Γ(d/2)Γ(2 − d/2) (m ) dxx(1 − x), (4.187) (2π) 0 so to order e2 2 Z 1 4e Ad π(q) = − 4 Γ(d/2)Γ(2 − d/2) dxx(1 − x) (2π) 0 (4.188) h i × [m2 + x(1 − x) q2]d/2−2 − (m2)d/2−2 .

As we let d → 4, the tricky term is i [m2 + x(1 − x) q2]d/2−2 − (m2)d/2−2 = exp (d/2 − 2) ln[m2 + x(1 − x) q2] i − exp (d/2 − 2) ln(m2)] = (d/2 − 2) ln[1 + x(1 − x) q2/m2]. (4.189) The ﬁnal result is 2 Z 1 2 e 2 2 π(q ) = 2 x(1 − x) ln 1 + x(1 − x) q /m dx. (4.190) 2π 0

The scattering of two particles of charges e1 and e2 that exchange mo- 0 0 mentum q = p1 − p1 = p2 − p2 is −ie e S = 1 2 δ4(p0 + p0 − p − p )¯u0 γau u¯0 γbu 4π2 1 2 1 2 1 1 2 2 (4.191) η [q2η − q q ]π(q2) × ab + ab a b . q2 (q2)2 The spinors obey the Dirac equation in momentum space (1.264)

c (i p γc + m)u(p, s) = 0 (4.192) so 0 a 0 a p1au¯1γ u1 = imu¯1γ u1. (4.193) The adjoint of Dirac’s equation is

† c † u (p, s)(−i p γc + m) = 0. (4.194) 122 Quantum Electrodynamics The spatial gammas are hermitian while γ0 is antihermitian, so

† c † 0 † 0 c c iu (p, s)(−i p γc + m)γ = iu (p, s)γ (i p γc + m) =u ¯(p, s)(i p γc + m) = 0. Thus 0 0 a 0 a p1au¯1γ u1 = −imu¯1γ u1. (4.195) Subtracting, we get

0 a 0 0 a 0 a qau¯1γ u1 = (p1a − p1a)¯u1γ u1 = (−im − (−im))¯u1γ u1 = 0. (4.196) So the amplitude (4.191) simpliﬁes to −ie e S = 1 2 [1 + π(q2)]δ4(p0 + p0 − p − p )¯u0 γau u¯0 γ u . (4.197) 4π2q2 1 2 1 2 1 1 2 a 2

For small ~p1 and ~p2, 0 0 0† 0 2 0† 0† 0 2 u¯ γ u = iu (γ ) u = −iu u ≈ −iδ 0 and iu (γ ) u ≈ −iδ 0 1 1 1 1 1 1 s1s1 1 1 s2s2 while 0 0 u¯1~γu1 = 0 andu ¯2~γu2 = 0. So the amplitude is approximately

−ie1e2 2 4 0 0 S = [1 + π(~q )]δ (p + p − p1 − p2) δs0 s δs0 s . (4.198) 4π2~q2 1 2 1 1 2 2 The lowest, Born, approximation to the nonrelativistic scattering due to a spin-independent potential V (|~x1 − ~x2|) which is 0 0 SB = −2πiδ(E1 + E2 − E1 − E2) TB (4.199) where

δ 0 δ 0 Z 0 0 s1s1 s2s2 3 3 −i~p ·~x −i~p ·~x i~p ·~x i~p ·~x T = d x d x V (|~x − ~x |)e 1 1 e 2 2 e 1 1 e 2 2 . B (2π)6 1 2 1 2

We set ~r = ~x1 − ~x2 and get Z −i 4 0 0 −i~q·~r 3 SB = δ (p + p − p1 − p2)δs0 s δs0 s V (r)e d r. (4.200) 4π2 1 2 1 1 2 2 Comparing this formula with (4.198), we have Z 1 + π(~q2) V (r)e−i~q·~r d3r = e e 1 2 ~q2 or e e Z 1 + π(~q2) V (r) = 1 2 ei~q·~r . (4.201) (2π)3 ~q2 4.11 One-loop QED 123 This is the potential that the extended charge distribution 1 Z c(~r) = δ3(~r) + π(~q2) ei~q·~r d3q, (4.202) 2(2π)3 which is normalized to unity Z π(0) c(~r) d3r = 1 + = 1, (4.203) 2 would produce if separated from itself by ~r Z Z c(~x)c(~y) V (r) = e e d3x d3y . (4.204) 1 2 4π|~x − ~y + ~r| Apart from a delta function, which we may write as 1 Z ∞ δ3(~r) = eiqr cos θ 2πq2dqd cos θ (2π)3 0 (4.205) 1 Z ∞ eiqr − eiqr = 2 qdq, (2π) 0 ir the isotropic charge distribution c0(r) = c(r) − δ3(~r) is Z ∞ 0 1 2 iqr cos θ 2 c (r) = 3 π(q ) e 2πq dqd cos θ 2(2π) 0 1 Z = π(q2) sin(qr) q dq (2π)2r (4.206) e2 Z ∞ Z 1 x(1 − x) q2 = 4 qdq dxx(1 − x) ln 1 + 2 sin(qr) 8π r 0 0 m 2 Z ∞ Z 1 2 −ie iqr x(1 − x) q = 4 e qdq dxx(1 − x) ln 1 + 2 . 16π r −∞ 0 m Vacuum polarization (4.190) shifts the energy of an atomic state with wave function ψ(r) by Z ∆E = ∆V (r)|ψ(r)|2 d3r (4.207) where (SW 11.2.28) e e Z π(q2) ∆V (r) = ∆V (r) = 1 2 d3qeiq·r . (4.208) (2π)3 q2 For hydrogen-like electrons in states with ` = 0 and wave function at r = 0

2 Zαm3/2 ψ(0) = √ (4.209) 4π n 124 Quantum Electrodynamics the shift is (SW 11.2.42) 4Z4α5m ∆E = = − (4.210) 15πn3 and is called the Uehling eﬀect. In the 2s state of hydrogen it is −1.122×10−7 eV.

4.12 Magnetic moment of the electron The one-loop correction to the interaction of an electron of momentum p with a photon of momentum p0 − p (SW’s ﬁgure 11.4(d)) is the logarithmically divergent integral (SW 11.3.1) Z h −i −i(p/0 − k/) + m i Γa(p0, p) = d4keγb(2π)4 γa (4.211) (2π)4 (p0 − k)2 + m2 − i h −i −i(p/ − k/) + m i h −i 1 i × eγ (2π)4 . (2π)4 (p − k)2 + m2 − i b (2π)4 k2 − i The ﬁrst step is to use Feynman’s trick 1 Z 1 Z x 1 = 2 dx dy (4.212) 3 ABC 0 0 Ay + B(x − y) + C(1 − x) to write the three denominators as 1 1 1 Z 1 Z x 0 2 2 2 2 2 = 2 dx dy (4.213) (p − k) + m − i (p − k) + m − i k − i 0 0 1 × h i3 [(p0 − k)2 + m2 − i]y + [(p − k)2 + m2 − i](x − y) + (k2 − i)(1 − x) Z 1 Z x 1 = 2 dx dy . h i3 0 0 [k − p0y − p(x − y)]2 + m2x2 + q2y(x − y) − i

We now assume that we have made the integral suﬃciently ﬁnite that we can shift k to k+p0y+p(x−y). The integral (4.211) for Γ(p0, p) then becomes SW’s (11.3.4) 2ie2 Z 1 Z x Z h i a 0 4 b /0 / Γ (p , p) = 4 dx dy d k γ − i[p (1 − y) − k − p/(x − y)] + m (2π) 0 0 ah 0 i × γ − i[p/(1 − x + y) − k/ − p/y] + m γb

h i3 k2 + m2x2 + q2y(x − y) − i . (4.214) 4.12 Magnetic moment of the electron 125 The next step is a Wick rotation that sets k0 = ik4 and makes the denominator a function of k2 = k2 + (k4)2. The O(4) symmetry of the denominator lets us drop terms in the numerator that are odd under reﬂections. Also since Γ(n) = (n − 1)!, our formula (4.161) for the area of the unit sphere in d dimensions gives for d = 4

2π4/2 A = = 2π2. (4.215) 4 Γ(4/2)

4 4 2 3 The integral (4.214) in terms of d k = id ke = 2iπ k dk now is 2 2 Z 1 Z x Z ∞ 2 b c a a 0 −4π e 3 k γ γ γ γcγb Γ (p , p) = 4 dx dy k dk − (2π) 0 0 0 4 h i + γb − i[p/0(1 − y) − p/(x − y)] + m γa h 0 i × − i[p/(1 − x + y) − p/y] + m γb

h i3 k2 + m2x2 + q2y(x − y) . (4.216)

Dirac’s spinors obey his equation (1.264) in momentum space

u¯0(ip/0 + m) = 0 and (ip/ + m)u = 0. (4.217)

So we may (eventually) write the vertex correction (4.216) as

2 2 Z 1 Z x Z ∞ 0 a 0 −4π e 3 u¯ Γ (p , p)u = 4 dx dy k dk (2π) 0 0 0 h i × u¯0 γa − k2 + 2m2(x2 − 4x + 2) + 2q2[y(x − y) + 1 − x] + 4imp0a(y − x + xy) + 4impa(x2 − xy − y) u

h i3 k2 + m2x2 + q2y(x − y) . (4.218)

Under the reﬂection y → x − y, the factor y − x + xy, which multiplies p0a, becomes x − y − x + x(x − y) = x2 − xy − y, which multiplies pa. This reﬂection maps the lower half of the triangle {(x, y)|0 ≤ x ≤ 1; 0 ≤ y ≤ x} into its upper half. The jacobian is unity. So we can replace each factor by the average of the two factors 1 1 − x(1 − x) = (y − x + xy + x2 − xy − y). (4.219) 2 2 126 Quantum Electrodynamics The vertex (4.218 then reduces to 2 2 Z 1 Z x Z ∞ 0 a 0 −4π e 3 u¯ Γ (p , p)u = 4 dx dy k dk (4.220) (2π) 0 0 0 h i × u¯0 γa − k2 + 2m2(x2 − 4x + 2) + 2q2[y(x − y) + 1 − x]

h i3 + −2im(p0a + pa)x(1 − x) u k2 + m2x2 + q2y(x − y) .

0 0 a This term is a correction to a term that involves hp , s |Aa(x)J (x)|p, si and we ignore the electromagnetic ﬁeld Aa(x) except for its momentum. So we are dealing with a matrix element of a current J a(x)

0 hp0, s0|J a(x)|p, si = hp0, s0|e−iP ·xJ a(0)eiP ·x|p, si = ei(p−p )·xhp0, s0|J a(0)|p, si. (4.221) Since the current J a(x) is conserved a 0 = ∂aJ (x) (4.222) it follows that 0 0 a 0 0 a 0 = hp , s |∂aJ (x)|p, si = ∂ahp , s |J (x)|p, si i(p−p0)·x 0 0 a 0 0 0 a = ∂ae hp , s |J (0)|p, si = i(p − p )ahp , s |J (0)|p, si. (4.223) Does the amplitude (4.220) obey this rule of current conservation? To check this, we note that since Dirac’s spinors obey his equation (4.217 or 4.237), we see that the γa term does 0 0 a 0 0 0 i(p − p )au¯ γ u =u ¯ (ip/ − ip/)u =u ¯ (m − m)u = 0. (4.224) So does the p0 − p term: 0 0 0a a 0 2 2 (p − p )au¯ (p − p )u =u ¯ (−m + m )u = 0. (4.225) If we integrate the charge density J 0(x) over all space, we get the charge operator Q Z Q = J 0(x, t) d3x. (4.226)

Because the current is conserved (4.222), the charge operator Q has a vanishing time derivative ∂Q Z Z = ∂ J 0(x, t) d3x = − ∇ · J(x, t) d3x ∂t 0 Z = − J(x, t) · dσ = 0 (4.227) 4.12 Magnetic moment of the electron 127 as long as the integral over the surface σ vanishes. (Massless scalar ﬁelds could pose a problem here.) The matrix element of the charge operator is an integral of the matrix element of the charge density, which is the time component of the current (4.221), Z hp0, s0|Q|p, si = hp0, s0| J 0(x, t) d3x|p, si

Z 0 = ei(p−p )·xd3xhp0, s0|J 0(0) d3x|p, si

= (2π)3δ3(p − p0)hp0, s0|J 0(0) d3x|p, si. (4.228)

The state |p, si is a state of one electron (of momentum p and spin s in the z direction), so it has charge −e. Therefore

Q|p, si = −e|p, si, (4.229) and so

0 0 0 0 3 0 hp s |Q|p, si = − ehp s |p, si = −eδ (p − p )δss0 . (4.230)

Comparing this equation with the matrix element (4.228) of the charge operator, we ﬁnd that

0 0 0 3 e hp , s |J (0) d x|p, si = − δ 0 . (4.231) (2π)3 ss

SW writes the matrix element of a current J a of a particle of charge −e as (10.6.9) ie hp0, s0|J a(0)|p, si = − u¯(p0, s0)Γa(p0, p)u(p, s). (4.232) (2π)3 Lorentz invariance requires the last factor here to be h i u¯(p0, s0)Γa(p0, p)u(p, s) =u ¯(p0, s0) γaF (q2) − (p + p0)aG(q2) 2m (p − p0)a i + H(q2) u(p, s). (4.233) 2m The hermiticity of the current implies that F, G, & H are real functions of q2 = (p0 − p)2. Charge conservation (4.223) requires that H(q2) = 0. So the vertex correction (4.220) is the sum of two form factors F (q2) and G(q2) h i i u¯0Γa(p0, p)u =u ¯0 γaF (q2) − (p + p0)aG(q2) u. (4.234) 2m 128 Quantum Electrodynamics Letting p0 → p in this equation and in the matrix element (4.232), we ﬁnd ie hp, s0|J a(0)|p, si = − u¯(p, s0)Γa(p, p)u(p, s) (2π)3 (4.235) ie h i i = − u¯(p, s0) γaF (0) − paG(0) u(p, s). (2π)3 m Since Dirac’s gamma matrices are deﬁned by the anticommutation relation {γa, γb} = 2ηab, it follows that

a a b ab a a a {γ , ip/ + m} = {γ , ipbγ + m} = 2ipbη + 2mγ = 2ip + 2mγ . (4.236) Dirac’s spinors obey his equation

u¯0(ip/0 + m) = 0 and (ip/ + m)u = 0. (4.237) So between spinors of the same momentum, the anticommutation relation (4.236) gives

u¯(p, s0){γa, ip/ + m}u(p, s) = 0 =u ¯(p, s0)(2ipa + 2mγa)u(p, s) or ipa u¯(p, s0)γau(p, s) = − u¯(p, s0)u(p, s). (4.238) m Using the formulas (1.273) for the spinors m − ip/ u(p, s) = u(0, s) p2p0(p0 + m) (4.239) m + ip/ v(p, s) = v(0, s), p2p0(p0 + m) we can compute the inner product

u¯(p, s0)u(p, s) = iu†(p, s0)γ0u(p, s)

i † 0 a† 0 = u (0, s )(m + ipaγ )γ (m − ip/)u(0, s) 2p0(p0 + m)

i † 0 0 a = u (0, s )γ (m − ipaγ )(m − ip/)u(0, s) 2p0(p0 + m) 1 = u¯(0, s0)(m − ip/)(m − ip/)u(0, s). (4.240) 2p0(p0 + m) The formula (4.238) for the terms linear in gamma tells us that

0 a a 0 0 0 u¯(0, s )paγ u(0, s) = −ipaδ0 u¯(0, s )u(0, s) = ip u¯(0, s )u(0, s). (4.241) 4.12 Magnetic moment of the electron 129 Also the anticommutation relation {γa, γb} = 2ηab tells us that

a b 1 a b ab 2 p/p/ = papbγ γ = papb{γ , γ } = papbη = −m . (4.242) 2 So we ﬁnd that the inner product is u¯(0, s0)u(0, s) m u¯(p, s0)u(p, s) = (2m2 + 2mp0) = u¯(0, s0)u(0, s). (4.243) 2p0(p0 + m) p0 The zero-momentum spinors are 1 0

1 1 0 1 1 1 u(~0, m = ) = √   and u(~0, m = − ) = √   , 2 2 1 2 2 0 0 1 (4.244)  0  −1

1 1  1  1 1  0  v(~0, m = ) = √   and v(~0, m = − ) = √   . 2 2  0  2 2  1  −1 0 So the inner product (4.243) is

0 m u¯(p, s )u(p, s) = δ 0 . (4.245) p0 ss Incidentally, SW’s pseudounitarity condition (1.209, SW’s 5.4.32) βD†(Λ)β = D−1(Λ) (4.246) leads to a much shorter derivation of this (4.245) result. Using the formulas (4.238 & 4.245), we can write the expression (4.235) for a matrix element of J a(0) as ie h ipa i i hp, s0|J a(0)|p, si = − u¯(p, s0) − F (0) − paG(0) u(p, s) (2π)3 m m epa h i = − u¯(p, s0) F (0) + G(0) u(p, s) (2π)3m a e p = − δ 0 F (0) + G(0) . (4.247) (2π)3 p0 ss Setting a = 0 and using the expression (4.231) for the matrix element of the charge density 0 0 0 3 e hp , s |J (0) d x|p, si = − δ 0 , (4.248) (2π)3 ss we get the normalization condition F (0) + G(0) = 1 (4.249) 130 Quantum Electrodynamics for these quantities when computed with counterterms. We can use Dirac’s equation (4.237) and the anticommutation relation (4.236) to get

u¯0[i(p + p0)a + 2mγa]u =u ¯0[i(p + p0)a − ip/0γa − iγap/]u 0 1 a 0 1 a 0 a a =u ¯ 2 i{γ , p/} + 2 i{γ , p/} − ip/γ − iγ p/ u 0 1 a b 0 =u ¯ 2 i[γ , γ ](p − p)b u. (4.250) We now can write the coeﬃcient of the ﬁrst term in the vertex (4.234) h i i u¯0Γa(p0, p)u =u ¯0 γaF (q2) − (p + p0)aG(q2) u (4.251) 2m as i h u¯0Γa(p0, p)u = − u¯0 (p + p0)a F (q2) + G(q2) 2m 1 i − [γa, γb](p0 − p) F (q2) u. (4.252) 2 b The generators of rotations and boosts in the Dirac representation of the Lorentz group are (1.175) for spatial values of i and j i i 0 σ 0 σ J ij = − [γi, γj] = − −i i , −i j 4 4 −σi 0 −σj 0 (4.253) 1 σk 0 = ijk 2 0 σk and i i 0 σ 0 1 J i0 = − [γi, γ0] = − −i i , −i 4 4 −σi 0 1 0 (4.254) i σ 0 = i . 2 0 −σi So their p = p0 = 0 matrix elements between the spinors (4.244) are

0 i j (1/2) u¯(0, s )[γ , γ ]u(0, s) = = 4iijk Jk = 2iijkσks0s s0s (4.255) u¯(0, s0)[γi, γ0]u(0, s) = 0.

So for small momenta and to lowest order in p and p0, the vertex (4.252) is

0 0 0 1 h 0 (1/2)i u¯(p , s )Γ(p , p)u(p, s) ≈ (p − p ) × J k,s0s F (0) m (4.256) 1 0 = (p − p ) × σ 0 F (0). 2m s s 4.12 Magnetic moment of the electron 131 In weak, static ﬁelds, the interaction hamiltonian is

Z H0 = − J(x) · A(x) d3x. (4.257)

So the interaction is

Z 0 0 0 ieF (0) i(p−p0)·x 0 3 hp , s |H |p, si = e A(x) · (p − p ) × σ 0 d x (2π)32m s s Z eF (0) i(p−p0)·x 3 = e B(x) · σ 0 d x (4.258) (2π)32m s s where B(x) = ∇ × A(x) is the magnetic ﬁeld. So for a slowly varying magnetic ﬁeld,

0 0 0 eF (0) 3 0 hp , s |H |p, si = B · σ 0 δ (p − p ). (4.259) 2m s s

So the magnetic moment is

e µ = − F (0). (4.260) 2m

The form factor G(q2) is

4π2e2 Z 1 Z x Z ∞ 4m2x(1 − x) G(q2) = − dx dy k3dk . (2π)4 h i3 0 0 0 k2 + m2x2 + q2y(x − y) (4.261) The k integral is ﬁnite:

2 2 Z 1 Z x 2 2 4π e 4m x(1 − x) G(q ) = − 4 dx dy h i (4.262) (2π) 0 0 4 m2x2 + q2y(x − y) e2m2 Z 1 Z x x(1 − x) = − 2 dx dy 2 2 2 . (4.263) 4π 0 0 m x + q y(x − y)

The magnetic moment of the electron to order e3 is

e µ = − 1 − G(0) (4.264) 2m 132 Quantum Electrodynamics where e2m2 Z 1 Z x x(1 − x) G(0) = − 2 dx dy 2 2 4π 0 0 m x e2m2 Z 1 Z x (1 − x) = − 2 dx dy 2 4π 0 0 m x e2m2 Z 1 (1 − x) e2 Z 1 = − 2 dx 2 = − 2 dx (1 − x) 4π 0 m 4π 0 e2 = − . (4.265) 8π2 So the absolute value of the magnetic moment of the electron to order e3 is e e2 e α e µ = 1 + = 1 + = 1.001161 (4.266) 2m 8π2 2m 2π 2m first calculated by Julian Schwinger in 1948. The 2018 experimental value by Gabrielse et al. (PRA 83, 052122 (2011)) is e µ = (1.00115965218073 ± 0.00000000000028) (4.267) exp 2m in which 73 could be 73 ± 28. Kinoshita and others have done QED calculations to 5 loops. Their best estimate (as of 2012) including all standard- model effects is (arXiv:1205.5368) e µ = 1.00115965218178 . (4.268) th 2m The standard model value of g − 2 agrees with experiment to 11 decimal digits e µ = µ = 1.00115965218 . (4.269) SM exp 2m Most of the theoretical uncertainty is due hadronic effects and to uncertainty in the experimental value of the fine-structure constant α.

4.13 Charge form factor of electron If we add in the contribution due to vacuum polaritzation, then the charge form factor of electron is 2 2 Z 1 Z x Z ∞ 2 2 4π e 3 F (q ) = π(q ) + 4 dx dy k dk (2π) 0 0 0 k2 − 2m2(x2 − 4x + 2) − 2q2[y(x − y) + 1 − x] × (4.270) h i3 k2 + m2x2 + q2y(x − y) 4.13 Charge form factor of electron 133 which diverges logarithmically. Let’s take the value of this quantity at q2 = 0 from experiment. Then the charge form factor is 2 2 Z 1 Z x Z ∞ 2 4π e 3 F (q ) = F (0) + 4 dx dy k dk (2π) 0 0 0 " k2 − 2m2(x2 − 4x + 2) − 2q2[y(x − y) + 1 − x] × h i3 k2 + m2x2 + q2y(x − y) # k2 − 2m2(x2 − 4x + 2) − . (4.271) h i3 k2 + m2x2

The k integral gives 2 2 Z 1 Z x 2 4π e F (q ) = F (0) + 4 dx dy (4.272) (2π) 0 0 − q2 − x3 + 2y2 − 2xy(1 + 2y) + x2(1 + 4y)

− m2x2 + q2x2(x − y)y ln(m2x2) + x2[m2x2 + q2(x − y)y] ln[m2x2 + q2(x − y)y] .h i 2x2(m2x2 + q2(x − y)y) (4.273) 5 Nonabelian gauge theory

5.1 Yang and Mills invent local nonabelian symmetry The action of a Yang-Mills theory is unchanged when a spacetime-dependent a unitary matrix U(x) = exp(−ita θ (x)) maps a vector ψ(x) of matter ﬁelds 0 † † † to ψi(x) = Uij(x)ψj(x). The symmetry ψ (x)U (x)U(x)ψ(x) = ψ (x)ψ(x) † i is obvious, but how can kinetic terms like ∂iψ ∂ ψ be made invariant? a Yang and Mills introduced matrices Ai = ta Ai of gauge ﬁelds in which the hermitian matrices ta are the generators of the Lie algebra of a compact gauge group and so obey the commutation relation

[ta, tb] = ifabctc (5.1) in which the fabc are real, totally antisymmetric structure constants. More importantly, they replaced ordinary derivatives ∂i by covariant derivatives a Diαβ = ∂iδαβ + Aiαβ ≡ ∂iδαβ + taαβ Ai (5.2) and required that covariant derivatives of fields transform like fields so that (Dψ)0 = UDψ or 0 0 0 0 D ψ = ∂i + Ai Uψ = ∂iU + U∂i + AiU ψ = U (∂i + Ai) ψ. (5.3) The nonabelian gauge field transforms as

0 † † Ai(x) = U(x)Ai(x)U (x) − (∂iU(x)) U (x). (5.4) In full detail, this is

0 † † Aiαδ(x) = Uαβ(x)Aiβγ(x)(U )γδ(x) − (∂iUαβ(x)) (U )βδ(x). (5.5) The nonabelian Faraday tensor is

Fik = [Di,Dk] = ∂iAk − ∂kAi + [Ai,Ak] (5.6) 5.2 SU(3) 135 in which both F and A are matrices in the Lie algebra. The nonabelian Faraday tensor transforms as

0 −1 −1 Fik(x) = U(x)FikU (x) = U(x)[Di,Dk]U (x). (5.7)

ik ik The trace Tr[FikF ] of the Lorentz-invariant product FikF is both Lorentz invariant and gauge invariant. The nonabelian generalization of the abelian action density 1 L = − F F jk − ψ¯[γk(∂ + ieA ) + m]ψ (5.8) 4 jk k k is 1 L = − TrF F jk − ψ¯[γk(∂ + ieA ) + m]ψ. (5.9) 4 jk k k In more detail, the Fermi action is ¯ k ¯ k a ψ[γ (∂k +ieAk)+m]ψ = ψ`α γ``0 (∂kδαβ +ietaαβAk)+mδ``0 δαβ ψ`0β. (5.10) Use of matrix notation and of summation conventions is necessary.

5.2 SU(3)

The gauge group of quantum chromodynamics is SU(3) which has eight generators. The Gell-Mann matrices are

0 1 0 0 −i 0 1 0 0 λ1 = 1 0 0 , λ2 = i 0 0 , λ3 = 0 −1 0 , 0 0 0 0 0 0 0 0 0 0 0 1 0 0 −i 0 0 0 λ4 = 0 0 0 , λ5 = 0 0 0  , λ6 = 0 0 1 , 1 0 0 i 0 0 0 1 0 0 0 0  1 0 0  1 λ7 = 0 0 −i , and λ8 = √ 0 1 0  . (5.11) 0 i 0 3 0 0 −2

The generators ta of the 3 × 3 deﬁning representation of SU(3) are these Gell-Mann matrices divided by 2

ta = λa/2 (5.12)

(Murray Gell-Mann, 1929–). 136 Nonabelian gauge theory

The eight generators ta are orthogonal with k = 1/2 1 Tr (t t ) = δ (5.13) a b 2 ab and satisfy the commutation relation

[ta, tb] = ifabc tc. (5.14) The trace formula i f c = − Tr [t , t ] t† . (5.15) ab k a b c gives us the SU(3) structure constants as

fabc = − 2i Tr ([ta, tb]tc) . (5.16) √ They are real and totally antisymmetric with f123 = 1, f458 = f678 = 3/2, and f147 = −f156 = f246 = f257 = f345 = −f367 = 1/2. While no two generators of SU(2) commute, two generators of SU(3) do. In the representation (5.11,5.12), t3 and t8 are diagonal and so commute

[t3, t8] = 0. (5.17) They generate the Cartan subalgebra of SU(3). The generators deﬁned by Eqs.(5.12 & 5.11) give us the 3 × 3 representation

D(α) = exp (iαata) (5.18) in which the sum a = 1, 2,... 8 is over the eight generators ta. This representation acts on complex 3-vectors and is called the 3. Note that if

D(α1)D(α2) = D(α3) (5.19) then the complex conjugates of these matrices obey the same multiplication rule ∗ ∗ ∗ D (α1)D (α2) = D (α3) (5.20) and so form another representation of SU(3). It turns out that (unlike in SU(2)) this representation is inequivalent to the 3; it is the 3. There are three quarks with masses less than about 100 MeV/c2—the u, d, and s quarks. The other three quarks c, b, and t are more massive; mc = 1.28 GeV, mb = 4.18 GeV, and mt = 173.1 GeV. Nobody knows why. Gell-Mann and Zweig suggested that the low-energy strong interactions were approximately invariant under unitary transformations of the three light quarks, which they represented by a 3, and of the three light antiquarks, which they represented by a 3. They imagined that the eight 5.2 SU(3) 137 light pseudoscalar mesons, that is, the three pions π−, π0, π+, the neutral 0 η, and the four kaons K0,K+,K− K , were composed of a quark and an antiquark. So they should transform as the tensor product 3 ⊗ 3 = 8 ⊕ 1. (5.21) They put the eight pseudoscalar mesons into an 8. They imagined that the eight light baryons — the two nucleons N and P , the three sigmas Σ−, Σ0, Σ+, the neutral lambda Λ, and the two cascades Ξ− and Ξ0 were each made of three quarks. They should transform as the tensor product 3 ⊗ 3 ⊗ 3 = 10 ⊕ 8 ⊕ 8 ⊕ 1. (5.22) They put the eight light baryons into one of these 8’s. When they were writing these papers, there were nine spin-3/2 resonances with masses somewhat heavier than 1200 MeV/c2 — four ∆’s, three Σ∗’s, and two Ξ∗’s. They put these into the 10 and predicted the tenth and its mass. In 1964, a tenth spin-3/2 resonance, the Ω−, was found with a mass close to their prediction of 1680 MeV/c2, and by 1973 an MIT-SLAC team had discovered quarks inside protons and neutrons. (George Zweig, 1937–) For a given quark, say the up quark, the action of quantum chromodynamics is 1 L = − TrF F jk − ψ¯[γk(∂ + ieA ) + m]ψ (5.23) 4 jk k k in which the index c on ψc takes on the values 1, 2, 3 which we may think of as red, green, and blue. The Faraday tensor is a 3 × 3 matrix

Fik = [Di,Dk] = ∂iAk − ∂kAi + [Ai,Ak]. (5.24) The trace term in the QCD action (5.23) is

jk h i k k i i k i Tr FjkF = Tr ∂iAk −∂kAi +[Ai,Ak] ∂ A −∂ A +[A ,A ] (5.25) Quarks and gluons are conﬁned in color-singlet particles. This eﬀect is robust and mysterious. 6 Standard model

6.1 Quantum chromodynamics If to a pure SU(3) gauge theory we add massless quarks in the fundamental or defining representation, then we get a theory of the strong interactions called quantum chromodynamics or QCD. Thus, let ψ be a complex 3-vector of Dirac fields   ψr ψ = ψg (6.1) ψb (so 12 fields in all). This complex 12-vector could represent u or “up” quarks. In terms of the covariant derivative 8 X b b Dµ = I∂µ + Aµ(x) = I∂µ + ig t Aµ(x), (6.2) b=1 the action density is 1 L = Tr [F F µν] − ψ (γµD + m) ψ (6.3) 2g2 µν µ where

Fµν = [Dµ,Dν]. (6.4) Nonperturbative effects are supposed to “confine” the quarks and massless gluons. There are 6 known “flavors” of quarks—u, d, c, s, t, b.

6.2 Abelian Higgs mechanism √ Suppose φ = (φ1 + iφ2)/ 2 is a complex scalar ﬁeld with action density ∗ a 2 ∗ ∗ 2 L = − ∂aφ ∂ φ + µ φ φ − λ(φ φ) . (6.5) 6.2 Abelian Higgs mechanism 139 The minimum of the potential V = − µ2φ∗φ + λ(φ∗φ)2 is given by the equation ∂V 0 = = −µ2 + 2λ|φ|2. (6.6) ∂|φ|2 It is a circle with µ2 v2 |φ|2 = ≡ . (6.7) 2λ 2 √ As is customary, we pick one asymmetrical minimum at φ = µ/ 2λ, so the mean values in the vacuum are r µ2 h0|φ |0i = v = and h0|φ |0i = 0. (6.8) 1 λ 2 The U(1) symmetry has been spontaneously broken. So v φ φ φ = √ + √1 + i√2 (6.9) 2 2 2 where now φ1 is the departure of the real part of the ﬁeld from its mean value v. So the potential V is V = − µ2φ∗φ + λ(φ∗φ)2

1 λ 2 = − µ2 (v + φ )2 + (φ )2 + (v + φ )2 + (φ )2 2 1 2 4 1 2 (6.10) µ4 = − + µ2φ2 + ... 4λ 1 where the dots denote cubic and quartic in φ1 and φ2. The quadratic part of the action determines the properties of the particles of the physical fields. The cubic and higher-order terms tell us how the particles interact with each other. The first part of the last homework problem is to find the masses of the particles of this theory. Scalar particles of zero mass that arise from spontaneous symmetry breaking are called Goldstone bosons. Now we give the theory a local U(1) symmetry by adding an abelian gauge field to the action density 1 L = − (D φ)∗Daφ + µ2φ∗φ − λ(φ∗φ)2 − F F ab (6.11) a 4 ab in which

Da = ∂a + ieAa and Fab = [Da,Db]. (6.12) 140 Standard model Once again the field assumes a mean value in the vacuum, let us say (6.9). But we write the field φ as 1 φ = √ (v + φ1 + iφ2). (6.13) 2 ∗ a The term −(Daφ) D φ then is ∗ a ∗ ∗ a a −(Daφ) D φ = − (∂aφ − ieAaφ )(∂ φ + ieA φ) 1 = − [∂ φ − i∂ φ − ieA (v + φ − iφ )] 2 a 1 a 2 a 1 2 a a a × [(∂ φ1 + i∂ φ2 + ieA (v + φ1 + iφ2)] (6.14) 1 = − (∂aφ − eAaφ )(∂ φ − eA φ ) 2 1 2 a 1 a 2 1 − (∂aφ + eAa(v + φ ))(∂ φ + eA (v + φ )). 2 2 1 a 2 a 1 So we set 1 1 B = A + ∂ φ or A = B − ∂ φ (6.15) a a ev a 2 a a ev a 2 and find that 1 φ φ −(D φ)∗Daφ = − (∂aφ − eBaφ + 2 ∂aφ )(∂ φ − eB φ + 2 ∂ φ ) a 2 1 2 v 2 a 1 a 2 v a 2 1 φ φ − (evBa + eBaφ − 1 ∂aφ )(evB + eB φ − 1 ∂ φ ) 2 1 v 2 a a 1 v a 2 1 − ∂aφ ∂ φ + e2v2BaB + ... (6.16) 2 1 a 1 a in which the dots denote cubic and quartic terms. The second part of the last homework problem is to find the masses of the particles of this theory.

6.3 Higgs’s mechanism The local gauge group of the Glashow-Salam-Weinberg electroweak theory is SU(2)` × U(1). What’s weird is that the SU(2)` symmetry applies only to the left-handed quarks and leptons and to the Higgs boson, a complex doublet (or 2-vector) of scalar fields H. Let’s leave out the fermions for the moment, and focus just on the Higgs and the gauge fields. The gauge transformation is H0(x) = U(x)H(x) (6.17) 0 † † Aµ(x) = U(x)Aµ(x)U (x) + (∂µU(x))U (x) 6.3 Higgs’s mechanism 141 in which the 2 × 2 unitary matrix U(x) is σa YI U(x) = exp i g αa(x) + i g0 β(x) . (6.18) 2 2 The generators here are the 3 Pauli matrices and the matrix I/2, where I is the 2 × 2 identity matrix. The action density of the theory (without the fermions) is 1 L = − (D H)†DµH + Tr(F F µν) + m2|H|2 − λ|H|4 (6.19) µ 4k µν in which the covariant derivative for the Higgs doublet is σa YI D H = I∂ + i g Aa + i g0 B H. (6.20) µ µ 2 µ 2 µ The potential energy of the Higgs field is V = − m2|H|2 + λ|H|4. (6.21) Its minimum is where ∂V 0 = = 2λ|H|2 − m2. (6.22) ∂|H|2 So m v |H| = √ = √ . (6.23) 2λ 2

By making an SU(2)` × U(1) gauge transformation, we can transform this mean value to 1 0 H0 = h0|H(x)|0i = √ . (6.24) 2 v This transformation is called “going to unitary gauge.” In this gauge, the Higgs potential is 1 1 V (v) = − m2v2 + λv4, (6.25) 2 4 and its second derivative is 00 2 2 2 2 V (v) = 3λv − m = 2m = mH . (6.26) The mass of the Higgs then is √ √ mH = 2 m = 2 λ v. (6.27) Experiments at LEP2 and at the LHC have revealed value of v to be v = 246.22 GeV. (6.28) 142 Standard model In 2012, the Higgs boson was discovered at the LHC. Its mass has been measured to be

mH = 125.18 ± 0.16 GeV. (6.29) The self coupling λ therefore is m2 1 125.182 λ = H = = 0.12924. (6.30) 2 v2 2 246.22 After spontaneous symmetry breaking, mass terms for the gauge bosons † µ emerge from the kinetic action of the Higgs doublet − (DµH) D H. Since the generators of a compact group like SU(2)` × U(1) are hermitian, the part of the kinetic action that contains the mass terms is σa YI σa YI L = − H† −i g Aa − i g0 B i g Aa + i g0 B H. m 2 µ 2 µ 2 µ 2 µ (6.31) In unitary gauge (6.24), these mass terms are 1 σa I σa I 0 L = − (0, v) g Aa + g0 B g Aµ + g0 Bµ m 2 2 µ 2 µ 2 a 2 v 3 0 1 2 1 gAµ + g Bµ g(Aµ − iAµ) = − (0, v) 1 2 3 0 8 g(Aµ + iAµ) −gAµ + g Bµ µ 0 µ µ µ gA3 + g B g(A1 − iA2 ) 0 × µ µ µ 0 µ g(A1 + iA2 ) −gA3 + g B v 2 µ µ v 1 2 3 0 g(A1 − iA2 ) = − g(Aµ + iAµ), −gAµ + g Bµ µ 0 µ 8 −gA3 + g B v2 = − g2 A1 Aµ + A2 Aµ + −gA3 + g0B −gAµ + g0Bµ . 8 µ 1 µ 2 µ µ 3 (6.32) The normalized complex, charged gauge bosons are 1 W ± = √ A1 ∓ iA2 (6.33) µ 2 µ µ and the normalized neutral ones are

1 3 0 Zµ = gAµ − g Bµ (6.34) pg2 + g02 and the photon, which is the normalized linear combination orthogonal to Zµ 1 0 3 Aµ = g Aµ + g Bµ . (6.35) pg2 + g02 6.3 Higgs’s mechanism 143 It remains massless. In terms of these properly normalized fields, the mass terms are g2v2 (g2 + g02)v2 L = − W −W +µ − Z Zµ. (6.36) M 4 µ 8 µ So the W + and the W − get the same mass v M = g = 80.370 ± 0.012 GeV/c2. (6.37) W 2 while the Z (also called the Z0) has mass p v M = g2 + g02 = 91.1876 ± 0.0021 GeV/c2. (6.38) Z 2 The accuracy and precision of the mass of the Z boson is an order of magnitude higher than that of the W boson because of measurements made in the Large Electron Collider. It is the photon. The fine-structure constant is e2 α = = 1/137.035 999 139(31) ≈ 1/137.036. (6.39) 4π~c Why do three gauge bosons become massive? Because there are three Goldstone bosons corresponding to three ways of moving h0|H|0i without changing the Higgs potential. Why does one gauge boson stay massless? Because one linear combination of the generators of SUL(2) ⊗ U(1) maps h0|H|0i to zero. In terms of these mass eigenstates, the original gauge bosons are 1 A1 = √ W + + W − µ 2 µ µ 1 A2 = √ W − − W + µ i 2 µ µ (6.40) 3 1 0 Aµ = g Aµ + gZµ pg2 + g02 1 0 Bµ = gAµ − g Zµ . pg2 + g02 The 4 × 4 matrices 1 1 0 1 0 0 P = 1 + γ5 = and P = 1 − γ5 = (6.41) ` 2 0 0 r 2 0 1 project out the upper and lower two components of a 4-component Dirac field. The SU(2)` gauge fields A~µ interact only with the upper two components ψ` = P`ψ, while the U(1)Y gauge field Bµ interacts with all four components of a Dirac field. 144 Standard model Thus the covariant derivative for a fermion of coupling g0 to the U(1) gauge field Bµ with hypercharge Y and coupling g to the SU(2)` gauge fields is σa D = I∂ + i g Aa P + i g0 YIB µ µ 2 µ ` µ σ1 1 + − σ2 1 − + = I∂µ + i g √ W + W + √ W − W (6.42) 2 2 µ µ 2 i 2 µ µ # σ3 1 0 0 I 0 + g Aµ + gZµ P` + i g Y gAµ − g Zµ . 2 pg2 + g02 pg2 + g02

The SU(2)` generators are 1 1 σ T~ = ~σ, T ± = (σ ± iσ ) and T 3 = 3 . (6.43) 2 2 1 2 2 In terms of them, the covariant derivative is

g + + − − 1 2 3 02 Dµ = I∂µ + i √ Wµ T + Wµ T P` + i Zµ g T P` − g YI 2 pg2 + g02 0 gg 3 + i Aµ T P` + YI (6.44) pg2 + g02 Equivalently, the left and right covariant derivatives are

h g + + − − 1 2 3 02 Dµ` = I∂µ + i √ Wµ T + Wµ T + i Zµ g T − g YI 2 pg2 + g02 0 gg 3 i + i Aµ T + YI P` (6.45) pg2 + g02 and h g02YI gg0YI i Dµr = I∂µ − i Zµ + i Aµ Pr. (6.46) pg2 + g02 pg2 + g02

The interaction strength or coupling constant of the photon Aµ is gg0 e = > 0. (6.47) pg2 + g02 The charge operator is 3 Q = T` + YI. (6.48) When acting on the doublet ν E = e (6.49) ` e 6.3 Higgs’s mechanism 145 to which we assign Y = −1/2, the charge Q gives 0 as the charge of the neutrino and − 1 as the charge of the electron. The photon-lepton term then is 0 gg 3 3 Aµ T + YI E` = e Aµ T + YI E` pg2 + g02 (6.50) 0 0 ν = e A e = −e A e µ 0 −1 e µ in which the first e is the absolute value of the charge (6.47) of the electron and the second is the field of the electron. The weak mixing angle θw is defined by Z cos θ − sin θ A3 = w w . (6.51) A sin θw cos θw B Equations (6.34 and 6.35) identify these trigonometric values as g g0 cos θw = and sin θw = . (6.52) pg2 + g02 pg2 + g02 Since the charge is Q = T 3 + YI, the hypercharge YI = Q − T 3, and so fields couple to the Z as

1 2 3 02 1 2 02 3 02 Zµ g T − g Y = Zµ (g + g )T − g Q (6.53) pg2 + g02 pg2 + g02 and to the photon as 0 0 gg 3 gg Aµ T + Y = Aµ Q. (6.54) pg2 + g02 pg2 + g02 We also have 2 02 g + g p 2 02 g p = g + g = (6.55) g2 + g02 cos θw and 02 02 g p 2 02 g g 2 p = g + g 2 02 = sin θw. (6.56) g2 + g02 g + g cos θw So the coupling to the Z is

1 2 02 3 02 g 3 2 p Zµ (g + g )T − g Q = Zµ T − sin θw Q (6.57) g2 + g02 cos θw and, if we set

e = g sin θw, (6.58) 146 Standard model then the coupling to the photon A is gg0 Aµ Q = g sin θw Aµ Q = e Aµ Q. (6.59) pg2 + g02 In these terms, the covariant derivative (6.44) is

g + + − − 1 2 3 02 Dµ = I∂µ + i √ Wµ T + Wµ T P` + i Zµ g T P` − g Y 2 pg2 + g02 0 gg 3 + i Aµ T P` + Y pg2 + g02 g + + − − g 3 2 = I∂µ + i √ Wµ T + Wµ T P` + i Zµ T P` − sin θw Q 2 cos θw

+ ie Aµ Q (6.60) in which the generators T ± and T 3 are those of the representation to which the ﬁelds they act on belong. When acting on left-handed fermions, they 1 are half the Pauli matrices, T = 2 σ. When acting on right-handed fermions, they are zero, T = 0, and so the explicit appearance of P` is unnecessary. Since g = e/ sin θw, the couplings involve one new parameter θw. Our mass formulas (6.37 and 6.38) for the W and the Z show that their masses are related by

v g p 2 02 v p 2 02 v MW = g = g + g = cos θw g + g = cos θw MZ . 2 pg2 + g02 2 2 (6.61) Experiments have determined the masses and shown that at the mass of the Z and in the MS convention 2 sin θw = 0.23122(4) or θw = 0.50163 (6.62) and that v = 246.22 GeV. (6.63)

6.4 Quark and lepton interactions

The right-handed fermions ur, dr, er, and νe,r are singlets under SUL(2) ⊗ 3 UY (1). So they have T = 0. The deﬁnition (6.48) of the charge Q Q = T 3 + YI (6.64) then implies that

Yr = Qr. (6.65) 6.4 Quark and lepton interactions 147

That is, Yνe,r = 0, Yer = −1, Yur = 2/3, and Ydr = −1/3. The left-handed fermions are in doublets ν u E = e and Q = (6.66) ` e− ` d

3 with T = ±1/2. So the choices YE = −1/2 and YQ = 1/6 and the deﬁnition (6.48) of the charge Q give the right charges:

ν 0 u 2u/3 QE = Q e = and QQ = Q = . (6.67) ` e− −e− ` d −d/3

Fermion-gauge-boson interactions are due to the covariant derivative (6.60) acting on either the left- or right-handed ﬁelds. On right-handed fermions, the covariant derivative is just

r g 2 Dµ = I∂µ + i Zµ − sin θw Q + ie Aµ Q cos θw 2 sin θw = I∂µ − ig Zµ Q + ie Aµ Q cos θw sin θw = I∂µ − ie Zµ Q + ie Aµ Q = I∂µ − ie tan θw Zµ Q + ie Aµ Q. cos θw (6.68) So the covariant derivative of a neutral right-handed fermion is just the ordinary derivative. On left-handed fermions, the covariant derivative is

` g + + − − g 3 2 Dµ = I∂µ + i√ Wµ T + Wµ T + i Zµ T − sin θw Q + ieAµQ 2 cos θw 3 e + + − − e T = I∂µ + i √ Wµ T + Wµ T + i Zµ − sin θw Q 2 sin θw cos θw sin θw + ie Aµ Q. (6.69) For the ﬁrst family or generation of quarks and leptons, the kinetic action density is

` r ` r Lk = − E`D/ E` − ErD/ Er − Q`D/ Q` − QrD/ Qr (6.70)

µ 5 in which D/ ≡ γ Dµ. The 4 × 4 matrix γ5 = γ plays the role of a ﬁfth 4 (spatial) gamma matrix γ = γ5 in 5-dimensional space-time in the sense that the anticommutator {γa, γb} = 2ηab (6.71) 148 Standard model in which η is the 5×5 diagonal matrix with η00 = −1 and ηaa = 1 for a = 1, 2, 3, 4. In Weinberg’s notation, γ5 is 1 0 γ = . (6.72) 5 0 −1 The combinations 1 1 P = (1 + γ ) and P = (1 − γ ) (6.73) ` 2 5 r 2 5 are projection operators onto the left- and right-handed ﬁelds. That is,

P`Q = Q` and P`Q` = Q` (6.74) with a similar equation for Pr. We can write Lk as ` r ` r Lk = − ED/ P`E − ED/ PrE − QD/ P`Q − QD/ PrQ h ` r ` r i (6.75) = − ED/ P`E + ED/ PrE + QD/ P`Q + QD/ PrQ . Homework set 4, problem 1: Show that / ` 1 † 0 / ` 1 / ` 1 E`D E` = [ 2 (1 + γ5)E] iγ D 2 (1 + γ5)E = ED 2 (1 + γ5)E. (6.76) Recall that in Weinberg’s notation 0 I ψ = ψ†iγ0 = ψ†β = ψ† (6.77) I 0 in which I is the 2 × 2 identity matrix.

6.5 Quark and Lepton Masses The Higgs mechanism also gives masses to the fermions, but somewhat ar- bitrarily. Dirac’s action density (4.56) has as its mass term † 0 † 0 † 0 2 2 −m ψψ = − im ψ γ ψ = −im ψ γ (P` + Pr)ψ = −im ψ γ (P` + Pr )ψ. (6.78) Since {γ0, γ5} = 0, this mass term is † 0 † 0 −m ψψ = − im ψ Prγ P`ψ − im ψ P`γ Prψ † 0 † 0 = − im (Prψ) γ P`ψ − im (P`ψ) γ Prψ (6.79) † 0 † 0 = − im ψrγ ψ` − im ψ` γ ψr = −m ψrψ` − m ψ`ψr.

Incidentally, because the fields ψ` and ψr are independent, we can redefine them as 0 iθ 0 iφ ψ` = e ψ` and ψr = e ψr (6.80) 6.5 Quark and Lepton Masses 149 at will. Such a redefinition changes the mass term to

0 0∗ i(θ−φ) −i(θ−φ) − m ψrψ` − m ψ`ψr = −m e ψrψ` − m e ψ`ψr. (6.81) So the phase of a Dirac mass term has no signiﬁcance. The deﬁnition (6.77) of ψ shows that the Dirac mass term is † † † 0 I ψ` † † − mψψ = −m ψ βψ = −m ψ` , ψr = −m ψ` ψr + ψrψ` . I 0 ψr (6.82) These mass terms are invariant under the Lorentz transformations 0 0 ∗ ψ = exp(−z · σ) ψ` and ψ = exp(z · σ) ψr ` r (6.83) because 0† 0 † ∗ ∗ † ψ ψ = ψ exp(−z · σ) exp(z · σ) ψr = ψ ψr ` r ` ` (6.84) 0† 0 † † ψr ψ` = ψr exp(z · σ) exp(−z · σ) ψ` = ψr ψ`.

They are not invariant under rigid, let alone local, SU(2)` transformations. But we can make them invariant by using the Higgs ﬁeld H(x). For † instance, the quantity Q`Hdr is invariant under local SU(2)` transformations. In unitary gauge, its mean value in the vacuum is

† 1 † h0|Q Hdr|0i = √ v h0|d dr|0i. (6.85) ` 2 ` So the term † ∗ † † − cd Q`Hdr − cd drH Q` (6.86) is invariant, and in the vacuum it is

† ∗ † † 1 † ∗ † h0| − cd Q Hdr − c d H Q`|0i = √ v h0| − cd d dr − c d d`|0i (6.87) ` d r 2 ` d r which gives to the d quark the mass

|cd| md = √ v. (6.88) 2

Note that we must add one new parameter cd to get one new mass md. This parameter cd is complex in general, but the mass md depends only upon the absolute value and not upon its phase of cd. Similarly, the term

† ∗ † † − ce E` Her − ce erH E` (6.89) 150 Standard model is invariant, and in the vacuum it is

† ∗ † † 1 † ∗ † h0| − ce E Her − c e H Q`|0i = √ v h0| − ce e er − c e e`|0i (6.90) ` e r 2 ` e r which gives to the electron the mass

|ce| me = √ v. (6.91) 2

Again, we must add one new (complex) parameter ce to get one new mass me. The mass of the up quark requires a new trick. The Higgs ﬁeld H transforms under SU(2)` × U(1) as σa I H0(x) = exp i g αa(x) + i g0 β(x) H(x). (6.92) 2 2 If for clarity, we leave aside the U(1) part for the moment, then the Higgs ﬁeld H transforms under SU(2)` as σa H0(x) = exp i g αa(x) H(x). (6.93) 2

∗ † Let us use H to be the complex column vector whose components are H1 † ∗ and H2. How does σ2H transform under SU(2)`? Suppressing our explicit mention of the space-time dependence and using the asterisk to mean hermitian conjugation when applied to operators but complex conjugation when 2 applied to matrices and vectors, we have, since σ2 is imaginary with σ2 = I while σ1 and σ3 are real, σa ∗ σ∗ (σ H∗)0 = σ exp i g αa H = σ exp −i g a αa H∗ 2 2 2 2 2 (6.94) σ∗ σa = σ exp −i g a αa σ σ H∗ = exp i g αa σ H∗. 2 2 2 2 2 2 Thus, the term † ∗ ∗ † T − cuQ`σ2H ur − cuurH σ2Q` (6.95) is invariant under SU(2)`. In the vacuum of the unitary gauge, it is

|cu| mu = √ v. (6.97) 2 But there are three families of generations of quarks and leptons on which the gauge fields act simply: 0 0 0  u   c   t   d   s   b  F1 =   ,F2 =   , and F3 =   . (6.98) νe νµ ντ  e µ τ The quark and lepton flavor families are 0 0 0 u c t Q0 = ,Q0 = , and Q0 = ; 1 d 2 s 3 b 0 0 0 (6.99) ν ν ν E0 = e ,E0 = µ , and E0 = τ . 1 e 2 µ 3 τ These are called the flavor eigenstates or more properly flavor eigenfields, designated here with primes. They are the ones on which the W ± act simply. − − 0 0 0 0 The weak interactions use Wµ T to map the flavor up fields u1 = u , u2 = c , 0 0 0 0 0 0 0 0 + + u3 = t into the flavor down fields d1 = d , d2 = s , d3 = b , and Wµ T to 0 0 map the flavor down fields di into the flavor up fields ui. The action density

3 X 0† 0 ∗ 0† † 0 −cdij Q`iHdrj − cdij drjH Q`i (6.100) i,j=1 gives for the d0, s0, and b0 quarks the mixed mass terms

3 3 v X 0† 0 ∗ 0† 0 v X 0† 0 ∗ 0† 0 √ −cdij dìdrj − cdij drjdì = √ −cdij dìdrj − cdij drjdì. 2 i,j=1 2 i,j=1 (6.101)

The 3 × 3 mass matrix Md with entries v [Md]ij = √ cdij (6.102) 2 need have no special properties. It need not be hermitian because for each i and j, the term (6.101) is hermitian. But every 3 × 3 complex matrix has a singular-value decomposition † Md = LdΣdRd (6.103) 152 Standard model in which Ld and Rd are 3 × 3 unitary matrices, and Σd is a 3 × 3 diagonal matrix with nonincreasing positive singular values on its main diagonal. The singular value decomposition works for any N × M (real or) complex matrix. Every complex M × N rectangular matrix A is the product of an M × M unitary matrix U, an M × N rectangular matrix Σ that is zero except on its main diagonal which consists of its nonnegative singular values † Sk, and an N × N unitary matrix V A = U Σ V †. (6.104) This singular-value decomposition is a key theorem of matrix algebra. One can use the Matlab command “[U,S,V] = svd(A)” to perform the svd A = USV †. The singular values of Σd are the masses mb, ms, and md:   mb 0 0 Σd =  0 ms 0  . (6.105) 0 0 md So the mass eigenﬁelds of the left and right down-quark ﬁelds are

† 0 † 0† † dri = Rdijdrj and d`i = d`jLdji or d`i = Ldijd`j. (6.106) The inverse relations are

0 0† † † 0 dri = Rdijdrj and d`i = d`jLdji or d`i = Ldijd`j (6.107) or in matrix notation

0 0† † † 0† † † 0 dr = Rd dr, dr = dr Rd, d` = d`Ld, and d` = Ld d` (6.108) in which b d` = s (6.109) d ` are the down-quark ﬁelds of deﬁnite masses. Similarly, the up quark action density

3 X 0† ∗ 0 ∗ 0† T 0 −cuijQ`iσ2H urj − cuijurjH σ2Q`i (6.110) i,j=1 gives for the three known families the mixed mass terms

3 3 iv X 0† 0 ∗ 0† 0 iv X 0† 0 ∗ 0† 0 √ cuijuì urj − cuijurj uì = √ cuijuì urj − cuijurj uì. (6.111) 2 i,j=1 2 i,j=1 6.6 CKM matrix 153

The 3 × 3 mass matrix Mu with entries iv [Mu]ij = √ cuij (6.112) 2 need have no special properties. It need not be hermitian because for each i and j, the term (6.111) is hermitian. But every 3 × 3 complex matrix Mu has a singular value decomposition

† Mu = LuΣuRu (6.113) in which Lu and Ru are 3 × 3 unitary matrices, and Σu is a 3 × 3 diagonal matrix with nonincreasing positive singular values on its main diagonal. These singular values are the masses mt, mc, and mu:   mt 0 0 Σu =  0 mc 0  . (6.114) 0 0 mu So the mass eigenﬁelds of the left and right up-quark ﬁelds are

† 0 † 0† † 0 uri = Ruijurj and u`i = u`jLuji or u`i = Luiju`j. (6.115) The inverse relations are

0 0† † † 0 uri = Ruijurj and u`i = u`jLuji or u`i = Luiju`j (6.116) or in matrix notation

0 0† † † 0† † † 0 ur = Ru ur, ur = ur Ru u` = u`Lu, and u` = Luu` (6.117) in which t u` = c (6.118) u ` are the up-quark ﬁelds of deﬁnite masses.

6.6 CKM matrix We will use the labels u, c, t and d, s, b for the states that are eigenstates of the quadratic part of the hamiltonian after the Higgs mechanism has given a mean value to the real part of the neutral Higgs boson in the unitary gauge. The u, c, t quarks have the same charge 2e/3 > 0 and the same T 3 = 1/2, so they all have the same electroweak interactions. Similarly, the d, s, b quarks have the same charge −e/3 < 0 and the same T 3 = −1/2, so they also all have the same electroweak interactions. 154 Standard model The right-handed covariant derivative (6.68)

r Dµ = I∂µ − ie tan θw Zµ Q + ie Aµ Q (6.119) just sends the fields of these mass eigenstates into themselves multiplied by their charge and either a Z or a photon. That is, 0† r 0 † † r † r ur Dµur = urRuDµRuur = urDµur (6.120) 0† r 0 † † r † r dr Dµdr = drRdDµRddr = drDµdr In these terms, the interactions of the Z and the photon with the right- handed fields are diagonal both in mass and in flavor. But the left-handed covariant derivative (6.69) is

` e + + − − Dµ = I∂µ + i √ Wµ T + Wµ T 2 sin θw (6.121) e T 3 +i Zµ − sin θw Q + ie Aµ Q. cos θw sin θw So we have 0 0† 0† ` u` † † † † ` Lu u` u` d` Dµ 0 = u`Lu d`Ld Dµ . (6.122) d` Ld d` † † Some of the unitary matrices just give unity, LuLu = I and LdLd = I like † † RuRu = I and RdRd = I in the right-handed covariant derivatives (6.120). Thus the interactions of the Z and the photon with the both the right- handed fields and with the left-handed fields are diagonal both in mass and in flavor. The Z and the photon do not mediate top-to-charm or charm-to- up or µ− → e− + γ decays. Also, the Higgs mass terms are diagonal, so the neutral Higgs boson can’t mediate such processes. Thus, in the standard model, there are no flavor-changing neutral-currents. The only changes are in the nonzero parts of T ± which become † + 0 LuLd 0 V − 0 0 0 0 Tckm = ≡ and Tckm = † ≡ † 0 0 0 0 LdLu 0 V 0 (6.123) † in which the unitary matrix V = LuLd is the CKM matrix (Nicola Cabibbo, Makoto Kobayashi, and Toshihide Maskawa). The left-handed covariant derivative on the mass eigenfields then is

` e + + − − Dµ = I∂µ + i √ Wµ Tckm + Wµ Tckm 2 sin θw (6.124) e T 3 +i Zµ − sin θw Q + ie Aµ Q. cos θw sin θw 6.6 CKM matrix 155 It has a second part that acts more or less like the right-handed covariant − − derivative, but the first part uses Wµ T to map the up fields u, c, t into + + linear combinations of the down fields d, s, b and Wµ T to map the down ± fields into linear combinations of the up fields. The Wµ terms are sensitive † to the CKM matrix V = LuLd. We write them suggestively as   u c   +   † 0 VWµ t u c t d s b † −   (6.125) V Wµ 0 d   s b   u †  c   d   +   0 Wµ  t  = u c t V s −    . Wµ 0  d  b   V s b (6.126)

By choosing the phases of the six quark fields, that is, u(x) → eiθu u(x) iθ † ... b(x) → e b b(x), one may make the CKM matrix LuLd real apart from a single phase. The existence of that phase probably is the cause of most of the breakdown of CP invariance that Fitch and Cronin and others have observed since 1964. The magnitudes of the elements of the CKM matrix V are     |Vud| |Vus| |Vub| 0.97427 0.22536 0.00355 V = |Vcd| |Vcs| |Vcb| = 0.22522 0.97343 0.0414  . (6.127) |Vtd| |Vts| |Vtb| 0.00886 0.0405 0.99914 Although there is only one phase exp(iδ) in the CKM matrix V , the experimental constraints on this phase often are expressed in terms of the angles α, β, and γ defined as ∗ ∗ α = arg [−VtdVtb/ (VudVub)] ∗ ∗ β = arg [−VcdVcb/ (VtdVtb)] (6.128) ∗ ∗ γ = arg [−VudVub/ (VcdVcb)] . If V is unitary, then α + β + γ = 180◦. From B → ππ, ρπ, and ρρ decays, the limits on the angle α are roughly α = (85.4 ± 4)◦. (6.129) 156 Standard model From B± → DK± decays, the limits on the angle γ are roughly γ = (68.0 ± 8)◦. (6.130) So the angle β is about 26.6◦. One of the quark-Higgs interactions is √ √ 0† 0 2 0† 0 2 0† † 0 −cdijQìHdrj = − Q` MddrH = − Q` LdΣdRddrH √v √v 2 2 0 = − Q†Σ d H = − Q† √ Σ d (6.131) v ` d r v ` (v + h)/ 2 d r h h = − d† 1 + Σ d = −m d† 1 + d . ` v d r di ì v ri A similar term describes the coupling of the up quarks to the Higgs h − m u† 1 + u . (6.132) ui ì v ri Thus, the rate of quark-antiquark to Higgs is proportional to the mass of the quark in the standard model.

6.7 Lepton Masses We can treat the leptons just like the quarks. The up leptons are the flavor 0 0 0 neutrinos νe, νµ, and ντ , and the down leptons are the flavor charged leptons e0, µ0, and τ 0. The action density 3 X 0† 0 ∗ 0† † 0 −ceij EìHerj − ceij erjH Eì (6.133) i,j=1 gives for the e0, µ0, and τ 0 the mixed mass terms 3 v X 0† 0 ∗ 0† 0 √ −ceij eìerj − ceij erjeì. (6.134) 2 i,j=1

The 3 × 3 mass matrix Me with entries v [Me]ij = √ ceij (6.135) 2 has a singular value decomposition † Me = LeΣeRe (6.136) in which Le and Re are 3 × 3 unitary matrices, and Σe is a 3 × 3 diagonal matrix with nonincreasing positive singular values mτ , mµ, and me on its main diagonal. 6.8 Before and after symmetry breaking 157 6.8 Before and after symmetry breaking Before spontaneous symmetry breaking, all the fields of the standard model are massless, and the local symmetry under SU(2)` ⊗ U(1) is exact. Under these gauge transformations, the left-handed electron and neutrino fields are 0 0 rotated among themselves. If e` is a linear combination of itself and of νe`, 0 0 then these two fields, e` and νe`, must be of the same kind. The left-handed electron field is a Dirac field. Thus, the left-handed neutrino field also must be a Dirac field. This makes sense because before symmetry breaking, all the fields are massless, and so there is no problem combining two Majorana fields of the same mass, namely zero, into one Dirac field. Thus, there are 0 0 0 three flavor Dirac neutrino fields νe`, νµ`, and ντ`. 0 A massless left-handed neutrino field ν` satisfies the two-component Dirac equation 0 (∂0I − ∇ · σ) ν`(x) = 0 (6.137) which in momentum space is 0 (E + p · σ) ν`(p) = 0. (6.138) Since the angular momentum is J = σ/2, and E = |p|, we have 1 pˆ · J ν0(p) = − ν0(p). (6.139) ` 2 ` 0 The left-handed neutrino field ν` annihilates neutrinos of negative helicity and creates antineutrinos of positive helicity. Since the neutrinos are massive, there may be right-handed neutrino fields. As for the up quarks, we can use them to make an action density 3 X 0† ∗ 0 ∗ 0† T 0 −cνijEìσ2H νrj − cνijνrjH σ2Eì (6.140) i,j=1 that is invariant under SU(2)` ⊗ U(1) and that gives for the neutrinos the mixed mass terms 3 X iv 0† 0 ∗ 0† 0 √ (cνijνìνrj − cνijνrjνì). (6.141) i,j=1 2

The 3 × 3 mass matrix Mν with entries iv [Mν]ij = √ cνij (6.142) 2 has a singular value decomposition † Mν = LνΣνRν (6.143) 158 Standard model in which Lν and Rν are 3 × 3 unitary matrices, and Σν is a 3 × 3 diagonal matrix with nonincreasing positive singular values mντ , mνµ , and mνe on its main diagonal (here, I have assumed that the neutrino masses mimic those of the charged leptons and quarks, rising with family number). The neutrino † CKM matrix then would be Lν Le, but since we are accustomed to treating the charged leptons as flavor and mass eigenfields, we apply the neutrino CKM matrix to the neutrinos rather than to the charged leptons. Thus the neutrino CKM matrix is † Vν = Le Lν. (6.144) By choosing the phases of the six lepton fields, we can make the neutrino CKM matrix real except for CP -breaking phases. If the neutrinos are Dirac fields, then there is one such phase; if not, there are three. So far, I have assumed that the mass terms for the neutrinos are the usual 0 Dirac mass terms. However, the right-handed Majorana neutrino fields νr are not affected by the SU(2)` ⊗ U(1). Note that a gauge transformation between e and νe rotates the operators a(p, s, e) and a(p, s, νe) into each other. This rotation makes sense only when the two particles have the same mass. In the standard model, such a gauge transformation makes sense only before symmetry breaking when all the particles are massless. Moreover, only when the particles are massless can one say that they are left- or right-handed. While the particles are massless, the operator a(p, −) annihilates a particle of negative helicity and occurs only in a left-handed field, while the operator a(p, +) annihilates a particle of positive helicity and occurs only in a right-handed field. But when the 1 particles are massive, the operator a(p, 2 ) annihilates a particle that is spin up and occurs in both left-handed and right-handed fields. So a symmetry 1 transformation that acted on the operator a(p, 2 ) would change both left- handed and right-handed fields. The left-handed fields of the neutrino and electron are Z a1(p, −, νe) + ia2(p, −, νe) ipx νe,`(x) = u(p, −) √ e (6.145) 2 a†(p, +, ν ) + ia†(p, +, ν ) d3p +v(p, +) 1 e √ 2 e e−ipx (6.146) 2 (2π)3/2 Z a1(p, −, e) + ia2(p, −, e) ipx e`(x) = u(p, −) √ e (6.147) 2 a†(p, +, e) + ia†(p, +, e) d3p +v(p, +) 1 √ 2 e−ipx (6.148) 2 (2π)3/2 where (p, −, νe) means momentum p, spin down, and electron flavor, and 6.8 Before and after symmetry breaking 159

(p, +, νe) means momentum p, spin up, and electron flavor. These fields satisfy equations like (6.137–6.139) apart from their interactions with other fields. Since a gauge transformation maps the fields νe,`(x) and e`(x) into each other, we know that when all the fields are massless, before symmetry breaking, there are (for each momentum) at least two neutrino and antineutrino states

1 h † † i √ a (p, −, νe) − ia (p, −, νe) |0i (6.149) 2 1 2 1 h † † i √ a (p, +, νe) + ia (p, +, νe) |0i (6.150) 2 1 2 for each of the three flavors, f = e, µ, τ. So there are at least six neutrino (and antineutrino) states. The right-handed electron field exists and interacts with gauge bosons and † other fields. So there are 12 electron states ai (p, ±, ef )|0i for i = 1 and 2 and for the three flavors, f = e, µ, τ. We don’t know yet whether a right-handed neutrino field exists or interacts with other fields. So there may be only 6 neutrino states or as many as 12. Neutrino oscillations tell us that neutrinos have masses. If there are 12 neutrino states, then there can be three massive Dirac neutrinos analagous to the e, µ, and τ or six massive Majorana neutrinos or some intermediate combination. If there are only 6, then there can be 3 massive Majorana neutrinos. The Majorana mass terms for the right-handed neutrino fields are 6 X 1 h i im ν0T σ ν0 + (im ν0T σ ν0 )† . (6.151) 2 ij ri 2 rj ij ri 2 rj ij=1 They are Lorentz invariant because under the Lorentz transformations (6.83) ν00T σ ν00 = ν0T exp(z∗ · σT)σ exp(z∗ · σ) ν0 r 2 r r 2 r (6.152) 0T ∗ ∗ 0 0T 0 = ν r σ2 exp(−z · σ) exp(z · σ) νr = ν r σ2 νr. The Majorana mass terms (6.151) are unrelated to the scale v of the Higgs field’s mean value. One can show that the complex matrix mij is symmetric. One then must combine the mass matrix in (6.151) with the mass matrix Mν in (6.142). The resulting mass matrix will have a singular-value decomposition with six singular values that would be the masses of the “physical” neutrinos. If these six masses are equal in pairs, then the three pairs would form three Dirac neutrinos. Whether or not there are right-handed neutrinos, we can make Majorana T mass terms like ν` σ2ν`, which are Lorentz invariant but not invariant under 160 Standard model

SU`(2) or UY (1). We can make them gauge invariant by using a triplet ~ ~0 ~ † φ = σiφi of Higgs ﬁelds that transforms as φ · ~σ = g(φ · ~σ)g for g ∈ SU`(2) 0 and that carries a value of Y = −1. Then if σ2 has Lorentz indices and σ2 has SU`(2) indices, the term

T 0 ~ E` σ2σ2(φ · ~σ)E` (6.153) is both Lorentz invariant and gauge invariant. If the potential V (φ~) has minima at φ~ 6= 0, then this term violates lepton number and gives a Ma- jorana mass to the neutrino. Another possibility is to say that at higher energies a theory with new fields of very high mass Λ plays a role, and that when one path-integrates over these heavy fields, one is left with an effective, nonrenormalizable term in the action

(H†E )2 − ` (6.154) Λ

which gives a Majorana mass to ν`. Models with both right-handed and left-handed neutrinos are easier to think about, but only experiments can tell us whether right-handed neutrinos exist. What is known experimentally is that there are at least three masses that satisfy

2 2 2 −5 2 ∆m21 ≡ m2 − m1 = (7.53 ± 0.18) × 10 eV 2 2 2 −3 2 ∆m32 ≡ m3 − m2 = (2.44 ± 0.06) × 10 eV normal mass hierarchy 2 2 2 −3 2 ∆m32 ≡ m3 − m2 = (2.52 ± 0.07) × 10 eV inverted mass hierarchy. (6.155)

If the neutrinos are Dirac particles, then they have a CKM matrix like that of the quarks with one CP -violating phase. But whereas one chooses to make the mass and flavor eigenfields the same for the up quarks u, c, t, for the leptons one makes the mass and flavor eigenfields the same for the down or charged leptons e, µ, τ. So the neutrino CKM matrix actually is † V = LeLν. If they are three Majorana particles, then their CKM matrix has two extra CP -violating phases α12 and α31. A common convention for the 6.9 The Seesaw Mechanism 161 neutrino CKM matrix is    −iδ 1 0 0 cos θ13 0 sin θ13e V = 0 cos θ23 sin θ23   0 1 0  (6.156) iδ 0 − sin θ23 cos θ23 − sin θ13e 0 cos θ13     cos θ12 sin θ12 0 1 0 0 iα /2 × − sin θ12 cos θ12 0 0 e 12 0  . (6.157) 0 0 1 0 0 eiα31/2 This convention without the last 3 × 3 matrix also is used for the quark CKM matrix. The current estimates are

2 sin (2θ12) = 0.846 ± 0.021 (6.158) +0.001 sin2(2θ ) = 0.999 normal mass hierarchy (6.159) 23 −0.018 +0.000 sin2(2θ ) = 1.000 inverted mass hierarchy (6.160) 23 −0.017 2 sin (2θ13) = 0.093 ± 0.008. (6.161)

Two of these are big angles: 2θ12 ≈ 2θ23 = π/2±nπ. In the normal hierarchy, the lightest neutrino is about 2/3 electron, 1/6 muon, and 1/6 tau; the very slightly heavier neutrino is about 1/3 electron, 1/3 muon, and 1/3 tau; and the much heavier heavier neutrino is about 1/6 electron, 5/12 muon, and 5/12 tau.

6.9 The Seesaw Mechanism Why are the neutrino masses so light? Suppose we wish to find the eigenvalues of the real, symmetric mass matrix 0 m M = (6.162) m M in which m is an ordinary mass and M is a huge mass. The eigenvalues µ of this hermitian mass matrix satisfy det (M − µI) = µ(µ − M) − m2 = 0 with √ 2 2 2 solutions µ± = M ± M + 4m /2. The larger mass µ+ ≈ M +m /M is 2 approximately the huge mass M and the smaller mass µ− ≈ −m /M is tiny. The physical mass of a fermion is the absolute value of its mass parameter, here m2/M. 2 The product of the two eigenvalues is the constant µ+µ− = det M = −m so as µ− goes down, µ+ must go up. In 1975, Gell-Mann, Ramond, Slansky, and Jerry Stephenson invented this “seesaw” mechanism as an explanation 162 Standard model of why neutrinos have such small masses, less than 1 eV/c2. If mc2 = 10 2 MeV, and µ−c ≈ 0.01 eV, which is a plausible light-neutrino mass, then the rest energy of the huge mass would be Mc2 = 107 GeV. This huge mass would be one of the six neutrino masses and would point at new physics, beyond the standard model. Yet the small masses of the neutrinos may be related to the weakness of their interactions. Before leaving the subject of fermion masses, let’s look more closely at Dirac and Majorana mass terms. A Dirac field is a linear combination of two Majorana fields of the same mass 1 L + i` ψ = √ (6.163) 2 R + ir in which L and ` are two-component left-handed spinors, and R and r are two-component right-handed spinors. The Dirac mass term 0 I m ψψ = i m ψ†γ0ψ = m ψ† ψ I 0 1 0 I L + i` = m L† − i`†,R† − ir† 2 I 0 R + ir 1 R + ir = m L† − i`†,R† − ir† 2 L + i` 1 h i = m L† − i`† (R + ir) + R† − ir† (L + i`) (6.164) 2 1 = m R† − ir† (L + i`) + h.c., 2 in which h.c. means hermitian conjugate, gives mass m to the particle and antiparticle of the Dirac field ψ. We may set

∗ ∗ R = iσ2 L ⇐⇒ L = −iσ2 R (6.165) ∗ ∗ r = iσ2 ` ⇐⇒ ` = −iσ2 r (6.166) or equivalently

† ! †! R1 L2 L1 −R2 = † ⇐⇒ = † (6.167) R2 −L1 L2 R1 † ! †! r1 `2 `1 −r2 = † ⇐⇒ = † (6.168) r2 −`1 `2 r1

† T which are the Majorana conditions (1.252 & 1.253). Since R = −iL σ2, we 6.10 Neutrino Oscillations 163 can write the Dirac mass term (6.164) in terms of left-handed fields as 1 m ψψ = m −i LT − `T σ (L + i`) + h.c. (6.169) 2 2 1 = m LT − i`T (−i σ )(L + i`) + h.c. (6.170) 2 2 1 0 −1 L1 + i`1 = m L1 − i`1,L2 − i`2 + h.c. (6.171) 2 1 0 L2 + i`2 1 −L2 − i`2 = m L1 − i`1,L2 − i`2 + h.c. (6.172) 2 L1 + i`1 1 = m (L − i` )(−L − i` ) + (L − i` )(L + i` ) + h.c. (6.173) 2 1 1 2 2 2 2 1 1 The fermion fields anticommute, so the Dirac mass term is 1 m ψψ = m (−2L L − 2` ` ) + h.c. = −m (L L + ` ` ) + h.c., (6.174) 2 1 2 1 2 1 2 1 2 and it says that the fields L and ` have the same mass m, as they must if they are to form a Dirac field. † T Since L = iR σ2, we also can write the Dirac mass term in terms of the right-handed fields as 1 m ψψ = m RT − irT iσ (R + ir) + h.c. (6.175) 2 2 1 0 1 R1 + ir1 = m R1 − ir1,R2 − ir2 + h.c. (6.176) 2 −1 0 R2 + ir2

= m (R1R2 + r1r2) + h.c. (6.177) So the fields R and r have the same mass m, as they must if they are to form a Dirac field. The Majorana mass term for a right-handed field r of mass m evidently is

m r1 r2 + h.c. (6.178)

6.10 Neutrino Oscillations The phase diﬀerence ∆φ between two highly relativistic neutrinos of momentum p going a distance L in a time t ≈ L varies with their masses m1 and m2 as LE LE q q ∆φ = t ∆E = ∆E = p2 + m2 − p2 + m2 (6.179) p p 1 2 164 Standard model in natural units. We can approximate this phase by using the ﬁrst two terms 2 2 of the binomial expansion of the square roots with y = 1 and x = mi /p q q LE∆m2 L∆m2 ∆φ = LE 1 + m2/p2 − 1 + m2/p2 ≈ ≈ (6.180) 1 2 p2 E 2 3 or in ordinary units ∆φ ≈ L∆m c /(~E). 7 Path integrals

7.1 Path integrals and Richard Feynman Since Richard Feynman invented them over 70 years ago, path integrals have been used with increasing frequency in high-energy and condensed- matter physics, in optics and biophysics, and even in ﬁnance. Feynman used them to express the amplitude for a process as a sum of all the ways the process could occur each weighted by an exponential of its classical action exp(iS/~). Others have used them to compute partition functions and to study the QCD vacuum. (Richard Feynman, 1918–1988)

7.2 Gaussian integrals and Trotter’s formula

Path integrals are based upon the gaussian integral (??) which holds for real a 6= 0 and real b ∞ r Z 2 iπ 2 eiax +2ibx dx = e−ib /a (7.1) −∞ a and upon the gaussian integral (??) ∞ r Z 2 π 2 e−ax +2ibx dx = e−b /a (7.2) −∞ a which holds both for Re a > 0 and also for Re a = 0 with b real and Im a 6= 0. The extension of the integral formula (7.1) to any n × n real symmetric nonsingular matrix sjk and any real vector cj is (exercises ?? & ??)

∞ r n Z (iπ) −1 isjkxj xk+2icj xj −icj (s )jkck e dx1 . . . dxn = e (7.3) −∞ det s 166 Path integrals in which det a is the determinant of the matrix a, a−1 is its inverse, and sums over the repeated indices j and k from 1 to n are understood. One may similarly extend the gaussian integral (7.2) to any positive symmetric n × n matrix sjk and any vector cj (exercises ?? & ??)

∞ r n Z π −1 −sjkxj xk+2icj xj −cj (s )jkck e dx1 . . . dxn = e . (7.4) −∞ det s Path integrals also are based upon Trotter’s product formula (Trotter, 1959; Kato, 1978) n ea+b = lim ea/n eb/n (7.5) n→∞ both sides of which are symmetrically ordered and obviously equal when ab = ba. Separating a given hamiltonian H = K + V into a kinetic part K and a potential part V , we can use Trotter’s formula to write the time-evolution operator e−itH/~ as n e−it(K+V )/~ = lim e−itK/(n~) e−itV/(n~) (7.6) n→∞ and the Boltzmann operator e−βH as n e−β(K+V ) = lim e−βK/n e−βV/n . (7.7) n→∞

7.3 Path integrals in quantum mechanics

Path integrals can represent matrix elements of the time-evolution operator exp( − i(tb − ta)H/~) in which H is the hamiltonian. For a particle of mass m moving nonrelativistically in one dimension in a potential V (q), the hamiltonian is p2 H = + V (q). (7.8) 2m The position and momentum operators q and p obey the commutation rela- 0 0 0 0 tion [q, p] = i~. Their eigenstates |q i and |p i have eigenvalues q and p for all real numbers q0 and p0

q |q0i = q0 |q0i and p |p0i = p0 |p0i. (7.9)

These eigenstates are complete. Their outer products |q0ihq0| and |p0ihp0| 7.3 Path integrals in quantum mechanics 167 provide expansions for the identity operator I and have inner products (??) that are phases

0 0 Z ∞ Z ∞ eiq p /~ I = |q0ihq0| dq0 = |p0ihp0| dp0 and hq0|p0i = √ . (7.10) −∞ −∞ 2π~ Setting = (tb − ta)/n and writing the hamiltonian (7.8) over ~ as H/~ = 2 p /(2m~) + V/~ = k + v, we can write Trotter’s formula (7.6) for the time- evolution operator as the limit as n → ∞ of n factors of e−ike−iv

e−i(tb−ta)(k+v) = e−ik e−iv e−ik e−iv ··· e−ik e−iv e−ik e−iv. (7.11) The advantage of using Trotter’s formula is that we now can evaluate the −ik −iv matrix element hq1|e e |qai between eigenstates |qai and |q1i of the position operator q by inserting the momentum-state expansion (7.10) of the identity operator I between these two exponentials ∞ 2 Z −ik −iv −ip /(2m~) 0 0 0 −i V (q)/~ hq1|e e |qai = hq1| e |p ihp | dp e |qai (7.12) −∞ and using the eigenvalue formulas (7.9) ∞ Z 02 −ik −iv −ip /(2m~) 0 −iV (qa)/~ 0 0 hq1|e e |qai = e hq1|p i e hp |qai dp . (7.13) −∞ 0 Now using the formula (7.10) for the inner product hq1|p i and the complex 0 conjugate of that formula for hp |qai, we get ∞ 0 Z 02 0 dp −ik −iv −i V (qa)/~ −ip /(2m~) i(q1−qa)p /~ hq1|e e |qai = e e e . (7.14) −∞ 2π~ In this integral, the momenta that are important are very high, being of p order m~/ which diverges as → 0; nonetheless, the integral converges. If we adopt the suggestive notation q1 − qa = q˙a and use the gaussian integral (7.1) with a = −/(2m~), x = p, and b = q/˙ (2~) Z ∞ p2 q˙ p dp r m mq˙2 exp − i + i = exp i , (7.15) −∞ 2m~ ~ 2π~ 2πi~ ~ 2 then we ﬁnd Z ∞ 02 0 −ik −iv 1 − V (qa) p q˙a p 0 hq1|e e |qai = e exp − i + i dp 2π~ −∞ 2m~ ~ 1/2 2 m mq˙a = exp i − V (qa) . (7.16) 2πi~ ~ 2 −ik −iv The dependence of the amplitude hq1|e e |qai upon q1 is hidden in the formulaq ˙a = (q1 − qa)/. 168 Path integrals The next step is to use the position-state expansion (7.10) of the identity operator to link two of these matrix elements together Z ∞ −ik −iv2 −ik −iv −ik −iv hq2| e e |qai = hq2| e e |q1ihq1| e e |qai dq1 −∞ (7.17) Z ∞ 2 2 m mq˙1 mq˙a = exp i − V (q1) + − V (qa) dq1 2πi~ −∞ ~ 2 2 where nowq ˙1 = (q2 − q1)/. By stitching together n = (tb − ta)/ time intervals each of length and letting n → ∞, we get Z −niH/~ −ik −iv −ik −iv hqb|e |qai = hqb|e e |qn−1i · · · hq1|e e |qai dqn−1 ··· dq1 Z " n−1 2 # m n/2 X mq˙j = exp i − V (q ) dq ··· dq 2πi 2 j n−1 1 ~ ~ j=0  n−1  m n/2Z X = exp i L dq ··· dq (7.18) 2πi  j n−1 1 ~ ~ j=0

2 in which Lj = mq˙j /2 − V (qj) is the lagrangian of the jth interval, and the qj integrals run from − ∞ to ∞. In the limit → 0 with n = (tb − ta)/, this multiple integral is an integral over all paths q(t) that go from qa, ta to qb, tb Z −i (tb−ta) H/~ iS[q]/~ hqb|e |qai = e Dq (7.19) in which each path is weighted by the phase of its classical action

Z tb Z tb mq˙(t)2 S[q] = L(q, ˙ q) dt = − V (q(t)) dt (7.20) ta ta 2

n/2 in units of ~ and Dq = (mn/(2πi~(tb − ta))) dqn−1 . . . dq1. −i (t −ta) H/ If we multiply the path-integral (7.19) for hqb|e b ~|qai from the left by |qbi and from the right by hqa| and integrate over qa and qb as in the resolution (7.10) of the identity operator, then we can write the time- evolution operator as an integral over all paths from ta to tb Z −i (tb−ta) H/~ iS[q]/~ e = |qbi e hqa| Dq dqa dqb (7.21)

n/2 with Dq = (mn/(2πi~(tb − ta))) dqn−1 . . . dq1 and S[q] the action (7.20). 7.3 Path integrals in quantum mechanics 169 The path integral for a particle moving in three-dimensional space is Z i Z tb −i(tb−ta)H/~ 1 2 hqb|e |qai = exp 2 m q˙ (t) − V (q(t)) dt Dq (7.22) ~ ta 3n/2 where Dq = (mn/(2πi~(tb − ta))) dqn−1 . . . dq1. Let us ﬁrst consider macroscopic processes whose actions are large compared to ~. Apart from the factor Dq, the amplitude (7.22) is a sum of iS[q]/~ phases e one for each path from qa, ta to qb, tb. When is this amplitude big? When is it small? Suppose there is a path qc(t) from qa, ta to qb, tb that obeys the classical equation of motion (??–??)

δS[qc] 0 = mq¨jc + V (qc) = 0. (7.23) δqjc

Its action may be minimal. It certainly is stationary: a path qc(t)+δq(t) that differs from qc(t) by a small detour δq(t) has an action S[qc +δq] that differs from S[qc] only by terms of second order and higher in δq. Thus a classical path has infinitely many neighboring paths whose actions differ only by integrals of (δq)n, n ≥ 2, and so have the same action to within a small fraction of ~. These paths add with nearly the same phase to the path integral (7.22) −i(tb−ta)H/~ and so make a huge contribution to the amplitude hqb|e |qai. But if no classical path goes from qa, ta to qb, tb, then the nonclassical, nonsta- tionary paths that go from qa, ta to qb, tb have actions that differ from each other by large multiples of ~. These amplitudes cancel each other, and their sum, which is amplitude for going from qa, ta to qb, tb, is small. Thus the path-integral formula for an amplitude in quantum mechanics explains why macroscopic processes are described by the principle of stationary action (section ??). What about microscopic processes whose actions are tiny compared to ~? The path integral (7.22) gives large amplitudes for all microscopic processes. On very small scales, anything can happen that doesn’t break a conservation law.

The path integral for two or more particles {q} = {q1,..., qk} interacting with a potential V ({q}) is Z −i(tb−ta)H/~ iS[{q}]/~ h{q}b|e |{q}ai = e D{q} (7.24) where Z tb m q˙2(t) m q˙2(t) S[{q}] = 1 1 + ··· + k k − V ({q(t)}) dt (7.25) ta 2 2 and D{q} = Dq1 ··· Dqk. 170 Path integrals Example 7.1 (A free particle) For a free particle, the potential is zero, and the path integral (7.19, 7.20) is the → 0, n → ∞ limit of

m n/2 −i t H/~ hqb|e |qai = (7.26) 2πi~ Z 2 2 m (qb − qn−1) (q1 − qa) × exp i + ··· + dqn−1 ··· dq1. 2~ 2 2

The q1 integral is by the gaussian formula (7.1) r m Z 2 2 m 2 im[(q2−q1) +(q1−qa) ]/(2~) im(q2−qa) /(2~2) e dq1 = e . (7.27) 2πi~ 2πi~2

The q2 integral is (exercise ??) r m Z 2 2 m 2 im[(q3−q2) +(q2−qa) /2]/(2~) im(q3−qa) /(2~3) √ e dq2 = e . 2 2 πi~ 2πi~3 (7.28)

Doing all n − 1 integrals (7.26) in this way and setting n = tb − ta, we get

r 2 m im(qb − qa) hq | e−i (tb−ta) H/~ |q i = exp b a 2πi n 2 n ~ ~ (7.29) r m im(q − q )2 = exp b a . 2πi~(tb − ta) 2~(tb − ta) The path integral (7.26) is perfectly convergent even though the velocities p q˙j = (qj+1 −qj)/ that are important are very high, being of order ~/(m). It is easier to compute this amplitude (7.29) by using the outer products (7.10) (exercise ??).

In three dimensions, the amplitude to go from qa, ta to qb, tb is

m 3/2 im(q − q )2 −i(tb−ta)H/~ b a hqt|e |q0i = exp . (7.30) 2πi~(tb − ta) 2~(tb − ta)

7.4 Path integrals for quadratic actions

If a path q(t) = qc(t) + x(t) differs from a classical path qc(t) by a detour x(t) that vanishes at the endpoints x(ta) = 0 = x(tb) so that both paths go from qa, ta to qb, tb, then the difference S[qc + x] − S[qc] in their actions vanishes to first order in the detour x(t) (section ??). Thus the actions of 7.4 Path integrals for quadratic actions 171 the two paths differ by a time integral of quadratic and higher powers of the detour x(t)

Z tb 1 2 S[qc + x] = 2 mq˙(t) − V (q(t)) dt ta Z tb 1 2 = 2 m (q ˙c(t) +x ˙(t)) − V (qc(t) + x(t)) dt ta Z tb 00 m 2 m 2 0 V (qc) 2 = q˙c + mq˙cx˙ + x˙ − V (qc) − V (qc)x − x ta 2 2 2 V 000(q ) V 0000(q ) − c x3 − c x4 − ... dt (7.31) 6 24 Z tb Z tb 00 hm 2 i m 2 V (qc) 2 = q˙c − V (qc) dt + x˙ − x ta 2 ta 2 2 V 000(q ) V 0000(q ) − c x3 − c x4 − ... dt 6 24

= S[qc] + ∆S[qc, x] in which S[qc] is the action of the classical path, and the detour x(t) is a loop that goes from x(ta) = 0 to x(tb) = 0. If the potential V (q) is quadratic in the position q, then the third V 000 and higher derivatives of the potential vanish, and the second derivative is 00 00 a constant V (qc(t)) = V . In this quadratic case, the correction ∆S[qc, x] 00 depends only on the time interval tb − ta and on ~, m, and V

Z tb 1 2 1 00 2 ∆S[qc, x] = ∆S[x] = 2 m x˙ (t) − 2 V x (t) dt. (7.32) ta

It is independent of the classical path. Thus for quadratic actions, the path integral (7.19) is an exponential of the 00 action S[qc] of the classical path multiplied by a function f(tb − ta, ~, m, V ) 00 of the time interval tb − ta and of ~, m, and V Z Z −i(tb−ta)H/~ iS[q]/~ i(S[qc]+∆S[x])/~ hqb|e |qai = e Dq = e Dq Z = eiS[qc]/~ ei∆S[x])/~ Dx (7.33)

00 iS[qc]/~ = f(tb − ta, ~, m, V ) e .

00 The function f = f(tb − ta, ~, m, V ) is the limit as n → ∞ of the (n − 1)- 172 Path integrals dimensional integral mn n/2Z i∆S[x])/~ f = e dxn−1 . . . dx1 (7.34) 2πi~(tb − ta) where n 2 tb − ta X 1 (xj − xj−1) 1 ∆S[x] = m − V 00 x2 (7.35) n 2 [(t − t )/n]2 2 j j=1 b a and xn = 0 = x0. More generally, the path integral for any quadratic action of the form

Z tb S[q] = u q˙2(t) + v q(t)q ˙(t) + w q2(t) + s(t)q ˙(t) + j(t) q(t) dt (7.36) ta is (exercise ??)

−i(tb−ta)H/~ iS[qc]/~ hqb|e |qai = f(ta, tb, ~, u, v, w) e . (7.37) The dependence of the amplitude upon s(t) and j(t) is contained in the classical action S[qc]. These formulas (7.33–7.37) may be generalized to any number of particles with coordinates {q} = {q1,..., qk} moving nonrelativistically in a space of multiple dimensions as long as the action is quadratic in the {q}’s and their velocities {q˙}. The amplitude is then an exponential of the action S[{q}c] of the classical path multiplied by a function f(ta, tb, ~,... ) that is independent of the classical path qc

−i(tb−ta)H/~ iS[{q}c]/~ h{q}b|e |{q}ai = f(ta, tb, ~,... ) e . (7.38) Example 7.2 (A free particle) The classical path of a free particle going from qa at time ta to qb at time tb is

t − ta qc(t) = qa + (qb − qa). (7.39) tb − ta Its action is Z tb 2 1 2 m(qb − qa) S[qc] = m q˙c dt = (7.40) ta 2 2(tb − ta) and for this case our quadratic-potential formula (7.38) is m(q − q )2 −i(tb−ta)H/~ b a hqb|e |qai = f(tb − ta, ~, m) exp i (7.41) 2~(tb − ta) which agrees with our explicit calculation (7.30) when f(tb − ta, ~, m) = 3/2 [m/(2πi~(tb − ta))] . 7.4 Path integrals for quadratic actions 173 Example 7.3 (Bohm-Aharonov eﬀect) From the formula (??) for the action of a relativistic particle of mass m and charge e, it follows (exercise ??) that the action a nonrelativistic particle in an electromagnetic ﬁeld with no scalar potential is

Z tb Z qb 1 2 1 S = 2 mq˙ + eA · q˙ dt = 2 mq˙ + e A · dq . (7.42) ta qa Since this action is quadratic in q˙, the amplitude for a particle to go from qa at ta to qb at tb is an exponential of the classical action

−i(tb−ta)H/~ iS[qc]/~ hqb|e |qai = f(tb − ta, ~, m, e) e (7.43) multiplied by a function f(tb −ta, ~, m, e) that is independent of the path qc. A beam of such particles goes horizontally past but not through a vertical pipe in which a vertical magnetic field is confined. The particles can go both ways around the pipe of cross-sectional area S but do not enter it. The difference in the phases of the amplitudes for the two paths is a loop integral from the source to the detector around the pipe and back to the source I mq˙ dq I mq˙ · dq e Z I mq˙ · dq eΦ + e A · = + B · dS = + (7.44) 2 ~ 2~ ~ S 2~ ~ in which Φ is the magnetic flux through the cylinder.

Example 7.4 (Harmonic oscillator) The action

Z tb 1 1 S = mq˙2(t) − mω2q2(t) dt (7.45) ta 2 2 of a harmonic oscillator is quadratic in q andq ˙. So apart from a factor f, its path integral (7.33–7.35) is an exponential

−i(tb−ta)H/~ iS[qc]/~ hqb|e |qai = f e (7.46) of the action S[qc] (exercise ??) 2 2 mω qa + qb cos(ω(tb − ta)) − 2qaqb S[qc] = (7.47) 2 sin(ω(tb − ta)) of the classical path

qb − qa cos ω(tb − ta) qc(t) = qa cos ω(t − ta) + sin ω(t − ta) (7.48) sin ω(tb − ta) that runs from qa, ta to qb, tb and obeys the classical equation of motion 2 mq¨c(t) = − ω qc(t). 2 The factor f is a function f(tb −ta, ~, m, mω ) of the time interval and the 174 Path integrals parameters of the oscillator. It is the n → ∞ limit of the (n−1)-dimensional integral (7.34)

mn n/2 Z i∆S[x])/~ f = e dxn−1 . . . dx1 (7.49) 2πi~(tb − ta) over all loops that run from 0 to 0 in time tb − ta in which the quadratic correction to the classical action is (7.35)

n 2 tb − ta X 1 (xj − xj−1) 1 ∆S[x] = m − mω2 x2, (7.50) n 2 [(t − t )/n]2 2 j j=1 b a and xn = 0 = x0. Setting tb − ta = T , we use the many-variable imaginary gaussian integral (7.3) to write f as r h mn in/2Z h mn in/2 (iπ)n−1 iajkxj xk f = e dxn−1 . . . dx1 = (7.51) 2πi~T 2πi~T det a in which the quadratic form ajkxjxk is

n nm X 1 (ωT )2 − x x + (x2 + x2 ) − x2 (7.52) T j j−1 2 j j−1 2n2 j ~ j=1 which has no linear term because x0 = xn = 0. The (n − 1)-dimensional square matrix a is a tridiagonal Toeplitz matrix

 y −1 0 0 ··· −1 y −1 0 ··· nm    0 −1 y −1 ··· a =   . (7.53) 2~T  0 0 −1 y ···  . . . . .  ......

Apart from the factor nm/(2~T ), the matrix a = (nm/(2~T )) Cn−1(y) is a tridiagonal matrix Cn−1(y) whose oﬀ-diagonal elements are −1 and whose diagonal elements are (ωT )2 y = 2 − . (7.54) n2

Their determinants |Cn(y)| = det Cn(y) obey (exercise ??) the recursion relation

|Cn+1(y)| = y |Cn(y)| − |Cn−1(y)| (7.55) 7.5 Path integrals in statistical mechanics 175

2 and have the initial values |C1(y)| = y and |C2(y)| = y − 1. The trigonometric functions Un(y) = sin[(n + 1)θ]/ sin θ with y = 2 cos θ obey the same recursion relation and have the same initial values (exercise ??), so sin(n + 1)θ |C (y)| = . (7.56) n sin θ Since for large n ω2t2 ωT θ = arccos(y/2) = arccos 1 − ≈ , (7.57) 2n2 n the determinant of the matrix a is nm n−1 nm n−1 sin nθ det a = |C (y)| = 2 T n−1 2 T sin θ ~ ~ (7.58) nm n−1 sin(ωT ) nm n−1 n sin ωT ≈ ≈ . 2~T sin(ωT/n) 2~T ωT Thus the factor f is r s h mn in/2 (iπ)n−1 h mn in/2 2πi T n−1 ωT f = = ~ 2πi~T det a 2πi~T nm n sin ωT r mω = . (7.59) 2πi~ sin ωT

The amplitude (7.46) is then an exponential of the action S[qc] (7.47) of the classical path (7.48) multiplied by this factor f r mω −i(tb−ta)H/~ hqb|e |qai = (7.60) 2πi~ sin ω(tb − ta) ( 2 2 ) i mω q + q cos(ω(t − ta)) − 2qaq × exp a b b b . ~ 2 sin(ω(tb − ta))

As these examples (7.2 & 7.4) suggest, path integrals are as mathemati- cally well deﬁned as ordinary integrals.

7.5 Path integrals in statistical mechanics

At the imaginary time t = −i~β = −i~/(kT ), the time-evolution operator e−itH/~ becomes the Boltzmann operator e−βH whose trace is the 176 Path integrals partition function Z(β) at inverse energy β = 1/(kT ) X Z(β) = Tre−βH = hn|e−βH |ni (7.61) n in which the states |ni form a complete orthonormal set, k = 8.617 × 10−5 eV/K is Boltzmann’s constant, and T is the absolute temperature. Partition functions are important in statistical mechanics and quantum ﬁeld theory. Since the Boltzmann operator e−βH is the time-evolution operator e−itH/~ at the imaginary time t = − i~β, we can write it as a path integral by imitating the derivation of the preceding section (7.3). We will use the same hamiltonian H = p2/(2m) + V (q) and the operators q and p which have complete sets of eigenstates (7.9) that satisfy (7.10). Changing our deﬁnitions of , k, and v to = β/n, k = βp2/(2m), and v = βV (q), we can write Trotter’s formula (7.7) for the Boltzmann operator as the n → ∞ limit of n factors of e−k e−v

e−βH = e−k e−v e−k e−v ··· e−k e−v e−k e−v. (7.62)

If we adopt the suggestive notation q1 − qa = ~ q˙a and use the gaussian integral (7.2) with a = /(2m), x = p, and b = q/˙ 2

Z ∞ p2 dp r m mq˙2 exp − + i q˙ p = 2 exp − , (7.64) −∞ 2m 2π~ 2π~ 2 then we ﬁnd Z ∞ 02 0 −k −v − V (qa) p 0 dp hq1|e e |qai = e exp − + i p q˙a −∞ 2m 2π~ 1/2 2 m m q˙a = exp − + V (qa) (7.65) 2π~2 2 in which q1 is hidden in the formula q1 − qa = ~ q˙a. 7.5 Path integrals in statistical mechanics 177 The next step is to link two of these matrix elements together Z ∞ −k −v2 −k −v −k −v hq2| e e |qai = hq2|e e |q1ihq1|e e |qaidq1 (7.66) −∞ Z ∞ 2 2 m m q˙1 m q˙0 = 2 exp − + V (q1) + + V (qa) dq1. 2π~ −∞ 2 2 Passing from 2 to n and suppressing some integral signs, we get

−nH RRR ∞ −k −v −k −v hqb|e |qai = −∞hqb|e e |qn−1i · · · hq1|e e |qai dqn−1 . . . dq1 m q˙2 m n/2RRR ∞ Pn−1 j = 2 exp − + V (qj) dqn−1 . . . dq1. 2π~ −∞ j=0 2

Setting du = ~ = ~β/n and taking the limit n → ∞, we ﬁnd that the −βH matrix element hqb|e |qai is the path integral Z −βH −Se[q]/~ hqb|e |qai = e Dq (7.67) in which each path is weighted by its euclidian action Z ~β mq˙2(u) Se[q] = + V (q(u)) du, (7.68) 0 2 q˙ is the derivative of the coordinate q(u) with respect to euclidian time 2 n/2 u = ~β, and Dq ≡ (n m/2π ~ β) dqn−1 . . . dq1. A derivation identical to the one that led from (7.62) to (7.68) leads in a more elaborate notation to Z −(βb−βa)H −Se[q]/~ hqb|e |qai = e Dq (7.69) in which each path is weighted by its euclidian action

Z ~βb mq˙2(u) S [q] = + V (q(u)) du, (7.70) e 2 ~βa andq ˙ and Dq are the same as in (7.68). If we multiply the path integral (7.69) from the left by |qbi and from the right by hqa| and integrate over qa and qb as in the resolution (7.10) of the identity operator, then we can write the Boltzmann operator as an integral over all paths from ta to tb Z −(βb−βa) H −Se[q]/~ e = |qbi e hqa| Dq dqa dqb (7.71)

n/2 with Dq = (mn/(2πi~(tb − ta))) dqn−1 . . . dq1 and Se[q] the action (7.70). 178 Path integrals

To get the partition function Z(β), we set qb = qa ≡ qn and integrate over all n q’s letting n → ∞ Z −βH −βH Z(β) = Tr e = hqn|e |qni dqn (7.72) Z 1 Z ~β mq˙2(u) = exp − + V (q(u)) du Dq ~ 0 2 2 n/2 where Dq ≡ (n m/2π ~ β) dqn . . . dq1. We sum over all loops q(u) that go from q(0) = qn at euclidian time 0 to q(~β) = qn at euclidian time ~β. In the low-temperature limit, T → 0 and β → ∞, the Boltzmann operator exp(−βH) projects out the ground state |E0i of the system

−βH X −βEn −βE0 lim e = lim e |EnihEn| = e |E0ihE0|. (7.73) β→∞ β→∞ n The maximum-entropy density operator (section ??, example ??) is the Boltzmann operator e−βH divided by its trace Z(β) e−βH e−βH ρ = = . (7.74) Tr(e−βH ) Z(β) Its matrix elements are matrix elements of Boltzmann operator (7.68) divided by the partition function (7.72) hq |e−βH |q i hq |ρ|q i = b a . (7.75) b a Z(β)

In three dimensions with q˙(u) = dq(u)/du, the qa, qb matrix element of the Boltzmann operator is the analog of equation (7.68) (exercise ??)

Z Z ~β 2 −βH 1 mq˙ (u) hqb|e |qai = exp − + V (q(u)) du Dq (7.76) ~ 0 2 2 3n/2 where Dq ≡ (n m/2π ~ β) dqn−1 . . . dq1, and the partition function is the integral over all loops that go from q0 = qn to qn in time ~β Z 1 Z ~β mq˙2(u) Z(β) = exp − + V (q(u)) du Dq (7.77) ~ 0 2 2 3n/2 where now Dq ≡ (n m/2π ~ β) dqn . . . dq1. Because the Boltzmann operator e−βH is the time-evolution operator −itH/ e ~ at the imaginary time t = − iu = −i~β = −i~/(kT ), the path integrals of statistical mechanics are called euclidian path integrals. 7.5 Path integrals in statistical mechanics 179 Example 7.5 (Density operator for a free particle) For a free particle, the matrix element of the Boltzmann operator e−βH is the n = β/ → ∞ limit of the integral n/2 −βH m hqb|e |qai = (7.78) 2π~2 Z 2 2 m(qb − qn−1) (q1 − qa) × exp − · · · − dqn−1 ··· dq1. 2~2 2~2 (7.79)

The formula (7.2) gives for the q1 integral

2 2 −m(q −qa) /(2 2) m 1/2Z 2 2 2 e 2 ~ −[m(q2−q1) +m(q1−qa) ]/(2~ ) e dq1 = √ . (7.80) 2π~2 2

The q2 integral is (exercise ??)

2 2 −m(q −qa) /(2 3) m 1/2Z 2 2 2 2 e 3 ~ −m(q3−q2) /(2~ )−m(q2−qa) ]/(4~ ) e dq2 = √ . 4π~2 3 (7.81) All n − 1 integrations give

2 2 r −m(q −qa) /(2 n) m e b ~ r m 2 2 −βH −m(qb−qa) /(2~ β) hqb|e |qai = √ = e . 2π~2 n 2π~2β (7.82) The partition function is the integral of this matrix element over qa = qb m 1/2 Z m 1/2 Z(β) = dqa = L (7.83) 2π~2β 2π~2β where L is the (inﬁnite) 1-dimensional volume of the system. The qb, qa matrix element of the maximum-entropy density operator is

2 2 e−m(qb−qa) /(2~ β) hq |ρ|q i = . (7.84) b a L In 3 dimensions, equations (7.82 & 7.83) are 3/2 3/2 mkT 2 2 mkT −βH −m(qb−qa) /(2~ β) 3 hqb|e |qai = e and Z(β) = L . 2π~2 2π~2 (7.85)

Example 7.6 (Partition function at high temperatures) At high temperatures, the time ~β = ~/(kT ) is very short, and the density operator (7.85) 180 Path integrals for a free particle√ shows that free paths are damped and limited to distances of order ~/ mkT . We thus can approximate the path integral (7.77) for the partition function by replacing the potential V (q(u)) by V (qn) and then using the free-particle matrix element (7.85)

Z 1 Z ~β mq˙2(u) Z(β) ≈ e−βV (qn) exp − du Dq (7.86) ~ 0 2 Z 3/2 Z −βV (qn) −βH mkT −βV (qn) = e hqn|e |qni dqn = e dqn. 2π~2

7.6 Boltzmann path integrals for quadratic actions 2 n/2 Apart from the factor Dq ≡ (n m/2π ~ β) dqn−1 . . . dq1, the euclidian path integral

Z Z ~β 2 −βH 1 mq˙ (u) hqb|e |qai = exp − + V (q(u)) du Dq (7.87) ~ 0 2

−Se[q]/ is a sum of positive terms e ~ one for each path from qa, 0 to qb, β. If a path from qa, 0 to qb, β obeys the classical euclidian equation of motion d2q m ce = m q¨ = V 0(q ) (7.88) du2 ce ce then its euclidian action Z ~β m q˙2(u) Se[q] = + V (u) du (7.89) 0 2 is stationary and may be minimal. So we can approximate the euclidian action Se[qce + x] as we approximated the action S[qc + x] in section 7.4. The euclidian action Se[qce +x] of an arbitrary path from qa, 0 to qb, β is the stationary euclidian action Se[qce] plus a u-integral of quadratic and higher powers of the detour x which goes from x(0) = 0 to x(~β) = 0

Z ~β Z ~β 00 hm 2 i m 2 V (qce) 2 Se[qce + x] = q˙ce + V (qce) du + x˙ + x 0 2 0 2 2 V 000(q ) V 0000(q ) + ce x3 + ce x4 + ... du 6 24

= Se[qce] + ∆Se[qce, x], (7.90) 7.6 Boltzmann path integrals for quadratic actions 181

−βH and the path integral for the matrix element hqb|e |qai is Z −βH −Se[qce]/~ −∆Se[qce,x]/~ hqb|e |qai = e e Dx (7.91)

2 n/2 as n → ∞ where Dx = (n m/2π ~ β) dqn−1 . . . dq1 in the limit n → ∞. If the action is quadratic in q andq ˙, then the integral ∆Se[qce, x] over the detour x is a gaussian path integral that is independent of the path qce and 00 so is a function f only of the parameters β, m, ~, and V Z −βH −Se[qce]/~ −∆Se[x]/~ hqb|e |qai = e e Dx (7.92) 00 −S [q ]/ = f(β, ~, m, V ) e e ce ~ where mn n/2Z 00 −∆Se[x]/~ f(β, ~, m, V ) = e dxn−1 . . . dx1, 2π~2β n 2 β X m (xj − xj−1) 1 ∆S [x] = ~ + V 00 x2, (7.93) e n 2 2 (β/n)2 2 j j=1 ~ and xn = 0 = x0.

Example 7.7 (Density operator for the harmonic oscillator) The path qce(β) that satisﬁes the classical euclidian equation of motion (7.88) d2q (u) q¨ (u) = ce = ω2q (u) (7.94) ce du2 ce and goes from qa, 0 to qb, ~β is

sinh(ωu) qb + sinh[ω(~β − u)] qa qce(u) = . (7.95) sinh(~ωβ) Its euclidian action is (exercise ??)

Z ~β 2 2 2 mq˙ce(u) mω qce(u)) Se[qce] = + du 0 2 2 (7.96) mω 2 2 = cosh(~ωβ)(qa + qb ) − 2qaqb . 2~ sinh(~ωβ) Since V 00 = mω2, our formulas (7.92 & 7.93) for quadratic actions give as the matrix element

−βH 2 −Se[qce]/~ hqb|e |qai = f(β, ~, m, mω ) e (7.97) 182 Path integrals in which mn n/2Z 2 −∆Se[x]/~) f(β, ~, m, mω ) = e dxn−1 . . . dx1, 2π~2β n 2 2 2 (7.98) β X m (xj − xj−1) mω xj ∆S [x] = ~ + , e n 2 2 (β/n)2 2 j=1 ~ and xn = 0 = x0. We can do this integral by using the formula (7.4) for a many variable real gaussian integral r h mn in/2Z h mn in/2 (π)n−1 −ajkxj xk f = e dxn−1 . . . dx1 = (7.99) 2π~2B 2π~2B det a in which the positive quadratic form ajkxjxk is n nm X ( ωB)2 − 2x x + x2 + x2 + ~ x2 (7.100) 2 2B j j−1 j j−1 n2 j ~ j=1 which has no linear term because x0 = xn = 0. 2 The matrix a is (nm/(2~ B)) Cn−1(y) in which Cn−1(y) is a square, tridiagonal, (n − 1)-dimensional matrix whose oﬀ-diagonal elements are −1 and 2 2 whose diagonal elements are y = 2 + (~ωβ) /n . The determinants |Cn(y)| obey the recursion relation |Cn+1(y)| = y |Cn(y)| − |Cn−1(y)| and have the 2 initial values C1(y) = y and C2(y) = y − 1. So do the hyperbolic functions sinh(n+1)θ/ sinh θ with y = 2 cosh θ. So we set Cn(y) = sinh(n+1)θ/ sinh θ with θ = arccosh(y/2). We then get as the matrix element (7.97) r 2 2 −βH mω mω[cosh(~ωβ)(qa + qb ) − 2qaqb] hqb|e |qai = exp − . 2π~ sinh(~ωβ) 2~ sinh(~ωβ) (7.101) The partition function is the integral over qa of this matrix element for qb = qa r Z 2 mω mω[cosh(~ωβ) − 1]qa Z(β) = exp − dqa 2π~ sinh(~ωβ) ~ sinh(~ωβ) 1 = p . (7.102) 2[cosh(~ωβ) − 1] The matrix elements of the maximum-entropy density operator (7.74) are hq |e−βH |q i hq |ρ|q i = b a (7.103) b a Z(β) s mω[cosh( ωβ) − 1] mω[cosh( ωβ)(q2 + q2) − 2q q ] = ~ exp − ~ a b a b π~ sinh(~ωβ) 2~ sinh(~ωβ) 7.7 Mean values of time-ordered products 183 which reveals the ground-state wave functions r mω 2 2 −mω(qa+q )/(2~) lim hqb|ρ|qai = hqb|0ih0|qai = e b . (7.104) β→∞ π~ The partition function gives us the ground-state energy 1 −βE0 −β~ω/2 lim Z(β) = lim p = e = e . (7.105) β→∞ β→∞ 2[cosh(~ωβ) − 1]

7.7 Mean values of time-ordered products In the Heisenberg picture, the position operator at time t is

q(t) = eitH/~q e−itH/~ (7.106) in which q = q(0) is the position operator at time t = 0 or equivalently the position operator in the Schr¨odingerpicture. The position operator q at the imaginary time t = −iu = −i~β = −i~/(kT ) is the euclidian position operator

uH/~ −uH/~ qe(u) = qe(~β) = e q e . (7.107) The time-ordered product of two position operators is q(t1) q(t2) if t1 ≥ t2 T [q(t1)q(t2)] = = q(t> ) q(t< ) (7.108) q(t2) q(t1) if t2 ≥ t1 in which t> is the later and t< the earlier of the two times t1 and t2. Similarly, the time-ordered product of two euclidian position operators at euclidian times u1 = ~β1 and u2 = ~β2 is qe(u1) qe(u2) if u1 ≥ u2 T [qe(u1)qe(u2)] = = qe(u> ) qe(u< ). (7.109) qe(u2) qe(u1) if u2 ≥ u1 The matrix element of the time-ordered product (7.108) of two position operators and two exponentials e−itH/~ between states |ai and |bi is

in which the integrations are over all paths that go from −t ≤ t< to t ≥ t> . The mean value of the time-ordered product of k position operators is Z iS[q]/ hn|qbiq(t1) ··· q(tk)e ~hqa|ni Dq hn|T [q(t1) ··· q(tk)]|ni = Z (7.115) iS[q]/ hn|qbie ~hqa|ni Dq in which the integrations are over all paths that go from some time before t1, . . . , tk to some time them. We may perform the same operations on the euclidian position operators by replacing t by −iu = −i~β. A matrix element of the euclidian time- ordered product (7.109) between two states is

As u → ∞, the exponential e−uH/~ projects (7.73) states in onto the ground state |0i which is an eigenstate of H with energy E0. So we replace the 7.8 Quantum ﬁeld theory 185 arbitrary states in (7.116) with the ground state and use the path-integral formula (7.71 for the last three exponentials of (7.116) Z −2uE0/~ −Se[q]/~ e h0|T [qe(u1)qe(u2)]|0i = h0|qbiq(u1)q(u2)e hqa|0iDq. (7.117) The same equation without the time-ordered product is Z −2uE0/~ −2uE0/~ −Se[q]/~ e h0|0i = e = h0|qbie hqa|0iDq. (7.118)

7.8 Quantum ﬁeld theory

Quantum mechanics imposes upon n coordinates qi and conjugate momenta pk the equal-time commutation relations

[qi, pk] = i ~ δi,k and [qi, qk] = [pi, pk] = 0. (7.121)

In the theory of a single spinless quantum ﬁeld, a coordinate qx ≡ φ(x) and a conjugate momentum px ≡ π(x) are associated with each point x of space. The operators φ(x) and π(x) obey the commutation relations 0 0 [φ(x), π(x )] = i ~ δ(x − x ) (7.122) [φ(x), φ(x0)] = [π(x), π(x0)] = 0 inherited from quantum mechanics. To make path integrals, we will replace space by a 3-dimensional lattice of 186 Path integrals points x = a(i, j, k) = (ai, aj, ak) and eventually let the distance a between adjacent points go to zero. On this lattice and at equal times t = 0, the ﬁeld operators obey discrete forms of the commutation relations (7.122)

[φ(a(i, j, k)), π(a(`, m, n))] = i ~ δ δ δ a3 i,` j,m k,n (7.123) [φ(a(i, j, k)), φ(a(`, m, n))] = [π(a(i, j, k)), π(a(`, m, n))] = 0. The vanishing commutators imply that the ﬁeld and the momenta have “simultaneous” eigenvalues φ(a(i, j, k))|φ0i = φ0(a(i, j, k))|φ0i and π(a(i, j, k))|π0i = π0(a(i, j, k))|π0i (7.124) for all lattice points a(i, j, k). Their inner products are

r 3 Y a 3 0 0 hφ0|π0i = eia φ (a(i,j,k))π (a(i,j,k))/~. (7.125) 2π i,j,k ~ These states are complete Z Y Z Y |φ0ihφ0| dφ0(a(i, j, k)) = I = |π0ihπ0| dπ0(a(i, j, k)) (7.126) i,j,k i,j,k and orthonormal Y hφ0|φ00i = δ(φ0(a(i, j, k)) − φ00(a(i, j, k))) (7.127) i,j,k with a similar equation for hπ0|π00i. The hamiltonian for a free ﬁeld of mass m is Z m2c4 a3 X m2c4 H = 1 π2 + c2(∇φ)2 + φ2 d3x = π2 + c2(∇φ )2 + φ2 2 2 2 v v 2 v ~ v ~ (7.128) where v = a(i, j, k), πv = π(a(i, j, k)), φv = φ(a(i, j, k)), and the square of 2 the lattice gradient (∇φv) is (φ(a(i + 1, j, k)) − φ(a(i, j, k)))2 + (φ(a(i, j + 1, k)) − φ(a(i, j, k)))2 + (φ(a(i, j, k + 1)) − φ(a(i, j, k)))2/a2. (7.129)

3 4 Other ﬁelds or terms, such as c φ /~, can be added to this hamiltonian. To simplify the appearance of the equations in the rest of this chapter, I will mostly use natural units in which ~ = c = 1. To convert the value of a physical quantity from natural units to universal units, one multiplies or divides its natural-unit value by suitable factors of ~ and c until one gets the right dimensions. For instance, if V = 1/m is the value of a time in 7.8 Quantum ﬁeld theory 187

2 natural units, where m is a mass, then the time you want is T = ~/(mc ). If V = 1/m is supposed to be a length, then the needed length is L = ~/(mc). 3 P 2 3 P 2 2 2 We set K = a v πv/2 and V = (a /2) v(∇φv) + m φv + P (φv) in which P (φv) represents the self-interactions of the ﬁeld. With = (tb −ta)/n, Trotter’s product formula (7.6) is the n → ∞ limit of n n e−i(tb−ta)(K+V ) = e−i(tb−ta)K/ne−i(tb−ta)V/n = e−iK e−iV . (7.130)

We again adopt the suggestive notation φ˙a = (φ1 − φa)/ and use the gaussian integral (7.1) to ﬁnd

" 3 1/2 # Y a 3 ˙2 2 2 2 hφ |e−iK e−iV |φ i = eia [φav−(∇φav) −m φav−P (φv)]/2 . 1 a 2πi v (7.134)

The product of n = (tb − ta)/ such time intervals is " n/2 # Y a3n Z hφ |e−i(tb−ta)H |φ i = eiSv Dφ (7.135) b a 2πi(t − t ) v v b a in which 3 n−1 tb − ta a X h i S = φ˙2 − (∇φ )2 − m2φ2 − P (φ ) (7.136) v n 2 jv jv jv v j=0 ˙ φjv = n(φj+1,v − φj,v)/(tb − ta), and Dφv = dφn−1,v ··· dφ1,v. −i(t −ta)H The amplitude hφb|e b |φai is the integral over all ﬁelds that go from φa(x) at ta to φb(x) at tb each weighted by an exponential Z −i(tb−ta)H iS[φ] hφb|e |φai = e Dφ (7.137) 188 Path integrals of its action Z tb Z 1 h i S[φ] = dt d3x φ˙2 − (∇φ)2 − m2φ2 − P (φ) (7.138) ta 2 in which Dφ is the n → ∞ limit of the product over all spatial vertices v

" n/2 # Y a3n Dφ = dφ ··· dφ . (7.139) 2πi(t − t ) n−1,v 1,v v b a Equivalently, the time-evolution operator is Z −i(tb−ta)H iS[φ] e = |φbie hφa| Dφ DφaDφb (7.140)

Q in which DφaDφb = v dφa,vdφb,v is an integral over the initial and final states. As in quantum mechanics (section 7.4), the path integral for an action that is quadratic in the fields is an exponential of the action of a stationary process times a function of the times and of the other parameters in the action Z −i(tb−ta)H iS[φ] iS[φc] hφb|e |φai = e Dφ = f(ta, tb,... ) e (7.141) in which S[φc] is the action of the process that goes from φ(x, ta) = φa(x) to φ(x, tb) = φb(x) and obeys the classical equations of motion, and the function f is a path integral over all fields that go from φ(x, ta) = 0 to φ(x, tb) = 0. Example 7.8 (A stationary process) The field Z φ(x, t) = eik·x[a(k) cos ωt + b(k) sin ωt] d3k (7.142)

p with ω = k2 + m2 makes the action (7.138) for P = 0 stationary because it is a solution of the equation of motion ∇2φ − φ¨ − m2φ = 0. In terms of the Fourier transforms Z d3x Z d3x φ˜(k, t ) = e−ik·x φ(x, t ) and φ˜(k, t ) = e−ik·x φ(x, t ) , a a (2π)3 b b (2π)3 (7.143) the solution that goes from φ(x, ta) to φ(x, tb) is

Z sin ω(t − t) φ˜(k, t ) + sin ω(t − t ) φ˜(k, t ) φ(x, t) = eik·x b a a b d3k. (7.144) sin ω(tb − ta) 7.9 Finite-temperature ﬁeld theory 189

The solution that evolves from φ(x, ta) and φ˙(x, ta) is Z sin ω(t − ta) ˜ φ(x, t) = eik·x cos ω(t − t ) φ˜(k, t ) + φ˙(k, t ) d3k a a ω a (7.145) ˜ in which the Fourier transform φ˙(k, ta) is deﬁned as in (7.143).

Like a position operator (7.106), a field at time t is defined as φ(x, t) = eitH/~φ(x)e−itH/~ (7.146) in which φ(x) = φ(x, 0) is the field at time zero, which obeys the commutation relations (7.122). The time-ordered product of several fields is their product with newer (later time) fields standing to the left of older (earlier time) fields as in the definition (7.108). The logic (7.110–7.115) of the derivation of the path-formulas for time-ordered products of position operators applies directly to field operators. One finds (exercise ??) for the mean value of the time-ordered product of two fields in an energy eigenstate |ni Z iS[φ]/ hn|φbiφ(x1)φ(x2)e ~hφa|ni Dφ hn|T [φ(x1)φ(x2)]|ni = Z (7.147) iS[φ]/ hn|φbie ~hφa|ni Dφ in which the integrations are over all paths that go from before t1 and t2 to after both times. The analogous result for several fields is (exercise ??) Z iS[φ]/ hn|φbiφ(x1) ··· φ(xk)e ~hφa|ni Dφ hn|T [φ(x1) ··· φ(xk)]|ni = Z iS[φ]/ hn|φbie ~hφa|ni Dφ (7.148) in which the integrations are over all paths that go from before the times t1, . . . , tk to after them.

7.9 Finite-temperature ﬁeld theory

Since the Boltzmann operator e−βH = e−H/(kT ) is the time evolution −itH/ operator e ~ at the imaginary time t = − i~β = −i~/(kT ), the formulas of finite-temperature field theory are those of quantum field theory with t replaced by −iu = −i~β = −i~/(kT ). 190 Path integrals If as in section 7.8, we use as our hamiltonian H = K + V where K and V are sums over all lattice vertices v = a(i, j, k) = (ai, aj, ak) of the cubes of volume a3 times the squared momentum and potential terms

a3 X a3 X H = K + V = π2 + (∇φ )2 + m2φ2 + P (φ ). (7.149) 2 v 2 v v v v v A matrix element of the ﬁrst term of the Trotter product formula (7.7) n e−β(K+V ) = lim e−βK/n e−βV/n (7.150) n→∞ is the imaginary-time version of (7.133) with = ~β/n Z 3 0 Y a dπ 3 2 0 hφ |e−K e−V |φ i = e−V (φa) v ea [−πv/2+i(φ1v−φav)πv] . 1 a 2π v (7.151)

Setting φ˙av = (φ1v − φav)/, we ﬁnd, instead of (7.134)

" 3 1/2 # Y a 3 ˙2 2 2 2 hφ |e−K e−V |φ i = e−a [φav+(∇φav) +m φav+P (φv)]/2 . 1 a 2π v (7.152) The product of n = ~β/ such inverse-temperature intervals is

" n/2 # Y a3n Z hφ |e−βH |φ i = e−Sev Dφ (7.153) b a 2πβ v v in which the euclidian action is

n−1 β a3 X h i S = φ˙2 + (∇φ )2 + m2φ2 + P (φ ) (7.154) ev n 2 jv jv jv v j=0

φ˙jv = n(φj+1,v − φj,v)/β, and Dφv = dφn−1,v ··· dφ1,v. −(β −βa)H The amplitude hφb|e b |φai is the integral over all ﬁelds that go from φa(x) at βa to φb(x) at βb each weighted by an exponential Z −(βb−βa)H −Se[φ] hφb|e |φai = e Dφ (7.155) of its euclidian action

Z βb Z 3 1 h 2 2 2 2 i Se[φ] = du d x φ˙ + (∇φ) + m φ + P (φ) (7.156) βa 2 7.9 Finite-temperature ﬁeld theory 191 in which Dφ is the n → ∞ limit of the product over all spatial vertices v

" n/2 # Y a3n Dφ = dφ ··· dφ . (7.157) 2π(β − β ) n−1,v 1,v v b a Equivalently, the Boltzmann operator is Z −(βb−βa)H −Se[φ] e = |φbie hφa| Dφ DφaDφb (7.158) Q in which DφaDφb = v dφa,vdφb,v is an integral over the initial and final states. The trace of the Boltzmann operator is the partition function Z Z −βH −Se[φ] −Se[φ] Z(β) = Tr(e ) = e hφa|φbi Dφ DφaDφb = e Dφ Dφa (7.159) which is an integral over all fields that go back to themselves in euclidian time β. Like a position operator (7.107), a field at an imaginary time t = −iu = −i~β is defined as uH/~ −uH/~ φe(x, u) = φe(x, ~β) = e φ(x) e . (7.160) in which φ(x) = φ(x, 0) = φe(x, 0) is the field at time zero, which obeys the commutation relations (7.122). The euclidian-time-ordered product of several fields is their product with newer (higher u = ~β) fields standing to the left of older (lower u = ~β) fields as in the definition (7.109). The euclidian path integrals for the mean values of euclidian-time-ordered- products of fields are similar to those (7.161 & 7.148) for ordinary time- ordered-products. The euclidian-time-ordered-product of the fields φ(xj) = φ(xj, uj) is the path integral Z −Se[φ]/ hn|φbiφ(x1)φ(x2)e ~hφa|ni Dφ hn|T [φe(x1)φe(x2)]|ni = Z (7.161) −S [φ]/ hn|φbie e ~hφa|ni Dφ in which the integrations are over all paths that go from before u1 and u2 to after both euclidian times. The analogous result for several fields is Z −Se[φ]/ hn|φbiφ(x1) ··· φ(xk)e ~hφa|ni Dφ hn|T [φe(x1) ··· φe(xk)]|ni = Z −S [φ]/ hn|φbie e ~hφa|ni Dφ (7.162) 192 Path integrals in which the integrations are over all paths that go from before the times u1, . . . , uk to after them. A distinctive feature of these formulas is that in the low-temperature β = 1/(kT ) → ∞ limit, the Boltzmann operator is a multiple of an outer product |0ih0| of the ground-state kets, e−βH → e−βE0 |0ih0|. In this limit, the integrations are over all fields that run from u = −∞ to u = ∞ and the energy eigenstates are the ground state of the theory Z −Se[q]/ h0|φbiφ(x1) ··· φ(xk)e ~hφa|0i Dφ h0|T [φe(x1) ··· φe(xk)]|0i = Z . −S [q]/ h0|φbie e ~hφa|0i Dφ (7.163) Formulas like this one are used in lattice gauge theory.

7.10 Perturbation theory Field theories with hamiltonians that are quadratic in their ﬁelds like Z 1 h 2 2 2 2 i 3 H0 = 2 π (x) + (∇φ(x)) + m φ (x) d x (7.164) are soluble. Their ﬁelds evolve in time as

φ(x, t) = eitH0 φ(x, 0)e−itH0 . (7.165)

The mean value in the ground state of H0 of a time-ordered product of these fields is a ratio (7.148) of path integrals Z iS0[φ] h0|φbi φ(x1) . . . φ(xn) e hφa|0i Dφ h0|T [φ(x1) . . . φ(xk)] |0i = Z (7.166) iS0[φ] h0|φbi e hφa|0i Dφ in which the action S0[φ] is quadratic in the field φ Z 1 h ˙2 2 2 2 i 4 S0[φ] = 2 φ (x) − (∇φ(x)) − m φ (x) d x Z (7.167) 1 a 2 2 4 = 2 − ∂aφ(x)∂ φ(x) − m φ (x) d x and the integrations are over all fields that run from φa at a time before the times t1, . . . , tk to φb at a time after t1, . . . , tk. The path integrals in the ratio (7.166) are gaussian and doable. 7.10 Perturbation theory 193 The Fourier transforms Z Z d4p φ˜(p) = e−ipxφ(x) d4x and φ(x) = eipxφ˜(p) (7.168) (2π)4 turn the spacetime derivatives in the action into a quadratic form Z d4p S [φ] = − 1 |φ˜(p)|2 (p2 + m2) (7.169) 0 2 (2π)4 in which p2 = p2 − p02 and φ˜(−p) = φ˜∗(p) by (??) since the field φ is real. The initial hφa|0i and final h0|φbi wave functions produce the i in the Feynman propagator (2.24). Although its exact form doesn’t matter here, the wave function hφ|0i of the ground state of H0 is the exponential (??)

1 Z p d3p hφ|0i = c exp − |φ˜(p)|2 p2 + m2 (7.170) 2 (2π)3 in which φ˜(p) is the spatial Fourier transform of the eigenvalue φ(x) Z φ˜(p) = e−ip·x φ(x) d3x (7.171) and c is a normalization factor that will cancel in ratios of path integrals. Apart from −2i ln c which we will not keep track of, the wave functions hφa|0i and h0|φbi add to the action S0[φ] the term i Z p d3p ∆S [φ] = p2 + m2 |φ˜(p, t)|2 + |φ˜(p, −t)|2 (7.172) 0 2 (2π)3 in which we envision taking the limit t → ∞ with φ(x, t) = φb(x) and φ(x, −t) = φa(x). The identity (Weinberg, 1995, pp. 386–388) Z ∞ f(+∞) + f(−∞) = lim f(t) e−|t| dt (7.173) →0+ −∞

(exercise ??) allows us to write ∆S0[φ] as iZ Z ∞ d3p p 2 2 ˜ 2 −|t| ∆S0[φ] = lim p + m |φ(p, t)| e dt 3 . (7.174) →0+ 2 −∞ (2π) To first order in , the change in the action is (exercise ??) iZ Z ∞ d3p p 2 2 ˜ 2 ∆S0[φ] = lim p + m |φ(p, t)| dt 3 →0+ 2 −∞ (2π) iZ p d4p = lim p2 + m2 |φ˜(p)|2 . (7.175) →0+ 2 (2π)4 194 Path integrals Thus the modified action is 1 Z p d4p S [φ, ] = S [φ] + ∆S [φ] = − |φ˜(p)|2 p2 + m2 − i p2 + m2 0 0 0 2 (2π)4 1 Z d4p = − |φ˜(p)|2 p2 + m2 − i (7.176) 2 (2π)4 since the square root is positive. In terms of the modified action, our formula (7.166) for the time-ordered product is the ratio Z iS0[φ,] φ(x1) . . . φ(xn) e Dφ h0|T [φ(x1) . . . φ(xn)] |0i = Z . (7.177) eiS0[φ,] Dφ

We can use this formula (7.177) to express the mean value in the vacuum |0i of the time-ordered exponential of a spacetime integral of j(x)φ(x), in which j(x) is a classical (c-number, external) current, as the ratio Z 4 Z0[j] ≡ h0| T exp i j(x) φ(x) d x |0i Z Z exp i j(x) φ(x) d4x eiS0[φ,] Dφ (7.178) = Z . eiS0[φ,] Dφ

Since the state |0i is normalized, the mean value Z0[0] is unity,

Z0[0] = 1. (7.179) If we absorb the current into the action Z 4 S0[φ, , j] = S0[φ, ] + j(x) φ(x) d x (7.180) then in terms of the current’s Fourier transform Z ˜j(p) = e−ipx j(x) d4x (7.181) the modiﬁed action S0[φ, , j] is (exercise ??) Z h i d4p S [φ, , j] = − 1 |φ˜(p)|2 p2 + m2 − i − ˜j∗(p)φ˜(p) − φ˜∗(p)˜j(p) . 0 2 (2π)4 (7.182) Changing variables to

ψ˜(p) = φ˜(p) − ˜j(p)/(p2 + m2 − i) (7.183) 7.10 Perturbation theory 195 we write the action S0[φ, , j] as (exercise ??) Z ˜j∗(p)˜j(p) d4p S [φ, , j] = − 1 |ψ˜(p)|2 p2 + m2 − i − 0 2 (p2 + m2 − i) (2π)4 Z ˜j∗(p)˜j(p) d4p = S [ψ, ] + 1 . (7.184) 0 2 (p2 + m2 − i) (2π)4 And since Dφ = Dψ, our formula (7.178) gives simply (exercise ??) i Z |˜j(p)|2 d4p Z [j] = exp . (7.185) 0 2 p2 + m2 − i (2π)4 Going back to position space, one ﬁnds (exercise ??) i Z Z [j] = exp j(x) ∆(x − x0) j(x0) d4x d4x0 (7.186) 0 2 in which ∆(x − x0) is Feynman’s propagator (2.24)

Z eip(x−x0) d4p ∆(x − x0) = . (7.187) p2 + m2 − i (2π)4

The functional derivative (chapter ??) of Z0[j], deﬁned by (7.178), is 1 δZ [j] Z 0 = h0| T φ(x) exp i j(x0)φ(x0)d4x0 |0i (7.188) i δj(x) while that of equation (7.186) is 1 δZ [j] Z 0 = Z [j] ∆(x − x0) j(x0) d4x0. (7.189) i δj(x) 0

Thus the second functional derivative of Z0[j] evaluated at j = 0 gives 2 0 1 δ Z0[j] 0 h0| T φ(x)φ(x ) |0i = 2 0 = −i ∆(x − x ). (7.190) i δj(x)δj(x ) j=0 Similarly, one may show (exercise ??) that 4 1 δ Z0[j] h0| T [φ(x1)φ(x2)φ(x3)φ(x4)] |0i = 4 i δj(x1)δj(x2)δj(x3)δj(x4) j=0 = − ∆(x1 − x2)∆(x3 − x4) − ∆(x1 − x3)∆(x2 − x4)

− ∆(x1 − x4)∆(x2 − x3). (7.191) Suppose now that we add a potential V = P (φ) to the free hamiltonian (7.164). Scattering amplitudes are matrix elements of the time-ordered exponential T exp −i R P (φ) d4x (Weinberg, 1995, p. 260) Our formula (7.177) 196 Path integrals for the mean value in the ground state |0i of the free hamiltonian H0 of any time-ordered product of ﬁelds leads us to Z Z 4 iS0[φ,] Z exp −i P (φ) d x e Dφ 4 h0|T exp −i P (φ) d x |0i = Z . eiS0[φ,] Dφ (7.192) Using (7.190 & 7.191), we can cast this expression into the magical form Z Z 4 δ 4 h0|T exp −i P (φ) d x |0i = exp −i P d x Z0[j] . iδj(x) j=0 (7.193) The generalization of the path-integral formula (7.177) to the ground state |Ωi of an interacting theory with action S is Z iS[φ,] φ(x1) . . . φ(xn) e Dφ hΩ|T [φ(x1) . . . φ(xn)] |Ωi = Z (7.194) eiS[φ,] Dφ in which a term like iφ2 is added to make the modiﬁed action S[φ, ]. These are some of the techniques one uses to make states of incoming and outgoing particles and to compute scattering amplitudes (Weinberg, 1995, 1996; Srednicki, 2007; Zee, 2010).

7.11 Application to quantum electrodynamics In the Coulomb gauge ∇ · A = 0, the QED hamiltonian is Z 1 2 1 2 3 H = Hm + 2 π + 2 (∇ × A) − A · j d x + VC (7.195) in which Hm is the matter hamiltonian, and VC is the Coulomb term 1 Z j0(x, t) j0(y, t) V = d3x d3y. (7.196) C 2 4π|x − y| The operators A and π are canonically conjugate, but they satisfy the Coulomb-gauge conditions

∇ · A = 0 and ∇ · π = 0. (7.197)

One may show (Weinberg, 1995, pp. 413–418) that in this theory, the 7.11 Application to quantum electrodynamics 197 analog of equation (7.194) is Z iSC O1 ... On e δ[∇ · A] DA Dψ hΩ|T [O1 ... On] |Ωi = Z (7.198) eiSC δ[∇ · A] DA Dψ in which the Coulomb-gauge action is Z Z 1 ˙ 2 1 2 4 SC = 2 A − 2 (∇ × A) + A · j + Lm d x − VC dt (7.199) and the functional delta function Y δ[∇ · A] = δ(∇ · A(x)) (7.200) x enforces the Coulomb-gauge condition. The term Lm is the action density of the matter ﬁeld ψ. Tricks are available. We introduce a new ﬁeld A0(x) and consider the factor

Z Z 1 2 F = exp i ∇A0+∇4−1j0 d4x DA0 (7.201) 2 which is just a number independent of the charge density j0 since we can cancel the j0 term by shifting A0. By 4−1, we mean − 1/4π|x − y|. By integrating by parts, we can write the number F as (exercise ??) Z Z 1 02 0 0 1 0 −1 0 4 0 F = exp i 2 ∇A − A j − 2 j 4 j d x DA Z Z Z (7.202) 1 02 0 0 4 0 = exp i 2 ∇A − A j d x + i VC dt DA .

So when we multiply the numerator and denominator of the amplitude (7.198) by F , the awkward Coulomb term VC cancels, and we get Z iS0 O1 ... On e δ[∇ · A] DA Dψ hΩ|T [O1 ... On] |Ωi = (7.203) Z 0 eiS δ[∇ · A] DA Dψ where now DA includes all four components Aµ and Z 0 1 ˙ 2 1 2 1 02 0 0 4 S = 2 A − 2 (∇ × A) + 2 ∇A + A · j − A j + Lm d x. (7.204)

Since the delta-functional δ[∇ · A] enforces the Coulomb-gauge condition, we can add to the action S0 the term (∇ · A˙) A0 which is − A˙ · ∇A0 after 198 Path integrals we integrate by parts and drop the surface term. This extra term makes the action gauge invariant Z 1 ˙ 0 2 1 2 0 0 4 S = 2 (A − ∇A ) − 2 (∇ × A) + A · j − A j + Lm d x Z (7.205) 1 ab b 4 = − 4 Fab F + A jb + Lm d x.

Thus at this point we have Z iS O1 ... On e δ[∇ · A] DA Dψ hΩ|T [O1 ... On] |Ωi = Z (7.206) eiS δ[∇ · A] DA Dψ in which S is the gauge-invariant action (7.205), and the integral is over all ﬁelds. The only relic of the Coulomb gauge is the gauge-ﬁxing delta functional δ[∇ · A]. We now make the gauge transformation

0 0 iqΛ(x) Ab(x) = Ab(x) + ∂bΛ(x) and ψ (x) = e ψ(x) (7.207) in the numerator and also, using a different gauge transformation Λ0, in the denominator of the ratio (7.206) of path integrals. Since we are integrating over all gauge fields, these gauge transformations merely change the order of integration in the numerator and denominator of that ratio. They are like R ∞ R ∞ replacing −∞ f(x) dx by −∞ f(y) dy. They change nothing, and so 0 hΩ|T [O1 ... On] |Ωi = hΩ|T [O1 ... On] |Ωi (7.208) in which the prime refers to the gauge transformations (7.207) Λ and Λ0. We’ve seen that the action S is gauge invariant. So is the measure DA Dψ. We now restrict ourselves to operators O1 ... On that are gauge invariant. So in the right-hand side of equation (7.208), the replacement of the fields by their gauge transforms affects only the term δ[∇ · A] that enforces the Coulomb-gauge condition Z iS O1 ... On e δ[∇ · A + 4Λ] DA Dψ hΩ|T [O1 ... On] |Ωi = Z . (7.209) eiS δ[∇ · A + 4Λ0] DA Dψ

We now have two choices. If we integrate over all gauge functions Λ(x) and Λ0(x) in both the numerator and the denominator of this ratio (7.209), then apart from over-all constants that cancel, the mean value in the vacuum of 7.11 Application to quantum electrodynamics 199 the time-ordered product is the ratio Z iS O1 ... On e DA Dψ hΩ|T [O1 ... On] |Ωi = Z (7.210) eiS DA Dψ in which we integrate over all matter fields, gauge fields, and gauges. That is, we do not fix the gauge. The analogous formula for the euclidian time-ordered product is Z −Se O1 ... On e DA Dψ hΩ|T [Oe,1 ... Oe,n] |Ωi = Z (7.211) e−Se DA Dψ in which the euclidian action Se is the spacetime integral of the energy density. This formula is quite general; it holds in nonabelian gauge theories and is important in lattice gauge theory. Our second choice is to multiply the numerator and the denominator of the 1 R 2 4 ratio (7.209) by the exponential exp[−i 2 α (4Λ) d x] and then integrate over Λ(x) in the numerator and over Λ0(x) in the denominator. This oper- ation just multiplies the numerator and denominator by the same constant factor, which cancels. But if before integrating over all gauge transformations, we shift Λ so that 4Λ changes to 4Λ−A˙ 0, then the exponential factor 1 R ˙ 0 2 4 is exp[−i 2 α (A − 4Λ) d x]. Now when we integrate over Λ(x), the delta function δ(∇ · A + 4Λ) replaces 4Λ by − ∇ · A in the inserted exponen- 1 R ˙ 0 2 4 tial, converting it to exp[−i 2 α (A + ∇ · A) d x]. This term changes the gauge-invariant action (7.205) to the gauge-fixed action Z 1 α S = − F F ab − (∂ Ab)2+Abj + L d4x. (7.212) α 4 ab 2 b b m This Lorentz-invariant, gauge-fixed action is much easier to use than the Coulomb-gauge action (7.199) with the Coulomb potential (7.196). We can use it to compute scattering amplitudes perturbatively. The mean value of a time-ordered product of operators in the ground state |0i of the free theory is Z iSα O1 ... On e DA Dψ h0|T [O1 ... On] |0i = Z . (7.213) eiSα DA Dψ

By following steps analogous to those the led to (7.187), one may show 200 Path integrals (exercise ??) that in Feynman’s gauge, α = 1, the photon propagator is Z η d4q h0|T [A (x)A (y)] |0i = − i4 (x − y) = − i µν eiq·(x−y) . µ ν µν q2 − i (2π)4 (7.214)

7.12 Fermionic path integrals In our brief introduction (??–??) and (??–??), to Grassmann variables, we learned that because θ2 = 0 (7.215) the most general function f(θ) of a single Grassmann variable θ is f(θ) = a + b θ. (7.216) So a complete integral table consists of the integral of this linear function Z Z Z Z f(θ) dθ = a + b θ dθ = a dθ + b θ dθ. (7.217)

This equation has two unknowns, the integral R dθ of unity and the integral R θ dθ of θ. We choose them so that the integral of f(θ + ζ) Z Z Z Z f(θ + ζ) dθ = a + b (θ + ζ) dθ = (a + b ζ) dθ + b θ dθ (7.218) is the same as the integral (7.217) of f(θ). Thus the integral R dθ of unity must vanish, while the integral R θ dθ of θ can be any constant, which we choose to be unity. Our complete table of integrals is then Z Z dθ = 0 and θ dθ = 1. (7.219)

The anticommutation relations for a fermionic degree of freedom ψ are {ψ, ψ†} ≡ ψ ψ† + ψ†ψ = 1 and {ψ, ψ} = {ψ†, ψ†} = 0. (7.220) Because ψ has ψ†, it is conventional to introduce a variable θ∗ = θ† that anti-commutes with itself and with θ {θ∗, θ∗} = {θ∗, θ} = {θ, θ} = 0. (7.221) The logic that led to (7.219) now gives Z Z dθ∗ = 0 and θ∗ dθ∗ = 1. (7.222) 7.12 Fermionic path integrals 201 We deﬁne the reference state |0i as |0i ≡ ψ|si for a state |si that is not annihilated by ψ. Since ψ2 = 0, the operator ψ annihilates the state |0i

ψ|0i = ψ2|si = 0. (7.223)

The eﬀect of the operator ψ on the state

† 1 ∗ † 1 ∗ |θi = exp ψ θ − 2 θ θ |0i = 1 + ψ θ − 2 θ θ |0i (7.224) is

† 1 ∗ θ|θi = θ(1 + ψ θ − 2 θ θ)|0i = θ|0i. (7.226) The state |θi therefore is an eigenstate of ψ with eigenvalue θ

ψ|θi = θ|θi. (7.227)

Z ∗ Z Z ec θ θ dθ∗dθ = (1 + c θ∗θ) dθ∗dθ = (1 − c θ θ∗) dθ∗dθ = −c. (7.230)

The identity operator for the space of states

c|0i + d|1i ≡ c|0i + dψ†|0i (7.231) 202 Path integrals is (exercise ??) the integral Z I = |θihθ| dθ∗dθ = |0ih0| + |1ih1| (7.232) in which the diﬀerentials anti-commute with each other and with other fermionic variables: {dθ, dθ∗} = 0, {dθ, θ} = 0, {dθ, ψ} = 0, and so forth. The case of several Grassmann variables θ1, θ2, . . . , θn and several Fermi operators ψ1, ψ2, . . . , ψn is similar. The θk anticommute among themselves

∗ ∗ ∗ {θi, θj} = {θi, θj } = {θi , θj } = 0 (7.233) while the ψk satisfy

† † † {ψk, ψ` } = δk` and {ψk, ψl} = {ψk, ψ` } = 0. (7.234)

The reference state |0i is

n ! Y |0i = ψk |si (7.235) k=1 in which |si is any state not annihilated by any ψk (so the resulting |0i isn’t zero). The direct-product state

n ! " n # X 1 Y 1 |θi ≡ exp ψ†θ − θ∗θ |0i = 1 + ψ†θ − θ∗θ |0i k k 2 k k k k 2 k k k=1 k=1 (7.236) is (exercise ??) a simultaneous eigenstate of each ψk

ψk|θi = θk|θi. (7.237)

It follows that

θ`θk|θi = ψ`ψk|θi = −ψkψ`|θi = −θkθ`|θi (7.239)

" n #" n # Y 1 Y 1 hζ|θi = h0| (1 + ζ∗ψ − ζ∗ζ ) (1 + ψ†θ − θ∗θ ) |0i k k 2 k k ` ` 2 ` ` k=1 `=1 " n # X 1 † † † = exp ζ∗θ − (ζ∗ζ + θ∗θ ) = eζ θ−(ζ ζ+θ θ)/2. (7.240) k k 2 k k k k k=1

The identity operator is

Z n Y ∗ I = |θihθ| dθkdθk. (7.241) k=1

Example 7.10 (Gaussian Grassmann Integral) For any 2 × 2 matrix A, we may compute the gaussian integral

Z −θ†Aθ ∗ ∗ g(A) = e dθ1dθ1dθ2dθ2 (7.242) by expanding the exponential. The only terms that survive are the ones that ∗ ∗ have exactly one of each of the four variables θ1, θ2, θ1, and θ2. Thus, the integral is the determinant of the matrix A

Z 1 g(A) = (θ∗A θ )2 dθ∗dθ dθ∗dθ 2 k k` ` 1 1 2 2 Z ∗ ∗ ∗ ∗ ∗ ∗ = (θ1A11θ1 θ2A22θ2 + θ1A12θ2 θ2A21θ1) dθ1dθ1dθ2dθ2

= A11A22 − A12A21 = det A. (7.243)

The natural generalization to n dimensions

Z n −θ†Aθ Y ∗ e dθkdθk = det A (7.244) k=1 is true for any n × n matrix A. If A is invertible, then the invariance of 204 Path integrals Grassmann integrals under translations implies that Z n Z n −θ†Aθ+θ†ζ+ζ†θ Y ∗ −θ†A(θ+A−1ζ)+θ†ζ+ζ†(θ+A−1ζ) Y ∗ e dθkdθk = e dθkdθk k=1 k=1 Z n −θ†Aθ+ζ†θ+ζ†A−1ζ Y ∗ = e dθkdθk k=1 Z n −(θ†+ζ†A−1)Aθ+ζ†θ+ζ†A−1ζ Y ∗ = e dθkdθk k=1 Z n −θ†Aθ+ζ†A−1ζ Y ∗ = e dθkdθk k=1 † −1 = det A eζ A ζ . (7.245) The values of θ and θ† that make the argument −θ†Aθ+θ†ζ+ζ†θ of the exponential stationary are θ = A−1ζ and θ† = ζ†A−1. So a gaussian Grassmann integral is equal to its exponential evaluated at its stationary point, apart from a prefactor involving the determinant det A. This result is a fermionic echo of the bosonic result (??).

One may further extend these definitions to a Grassmann field χm(x) and an associated Dirac field ψm(x). The χm(x)’s anticommute among themselves and with all fermionic variables at all points of spacetime

0 ∗ 0 ∗ ∗ 0 {χm(x), χn(x )} = {χm(x), χn(x )} = {χm(x), χn(x )} = 0 (7.246) and the Dirac field ψm(x) obeys the equal-time anticommutation relations † 0 0 {ψm(x, t), ψ (x , t)} = δmn δ(x − x )(n, m = 1,..., 4) n (7.247) 0 † † 0 {ψm(x, t), ψn(x , t)} = {ψm(x, t), ψn(x , t)} = 0. As in (7.235), we use eigenstates of the field ψ at t = 0. If |0i is defined in terms of a state |si that is not annihilated by any ψm(x, 0) as " # Y |0i = ψm(x, 0) |si (7.248) m,x then (exercise ??) the state ! Z X 1 |χi = exp ψ† (x, 0) χ (x) − χ∗ (x)χ (x) d3x |0i m m 2 m m m Z † 1 † 3 = exp ψ χ − 2 χ χ d x |0i (7.249) 7.12 Fermionic path integrals 205 is an eigenstate of the operator ψm(x, 0) with eigenvalue χm(x)

ψm(x, 0)|χi = χm(x)|χi. (7.250)

The inner product of two such states is (exercise ??)

Z 1 1 hχ0|χi = exp χ0†χ − χ0†χ0 − χ†χ d3x . (7.251) 2 2 The identity operator is the integral Z I = |χihχ| Dχ∗Dχ (7.252) in which ∗ Y ∗ Dχ Dχ ≡ dχm(x)dχm(x). (7.253) m,x

The hamiltonian for a free Dirac ﬁeld ψ of mass m is the spatial integral Z 3 H0 = ψ (γ · ∇ + m) ψ d x (7.254) in which ψ ≡ iψ†γ0 and the gamma matrices (??) satisfy

{γa, γb} = 2 ηab (7.255) where η is the 4 × 4 diagonal matrix with entries (−1, 1, 1, 1). Since ψ|χi = 0 † 0 0† 0 χ|χi and hχ |ψ = hχ |χ , the quantity hχ | exp( − iH0)|χi is by (7.251) Z hχ0|e−iH0 |χi = hχ0|χi exp − i χ0 (γ · ∇ + m) χ d3x (7.256) Z 1 0† † 1 0† 0 0 3 = exp 2 (χ − χ )χ − 2 χ (χ − χ) − iχ (γ·∇ + m)χd x Z h 1 † 1 0† 0 i 3 = exp 2 χ˙ χ − 2 χ χ˙ − iχ (γ · ∇ + m) χ d x in which χ0† − χ† = χ˙ † and χ0 − χ = χ˙. Everything within the square brackets is multiplied by , so we may replace χ0† by χ† and χ0 by χ so as to write to ﬁrst order in Z 0 −iH0 1 † 1 † 3 hχ |e |χi = exp 2 χ˙ χ − 2 χ χ˙ − iχ (γ · ∇ + m) χ d x (7.257) in which the dependence upon χ0 is through the time derivatives. 206 Path integrals Putting together n = 2t/ such matrix elements, integrating over all intermediate-state dyadics |χihχ|, and using our formula (7.252), we ﬁnd Z Z −2itH0 1 † 1 † 4 ∗ hχt|e |χ−ti = exp 2 χ˙ χ − 2 χ χ˙ − iχ (γ·∇ + m) χd x Dχ Dχ. (7.258) Integratingχ ˙ †χ by parts and dropping the surface term, we get Z Z −2itH0 † 4 ∗ hχt|e |χ−ti = exp − χ χ˙ − iχ (γ·∇ + m) χ d x Dχ Dχ. (7.259)

Since − χ†χ˙ = − iχγ0χ˙, the argument of the exponential is Z Z 0 4 µ 4 i − χγ χ˙ − χ (γ · ∇ + m) χ d x = i − χ (γ ∂µ + m) χ d x. (7.260)

We then have Z Z −2itH0 4 ∗ hχt|e |χ−ti = exp i L0(χ) d x Dχ Dχ (7.261)

µ in which L0(χ) = − χ (γ ∂µ + m) χ is the action density (4.56) for a free Dirac ﬁeld. Thus the amplitude is a path integral with phases given by the classical action S0[χ]

Z R 4 Z −2itH0 i L0(χ) d x ∗ iS0[χ] ∗ hχt|e |χ−ti = e Dχ Dχ = e Dχ Dχ (7.262) and the integral is over all ﬁelds that go from χ(x, −t) = χ−t(x) to χ(x, t) = χt(x). Any normalization factor will cancel in ratios of such integrals. Since Fermi ﬁelds anticommute, their time-ordered product has an extra minus sign

0 0 0 0 T ψ(x1)ψ(x2) = θ(x1 − x2) ψ(x1) ψ(x2) − θ(x2 − x1) ψ(x2) ψ(x1). (7.263)

The logic behind our formulas (7.148) and (7.166) for the time-ordered product of bosonic fields now leads to an expression for the time-ordered product of 2n Dirac fields (with Dχ00 and Dχ0 and so forth suppressed) Z 00 iS0[χ] 0 ∗ h0|χ i χ(x1) . . . χ(x2n) e hχ |0i Dχ Dχ h0|T ψ(x1) . . . ψ(x2n) |0i = Z . h0|χ00i eiS0[χ]hχ0|0i Dχ∗Dχ (7.264) As in (7.177), the effect of the inner products h0|χ00i and hχ0|0i is to insert 7.12 Fermionic path integrals 207 -terms which modify the Dirac propagators Z iS0[χ,] ∗ χ(x1) . . . χ(x2n) e Dχ Dχ h0|T ψ(x1) . . . ψ(x2n) |0i = Z . (7.265) eiS0[χ,] Dχ∗Dχ

Imitating (7.178), we introduce a Grassmann external current ζ(x) and deﬁne a fermionic analog of Z0[j]

Z R 4 e ζχ+χζ d xeiS0[χ,]Dχ∗Dχ h R ζψ+ψζ d4xi Z0[ζ] ≡ h0| T e |0i = Z . (7.266) eiS0[χ,]Dχ∗Dχ

Example 7.11 (Feynman’s fermion propagator) Since

Z d4p −i (−iγνp + m) i (γµ∂ + m) ∆(x − y) ≡ i (γµ∂ + m) eip(x−y) ν µ µ (2π)4 p2 + m2 − i Z d4p (−iγνp + m) = eip(x−y) (iγµp + m) ν (2π)4 µ p2 + m2 − i Z d4p p2 + m2 = eip(x−y) = δ4(x − y), (2π)4 p2 + m2 − i (7.267)

µ the function ∆(x − y) is the inverse of the diﬀerential operator i(γ ∂µ + m). Thus the Grassmann identity (7.245) implies that Z0[ζ] is

Z R µ 4 e [ζχ+χζ−i χ(γ ∂µ+m)χ]d xDχ∗Dχ h R ζψ+ψζ d4xi h0| T e |0i = Z eiS0[χ,]Dχ∗Dχ (7.268) Z = exp ζ(x)∆(x − y)ζ(y) d4xd4y .

Diﬀerentiating we get

Z d4p −iγνp + m h0|T ψ(x)ψ(y) |0i = ∆(x − y) = −i eip(x−y) ν . (7.269) (2π)4 p2 + m2 − i 208 Path integrals 7.13 Application to nonabelian gauge theories The action of a generic non-abelian gauge theory is Z 1 µν µ 4 S = − 4 FaµνFa − ψ (γ Dµ + m) ψ d x (7.270) in which the Maxwell ﬁeld is

Faµν ≡ ∂µAaν − ∂νAaµ + g fabc Abµ Acν (7.271) and the covariant derivative is

Dµψ ≡ ∂µψ − ig ta Aaµ ψ. (7.272)

Here g is a coupling constant, fabc is a structure constant (??), and ta is a generator (??) of the Lie algebra (section ??) of the gauge group. One may show (Weinberg, 1996, pp. 14–18) that the analog of equation (7.206) for quantum electrodynamics is Z iS O1 ... On e δ[Aa3] DA Dψ hΩ|T [O1 ... On] |Ωi = Z (7.273) iS e δ[Aa3] DA Dψ in which the functional delta function Y δ[Aa3] ≡ δ(Aa3(x)) (7.274) x,b enforces the axial-gauge condition, and Dψ stands for Dψ∗Dψ. Initially, physicists had trouble computing nonabelian amplitudes beyond the lowest order of perturbation theory. Then DeWitt showed how to compute to second order (DeWitt, 1967), and Faddeev and Popov, using path integrals, showed how to compute to all orders (Faddeev and Popov, 1967).

7.14 Faddeev-Popov trick The path-integral tricks of Faddeev and Popov are described in (Weinberg, 1996, pp. 19–27). We will use gauge-fixing functions Ga(x) to impose a gauge a condition on our non-abelian gauge fields Aµ(x). For instance, we can use 3 µ Ga(x) = Aa(x) to impose an axial gauge or Ga(x) = i∂µAa (x) to impose a Lorentz-invariant gauge. Under an infinitesimal gauge transformation (??)

λ Aaµ = Aaµ − ∂µλa − g fabc Abµ λc (7.275) 7.14 Faddeev-Popov trick 209 the gauge fields change, and so the gauge-fixing functions Gb(x), which depend upon them, also change. The jacobian J of that change at λ = 0 is λ λ δGa(x) DG J = det ≡ (7.276) δλb(y) λ=0 Dλ λ=0 and it typically involves the delta function δ4(x − y). Let B[G] be any functional of the gauge-fixing functions Gb(x) such as Y Y 3 B[G] = δ(Ga(x)) = δ(Aa(x)) (7.277) x,a x,a in an axial gauge or i Z i Z B[G] = exp (G (x))2 d4x = exp − (∂ Aµ(x))2 d4x (7.278) 2 a 2 µ a in a Lorentz-invariant gauge. We want to understand functional integrals like (7.273) Z iS O1 ... On e B[G] J DA Dψ hΩ|T [O1 ... On] |Ωi = Z (7.279) eiS B[G] J DA Dψ in which the operators Ok, the action functional S[A], and the differentials DADψ (but not the gauge-fixing functional B[G] or the Jacobian J) are gauge invariant. The axial-gauge formula (7.273) is a simple example in which B[G] = δ[Aa3] enforces the axial-gauge condition Aa3(x) = 0 and the determinant J = det (δab∂3δ(x − y)) is a constant that cancels. If we translate the gauge fields by gauge transformations Λ and Λ0, then the ratio (7.279) does not change Z Λ Λ iSΛ Λ Λ Λ Λ O1 ... On e B[G ] J DA Dψ hΩ|T [O1 ... On] |Ωi = (7.280) Z Λ0 0 0 0 0 eiS B[GΛ ] J Λ DAΛ DψΛ R R any more than f(y) dy is different from f(x) dx. Since the operators Ok, the action functional S[A], and the differentials DADψ are gauge invariant, most of the Λ-dependence goes away Z iS Λ Λ O1 ... On e B[G ] J DA Dψ hΩ|T [O1 ... On] |Ωi = . (7.281) Z 0 0 eiS B[GΛ ] J Λ DA Dψ

Let Λλ be a gauge transformation Λ followed by an inﬁnitesimal gauge 210 Path integrals transformation λ. The jacobian J Λ is a determinant of a product of matrices which is a product of their determinants

Λλ Z Λλ Λ δGa (x) δGa (x) δΛλc(z) 4 J = det = det d z δλb(y) λ=0 δΛλc(z) δλb(y) λ=0 Λλ δGa (x) δΛλc(z) = det det δΛλc(z) λ=0 δλb(y) λ=0 Λ Λ δGa (x) δΛλc(z) DG DΛλ = det det ≡ . (7.282) δΛc(z) δλb(y) λ=0 DΛ Dλ λ=0 Now we integrate over the gauge transformations Λ (and Λ0) with weight −1 function ρ(Λ) = (DΛλ/Dλ|λ=0) and ﬁnd, since the ratio (7.281) is Λ- independent

Z DGΛ O ... O eiS B[GΛ] DΛ DA Dψ 1 n DΛ hΩ|T [O ... O ] |Ωi = 1 n Λ0 Z 0 DG eiS B[GΛ ] DΛ0 DA Dψ DΛ0 Z iS Λ Λ O1 ... On e B[G ] DG DA Dψ = Z eiS B[GΛ] DGΛ DA Dψ Z iS O1 ... On e DA Dψ = Z . (7.283) eiS DA Dψ

Thus the mean-value in the vacuum of a time-ordered product of gauge- invariant operators is a ratio of path integrals over all gauge fields without any gauge fixing. No matter what gauge condition G or gauge-fixing functional B[G] we use, the resulting gauge-fixed ratio (7.279) is equal to the ratio (7.283) of path integrals over all gauge fields without any gauge fixing. All gauge-fixed ratios (7.279) give the same time-ordered products, and so we can use whatever gauge condition G or gauge-fixing functional B[G] is most convenient. The analogous formula for the euclidian time-ordered product is Z −Se O1 ... On e DA Dψ hΩ|Te [O1 ... On] |Ωi = Z (7.284) e−Se DA Dψ 7.15 Ghosts 211 where the euclidian action Se is the spacetime integral of the energy density. This formula is the basis for lattice gauge theory. The path-integral formulas (7.210 & 7.211) derived for quantum electrodynamics therefore also apply to nonabelian gauge theories.

7.15 Ghosts Faddeev and Popov showed how to do perturbative calculations in which one does ﬁx the gauge. To continue our description of their tricks, we return to the gauge-ﬁxed expression (7.279) for the time-ordered product Z iS O1 ... On e B[G] J DA Dψ hΩ|T [O1 ... On] |Ωi = Z (7.285) eiS B[G] J DA Dψ

µ set Gb(x) = −i∂µAb (x) and use (7.278) as the gauge-ﬁxing functional B[G] i Z i Z B[G] = exp (G (x))2 d4x = exp − (∂ Aµ(x))2 d4x . (7.286) 2 a 2 µ a

µ 2 This functional adds to the action density the term −(∂µAa ) /2 which leads to a gauge-field propagator like the photon’s (7.214) h i Z η δ d4q h0|T Aa (x)Ab (y) |0i = − iδ 4 (x − y) = − i µν ab eiq·(x−y) . µ ν ab µν q2 − i (2π)4 (7.287) What about the determinant J? Under an infinitesimal gauge transformation (7.275), the gauge field becomes

λ Aaµ = Aaµ − ∂µλa − g fabc Abµ λc (7.288) λ µ λ and so Ga(x) = i∂ Aaµ(x) is Z λ µ µ 4 4 Ga(x) = i∂ Aaµ(x) + i∂ [−δac∂µ − g fabcAbµ(x)] δ (x − y)λc(y) d y. (7.289) The jacobian J then is the determinant (7.276) of the matrix λ δGa(x) 4 ∂ µ 4 = −iδac 2 δ (x−y)−ig fabc µ Ab (x)δ (x − y) (7.290) δλc(y) λ=0 ∂x that is ∂ J = det −iδ 2 δ4(x − y) − ig f Aµ(x)δ4(x − y) . (7.291) ac abc ∂xµ b 212 Path integrals But we’ve seen (7.244) that a determinant can be written as a fermionic path integral Z n −θ†A θ Y ∗ det A = e dθkdθk. (7.292) k=1 So we can write the jacobian J as Z Z ∗ ∗ µ 4 ∗ J = exp iωa2ωa + igfabcωa∂µ(Ab ωc) d x Dω Dω (7.293)

∗ µ which contributes the terms −∂µωa∂ ωa and ∗ µ ∗ µ − ∂µωa g fabc Ab ωc = ∂µωa g fabc Ac ωb (7.294) to the action density. Thus we can do perturbation theory by using the modiﬁed action density

0 1 µν 1 µ 2 ∗ µ ∗ µ L = − 4 FaµνFa − 2 (∂µAa ) − ∂µωa∂ ωa + ∂µωa g fabc Ac ωb − ψ (D6 + m) ψ (7.295) µ µ a in which D6 ≡ γ Dµ = γ (∂µ−igt Aaµ). The ghost ﬁeld ω is a mathematical device, not a physical ﬁeld describing real particles, which would be spinless fermions violating the spin-statistics theorem (example ??).

7.16 Integrating over the momenta When a hamiltonian is quadratic in the momenta like (7.128 & 10.8), one easily integrates over its momentum and converts it into its lagrangian. If, however, the hamiltonian is a more complicated function of the momenta, one usually can’t path-integrate over the momenta analytically. The partition function is then a path integral over both φ and π Z Z β Z h i Z(β) = exp iφ˙(x)π(x) − H(φ, π) dtd3x Dφ Dπ. (7.296) 0 This exponential is not positive, and so is not a probability distribution for φ and π. The Monte Carlo methods of chapter ?? are designed for probability distributions, not for distributions that assume negative or complex values. This is one aspect of the sign problem. The integral over the momentum Z Z β Z h i P [φ] = exp iφ˙(x)π(x) − H(φ, π) dtd3x Dπ (7.297) 0 is positive, however, and is a probability distribution. So one can numerically integrate over the momentum, make a look-up table for P [φ], and then apply 7.16 Integrating over the momenta 213 the usual Monte Carlo method to the probability functional P [φ] (Amdahl and Cahill, 2016). Programs that do this are in the repository Path integrals at github.com/kevinecahill. 8 Landau theory of phase transitions

8.1 First- and second-order phase transitions In a first-order phase transition, a physical quantity, called an order parameter, changes discontinuously. Often there is a point where the dis- continuous phase transition becomes continuous. Such a point is called a critical point. The continuous phase transition is called a second-order phase transition. Binary alloys, fluids, and helium-4 exhibit phase transitions. The archetypical example is the ferromagnet. In an idealized ferromagnet with spins parallel or antiparallel to a given axis, the order parameter is the total magnetization M along this axis. At low temperatures, an external magnetic field H will favor one spin direction. At H = 0 the states will be in equilibrium, but small changes in H will make M change discontinuously. There is a first-order phase transition at H = 0 at low temperatures. At slightly higher temperatures, the spins fluctuate and the mean value of |M| decreases. At the critical temperature Tc, tiny changes in H no longer make M change discontinuously. At T > Tc, the phase transition is continuous and second order. Landau used a Gibbs free-energy potential G which obeys

∂G = − H. (8.1) ∂M T

He worked at T ≈ Tc and M ≈ 0 and expanded G(M) at H = 0 as

G(M) = A(T ) + B(T )M 2 + C(T )M 4 + ... (8.2) keeping only even terms because of the symmetry under M → −M. So at 8.1 First- and second-order phase transitions 215 H = 0 equation (8.1) gives

∂G 3 0 = − H = = 2B(T )M + 4C(T )M . (8.3) ∂M T If B and C are positive, the only solution is M = 0. But if C > 0 and if for T < Tc it happens that B < 0, then we get

2CM 2 + B = 0 or M = p−B/2C. (8.4)

The approximations

B(T ) ≈ b(T − Tc) and C(T ) = c (8.5) give p M = ± (b/2c)(Tc − T ) for T < T c (8.6) while M = 0 for T > Tc. The minimum of the potential

G(M,H) = A(T ) + B(T )M 2 + C(T )M 4 − HM (8.7) as a function of M at constant T gives equation (8.3). One may represent the magnetization in terms of a spin density s(x) as Z M = s(x) d3x. (8.8)

Gibb’s free energy then becomes Z 1 G = (∇s)2 + b(T − T )s2 + cs4 − Hs d3x. (8.9) 2 c The spin density that minimizes the free energy must make G stationary δG 0 = = −∇2s(x) + 2b(T − T )s(x) + 4cs3(x) − H(x). (8.10) δs(x) c

For T > Tc, the macroscopic magnetization vanishes and so the spin density s is small. So we drop the nonlinear term and get

2 −∇ + 2b(T − Tc) s(x) = H(x). (8.11)

3 The spin density D(x) generated by H(x) = H0δ (x) is the Green’s function of the diﬀerential equation

2 −∇ + 2b(T − Tc) D(x) = H0 δ(x). (8.12) 216 Landau theory of phase transitions

Expressing both sides as Fourier transforms, we get for T > Tc Z d3k H eik·x D(x) = 0 (2π)3 2 k + 2b(T − Tc) (8.13) H = 0 e−r/ξ 4πr where the correlation length −1/2 ξ = [2b(T − Tc)] (8.14) which diverges as T → Tc indicating the onset of large-scale fluctuations. Different classes of systems have different critical exponents. For example, when T < Tc, the magnetization can go as β M ≈ (Tc − T ) (8.15) in which instead of β = 1/2 one has β = 0.313 for some fluids and some 3D magnets. −γ The susceptibility χ = (∂M/∂H)|H = 0 diverges as |T − Tc| . We differentiate the equation (8.3) with respect to H

∂G 3 0 = − H = = 2B(T )M + 4C(T )M (8.16) ∂M T getting − 1 = 2Bχ + 4C3M 2χ. (8.17) So −1 χ = . (8.18) 2B + 12CM 2 p Setting B = b(T − Tc), C = c, and M = ± (b/2c)(Tc − T ), we get −1 −1 1 χ = = = (8.19) 2b(T − Tc) + 6b(Tc − T ) 4b(Tc − T ) 4b(T − Tc) which says γ = 1. 9 Eﬀective ﬁeld theories and gravity

9.1 Effective field theories Suppose a field φ whose mass M is huge compared to accessible energies interacts with a field ψ of a low-energy theory such as the standard model 1 a 1 2 2 Lφ = − 2 ∂aφ(x) ∂ φ(x) − 2 M φ (x) + g ψ(x)ψ(x)φ(x). (9.1) 2 a Compared to the mass term M , the derivative terms ∂aφ ∂ φ contribute little to the low-energy path integral. So we drop them and represent the 1 2 2 effect of the heavy field φ as L0 = − 2 M φ + g ψψφ. Completing the square g 2 g2 L = − 1 M 2 φ − ψψ + (ψψ)2 (9.2) 0 2 M 2 2M 2 and shifting φ by gψψ/M 2, we evaluate the path integral over φ as Z Z g2 Z g2 exp i − 1 M 2φ2 + (ψψ)2 d4x Dφ = exp i (ψψ)2 d4x 2 2M 2 2M 2 apart from a field-independent factor. The net effect of heavy field φ is thus to add to the low-energy theory a new interaction g2 L (ψ) = (ψψ)2 (9.3) φ 2M 2 which is small because M 2 is large. If a gauge boson Ya of huge mass M interacts with a spin-one-half field ψ as 1 2 a a L0 = − 2 M YaY + igψγ ψYa (9.4) ig ig g2 = − 1 M 2 Y − ψγ ψ Y a − ψγaψ − ψγaψ ψγ ψ 2 a M 2 a M 2 2M 2 a 218 Effective field theories and gravity then the new low-energy interaction is g2 S = − ψγaψ ψγ ψ. (9.5) Y 2M 2 a The low-energy theory of the weak interactions by Fermi and by Gell-Mann and Feynman is an example, but with γa → γa(1 + γ5). One way of explaining the lightness of the masses of the (known) neutrinos is to say that new a field of very high mass M plays a role, and that when one path-integrates over this heavy field, one is left with an effective, nonrenormalizable term in the action g2 Lσ H∗HTσ L (9.6) M 2 2 which gives a tiny mass to ν`. Here’s how this can work: take as part of the action density of the high-energy theory

T ∗ Lψ = −ψ(∂/ + M)ψ + g ψH σ2L + g Lσ2H ψ (9.7) where M is huge. Drop ∂/ and complete the square: g g g2 L = −M ψ − Lσ H∗ ψ − HTσ L + Lσ H∗HTσ L. (9.8) ψ0 M 2 M 2 M 2 2 The path integral over the field ψ of mass M yields a field-independent constant and leaves in the action the term g2 g2 L = Lσ H∗HTσ L = − (¯νh∗0 − eh¯ −)(h+e − h0ν). (9.9) ν M 2 2 M The mean value of the Higgs field in the vacuum yields the tiny mass term g2v2 L = νν. (9.10) νm 2M If the neutrino is a Dirac field, then this is a Dirac mass term. If the neutrino is a Majorana field, then this is a Majorana mass term. In both cases, the mass is intrinsically tiny g2v2 m ∼ . (9.11) ν 2M

9.2 Is gravity an effective field theory? The action of the gravitational field is c3 Z √ c3 Z √ S = R g d4x = gik R g d4x. (9.12) E 16πG 16πG ik 9.2 Is gravity an effective field theory? 219 The action of fields whose energy-momentum tensor is T ik interacting with gravity is c3 Z √ 1 Z √ S = gik R g d4x + T ikg g d4x. (9.13) 16πG ik 2c ik The action of a free scalar field of mass m in flat spacetime is Z 2 4 1 ˙2 2 2 m c 2 4 Sφf = φ − c (∇φ) − φ d x (9.14) 2 ~2 where d4x = cdtd3x and a dot means a time derivative. In curved spacetime and in natural units, the action is 1 Z √ S = − ∂iφ∂kφ g + m2φ2 g d4x. (9.15) φ 2 ik The total action is c3 Z √ 1 Z √ S = R g d4x − ∂iφ∂kφ g + m2φ2 g d4x. (9.16) 16πG 2 ik √ p The change in g ≡ | det gik| is √ 1 δ det g 1√ 1 √ δ g = − √ ik = g gikδg = − g g δgik. (9.17) 2 g 2 ki 2 ik The change in the inverse of the metric is

ik ij `k δg = − g g δgj`. (9.18) So when the metric changes, the action (9.15 changes by 1 Z h √ √ √ i δS = − ∂iφ∂kφ ( g δg + g δ g) + m2φ2 δ g d4x φ 2 ik ik 1 Z h √ √ √ i = − ∂iφ∂kφ g δg + g 1 g gj`δg + m2φ2 1 g gikδg d4x 2 ik ik 2 `j 2 ki 1 Z h i √ = − ∂iφ∂kφ δg + g 1 gj`δg + m2φ2 1 gikδg g d4x 2 ki ik 2 `j 2 ki 1 Z h i √ = − ∂iφ∂kφ + 1 ∂jφ∂`φ g + 1 m2φ2gik δg g d4x 2 2 j` 2 ki 1 Z h i √ 1 Z √ = Lgik − ∂iφ∂kφ δg g d4x = T ik δg g d4x (9.19) 2 ki 2 ki in which the energy-momentum tensor is

T ik = Lgik − ∂iφ ∂kφ. (9.20) 220 Effective field theories and gravity Equivalently, in terms of the change in the inverse metric (9.18), the change in the scalar action is 1 Z √ δS = − T δgki g d4x (9.21) φ 2 ik The variation of the gravitational action is c3 Z 1 √ δS = R − g R g δgik d4x. (9.22) E 16πG ik 2 ik The variation of the whole action must vanish c3 Z 1 √ 1 Z √ 0 = R − g R g δgik d4x − T δgki g d4x. (9.23) 16πG ik 2 ik 2c ik Thus Einstein’s equations are 1 8πG R − g R = T . (9.24) ik 2 ik c4 ik p In terms of the Planck mass mP = ~c/G, the way the world works is 1 8π~ Rik − gik R = 3 2 Tik (9.25) 2 c mP or in natural units 1 8π Rik − gik R = 2 Tik. (9.26) 2 mP The total action is proportional to Z Z √ 4 i k 2 2 √ 4 Sˆ = R g d x − 8πG ∂ φ∂ φ gik + m φ g d x Z (9.27) h 8π i k 2 2 i√ 4 = R − 2 ∂ φ ∂ φ gik + m φ g d x. mP For a massless scalar field, this is simply Z ˆ 8π i k √ 4 S0 = R − 2 ∂ φ ∂ φ gik g d x. (9.28) mP

The second term in this action Sˆ0 could arise in a path integral over a gauge ﬁeld Ai whose mass is M = mP the Planck mass Z Z √ √ h 1 2 4 π i 4 π i exp i − mP Ai − 2 ∂iφ A − 2 ∂ φ 2 mP mP 8π i k i√ 4 + 2 ∂ φ ∂ φ gik g d x DA (9.29) mP 9.3 Quantization of Fields in Curved Space 221 which might arise from an interaction such as

1 2 i √ i L = − 2 mP AiA + 4 πAi∂ φ. (9.30) Let’s try this again. The usual action is Z 3 √ 4 c 1 ,i 2 2 S = g d x R − φ φ,i + m φ (9.31) 16πG 2

2 where 1/G = M /(~c) in which M = mP is the Planck mass. So Z 2 2 √ 4 c M 1 ,i 2 2 S = g d x R − φ φ,i + m φ . (9.32) 16π~ 2

9.3 Quantization of Fields in Curved Space Some good references for these ideas are: Einstein’s Gravity in a Nutshell by Zee Quantum Fields in Curved Space by Birrell and Davies, Conformal Field Theory by Di Francesco, Mathieu, and Sénéchal Let’s focus on scalar fields for simplicity. We usually expand a scalar field in flat space as Fourier might have (1.56) Z d3p h i φ(x) = a(p) eip·x + a†(p) e−ip·x . (9.33) p(2π)32p0 The field φ obeys the Klein-Gordon equation

2 2 2 2 (∇ − ∂0 − m ) φ(x) ≡ (2 − m ) φ(x) = 0 (9.34) because the ﬂat-space modes, which have p2 = −m2,

ip·x fp(x) = e (9.35) do 2 2 2 2 (∇ − ∂0 − m ) fp(x) ≡ (2 − m ) fp(x) = 0. (9.36)

In terms of these functions fp(x), the ﬁeld is Z 3 d p h † ∗ i φ(x) = a(p) fp(x) + a (p) fp (x) . (9.37) p(2π)32p0

In terms of a discrete set of modes fn, it is

X h † ∗ i φ(x) = an(p) fn(x) + an fn(x) . (9.38) n 222 Eﬀective ﬁeld theories and gravity

The action for a scalar ﬁeld in a space described by the metric gik is Z 1 √ 4 h ik 2 2i S = − g d x g φ,iφ,k + m + ξR φ (9.39) 2 in which R is the scalar curvature, ξ is a constant, commas denote derivatives ij as in φ,k = ∂kφ, g is the inverse of the metric gij, and g is the absolute value of the determinant of the metric gij. If the spacetime metric is gij, then instead of (9.34), the ﬁeld φ obeys the covariant Klein-Gordon equation

h√ ij i 2 √ ∂i g g ∂j φ(x) − m + ξR g φ(x) = 0. (9.40) However, to simplify what follows, we will now set ξ = 0. To quantize in the new coordinates or in the gravitational field of the 0 metric gij, we need solutions fn(x) of the curved-space equation (9.40) h√ ij 0 i 2√ 0 ∂i g g ∂j fn(x) − m g fn(x) = 0 (9.41) which we label with primes to distinguish them from the flat-space solutions (9.35). We use these solutions to expand the field in terms of curved-space annihilation and creation operators, which we also label with primes

X h 0 0 0† 0∗ i φ(x) = an(p) fn(x) + an fn (x) . (9.42) n The ﬂat-space modes obey the orthonormality relations Z 3 h ∗ ∗ i (fp, fq) = i d x fp (x) ∂tfq(x) − ∂tfp (x) fq(x) Z h i = i d3x e−ipx (−iq0)eiqx − ip0e−ipx eiqx (9.43)

= 2p0(2π)3δ3(p − q),

∗ ∗ 0 3 3 ∗ (fp , fq ) = − 2p (2π) δ (p − q) and (fp, fq ) = 0. (9.44) In terms of discrete modes, the ﬂat-space scalar product is Z 3 h ∗ ∗ i (fn, fm) = i d x fn(x) ∂tfm(x) − ∂tfn(x) fm(x) = δnm (9.45) and its orthonormality relations are

∗ ∗ ∗ (fn, fm) = δnm, (fn, fm) = − δnm and (fn, fm) = 0. (9.46) The scalar product (9.45) is a special case of more general curved-space scalar product Z √ 3 ah ∗ ∗ i (f, g) = i gS d S v f (x) ∂ag(x) − ∂af (x) g(x) (9.47) S 9.3 Quantization of Fields in Curved Space 223 in which the integral is over a spacelike surface S with a future-pointing a timelike vector v , and gS is the absolute value of the spatial part of the metric gik. This more general scalar product (9.47) is hermitian

(f, g)∗ = (g, f). (9.48)

It also satisﬁes the rule

(f, g)∗ = − (f ∗, g∗). (9.49)

One may use Gauss’s theorem to show (Hawking and Ellis, 1973, section 2.8) that this inner product is independent of S and v. The curved-space modes fn(x) obey orthonormality relations

0 0 0∗ 0∗ 0 0∗ (fn, fm) = δnm, (fn , fm) = − δnm, and (fn, fm) = 0 (9.50) like those (9.46) of the flat-space modes. ipnx The flat-space modes fn(x) = e naturally describe particles of momentum pn in flat Minkowski space. In curved space, however, there are in general no equally natural modes. We can consider other complete sets of 00 modes fn (x) that are solutions of the curved-space Klein-Gordon equation (9.41) and obey the orthonormality relations (9.50). We can use any of these complete sets of mode functions to expand a scalar field φ(x)

X † ∗ φ(x) = anfn(x) + anfn(x) n X 0 0 0† 0∗ φ(x) = anfn(x) + an fn (x) (9.51) n X 00 00 00† 00∗ φ(x) = anfn (x) + an fn (x). n

The curved-space orthonormality relations (9.50) imply that

X † ∗ (f`, φ) = an(f`, fn) + an(f`, fn) = a` n 0 X 0 0 0 0† 0 0∗ 0 (f`, φ) = an(f`, fn) + an (f`, fn ) = a` (9.52) n 00 X 00 00 00 00† 00 00∗ 00 (f` , φ) = an(f` , fn ) + an (f` , fn ) = a` n 224 Effective field theories and gravity and that ∗ X ∗ † ∗ ∗ † (f` , φ) = an(f` , fn) + an(f` , fn) = −a` n 0∗ X 0 0∗ 0 0† 0∗ 0∗ 0† (f` , φ) = an(f` , fn) + an (f` , fn ) = −a` (9.53) n 00∗ X 00 00∗ 00 00† 00∗ 00∗ 00† (f` , φ) = an(f` , fn ) + an (f` , fn ) = −a` . n The completeness of the mode functions lets us expand them in terms of each other. Suppressing the spacetime argument x, we have 0 X ∗ fj = αjifi + βji fi . (9.54) i The curved-space orthonormality relations (9.50) let us identify these Bo- goliubov coefficients

0 X h ∗ i (f`, fj) = αji(f`, fi) + βji (f`, fi ) = αj` i (9.55) ∗ 0 X h ∗ ∗ ∗ i (f` , fj) = αji(f` , fi) + βji (f` , fi ) = −βj`. i To ﬁnd the inverse relations, we use the completeness of the mode functions 0 fi to expand the fj’s X 0 0∗ fj = cjifi + dji fi (9.56) i and then use the orthonormality relations (9.50) to form the inner products

0 X h 0 0 0 0∗ i (f`, fj) = cji(f`, fi ) + dji (f`, fi ) = cj` i (9.57) 0∗ X h 0∗ 0 0∗ 0∗ i (f` , fj) = cji(f` , fi ) + dji (f` , fi ) = −dj`. i

The hermiticity (9.48) of the scalar product tells us that the cj`’s are related to the α’s 0 ∗ ∗ cj` = (fj, f`) = α`j. (9.58) The hermiticity (9.48) of the scalar product and the rule (9.49) show that 0∗ 0∗ ∗ ∗ 0 dj` = − (f` , fj) = − (fj, f` ) = (fj , f`) = −β`j. (9.59) So the inverse relation (9.56) is

X ∗ 0 0∗ fj = αijfi − βij fi . (9.60) i 9.3 Quantization of Fields in Curved Space 225 The formulas (9.54 & 9.60) that relate the mode functions of different metrics are known as Bogoliubov transformations. The vacuum state for a given metric is the state that is mapped to zero by all the annihilation operators. Our formulas (9.53) for the annihilation and creation operators 0 0 a` = (f`, φ) and a` = (f`, φ) (9.61) † ∗ 0† 0∗ a` = − (f` , φ) and a` = − (f` , φ) let us express the annihilation and creation operators for one metric in terms of those for a different metric. Thus, using our formula (9.54) for the f 0’s in terms of the f’s, we find

0 0 X h ∗ i X † aj = (fj, φ) = αji (fi, φ) + βji (fi , φ) = αji ai − βji ai . (9.62) i i Our formula (9.60) for the f’s in terms of the f 0’s gives us the inverse relation

X h ∗ 0 0∗ i X ∗ 0 0† aj = (fj, φ) = αij (fi , φ) − βij (fi , φ) = αij ai + βij ai . (9.63) i i 0 0 The annihilation operators aj deﬁne the vacuum state |0i by the rules 0 0 aj |0i = 0 (9.64) for all modes j. Thus our formula (9.63) for aj says that

0 X ∗ 0 0† 0 X 0† 0 aj |0i = αij ai + βij ai |0i = βij ai |0i (9.65) i i The adjoint of this equation is

8 † 8 X 0 ∗ h0|aj = h0| aiβij. (9.66) i

† 0 Thus the mean value of the number operator ajaj in the |0i vacuum is

8 † 0 8 X 0 ∗ X 0† 0 h0|ajaj |0i = h0| aiβij βkj ak |0i . (9.67) i k The commutation relations

0 0† [ai, ak ] = δik (9.68) 0 0 0 and the deﬁnition aj |0i = 0 (9.64) of the vacuum |0i imply that the average number (9.67) of particles of mode j in the (normalized) vacuum |0i0 is

8 † 0 8 X ∗ 0 X ∗ h0|ajaj |0i = h0| βijβkj δik|0i = βijβij. (9.69) ik i 226 Eﬀective ﬁeld theories and gravity It follows that the vacuum of one metric contains particles of the other metric unless the Bogoliubov matrix

∗ 0 βj` = − (f` , fj) (9.70) vanishes. The value of βj` in the Minkowski-space scalar product (9.43) is Z ∗ 0 3 h 0 0 i βj` = − (f` , fj) = −i d x f`(x) ∂tfj(x) − ∂tf`(x) fj(x) . (9.71)

0 This integral for βj` is nonzero, for example, when the functions f` and fj have diﬀerent frequencies but are not spatially orthogonal. An example due to Rindler. Let us consider the two metrics ds2 = dx2 + dy2 + dz2 − dt2 = dr2 + dy2 + dz2 − r2du2 (9.72) in which y and z are the same, r plays the role of x and u that of time. The ﬁrst metric has gik = ηik and g = | det(η)| = 1. The second metric has −r2 0 0 0  0 1 0 0 2 gik =   and g = r . (9.73)  0 0 1 0 0 0 0 1 The inverse metric is −r−2 0 0 0 ik  0 1 0 0 g =   . (9.74)  0 0 1 0 0 0 0 1 The equation of motion is

h√ ik i 2√ ∂i gg ∂kf = m gf. (9.75) So we must solve

−1 2 ∂u(−r ∂uf) + ∂r(r∂rf) + ∂y(r∂yf) + ∂z(r∂zf) = m rf. (9.76) Now u, r, y, z are independent coordinates, so this equation of motion is

−1 2 2 2 2 − r ∂uf + ∂r(r∂rf) + r∂y f + r∂z f = m rf (9.77) or −2 2 −1 2 2 2 2 − r ∂uf + r ∂rf + ∂r f + ∂y f + ∂z f = m f. (9.78) Let’s make f a plane wave in the y and z directions

f(u, r, y, z) = f(u, r)eiypy+izPz . (9.79) 9.3 Quantization of Fields in Curved Space 227 If we also set 2 2 2 2 M = m + Py + Pz , (9.80) then we must solve

−2 2 −1 2 2 − r ∂uf + r ∂rf + ∂r f = M f (9.81) where f = f(u, r). We now make the Daniel-Middleton transformation, separating the dependence of f upon r and u

f(u, r) = a(u)b(r). (9.82)

Our diﬀerential equation (9.81) reduces to

− r−2ab¨ + r−1ab0 + ab00 = M 2ab (9.83) in which dots denote ∂u and primes ∂r. Dividing by a, we get a¨ b b0 − + + b00 − M 2b = 0. (9.84) a r2 r As a(u), we choose a¨ a(u) = eiωu and = − ω2. (9.85) a Then our equation (9.84) for b(r) is

b0 ω2 b00 + − M 2 − b = 0 (9.86) r r2 or equivalently r2b00 + rb0 − M 2r2 − ω2 b = 0. (9.87)

Yi’s solution is a modiﬁed Bessel function Iν(z) for imaginary ν. Bogoliubov’s β matrix is

∗ 0 βj` = − (f` , fj). (9.88) The scalar product (9.47) uses the metric of one, not both, of the solutions. If we choose the ﬂat-space metric, then we need to write the solution 0 fj(u, r, y, z) in terms of the ﬂat-space coordinates t, x, y, z. The mean occu- pation number (9.69) is the sum

X 8 † 0 X 2 X ∗ 0 2 h0|ajaj |0i = |βij| = |(fj , fi )| . (9.89) j ij ij 228 Eﬀective ﬁeld theories and gravity 9.4 Accelerated coordinate systems We recall the Lorentz transformations t0 = γ(t − vx), x0 = γ(x − vt), y0 = y, and z0 = z (9.90) t = γ(t0 + vx0), x = γ(x0 + vt0), y = y0, and z = z0 √ in which γ = 1/ 1 − v2. The velocities (in the x direction) are

dx dx0 γ(dx − vdt) (u − v) u = and u0 = = = . (9.91) dt dt0 γ(dt − vdx) (1 − vu)

So the accelerations (in the x direction) are

du a = dt du0 h (u − v) i a0 = = d γ(dt − vdx) dt0 (1 − vu) h du (u − v)vdui = + γ(dt − vdx) (1 − vu) (1 − vu)2 hdu(1 − vu) (u − v)vdui = + γ(dt − vdx) (1 − vu)2 (1 − vu)2 (9.92) du(1 − v2) = γ(1 − vu)2(dt − vdx) (1 − v2)a = γ(1 − vu)2(1 − vu) (1 − v2)a a = = . γ(1 − vu)3 γ3(1 − vu)3

Now we let the acceleration a0 be a constant. That is, the acceleration in the (instantaneous) rest frame of the frame moving instantaneously at u = v in the laboratory frame is a constant, a0 = α. In this case, since u = v, we get an equation (Rindler, 2006, sec. 3.7) a a a a0 = α = = = γ3(1 − vu)3 γ3(1 − v2)3 (1 − v2)3/2 (9.93) 1 du 1 du d u = = = √ (1 − v2)3/2 dt (1 − u2)3/2 dt dt 1 − u2 that we can integrate u √ = α (t − t0). (9.94) 1 − u2 9.5 Scalar ﬁeld in an accelerating frame 229 Squaring and solving for u, we ﬁnd dx α(t − t ) u = = 0 (9.95) p 2 2 dt 1 + α (t − t0) which we can integrate to

Z 2 p 2 2 −1 α (t − t0) 1 + α (t − t0) − 1 x = α dt = + x0. (9.96) p 2 2 1 + α (t − t0) α

0 0 The proper time of the accelerating frame√ is τ = t . An interval√ dt of proper time is one at which dx0 = 0. So dt0 = 1 − v2dt or dtτ = 1 − u2dt where u(t) is the velocity (9.95). Integrating the equation s p α2(t − t )2 dt dt0 = dτ = 1 − u2dt = 1 − 0 dt = , 2 2 p 2 2 1 + α (t − t0) 1 + α (t − t0) (9.97) we get Z 0 0 α dt α(t − t ) = α(τ − τ0) = = arcsinh(α(t − t0)). (9.98) 0 p 2 2 1 + α (t − t0) So

α(t − t0) = sinh(α(τ − τ0)). (9.99)

The formula (9.96) for x now gives

α(x − x0) = cosh(α(τ − τ0)). (9.100)

9.5 Scalar ﬁeld in an accelerating frame For simplicity, we’ll work with a real scalar ﬁeld (1.54)

Z d3p h i φ(x) = a(p) eip·x + a†(p) e−ip·x . (9.101) p(2π)32p0

If the field is quantized in a box of volume V , then the expansion of the field is X 1 h i φ(x) = √ a(k) eik·x + a†(k) e−ik·x . (9.102) 2ω V k k For the scalar field (9.101), the zero-temperature correlation function is 230 Effective field theories and gravity the mean value in the vacuum Z d3p h i h0|φ(x, t)φ(x0, t0)|0i = h0| a(p) eip·x + a†(p) e−ip·x p(2π)32p0 3 0 Z d p h 0 0 0 0 i × a(p0) eip ·x + a†(p0) e−ip ·x |0i p(2π)32p00 3 3 0 Z d pd p 0 0 = h0| a(p)a†(p0) eip·x−ip ·x |0i (2π)3p2p02p00 3 3 0 Z d pd p 0 0 = δ3(p − p0)eip·x−ip ·x (2π)3p2p02p00 3 Z d p 0 = eip·(x−x ). (9.103) (2π)32p0

This integral is simpler for massless ﬁelds. Setting p = |p| and r = |x − x0|, we add a small imaginary part to the exponential and ﬁnd

3 Z d p 0 h0|φ(x, t)φ(x0, t0)|0i = eipr cos θ−ip(t−t −i) (2π)32p

Z pdp dcos θ 0 = eipr cos θ−ip(t−t −i) (2π)22 ipr −ipr Z pdp e − e 0 = e−ip(t−t −i) (2π)22 ipr Z ∞ dp ipr −ipr −ip(t−t0−i) = 2 e − e e 0 (2π) 2ir 1 1 1 = − − (2π)22ir i(r − (t − t0)) i(r + (t − t0)) 1 1 1 = + (2π)22r r − (t − t0) r + (t − t0) 1 = . (9.104) (2π)2[(x − x0)2 − (t − t0)2] This is the zero-temperature correlation function. In the instantaneous rest frame of an accelerated observer moving in the x direction with coordinates (9.99 & 9.100), this two-point function is

α2/(2π)2 h0|φ(x, t)φ(x0, t0)|0i = cosh(ατ) − cosh(ατ 0)2 − sinh(ατ) − sinh(ατ 0)2 α2 = − (9.105) (4π)2 sinh2[α(τ − τ 0)/2] 9.5 Scalar ﬁeld in an accelerating frame 231 in which the identities

cosh(ατ) cosh(ατ 0) − sinh(ατ) sinh(ατ 0) = cosh(α(τ − τ 0)) (9.106) and cosh(α(τ − τ 0)) − 1 = 2 sinh(α(τ − τ 0)/2) (9.107) as well as cosh2 ατ − sinh2 ατ = 1 were used. Now let’s compute the same correlation function at a ﬁnite inverse temperature β = 1/(kBT )

0 −βH . −βH hφ(0, τ)φ(0, τ )iβ = Tr φ(0, 0)φ(0, t)e Tr e . (9.108)

The mean value of the number operator for a given momentum k is

† † −βH . −βH ha (k) a(k)iβ = nk = Tr a (k) a(k)e Tr e (9.109) in which X † 1 H0 = ωk a (k) a(k) + 2 . (9.110) k The trace over all momenta with k0 6= k is unity, and we are left with

† . † † † −βωka a −βωka a ha aiβ = nk = Tr a ae Tr e

1 ∂ † (9.111) = − Tre−βωka a −βω a†a ωkTr e k ∂β in which a†a ≡ a†(k) a(k) and the 1/2 terms have cancelled. The trace is

† X 1 Tre−βωka a = e−βnωk = . (9.112) 1 − e−βωk n So we have

−βωk −βωk † 1 − e (−ωke ) 1 ha aiβ = − = . (9.113) 2 βωk ωk 1 − e−βωk e − 1 In the trace (9.108), only terms that don’t change the number of quanta in each mode contribute. So the mean value of the correlation function at inverse temperature β = 1/(kBT ) is X 1 −1 0 hφ(0, τ)φ(0, τ 0)i = e~ωk/kB T − 1 eiωk(τ−τ ) β 2kV k (9.114) −1 0 + e~ωk/kB T − 1 + 1 e−iωk(τ−τ ) . 232 Eﬀective ﬁeld theories and gravity In the continuum limit, this τ, τ 0 correlation function is two integrals. For massless particles, the simpler one requires some regularization because it involves zero-point energies Z 3 Z 2 X 1 0 d k 0 k dk 0 A = e−ik(τ−τ ) = e−ik(τ−τ ) = e−ik(τ−τ ) β 2kV (2π)32k (2π)2k k Z ∞ kdk −ik(τ−τ 0) = 2 e . 0 (2π) (9.115)

We send τ − τ 0 → τ − τ 0 + i and ﬁnd Z ∞ Z ∞ kdk −ik(τ−τ 0+i) d dk −ik(τ−τ 0+i) Aβ = 2 e = i 2 e 0 (2π) dτ 0 (2π) Z ∞ d dk −ik(τ−τ 0+i) 1 d 1 = i 2 e = 2 0 (9.116) dτ 0 (2π) (2π) dτ (τ − τ ) 1 1 = − (2π)2 (τ − τ 0)2 as → 0. The second integral is

Z ∞ −1 kdk βk 0 Bβ = 2 e − 1 2 cos(k(τ − τ )) (9.117) 0 (2π) which Mathematica says is 1 csch2(π(τ − τ 0)/β) B = − . (9.118) β (2π)2(τ − τ 0)2 4β2 Thus the ﬁnite-temperature correlation function is

0 1 hφ(0, τ)φ(0, τ )iβ = Aβ + Bβ = − . (9.119) 4β2 sinh2(π(τ − τ 0)/β) Equating this formula to the zero–temperature correlation function (9.105) in the accelerating frame, we ﬁnd 1 α2 = (9.120) 4β2 sinh2(π(τ − τ 0)/β) (4π)2 sinh2[α(τ − τ 0)/2] which says redundantly α α k T = and πk T = . (9.121) B 2π B 2 So a detector accelerating uniformly with acceleration α in the vacuum feels 9.6 Maximally symmetric spaces 233 a nonzero temperature

α T = ~ . (9.122) 2πckB

This result (Davies, 1975) is equivalent to the ﬁnding (Hawking, 1974) that a gravitational ﬁeld of local acceleration g makes empty space radiate at a temperature

g T = ~ . (9.123) 2πckB

Black holes are not black.

9.6 Maximally symmetric spaces The spheres S2 and S3 and the hyperboloids H2 and H3 are maximally sym- 0 0 0 0 metric spaces. A transformation x → x is an isometry if gik(x ) = gik(x ) i k 0 0 0i 0k 0 0i 0k in which case the distances gik(x)dx dx = gik(x )dx dx = gik(x )dx dx are the same. To see what this symmetry condition means, we consider the inﬁnitesimal transformation x0` = x` + y`(x) under which to lowest order 0 ` 0i i i j gik(x ) = gik(x) + gik,`y and dx = dx + y,jdx . The symmetry condition requires

i k ` i i j k k m gik(x)dx dx = (gik(x) + gik,`y )(dx + y,jdx )(dx + y,mdx ) (9.124) or

` m j 0 = gik,` y + gim y,k + gjk y,i. (9.125)

The vector ﬁeld yi(x) must satisfy this condition if x0i = xi + yi(x) is to be a symmetry of the metric gik(x). Since the covariant derivative of the metric tensor vanishes, gik;` = 0, we may write the condition on the symmetry vector y`(x) as

0 = yi;k + yk;i. (9.126) 234 Eﬀective ﬁeld theories and gravity

` In more detail, we subtract gik;`y , which vanishes, from the condition (9.125) that y be a Killing vector:

` m j ` 0 = gik,`y + gimy,k + gjky,i − gik;`y ` m j ` m ` j ` = gik,`y + gimy,k + gjky,i − gik,`y + gimΓ `ky + gkjΓ ìy m j m ` j ` = gimy,k + gjky,i + gimΓ `ky + gkjΓ ìy (9.127) m m ` j j ` = gim(y,k + Γ `ky ) + gjk(y,i + Γ ìy ) m j = gimy;k + gjky;i = yi;k + yk;i. Students leery of the last step may use the vanishing of the covariant derivative of the metric tensor and the derivation property

(AB);k = A;kB + AB;k (9.128) of covariant derivatives, to show that

m j 0 = (gimy );k + (gjky );i = yi;k + yk;i. (9.129) The symmetry vector y` is a Killing vector (Wilhelm Killing, 1847–1923). We may use symmetry condition (9.125) or (9.126) either to find the symmetries of a space with a known metric or to find a metric with given symmetries. Example 9.1 (Killing vectors of the sphere S2) The first Killing vector θ φ is (y1, y1 ) = (0, 1). Since the components of y1 are constants, the symmetry condition (9.125) says gik,φ = 0 which tells us that the metric is independent θ φ of φ. The other two Killing vectors are (y2, y2 ) = (sin φ, cot θ cos φ) and θ φ (y3, y3 ) = (cos φ, − cot θ sin φ). The symmetry condition (9.125) for i = k = θ and Killing vectors y2 and y3 tell us that gθφ = 0 and that gθθ,θ = 0. So gθθ is a constant, which we set equal to unity. Finally, the symmetry condition (9.125) for i = k = φ and the Killing vectors y2 and y3 tell us that 2 gφφ,θ = 2 cot θgφφ which we integrate to gφφ = sin θ. The 2-dimensional space with Killing vectors y1, y2, y3 therefore has the metric 2 gθθ gθφ eθ · eθ eθ · eφ R 0 gik = = = 2 2 (9.130) gφθ gφφ eφ · eθ eφ · eφ 0 R sin θ of the sphere S2. Example 9.2 (Killing vectors of the hyperboloid H2) The metric 1 0 (g ) = R2 . (9.131) ij 0 sinh2 θ 9.6 Maximally symmetric spaces 235

2 2 2 2 of the hyperboloid H is diagonal with gθθ = R and gφφ = R sinh θ. θ φ The Killing vector (y1, y1 ) = (0, 1) satisﬁes the symmetry condition (9.125). Since gθθ is independent of θ and φ, the θθ component of (9.125) implies θ 2 2 that y,θ = 0. Since gφφ = R sinh θ, the φφ component of (9.125) says φ θ θ that y,φ = − coth θ y . The θφ and φθ components of (9.125) give y,φ = 2 φ θ φ − sinh θ y,θ. The vectors y2 = (y2, y2 ) = (sin φ, coth θ cos φ) and y3 = θ φ (y3, y3 ) = (cos φ, − coth θ sin φ) satisfy both of these equations.

The Lie derivative Ly of a scalar field A is defined in terms of a vector ` ` field y (x) as LyA = y A,`. The Lie derivative Ly of a contravariant vector F i is i ` i ` i ` i ` i LyF = y F,` − F y,` = y F;` − F y;` (9.132) ` i k ` i k in which the second equality follows from y Γ`kF = F Γ`ky . The Lie derivative Ly of a covariant vector Vi is ` ` ` ` LyVi = y Vi,` + V`y,i = y Vi;` + V`y;i. (9.133)

Similarly, the Lie derivative Ly of a rank-2 covariant tensor Tik is ` ` ` LyTik = y Tik,` + T`ky,i + Ti`y,k. (9.134) We see now that the condition (9.125) that a vector ﬁeld y` be a symmetry of a metric gjm is that its Lie derivative ` m j Lygik = gik,` y + gim y,k + gjk y,i = 0 (9.135) must vanish. A maximally symmetric space (or spacetime) in d dimensions has d translation symmetries and d(d − 1)/2 rotational symmetries which gives a total of d(d + 1)/2 symmetries associated with d(d + 1)/2 Killing vectors. Thus for d = 2, there is one rotation and two translations. For d = 3, there are three rotations and three translations. For d = 4, there are six rotations and four translations. A maximally symmetric space has a curvature tensor that is simply related to its metric tensor

Rijk` = c (gikgj` − gi`gjk) (9.136) ki k where c is a constant (Zee, 2013, IX.6). Since g gik = gk = d is the number of dimensions of the space(time), the Ricci tensor and the curvature scalar of a maximally symmetric space are

ki `j Rj` = g Rijk` = c (d − 1) gj` and R = g Rj` = c d(d − 1). (9.137) 236 Eﬀective ﬁeld theories and gravity

9.7 Conformal algebra 0 0 0 An isometry requires that gik(x ) = gik(x ). The looser condition 0 0 0 0 0 gik(x ) = Ω(x ) gik(x ) with Ω(x ) > 0 (9.138) says that two metrics are conformally related. An equivalent condition is 0 0 ` m 0 0 gik(x ) = x,i0 x,k0 g`m(x) = Ω(x )gik(x ). (9.139) This condition guarantees that angles do not change: 0 gikuivk gikuivk q q = √ √ . (9.140) 0 0 gijuiuj gijvivj gijuiuj gijvivj For the inﬁnitesimal transformation x0` = x` + y`, (9.141) we approximate the factor Ω as Ω2(x0) ≈ 1 + ω(x0). (9.142) The conformal condition (9.139) then says that ` ` m m ` g`m(δi − y,i)(δk − y,k) = (1 + ω)(gik + y gik,`) (9.143) or m ` ` gimy,k + g`ky,i + y gik,` + ωgik = 0. (9.144) This is Killing’s conformal condition. Multiplication by gik and summing over i and k gives k m i ` ik ` δmy,k + δ`y,i + g y gik,` + 4ω = 0 (9.145) or k ` ik 2y,k + y g gik,` + 4ω = 0. (9.146) The function ω(x) therefore

1 i ` ik ω = − 2y + y g gik,` . (9.147) 4 ,i Substituting this formula for ω into Killing’s conformal condition (9.144) gives a condition on the metric g and on the vector y m ` ` 1 j 1 ` jn gimy + g`ky + y gik,` − y + y g gjn,` gik = 0. (9.148) ,k ,i 2 ,j 4 9.8 Conformal algebra in ﬂat space 237 When ω = 0, Killing’s conformal condition (9.144)

m ` ` gimy,k + g`ky,i + y gik,` = 0 (9.149) is the same as his isometry condition (9.125) . Using the vanishing of the covariant derivative of the metric tensor as in (9.127), we can write Killing’s conformal condition (9.144) more succinctly as

yi;k + yk;i + ω gik = 0 (9.150) or as

Lygik = −ω gik. (9.151)

9.8 Conformal algebra in ﬂat space

To find Killing’s conformal vectors y in flat space, we set gik = ηik and use the formula (9.147) for ω to find 1 ω = − yi (9.152) 2 ,i i which in d dimensions of spacetime is ω = −2y,i/d. So in flat-space, Killing’s conformal condition (9.144) is

1 ` 2 ` yi,k + yk,i = ηik y or yi,k + yk,i = ηik y . (9.153) 2 ,` d ,` Inﬁnitesimal transformations x0i = xi + yi (9.154) that obey this rule generate the conformal algebra of Minkowski space. We already know some of these vectors. The vector

` ` ` k y = a + b kx (9.155) i ` with a and b k constant and b antisymmetric, bik = − bki, has a vanishing ` divergence y,` = 0 and satisﬁes Killing’s conformal condition (9.153)

2 ` yi,k + yk,i = bik + bki = ηik y = 0. (9.156) d ,` i These y’s generate the translations a and the Lorentz transformations bik. A conformal transformation x → x0 changes the metric by no more than an overall factor. That means that

0 0i 0k i k Ω(x ) ηikdx dx = ηikdx dx . (9.157) 238 Effective field theories and gravity So if space is just stretched by a constant factor σ, so that x0i = σxi, then dx0i = σdxi, and the stretch condition (9.157) is satisfied with Ω(x0) = σ−2. This kind of conformal transformation is a dilation, which some call a dilatation. The Killing vector y is

yi = σxi (9.158) with σ a constant. This vector obeys Killing’s ﬂat-space conformal condition ` (9.153) because y,` = c d in d dimensions, and so

j j 2 ` yi,k + yk,i = ηij c δ + ηkj c δ = 2ηik c = ηik y . (9.159) k i d ,`

The inversion r2 xi x0i = (9.160) x2

2 k where r is a length and x = x xk also obeys Killing’s conformal condition (9.153). The change in x0i is

r2 2xix dx0i = δi − j dxj (9.161) x2 j x2 so with Ω ≡ Ω(x0)

r2 2 2xix 2xkx Ωη dx0idx0k = Ωη δi − j dxj δk − ` dx` ik ik x2 j x2 ` x2 r2 2 2η xkx 2η xix 4x2x x = Ω η − jk ` − i` j + j ` dxjdx` x2 j` x2 x2 (x2)2 r2 2 = Ω η dxjdx`. (9.162) x2 j`

So the stretch condition (9.157) is satisﬁed with

x2 2 Ω(x0) = . (9.163) r2

Since both inversions and translations obey Killing’s conformal condition (9.153), we may combine them so as to shrink an inversion to its infinitesimal form. We invert, translate by a tiny vector a, and re-invert and so find as 9.8 Conformal algebra in flat space 239 the legitimately infinitesimal form of an inversion the transformation xi xi xi xj xk xi → → + ai → + ai η + aj + ak x2 x2 x2 jk x2 x2 xi 1 2a · x = + ai η + + a2 (9.164) x2 jk x2 x2 xi = x2 + ai 1 + 2a · x + a2x2 ' xi + aix2 (1 − 2a · x) x2 i ik 2 i k = x + ak(η x − 2x x ) (9.165) which is a conformal transformation. If we use ∂i to differentiate Killing’s conformal condition (9.153), then we get

i i 2 i ` ∂ yi,k + ∂ yk,i = ηik ∂ y (9.166) d ,` 2 i i or, with ∂ = ∂ ∂i and ∂ · y = ∂iy , 2 1 − ∂ ∂ · y = − ∂2y (9.167) d k k which says that in two dimensions (d = 2) every Killing vector is a solution of Laplace’s equation:

2 ∂ yk = 0 if d = 2. (9.168) There are inﬁnitely many Killing vectors in two-dimensions, a fact exploited in string theory. Diﬀerentiating the same equation (9.167) again, we get

2 (d − 2) ∂i∂k ∂ · y = − d ∂ yk,i (9.169)

2 2 2 which says that ∂ yk,i = ∂ yi,k. Applying this symmetry and ∂ to Killing’s conformal condition (9.153), we get

2 2 ηik ∂ ∂ · y = d ∂ yk,i. (9.170) Comparing the right-hand sides of the last two equations (9.169 & 9.170), we ﬁnd 2 (d − 2) ∂i∂k + ηik∂ ∂ · y = 0. (9.171) Diﬀerentiating the same equation (9.167) a third time, we have

k 2 k ∂ (d − 2) ∂k ∂ · y = − d ∂ ∂ yk (9.172) 240 Effective field theories and gravity or (d − 1) ∂2 ∂ · y = 0. (9.173) In the case of one dimension (d = 1), Killing’s conformal condition (9.153) reduces to 2∂0y0 = 2∂0y0 and so places no restriction on Killing’s vector, as reflected equations (9.167 & 9.173). Every function y0(x0) ≡ y(t) is a conformal Killing vector. The condition (9.157) for a finite one-dimensional conformal transformation is dx0 = Ω−1(x0). (9.174) dx This condition constrains the transformation x → x0 = x0(x) only by requiring that the derivative (9.174) be positive. The vanishing (9.168) of the laplacian of the Killing vector in d = 2 dimensions means that every Killing vector is a solution of Laplace’s equation: ∂2yk = 0 if d = 2. (9.175) In two-dimensional euclidian space, we let the d = 2 space be the complex plane and let y0 and y1 be the real and imaginary parts of an analytic (or antianalytic) function: z = x0 + ix1 and f(z) = y0 + iy1. (9.176) Then the Cauchy-Riemann conditions 0 1 0 1 y,0 = y,1 and y,1 = − y,0 (9.177) imply that the real and imaginary parts of every analytic function f(z) are harmonic 0 0 1 1 y,00 + y,11 = 0 and y,00 + y,11 = 0 (9.178) 0 1 0 1 0 1 because y,00 = y10 = −y,11 and y,00 = −y,10 = −y,11. The real and imaginary parts of every antianalytic function f(z∗) also are harmonic, and so satisfy Killing’s condition (9.175). There are infinitely many solutions for the Killing vectors of an infinitesimal conformal transformation in two- dimensional euclidian space. What about finite conformal transformations in two-dimensional euclidian space? The definition (9.157) of a conformal transformation in flat space is 0 0 ηik = Ω(x )ηik. (9.179) But metrics transform like this: ∂xj ∂x` ∂x0i ∂x0k η0 = η and η0ik = ηj`. (9.180) ik ∂x0i ∂x0k j` ∂xj ∂x` 9.8 Conformal algebra in flat space 241 In two-dimensional euclidian space, η is the 2×2 matrix η is the 2×2 matrix 1 0 η = , (9.181) 0 1 so these equations say that

∂x00 ∂x00 ∂x00 ∂x00 ∂x00 2 ∂x00 2 Ω = Ωη00 = η00 + η11 = + ∂x0 ∂x0 ∂x1 ∂x1 ∂x0 ∂x1 ∂x01 ∂x01 ∂x01 ∂x01 ∂x01 2 ∂x01 2 Ω = Ωη11 = η00 + η11 = + (9.182) ∂x0 ∂x0 ∂x1 ∂x1 ∂x0 ∂x1 ∂x00 ∂x01 ∂x00 ∂x01 ∂x00 ∂x01 ∂x00 ∂x01 0 = Ωη01 = η00 + η11 = + . ∂x0 ∂x0 ∂x1 ∂x1 ∂x0 ∂x0 ∂x1 ∂x1 More succinctly, these conditions are

∂x00 2 ∂x00 2 ∂x01 2 ∂x01 2 0 + 1 = 0 + 1 ∂x ∂x ∂x ∂x (9.183) ∂x00 ∂x01 ∂x00 ∂x01 = − . ∂x0 ∂x0 ∂x1 ∂x1 These conditions are equivalent either to ∂x00 ∂x01 ∂x01 ∂x00 = and = − (9.184) ∂x0 ∂x1 ∂x0 ∂x1 which are the Cauchy-Riemann conditions ux = vy and vx = − uy for f = u + iv to be an analytic function of z = x + iy or to ∂x00 ∂x01 ∂x01 ∂x00 = − and = (9.185) ∂x0 ∂x1 ∂x0 ∂x1 which are the Cauchy-Riemann conditions ux = − vy and vx = uy for f(¯z) = u + iv to be an analytic function ofz ¯ = x − iy.

9.8.1 Angles and analytic functions

An analytic function f(z) maps curves in the z plane into curves in the f(z) plane. In general, this mapping preserves angles. To see why, we consider the angle dθ between two tiny complex lines dz = exp(iθ) and dz0 = exp(iθ0) that radiate from the same point z. This angle dθ = θ0 − θ is the phase of the ratio 0 iθ0 dz e 0 = = ei(θ −θ). (9.186) dz eiθ 242 Eﬀective ﬁeld theories and gravity Let’s use w = ρeiφ for f(z). Then the analytic function f(z) maps dz into

dw = f(z + dz) − f(z) ≈ f 0(z) dz (9.187) and dz0 into dw0 = f(z + dz0) − f(z) ≈ f 0(z) dz0. (9.188)

The angle dφ = φ0 − φ between dw and dw0 is the phase of the ratio

0 iφ0 0 0 0 iθ0 dw e f (z) dz dz e 0 = = = = = ei(θ −θ). (9.189) dw eiφ f 0(z) dz dz eiθ So as long as the derivative f 0(z) does not vanish, the angle in the w-plane is the same as the angle in the z-plane

dφ = dθ. (9.190)

Analytic functions preserve angles. They are conformal maps. What if f 0(z) = 0? In this case, dw ≈ f 00 (z) dz2/2 and dw0 ≈ f 00 (z) dz02/2, and so the angle dφ = dφ0 − dφ between these two tiny complex lines is the phase of the ratio

0 iφ0 00 02 02 dw e f (z) dz dz 0 = = = = e2i(θ −θ). (9.191) dw eiφ f 00 (z) dz2 dz2 So angles are doubled, dφ = 2dθ. In general, if the ﬁrst nonzero derivative is f (n)(z), then

0 iφ0 (n) 0n 0n dw e f (z) dz dz 0 = = = = eni(θ −θ) (9.192) dw eiφ f (n)(z) dzn dzn and so dφ = ndθ. The angles increase by a factor of n. Example 9.3 (zn) The function f(z) = zn has only one nonzero derivative (k) n f (0) = n! δnk at the origin z = 0. So at z = 0 the map z → z scales angles by n, dφ = n dθ, but at z 6= 0 the ﬁrst derivative f (1)(z) = nzn−1 is not equal to zero. So zn is conformal except at the origin.

Example 9.4 (M¨obiustransformation) The function az + b f(z) = (9.193) cz + d maps (straight) lines into lines and circles and circles into circles and lines, unless ad = bc in which case it is the constant b/d. 9.8 Conformal algebra in ﬂat space 243

But if we require that f(z) be entire and invertible on the Riemann sphere (the complex plane with the “point” z = ∞ mapped onto the north pole) so that the transformations form the special conformal group, then the function f(z) cannot vanish at more than one value of z or f −1(0) would not be unique. Similarly, f(z) can’t have more than one pole. So the only functions f(z) that are entire and invertible on the Riemann sphere are those of the projective transformation az + b f(z) = with ad − bc = 1 (9.194) cz + d and a, b, c, and d complex, which is a special case of the fractional linear or Möbiustransformation (9.193). These matrices form the group SL(2, C) which is isomorphic to the Lorentz group SO(3, 1). So the requirement that the finite transformations form a group has reduced the symmetry of the conformal group to that of the Lorentz group. In two-dimensional Minkowski space, we must solve Killing’s equation for the vector of an infinitesimal transformation

k k y,11 = y,00. (9.195) So we use Dirac’s light-cone coordinates. We set

x± = t ± x. (9.196)

Then 2 2 2 i k ds = − dt + dx = −(dt + dx)(dt − dx) = ηikdx dx − − + + + − − + (9.197) = η−−dx dx + η++dx dx + η−+dx dx + η+−dx dx

1 with η−+ = η+− = − 2 and η−− = η++ = 0. So the inverse metric has η−+ = η+− = −2. The light-cone derivatives are 1 ∂ = (∂ ± ∂ ) . (9.198) ± 2 t x They act like this:

+ − ∂+x = 1 and ∂−x = 1 − + (9.199) ∂+x = 0 and ∂−x = 0. In light-cone coordinates, the equation (9.195) for the d = 2 Killing vector is k k ∂−∂+y = ∂+∂−y = 0. (9.200) 244 Effective field theories and gravity Any vector yk(x+) satisfies this equation because

k + 0k + ∂−y (x ) = y ∂−x = 0. (9.201) Similarly, any function yk(x−) obeys the same equation. So there are two sets of infinitely many solutions yk(x+) and yk(x−) of Killing’s equation (9.200). We let k + − k + k − y (x , x ) = y+(x ) + y−(x ). (9.202) To understand finite conformal transformations in 2-d Minkowski space, we return to the definition (9.179)

0 0 ηik = Ω(x )ηik. (9.203) of a conformal transformation in ﬂat space and recall how (9.204) metrics transform ∂xj ∂x` ∂x0i ∂x0k η0 = η and η0ik = ηj`. (9.204) ik ∂x0i ∂x0k j` ∂xj ∂x` In two-dimensional Minkowski space, η is the 2 × 2 matrix −1 0 η = , (9.205) 0 1 so these equations say that

∂x00 ∂x00 ∂x00 ∂x00 ∂x00 2 ∂x00 2 − Ω = Ωη00 = η00 + η11 = − + ∂x0 ∂x0 ∂x1 ∂x1 ∂x0 ∂x1 ∂x01 ∂x01 ∂x01 ∂x01 ∂x01 2 ∂x01 2 Ω = Ωη11 = η00 + η11 = − + ∂x0 ∂x0 ∂x1 ∂x1 ∂x0 ∂x1 ∂x00 ∂x01 ∂x00 ∂x01 ∂x00 ∂x01 ∂x00 ∂x01 0 = Ωη01 = η00 + η11 = − + . ∂x0 ∂x0 ∂x1 ∂x1 ∂x0 ∂x0 ∂x1 ∂x1 (9.206) More succinctly, these conditions are

∂x00 2 ∂x00 2 ∂x01 2 ∂x01 2 0 − 1 = − 0 + 1 ∂x ∂x ∂x ∂x (9.207) ∂x00 ∂x01 ∂x00 ∂x01 = . ∂x0 ∂x0 ∂x1 ∂x1 In terms of light-cone coordinates, these conditions are

00 00 01 01 ∂+x ∂−x = − ∂+x ∂−x 00 01 00 01 (9.208) ∂+x ∂−x = − ∂−x ∂+x . 9.8 Conformal algebra in ﬂat space 245 We can satisfy them by having x0k be functions either of x+ or of x−: x0k(t, x) = xk(x+) or x0k(t, x) = xk(x−) (9.209) but not both as would be okay (9.202) for inﬁnitesimal transformations. Example 9.5 (Liouville’s equation) Liouville’s equation is φ φtt − φxx + 2e = 0 (9.210) or 1 φ ∂+ ∂−φ = − 2 e . (9.211) It has as a solution 4f 0(x+)g0(x−) eφ = (9.212) (f(x+) − g(x−))2 in which f 0 g0 > 0. In more than two dimensions (d ≥ 3), the vanishing (9.173) of the laplacian of the divergence ∂ · y and the third-order equation (9.171) imply that the all the second derivatives of the divergence vanish

∂i ∂k (∂ · y) = 0. (9.213) So the divergence of Killing’s vector must be at most linear in the coordinates i ∂ · y = a + bix for d ≥ 3. (9.214) To see what this implies, we differentiate Killing’s conformal condition (9.153) 2 yi,k` + yk,i` = ηik(∂ · y),`, (9.215) d add to it the same equation with k and ` interchanged 2 yi,`k + y`,ik = ηi`(∂ · y),k, (9.216) d and then subtract the same equation with i, k, ` permuted 2 yk,ì + y`,ki = ηk`(∂ · y),i (9.217) d so as to get 2 2yi,k` = ηik(∂ · y),` + ηi`(∂ · y),k − ηk`(∂ · y),i . (9.218) d So if we substitute the at-most-linear condition (9.214) into this equation (9.218), then we find 2 2yi,k` = [ηikb` + ηi`bk − ηk`bi] (9.219) d 246 Effective field theories and gravity which says that yi can be at most quadratic in the coordinates. j j k yi = αi + βijx + γijkx x (9.220) in which γijk = γikj. Killing’s conformal condition (9.153) must hold for all x’s and is linear in the vector y, so we can impose it successively on ai, bij, and cijk by differentiating with respect to the variables x` and setting them equal to i zero. The vector α is an arbitrary translation. The condition on βij is 2 β + β = η β` (9.221) ik ki d ik ` which says that βij is the sum

βij = bij + c ηij (9.222) of an antisymmetric part bij = −bji that generates a Lorentz transformation and a term proportional to ηij that generates a scale transformation. The conformal condition (9.153) makes the quadratic term satisfy the relations

γik` + γki` = 2ηikd` (9.223)

γi`k + γ`ik = 2ηi`dk (9.224)

γkì + γ`ki = 2ηk`di (9.225) in which 1 d = γr . (9.226) ` d r` Subtracting the last equation (9.225) from the sum of the first two (9.223 & 9.224), we find

γik` = ηikd` + ηi`dk − ηk`di. (9.227) The corresponding infinitesimal transformation is x0i = xi + 2(x · d)xi − dix2 (9.228) which is called a special conformal transformation whose exponential or finite form is xi − dix2 x0i = . (9.229) 1 − 2d · x + d2x2 The general Killing vector and its infinitesimal transformation are i i i k i ik 2 i k y = a + b kx + cx + dk(η x − 2x x ) (9.230) and 0i i i i k i ik 2 i k x = x + a + b kx + cx + dk(η x − 2x x ) (9.231) 9.9 Maxwell’s action is conformally invariant for d = 4 247 in which we see a translation, a Lorentz transformation, a dilation, and a special conformal transformation. The differential forms of the Poincarétransformations and the dilation D and the inversion k are

Pi = ∂i,Jik = (xi ∂k − xk ∂i), (9.232) i i ik j i k D = x ∂i, and K = (η x xj − 2x x ) ∂k. The commutators of the conformal Lie algebra are [P i,P j] = 0, [Ki,Kj] = 0, i i i i [D,P ] = − P , [D,Jij] = 0, [D,K ] = K , [J ij,P `] = − ηi`P j + ηj`P i, [J ij,K`] = −ηi`Kj + ηj`Ki (9.233) [J ij,J `m] = − ηi`J jm − ηjmJ i` + ηj`J im + ηimJ j`, [Ki,P j] = 2J ij + 2ηijD. These commutation relations say that D is a Lorentz scalar, and that Ki transforms as a vector. There are d P ’s, d K’s, d(d − 1)/2 J’s, and one D. That’s (d + 2)(d + 1)/2 generators, which is the same number of generators as SO(d + 2). In fact, Minkowski space in d dimensions has SO(d − 1, 1) as its Lorentz group and SO(d, 2) as its conformal algebra. The “Lorentz group” of a spacetime with d spatial dimensions and 2 time dimensions is also SO(d, 2).

9.9 Maxwell’s action is conformally invariant for d = 4 0 Under a change of coordinates from x to x , a covariant vector field Ai(x) changes this way k 0 0 k ∂x A (x ) = x 0 A (x) = A (x). (9.234) i ,i k ∂x0i k So under the tiny change (with a convenient minus sign) x0 = x − y or x = x0 + y (9.235) the change in Ai is 0 δAi(x) ≡ Ai(x) − Ai(x) 0 0 0 0 0 = Ai(x) − Ai(x ) + Ai(x ) − Ai(x) j 0 0 j j (9.236) = y Ai,j(x ) + δi + y,i0 Aj(x) − Ai(x) j 0 0 j = y Ai,j(x ) + y,i0 Aj(x). 248 Effective field theories and gravity To lowest order in y, this is j j δAi(x) = y Ai,j(x) + y,i Aj(x). (9.237) So the change in the Maxwell-Faraday tensor is

δFik = δ (∂iAk − ∂kAi) = ∂i (δAk) − ∂k (δAi) j j j j = ∂i y Ak,j(x) + y,k Aj(x) − ∂k y Ai,j(x) + y,i Aj(x) j j j j j j j = y ∂jFik + y,i Ak,j − y,k Ai,j + y,ki Aj − y,ik Aj + y,k ∂iAj − y,i ∂kAj j j j = y ∂jFik + y,i Fjk + y,k Fij. (9.238) Thus the change in Maxwell’s action density is

ik ik ik j j j δ F Fik = 2 F δFik = 2 F y ∂jFik + y,i Fjk + y,k Fij j ik ik j ik j ik = ∂j y F Fik − (∂ · y)F Fik + 2y,i F Fjk + 2y,k F Fij j ik ik ik j ik j = ∂j y F Fik − (∂ · y)F Fik + 2yj,i F F k + 2yj,k F Fi j ik ik ik j = ∂j y F Fik − (∂ · y)F Fik + 4yj,i F F k (9.239) j ik ik ij k = ∂j y F Fik − (∂ · y)F Fik + 4yk,i F F j. We can rewrite this as ik j ik ij k 1 δ F Fik = ∂j y F Fik + 2F F j yk,i + yi,k − 2 ηik∂ · y . (9.240) But if y is a conformal Killing vector obeying (9.153), then this change in the Maxwell action density is 2 1 δ F ikF = ∂ yjF ikF + 2F ijF k η y` − η ∂ · y . (9.241) ik j ik j d ik ,` 2 ik Iﬀ d = 4, this change is a total derivative

ik j ik δ F Fik = ∂j y F Fik ⇐⇒ d = 4. (9.242) The action density of pure (only gauge ﬁelds), classical Yang-Mills theory also is conformally invariant in four dimensions (d = 4). Quantum eﬀects introduce a mass scale, however, and break conformal invariance.

9.10 Massless scalar field theory is conformally invariant if d = 2 Under a tiny change of coordinates x = x0 + y, the change in a scalar field φ is j ,j δφ = y φ,j = yjφ . (9.243) 9.11 Christoffel symbols as nonabelian gauge fields 249 So the change in the action density is 1 ,i ,i ,i ,i ,j δ φ φ,i = φ δφ,i = φ ∂iδφ = φ ∂i yjφ 2 ,i ,j ,i ,j = φ φ yj,i + φ yjφ,i (9.244) 1 ,i ,j ,i ,j = φ φ yj,i + yi,j + φ φ yj. 2 ,i If y obeys the condition (9.153) for a conformal Killing vector, then the change (9.244) in the action density is a total divergence in two dimensions 1 ,i 1 ,i ,j 2 ` ,i ,j δ φ φ,i = φ φ ηijy + φ φ yj 2 2 d ,` ,i 1 ,i ` ,i ` 1 ,i ` 1 ,i ` = φ φ,i y + φ φ,i` y = φ φ,i y + φ φ,i y d ,` d ,` 2 ,` 1 ,i ` = ∂` φ φ,i y ⇐⇒ d = 2. (9.245) 2 Thus the classical theory of a free massless scalar field is conformally invariant in two-dimensional spacetimes.

The rest of this chapter is at best a ﬁrst draft, not ready for human consumption.

9.11 Christoffel symbols as nonabelian gauge fields i 0i 0i k A contravariant vector V transforms like dx = x,kdx as ∂x0i V 0i(x0) = V k(x) = x0i V k(x) = Ei (x) V k(x). (9.246) ∂xk ,k k i 0i The 4 × 4 matrix E k(x) = x,k depends upon the spacetime point x and is a member of the huge noncompact group GL(4, R). The insight of Yang and Mills (section 5.1) lets us define a covariant i derivative D` = ∂` + A` of a contravariant vector V

k k k j (D`V ) = (∂`δj + A`j)V (9.247) that transforms as ∂xi ∂x0k (D V )k0 = (D V )m. (9.248) ` ∂x0` ∂xm i 250 Eﬀective ﬁeld theories and gravity We need k k j0 0 k 0k 0j 0k 0k 0j (∂`δj + A`j)V = (∂`δj + A`j)V = V,`0 + A`jV i 0k m m n i 0k m m n = x,`0 x,m(∂iδn + Ain)V = x,`0 x,m(V,i + AinV ) (9.249) or 0k 0k 0j i 0k m m n V,`0 + A`jV = x,`0 x,m(V,i + AinV ). (9.250) Decoding the left-hand side, we get

0k m 0k 0j i 0k m m n ∂`0 (x,mV ) + A`jV = x,`0 x,m(V,i + AinV ) (9.251) or i 0k m 0k 0j i 0k m m n x,`0 ∂i(x,mV ) + A`jV = x,`0 x,m(V,i + AinV ) (9.252) or

i 0k m i 0k m 0k 0j i 0k m m n x,`0 x,mV,i + x,`0 x,miV + A`jV = x,`0 x,m(V,i + AinV ). (9.253) After a cancelation, we get

n 0k m 0k 0j i 0k m n x,`0 x,mnV + A`jV = x,`0 x,mAinV (9.254) or 0k 0j p 0k m n n 0k m A`jV = x,`0 x,mApnV − x,`0 x,mnV (9.255) which is 0k 0j m p 0k m n n 0k m A`jx,mV = x,`0 x,mApnV − x,`0 x,mnV (9.256) or after the interchange of m and n in the second term

0k 0j m p 0k n m n 0k m A`jx,mV = x,`0 x,nApmV − x,`0 x,mnV . (9.257) Since the vector V m is arbitrary, we can promote this to an equation with three free indexes

0k 0j p 0k n n 0k A`jx,m = x,`0 x,nApm − x,`0 x,mn (9.258) 0j and then multiply by the inverse of x,m 0k 0j m p 0k n m n 0k m A`jx,mx,i0 = x,`0 x,nApmx,i0 − x,`0 x,mnx,i0 (9.259) so as to get 0k m 0k p n p 0k n Aì = x,`0 x,pAnmx,i0 − x,`0 x,npx,i0 . (9.260) 0k 0k Partial derivatives of the same kind commute, so x,np = x,pn. It follows 9.11 Christoffel symbols as nonabelian gauge fields 251 that the inhomogeneous term in the transformation law (9.260) for A is symmetric in ` and i

p 0k n n 0k p p 0k n p 0k n − x,`0 x,npx,i0 = − x,i0 x,npx,`0 = − x,i0 x,pnx,`0 = − x,i0 x,npx,`0 . (9.261)

m 0k p n The homogeneous term x,`0 x,pAnmx,i0 in the transformation law (9.260) for p p p A is symmetric in ` and i if Anm is symmetric in n and m, Anm = Amn. Thus we can use gauge ﬁelds that are symmetric in their lower indexes, p p Anm = Amn. The transformations are

0k 0k j k j 0 i −1T i −1i V = x,jV ≡ E jV and V` = x,`0 Vi ≡ E ` Vi = E `Vi (9.262) and 0k m 0k p n p 0k n A`i = x,`0 x,pAnmx,i0 − x,`0 x,npx,i0 −1m k p −1n p 0k n = E `E pAnmE i − x,`0 x,npx,i0 . We decode the inhomogeneous term

p 0k n 0k n k n k −1n −x,`0 x,npx,i0 = − (∂`0 x,n)x,i0 = −(∂`0 E n)x,i0 = −(∂`0 E n)E i (9.263) −1n k k −1n 0k n = − E i(∂`0 E n) = E n(∂`0 E i) = x,n(∂`0 x,i0 ) So on a contravariant vector V 0 = E−1V , A0 is

0 −1m −1 −1 A` = E ÈAmE + E∂`0 E (9.264) or 0k −1m k p −1n k −1n Aì = E Èp AmnEi + En∂`0 Ei . (9.265) This gauge field is the affine connection, aka, Christoffel symbol (of the k k second kind) A ì = Γ ì. The usual formula for the covariant derivative of a contravariant vector is

k k k j k k j k k j (D`V ) = (∂`δj + A`j)V = (∂`δj + Γ `j)V = ∂`V + Γ `jV . (9.266)

We can write the chain rule identities ∂xi ∂x0k ∂x0i ∂xk = δi = (9.267) ∂x0k ∂x` ` ∂xk ∂x0` k 0k in terms of the matrix E ` = x,` as ∂xi ∂xk Ek = δi = Ei (9.268) ∂x0k ` ` k ∂x0` 252 Eﬀective ﬁeld theories and gravity

k i which show that the inverse of the matrices E ` and E k are i k −1i ∂x i −1k ∂x k E = = x 0 and E = = x 0 . (9.269) k ∂x0k ,k ` ∂x0` ,` The very general identity

−1 −1 −1 0 = ∂iI = ∂i(CC ) = C,i C + CC,i (9.270) will soon be useful. So we can write n n m −1n m ∂x m ∂x E E 0 = E ∂`0 = E ∂k0 n k,` n ∂x0k n ∂x0` (9.271) m −1n m −1n = E n E `,k0 = −E n,k0 E `.

A covariant vector Vj transforms like a derivative ∂j k 0 0 ∂x k −1k V (x ) = V (x) = x 0 V (x) ≡ E V . (9.272) j ∂x0j k ,j k j k

The Yang-Mills trick now requires that we ﬁnd a gauge ﬁeld B` so that the covariant derivative of a covariant vector j j (D`V )k = (∂`δk + B`k)Vj (9.273) transforms as ∂xi ∂xn (D V ) 0 = (D V ) . (9.274) ` k ∂x0` ∂x0k i n We need j j 0 0 j 0j 0 0 0j 0 (∂`δ +B )Vj = (∂`δ + B )Vj = Vk,`0 + B Vj k `k k `k `k (9.275) i n p p i n p = x,`0 x,k0 (∂iδn + Bin)Vp = x,`0 x,k0 (Vn,i + BinVp) or 0 0j 0 i n p Vk,`0 + B`kVj = x,`0 x,k0 (Vn,i + BinVp). (9.276) Decoding the left-hand side, we get

n 0j n n n 0j n (x,k0 Vn),`0 + B x,j0 Vn = x,k0`0 Vn + x,k0 Vn,`0 + B x,j0 Vn `k `k (9.277) n n i 0j n = x,k0`0 Vn + x,k0 Vn,i x,`0 + B`k x,j0 Vn. So combining the last two equations, we ﬁnd

n n i 0j n i n i n p x,k0`0 Vn + x,k0 Vn,i x,`0 + B`k x,j0 Vn = x,`0 x,k0 Vn,i + x,`0 x,k0 BinVp (9.278) or n 0j n i n p x,k0`0 Vn + B`k x,j0 Vn = x,`0 x,k0 BinVp. (9.279) 9.11 Christoﬀel symbols as nonabelian gauge ﬁelds 253 Interchanging p and n in the third term

n 0j n i p n x,k0`0 Vn + B`k x,j0 Vn = x,`0 x,k0 Bip Vn (9.280) and then using the arbitrariness of Vn, we get

n 0j n i p n x,k0`0 + B`k x,j0 = x,`0 x,k0 Bip. (9.281) 0m Multiplying by x,n gives

n 0m 0j n 0m n 0m 0m i p n 0m x,k0`0 x,n + B`k x,j0 x,n = x,k0`0 x,n + B`k = x,`0 x,k0 Bip x,n (9.282) or 0m i p n 0m n 0m B`k = x,`0 x,k0 Bip x,n − x,k0`0 x,n . (9.283) n 0m The inhomogeneous term −x,k0`0 x,n is symmetric in k and `. The homoge- i p n 0m neous term x,`0 x,k0 Bip x,n also is symmetric in k and `. Thus we can use n n gauge ﬁelds that are symmetric in their lower indexes, Bip = Bpi. In terms of the matrices k k 0k −1k ∂x k E = x and E = = x 0 (9.284) ` ,` ` ∂x0` ,` and in view of the identities (9.270) and (9.271), the transformation rule (9.283) for B has the equivalent form

0m −1T i m n −1p m −1n B`k = (E )` E nBipE k + E n,`0 E k. (9.285)

The transformation law for A` is

0k −1m −1n p k −1n k A`i = E `E iAnmE p − E i∂`0 E n. (9.286) We write the formula (9.285) for B0 as

0m −1T i m n −1p −1n m − B`k = − (E )` E nBipE − E E n,`0 k k (9.287) −1T i m n −1p −1n m = − (E )` E nBipE k − E ` E n,k0 and do the substitutions k → i, i → m, and m → k

0k −1m k n −1p −1n k − B`i = − E `E nBmpE i − E i ∂`0 E n. (9.288) The inhomogeneous terms in the equations for A0 and B0 are the same. We interchange p and n in the formula for B0

0k −1m k p −1n −1n k − B = − E E BmnE − E ∂`0 E ì ` p i i n (9.289) −1m −1n p k −1n k = − E È iBmnE p − E i ∂`0 E n. 254 Effective field theories and gravity Since the gauge fields are symmetric in their lower indexes, we can write this as

0k −1m −1n p k −1n k − Bì = − E È iBnmE p − E i ∂`0 E n. (9.290) Since the transformation laws are the same, we can identify the two gauge fields apart from a minus sign

k k k Aì = − Bì = Γ i`. (9.291) The gauge fields A and B correspond to the Christoffel symbols, the Γ’s of the usual notation,

i i i i i k i i k A`k : D`V = V;` = V,` + Γ k`V = V,` + e · ek,`V (9.292) k i k k B`i : D`Vi = Vi;` = V,` − Γ i` Vk = Vi,` − e · ei,` Vk

9.12 Spin connection a The Lorentz index a on a tetrad ci is subject to local Lorentz transformations. A spin-one-half ﬁeld ψb also has an index b that is subject to local Lorentz transformations. We can use the insight of Yang and Mills to in- ab troduce a gauge ﬁeld, called the spin connection ωi which multiplies the generators Jab of the Lorentz group ab ωi = ωi Jab. (9.293)

Since the generators Jab are antisymmetric Jab = − Jba, the spin connection is also antisymmetric ab ba ωi = − ωi . (9.294) ab So for each spacetime index i, there are six independent ωi ’s. The spin connection is

a k a j a a k j a k ω b ` = − cb ck,` − Γk` cj = cj cb Γ k` − ck,` cb . (9.295) Under a general coordinate transformation and a local Lorentz transformation, the spin connection (9.295) transforms as ∂xi h i ω0a = La ωd − (∂ La ) L−1e . (9.296) b ` ∂x0` d e i i e b This example is a work in progress.... Example 9.6 (Relativistic particle) The lagrangian of a relativistic particle of mass m is L = −mp1 − q˙2 in units with c = 1. The hamiltonian is 9.12 Spin connection 255 H = pp2 + m2. The path integral for the ﬁrst step (7.63) toward Z(β) is ∞ √ 0 Z 02 2 0 dp −H − p +m ip (q1−qa)/~ hq1|e |qai = e e (9.297) −∞ 2π~ which is a nontrivial integral. In the m → 0 limit, it is ∞ 0 Z 0 0 dp −H −|p | ip (q1−qa)/~ hq1|e |qai = e e −∞ 2π~ ∞ 0 0 Z 0 p (q − q ) dp = 2 e−p cos 1 a 0 ~ 2π~ ∞ 0 Z 0 0 0 dp = e−p +ip (q1−qa)/~ + e−p−ip (q1−qa)/~ 0 2π~ 1 1 1 = + − i(q1 − qa)/~ + i(q1 − qa)/~ 2π~ 2 1 = 2 2 2 + (q1 − qa) /~ 2π~ 1 1 = 2 2 . (9.298) π~ 1 + (q1 − qa) /(~) So the partition function is n → ∞ limit of the product of the n integrals n Z ∞ 1 dqn dq1 Z(β) = 2 2 ··· 2 2 π~ −∞ 1 + (qn − qn−1) /(~) 1 + (q1 − qn) /(~) in which qn = qa. n 1 n Z(β) = (π~) . (9.299) π~ In one space dimension, quantum mechanics gives the result L Z(β) = (9.300) π~βc in which c has reappeared. and we must take the L → ∞ limit. In two space dimensions, it gives 2 Z ∞ √ 2 Z ∞ √ L −βc m2c2+p2 L −βc m2c2+u Z(β) = 2 e 2πpdp = 2 e du (2π~) 0 4π~ 0 2 Z ∞ √ 2 2 (Lmc) −βmc2 1+v (Lmc) 2(1 + βmc ) −βmc2 = 2 e dv = 2 2 2 e (9.301) 4π~ 0 4π~ (βmc ) 2 2 L (1 + βmc ) 2 = e−βmc . 2π~2 (βc)2 10 Field Theory on a Lattice

10.1 Scalar Fields To represent a hermitian ﬁeld ϕ(x), we put a real number ϕ(i, j, k, `) at each vertex of the lattice and label a lattice of spacetime points x by a 4-vector of integers as x = a(i, j, k, `) = as (10.1) where a is the lattice spacing and

s = (i, j, k, `) (10.2) is a 4-vector of lattice sites. The derivative ∂iϕ(x) is approximated as ϕ(x + î) − ϕ(x) ∂ ϕ(x) ≈ (10.3) i a in which x is the discrete position 4-vector and î is a unit 4-vector pointing in the i direction. So the euclidian action is the sum over all lattice sites of X 1 1 S = (∂ ϕ(x))2 a4 + m2 ϕ2(x) a4 e 2 i 2 x (10.4) X 1 2 1 = ϕ(x + î) − ϕ(x) a2 + m2 ϕ2(x) a4 2 2 x We switch to dimensionless variables

φ = a ϕ and µ = a m. (10.5)

In these terms, the action density is 1 X 1 S = − φ(x + î)φ(x) + 8 + µ2 φ2(x) e 2 2 (10.6) xi 10.2 Finite-temperature field theory 257 or if the self interaction happens to be quartic 1 1 λ S = (∂ φ(x))2 a4 + m2 φ2(x) a4 + φ4(x) a4 e 2 i 2 4 1 X 1 λ (10.7) = − φ(x + î)φ(x) + 8 + µ2 φ2(x) + φ4(x). 2 2 4 xi

10.2 Finite-temperature field theory Since the Boltzmann operator e−βH = e−H/(kT ) is the time evolution op- −itH/ erator e ~ at the imaginary time t = − i~β = −i~/(kT ), the formulas of finite-temperature field theory are those of quantum field theory with t replaced by −iu = −i~β = −i~/(kT ). Our hamiltonian is H = K +V where K and V are sums over the vertices of the spatial lattice v = a(i, j, k) = (ai, aj, ak) of the cubes of volume a3 times the squared momentum and the potential energy a4 X a4 X aH = aK + aV = $2 + (∇ϕ )2 + m2ϕ2 + P (ϕ ) 2 v 2 v v v v v (10.8) 1 X 1 X = π2 + (∇φ )2 + µ2φ2 + P (φ ). 2 v 2 v v v v v We are assuming that the action if quadratic in the time derivatives of the fields. A matrix element of the first term of the Trotter product formula (7.7) n e−β(K+V ) = lim e−βK/n e−βV/n (10.9) n→∞ is the imaginary-time version of (7.133) with = ~β/n Z 3 0 Y a dπ 3 2 0 hφ |e−K e−V |φ i = e−V (φa) v ea [−πv/2+i(φ1v−φav)πv] . 1 a 2π v

(10.10)

Setting φ˙av = (φ1v − φav)/, we ﬁnd, instead of (7.134) " 3 1/2 # Y a 3 ˙2 2 2 2 hφ |e−K e−V |φ i = e−a [φav+(∇φav) +m φav+P (φv)]/2 . 1 a 2π v (10.11) The product of n = ~β/ such inverse-temperature intervals is " n/2 # Y a3n Z hφ |e−βH |φ i = e−Sev Dφ (10.12) b a 2πβ v v 258 Field Theory on a Lattice in which the euclidian action (10.4 or 10.7) is a sum over all the vertices x = a(i, j, k, `) of the spacetime lattice

n−1 β a3 X h i S = φ˙2 + (∇φ )2 + m2φ2 + P (φ ) (10.13) ev n 2 jv jv jv v j=0 where φ˙jv = n(φj+1,v − φj,v)/β and Dφv = dφn−1,v ··· dφ1,v. −(β −βa)H The amplitude hφb|e b |φai is the integral over all ﬁelds that go from φa(x) at βa to φb(x) at βb each weighted by an exponential Z −(βb−βa)H −Se[φ] hφb|e |φai = e Dφ (10.14) of its euclidian action

Z βb Z 3 1 h 2 2 2 2 i Se[φ] = du d x φ˙ + (∇φ) + m φ + P (φ) (10.15) βa 2 in which Dφ is the n → ∞ limit of the product over all spatial vertices v

" n/2 # Y a3n Dφ = dφ ··· dφ . (10.16) 2π(β − β ) n−1,v 1,v v b a Equivalently, the Boltzmann operator is Z −(βb−βa)H −Se[φ] e = |φbie hφa| Dφ DφaDφb (10.17) Q in which DφaDφb = v dφa,vdφb,v is an integral over the initial and final states. The trace of the Boltzmann operator is the partition function Z Z −βH −Se[φ] −Se[φ] Z(β) = Tr(e ) = e hφa|φbi Dφ DφaDφb = e Dφ Dφa (10.18) which is an integral over all fields that go back to themselves in euclidian time β. So we must require that the field ϕ is periodic in euclidian time. If na is the extend of the lattice in euclidian time, then ϕ(i, j, k, ` + n) = ϕ(i, j, k, `). (10.19) As a convenience, in scalar field theoeries one usually also imposes periodic boundary conditions in the spatial directions. One then has full periodicity: for any four multiples of the lengths Li of the lattice in the four directions ni = nLi, one has

ϕ(i + n1, j + n2, k + n3, ` + n4) = ϕ(i, j, k, `). (10.20) 10.2 Finite-temperature ﬁeld theory 259

Like a position operator (7.107), a ﬁeld at an imaginary time t = −iu = −i~β is deﬁned as

uH/~ −uH/~ φe(x, u) = φe(x, ~β) = e φ(x, 0) e (10.21) in which φ(x) = φ(x, 0) = φe(x, 0) is the field at time zero, which obeys the commutation relations (7.122). The euclidian-time-ordered product of several fields is their product with newer (higher u = ~β) fields standing to the left of older (lower u = ~β) fields as in the definition (7.109). We set ~ = 1 to avoid clutter and irrelevant thinking. The euclidian path integrals for the mean values of euclidian-time-ordered-products of fields are similar to those (7.161 & 7.148) for ordinary time-ordered-products. The euclidian-time-ordered-product of the fields φ(xj) = φ(xj, uj) is the path integral Z −Se[φ] hn|φbiφ(x1)φ(x2)e hφa|ni Dφ hn|T [φe(x1)φe(x2)]|ni = Z (10.22) −S [φ] hn|φbie e hφa|ni Dφ in which the integrations are over all paths that go from before u1 and u2 to after both euclidian times. The analogous result for several fields is Z −Se[φ]/ hn|φbiφ(x1) ··· φ(xk)e ~hφa|ni Dφ hn|T [φe(x1) ··· φe(xk)]|ni = Z −S [φ]/ hn|φbie e ~hφa|ni Dφ (10.23) in which the integrations are over all paths that go from before the times u1, . . . , uk to after them. In the low-temperature β = 1/(kT ) → ∞ limit, the Boltzmann operator is proportional to the outer product |0ih0| of the ground-state kets, e−βH → e−βE0 |0ih0|. In this limit, the integrations are over all fields that run from u = −∞ to u = ∞, and the only energy eigenstate |ni that contributes is the ground state |0i of the theory Z −Se[q]/ h0|φbiφ(x1) ··· φ(xk)e ~hφa|0i Dφ h0|T [φe(x1) ··· φe(xk)]|0i = Z . −S [q]/ h0|φbie e ~hφa|0i Dφ (10.24) Formulas like this one are used in lattice gauge theory. For a free scalar field 260 Field Theory on a Lattice theory, they take the form of a multiple integral over lattice sites s

−βH0 Y −S0e Z0(β) = Tr(e ) = dφse (10.25) s in which the free euclidian action (10.6) is a sum over the spacetime lattice with nearest-neighbor coulings 1 X 1 S = − φ(x + ˆi)φ(x) + 8 + µ2 φ2(x) 0e 2 2 xi (10.26) 1 X = φ K φ . 2 n nm m n,m

The matrix K is n4 × n4. Its elements are

X 2 Ksu = − [δs+i,u + δs−i,u − 2δsu] + µ δsu (10.27) i>0 in which i = (1, 0, 0, 0) etc. The integral is a gaussian. If the lattice is four dimensional with n4 sites, then Z(β) is   Y 1 X Z0(β) = dφs exp − φsKss0 φs0  2 0 s s,s (10.28) s (2π)n4 = det K in which β = na. In the presence of a classical current J, the otherwise free partition function becomes   Y 1 X X Z (β, J) = dφ exp − φ K 0 φ 0 + J φ (10.29) 0 s  2 s ss s s s s s,s0 s

We shift the ﬁeld to φ0 = φ + K−1J (10.30) so that 1 1 − φ0Kφ0 + J · φ0 = − φ + JK−1 K φ + K−1J + J · φ + K−1J 2 2 1 1 = − φKφ + J · K−1J. 2 2 (10.31) 10.3 The Propagator 261 and   Y 1 X X −1 Z (β, J) = dφ exp − φ K 0 φ 0 + J K J 0 . 0 s  2 s ss s s s,s0 s  s s,s0 s,s0 (10.32) s   n4 (2π) 1 X −1 = exp J K J 0 . det K 2 s s,s0 s  s,s0 The formula (10.29) now gives us the propagator as 2 −1 ∂ −1 hφsφs0 i = Z0(β, J) Z0(β, J) = Kss0 . (10.33) ∂Js∂Js0 J=0 −1 To ﬁnd Kss0 , we use X Z π d4k K K−1 = δ = eik·(s−u). (10.34) st tu su (2π)4 t −π We also use this formula to write K as Z π 4 d k ik·(s−u) Ksu = 4 K(k) e (10.35) −π (2π) where 4 X 2 K(k) = 4 sin (kj/2) + µ. (10.36) j=1

10.3 The Propagator 10.4 Pure Gauge Theory

The gauge-covariant derivative is deﬁned in terms of the generators ta of a compact Lie algebra

[ta, tb] = ifabctc (10.37) b and a gauge-ﬁeld matrix Ai = igAi tb as b Di = ∂i − Ai = ∂i − igAi tb (10.38) summed over all the generators, and g is a coupling constant. Since the group is compact, we may raise and lower group indexes without worrying about factors or minus signs. The Faraday matrix is

Fij = [Di,Dj] = [I∂i−Ai(x), I∂j −Aj(x)] = −∂iAj +∂jAi+[Ai,Aj] (10.39) 262 Field Theory on a Lattice in matrix notation. With more indices exposed, it is

(Fij)cd = (−∂iAj + ∂jAi + [Ai,Aj])cd b b b b b e e (10.40) = − igtcd ∂iAj − ∂jAi + [igt Ai , igt Aj] . cd Summing over repeated indices, we get

b b b 2 b e b e (Fij)cd = − igtcd ∂iAj − ∂jAi − g Ai Aj [t , t ] cd b b b 2 b e f = − igtcd ∂iAj − ∂jAi − g Ai Ajifbef tcd b b b 2 b e f = − igtcd ∂iAj − ∂jAi − ig Ai Ajfbef tcd (10.41) b b b 2 f e b = − igtcd ∂iAj − ∂jAi − ig Ai Ajffebtcd b b b f e = − igtcd ∂iAj − ∂jAi + gAi Ajffeb b b b f e b b = − igtcd ∂iAj − ∂jAi + gfbfe Ai Aj = − igtcd Fij where b b b f e Fij = ∂iAj − ∂jAi + gfbfe Ai Aj (10.42) is the Faraday tensor. The action density of this tensor is 1 L = − F b F ij. (10.43) F 4 ij b The trace of the square of the Faraday matrix is

ij h b b c iji Tr FijF = Tr igt Fij igt Fc 2 b ij b c 2 b ij (10.44) = − g Fij Fc Tr(t t ) = −g Fij Fc kδbc 2 b ij = − kg Fij Fb . So the Faraday action density is 1 1 1 L = − F b F ij = Tr F F ij = Tr F F ij . (10.45) F 4 ij b 4kg2 ij 2g2 ij The theory described by this action density, without scalar or spinor fields, is called pure gauge theory. The purpose of a gauge field is to make the gauge-invariant theories. So if a field ψa is transformed by the group element g(x)

0 ψa(x) = g(x)abψb(x), (10.46) 10.4 Pure Gauge Theory 263 then we want the covariant derivative of the ﬁeld to go as

0 0 (Diψ(x)) = ∂i − A (x) g(x)ψ(x) i (10.47) = g(x)Diψ(x) = g(x) [(∂i − Ai(x)) ψ(x)] .

So the gauge ﬁeld must go as

0 ∂ig − Aig = − gAi (10.48) or as

0 −1 −1 Ai(x) = g(x)Ai(x)g (x) + (∂ig(x))g (x). (10.49)

So if g(x) = exp(−iθa(x)ta), then a gauge transforms as

0 −iθa(x)ta iθa(x)ta −iθa(x)ta iθa(x)ta Ai(x) = e Ai(x) e + (∂ie ) e . (10.50)

How does an exponential of a path-ordered, very short line integral of gauge ﬁelds go? We will evaluate how the path-ordered exponential in which g0 is a coupling constant

i0 h −1 −1 i i P exp g0Ai(x)dx = P exp g(x)g0Ai(x)g (x) + (∂ig(x))g (x) dx

h −1 i i = P exp g(x)g0Ai(x)g (x) + ∂i log g(x) dx (10.51) changes under the gauge transformation (10.50) in the limitt dxi → 0. We ﬁnd

h i i0 g0Ai(x)dx −1 P exp g0Ai(x)dx = P g(x)e g (x)

i i i ×elog g(x+dx /2)−log g(x)+log g(x)−log g(x−dx /2)

h i = P g(x)eg0Ai(x)dx g−1(x) (10.52) i ×g(x + dxi/2)g−1(x)g(x)g−1(x − dxi/2)

i = g(x + dxi/2)eg0Ai(x)dx g−1(x − dxi/2).

Putting together a chain of such inﬁnitesimal links, we get

Z x 0 Z x 0 0i 0 0i −1 P exp g0Ai(x )dx = g(x)P exp g0Ai(x )dx g (y). (10.53) y y 264 Field Theory on a Lattice In particular, this means that the trace of a closed loop is gauge invariant I 0 I i i −1 TrP exp g0Ai(x)dx = Trg(x)P exp g0Ai(x)dx g (x) I i = TrP exp g0Ai(x)dx . (10.54)

10.5 Pure Gauge Theory on a Lattice Wilson’s lattice gauge theory is inspired by these last two equations. Another source of inspiration is the approximation for a loop of tiny area dx ∧ dy which in the joint limit dx → 0 and dy → 0 is I i h i W = P exp g0Ai(x)dx = exp g0 Ay,x − Ax,y + g0[Ax,Ay] dxdy = exp − g0Fxy dxdy . (10.55)

To derive this formula, we will ignore the bare coupling constant g0 for the moment and apply the Baker-Campbell-Hausdorf identity 1 1 1 eA eB = exp A + B + [A, B] + [A, [A, B]] + [B, [B,A]] + ... . 2 12 12 (10.56) to the product W = exp Ax dx exp Ay dy + Ay,x dxdy (10.57) × exp − Ax dx − Ax,y dxdy exp − Ay dy .

We get 1 W = exp Ax dx + Ay dy + Ay,x dxdy + [Ax dx, (Ay + Ay,x dx)dy] 2 1 × exp − Ax dx − Ay dy − Ax,y dxdy + [(Ax + Ax,y dy)dx, Ay dy] . 2 (10.58) Applying again the BCH identity, we get h i W = exp Ay,x − Ax,y + [Ax,Ay] dxdy = exp Fxy dxdy , (10.59) an identity that is the basis of lattice gauge theory. 10.5 Pure Gauge Theory on a Lattice 265

Restoring g0, we divide the trace of W by the dimension n of the matrices ta and subtract from unity 1 I 1 h i 1 − Tr P exp g A (x)dxi = 1 − Tr exp g F dxdy n 0 i n 0 xy (10.60) 1 h 1 2i = − Tr g F dxdy + g F dxdy . n 0 xy 2 0 xy The generators of SU(n), SO(n), and Sp(2n) are traceless, so the ﬁrst term vanishes, and we get 1 I 1 2 1 − Tr P exp g A (x)dxi = − Tr g F dxdy . (10.61) n 0 i 2n 0 xy Recalling the more explicit form (10.41) of the Faraday matrix, we have 1 g2 h i kg2 1 − Tr(W ) = 0 Tr t F a t F b dxdy2 = 0 F a dxdy2 (10.62) n 2n a xy b xy 2n xy in which k is the constant of the normalization Tr(tatb) = kδab. For SU(2) with ta = σa/2 and for SU(3) with ta = λa/2, this constant is k = 1/2. The Wilson action is a sum over all the smallest squares of the lattice, called the plaquettes, of the quantity n 1 1 a 2 4 S2 = 2 1 − Tr(W ) = Fij a (10.63) 2kg0 n 4 in which a is the lattice spacing. The full Wilson action is the sum of this quantity over all the elementary squares of the lattice and over i, j = 1, 2, 3, 4. References

Amdahl, David, and Cahill, Kevin. 2016. Path integrals for awkward actions. arXiv, 1611.06685. Davies, P. C. W. 1975. Scalar particle production in Schwarzschild and Rindler metrics. J. Phys., A8, 609–616. DeWitt, Bryce S. 1967. Quantum Theory of Gravity. II. The Manifestly Covariant Theory. Phys. Rev., 162(5), 1195–1239. Faddeev, L. D., and Popov, V. N. 1967. Feynman diagrams for the Yang-Mills ﬁeld. Phys. Lett. B, 25(1), 29–30. Hawking, S. W. 1974. Black hole explosions. Nature, 248, 30–31. Kato, T. 1978. Topics in Functional Analysis. Academic Press. Pages 185–195. Rindler, Wolfgang. 2006. Relativity: Special, General, and Cosmological. 2d edn. Oxford University Press. Srednicki, Mark. 2007. Quantum Field Theory. Cambridge University Press. Trotter, H. F. 1959. On the product of semi-groups of operators. Proc. Ann. Math., 10, 545–551. Weinberg, Steven. 1995. The Quantum Theory of Fields. Vol. I Foundations. Cambridge University Press. Weinberg, Steven. 1996. The Quantum Theory of Fields. Vol. II Modern Applica- tions. Cambridge, UK: Cambridge University Press. Zee, A. 2013. Einstein Gravity in a Nutshell. New Jersey: Princeton University Press. Zee, Anthony. 2010. Quantum Field Theory in a Nutshell. 2d edn. Princeton University Press. Index

complex-variable theory and nonabelian gauge theories, 200–204 conformal mapping, 233–235 ghosts, 203–204 conformal algebra, 228–241 the method of Faddeev and Popov, conformal mapping, 233–235 200–204 conjugate momenta, 81 and perturbative field theory, 184–205 conserved quantities, 79–84 and quantum electrodynamics, 188–192 Covariant derivatives and the Bohm-Aharonov effect, 164–165 in Yang-Mills theory, 126 and The principle of stationary action, 161 decuplet of baryon resonances, 129 density operator for a free particle, 170–171 effective field theories, 209–210 density operator for harmonic oscillator, energy density, 79–84 173–175 euclidian, 167–170 Feynman’s propagator fermionic, 192–199 as a Green’s function, 57 finite temperature, 167–170 fractional linear transformation, 234 for harmonic oscillator in imaginary time, gauge theory, 104–156 173–175 gaussian integrals, 157–158 for partition function, 167–170 Gell-Mann’s SU(3) matrices, 127 for the harmonic oscillator, 165–167 Grassmann variables, 192–199 in finite-temperature field theory, 181–184, groups 249–253 SU(3), 127–129 in imaginary time, 167–170 SU(3) structure constants, 128 in quantum mechanics, 158–167 and Yang-Mills gauge theory, 104–156 in real time, 158–167 Gell-Mann matrices, 127–129 Minkowski, 158–167 Lie groups of fields, 177–205 SU(3), 127–129 SU(3) generators, 127 of fields in euclidian space, 181–184, 249–253 SU(3) structure constants, 128 of fields in imaginary time, 181–184, 249–253 hamiltonian density of scalar fields, 81 of fields in real time, 177–181 Lagrange’s equation partition function for a free particle, in field theory, 77–79 170–171 maximally symmetric spaces, 225–228 principle of stationary action Möbiustransformation, 234 in field theory, 77–79 natural units, 178 sign problem, 204 octet of baryons, 129 symmetries, 79–84 octet of pseudoscalar mesons, 128 tensors path integrals, 157–205 maximally symmetric spaces, 225–228 and gaussian integrals, 157–158 time-ordered products, 175–177, 181, 183–184, and lattice gauge theories, 203 251–253 268 Index

of euclidian ﬁeld operators, 183–184, 251–253 of ﬁeld operators, 181 of position operators, 175–177 Yang-Mills theory, 104–156