Phenomenology of Particle Physics NIU Spring 2002 PHYS586 Lecture Notes

June 23, 2005

Diligent eﬀorts have been made to eliminate all misteaks. – Anonymous

Contents

1 Special Relativity and Lorentz Transformations 4

2 Relativistic quantum mechanics of single particles 10 2.1 Klein-GordonandDiracequations ...... 10 2.2 SolutionsoftheDiracEquation ...... 16 2.3 TheWeylequation ...... 23

3 Maxwell’s equations and electromagnetism 26

4 Field Theory and Lagrangians 28 4.1 The field concept and Lagrangian dynamics ...... 28 4.2 Quantization of free scalar field theory ...... 35 4.3 Quantization of free Dirac fermion field theory ...... 40

5 Interacting scalar field theories 43 5.1 Scalar field with φ4 coupling...... 43 5.2 Scattering processes and cross sections ...... 48 5.3 Scalar field with φ3 coupling...... 55 5.4 Feynmanrules ...... 62

1 6 Quantum Electro-Dynamics (QED) 69 6.1 QEDLagrangianandFeynmanrules ...... 69 6.2 e e+ µ µ+ ...... 77 − → − 6.3 e e+ ff...... 82 − −→ 6.4 Helicities in e e+ µ µ+ ...... 86 − → − 6.5 Bhabha scattering (e e+ e e+) ...... 92 − → − 6.6 Crossingsymmetry...... 98 6.7 e µ+ e µ+ ...... 99 − → − 6.8 Moller scattering (e e e e )...... 100 − − → − − 6.9 Gauge invariance in Feynman diagrams ...... 101 6.10 Compton scattering (γe γe )...... 103 − → − 6.11 e+e γγ ...... 110 − → 7 Decay processes 113 7.1 Decayratesandpartialwidths ...... 113 7.2 Two-bodydecays...... 114 7.3 Scalar decays to fermion-antifermion pairs; Higgs decay ...... 116 7.4 Three-bodydecays ...... 119

8 Fermi theory of weak interactions 122 8.1 Weaknucleardecays ...... 122 8.2 Muondecay...... 124 8.3 Correctionstomuondecay ...... 134 8.4 Inverse muon decay (e ν ν µ )...... 136 − µ → e − 8.5 e ν µ ν ...... 138 − e → − µ 8.6 Charged currents and π± decay...... 139 8.7 Unitarity, renormalizability, and the W boson...... 145

9 Gauge theories 150 9.1 Groupsandrepresentations ...... 150 9.2 The Yang-Mills Lagrangian and Feynman rules ...... 160

10 Quantum Chromo-Dynamics (QCD) 167 10.1 QCDLagrangianandFeynmanrules ...... 167 10.2 Quark-quark scattering (qq qq)...... 169 → 10.3 Renormalization ...... 171 10.4 Gluon-gluon scattering (gg gg)...... 180 → 10.5 The parton model, parton distribution functions, and hadron-hadron scattering . . . . . 182 10.6 Top-antitop production in pp collisions...... 191

2 10.7 Top-antitop production in pp collisions...... 193 10.8 Kinematics in hadron-hadron scattering ...... 194 + 10.9 Drell-Yan scattering (` `− production in hadron collisions) ...... 196

11 Spontaneous symmetry breaking 199 11.1 Globalsymmetrybreaking...... 199 11.2 Local symmetry breaking and the Higgs mechanism ...... 202 11.3 Goldstone’s Theorem and the Higgs mechanism in general ...... 206

12 The standard electroweak model 208 12.1 SU(2) U(1) representationsandLagrangian ...... 208 L × Y 12.2 The Standard Model Higgs mechanism ...... 212 12.3 Fermion masses and Cabibbo-Kobayashi-Maskawa mixing ...... 217

Homework Problems 225

Appendix A: Natural units and conversions 241

Appendix B: Dirac Spinor Formulas 242

Index 243

3 1 Special Relativity and Lorentz Transformations

A successful description of elementary particles must be consistent with the two pillars of modern physics: special relativity and quantum mechanics. Let us begin by establishing some important features of special relativity. Spacetime has four dimensions. For any given event (for example, a ﬁrecracker explodes, or a particle decays to two other particles) one can assign a four-vector position:

(ct,x,y,z) = (x0,x1,x2,x3)= xµ (1.1)

The Greek indices µ,ν,ρ,... run over the values 0, 1, 2, 3, and c is the speed of light in vacuum. As a matter of terminology, xµ is an example of a contravariant four-vector. It is often useful to change our coordinate system (or “inertial reference frame”) according to

µ µ µ ν x x0 = L x . (1.2) → ν Such a change of coordinates is called a Lorentz transformation. Here Lµ is a 4 4 real matrix which ν × parameterizes the Lorentz transformation. It is not arbitrary, however, as we will soon see. The laws of physics should not depend on what coordinate system we use; this is a guiding principle in making a sensible theory. As a simple example of a Lorentz transformation, suppose we rotate our coordinate system about the z-axis by an angle α. Then in the new coordinate system:

µ x0 = (ct0,x0, y0,z0) (1.3) where

ct0 = ct

x0 = x cos α + y sin α

y0 = x sin α + y cos α − z0 = z. (1.4)

Alternatively, we could go to a frame moving with respect to the original frame with velocity v along the z direction, with the origins of the two frames coinciding at time t = t0 = 0. Then:

ct0 = γ(ct βz) − x0 = x

y0 = y

z0 = γ(z βct). (1.5) − where

β = v/c; γ = 1/ 1 β2 (1.6) − q 4 Another way of rewriting this is to deﬁne the rapidity η by β = tanh η, so that γ = cosh η and βγ = sinh η. Then we can rewrite eq. (1.5),

0 0 3 x0 = x cosh η x sinh η − 1 1 x0 = x 2 2 x0 = x 3 0 3 x0 = x sinh η + x cosh η. (1.7) − This is an example of a boost. Another example of a contravariant four-vector is given by the 4-momentum formed from the energy E and spatial momentum ~p of a particle:

pµ = (E/c, ~p). (1.8)

All contravariant four-vectors transform the same way under a Lorentz transformation:

µ µ ν a0 = L νa . (1.9)

A key property of special relativity is that for any two events one can deﬁne a proper interval which is independent of the Lorentz frame, and which tells us how far apart the two events are in a coordinate-independent sense. So, consider two events occurring at xµ and xµ + dµ, where dµ is some four-vector displacement. The proper interval between the events is

(∆τ)2 = (d0)2 (d1)2 (d2)2 (d3)2 = g dµdν (1.10) − − − µν where 10 0 0 0 1 0 0 gµν =  −  (1.11) 0 0 1 0  −   0 0 0 1   −  is known as the metric tensor. Here, and from now on, we adopt the Einstein summation convention, in which repeated indices µ,ν,... are taken to be summed over. It is an assumption of special relativity that gµν is the same in every inertial reference frame. The existence of the metric tensor allows us to deﬁne covariant four-vectors by lowering an index:

x = g xν = (ct, x, y, z), (1.12) µ µν − − − p = g pν = (E/c, p , p , p ). (1.13) µ µν − x − y − z Furthermore, one can deﬁne an inverse metric gµν so that

µν µ g gνρ = δρ , (1.14)

5 µ where δν =1 if µ = ν, and otherwise = 0. It follows that 10 0 0 0 1 0 0 gµν =  −  . (1.15) 0 0 1 0  −   0 0 0 1   −  Then one has, for any vector aµ,

ν µ µν aµ = gµνa ; a = g aν. (1.16)

It follows that covariant four-vectors transform as

ν aµ0 = Lµ aν (1.17) where (note the positions of the indices!)

ν νσ ρ Lµ = gµρg L σ. (1.18)

Because one can always use the metric to go between contravariant and covariant four-vectors, people often use a harmlessly sloppy terminology and neglect the distinction, simply referring to them as four-vectors. If aµ and bµ are any four-vectors, then

aµbνg = a b gµν = a bµ = aµb a b (1.19) µν µ ν µ µ ≡ · is a scalar quantity. For example, if pµ and qµ are the four-momenta of any two particles, then p q is a · Lorentz-invariant; it does not depend on which inertial reference frame it is measured in. In particular, a particle with mass m satisﬁes the on-shell condition

p2 = pµp = E2/c2 ~p2 = m2c2. (1.20) µ − This equation, plus the conservation of four-momentum, is enough to solve most problems in relativistic kinematics. The Lorentz-invariance of equation (1.19) implies that, if aµ and bµ are constant four-vectors, then

µ ν µ ν gµνa0 b0 = gµνa b , (1.21) so that

µ ν ρ σ ρ σ gµνL ρL σa b = gρσa b . (1.22)

Since aµ and bν are arbitrary, it must be that:

µ ν gµνL ρL σ = gρσ. (1.23)

6 This is the fundamental constraint that a Lorentz transformation matrix must satisfy. In matrix form, it could be written as LT gL = g. If we contract eq. (1.23) with gρκ, we obtain

κ ν κ Lν L σ = δσ (1.24)

Applying this to eqs. (1.2) and (1.17), we ﬁnd that the inverse Lorentz transformation of any four-vector is

ν µ ν a = a0 Lµ (1.25) µ aν = aµ0 L ν (1.26)

Let us now consider some particular Lorentz transformations. To begin, we note that as a matrix, det(L)= 1. (See homework problem.) An example of a “large” Lorentz transformation with det(L)= ± 1 is: − 1 0 0 0 − 0 100 Lµ =   . (1.27) ν 0 010    0 001    This just ﬂips the sign of the time coordinate, and is therefore known as time reversal:

0 0 1 1 2 2 3 3 x0 = x x0 = x x0 = x x0 = x . (1.28) − Another “large” Lorentz transformation is parity, or space inversion:

10 0 0 µ 0 1 0 0 L ν =  −  , (1.29) 0 0 1 0  −   0 0 0 1   −  so that:

0 0 1 1 2 2 3 3 x0 = x x0 = x x0 = x x0 = x . (1.30) − − − It was once thought that the laws of physics have to be invariant under these operations. However, it was shown experimentally in the 1950’s that parity is violated in the weak interactions, speciﬁcally in 60 the weak decays of the Co nucleus and the K± mesons. Likewise, experiments in the 1960’s on the decays of K0 mesons showed that time-reversal invariance is violated (at least if very general properties of quantum mechanics and special relativity are assumed). However, all experiments up to now are consistent with invariance of the laws of physics under Lorentz transformations that are continuously connected to the identity; these are known as “proper” Lorentz transformations and have det(L) = +1. They can be built up out of inﬁnitesimal Lorentz transformations:

Lµ = δµ + ωµ + (ω2), (1.31) ν ν ν O

7 µ where we agree to drop everything with more than one ω ν. Then, according to eq. (1.23),

µ µ ν ν gµν(δρ + ω ρ + ...)(δσ + ω σ + ...)= gρσ, (1.32) or

gρσ + ωσρ + ωρσ + ... = gρσ. (1.33)

Therefore

ω = ω (1.34) σρ − ρσ is an antisymmetric 4 4 matrix, with 4 3/2 1 = 6 independent entries. These correspond to 3 × · · rotations (ρ,σ = 1, 2 or 1,3 or 2,3) and 3 boosts (ρ,σ = 0, 1 or 0,2 or 0,3). It is a mathematical fact that any Lorentz transformation can be built up out of repeated infinitesimal boosts and rotations, plus the operations of time-reversal and space inversion. Lorentz transformations obey the mathematical properties of a group, known as the Lorentz group. The subset of Lorentz transformations that can be built out of repeated infinitesimal boosts and rotations form a smaller group, called the proper Lorentz group. In the Standard Model of particle physics and generalizations of it, all interesting objects, including operators, states, particles, and fields, live in representations of the Lorentz group. We’ll study these group representations in more detail later. So far we have considered constant four-vectors. However, one can also consider four-vectors that depend on position in spacetime. For example, suppose that F (x) is a scalar function of xµ. [It is usual to leave the index µ off of xµ when it is used as the argument of a function, so F (x) really means F (x0,x1,x2,x3).] Under a Lorentz transformation from coordinates xµ x µ, at a given fixed point → 0 in spacetime the value of the function F 0 reported by an observer using the primed coordinate system is taken to be equal to the value of the original function F in the original coordinates:

F 0(x0)= F (x). (1.35)

Then ∂F 1 ∂F ∂ F = ( , ~ F ) (1.36) µ ≡ ∂xµ c ∂t ∇ is a covariant four-vector. This is because: ν ∂ ∂x ∂ ν (∂µF )0(x0) µ F 0(x0)= µ ν F (x)= Lµ (∂νF )(x), (1.37) ≡ ∂x0 ∂x0 ∂x showing that it transforms according to eq. (1.17). [The second equality uses the chain rule and eq. (1.35); the last equality uses eq. (1.25) with aµ = xµ.] By raising the index, one obtains a contravariant four-vector function 1 ∂F ∂µF = gµν∂ F = ( , ~ F ). (1.38) ν c ∂t −∇

8 One can obtain another scalar function by acting twice with the 4-dimensional derivative operator on F , contracting the indices on the derivatives:

1 ∂2F ∂µ∂ F = 2F. (1.39) µ c2 ∂t2 − ∇ The object ∂µ∂ F is a 4-dimensional generalization of the Laplacian. − µ A tensor is an object which can carry an arbitrary number of spacetime vector indices, and trans- µν µ forms appropriately when one goes to a new reference frame. The objects g and gµν and δν are constant tensors. Four-vectors and scalar functions and 4-derivatives of them are also tensors. In

µ1µ2... general, the deﬁning characteristic of a tensor function Tν1ν2... (x) is that under a change of reference frame, it transforms so that in the primed coordinate system, the corresponding tensor T 0 is:

µ1µ2... µ1 µ2 σ1 σ2 ρ1ρ2... T 0 (x0)= L L L L T (x). (1.40) ν1ν2... ρ1 ρ2 ··· ν1 ν2 ··· σ1σ2... A special and useful constant tensor is the totally antisymmetric Levi-Civita tensor:

+1 if µνρσ is an even permutation of 0123 ² = 1 if µνρσ is an odd permutation of 0123 (1.41) µνρσ  −  0 otherwise

One use for the Levi-Civita tensor is in understanding the Lorentz invariance of four-dimensional integration. Define 4 four-vectors which in a particular frame are given by the infinitesimal differentials

Aµ = (cdt, 0, 0, 0); (1.42) Bµ = (0,dx, 0, 0); (1.43) Cµ = (0, 0,dy, 0); (1.44) Dµ = (0, 0, 0,dz). (1.45)

Then the 4-dimensional volume element

d4x dx0dx1dx2dx3 = AµBνCρDσ² (1.46) ≡ µνρσ is Lorentz-invariant, since in the last expression it has no uncontracted four-vector indices. It follows that if F (x) is a Lorentz scalar function of xµ, then the integral

I[F ]= d4x F (x) (1.47) Z is invariant under Lorentz transformations. This is good because eventually we will learn to deﬁne theories in terms of such an integral, known as the action.

9 2 Relativistic quantum mechanics of single particles

2.1 Klein-Gordon and Dirac equations

Any realistic theory must be consistent with quantum mechanics. In this section, we consider how to formulate a theory of quantum mechanics that is consistent with special relativity. Suppose that Φ(x) is the wavefunction of a free particle in 4-dimensional spacetime. A fundamental principle of quantum mechanics is that the time dependence of Φ is determined by a Hamiltonian operator, according to: ∂ HΦ= i¯h Φ (2.48) ∂t Now, the three-momentum operator is given by

P~ = i¯h~ . (2.49) − ∇ Since H and P~ commute, one can always take Φ to be one of the basis of eigenstates with energy and momentum eigenvalues E and ~p respectively:

HΦ= EΦ; P~ Φ= ~pΦ. (2.50)

One can now turn this into a relativistic Schrodinger wave equation for free particle states, by using the fact that special relativity implies:

E = m2c4 + ~p2c2, (2.51) q where m is the mass of the particle. To make sense of this as an operator equation, we could try expanding it in an inﬁnite series, treating ~p2 as small compared to m2c2:

∂ ~p2 ~p4 i¯h Φ = mc2(1 + + ...)Φ (2.52) ∂t 2m2c2 − 8m4c4 ¯h2 ¯h4 = [mc2 2 ( 2)2 + ...]Φ (2.53) − 2m∇ − 8m3c2 ∇ If we keep only the first two terms, then we recover the standard non-relativistic quantum mechanics of a free particle; the first term mc2 is an unobservable constant contribution to the Hamiltonian, proportional to the rest energy, and the second term is the usual non-relativistic kinetic energy. However, the presence of an infinite number of derivatives leads to horrible problems, including apparently non-local effects. Instead, one can consider the operator H2 acting on Φ, avoiding the square root. It follows that

H2Φ= E2Φ = (c2~p2 + m2c4)Φ = (c2P~ 2 + m2c4)Φ, (2.54) so that: ∂2Φ = 2Φ+ m2Φ. (2.55) − ∂t2 −∇

10 As an aside, here and from now on we have set c = 1 andh ¯ = 1 by a choice of units. This convention means that mass, energy, and momentum all have the same units (GeV), while time and distance have 1 units of GeV− , and velocity is dimensionless. These conventions greatly simplify the equations of particle physics. One can always recover the usual metric system units using the following conversion table for energy, mass, distance, and time, respectively:

3 (1 GeV) = 1.6022 10− erg, (2.56) × 2 24 (1 GeV)/c = 1.7827 10− g, (2.57) × 1 14 (1 GeV)− (¯hc) = 1.9733 10− cm, (2.58) × 1 25 (1 GeV)− ¯h = 6.5822 10− sec. (2.59) × Using eq. (1.39), the wave-equation eq. (2.55) can be rewritten in a manifestly Lorentz-invariant way as

µ 2 (∂ ∂µ + m )Φ = 0 (2.60)

This relativistic generalization of the Schrodinger equation is known as the Klein-Gordon equation. It is easy to guess the solutions of the Klein-Gordon equation. If we try:

ik x Φ(x) = Φ0e− · , (2.61) where Φ is a constant and kµ is a four-vector, then ∂ Φ= ik Φ and so 0 µ − µ ∂µ∂ Φ= kµk Φ= k2Φ (2.62) µ − µ − Therefore, we only need to impose k2 = m2 to have a solution. It is then easy to check that this is an eigenstate of H and P~ with energy E = k0 and three-momentum ~p = ~k, satisfying E2 = ~p2 + m2. However, there is a big problem with this. If kµ = (E, ~p) gives a solution, then so does kµ = ( E, ~p). − By increasing ~p , one can have E arbitrarily large. This is a disaster, because the energy is not | | | | bounded from below. If the particle can interact, it will make transitions from higher energy states to lower energy states. Nothing can stop this, leading, it would seem, to the release of an inﬁnite amount of energy as the particle acquires a larger and larger three-momentum! If only it were true, this phenomenon would be very interesting to Exxon. In 1927, Dirac suggested an alternative, based on the observation that the problem with the Klein- Gordon equation seems to be that it is quadratic in H or equivalently ∂/∂t; this leads to the sign ambiguity for E. Dirac could also have been† motivated by the fact that particles like the electron have spin; since they have more than one intrinsic degree of freedom, trying to explain them with a single wavefunction Φ is doomed to failure. Instead, Dirac proposed to write a relativistic Schrodinger equation, for a multi-component wavefunction Ψa, where the spinor index a = 1, 2,...,n runs over the components. The wave equation should be linear in ∂/∂t; since relativity places t on the same footing

†Apparently, he realized this only in hindsight.

11 as x,y,z, it should also be linear in derivatives of the spatial coordinates. Therefore, the equation ought to take the form ∂ i Ψ= HΨ = (~α P~ + βm)Ψ (2.63) ∂t · where α , α , α , and β are n n matrices acting in “spinor space”. x y z × To determine what ~α and β have to be, consider H2Ψ. There are two ways to evaluate the result. First, by exactly the same reasoning as for the Klein-Gordon equation, one ﬁnds

∂2 Ψ = ( 2 + m2)Ψ. (2.64) −∂t2 −∇ On the other hand, expressing H in terms of the right-hand side of eq. (2.63), we ﬁnd:

∂2 3 ∂ ∂ ∂ Ψ= α α im (α β + βα ) + β2m2 Ψ (2.65) −∂t2 − j k ∂xj ∂xk − j j ∂xj  j,kX=1 Xj   Since partial derivatives commute, one can write:

3 ∂ ∂ 1 3 ∂ ∂ α α = (α α + α α ) (2.66) j k ∂xj ∂xk 2 j k k j ∂xj ∂xk j,kX=1 j,kX=1 Then comparing eqs. (2.64) and (2.65), one ﬁnds that the two agree if, for j, k = 1, 2, 3:

β2 = 1; (2.67)

αjβ + βαj = 0; (2.68)

αjαk + αkαj = 2δjk. (2.69)

The simplest solution occurs turns out to require n = 4 spinor indices. This may be somewhat surprising, since naively one only needs n = 2 to describe a spin-1/2 particle like the electron. As we will see, the Dirac equation automatically describes positrons as well as electrons, accounting for the doubling. It is easiest to write the solution in terms of 2 2 Pauli matrices: × 0 1 0 i 1 0 1 0 σ1 = ; σ2 = − ; σ3 = ; and σ0 = . 1 0 i 0 0 1 0 1 µ ¶ µ ¶ µ − ¶ µ ¶ Then one can check that the 4 4 matrices × 0 σ0 σj 0 β = ; αj = − ; (j = 1, 2, 3) (2.70) σ0 0 0 σj µ ¶ µ ¶ obey the required conditions. The matrices β, α are written in 2 2 block form, so “0” actually j × denotes a 2 2 block of 0’s. Equation (2.63) is known as the Dirac equation, and the 4-component × object is known as a Dirac spinor. Note that the fact that Dirac spinor space is 4-dimensional, just like ordinary spacetime, is really just a coincidence.‡ One must be careful not to confuse the two types of 4-dimensional spaces!

‡For example, if we lived in 10 dimensional spacetime, it turns out that Dirac spinors would have 32 components.

12 It is convenient and traditional to rewrite the Dirac equation in a nicer way by multiplying it on the left by the matrix β, and deﬁning

0 j γ = β; γ = βαj; (j = 1, 2, 3). (2.71)

The result is ∂ ∂ ∂ ∂ i(γ0 + γ1 + γ2 + γ3 ) m Ψ = 0, (2.72) ∂x0 ∂x1 ∂x2 ∂x3 · − ¸ or, even more nicely:

(iγµ∂ m)Ψ = 0. (2.73) µ − The γµ matrices are explicitly given, in 2 2 blocks, by: × 0 σ0 0 σj γ0 = ; γj = ; (j = 1, 2, 3). (2.74) σ0 0 σj 0 µ ¶ µ − ¶ [It should be noted that the solution found above for the γµ is not unique. To see this, suppose U is any unitary 4 4 matrix satisfying U U = 1. Then the Dirac equation implies: × † µ U(iγ ∂ m)U †UΨ = 0, (2.75) µ − µ µ from which it follows that, writing γ0 = Uγ U †, and Ψ0 = UΨ,

µ (iγ0 ∂ m)Ψ0 = 0. (2.76) µ − µ µ So, the new γ0 matrices together with the new spinor Ψ0 are just as good as the old pair γ , Ψ; there are an inﬁnite number of diﬀerent, equally valid choices. The set we’ve given above is called the chiral or Weyl representation. Another popular choice used by some textbooks (but not here) is the Pauli-Dirac representation.] Many problems involving fermions in high-energy physics involve many gamma matrices dotted into partial derivatives or momentum four-vectors. To keep the notation from getting too bloated, it is often useful to use the Feynman slash notation:

µ γ aµ = a/ (2.77) for any four-vector aµ. Then the Dirac equation takes the even more compact form:

(i∂/ m)Ψ = 0. (2.78) − Some important properties of the γµ matrices are:

0 0 j j γ † = γ ; γ † = γ (j = 1, 2, 3) (2.79) − 0 µ 0 µ γ γ †γ = γ (2.80)

Tr(γµγν) = 4gµν (2.81) µ γ γµ = 4 (2.82) γ γ + γ γ = γ ,γ = 2g (2.83) µ ν ν µ { µ ν} µν

13 Note that on the right-hand sides of each of eqs. (2.82) and (2.83), there is an implicit 4 4 unit × matrix. It turns out that one almost never needs to know the explicit form of the γµ. Instead, the equations above can be used to derive identities needed in practical work. (You will get some practice with this from the homework.)

How does a Dirac spinor Ψa(x) transform under a Lorentz transformation? It carries no vector index, so it is not a tensor. On the other hand, the fact that the Hamiltonian “mixes up” the components of Ψa(x) is a clue that it doesn’t transform like an ordinary scalar function either. Instead, we might expect that the spinor reported by an observer in the primed frame is given by

Ψ0(x0) = ΛΨ(x), (2.84) where Λ is a 4 4 matrix that depends on the Lorentz transformation matrix Lµ . In fact, you will × ν µ µ µ show for homework that for an inﬁnitesimal Lorentz transformation L ν = δν + ω ν, one has:

i µν Ψ0(x0) = (1 ω S )Ψ(x), (2.85) − 2 µν where i Sµν = [γµ,γν]. (2.86) 4 To obtain the result for a ﬁnite Lorentz transformation, we can apply the same inﬁnitesimal transformation a large number of times N, with N . Letting Ω = Nω , we obtain, after N iterations → ∞ µν µν µ of the Lorentz transformation parameterized by ω ν:

Lµ = (δµ + Ωµ /N)N [exp(Ω)]µ (2.87) ν ν ν → ν as N . Here we are using the identity: → ∞ lim (1 + x/N)N = exp(x), (2.88) N →∞ with the exponential of a matrix to be interpreted in the power series sense:

exp(M)=1+ M + M 2/2+ M 3/6+ .... (2.89)

For the Dirac spinor, one has in the same way:

N i µν i µν Ψ0(x0) = 1 Ω S /N Ψ(x) exp Ω S Ψ(x). (2.90) 2 µν 2 µν µ − ¶ → µ− ¶ µ So, we have found the Λ that appears in eq. (2.84) corresponding to the L ν that appears in eq. (2.87): i Λ = exp Ω Sµν . (2.91) 2 µν µ− ¶ As an example, consider a boost in the z direction: 0 0 0 η − 0 00 0 Ωµ =   . (2.92) ν 0 00 0    η 0 0 0   −  14 Then η2 0 0 0 0 0 0 η3 − 0 00 0 0 00 0 Ω2 =   , Ω3 =   , etc. (2.93) 0 00 0 0 00 0  2   3   0 0 0 η   η 0 0 0     −  so that 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 Lµ = (1+ η2/2+ ...)   (η + η3/6+ ...)   ν 0 0 0 0 − 0 0 0 0      0 0 0 1   1 0 0 0  cosh η 0 0 sinh η    − 0 00 0 =   , (2.94) 0 00 0    sinh η 0 0 cosh η   −  in agreement with eq. (1.7). Meanwhile, Ω = Ω = η, so 03 − 30 − i η η σ3 0 Ω Sµν = [γ0,γ3]= (2.95) −2 µν −4 2 0 σ3 µ − ¶ in 2 2 blocks. Since this matrix is diagonal, it is particularly easy to exponentiate: × η/20 0 0 eη/2 0 0 0 η/2 0 η/2 0 0 0 e− 0 0 Λ = exp  −  =  η/2  . (2.96) 0 0 η/2 0 0 0 e− 0  −   η/2   0 0 0 η/2   0 0 0 e      This, then is the matrix which boosts a Dirac spinor in the z direction with rapidity η, in eq. (2.84). Since Ψ is not a scalar, it is natural to ask whether one can use it to construct a scalar quantity. A natural guess is to get rid of all the pesky spinor indices by

4 Ψ†Ψ(x) Ψ† Ψ . (2.97) ≡ a a aX=1

However, under a Lorentz transformation, Ψ0(x0) = ΛΨ(x) and Ψ0†(x0)=Ψ†(x)Λ†, so:

Ψ0†Ψ0(x0)=Ψ†Λ†ΛΨ(x). (2.98)

This will therefore be a scalar function if Λ†Λ = 1, in other words if Λ is a unitary matrix. However, this is not true, as the example of eq. (2.96) clearly shows! Instead, with amazing foresight, let us consider the object

0 Ψ†γ Ψ. (2.99)

Under a Lorentz transformation:

0 0 Ψ0†γ Ψ0(x0)=Ψ†Λ†γ ΛΨ(x). (2.100)

15 0 Therefore, Ψ†γ Ψ will transform as a scalar if:

0 0 Λ†γ Λ= γ . (2.101)

One can check that this is indeed true for the special case of eq. (2.96). More importantly, it can be proved for any i Λ = 1 ω Sµν (2.102) − 2 µν that is infinitesimally close to the identity, using eqs. (2.79) and (2.80). Therefore, it is true for any proper Lorentz transformation built out of infinitesimal ones. Motivated by this, one defines, for any Dirac spinor Ψ,

0 Ψ Ψ†γ . (2.103) ≡ One should think of Ψ as a column vector in spinor space, and Ψ as a row vector. Then their inner product,

ΨΨ (2.104) with all spinor indices contracted, transforms as a scalar function under proper Lorentz transformations. Similarly, one can show that

ΨγµΨ (2.105) transforms as a four-vector. One should think of eq. (2.105) as a (row vector) (matrix) (column × × vector) in spinor-index space, with a spacetime vector index µ hanging around.

2.2 Solutions of the Dirac Equation

Our next task is to construct solutions to the Dirac equation. Let us separate out the xµ-dependent part as a plane wave, by trying

ip x Ψ(x)= u(p, s)e− · . (2.106)

Here pµ is a four-vector momentum, with p0 = E > 0. A solution to the Dirac equation must also satisfy the Klein-Gordon equation, so p2 = E2 ~p2 = m2. The object u(p, s) is a spinor, labeled by − the 4-momentum p and s. For now s just distinguishes between distinct solutions, but it will turn out to be related to the spin. Plugging this into the Dirac equation, we obtain a 4 4 eigenvalue equation: × (p/ m)u(p, s) = 0. (2.107) − To simplify things, ﬁrst consider this equation in the rest frame of the particle, where pµ = (m, 0, 0, 0). In that frame,

m(γ0 1)u(p, s) = 0. (2.108) −

16 Using the explicit form of γ0, we can therefore write in 2 2 blocks: × 1 1 − u(p, s) = 0, (2.109) 1 1 µ − ¶ where each “1” means a 2 2 unit matrix. The solutions are clearly × χ u(p, s)= √m s , (2.110) χ µ s ¶ where χs can be any 2-vector, and the √m normalization is a convention. In practice, it is best to choose the χs orthonormal, satisfying χs†χr = δrs for r, s = 1, 2. A particularly nice choice is: 1 0 χ = ; χ = . (2.111) 1 0 2 1 µ ¶ µ ¶ As we will see, these just correspond to spin eigenstates S = 1/2 and 1/2. z − Now to construct the corresponding solution in any other frame, one can just boost the spinor using eqs. (2.91). For example, consider the solution 1 ip x0 0 imt0 Ψ0(x0)= u(p, 1)e− · = √m   e− . (2.112) 1    0    in a frame where the particle is at rest; we have called it the primed frame for convenience. We suppose the primed frame is moving with respect to the unprimed frame with rapidity η in the z direction. Thus, the particle has, in the unprimed frame:

0 3 E = p = m cosh η; pz = p = m sinh η. (2.113)

1 Now, Ψ(x)=Λ− Ψ0(x0) from eq. (2.84), so using the inverse of eq. (2.96): η/2 e− 0 ip x Ψ(x)= √m   e− · . (2.114) eη/2    0    We can rewrite this, noting that from eq. (2.113),

η/2 η/2 √me = E + p ; √me− = E p . (2.115) z − z p p Therefore, one solution of the Dirac equation for a particle moving in the z direction, with energy E and three-momentum p = √E2 m2, is: z − √E p − z 0 ip x Ψ(x)=   e− · , (2.116) √E + pz    0    so that √E p − z 0 u(p, 1) =   (2.117) √E + pz    0    17 in this frame. 0 Similarly, if we use instead χ = in eq. (2.110) in the rest frame, and apply the same procedure, 2 1 we find a solution: µ ¶ 0 0 η/2 e ip x √E + pz ip x Ψ(x)= √m   e− · =   e− · (2.118) 0 0  η/2     e−   √E pz     −  so that 0 √E + pz u(p, 2) =   (2.119) 0    √E pz   −  in this frame. Note that pz in eqs. (2.116) and (2.118) can have either sign, corresponding to the wavefunction for a particle moving in either the +z or z directions. − In order to make a direct connection between spin and the various components of a Dirac spinor, let us now consider how to construct the spin operator S~. To do this, recall that by definition, spin is the difference between the total angular momentum operator J~ and the orbital angular momentum operator L~ :

J~ = L~ + S.~ (2.120)

Now,

L~ = ~x P,~ (2.121) × where ~x and P~ are the three-dimensional position and momentum operators. The total angular momentum must be conserved, or in other words it must commute with the Hamiltonian:

[H, J~] = 0. (2.122)

Using the Dirac Hamiltonian given in eq. (2.63), we have

[H, L~ ]=[~α P~ + βm, ~x P~ ]= i~α P,~ (2.123) · × − × where we have used the canonical commutation relation (with ¯h =1) [P ,x ]= iδ . So, comparing j k − jk eqs. (2.120), (2.122) and (2.123), it must be true that:

~σ P~ 0 [H, S~]= i~α P~ = i − × . (2.124) × 0 ~σ P~ µ × ¶ One can now observe that the matrix: 1 ~σ 0 S~ = (2.125) 2 0 ~σ µ ¶ obeys eq. (2.124). So, it must be the spin operator acting on Dirac spinors.

18 In particular, the z-component of the spin operator for Dirac spinors is given by the diagonal matrix:

1 0 0 0 1 0 1 0 0 Sz =  −  (2.126) 2 0 0 1 0    0 0 0 1   −  Therefore, the solutions in eqs. (2.116) and (2.118) can be identified to have spin eigenvalues Sz = +1/2 and S = 1/2, respectively. In general, a Dirac spinor eigenstate with S = +1/2 will have only the z − z first and third components non-zero, and one with S = 1/2 will have only the second and fourth z − components non-zero, regardless of the direction of the momentum. Note that, as promised, Sz = 1/2 ( 1/2) exactly corresponds to the use of χ (χ ) in eq. (2.110). − 1 2 The helicity operator gives the relative orientation of the spin of the particle and its momentum. It is defined to be:

~p S~ h = · . (2.127) ~p | | Like S , helicity has possible eigenvalues 1/2 for a spin-1/2 particle. For example, if p > 0, then z ± z eqs. (2.116) and (2.118) represent states with helicity +1/2 and 1/2 respectively. The helicity is not − invariant under Lorentz transformations for massive particles. This is because one can always boost to a different frame in which the 3-momentum is flipped but the spin remains the same. (Also, note that unlike Sz, helicity is not even well-defined for a particle exactly at rest.) However, a massless particle moves at the speed of light in any inertial frame, so one can never boost to a frame in which its 3-momentum direction is flipped. This means that for massless (or very energetic, so that E m) À particles, the helicity is fixed and invariant under Lorentz transformations. In any frame, a particle with ~p and S~ parallel has helicity h = 1/2, and a particle with ~p and S~ antiparallel has helicity h = 1/2. − Helicity is particularly useful in the high-energy limit. For example, we can consider four solutions obtained from the E,p m limits of eqs. (2.116) and (2.118), so that p = E: z À | z| 0 0 iE(t z) Ψ =   e− − [~p , S~ , h = +1/2] (2.128) pz>0,Sz=+1/2 √2E ↑ ↑    0   0  √ 2E iE(t z) ~ Ψpz>0,Sz= 1/2 =   e− − [~p , S , h = 1/2] (2.129) − 0 ↑ ↓ −    0   √2E  0 iE(t+z) Ψ =   e− [~p , S~ , h = 1/2] (2.130) pz<0,Sz=+1/2 0 ↓ ↑ −    0   0  0 iE(t+z) ~ Ψpz<0,Sz= 1/2 =   e− [~p , S , h = +1/2] (2.131) − 0 ↓ ↓  √   2E    19 In this high-energy limit, a Dirac spinor with h = +1/2 is called right-handed (R) and one with h = 1/2 is called left-handed (L). Notice that a high-energy L state is one that has the last two − entries zero, while a high-energy R state always has the first two entries zero. It is useful to define matrices which project onto L and R states in the high-energy or massless limit. In 2 2 blocks: × 1 0 0 0 P = ; P = , (2.132) L 0 0 R 0 1 µ ¶ µ ¶ where 1 and 0 mean the 2 2 unit and zero matrices, respectively. Then P acting on any Dirac spinor × L gives back a left-handed spinor, by just killing the last two components. The projectors obey the rules:

2 2 PL = PL; PR = PR; PRPL = PLPR = 0. (2.133)

It is traditional to write PL and PR in terms of a “ﬁfth” gamma matrix, which in our conventions is given in 2 2 blocks by: × 1 0 γ5 = − . (2.134) 0 1 µ ¶ Then 1 γ 1+ γ P = − 5 ; P = 5 . (2.135) L 2 R 2

The matrix γ5 satisﬁes the equations:

2 γ = 1; γ† = γ ; γ ,γ = 0. (2.136) 5 5 5 { 5 µ} So far, we have been considering Dirac spinor wavefunction solutions of the form

ip x Ψ(x)= u(p, s)e− · (2.137) with p0 = E > 0. We have successfully interpreted these solutions in terms of a spin-1/2 particle, say, the electron. However, there is nothing mathematically wrong with these solutions for pµ with p0 < 0 and p2 = m2. So, like the Klein-Gordon equation, the Dirac equation has the embarrassment of negative energy solutions. Dirac proposed to get around the problem of negative energy states by using the fact that spin-1/2 particle are fermions. The Pauli exclusion principle dictates that two fermions cannot occupy the same quantum state. Therefore, Dirac proposed that all of the negative energy states are occupied. This prevents electrons with positive energy from making disastrous transitions to the E < 0 states. The infinite number of filled E < 0 states is called the Dirac sea. Now, if one of the states in the Dirac sea becomes unoccupied, it leaves behind a “hole”. Since a hole is the absence of an E < 0 state, it effectively has energy E > 0. An electron has charge e, − § − so the hole corresponding to its absence effectively has the opposite charge, +e. Since both electrons

§Here, e is always deﬁned to be positive, so that the electron has charge e. (Some references deﬁne e to be negative.) −

20 and holes obey p2 = m2, they have the same mass. Dirac’s proposal therefore predicts the existence of “anti-electrons” or positrons, with positive energy and positive charge. The positron was indeed discovered in 1932 in cosmic ray experiments. Feynman and St¨uckelberg noted that one can reinterpret the positron as a negative energy electron moving backwards in time, so that pµ pµ and S~ S~. According to this interpretation, the → − → − wavefunction for a positron with 4-momentum pµ with p0 = E > 0 is

ip x Ψ(x)= v(p, s)e · . (2.138)

Now, using the Dirac equation (2.78), v(p, s) must satisfy the eigenvalue equation:

(p/ + m)v(p, s) = 0. (2.139)

We can now construct solutions to this equation just as before. First, in the rest (primed) frame of the particle, we have in 2 2 blocks: × m m v(p, s) = 0. (2.140) m m µ ¶ So, the solutions are

ξ 0 Ψ (x )= √m s eimt (2.141) 0 0 ξ µ − s ¶ for any two-vector ξs. One must be careful in interpreting the quantum numbers of the positron solutions to the Dirac equation. This is because the Hamiltonian, 3-momentum, and spin operators of a positron described by the wavefunction eq. (2.138) are all given by the negative of the expressions one would use for an electron wavefunction. Thus, acting on a positron wavefunction, one has ∂ HΨ = i Ψ (2.142) − ∂t P~ Ψ = i~ Ψ (2.143) ∇ 1 ~σ 0 S~Ψ = Ψ, (2.144) 2 0 ~σ − µ ¶ where H, P~ , and S~ are the operators whose eigenvalues are to be interpreted as the energy, 3- momentum, and spin of the positive-energy positron antiparticle. Therefore, to describe a positron with spin S = +1/2 or 1/2, one should use, respectively, z − 0 1 ξ = ; or ξ = , (2.145) 1 1 2 0 µ ¶ µ ¶ in eq. (2.141). Now we can boost to the unprimed frame just as before, yielding the solutions: 0 √E + pz v(p, 1) =   , (2.146) 0    √E pz   − −  21 √E p − z 0 v(p, 2) =   (2.147) √E + pz  −   0    Here v(p, 1) corresponds to a positron moving in the +z direction with 3-momentum pz and energy 2 2 E = pz + m and Sz = +1/2, hence helicity h = +1/2 if pz > 0. Similarly, v(p, 2) corresponds to a positronp with the same energy and 3-momentum, but with S = 1/2, and therefore helicity h = 1/2 z − − if pz > 0.

Note that for positron wavefunctions, PL projects onto states that describe right-handed positrons in the high-energy limit, and PR projects onto states that describe left-handed positrons in the high- energy limit. If we insist that PL projects on to left-handed spinors, and PR projects on to right-handed spinors, then we must simply remember that a right-handed positron is described by a left-handed spinor (annihilated by PR), and vice versa! The Dirac spinors given above only describe electrons and positrons with both momentum and spin aligned along the z direction. More generally, we could construct u(p, s) and v(p, s) for states ± describing electrons or positrons with any ~p and spin. However, in general that is quite a mess, and it turns out to be not particularly useful. For a more useful way of encoding information, consider the Dirac row spinors:

0 0 u(p, s)= u(p, s)†γ v(p, s)= v(p, s)†γ . (2.148)

Then the quantities u(p, s)u(k,r) and u(p, s)v(k,r) and v(p, s)u(k,r) and v(p, s)v(k,r) are all Lorentz scalars. For example, in the rest frame of a particle with mass m and spin Sz = +1/2, one has 0 σ0 u(p, 1) = u(p, 1) γ0 = √m (1 0 1 0) = ( √m 0 √m 0 ) , (2.149) † σ0 0 µ ¶ so that √m 0 u(p, 1)u(p, 1) = ( √m 0 √m 0 )   = 2m. (2.150) √m    0    Since this quantity is a scalar, it must be true that u(p, 1)u(p, 1) = 2m in any Lorentz frame, in other words, for any pµ. More generally, if s, r = 1, 2 represent orthonormal spin state labels, then the u and v spinors obey:

u(p, s)u(p,r) = 2mδsr, (2.151) v(p, s)v(p,r)= 2mδ , (2.152) − sr v(p, s)u(p,r)= u(p, s)v(p,r) = 0. (2.153)

µ Similarly, one can show that u(p, s)γ u(p,r) = 2(m,~0)δsr in the rest frame. Since it is a four-vector, it must be that in any frame:

µ µ u(p, s)γ u(p,r) = 2p δsr. (2.154)

22 But by far the most useful identities that we will use later on are the spin-sum equations:

2 u(p, s)u(p, s) = p/ + m; (2.155) sX=1 2 v(p, s)v(p, s) = p/ m. (2.156) − sX=1 Here the spin state label s is summed over. These equations are to be interpreted in the sense of a row vector times a column vector giving a 4 4 matrix, like: × a1 a1b1 a1b2 a1b3 a1b4 a2 a2b1 a2b2 a2b3 a2b4   ( b1 b2 b3 b4 )=   . (2.157) a3 a3b1 a3b2 a3b3 a3b4      a4   a4b1 a4b2 a4b3 a4b4      We will use eqs. (2.155) and (2.156) often when calculating cross sections and decay rates involving fermions. As a check, note that if we act on the left of eq. (2.155) with p/ m, the left hand side vanishes − because of eq. (2.107), and the right hand side vanishes because of

(p/ m)(p/ + m)= p/p/ m2 = 0. (2.158) − − This relies on the identity

p/p/ = p2, (2.159) which follows from

µ ν µ ν 2 p/p/ = p p γµγν = p p ( γνγµ + 2gµν)= p/p/ + 2p . (2.160) − − A similar consistency check works if we act on the left of eq. (2.156) with p/ + m.

2.3 The Weyl equation

It turns out that the Dirac equation can be replaced by something simpler and more fundamental if we examine the special case m = 0. If we go back to Dirac’s guess for the Hamiltonian, we now have just

H = ~α P,~ (2.161) · and there is no need for the matrix β. Therefore, eqs. (2.67) and (2.68) are not applicable, and we have only the one requirement:

αiαj + αkαj = 2δjk. (2.162)

Now there are two distinct solutions involving 2 2 matrices, namely ~α = ~σ or ~α = ~σ. So we have × − two possible quantum mechanical wave equations: ∂ i ψ = i~σ ~ ψ. (2.163) ∂t ± · ∇

23 If we now deﬁne

σµ = (σ0,σ1,σ2,σ3), (2.164) σµ = (σ0, σ1, σ2, σ3), (2.165) − − − then we can write the two possible equations in the form:

µ iσ ∂µψL = 0, (2.166) µ iσ ∂µψR = 0. (2.167)

Here I have attached labels L and R because the solutions to these equations turn out to have left and right helicity, respectively, as we will see in a moment. Each of these equations is called a Weyl equation. They are similar to the Dirac equation, but only apply to massless spin-1/2 particles, and are 2 2 matrix equations rather than 4 4. The two-component objects ψ and ψ are called Weyl × × L R spinors. We can understand the relationship of the Dirac equation to the Weyl equations if we notice that the γµ matrices can be written as 0 σµ γµ = . (2.168) σµ 0 µ ¶ [Compare eq. (2.74).] If we now write a Dirac spinor in its L and R helicity components,

Ψ Ψ= L , (2.169) Ψ µ R ¶ then the Dirac equation becomes: 0 σµ Ψ Ψ i ∂ L = m L , (2.170) σµ 0 µ Ψ Ψ µ ¶ µ R ¶ µ R ¶ or

µ iσ ∂µΨR = mΨL, (2.171) µ iσ ∂µΨL = mΨR. (2.172)

Comparing with eqs. (2.166) and (2.167) when m = 0, we can indeed identify ΨR as a right-handed helicity Weyl fermion, and ΨL as a left-handed helicity Weyl fermion.

Note that if m = 0, one can consistently set ΨR = 0 as an identity in the Dirac spinor without violating eqs. (2.171) and (2.172). Then only the left-handed helicity fermion exists. This is how neutrinos appear in the Standard Model; a massless neutrino corresponds to a left-handed Weyl fermion. (Recent evidence strongly suggests that neutrinos do have mass, so that this discussion has to be modiﬁed slightly. However, for experiments in which neutrino masses can be neglected, it is still proper to treat a neutrino as either a Weyl fermion, or equivalently as the left-handed part of a Dirac neutrino with no right-handed part.)

24 The two versions of the Weyl equation, eqs. (2.166) and (2.167), are actually not distinct. To see this, suppose we take the Hermitian conjugate of eq. (2.167). Since (~ ) = ~ and (σµ) = σµ, we ∇ † −∇ † obtain:

0 ∂ i(σ ~σ ~ )ψ† = 0 (2.173) ∂t − · ∇ R or:

µ iσ ∂µψR† = 0. (2.174)

In other words, a right-handed Weyl spinor is the Hermitian conjugate of a left-handed Weyl spinor, and vice versa. The modern point of view is that the Weyl equation is fundamental, and all 4-component Dirac fermions can be thought of as consisting of two 2-component Weyl fermions coupled together by a mass. This is because in the Standard Model, all fermions are massless until they are provided with a mass term by the spontaneous symmetry breaking. Since a right-handed Weyl fermion is the Hermitian conjugate of a left-handed Weyl fermion, all one really needs is the single Weyl equation (2.166). A Dirac fermion is then thought of as

χ Ψ= , (2.175) ξ µ † ¶ where χ and ξ are independent left-handed two-component Weyl fermions. All fermion degrees of freedom can thus be thought of as left-handed Weyl fermions.

25 3 Maxwell’s equations and electromagnetism

In the previous section, we worked out the relativistic wave equations governing the quantum mechanics of scalars and fermions. A more familiar set of relativistic wave equations are those governing electromagnetic ﬁelds. Recall that Maxwell’s equations can be written in the form:

~ E~ = eρ (3.1) ∇· ∂E~ ~ B~ = eJ~ (3.2) ∇× − ∂t ~ B~ = 0 (3.3) ∇· ∂B~ ~ E~ + = 0, (3.4) ∇× ∂t where ρ is the local charge density and J~ is the current density, with the magnitude of the electron’s charge, e, factored out. These equations can be rewritten in a manifestly relativistic form, using the following observations. First, suppose we add together the equations obtained by taking ∂/∂t of eq. (3.1) and ~ of eq. (3.2). Since the divergence of a curl vanishes identically, this yields the Law of ∇· Local Conservation of Charge: ∂ρ + ~ J~ = 0. (3.5) ∂t ∇· To put this into a Lorentz-invariant form, we can form a four-vector:

J µ = (ρ, J~), (3.6) so that eq. (3.5) becomes

µ ∂µJ = 0. (3.7)

Furthermore, we can write the electric and magnetic ﬁelds as derivatives of the electric and magnetic potentials V and A~:

∂A~ E~ = ~ V , (3.8) −∇ − ∂t B~ = ~ A.~ (3.9) ∇× Now if we assemble the potentials into a four-vector potential:

Aµ = (V, A~), (3.10) then we can write the electric and magnetic ﬁelds as components of an antisymmetric tensor:

F = ∂ A ∂ A (3.11) µν µ ν − ν µ 0 Ex Ey Ez Ex 0 Bz By =  − −  (3.12) Ey Bz 0 Bx  − −   Ez By Bx 0   − −  26 Now the Maxwell equations (3.1) and (3.2) correspond to the relativistic wave equation

µν ν ∂µF = eJ , (3.13) or equivalently,

∂ ∂µAν ∂ν∂ Aµ = eJ ν. (3.14) µ − µ The remaining Maxwell equations (3.3) and (3.4) are equivalent to the identity:

∂ρFµν + ∂µFνρ + ∂νFρµ = 0, (3.15) for µ,ν,ρ = any of 0, 1, 2, 3. Note that this equation is automatically true because of eq. (3.11). Also, µν µν because F is antisymmetric and partial derivatives commute, ∂µ∂νF = 0, so that the Law of Local Conservation of Charge eq. (3.7) follows from eq. (3.13). The object Aµ(x) is fundamental, and the objects E~ and B~ are derived from it. The theory of electromagnetism as described by Aµ(x) is subject to a redundancy known as gauge invariance. To see this, we note that eq. (3.14) is unchanged if we do the transformation

Aµ(x) Aµ(x)+ ∂µλ(x), (3.16) → where λ(x) is any function of position in spacetime. In components, this amounts to:

∂λ V V + (3.17) → ∂t A~ A~ ~ λ (3.18) → − ∇ This transformation leaves F µν (or equivalently E~ and B~ ) unchanged. Therefore, the new Aµ is just as good as the old Aµ for the purposes of describing a particular physical situation.

27 4 Field Theory and Lagrangians

4.1 The ﬁeld concept and Lagrangian dynamics

It is now time to make a conceptual break with our earlier treatment of relativistic quantum-mechanical wave equations for scalar and Dirac particles. There are two reasons for doing this. First, the existence of negative energy solutions lead us to the concept of antiparticles. Now, a hole in the Dirac sea, representing a positron, can be removed if the state is occupied by a positive energy electron, releasing an energy of at least 2m. This forces us to admit that the total number of particles is not conserved. The Klein-Gordon wavefunction φ(x) and the Dirac wavefunction Ψ(x) were designed to describe single- particle probability amplitudes, but the correct theory of nature evidently must describe a variable number of particles. Secondly, we note that in the electromagnetic theory, Aµ(x) are not just quantum mechanical wavefunctions; rather, they exist classically too. If we follow this example with scalar and spinor particles, we are lead to abandon φ(x) and Ψ(x) as quantum wavefunctions representing single-particle states, and reinterpret them as fields that have meaning even classically. Specifically, a scalar particle is described by a field φ(x). Classically, this just means that for every point xµ, the object φ(x) returns a number. Quantum mechanically, φ(x) becomes an operator (rather than a state or wavefunction). There is a distinct operator for each xµ. Therefore, we no longer have a position operator xµ; rather, it is just an ordinary number label which tells us which operator φ we are talking about. If φ(x) is now an operator, what states will it act on? The answer is that we can start with a vacuum state

0 (4.1) | i which describes an empty universe with no particles in it. If we now act with our ﬁeld operator, we obtain a state:

φ(x) 0 (4.2) | i which, at the time t = x0, contains one particle at ~x. (What this state describes at other times is a much more complicated question!) If we act again with our ﬁeld operator at a diﬀerent point yµ, we get a state

φ(y)φ(x) 0 (4.3) | i which in general can be a linear combination of states containing any number of particles. [The operator φ(y) can either add another particle, or remove the particle added to the vacuum state by φ(x). But, in addition, the particles can interact to change their number.] In general, the ﬁeld operator φ(x) acts on any state by adding or subtracting particles. So this is the right framework to describe the quantum mechanics of a system with a variable number of particles.

28 Similarly, to describe spin-1/2 particles and their antiparticles, like the electron and the positron, or quarks and antiquarks, we will want to use a Dirac field Ψ(x). Classically, Ψ(x) is a set of four functions (one for each of the Dirac spinor components). Quantum mechanically, Ψ(x) is a set of four operators. We can build up any state we want, containing any number of electrons and positrons, by acting on the vacuum state 0 enough times with the fields Ψ(x) and Ψ(x)=Ψ γ0. | i † The vector field Aµ(x) associated with electromagnetism is already very familiar in the classical theory, as the electrical and vector potentials. In the quantum theory of electromagnetism, Aµ(x) becomes an operator which can add or subtract photons from the vacuum. This way of dealing with theories of multi-particle physics is called field theory. In order to describe how particles in a field theory evolve and interact, we need to specify a single object called the action. Let us first review how the action principle works in a simple, non-field-theory setting. Let the variables qn(t) describe the configuration of a physical system. (For example, a single q(t) could describe the displacement of a harmonic oscillator from equilibrium, as a function of time.) Here n is just a label distinguishing the different configuration variables. Classically, we could specify the equations of motion for the qn, which we could get from knowing what forces were acting on it. A clever way of summarizing this information is to specify the action

tf S = L(qn, q˙n) dt. (4.4) Zti

Here ti and tf are ﬁxed initial and ﬁnal times, and L is the Lagrangian. It is given in simple systems by

L = T V (4.5) − where T is the total kinetic energy and V is the total potential energy. Thus the action S is a functional of qn; if you specify a particular trajectory qn(t), then the action returns a single number.

The Lagrangian is a function of qn(t) and its ﬁrst derivative.

The usefulness of the action is given by Hamilton’s principle, which states that if qn(t) satisﬁes the equations of motion, and qn(ti) and qn(tf ) are held ﬁxed as boundary conditions, then S is minimized.

Since S is at an extremum, this means that any small variation in qn(t) will lead to no change in S, so that q (t) q (t)+ δq (t) implies δS = 0, provided that q (t) obeys the equations of motion. Here n → n n n δqn(t) is any small function of t which vanishes at both ti and tf , as shown in the ﬁgure.

29 q(t2)

q(t)+δq(t) q(t) q(t)

q(t1)

t1 t2 t

So, let us compute δS. First, note that by the chain rule, we have: ∂L ∂L δL = δqn + δq˙n (4.6) n ∂qn ∂q˙n X µ ¶ Now, since d δq˙ = (δq ), (4.7) n dt n we have

tf ∂L d ∂L δS = δqn + (δqn) dt (4.8) n ti ∂qn dt ∂q˙n X Z · ¸ Now integrating by parts yields:

t f ∂L d ∂L ∂L t=tf δS = δqn dt + δqn . (4.9) t ∂qn − dt ∂q˙n ∂q˙n t=ti n Z i · µ ¶¸ n ¯ X X ¯ ¯ The last term vanishes because of the boundary conditions on δqn. Since the variation δS is supposed to vanish for any δqn(t), it must be that ∂L d ∂L = 0 (4.10) ∂q dt ∂q˙ n − µ n ¶ is the equation of motion for qn(t), for each n.

As a simple example, suppose there is only one qn(t)= x(t), the position of some particle moving in one dimension in a potential V (x). Then 1 L = T V = mx˙ 2 V (x), (4.11) − 2 −

30 from which there follows: ∂L ∂V = = F (4.12) ∂x − ∂x which we recognize as the Newtonian force, and d ∂L d = (mx˙)= mx.¨ (4.13) dt ∂x˙ dt µ ¶ So the equation of motion is Newton’s second law:

F = mx.¨ (4.14)

Everything we need to know about a physical system is encoded in a precise way in the action S, or equivalently the Lagrangian L. In quantum field theory, it tells us what the particle masses are, how they interact, how they decay, and what symmetries provide selection rules on their behavior. As a first example of an action for a relativistic field theory, consider a scalar field φ(x) = φ(t, ~x).

Then, in the previous discussion, we identify the label n with the spatial position ~x, and qn(t)= φ(t, ~x). The action is obtained by summing contributions from each x:

tf tf S = dtL(φ, φ˙)= dt d3~x (φ, φ˙). (4.15) L Zti Zti Z Now, since this expression depends on φ˙, it must also depend on ~ φ in order to be Lorentz invariant. ∇ This just means that the form of the Lagrangian allows it to depend on the differences between the field evaluated at infinitesimally nearby points. So, a better way to write the action is:

S = d4x (φ, ∂ φ). (4.16) L µ Z The object is known as the Lagrangian density. Specifying a particular form for deﬁnes the theory. L L

To find the classical equations of motion for the field, we must find φ(x) so that S is extremized; in other words, for any small variation φ(x) φ(x)+δφ(x), we must have δS = 0. By a similar argument → as above, this implies the equations of motion:

δ δ L ∂µ L = 0. (4.17) δφ − Ãδ(∂µφ)!

δ Here, we use L to mean the partial derivative of with respect to φ; δ is used rather than ∂ to avoid δφ L confusing between derivatives with respect to φ and spacetime partial derivatives with respect to xµ. δ Likewise, L means a partial derivative of with respect to the object ∂µφ, upon which it depends. δ(∂µφ) L As an example, consider the choice: 1 1 = ∂ φ∂µφ m2φ2 (4.18) L 2 µ − 2 It follows that: δ L = m2φ (4.19) δφ −

31 and δ δ 1 1 1 L = gαβ∂ φ∂ φ = gµβ∂ φ + gαµ∂ φ = ∂µφ (4.20) δ(∂ φ) δ(∂ φ) 2 α β 2 β 2 α µ µ · ¸ Therefore,

δ µ ∂µ L = ∂µ∂ φ, (4.21) Ãδ(∂µφ)! and so the equation of motion following from eq. (4.17) is:

µ 2 ∂µ∂ φ + m φ = 0 (4.22)

This we recognize as the Klein-Gordon wave equation for a scalar particle of mass m; compare to eq. (2.60). This equation was originally introduced with the interpretation as the equation governing the quantum wavefunction of a single scalar particle. Now it has reappeared with a totally different interpretation, as the classical equation of motion for the scalar field. The previous discussion for scalar fields can be extended to other types of fields as well. Consider a general list of fields Φj, which could include scalar fields φ(x), Dirac fields Ψ(x) with four components, Weyl fields ψ(x) with two components, or vector fields Aµ(x) with four components, or several copies of any of these. The Lagrangian density (Φ ,∂ Φ ) determines the classical equations of motion through L j µ j the principle that S = d4x should be stationary to first order when a deviation Φ (x) Φ (x)+ L j → j δΦj(x) is made, with boundaryR conditions that δΦj(x) vanishes on the boundary of the spacetime region on which S is evaluated. (Typically one evaluates S between two times ti and tf , and over all of space, so this means that δΦ(x) vanishes very far away from some region of interest.) Similarly to eq. (4.6), we have by the chain rule:

4 δ δ δS = d x δΦj L + δ(∂µΦj) L (4.23) " δΦj Ãδ(∂µΦj)!# Z Xj

Now using δ(∂µΦj) = ∂µ(δΦj), and integrating the second term by parts (this is where the boundary conditions come in!), one obtains

4 δ δ δS = d x δΦj L ∂µ L . (4.24) "δΦj − Ãδ(∂µΦj)!# Z Xj

If we require this to vanish for each and every arbitrary variation δΦj, we obtain the Euler-Lagrange equations of motion:

δ δ L ∂µ L = 0, (4.25) δΦj − Ãδ(∂µΦj)! for each j. The Lagrangian density should be a real quantity that transforms under proper Lorentz L transformations as a scalar. For example, let us consider how to make a Lagrangian density for a Dirac ﬁeld Ψ(x). Under Lorentz transformations, Ψ(x) transforms exactly like the wavefunction solution to the Dirac equation.

32 But now it is interpreted instead as a field; classically it is a function on spacetime, and quantum µ mechanically it is an operator for each point x . Now, Ψ(x) is a complex 4-component object, so Ψ†(x) is also a field. One should actually treat Ψ(x) and Ψ†(x) as independent fields, in the same way that in complex analysis one often treats z = x + iy and z = x iy as independent variables. As we found in ∗ − 0 section 2.1, if we want to build Lorentz scalar quantities, it is useful to use Ψ(x)=Ψ†γ as a building block. A good guess for the Lagrangian for a Dirac field is:

= iΨγµ∂ Ψ mΨΨ (4.26) L µ − which can also be written as:

0 µ 0 = iΨ†γ γ ∂ Ψ mΨ†γ Ψ. (4.27) L µ − This is a Lorentz scalar, so that when integrated d4x it will give a Lorentz-invariant number. Let us now compute the equations of motion that followR from it. First, let us ﬁnd the equations of motion obtained by varying with respect to Ψ†. For this, we need:

δ 0 µ 0 L = iγ γ ∂µΨ mγ Ψ (4.28) δΨ† − δ L = 0. (4.29) δ(∂µΨ†) The second equation just reﬂects the fact that the Lagrangian only contains the derivative of Ψ, not the derivative of Ψ†. So, by plugging in to the general result eq. (4.25), the equation of motion is simply:

iγµ∂ Ψ mΨ = 0, (4.30) µ −

δ 0 0 2 where we have multiplied δΨL† = 0 on the left by γ and used the fact that (γ ) = 1. This is, of course, the Dirac equation. We can also ﬁnd the equations of motion obtained from the Lagrangian by varying with respect to Ψ. For that, we need: δ L = mΨ, (4.31) δΨ − δ L = iΨγµ. (4.32) δ(∂µΨ)

Plugging these into eq. (4.25), we obtain:

i∂ Ψγµ mΨ = 0. (4.33) − µ − However, this is nothing new; it is just the Hermitian conjugate of eq. (4.30). [The proof is very similar to Homework Set 2, Problem 3.]

33 The Lagrangian density for an electromagnetic ﬁeld is: 1 1 = F F µν = (∂ A ∂ A )(∂µAν ∂νAµ). (4.34) LEM −4 µν −4 µ ν − ν µ − To ﬁnd the equations of motion that follow from this Lagrangian, we compute: δ LEM = 0, (4.35) δAν since Aµ doesn’t appear in the Lagrangian without a derivative acting on it, and δ δ 1 LEM = (∂ A ∂ A )(∂ A ∂ A )gαρgβσ δ(∂ A ) δ(∂ A ) 4 α β β α ρ σ σ ρ µ ν µ ν ·− − − ¸ 1 = (∂ A ∂ A )gµρgνσ (∂ A ∂ A )gνρgµσ −4 ρ σ − σ ρ − ρ σ − σ ρ h +(∂ A ∂ A )gµαgβσ (∂ A ∂ A )gναgµβ α β − β α − α β − α β = ∂µAν + ∂νAµ i − = F µν. (4.36) − So, the equations of motion,

δ EM δ EM L ∂µ L = 0 (4.37) δAν − Ãδ(∂µAν)! reduce to

µν ∂µF = 0, (4.38) which we recognize as Maxwell’s equations in the case of vanishing J µ [see eq. (3.13)]. In order to include the eﬀects of a 4-current J µ, we can simply add a term

= eJ µA (4.39) Lcurrent − µ to the Lagrangian density. Then, since δ Lcurrent = eJ ν; (4.40) δAν − δ Lcurrent = 0, (4.41) δ(∂µAν) the classical equations of motion for the vector ﬁeld become

∂ F µν eJ ν = 0, (4.42) µ − in agreement with eq. (3.13). The current density J µ may be regarded as an external source for the electromagnetic field of unspecified origin; or it can be built out of the fields for charged particles, as we will see.

34 4.2 Quantization of free scalar ﬁeld theory

Let us now turn to the question of quantizing a field theory. To begin, let us recall how one quantizes a simple generic system based on variables qn(t), given the Lagrangian L(qn, q˙n). First, one defines the canonical momenta conjugate to each qn: ∂L pn (4.43) ≡ ∂q˙n The classical Hamiltonian is then defined by

H(qn,pn) pnq˙n L(qn, q˙n), (4.44) ≡ n − X where theq ˙n are to be eliminated using eq. (4.43). To go to the corresponding quantum theory, the qn and pn and H are reinterpreted as Hermitian operators acting on a Hilbert space of states. The operators obey canonical equal-time commutation relations:

[pn,pm] = 0; (4.45)

[qn, qm] = 0; (4.46) [p , q ] = i¯hδ , (4.47) n m − nm and the time evolution of the system is determined by the Hamiltonian operator H. Let us now apply this to the theory of a scalar field φ(x) with the Lagrangian density given by eq. (4.18). Then φ(x)= φ(t, ~x) plays the role of qn(t), with ~x playing the role of the label n; there is a different field at each point in space. The momentum conjugate to φ is: δ δ 1 π(~x) L = φ˙2 + ... = φ.˙ (4.48) ≡ δφ˙ δφ˙ 2 h i It should be emphasized that π(~x), the momentum conjugate to the field φ(~x), is not in any way the mechanical momentum of the particle. Notice that π(~x) is a scalar function, not a three-vector or a four-vector! The Hamiltonian is obtained by summing over the fields at each point ~x:

H = d3~xπ(~x)φ˙(x) L (4.49) − Z 1 = d3~xπ(~x)φ˙(x) d3~x [φ˙2 (~ φ)2 m2φ2] (4.50) − 2 − ∇ − Z Z 1 = d3~x [π2 + (~ φ)2 + m2φ2]. (4.51) 2 ∇ Z Notice that a nice feature has emerged: this Hamiltonian is a sum of squares, and is therefore always 0. There are no dangerous solutions with arbitrarily large negative energy, unlike the case of the ≥ single-particle Klein-Gordon wave equation.

35 At any given ﬁxed time t, the ﬁeld operators φ(~x) and their conjugate momenta π(~x) are Hermitian operators satisfying commutation relations exactly analogous to eqs. (4.45)-(4.47):

[φ(~x), φ(~y)] = 0; (4.52) [π(~x),π(~y)] = 0; (4.53) [π(~x), φ(~y)] = i¯hδ(3)(~x ~y). (4.54) − − As we will see, it turns out to be profitable to analyze the system in a way similar to the way one treats the harmonic oscillator in one-dimensional nonrelativistic quantum mechanics. In that system, one defines “creation” and “annihilation” (or “raising” and “lowering”) operators a† and a as complex linear combinations of the position and momentum operators x and p. Similarly, here we will define:

3 i~p ~x a~p = d ~xe− · E~pφ(~x)+ iπ(~x) , (4.55) Z £ ¤ where

2 2 E~p = ~p + m , (4.56) q with the positive square root always taken. The coefficients in front of φ and π in this expression reflect an arbitrary choice (and in fact they are chosen in different ways by various textbooks). Equation (4.55) defines a distinct annihilation operator for each three-momentum ~p. Taking the Hermitian conjugate yields:

3 i~p ~x a† = d ~xe · E φ(~x) iπ(~x) . (4.57) ~p ~p − Z £ ¤ The usefulness of these deﬁnitions appears if we compute the commutator [a ,a† ]. This gives: ~p ~k

3 3 i~p ~x i~k ~y [a ,a† ] = d ~x d ~ye− · e · [E φ(~x)+ iπ(~x) , E φ(~y) iπ(~y)] (4.58) ~p ~k ~p ~k − Z Z 3 3 i~p ~x i~k ~y (3) = d ~x d ~ye− · e · (E + E ) δ (~x ~y) (4.59) ~k ~p − Z Z where eqs. (4.52)-(4.54) have been used. Now performing the ~y integral using the deﬁnition of the delta function, one obtains:

3 i(~k ~p) x [a ,a† ] = (E + E ) d ~xe − · . (4.60) ~p ~k ~k ~p Z This can be further reduced by using the important identity:

3 i~q ~x 3 (3) d ~xe · = (2π) δ (~q), (4.61) Z valid for any 3-vector ~q, to obtain the ﬁnal result:

3 (3) [a ,a† ] = (2π) 2E δ (~k ~p). (4.62) ~p ~k ~p −

36 ~ Here we have put E~k = E~p, using the fact that the delta function vanishes except when k = ~p. Ina similar way, one can check the commutators:

[a ,a ]=[a†,a† ] = 0. (4.63) ~p ~k ~p ~k Up to a constant factor on the right-hand side, these results have the same form as the harmonic oscillator algebra [a,a†]=1;[a,a]=[a†,a†] = 0, that should be familiar from non-relativistic quantum mechanics. Therefore, the quantum mechanics of a scalar field behaves like an infinite collection of harmonic oscillators, one associated with each three-momentum ~p. Now we can understand the Hilbert space of states for the quantum field theory. We start with the vacuum state 0 which is taken to be annihilated by all of the lowering operators: | i a 0 = 0. (4.64) ~k| i Acting on the vacuum state with any raising operator produces

a† 0 = ~k , (4.65) ~k| i | i which describes a state with a single particle with three-momentum ~k (and no deﬁnite position). Acting multiple times with raising operators produces a state with multiple particles. So

~ ~ ~ a~† a~† ...a~† 0 = k1, k2,..., kn (4.66) k1 k2 kn | i | i describes a state of the universe with n φ-particles with momenta ~k1,~k2,...,~kn. Note that these states are automatically symmetric under interchange of any two of the momentum labels ~ki, because of the identity [a† ,a† ] = 0. This is another way of saying that the multiparticle states obey Bose-Einstein ~ki ~kj statistics for identical particles with integer (in this case, 0) spin. In order to check our interpretation of the quantum theory, let us now evaluate the Hamiltonian operator in terms of the raising and lowering operators, and then determine how H acts on the space of states. First, let us invert the deﬁnitions eqs. (4.55) and (4.57) to ﬁnd φ(~x) and π(~y) in terms of the a†~p and a~p. We begin by noting that:

3 i~p ~x a~p + a† ~p = d ~xe− · 2E~pφ(~x). (4.67) − Z Now we act on both sides by:

i~p ~y dp˜ e · , (4.68) Z where we have introduced the very convenient shorthand notation: d3~p dp˜ (4.69) ≡ (2π)32E Z Z ~p used often from now on. The result is that eq. (4.67) becomes

3 i~p ~y 3 d ~p i~p (~y ~x) dp˜ e · (a + a† )= d ~xφ(~x) e · − (4.70) ~p ~p (2π)3 Z − Z (Z )

37 Since the ~p integral in braces is equal to δ(3)(~y ~x) [see eq. (4.61)], one obtains after performing the ~x − integral:

i~p ~y φ(~y)= dp˜ e · (a~p + a† ~p), (4.71) Z − or, renaming ~y ~x, and ~p ~p in the second term on the right: → →− i~p ~x i~p ~x φ(~x)= dp˜ (e · a~p + e− · a†~p). (4.72) Z This expresses the original field in terms of raising and lowering operators. Similarly, for the conjugate momentum field, one finds:

i~p ~x i~p ~x π(~x)= i dpE˜ (e · a e− · a†). (4.73) − ~p ~p − ~p Z Now we can plug the results of eqs. (4.72) and (4.73) into the expression eq. (4.51) for the Hamil- tonian. The needed terms are:

1 3 2 1 3 2 i~k ~x i~k ~x i~p ~x i~p ~x d ~xπ(x) = d ~x dk˜ dp˜ ( i) E E e · a e− · a† e · a e− · a† (4.74) 2 2 − ~k ~p ~k − ~k ~p − ~p Z Z Z Z ³ ´ ³ ´ 1 3 2 1 3 i~k ~x i~k ~x i~p ~x i~p ~x d ~x (~ φ) = d ~x dk˜ dp˜ i~ke · a i~ke− · a† i~pe · a i~pe− · a† (4.75) 2 ∇ 2 ~k − ~k · ~p − ~p 2Z Z2 Z Z ³ ´ ³ ´ m 3 2 m 3 i~k ~x i~k ~x i~p ~x i~p ~x d ~xφ(x) = d ~x dk˜ dp˜ e · a + e− · a† e · a + e− · a† . (4.76) 2 2 ~k ~k ~p ~p Z Z Z Z ³ ´ ³ ´ Adding up the pieces, one ﬁnds:

1 ˜ 3 2 ~ i(~k+~p) ~x i(~k+~p) ~x H = dk dp˜ d ~x (m k ~p E~kE~p) a~ka~pe · + a~† a†~pe− · 2 ( − · − k Z Z Z h i 2 ~ i(~p ~k) ~x i(~k ~p) ~x +(m + k ~p + E~kE~p) a~† a~pe − · + a~ka†~pe − · (4.77) · k ) h i Now one can do the ~x integration using:

3 i(~k+~p) ~x 3 (3) d ~xe± · = (2π) δ (~k + ~p) (4.78) Z 3 i(~k ~p) ~x 3 (3) d ~xe± − · = (2π) δ (~k ~p). (4.79) − Z As a result, the coeﬃcient (m2 ~k ~p E E ) of the aa and a a terms vanishes after plugging in − · − ~k ~p † † ~k = ~p as enforced by the delta function. Meanwhile, for the aa and a a terms one has ~k = ~p from − † † the delta function, so that

(m2 + ~k ~p + E E )δ(3)(~k ~p) = 2E2δ(3)(~k ~p). (4.80) · ~k ~p − ~p − Now performing the ~k integral in eq. (4.77), one ﬁnds: 1 H = dpE˜ a†a + a a† . (4.81) 2 ~p ~p ~p ~p ~p Z ³ ´

38 Finally, we can rearrange the second term, using

3 (3) a a† = a†a + (2π) 2E δ (~p ~p) (4.82) ~p ~p ~p ~p ~p − from eq. (4.62). The last term is inﬁnite, so

H = dpE˜ a†a + , (4.83) ~p ~p ~p ∞ Z where “ ” means an infinite, but constant, contribution to the energy. Since a uniform constant ∞ contribution to the energy of all states is unobservable and commutes will all other operators, we are free to ignore it. This is a simple example of the process known as renormalization. (In a more careful treatment, one could “regulate” the theory by quantizing the theory confined to a box of finite volume, and neglecting all contributions coming from momenta greater than some cutoff ~p . Then | |max the infinite constant would be rendered finite. Since we are going to ignore the constant anyway, we won’t bother doing this.) So, from now on,

H = dpE˜ ~p a†~pa~p (4.84) Z is the Hamiltonian operator. Acting on the vacuum state,

H 0 = 0, (4.85) | i since all a~p annihilate the vacuum. This shows that the infinite constant we dropped from H is actually the infinite energy density associated with an infinite universe of empty space, filled with the zero-point energies of an infinite number of oscillators, one for each possible momentum 3-vector ~p. But we’ve already agreed to ignore it, so let it go. Now, you will show for homework that:

[H,a† ]= E a† . (4.86) ~k ~k ~k

Acting with H on a one-particle state, we therefore obtain

H ~k = Ha† 0 =[H,a† ] 0 = E a† 0 = E ~k . (4.87) | i ~k| i ~k | i ~k ~k| i ~k| i

~ ~ 2 2 This proves that the one-particle state with 3-momentum k has energy eigenvalue E~k = k + m , as expected. More generally, a multi-particle state q

~ ~ ~ k1, k2,..., kn = a~† a~† ...a~† 0 (4.88) | i k1 k2 kn | i is easily shown to be an eigenstate of H with eigenvalue E + E + ... + E . Note that it is not ~k1 ~k2 ~kn possible to construct a state with a negative energy eigenvalue!

39 4.3 Quantization of free Dirac fermion ﬁeld theory

Let us now apply the wisdom obtained by the quantization of a scalar field in the previous subsection to the problem of quantizing a Dirac fermion field that describes electrons and positrons. A sensible strategy is to expand the fields Ψ and Ψ† in terms of operators that act on states by creating and destroying particles with a given 3-momentum. Now, since Ψ is a spinor with four components, one must expand it in a basis for the four-dimensional spinor space. A convenient such basis is the solutions we found to the Dirac equation, u(p, s) and v(p, s). So we expand the Dirac field, at a given fixed time t, as:

2 i~p ~x i~p ~x Ψ(~x)= dp˜ u(p, s)e · b~p,s + v(p, s)e− · d†~p,s (4.89) sX=1 Z h i Here s labels the two possible spin states in some appropriate basis (for example, S = 1/2). The op- z ± erator b~p,s will be interpreted as an annihilation operator, which removes an electron with 3-momentum

~p and spin state s from whatever state it acts on. The operator d†~p,s is a creation operator, which adds a positron to whatever state it acts on. We are using b, b† and d,d† rather than a,a† in order to distinguish the fermion and antifermion creation and annihilation operators from the scalar ﬁeld versions. Taking the Hermitian conjugate of eq. (4.89), and multiplying by γ0 on the right, we get:

2 i~p ~x i~p ~x Ψ(~x)= dp˜ u(p, s)e− · b†~p,s + v(p, s)e · d~p,s (4.90) sX=1 Z h i

The operators b†~p,s (d~p,s) create (destroy) an electron (positron) with the corresponding 3-momentum and spin. More generally, if the Dirac ﬁeld describes some fermions other than the electron-positron system, then you can substitute “particle” for electron and “antiparticle” for positron. So b†, b act on states to create and destroy particles, while d†,d create and destroy antiparticles. Just as in the case of a scalar ﬁeld, we assume the existence of a vacuum state 0 , which describes | i a universe of empty space with no electrons or positrons present. The annihilation operators yield 0 when acting on the vacuum state:

b 0 = d 0 = 0 (4.91) ~p,s| i ~p,s| i for all ~p and s. To make a state describing a single electron with 3-momentum ~p and spin state s, just act on the vacuum with the corresponding creation operator:

b† 0 = e− . (4.92) ~p,s| i | ~p,si Similarly,

b† b† 0 = e− ; e− (4.93) ~k,r ~p,s| i | ~k,r ~p,si is a state containing two electrons, etc.

40 Now, electrons are fermions; they must obey Fermi-Dirac statistics for identical particles. This means that if we interchange the momentum and spin labels for the two electrons in the state of eq. (4.93), we must get a minus sign:

e− ; e− = e− ; e− . (4.94) | ~k,r ~p,si −| ~p,s ~k,ri A corollary of this is the Pauli exclusion principle, which states that two electrons cannot be in exactly the same state. In the present case, that means that we cannot add to the vacuum two electrons with exactly the same 3-momentum and spin. Taking ~k = ~p and r = s in eq. (4.94):

e− ; e− = e− ; e− (4.95) | ~p,s ~p,si −| ~p,s ~p,si which can only be true if e− ; e− = 0, in other words there is no such state. | ~p,s ~p,si Writing eq. (4.94) in terms of creation operators, we have:

b† b† 0 = b† b† 0 . (4.96) ~k,r ~p,s| i − ~p,s ~k,r| i So, instead of the commutation relation

[b† , b† ] = 0 (Wrong!) (4.97) ~k,r ~p,s that one might expect from comparison with the scalar ﬁeld, we must have an anticommutation relation:

b† b† + b† b† = b† , b† = 0. (4.98) ~k,r ~p,s ~p,s ~k,r { ~k,r ~p,s} Taking the Hermitian conjugate, we must also have:

b , b = 0. (4.99) { ~k,r ~p,s} Similarly, applying the same thought process to identical positron fermions, one must have:

d† ,d† = d ,d = 0. (4.100) { ~k,r ~p,s} { ~k,r ~p,s} Note that in the classical limit,h ¯ 0, these equations are unaﬀected, since ¯h doesn’t appear → anywhere. So it must be true that b, b†, d, and d† anticommute even classically. So, as classical ﬁelds, one must have

Ψ(x), Ψ(y) = 0 (4.101) { } Ψ†(x), Ψ†(y) = 0 (4.102) { } Ψ(x), Ψ†(y) = 0. (4.103) { } Evidently, the Dirac ﬁeld is not a normal number, but rather an anticommuting or Grassman number. Interchanging the order of any two Grassman numbers results in an overall minus sign.

41 In order to discover how the classical equations (4.101) and (4.103) are modiﬁed when one goes to the quantum theory, let us construct the momentum conjugate to Ψ(~x). It is:

δ 0 0 0 (~x)= L = iΨγ = iΨ†γ γ = iΨ†. (4.104) P δ∂0Ψ So the momentum conjugate to the Dirac spinor ﬁeld is just i times its Hermitian conjugate. Now, naively following the path of canonical quantization, one might expect the equal-time commutation relation:

[ (~x), Ψ(~y)] = i¯hδ(3)(~x ~y). (Wrong!) (4.105) P − − However, this clearly cannot be correct, since these are anticommuting ﬁelds; in the classical limit ¯h 0, eq. (4.105) disagrees with eq. (4.103). So, instead we postulate a canonical anticommutation → relation for the Dirac ﬁeld operator and its conjugate momentum operator:

(~x), Ψ(~y) = i¯hδ(3)(~x ~y). (4.106) {P } − − Now just rewriting = iΨ , this becomes: P † (3) Ψ†(~x), Ψ(~y) = ¯hδ (~x ~y). (4.107) { } − − From this, using a strategy similar to that used for scalar ﬁelds, one can obtain:

3 (3) b , b† = d ,d† = (2π) 2E δ (~p ~k) δsr, (4.108) { ~p,s ~k,r} { ~p,s ~k,r} ~p − and all other anticommutators of b, b†,d,d† operators vanish. One also can check that the Hamiltonian is 2

H = dpE˜ ~p (b†~p,sb~p,s + d†~p,sd~p,s) (4.109) sX=1 Z in a way very similar to the way we found the Hamiltonian for a scalar ﬁeld in terms of a,a† operators. In doing so, one must again drop an inﬁnite constant contribution (negative, this time) which is unobservable because it is the same for all states. Note that H again has energy eigenvalues that are 0. For homework, you will show that: ≥

[H, b† ] = E b† (4.110) ~k,s ~k ~k,s

[H,d† ] = E d† . (4.111) ~k,s ~k ~k,s (Note that these equations are commutators rather than anticommutators!) It follows that the eigenstates of energy and 3-momentum are given in general by:

b†~p ,s ...b†~p ,s d~† ...d~† 0 , (4.112) 1 1 n n k1,r1 km,rm | i which describes a state with n electrons (particles) and m positrons (antiparticles) with the obvious 3-momenta and spins, and total energy E + ... + E + E + ... + E . ~p1 ~pn ~k1 ~km

42 5 Interacting scalar ﬁeld theories

5.1 Scalar ﬁeld with φ4 coupling

So far, we have been dealing with free field theories. These are theories in which the Lagrangian density is quadratic in the fields, so that the Euler-Lagrange equations obtained by varying are linear wave L equations in the fields, with exact solutions that are not too hard to find. At the quantum level, this nice feature shows up in the simple time evolution of the states. In field theory, as in any quantum system, the time evolution of a state X is given in the Schrodinger picture by | i d i¯h X = H X . (5.1) dt| i | i So, in the case of multiparticle states with an energy eigenvalue E as described above, the solutions are just

i(t t0)H i(t t0)E X(t) = e− − X(t ) = e− − X(t ) . (5.2) | i | 0 i | 0 i

In other words, the state at some time t is just the same as the state at some previous time t0, up to a phase. So nothing ever happens to the particles in a free theory; their number does not change, and their momenta and spins remain the same. We are interested in describing a more interesting situation where particles can scatter off each other, perhaps inelastically to create new particles, and in which some particles can decay into other sets of particles. To describe this, we need a Lagrangian density which contains terms with more than two fields. At the classical level, this will lead to non-linear equations of motion that have to be solved approximately. At the quantum level, finding exact energy eigenstates of the Hamiltonian is not possible, so one usually treats the non-quadratic part of the Hamiltonian as a perturbation on the quadratic part. This gives only approximate answers, which is unfortunate, but makes life interesting. As an example, consider the free Lagrangian for a scalar field φ, as given in eq. (4.18), and add to it an interaction term:

= + ; (5.3) L L0 Lint λ = φ4. (5.4) Lint −24 Here λ is a dimensionless number, a parameter of the theory known as a coupling. It governs the strength of interactions; if we set λ = 0, we would be back to the free theory in which nothing interesting ever happens. The factor of 1/4! = 1/24 is a convention, and the reason for it will be apparent later. Now canonical quantization can proceed as before, except that now the Hamiltonian is

H = H0 + Hint, (5.5) where λ H = d3~x = d3~xφ(~x)4. (5.6) int − Lint 24 Z Z 43 Let us write this in terms of creation and annihilation operators, using eq. (4.72):

λ 3 i~q1 ~x i~q1 ~x i~q2 ~x i~q2 ~x Hint = d ~x dq˜1 dq˜2 dq˜3 dq˜4 a~q e · + a† e− · a~q e · + a† e− · 24 1 ~q1 2 ~q2 Z Z Z Z Z ³ ´ ³ ´ i~q3 ~x i~q3 ~x i~q4 ~x i~q4 ~x a e · + a† e− · a e · + a† e− · (5.7) ~q3 ~q3 ~q4 ~q4 ³ ´ ³ ´ Now we can perform the d3~x integration, using eq. (4.61). The result is:

λ 3 (3) Hint = (2π) dq˜1 dq˜2 dq˜3 dq˜4 a† a† a† a† δ (~q1 + ~q2 + ~q3 + ~q4) 24 ~q1 ~q2 ~q3 ~q4 Z Z Z Z " (3) +4a† a† a† a~q δ (~q1 + ~q2 + ~q3 ~q4) ~q1 ~q3 ~q3 4 − (3) +6a† a† a~q a~q δ (~q1 + ~q2 ~q3 ~q4) ~q1 ~q2 3 4 − − (3) +4a† a~q a~q a~q δ (~q1 ~q3 ~q3 ~q4) ~q1 2 3 4 − − − (3) +a~q1 a~q2 a~q3 a~q4 δ (~q1 + ~q2 + ~q3 + ~q4) (5.8) #

Here we have combined several like terms, by relabeling the momenta, giving rise to the factors of 4, 6, and 4. This involves reordering the a’s and a†’s. In doing so, we have ignored the fact that a’s do not commute with a†’s when the 3-momenta are exactly equal. This should not cause any worry, because it just corresponds to the situation where a particle is “scattered” without changing its momentum at all, which is the same as no scattering, and therefore not of interest. To see how to use the interaction Hamiltonian, it is useful to tackle a speciﬁc process. For example, consider a scattering problem in which we have two scalar particles with 4-momenta pa,pb which interact, producing two scalar particles with 4-momenta k1, k2:

p p k k . (5.9) a b → 1 2 We will work in the Schrodinger picture of quantum mechanics, in which operators are time-independent and states evolve in time according to eq. (5.1). Nevertheless, in the far past, we assume that the incoming particles were far apart, so the system is accurately described by the free Hamiltonian and its energy eigenstates. The same applies to the outgoing particles in the far future. So, we can pretend that Hint is “turned oﬀ” in both the far past and the far future. The states that are simple two-particle states are

pa,pb IN = a† a† 0 , (5.10) | i ~pa ~pb | i in the far past, and

k1, k2 OUT = a~† a~† 0 , (5.11) | i k1 k2 | i in the far future. These are built out of creation and annihilation operators just as before, so they are eigenstates of the free Hamiltonian H0, but not of the full Hamiltonian. Now, we are interested in

44 computing the probability amplitude that the state p ,p evolves to the state k , k . According | a biIN | 1 2iOUT to the rules of quantum mechanics this is given by their overlap at a common time, say in the far future:

k , k p ,p (5.12) OUTh 1 2| a biOUT The state p ,p is the time evolution of p ,p from the far past to the far future: | a biOUT | a biIN iT H p ,p = e− p ,p , (5.13) | a biOUT | a biIN where T is the long time between the far past time when the initial state was created and the far future time at which the overlap is computed. So we have:

iT H k , k p ,p = k , k e− p ,p . (5.14) OUTh 1 2| a biOUT OUTh 1 2| | a biIN The states appearing on the right-hand side are simple; see eqs. (5.10) and (5.11). The complications iT H are hidden in the operator e− . iT H In general, e− cannot be written exactly in a useful way in terms of creation and annihilation operators. However, we can do it perturbatively, order by order in the coupling λ. For example, let us consider the contribution linear in λ. We use the deﬁnition of the exponential to write:

iT H N N e− = [1 iHT/N] = [1 i(H + H )T/N] , (5.15) − − 0 int for N . Now, the part of this that is linear in H can be expanded as: → ∞ int N 1 iT H − N n 1 n e− = [1 iH T/N] − − ( iH T/N) [1 iH T/N] (5.16) − 0 − int − 0 nX=0

iT H0 (Here we have dropped the 0th order part, e− as uninteresting; it just corresponds to the particles evolving as free particles.) We can now turn this discrete sum into an integral, by letting t = nT/N, and dt = T/N in the limit of large N:

T iT H i(T t)H0 itH0 e− = i dt e− − H e− . (5.17) − int Z0

Next we can use the fact that we know what H0 is when acting on the simple states of eqs. (5.10) and (5.11):

itH0 itEi e− p ,p = e− p ,p , (5.18) | a biIN | a biIN i(T t)H0 i(T t)E k , k e− − = k , k e− − f , (5.19) OUTh 1 2| OUTh 1 2| where

Ei = E~pa + E~pb ; (5.20) E = E + E (5.21) f ~k1 ~k2

45 are the energies of the initial and ﬁnal states, respectively. So we have:

T iT H i(T t)E itEi k , k e− p ,p = i dt e− − f e− k , k H p ,p . (5.22) OUTh 1 2| | a biIN − OUTh 1 2| int| a biIN Z0 First let us do the t integral:

T T/2 0 i(T t)E itEi i(E +Ei)T/2 it (E Ei) dt e− − f e− = e− f dt0 e f − (5.23) 0 T/2 Z Z− where we have redeﬁned the integration variable by t = t + T/2. As we take T , we can use the 0 → ∞ integral identity

∞ dx eixA = 2πδ(A) (5.24) Z−∞ to obtain:

T i(T t)E itEi i(E +Ei)T/2 dt e− − f e− = 2πδ(E E ) e− f . (5.25) f − i Z0

i(E +Ei)T/2 This tells us that energy conservation will be enforced, and (ignoring the phase factor e− f which will just give 1 when we take the complex square of the probability amplitude):

k , k p ,p = i 2πδ(E E ) k , k H p ,p . (5.26) OUTh 1 2| a biOUT − f − i OUTh 1 2| int| a biIN

Now we are ready to use our expression for Hint in eq. (5.8). The action of Hint on eigenstates of the free Hamiltonian can be read off from the different types of terms. The aaaa-type term will remove four particles from the state. Clearly we don’t have to worry about that, because there were only two particles in the state to begin with! The same goes for the a†aaa-type term. The terms of type a†a†a†a† and a†a†a†a create more than the two particles we know to be in the final state, so we can ignore them too. Therefore, the only term that will play a role in this example is the a†a†aa contribution:

λ 3 (3) Hint = (2π) dq˜1 dq˜2 dq˜3 dq˜4 δ (~q1 + ~q2 ~q3 ~q4) a† a† a~q a~q , (5.27) 4 − − ~q1 ~q2 3 4 Z Z Z Z Therefore, we have: λ k , k p ,p = i(2π)4 dq˜ dq˜ dq˜ dq˜ δ(4)(~q + ~q ~q ~q ) OUTh 1 2| a biOUT − 4 1 2 3 4 1 2 − 3 − 4 Z Z Z Z OUT k1, k2 a† a† a~q a~q pa,pb IN. (5.28) h | ~q1 ~q2 3 4 | i Here we have combined the three-momenta delta function from eq. (5.27) with the energy delta function from eq. (5.26) to give a 4-momentum delta function. It remains to evaluate:

OUT k1, k2 a† a† a~q a~q pa,pb IN, (5.29) h | ~q1 ~q2 3 4 | i which, according to eqs. (5.10) and (5.11), is equal to

0 a~ a~ a† a† a~q a~q a† a† 0 . (5.30) h | k1 k2 ~q1 ~q2 3 4 ~pa ~pb | i

46 This can be done using the commutation relations of eqs. (4.62) and (4.63). The strategy is to commute a~q and a~q to the right, so they can give 0 when acting on 0 , and commute a† and a† to the left 3 4 | i ~q1 ~q2 so they can give 0 when acting on 0 . Along the way, one picks up delta functions whenever the h | 3-momenta of an a and a† match. One contribution occurs when ~q3 = ~pa and ~q4 = ~pb and ~q1 = ~k1 and

~q2 = ~k2. It yields:

(2π)32E δ(3)(~q ~p ) (2π)32E δ(3)(~q ~p ) (2π)32E δ(3)(~q ~k ) (2π)32E δ(3)(~q ~k ). (5.31) ~q3 3 − a ~q4 4 − b ~q1 1 − 1 ~q2 2 − 2 There are 3 more similar terms, which you can check give equal contributions when put into eq. (5.28), 3 after relabeling momenta; this cancels the factor of 1/4 in eq. (5.28). Now, the factors of (2π) 2E~q3 etc. all neatly cancel against the corresponding factors in the denominator of dq˜3, etc. The three- 3 3 3 3 momentum delta functions then make the remaining d ~q1, d ~q2, d ~q3, and d ~q4 integrations trivial; they just set the four-vectors q3 = pa, q4 = pb, q1 = k1, and q2 = k2 in the remaining 4-momentum delta function that was already present in eq. (5.28). Putting it all together, we are left with the remarkably simple result:

k , k p ,p = iλ(2π)4δ(4)(k + k p p ). (5.32) OUTh 1 2| a biOUT − 1 2 − a − b Rather than go through this whole messy procedure every time we invent a new interaction term for the Lagrangian density, or every time we think of a new scattering process, one can instead summarize the procedure with a simple set of diagrammatic rules. These rules, called Feynman rules, are useful both as a precise summary of a matrix element calculation, and as a heuristic guide to what physical process the calculation represents. In the present case, the Feynman diagram for the process is:

initial state ﬁnal state

Here the two lines coming from the left represent the incoming state scalar particles, which get “destroyed” by the annihilation operators in Hint. The vertex where the four lines meet represents the interaction itself, and is associated with the factor iλ. The two lines outgoing to the right represent − the two ﬁnal state scalar particles, which are resurrected by the two creation operators in Hint. This is just the simplest of many Feynman diagrams one could write down for the process of two particle scattering in this theory. But all other diagrams represent contributions that are higher order in λ, so if λ is small we can ignore them.

47 5.2 Scattering processes and cross sections

In subsection 5.1, we learned that the matrix element corresponding to 2 particle to 2 particle scattering in a scalar ﬁeld theory with interaction Lagrangian λ φ4 is: − 24 k k p p = iλ(2π)4δ(4)(k + k p p ) (5.33) OUTh 1 2| a biOUT − 1 2 − a − b Now we would like to learn how to translate this information into something physically meaningful that could in principle be measured in an experiment. The matrix element itself is inﬁnite whenever 4-momentum is conserved, and zero otherwise. So clearly we must do some work to relate it to an appropriate physically observable quantity, namely the cross section.

The cross section relates the expected number of scattering events NS that will occur if two large sets of particles are allowed to collide. Suppose that we have Na particles of type a and Nb of type b, formed into large packets of uniform density which move completely through each other as shown.

~v ~v a −→ ←− b

The total number of scattering events occurring while the packets move through each other should be proportional to each of these numbers, and inversely proportional to the area A that is perpendicular to their relative velocity ~v ~v . The equation a − b N N N = a b σ (5.34) S A defines the cross section σ. 24 2 The dimensions for cross section are the same as area, and the official unit is 1 barn = 10− cm 2 = 2568 GeV− . However, a barn is a very large cross section,† so more commonly-used units are obtained by using the prefixes nano-, pico-, and femto-:

33 2 6 2 1 nb = 10− cm = 2.568 10− GeV− (5.35) × 36 2 9 2 1 pb = 10− cm = 2.568 10− GeV− (5.36) × 39 2 12 2 1 fb = 10− cm = 2.568 10− GeV− . (5.37) × As an example, the Tevatron collides protons (p) and antiprotons (p) with a center-of-momentum

(CM) energy of ECM = 2000 GeV. The design goal for Run IIa is to achieve:

NpNp 1 = 2 fb− (5.38) A †The joke is that achieving an event with such a cross section is “as easy as hitting the broad side of a barn”.

48 in about 2 years of running time; this quantity is called the integrated luminosity. Here “integrated” means integrated over time; the instantaneous rate of increase of NpNp/A is called the luminosity. In the case of the Tevatron, the design-goal peak luminosity in Run IIa is about 2 1032 cm 2 sec 1. × − − To ﬁgure out how many scattering events one can expect at the collider, one needs to know the corresponding cross section for that type of event, which depends on the ﬁnal state. The total cross section for any type of scattering at the Tevatron is quite large, approximately

σ(pp anything) = 0.075 barns. (5.39) → However, the vast majority of these scattering events are extremely uninteresting, featuring ﬁnal states of well-known and well-understood hadrons. A more interesting ﬁnal state would be anything involving a top quark (t)and anti-top quark (t) pair, for which the Tevatron cross section is about:

σ(pp tt + anything) = 6 pb. (5.40) → This means that one can expect to produce about 12000 top pairs in Run IIa. However, only a fraction of these will be identiﬁed as such. A more speculative possibility is that the Tevatron will produce a

Higgs boson h. If the mass of the Higgs boson is mh = 130 GeV, then the Tevatron cross section to produce it in association with a Z0 boson is roughly:

σ(pp hZ0 + anything) = 75 fb. (5.41) → This implies 150 or so hZ0 events in Run IIa. (Unfortunately, there is also a significant background from similar-looking events with a more mundane origin.) In general, our task is to figure out how to relate the matrix element for a collision process to the corresponding cross section. Let us consider a general situation in which two particles a and b with masses ma and mb and 4-momenta pa and pb collide, producing n final-state particles with masses mi and 4-momenta ki, where i = 1, 2,...,n: p , m a a k1, m1 ......

pb, mb kn, mn

As an abbreviation, we can call the initial state i = p , m ; p , m and the final state f = | i | a a b biIN | i k , m ; ... ; k , m . In general, all of the particles could be different, so that a different species | 1 1 n niOUT of creation and annihilation operators might be used for each. They can be either fermions or bosons, provided that the process conserves angular momentum. If the particles are not scalars, then i and | i

49 f should also carry labels that specify the spin of each particle. Now, because of four-momentum | i conservation, we can always write:

n 4 (4) f i = i f (2π) δ (pa + pb ki). (5.42) h | i M → − Xi=1

Here i f is called the reduced matrix element for the process. In the example in subsection 5.1, the M → reduced matrix element we found to ﬁrst order in the coupling λ was simply a constant: φφ φφ = iλ. M → − However, in general, can be a non-trivial Lorentz-scalar function of the various 4-momenta and spin M eigenvalues of the particles in the problem. It is computed order-by-order in perturbation theory, so it is only known approximately. According to the postulates of quantum mechanics, the probability of a transition from the state i to the state f is: | i | i f i 2 i f = |h | i| . (5.43) P → f f i i h | ih | i Here we have divided the matrix element by the norms of the states, which are not unity; they will be computed below. Now, the total number of scattering events expected to occur is:

NS = NaNb i f (5.44) P → Xf

Here NaNb represents the total number of initial states, one for each possible pair of incoming particles.

The sum f is over all possible ﬁnal states f. To evaluate this, recall that if one puts particles in a large boxP of volume V , then the density of one-particle states with 3-momentum ~k is

d3~k density of states = V . (5.45) (2π)3

So, including a sum over the n ﬁnal-state particles implies

n d3~k V i . (5.46) → " (2π)3 # Xf iY=1 Z Putting this into eq. (5.44) and comparing with the deﬁnition eq. (5.34), we have for the diﬀerential contribution to the total cross section:

n 3 d ~ki dσ = i f A V . (5.47) P → " (2π)3 # iY=1 Let us now suppose that each packet of particles consists of a cylinder with a large volume V . Then the total time T over which the particles can collide is given by the time it takes for the two packets to move through each other: V T = . (5.48) A ~v ~v | a − b|

50 (Assume that V and A are very large compared to the cube and square of the particles’ Compton wavelengths.) It follows that the diﬀerential contribution to the cross section is:

n 3 V d ~ki dσ = i f V 3 . (5.49) P → T ~va ~vb " (2π) # | − | iY=1 The total cross section is obtained by integrating over 3-momenta of the final state particles. Note that the differential cross section dσ depends only on the collision process being studied. So in the following we expect that the arbitrary volume V and packet collision time T should cancel out. To see how that happens in the example of φφ φφ scattering with a φ4 interaction, let us first → compute the normalizations of the states appearing in i f . For the initial state i of eq. (5.10), one P → | i has:

i i = 0 a~p a~p a† a† 0 . (5.50) h | i h | b a ~pa ~pb | i

To compute this, one can commute the a† operators to the left:

i i = 0 a~p [a~p a† ]a† 0 + 0 a~p a† a~p a† 0 . (5.51) h | i h | b a ~pa ~pb | i h | b ~pa a ~pb | i

Now the incoming particle momenta ~p and ~p are always different, so in the last term the a† just a b ~pa commutes with a according to eq. (4.62), yielding 0 when acting on 0 . The first term can be ~pb h | simplified using the commutator eq. (4.62):

3 (3) i i = 0 a~p a† 0 (2π) 2Eaδ (~pa ~pa). (5.52) h | i h | b ~pb | i −

Commuting a† to the left in the same way yields the norm of the state i : ~pb | i i i = (2π)6 4E E δ(3)(~p ~p )δ(3)(~p ~p ). (5.53) h | i a b a − a b − b This result is doubly inﬁnite, since the arguments of each delta function vanish! In order to successfully interpret it, recall the origin of the 3-momentum delta functions. One can write

3 (3) 3 i~0 x (2π) δ (~p ~p)= d ~xe · = V, (5.54) − Z showing that a 3-momentum delta-function with vanishing argument corresponds to V/(2π)3, where V is the volume that the particles occupy. So we obtain

i i = 4E E V 2 (5.55) h | i a b for the norm of the incoming state. Similarly, the norm of the state f is: | i n n f f = (2π)32E δ(3)(~k ~k )= (2E V ). (5.56) h | i i i − i i iY=1 iY=1

51 In doing this, there is one subtlety; unlike the colliding particles, it could be that two identical outgoing particles have exactly the same momentum. This seemingly could produce “extra” contributions when we commute a† operators to the left. However, one can usually ignore this, since the probability that two outgoing particles will have exactly the same momentum is vanishingly small in the limit V . → ∞ Next we turn to the square of the matrix element:

2 2 2 4 (4) f i = i f (2π) δ (pa + pb ki) . (5.57) |h | i| |M → | − h X i This also is apparently the square of an inﬁnite quantity. To interpret it, we again recall the origin of the delta functions:

T/2 it(Ef Ei) 2πδ(Ef Ei)= dt e − = T (for Ef = Ei) (5.58) − T/2 Z− 3 3 i~x ( ~k ~p) (2π) δ( ~k ~p)= d ~xe · − = V (for ~k = ~p). (5.59) − X X Z P P X X So we can write: n (2π)4δ(4)(p + p k ) = TV. (5.60) a b − i Xi=1 Now if we use this to replace one of the two 4-momentum delta functions in eq. (5.57), we have:

n 2 2 4 (4) f i = i f (2π) TVδ (pa + pb ki). (5.61) |h | i| |M → | − Xi=1 Plugging the results of eqs. (5.55), (5.56) and (5.61) into eq. (5.43), we obtain an expression for the transition probability:

n n 2 4 (4) T 1 i f = i f (2π) δ (pa + pb ki) . (5.62) P → |M → | − 4EaEbV 2EiV Xi=1 iY=1 µ ¶ Putting this in turn into eq. (5.49), we ﬁnally obtain:

2 i f dσ = |M → | dΦ (5.63) 4E E ~v ~v n a b| a − b| where dΦn is a standard object known as the n-body diﬀerential Lorentz-invariant phase space:

n n 3~ 4 (4) d ki dΦn = (2π) δ (pa + pb ki) 3 . (5.64) − Ã(2π) 2Ei ! Xi=1 iY=1 As expected, all factors of T and V have canceled out of the formula eq. (5.63) for the diﬀerential cross section. In the formula eq. (5.63), one can write for the velocities:

~pa ~pb ~va = ; ~vb = . (5.65) Ea Eb

52 Now, assuming that the collision is head-on so that ~vb is opposite to ~va (or 0), the denominator in eq. (5.63) is:

4E E ~v ~v = 4(E ~p + E ~p ) (5.66) a b| a − b| a| b| b| a| The most common case one encounters is two-particle scattering to a ﬁnal state with two particles. In the center-of-momentum frame, ~p = ~p , so that the 2-body Lorentz-invariant phase space becomes: b − a 3~ 3~ 4 (3) ~ ~ d k1 d k2 dΦ2 = (2π) δ (k1 + k2)δ(Ea + Eb E1 E2) 3 3 (5.67) − − (2π) 2E1 (2π) 2E2 (3) ~ ~ ~ 2 2 ~ 2 2 δ (k1 + k2) δ( k1 + m1 + k2 + m2 ECM) 3 3 = − d ~k1 d ~k2 (5.68) 2q ~ 2 2 ~q2 2 16π k1 + m1 k2 + m2 q q where

E E + E (5.69) CM ≡ a b is the center-of-momentum energy of the process. Now one can do the ~k2 integral; the 3-momentum delta function just sets ~k = ~k (as it must be in the CM frame). If we deﬁne 2 − 1 K ~k (5.70) ≡| 1| for convenience, then

3 2 2 d ~k1 = K dK dΩ = K dK dφd(cos θ), (5.71) where Ω = (θ, φ) are the spherical coordinate angles for ~k1. So

δ( K2 + m2 + K2 + m2 E ) 1 2 CM 2 ... dΦ2 = ... − K dK dφd(cos θ), (5.72) q 2 q2 2 2 Z Z K + m1 K + m2 q q where ... represents any quantity. To do the remaining dK integral, it is convenient to change variables to the argument of the delta function. So, deﬁning

W = K2 + m2 + K2 + m2 E , (5.73) 1 2 − CM q q we ﬁnd

2 2 2 2 K + m1 + K + m2 dW = KdK. (5.74) q 2 2 q 2 2 K + m1 K + m2 q q 2 2 2 2 Noticing that the delta function predestines K + m1 + K + m2 to be replaced by ECM, we can write: q q K2dK KdW = . (5.75) 2 2 2 2 ECM K + m1 K + m2 q q 53 Using this in eq. (5.72), and integrating dW using the delta function δ(W ), we obtain: K dΦ2 = 2 dφd(cos θ) (5.76) 16π ECM for the Lorentz-invariant phase space of a two-particle ﬁnal state. Meanwhile, in the CM frame, eq. (5.66) simpliﬁes according to:

4E E ~v ~v = 4(E ~p + E ~p ) = 4 ~p (E + E ) = 4 ~p E . (5.77) a b| a − b| a| b| b| a| | a| a b | a| CM Therefore, using the results of eqs. (5.76) and (5.77) in eq. (5.63), we have: ~ 2 k1 dσ = i f | | dΩ (5.78) |M → | 64π2E2 ~p CM| a| for the differential cross section for two-particle scattering to two particles. The matrix element is quite often symmetric under rotations about the collision axis determined by the incoming particle momenta. If so, then everything is independent of φ, and one can do dφ = 2π, leaving: R ~ 2 k1 dσ = i f | | d(cos θ). (5.79) |M → | 32πE2 ~p CM| a| If the particle masses satisfy ma = m1 and mb = m2 (or they are very small), then one has the further simplification ~k = ~p , so that | 1| | a| 2 1 dσ = i f 2 d(cos θ). (5.80) |M → | 32πECM The formulas eqs. (5.78)-(5.79) will be used often in the rest of these notes. We can finally interpret the meaning of the result eq. (5.32) that we obtained for scalar φ4 theory.

Since we found φφ φφ = iλ, the differential cross section in the CM frame is: M → − λ2 dσφφ φφ = 2 d(cos θ). (5.81) → 32πECM 1 Now we can integrate over θ using 1 d(cos θ) = 2. However, there is a double-counting problem that − we must take into account. The anglesR (θ, φ) that we have integrated over represent the direction of the 3-momentum of one of the final state particles. The other particle must then have 3-momentum in the opposite direction (π θ, φ). The two possible final states with ~k along those two opposite − − 1 directions are therefore actually the same state, because the two particles are identical. So we have actually counted each state twice when integrating over all dΩ. To take this into account, we have to divide by 2, arriving at the result for the total cross section: λ2 σφφ φφ = 2 . (5.82) → 32πECM In the system of units with c =h ¯ = 1, energy has the same units as 1/distance. Since λ is dimensionless, it checks that σ indeed has units of area. This is a very useful thing to check whenever one has found a cross section!

54 5.3 Scalar ﬁeld with φ3 coupling

For our next example, let us consider a theory with a single scalar field as before, but with an interaction Lagrangian that is cubic in the field: µ = φ3, (5.83) Lint − 6 instead of eq. (5.4). Here µ is a coupling which has the same dimensions as mass. As before, let us compute the matrix element for 2 particle to particle scattering, φφ φφ. → The definition and quantization of the free Hamiltonian proceeds exactly as before, with equal time commutators given by eqs. (4.62) and (4.63), and the free Hamiltonian by eq. (4.84). The interaction part of the Hamiltonian can be obtained in exactly the same way as the discussion leading up to eq. (5.8), yielding:

µ 3 (3) Hint = (2π) dq˜1 dq˜2 dq˜3 a† a† a† δ (~q1 + ~q2 + ~q3) 6 ~q1 ~q2 ~q3 Z Z Z " (3) +3a† a† a~q δ (~q1 + ~q2 ~q3) ~q1 ~q2 3 − (3) +3a† a~q a~q δ (~q1 ~q2 ~q3) ~q1 2 3 − − (3) +a~q1 a~q2 a~q3 δ (~q1 + ~q2 + ~q3) . (5.84) #

As before, we want to calculate:

iT H ~k ,~k ~p ~p = ~k ,~k e− ~p , ~p (5.85) OUTh 1 2| a biOUT OUTh 1 2| | a biIN

iT H where H = H0 + Hint is the total Hamiltonian. However, this time if we expand e− only to ﬁrst order in Hint, the contribution is clearly zero, because the net number of particles created or destroyed by Hint is always odd. Therefore we must work to second order in Hint, or equivalently in the coupling µ. iT H Expanding the operator e− to second order in Hint, one has in the large-N limit [compare to eq. (5.15) and the surrounding discussion]:

N iT H T e− = 1 i(H + H ) (5.86) 0 int N · − ¸ N 2 N n 2 N n m 2 m n − − − T T T T T = 1 iH − − − ( iH ) 1 iH ( iH ) 1 iH , − 0 N − int N − 0 N − int N − 0 N nX=0 mX=0 · ¸ · ¸ · ¸ (5.87)

2 where we have kept only terms that are of order Hint. Now we can convert the discrete sums into integrals over the variables t = Tn/N and t0 = T m/N + t with ∆t = ∆t0 = T/N. The result is:

T T 0 0 iT H iH0(T t ) iH0(t t) iH0t e− = dt dt0 e− − ( iHint) e− − ( iHint) e− . (5.88) 0 T t − − Z Z −

55 When we sandwich this between the states ~k ,~k and ~p , ~p , we can substitute h 1 2| | a bi 0 0 iH0(T t ) iE (T t ) e− − e− f − ; (5.89) → iH0t iEit e− e− , (5.90) → where Ei = E + E is the initial state energy eigenvalue and E = E + E is the ﬁnal state energy ~pa ~pb f ~k1 ~k2 eigenvalue. Now, ( iH ) ~p , ~p will consist of a linear combination of eigenstates X of H with − int | a bi | i 0 diﬀerent energies EX . So we can also substitute

0 0 iH0(t t) iEX (t t) e− − e− − (5.91) → provided that in place of EX we will later put in the appropriate energy eigenvalue of the state created by each particular term in iH acting on the initial state. So, we have: − int iT H e− = ( iH ) I ( iH ) (5.92) − int − int where

T T 0 0 iE (T t ) iEX (t t) iEit I = dt dt0 e− f − e− − e− . (5.93) 0 T t Z Z − To evaluate the integral I, we can ﬁrst redeﬁne t t T/2 and t t T/2, so that → − 0 → 0 −

T/2 T/2 0 iT (Ei+E )/2) it(EX Ei) it (E EX ) I = e− f dt e − dt0 e f − . (5.94) T/2 t Z− Z

iT (Ei+E )/2) Now e− f is just a constant phase which will go away when we take the complex square of the matrix element, so we ignore it. Taking the limit of a very long time T : → ∞ 0 ∞ it(EX E ) ∞ it (E EX +i²) I = dt e − f dt0 e f − . (5.95) t Z−∞ Z The t0 integral does not have the form of a delta function because its lower limit of integration is t. 0 Therefore we have inserted an inﬁnitesimal factor e ²t so that the integral makes sense for t ; we − 0 → ∞ will take ² 0 at the end. Performing the t integration, we get: → 0

i ∞ it(E Ei) I = dt e f − (5.96) ÃEf EX + i²! − Z−∞ i = 2πδ(E E ) . (5.97) f − i E E + i² Ã f − X ! As usual, energy conservation between the initial and ﬁnal states is thus automatic. Putting together the results above, we have so far:

~ ~ i OUT k1, k2 ~pa, ~pb OUT = 2πδ(Ef Ei) 0 a~ a~ ( iHint) ( iHint)a† a† 0 (5.98) h | i − h | k1 k2 − E E + i² − ~pa ~pb | i Ã f − X ! Let us now evaluate the matrix element eq. (5.98). To do this, we can divide the calculation up into pieces, depending on how many a and a† operators are contained in each factor of Hint. First, let us

56 consider the contribution when the right Hint contains a†aa terms acting on the initial state, and the left Hint contains a†a†a terms. Taking these pieces from eq. (5.84), the contribution from eq. (5.98) is:

OUT ~k1,~k2 ~pa, ~pb OUT = h | i (a†a†a)(a†aa) part ¯ µ 2¯ i (2π)3 ¯ 2πδ(E E ) dr˜ dr˜ dr˜ dq˜ dq˜ dq˜ − 2 f − i 1 2 3 1 2 3 · ¸ Z Z Z Z Z Z δ(3)(~r + ~r ~r ) δ(3)(~q ~q ~q ) 1 2 − 3 1 − 2 − 3 i 0 a~ a~ a† a† a~r a† a~q a~q a† a† 0 . (5.99) h | k1 k2 ~r1 ~r2 3 E E ~q1 2 3 ~pa ~pb | i Ã f − X !

The factor involving EX is left inserted within the matrix element to remind us that EX should be replaced by the eigenvalue of the free Hamiltonian H0 acting on the state to its right. The last line in eq. (5.99) can be calculated using the following general strategy. We commute a’s to the right and a†’s to the left, using eqs. (4.62) and (4.63). In doing so, we will get a non-zero contribution with a delta function whenever the 3-momentum of an a equals that of an a† with which it is commuted, removing that a,a† pair. In the end, every a must “contract” with some a† in this way (and vice versa), because an a acting on 0 or an a acting on 0 vanishes. | i † h | This allows us to identify what E is. The a and a operators must be contracted with a† X ~q2 ~q3 ~pa and a† if a non-zero result is to be obtained. There are two ways to do this: either pair up [a ,a† ] ~pb ~q2 ~pa and [a ,a† ], or pair up [a ,a† ] and [a ,a† ]. In both cases, the result can be non-zero only if ~q3 ~pb ~q2 ~pb ~q3 ~pa ~q + ~q = ~p + ~p . The delta-function δ(3)(~q ~q ~q ) then insures that there will be a non-zero 2 3 a b 1 − 2 − 3 contribution only when

~q = ~p + ~p Q.~ (5.100) 1 a b ≡

So the energy eigenvalue of the state a† a~q a~q a† a† 0 must be replaced by ~q1 2 3 ~pa ~pb | i

E = E = ~p + ~p 2 + m2 (5.101) X Q~ | a b| q whenever there is a non-zero contribution. Evaluating the quantity

0 a~ a~ a† a† a~r a† a~q a~q a† a† 0 (5.102) h | k1 k2 ~r1 ~r2 3 ~q1 2 3 ~pa ~pb | i now yields four distinct non-zero terms, corresponding to the following ways of contracting a’s and a†’s:

[a ,a† ] [a ,a† ] [a ,a† ] [a ,a† ] [a ,a† ] (5.103) ~k1 ~r1 ~k2 ~r2 ~q2 ~pa ~q3 ~pb ~r3 ~q1 [a ,a† ] [a ,a† ] [a ,a† ] [a ,a† ] [a ,a† ] (5.104) ~k1 ~r2 ~k2 ~r1 ~q2 ~pa ~q3 ~pb ~r3 ~q1 [a ,a† ] [a ,a† ] [a ,a† ] [a ,a† ] [a ,a† ] (5.105) ~k1 ~r1 ~k2 ~r2 ~q2 ~pb ~q3 ~pa ~r3 ~q1 [a ,a† ] [a ,a† ] [a ,a† ] [a ,a† ] [a ,a† ] (5.106) ~k1 ~r2 ~k2 ~r1 ~q2 ~pb ~q3 ~pa ~r3 ~q1

57 For the ﬁrst of these contributions to eq. (5.102), we get:

(2π)32E δ(3)(~r ~k ) (2π)32E δ(3)(~r ~k ) (2π)32E δ(3)(~q ~p ) ~r1 1 − 1 ~r2 2 − 2 ~q2 2 − a (2π)32E δ(3)(~q ~p ) (2π)32E δ(3)(~r ~q ) (5.107) ~q3 3 − b ~r3 3 − 1 2 Now, the various factors of (2π) 2E just cancel the factors in the denominators of the deﬁnition of dq˜i and dr˜i. One can do the ~q1, ~q2, ~q3, ~r1, ~r2, and ~r3 integrations trivially, using the 3-momentum delta functions, resulting in the following contribution to eq. (5.99):

µ 2 i 1 i (2π)4δ(E E )δ(3)(~k + ~k ~p ~p ). (5.108) − 2 E E 2E f − i 1 2 − a − b µ ¶ Ã f − Q~ ! Q~ The two delta functions can be combined into δ(4)(k + k p p ). Now, the other three sets of 1 2 − a − b contractions listed in eqs. (5.104)-(5.106) are exactly the same, after a relabeling of momenta. This gives a factor of 4, so (replacing E E , as allowed by the delta function) we have: f → i ~ ~ 2 i 4 (4) OUT k1, k2 ~pa, ~pb OUT † † † = ( iµ) (2π) δ (k1 + k2 pa pb). (5.109) h | i (a a a)(a aa) part − (Ei E ~ )(2E ~ ) − − ¯ − Q Q ¯ One can draw a simple¯ picture illustrating what has happened in the preceding formulas:

a†aa a†a†a initial state ﬁnal state

The initial state contains two particles, denoted by the lines on the left. Acting with the first factor of Hint (on the right in the formula, and represented by the vertex on the left in the figure) destroys the two particles and creates a virtual particle in their place. The second factor of Hint destroys the virtual particle and creates the two final state particles, represented by the lines on the right. The three-momentum carried by the intermediate virtual particle is Q~ = ~pa + ~pb = ~k1 + ~k2, so momentum is conserved at the two vertices.

However, there are other contributions which must be included. Another one occurs if the Hint on the right in eq. (5.98) contains a†a†a† operators, and the other Hint contains aaa operators. The corresponding picture is this:

a†a†a†

initial state ﬁnal state

aaa

58 Here the Hint carrying a†a†a† (the rightmost one in the formula) is represented by the upper left vertex in the ﬁgure, and the one carrying aaa is represented by the lower right vertex. The explicit formula corresponding to this picture is:

OUT ~k1,~k2 ~pa, ~pb OUT = h | i (aaa)(a†a†a†) part ¯ µ 2¯ i (2π)3 ¯ 2πδ(E E ) dr˜ dr˜ dr˜ dq˜ dq˜ dq˜ − 6 f − i 1 2 3 1 2 3 · ¸ Z Z Z Z Z Z (3) (3) δ (~r1 + ~r2 + ~r3) δ (~q1 + ~q2 + ~q3) i 0 a~ a~ a~r a~r a~r a† a† a† a† a† 0 (5.110) h | k1 k2 1 2 3 E E ~q1 ~q2 ~q3 ~pa ~pb | i Ã f − X !

As before, we can calculate this by commuting a’s to the right and a†’s to the left. In the end, non-zero contributions arise only when each a is contracted with some a†. In doing so, we should ignore any terms which arise whenever a ﬁnal state state a~k is contracted with an initial state a†~p. That would correspond to a situation with no scattering, since the initial state particle and the ﬁnal state particle would be exactly the same. The result contains 36 distinct contributions, corresponding to the 6 ways of contracting a and ~k1 a with any two of a† , a† , and a† , times the 6 ways of contracting a† and a† with any two of ~k2 ~q1 ~q2 ~q3 ~pa ~pb a~r1 , a~r2 , a~r3 . However, all 36 of these contributions are identical under relabeling of momentum, so we can just calculate one of them and multiply the answer by 36. This will neatly convert the factor of ( iµ/6)2 to ( iµ)2. We also note that E must be replaced by the free Hamiltonian energy eigenvalue − − X of the state a† a† a† a† a† 0 , namely: ~q1 ~q2 ~q3 ~pa ~pb | i

EX = E~q1 + E~q2 + E~q3 + E~pa + E~pb . (5.111)

For example consider the term obtained from the following contractions of a’s and a†’s:

[a ,a† ] [a ,a† ] [a ,a† ] [a ,a† ] [a ,a† ]. (5.112) ~k1 ~q1 ~k2 ~q2 ~r1 ~pa ~r2 ~pb ~r3 ~q3

This leads to factors of (2π)32E and momentum delta functions just as before. So we can do the 3- momentum ~q1,2 and ~r1,2,3 integrals using the delta functions, in the process setting ~q1 = ~k1 and ~q2 = ~k2 and ~r1 = ~pa and ~r2 = ~pb and ~r3 = ~q3. Finally, we can do the ~q3 integral using one of the delta functions already present in eq. (5.110), resulting in ~q = ~p ~p = ~k ~k = Q~ , with Q~ the same as was 3 − a − b − 1 − 2 − deﬁned in eq. (5.100). This allows us to identify in this case:

E = E + E + E + E + E = 2Ei + E . (5.113) X ~k1 ~k2 ~pa ~pb Q~ Q~

The end result is:

~ ~ 2 i 4 (4) OUT k1, k2 ~pa, ~pb OUT † † † = ( iµ) − (2π) δ (k1 + k2 pa pb). (5.114) h | i (aaa)(a a a ) part − (Ei + E ~ )(2E ~ ) − − ¯ Q Q ¯ It is now proﬁtable to¯ combine the two contributions we have found. One hint that this is a good idea is the fact that the two cartoon ﬁgures we have drawn for them are topologically the same; the

59 second one just has a line that moves backwards. So if we just ignore the distinction between internal lines that move backwards and those that move forwards, we can draw a single Feynman diagram to represent both results combined:

pa k1

pa + pb

pb k2 The initial state is on the left, and the ﬁnal state is on the right, and the ﬂow of 4-momentum is indicated by the arrows, with 4-momentum conserved at each vertex. The result of combining these two contributions is called the s-channel contribution, to distinguish it from still more contributions that we will get to soon. Using a common denominator for (E E ) i − Q~ and (Ei + EQ~ ), we get:

2 i 4 (4) OUT ~k1,~k2 ~pa, ~pb OUT = ( iµ) (2π) δ (k1 + k2 pa pb). (5.115) h | i s channel − E2 E2 − − ¯ − i − Q~ ¯ ¯ If we now consider the four-vector

pµ + pµ = ( ~p 2 + m2 + ~p 2 + m2 , ~p + ~p ) = (E , Q~ ), (5.116) a b | a| | b| a b i q q then we recognize that

(p + p )2 = E2 Q~ 2 = E2 E2 + m2. (5.117) a b i −| | i − Q~ µ µ ~ [Note that pa + pb is not equal to (EQ~ , Q).] So we can rewrite the term i i 2 2 = 2 2 , (5.118) E E (pa + p ) m i − Q~ b − The ﬁnal result is that the s-channel contribution to the matrix element is:

~ ~ 2 i 4 (4) OUT k1, k2 ~pa, ~pb OUT = ( iµ) 2 2 (2π) δ (k1 + k2 pa pb). (5.119) h | i s channel − (pa + pb) m − − ¯ − − ¯ ¯ 2 2 Note that one could just as well have put (k1 + k2) in place of (pa + pb) in this expression, because of the delta function. Now one can go through the same whole process with contributions that come from the rightmost

Hint (acting ﬁrst on the initial state) consisting of a†a†a terms, and the leftmost Hint containing a†aa terms. One can draw Feynman diagrams which represent these terms, which look like:

60 p pa k1 a k1

p k pa k2 a − 1 −

pb k2 pb k2

These are referred to as the t-channel and u-channel contributions respectively. Here we have combined all topologically-identical diagrams. This is a standard procedure which is always followed; the diagrams we have drawn with dashed lines for the scalar ﬁeld are the Feynman diagrams for the process. After much juggling of factors of (2π)3 and doing 3-momentum integrals using delta functions, but using no new concepts, the contributions of the t-channel and u-channel Feynman diagrams can be found to be simply:

~ ~ 2 i 4 (4) OUT k1, k2 ~pa, ~pb OUT = ( iµ) 2 2 (2π) δ (k1 + k2 pa pb), (5.120) h | i t channel − (pa k1) m − − ¯ − − − ¯ and ¯

~ ~ 2 i 4 (4) OUT k1, k2 ~pa, ~pb OUT = ( iµ) 2 2 (2π) δ (k1 + k2 pa pb). (5.121) h | i u channel − (pa k2) m − − ¯ − − − ¯ ¯ 4 (4) The reduced matrix element can now be obtained by just stripping oﬀ the factors of (2π) δ (k1 + k p p ), as demanded by the deﬁnition eq. (5.42). So the total reduced matrix element, suitable 2 − a − b for plugging into the formula for the cross section, is:

φφ φφ = s + t + u (5.122) M → M M M where: i = ( iµ)2 , (5.123) Ms − (p + p )2 m2 a b − i = ( iµ)2 , (5.124) Mt − (p k )2 m2 a − 1 − i = ( iµ)2 . (5.125) Mu − (p k )2 m2 a − 2 −

The reason for the terminology s, t, and u is because of the standard kinematic variables for 2 2 → scattering known as Mandelstam variables:

2 2 s = (pa + pb) = (k1 + k2) , (5.126) t = (p k )2 = (k p )2, (5.127) a − 1 2 − b u = (p k )2 = (k p )2. (5.128) a − 2 1 − b

61 (You will work out some of the properties of these kinematic variables for homework.) The s-, t-, and u-channel diagrams are simple functions of the corresponding Mandelstam variables: i = ( iµ)2 , (5.129) Ms − s m2 −i = ( iµ)2 , (5.130) Mt − t m2 −i = ( iµ)2 . (5.131) Mu − u m2 − Typically, if one instead scatters fermions or vector particles or some combination of them, the s- t- and u- channel diagrams will have a similar form, with m always the mass of the particle on the internal line, but with more junk in the numerators coming from the appropriate reduced matrix elements.

5.4 Feynman rules

It is now possible to abstract what we have found, to obtain the general Feynman rules for calculating reduced matrix elements in a scalar ﬁeld theory. Evidently, the reduced matrix element is the sum M of contributions from each topologically distinct Feynman diagram, with external lines corresponding to each initial state or ﬁnal state particle. For each term in the interaction Lagrangian y = φn, (5.132) Lint −n! with coupling y, one can draw a vertex at which n lines meet. At each vertex, 4-momentum must be conserved. Then: For each vertex appearing in a diagram, we should put a factor of iy. For the examples we have • − done with y = λ and y = µ, the Feynman rules are just:

iµ ←→ −

iλ ←→ −

Note that the conventional factor of 1/n! in the Lagrangian eq. (5.132) makes the corresponding Feynman rule simple in each case. For each internal scalar ﬁeld line carrying 4-momentum pµ, we should put a factor of • i/(p2 m2 + i²): −

i ←→ p2 m2 + i² p −

This factor associated with internal scalar ﬁeld lines is called the Feynman propagator. Here we have added an imaginary inﬁnitesimal term i², with the understanding that ² 0 at the end of → the calculation; this turns out to be necessary for cases in which p2 become very close to m2. This

62 corresponds to the particle on the internal line being nearly “on-shell”, because p2 = m2 is the equation satisﬁed by a free particle in empty space. Typically, the i² only makes a diﬀerence for propagators involved in closed loops (discussed below). For each external line, we just have a factor of 1: •

or 1 ←→

It is useful to mention this rule simply because in the cases of fermions and vector fields, external lines will turn out to carry non-trivial factors not equal to 1. (Here the gray blobs represent the rest of the Feynman diagram.) Those are all the rules one needs to calculate reduced matrix elements for Feynman diagrams without closed loops, also known as tree diagrams. The resulting calculation is said to be a tree- level. There are also Feynman rules which apply to diagrams with closed loops (loop diagrams) which have not arisen explicitly in the preceding discussion, but could be inferred from more complicated calculations. For them, one needs the following additional rules: For each closed loop in a Feynman diagram, there is an undetermined 4-momentum `µ. These • loop momenta should be integrated over according to: d4` . (5.133) (2π)4 Z Loop diagrams quite often diverge because of the integration over all `µ, because of the contribution from very large `2 . This can be fixed by introducing a cutoff `2 in the integral, or by other slimy | | | |max tricks, which can make the integrals finite. The techniques of getting physically meaningful answers out of this are known as regularization (making the integrals finite) and renormalization (redefining coupling constants and masses so that the physical observables don’t depend explicitly on the unknown cutoff). If a Feynman diagram with one or more closed loops can be transformed into an exact copy of • itself by interchanging any number of internal lines through a smooth deformation, without moving the external lines, then there is an additional factor of 1/N, where N is the number of distinct permutations of that type. (This is known as the “symmetry factor” for the loop diagram.) Some examples might be useful. In the φ3 theory, there are quite a few Feynman diagrams that will describe the scattering of 2 particles to 3 particles. One of them is shown below:

pa k1 + k2

pa + pb k2

pb k3

63 For this diagram, according to the rules, the contribution to the reduced matrix element is just i i = ( iµ)3 . (5.134) M − (p + p )2 m2 (k + k )2 m2 · a b − ¸ · 1 2 − ¸ Imagine having to calculate this starting from scratch with creation and annihilation operators, and tremble with fear! Feynman rules are good. An example of a Feynman diagram with a closed loop in the φ3 theory is:

pa ` k1

pa + pb pa + pb

pb ` p p k − a − b 2 There is a symmetry factor of 1/2 for this diagram, because one can smoothly interchange the two lines carrying 4-momenta `µ and `µ pµ pµ to get back to the original diagram, without moving the − a − b external lines. So the reduced matrix element for this diagram is:

1 i 2 d4` i i = ( iµ)4 . (5.135) M 2 − (p + p )2 m2 (2π)4 (` p p )2 m2 + i² `2 m2 + i² · a b − ¸ Z · − a − b − ¸ · − ¸ Again, deriving this result starting from the creation and annihilation operators is possible, but ex- traordinarily unpleasant! In the future, we will simply guess the Feynman rules for any theory from staring at the Lagrangian density. The general procedure for doing this is rather simple, and is outlined below. A Feynman diagram is a precise representation of a contribution to the reduced matrix element M for a given physical process. The diagrams are built out of three types of building blocks:

vertices interactions ←→ internal lines free virtual particle propagation ←→ external lines initial, ﬁnal states. ←→ The Feynman rules specify a mathematical expression for each of these objects. They follow from the

Lagrangian density, which deﬁnes a particular theory.‡

To generalize what we have found for scalar fields, let us consider a set of generic fields Φi, which can include both commuting bosons and anticommuting fermions. They might include real or complex scalars, Dirac or Weyl fermions, and vector fields of various types. The index i runs over a list of all the fields, and over their spinor or vector indices. Now, it is always possible to obtain the Feynman rules by writing an interaction Hamiltonian and computing matrix elements. Alternatively, one can use powerful path integral techniques that are beyond the scope of this course to derive the Feynman

‡It is tempting to suggest that the Feynman rules themselves should be taken as the deﬁnition of the theory. However, this would only be suﬃcient to describe phenomena that occur in a perturbative weak-coupling expansion.

64 rules. However, in the end the rules can be summarized very simply in a way that could be guessed from the examples of real scalar field theory that we have already worked out. In these notes, we will follow the technique of guessing; more rigorous derivations can be found in field theory textbooks. For interactions, we have now found in two cases that the Feynman rule for n scalar lines to meet at a vertex is equal to i times the coupling of n scalar fields in the Lagrangian with a factor of 1/n!. − More generally, consider an interaction Lagrangian term: X = i1i2...iN Φ Φ ... Φ , (5.136) Lint − P i1 i2 iN where P is the product of n! for each set of n identical fields in the list Φi1 , Φi2 ,..., ΦiN , and Xi1i2...iN is the coupling constant which determines the strength of the interaction. The corresponding Feynman rule attaches N lines together at a vertex. Then the mathematical expression assigned to this vertex is iX . The lines for distinguishable fields among i ,i ,...,i should be labeled as such, or − i1i2...iN 1 2 N otherwise distinguished by drawing them differently from each other. For example, consider a theory with two real scalar fields φ and ρ. If the interaction Lagrangian includes terms, say, λ λ = 1 φ2ρ2 2 φ3ρ, (5.137) Lint − 4 − 6 then there are Feynman rules:

iλ ←→ − 1

iλ ←→ − 2

Here the longer-dashed lines correspond to the field φ, and the shorter-dashed lines to the field ρ. As another example, consider a theory in which a real scalar field φ couples to a Dirac fermion Ψ according to:

= yφΨΨ. (5.138) Lint − 0 In this case, we must distinguish between lines for all three ﬁelds, because Ψ=Ψ†γ is independent of Ψ. For Dirac fermions, one draws solid lines with an arrow coming in to a vertex representing Ψ in

Hint, and an arrow coming out representing Ψ. So the Feynman rule for this interaction is: a

iyδ b ←→ − a b

65 Note that this Feynman rule is proportional to a 4 4 identity matrix in Dirac spinor space. This is × a because the interaction Lagrangian can be written yδ bφΨ Ψ , where a is the Dirac spinor index for − a b Ψ and b for Ψ. Often, one just suppresses the spinor indices, and writes simply iy for the Feynman − rule, with the identity matrix implicit. The interaction Lagrangian eq. (5.138) is called a Yukawa coupling. This theory has an actual physical application: it is precisely the type of interaction that applies between the Standard Model Higgs boson φ = h and each Dirac fermion Ψ, with the coupling y proportional to the mass of that fermion. We will return to this interaction when we discuss the decays of the Higgs boson into fermion- antifermion pairs. Let us turn next to the topic of internal lines in Feynman diagrams. These are determined by the free (quadratic) part of the Lagrangian density. Recall that for a scalar ﬁeld, we can write the free Lagrangian after integrating by parts as: 1 = φ( ∂ ∂µ m2)φ. (5.139) L0 2 − µ − This corresponded to a Feynman propagator rule for internal scalar lines i/(p2 m2 + i²). So, up to − the i² factor, the propagator is just proportional to i divided by the inverse of the coeﬃcient of the quadratic piece of the Lagrangian density, with the replacement

∂ ip . (5.140) µ −→ − µ

The free Lagrangian density for generic ﬁelds Φi can always be put into either the form 1 = Φ P Φ , (5.141) L0 2 i ij j Xi,j for real ﬁelds, or the form

= (Φ†) P Φ , (5.142) L0 i ij j Xi,j for complex ﬁelds (including, for example, Dirac spinors). To accomplish this, one may need to integrate the action by parts, throwing away a total derivative in which will not contribute to S = d4x . L0 L Here Pij is a matrix that involves spacetime derivatives and masses. Then it turns out that the FeynmanR propagator can be found by making the replacement eq. (5.140) and taking i times the inverse of the matrix Pij:

1 i(P − )ij. (5.143)

This corresponds to an internal line in the Feynman diagram labeled by i at one end and j at the other. As an example, consider the free Lagrangian for a Dirac spinor Ψ, as given by eq. (4.26). According to the prescription of eqs. (5.142) and (5.143), the Feynman propagator connecting vertices with spinor indices a and b should be:

1 b i[(p/ m)− ] . (5.144) − a

66 In order to make sense of the inverse matrix, we can write it as a fraction, then multiply numerator and denominator by (p/ + m), and use the fact that p/p/ = p2 from eq. (2.159):

i i(p/ + m) i(p/ + m) = = . (5.145) p/ m (p/ m)(p/ + m) p2 m2 + i² − − − In the last line we have put in the i² factor needed for loop diagrams as a prescription for handling the possible singularity at p2 = m2. So the Feynman rule for a Dirac fermion internal line is:

p b b b → a i([p/]a + mδa ) ←→ p2 m2 + i² −

Here the arrow direction on the fermion line distinguishes the direction of particle flow, with particles (anti-particles) moving with (against) the arrow. For electrons and positrons, this means that the arrow on the propagator points in the direction of the flow of negative charge. As indicated, the 4- momentum pµ appearing in the propagator is also assigned to be in the direction of the arrow on the internal fermion line. Next we turn to the question of Feynman rules for external particle and anti-particle lines. At a fixed time t = 0, a generic field Φ is written as an expansion of the form:

i~p ~x i~p ~x Φ(~x)= dp˜ i(~p, n) e · a~p,n + f(~p, n) e− · b†~p,n , (5.146) n X Z h i where a~p,n and b†~p,n are annihilation and creation operators (which may or may not be Hermitian conjugates of each other); n is an index running over spins and perhaps other labels for diﬀerent particle types; and i(~p, n) and f(~p, n) are expansion coeﬃcients. In general, we build an interaction

Hamiltonian out of the fields Φ. When acting on an initial state a† 0 on the right, Hint will therefore ~k,m| i ~ produce a factor of i(k, m) after commuting (or anticommuting, for fermions) the a~p,n operator in Φ to the right, removing the a† . Likewise, when acting on a final state 0 b on the left, the interaction ~k,m h | ~k,m Hamiltonian will produce a factor of f(~k, m). Therefore, initial and final state lines just correspond to the appropriate coefficient of annihilation and creation operators in the Fourier mode expansion for that field. For example, comparing eq. (5.146) to eqs. (4.72), (4.89) and (4.90); in the scalar case, we find that i(~p, n) and f(~p, n) are both just equal to 1. For Dirac fermions, we see from eq. (4.89) that the coefficient for an initial state particle (electron) µ carrying 4-momentum p and spin state s is u(p, s)a, where a is a spinor index. Similarly, the coefficient for a final state antiparticle (positron) is v(p, s)a. So the Feynman rules for these types of external particle lines are:

p initial state electron: → u(p, s)a a ←→

67 p ﬁnal state positron: → v(p, s) a ←→ a

Here the blobs represent the rest of the Feynman diagram in each case. Similarly, considering the expansion of the field Ψ in eq. (4.90), we see that the coefficient for an initial state antiparticle (positron) is v(p, s)a and that for a final state particle (electron) is u(p, s)a. So the Feynman rules for these external states are:

a p initial state positron: → v(p, s)a ←→

p a ﬁnal state electron: → u(p, s)a ←→

Note that in these rules, the pµ label of an external state is always the physical 4-momentum of that particle or anti-particle; this means that with the standard convention of initial state on the left and ﬁnal state on the right, the pµ associated with each of u(p, s), v(p, s), u(p, s) and v(p, s) is always taken to be pointing to the right. For v(p, s) and v(p, s), this is in the opposite direction to the arrow on the fermion line itself.

68 6 Quantum Electro-Dynamics (QED)

6.1 QED Lagrangian and Feynman rules

Let us now see how all of these general rules apply in the case of Quantum Electrodynamics. This is the quantum field theory governing photons (quantized electromagnetic waves) and charged fermions and antifermions. The fermions in the theory are represented by Dirac spinor fields Ψ carrying electric charge Qe, where e is the magnitude of the charge of the electron. Thus Q = 1 for electrons and − positrons, +2/3 for up, charm and top quarks and their anti-quarks, and 1/3 for down, strange and − bottom quarks and their antiquarks. (Recall that a single Dirac field, assigned a single value of Q, is used to describe both particles and their anti-particles.) The free Lagrangian for the theory is: 1 = F µνF + Ψ(iγµ∂ m)Ψ. (6.1) L0 −4 µν µ − Now, in section 3 we found that the electromagnetic field Aµ couples to the 4-current density J µ = (ρ, J~) by a term in the Lagrangian eJ µA [see eq. (4.39)]. Since J µ must be a four-vector built out − µ of the charged fermion fields Ψ and Ψ, we can guess that:

J µ = QΨγµΨ. (6.2)

The interaction Lagrangian density for a fermion with charge Qe and electromagnetic ﬁelds is therefore:

= eQΨγµΨA . (6.3) Lint − µ The value of e is determined by experiment. However, it is a running coupling constant, which means that its value has a logarithmic dependence on the characteristic energy of the process. For very low energy experiments, the numerical value is e 0.3028, corresponding to the experimental result for ≈ the ﬁne structure constant: e2 α 1/137.036. (6.4) ≡ 4π ≈ For experiments done at energies near 100 GeV, the appropriate value is a little larger, more like e 0.313. ≈ Let us take a small detour to check that eq. (6.2) really has the correct form and normalization to be the electromagnetic current density. Consider the total charge operator:

3 3 0 3 0 3 Q = d ~xρ(~x)= d ~xJ (~x)= Q d ~x Ψγ Ψ= Q d ~x Ψ†Ψ. (6.5) Z Z Z Z Plugging in eqs. (4.89)b and (4.90), and doing the ~x integration, and one of the momentum integrations using the resulting delta function, one ﬁnds:

2 2 1 Q = Q dp˜ [u†(p, s)b†~p,s + v†(p, s)d~p,s][u(p,r)b~p,r + v(p,r)d†~p,r]. (6.6) 2E~p sX=1 rX=1 Z b

69 This can be simpliﬁed using some spinor identities:

u†(p, s)u(p,r)= v†(p, s)v(p,r) = 2E~p δsr; (6.7)

u†(p, s)v(p,r)= v†(p, s)u(p,r) = 0. (6.8)

Taking into account that the operators d,d† satisfy the anticommutation relation eqs. (4.108), the result is:

2 Q = Q dp˜ b† b d† d . (6.9) ~p,s ~p,s − ~p,s ~p,s sX=1 Z h i b Here we have dropped an inﬁnite contribution, much like we had to do in getting to eqs. (4.84) and eqs. (4.109). This represents a uniform and constant (and therefore unobservable) inﬁnite charge density throughout all space. The result is that the total charge eigenvalue of the vacuum vanishes:

Q 0 = 0. (6.10) | i One can now check that b

[Q, b† ] = Q b† (6.11) ~k,r ~k,r [Q,b d† ] = Qd† . (6.12) ~k,r − ~k,r Therefore, b

Q b† 0 = Q b† 0 (6.13) ~k,r| i ~k,r| i for single-particle states, and b

Qd† 0 = Qd† 0 (6.14) ~k,r| i − ~k,r| i for single anti-particle states. More generally,b the eigenvalue of Q acting on a state with N particles and N antiparticles is (N N)Q. From eq. (6.5), this veriﬁes that the charge density ρ is indeed the − b time-like component of the four-vector QΨγµΨ, which must therefore be equal to J µ. The full QED Lagrangian is invariant under gauge transformations: 1 A (x) A (x) ∂ θ(x), (6.15) µ → µ − e µ Ψ(x) eiQθ Ψ(x), (6.16) → iQθ Ψ(x) e− Ψ(x), (6.17) → where θ(x) is an arbitrary function of spacetime [equal to eλ(x) in eq. (3.16)]. A nice way to see this − invariance is to deﬁne the covariant derivative:

D Ψ (∂ + iQeA )Ψ. (6.18) µ ≡ µ µ

70 Here the term “covariant” refers to the gauge transformation symmetry, not the Lorentz transformation symmetry as it did when we introduced covariant four-vectors. Note that the covariant derivative actually depends on the charge of the ﬁeld it acts on. Now one can write the full Lagrangian density as 1 = + = F µνF + iΨγµD Ψ mΨΨ. (6.19) L L0 Lint −4 µν µ − The ordinary derivative of the spinor transforms under the gauge transformation with an “extra” term:

∂ Ψ ∂ (eiQθΨ) = eiQθ∂ Ψ+ iQΨ∂ θ. (6.20) µ → µ µ µ The point of the covariant derivative is that it transforms under the gauge transformation the same way Ψ does, by acquiring a phase:

D Ψ eiQθD Ψ. (6.21) µ → µ

Here the contribution from the transformation of Aµ in Dµ cancels the extra term in eq. (6.20). Using eqs. (6.15)-(6.17) and (6.21), it is easy to see that is invariant under the gauge transformation, since L the multiplicative phase factors just cancel. Returning to the interaction term eq. (6.3), we can now identify the Feynman rule for QED interactions, by following the general prescription outlined with eq. (5.136): a µ iQe(γµ) b ←→ − a b This one interaction vertex governs all physical processes in QED. To find the Feynman rules for initial and final state photons, consider the Fourier expansion of the vector field at a fixed time:

3 i~p ~x i~p ~x Aµ = dp˜ ²µ(p, λ)e · a~p,λ + ²µ∗ (p, λ)e− · a†~p,λ . (6.22) λX=0 Z h i

Here ²µ(p, λ) is a basis of four polarization four-vectors labeled by λ = 0, 1, 2, 3. They satisfy the orthonormality condition:

1 for λ = λ0 = 1, 2, or 3 µ − ² (p, λ)²µ∗ (p, λ0) =  +1 for λ = λ0 = 0 (6.23)  0 for λ = λ0 6  The operators a†~p,λ and a~p,λ act on the vacuum state by creating and destroying photons with momentum ~p and polarization vector ²µ(p, λ). However, not all of the four degrees of freedom labeled by λ can be physical. From classical electromagnetism, we know that electromagnetic waves are transversely polarized. This means that

71 the electric and magnetic ﬁelds are perpendicular to the 3-momentum direction of propagation. In terms of the potentials, it means that one can always choose a gauge in which A0 = 0 and the Lorenz µ gauge condition ∂µA = 0 is satisﬁed. Therefore, physical electromagnetic wave quanta corresponding µ µ ip x 2 to the classical solutions to Maxwell’s equations A = ² e− · with p = 0 can be taken to obey:

²µ = (0,~²), (6.24) µ pµ² = 0 (6.25)

(or, equivalent to the last condition, ~² ~p = 0). After imposing these two conditions, only two of the · four λ’s will survive as valid initial or ﬁnal states for any given pµ. For example, suppose that a state contains a photon with 3-momentum ~p = pzˆ, so pµ = (p, 0, 0,p). Then we can choose a basis of transverse linearly-polarized vectors with λ = 1, 2:

²µ(p, 1) = (0, 1, 0, 0) x-polarized; (6.26) ²µ(p, 2) = (0, 0, 1, 0) y-polarized. (6.27)

However, in high-energy physics it is usually more useful to instead use a basis of left- and right-handed circular polarizations that carry deﬁnite helicities λ = R,L: 1 ²µ(p, R) = (0, 1,i, 0) right-handed (6.28) √2 1 ²µ(p,L) = (0, 1, i, 0) left-handed. (6.29) √2 −

In general, incoming photon lines have a Feynman rule ²µ(p, λ) and outgoing photon lines have a

Feynman rule ²µ∗ (p, λ), where λ = 1, 2 in some convenient basis of choice. Often, we will sum or average over the polarization labels λ, so the ²µ(p, λ) will not need to be listed explicitly for a given momentum.

Let us next construct the Feynman propagator for photon lines. The free Lagrangian density given in eq. (4.34): can be rewritten as: 1 = Aµ (g ∂ ∂ρ ∂ ∂ ) Aν, (6.30) L0 2 µν ρ − µ ν where we have dropped a total derivative. (The action, obtained by integrating , does not depend on L0 total derivative terms.) Therefore, following the prescription of eq. (5.143), it appears that we ought to ﬁnd the propagator by ﬁnding the inverse of the 4 4 matrix × P = p2g + p p . (6.31) µν − µν µ ν Unfortunately, however, this matrix is not invertible. The reason for this can be traced to the gauge invariance of the theory; not all of the physical states we are attempting to propagate are really physical.

72 This problem can be avoided using a trick, due to Fermi, called “gauge fixing”. As long as we agree µ to stick to Lorenz gauge, ∂µA = 0, we can add a term to the Lagrangian density proportional to µ 2 (∂µA ) : 1 (ξ) = (∂ Aµ)2. (6.32) L0 L0 − 2ξ µ In Lorenz gauge, not only does the extra term vanish, but also its contribution to the equations of motion vanishes. Here ξ is an arbitrary new gauge-fixing parameter; it can be picked at will. Physical results should not depend on the choice of ξ. The new term in the modified free Lagrangian (ξ) is L0 called the gauge-fixing term. Now the matrix to be inverted is: 1 P = p2g + 1 p p . (6.33) µν µν ξ µ ν − µ − ¶ 1 νρ To find the inverse, one notices that as a tensor, (P − ) can only be a linear combination of terms proportional to gνρ and to pνpρ. So, making a guess:

1 νρ νρ ν ρ (P − ) = C1 g + C2 p p (6.34) and requiring that

1 νρ ρ Pµν(P − ) = δµ, (6.35) one ﬁnds the solution 1 1 ξ C = ; C = − . (6.36) 1 −p2 2 (p2)2

It follows that the desired Feynman propagator for a photon with momentum pµ is: i p p g + (1 ξ) µ ν (6.37) p2 + i² µν p2 ·− − ¸ (Here we have put in the i² factor in the denominator as usual.) In a Feynman diagram, this propagator corresponds to an internal wavy line, labeled by µ and ν at opposite ends, and carrying 4-momentum p. The gauge-ﬁxing parameter ξ can be chosen at the convenience or whim of the person computing the Feynman diagram. The most popular choice for simple calculations is ξ = 1, called Feynman gauge. Then the Feynman propagator for photons is simply: ig − µν (Feynman gauge). (6.38) p2 + i² Another common choice is ξ = 0, known as Landau gauge, for which the Feynman propagator is: i p p − g µ ν (Landau gauge) (6.39) p2 + i² µν p2 · − ¸ [Comparing to eq. (6.32), we see that this is really obtained as a formal limit ξ 0.] The Landau → gauge photon propagator has the nice property that it vanishes when contracted with either pµ or pν,

73 which can make some calculations simpler (especially certain loop diagram calculations). Sometimes it is useful to just leave ξ unspecified. Even though this means having to calculate more terms, the payoff is that in the end one can demand that the final answer for the reduced matrix element is independent of ξ, providing a consistency check. We have now encountered most of the Feynman rules for QED. There are some additional rules having to do with minus signs because of Fermi-Dirac statistics; these can be understood by carefully considering the effects of anticommutation relations for fermionic operators. Also, in practice one usually does not write out the spinor indices explicitly. All the rules are summarized in the following two pages in a cookbook form. Of course, the best way to understand how the rules work is to do some examples. That will be the subject of the next few subsections.

74 Feynman rules for QED

To ﬁnd the contributions to the reduced matrix element for a physical process involving charged M Dirac fermions and photons: 1. Draw all topologically distinct Feynman diagrams, with wavy lines representing photons, and solid lines with arrows representing fermions, using the rules below for external lines, internal lines, and interaction vertices. The arrow direction is preserved when following each fermion line. Enforce four- momentum conservation at each vertex. 2. For external lines, write (with 4-momentum pµ always to the right, and spin polarization s or λ as appropriate):

initial-state fermion: u(p, s) ←→

initial-state antifermion: v(p, s) ←→

ﬁnal-state fermion: u(p, s) ←→

ﬁnal-state antifermion: v(p, s) ←→

µ initial-state photon: ² (p, λ) ←→ µ

µ ﬁnal-state photon: ² (p, λ) ←→ µ∗

3. For internal fermion lines, write Feynman propagators:

i(p/ + m) ←→ p2 m2 + i² − with 4-momentum pµ along the arrow direction, and m the mass of the fermion. For internal photon lines, write:

µ ν i p p g + (1 ξ) µ ν p2 + i² µν p2 ←→ ·− − ¸ with 4-momentum pµ along either direction in the wavy line. (Use ξ = 1 for Feynman gauge and ξ = 0 for Landau gauge.)

4. For interaction vertices, write:

75 µ iQeγµ ←→ −

The vector index µ is to be contracted with the corresponding index on the photon line to which it is connected. This will be either an external photon line factor of ²µ or ²µ∗ , or on an internal photon line propagator index.

5. For each loop momentum `µ that is undetermined by four-momentum conservation with ﬁxed external-state momenta, perform an integration

d4` . (2π)4 Z Getting a ﬁnite answer from these loop integrations often requires that they be regularized by introducing a cutoﬀ or some other trick.

6. Put a factor of ( 1) for each closed fermion loop. − 7. To take into account suppressed spinor indices on fermion lines, write terms involving spinors as follows. For fermion lines that go all the way through the diagram, start at the end of each fermion line (as deﬁned by the arrow direction) with a factor u or v, and write down factors of γµ or (p/ + m) consecutively, following the line backwards until a u or v spinor is reached. For closed fermion loops, start at an arbitrary vertex on the loop, and follow the fermion line backwards until the original point is reached; take a trace over the gamma matrices in the closed loop.

8. If a Feynman diagram with one or more closed loops can be transformed into an exact copy of itself by interchanging any number of internal lines through a smooth deformation without moving the external lines, then there is an additional symmetry factor of 1/N, where N is the number of distinct permutations of that type.

9. After writing down the contributions from each diagram to the reduced matrix element ac- M cording to the preceding rules, assign an additional relative minus sign between diﬀerent diagram contributions whenever the written ordering of external state spinor wavefunctions u,v, u, v diﬀers by an odd permutation.

76 + + 6.2 e−e µ−µ → In the next few subsections we will study some of the basic scattering processes in QED, using the Feynman rules found in the previous section, and the general discussion of cross sections given in section 5.2. These calculations will involve several thematic tricks which are common to many Feynman diagram evaluations. We begin with electron-positron annihilation into a muon-antimuon pair:

+ + e−e µ−µ . → We will calculate the differential and total cross sections for this process in the center-of-momentum frame, to leading order in the coupling e. Since the mass of the muon (and the anti-muon) is about mµ = 105.66 MeV, this process requires a center-of-momentum energy of at least √s = 211.3 MeV. By contrast, the mass of the electron is only about me = 0.511 MeV, which we can therefore safely neglect. The error made in doing so is far less than the error made by not including higher-order corrections. A good first step is to label the momentum and spin data for the initial state electron and positron and the final state muon and anti-muon: Particle Momentum Spin Spinor e− pa sa u(pa, sa) + e pb sb v(pb, sb) (6.40) µ− k1 s1 u(k1, s1) + µ k2 s2 v(k2, s2)

At order e2, there is only one Feynman diagram for this process. Here it is:

pa k1

pa + pb

µ ν

pb k2

Applying the rules for QED to turn this picture into a formula for the reduced matrix element, we ﬁnd: ig = [v(p , s ) (ieγµ) u(p , s )] − µν [u(k , s ) (ieγν) v(k , s )] . (6.41) b b a a (p + p )2 1 1 2 2 M · a b ¸ µ The v(pb, sb) (ieγ ) u(pa, sa) part is obtained by starting at the end of the electron-positron line with the positron external state spinor, and following it back to its beginning. The interaction vertex is iQeγµ = ieγµ, since the charge of the electron and muon is Q = 1. Likewise, the − − ν u(k1, s1) (ieγ ) v(k2, s2) part is obtained by starting at the end of the muon-antimuon line with the muon external state spinor, and following it backwards. The photon propagator is written in Feyn- man gauge, for simplicity, and carries indices µ and ν which connect to the two fermion lines at their respective interaction vertices.

77 We can write this result more compactly by using abbreviations v(p , s ) = v and u(p , s ) u , b b b a a ≡ a 2 etc. Writing the denominator of the photon propagator as the Mandelstam variable s = (pa + pb) = 2 (k1 + k2) , and using the metric in the photon propagator to lower the index on one of the gamma matrices, we get:

e2 = i (v γ u )(u γµv ). (6.42) M s b µ a 1 2 The diﬀerential cross section involves the complex square of : M 4 2 e µ ν = (v γ u )(u γ v )(v γ u )∗(u γ v )∗. (6.43) |M| s2 b µ a 1 2 b ν a 1 2 Evaluating the complex conjugated terms in parentheses can be done systematically by taking the Hermitian conjugate of the Dirac spinors and matrices they are made of, taking care to write them in the reverse order. So, for example,

0 0 0 (vbγνua)∗ = (vb†γ γνua)∗ = ua† γν†γ vb = ua† γ γνvb = uaγνvb. (6.44)

0 0 The third equality follows from the identity (B.4), which implies γν†γ = γ γν. Similarly,

ν ν (u1γ v2)∗ = v2γ u1. (6.45)

[Following the same strategy, one can show that in general,

(xγµγν ...γρy)∗ = yγρ ...γνγµx (6.46) where x and y are any u and v spinors.] Therefore we have:

e4 2 = (v γ u )(u γ v ) (u γµv )(v γνu ). (6.47) |M| s2 b µ a a ν b 1 2 2 1

At this point, we could work out explicit forms for the external state spinors and plug eq. (6.47) into eq. (5.78) to find the differential cross section for any particular set of spins. However, this is not very convenient, and fortunately it is not necessary either. In a real experiment, the final state spins of the muon and anti-muon are typically not measured. Therefore, to find the total cross section for all possible final states, we should sum over s1 and s2. Also, if the initial-state electron spin states are unknown, we should average over sa and sb. (One must average, not sum, over the initial-state spins, because sa and sb cannot simultaneously take on both spin-up and spin-down values; there is only one initial state, even if it is unknown.) These spin sums and averages will allow us to exploit the identities (B.27) and (B.28), so that the explicit forms for the spinors are never needed. After doing the spin sum and average, the differential cross section must be symmetric under rotations about the collision axis. This is because the only special directions in the problem are the momenta of the particles, so that the cross section can only depend on the angle θ between the collision

78 axis determined by the two initial-state particles and the scattering axis determined by the two ﬁnal- state particles. So, we can apply eq. (5.79) to obtain:

dσ 1 1 ~k = 2 | 1| . (6.48) d(cos θ) 2 s 2 s s s |M| 32πs ~pa Xa Xb X1 X2 | | in the center-of-momentum frame, with the eﬀects of initial state spin averaging and ﬁnal state spin summing now included. We can now use eqs. (2.155) and (2.156), which in the present situation imply

uaua = p/a + me, (6.49) s Xa v2v2 = k/2 mµ. (6.50) s − X2 Neglecting me as promised earlier, we obtain: 1 1 e4 2 µ / ν = 2 (vbγµp/aγνvb) (u1γ [k2 mµ]γ u1) (6.51) 2 s 2 s s s |M| 4s s s − Xa Xb X1 X2 Xb X1 Now we apply another trick. A dot product of two vectors is equal to the trace of the vectors multiplied in the opposite order to form a matrix:

b1 b1 b1a1 b1a2 b1a3 b1a4 b2 b2 b2a1 b2a2 b2a3 b2a4 ( a1 a2 a3 a4 )   = Tr   ( a1 a2 a3 a4 ) Tr   . (6.52) b3 b3 ≡ b3a1 b3a2 b3a3 b3a4   h  i    b4   b4   b4a1 b4a2 b4a3 b4a4        Applying this to each expression in parentheses in eq. (6.51), we move the barred spinor (thought of as a row vector) to the end and take the trace over the resulting 4 4 Dirac spinor matrix. So: ×

vbγµp/aγνvb = Tr[γµp/aγνvbvb] (6.53) u γµ[k/ m ]γνu = Tr[γµ(k/ m )γνu u ], (6.54) 1 2 − µ 1 2 − µ 1 1 so that 1 e4 2 µ / ν = 2 Tr[γµp/aγνvbvb] Tr[γ (k2 mµ)γ u1u1]. (6.55) 4 |M| 4s s s − spinsX Xb X1 The reason this trick of rearranging into a trace is useful is that now we can once again exploit the spin-sum identities, this time in the form:

vbvb = p/b me (6.56) s − Xb u1u1 = k/1 + mµ. (6.57) s X1 The result, again neglecting me, is: 4 1 2 e µ ν = Tr[γµp/ γνp/ ] Tr[γ (k/ mµ)γ (k/ + mµ)]. (6.58) 4 |M| 4s2 a b 2 − 1 spinsX

79 Next we must evaluate the traces. First,

α β Tr[γµp/aγνp/b] = pa pb Tr[γµγαγνγβ] (6.59) = pαpβ(4g g 4g g + 4g g ) (6.60) a b µα νβ − µν αβ µβ να = 4p p 4g p p + 4p p , (6.61) aµ bν − µν a · b bµ aν where we’ve used the general result for the trace of four gamma matrices found in Homework Set 1. Similarly, making use of the fact that the trace of an odd number of gamma matrices is zero:

Tr[γµ(k/ m )γν(k/ + m )] = Tr[γµk/ γνk/ ] m2 Tr[γµγν] (6.62) 2 − µ 1 µ 2 1 − µ = 4kµkν 4gµνk k + 4kµkν 4gµνm2 . (6.63) 2 1 − 1 · 2 1 2 − µ Taking the product of the two traces, one ﬁnds that the answer reduces to simply: 1 e4 2 = 8 (p k )(p k ) + (p k )(p k ) + (p p )m2 (6.64) 4 |M| s2 a · 2 b · 1 a · 1 b · 2 a · b µ spinsX h i

Our next task is to work out the kinematic quantities appearing in eq. (6.64). Let us call P and K the magnitudes of the 3-momenta of the electron and the muon, respectively. We assume that the electron is initially moving in the +z direction, and the muon makes an angle θ with respect to the positive z axis, within the yz plane. The on-shell conditions for the particles are:

p2 = p2 = m2 0 (6.65) a b e ≈ 2 2 2 k1 = k2 = mµ. (6.66) Then we have:

pa = (P, 0, 0, P ) (6.67) p = (P, 0, 0, P ) (6.68) b − 2 2 k1 = ( K + mµ, 0, K sin θ, K cos θ) (6.69) q k = ( K2 + m2 , 0, K sin θ, K cos θ). (6.70) 2 µ − − 2 2 q Since (pa + pb) = (k1 + k2) = s, it follows that s P = (6.71) r4 s 2 K = mµ, (6.72) r4 − and s p p = 2P 2 = (6.73) a · b 2 s 4m2 p k = p k = P K2 + m2 P K cos θ = 1 cos θ 1 µ (6.74) a · 1 b · 2 µ − 4  − s − s  q   s 4m2 p k = p k = P K2 + m2 + P K cos θ = 1 + cos θ 1 µ . (6.75) a · 2 b · 1 µ 4  s − s  q   80 Putting these results into eq. (6.64), we get:

1 8e4 s 2 4m2 2 s 2 4m2 2 sm2 2 = 1 + cos θ 1 µ + 1 cos θ 1 µ + µ 4 |M| s2 4 s − s 4 − s − s 2 spinsX ½µ ¶ · ¸ µ ¶ · ¸ ¾ 4m2 4m2 = e4 1+ µ + 1 µ cos2 θ . (6.76) " s Ã − s ! # Finally we can plug this into eq. (6.48): dσ 1 K = 2 (6.77) d(cos θ) 4 |M| 32πsP spinsX e4 4m2 4m2 4m2 = 1 µ 1+ µ + 1 µ cos2 θ , (6.78) 32πss − s " s Ã − s ! # or, rewriting in terms of the ﬁne structure constant α = e2/4π,

dσ πα2 4m2 4m2 4m2 = 1 µ 1+ µ + 1 µ cos2 θ . (6.79) d(cos θ) 2s s − s " s Ã − s ! #

1 1 2 Doing the integral over cos θ using 1 d(cos θ) = 2 and 1 cos θd(cos θ) = 2/3, we ﬁnd the total cross − − section: R R

4πα2 4m2 2m2 σ = 1 µ 1+ µ . (6.80) 3s s − s Ã s !

2 It is a useful check that the cross section has units of area; recall that when c =h ¯ = 1, then s = ECM 2 2 has units of mass or length− . Equations (6.79) and (6.80) have been tested in many experiments, and correctly predict the rate of production of muon-antimuon pairs at electron-positron colliders. Let us examine some special limiting + cases. Near the energy threshold for µ µ− production, one may expand in the quantity

∆E = √s 2m (6.81) − µ To leading order in small ∆E, eq. (6.80) becomes

πα2 ∆E σ 2 . (6.82) ≈ 2mµ s mµ The cross section therefore rises like the square root of the energy excess over the threshold. However, going to increasing energy, σ quickly levels oﬀ because of the 1/s factors in eq. (6.80). Maximizing with respect to s, one ﬁnds that the largest cross section in eq. (6.80) is reached for √s = 1+ √21 m µ ≈ q 2.36mµ, and is about

2 2 σmax = 0.54α /mµ = 1000 nb. (6.83)

In the high energy limit √s m , the cross section decreases proportional to 1/s: À µ 4πα2 σ . (6.84) ≈ 3s

81 However, this formula is not good for arbitrarily high values of the center-of-momentum energy, because there is another diagram in which the photon is replaced by a Z0 boson. This eﬀect becomes important when √s is not small compared to mZ = 91.19 GeV.

+ 6.3 e−e ff. −→ + In the last subsection, we calculated the cross section for producing a muon-antimuon pair in e−e collisions. We can easily generalize this to the case of production of any charged fermion f and antifermion f. The Feynman diagram for this process is obtained by simply replacing the muon-antimuon line by an f-f line:

e− f

e+ f

The reduced matrix element for this process has exactly the same form as for e e+ ff, except − −→ + that the photon-µ−-µ vertex is replaced by a photon-f-f vertex, with:

ieγν iQ eγν, (6.85) −→ − f where Qf is the charge of the fermion f. In the case of quarks, there are three indistinguishable colors for each flavor (up, down, strange, charm, bottom, top). The photon-quark-antiquark vertex is diagonal in color, so the three colors are simply summed over in order to find the total cross section for a given flavor. In general, if we call nf the number of colors (or perhaps other non-spin degrees of freedom) of the fermion f, then we have:

2 2 2 e−e+ ff = nf Qf e−e+ µ−µ+ , (6.86) |M → | |M → | 2 where it is understood that mµ should be replaced by mf everywhere in e−e+ ff . It follows that |M → | the diﬀerential and total cross sections for e e+ ff are obtained from eqs. (6.79) and (6.80) by − −→ just multiplying by n Q2 and replacing m m : f f µ → f

2 2 2 2 4πα 4mf 2mf σe−e+ ff = nf Qf 1 1+ . (6.87) → 3s s − s Ã s !

In the high-energy limit √s m , we have: À f 2 2 4πα σe−e+ ff = nf Qf . (6.88) → 3s The following graph compares the total cross section for e+e ff (solid line) as given by eq. (6.87) − → to the asymptotic approximation (dashed line) given by eq. (6.88).

82 2α2 2 0.54 nfQf /mf

0 0 1 2 3 4 5 ECM/mf

We see that the true cross-section is always less than the asymptotic approximation, but the two already agree fairly well when √s > 2.5mf . This means that when several fermions contribute, the ∼ 2 total cross section well above threshold is just equal to the sum of nf Qf for the available states times 2 a factor 4πα /3s. For example, the up quark has charge Qu = +2/3, and there are three colors, so the prefactor indicated above is 3(2/3)2 = 4/3. The prefactors for all the fundamental charged fermion types with masses less than mZ are:

up, charm quarks: Q = 2/3, n = 3 n Q2 = 4/3 f f −→ f f down, strange, bottom quarks: Q = 1/3, n = 3 n Q2 = 1/3 (6.89) f − f −→ f f muon, tau leptons: Q = 1, n = 1 n Q2 = 1. f − f −→ f f However, free quarks are not seen in nature because the QCD color force confines them within color-singlet hadrons. This means that the quark-antiquark production process e e+ QQ cross − −→ section cannot easily be interpreted in terms of specific particles in the final state. Instead, one should view the quark production as a microscopic process, occurring at a distance scale much smaller than a typical hadron. Before we “see” them in macroscopic-sized detectors, the produced quarks then undergo further strong interactions which end up producing hadronic jets of particles with momenta close to those of the original quarks. This always involves at least the further production of a quark- antiquark pair in order to make the final state hadrons color singlets. A Feynman-diagram cartoon of the situation might looks as follows:

83 hadronic jet

e−

hadronic jet

Because the hadronic interactions are most important at the strong-interaction energy scale of a few hundred MeV, the calculation of the cross section can only be trusted for energies that are significantly higher than this. When √s 1 GeV, one can make the approximation: À + + σ(e−e hadrons) σ(e−e qq). (6.90) −→ ≈ q −→ X The final state can be quite complicated, so to test QED production of quarks, one can just measure the total cross section for producing hadrons. The traditional measure of the total hadronic cross section is the variable Rhadrons, defined as the ratio: σ(e e+ hadrons) R = − −→ . (6.91) hadrons σ(e e+ µ µ+) − −→ − When the approximation is valid, one can always produce up, down and strange quarks, which all have masses < 1 GeV. The threshold to produce charm-anticharm quarks occurs roughly when √s> 2m 3 c ≈ GeV, and that to produce bottom-antibottom quarks is at roughly √s > 2m 10 GeV. As each of b ≈ these thresholds is passed, one gets a contribution to Rhadrons which is approximately a constant 2 proportional to nf Qf . So, for √s< 3 GeV, one has 4 1 1 R = + + = 2. (u,d,s quarks) (6.92) hadrons 3 3 3 For 3 GeV< √s< 10 GeV, the charm quark contributes, and the ratio is 4 1 1 4 10 R = + + + = . (u,d,s,c quarks) (6.93) hadrons 3 3 3 3 3 Finally, for √s> 10 GeV, we get 4 1 1 4 1 11 R = + + + + = . (u,d,s,c,b quarks). (6.94) hadrons 3 3 3 3 3 3

84 Besides these “continuum” contributions to Rhadrons, there are resonant contributions that come from e e+ hadronic bound states. These bound states tend to have very large, but very narrow, − → production cross sections when √s is in just the right energy range to produce them. For example, when the bound state consists of a charm and anticharm quark, one gets the J/ψ particle resonance at √s = 3.09687 GeV, with a width of 0.00087 GeV. These resonances contribute very sharp peaks to the measured Rhadrons. Experimentally, Rhadrons is quite hard to measure, being plagued by systematic detector eﬀects. Many of the older experiments at lower energy tended to underestimate the systematic uncertainties. Here is a plot of the data, from the Review of Particle Properties, [D.E. Groom et al., The European Physical Journal C15 (2000) 1.]

5 γγ 2 MARK I {MARK I/LGW 4 MEA R 3

2 J/ψ(1S) ψ(2S)

1 2 3 4 5 6 7

4 R

3 AMY CRYSTAL BALL JADE MARK J TASSO CELLO CUSB LENA MD−1 TOPAZ (nS) VENUS n=1234 CLEO DASP II MAC PLUTO

2 10 15 20 25 30 35 40 45 5055 60

Ecm (GeV)

The approximate agreement with the predictions for Rhadrons in eqs. (6.92)-(6.94) provides a crucial test of the quark model of hadrons, including the charges of the quarks and the number of colors. [This plot is actually somewhat old (1992), and it is now believed that the error bars on many of the data points should actually be larger than shown, and some points perhaps discarded. A more up-to-date plot is found in Peskin and Schroeder, Figure 5.3.]

85 + + 6.4 Helicities in e−e µ−µ → Up to now, we have computed the cross section by averaging over the unknown spins of the initial state + electron and positron. However, some e−e colliders can control the initial spin states, using polarized beams. This means that the beams are arranged to have an excess of either L-handed or R-handed helicity electrons and positrons. Practical realities make it impossible to achieve 100%-pure polarized beams, of course. At a proposed future Linear Collider, it is a very important part of the experimental program to be able to run with at least the electron beam polarized. Present estimates are that one might be able to get 90% or 95% pure polarization for the electron beam (either L or R), with perhaps 60% polarization for the positron beam. This terminology means that when the beam is operating in R mode, then a polarization of X% implies that

= X% (6.95) PR − PL where and are the probabilities of measuring the spin pointing along and against the 3- PR PL momentum direction, respectively. This experimental capability shows that one needs to be able to calculate cross sections without assuming that the initial spin state is random and can be averaged over. We could redo the calculation of the previous subsections with particular spinors u(p, s) and v(p, s) for the desired speciﬁc spin states s of initial state particles. However, then we would lose our precious trick of evaluating s uu and s vv. A nicer way is to keep the sum over spins, but eliminate the “wrong” polarizationP from the sumP using a projection matrix. So, for example, we can use

P u(p, s) L initial-state particle (6.96) L ←→ P u(p, s) R initial-state particle (6.97) R ←→ in place of the usual Feynman rules for an initial state particle. Summing over the spin s will not change the fact that the projection matrix allows only L- or R-handed electrons to contribute to the cross section. Now our traces over gamma matrices will involve γ5, because of the explicit expressions for PL and PR [see eq. (2.135)]. To get the equivalent rules for an initial state antiparticle, we must remember that the spin operator acting on v(p, s) spinors is the opposite of the spin operator acting on u(p, s) spinors. Therefore, PL acting on a v(p, s) spinor projects onto a R-handed antiparticle. So if we form the object v(p, s)PR = 0 0 v†(p, s)γ PR = v†(p, s)PLγ , the result must describe a R-handed positron; in this case, the bar on the spinor for an antiparticle “corrects” the handedness. So, for an initial state antiparticle with either L or R polarization, we can use:

v(p, s)P L initial-state anti-particle (6.98) L ←→ v(p, s)P R initial-state anti-particle (6.99) R ←→

86 In some cases, one can also measure the polarizations of outgoing particles, for example by observing their decays. Tau leptons and anti-taus decay by the weak interaction processes:

τ − e−ν ν (6.100) −→ τ e τ + e+ν ν , (6.101) −→ τ e with the angular distributions of the ﬁnal state directions depending on the spin of the µ. So, in general we also need to use:

u(p, s)P L final-state particle, (6.102) R ←→ u(p, s)P R final-state particle, (6.103) L ←→ P v(p, s) L final-state anti-particle, (6.104) R ←→ P v(p, s) R final-state anti-particle, (6.105) L ←→ in order to be able to calculate cross-sections to final states with specific L or R spin polarization states. (Note that as a result of the barred spinor notation, the general rule is that the projection matrix in an initial state has the same handedness as the incoming particle or antiparticle, while the projection matrix in a final state has the opposite handedness of the particle being produced.) As an example, let us consider the process:

+ + e−e µ−µ , (6.106) R R −→ where the helicities of the initial state particles are now assumed to be known perfectly. The reduced matrix element for this process, following from the same Feynman diagram as before, is: e2 = i (v P γµP u ) (u γ v ). (6.107) M s b R R a 2 µ 1 As you should have discovered in Homework Set 2, a projection matrix can be moved through a gamma matrix by changing L R: ↔ µ µ PRγ = γ PL (6.108) µ µ PLγ = γ PR (6.109)

µ This follows from the fact that γ5 anticommutes with γ : 1+ γ 1 γ P γµ = 5 γµ = γµ − 5 = γµP . (6.110) R 2 2 L µ ¶ µ ¶ Therefore, eq. (6.107) simply vanishes, because

µ µ PRγ PR = γ PLPR = 0. (6.111)

+ + + + Therefore, the process e−e µ µ does not occur in QED. Similarly, e−e µ µ does not R R −→ − L L −→ − + occur in QED. By the same type of argument, µ− and µ in the ﬁnal state must have opposite L,R polarizations from each other in QED.

87 To study a non-vanishing reduced matrix element, let us therefore consider the process:

+ + e−e µ−µ , (6.112) L R −→ L R in which we have now assumed that all helicities are perfectly known. To simplify matters, we will assume the high energy limit √s m . The reduced matrix element can be simply obtained from À µ eq. (6.42) by just putting in the appropriate L and R projection matrices acting on each external state spinor:

e2 = i (v P γµP u )(u P γ P v ). (6.113) M s b R L a 1 R µ L 2 This can be simpliﬁed slightly by using the properties of the projections matrices:

µ µ µ PRγ PL = γ PLPL = γ PL, (6.114) so that e2 = i (v γµP u )(u γ P v ), (6.115) M s b L a 1 µ L 2 and so 4 2 e µ ν = (v γ P u )(u γ P v )(v γ P u )∗(u γ P v )∗. (6.116) |M| s2 b L a 1 µ L 2 b L a 1 ν L 2 To evaluate this, we compute:

ν 0 ν ν 0 (vbγ PLua)∗ = (vb†γ γ PLua)∗ = ua† (PL)†(γ )†γ vb (6.117) 0 ν = ua† PLγ γ vb (6.118) 0 ν = ua† γ PRγ vb (6.119) ν = uaPRγ vb. (6.120)

ν 0 0 ν 0 0 0 Here we have used the facts that (PL)† = PL, and (γ )†γ = γ γ , and PLγ = γ PR, and ua† γ = ua. In a similar way,

(u1γνPLv2)∗ = v2PRγνu1. (6.121)

Therefore,

e4 2 = (v γµP u )(u P γνv )(u γ P v )(v P γ u ). (6.122) |M| s2 b L a a R b 1 µ L 2 2 R ν 1 Because the spin projection matrices will only allow the speciﬁed set of spins to contribute anyway, we are free to sum over the spin labels sa, sb, s1, and s2. Let us do so, since it allows us to apply the tricks

uaua = p/a + me (6.123) s Xa v2v2 = k/2 mµ. (6.124) s − X2 88 Neglecting the masses because of the high-energy limit, we therefore get

e4 2 µ ν / = 2 (vbγ PLp/aPRγ vb) (u1γµPLk2PRγνu1) (6.125) |M| s s s X1 Xb This can be simpliﬁed by eliminating excess projection matrices, using:

PLp/aPR = p/aPRPR = p/aPR, (6.126)

PLk/2PR = k/2PRPR = k/2PR, (6.127) to get

e4 2 µ ν / = 2 (vbγ p/aPRγ vb) (u1γµk2PRγνu1). (6.128) |M| s s s X1 Xb Again using the trick of putting the barred spinor at the end and taking the trace [see the discussion around eq. (6.52)] for each quantity in parentheses, this becomes:

e4 2 µ ν / = 2 Tr[γ p/aPRγ vbvb] Tr[γµk2PRγνu1u1]. (6.129) |M| s s s X1 Xb

Doing the sums over s1 and sb using the usual trick gives:

4 2 e µ ν = Tr[γ p/ PRγ p/ ] Tr[γµk/ PRγνk/ ]. (6.130) |M| s2 a b 2 1 Now it is time to evaluate the traces. We have

µ ν µ 1+ γ5 ν Tr[γ p/ PRγ p/ ] = Tr[γ p/ γ p/ ] (6.131) a b a 2 b µ ¶ 1 µ ν 1 µ ν = Tr[γ p/ γ p/ ]+ Tr[γ p/ γ p/ γ5], (6.132) 2 a b 2 a b where we have used the fact that γ5 anticommutes with any gamma matrix to rearrange the order in the last term. The ﬁrst of these traces was evaluated in section 6.2. The trace involving γ5 is, from eq. (B.17):

µ ν µ α ν β Tr[γ p/aγ p/bγ5] = paαpbβTr[γ γ γ γ γ5] (6.133) µανβ = paαpbβ(4i² ). (6.134) where ²µανβ is the totally antisymmetric Levi-Civita tensor deﬁned in eq. (1.41). Putting things together:

µ ν µ ν µν µ ν µανβ Tr[γ p/ PRγ p/ ] = 2 p p g (pa pb)+ p p + ipaαpbβ² . (6.135) a b a b − · b a h i In exactly the same way, we get

Tr[γ k/ P γ k/ ]=2[k k g (k k )+ k k + ikρkσ² ] . (6.136) µ 2 R ν 1 2µ 1ν − µν 2 · 1 1µ 2ν 2 1 µρνσ

89 Finally, we have to multiply these two traces together, contracting the indices µ and ν. Note that the cross-terms containing only one ² tensor vanish, because the epsilon tensors are antisymmetric under µ ν, while the other terms are symmetric. The term involving two epsilon tensors can be evaluated ↔ using the useful identity

²µανβ² = 2δαδβ + 2δαδβ, (6.137) µρνσ − ρ σ σ ρ which you can verify by brute force substitution of indices. The result is simply:

µ ν Tr[γ p/ PRγ p/ ] Tr[γµk/ PRγνk/ ] = 16(pa k2)(pb k1), (6.138) a b 2 1 · · so that 16e4 2 = (p k )(p k ). (6.139) |M| s2 a · 2 b · 1 This result should be plugged in to the formula for the diﬀerential cross section:

dσ − + − + e e µ µ ~k1 L R→ L R = 2 | | (6.140) d(cos θ) |M| 32πs ~p | a| (Note that one does not average or sum over spins in this case, because they have already been fixed!) The kinematics is of course not affected by the fact that we have fixed the helicities, and so can be taken from the discussion in 6.2 with mµ replaced by 0. It follows that:

− − 4 dσe e+ µ µ+ e L R→ L R = (1 + cos θ)2 (6.141) d(cos θ) 32πs πα2 = (1 + cos θ)2. (6.142) 2s The angular dependence of this result can be understood from considering the conservation of angular momentum in the event. Drawing a short arrow to represent the direction of the spin:

µL−

+ eL− θ eR

+ µR

This shows that the total spin angular momentum of the initial state is S = 1 (taking the electron to zˆ − be moving in the +z direction). The total spin angular momentum of the ﬁnal state is S = 1, where nˆ − nˆ is the direction of the µ . This explains why the cross section vanishes if cos θ = 1; that corresponds − − to a ﬁnal state with the total spin angular momentum in the opposite direction from the initial state. The quantum mechanical overlap for two states with measured angular momenta in exactly opposite

90 directions must vanish. If we describe the initial and ﬁnal states as eigenstates of angular momentum with J = 1:

2 dσe−e+ µ−µ+ πα R L → R L = (1 + cos θ)2, (6.146) d(cos θ) 2s corresponding to the picture: µR−

+ eR− θ eL

+ µL with all helicities reversed compared to the previous case. If we compute the cross sections for the ﬁnal state muon to have the opposite helicity from the initial state electron, we get

2 dσe−e+ µ−µ+ dσe−e+ µ−µ+ πα L R→ R L = R L → L R = (1 cos θ)2, (6.147) d(cos θ) d(cos θ) 2s − corresponding to the pictures: µR− µL−

+ + eL− θ eR eR− θ eL

+ + µL µR

These are 4 of the possible 24 = 16 possible helicity conﬁgurations for e e+ µ µ+. However, as we − → − have already seen, the other 12 possible helicity combinations all vanish, because they contain either + + e− and e with the same helicity, or µ− and µ with the same helicity. If we take the average of the initial state helicities, and the sum of the possible ﬁnal state helicities, we get:

1 dσe−e+ µ−µ+ dσe−e+ µ−µ+ dσe−e+ µ−µ+ dσe−e+ µ−µ+ L R→ L R + R L → R L + L R→ R L + R L → L R + 12 0 4 " d(cos θ) d(cos θ) d(cos θ) d(cos θ) · # 1 πα2 = 2(1 + cos θ)2 + 2(1 cos θ)2 (6.148) 4 Ã 2s ! − h i πα2 = (1 + cos2 θ), (6.149) 2s

91 in agreement with the √s m limit of eq. (6.79). À µ + + The vanishing of the cross sections for eL−eL and eR−eR in the above process can be generalized beyond this example and even beyond QED. Consider any ﬁeld theory in which interactions are given by a fermion-antifermion-vector vertex with a Feynman rule proportional to a gamma matrix γµ. If an initial state fermion and antifermion merge into a vector, or a vector splits into a ﬁnal state fermion and antifermion:

then by exactly the same argument as before, the fermion and antifermion must have opposite helicities, µ µ µ µ because of vPLγ PLu = vPRγ PRu = 0 and uPLγ PLv = uPRγ PRv = 0 and the rules of eqs. (6.96)- (6.99) and (6.102)-(6.105). Moreover, if an initial state fermion (or anti-fermion) interacts with a vector and emerges as a ﬁnal state fermion (or anti-fermion):

µ then the fermions (or anti-fermions) must have the same helicity, because of the identities uPLγ PLu µ µ µ = uPRγ PRu = 0 and vPLγ PLv = vPRγ PRv = 0. This is true even if the interaction with the vector changes the fermion from one type to another. These rules embody the concept of helicity conservation in high energy scattering. They are obviously useful when the helicities of the particles are controlled or measured by the experimenter. They are also useful because, as we will see, the weak interactions only aﬀect fermions with L helicity and antifermions with R helicity. The conservation of angular momentum together with helicity conservation often allows one to know in which direction a particle is most likely to emerge in a scattering or decay experiment, and in what cases one may expect the cross section to vanish or be enhanced.

+ + 6.5 Bhabha scattering (e−e e−e ) → In this subsection we consider the process of Bhabha scattering:

+ + e−e e−e . (6.150) →

92 For simplicity we will only consider the case of high-energy scattering, with √s = E m , and CM À e we will consider all spins to be unknown (averaged over in the initial state, summed over in the ﬁnal state). Label the momentum and spin data as follows:

Particle Momentum Spin Spinor e− pa sa u(pa, sa) + e pb sb v(pb, sb) (6.151) e− k1 s1 u(k1, s1) + e k2 s2 v(k2, s2)

At order e2, there are two Feynman diagram for this process:

pa k1 pa k1 µ pa + pb and p k µ ν a − 1 ν pb k2 pb k2

The first of these is called the s-channel diagram; it is exactly the same as the one we drew for e e+ µ µ+. The second one is called the t-channel diagram. Using the QED Feynman rules listed − → − at the end of subsection 6.1, the corresponding contributions to the reduced matrix element for the process are: ig = [v (ieγµ)u ] − µν [u (ieγν)v ] , (6.152) s b a (p + p )2 1 2 M · a b ¸ and ig = ( 1) [u (ieγµ)u ] − µν [v (ieγν)v ] . (6.153) Mt − 1 a (p k )2 b 2 · a − 1 ¸ The additional ( 1) factor in is due to Rule 9 in the QED Feynman rules. It arises because the − Mt order of spinors in the written expression for is b,a, 1, 2, but that in is 1,a,b, 2, and these differ Ms Mt from each other by an odd permutation. We could have just as well assigned the minus sign to Ms instead; only the relative phases of terms in the matrix element are significant. Therefore the full reduced matrix element for Bhabha scattering, written in terms of the Mandelstam variables s = (p + p )2 and t = (p k )2, is: a b a − 1 1 1 = + = ie2 (v γ u )(u γµv ) (u γ u )(v γµv ) . (6.154) s t s b µ a 1 2 t 1 µ a b 2 M M M ½ − ¾

93 Taking the complex conjugate of this gives:

∗ = ∗ + ∗ (6.155) M Ms Mt 2 1 ν 1 ν = ie (v γ u )∗(u γ v )∗ (u γ u )∗(v γ v )∗ (6.156) s b ν a 1 2 t 1 ν a b 2 − ½ − ¾ 1 1 = ie2 (u γ v )(v γνu ) (u γ u )(v γνv ) . (6.157) s a ν b 2 1 t a ν 1 2 b − ½ − ¾ The complex square of the reduced matrix element, 2 = , contains a pure s-channel piece |M| M∗M proportional to 1/s2, a pure t-channel piece proportional to 1/t2, and an interference piece proportional to 1/st. For organizational purposes, it is useful to calculate these pieces separately. The pure s-channel contribution calculation is exactly the same as what we did before for e e+ − → µ µ+, except that now we can substitute m m 0. Therefore, plagiarizing the result of eq. (6.64), − µ → e → we have: 1 8e4 2 = [(p k )(p k ) + (p k )(p k )] . (6.158) 4 |Ms| s2 a · 2 b · 1 a · 1 b · 2 spinsX The pure t-channel contribution can be calculated in a very similar way. We have:

e4 2 = (v γ v )(v γ v )(u γµu )(u γνu ). (6.159) |Mt| t2 2 ν b b µ 2 1 a a 1 Taking the average of initial state spins and the sum over ﬁnal state spins allows us to use the identities u u = p and v v = p (neglecting m ). The result is: sa a a /a sa b b /b e

P P 4 1 2 e µ ν t = 2 (v2γνp/bγµv2)(u1γ p/aγ u1) (6.160) 4 |M | 4t s ,s spinsX X1 2 4 e µ ν = 2 Tr[γνp/bγµv2v2]Tr[γ p/aγ u1u1], (6.161) 4t s ,s X1 2 in which we have turned the quantity into a trace by moving the u1 to the end. Now performing the sums over s1, s2 gives:

4 1 2 e µ ν t = Tr[γνp/ γµk/ ]Tr[γ p/ γ k/ ] (6.162) 4 |M | 4t2 b 2 a 1 spinsX 4 e µ ν = Tr[γµk/ γνp/ ]Tr[γ p/ γ k/ ]. (6.163) 4t2 2 b a 1 In the second line, the ﬁrst trace has been rearranged using the cyclic property of traces. The point of doing so is that now these traces have exactly the same form that we encountered in eq. (6.58), but with p k and m 0. Therefore we can obtain 1 2 by simply making the same a ↔ 2 µ → 4 spins |Mt| replacements pa k2 and mµ 0 in eq. (6.64): P ↔ → 1 8e4 2 = [(p k )(p k ) + (k k )(p p )] . (6.164) 4 |Mt| t2 a · 2 b · 1 2 · 1 b · a spinsX

94 Next, consider the interference term:

4 1 e ν µ ∗ = (v γ u )(u γ u )(u γ v )(v γ u ). (6.165) 4 Mt Ms −4st b µ a a 1 1 2 2 ν b spinsX spinsX We chose to write the factors parentheses in that order, so that now we can use the tricks u u = p , sa a a /a / / and s1 u1u1 = k1, and s2 v2v2 = k2, to obtain: P P P1 e4 ν / µ/ t∗ s = (vbγµp/aγ k1γ k2γνvb) (6.166) 4 M M −4st s spinsX Xb which can now be converted into a trace by the usual trick of moving the vb to the end: 1 e4 ν / µ/ t∗ s = Tr[γµp/aγ k1γ k2γνvbvb] (6.167) 4 M M −4st s spinsX Xb 4 e ν µ = Tr[γµp/ γ k/ γ k/ γνp/ ]. (6.168) −4st a 1 2 b Now we are faced with the task of computing the trace of 8 gamma matrices. In principle, the trace of any number of gamma matrices can be performed with the algorithm of eq. (B.10). The procedure is to replace the trace over 2n gamma matrices by a sum over traces of 2n 2 gamma matrices, and − repeat until all traces are short enough to evaluate using eq. (B.8). However, in our present case it is easier to simplify the contents of the trace before evaluating it. So, we use eq. (B.21) to write:

ν µ ν γµp/ γ k/ γ = 2k/ γ p/ , (6.169) a 1 − 1 a which implies that

ν µ ν Tr[γµp/ γ k/ γ k/ γνp/ ]= 2Tr[k/ γ p/ k/ γνp/ ]. (6.170) a 1 2 b − 1 a 2 b This can be further simpliﬁed by now using eq. (B.20) to write:

ν γ p/ k/ γν = 4pa k2, (6.171) a 2 · so that:

ν µ Tr[γµp/ γ k/ γ k/ γνp/ ] = 8(pa k2)Tr[k/ p/ ] (6.172) a 1 2 b − · 1 b = 32(p k )(k p ), (6.173) − a · 2 1 · b in which the trace has ﬁnally been performed using eq. (B.8). So, from eq. (6.168)

1 8e4 ∗ = (p k )(p k ). (6.174) 4 Mt Ms st a · 2 b · 1 spinsX Taking the complex conjugate of both sides, we also have:

1 8e4 ∗ = (p k )(p k ). (6.175) 4 MsMt st a · 2 b · 1 spinsX

95 This completes the contributions to the total

1 2 1 2 2 = + + ∗ + ∗ . (6.176) 4 |M| 4 |Ms| |Mt| Mt Ms MsMt spinsX spinsX h i It remains to identify the dot products of momenta appearing in the above formulas. This can be done by carrying over the kinematic analysis for the case e e+ µ µ+ as worked out in eqs. (6.65)-(6.75), − → − with of course m m . Letting θ be the angle between the 3-momenta directions of the initial state µ → e electron and the ﬁnal state electron, we have:

p p = k k = s/2, (6.177) a · b 1 · 2 p k = p k = t/2, (6.178) a · 1 b · 2 − p k = p k = u/2, (6.179) a · 2 b · 1 − with s t = (1 cos θ) (6.180) −2 − s u = (1 + cos θ). (6.181) −2 It follows from eqs. (6.158), (6.164), (6.174), and (6.175) that:

1 2e4 2 = (u2 + t2) (6.182) 4 |Ms| s2 spinsX 1 2e4 2 = (u2 + s2) (6.183) 4 |Mt| t2 spinsX 4 1 4e 2 ( ∗ + ∗ ) = u . (6.184) 4 Mt Ms MsMt st spinsX Putting this into eq. (5.80), since ~p = ~k , we obtain the spin-averaged diﬀerential cross-section | a| | 1| for Bhabha scattering:

dσ e4 u2 + t2 u2 + s2 2u2 = + + (6.185) d(cos θ) 16πs ( s2 t2 st ) πα2 u2 + t2 u2 + s2 2u2 = + + , (6.186) s ( s2 t2 st ) This result actually diverges for cos θ 1, because of the t’s in the denominator. In that limit, the → diﬀerential cross section blows up quadratically,

dσ 8πα2 . (6.187) d(cos θ) ∼ s(1 cos θ)2 − This is not an integrable singularity:

1 dσ d(cos θ) . (6.188) 1 d(cos θ) −→ ∞ Z− 96 The infinite total cross section corresponds to the infinite range of the Coulomb potential between two charged particles. It arises entirely from the t-channel diagram, in which the electron and positron scatter off of each other in the forward direction (θ 0). It simply reflects that an infinite-range ≈ interaction will always produce some deflection, although it may be extremely small. This result is the relativistic generalization of the non-relativistic, classical Rutherford scattering problem, in which an electron or alpha particle (or some other light charged particle) scatters off of the classical electric field of a heavy nucleus. As worked out in many textbooks on classical physics (for example, H. Goldstein’s Classical Mechanics, J.D. Jackson’s Classical Electrodynamics), the differential cross section for a non- relativistic light particle with charge QA and a heavy particle with charge QB, with center-of-momentum energy ECM to scatter through their Coulomb interaction is:

dσ πQ2 Q2 α2 Rutherford = A B . (6.189) d(cos θ) 2E2(1 cos θ)2 − (Here one must be careful in comparing results, because the charge e used by Goldstein and Jackson diﬀers from the one used here by a factor of √4π.) Comparing the non-relativistic Rutherford result to the relativistic Bhabha result, we see that in both cases the small-angle behavior scales like 1/θ4, and does not depend on the signs of the charges of the particles. In a real experiment, there is always some minimum scattering angle that can be resolved. In a colliding-beam experiment, this is usually dictated by the fact that detectors cannot be placed within or too close to the beamline. In other experiments, one is limited by the angular resolution of detectors. Therefore, the true observable quantity is typically something more like:

cos θmin dσ σexperiment = . (6.190) cos θ d(cos θ) Z− min Of course, in the Real World, the minimum angle is just one of many practical factors which have to be included. In terms of the Feynman diagram interpretation, the divergence for small θ corresponds to the photon propagator going on-shell; in other words, the situation where the square of the t-channel virtual photon’s 4-momentum is nearly equal to 0, the classical value for a real massless photon. For any scattering angle θ> 0, one has s (p k )2 = t = (1 cos θ) > 0, (6.191) a − 1 2 − so that the virtual photon is said to be off-shell. In general, any time that a virtual (internal line) particle can go on-shell, there will be a divergence in the cross section due to the denominator of the Feynman propagator blowing up. Sometimes this is a real divergence with a physical interpretation, as in the case of Bhabha scattering. In other cases, the divergence is removed by higher-order effects, such as the finite life-time of the virtual particle, which will give an imaginary part to its squared mass, removing the singularity in the Feynman propagator.

97 6.6 Crossing symmetry

Consider the two completely diﬀerent processes:

+ + e−e µ−µ (6.192) −→ + + e−µ e−µ . (6.193) −→ The two Feynman diagrams are very similar:

e− µ− e− e−

and

e+ µ+

µ+ µ+

In fact, by stretching and twisting, one can turn the ﬁrst process into the second by the transformation:

+ initial e final e− (6.194) → + final µ− initial µ , (6.195) → Two processes related to each other by exchanging some initial state particles with their antiparticles in the final state are said to be related by crossing. Not surprisingly, the reduced matrix elements for these processes are also quite similar. For example, in the high-energy limit √s m , À µ 2 2 1 2 4 u + t e−e+ µ−µ+ = 2e (6.196) 4 |M → | Ã s2 ! spinsX 2 2 1 2 4 u + s e−µ+ e−µ+ = 2e . (6.197) 4 |M → | Ã t2 ! spinsX This similarity is generalized and made more precise by the following theorem.

Crossing Symmetry Theorem: Suppose two Feynman diagrams with reduced matrix elements and are related by the exchange (“crossing”) of some initial state particles and M M0 antiparticles for the corresponding ﬁnal state antiparticles and particles. If the crossed particles have 4-momenta P µ,...P µ in , then 2 can be obtained by substituting 1 n M spins |M| µ µ 2 P 0 = P into the mathematical expression forP 0 : i − i spins |M | P µ µ 2 F µ µ 2 (P1 ,...,Pn ,...) = ( 1) 0(P10 ,...,Pn0 ,...) , (6.198) spins |M | − spins |M | ¯ 0µ µ ¯Pi Pi X X ¯ →− ¯ with the other (uncrossed) particle 4-momenta unaﬀected. Here F is the¯ number of fermion lines that were crossed.

98 If pµ = (E, ~p) is a valid physical 4-momentum, then p µ = ( E, ~p) is never a physical 4-momentum, 0 − − since it has negative energy. So the relation between the two diagrams is a formal one rather than a relation between physical reduced matrix elements that would actually be measured; when one of the matrix elements involves the 4-momenta appropriate for a physical process, the other one does not. However, it is still perfectly valid as a mathematical relation, and therefore extremely useful. In general, if one has calculated the cross section or reduced matrix element for one process, one can obtain the results for several “crossed” processes by merely substituting momenta, with no additional labor. Equation (6.198) might look dubious at ﬁrst, since it may look like the right-hand side is negative for odd F . However, the point is that when the expression for 2 is analytically continued to unphysical |M0| momenta, it always ﬂips sign for odd F . We will see this in some examples later.

+ + 6.7 e−µ e−µ → Let us apply the crossing symmetry to the example of eqs. (6.192) and (6.193) by assigning primed momenta to the Feynman diagram for e e+ µ µ+: − → − e p initial state − ↔ a0 e+ p ( ↔ b0 µ k ﬁnal state − ↔ 10 µ+ k ( ↔ 20 Label the momenta in the crossed Feynman diagram (for e µ+ e µ+) without primes: − → − e p initial state − ↔ a µ+ p ( ↔ b e k ﬁnal state − ↔ 1 µ+ k . ( ↔ 2 Then the Crossing Symmetry Theorem tells us that we can get the reduced matrix element for the process e µ+ e µ+ as a function of physical momenta p ,p , k , k by substituting unphysical − → − a b 1 2 momenta

p0 = p ; p0 = k ; k0 = p ; k0 = k , (6.199) a a b − 1 1 − b 2 2 into the formula for the reduced matrix element for the process e e+ µ µ+. This means that we − → − can identify:

2 2 s0 = (p0 + p0 ) = (p k ) = t (6.200) a b a − 1 2 2 t0 = (p0 k0 ) = (p + p ) = s (6.201) a − 1 a b 2 2 u0 = (p0 k0 ) = (p k ) = u. (6.202) a − 2 a − 2 In other words, crossing symmetry tells us that the formulas for the reduced matrix elements for these two processes are just related by the exchange of s and t, as illustrated in the high-energy limit in eqs. (6.196) and (6.197).

99 We can carry this further by considering another related process:

e−µ− e−µ−, (6.203) → with physical momenta: e p initial state − ↔ a µ p ( − ↔ b e k ﬁnal state − ↔ 1 . µ k ( − ↔ 2 This time, the Crossing Symmetry Theorem tells us that we can identify the matrix element by again starting with the diagram for e e+ µ µ+ and replacing: − → −

p0 = p ; p0 = k ; k0 = k ; k0 = p , (6.204) a a b − 1 1 2 2 − b so that:

2 2 s0 = (p0 + p0 ) = (p k ) = t, (6.205) a b a − 1 2 2 t0 = (p0 k0 ) = (p k ) = u, (6.206) a − 1 a − 2 2 2 u0 = (p0 k0 ) = (p + p ) = s. (6.207) a − 2 a b Here the primed Mandelstam variable are the unphysical ones for the e e+ µ µ+ process, and the − → − unprimed ones are for the desired process e µ e µ . We can therefore infer, from eq. (6.196), − − → − − that 2 2 1 2 4 s + u e−µ− e−µ− = 2e , (6.208) 4 |M → | Ã t2 ! spinsX by doing the substitutions indicated by eqs. (6.205)-(6.207).

6.8 Moller scattering (e−e− e−e−) → As another example, let us consider the case of Moller scattering:

e−e− e−e−. (6.209) → There are two Feynman diagrams for this process:

e− e− e− e−

and

e− e− e− e−

100 Now, nobody can stop us from getting the result for this process by applying the Feynman rules to get the reduced matrix element, taking the complex square, summing and averaging over spins, and computing the Dirac traces. However, an easier way is to note that this is a crossed version of Bhabha scattering, which we studied last time. Making a table of the momenta:

Bhabha Moller

e− p0 e− p ↔ a ↔ a + e p0 e− p ↔ b ↔ b e− k0 e− k ↔ 1 ↔ 1 + e k0 e− k , ↔ 2 ↔ 2 we see that crossing symmetry allows us to compute the Moller scattering by identifying the (initial state positron, ﬁnal state positron) in Bhabha scattering with the (ﬁnal state electron, initial state electron) in Moller scattering, so that:

p0 = p p0 = k k0 = k k0 = p . (6.210) a a b − 2 1 1 2 − b So, the Moller scattering reduced matrix element is obtained by putting

2 2 s0 = (p0 + p0 ) = (p k ) = u (6.211) a b a − 2 2 2 t0 = (p0 k0 ) = (p k ) = t (6.212) a − 1 a − 1 2 2 u0 = (p0 k0 ) = (p + p ) = s (6.213) a − 2 a b into the corresponding result for Bhabha scattering. Using the results of eq. (6.182)-(6.184), we therefore get:

2 2 2 2 2 1 2 2 s + t s + u 2s e−e− e−e− = 2e + + . (6.214) 4 |M → | ( u2 t2 ut ) spinsX Applying eq. (5.80), we ﬁnd the diﬀerential cross section:

dσe−e− e−e− 1 1 2 → = e−e− e−e− (6.215) d(cos θ) 32πs 4 |M → |  spinsX πα2 s2 + t2 s2 + u2 2s2 = + + . (6.216) s ( u2 t2 ut ) Note that kinematic factors involved in turning the squared matrix element into a cross section, like the 1/32πs in this case, are not determined by crossing.

6.9 Gauge invariance in Feynman diagrams

Let us now turn to the issue of gauge invariance as it is manifested in QED Feynman diagrams. Recall that when we found the Feynman propagator for a photon, it contained a term which depended on

101 an arbitrary parameter ξ. We have been working with ξ = 1 (Feynman gauge). Consider what the matrix element for the process e e+ µ µ+ would be if instead we let ξ remain unﬁxed. Instead of − → − eq. (6.41), we would have ie2 (p + p ) (p + p ) = (v γµu )(u γνv ) g + (1 ξ) a b µ a b ν . (6.217) b a 1 2 (p + p )2 µν (p + p )2 M a b ·− − a b ¸ If the answer is to be independent of ξ, then it must be true that the new term proportional to (1 ξ) − gives no contribution. This can be easily proved by observing that it contains the factor

µ (vbγ ua)(pa + pb)µ = vbp/ ua + vbp/ ua = mvbua mvbua = 0. (6.218) a b −

Here we have applied the Dirac equation, as embodied in eqs. (B.22) and (B.23), to write p/aua = mua and vbp/ = mp/ . For any photon propagator connected (at either end) to an external fermion line, the b − b proof is similar. And, in general, one can show that the 1 ξ term will always cancel when one includes − all Feynman diagrams contributing to a particular process. So we can choose the most convenient value of ξ, which is usually ξ = 1. Another aspect of gauge invariance involves a feature that we have not explored in an example so µ µ far: external state photons. Recall that the Feynman rules associate factors of ² (p, λ) and ²∗ (p, λ) to initial or ﬁnal state photons, respectively. Now, making a gauge transformation on the photon ﬁeld results in:

Aµ Aµ + ∂µλ (6.219) → where λ is any function. In momentum space, the derivative ∂µ is proportional to pµ. So, in terms of the polarization vector for the electromagnetic ﬁeld, a gauge transformation is

²µ(p, λ) ²µ(p, λ)+ apµ (6.220) → where a is any quantity. [As a consistency check, note that the original polarization vector and momentum for a physical photon satisfy ²2 = 1 and ² p = 0 and p2 = 0; these are also obeyed by − · the polarization vectors after the gauge transformation.] Gauge invariance therefore implies that the reduced matrix element should be unchanged after the substitution in eq. (6.220). The reduced matrix element for a process with an external state photon can always be written in the form:

= ²µ(p, λ), (6.221) M Mµ which deﬁnes . Since must be invariant under a gauge transformation, it follows from eq. (6.220) Mµ M that

pµ = 0. (6.222) Mµ This relation is known as the Ward identity for QED. It says that if we replace the polarization vector for any photon by the momentum of that photon, then the reduced matrix element should become 0. This is a nice consistency check on calculations.

102 Another nice consequence of the Ward identity is that it provides for a simpliﬁed way to sum or average over unmeasured photon polarization states. Consider a photon with momentum taken to be along the positive z axis, with pµ = (P, 0, 0, P ). Summing over the two polarization vectors in eqs. (6.26)-(6.27), we have:

2 2 2 µ ν 2 2 = ∗² (p, λ)²∗ (p, λ)= + . (6.223) |M| MµMν |M1| |M2| λX=1 λX=1 The Ward identity implies that P + P = 0, so 2 = 2. Therefore, we can write: M0 M3 |M0| |M3| 2 2 2 2 2 2 µν = + + + = g ∗. (6.224) |M| −|M0| |M1| |M2| |M3| − MµMν λX=1 The last expression is written as a Lorentz scalar, so it is true for any momentum, not just momenta oriented along the z direction. Gauge invariance, as expressed by the Ward identity, therefore implies that we can always sum over a photon’s polarization states by the rule: 2 µ ν µν µν ² (p, λ) ²∗ (p, λ) = g + (irrelevant) , (6.225) − λX=1 as long as we are taking the sum of the complex square of a reduced matrix element. Although the (irrelevant)µν part is non-zero, it must vanish when contracted with , according to eq. (6.224). MµMν∗

6.10 Compton scattering (γe− γe−) → As our ﬁrst example of a process with external-state photons, consider Compton scattering:

γe− γe−. (6.226) → First let us assign the following labels to the external states: µ initial γ ² (pa, λa) initial e− u(pb, sb) ν ﬁnal γ ² ∗(k1, λ1) ﬁnal e− u(k2, s2). There are two Feynman diagrams for this process:

and

which are s-channel and u-channel, respectively. Applying the QED Feynman rules, we obtain:

ν i(p/a + p/b + m) µ = u (ieγ ) (ieγ )u ²∗ ² (6.227) Ms 2 (p + p )2 m2 b 1ν aµ " a b − # 2 e ν µ = i u2γ (p/ + p/ + m)γ ub ²∗ ²aµ (6.228) − s m2 a b 1ν − h i 103 and / µ i(p/b k1 + m) ν = u (ieγ ) − (ieγ )u ²∗ ² (6.229) Mu 2 (p k )2 m2 b 1ν aµ " b − 1 − # 2 e µ ν = i u2γ (p/ k/ + m)γ ub ²∗ ²aµ. (6.230) − u m2 b − 1 1ν − h i Before squaring the total reduced matrix element, it is useful to simplify. So we note that:

µ µ µ (p/ + m)γ ub = p/ ,γ ub γ (p/ m)ub (6.231) b { b } − b − = p γσ,γµ u + 0 (6.232) bσ{ } b σµ = 2pbσg ub, (6.233) µ = 2pb ub, (6.234) where we have used eq. (B.22). So the reduced matrix element is:

2 1 ν µ µ ν 1 µ ν ν µ = ie ²∗ ²aµ u2 γ p/ γ + 2p γ + ( γ k/ γ + 2p γ ) ub. (6.235) M − 1ν s m2 a b u m2 − 1 b · − ³ ´ − ¸

Taking the complex conjugate of eq. (6.235) gives:

2 1 ρ σ ρ σ 1 σ ρ σ ρ ∗ = ie ²1σ²∗ ub (γ p/ γ + 2p γ )+ ( γ k/ γ + 2p γ ) u2. (6.236) M aρ s m2 a b u m2 − 1 b · − − ¸ Now we multiply together eqs. (6.235) and (6.236), and average over the initial photon polarization λa and sum over the ﬁnal photon polarization λ1, using 1 2 1 ² ²∗ = g + irrelevant, (6.237) 2 aµ aρ 2 µρ λXa=1 2 ²1∗ν²1σ = gνσ + irrelevant, (6.238) λX1=1 to obtain: 4 1 2 e 1 ν µ µ ν 1 µ ν ν µ = u2 (γ p/ γ + 2p γ )+ ( γ k/ γ + 2p γ ) ub 2 |M| 2 s m2 a b u m2 − 1 b λXa,λ1 ½ · − − ¸ ¾ 1 1 ub (γµp/ γν + 2pbµγν)+ ( γνk/ γµ + 2pbνγµ) u2 . (6.239) s m2 a u m2 − 1 ½ · − − ¸ ¾ Next we can average over sb, and sum over s2, using the usual tricks:

ubub = p/b + m, (6.240) s Xb 1 u2 ...u2 = Tr[... (k/2 + m)]. (6.241) 2 s X2 The result is a single spinor trace: 4 1 e 1 µ 1 2 = Tr (γνp/ γµ + 2p γν)+ ( γµk/ γν + 2pνγµ) (p/ + m) 4 |M| 4 s m2 a b u m2 − 1 b b spinsX ·½ − − ¾ 1 1 (γµp/ γν + 2pbµγν)+ ( γνk/ γµ + 2pbνγµ) (k/ + m) . (6.242) s m2 a u m2 − 1 2 ½ − − ¾ ¸ 104 Doing this trace requires a little patience and organization, but it is not impossible. The end result can be written most compactly in terms of s m2 p p = − , (6.243) a · b 2 m2 u p k = − . (6.244) b · 1 2 After some calculation, one ﬁnds:

1 p p p k 1 1 1 1 2 2 = 2e4 a · b + b · 1 + 2m2 + m4 . (6.245) 4 |M| "pb k1 pa pb pa pb − pb k1 pa pb − pb k1 # spinsX · · µ · · ¶ µ · · ¶

Equation (6.245) is a Lorentz scalar. We can now find the differential cross section after choosing a reference frame. We will do this first in the center-of-momentum frame, and then redo it in the “lab” frame in which the initial electron is at rest. In the center-of-momentum frame, the kinematics is just like in the case eµ eµ studied in → Homework Set 5. Call the magnitude of the 3-momentum of the photon in the initial state P . Then 2 2 2 2 2 using four-momentum conservation and the on-shell conditions pa = k1 = 0 and pb = k2 = m , and taking the initial state photon momentum to be in the +z direction and the final state photon momentum to make an angle θ with the z-axis, we have:

µ pa = (P, 0, 0, P ) (6.246) pµ = ( P 2 + m2, 0, 0, P ) (6.247) b − µ p k1 = (P, 0, P sin θ, P cos θ) (6.248) kµ = ( P 2 + m2, 0, P sin θ, P cos θ). (6.249) 2 − − p 1

θ a b

The initial and final state photons have the same energy, as do the initial and final state electrons, so define:

Eγ = P, (6.250) 2 2 Ee = P + m . (6.251) p Then we have:

2 s = (Ee + Eγ) , (6.252)

105 p p = E (E + E ), (6.253) a · b γ e γ p k = E (E + E cos θ), (6.254) b · 1 γ e γ ~k | 1| = 1. (6.255) ~p | a| So, applying eq. (5.79), we obtain:

dσ πα2 E + E E + E cos θ cos θ 1 = e γ + e γ + 2m2 − d(cos θ) s "Ee + Eγ cos θ Ee + Eγ Ã(Ee + Eγ)(Ee + Eγ cos θ)! 2 cos θ 1 +m4 − . (6.256) Ã(Ee + Eγ)(Ee + Eγ cos θ)!   In a typical Compton scattering situation, the energies in the center-of-momentum frame are much larger than the electron’s mass. So consider the high-energy limit E m. Naively, we can set E = E γ À e γ and m = 0 in that limit. However, this requires some care for cos θ 1, since then the denominator ≈− factor Ee + Eγ cos θ can become large. This is most important for the ﬁrst term in eq. (6.256). In fact, this term gives the dominant contribution to the total cross section, coming from the cos θ 1 ≈− (back-scattering) region. Integrating with respect to cos θ, we ﬁnd:

1 E + E E + E E + E e γ d(cos θ) = e γ ln e γ . (6.257) 1 Ee + Eγ cos θ Eγ ÃEe Eγ ! Z− − Now, expanding in the small mass of the electron,

2 2 2 2 m m Ee Eγ = Eγ 1+ m /Eγ Eγ = Eγ 1+ 2 + ... 1 , (6.258) − − Ã 2Eγ − ! ≈ 2Eγ q E + E 2E . (6.259) e γ ≈ γ Therefore,

1 Ee + Eγ 2 2 2 d(cos θ) 2 ln(4Eγ/m ) 2 ln(s/m ), (6.260) 1 Ee + Eγ cos θ ≈ ≈ Z− with the dominant contribution coming from cos θ near 1, where the denominator of the integrand − becomes small. Integrating the second term in eq. (6.256), one ﬁnds:

1 E + E cos θ 2E e γ d(cos θ)= e 1. (6.261) 1 Ee + Eγ Ee + Eγ ≈ Z− The remaining two terms vanish as m2/s 0. So, for s m2 we have: → À 1 dσ πα2 σ = d(cos θ)= 2 ln(s/m2) + 1 (6.262) 1 d(cos θ) s Z− h i plus terms which vanish like (m2/s2)ln(s/m2) as m2/s 0. The cross section falls at high energy → like 1/s, but with a logarithmic enhancement coming from back-scattered photons with angles θ π. ≈ The origin of this enhancement can be traced to the u-channel propagator, which becomes large when

106 u m2 = 2p k = 2E (E + E cos θ) becomes small. This corresponds to the virtual electron in the − b · 1 γ e γ u-channel Feynman diagram going nearly on-shell. (Notice that s m2 can never become small when − s m2.) À Just for fun, let us redo the analysis of Compton scattering, starting from eq. (6.245), but now working in the lab frame in which the initial state electron is at rest. Let us call the energy of the initial state photon ω and that of the ﬁnal state photon ω0. (Recall thath ¯ = 1 in our units, so the energy of a photon is equal to its angular frequency.)

θ a b

Then, in terms of the lab photon scattering angle θ, the 4-momenta are:

pa = (ω, 0, 0, ω) (6.263)

pb = (m, 0, 0, 0) (6.264)

k1 = (ω0, 0, ω0 sin θ, ω0 cos θ) (6.265)

k = (ω + m ω0, 0, ω0 sin θ, ω ω0 cos θ). (6.266) 2 − − −

2 2 2 2 Here we have used four-momentum conservation and the on-shell conditions pa = k1 = 0 and pb = m . 2 2 Applying the last on-shell condition k2 = m now leads to:

2m(ω ω0) + 2ωω0(cos θ 1) = 0. (6.267) − −

This can be used to solve for ω0 in terms of cos θ, or vice versa: ω ω0 = ; (6.268) 1+ ω (1 cos θ) m − m(ω ω ) cos θ = 1 − 0 . (6.269) − ωω0 In terms of lab-frame variables, one has:

p p = ωm; (6.270) a · b p k = ω0m, (6.271) b · 1 so that, from eq. (6.245)

1 ω ω 1 1 1 1 2 2 = 2e4 + 0 + 2m + m2 . (6.272) 4 |M| "ω0 ω ω − ω0 ω − ω0 # spinsX µ ¶ µ ¶

107 This can be simpliﬁed slightly by using 1 1 m = cos θ 1, (6.273) ω ω µ − 0 ¶ − which follows from eq. (6.267), so that

1 ω ω 2 = 2e4 + 0 sin2 θ . (6.274) 4 |M| ω0 ω − spinsX · ¸ Now to ﬁnd the diﬀerential cross-section, we must apply eq. (5.63): 1 1 dσ = 2 dΦ (6.275) 4E E ~v ~v 4|M| 2 a b| a − b| µ ¶

where dΦ2 is the two-body Lorentz-invariant phase space, as deﬁned in eq. (5.64), for the ﬁnal state particles in the lab frame. Evaluating the prefactors for the case at hand:

Ea = ω, (6.276)

Eb = m, (6.277) ~v ~v = 1 0 = 1. (6.278) | a − b| − (Recall that the speed of the photon is c = 1.) The two-body phase space in the lab frame is:

3~ 3~ 4 (4) d k1 d k2 dΦ2 = (2π) δ (k1 + k2 pa pb) 3 3 (6.279) − − (2π) 2E1 (2π) 2E2 3~ (3) ~ ~ d k1 3~ = δ (k1 + k2 ωz) δ(ω0 + E2 ω m) 2 d k2, (6.280) − − − 16π ω0E2 b where ω is now deﬁned to be equal to E = ~k and E is deﬁned to be ~k 2 + m2. Performing the 0 1 | 1| 2 | 2| ~k integral using the 3-momentum delta function just sets ~k = ωz ~k ,q resulting in: 2 2 − 1 3~ d k1 b dΦ2 = δ(ω0 + E2 ω m) 2 , (6.281) − − 16π ω0E2 where now

E = ω 2 2ωω cos θ + ω2 + m2. (6.282) 2 0 − 0 p The phase space for the ﬁnal state photon can be simpliﬁed by writing it in terms of angular coordinates and doing the integral dφ = 2π: R 3 2 2 d ~k1 = dφ d(cos θ) ω0 dω0 = 2πd(cos θ) ω0 dω0. (6.283)

Therefore,

d(cos θ) ω0dω0 dΦ2 = δ(ω0 + E2 ω m) . (6.284) − − 8πE2

108 In order to do the ω0 integral, it is simplest, as usual, to make a change of integration variables to the argument of the delta function. So, deﬁning

K = ω0 + E ω m, (6.285) 2 − − we have dK ω ω cos θ =1+ 0 − . (6.286) dω0 E2 Therefore, ω dω ω dK 0 0 = 0 . (6.287) E E + ω ω cos θ 2 2 0 − Performing the dK integration sets K = 0, so E = ω + m ω . Using eq. (6.287) in eq.(6.284) gives 2 − 0 ω d(cos θ) ω 2 dΦ = 0 = 0 d(cos θ), (6.288) 2 m + ω(1 cos θ) 8π 8πmω − where eq. (6.268) has been used to simplify the denominator. Finally using this in eq. (6.275) yields:

1 ω 2 dσ = 2 0 d(cos θ), (6.289) 4 |M|  32πmω2 spinsX   so that, putting in eq. (6.274),

dσ πα2 ω 2 ω ω = 0 0 + sin2 θ . (6.290) d(cos θ) m2 ω ω ω µ ¶ · 0 − ¸ This result is the Klein-Nishina formula. As in the center-of-momentum frame, the largest differential cross-section is found for back-scattered photons with cos θ = 1, which corresponds to the smallest possible final-state photon energy ω . − 0 Notice that, according to eq. (6.268), when ω m, one gets very low energies ω when the photon is À 0 1 back-scattered. One can now integrate the lab-frame differential cross section 1 d(cos θ) to get the − total cross section, using the dependence of ω0 on cos θ as given in eq. (6.268). TheR result is: 1 2 2m 2ω 4 2(m + ω) σ = πα2 ln 1+ + + . (6.291) ωm ω2 ω3 m ω2 m(m + 2ω)2 ·µ − − ¶ µ ¶ ¸ In the high-energy (small m) limit, this becomes: 1 2ω 1 σ = πα2 ln + + ... . (6.292) ωm m 2ωm · µ ¶ ¸ We can re-express this in terms of the Mandelstam variable s = (ω + m)2 ω2 = 2ωm + m2 2ωm, − ≈ πα2 σ = 2ln(s/m2)+1+ ... . (6.293) s h i Equation (6.293) is the same result that we found in the center-of-momentum frame, eq. (6.262). This is an example of a general fact: the total cross section does not depend on the choice of reference frame

109 as long as one boosts along a direction parallel to the collision axis. To see why, one need only look at the deﬁnition of the total cross-section given in eq. (5.34). The numbers of particles NS, Na, and Nb can be simply counted, and so certainly do not depend on any choice of inertial reference frame, while the area A is invariant under Lorentz boosts along the collision axis. The low-energy Thomson scattering limit is also interesting. In the lab frame, ω m implies ¿ ω0/ω = 1, so that dσ πα2 = (1 + cos2 θ), (6.294) d(cos θ) m2 which is independent of energy. The total cross section in this limit is then

8πα2 σ = . (6.295) 3m2 Unlike the case of high-energy Compton scattering, Thomson scattering is symmetric under θ π θ, → − with a factor of 2 enhancement in the forward (θ = 0) and backward (θ = π) directions compared to right-angle (θ = π/2) scattering.

+ 6.11 e e− γγ → As our ﬁnal example of a QED process let us consider e+e γγ in the high-energy limit. We could − → always compute the reduced matrix element starting from the Feynman rules. However, this process is obtained by “crossing” Compton scattering. Labeling the physical momenta for Compton scattering now with primes, we can make the following comparison table:

+ Compton e e− γγ → + γ p0 e p ↔ a ↔ a e− p0 e− p ↔ b ↔ b γ k0 γ k ↔ 1 ↔ 1 e− k0 γ k . ↔ 2 ↔ 2 + For convenience, we have chosen e to be labeled by “a”, so that the initial-state e− can have the same label “b” in both processes. Then according to the Crossing Symmetry Theorem, we can obtain 2 spins e+e− γγ by making the replacements |M → | P p0 = k ; p0 = p ; k0 = k ; k0 = p (6.296) a − 2 b b 1 1 2 − a in the Compton scattering result (for small m):

2 4 pa0 pb0 pb0 k10 γe− γe− = 8e · + · (6.297) |M → | Ãpb0 k10 pa0 pb0 ! spinsX · · obtained from eq. (6.245). Because the crossing involves one fermion (a ﬁnal state electron changes into an initial state positron), there is also a factor of ( 1)1 = 1, according to eq. (6.198). So, the − −

110 result is:

2 4 k2 pb pb k1 e+e− γγ = ( 1)8e − · + · (6.298) |M → | − pb k1 k2 pb spinsX µ · − · ¶ k p p k = 8e4 2 · b + b · 1 . (6.299) p k k p µ b · 1 2 · b ¶ The situation here is 2 2 massless particle scattering, so we can steal the kinematics information → directly from eqs. (6.177)-(6.181). The relevant facts for the present case are: u s p k = = (1 + cos θ) (6.300) b · 1 − 2 4 t s p k = = (1 cos θ). (6.301) b · 2 −2 4 − Therefore,

2 1 2 4 t u 4 1 + cos θ e+e− γγ = 2e + = 4e 2 . (6.302) 4 |M → | u t Ã sin θ ! spinsX µ ¶ It is a useful check, and a vindication of the ( 1)F factor in the Crossing Symmetry Theorem, that − this is positive! Now plugging this into the formula eq. (5.80) for the diﬀerential cross section, we get:

dσ 2πα2 1 + cos2 θ = 2 . (6.303) d cos(θ) s Ã sin θ ! This cross section is symmetric as θ π θ. That was inevitable, since the two final state photons are → − identical; the final state in which a photon is observed coming out at a (θ, φ) is actually the same as a state in which a photon is observed at an angle (π θ, φ). When we find the total cross section, we − − should therefore divide by 2 (or just integrate over half of the range for cos θ) to avoid overcounting. However, the integrand diverges for sin θ = 0. This is because we have set m = 0 in the kinematics, which is not valid for scattering very close to the collision axis. If you put back non-zero m, you will find that instead of diverging, the total cross section features a logarithmic enhancement ln(s/m2) coming from small sin θ. In a real e e+ γγ experiment, however, photons very close to the electron and − → positron beams will not be seen by any detector. The observable cross section in one of these colliding beam experiments is something more like

1 cos θmax dσ σcut = d(cos θ) (6.304) 2 cos θmax d(cos θ) Z− 2πα2 1 + cos θ = ln max cos θ . (6.305) s 1 cos θ − max · µ − max ¶ ¸ (You can easily check that this is a positive and increasing function of cos θmax.) On the other hand, if you are interested in the total cross section for electron-positron annihilation with no cuts applied on the angle, then you must take into account the non-zero electron mass. Redoing everything with m √s but non-zero, you can show: ¿ 2πα2 s σ = ln 1 . (6.306) s 2m2 · µ ¶ − ¸ 111 The logarithmic enhancement at large s in this formula comes entirely from the sin θ 0 region. Note ≈ that this formula is just what you would have gotten by plugging in

cos θ = 1 4m2/s (6.307) max − into eq. (6.305), for small m. In this sense, the ﬁnite mass of the electron “cuts oﬀ” the would-be logarithmic divergence of the cross section for small sin θ.

112 7 Decay processes

7.1 Decay rates and partial widths

So far, we have studied only 2 2 scattering processes. This is because in QED as a stand-alone → theory, all particles are stable. Photons cannot decay into anything else, (in any theory) since they are massless. In QED, the photon interaction vertex with a fermion line does not change the type of fermion. So if there is only a muon in the initial state, you can easily convince yourself that there must be at least one muon in the final state. (In general, there could be n muons and n 1 antimuons, where − n> 0.) Going to the rest frame of the initial muon, the total energy of the process is simply mµ, but the energy of the final state would have to be larger than this. The weak interactions get around this by allowing interaction vertices that change the fermion type. For any type of unstable particle, the probability that a decay will occur in a short interval of time ∆t should be proportional to ∆t. So we can define the decay rate or decay width Γ as the constant of proportionality:

(Probability of decay in time ∆t) = Γ∆t. (7.1)

Suppose we observe the decays of a large sample of particles of this type. If we start with N(t) particles at time t, the number of particles remaining a short time later is therefore:

N(t + ∆t) = (1 Γ∆t) N(t). (7.2) − It follows that dN = Γ, (7.3) dt − so that

Γt N(t)= e− N(0) (7.4)

When one has computed or measured Γ for some particle, it is traditional to quote the result as measured in the rest frame of the particle. If the particle is moving with velocity β, then because of relativistic time dilation, the survival probability for a particular particle as a function of the laboratory time t is:

Γt√1 β2 (Probability of particle survival) = e− − . (7.5)

The quantity

τ = 1/Γ (7.6) is also known as the mean lifetime of the particle at rest. There is often more than one ﬁnal state available for a decaying particle. One can then compute or measure the decay rates into particular

113 final states. The rate Γ for a particular final state or class of final states is called a partial width. The sum of all exclusive partial widths should add up to the total decay rate, of course. Consider a process in which a particle at rest with 4-momentum

pµ = (M, 0, 0, 0) (7.7) decays to several particles with 4-momenta k and masses m . Given the reduced matrix element for i i M this process, one can show by arguments similar to those in 5.2 for cross sections that the differential decay rate is: 1 dΓ = 2 dΦ , (7.8) 2M |M| n where n 3~ 4 (4) d ki dΦn = (2π) δ (p ki) 3 (7.9) − Ã(2π) 2Ei ! Xi iY=1 is the Lorentz-invariant n-body phase space. [Compare this to eq. (5.64); you will see that the only difference is that the pa + pb in the 4-momentum delta function for a scattering process has been replaced by p for a decay process.] To find the contribution to the decay rate for particles with 3- momenta restricted to be in some ranges, we should integrate dΓ over those ranges. To find the total decay rate Γ, we should integrate the 3-momenta over all available ~ki. The energies in this formula are defined by

~ 2 2 Ei = ki + mi . (7.10) q 7.2 Two-body decays

Most of the decay processes that one encounters in high-energy physics are two-particle or three-particle final states. As the number of particles in the final state increases, the decay rate for that final state tends to decrease, so a particle will typically decay into few-particle states if it can. If a two-particle final state is available, it is usually a very good bet that three-particle final states will lose. Let us simplify the formula eq. (7.8) for the case of two-particle final states with arbitrary masses. The evaluation of the two-particle final-state phase space is exactly the same as in eqs. (5.67)-(5.76), with the simple replacement E M. Therefore, CM → K dΦ = dφ d(cos θ ), (7.11) 2 16π2M 1 1 and K dΓ= 2 dφ d(cos θ ). (7.12) 32π2M 2 |M| 1 1 It remains to solve for K. Energy conservation requires that

2 2 2 2 E1 + E2 = K + m1 + K + m2 = M. (7.13) q q 114 One can solve this by writing it as

E M = E2 m2 + m2 (7.14) 1 − 1 − 1 2 q and squaring both sides. The solution is: M 2 + m2 m2 E = 1 − 2 (7.15) 1 2M M 2 + m2 m2 E = 2 − 1 (7.16) 2 2M 2 2 2 λ(M , m1, m2) K = (7.17) q 2M where

λ(x,y,z) x2 + y2 + z2 2xy 2xz 2yz. (7.18) ≡ − − − is known as the triangle function.† It is useful to tabulate results for some common special cases: If the ﬁnal-state masses are equal, m = m = m, then the ﬁnal state particles share the energy • 1 2 equally in the rest frame:

E1 = E2 = M/2, (7.19) M K = 1 4m2/M 2, (7.20) 2 − q and 2 dΓ= |M| 1 4m2/M 2 dφ d(cos θ ). (7.21) 64π2M − 1 1 q If one of the final-state particle is massless, m = 0, then: • 2 M 2 + m2 E = 1 , (7.22) 1 2M M 2 m2 E = K = − 1 . (7.23) 2 2M Notice that since the final state particles have equal 3-momentum magnitudes, the heavier particle gets more energy. In this case, 2 dΓ= |M| 1 m2/M 2 dφ d(cos θ ). (7.24) 64π2M − 1 1 1 ³ ´ If the decaying particle has spin 0, or if its spin is not measured, then there can be no special direction in the decay, so the final state particles must be distributed isotropically in the center-of- momentum frame. One then obtains the total decay rate from

dφ d(cos θ ) 4π, (7.25) 1 1 → provided that the two ﬁnal state particles are distinguishable. There is an extra factor of 1/2 if they are identical, to avoid counting each ﬁnal state twice [see the discussion leading to eq. (5.82)].

†It is so-named because, if each of √x, √y, √z is less than the sum of the other two, then λ(x, y, z) is 16 times − the square of the area of a triangle with sides √x, √y, √z. Note, however, that M, m1, m2 never form a triangle; if M < m1 + m2, then the decay is forbidden.

115 7.3 Scalar decays to fermion-antifermion pairs; Higgs decay

Let us now consider a simple and very important decay process, namely a scalar particle φ decaying to a fermion-antifermion pair. As a model, let us consider the Lagrangian already mentioned in section 5.4:

= yφΨΨ. (7.26) Lint − This type of interaction is called a Yukawa interaction, and y is a Yukawa coupling. The corresponding Feynman rule was argued to be equal to iy times an identity matrix in Dirac spinor space, as shown − in the picture immediately after eq. (5.138). Consider an initial state containing a single φ particle of mass M and 4-momentum p, and a ﬁnal state containing a fermion and antifermion each with mass m and with 4-momenta and spins k1, s1 and k2, s2 respectively. Then the matrix element will be of the form

k , s ; k , s p = i (2π)4δ(4)(k + k p) (7.27) OUTh 1 1 2 2| iOUT − M 1 2 − The Feynman rules for fermion external states don’t depend on the choice of interaction vertex, so they are the same as for QED. Therefore we can draw the Feynman diagram: 1

2 and immediately write down the reduced matrix element for the decay:

= iy u(k , s ) v(k , s ). (7.28) M − 1 1 2 2 To turn into a physically observable decay rate, we need to compute the squared reduced matrix M element summed over ﬁnal state spins. From eq. (7.28),

∗ = iy v u , (7.29) M 2 1 so

2 = y2(u v )(v u ). (7.30) |M| 1 2 2 1

Summing over s2, we have:

2 2 = y u1(k/2 m)u1 (7.31) s |M| − X2 = y2Tr[(k/ m)u u ]. (7.32) 2 − 1 1

116 Now summing over s1 gives:

2 2 = y Tr[(k/2 m)(k/1 + m)] (7.33) s ,s |M| − X1 2 = y2 Tr[k/ k/ ] m2Tr[1] (7.34) 2 1 − = 4y2³(k k m2), ´ (7.35) 1 · 2 − where we have used the fact that the trace of an odd number of gamma matrices vanishes, and eqs. (B.7) and (B.8). The fermion and antifermion have the same mass m, so

M 2 = (k + k )2 = k2 + k2 + 2k k = m2 + m2 + 2k k (7.36) 1 2 1 2 1 · 2 1 · 2 implies that

M 2 k k m2 = 2m2. (7.37) 1 · 2 − 2 − Therefore,

4m2 2 = 2y2M 2 1 , (7.38) |M| Ã − M 2 ! spinsX and, using eq. (7.21),

3/2 y2M 4m2 dΓ= 1 dφ1 d(cos θ1). (7.39) 32π2 Ã − M 2 ! Doing the angular integrals ﬁnally gives the total decay rate:

3/2 y2M 4m2 Γ= 1 . (7.40) 8π Ã − M 2 ! In the Standard Model, the Higgs boson h plays the role of φ, and couples to each fermion f with a Lagrangian that is exactly of the form given above:

= y hΨ Ψ . (7.41) Lint − f f f Xf The Yukawa coupling for each fermion is approximately proportional to its mass: m y = f . (7.42) f 174 GeV

Actually, the mf appearing in this formula is not quite equal to the mass, because of higher-order corrections. For quarks, these corrections are quite large, and mf tends to come out considerably smaller than the inertial mass of the quark. At the Fermilab Tevatron Run II, one of the major goals is to look for a Higgs boson through its decay modes. Suppose for example that the mass of the Higgs is not too much larger than the current experimental limit of about 114 GeV. Then, since the top quark has a mass of about 175 GeV, the decay h tt is clearly forbidden. The next-lightest fermions in the →

117 Standard Model are the bottom quark, charm quark, and tau lepton, so we expect decays h bb and → h τ τ + and h cc. For quarks, the sum in eq. (7.41) includes a summation over 3 colors, leading → − → to an extra factor of n = 3 in the decay rate. Since the kinematic factor (1 4m2 /M 2)3/2 is close to 1 f − f h for all allowed fermion-antifermion ﬁnal states, the decay rate to a particular fermion is approximately:

2 nf y Mh Γ = f n m2 . (7.43) 8π ∝ f f

Estimates of the mf from present experimental data are:

m 3 GeV (7.44) b ≈ m 1.77 GeV (7.45) τ ≈ m 0.68 GeV, (7.46) c ≈ for a Higgs with mass of order Mh = 125 GeV. Even though the charm quark is heavier than the tau lepton, it turns out that mτ > mc because of the large higher-order corrections for the charm quark. Now we can predict:

+ 2 2 2 Γ(h bb):Γ(h τ −τ ):Γ(h cc) :: 3(m ) : (m ) : 3(m ) (7.47) → → → b τ c :: 1:0.12 : 0.05 (7.48)

One can deﬁne the branching fraction to be the partial decay rate into a particular ﬁnal state, divided by the total decay rate, so

Γ(h bb) BR(h bb)= → . (7.49) → Γtotal Then, if no other decays contribute signiﬁcantly,

BR(h bb) 0.86, (7.50) → ≈ + BR(h τ −τ ) 0.10, (7.51) → ≈ BR(h cc) 0.04. (7.52) → ≈ Detailed understanding of the higher-order corrections to these ratios, especially in certain variations of the Standard Model like supersymmetry, remain a subject of research. But the main lesson is that we expect bb final states, which will manifest themselves in the detector as b-jets, should dominate by a factor of 10 or so over the other available fermion-antifermion final states. Later we will discuss the production processes for the Higgs boson, to see what the whole events are expected to look like. The assumption that there are no other decays kinematically available to the Higgs boson applies in the Standard Model if Mh is in the range over which the Tevatron Run II could discover the Higgs. If the Higgs boson is heavier than about 150 GeV, then other final states become kinematically allowed, like decays to W and Z0 vector bosons: h W W +, and h Z0Z0. Since the squared Yukawa ± → − → coupling y2 .00015 is so small, the decay h bb will lose out to these other possibilities if they are b ≈ →

118 kinematically open. Moreover, in some extensions of the Standard Model, the Higgs boson could decay into some new exotic particles. This could also decrease the branching fraction into bb. Finally, let us remark on the helicities for this process. If we demanded that the ﬁnal states have particular helicities, then we would have obtained for the matrix element, using PR, PL projection matrices:

R-fermion, R-antifermion: = iy u P P v = iy u P v = 0 (7.53) M − 2 L L 1 − 2 L 1 6 L-fermion, L-antifermion: = iy u P P v = iy u P v = 0 (7.54) M − 2 R R 1 − 2 R 1 6 R-fermion, L-antifermion: = iy u P P v = 0 (7.55) M − 2 L R 1 L-fermion, R-antifermion: = iy u P P v = 0 (7.56) M − 2 R L 1 [Recall, from eqs. (6.102)-(6.105), that to get a L fermion or antifermion in the final state, one puts in a PR next to the spinor, while to get a R fermion or antifermion in the final state, one puts in a PL.] So, the rule here is that helicity is always violated by the scalar-fermion-antifermion vertex, since the scalar must decay to a state with equal helicities. One may understand this result from angular momentum conservation. The initial state had no spin and no orbital angular momentum. In the final state, the spins of the outgoing particles must therefore have opposite directions. Since they have momentum in opposite directions, this means they must also have the same helicity. Drawing a short arrow to represent the spin, the allowed cases of RR helicities and LL helicities look like:

R fermionh R anti-fermion

L fermionh L anti-fermion

Since the helicities of tau leptons can be (statistically) measured from their decays, this eﬀect may eventually be measured when a sample of h τ τ + decay events is obtained. → − 7.4 Three-body decays

Let us consider a generic three-body decay process in which a particle of mass M decays to three lighter particles with masses m1, m2, and m3. We will work in the rest frame of the decaying particle, so µ its four-vector is p = (M, 0, 0, 0), and the four-vectors of the remaining particles are k1, k2, and k3 respectively:

119 m1, k1

m3, k3 m2, k2

In general, the formula for a three-body differential decay rate is 1 dΓ= 2dΦ (7.57) 2M |M| 3 where 3~ 3~ 3~ 4 (4) d k1 d k2 d k3 dΦ3 = (2π) δ (p k1 k2 k3) 3 3 3 (7.58) − − − (2π) 2E1 (2π) 2E2 (2π) 2E3 is the Lorentz-invariant phase space. Since there are 9 integrals to do, and 4 delta functions, the result for dΓ is a differential with respect to 5 remaining variables. The best choice of 5 variables depends on the problem at hand, so there are several ways to present the result. Two of the 5 variables can be chosen to be the energies E1 and E2 of two of the final-state particles; then the energy of the third particle E = M E E is also known from energy conservation. In the rest frame of the 3 − 1 − 2 decaying particle, the three final-state particle 3-momenta must lie in a plane, because of momentum conservation. Specifying E1 and E2 also uniquely fixes the angles between the three particle momenta within this decay plane. The remaining 3 variables just correspond to the orientation of the decay plane with respect to some fixed coordinate axis. If we think of the three 3-momenta within the decay plane as describing a rigid body, then the relative orientation can be described using three Euler angles.

These can be chosen to be the spherical coordinate angles φ1 and θ1 for particle 1, and an angle α2 which measures the rotation of the 3-momentum direction of particle 2 as measured about the axis of the momentum vector of particle 1. Then one can show: 1 dΦ = dE dE dφ d(cos θ ) dα . (7.59) 3 256π5 1 2 1 1 2 The choice of which particles to label as 1 and 2 is arbitrary, and should be made to maximize convenience. If the initial state particle spin is averaged over, or if it is spinless, then there is no special direction to measure the orientation of the ﬁnal state decay plane with respect to. In that case, for particular

E1 and E2, the reduced matrix element cannot depend on the angles φ1, θ1 or α2, and one can do the integrals

2π 1 2π 2 dφ1 d(cos θ1) dα2 = (2π)(2)(2π) = 8π . (7.60) 0 1 0 Z Z− Z Then, 1 dΦ = dE dE , (7.61) 3 32π3 1 2

120 and so, for spinless or spin-averaged initial states, 1 dΓ= 2 dE dE . (7.62) 64π3M |M| 1 2 To do the remaining energy integrals, one must ﬁnd the limits of integration. If one resolves to do the

E2 integral ﬁrst, then by doing the kinematics one can show for any particular E1 that 1 Emax,min = (M E )(m2 + m2 m2) (E2 m2) λ(m2 , m2, m2) , (7.63) 2 2m2 − 1 23 2 − 3 ± 1 − 1 23 2 1 23 · q ¸ where

m2 = (k + k )2 = (p k )2 = M 2 2E M + m2 (7.64) 23 2 3 − 1 − 1 1 is the invariant (mass)2 of the combination of particles 2 and 3. Then the limits of integration for the

ﬁnal E1 integral are:

M 2 + m2 (m + m )2 m

In the special case that all ﬁnal state particles are massless (or small) m1 = m2 = m3 = 0, then these limits of integration simplify to: M M E < E < , (7.66) 2 − 1 2 2 M 0 < E < . (7.67) 1 2

121 8 Fermi theory of weak interactions

8.1 Weak nuclear decays

In nuclear physics, the weak interactions are responsible for decays of long-lived isotopes. A nucleus with Z protons and A Z neutrons, so A nucleons in all, is denoted by AZ. If kinematically allowed, − one can observe decays:

A A Z (Z +1)+ e−ν . (8.1) → e

The existence of the antineutrino νe was inferred by Pauli as a “desperate remedy” to save the principle of energy conservation. The simplest example of this is the decay of the neutron into a proton, electron, and antineutrino:

+ n p e−ν (8.2) → e with a mean lifetime of†

τ = 887 2 sec. (8.3) ± Decays of heavier nuclei can be thought of as involving the subprocess:

+ “n” “p ”e−ν , (8.4) → e where the quotes indicated that the neutron and proton are really not separate entities, but part of the nuclear bound states. So for example, tritium decays according to

3 3 8 H He + e−ν (τ = 5.6 10 sec = 17.7 years), (8.5) → e × and carbon-14 decays according to

14 14 11 C N+ e−ν (τ = 1.8 10 sec = 8280 years). (8.6) → e ×

Nuclear physicists usually quote the half life t1/2 rather than the mean lifetime τ. They are related by

t1/2 = τln(2), (8.7) so that t1/2 = 5730 years for Carbon-14, making it ideal for dating dead organisms. In the upper atmosphere, 14C is constantly being created by cosmic rays which produce energetic neutrons, which in turn convert 14N nuclei into 14C. Carbon-dioxide-breathing organisms, or those that eat them, maintain an equilibrium with the carbon content of the atmosphere, at a level of roughly 14C/12C 10 12. ≈ − However, this ratio is not constant; it dropped in the early 20th century as more ordinary 12C entered the atmosphere because of the burning of fossil fuels containing the carbon of organisms that have been

†The lifetime of the neutron is an infamous example of an experimental measurement which has shifted dramatically over time. As recently as the late 1960’s, it was thought that τn = 1010 30 seconds, and as late as 1980, τn = 920 20 ± ± seconds.

122 dead for very long time. The relative abundance 14C/12C 10 12 then doubled after 1954 because of ≈ − nuclear weapons testing, reaching a peak in the mid 1960’s from which it has since declined. In any case, dead organisms lose half of their 14C every 5730 30 years, and certainly do not regain it by ± 14 breathing or eating. So, by measuring the rate of e− beta rays consistent with C decay produced by a sample, and determining the atmospheric 14C/12C ratio as a function of time with control samples or by other means, one can date the death of a sample of organic matter. One can also have decays which release a positron and neutrino:

AZ A(Z 1) + e+ν . (8.8) → − e These can be thought of as coming from the subprocess

“p+” “n”e+ν . (8.9) → e

In free space, the proton cannot decay, simply because mp < mn, but under the right circumstances it is kinematically allowed when the proton and neutron are parts of nuclear bound states. An example is

14O 14N+ e+ν (τ = 71 sec). (8.10) → e The long lifetimes of these decays are what originally gave rise to the name “weak” interactions. Charged pions also decay through the weak interactions, with a mean lifetime of

8 τ ± = 2.2 10− sec. (8.11) π × This corresponds to a decay length of cτ = 7.8 meters. The probability that a charged pion with velocity β will travel a distance L in empty space before decaying is therefore

(L/7.8 m)√1 β2/β P = e− − . (8.12)

This means that a relativistic pion will typically travel several meters before decaying, unless it interacts (which it usually will in a collider detector). The main decay mode is

π− µ−ν (8.13) → µ with a branching fraction of 0.99988. (This includes also submodes in which an additional photon is radiated away.) The only other signiﬁcant decay mode is

π− e−ν . (8.14) → e with a branching fraction 1.2 10 4. This presents a puzzle: since the electron is lighter, there is × − more kinematic phase space available for the second decay, yet the ﬁrst decay dominates by almost a factor of 104. We will calculate the reason for this later.

123 8.2 Muon decay

The muon decays according to

6 µ− e−ν ν (τ = 2.2 10− sec). (8.15) → e µ × This corresponds to a decay width of

19 Γ = 3.0 10− GeV, (8.16) × implying a proper decay length of cτ = 659 meters. Muons do not undergo hadronic interactions like pions do, so that relativistic muons will usually penetrate at least the inner layers of particle detectors with a very high probability. The Feynman diagram for muon decay can be drawn as:

e−

µ− νe

νµ

This involves a 4-fermion interaction vertex.‡ Because of the correspondence between interactions and terms in the Lagrangian, we therefore expect that the Lagrangian should contain terms schematically of the form

= (ν ...µ) (e...ν ) or (ν ...ν ) (e...µ), (8.17) Lint µ e µ e where the symbols µ, νµ,e,νe mean the Dirac spinor ﬁelds for the muon, muon neutrino, electron, and electron neutrino, and the ellipses mean matrices in Dirac spinor space. To be more precise about the interaction Lagrangian, one needs clues from experiment. One clue is the fact that there are three quantum numbers, called lepton numbers, that are additively conserved to a high accuracy. They are assigned as:

+1 for e−,νe L = 1 for e+, ν (8.18) e  − e  0 for all other particles

 +1 for µ−,νµ L = 1 for µ+, ν (8.19) µ  − µ  0 for all other particles

 +1 for τ −,ντ L = 1 for τ +, ν (8.20) τ  − τ  0 for all other particles

‡We will eventually find out that this is not a true fundamental interaction of the theory, but rather an “effective” interaction that is derived from the low-energy effects of the W − boson.

124 So, for example, in the nuclear decay examples above, one always has Le = 0 in the initial state, and L = 1 1 = 0 in the ﬁnal state, with L = L = 0 trivially in each case. The muon decay mode e − µ τ in eq. (8.15) has (Le,Lµ) = (0, 1) in both the initial and ﬁnal states. If lepton numbers were not conserved, then one might expect that decays like

µ− e−γ (8.21) → would be allowed. However, this decay has never been observed, and the most recent limit from the LAMPF experiment at Los Alamos National Lab is

11 BR(µ− e−γ) < 1.2 10− . (8.22) → × This is a remarkably strong constraint, since this decay only has to compete with the already weak mode in eq. (8.15). It implies that

30 Γ(µ− e−γ) < 3.6 10− GeV. (8.23) → × The CLEO experiment has put similar (but not as stringent) bounds on tau lepton number non- conservation:

6 BR(τ − e−γ) < 2.7 10− , (8.24) → × 6 BR(τ − µ−γ) < 1.1 10− , (8.25) → × 0 6 BR(τ − e−π ) < 4 10− . (8.26) → × On the other hand, SuperKamiokande announced in 1998 that they had evidence for muon neutrino oscillations:

ν something else. (8.27) µ ↔

Here the “something else” could be either a ντ , or a new neutrino that does not have the usual weak interactions and is not part of the Standard Model. So it appears that at least Lµ may be violated after all. Fortunately, regardless of the outcome of this and other experiments probing lepton number violation in the neutrino sector, this is a small eﬀect for us, and we can ignore it. The conservation of lepton numbers suggests that the interaction Lagrangian for the weak interactions can always be written in terms of fermion bilinears involving one barred and one unbarred Dirac spinor from each lepton family. So, we will write the weak interactions for leptons in terms of building blocks with net Le = Lµ = Lτ = 0, for example, like the ﬁrst term in eq. (8.17) but not the second. More generally, we will want to use building blocks:

(`...ν`) or (ν` . . . `), (8.28) where ` is any of e, µ, τ. Now, since each Dirac spinor has 4 components, a basis for fermion bilinears involving any two ﬁelds Ψ and Ψ will have 4 4 = 16 elements. The can be classiﬁed by their 1 2 ×

125 transformation properties under the proper Lorentz group and the parity transformation ~x ~x, as → − follows:

Term Number Parity (~x ~x) Type →− Ψ1Ψ2 1 +1 Scalar= S Ψ γ Ψ 1 1 Pseudo-scalar = P 1 5 2 − Ψ γµΨ 4 ( 1)µ Vector = V 1 2 − Ψ γµγ Ψ 4 ( 1)µ Axial-vector = A 1 5 2 − − i Ψ [γµ,γν]Ψ 6 ( 1)µ( 1)ν Tensor = T 2 1 2 − − The entry under Parity indicates the multiplicative factor under which each of these terms transforms when ~x ~x, with →− +1 for µ = 0 ( 1)µ = (8.29) − 1 for µ = 1, 2, 3 ( − The weak interaction Lagrangian for leptons could be formed out of any product of such terms with

Ψ1, Ψ2 = `, ν`. Fermi originally proposed that the weak interaction fermion building blocks were of the type V , so that muon decays would be described by

V = G(ν γρµ)(eγ ν )+c.c. (8.30) Lint − µ ρ e Here “c.c.” means complex conjugate; this is necessary since the Dirac spinor ﬁelds are complex. Some other possibilities could have been that the building blocks were of type A:

A = G(ν γργ µ)(eγ γ ν )+c.c. (8.31) Lint − µ 5 ρ 5 e or some combination of V and A, or some combination of S and P , or perhaps even T . Fermi’s original proposal of V for the weak interactions turned out to be wrong. The most important clue for determining the correct answer for the proper Lorentz and parity structure of the weak interaction building blocks came from an experiment on polarized 60Co decay by Wu in 1957. The 60Co nucleus has spin J = 5, so that when cooled and placed in a magnetic ﬁeld, the nuclear spins align with B~ . Wu then measured the angular dependence of the electron spin from the decay

60 60 Co Ni + e−ν (8.32) → e The nucleus 60Ni has spin J = 4, so the net angular momentum carried away by the electron and antineutrino is 1. The observation was that the electron is emitted preferentially in the direction opposite to the original spin of the 60Co nucleus. This can be explained consistently with angular momentum conservation if the electron is always left-handed and the antineutrino is always right- handed. Using short arrows to designate spin directions, the most favored conﬁguration is:

126 J = 5 J = 4 −→ −→ −→ −→ e− νe 60Co 60Ni

Before After

The importance of this experiment and others was that right-handed electrons and left-handed antineutrinos do not seem to participate in the weak interactions. This means that when writing the interaction Lagrangian for weak interactions, we can always put a PL to the left of the electron’s Dirac

ﬁeld, and a PR to the right of a νe ﬁeld. This helped establish that the correct form for the fermion bilinear is V A: − 1 eP γρν = eγρP ν = eγρ(1 γ )ν . (8.33) R e L e 2 − 5 e Since this is a complex quantity, and the Lagrangian density must be real, one must also have terms involving the complex conjugate of eq. (8.33):

ρ ρ νePRγ e = νeγ PLe (8.34)

The feature that was most surprising at the time was that right-handed Dirac fermions and left-handed

Dirac barred fermion ﬁelds PRe, PRνe, ePL, and νePL never appear in any part of the weak interaction Lagrangian. For muon decay, the relevant four-fermion interaction Lagrangian is:

= 2√2G (ν γρP µ)(eγ P ν )+c.c. (8.35) Lint − F µ L ρ L e

2 Here GF is a coupling constant with dimensions of [mass]− , known as the Fermi constant. Its numerical value is most precisely determined from muon decay. The factor of 2√2 is a historical convention. Using the correspondence between terms in the Lagrangian and particle interactions, we therefore have two Feynman rules: µ, a e,c

i2√2G (γρP ) (γ P ) ←→ − F L ba ρ L cd

νµ, b νe,d

127 µ, a e, c

i2√2G (γρP ) (γ P ) ←→ − F L ab ρ L dc

νµ, b νe,d

These are related by reversing of all arrows, corresponding to the complex conjugate in eq. (8.35). The slightly separated dots in the Feynman rule picture are meant to indicate the Dirac spinor structure. The Feynman rules for external state fermions and antifermions are exactly the same as in QED, with neutrinos treated as fermions and antineutrinos as antifermions. This weak interaction Lagrangian for muon decay violates parity maximally, since it treats left-handed fermions differently from right-handed fermions. However, helicity is conserved by this interaction Lagrangian, just as in QED, because of the presence of one gamma matrix in each fermion bilinear. We can now derive the reduced matrix element for muon decay, and use it to compute the differential decay rate of the muon. Comparing this to the experimentally measured result will allow us to find the numerical value of GF , and determine the energy spectrum of the final state electron. At lowest order, the only Feynman diagram for µ e ν ν is: − → − e µ e−

µ− νe

νµ using the ﬁrst of the two Feynman rules above. Let us label the momenta and spins of the particles as follows: Particle Momentum Spin Spinor µ− pa sa u(pa, sa)= ua e− k1 s1 u(k1, s1)= u1 (8.36) νe k2 s2 v(k2, s2)= v2 νµ k3 s3 u(k3, s3)= u3 The reduced matrix element is obtained by starting at the end of each fermion line with a barred spinor and following it back (moving opposite the arrow direction) to the beginning. In this case, that means starting with the muon neutrino and electron barred spinors. The result is:

= i2√2G (u γρP u )(u γ P v ). (8.37) M − F 3 L a 1 ρ L 2

This illustrates a general feature; in the weak interactions, there should be a PL next to each unbarred spinor in a matrix element, or equivalently a PR next to each barred spinor. (The presence of the

128 gamma matrix ensures the equivalence of these two statements, since P P when moved through L ↔ R a gamma matrix.) Taking the complex conjugate, we have:

σ ∗ = i2√2G (u P γ u )(v P γ u ). (8.38) M F a R 3 2 R σ 1 Therefore,

2 = 8G2 (u γρP u )(u P γσu ) (u γ P v )(v P γ u ). (8.39) |M| F 3 L a a R 3 1 ρ L 2 2 R σ 1

In the following, we can neglect the mass of the electron me, since me/mµ < 0.005. Now we can average over the initial-state spin sa and sum over the ﬁnal-state spins s1, s2, s3 using the usual tricks: 1 1 uaua = (p/a + mµ), (8.40) 2 s 2 Xa u1u1 = k/1, (8.41) s X1 v2v2 = k/2, (8.42) s X2 u3u3 = k/3, (8.43) s X3 to turn the result into a product of traces:

1 2 2 ρ σ = 4G Tr[γ PL(p/ + mµ)PRγ k/ ]Tr[γρPLk/ PRγσk/ ] (8.44) 2 |M| F a 3 2 1 spinsX 2 σ ρ/ / / = 4GF Tr[γ p/aPRγ k3]Tr[γσk2PRγρk1] (8.45)

Fortunately, we have already seen a product of traces just like this one, in eq. (6.138), so that by substituting in the appropriate 4-momenta, we immediately get: 1 2 = 64G2 (p k )(k k ). (8.46) 2 |M| F a · 2 1 · 3 spinsX Our next task is to turn this reduced matrix element into a diﬀerential decay rate.

Applying the results of subsection 7.4 to the example of muon decay, with M = mµ and m1 = m = m = 0. According to our result of eq. (8.46), we need to evaluate the dot products p k and 2 3 a · 2 k k . Since these are Lorentz scalars, we can evaluate them by rotating to a frame where ~k is along 1 · 3 2 the z-axis. Then

pa = (mµ, 0, 0, 0) (8.47)

k2 = (Eνe , 0, 0,Eνe ). (8.48)

Therefore,

p k = m E . (8.49) a · 2 µ νe

129 Also, k k = 1 [(k + k )2 k2 k2]= 1 [(p k )2 0 0] = m2 /2, so 1 · 3 2 1 3 − 1 − 3 2 a − 2 − − 23 1 k k = (m2 2m E ). (8.50) 1 · 3 2 µ − µ νe Therefore, from eq. (8.46), 1 2 = 32G2 (m3 E 2m2 E2 ). (8.51) 2 |M| F µ νe − µ νe spinsX

Plugging this into eq. (7.62) with M = mµ, and choosing E1 = Ee and E2 = Eνe , we obtain:

G2 dΓ = dE dE F (m2 E 2m E2 ). (8.52) e νe 2π3 µ νe − µ νe

Doing the dEνe integral using the limits of integration of eq. (7.66), we obtain:

mµ 2 2 2 2 3 2 G G m E m E dΓ = dE dE F (m2 E 2m E2 ) = dE F µ e µ e . (8.53) e mµ νe 3 µ νe µ νe e 3 Ee 2π − π Ã 4 − 3 ! Z 2 − We have obtained the diﬀerential decay rate for the energy of the ﬁnal state electron:

2 2 dΓ GF mµ 2 4Ee = 3 Ee 1 . (8.54) dEe 4π Ã − 3mµ !

The shape of this distribution is shown below as the solid line:

−, ν e µ ν e

dΓ/dE

0 0 0.1 0.2 0.3 0.4 0.5

E/mµ

130 We see that the electron energy is peaked near its maximum value of mµ/2. This corresponds to the situation where the electron is recoiling directly against both the neutrino and antineutrino which are collinear; for example, k = (m /2, 0, 0, m /2), and kµ = kµ = (m /4, 0, 0, m /4): 1 µ − µ 2 3 µ µ

e µ νe − −→ − −→ νµ ←−

The helicity of the initial state is undeﬁned, since the muon is at rest. However, we know that the

ﬁnal state e−, νµ, and νe have well-deﬁned L, L, and R helicities respectively, as shown above, since this is dictated by the weak interactions. In the case of maximum Ee, therefore, the spins of νµ and

νe must be in opposite directions. The helicity of the electron is L, so its spin must be opposite to its 3-momentum direction. By momentum conservation, this tells us that the electron must move in the opposite direction to the initial muon spin in the limit that Ee is near the maximum. The smallest possible electron energies are near 0, which occurs when the neutrino and antineutrino move in nearly opposite directions, so that the 3-momentum of the electron recoiling against them is very small. We have done the most practically sensible thing by plotting the diﬀerential decay rate in terms of the electron energy, since that is what is directly observable in an experiment. Just for fun, however, let us pretend that we could directly measure the νµ and νe energies, and compute the distributions for them. To ﬁnd dΓ/dEνµ , we can take E2 = Eνe and E1 = Eνµ in eqs.(7.62) and (7.66)-(7.67), with the reduced matrix element from eq. (8.51). Then

G2 dΓ = dE dE F (m2 E 2m E2 ), (8.55) νµ νe 2π3 µ νe − µ νe and the range of integration for Eνe is now: m m µ E < E < µ , (8.56) 2 − νµ νe 2 so that

mµ 2 2 2 2 3 2 G G mµEν mµEν dΓ = dE dE F (m2 E 2m E2 ) = dE F µ µ . (8.57) νµ mµ νe 3 µ νe µ νe e 3 Eνµ 2π − π Ã 4 − 3 ! Z 2 −

Therefore, the Eνµ distribution of ﬁnal states has the same shape as the Ee distribution:

2 2 dΓ GF mµ 2 4Eνµ = 3 Eνµ 1 . (8.58) dEνµ 4π Ã − 3mµ !

Finally, we can ﬁnd dΓ/dEνe , by choosing E2 = Ee and E1 = Eνe in eqs. (7.62) and (7.66)-(7.67) with eq. (8.51). Then:

mµ 2 2 2 G G dΓ = dE dE F (m2 E 2m E2 ) = dE F m2 E2 2m E3 , (8.59) νe m e 3 µ νe µ νe νe 3 µ νe µ νe µ E 2π − 2π − Z 2 − νe ³ ´ 131 so that

2 2 dΓ GF mµ 2 2Eνe = 3 Eνe 1 . (8.60) dEνe 2π Ã − mµ !

This distribution is plotted as the dashed line in the previous graph. Unlike the distributions for Ee and mµ Eν, we see that dΓ/dEνe vanishes when Eνe approaches its maximum value of 2 . We can understand this by noting that when Eνe is maximum, the νe must be recoiling against both e and νe moving in the opposite direction, so the L, L, R helicities of e, νµ, and νe tell us that the total spin of the ﬁnal state is 3/2:

νe µ e ←− − ←− νµ ←−

Since the initial-state muon only had spin 1/2, the quantum states have 0 overlap, and the rate must vanish in that limit of maximal Eνe .

The total decay rate for the muon is found by integrating either eq. (8.54) with respect to Ee, or eq. (8.58) with respect to Eνµ , or eq. (8.60) with respect to Eνe . In each case, we get:

mµ/2 dΓ mµ/2 dΓ mµ/2 dΓ Γ = dE = dE = dE (8.61) dE e dE νµ dE νe Z0 µ e ¶ Z0 Ã νµ ! Z0 µ νe ¶ G2 m5 = F µ . (8.62) 192π3 It is a good check that the ﬁnal result does not depend on the choice of the ﬁnal energy integration 2 4 4 5 variable. It is also good to check units: GF has units of [mass]− or [time] , while mµ has units of 5 5 1 [mass] or [time]− , so Γ indeed has units of [mass] or [time]− . Experiments tell us that

19 Γ(µ− e−ν ν ) = 3.00 10− GeV, m = 0.106 GeV, (8.63) → µ e × µ so we obtain the numerical value of Fermi’s constant from eq. (8.62):

5 2 G = 1.16 10− GeV− . (8.64) F ×

The 4-fermion weak interaction Lagrangian of eq. (8.35) describes several other processes besides the decay µ e ν ν that we studied in subsection 8.2. As the simplest example, we can just replace − → − e µ each particle in the process by its anti-particle:

µ+ e+ν ν , (8.65) → e µ for which the Feynman diagram is just obtained by changing all of the arrow directions:

132 e+

µ+ νe

νµ

The evaluations of the reduced matrix element and the diﬀerential and total decay rates for this decay are very similar to those for the µ e ν ν . For future reference, let us label the 4-momenta for − → − e µ this process as follows:

Particle Momentum Spinor + µ pa0 va + e k10 v1 (8.66) νe k20 u2 νµ k30 v3.

The reduced matrix element, following from the “+c.c.” term in eq. (8.35), is then

= i2√2G (v γρP v )(u γ P v ). (8.67) M − F a L 3 2 ρ L 1 As one might expect, the result for the spin-summed squared matrix element,

2 2 = 128G (p0 k0 )(k0 k0 ), (8.68) |M| F a · 2 1 · 3 spinsX is exactly the same as obtained in eq. (8.46), with the obvious substitution of primed 4-momenta. The differential and total decay rates that follow from this are, of course, exactly the same as for µ− decay. This is actually a special case of a general symmetry relation between particles and anti-particles, which holds true in any local quantum field theory, and is known as the CPT Theorem. The statement of the theorem is that the laws of physics, as specified by the Lagrangian, are left unchanged after one performs the combined operations of replacing each particle by its antiparticle, or charge conjugation (C); parity ~x ~x, (P); and time reversal t t (T). It turns out to be impossible to write down a →− →− Lagrangian that fails to obey these rules, as long as it is invariant under proper Lorentz transformations and contains a finite number of spacetime derivatives and obeys some other technical assumptions. Among other things, the CPT Theorem implies that the mass and the total decay rate of a particle must each be equal to the same quantities for the corresponding anti-particle. (It does not say that the differential decay rate to a particular final state configuration necessarily has to be equal to the anti-particle differential decay rate to the same configuration of final-state anti-particles; that stronger result holds only if the theory is invariant under T. The four-fermion Fermi interaction for leptons does respect invariance under T, but it is violated by a tiny amount in the weak interactions of quarks.) We will study some other processes implied by the Fermi weak interaction Lagrangians in subsections 8.4, 8.5, and 8.6 below.

133 8.3 Corrections to muon decay

In the previous subsection, we derived the µ− decay rate in terms of Fermi’s four-fermion weak interaction coupling constant GF . Since this decay process is actually the one which is used to experimentally determine GF most accurately, it is worthwhile to note the leading corrections to it.

First, there is the dependence on me, which we neglected, but could have included at the cost of a more complicated phase space integration. Taking this into account using correct kinematics for m = 0 e 6 and the limits of integration in eq. (7.63)-(7.65), one ﬁnds that the decay rate must be multiplied by 2 2 a correction factor Fkin(me/mµ), where

F (x) = 1 8x + 8x3 x4 12x2lnx. (8.69) kin − − −

2 2 Numerically, Fkin(me/mµ) = 0.999813. There are also corrections coming from two types of QED eﬀects. First, there are loop diagrams involving virtual photons:

e− e−

µ− νe µ− νe

νµ νµ

e−

µ− νe

νµ

Evaluating these diagrams is beyond the scope of this course. But it should be clear that they give 2 contributions to the reduced matrix element proportional to e GF , since each contains two photon interaction vertices. These contributions to the reduced matrix element actually involve divergent loop integrals, which must be “regularized” by using a high-momentum cutoff. The logarithmic dependence on the cutoff is then absorbed into a redefinition of the mass and coupling parameters of the model, by the systematic process of renormalization. The interference of the loop diagrams with the original lowest-order diagram then gives a contribution to the decay rate proportional to α. There are also QED contributions from diagrams with additional photons in the final state:

134 γ γ

e− e−

µ− νe µ− νe

νµ νµ The QED diagrams involving an additional photon contribute to a 4-body final state, with a reduced matrix element proportional to eGF . After squaring, summing over final spins, and averaging over the initial spin, and integrating over the 4-body phase space, the contribution to the decay rate is again 2 proportional to αGF . Much of this contribution actually comes from very soft (low-energy) photons, which are difficult or impossible to resolve experimentally. Therefore, one usually just combines the two types of QED contributions into a total inclusive decay rate with one or more extra photons in the final state. After a heroic calculation, one finds that the QED effect on the total decay rate is to multiply by a correction factor

α 25 π2 α 2 F (α)=1+ + C + ... (8.70) QED π 8 2 π 2 Ã − ! µ ¶ where the C2 contribution refers to even higher-order corrections from: the interference between Feyn- man diagrams with two virtual photons and the original Feynman diagram; the interference between Feynman diagrams with one virtual photons plus one final state photon and the original Feynman diagram; the square of the reduced matrix element for a Feynman diagram involving one virtual photon; and two photons in the final state. A very complicated calculation shows that C 6.74. Because 2 ≈ of the renormalization procedure, the QED coupling α actually is dependent on the energy scale, and should be evaluated at the energy scale of interest for this problem, which is naturally mµ. At that scale, α 1/136, so numerically ≈ F (α) 0.995727. (8.71) QED ≈ Finally, there are corrections involving the fact that the point-like four-fermion interaction is actually replaced by the effects of a W − boson. This gives a correction factor

2 3mµ FW =1+ 2 1.000001, (8.72) 5MW ≈ using MW = 80.4 GeV. The predicted decay rate deﬁning GF experimentally including all these higher- order eﬀects is 2 5 GF mµ Γ − = F F F . (8.73) µ 192π3 kin QED W

The dominant remaining uncertainty in GF comes from the experimental input of the muon lifetime.

135 8.4 Inverse muon decay (e−ν ν µ−) µ → e Consider the process of muon-neutrino scattering on an electron:

e−ν ν µ−. (8.74) µ → e The Feynman diagram for this process is:

e−,pa νe, k1

νµ,pb µ−, k2 in which we see that the following particles have been crossed from the previous diagram for µ+ decay:

+ initial µ final µ− (8.75) → + final e initial e− (8.76) → final ν initial ν . (8.77) µ → µ In fact, this scattering process is often known as inverse muon decay. To apply the Crossing Symme- try Theorem stated in section 6.6, we can assign momentum labels pa, pb, k1, k2 to e−, νµ, νe, µ− respectively, as shown in the figure, and then make the following comparison table:

µ+ e+ν ν e ν ν µ → e µ − µ → e − + µ ,pa0 µ−, k2 + e , k10 e−,pa

νe, k20 νe, k1

νµ, k30 νµ,pb Therefore, we obtain the spin-summed, squared matrix element for e ν ν µ by making the − µ → e − replacements

p0 = k ; k0 = p ; k0 = k ; k0 = p (8.78) a − 2 1 − a 2 1 3 − b in eq. (8.68), and then multiplying by ( 1)3 for three crossed fermions: − 2 2 − − e νµ νeµ = 128GF (k2 k1)(pa pb). (8.79) |M → | · · spinsX Let us evaluate this result in the limit of high-energy scattering, so that mµ can be neglected, and in the center-of-momentum frame. In that case, all four particles being treated as massless, we can take the kinematics results from eqs. (6.177)-(6.181), so that p p = k k = s/2, and a · b 1 · 2 2 2 2 − − e νµ νeµ = 32GF s . (8.80) |M → | spinsX

136 Including a factor of 1/2 for the average over the initial-state electron spin,§ and using eq. (5.80),

dσ G2 s = F . (8.81) d(cos θ) 2π

1 This diﬀerential cross-section is isotropic (independent of θ), so it is trivial to integrate 1 d(cos θ) = 2 − to get the total cross-section: R

2 GF s − − σe νµ νeµ = . (8.82) → π Numerically, we can evaluate this using eq. (8.64):

2 √s − − σe νµ νeµ = 16.9 fb . (8.83) → ÃGeV!

In a typical experimental setup, the electrons will be contained in a target of ordinary material at rest in the lab frame. The muon neutrinos might be produced from a beam of decaying µ−, which are in turn produced by decaying pions, as discussed later. If we call the νµ energy in the lab frame Eνµ , then the center-of-momentum energy is given by

√s = E = 2E m + m2 2E m . (8.84) CM νµ e e ≈ νµ e q q Substituting this into eq. (8.83) gives:

2 Eνµ − − σe νµ νeµ = 1.7 10− fb . (8.85) → GeV × µ ¶ This is a very small cross-section for typical neutrino energies encountered in present experiments, but it does grow with Eνµ . The isotropy of e ν ν µ scattering in the center-of-momentum frame can be understood from − µ → e − considering what the helicities dictated by the weak interactions tell us about the angular momentum. Since this is a weak interaction process involving only fermions and not anti-fermions, they are all L helicity.

µ−

ν e− θ µ

νe

We therefore see that the initial and ﬁnal states both have total spin 0, so that the process is s-wave, and therefore necessarily isotropic.

§In the Standard Model, all neutrinos are left-handed, and all antineutrinos are right-handed. Since there is only one possible νµ helicity, namely L, it would be incorrect to average over the νµ spin. This is a general feature; one should never average over initial-state neutrino or antineutrino spins, as long as they are being treated as massless.

137 8.5 e−ν µ−ν e → µ As another example, consider the process of antineutrino-electron scattering:

e−ν µ−ν . (8.86) e → µ This process can again be obtained by crossing µ+ e+ν ν according to → e µ + initial µ final µ− (8.87) → + final e initial e− (8.88) → final ν initial ν , (8.89) e → e as can be seen from the Feynman diagram:

e−,pa µ−, k1

νe,pb νµ, k2

Therefore, we obtain the spin-summed squared matrix element for e ν µ ν by making the − e → − µ substitutions

p0 = k ; k0 = p ; k0 = p ; k0 = k (8.90) a − 1 1 − a 2 − b 3 2 in eq. (8.68), and multiplying again by ( 1)3 because of the three crossed fermions. The result this − time is:

2 = 128G2 (k p )(p k ) = 32G2 u2 = 8G2 s2(1 + cos θ)2, (8.91) |M| F 1 · b a · 2 F F spinsX where eqs. (6.179) and (6.181) for 2 2 massless kinematics have been used. Here θ is the angle between → the incoming e− and the outgoing µ− 3-momenta. Substituting this result into eq. (5.80), with a factor of 1/2 to account for averaging over the initial e− spin, we obtain:

− − 2 dσe νe µ νµ G s → = F (1 + cos θ)2. (8.92) d(cos θ) 8π Performing the d(cos θ) integration gives a total cross section of:

2 GF s − − σe νe µ νµ = . (8.93) → 3π

This calculation shows that in the center-of-momentum frame, the µ− tends to keep going in the same direction as the original e−. This can be understood from the helicity-spin-momentum diagram:

138 µ−

e− θ νe

νµ

Since the helicities of e−, νe, µ−, νµ are respectively L, R, L, R, the total spin of the initial state must be pointing in the direction opposite to the e− 3-momentum, and the total spin of the ﬁnal state must be pointing opposite to the µ− direction. The overlap between these two states is therefore maximized when the e− and µ− momenta are parallel, and vanishes when the µ− tries to come out in the opposite direction to the e−. Of course, this reaction usually occurs in a laboratory frame in which the initial e− was at rest, so one must correct for this when interpreting the distribution in the lab frame. The total cross section for this reaction is 1/3 of that for the reaction e ν µ ν . This is because − e → − µ the former reaction is an isotropic s-wave (angular momentum 0), while the latter is a p-wave (angular momentum 1), which can only use one of the three possible J = 1 ﬁnal states, namely, the one with J~ pointing along the νµ direction.

8.6 Charged currents and π± decay

The interaction Lagrangian term which is responsible for muon decay and the cross-sections discussed above is just one term in the weak-interaction Lagrangian. More generally, we can write the Lagrangian + as a product of a weak-interaction charged current Jρ− and its complex conjugate Jρ :

+ ρ = 2√2G J J − . (8.94) Lint − F ρ The weak-interaction charged current is obtained by adding together terms for pairs of fermions, with the constraint that the total charge of the current is 1, and all Dirac fermion ﬁelds involved in the − current are left-handed, and all barred ﬁelds are right-handed:

Jρ− = νeγρPLe + νµγρPLµ + ντ γρPLτ + uγρPLd0 + cγρPLs0 + tγρPLb0. (8.95)

Notice that we have included contributions for the quarks. The quark ﬁelds d0, s0, and b0 appearing here are actually not quite mass eigenstates, because of mixing; this is the reason for the primes. The complex conjugate of Jρ− has charge +1, and is given by:

+ ρ ρ ρ Jρ = (Jρ−)∗ = eγρPLνe + µγρPLνµ + τγρPLντ + d0γ PLu + s0γ PLc + b0γ PLt. (8.96)

Both J ρ and J +ρ transform under proper Lorentz transformations as four-vectors, and are V A − − fermion bilinears. Unfortunately, there is an obstacle to a detailed, direct testing of the (V A)(V A) form of the − − weak interaction Lagrangian for quarks. This is because the quarks are bound into hadrons by strong interactions, so that cross-sections and decays involving the weak interactions are subject to very large

139 and very complicated corrections. However, one can still do a quantitative analysis of some aspects of weak decays involving hadrons, by a method of parameterizing our ignorance.

To see how this works, consider the process of charged pion decay. A π− consists of a bound state made from a valence d quark and u antiquark, together with many virtual gluons and quark-antiquark pairs. The Feynman diagram for π µ ν decay following from the current-current Lagrangian − → − µ therefore looks like:

µ−, k1 d

π−

νµ, k2

Unfortunately, the left-part of this Feynman diagram, involving the complications of the π− bound state, is just a cartoon for the strong interactions, which are not very amenable to calculation. The u antiquark and d quark do not even have ﬁxed momenta in this diagram, since they can exchange energy and momentum with each other and with the virtual gluons and quark-anti-quark pairs. In principle, one can ﬁnd some distribution for the u and d momenta, and try to average over that distribution, but the strong interactions are very complicated so this is not very easy to do from the theoretical side. However, by considering what we do know about the current-current Lagrangian, we can write down the general form of the reduced matrix element. First, we know that the external state spinors for the fermions are: Particle Momentum Spinor µ− k1 u1 (8.97) νµ k2 v2 In terms of these spinors, we can write:

= i√2G f p (u γρP v ). (8.98) M − F π ρ 1 L 2 ρ In this formula, the factor (u1γ PLv2) just reﬂects the fact that leptons are immune from the complications of the strong interactions. The factor fπpρ takes into account the part of the reduced matrix ρ element involving the π−; here p is the 4-momentum of the pion. The point is that whatever the pion factor in the reduced matrix element is, we know that it is a four-vector in order to contract with the lepton part, and it must be proportional to pρ, since there is no other vector quantity in the problem that it can depend on. (Recall that pions are spinless, so there is no spin dependence.) So we are simply parameterizing all of our ignorance of the bound-state properties of the pion in terms of a single constant fπ, called the pion decay constant. It is a quantity with dimensions of mass. In principle we could compute it if we had perfect ability to calculate with the strong interactions. In practice, fπ is an experimentally measured quantity, with its value following most accurately from the π− lifetime

140 that we will compute below. The factor √2GF is another historical convention; it could have been absorbed into the deﬁnition of fπ. But it is useful to have the GF appear explicitly as a sign that this is a weak interaction; then fπ is entirely a strong-interaction parameter. Let us now compute the decay rate for π µ ν . Taking the complex square of the reduced − → − µ matrix element eq. (8.98), we have:

2 = 2G2 f 2p p (u γρP v )(v P γσu ). (8.99) |M| F π ρ σ 1 L 2 2 R 1 Now summing over ﬁnal state spins in the usual way gives:

2 2 2 ρ σ 2 2 = 2G f pρpσTr[γ PLk/ PRγ (k/ + mµ)] = 2G f Tr[p/k/ PRp/k/ ]. (8.100) |M| F π 2 1 F π 2 1 spinsX

Note that we do not neglect the mass of the muon, since mµ/mπ± = 0.1056 GeV/0.1396 GeV = 0.756 is not a small number. However, the term in eq. (8.100) which explicitly involves mµ does not contribute, since the trace of 3 gamma matrices (with or without a PR) vanishes. Evaluating the trace, we have:

2 Tr[p/k/ PRp/k/ ] = 4(p k1)(p k2) 2p (k1 k2). (8.101) 2 1 · · − · The decay kinematics tells us that:

2 2 2 2 2 2 p = mπ± ; k1 = mµ; k2 = mνµ = 0; (8.102) 1 2 2 2 1 2 2 p k = [ (p k ) + p + k ]= (m ± + m ); (8.103) · 1 2 − − 1 1 2 π µ 1 2 2 2 1 2 2 p k = [ (p k ) + p + k ]= ( m + m ± ); (8.104) · 2 2 − − 2 2 2 − µ π 1 2 2 2 1 2 2 k k = [(k + k ) k k ]= (m ± m ), (8.105) 1 · 2 2 1 2 − 1 − 2 2 π − µ 2 2 2 Therefore, Tr[p/k/ PRp/k/ ]= m (m ± m ), and 2 1 µ π − µ 2 2 2 2 2 2 mµ = 2GF fπ mπ± mµ 1 2 , (8.106) |M| Ã − mπ± ! spinsX so that using eq. (7.24), we get:

2 2 2 2 2 GF fπ mπ± mµ mµ dΓ= 2 1 2 dφd(cos θ), (8.107) 32π Ã − mπ± ! where (θ, φ) are the angles for the µ− three-momentum. Of course, since the pion is spinless, the diﬀerential decay rate is isotropic, so the angular integration trivially gives dφd(cos θ) 4π, and: → 2 2 2 2 2 GF fπ mπ± mµ mµ Γ(π− µ−νµ)= 1 2 . (8.108) → 8π Ã − mπ± ! The charged pion can also decay according to π e ν . The calculation of this decay rate is − → − e identical to the one just given, except that me is substituted everywhere for mµ. Therefore, we have:

2 2 2 2 2 GF fπ mπ± me me Γ(π− e−νe)= 1 2 , (8.109) → 8π Ã − mπ± !

141 and the ratio of branching fractions is predicted to be:

2 2 2 2 BR(π− e−νe) Γ(π− e−νe) me mπ± me 4 → = → = − = 1.2 10− . (8.110) BR(π µ ν ) Γ(π µ ν ) m2 m2 m2 × − → − µ − → − µ Ã µ !Ã π± − µ !

The dependence on fπ has canceled out of the ratio eq. (8.110), which is therefore a robust prediction of the theory. Since there are no other kinematically-possible two-body decay channels open to π−, it should decay to µ−νµ almost always, with a rare decay to e−νe occurring 0.012% of the time. This has been conﬁrmed experimentally. We can also use the measurement of the total lifetime of the π− to ﬁnd fπ numerically, using eq. (8.108). The result is:

fπ = 0.128 GeV. (8.111)

It is not surprising that this value is of the same order-of-magnitude as the mass of the pion. The most striking feature of the π decay rate is that it is proportional to m2 , with proportional − µ M to mµ. This is what leads to the strong suppression of decays to e−νe. We found this result just by calculating. To understand it better, we can draw a momentum-helicity-spin diagram, using the fact that the `− and ν` produced in the weak interactions are L and R respectively:

J = 0 −→ −→ e νe π−

The π− has spin 0, but the ﬁnal state predicted by the weak interaction helicities unambiguously has spin 1. Therefore, if helicity were exactly conserved, the π− could not decay at all! However, helicity conservation only holds in the high-energy limit in which we can treat all fermions as massless.

This decay is said to be helicity-suppressed, since the only reason it can occur is because mµ and me are non-zero. In the limit m 0, we recover exact helicity conservation and the reduced matrix ` → 2 element and the decay lifetime vanish. This explains why they should be proportional to m` and m` respectively. The helicity suppression of this decay is therefore a good prediction of the rule that the weak interactions aﬀect only L fermions and R antifermions. In the ﬁnal state, the charged lepton

µ− or e− is said to undergo a helicity flip, meaning that the L-helicity fermion produced by the weak interactions has an amplitude to appear in the final state as a R-fermion. In general, a helicity flip for a fermion entails a suppression in the reduced matrix element proportional to the mass of the fermion divided by its energy. Having computed the decay rate following from eq. (8.98), let us find a Lagrangian that would give rise to it involving a quantum field for the pion. Although the pion is a composite, bound-state particle, we can still invent a quantum field for it, in an approximate, “effective” description. The π− is a charged spin-0 field. Previously, we studied spin-0 particles described by a real scalar field. However, the particle and antiparticle created by a real scalar field turned out to be the same thing. Here we

142 + want something different; since the π− is charged, its antiparticle π is clearly a different particle. This means that the π− particle should be described by a complex scalar field.

Let us therefore deﬁne π−(x) to be a complex scalar ﬁeld, with its complex conjugate given by

+ π (x) (π−(x))∗. (8.112) ≡ We can construct a real free Lagrangian density from these complex ﬁelds as follows:

+ µ 2 + = ∂ π ∂ π− m ± π π−. (8.113) L µ − π [Compare to the Lagrangian density for a real scalar field, eq. (4.18).] This Lagrangian density describes + free pion fields with mass mπ± . At any fixed time t = 0, the π and π− fields can be expanded in creation and annihilation operators as:

i~p ~x i~p ~x π−(~x)= dp˜ (e · a~p, + e− · a†~p,+); (8.114) − Z + i~p ~x i~p ~x π (~x)= dp˜ (e · a~p,+ + e− · a†~p, ). (8.115) Z − Note that these ﬁelds are indeed complex conjugates of each other, and that they are each complex since a~p, and a~p,+ are taken to be independent. The operators a~p, and a†~p, act on states by destroying − − − and creating a π− particle with 3-momentum ~p. Likewise, the operators a~p,+ and a†~p,+ act on states by destroying and creating a π+ particle with 3-momentum ~p. In particular, the single particle states are:

a†~p, 0 = π−; ~p (8.116) −| i | i + a† 0 = π ; ~p (8.117) ~p,+| i | i One can now carry through canonical quantization as usual. Given an interaction Lagrangian, one can derive the corresponding Feynman rules for the propagator and interaction vertices. Since a π− moving forward in time is a π+ moving backwards in time, and vice versa, there is only one propagator for π± fields. It differs from the propagator for an ordinary scalar in that it carries an arrow indicating the direction of the flow of charge:

i ←→ p2 m2 + i² − π±

+ The external state pion lines also carry an arrow direction telling us whether it is a π− or a π particle. A pion line entering from the left with an arrow pointing to the right means a π+ particle in the initial state, while a line entering from the left with an arrow pointing back to the left means a π− particle in the initial state. Similarly, if a pion line leaves the diagram to the right, it represents a ﬁnal state pion, + with an arrow to the right meaning a π and an arrow to the left meaning a π−. We can summarize this with the following mnemonic ﬁgures:

143 initial state π+:

initial state π−:

ﬁnal state π+:

ﬁnal state π−:

In each case the Feynman rule factor associated with the initial- or ﬁnal-state pion is just 1. Returning to the reduced matrix element of eq. (8.98), we can interpret this as coming from a pion-lepton-antineutrino interaction vertex. When we computed the decay matrix element, the pion was on-shell, but in general this need not be the case. The pion decay constant fπ must therefore be generalized to a function f(p2), with

2 f(p ) p2=m2 = fπ (8.118) | π± 2 when the pion is on-shell. The momentum-space factor f(p )pρ can be interpreted by identifying the 4-momentum as a diﬀerential operator acting on the pion ﬁeld, using:

pρ i∂ρ. (8.119) ↔ Then reversing the usual procedure of inferring the Feynman rule from a term in the interaction

Lagrangian, we conclude that the eﬀective interaction describing π− decay is:

ρ 2 − = (µγ P ν ) f( ∂ )∂ π−. (8.120) Lint,π µνµ − L µ − ρ Here f( ∂2) can in principle be defined in terms of its power-series expansion in the differential operator − ∂2 = ~ 2 ∂2 acting on the pion field. In practice, one usually just works in momentum space where − ∇ − t it is f(p2). Since the Lagrangian must be real, we must also include the complex conjugate of this term:

ρ 2 + + = (ν γ P µ) f( ∂ )∂ π . (8.121) Lint,π µνµ − µ L − ρ The Feynman rules for these eﬀective interactions are: µ,a

π− √2G f(p2)p (γρP ) ←→ F ρ L ab

νµ, b

144 and

νµ,a

π+ √2G f(p2)p (γρP ) ←→ F ρ L ab

µ, b

In each Feynman diagram, the arrow on the pion line describes the direction of ﬂow of charge, and the 4-momentum pρ is taken to be ﬂowing in to the vertex. When the pion is on-shell, one can replace 2 f(p ) by the pion decay constant fπ.

Other charged mesons made out of a quark and antiquark, like the K±, D±, and Ds±, have their own decay constants fK , fD, and fDs , and their decays can be treated in a similar way.

8.7 Unitarity, renormalizability, and the W boson

An important feature of weak-interaction 2 2 cross sections following from Fermi’s four-fermion → interaction is that they grow proportional to s for very large s; see eqs. (8.82) and (8.93). This had to be true on general grounds just from dimensional analysis. Any reduced matrix element that contains one four-fermion interaction will be proportional to GF , so the cross-section will have to be proportional 2 4 2 to GF . Since this has units of [mass]− , and cross-sections must have dimensions of [mass]− , it must be that the cross-section scales like the square of the characteristic energy of the process, s, in the high- energy limit in which all other kinematic mass scales are comparatively unimportant. This behavior of σ s is not acceptable for arbitrarily large s, since the cross-section is bounded by the fact that ∝ the probability for any two particles to scatter cannot exceed 1. In quantum mechanical language, the iHt constraint is on the unitarity of the time-evolution operator e− . If the cross-section grows too large, then our perturbative approximation e iHt = 1 iHt represented by the lowest-order Feynman diagram − − must break down. The reduced matrix element found from just including this Feynman diagram will have to be compensated somehow by higher-order diagrams, or by changing the physics of the weak interactions at some higher energy scale. Let us develop the dimensional analysis of ﬁelds and couplings further. We know that the La- grangian must have the same units as energy. In the standard system in which c =h ¯ = 1, this is equal 3 3 to units of [mass]. Since d ~x has units of length, or [mass]− , and

L = d3~x , (8.122) L Z it must be that has units of [mass]4. This fact allows us to evaluate the units of all fields and couplings L in a theory. For example, a spacetime derivative has units of inverse length, or [mass]. Therefore, from the kinetic terms for scalars, fermions, and vector fields found for example in eqs. (4.18), (4.26), and (4.34), we find that these types of fields must have dimensions of [mass], [mass]3/2, and [mass]

145 respectively. This allows us to evaluate the units of various possible interaction couplings that appear in the Lagrangian density. For example, a coupling of n scalar fields, λ = n φn (8.123) Lint − n! 4 n implies that λn has units of [mass] − . A vector-fermion-fermion coupling, like e in QED, is dimensionless. The effective coupling fπ for on-shell pions has dimensions of [mass], because of the presence of a spacetime derivative together with a scalar field and two fermion fields in the Lagrangian. Summarizing this information for the known types of fields and couplings that we have encountered so far:

Object Dimension Role [mass]4 Lagrangian density L ∂µ [mass] derivative φ [mass] scalar field Ψ [mass]3/2 fermion field Aµ [mass] vector field 3 λ3 [mass] scalar coupling 0 4 λ4 [mass] scalar coupling y [mass]0 scalar-fermion-fermion (Yukawa) coupling e [mass]0 photon-fermion-fermion coupling 2 4 GF [mass]− fermion coupling 2 fπ [mass] fermion -scalar-derivative coupling u,v, u, v [mass]1/2 external-state spinors

4 Ni Nf Ni N [mass] − − reduced matrix element for Ni Nf particles M → f → 2 σ [mass]− cross section Γ [mass] decay rate.

It is a general fact that theories with couplings with negative mass dimension, like GF , or λn for n 5, always suffer from a problem known as non-renormalizability. In a renormalizable theory, ≥ ¶ the divergences that occur in loop diagrams due to integrating over arbitrarily large 4-momenta for virtual particles can be regularized by introducing a cutoff, and then the resulting dependence on the unknown cutoff can be absorbed into a redefinition of the masses and coupling constants of the theory. In contrast, in a non-renormalizable theory, one finds that this process requires introducing an infinite number of different couplings, each of which must be redefined in order to absorb the momentum-cutoff dependence. This dependence on an infinite number of different coupling constants makes non-renormalizable theories non-predictive, although only in principle. We can always use non-renormalizable theories as effective theories at low energies, as we have done in the case of the

¶The converse is not true; just because a theory has only couplings with positive or zero mass dimension does not guarantee that it is renormalizable. It is a necessary, but not suﬃcient, condition.

146 four-fermion theory of the weak interactions. However, when probed at sufficiently high energies, a non- renormalizable theory will encounter related problems associated with the apparent failure of unitarity (cross sections that grow uncontrollably with energy) and non-renormalizability (an uncontrollable dependence on more and more unknown couplings that become more and more important at higher energies). For this reason, we should always try to describe physical phenomena using renormalizable theories if we can. An example of a useful non-renormalizable theory is gravity. The effective coupling constant for Feynman diagrams involving gravitons is 1/M 2 , where M = 2.4 1018 GeV is the “reduced Planck Planck × Planck mass”. Like GF , this coupling has negative mass dimension, so tree-level cross-sections for 2 2 scattering involving gravitons grow with energy like s. This is not a problem as long as we stick → to scattering energies M , which corresponds to all known experiments and directly measured ¿ Planck phenomena. However, we do not know how unitarity is restored in gravitational interactions at energies comparable to MPlanck or higher. Unlike the case of the weak interactions, it is hard to conceive of an experiment with present technologies that could test competing ideas. To find a renormalizable “fix” for the weak interactions, we note that the (V A)(V A) current- − − current structure of the four-fermion coupling could come about from the exchange of a heavy vector particle. To do this, we imagine “pulling apart” the two currents, and replacing the short line segment by the propagator for a virtual vector particle. For example, one of the current-current terms is: µ, a e,c µ, a e, c

W ←→

νµ, b νe,d νµ, b νe,d

Since the currents involved have electric charges 1 and +1, the vector boson must carry charge 1 − − to the right. This is the W − vector boson. By analogy with the charged pion, W ± are complex vector ﬁelds, with an arrow on its propagator indicating the direction of ﬂow of charge.

The Feynman rule for the propagator of a charged vector W ± boson carrying 4-momentum p turns out to be:

ρ σ i p p g + ρ σ ←→ p2 m2 + i² − ρσ m2 − W " W #

In the limit of low energies, pρ m , this propagator just becomes a constant: | | ¿ W i p p g g + ρ σ i ρσ . (8.124) p2 m2 + i² − ρσ m2 −→ m2 − W " W # W In order to complete the correspondence between the eﬀective four-fermion interaction and the more

147 fundamental version involving the vector boson, we need W -fermion-antifermion vertex Feynman rules of the form: `, a

W, ρ g ρ i (γ PL) ←→ − √2 ab

ν`, b

ν`,a

W, ρ g ρ i (γ PL) ←→ − √2 ab

`, b

Here g is a fundamental coupling of the weak interactions, and the 1/√2 is a standard convention. These are the Feynman rules involving W ± interactions with leptons; there are similar rules for interactions + with the quarks in the charged currents Jρ and Jρ− given earlier in eqs. (8.95) and (8.96). The interaction Lagrangian for W bosons with standard model fermions corresponding to these Feynman rules is:

g +ρ ρ + int = W J − + W − J . (8.125) L −√2 ρ ρ ³ ´ Comparing the four-fermion vertex to the reduced matrix element from W -boson exchange, we ﬁnd that we must have: ig 2 i i2√2G = − , (8.126) F √ 2 − µ 2 ¶ ÃmW ! so that g2 G = . (8.127) F 2 4√2mW

The W ± boson has been discovered, with a mass mW = 80.4 GeV, so we conclude that

g 0.65. (8.128) ≈ Since this is a dimensionless coupling, there is at least a chance to make this into a renormalizable theory which is unitary in perturbation theory. At very high energies, the W ± propagator will behave 2 2 like 1/p , rather than the 1/mW that is encoded in GF in the four-fermion approximation. This “softens” the weak interactions at high energies, leading to cross sections which fall, rather than rise, at very high √s. When a massive vector boson appears in a ﬁnal state, it has a Feynman rule given by a polarization vector ²µ(p, λ), just like the photon did. The diﬀerence is that a massive vector particle V has three physical polarization states λ = 1, 2, 3, satisfying

ρ p ²ρ(p, λ)=0 (λ = 1, 2, 3). (8.129)

148 One can sum over these polarizations for an initial or ﬁnal state in a squared reduced matrix element, with the result:

3 pρpσ ²ρ(p, λ)²σ∗ (p, λ)= gρσ + 2 . (8.130) − mV λX=1 Summarizing the propagator and external state Feynman rules for a generic massive vector boson for future reference:

ρ σ i p p g + ρ σ ←→ p2 m2 + i² − ρσ m2 − V " V #

µ initial state vector: ²µ(p, λ) ←→

µ ﬁnal state vector: ²∗ (p, λ) ←→ µ

If the massive vector is charged, like the W ± bosons, then an arrow is added to each line to show the direction of flow of charge. The weak interactions and the strong interactions are invariant under non-Abelian gauge transformations, which involve a generalization of the type of gauge invariance we have already encountered in the case of QED. This means that the gauge transformations not only multiply fields by phases, but can mix the fields. In the next section we will begin to study the properties of field theories, known as Yang-Mills theories, which have a non-Abelian gauge invariance. This will enable us to get a complete theory of the weak interactions.

149 9 Gauge theories

9.1 Groups and representations

In this section, we will seek to generalize the idea of gauge invariance. Recall that in QED the Lagrangian is deﬁned in terms of a covariant derivative

Dµ = ∂µ + iQeAµ (9.1) and a ﬁeld strength

F = ∂ A ∂ A (9.2) µν µ ν − ν µ as 1 = F µνF + iΨD/Ψ mΨΨ. (9.3) L −4 µν − This Lagrangian is designed to be invariant under the local gauge transformation 1 A A0 = A ∂ θ (9.4) µ → µ µ − e µ iQθ Ψ Ψ0 = e Ψ, (9.5) → where θ(x) is any function of spacetime, called a gauge parameter. Now, the result of doing one gauge transformation θ1 followed by another gauge transformation θ2 is always a third gauge transformation parameterized by the function θ1 + θ2: 1 1 1 A (A ∂ θ ) ∂ θ = A ∂ (θ + θ ); (9.6) µ → µ − e µ 1 − e µ 2 µ − e µ 1 2 Ψ eiQθ2 (eiQθ1 Ψ) = eiQ(θ1+θ2)Ψ. (9.7) → Mathematically, these gauge transformations are an example of a group.

A group is a set of elements G = (I, g1, g2,...) and a rule for multiplying them, with the properties:

1) Closure: If gi and gj are elements of the group G, then the product gigj is also an element of G.

2) Associativity: gi(gjgk) = (gigj)gk.

3) Existence of an Identity: There is a unique element I of the group, such that for all gi in G,

Igi = giI = gi. 1 1 1 4) Inversion: For each gi, there is a unique inverse element (gi)− satisfying (gi)− gi = gi(gi)− = I.

It may or may not be also true that the group also satisﬁes the commutativity property:

gigj = gjgi. (9.8)

If this is satisﬁed, then the group is commutative or Abelian. Otherwise it is non-commutative or non-Abelian.

150 In trying to generalize the QED Lagrangian, we will be interested in continuous Lie groups. A continuous group has an uncountably infinite number of elements labeled by one or more continuously varying parameters, which turn out to be nothing other than the generalizations of the gauge parameter θ in QED. A Lie group is a continuous group that also has the desirable property of being differentiable with respect to the gauge parameters. The action of group elements on physics states or fields can be represented by a set of complex n n matrices acting on n-dimensional complex vectors. This association of group elements with n n × × matrices and the states or fields that they act on is said to form a representation of the group. The matrices obey the same rules as the group elements themselves. For example, the group of QED gauge transformations is the Abelian Lie group U(1). According to eq. (9.5), the group is represented on Dirac fermion fields by complex 1 1 matrices: × iQθ UQ(θ)= e . (9.9)

Here θ labels the group elements, and the charge Q labels the representation of the group. So we can say that the electron, muon, and tau Dirac fields live in three copies of a representation of the group U(1) with charge Q = 1; the Dirac fields for up, charm, and top quarks each live in different − representations with charge Q = 2/3; and the Dirac fields for down, strange and bottom quarks each live in a representation with Q = 1/3. We can read off the charge of any field if we know how it − transforms under the gauge group. A barred Dirac field transforms with the opposite phase from the original Dirac field of charge Q, and therefore has charge Q. − Objects which transform into themselves with no change are said to be in the singlet representation. In general, the Lagrangian should be invariant under gauge transformations, and therefore must be in the singlet representation. For example, each term of the QED Lagrangian carries no charge, and so is a singlet of U(1). The photon field Aµ has charge 0, and is therefore usually said (by a slight abuse of language) to transform as a singlet representation of U(1). [Technically, it does not really transform under gauge transformations as any representation of the group U(1), because of the derivative term in eq. (9.4), unless θ is a constant function so that one is making the same transformation everywhere in spacetime.] Let us now generalize to non-Abelian groups, which always involve representations containing more than one field or state. Let ϕi be a set of objects that together transform in some representation R of the group G. The number of components of ϕi is called the dimension of the representation, dR, so that i = 1,...,dR. Under a group transformation,

j ϕ ϕ0 = U ϕ (9.10) i → i i j j where Ui is a representation matrix. We are especially interested in transformations which are represented by unitary matrices, so that their action can be realized on the quantum Hilbert space by a unitary operator. Consider the subset of group elements which are inﬁnitesimally close to the identity

151 element. We can write these in the form:

j a a j U(²)i = (1+ i² T )i . (9.11)

aj Here the Ti are a basis for all the possible inﬁnitesimal group transformations. The number of matrices a a T is called the dimension of the group, dG, and there is an implicit sum over a = 1,...,dG. The ² are a set of dG inﬁnitesimal gauge parameters (analogous to θ in QED) which tell us how much of each is included in the transformation represented by U(²). If U(²) is to be unitary, then

1 a a U(²)† = U(²)− = 1 i² T , (9.12) − from which it follows that the matrices T a must be Hermitian. a a Consider two group transformations g² and gδ parameterized by ² and δ . By the closure property we can then form a new group element

1 1 g²gδg²− gδ− . (9.13)

Working with some particular representation, this corresponds to

1 1 a a b b c c d d U(²)U(δ)U(²)− U(δ)− = (1+ i² T )(1 + iδ T )(1 i² T )(1 iδ T ) (9.14) − − = 1 ²aδb[T a, T b]+ ... (9.15) − The closure property requires that this is a representation of the group element in eq. (9.13), which must also be close to the identity. It follows that

[T a, T b]= if abcT c (9.16) for some set of numbers f abc, called the structure constants of the group. In practice, one often picks a particular representation of matrices T a as the defining or fundamental representation. This determines the structure constants f abc once and for all. The set of matrices T a for all other representations are then required to reproduce eq. (9.16), which fixes their overall normalization. Equation (9.16) defines the Lie algebra corresponding to the Lie group, and the hermitian matrices T a are said to be generators of the Lie algebra for the corresponding representation. Physicists have a bad habit of using the words “Lie group” and “Lie algebra” interchangeably, because we often only care about the subset of gauge transformations that are close to the identity. For any given representation, one can always choose the generators so that:

a b ab Tr(TRTR)= I(R)δ . (9.17)

The number I(R) is called the index of the representation. A standard choice is that the index of the fundamental representation is 1/2. (This can always be achieved by rescaling the T a, if necessary.) From eqs. (9.16) and (9.17), one obtains for any other representation R:

abc a b c iI(R)f = Tr([TR, TR]TR) (9.18)

152 It follows, from the cyclic property of the trace, that f abc is totally antisymmetric under interchange of any two of a,b,c. By using the Jacobi identity,

[T a, [T b, T c]]+[T b, [T c, T a]]+[T c, [T a, T b]]=0, (9.19) which holds for any three matrices, one also ﬁnds the useful result:

f adef bce + f cdef abe + f bdef cae = 0. (9.20)

Two representations R and R0 are said to be equivalent if there exists some ﬁxed matrix X such that:

a 1 a XTRX− = TR0 , (9.21) for all a. Obviously, this requires that R and R0 have the same dimension. From a physical point of view, equivalent representations are indistinguishable from each other. A representation R of a Lie algebra is said to be reducible if it is equivalent to a representation in a block-diagonal form; in other words, if there is some matrix X that can be used to put all of the TR simultaneously in a block-diagonal form: a tr1 0 ... 0 0 ta ... 0 a 1  r2  XTRX− = . . . . for all a. (9.22) . . .. .    a   0 0 ... trn    a Here the tri are representation matrices for smaller representations ri. One calls this a direct sum, and writes it as

R = r r ... r . (9.23) 1 ⊕ 2 ⊕ ⊕ n A representation which is not equivalent to a direct sum of smaller representations in this way is said to be irreducible. Heuristically, reducible representations are those which can be chopped up into smaller pieces that can be treated individually. In physics, ﬁelds and states transform as irreducible representation of symmetry groups. With the above conventions on the Lie algebra generators, one can show that for each irreducible representation R:

a a j j (TRTR)i = C(R)δi (9.24)

(with an implicit sum over a = 1,...,dG), where C(R) is another characteristic number of the representation R, called the quadratic Casimir invariant. If we take the sum over a of eq. (9.17), it is equal to the trace of eq. (9.24). It follows that for each irreducible representation R, the dimension, the index, and the Casimir invariant are related to the dimension of the group by:

dGI(R)= dRC(R). (9.25)

153 The simplest irreducible representation of any Lie algebra is just:

aj Ti = 0. (9.26)

This is called the singlet representation. aj Suppose that we have some representation with matrices Ti . Then one can show that the matrices (T aj) also form a representation of the algebra eq. (9.16). This is called the complex conjugate of − i ∗ the representation R, and is often denoted R:

a a T = T ∗. (9.27) R − R If T a is equivalent to T a, so that there is some X such that R R

a 1 a XTRX− = TR, (9.28) then the representation R is said to be a real representation,† and otherwise R is said to be complex.

One can also form the tensor product of any two representations R, R0 of the Lie algebra to get another representation:

a j y j a y a j,y (TR)i δx + δi (TR0 )x = (TR R0 )i,x (9.29) ⊗

The representation R R has dimension d d 0 , and typically is reducible: ⊗ 0 R R

R R0 = R ... R (9.30) ⊗ 1 ⊕ ⊕ n with

0 dRdR = dR1 + ... + dRn . (9.31)

This is a way to make larger representations out of smaller ones. One can check from the identity eq. (9.20) that the matrices

(T a) c = if abc (9.32) b − form a representation, called the adjoint representation, with the same dimension as the group G. As a matter of terminology, the quadratic Casimir invariant of the adjoint representation is also called the Casimir invariant of the group, and given the symbol C(G). Note that, from eq. (9.25), the index of the adjoint representation is equal to its quadratic Casimir invariant:

C(G) C(adjoint) = I(adjoint). (9.33) ≡

I now list, without proof, some further group theory facts regarding Lie algebra representations:

†Real representations can be divided into two sub-cases, “positive-real” and “pseudo-real”, depending on whether the matrix X can or cannot be chosen to be symmetric. In a pseudo-real representation, the T a cannot all be made antisymmetric and imaginary; in a positive-real representation, they can.

154 The number of inequivalent irreducible representations of a Lie group is always inﬁnite. • Unlike group element multiplication, the tensor product multiplication of representations is both • associative and commutative:

(R R ) R = R (R R ) (9.34) 1 ⊗ 2 ⊗ 3 1 ⊗ 2 ⊗ 3 R R = R R (9.35) 1 ⊗ 2 2 ⊗ 1 The tensor product of any representation with the singlet representation just gives the original • representation back:

1 R = R 1= R. (9.36) ⊗ ⊗

The tensor product of two real representations R and R is always a direct sum of representations • 1 2 that are either real or appear in complex conjugate pairs.

The adjoint representation is always real. • The tensor product of two representations contains the singlet representation if and only if they • are complex conjugates of each other:

R R = 1 ... R = R (9.37) 1 ⊗ 2 ⊕ ←→ 2 1 It follows that if R is real, then R R contains a singlet. × The tensor product of a representation and its complex conjugate always contains the adjoint • representation:

R R = Adjoint .... (9.38) ⊗ ⊕ As a corollary of the preceding three rules, the tensor product of the adjoint representation with • itself always contains both the singlet and the adjoint:

Adjoint Adjoint = 1 Adjoint .... (9.39) ⊗ S ⊕ A ⊕ Here the S and A mean that the indices of the two adjoints on the left are combined symmetrically and antisymmetrically respectively.

If the tensor product of two representations contains a third, then the tensor product of the ﬁrst • representation with the conjugate of the third representation contains the conjugate of the second representation:

R R = R ..., R R = R ..., (9.40) 1 ⊗ 2 3 ⊕ ←→ 1 ⊗ 3 2 ⊕ R R R = 1 ..., R R = R .... (9.41) 1 ⊗ 2 ⊗ 3 ⊕ ←→ 1 ⊗ 2 3 ⊕ This is sometimes known as “crossing symmetry” for Lie algebra representations.

155 If R R = r ... r , then the indices satisfy the following rule: • 1 ⊗ 2 1 ⊕ ⊕ n n

I(R1)dR2 + I(R2)dR1 = I(ri). (9.42) Xi=1 Let us recall how Lie algebra representations work in the example of SU(2), the group of unitary 2 2 matrices (that’s the “U(2)” part of the name) with determinant 1 (that’s the “S”, for special, × part of the name). This group is familiar from the study of angular momentum in quantum mechanics, and the deﬁning or fundamental representation is the familiar spin-1/2 one with ϕi with i = 1, 2 or up,down. The Lie algebra generators in the fundamental representation are: σa T a = (a = 1, 2, 3) (9.43) 2 where the σa are the three Pauli matrices (see, for example, eq. (2.70). One ﬁnds that the structure constants are

+1 if a,b,c = 1, 2, 3 or 2, 3, 1 or 3, 1, 2 f abc = ²abc = 1 if a,b,c = 1, 3, 2 or 3, 2, 1 or 2, 1, 3 (9.44)  −  0 otherwise

Irreducible representations exist for any “spin” j = n/2, where n is an integer, and have dimension 2j + 1. The representation matrices J a in the spin-j representation satisfy the SU(2) Lie algebra:

[J a, J b]= i²abcJ c. (9.45)

3m0 Jm ϕm0 = mϕm (9.48) a a m0 (J J )m ϕm0 = j(j + 1)ϕm. (9.49)

We therefore recognize from eq. (9.24) that the quadratic Casimir invariant of the spin-j representation of SU(2) is C(Rj)= j(j + 1). It follows from eq. (9.25) that the index of the spin-j representation is

I(Rj)= j(j + 1)(2j + 1)/3. The j = 1/2 representation is real, because

a a σ ∗ 1 σ X( )X− = , (9.50) − 2 2 where 0 i X = . (9.51) i 0 µ − ¶

156 More generally, one can show that all representations of SU(2) are real. Making a table of the representations of SU(2):

spin dimension I(R) C(R) real? singlet 0 1 0 0 yes fundamental 1/2 2 1/2 3/4 yes adjoint 1 3 2 2 yes 3/2 4 5 15/4 yes ...... j 2j + 1 j(j + 1)(2j + 1)/3 j(j + 1) yes

The tensor product of any two representation of SU(2) is reducible to a direct sum, as:

j j = j j j j + 1 ... (j + j ) (9.52) 1 ⊗ 2 | 1 − 2|⊕| 1 − 2 |⊗ ⊗ 1 2 in which j is used to represent the representation of spin j. The group SU(2) has many applications in physics. First, it serves as the Lie algebra of angular momentum operators. The strong force is approximately (but not exactly) invariant under a different SU(2) isospin symmetry, under which the up and down quarks transform as a j = 1/2 doublet. Isospin is a global symmetry, meaning that the same symmetry transformation must be made simultaneously everywhere: u u exp(iθaσa/2) , (9.53) d d µ ¶ → µ ¶ with a constant θa that does not depend on position in spacetime. The weak interactions involve a still different SU(2), known as weak isospin or SU(2)L. Weak isospin is a gauge symmetry that acts only on left-handed fermion fields. The irreducible j = 1/2 representations of SU(2)L are composed of the pairs of fermions that couple to a W ± boson, namely: ν ν ν eL ; µL ; τL ; (9.54) e µ τ µ L ¶ µ L ¶ µ L ¶ u c t L ; L ; L . (9.55) d s b µ L0 ¶ µ L0 ¶ µ L0 ¶ Here eL means PLe, etc., and the primes mean that these are not quark mass eigenstates. When one makes an SU(2)L gauge transformation, the transformation can be different at each point in spacetime. However, one must make the same transformation simultaneously on each of these representations. We will come back to study the SU(2)L symmetry in more detail later, and see more precisely how it ties into the weak interactions and QED. One can generalize the SU(2) group to non-Abelian groups SU(N) for any integer N 2. The ≥ Lie algebra generators of SU(N) in the fundamental representation are Hermitian traceless N N × matrices. In general, a basis for the complex N N matrices is 2N 2 dimensional, since each matrix has × N 2 entries with a real and imaginary part. The condition that the matrices are Hermitian removes half of these, since each entry in the matrix is required to be the complex conjugate of another entry. Finally, the single condition of tracelessness removes one from the basis. That leaves d = N 2 1 as the SU(N) −

157 dimension of the group and of the adjoint representation. For N 3, the fundamental representation ≥ is complex. For example, Quantum Chromo-Dynamics (QCD), the theory of the strong interactions, is a gauge theory based on the group SU(3), with dimension dG = 8. In the fundamental representation, the generators of the Lie algebra are given by: 1 T a = λa (a = 1,..., 8) (9.56) 2 where the λa are known as the Gell-Mann matrices: 0 1 0 0 i 0 − λ1 = 1 0 0 ; λ2 = i 0 0 ; (9.57)     0 0 0 0 0 0 1 0 0  0 0 1 λ3 = 0 1 0 ; λ4 = 0 0 0 ; (9.58)  −    0 0 0 1 0 0  0 0 i  0 0 0  − λ5 = 0 0 0 ; λ6 = 0 0 1 ; (9.59)     i 0 0 0 1 0 0 0 0   1 0 0 λ7 = 0 0 i ; λ8 = 1 0 1 0 . (9.60)  −  √3   0 i 0 0 0 2    −  Note that each of these matrices is Hermitian and traceless, as required. They have also been engineered to satisfy Tr(λaλb) = 2δab, so that

Tr(T aT b) = 1/2, (9.61) and therefore

I(F ) = 1/2. (9.62)

One can also check that 4 (T aT a) j = δj, (9.63) i 3 i so that

C(F ) = 4/3. (9.64)

By taking commutators of each pair of generators, one ﬁnds that the non-zero structure constants of SU(3) are:

f 123 = 1; (9.65) f 147 = f 156 = f 246 = f 257 = f 345 = f 367 = 1/2; (9.66) − − f 458 = f 678 = √3/2. (9.67)

158 and those related to the above by permutations of indices, following from the condition that the f abc are totally antisymmetric. From these one can ﬁnd the adjoint representation matrices using eq. (9.32), with, for example: 0000 0 0 00 0010 0 0 00   0 100 0 0 00  −  1  0000 0 01/2 0  T = i   , (9.68) adjoint −  000 0 0 1/2 0 0   −   000 0 1/2 0 00     0 0 0 1/20 0 00     −   0000 0 0 00    etc. However, it is almost never necessary to actually use the explicit form of any matrix representation larger than the fundamental. Instead, one relies on group-theoretic identities. For example, calculations of Feynman diagrams often involve the index or Casimir invariant of the fundamental representation, and the Casimir invariant of the group. One can easily compute the latter by using eq. (9.24) and (9.32):

f abcf abd = C(G)δcd, (9.69) with the result C(G)=3. Following is a table of the smallest few irreducible representations of SU(3): dimension I(R) C(R) real? singlet 1 0 0 y fundamental 3 1/2 4/3 n anti-fundamental 3 1/2 4/3 n 6 5/2 10/3 n 6 5/2 10/3 n adjoint 8 3 3 y 10 15/2 6 n 10 15/2 6 n ...

It is usual to refer to each representation by its dimension in boldface. In general, the representations can be classiﬁed by two non-negative integers α and β. The dimension of the representation labeled by α,β is

dα,β = (α + 1)(β + 1)(α + β + 2)/2. (9.70)

Some tensor products involving these representations are:

3 3 = 1 8 (9.71) ⊗ ⊕ 3 3 = 3 6 (9.72) ⊗ A ⊕ S 3 3 = 3 6 (9.73) ⊗ A ⊕ S 3 3 3 = 1 8 8 10 (9.74) ⊗ ⊗ A ⊕ M ⊕ M ⊕ S 8 8 = 1 8 8 10 10 27 . (9.75) ⊗ S ⊕ A ⊕ S ⊕ A ⊕ A ⊕ S

159 Here, when we take the tensor product of two or more identical representations, the irreducible representations on the right side are labeled as A, S, or M depending on whether they involve an antisymmetric, symmetric, or mixed symmetry combination of the indices of the original representations on the left side. In SU(N), one can build any representation out of objects that carry only indices transforming under the fundamental N and anti-fundamental N representations. It is useful to employ lowered indices for the fundamental, and raised indices for the antifundamental. Then an object carrying n lowered and m raised indices:

j1...jm ϕi1...in (9.76) transforms under the tensor product representation

N ... N N ... N . (9.77) ⊗ ⊗ ⊗ ⊗ ⊗ n times m times Of course, this is always reducible.| To reduce{z } it,| one can{z dec}ompose ϕ into parts that have different symmetry and trace properties. So, for example, we can take an object that transforms under SU(3) as N N, and write it as: × 1 1 ϕj = (ϕj δjϕk) + ( δjϕk). (9.78) i i − N i k N i k The first term in parentheses transforms as an adjoint representation, and the second as a singlet, under SU(N). For SU(3), this corresponds to the rule of eq. (9.71). Similarly, an object that transforms under SU(N) as N N can be decomposed as × 1 1 ϕ = (ϕ + ϕ )+ (ϕ ϕ ). (9.79) ij 2 ij ji 2 ij − ji The two terms on the right-hand side correspond to an N(N + 1)/2-dimensional symmetric tensor, and an N(N 1)/2-dimensional antisymmetric tensor, irreducible representations. For N = 3, these − are the 6 and 3 representations, respectively, and this decomposition corresponds to eq. (9.72). By using this process of taking symmetric and anti-symmetric parts and removing traces, one can find all necessary tensor-product representation rules for any SU(N) group.

9.2 The Yang-Mills Lagrangian and Feynman rules

In this subsection, we will construct the Lagrangian and Feynman rules for a theory of Dirac fermions and gauge bosons transforming under a non-Abelian gauge group, called a Yang-Mills theory.

Let the Dirac fermion ﬁelds be given by Ψi, where i is an index in some representation of the gauge aj group with generators Ti . Here i = 1,...,dR and a = 1,...,dG. Under a gauge transformation, we have:

Ψ U jΨ (9.80) i → i j

160 where

U = exp(iθaT a). (9.81)

Specializing to the case of an inﬁnitesimal gauge transformation θa = ²a, we have

Ψ (1 + i²aT a) jΨ . (9.82) i → i j Our goal is to build a Lagrangian that is invariant under this transformation. First let us consider how barred spinors transform. Taking the Hermitian conjugate of eq. (9.82), we ﬁnd

i j a a i Ψ† Ψ† (1 i² T ) , (9.83) → − j where we have used the fact that T a are Hermitian matrices. (Notice that taking the Hermitian conjugate changes the heights of the representation indices, and in the case of matrices, reverses their order. So the Dirac spinor carries a lowered representation index, while the Hermitian conjugate spinor carries a raised index.) Now we can multiply on the right by γ0. The Dirac gamma matrices are completely separate from the gauge group representation indices, so we get the transformation rule for the barred Dirac spinors:

i j Ψ Ψ (1 i²aT a) i. (9.84) → − j We can rewrite this in a slightly diﬀerent way by noting that

ai ai ai Tj = (Tj )† = (T j)∗, (9.85) so that

i a a i j Ψ (1 + i² [ T ∗]) Ψ . (9.86) → − j i Comparing with eq. (9.27), this establishes that Ψ transforms in the complex conjugate of the representation carried by Ψi. i Since Ψ and Ψi transform as complex conjugate representations of each other, their tensor product must be a direct sum of representations that includes a singlet. The singlet is obtained by summing over the index i:

i Ψ Ψi. (9.87)

As a check of this, under an inﬁnitesimal gauge transformation, this term becomes:

i j k Ψ Ψ Ψ (1 i²aT a) i (1 + i²bT b) Ψ (9.88) i → − j i k i = Ψ Ψ + (²2), (9.89) i O

161 where the terms linear in ²a have indeed canceled. Therefore, we can include a fermion mass term in the Lagrangian:

i = mΨ Ψ (9.90) Lm − i

This shows that each component of the ﬁeld Ψi must have the same mass. Next we would like to include a derivative kinetic term for the fermions. Just as in QED, the term

i µ iΨ γ ∂µΨi (9.91) is not acceptable by itself, because ∂µΨi does not gauge transform in the same way that Ψi does. The problem is that the derivative can act on the gauge-parameter function ²a, giving an extra term:

∂ Ψ (1 + i²aT a) j∂ Ψ + i(∂ ²a)T ajΨ . (9.92) µ i → i µ j µ i j By analogy with QED, we can ﬁx this by writing a covariant derivative involving vectors ﬁelds, which will also transform in such a way as to cancel the last term in eq. (9.92):

a aj DµΨi = ∂µΨi + igAµTi Ψj (9.93)

a The vector boson fields Aµ are known as gauge fields. They carry an adjoint representation index a in addition to their spacetime vector index µ. The number of such fields is equal to the number of a generator matrices T , which we recall is the dimension of the gauge group dG. The quantity g is a coupling, known as a gauge coupling. It is dimensionless, and is the direct analog of the coupling e in aj QED. The entries of the matrix Ti play the role played by the charges q in QED. Notice that the definition of the covariant derivative depends on the representation matrices for the fermions, so there is really a different covariant derivative depending on which fermion representation one is acting on. Now one can check that the Lagrangian

i i = iΨ γµD Ψ mΨ Ψ (9.94) Lfermions µ i − i is invariant under infinitesimal gauge transformations, provided that the gauge field is taken to transform as: 1 Aa Aa ∂ ²a f abc²bAc . (9.95) µ → µ − g µ − µ The term with a derivative acting on ²a is the direct analog of a corresponding term in QED, see eq. (9.4). The last term vanishes for Abelian groups like QED, but it is necessary to ensure that the covariant derivative of a Dirac field transforms in in the same way as the field itself:

D Ψ (1 + i²aT a) jD Ψ . (9.96) µ i → i µ j a If we limit ourselves to constant gauge parameters ∂µ² = 0, then the transformation in eq. (9.95) is of the correct form for a ﬁeld in the adjoint representation.

162 Having introduced a gauge field for each Lie algebra generator, we must now include kinetic terms for them. By analogy with QED, the gauge field Lagrangian has the form: 1 gauge = F µνaF a , (9.97) L −4 µν a with an implied sum on a, where Fµν is an antisymmetric field strength tensor for each Lie algebra generator. However, we must also require that this Lagrangian is invariant under gauge transformations. This is accomplished if we choose:

F a = ∂ Aa ∂ Aa gf abcAb Ac . (9.98) µν µ ν − ν µ − µ ν Then one can check, using eq. (9.95), that

F a F a f abc²bF c . (9.99) µν → µν − µν From this it follows that gauge transforms as: L 1 1 1 F µνaF a F µνaF a F µνaf abc²bF c + (²2). (9.100) −4 µν → −4 µν − 2 µν O The extra term linear in ² vanishes, because the part F µνaF c is symmetric under interchange of a c, µν ↔ but its gauge indices are contracted with f abc which is antisymmetric under the same interchange. Therefore, eq. (9.97) is a gauge singlet. Putting the above results together, we have found a gauge-invariant Lagrangian for Dirac fermions coupled to a non-Abelian gauge ﬁeld:

= gauge + . (9.101) LYang-Mills L Lfermions Now we can ﬁnd the Feynman rules for this theory in the usual way. First we identify the kinetic terms that are quadratic in the ﬁelds. That part of is: LYang-Mills

1 i i = (∂ Aa ∂ Aa )(∂µAaν ∂νAaµ)+ iΨ ∂/Ψ mΨ Ψ . (9.102) Lkinetic −4 µ ν − ν µ − i − i

These terms have exactly the same form as in the QED Lagrangian, but with a sum over dG copies of the vector fields, labeled by a, and over dR copies of the fermion field, labeled by i. Therefore we can immediately obtain the Feynman rules for vector and fermion propagators. For gauge fields, we have:

µ, a ν, b i p p δab g + (1 ξ) µ ν p2 + i² µν p2 ←→ ·− − ¸ were pµ is the 4-momentum along either direction in the wavy line, and one can take ξ = 1 for Feynman gauge and ξ = 0 for Landau gauge. The δab in the Feynman rule just means that a gauge ﬁeld does not change to a diﬀerent type as it propagates. Likewise, the Dirac fermion propagator is:

j i i(p/ + m) δj ←→ i p2 m2 + i² −

163 with 4-momentum pµ along the arrow direction, and m the mass of the Dirac fermion. Again the factor j of δi means the fermion does not change its identity as it propagates. The interaction Feynman rules follow from the remaining terms in . First, there is a LYang-Mills fermion-vector-vector interaction coming from the covariant derivative in . Identifying the Feynman LΨ rule as i times the term in the Lagrangian,

= gAa ΨγµT aΨ (9.103) LAΨΨ − µ we get: i µ, a igT ajγµ ←→ − i j

This rule says that the coupling of a gauge field to a fermion line is proportional to the corresponding Lie algebra generator matrix. Since the matrices T a are not diagonal for non-Abelian groups, this interaction can change one fermion into another. The Lagrangian density gauge contains three- L gauge-field and four-gauge-field couplings, proportional to g and g2 respectively. After combining some terms using the antisymmetry of the f abc symbol, they can be written as

= gf abc(∂ Aa)AµbAνc (9.104) LAAA µ ν g2 = f abef cdeAa Ab AµcAνd. (9.105) LAAAA − 4 µ ν The three-gauge-field couplings involve a spacetime derivative, which we can treat according to the rule of eq. (8.119), as i times the momentum of the field it acts on (in the direction going into the vertex). Then the Feynman rule for the interaction of three gauge bosons with (spacetime vector, gauge) indices µ, a and ν, b and ρ,c is obtained by taking functional derivatives of the Lagrangian density with respect to the corresponding fields:

δ3 i a b c AAA, (9.106) δAµδAνδAρ L resulting in: ν, b q p µ, a gf abc[gµν(p q)ρ + gνρ(q k)µ + gρµ(k p)ν] ←→ − − − − k ρ,c

164 where pµ, qµ, and kµ are the gauge boson 4-momenta ﬂowing into the vertex. Likewise, the Feynman rule for the coupling of four gauge bosons with indices µ, a and ν, b and ρ,c and σ, d is:

δ4 i a b c d AAAA, (9.107) δAµδAνδAρδAσ L leading to:

µ, a ν, b ig2 f abef cde(gµρgνσ gµσgνρ) − − h+ f acef bde(gµνgρσ gµσgνρ) ←→ − + f adef bce(gµνgρσ gµρgνσ) − σ, d ρ,c i

There are more terms in these Feynman rules than in the corresponding Lagrangian, since the functional derivatives have a choice of several fields on which to act. Notice that these fields are invariant under the simultaneous interchange of all the indices and momenta for any two vector bosons, for example (µ,a,p) (ν, b, q). The above Feynman rules are all that is needed to calculated tree-level Feynman ↔ diagrams in a Yang-Mills theory with Dirac fermions. External state fermions and gauge bosons are assigned exactly the same rules as for fermions and photons in QED. The external state particles carry a representation or gauge index which is just determined by the interaction vertex to which that line is attached. [However, this is not quite the end of the story if one needs to compute loop diagrams. In that case, one must take into account that not all of the gauge fields that can propagate in loops are actually physical. One way to fix this problem is by introducing “ghost” fields that only appear in loops, and in particular never appear in initial or final states. The ghost fields do not create and destroy real particles; they are really just book-keeping devices that exist only to cancel the unphysical contributions of gauge fields in loops. We will not do any loop calculations in this course, so we will not go into more detail on that issue.] The Yang-Mills theory we have constructed makes several interesting predictions. One is that the gauge fields are necessarily massless. If one tries to get around this by introducing a mass term for the vector gauge fields, like:

2 a aµ A-mass = m A A , (9.108) L V µ then one ﬁnds that this term is not invariant under the gauge transformation of eq. (9.95). Therefore, if we put in such a term, we necessarily violate the gauge invariance of the Lagrangian, and the gauge symmetry will not be a symmetry of the theory. This sounds like a serious problem, because there is only one known freely-propagating, non-composite, massless vector ﬁeld, the photon. In particular, the massive W ± boson cannot be described by the Yang-Mills theory that we have so far. One way

165 to proceed would be to simply keep the term in eq. (9.108), and accept that the theory is not fully invariant under the gauge symmetry. The only problem with this is that the theory would be non- renormalizable in that case; as a related problem, unitarity would be violated in scattering at very high energies. Instead, we can explain the non-zero mass of the W ± boson by enlarging the theory to include scalar fields, leading to a spontaneous breakdown in the gauge symmetry. Another nice feature of the Yang-Mills theory is that several different couplings are predicted to be related to each other. Once we have picked a gauge group G, a set of irreducible representations for the fermions, and the gauge coupling g, then the interaction terms are all fixed. In particular, if we know the coupling of one type of fermion to the gauge fields, then we know g. This in turn allows us to predict, as a consequence of the gauge invariance, what the couplings of other fermions to the gauge fields should be (as long as we know their representations), and what the three-gauge-boson and four-gauge-boson vertices should be.

166 10 Quantum Chromo-Dynamics (QCD)

10.1 QCD Lagrangian and Feynman rules

The strong interactions are based on a Yang-Mills theory with gauge group SU(3)c, with quarks transforming in the fundamental 3 representation. The subscript c is to distinguish this as the group of invariances under transformations of the color degrees of freedom. As far as we can tell, this is an exact symmetry of nature. (There is also an approximate SU(3)flavor symmetry under which the quark flavors u,d,s transform into each other; isospin is an SU(2) subgroup of this symmetry.) Each of the quark Dirac fields u,d,s,c,b,t transforms separately as a 3 of SU(3)c, and each barred Dirac field u, d, s, c, b, t therefore transforms as a 3, as we saw on general grounds in subsection 9.2. For example, an up quark is created in an initial state by any one of the three color component fields:

ured u1 u = u = u . (10.1)  blue   2  ugreen u3     while an anti-up quark is created in an initial state by any of the ﬁelds:

u = ( ured ublue ugreen ) = ( u1 u2 u3 ) . (10.2)

Since SU(3)c is an exact symmetry, no experiment can tell the difference between a red quark and a blue quark, so the labels are intrinsically arbitrary. In fact, we can do a different SU(3)c transformation at each point in spacetime, but simultaneously on each quark flavor, so that:

a a a a a a u eiθ (x)T u, d eiθ (x)T d, s eiθ (x)T s, etc. (10.3) → → → a λa a where T = 2 with a = 1,..., 8, and the θ (x) are any gauge parameter functions of our choosing. This symmetry is in addition to the U(1)EM gauge transformations:

u eiθ(x)Qu u, d eiθ(x)Qd d, s eiθ(x)Qs s, etc. (10.4) → → → where Q = Q = Q = 2/3 and Q = Q = Q = 1/3. For each of the 8 generator matrices T a of u c t d s b − a SU(3)c, there is a corresponding gauge vector boson called a gluon, represented by a ﬁeld Gµ carrying both a spacetime vector index and an SU(3)c adjoint representation index. One says that the unbroken gauge group of the Standard Model is SU(3) U(1) , with the c × EM fermions and gauge bosons transforming as: SU(3) U(1) spin c × EM 2 1 u,c,t (3, + 3 ) 2 d, s, b (3, 1 ) 1 − 3 2 e, µ, τ (1, 1) 1 − 2 1 νe,νµ,ντ (1, 0) 2 γ (1, 0) 1 gluon (8, 0) 1

167 The gluon appears together with the photon ﬁeld Aµ in the full covariant derivative for quark ﬁelds. Using an index i = 1, 2, 3 to run over the color degrees of freedom, the covariant derivatives are: 2 D u = ∂ u + ig Ga T aju + ieAµ( )u , (10.5) µ i µ i 3 µ i j 3 i 1 D d = ∂ d + ig Ga T ajd + ieAµ( )d . (10.6) µ i µ i 3 µ i j −3 i

Here g3 is the coupling constant associated with the SU(3)c gauge interactions. The strength of the strong interactions comes from the fact that g e. 3 À Using the general results for a gauge theory in subsection 9.2, we know that the propagator for the gluons is that of a massless vector ﬁeld just like the photon:

µ, a ν, b i p p δab g + (1 ξ) µ ν p2 + i² µν p2 ←→ ·− − ¸

Note that it is traditional, in QCD, to use “springy” lines for gluons, to easily distinguish them from wavy photon lines. There are also quark-gluon interaction vertices for each ﬂavor of quark: i µ, a ig T ajγµ ←→ − 3 i j

Here the quark line can be any of u, d, s, c, b, t. The gluon interaction changes the color of the quarks when T a is non-diagonal, but never changes the ﬂavor of the quark line, so an up quark remains an up quark, a down quark remains a down quark, etc. The Lagrangian density also contains a “pure glue” part: 1 = F µνaF a (10.7) Lglue −4 µν F a = ∂ Ga ∂ Ga g f abcGb Gc , (10.8) µν µ ν − ν µ − 3 µ ν abc where f are the structure constants for SU(3)c given in eqs. (9.65)-(9.67). In addition to the propagator, this implies that there are three-gluon and four-gluon interactions:

The spacetime- and gauge-index structure are as given in section 9.2 in the general case, with g g . → 3

168 10.2 Quark-quark scattering (qq qq) → To see how the Feynman rules for QCD work in practice, let us consider the example of quark-quark scattering. This is not a directly observable process, because the quarks in both the initial state and ﬁnal state are part of bound states. However, it does form the microscopic part of a calculation for the observable process hadron+hadron jet+jet. We will see how to use the microscopic cross section → result to obtain the observable cross section later, in subsection 10.5. To be speciﬁc, let us consider the process of an up-quark and down-quark scattering from each other:

ud ud. (10.9) → Let us assign momenta, spin, and color to the quarks as follows: Particle Momentum Spin Spinor Color initial u p s1 u(p, s1)= u1 i initial d p0 s2 u(p0, s2)= u2 j (10.10) ﬁnal u k s3 u(k, s3)= u3 l ﬁnal d k0 s4 u(k0, s4)= u4 m

At leading order in an expansion in g3, there is only one Feynman diagram:

u p,i k,l µ, a

p k − ν, b p0, j k0, m d

The reduced matrix element can now be written down by the same procedure as in QED. One obtains, using Feynman gauge (ξ = 1):

µν ab ai bj ig δ = u3( ig3γµTl )u1 u4( ig3γνTm )u2 − (10.11) M − − " (p k)2 # h ih i − 2 ai aj µ = ig3Tl Tm [u3γµu1][u4γ u2] /t (10.12) where t = (p k)2. This matrix element is exactly what one ﬁnds in QED for e µ e µ , but with − − − → − − the QED squared coupling replaced by a product of matrices depending on the color combination:

e2 g2T aiT aj. (10.13) → 3 l m ai This illustrates that the “color charge matrix” g3Tl is analogous to the electric charge eQf . There are 34 = 81 color combinations for quark-quark scattering. In order to ﬁnd the diﬀerential cross section, we continue as usual by taking the complex square of the reduced matrix element:

2 4 ai aj bi bj 2 = g (T T )(T T )∗ , (10.14) |M| 3 l m l m |M| c 169 for each i,j,l,m (with no implied sum yet), and

=[u γ u ][u γµu ] /t. (10.15) M 3 µ 1 4 2 It is not possible, even in principle,c to distinguish between colors. However, one can always imagine fixing, by an arbitrary choice, that the incoming u-quark has color red= 1; then the colors of the other quarks can be distinguished up to SU(3)c rotations that leave the red component fixed. In practice, one does not measure the colors of quarks in an experiment, even with respect to some arbitrary choice, so we will sum over the colors of the final state quarks and average over the colors of the initial state quarks: 1 1 2. (10.16) 3 3 m |M| Xi Xj Xl X

To do the color sum/average most easily, we note that, because the gauge group generator matrices are Hermitian,

bi bj bl bm (Tl Tm )∗ = (Ti Tj ). (10.17)

Therefore, the color factor is

1 1 ai aj bi bj 1 ai bl aj bm (Tl Tm )(Tl Tm )∗ = (Tl Ti )(Tm Tj ) (10.18) 3 3 m 9 Xi Xj Xl X i,j,l,mX 1 = Tr(T aT b)Tr(T aT b) (10.19) 9 1 = I(3)δabI(3)δab (10.20) 9 1 1 = ( )2d (10.21) 9 2 G 2 = (10.22) 9 In doing this, we have used the definition of the index of a representation eq. (9.17); the fact that the index of the fundamental representation is 1/2; and the fact that the sum over a, b of δabδab just counts the number of generators of the Lie algebra dG, which is 8 for SU(3)c. Meanwhile, the rest of 2, including a sum over final state spins and an average over initial state |M| spins, can be taken directly from the corresponding result for e µ e µ in QED, which we found − − → − − by crossing symmetry in eq. (6.208). Stripping off the factor e4 associated with the QED charges, we find in the high energy limit of negligible quark masses,

2 2 1 1 2 s + u = 2 2 . (10.23) 2 s 2 s s s |M| Ã t ! X1 X2 X3 X4 c 4 Putting this together with the factor of g3 and the color factor above, we have 1 1 4g4 s2 + u2 2 2 = 3 . (10.24) |M| ≡ 9 4 |M| 9 Ã t2 ! colorsX spinsX

170 The notation 2 is a standard notation, which for a general process implies the appropriate sum/average |M| over spin and color. The diﬀerential cross section for this process is therefore:

dσ 1 2πα2 s2 + u2 = 2 = s , (10.25) d(cos θ) 32πs|M| 9s Ã t2 ! where g2 α = 3 (10.26) s 4π is the strong-interaction analog of the ﬁne structure constant. Since we are neglecting quark masses, the kinematics for this process is the same as in any massless 2 2 process, for example as found in → eqs. (6.177)-(6.181). Therefore, can replace cos θ in favor of the Mandelstam variable t, using 2dt d(cos θ)= , (10.27) s so dσ 4πα2 s2 + u2 = s . (10.28) dt 9s2 Ã t2 !

10.3 Renormalization

Since the strong interactions involve a coupling that is not small, we should worry about higher-order corrections to the treatment of quark-quark scattering in the previous subsection. Let us discuss this issue in a more general framework than just QCD. In a general gauge theory, the Feynman diagrams contributing to the reduced matrix element at one-loop order in fermion+fermion fermion+fermion 0 → 0 scattering are the following:

171 In each of these diagrams, there is a loop momentum `µ which is unfixed by the external 4-momenta, and must be integrated over. Only the first two diagrams give a finite answer when one naively integrates d4`. This is not surprising; we do not really know what physics is like at very high energy and momentum scales, so we have no business in integrating over them. Therefore, one must introduce a very high cutoff mass scale M, and replace the loop-momentum integral by one which kills the contributions to the reduced matrix element from `µ M. Physically, M should be the mass scale | | ≥ at which some as-yet-unknown new physics enters in to alter the theory. It is generally thought that the highest this cutoff is likely to be is about M = 2.4 1018 GeV (give or take an order of Planck × magnitude), but it could conceivably be much lower. As an example of what can happen, consider the next-to-last Feynman diagram given above. Let us call qµ = pµ kµ the 4-momentum flowing through either of the vector-boson propagators. Then − the part of the reduced matrix element associated with the fermion loop is: / 4 aj µ i(` + /q +m ˆ f ) bi ν i(/` +m ˆ f ) ( 1) d ` Tr igˆ(Tf ) γ igˆ(Tf )j γ . µ i 2 2 2 2 − ` M ( − "(` + q) mˆ f + i²# − "` mˆ f + i²#) Xf Z| |≤ h i − h i − (10.29)

This involves a sum over all fermions that can propagate in the loop, and a trace over the spinor indices of the fermion loop. For reasons that will become clear shortly, we are calling the gauge coupling of the theoryg ˆ and the mass of each fermion speciesm ˆ f . We are being purposefully vague about what 4 `µ M d ` means, in part because there are actually several different ways to cutoff the integral at large | |≤ MR . (A straightforward step-function cutoff will work, but is clumsy to carry out and even clumsier to interpret.) The d4` factor can be written as an angular part times a radial part ` 3d ` . Now there are up to | | | | five powers of ` in the numerator (three from the d4`, and two from the propagators), and four powers | |

172 of ` in the denominator from the propagators. So naively, one might expect that the result of doing | | the integral will scale like M 2 for a large cutoﬀ M. However, there is a conspiratorial cancellation, so that the large-M behavior is only logarithmic. The result is proportional to:

gˆ2(q2gµν qµqν) Tr(T aT b) [ln(M/m)+ ...] (10.30) − f f Xf where the ... represents a contribution that does not get large as M gets large. The m is a characteristic µ mass scale of the problem; it is something with dimensions of mass built out of q and them ˆ f . It must appear in the formula in the way it does in order to make the argument of the logarithm dimensionless. Any small arbitrariness made in the precise deﬁnition of m can be absorbed into the “...”. When one uses eq. (10.30) in the rest of the Feynman diagram, it is clear that the entire contribution must be proportional to:

gˆ4 I(R )ln(M/m)+ .... (10.31) Mfermion loop in gauge prop. ∝ f Xf What we are trying to keep track of here is just the number of powers ofg ˆ, the group-theory factor, and the large-M dependence on ln(M/m). A similar sort of calculation applies to the last diagram involving a gauge vector boson loop. Each of the three-vector couplings involves a factor of f abc, with two of the indices contracted because of the propagators. So it must be that the loop part of the diagram make a contribution proportional to f acdf bcd = C(G)δab. It is again logarithmically divergent, so that

gˆ4C(G)ln(M/m)+ .... (10.32) Mgauge loop in gauge prop. ∝ Doing everything carefully, one ﬁnds that the contributions to the diﬀerential cross section is given by:

gˆ2 11 4 dσ = dσ (ˆg) 1+ C(G) I(R ) ln(M/m)+ ... (10.33) tree  4π2  3 − 3 f    Xf    where dσtree(ˆg) is the tree-level result (which we have already worked out in the special case of QCD), considered to be a function ofg ˆ. To be specific, it is proportional tog ˆ4. Let us ignore all the other diagrams for now; the justification for this will be revealed soon. The cutoff M may be quite large. Furthermore, by definition, we do not know what it is, or what the specific very-high-energy physics associated with it is. (If we did, we could just redo the calculation with that physics included, and a higher cutoff.) Therefore, it is convenient to absorb our ignorance of M into a redefinition of the coupling. Specifically, inspired by eq. (10.33), one defines a renormalized or running coupling g(Q) by writing:

(g(Q))2 11 4 gˆ = g(Q) 1 C(G) I(R ) ln(M/Q) , (10.34)  − 16π2  3 − 3 f    Xf     

173 Here Q is a new mass scale, called the renormalization scale, that we get to pick. The original coupling gˆ is called the bare coupling. One can invert this relation to write the renormalized coupling in terms of the bare coupling:

gˆ2 11 4 g(Q) =g ˆ 1+ C(G) I(R ) ln(M/Q)+ ... . (10.35)  16π2  3 − 3 f    Xf    where we are treating g(Q) as an expansion ing ˆ, dropping terms of orderg ˆ5 everywhere. 4 The reason for this strategic deﬁnition is that, since we know that dσtree is proportional tog ˆ , we can now write:

4 dσ = dσtree(g) (ˆg/g) (10.36) g2 11 4 = dσ (g) 1+ C(G) I(R ) ln(Q/m)+ ... . (10.37) tree  4π2  3 − 3 f    Xf    Here we are again dropping terms which go like g4; these are comparable to 2-loop contributions that we are neglecting anyway. The factor dσtree(g) is the tree-level differential cross-section, but computed with g(Q) in place ofg ˆ. This formula looks very much like eq. (10.33), but with the crucial difference that the unknown cutoff M has disappeared, and is replaced by a scale Q that we know, because we get to pick it. What should we pick Q to be? In principle we could pick it to be the cutoff M, except that we do not know what that is. Besides, the logarithm could then be very large, and perturbation theory would converge very slowly or not at all. For example, suppose that M = MPlanck, and the characteristic energy scale of the experiment we are doing is, say, m = 0.511 MeV or m = 1000 GeV. These choice might be appropriate for experiments involving a non-relativistic electron and a TeV-scale collider, respectively. Then

ln(M/m) 50 or 35. (10.38) ≈ This logarithm typically gets multiplied by 1/16π2 times g2 times a group-theory quantity, but is still large. This suggests that a really good choice for Q is to make the logarithm ln(Q/m) as small as possible, so that the correction term in eq. (10.37) is small. Therefore, one should choose

Q m. (10.39) ≈ Then, to a ﬁrst approximation, one can calculate using the tree-level approximation using a renormalized coupling g(Q), knowing that the one-loop correction from these diagrams is small. The choice of renormalization scale eq. (10.39) allows us to write:

dσ dσ (g(Q)) (10.40) ≈ tree Of course, this is only good enough to get rid of the large logarithmic one-loop corrections. If you really want all one-loop corrections, there is no way around calculating all the one-loop diagrams, keeping all the pieces, not just the ones that get large as M . → ∞

174 What about the remaining diagrams? If we isolate the M behavior, they fall into three → ∞ classes. First, there are diagrams which are not divergent at all (the ﬁrst two diagram). Second, there are diagrams (the third through sixth diagrams) which are individually divergent like ln(M/m), but sum up to a total which is not divergent. Finally, the seventh through tenth diagrams have a logarithmic divergence, but it can be absorbed into a similar redeﬁnition of the mass. A clue to this is that they all involve sub-diagrams:

The one-loop renormalized or running mass mf (Q) is deﬁned in terms of the bare massm ˆ f by

g2 mˆ f = mf (Q) 1 C(Rf )ln(M/Q) , (10.41) Ã − 2π2 ! or g2 mf (Q) =m ˆ f 1+ C(Rf )ln(M/Q)+ ... , (10.42) Ã 2π2 ! where C(Rf ) is the quadratic Casimir invariant of the representation carried by the fermion f. It is an amazing fact that the two redefinitions eqs. (10.34) and (10.41) are enough to remove the cutoff dependence of all cross sections in the theory up to and including one-loop order. In other words, one can calculate dσ for any process, and express it in terms of the renormalized mass m(Q) and the renormalized coupling g(Q), with no M-dependence. This is what it means for a theory to be renormalizable at one loop order. In Yang-Mills theories, one can show that by doing some redefinitions of the form:

L 2n gˆ = g(Q) 1+ bng pn(ln(M/Q)) , (10.43) " # nX=1 L 2n mˆ f = mf (Q) 1+ cng qn(ln(M/Q)) , (10.44) " # nX=1 one can simultaneously eliminate all dependence on the cutoff in any process up to L-loop order. Here pn(x) and qn(x) are polynomials of degree n, and bn,cn are some constants that depend on group theory invariants like the Casimir invariants of the group and the representations, and the index. At any finite loop order, what is left in the expression for any cross section after writing it in terms of the renormalized mass m(Q) and renormalized coupling g(Q) is a polynomial in ln(Q/m); these are to be made small by choosing Q m. This is what it means for a theory to be renormalizable at all loop † ≈ orders. Typically, the specifics of these redefinitions is only known at 2- or 3- or occasionally 4- loop order, except in some special theories. If a theory is non-renormalizable, it does not necessarily mean that the theory is useless; we saw that the four-fermion theory of the weak interactions makes reliable

†Of course, there might be more than one characteristic energy scale in a given problem, rather than a single m. If so, and if they are very diﬀerent from each other, then one may be stuck with some large logarithms, no matter what Q is chosen. This has to be dealt with by fancier methods.

175 predictions, and we still have no more predictive theory for gravity than Einstein’s relativity. It does mean that we expect the theory to have trouble making predictions about processes at high energy scales. We have seen that we can eliminate the dependence on the unknown cutoff of a theory by defining a renormalized running coupling g(Q) and mass mf (Q). When one does an experiment in high energy physics, the results are first expressed in terms of observable quantities like cross sections, decay rates, and physical masses of particles. Using this data, one extracts the value of the running couplings and running masses at some appropriately-chosen renormalization scale Q, using a theoretical prediction like eq. (10.37), but with the non-logarithmic corrections included too. (The running mass is not quite the same thing as the physical mass. The physical mass can be determined from the experiment by kinematics, the running mass is related to it by various corrections.) The running parameters can then be used to make predictions for other experiments. This tests both the theoretical framework, and the specific values of the running parameters. The bare coupling and the bare mass never enter into this process. If we measure dσ in an experiment, we see from eq. (10.33) that in order to determine the bare couplingg ˆ from the data, we would also need to know the cutoff M. However, by definition we do not know what M is. We could guess at it, but it would be a wild guess, devoid of practical significance. A situation which arises quite often is that one extracts running parameters from an experiment with a characteristic energy scale Q0, and one wants to compare with data from some other experiment which has a completely different characteristic energy scale Q. Here Q0 and Q each might be the mass of some particle that is decaying, or the momentum exchanged between particles in a collision, or some suitable average of particle masses and exchanged momenta. It would be unwise to use the same renormalization scale when computing the theoretical expectations for both experiments, because the loop corrections involved in at least one of the two cases will be unnecessarily large. What we need is a way of taking a running coupling as determined in the first experiment at a renormalization scale

Q0, and getting from it the running coupling at any other scale Q. The change of the choice of scale

Q is known as the renormalization group.‡ As an example, let us consider how g(Q) changes in a Yang-Mills gauge theory. Since the differential cross section dσ for fermion+fermion fermion+fermion is in principle an observable, it cannot 0 → 0 possibly depend on the choice of Q, which is an arbitrary one made by us. Therefore, we can require that eq. (10.37) is independent of Q. Remembering that dσ g4, we find: tree ∝ d 4 dg g2 11 4 1 0= (dσ) = (dσ ) + C(G) I(R ) + ... , (10.45) dQ tree g dQ 4π2  3 − 3 f  Q   Xf      4 where we are dropping all higher-loop-order terms that are proportional to (dσtree)g . The first term 4 in eq. (10.45) comes from the derivative acting on the g inside dσtree. The second term comes from the derivative acting on the lnQ one-loop correction term. The contribution from the derivative acting

‡The use of the word “group” is historical; this is not a group in the mathematical sense deﬁned earlier.

176 on the g2 in the one-loop correction term can be self-consistently judged, from the equation we are 5 about to write down, as proportional to (dσtree)g , so it is neglected as a higher-loop-order eﬀect in the expansion in g2. So, it must be true that:

dg g3 11 4 Q = C(G)+ I(R ) . (10.46) dQ 16π2 − 3 3 f  Xf   This differential equation, called the renormalization group equation or RG equation, tells us how to change the coupling g(Q) when we change the renormalization scale. An experimental result will provide a boundary condition at some scale Q0, and then we can solve the RG equation to find g(Q) at some other scale. Other experiments then test the whole framework. The right-hand side of the RG equation is known as the beta function for the running coupling g(Q), and is written β(g), so that: dg Q = β(g). (10.47) dQ In a Yang-Mills gauge theory, g3 g5 β(g)= b + b + ... (10.48) 16π2 0 (16π2)2 1 where we already know that 11 4 b = C(G)+ I(R ), (10.49) 0 − 3 3 f Xf and, just to give you an idea of how it goes, 34 20 b = C(G)2 + C(G) I(R ) + 4 C(R )I(R ), (10.50) 1 − 3 3 f f f Xf Xf etc. In practical calculations, it is usually best to work within an effective theory in which fermions much heavier than the scales Q of interest are ignored. The sum over fermions then includes only those satisfying mf < Q. The difference between this effective theory and the more complete theory with all ∼ known fermions included can be absorbed into a redefinition of running parameters. The advantage of doing this is that perturbation theory will converge more quickly and reliably if heavy fermions (that are, after all, irrelevant to the process under study) are not included. In the one-loop order approximation, one can solve the RG equation explicitly. Writing dg2 b = 0 g4, (10.51) dlnQ 8π2 it is easy to check that 2 2 g (Q0) g (Q)= 2 . (10.52) 1 b0g (Q0) ln(Q/Q ) − 8π2 0 To see how this works in QCD, let us examine the one-loop beta function. In SU(3), C(G) = 3, and each quark flavor is in a fundamental 3 representation with I(3) = 1/2. Therefore, 2 b = 11 + n (10.53) 0,QCD − 3 f

177 where nf is the number of “active” quarks in the effective theory, usually those with mass < Q. ∼ The crucial fact is that since there are only 6 quark flavors known, b0,QCD is definitely negative for all accessible scales Q, and so the beta function is definitely negative. For an effective theory with n = (3, 4, 5, 6) quark flavors, b = ( 9, 25/3, 23/3, 7). Writing the solution to the RG equation, f 0 − − − − eq. (10.52), in terms of the running αs, we have:

αs(Q0) αs(Q)= . (10.54) 1 b0αs(Q0) ln(Q/Q ) − 2π 0 Since b0 is negative, we can make αs blow up by choosing Q small enough. To make this more explicit, we can deﬁne a quantity

2π/b0αs(Q0) ΛQCD = Q0e− , (10.55) with dimensions of [mass], implying that 2π αs(Q)= . (10.56) b0ln(ΛQCD/Q)

This shows that ΛQCD is the scale at which the QCD gauge coupling is predicted to blow up, according to the 1-loop RG equation. A qualitative graph of the running of αs(Q) as a function of renormalization scale Q is shown below:

α S(Q)

Λ QCD Renormalization scale Q

Of course, once αs(Q) starts to get big, we should no longer trust the one-loop approximation, since two-loop effects are definitely big. The whole analysis has been extended to four-loop order, with significant numerical changes, but the qualitative effect remains: at any finite loop order, there is

178 some scale ΛQCD at which the gauge coupling is predicted to blow up in a theory with a negative beta function. This is not a sign that QCD is wrong. Instead, it is a sign that perturbation theory is not going to be able to make good predictions when we do experiments near Q = ΛQCD or lower energy scales. One can draw Feynman diagrams and make rough qualitative guesses, but the numbers cannot be trusted. On the other hand, we see that for experiments conducted at characteristic energies much larger than ΛQCD, the gauge coupling is not large, and is getting smaller as Q gets larger. This means that perturbation theory becomes more and more trustworthy at higher and higher energies. This nice property of theories with negative β functions is known as asymptotic freedom. The name refers to the fact that quarks in QCD are becoming free (since the coupling is becoming small) as we probe them at larger energy scales. Conversely, the fact that the QCD gauge coupling becomes non-perturbative in the infrared means that we cannot expect to describe free quarks at low energies using perturbation theory. This theoretical prediction goes by the name of infrared slavery. It agrees well with the fact that one does not observe free quarks outside of bound states. While it has not been proved mathematically that the infrared slavery of QCD necessarily requires the absence of free quarks, the two ideas are certainly compatible, and more complicated calculations show that they are plausibly linked. Heuristically, the growth of the QCD coupling means that at very small energies or large distances, the force between two free color charges is large and constant. In the early universe, after the temperature dropped below ΛQCD, all quarks and antiquarks and gluons arranged themselves into color-singlet bound states, and have remained that way ever since. An important feature of the renormalization of QCD is that we can actually trade the gauge coupling as a parameter of the theory for the scale ΛQCD. This is remarkable, since g3 (or equivalently αs) is a dimensionless coupling, while ΛQCD is a mass scale. If we want, we can specify how strong the QCD interactions are either by quoting what αs(Q0) is at some specified Q0, or by quoting what ΛQCD is. This trade of a dimensionless parameter for a mass scale in a gauge theory is known as dimensional transmutation. Working in a theory with five “active” quarks u,d,s,c,b (the top quark is treated as part of the unknown theory above the cutoff), one finds ΛQCD is about 200 MeV. One can also work in an effective theory with only four active quarks u,d,s,c, in which case ΛQCD is about 290 MeV. Alternatively, α (m ) = 0.118 0.003. This result holds when the details of the cutoff and the s Z ± renormalization are treated in the most popular way, called the MS scheme.§ We can contrast this situation with the case of QED. For a U(1) group, there is no non-zero structure constant, so C(G) = 0. Also, since the generator of the group in a representation of charge Qf is just the 1 1 matrix Q , the index for a fermion with charge Q is I(R )= Q2 . Therefore, × f f f f 4 16 4 4 b = 3n (2/3)2 + 3n (1/3)2 + n ( 1)2 = n + n + n , (10.57) 0,QED 3 u d ` − 9 u 9 d 3 ` h i §In this scheme, one cuts off loop momentum integrals by dimensional regularization, continuously varying the number of spacetime dimensions infinitesimally away from 4, rather than putting in a particular cutoff M.

179 where nu is the number of up-type quark ﬂavors (u,c,t), and nd is the number of down-type quark

ﬂavors (d, s, b), and n` is the number of charged leptons (e, µ, τ) included in the chosen eﬀective theory.

If we do experiments with a characteristic energy scale me < Q < mµ, then only the electron itself ∼ ∼ contributes, and b0,EM = 4/3, so:

de e3 4 = β = (m < Q < m ). (10.58) dlnQ e 16π2 3 e µ µ ¶ ∼ ∼ This corresponds to a very slow running. (Notice that the smaller a gauge coupling is, the slower it will run.) If we do experiments at characteristic energies that are much less than the electron mass, then the relativistic electron is not included in the eﬀective theory (virtual electron-positron pairs are less and less important at low energies), so b0,EM = 0, and the electron charge does not run at all: de = 0 (Q m ). (10.59) dlnQ ¿ e This means that QED is not quite “infrared free”, since the eﬀective electromagnetic coupling is perturbative, but does not get arbitrarily small, at very large distance scales. At extremely high energies, the coupling e could in principle become very large, because the QED beta function is always positive. Fortunately, this is predicted to occur only at energy scales far beyond what we can probe, because e runs very slowly. Furthermore, QED is embedded in a larger, more complete theory anyway at energy scales in the hundreds of GeV range, so the apparent blowing up of α much farther in the ultraviolet is just an illusion.

10.4 Gluon-gluon scattering (gg gg) → Let us now return to QCD scattering processes. Because there are three-gluon and four-gluon interaction vertices, one has the interesting process gg gg even at tree-level. (It is traditional to represent → the gluon particle name, but not its quantum ﬁeld, by g.) In homework set 7, problem 1, you should have noted that the corresponding QED process of γγ γγ does not happen at tree-level, but does → occur at one loop. In QCD, because of the three-gluon and four-gluon vertices, there are four distinct Feynman diagrams that contribute at tree-level:

The calculation of the differential cross-section from these diagrams is an important, but quite tedious, one. Just to get an idea of how this proceeds, let us write down the reduced matrix element for the first (“s-channel”) diagram, and then skip directly to the final answer.

180 Choosing polarization vectors and color indices for the gluons, we have:

Particle Polarization vector Color µ µ initial gluon ²1 = ² (p, λ1) a ν ν initial gluon ²2 = ² (p0, λ2) b (10.60) ρ ρ ﬁnal gluon ²3∗ = ² ∗(k, λ3) c σ σ ﬁnal gluon ²4∗ = ² ∗(k0, λ4) d

Let the internal gluon line carry (vector,gauge) indices (κ, e) on the left and (λ, f) on the right. Labeling the Feynman diagram in detail: µ, a ρ,c p k

p + p0

κ, e λ, f

p ν, b 0 k0 σ, d

The momenta flowing into the leftmost 3-gluon vertex are, starting from the upper-left incoming gluon and going clockwise, (p, p p ,p ). Also, the momenta flowing into the rightmost 3-gluon vertex are, − − 0 0 starting from the upper-right final-state gluon and going clockwise, ( k, k , k + k ). So we can use − − 0 0 the Feynman rules of 10.1 to obtain:

µ ν ρ σ aeb gg gg,s-channel = ²1 ²2²3∗²4∗ gf gµκ(2p + p0)ν + gκν( p 2p0)µ + gνµ(p0 p)κ M → − − − − cdf h i gf g (k0 k©) + g ( 2k0 k) + g (2k + k0) ª − ρσ − λ σλ − − ρ λρ σ h igκλδ©ef ªi − 2 . (10.61) " (p + p0) #

After writing down the reduced matrix elements for the other three diagrams, adding them together, taking the complex square, summing over final state polarizations and averaging over initial state polarizations, summing over final-state gluon colors and averaging over initial-state gluon colors according 1 1 to 8 a 8 b c d one finds:

P P P P 2 2 2 2 2 2 2 dσgg gg 9παs u + t s + u s + t → = + + + 3 . (10.62) dt 4s2 " s2 t2 u2 #

When we collide a proton with another proton or an antiproton, this process, and the process ud ud, → are just two of many possible subprocesses that can occur. There is no way to separate the proton into simpler parts, so we will have to deal with all of these possible subprocesses on an equal footing. We will consider the subprocesses of proton-(anti)proton scattering more systematically in the next subsection.

181 10.5 The parton model, parton distribution functions, and hadron-hadron scattering

The next 10 years or so of particle physics are likely to be dominated by hadron-hadron colliders. First, our own Tevatron will explore pp collisions at √s = 2 TeV; then the Large Hadron Collider will take over the frontier with pp collisions at √s = 14 TeV. It is virtually certain that the most exciting discoveries in high-energy physics in the next decade will occur at one or both of these colliders. Therefore, we will spend several subsections on the basic ideas behind hadron-hadron collisions. In general, a hadron is a QCD bound state of quarks, anti-quarks, and gluons. The characteristic size of a hadron, like the proton or antiproton, is always roughly 1/Λ 10 13 cm, since this is QCD ≈ − the scale at which the strongly-interacting particles are confined. In general, the point-like quark, antiquark, and gluon parts of a hadron are called partons, and the description of hadrons in terms of them is called the parton model. Suppose we scatter a hadron off of another particle (which might be another hadron or a lepton or photon) with a total momentum exchange much larger than ΛQCD. The scattering can be thought of as a factored into a “hard scattering” of one of the point-like partons, with the remaining partons as spectators, and “soft” QCD processes which involve exchanges and radiation of low energy virtual gluons. The hard scattering subprocess takes place on a time scale which is much shorter than 1/ΛQCD expressed in seconds. Because of asymptotic freedom, at higher scattering energies it becomes a better and better approximation to think of the partons as individual entities which move collectively before and after the scattering, but are free particles at the moment of scattering. As a first approximation, we can consider only the hard scattering processes, and later worry about adding on the various soft processes as part of the higher-order corrections. This way of thinking about things allows us to compute cross sections for hadron scattering by first calculating the partonic cross-sections leading to a desired final state, and then combining them with information about the multiplicity and momentum distributions of partons within the hadronic bound states. For example, suppose we want to calculate the scattering of a proton and antiproton. This can involve many different 2 2 partonic subprocesses, including: → qq qq, (10.63) → qq0 qq0, (10.64) → gg gg, (10.65) → qg qg, (10.66) → where q is any quark and g is a gluon. Let the center-of-mass energy of the proton and antiproton be called √s. Each parton only carries a fraction of the energy of the energy of the proton it belongs to, so the partonic center-of-mass energy, call it √sˆ, will be significantly less than √s. One often uses hatted Mandelstam variabless ˆ, tˆ andu ˆ for the partonic scattering event. If √sˆ Λ , then the À QCD two partons in the final state will be sufficiently energetic that they can usually escape from most or

182 all of the spectator quarks before hadronizing (forming bound states). However, before traveling a distance 1/ΛQCD, they must rearrange themselves into color-singlet combinations, possibly by creating quark-antiquark or gluon-antigluon pairs out of the vacuum. This hadronization process can be quite complicated, but will always result in a jet of hadronic particles moving with roughly the same 4- momentum as the parton that was produced. So, all of the partonic cross sections above contribute to the observable cross section for the process:

pp jet+jet+ X, (10.67) → where the X includes stray hadronic junk left over from the original proton and antiproton. Similarly, partonic scatterings like:

qq qqg, (10.68) → qg qgg, (10.69) → qg qgg, (10.70) → gg ggg, (10.71) → and so on, will contribute to an observable cross section for

pp jet+jet+jet+ X. (10.72) → In order to use the calculation of cross-sections for partonic processes like eqs. (10.63)-(10.66) to obtain measurable cross sections, we need to know how likely it is to have a given parton inside the initial-state hadron with a given 4-momentum. Since we are mostly interested in high-energy scattering problems, we can make things simple, and treat the hadron and all of its constituents as nearly massless. (For the proton, this means that we are assuming that √s m 1 GeV.) Suppose we therefore take À p ≈ the total 4-momentum of the hadron h in an appropriate Lorentz frame to be:

µ ph = (E, 0, 0,E). (10.73)

This is sometimes called the “inﬁnite momentum frame”, even though E is ﬁnite, since E m . À p Consider a partonic constituent A which carries a fraction x of the hadron’s momentum:

µ pA = x(E, 0, 0,E). (10.74)

The variable x is a standard notation, and is called Feynman’s x. It is older than QCD, dating back to a time when the proton was suspected to contain point-like partons with properties that were then obscure. In order to describe the partonic content of a hadron, one deﬁnes: Probability of ﬁnding a parton of type A with 4-momentum between xpµ and (x + dx)pµ = f h(x) dx (10.75)   A inside a hadron h with 4-momentum pµ   h The function fA(x) is called the parton distribution function or PDF for the parton A in the hadron h. The parton can be either one of the two or three valence quarks or antiquarks which are the nominal

183 constituents of the hadron, or one of an indeterminate number of virtual sea quarks and gluons. Either type of parton can participate in a scattering event. Hadronic collisions studied in laboratories usually involve protons or antiprotons, so the PDFs of the proton and antiproton are especially interesting. The proton is nominally a bound state of three valence quarks, namely two up quarks and one down quark, so we are certainly interested in the up-quark and down-quark distribution functions

p p fu(x) and fd (x).

The proton also contains virtual gluons, implying a gluon distribution function:

p fg (x). (10.76)

Furthermore, there are always virtual quark-antiquark pairs within the proton. This adds additional p p contributions to fu (x) and fd (x), and also means that there is a non-zero probability of ﬁnding antiup, antidown, or strange or antistrange quarks:

f p(x), f p(x), f p(x), f p(x). (10.77) u d s s

p These parton distribution functions are implicitly summed over color and spin. So fu (x) tells us the probability of finding an up quark with the given momentum fraction x and any color and spin. Since the antiparticle of the gluon is another gluon (it lives in the adjoint representation of the gauge group, p which is always a real representation), there is not a separate fg (x). [Although the charm, bottom and top quarks are heavier than the proton, virtual charm-anticharm, bottom-antibottom, and top-antitop pairs can exist as long as their total energy does not exceed mp. This can happen because, as virtual particles, they need not be on-shell. So, one can even talk about the parton distribution functions f p(x), f p(x), f p(x), f p(x), f p(x), f p(x). Fortunately, these are very c c b b t t small so one can usually neglect them.] Given the PDFs for the proton, the PDFs for the antiproton follow immediately from the fact that it is the proton’s antiparticle. The probability of finding a given parton in the proton with a given x is the same as the probability of finding the corresponding antiparton in the antiproton with the same x. Therefore, if we know the PDFs for the proton, there is no new information in the PDFs for the antiproton. We can just describe everything having to do with proton and antiproton collisions in terms of the proton PDFs. To simplify the notation, it is traditional to write the proton and antiproton PDFs as:

p p g(x) = fg (x)= fg (x), (10.78) p p u(x) = fu (x)= fu (x), (10.79) d(x) = f p(x)= f p(x), (10.80) d d p p u(x) = fu (x)= fu (x), (10.81)

184 d(x) = f p(x)= f p(x), (10.82) d d p p s(x) = fs (x)= fs (x), (10.83) p p s(x) = fs (x)= fs (x). (10.84)

Since we will be doing calculations in a renormalizable field theory, with a cutoff and renormalized running couplings and masses, the PDFs must also be treated as running functions. The running of PDFs implies a mild logarithmic dependence on Q. So they are really functions u(x, Q) etc., although the Q-dependence is often left implicit. In the proton, antiquarks are always virtual, and so must be accompanied by a quark with the same flavor. This implies that if we add up all the up quarks found in the proton, and subtract all the anti-ups, we must find a total of 2 quarks:

1 dx [u(x) u(x)] = 2. (10.85) − Z0 Similarly, summing over all x the probability of ﬁnding a down quark with a given x, and subtracting the same thing for anti-downs, one must have:

1 dx [d(x) d(x)] = 1. (10.86) − Z0

Because a virtual strange quark is always¶ accompanied by an antistrange, there follows the sum rule:

1 dx [s(x) s(x)] = 0. (10.87) − Z0 In fact, this rule can be made more speciﬁc, since the strange quarks in the proton come from the process of a virtual gluon splitting into a strange and anti-strange pair. Since the virtual gluon treats quarks and antiquarks on an equal footing, for every strange quark with a given x, there must be an equal probability of ﬁnding an antistrange with the same x:

s(x)= s(x). (10.88)

Similarly, we can divide the up-quark PDF into a contribution uv(x) from the two valence quarks, and a contribution us(x) from the sea quarks that are accompanied by an anti-up. So we have:

u(x)= uv(x)+ us(x); us(x)= u(x) (10.89)

d(x)= dv(x)+ ds(x); ds(x)= d(x). (10.90)

There is also a constraint that the total 4-momentum of all partons found in the proton must be equal to the 4-momentum of the proton that they form. This rule takes the form:

1 dx x[g(x)+ u(x)+ u(x)+ d(x)+ d(x)+ s(x)+ s(x)+ ...] = 1, (10.91) Z0 ¶Although QCD interactions do not change quark ﬂavors, there is a small strangeness violation in the weak interactions, so the following rule is not quite exact.

185 or more generally, for any hadron h made out of parton species A,

1 h dx xfA(x) = 1. (10.92) 0 XA Z h Each term xfA(x) represents the probability that a parton is found with a given momentum fraction x, multiplied by that momentum fraction. One of the ﬁrst compelling pieces of evidence that the gluons are actual vectors particles carrying real momentum and energy, and not just abstract group-theoretic constructs, was that if one excludes them from the sum rule eq. (10.91), only about half of the proton’s 4-momentum is accounted for:

1 dx x[u(x)+ u(x)+ d(x)+ d(x)+ s(x)+ s(x)+ ...] 0.5. (10.93) ≈ Z0

Consider the very naive model of the proton in which all gluons and sea quark-antiquark pairs are ignored. In that case, each of the two up and one down quarks would carry exactly 1/3 of the 4-momentum of the proton. Then the parton distribution functions would be proportional to δ(x 1/3): − u (x) = 2δ(x 1/3); (10.94) naive − d (x)= δ(x 1/3); (10.95) naive − gnaive(x)= unaive(x)= dnaive(x)= snaive(x)= snaive(x) = 0, (10.96) with no Q-dependence. The factor of 2 in eq. (10.94) is necessary because there are two up-quarks to find in a proton; if one integrates the PDFs over all possible x, one must find them both. However, this naive model is hopelessly wrong; the impact of virtual gluons and antiquarks in reality is very significant, and forces the valence up quarks and down quarks to share the 4-momentum. As we will see, this means that the distributions for quarks is considerably spread out, with the average value shifted to lower x. If we could solve the bound state problem for the proton in QCD, like one can solve the hydrogen atom in quantum mechanics, then we could derive the PDFs directly from the Hamiltonian. However, we saw in subsection 10.3 why this is not practical; perturbation theory in QCD is no good for studying low-energy problems like bound-state problems. Instead, the proton PDFs are determined by experiments in which charged leptons and neutrinos probe the proton, like ` p ` + X and νp ` + X. − → − → − Several collaborations perform fits of available data to determine the PDFs, and periodically publish updated result both in print and as computer code. One such collaboration is CTEQ, (Coordinated Theoretical-Experimental Project on QCD). Other popular PDF sets are GRV (Glück, Reya, Vogt) and MRS (Martin, Roberts, Stirling). In each case, the fits of the PDFs are to some template functions, with several tens or hundreds of numerical coefficients and exponents to be fit by data. Because of different techniques and weighting of the data, the PDFs from different groups are always somewhat different.

186 As an example, let us consider the recent CTEQ5L PDF set. (Here, the “5” says which update of the PDFs is being provided (this one in the year 2000), and the “L” stands for “lowest order”, which is the set appropriate when one only has the lowest-order formula for the partonic cross-section. There are also sets CTEQ5M and CTEQ5DIS which are slightly diﬀerent and are appropriate when non-leading order cross-sections are used.) At Q = 10 GeV, the PDFs for u(x), u(x), and the valence contribution u (x) u(x) u(x) are shown below: v ≡ − 1 CTEQ5L, Q=10 GeV

0.8 u(x) u(x) = us(x)

uv(x) = u(x) - u(x) 0.6 x f(x) 0.4

0.2

0 0 0.2 0.4 0.6 0.8 1 x

It is traditional to graph x times the PDF in each case, since they all tend to get large near x = 0. Just to give an idea of how these things are done, the up-quark PDF found in the CTEQ5L ﬁt is:

2 3 4 5 c u(x) = ec0+c1x+c2x +c3x +c4x +c5[1+c6ln(10 x)][ln(1/x)] 7 x(1 x)A(B x)C (10.97) − − where, with p 1.2 + ln(ln(Q/0.496096 GeV)), ≡− A = 4.13843 (10.98) B = 1.03990 (10.99) C = 0.763881 + 0.500831p 0.0923737p2 (10.100) − − c = 1.3325 0.370372p + 0.128864p2 (10.101) 0 − c = 3.6968 + 0.660393p 1.01787p2 (10.102) 1 − c = 13.2268 4.62515p + 1.92648p2 (10.103) 2 − − c = 16.4598 + 7.26123p 2.19899p2 (10.104) 3 − c = 8.6313 3.9812p + 0.697015p2 (10.105) 4 − − c = 0.754469 + 0.0325507p 0.0470668p2 (10.106) 5 − 2 c6 = 0.0237123 + 0.00537268p + 0.0011187p (10.107) 2 c7 = 1.39916 + 0.0197988p + 0.0177581p . (10.108)

187 We see from the graph that the valence up-quark distribution is peaked below x = 0.2, with a long tail for larger x (where an up-quark is found to have a larger fraction of the proton’s energy). There is even a signiﬁcant chance of ﬁnding that an up quark has more than half of the proton’s 4-momentum. In contrast, the sea quark distribution u(x) is strongly peaked near x = 0. This is a general feature of sea partons; the chance that a virtual particle can appear is greater when it carries a smaller energy, and thus a smaller fraction x of the proton’s total momentum. The solid curve shows the total up-quark PDF for this value of Q. For comparison, here are the corresponding graphs for the down quark and antiquark: 1 CTEQ5L, Q=10 GeV

0.8 d(x) d(x) = ds(x)

dv(x) = d(x) - d(x) 0.6 x f(x) 0.4

0.2

0 0 0.2 0.4 0.6 0.8 1 x

Notice that the sea distribution d(x) in this case is not very diﬀerent from that of the anti-up, but dv(x) is of course only about half as big as uv(x), since there is only one valence down quark to ﬁnd in the proton. Next, let us look at the strange and gluon PDFs:

188 1 CTEQ5L, Q=10 GeV

0.8 g(x) s(x) = s(x)

0.6 x f(x) 0.4

0.2

0 0 0.2 0.4 0.6 0.8 1 x

The parton distribution function for gluons grows very quickly as one moves towards x = 0. This is because there are 8 gluon color combinations available, and each virtual gluon can give rise to more virtual gluons because of the 3-gluon and 4-gluon vertex. This means that the chance of finding a gluon gets very large if one requires that it only have a small fraction of the total 4-momentum of the proton. The PDFs s(x)= s(x) are suppressed by the non-zero strange quark mass, since this imposes a penalty on making virtual strange and antistrange quarks. This explains why s(x) < d(x). The value of Q = 10 GeV corresponds roughly to the appropriate energy scale for many of the experiments which are actually used to fit for the PDFs. However, at the Tevatron and LHC, one will often be studying events with a much larger characteristic energy scale, like Q m for top events and ∼ t perhaps Q 1000 GeV for supersymmetry events at the LHC. Increasing Q corresponds to probing the ∼ proton at larger energy scales, or shorter distance scales. Heuristically, at smaller distance scales, the chances of finding a virtual sea parton increases, while the chances of finding a valence quark remain the same. So, for larger Q, one expects that the sea PDFs, especially g(x), will become relatively more important. Furthermore, there are more virtual partons to share the total momentum at larger Q, so the distribution for each species tends to shift slightly toward lower values of x. These effects can be made quantitative using a set of equations called the Altarelli-Parisi equations, which dictate how the PDFs change with Q. To give some idea, the next two graphs show the CTEQ5L PDFs for Q = 100 and 1000 GeV, roughly appropriate for the Tevatron and LHC respectively:

189 1 CTEQ5L, Q=100 GeV

0.8 g

0.6 x f(x) u 0.4

0.2 d d u 0 0 0.2 0.4 0.6 0.8 1 x 1 CTEQ5L, Q=1000 GeV

0.8 g

0.6 x f(x) u 0.4

0.2 d d u 0 0 0.2 0.4 0.6 0.8 1 x

Now suppose we have available a set of PDFs, and let us see how to use them to get a cross-section.

Consider scattering two hadrons h and h0, and let the partonic diﬀerential cross sections for the desired ﬁnal state X be

dσˆ(ab X) (10.109) → for any two partons a (to be taken from h) and b (from h0). The hat is used as a reminder that this is a partonic process. If X has two particles 1,2 in it, then one deﬁnes partonic Mandelstam variables:

2 sˆ = (pa + pb) ; (10.110) tˆ = (p k )2; (10.111) a − 1 uˆ = (p k )2. (10.112) a − 2

190 Let us work in the center-of-momentum frame, with approximately massless hadrons and partons, so that

µ ph = (E, 0, 0,E); (10.113) µ p 0 = (E, 0, 0, E). (10.114) h −

Then we can deﬁne a Feynman x for each of the partons, xa and xb, so that:

µ pa = xa(E, 0, 0,E); (10.115) pµ = x (E, 0, 0, E). (10.116) b b − 2 It follows that, with s = (ph + ph0 ) for the whole hadronic event,

sˆ = (x + x )2E2 (x x )E2 = 4x x E2 = sx x . (10.117) a b − a − b a b a b Here s is typically fixed by the experiment [(2000 GeV)2 at the Tevatron], whiles ˆ is different for each event. Now, to find the total cross-section to produce the final state X in h, h0 collisions, we should multiply the partonic cross section by the probabilities of finding in h a parton a with momentum fraction in the range xa to xa + dxa and the same probability for parton b in h0; then integrate over all possible xa and xb, and then sum over all the different parton species a and b. The result is:

1 1 h h0 dσ(hh0 X)= dxa dxb dσˆ(ab X) fa (xa) fb (xb). (10.118) → 0 0 → Xa,b Z Z This integration is typically done by computer, using PDFs with Q equal to some characteristic energy of the event (for example Q =s ˆ). The partonic diﬀerential cross-section dσˆ(ab X) depends on the → µ µ µ µ µ µ momentum fractions xa and xb through pa = xaph and pb = xbph0 , with ph and ph0 controlled or known by the experimenter.

10.6 Top-antitop production in pp collisions

As an example, let us consider top-antitop production at the Tevatron. The parton-level processes that can contribute to this are, with the ﬁrst parton taken to be from the proton and the second from the antiproton:

uu tt, (10.119) → dd tt, (10.120) → gg tt, (10.121) → dd tt, (10.122) → uu tt, (10.123) → ss tt, (10.124) → ss tt. (10.125) →

191 These are listed in the order of their numerical importance in contributing to the total cross section. Notice that the most likely thing is to find a quark in the proton and an antiquark in the anti-proton, but there is also a small but non-zero probability of finding an anti-quark in the proton, and a quark in the anti-proton. Ignoring the tiny contributions from charm, bottom and top partons, these are the only parton-level cross-sections that can contribute. All of the processes involving quark and antiquark in the initial state have the same dσˆ, namely dσˆ(qq tt) → , (10.126) dtˆ which you will compute in Homework Set 10. The gluon-gluon process has a partonic cross-section which is a little harder to obtain: dσˆ(gg tt) πα2 6(m2 tˆ)(m2 uˆ) m2(ˆs 4m2) → = s − − − dtˆ 8ˆs2 sˆ2 − 3(m2 tˆ)(m2 uˆ) · − − 4[(m2 tˆ)(m2 uˆ) 2m2(m2 + tˆ)] 4[(m2 tˆ)(m2 uˆ) 2m2(m2 +u ˆ)] + − − − + − − − 3(m2 tˆ)2 3(m2 uˆ)2 − − 3[(m2 tˆ)(m2 uˆ)+ m2(û tˆ)] 3[(m2 tˆ)(m2 uˆ)+ m2(tˆ uˆ)] − − − − − − , (10.127) − sˆ(m2 tˆ) − sˆ(m2 uˆ) − − ¸ where m is the mass of the top quark. Before plugging the partonic differential cross-sections into the general formula, one generally chooses to express things in terms of a more appropriate variable than tˆ. In the experimental Real World, final state particles very close to the beam are not observed, which leads to a high uncertainty in the momentum component along the beam direction. The component of the momentum transverse to the beam, pT , therefore makes a more useful variable. One can trade the dependence of dσ/dˆ tˆ on s,ˆ tˆ for a dependence ons, ˆ pT . It is also useful to note that eq. (10.117) implies sˆ dsˆ xb = ; dxb = . (10.128) xas xas

So instead of integrating over xb, we can integrate overs ˆ. The limits of integration ons ˆ are from 2 sˆmin = 4mt (the minimum required to make a top-antitop pair) tos ˆmax = s (the maximum available from the proton and antiproton, corresponding to xa = xb = 1). For a givens ˆ, the range of xa is from s/sˆ to 1. Relabeling xa as just x, we therefore have: dσ(pp tt) s 1 1 dσˆ(qq tt) → = dsˆ dx → u(x)u(ˆs/xs)+ d(x)d(ˆs/xs) dpT 4m2 s/sˆ xs dpT Z t Z ½ h +u(x)u(ˆs/xs)+ d(x)d(ˆs/xs) + 2s(x)s(ˆs/xs) dσˆ(gg tt) i + → g(x)g(ˆs/xs) . (10.129) dp T ¾ If only the total cross-section is needed, one can of course integrate both sides with respect to pT . Using the CTEQ5L PDFs and Q = mt = 175 GeV, one can compute the individual partonic contributions to the total cross-section, as a function of mt. Using √s = 2000 GeV, and αs(Q) = 0.1156, and working with the leading-order partonic cross sections, the results look like:

192 10000 total uu

1000 dd

gg t) [fb] t) (t σ 100

10 ss + ss uu

170 172 174 176 178 180 mtop [GeV}

This can be understood qualitatively from the PDFs as follows. To produce a top-antitop pair, we 2 must haves> ˆ 4mt , so according to eq. (10.117),

2 2 2 xaxb > 4mt /s = 4(175) /(2000) = 0.030625. (10.130)

So at least one of the Feynman x’s must be larger than 0.175. This means that the largest contributions come from the valence quarks. Since there are roughly twice as many valence up quarks as down quarks in the proton for a given x, and twice as many antiups as antidowns in the antiproton, the ratio of top-antitop events produced from up-antiup should be about 4 times that from down-antidown. The gluon-gluon contribution is suppressed in this case because most of the gluons are at small x and do not have enough energy to make a top-antitop pair. Finally, the contributions from sea partons are highly suppressed for the same reason. Unfortunately, the accuracy of the above results, obtained with only leading order cross sections, is not very high. The correction to the quark-antiquark processes is of order 10 to 20%, while the gluon-gluon process gets about a 70% correction when higher-order eﬀects are included. So accurate comparisons with experiment require a much more detailed and sophisticated treatment of the higher- order eﬀects.

10.7 Top-antitop production in pp collisions

In the previous subsection, we studied the production of top-antitop pairs at the Tevatron, a pp collider. One can do the same thing for the Large Hadron Collider, a pp collider, by taking into account the larger √s and the diﬀerent parton distribution function roles. Since both of the initial-state hadrons are protons, if one contributed a quark parton, the other must contribute an antiquark parton. The

193 formula for the total cross-section is now: s 1 1 σ(pp tt) = dsˆ dx σˆ(qq tt) 2u(x)u(ˆs/xs) + 2d(x)d(ˆs/xs) + 2s(x)s(ˆs/xs) → 4m2 s/sˆ xs → Z t Z ½ h i +ˆσ(gg tt) g(x)g(ˆs/xs) . (10.131) → ¾ The factors of 2 are because each proton can contribute either the quark, or the antiquark; then the contribution of the other proton is fixed. The gluon-gluon contribution has the same form as in pp collisions, because the gluon distribution is identical in protons and in antiprotons. If you numerically integrate the above formula with a computer with your favorite set of PDFs, you will find that now the gluon-gluon partonic contribution is dominant for the LHC. This is partly because all of the quark-antiquark contributions now require a sea antiquark PDF, but that is not the main reason. The real effect is that at very high energies like at the LHC, the top quark is almost massless (!) and so one can make them using partons with much lower x. At low x, we saw that the gluon distribution function is very large; one has plenty of gluons available with 1/10 of the protons’ total energy. This is actually a general feature, holding for any process with a final state that is not very heavy. This is why you sometimes hear people somewhat whimsically call the LHC a “gluon collider”; with so much energy available for the protons, almost everything is dominated by the large gluon PDF at low x. There are some processes that do not rely on gluons at all, however. We will see one example in subsection 10.9. Those processes are dominated by sea quarks at the LHC.

10.8 Kinematics in hadron-hadron scattering

Let us now consider the general problem of kinematics associated with hadron-hadron collisions with underlying 2 2 parton scattering. To make things simple, we will suppose all of the particles are → essentially massless, so what we are about to do does not quite work for tt in the ﬁnal state (but could be generalized to do so). After doing a sum/average over spins, colors, and any other unobserved degrees of freedom, we should be able to compute the diﬀerential cross-section for the partonic event from its Feynman diagrams as:

dσˆ(ab 12) → . (10.132) dtˆ As we learned in subsection 10.5, we can then write:

h h0 dσˆ(ab 12) dσ(hh0 12 + X) = f (xa)f (xb) → dtˆ dxa dxb. (10.133) → a b dtˆ Xa,b A picture of the scattering in real space might look like:

194 1

p p 0 h xaph xbph0 h → h →dσˆ ← h0 ← fa ˆ fb a dt b h h0

There are several different ways to choose the kinematic variables describing the final state. There are three significant degrees of freedom: two angles at which the final-state particles emerge with respect to the collision axis, and one overall momentum scale. (Once the magnitude of the momentum transverse to the beam for one particle is specified, the other is determined.) The angular dependence about the collision axis is trivial, so we can ignore it. For example, we can use the following three variables: momentum of particle 1 transverse to the collision axis, pT ; the total center-of-momentum energy of the final state partons, √sˆ; and the longitudinal rapidity of the two-parton system in the lab frame, defined by 1 Y = ln(x /x ). (10.134) 2 a b This may look like an obscure definition, but it is the rapidity (see section 1) needed to boost along the collision axis to get to the center-of-momentum frame for the two-parton system. It is equal to 0 if the final-state particles are back-to-back, which occurs in the unlikely case that the initial-state partons had the same energy in the lab frame. Instead of the variables (xa,xb, tˆ), we can use the more 2 directly observable variables (ˆs,pT ,Y ), or perhaps some subset of these with the others integrated over. Working in the center-of-momentum frame of the partons, we can write:

µ ˆ ˆ pa = (E, 0, 0, E); (10.135) pµ = (E,ˆ 0, 0, Eˆ); (10.136) b − kµ = (E,ˆ 0,p , Eˆ2 p2 ); (10.137) 1 T − T q kµ = (E,ˆ 0, p , Eˆ2 p2 ), (10.138) 2 − T − − T q with

2 sˆ = 4Eˆ = xaxbs, (10.139)

195 from which it follows that sˆ tˆ= (p k )2 = 1+ 1 4p2 /sˆ ; (10.140) a − 1 2 − T µ q ¶ sˆ uˆ = (p k )2 = 1 1 4p2 /sˆ , (10.141) a − 2 2 − − T µ q ¶ and so ˆ ˆ ˆ 2 tuˆ t(sxaxb + t) pT = = , (10.142) sˆ − sxaxb where the last equality usesu ˆ = sˆ tˆ = sx x tˆ for massless particles. Making the change of − − − a b − ˆ 2 variables (xa,xb, t) to (ˆs, pT ,Y ) for a diﬀerential cross section requires

2 dtdxˆ a dxb = Jdsdˆ (pT ) dY, (10.143) where J is the determinant of the Jacobian matrix of the transformation. Evaluating it, by taking the inverse of the determinant of its inverse, one ﬁnds:

1 ∂(ˆs, p2 ,Y ) − x x J = T = a b . (10.144) ¯ ∂(t,xˆ ,x ) ¯ sˆ + 2tˆ ¯ a b ¯ ¯ ¯ ¯ ¯ Therefore, eq. (10.133) becomes: ¯ ¯

dσ(hh0 12 + X) xaxb h h0 dσˆ(ab 12) →2 = fa (xa)fb (xb) → , (10.145) dsdˆ (pT ) dY sˆ + 2tˆ dtˆ Xa,b ˆ 2 where the variables xa,xb, t on the right-hand side are understood to be determined in terms ofs,p ˆ T and Y by eqs. (10.139), (10.142), and (10.134).

+ 10.9 Drell-Yan scattering (` `− production in hadron collisions)

As an example, let us consider the process of Drell-Yan scattering, which is the production of lepton pairs in hadron-hadron collisions through a virtual photon:

+ hh0 ` `−. (10.146) → This must involve QED, rather than QCD, partonic scattering, since the leptons are singlets under

SU(3)c color. However, it still depends on QCD, because to evaluate it we need to know the PDFs for the quarks inside the hadrons. Since gluons have no electric charge and do not couple to photons, the underlying partonic process is always:

+ qq ` `−. (10.147) → with the q coming from either h or h0. The cross section for this can be obtained by exactly the same methods as in 6.2 for e+e µ+µ ; we just need to remember to use the charge of the quark Q − → − q instead of the charge of the electron, and to average over initial-state colors. The latter eﬀect leads to

196 a suppression of 1/3; there is no reaction if the colors do not match. One ﬁnds that the diﬀerential partonic cross section is:

dσˆ(qq `+` ) 2πα2Q2(tˆ2 +u ˆ2) → − = q . (10.148) dtˆ 3ˆs4 Therefore, usingu ˆ = sˆ tˆ for massless scattering, and writing separate contributions from ﬁnding − − the quark, and the antiquark, in h:

+ 2 2 2 dσ(hh ` ` ) 2πα tˆ + (ˆs + tˆ) 0 0 0 → − = x x Q2 f h(x )f h (x )+ f h(x )f h (x ) . (10.149) 2 4 ˆ a b q q a q b q a q b dsdpˆ T dY 3 sˆ (ˆs + 2t) q X h i This can be used to make a prediction for the experimental distribution of events with respect to each ofs ˆ, pT , and Y . Alternatively, we can integrate the partonic diﬀerential cross-section with respect to tˆ ﬁrst. Since tˆ determines the scattering angle with respect to the collision axis, this will eliminate the integration 2 over pT . The total partonic cross section obtained by integrating eq. (10.148) is: 2 2 + 4πα Qq σˆ(qq ` `−)= . (10.150) → 9ˆs Therefore we get:

2 + 4πα 2 h h0 h h0 dσ(hh0 ` `−)= dxadxb Qq fq (xa)fq (xb)+ fq (xa)fq (xb) . (10.151) → 9ˆs q X h i

Now making the change of variables from (xa,xb) to (ˆs, Y ) requires:

1 ∂(ˆs, Y ) − dsdYˆ dxadxb = dsdYˆ = , (10.152) ¯∂(xa,xb)¯ s ¯ ¯ ¯ ¯ we therefore have: ¯ ¯ + 2 dσ(hh0 ` `−) 4πα 2 h h0 h h0 → = Qq fq (xa)fq (xb)+ fq (xa)fq (xb) . (10.153) dsdYˆ 9ssˆ q X h i Still another way to present the result is to leave onlys ˆ unintegrated, by again ﬁrst integrating the partonic diﬀerential cross-section with respect to tˆ, and then trading one of the Feynman-x variables fors ˆ, and do the remaining x-integration. This is how we wrote the top-antitop total cross-section. The Jacobian factor in the change of variables from (x ,x ) (x , sˆ) is now: a b → a 1 ∂sˆ − dxa dsˆ dxa dxb = dxa dsˆ = . (10.154) ¯∂xb ¯ xas ¯ ¯ ¯ ¯ Using this in eq. (10.151) now gives, after¯ replacing¯ xa by x and integrating: + 2 1 dσ(hh0 ` `−) 4πα 2 sˆ h h0 h h0 → = 2 Qq dx fq (x)fq (ˆs/xs)+ fq (x)fq (ˆs/xs) (10.155) dsˆ 3ˆs q s/sˆ xs X Z h i This version makes a nice prediction that is independent of the actual parton distribution functions. The right-hand side could have depended on boths ˆ and s in an arbitrary way, but to the extent that

197 the PDFs are independent of Q, we see that it is predicted to scale like 1/sˆ2 times some function of the ratios/s ˆ . Since the PDFs run very slowly with Q, this is a reasonably good prediction. Drell-Yan scattering has been studied in hh0 = pp, pp, π±p, and K±p scattering experiments, and in each case the results indeed satisfy the scaling law:

+ 2 dσ(hh0 ` `−) sˆ → = F 0 (ˆs/s) (10.156) dsˆ hh to a very good approximation. Furthermore, the functions Fhh0 gives information about the PDFs. Because of the relatively unambiguous signals of muons in particle detectors, the Drell-Yan process

+ pp µ−µ + X or (10.157) → + pp µ−µ + X (10.158) → is often one of the ﬁrst things one studies at a hadron collider to make sure everything is working correctly. It also provides a relatively clean test of the PDFs, especially at small x.

198 11 Spontaneous symmetry breaking

11.1 Global symmetry breaking

Not all of the symmetries of the laws of physics are evident in the state that describes a physical system, or even in the vacuum state with no particles. For example, in condensed matter physics, the ground state of a ferromagnetic system involves a magnetization vector that points in some particular direction, even though Maxwell’s equations in matter do not contain any special direction. This is because it is energetically favorable for the magnetic moments in the material to line up, rather than remaining randomized. The state with randomized magnetic moments is unstable to small perturbations, like a stick balanced on one end, and will settle in the more energetically-favored magnetized state. The situation in which the laws of physics are invariant under some symmetry transformations, but the vacuum state is not, is called spontaneous symmetry breaking. In this section we will study how this works in quantum field theory. There are two types of continuous symmetry transformations; global, in which the transformation does not depend on position in spacetime, and local (or gauge) in which the transformation can be different at each point. We will work out how spontaneous symmetry breaking works in each of these cases, using the example of a U(1) symmetry, and then guess the generalizations to non-Abelian symmetries. Consider a complex scalar field φ(x) with a Lagrangian density:

µ = ∂ φ∗∂ φ V (φ, φ∗) (11.1) L µ − where

2 2 V (φ, φ∗)= m φ∗φ + λ(φ∗φ) (11.2) is the potential energy. This Lagrangian is invariant under the global U(1) transformations:

iα φ(x) φ0(x)= e φ(x) (11.3) → where α is any constant. Classically, one can solve the equations of motion for φ and φ∗ following from , which are L µ δV ∂ ∂µφ + = 0 (11.4) δφ∗ µ δV ∂ ∂ φ∗ + = 0, (11.5) µ δφ

In particular, there is a solution with ∂µφ = ∂µφ∗ = 0, so that φ(x) is just equal to a constant which minimizes the potential. If m2 > 0 and λ> 0, then the minimum of the potential is clearly at φ = 0. Quantum mechanically, the ground state of the system will be one in which the expectation value of φ(x) vanishes:

0 φ(x) 0 = 0. (11.6) h | | i

199 The scalar particles created and destroyed by the field φ(x) correspond to quantized oscillations of φ(x) about the minimum of the potential. They have squared mass equal to m2, and interact with a four-scalar vertex proportional to λ. Let us now consider what happens if the signs of the parameters m2 and λ are different. If λ< 0, then the potential V (φ, φ ) is unbounded from below for arbitrarily large φ . This cannot lead to ∗ | | an acceptable theory. Classically there would be runaway solutions in which φ(x) , gaining | | → ∞ an infinite amount of kinetic energy. Quantum mechanically, the expectation value of φ(x) will grow without bound. However, there is nothing wrong with the theory if m2 < 0. (One should not think of m2 as necessarily the square of some mythical real number m, but simply as a parameter that appears in the Lagrangian density.) In that case, the potential V (φ, φ∗) has a “Mexican hat” shape, with a local maximum at φ = 0, and a degenerate set of minima with v2 φ 2 = , (11.7) | min| 2 where we have defined:

v = m2/λ. (11.8) − q The potential V does not depend on the phase of φ(x) at all, so it is impossible to unambiguously determine the phase of φ(x) at the minimum. However, by an arbitrary choice, we can make Im(φmin)= 0. In quantum mechanics, the system will have a ground state in which the expectation value of φ(x) is constant and equal to the classical minimum: v 0 φ(x) 0 = . (11.9) h | | i √2 The quantity v is a measurable property of the vacuum state, known as the vacuum expectation value, or VEV, of φ(x). If we now ask what the VEV of φ is after performing a U(1) transformation of the form eq. (11.3), we ﬁnd:

iα iα v v 0 φ0(x) 0 = e 0 φ(x) 0 = e = . (11.10) h | | i h | | i √2 6 √2 The VEV is not invariant under the U(1) symmetry operation acting on the fields of the theory; this reflects the fact that we had to make an arbitrary choice of phase. One cannot restore the invariance by defining the symmetry operation to also multiply 0 by a phase, since 0 will rotate by the opposite | i h | phase, canceling out of eq. (11.10). Therefore, the vacuum state must not be invariant under the global U(1) symmetry rotation, and the symmetry is spontaneously broken. The sign of the parameter m2 is what determines whether or not spontaneous symmetry breaking takes place in the theory. In order to further understand the behavior of this theory, it is convenient to rewrite the scalar field in terms of its deviation from its VEV. One way to do this is to write: 1 φ(x) = [v + R(x)+ iI(x)] (11.11) √2 1 φ∗(x) = [v + R(x) iI(x)] (11.12) √2 −

200 where R and I are each real scalar ﬁelds, representing the real and imaginary parts of φ. The derivative part of the Lagrangian can now be rewritten in terms of R and I, as: 1 1 = ∂µR∂ R + ∂µI∂ I. (11.13) L 2 µ 2 µ The potential appearing in the Lagrangian can be found in terms of R and I most easily by noticing that it can be rewritten as

2 2 4 V (φ, φ∗)= λ(φ∗φ v /2) λv /4. (11.14) − − Dropping the last term that does not depend on the fields, and plugging in eqs. (11.11),(11.12), this becomes: λ V (R,I) = [(v + R)2 + I2 v2]2 (11.15) 4 − λ = λv2R2 + λvR(R2 + I2)+ (R2 + I2)2. (11.16) 4 Comparing this expression with our previous discussion of real scalar fields in section 5, we can interpret the terms proportional to λv as RRR and RII interaction vertices, and the term proportional to λ as RRRR, RRII, and IIII interaction vertices. The first term proportional to λv2 is a mass term for R, but there is no term quadratic in I, so it corresponds to a massless real scalar field. Comparing to the Klein-Gordon Lagrangian density of eq. (4.18), we can identify:

m2 = 2λv2 = 2m2 (11.17) R − 2 mI = 0. (11.18)

It is useful to redo this analysis in a slightly different way, by writing 1 φ(x) = [v + h(x)]eiG(x)/v (11.19) √2 1 iG(x)/v φ∗(x) = [v + h(x)]e− (11.20) √2 instead of eq. (11.11),(11.12). Again h(x) and G(x) are two real scalar fields, related to R(x) and I(x) by a non-linear functional transformation. In terms of these fields, the potential is: λ V (h)= λv2h2 + λvh3 + h4. (11.21) 4 Notice that the field G does not appear in V at all. This is because G just corresponds to the phase of φ, and the potential was chosen to be invariant under U(1) phase transformations. However, G does have interactions coming from the part of the Lagrangian density containing derivatives. To find the derivative part of the Lagrangian, we compute:

1 iG/v i(v + h) ∂µφ = e [∂µh + ∂µG] (11.22) √2 v 1 iG/v i(v + h) ∂µφ∗ = e− [∂µh ∂µG], (11.23) √2 − v

201 so that: 1 1 h 2 = ∂µh∂ h + 1+ ∂µG∂ G. (11.24) derivatives 2 µ 2 v µ L µ ¶ The quadratic part of the Lagrangian, which determines the propagators for h and G, is 1 1 1 = ∂µh∂ h m2 h2 + ∂µG∂ G, (11.25) Lquadratic 2 µ − 2 h 2 µ with

2 2 mh = 2λv ; (11.26) 2 mG = 0. (11.27)

This conﬁrms the previous result that the spectrum of particles consists of a massive real scalar (h) and a massless one (G). The interaction part of the Lagrangian following from eqs. (11.21) and (11.24) is: 1 1 λ = h + h2 ∂µG∂ G λvh3 h4. (11.28) int v 2v2 µ 4 L µ ¶ − − The ﬁeld and particle represented by G is known as a Nambu-Goldstone boson (or sometimes just a Goldstone boson). The original U(1) symmetry acts on G by shifting it by a constant that depends on the VEV:

G G0 = G + αv; (11.29) → h h0 = h. (11.30) → This explains why G only appears in the Lagrangian with derivatives acting on it. In general, a broken global symmetry is always signaled by the presence of a massless Nambu-Goldstone boson with only derivative interactions. This is an example of Goldstone’s theorem, which we will state in a more general framework in subsection 11.3.

11.2 Local symmetry breaking and the Higgs mechanism

Let us now consider how things change if the spontaneously broken symmetry is local, or gauged. As a simple example, consider a U(1) gauge theory with a fermion ψ with charge q and gauge coupling g and a vector ﬁeld Aµ transforming according to:

ψ(x) eiqθ(x)ψ(x) (11.31) → iqθ(x) ψ(x) e− ψ(x) (11.32) → 1 A (x) A (x) ∂ θ(x). (11.33) µ → µ − g µ In order to make a gauge-invariant Lagrangian density, the ordinary derivative is replaced by the covariant derivative:

Dµψ = (∂µ + igqAµ)ψ. (11.34)

202 Now, in order to spontaneously break the gauge symmetry, we introduce a complex scalar ﬁeld φ with charge +1. It transforms under a gauge transformation like

φ(x) eiθ(x)φ(x) (11.35) → iθ(x) φ∗(x) e− φ∗(x). (11.36) →

To make a gauge-invariant Lagrangian, we again replace the ordinary derivative acting on φ, φ∗ by covariant derivatives:

Dµφ = (∂µ + igAµ)φ (11.37)

D φ∗ = (∂ igA )φ∗. (11.38) µ µ − µ The Lagrangian density for the scalar and vector degrees of freedom of the theory is:

µ 1 µν = D φ∗D φ V (φ, φ∗) F F , (11.39) L µ − − 4 µν where V (φ, φ∗) is as given before in eq. (11.2). Because the covariant derivative of the ﬁeld transforms like

D φ eiθD φ, (11.40) µ → µ this Lagrangian is easily checked to be gauge-invariant. If m2 > 0 and λ > 0, then this theory describes a massive scalar, with self-interactions with a coupling proportional to λ, and interaction with the massless vector field Aµ. However, if m2 < 0, then the minimum of the potential for the scalar field brings about a non-zero VEV 0 φ 0 = v/√2= m2/2λ, just as in the global symmetry case. Using the same decomposition h | | i − of φ into real fields h andp G as given in eq. (11.19), one finds: 1 1 D φ = ∂ h + ig(A + ∂ G)(h + v) eiG/v. (11.41) µ √ µ µ gv µ 2 · ¸ It is convenient to define a new vector field: 1 V = A + ∂ G, (11.42) µ µ gv µ since this is the object that appears in eq. (11.41). Then

1 iG/v Dµφ = [∂µh + igVµ(v + h)] e , (11.43) √2 1 iG/v Dµφ∗ = [∂µh igVµ(v + h)] e− . (11.44) √2 − Note also that since

∂ A ∂ A = ∂ V ∂ V , (11.45) µ ν − ν µ µ ν − ν µ

203 the vector field strength part of the Lagrangian is the same written in terms of the new vector Vµ as it was in terms of the old vector Aµ: 1 1 F µνF = (∂ V ∂ V )(∂µV ν ∂νV µ). (11.46) −4 µν −4 µ ν − ν µ − The complete Lagrangian density of the vector and scalar degrees of freedom is now: 1 1 = ∂µh∂ h + g2(v + h)2V µV F µνF λ(vh + h2/2)2. (11.47) L 2 µ µ − 4 µν − h i This Lagrangian has the very important property that the field G has completely disappeared! Reading off the part quadratic in h, we see that it has the same squared mass as in the global symmetry case, namely

2 2 mh = 2λv . (11.48)

There is also a term quadratic in the vector ﬁeld:

1 g2v2 = F µνF + V µV . (11.49) LVV −4 µν 2 µ This means that by spontaneously breaking the gauge symmetry, we have given a mass to the corresponding vector ﬁeld:

2 2 2 mV = g v . (11.50)

We can understand why the disappearance of the field G goes along with the appearance of the vector boson mass as follows. A massless spin-1 vector boson (like the photon) has only two possible polarization states, each transverse to its direction of motion. In contrast, a massive spin-1 vector boson has three possible polarization states; the two transverse, and one longitudinal (parallel) to its direction of motion. The additional polarization state degree of freedom had to come from somewhere, so one real scalar degree of freedom had to disappear. The words used to describe this are that the vector boson becomes massive by “eating” the would-be Nambu-Goldstone boson, which becomes its longitudinal polarization degree of freedom. This is called the Higgs mechanism. The original field φ(x) is called a Higgs field, and the surviving real scalar degree of freedom h(x) is called by the generic term Higgs boson. The Standard Model Higgs boson and the masses of the W ± and Z bosons result from a slightly more complicated version of this same idea, as we will see. An alternative way to understand what has just happened to the would-be Nambu-Goldstone boson G(x) is that it has been “gauged away”. Recall that 1 φ = (v + h)eiG/v (11.51) √2 behaves under a gauge transformation as:

φ eiθφ. (11.52) →

204 Normally we think of θ in this equation as just some ordinary function of spacetime, but since this is true for any θ, we can choose it to be proportional to the Nambu-Goldstone ﬁeld itself:

θ(x)= G(x)/v. (11.53) − This choice, known as “unitary gauge”, eliminates G(x) completely, just as we saw in eq. (11.47). In unitary gauge, 1 φ(x) = [v + h(x)]. (11.54) √2

Notice also that the gauge transformation eq. (11.53) gives exactly the term in eq. (11.42), so that Vµ is simply the unitary gauge version of Aµ. The advantage of unitary gauge is that the true particle content of the theory (a massive vector and real Higgs scalar) is more obvious than in the version of the Lagrangian written in terms of the original ﬁelds φ and Aµ. However, it turns out to be easier to prove that the theory is renormalizable if one works in a diﬀerent gauge in which the would-be Nambu-Goldstone bosons are retained. The physical predictions of the theory do not depend on which gauge one chooses, but the ease with which one can compute those results depends on picking the right gauge for the problem at hand. Let us catalog the propagators and interactions of this theory, in unitary gauge. The propagators of the Higgs scalar and the massive vector are:

µ ν i p p g + µ ν ←→ p2 m2 + i² − µν m2 − V " V # i ←→ p2 m2 + i² − h

From eq. (11.47), there are also hV V and hhV V interaction vertices: µ

2ig2vgµν ←→

2ig2v2gµν ←→

205 There are also hhh and hhhh self-interactions:

6iλv ←→ −

6iλ ←→ −

Finally, a fermion with charge q inherits the same interactions with Vµ that it had with Aµ, coming from the covariant derivative:

µ igqγµ ←→ −

This is a general way of making massive vector ﬁelds in gauge theories which interact with scalars and fermions, without ruining renormalizability.

11.3 Goldstone’s Theorem and the Higgs mechanism in general

Let us now state, without proof, how all of the considerations above generalize to arbitrary groups.

First, suppose we have scalar ﬁelds φi in some representation of a global symmetry group with gener- aj ators Ti . There is some potential

V (φi, φi∗) (11.55) which we presume has a minimum at which at least some of the φi are non-zero. This of course depends on the parameters and couplings appearing in V . The ﬁelds φi will then have VEVs which can be written:

vi 0 φi 0 = . (11.56) h | | i √2 Any group generators which satisfy

aj Ti vj = 0 (11.57)

206 correspond to unbroken symmetry transformations. In general, the unbroken global symmetry group is the one formed from the unbroken symmetry generators, and the vacuum state is invariant under this unbroken subgroup. The broken generators satisfy

T ajv = 0. (11.58) i j 6 Goldstone’s Theorem states that for every spontaneously broken generator, labeled by a, of a global symmetry group, there must be a corresponding Nambu-Goldstone boson. [The group U(1) has just one generator, so there was just one Nambu-Goldstone boson.] In the case of a local or gauge symmetry, each of the would-be Goldstone bosons is eaten by the vector field with the corresponding index a. The vector fields for the broken generators become massive, with squared masses that can be computed in terms of the VEV(s) and the gauge coupling(s) of the theory. There are also Higgs boson(s) for the uneaten components of the scalar fields that obtained VEVs. One might also ask whether it is possible for fields other than scalars to obtain vacuum expectation values. If one could succeed in concocting a theory in which a fermion spinor field or a vector field has a VEV:

0 Ψ 0 = 0 (?), (11.59) h | α| i 6 0 A 0 = 0 (?), (11.60) h | µ| i 6 then Lorentz invariance will necessarily be broken, since the alleged VEV carries an uncontracted spinor or vector index, and therefore transforms under the Lorentz group. This would imply that the broken generators would include Lorentz boosts and rotations, in contradiction with experiment. However, one can have vacuum expectation values for antifermion-fermion ﬁelds, since they can form a Lorentz scalar:

0 ΨΨ 0 = 0 (11.61) h | | i 6 In fact, in QCD one can show that the antiquark-quark ﬁelds do have vacuum expectation values:

0 uu 0 0 dd 0 0 ss 0 µ3 = 0, (11.62) h | | i ≈ h | | i ≈ h | | i≈ 6 where µ is a quantity with dimensions of [mass] which is set by the scale ΛQCD at which non-perturbative eﬀects become important. This is known as chiral symmetry breaking. The chiral symmetry is a global, approximate symmetry by which left-handed u,d,s quarks are rotated into each other and right-handed u,d,s quarks are rotated into each other. (The objects qq are color singlets, so these antifermion-fermion

VEVs do not break SU(3)c symmetry.) Chiral symmetry breaking is actually the mechanism which is responsible for most of the mass of the proton and the neutron, and therefore most of the mass of everyday objects. When the chiral symmetry is spontaneously broken, the Goldstone bosons which 0 arise include the pions π± and π . They are not exactly massless because the chiral symmetry was only approximate to begin with, but the Goldstone theorem successfully explains why they are much lighter than the proton. Unfortunately, these issues cannot be explored here.

207 12 The standard electroweak model

12.1 SU(2) U(1) representations and Lagrangian L × Y In this section, we will study the Higgs mechanism of the Standard Model, which is responsible for the masses of W ± and Z bosons, and the masses of the leptons and most of the masses of the heavier quarks.

The electroweak interactions are mediated by three massive vector bosons W ±, Z and the massless photon γ. The gauge group before spontaneous symmetry breaking must therefore have four generators. After spontaneous symmetry breaking, the remaining unbroken gauge group is electromagnetic gauge invariance. A viable theory must explain the qualitative experimental facts that the W ± bosons couple only to L-fermions (and R-antifermions), that the Z boson couples diﬀerently to L-fermions and R-fermions, but γ couples with the same strength to L-fermions and R-fermions. Also, there are very stringent quantitative experimental tests involving the relative strengths of fermion-antifermion- vector couplings and the ratio of the W and Z masses. The Standard Model (SM) of electroweak interactions of Glashow, Weinberg and Salam successfully incorporates all of these features and tests into a spontaneously broken gauge theory. In the SM, the gauge symmetry breaking is:

SU(2) U(1) U(1) . (12.1) L × Y → EM We will need to introduce a Higgs ﬁeld to produce this pattern of symmetry breaking.

The SU(2)L subgroup is known as weak isospin. Left-handed SM fermions are known to be doublets under SU(2)L:

ν ν ν u c t e , µ , τ , L , L , L . (12.2) e µ τ d s b µ L ¶ µ L ¶ µ L ¶ µ L ¶ µ L ¶ µ L ¶ Notice that the electric charge of the upper member of each doublet is always 1 greater than that of the lower member. The SU(2)L representation matrix generators acting on these ﬁelds are proportional to the Pauli matrices:

T a = σa/2, (a = 1, 2, 3) (12.3) with corresponding vector gauge boson ﬁelds:

a Wµ , (a = 1, 2, 3) (12.4) and a coupling constant g. The right-handed fermions

eR, µR, τR, uR, cR, tR, dR, sR, bR, (12.5) are all singlets under SU(2)L.

Meanwhile, the U(1)Y subgroup is known as weak hypercharge and has a coupling constant g0 and a vector boson Bµ, sometimes known as the hyperphoton. The weak hypercharge Y is a conserved

208 charge just like electric charge q, but it is diﬀerent for left-handed and right-handed fermions. Both members of an SU(2)L doublet must have the same weak hypercharge in order to satisfy SU(2)L gauge invariance. Following the general discussion of Yang-Mills gauge theories in section 9.2 [see eqs. (9.97) and (9.98)], the pure-gauge part of the electroweak Lagrangian density is: 1 1 = W a W aµν B Bµν, (12.6) Lgauge −4 µν − 4 µν where:

W a = ∂ W a ∂ W a g²abcW bW c (12.7) µν µ ν − ν µ − µ ν B = ∂ B ∂ B (12.8) µν µ ν − ν µ

abc 123 are the SU(2)L and U(1)Y field strengths. The totally antisymmetric ² (with ² = +1) are the structure constants for SU(2) . This provides for kinetic terms of the vector fields, and W a L Lgauge µ self-interactions. The interactions of the electroweak gauge bosons with fermions are determined by the covariant derivative. For example, the covariant derivatives acting on the lepton fields are:

νe a a νe Dµ = ∂µ + ig0BµY`L + igWµ T (12.9) eL eL µ ¶ h i µ ¶ DµeR = ∂µ + ig0BµY`R eR. (12.10) £ ¤ where Y`L and Y`R are the weak hypercharges of left-handed leptons and right-handed leptons, and 2 2 unit matrices are understood to go with the ∂ and B terms in eq. (12.9). A multiplicative × µ µ factor can always be absorbed into the deﬁnition of the coupling g0, so without loss of generality, it is traditional to set Y = Q = 1. The weak hypercharges of all other fermions are then ﬁxed. † `R ` − Using the explicit form of the SU(2)L generators in terms of Pauli matrices in eq. (12.3), the covariant derivative of left-handed leptons is:

3 1 2 νe νe Bµ 0 g Wµ Wµ iWµ νe Dµ = ∂µ + i g0Y + − (12.11) e e `L 0 B 2 W 1 + iW 2 W 3 e µ L ¶ µ L ¶ · µ µ ¶ µ µ µ − µ ¶¸ µ L ¶ Therefore, the covariant derivatives of the lepton ﬁelds can be summarized as:

g 3 g 1 2 D ν = ∂ ν + i g0Y B + W ν + i (W iW )e ; (12.12) µ e µ e `L µ 2 µ e 2 µ µ L µ ¶ − g 3 g 1 2 D e = ∂ e + i g0Y B W e + i (W + iW )ν ; (12.13) µ L µ L `L µ 2 µ L 2 µ µ e µ − ¶ D e = ∂ e ig0B e . (12.14) µ R µ R − µ L The covariant derivative of a ﬁeld must carry the same electric charge as the ﬁeld itself, in order for charge to be conserved. Evidently, then, W 1 iW 2 must carry electric charge +1 and W 1 + iW 2 must µ − µ µ µ †Some references use weak hypercharges Y which are all normalized to be a factor of 2 larger than here.

209 carry electric charge 1, so these must be identiﬁed with the W bosons of the weak interactions. − ± Consider the interaction Lagrangian following from

µ νe = i ( νe eL ) γ Dµ (12.15) L eL g µ ¶ g = ν γµe (W 1 iW 2) e γµν (W 1 + iW 2)+ .... (12.16) −2 e L µ − µ − 2 L e µ µ Comparing with eqs. (8.95), (8.96), and (8.125), we ﬁnd that to reproduce the weak-interaction La- grangian of muon decay, we must have: 1 W + (W 1 iW 2); (12.17) µ ≡ √2 µ − µ 1 1 2 W − (W + iW ). (12.18) µ ≡ √2 µ µ

The 1/√2 normalization agrees with our previous convention; the real reason for it is so that the kinetic terms for W have a standard normalization: = 1 (∂ W + ∂ W +)(∂µW ν ∂νW µ). ± L − 2 µ ν − ν µ − − − 3 The vector bosons Bµ and Wµ are both electrically neutral. As a result of spontaneous symmetry breaking, we will find that they mix. In other words, the fields with well-defined masses (“mass 3 eigenstates” or “mass eigenfields”) are not Bµ and Wµ , but are orthogonal linear combinations of these two gauge eigenstate fields. One of the mass eigenstates is the photon field Aµ, and the other is the massive Z boson vector field, Zµ. One can write the relation between the gauge eigenstate and mass eigenstate fields as a rotation in field space by an angle θW , known as the weak mixing angle or Weinberg angle: W 3 cos θ sin θ Z µ = W W µ , (12.19) B sin θ cos θ A µ µ ¶ µ − W W ¶ µ µ ¶ with the inverse relation: Z cos θ sin θ W 3 µ = W − W µ . (12.20) A sin θ cos θ B µ µ ¶ µ W W ¶ µ µ ¶

We now require that the resulting theory has the correct photon coupling to fermions, by requiring that the ﬁeld Aµ appears in the covariant derivatives in the way dictated by QED. The covariant derivative of the right-handed electron ﬁeld eq. (12.14) can be written:

D e = ∂ e ig0 cos θ A e + ig0 sin θ Z e . (12.21) µ R µ R − W µ R W µ R Comparing to D e = ∂ e ieA e from QED, we conclude that: µ R µ R − µ R

g0 cos θW = e. (12.22)

Similarly, one ﬁnds using eq. (12.19) that: g D e = ∂ e + i g0Y cos θ sin θ A e + ... (12.23) µ L µ L `L W 2 W µ L µ − ¶ 210 Again comparing to the prediction of QED that D e = ∂ e ieA e , it must be that: µ L µ L − µ L g sin θ g0Y cos θ = e (12.24) 2 W − `L W is the electromagnetic coupling. In the same way: g D ν = ∂ ν + i g0Y cos θ + sin θ A ν + ... (12.25) µ e µ e `L W 2 W µ e µ ¶ where the ... represent W and Z terms. However, we know that the neutrino has no electric charge, and therefore its covariant derivative cannot involve the photon. So, the coeﬃcient of Aµ in eq. (12.25) must vanish: g sin θ + g0Y cos θ = 0. (12.26) 2 W `L W Now combining eqs. (12.22), (12.24), and (12.26), we learn that:

Y = 1/2, (12.27) `L − gg e = 0 , (12.28) 2 2 g + g0 tan θW = gp0/g, (12.29) so that g g sin θ = 0 ; cos θ = . (12.30) W 2 2 W 2 2 g + g0 g + g0 These are requirements that will havep to be satisﬁed by the spontaneousp symmetry breaking mechanism. The present numerical values from experiment are approximately:

g = 0.652; (12.31)

g0 = 0.357; (12.32) e = 0.313; (12.33) 2 sin θW = 0.231. (12.34)

These are all renormalized, running parameters, evaluated at a renormalization scale Q = mZ = 91.19 GeV in the MS scheme. In a similar way, one can work out what the weak hypercharges of all of the other SM quarks and leptons have to be, in order to reproduce the correct electric charges appearing in the coupling to the photon ﬁeld Aµ from the covariant derivative. The results can be summarized in terms of the SU(3) SU(2) U(1) representations: c × L × Y ν ν ν e , µ , τ , (1, 2, 1 ) e µ τ 2 µ L ¶ µ L ¶ µ L ¶ ←→ − e , µ , τ , (1, 1, 1) R R R ←→ − u c t L , L , L , (3, 2, 1 ) (12.35) d s b 6 µ L ¶ µ L ¶ µ L ¶ ←→ u , c , t , (3, 1, 2 ) R R R ←→ 3 d , s , b , (3, 1, 1 ). R R R ←→ − 3

211 In general, the electric charge of any ﬁeld f is given in terms of the eigenvalue of the 3 component of weak isospin matrix, T 3, and the weak hypercharge Y , as:

3 Qf = Tf + Yf . (12.36)

Here T 3 is +1/2 for the upper component of a doublet, 1/2 for the lower component of a doublet, and f − 0 for an SU(2)L singlet. The couplings of the SM fermions to the Z boson then follow as a prediction. One ﬁnds for each SM fermion f:

= Zµfγ (gf P + gf P )f, (12.37) LZff − µ L L R R where

f 3 g = g cos θ T g0 sin θ Y ; (12.38) L W fL − W fL f g = g0 sin θ Y (12.39) R − W fR with coeﬃcients:

Fermion T 3 Y Y Q fL fL fR f ν ,ν ,ν 1 1 0 0 e µ τ 2 − 2 e, µ, τ 1 1 1 1 − 2 − 2 − − 1 1 2 2 u,c,t 2 6 3 3 d, s, b 1 1 1 1 − 2 6 − 3 − 3

12.2 The Standard Model Higgs mechanism

Let us now turn to the question of how to spontaneously break the electroweak gauge symmetry in a way that satisfies the above conditions. There is actually more that one way to do this, but we will follow the simplest possibility, which is to introduce a complex SU(2)L-doublet scalar Higgs field with weak hypercharge YΦ = +1/2: φ+ 1 Φ= (1, 2, ). (12.40) φ0 2 µ ¶ ←→ Each of the fields φ+ and φ0 is a complex scalar field; we know that they carry electric charges +1 and 0 respectively from eq. (12.36). Under gauge transformations, Φ transforms as:

iθa(x)σa/2 SU(2) : Φ(x) Φ0(x) = e Φ(x); (12.41) L → iθ(x)/2 U(1) : Φ(x) Φ0(x) = e Φ(x). (12.42) Y → The Hermitian conjugate ﬁeld transforms as:

iθaσa/2 SU(2) : Φ† Φ0† = Φ†e− ; (12.43) L → iθ/2 U(1) : Φ† Φ0† = Φ†e− . (12.44) Y →

212 It follows that the combinations

µ Φ†Φ and D Φ†DµΦ (12.45) are gauge singlets. We can therefore build a gauge-invariant potential:

2 2 V (Φ, Φ†)= m Φ†Φ+ λ(Φ†Φ) , (12.46) and the Lagrangian density for Φ is:

µ = D Φ†D Φ V (Φ, Φ†). (12.47) L µ −

0 Now, provided that m2 < 0, then Φ = is a local maximum, rather than a minimum, of the 0 potential. This will ensure the spontaneousµ symmetry¶ breaking that we demand. There are degenerate minima of the potential with

2 2 Φ†Φ= v /2; v = m /λ. (12.48) − q Without loss of generality, we can choose the VEV of Φ to be real, and entirely in the second (electrically neutral) component of the Higgs ﬁeld: 0 0 Φ 0 = . (12.49) v/√2 h | | i µ ¶

This is a convention, which can always be achieved if necessary by doing an SU(2)L gauge transformation on the ﬁeld Φ to make it so. By deﬁnition, the surviving U(1) gauge symmetry is U(1)EM, so the component of Φ obtaining a VEV must be the one assigned 0 electric charge. The U(1)EM gauge transformations acting on Φ are a combination of SU(2)L and U(1)Y transformations: 1 0 Φ Φ = exp iθ Φ, (12.50) 0 0 0 → · µ ¶¸ or, in components,

φ+ eiθφ+ (12.51) → φ0 φ0. (12.52) → Comparing with the QED gauge transformation rule of eq. (9.5), we see that indeed φ+ and φ0 have charges +1 and 0, respectively. The Higgs ﬁeld Φ has two complex, so four real, scalar ﬁeld degrees of freedom. Therefore, following the example of section 11.2, we can write it as:

iGa(x)σa/2v 0 Φ(x)= e v+h(x) (12.53) µ √2 ¶ where Ga(x) (a = 1, 2, 3) and h(x) are each real scalar ﬁelds. The Ga are would-be Nambu-Goldstone bosons, corresponding to the three broken generators in SU(2) U(1) U(1) . The would-be L × Y → EM

213 Nambu-Goldstone fields can be removed by going to unitary gauge, which means performing an SU(2)L gauge transformation of the form of eq. (12.41), with θa = Ga/v. This completely eliminates the Ga − from the Lagrangian, so that in the unitary gauge we have simply 0 Φ(x)= v+h(x) . (12.54) µ √2 ¶ The field h creates and destroys the physical Higgs particle, an electrically neutral real scalar that has yet to be discovered experimentally. We can now plug this into the Lagrangian density of eq. (12.47), to find interactions and mass terms for the remaining Higgs field h and the vector bosons. The covariant derivative of Φ in unitary gauge is:

1 0 i g g 0 D Φ = + 0 B + W aσa (12.55) µ √ ∂ h √ 2 µ 2 µ v + h 2 µ µ ¶ 2 · ¸ µ ¶ and its Hermitian conjugate is:

1 i g0 g a a D Φ† = ( 0 ∂ h ) ( 0 v + h ) B + W σ . (12.56) µ √ µ √ 2 µ 2 µ 2 − 2 · ¸ Therefore,

2 3 µ 1 µ (v + h) g0Bµ + gWµ √2gWµ− D Φ†D Φ= ∂ h∂ h + (0 1) µ 2 µ 8 √2gW + g B gW 3 µ µ 0 µ − µ ¶ g Bµ + gW 3µ √2gW µ 0 0 − (12.57) √2gW +µ g Bµ gW 3µ 1 µ 0 − ¶ µ ¶ or, after simplifying,

2 µ 1 µ (v + h) 2 + µ 1 3 3µ µ D Φ†D Φ= ∂ h∂ h + g W W − + (gW g0B )(gW g0B ) . (12.58) µ 2 µ 4 µ 2 µ µ · − − ¸ This can be further simpliﬁed using:

3 2 2 3 2 2 gW g0B = g + g cos θ W sin θ B = g + g Z , (12.59) µ − µ 0 W µ − W µ 0 µ q h i q where the ﬁrst equality uses eq. (12.30) and the second uses eq. (12.20). So ﬁnally we have:

2 µ 1 µ (v + h) 2 + µ 1 2 2 µ = D Φ†D Φ= ∂ h∂ h + g W W − + (g + g0 )Z Z . (12.60) Φ kinetic µ 2 µ 4 µ 2 µ L · ¸ 2 2 The parts of this proportional to v make up (mass) terms for the W ± and Z vector bosons, vindicating the earlier assumption of neutral vector boson mixing with the form that we took for the sine and cosine of the weak mixing angle. Since there is no such (mass)2 term for the photon ﬁeld Aµ, we have successfully shown that the photon remains massless, in agreement with the fact that U(1)EM gauge invariance remains unbroken. The speciﬁc prediction is:

g2v2 m2 = (12.61) W 4 (g2 + g 2)v2 m2 = 0 (12.62) Z 4

214 which agrees with the experimental values provided that the VEV is approximately:

v = 246 GeV. (12.63)

A non-trivial prediction of the theory is that

mW /mZ = cos θW . (12.64)

All of the above predictions are subject to small, but measurable, loop corrections. For example, the present experimental values m = 80.42 0.06 GeV and m = 91.1882 0.0022 GeV yield: W ± Z ± 2 on shell 2 2 sin θ − 1 m /m = 0.2223 0.0011, (12.65) W ≡ − W Z ± which is signiﬁcantly lower than the MS-scheme running value in eq. (12.34). The remaining terms in eq. (12.60) are Higgs-vector-vector and Higgs-Higgs-vector-vector couplings. This part of the Lagrangian density implies the following unitary gauge Feynman rules:

µ W ± ν µ Z ν

i p p i p p g + µ ν g + µ ν p2 m2 + i² − µν m2 p2 m2 + i² − µν m2 − W " W # − Z " Z #

W ν Zµ h h

W µ Zν g2v (g2 + g 2)v igµν igµν 0 2 2

h W µ h Zµ

h W ν h Zν g2 (g2 + g 2) igµν igµν 0 2 2

with the arrow direction on W ± lines indicating the direction of the ﬂow of positive charge. The ﬁeld- strength Lagrangian terms of eq. (12.6) provides the momentum part of the W , Z propagators above, and also yields 3-gauge-boson and 4-gauge-boson interactions:

215 q Wν p Aµ ie[gµν(p q)ρ + gνρ(q k)µ + gρµ(k p)ν] − − − −

k Wρ

q Wν p Zµ ig cos θ [gµν(p q)ρ + gνρ(q k)µ + gρµ(k p)ν] − W − − −

k Wρ W µ W ν Aµ W σ

W σ W ρ Aν W ρ

ig2Xµν,ρσ ie2Xµν,ρσ −

Zµ W ρ Aµ W ρ

Zν W σ Zν W σ

ig2 cos2 θ Xµν,ρσ ig2 sin θ cos θ Xµν,ρσ − W − W W where:

Xµν,ρσ = 2gµνgρσ gµρgνσ gµσgνρ. (12.66) − −

Finally, the Higgs potential V (Φ, Φ†) gives rise to a mass and self-interactions for h. In unitary gauge: λ V (h)= λv2h2 + λvh3 + h4, (12.67) 4 just as in the toy model studied in 11.2. Therefore, the Higgs boson has self-interactions with Feynman rules:

216 6iλv 6iλ − − and a mass

mh = √2λv, (12.68)

It would be great if we could evaluate this numerically using present data. Unfortunately, while we know what the Higgs VEV v is, there is no experiment which directly sheds light on λ. There are indirect effects of the Higgs mass in loops which affect some present experiments. These experiments suggest that mh cannot be too large, but these bounds are not very strong, especially if we consider that the SM is probably not the final story.

12.3 Fermion masses and Cabibbo-Kobayashi-Maskawa mixing

The gauge group representations for fermions in the Standard Model are chiral. This means that the left-handed fermions transform in a diﬀerent representation than the right-handed fermions. Chiral fermions have the property that they cannot have masses without breaking the symmetry that makes them chiral. For example, suppose we try to write down a mass term for the electron:

= m ee. (12.69) Lelectron mass − e The Dirac spinor for the electron can be separated into left- and right-handed pieces,

e = PLeL + PReR, (12.70) and the corresponding barred spinor as:

0 0 0 e = (eL† PL + eR† PR)γ = eL† γ PR + eR† γ PL = eLPR + eRPL, (12.71) where, to avoid any confusion between between (eL) and (e)PL, we explicitly deﬁne

0 0 e e† γ ; e e† γ . (12.72) L ≡ L R ≡ L Equation (12.69) can therefore be written:

= m (e e + e e ). (12.73) Lelectron mass − e L R R L

The point is that this is clearly not a gauge singlet. In the ﬁrst place, the eL part of each term transforms as a doublet under SU(2)L, and the eR is a singlet, so each term is an SU(2)L doublet.

217 Furthermore, the ﬁrst term has Y = Y Y = 1/2, while the second term has Y = 1/2. All terms eR − eL − in the Lagrangian must be gauge singlets in order not to violate the gauge symmetry, so the electron mass is disqualiﬁed from appearing in this form. More generally, for any Standard Model fermion f, the naive mass term

= m (f f + f f ) (12.74) Lf mass − f L R R L is not an SU(2)L singlet, and is not neutral under U(1)Y , and so is not allowed. Fortunately, fermion masses can still arise with the help of the Higgs ﬁeld. For the electron, there is a gauge-invariant term: φ+ = y ( ν e ) e +c.c. (12.75) electron Yukawa e e L φ0 R L − µ ¶

Here ye is a Yukawa coupling of the type we studied in 7.3. The ﬁeld ( νe eL ) carries weak hypercharge +1/2, as does the Higgs ﬁeld, and e carries weak hypercharge 1, so the whole term is a U(1) singlet, R − Y as required. Moreover, the doublets transform under SU(2)L as: + + φ iθaσa/2 φ e− (12.76) φ0 → φ0 µ ¶ µ a¶a ( ν e ) ( ν e ) e+iθ σ /2, (12.77) e L → e L so eq. (12.75) is also an SU(2)L singlet. Going to the unitary gauge of eq. (12.54), it becomes:

ye Yukawa = (v + h)(eLeR + eReL), (12.78) L −√2 or, reassembling the Dirac spinors without projection matrices:

ye Yukawa = (v + h)ee. (12.79) L −√2 This can now be interpreted as an electron mass, equal to

yev me = , (12.80) √2 and, as a bonus, an electron-positron-Higgs interaction vertex, with Feynman rule:

iy /√2 − e

Since we know the electron mass and the Higgs VEV already, we can predict the electron Yukawa coupling:

0.511 MeV 6 y = √2 = 2.94 10− . (12.81) e 246 GeV µ ¶ × 218 Unfortunately, this is so small that we can forget about ever observing the interactions of the Higgs particle h with an electron. Notice that although the neutrino participates in the Yukawa interaction, it disappears in unitary gauge from that term. Masses for all of the other leptons, and the down-type quarks (d, s, b) in the Standard Model arise in exactly the same way. For example, the bottom quark mass comes from the gauge-invariant Yukawa coupling: φ+ = y ( t b ) b +c.c., (12.82) b L L φ0 R L − µ ¶ implying that, in unitary gauge, we have a b-quark mass and an hbb vertex: y = b (v + h)bb. (12.83) L −√2 The situation is slightly diﬀerent for up-type quarks (u,c,t), because the complex conjugate of the ﬁeld

Φ must appear in order to preserve U(1)Y invariance. It is convenient to deﬁne

0 ˜ 0 1 φ ∗ Φ Φ∗ = + , (12.84) ≡ 1 0 φ ∗ µ − ¶ µ − ¶ which transforms as an SU(2)L doublet in exactly the same way that Φ does:

a a Φ eiθ σ /2Φ; (12.85) → a a Φ˜ eiθ σ /2Φ˜. (12.86) → Also, Φ˜ has weak hypercharge Y = 1/2. (The ﬁeld φ+ has a negative electric charge.) Therefore, − ∗ one can write a gauge-invariant Yukawa coupling for the top quark as:

0 φ ∗ = yt ( tL bL ) + tR +c.c. (12.87) L − φ ∗ µ − ¶ Going to unitary gauge, one ﬁnds that the top quark has a mass: y = t (v + h)tt. (12.88) L −√2 In all cases, the unitary-gauge version of the gauge-invariant Yukawa interaction is y = f (v + h)ff. (12.89) L −√2 The mass and the h-fermion-antifermion coupling obtained by each Standard Model fermion in this way are both proportional to yf . The Higgs mechanism not only explains the masses of the W ± and Z bosons, but also explains the masses of fermions. Notice that all of the particles in the Standard Model (except the photon and gluon, which must remain massless because of SU(3) U(1) gauge c × EM invariance) get a mass from spontaneous electroweak symmetry breaking of the form:

mparticle = kv, (12.90)

219 where k is some combination of dimensionless couplings. For the fermions, it is proportional to a

Yukawa coupling; for the W ± and Z bosons it depends on gauge couplings, and for the Higgs particle itself, it is the Higgs self-coupling λ. So all mass scales in the Standard Model are set by v. There are two notable qualiﬁcations for quarks. First, there are large corrections from gluon loops, which modify the tree-level prediction. For each quark, the physical mass measured from kinematics is y v 4α m = q 1+ s + ... . (12.91) q √ 3π 2 · ¸ The QCD gluon-loop correction increases mt by roughly 6%, and has an even larger eﬀect on mb because

αs is larger at the renormalization scale Q = mb than at Q = mt. The two-loop and higher corrections (indicated by ...) are smaller but still significant, and must be taken into account in precision work, for example when predicting the branching fractions of the Higgs boson decay. This was the reason for the mf notation that we used in 7.3. A second qualification is that there is actually another source of quark masses, coming from non-perturbative QCD effects. If the Higgs field did not break SU(2) U(1) , L × Y then these chiral symmetry breaking effects would do it anyway, using an antifermion-fermion VEV rather than a scalar VEV. This gives contributions to all quark masses that are roughly of order ΛQCD. For the top, bottom, and even charm quarks, this is relatively insignificant. However, for the up and down quarks, it is actually the dominant effect. They get only about 5 to 10 MeV of their mass from the Higgs field. Therefore, the most important source of mass in ordinary matter is really QCD, not the Standard Model Higgs field. The Standard Model fermions consist of three identical families. Therefore, the most general form of the Yukawa interactions is actually: + i φ j = νi ` ye ` +c.c., (12.92) e,µ,τ Yukawas L φ0 i Rj L − µ ¶ ¡ ¢ + i φ j = ui d yd d +c.c., (12.93) d,s,b Yukawas L L φ0 i Rj L − µ ¶ ¡ ¢ 0 i φ ∗ j i u u,c,t Yukawas = uL dL + y i uRj +c.c.. (12.94) L − φ ∗ µ − ¶ Here i, j are indices that run over the three¡ families,¢ so that:

eR eL νe ` = µ ; ` = µ ; ν = ν ; (12.95) Rj  R  Lj  L  i  µ  τR τL ντ       dL dR uL uR d = s ; d = s ; u = c ; u = c . (12.96) Lj  L  Rj  R  Lj  L  Rj  R  bL bR tL tR         In general, the Yukawa couplings

j j j yei , ydi , yui (12.97) are complex 3 3 matrices in family space. In unitary gauge, the Yukawa interaction Lagrangian can × be written as:

h i j i j i j = 1+ `0 me `0 + d0 md d0 + u0 mu u0 +c.c., (12.98) L − v L i Rj L i Rj L i Rj µ ¶ n o 220 where

j v j mf i = yf . (12.99) √2 i (The fermion ﬁelds are now labeled with a prime, to distinguish them from the basis we are about to introduce.) It therefore appears that the masses of Standard Model fermions are actually 3 3 complex × matrices. Its is most convenient to work in a basis in which the fermion masses are real and positive, so that the Feynman propagators are simple. This can always be accomplished, thanks to the following:

Mass Diagonalization Theorem. Any complex matrix M can be diagonalized by a biunitary transformation:

UL† MUR = MD (12.100)

where MD is diagonal with positive real entries, and UL and UR are unitary matrices.

To apply this in the present case, consider the following redeﬁnition of the lepton ﬁelds:

j j `Li0 = LLi `Lj; `Ri0 = LRi `Rj. (12.101) where L j and L j are unitary 3 3 matrices. The lepton mass term in the unitary gauge Lagrangian Li Ri × then becomes:

h i j = 1+ ` (L† meL ) ` . (12.102) v L L R i Rj L − µ ¶ Now, the theorem just stated assures us that we can choose the matrices LL and LR so that:

me 0 0 L† meL = 0 m 0 . (12.103) L R  µ  0 0 mτ   So we can write in terms of the unprimed (mass eigenstate) fields: h = 1+ (m ee + m µµ + m ττ). (12.104) v e µ τ L − µ ¶ In the same way, one can do unitary-matrix redefinitions of the quark fields:

j j dLi0 = DLi dLj; dRi0 = DRi dRj; (12.105) j j uLi0 = ULi uLj; uRi0 = URi uRj, (12.106) chosen in such a way that

md 0 0 D† mdD = 0 m 0 ; (12.107) L R  s  0 0 mb   mu 0 0 U † muU = 0 m 0 , (12.108) L R  c  0 0 mt   221 with real and positive entries. It might seem like worrying about the possibility of 3 3 Yukawa matrices was just a waste of time, × but one must now consider how these field redefinitions from primed (gauge eigenstate) to unprimed (mass eigenstate) fields affect the other terms in the Lagrangian. First, consider the derivative kinetic terms. For the leptons, contains L j j 0 j i`L∂`/ Lj0 = i(`LLL† ) ∂/(LL`L)j = i`L∂`/ Lj. (12.109)

This relies on the fact that the ﬁeld redeﬁnition matrix LL is unitary, LL† LL = 1. The same thing works for all of the other derivative kinetic terms, for example, for right-handed up-type quarks:

j j j iuR0∂u/ Rj0 = i(uRUR† ) ∂/(URuR)j = iuR∂u/ Rj, (12.110) which relies on UR† UR = 1. So the redefinition has no effect at all here; the form of the derivative kinetic terms is exactly the same for unprimed fields as for primed fields. There are also interactions of fermions with gauge bosons. For example, for the right-handed leptons, the QED Lagrangian contains a term

j µ j µ j µ eA ` 0γ `0 = eA (` L† ) γ (L ` ) = eA ` γ ` . (12.111) − µ R Rj − µ R R R R j − µ R Rj

Just as before, the unitary condition (this time LR† LR = 1) guarantees that the form of the Lagrangian term is exactly the same for unprimed fields as for primed fields. You can show quite easily that the same thing applies to interactions of all fermions with Zµ and the gluon fields. The unitary redefinition matrices for quarks just commute with the SU(3)c generators, since they act on different indices. There is one place in the Standard Model where the above argument does not work, namely the interactions of W ± vector bosons. This is because the W ± interactions involve two different types of fermions, with different unitary redefinition matrices. Consider first the interactions of the W ± with leptons. In terms of the original primed fields:

g + i µ = W ν0 γ `0 +c.c., (12.112) L −√2 µ Li so that

g + i µ = W ν0 γ (LL`L)i +c.c. (12.113) L −√2 µ Since the neutrinos did not have a mass term, we did not have to make a unitary redeﬁnition for them. But now we are free to do so; deﬁning νi in the same way as the corresponding charged leptons,

νi0 = LLνj, we get,

i i ν0 = (νLL† ) , (12.114) resulting in

g + i µ = W ν γ `Li +c.c. (12.115) L −√2 µ

222 So once again the interactions of W ± bosons have exactly the same form for mass-eigenstate leptons.

However, consider the interactions of W ± bosons with quarks. In terms of the original primed ﬁelds,

g + i µ = W u0 γ d0 +c.c. (12.116) L −√2 µ L Li which becomes:

g + i µ = W u γ (U † DLdL)i +c.c. (12.117) L −√2 µ L L

There is no reason why UL† DL should be equal to the unit matrix, and in fact it is not. So we have ﬁnally encountered a consequence of going to the mass eigenstate basis. The charged-current weak interactions contain a non-trivial matrix operating in quark family space,

V = UL† DL, (12.118) called the Cabibbo-Kobayashi-Maskawa matrix (or CKM matrix). The CKM matrix V is itself unitary, since V † = (UL† DL)† = DL† UL, implying that V †V = V V † = 1. But, it cannot be removed by going to some other basis using a further unitary matrix without ruining the diagonal quark masses. So we are stuck with it. One can think of V as just a unitary rotation acting on the left-handed down quarks. From eq. (12.117), we can deﬁne

j dLi0 = Vi dLj, (12.119) where the dLj are mass eigenstate quark ﬁelds, and the dLi0 are the quarks that interact in a simple way with W bosons:

g + i µ = W u γ d0 +c.c. (12.120) L −√2 µ L Li This is the entire eﬀect of the non-diagonal Yukawa matrices. The numerical entries of the CKM matrix are a subject of continuing experimental investigation. To a ﬁrst approximation, it turns out that the CKM mixing is just a rotation of down and strange quarks:

cos θc sin θc 0 V = sin θ cos θ 0 (12.121)  − c c  0 0 1   + where θc is called the Cabibbo angle. This implies that the interactions of W with the mass-eigenstate quarks are very nearly:

g + µ µ µ µ µ = W cos θc[uLγ dL + cLγ sL] + sin θc[uLγ sL cLγ dL]+ tLγ bL . (12.122) L −√2 µ − ³ ´ The terms proportional to sin θc are responsible for strangeness-changing decays. Numerically,

cos θ 0.975; sin θ 0.22. (12.123) c ≈ c ≈

223 Strange hadrons have long lifetimes because they decay through the weak interactions, and with 2 reduced matrix elements that are proportional to sin θc = 0.05. In general, the CKM matrix is:

Vud Vus Vub 0.975 0.22 0.004 V = V V V 0.22 0.975 0.04 , (12.124)  cd cs cb  ≈   Vtd Vts Vtb 0.003 0.04 0.999     where the numerical values given indicate rough estimates of the magnitude only (not the sign or phase).‡ Weak decays involving the W bosons allow the entries of the CKM matrix to be probed experimentally. For example, decays

B D`+ν , (12.125) → ` where B is a meson containing a bottom quark and D contains a charm quark, can be used to extract V . The very long lifetimes of B mesons are explained by the fact that V and V are very small. | cb| | ub| | cb| One of the ways of testing the Standard Model is to check that the CKM matrix is indeed unitary:

V †V = 1. (12.126)

This is an automatic consequence of the Standard Model, but if there is further unknown physics out there, then it could aﬀect the weak interactions in such a way as to appear to violate CKM unitarity.

‡In fact, the CKM matrix contains one phase that cannot be removed by redeﬁning phases of the fermion ﬁelds. This phase is the only source of CP violation in the Standard Model.

224 Physics 586 Homework Set 1. Due Jan. 24, 2002.

Problem 1. Prove that any Lorentz transformation matrix Lµ satisﬁes det(L)= 1. ν ± [Hint: Recall that det(AB) = det(A)det(B) for any matrices A, B.]

Problem 2. Suppose a particle A of mass M is at rest, and then decays to another particle B of mass m and a third massless particle C. Find the energies of the particles B and C, using conservation of 4-momentum.

Problem 3. Prove that the following statements are true, where N1, N2, N3, and N4 are certain integers that you will determine. [Hint: You do not need the explicit forms of the gamma matrices at all, just eqs. (2.81)-(2.83) in the lecture notes.] µ (a) γ γνγµ = N1γν. µ (b) γ γνγργµ = N2gνρ. (c) Tr(γ γ γ γ )= N (g g g g + g g ). µ ν ρ σ 3 µν ρσ − µρ νσ µσ νρ [Hint: use the cyclic property of the trace: Tr(ABCD) = Tr(BCDA).] (d) [γ , [γ ,γ ]] = N (g γ g γ ). ρ µ ν 4 ρµ ν − ρν µ [Hint: Write out the left side as four explicit terms, and then use eq. (2.83) in the lecture notes several times.]

Problem 4. In this problem, we will check the Lorentz invariance of the Dirac equation, and in the process determine the Lorentz transformation rule for Dirac spinors. Suppose that two coordinate systems are related by a Lorentz transformation

µ µ ν x0 = L νx .

The wavefunction Ψ0(x0) as reported by an observer in the primed frame should be related to that in the unprimed frame by

Ψ0(x0) = ΛΨ(x) where Λ is a 4 4 matrix. Now, the Dirac equation in the unprimed frame is × ∂ (iγµ m)Ψ(x) = 0, ∂xµ − and in the primed frame it is

µ ∂ (iγ µ m)Ψ0(x0) = 0. ∂x0 − (a) Show that these equations are consistent provided that

1 ρ µ µ Λ− γ Lρ Λ= γ .

225 µ µ µ µ (b) Now suppose that L ν = δν + ω ν with ω ν infinitesimal, as in eq. (1.31) of the lecture notes. Prove that the equation found in part (a) is satisfied if 1 Λ=1+ ωµν[γ ,γ ] 8 µ ν [Hints: Use the result you found in part (d) of Problem 2. Note that if ² is an infinitesimal matrix, then (1 + ²) 1 = 1 ². Also use the fact that ω is antisymmetric.] − − µν

226 Physics 586 Homework Set 2: Fun with Dirac spinors! Due Jan. 31, 2002.

Problem 1. Prove each of the following statements, where the numbers Ni are constants that you will identify: 2 (a) p/k/p/ = N1p k/ + N2(k p)p/ · 2 2 2 (b) Tr[p/k/p/k/]= N3p k + N4(p k) · 2 (c) Tr[p/k//qp/]= N5(p k)(q p)+ N6p (q k) · · ·

Problem 2. Simplify the following expressions by getting rid of one projection matrix:

(a) PLkP/ L

(b) PRkP/ L

(d) PRp/kP/ L

(e) PLa//b/cd//ek/p//q/rPL

Problem 3. By taking the Hermitian conjugates of the Dirac equations (p/ m)u(p, s) = 0 and − (p/ + m)v(p, s) = 0, prove that:

u(p, s)(p/ m) = 0 and − v(p, s)(p/ + m) = 0.

[Hint: use eqs. (2.79), (2.80) and (2.103).]

Problem 4. Suppose that u(p, s) and u(k,r) are solutions of the Dirac equation with mass m, with diﬀerent 4-momenta pµ, kµ and spin labels r, s. Prove the Gordon decomposition identity: 1 1 u(k,r)γµu(p, s) = (pµ + kµ) u(k,r)u(p, s)+ (p k ) u(k,r)[γµ,γν]u(p, s) 2m 2 ν ν ½ − ¾ [Hint: Start with 1 (p k )u(k,r)[γµ,γν]u(p, s), and rewrite the individual terms using eq. (2.83) so 2 ν − ν that you can make use of the Dirac equations pu/ (p, s)= mu(p, s) and u(k,r)k/ = mu(k,r).] Physics 586 Homework Set 3 Due Feb. 7, 2002.

Problem 1. In the theory of a free scalar ﬁeld φ as described in section 4.2, compute the commutators

[H,a† ] = E a† and ~k ~k ~k

[H,a~k] = ?

In the theory of a free Dirac fermion ﬁeld as described in section 4.3, use the anticommutation relations of b, b† to compute the commutators

[H, b† ] = ? and ~k,s

[H, b~k,s] = ?

Problem 2. Consider the operator:

P~ = d3~xπ(~x) ~ φ(~x), − ∇ Z in the case of a free real scalar ﬁeld φ as studied in 4.2. Rewrite this operator in terms of the a~p and a†~p operators, and show that the result is:

~ P = dp˜ ~pa†~pa~p Z [Hint: You will have to argue that certain terms vanish, including an apparently inﬁnite one, by carefully noting their behavior as ~p ~p.] What is P~ acting on the vacuum state 0 ? Compute the →− | i commutator:

[P,a~ † ] =? ~k

What is the eigenvalue of P~ acting on the state ~k = a† 0 ? | i ~k| i

Problem 3. (More review of relativistic kinematics.) A particle of mass M, with energy E, collides with another particle of mass M which is initially at rest. The result of the collision is two identical particles, each of mass m. (a) Find a Lorentz transformation to the center-of-momentum frame, and ﬁnd the energies of each of the two initial particle in that frame.

(b) Find the maximum and minimum possible energies Emax and Emin of the two ﬁnal state particles in the lab frame. (c) Find the energies of the two ﬁnal state particles, if one of them is emitted at a right angle to the initial direction of the incident particle in the lab frame. Physics 586 Homework Set 4 Due Feb. 21, 2002.

Problem 1. Consider a general 2 particle 2 particle scattering problem. The initial-state particles → have masses ma and mb and the ﬁnal state particles have masses m1 and m2. Their 4-momenta are 2 2 2 2 2 2 2 2 respectively pa, pb, k1, and k2. The particles are on-shell, so pa = ma, pb = mb , k1 = m1, and k2 = m2. Consider the Mandelstam variables deﬁned by eq. (5.126)-(5.128) in the lecture notes.

2 2 2 2 (a) Prove that s + t + u = ma + mb + m1 + m2. This means that one can always eliminate one of the Mandelstam variables in terms of the other two.

(b) Find expressions for each of these Lorentz-invariant quantities in terms of only s, t, ma, mb, m1, and m2: p p , p k , p k , p k , p k , k k . a · b a · 1 a · 2 b · 1 b · 2 1 · 2 (c) Suppose we are in the center-of-momentum frame, so that ~p = ~p . Also assume that ~p points b − a a along the positive z-axis (θ = 0), and that ~k1 points along a direction making an angle θ with respect to the positive z-axis. Find expressions for s, t, and u in terms of the total center-of-momentum energy

ECM, and cos θ and the masses ma, mb, m1, and m2.

(d) Under the same conditions as part (c), ﬁnd an expression for cos θ in terms of the Mandelstam variable t and ECM and the masses.

Problem 2. Consider the matrix element obtained for φφ φφ scattering in φ3 theory in section 5.3, → eqs. (5.122)-(5.125).

(a) Find the diﬀerential cross section

dσ d(cos θ) as a function of µ, ECM, m, and cos θ.

(b) Find the total cross section in terms of µ, ECM, m. [Hints: integrate directly in terms of the variable cos θ. Remember to be wary of tricky factors of 2 at the very end!] Problem 3. Consider a field theory consisting of two distinct real scalar fields Φ and φ. They have masses M and m, so that the free Lagrangian density is 1 1 1 1 = ∂ Φ∂µΦ M 2Φ2 + ∂ φ∂µφ m2φ2. L0 2 µ − 2 2 µ − 2 This means that one can do canonical quantization of each field separately just as discussed in class.

Call A†~p, A~p the creation and annihilation operators for the ﬁeld Φ, and a†~p,a~p the ones for the ﬁeld φ. (Each of A† and A always commutes with each of a† and a.) Now suppose we add to this theory an interaction Lagrangian density: µ = Φφ2. L − 2 where µ is a constant coupling.

(a) Find an expression like eq. (5.8) or eq. (5.84) for the interaction Hamiltonian for this theory in terms of A†, A and a†,a operators.

(b) Consider the transition between a state

~p = A† 0 | i ~p| i with one Φ particle carrying momentum ~p, and the state

~ ~ k1, k2 = a~† a~† 0 | i k1 k2 | i with two φ particles with momenta ~k1 and ~k2. Find the matrix element

iT H ~k ,~k ~p = ~k ,~k e− ~p OUTh 1 2| iIN OUTh 1 2| | iOUT to ﬁrst order in µ, using techniques like those in section 5. (Do NOT use Feynman rules.)

(c) Find the reduced matrix element Φ φφ, obtained from the deﬁnition M → 4 (4) OUT ~k1,~k2 ~p IN = Φ φφ (2π) δ (p k1 k2) h | i M → − − What do you conclude about the Feynman rule for the Φφφ interaction vertex in this theory?

(d) Now, draw the Feynman diagrams which correspond to φφ φφ scattering in this theory. Use → the Feynman rule you found above, together with the Feynman propagator for the ﬁeld Φ, to write the complete reduced matrix element for this process, to second order in µ. [Hint: There is more than one Feynman diagram!]

Problem 4. Consider the scalar φ4 theory discussed in lecture 5.1. Draw all Feynman diagrams corresponding to φφ φφφφ scattering, to order λ2. Clearly label the particle 4-momenta for each → external and internal line in your diagram. Write down the reduced matrix element for each diagram, using Feynman rules. Physics 586 Homework Set 5 Due Feb. 28, 2002.

Problem 1. Consider the process of antimuon scattering oﬀ of an electron:

+ + µ e− µ e− →

You may assume me is negligible, but keep mµ. Work in the center-of-momentum frame. Assign the incoming electron and antimuon 4-momenta pa and pb, and call the magnitude of their 3-momenta P .

Assign the ﬁnal-state electron and antimuon 4-momenta k1 and k2, and note that the magnitude of their 3-momenta is also P . (a) Draw the Feynman diagram(s) which contribute to the reduced matrix element for this process at order e2.

(b) Deﬁne θ to be angle between the initial-state e− and the ﬁnal state e− directions in the center-of- momentum frame. Work out all of the following quantities in terms of mµ, P , and cos θ:

2 2 2 2 pa, pb , k1, k2, (p p ), (p k ), (p k ), a · b a · 1 a · 2 (p k ), (p k ), (k k ), b · 1 b · 2 1 · 2 s = (p + p )2, t = (p k )2, u = (p k )2. a b a − 1 a − 2 (c) Write down the reduced matrix element . M (d) Take the complex square of the reduced matrix element, sum over final state spins, and average over initial state spins, and simplify. Write the result in terms of Mandelstam variables s,t,u, and then rewrite it in terms of P and the scattering angle θ. (e) Find the differential cross section. Simplify your answer as much as possible. (f) Now take m 0. What is the differential cross section? You should note something odd for a µ → particular value of cos θ. (g) Consider the 16 possible helicity processes:

+ + + + µ e− µ e−; µ e− µ e−; L L → L L L L → R L + + + + µ e− µ e−; µ e− µ e−; L L → L R L L → R R + + + + µ e− µ e−; µ e− µ e−; L R → L L L R → R L + + + + µ e− µ e−; µ e− µ e−; L R → L R L R → R R + + + + µ e− µ e−; µ e− µ e−; R L → L L R L → R L + + + + µ e− µ e−; µ e− µ e−; R L → L R R L → R R + + + + µ e− µ e−; µ e− µ e−; R R → L L R R → R L + + + + µ e− µ e−; µ e− µ e−. R R → L R R R → R R

Do not compute them. Instead, ﬁgure out which ones vanish by helicity conservation. Physics 586 Homework Set 6 Due March 7, 2002.

Problem 1. Consider the process of polarized Bhabha scattering:

+ + e−e e−e L L →

You may assume me is negligible, and that the ﬁnal state spins are unobserved. However, the initial state helicities are known, as indicated. Work in the center-of-momentum frame. Assign the incoming electron and positron 4-momenta pa and pb, and call the magnitude of their 3-momenta P . Assign the

ﬁnal-state electron and positron 4-momenta k1 and k2, and note that the magnitude of their 3-momenta is also P .

(a) Deﬁne θ to be angle between the initial-state e− and the ﬁnal state e− directions in the center-of- momentum frame. Work out all of the following quantities in terms of P , and cos θ:

2 2 2 2 pa, pb , k1, k2, (p p ), (p k ), (p k ), a · b a · 1 a · 2 (p k ), (p k ), (k k ), b · 1 b · 2 1 · 2 s = (p + p )2, t = (p k )2, u = (p k )2. a b a − 1 a − 2 (b) Draw the Feynman diagram(s) which contribute to the reduced matrix element for this process at order e2. Do any of them vanish because of helicity conservation? (c) Write down the reduced matrix element . M (d) Take the complex square of the reduced matrix element, sum over final state spins, and simplify. Write the result in terms of Mandelstam variables s,t,u, and then rewrite it in terms of P and the scattering angle θ. (e) Find the differential cross section. Simplify your answer as much as possible. (f) Redo steps (b) through (e), for the same process, but with different initial-state helicities:

+ + e−e e e−. L R → [Hint: It is always a good idea to simplify traces as much as possible using gamma matrix algebra before trying to compute them.] Physics 586 Homework Set 7 Due March 28, 2002.

Problem 1. Draw, but do not compute, the following Feynman diagrams in QED: (a) All tree-level diagrams contributing to e e+ µ µ+γ. − → − (b) All one-loop diagrams contributing to e e+ µ µ+. − → − (c) A representative one-loop diagram contributing to γγ γγ. →

Problem 2. In certain extensions of the Standard Model, including the Minimal Supersymmetric Standard Model (MSSM), there is predicted to be an electrically neutral spin-0 particle A0 called the “pseudo-scalar Higgs boson”. The interaction Lagrangian of this particle with Standard Model Dirac fermions has the form:

0 = y0A Ψγ Ψ Lint 5 where y0 is a coupling constant. 0 (a) Compute the partial decay rate for A into a fermion anti-fermion pair, as a function of y0, the mass of the pseudo-scalar MA0 , and the mass of the fermion mf . You should ﬁnd a result of the form:

2 p 2 4mf Γ= Ny0 MA0 1 2 , Ã − MA0 ! where N and p are numbers that you will compute. (Hint: p = 3/2.) 6 (b) In the MSSM, it has been calculated that the ratio of the couplings of A0 to top and bottom quarks is:

y0 0 m A bb = b tan2 β y m A0 0tt t where tan β is a parameter of the model which is believed to be in the range:

2 < tan β < 55.

Here, mt and mb diﬀer somewhat from the actual masses, because of higher-order corrections. Taking mt = 165 GeV and mb = 3 GeV and mt = 175 GeV and mb = 5 GeV and mA0 = 400 GeV, make a plot of

BR(A0 bb) → BR(A0 tt) → as a function of tan β, using representative points tan β = 2, 5, 10, 20, 30, 40, 50. For what values of tan β is the branching fraction into bb greater than for tt? Problem 3. Professor Eric Einform, the most brilliant scientist on the far-oﬀ planet Xany where particle accelerators have not yet been invented, has proposed two alternative theories for muon decays. His two proposed Lagrangians correspond to what we would call S and S P theories of the weak − interaction current:

S = G(ν µ)(eν )+c.c. Lint − µ e S P − = G(ν P µ)(eP ν )+c.c. Lint − µ L L e (a) For each theory, calculate

1 2 2 |M| spinsX for µ e ν ν decay. − → − µ e (b) Find the distribution for the electron energy produced in muon decay,

dΓ , dEe for each of these theories, and integrate to ﬁnd Γ in each case. Physics 586 Homework Set 8 Due April 4, 2002.

Problem 1. Consider the process of top-quark decay:

t bW +. → This is a weak interaction process, following from the following term in the interaction Lagrangian:

g µ = W −(bγ PLt)+c.c. L −√2 µ where t, b are the Dirac spinor ﬁelds for the top quark. The top quark decays very quickly, as you will discover below, so it does not form complicated bound states like the lighter quarks. Therefore, one can just use the simple weak-interaction Feynman rule implied by the above interaction Lagrangian.

µ + µ (a) Let the 4-momentum of the top quark be p , and that of the W boson be k1 , and that of the µ bottom quark be k2 . Find the kinematic quantities:

2 2 2 p ; k1; k2; p k ; p k ; k k · 1 · 2 1 · 2 in terms of mt and mW . Treat the bottom quark as massless. (This is a good approximation, since m 175 GeV, m = 80.4 GeV, and m 5 GeV.) t ≈ W b ≈ (b) Write down the reduced matrix element for top quark decay.

(c) Take the complex square of the reduced matrix element, and sum over the ﬁnal state polarizations of the W + boson, using eq. (8.130). Then average over the initial t spin, and sum over the initial spin of the b. (The quarks have 3 colors, but the color of the ﬁnal state bottom quark is constrained to be the same as that of the initial state top quark. Since one should average over the initial state quark color, the net color factor is just 1.)

(d) Compute the decay rate of the top quark. You should ﬁnd a result of the form:

2 3 2 2 N3 + g mt MW MW Γ(t bW )= 2 1+ N2 2 1 2 → N1πMW Ã mt !Ã − mt ! where N1, N2, and N3 are positive integers that you will ﬁnd.

(e) Find the numerical value of the top quark lifetime in seconds, and the decay width in GeV. Use g = 0.65, m = 175 GeV, m = 80.4 GeV. What is Γ(t bW +)/m ? t W → t Physics 586 Homework Set 9 Due April 11, 2002.

Problem 1. Consider the process of W − boson decay, through the interaction Lagrangian:

g +µ = W −J +c.c. L −√2 µ +µ µ µ µ µ µ µ J = bγ PLt + sγ PLc + dγ PLu + τγ PLντ + µγ PLνµ + eγ PLνe.

(Notice that we are ignoring the problem of mass eigenstates being not quite the same as the ﬁelds that couple to the W −.)

(a) The W − boson cannot decay into a ﬁnal state with a bottom quark; why? (This is a useful thing sometimes; if your experiment tags a bottom quark, chances are it didn’t come from a W − decay unless it was mis-tagged.) (b) Treating the electron as massless, compute the decay rate for W e ν , in terms of g and M . − → − e W [Draw the Feynman diagram, write down the reduced matrix element, take its complex square, average over the three possible initial polarizations of the W − boson using eq. (8.130), sum over all possible

final state spins, and apply eq. (7.24) with m1 = 0.] (c) From your answer to part (b), infer the results for: Γ(W µ ν ) and Γ(W τ ν ) and − → − µ − → − τ Γ(W du) and Γ(W sc), treating all of the final-state fermions as massless. Remember that − → − → each quark in the final state has 3 possible colors, which you must sum over. The antiquarks are constrained to have the opposite color, so once you have summed over the quark colors, you should not sum over the antiquark’s anticolors. Because the W − has a large mass and the decay happens quickly, you can assume that the strong interactions of the quarks and antiquarks are irrelevant until long after the decay has occurred. (d) From the above results, predict the total decay width of the W boson in GeV, its lifetime in seconds, and its branching fraction into each of the possible final states.

Problem 2. The Z-boson is a massive real vector particle which happens to interact with left-handed fermions diﬀerently than with right-handed fermions. Its interaction Lagrangian for any particular Dirac spinor fermion ﬁeld f can be written in the form

= Zµfγ (g P + g P )f, Lint − µ L L R R where gL and gR are certain coupling constants that we will identify more explicitly later in the course. (a) Write down the corresponding Feynman rule for Z boson interactions with f.

(b) Treating f as massless, compute the partial decay width for Z ff, in terms of g , g , and M . → L R Z Physics 586 Homework Set 10 Due April 18, 2002.

Problem 1. Consider the process of top quark production in QCD in proton-anti-proton collisions. The most important subprocesses contributing to this at the Tevatron collider are, in order of importance:

uu tt, → dd tt, → gg tt. → (a) Draw the tree-level Feynman diagrams contributing to these in QCD. You should ﬁnd a total of 5 diagrams.

(b) Find the reduced matrix element for the ﬁrst process only. Do not neglect the top-quark mass, mt.

(c) By the usual procedure, compute the spin sum/averaged and color sum/averaged diﬀerential total cross section dσ uu tt , d(cos→θ) which is identical to that for dσ dd tt . d(cos→θ)

You should ﬁnd that, apart from the color factor, this is extremely similar to a certain QED calculation that we did in class. Write your answers in terms of αs, mt, and the center-of-momentum energy for the quark-antiquark system, √sˆ.

(d) Integrate to ﬁnd the total cross section for uu tt or dd tt in QCD. → → (e) Consider the following VERY CRUDE model for the proton. Take it to consist of two up quarks and one down quark, each having exactly 1/5 of the proton’s 4-momentum. (The gluons share the rest.) Similarly, assume that the antiproton consists of two anti-up quarks and an anti-down quark, each having exactly 1/5 of the anti-proton’s 4-momentum. Convince yourself persuasively, in writing, that the total cross-section for producing a tt pair plus X =anything in proton-antiproton collisions in this model should be:

σ(pp tt + X) = 4σ(uu tt)+ σ(dd tt), (12.127) → → → where the cross-sections on the right-hand side are evaluated with center-of-momentum energy √sˆ =

√s/5, where √s is the center-of-momentum energy of the pp pair. Now plug in mt = 175 GeV and √s = 2000 GeV, and use a running QCD coupling α (m ) 0.11, to get a numerical answer for σ in s t ≈ femtobarns. (f) The last process, gg tt is less important, but still signiﬁcant, at the Tevatron. It is the dominant → source of top quarks at the future Large Hadron Collider. Write down a complete expression for the reduced matrix element for this process. You do not have to simplify it.

We will discuss a much more accurate model for doing part (e) in class.

If you are very ambitious, you may attempt to ﬁnd the diﬀerential cross section for gg tt using the → matrix element found in part (f). I will not try to stop you, but I will also not give you extra credit for it! Physics 586 Homework Set 11 Due April 25, 2002.

This problem involves very little calculation, but some thought.

Problem 1. Consider the following process at the Tevatron collider:

pp µ−ν + X. → µ Here X means any hadronic particles, but does not include any leptons, neutrinos, photons, or anything exotic.

(a) Identify all the tree-level 2 2 partonic subprocesses that can contribute to this process. You → should not include charm, bottom, or top quarks as initial-state partons. [Hint: what particle studied in this course has a coupling to a muon and muon antineutrino?]

(b) Assume that the total cross for each subprocess you identiﬁed in part (a) is known. Write a complete and explicit integral expression for the total cross-section

σ(pp µ−ν + X). → µ You do not need to evaluate it. It may involve the PDFs g(x),u(x),d(x), s(x), u(x), d(x), s(x).

pp µ−ν + X. → µ

(d) Repeat steps (a) and (b) for the process of single-top production at the Tevatron:

pp tb + X. → (Here X does not include a top or bottom quark.) Physics 586 Homework Set 12 Due May 7, 2002.

This problem set is OPTIONAL. If you turn it in and it is higher than your average, then it will be averaged into your ﬁnal grade. Otherwise it won’t be. Doing it will bring you 40 years of good luck. I am leaving on a trip starting May 8, so it must be turned in by May 7 at 5:00 PM in order to count.

Problem 1. In Homework Set 9, problem 2, you worked out the partial decay width for Z ff in f f → terms of couplings gL and gR which were then unknown. You should have gotten a result that was f 2 f 2 proportional to (gL) + (gR) ; only this information is needed for the present problem. In class, we have since found out what these couplings are; see eqs. (12.38) and (12.39). f f f 2 f 2 (a) Make a table of the numerical values of the coeﬃcients gL, gR, and (gL) + (gR) for all fermions in the Standard Model, using the values in eqs. (12.31)-(12.34) of the notes. (b) Using the results of part (a), ﬁnd the branching fractions for Z ff, for each Standard Model → fermion f, treating them all as massless except the top quark. Don’t forget to take color into account.

(c) Suppose a τ − lepton is observed coming from Z decay. In the lab frame, it will be very energetic, since m m . So it can have a well-deﬁned helicity. What is the probability that it is left-handed? Z À τ

Problem 2. The most important goal of high energy physics these days is to ﬁnd the Higgs boson (or whatever takes its place in non-standard models). + (a) At an e e− collider, the main process by which one can look for the Higgs boson is:

+ e e− hZ. → Draw the single Feynman diagram which makes the largest contribution to this process. Write down a complete and explicit expression for the reduced matrix element for it. (You do not have to simplify it.) (b) At the Tevatron, the processes by which one can make a Higgs boson include:

pp Zh + X → pp W +h + X → pp W −h + X. → Identify the parton-level subprocesses which can lead to the above processes, and draw a Feynman diagram for each one.

Have a good summer! Appendix A: Natural units and conversions

The speed of light andh ¯ are:

c = 2.99792458 1010 cm/sec (A.1) × 34 ¯h = 1.0546 10− J sec (A.2) × 25 = 6.5821 10− GeV sec. (A.3) × The value of c is exact, by definition. (Since October 1983, the official definition of 1 meter is the distance traveled by light in a vacuum in exactly 1/299792458 of a second.) In units with c =h ¯ = 1, some other conversion factors are:

3 1 GeV = 1.6022 10− erg (A.4) × 24 1 GeV = 1.7827 10− g (A.5) × 1 14 1 GeV− = 1.9733 10− cm (A.6) × Conversions of particle decay widths to mean lifetimes and vice versa are obtained using:

1 25 1 GeV− = 6.5822 10− sec (A.7) × 24 1 1 sec = 1.5193 10 GeV− (A.8) ×

2 Conversions of cross-sections in GeV− to barn units involve:

2 4 1 GeV− = 3.894 10− barns (A.9) × = 3.894 105 nb (A.10) × = 3.894 108 pb (A.11) × = 3.894 1011 fb, (A.12) × and in reverse:

33 2 6 2 1 nb = 10− cm = 2.568 10− GeV (A.13) × 36 2 9 2 1 pb = 10− cm = 2.568 10− GeV (A.14) × 39 2 12 2 1 fb = 10− cm = 2.568 10− GeV . (A.15) × Appendix B: Dirac Spinor Formulas

In the Weyl (or chiral) representation): 0 σµ γµ = (B.1) σµ 0 µ ¶ where 1 0 0 1 σ0 = σ0 = ; σ1 = σ1 = ; 0 1 1 0 µ ¶ − µ ¶ 0 i 1 0 σ2 = σ2 = − ; σ3 = σ3 = . − i 0 − 0 1 µ ¶ µ − ¶

0 0 0 2 γ † = γ ; (γ ) = 1 (B.2) j j γ † = γ (j = 1, 2, 3) (B.3) − 0 µ 0 µ γ γ †γ = γ (B.4) γ γ + γ γ = γ ,γ = 2g (B.5) µ ν ν µ { µ ν} µν [γ , [γ ,γ ]] = 4(g γ g γ ) (B.6) ρ µ ν ρµ ν − ρν µ The trace of an odd number of γµ matrices is 0.

Tr(1) = 4 (B.7)

Tr(γµγν) = 4gµν (B.8) Tr(γ γ γ γ ) = 4(g g g g + g g ) (B.9) µ ν ρ σ µν ρσ − µρ νσ µσ νρ Tr(γ γ ...γ ) = g Tr(γ γ ...γ ) g Tr(γ γ ...γ ) µ1 µ2 µ2n µ1µ2 µ3 µ4 µ2n − µ1µ3 µ2 µ4 µ2n ... + ( 1)kg Tr(γ γ ...γ γ ...γ )+ ... − µ1µk µ2 µ3 µk−1 µk+1 µ2n

+gµ1µ2n Tr(γµ2 γµ3 ...γµ2n−1 ) (B.10)

In the chiral (or Weyl) representation, in 2 2 block form: × 1 0 γ5 = − (B.11) 0 1 µ ¶ 1 γ 1 0 1+ γ 0 0 P = − 5 = ; P = 5 = (B.12) L 2 0 0 R 2 0 1 µ ¶ µ ¶ The matrix γ5 satisﬁes:

2 µ γ† = γ ; γ = 1; γ ,γ = 0 (B.13) 5 5 5 { 5 }

Tr(γµγ5) = 0 (B.14)

Tr(γµγνγ5) = 0 (B.15)

Tr(γµγνγργ5) = 0 (B.16)

Tr(γµγνγργσγ5) = 4i²µνρσ (B.17) µ γ γµ = 4 (B.18) γµγ γ = 2γ (B.19) ν µ − ν µ γ γνγργµ = 4gνρ (B.20) γµγ γ γ γ = 2γ γ γ (B.21) ν ρ σ µ − σ ρ ν

(p/ m)u(p, s) = 0; (p/ + m)v(p, s) = 0 (B.22) − u(p, s)(p/ m)=0; v(p, s)(p/ + m) = 0 (B.23) −

u(p, s)u(p,r) = 2mδsr (B.24) v(p, s)v(p,r)= 2mδ (B.25) − sr v(p, s)u(p,r)= u(p, s)v(p,r) = 0 (B.26)

u(p, s)u(p, s) = p/ + m (B.27) s X v(p, s)v(p, s) = p/ m (B.28) s − X Index Abelian (commutative) group, 150, 151 canonical commutation relations, 18 action, 9, 29, 31 canonical quantization active quarks, 178, 179 complex scalar field, 143 adjoint representation, 154, 155, 158–160, 162, Dirac fermion fields, 42 167 real scalar fields, 35, 36 angular momentum conservation, 90–92, 119, Carbon-14 dating, 122 126 Casimir invariant, 153 angular momentum operator, 18 charge conjugation, 133 angular resolution, 97 charge conservation, 26, 27 annihilation operator, 36, 40 charge density, 26 anticommutation relations, 41, 42, 74 charge operator, 69 antineutrino, 122, 123 charged current, 139, 148 antineutrino-electron scattering, 138 chiral (Weyl) representation of Dirac matrices, antiparticle, 21, 28, 29, 40, 67, 86, 87 13, 242 associativity property of group, 150 chiral representations (of gauge group), 217 asymptotic freedom, 179, 182 chiral symmetry breaking, 207, 220 CLEO experiment, 125 bare coupling, 174, 176 closure (of group), 152 bare mass, 175, 176 closure property of group, 150 barn (unit of cross section), 48, 49, 241 color sum, 170, 237 barred Dirac spinor, 16 commutation relations, 35–37 beta function, 177 commutative (Abelian) group, 150 QCD, 177 commutator of Lie algebra generators, 152 QED, 180 complex conjugate of spinor expressions, 78, 94, Bhabha scattering, 92 127–129, 139 biunitary transformation, 221 complex conjugate representation, 154 boost, 4, 5, 8, 14, 15 complex scalar field, 143 Bose-Einstein statistics, 37 Compton scattering, 103–110 branching fractions, 118 conservation of angular momentum, 90–92, 119, W boson, 236 126 Z boson, 240 conservation of charge, 26, 27 charged pion, 123, 142 conservation of energy, 46, 56, 122, 225 Higgs boson, 118, 220 conservation of helicity, 92, 142, 231, 232 Cabibbo angle, 223, 224 conservation of lepton number, 124, 125 Cabibbo-Kobayashi-Maskawa (CKM) mixing, 223, conservation of momentum, 6, 50, 75, 76, 225 224 contravariant four-vector, 4–6, 8

244 Coulomb potential, 97 dimensional analysis, 132, 145, 146 coupling dimensional regularization, 179 electromagnetic (e), 34, 69, 211 dimensional transmutation, 179 electroweak (g), 148 Dirac equation, 11–14

electroweak (g0), 208, 209 Dirac ﬁeld, 29, 32, 33, 40–42

Fermi (GF ), 127, 134 Dirac matrices, 13, 242 general, 65 Dirac spinor, 12

QCD (g3, αs), 168, 169, 171, 178 direct sum representation, 153 scalar3, 55 double-counting problem, identical particles in scalar4, 43 ﬁnal state, 54, 115 scalarn, 62, 65 Drell-Yan scattering, 196–198 Yang-Mills gauge, 162–166, 202 electroweak Standard Model (Glashow-Weinberg- Yukawa, 66, 116–118, 218–220, 222 Salam) gauge theory, 208–217 couplings energy conservation, 46, 56, 122, 225 Z boson, 212 ²µνρσ (Levi-Civita) tensor, 9 covariant derivative equal-time commutators, 35 QED, 70, 71, 150 Euler-Lagrange equations of motion, 29–33, 43 Yang-Mills, 162 covariant four-vector, 6 femtobarns (fb), 48

CPT Theorem, 133 Fermi constant (GF ), 127, 128 creation operator, 36 numerical value, 132

fermionic, 40 relation to mW , 148 cross section, 48, 49 Fermi weak interaction theory, 127 φφ φφ, 54 → Fermi-Dirac statistics, 41, 74 differential, 50, 51 Feynman diagram, 47, 60–64 two particle two particle, 54 → Feynman gauge, 73, 75, 102, 163 crossing, 98–101, 110, 111, 136, 138 Feynman propagator CTEQ, 186 charged massive vector fields, 147 current density, electromagnetic, 26, 34, 69 Dirac fields, 66, 67, 75 cutoff, 39, 63, 76, 134, 146, 172–176, 179 generic fields, 66 gluons, 168 decay rate (width, Γ), 113 Higgs boson, 205 defining representation, 152 massive vector fields, 149 density of states, 50 photon fields, 72–74 differential cross section, 50, 51 scalar, complex (π±), 143 dimension of group (dG), 152, 153 scalar, real, 62 dimension of representation (dR), 151 Feynman propagator dimension of tensor product representation, 154 W ± bosons, 147 Yang-Mills vector fields, 163 gauge eigenstate fields, 210, 221–223 Feynman rule gauge fixing, 73 Yukawa coupling, 66 gauge-fixing parameter (ξ), 73 Feynman rules, 47 gauge invariance, 27, 70, 71, 150 W -fermion-antifermion, 148 non-Abelian, 150, 151, 160–163 Fermi weak interaction theory, 127, 128 Yang-Mills, 164, 165 fermion external lines, 67, 68 Gell-Mann matrices, 158 fermions with known helicity, 86, 87 generators general, 64–67 SU(2), 156 massive vector bosons, 149 SU(3), 158 pion decay, 143–145 SU(N), 157 QCD, 168, 169 of Lie algebra, 152 QED, 71–76 GeV, 11 scalar φn theory, 62, 63, 65 global symmetry, 157, 199 standard electroweak gauge theory, 215–217 global symmetry breaking, 199–202 Yang-Mills, 163–165 gluon, 140, 167, 168 Yukawa coupling, 65 propagator, 168 Yukawa coupling, electron, 218 gluon collider, 194 Yukawa theory, 66 gluon-gluon scattering, 192–194 Feynman slash notation, 13 Goldstone boson, 201, 202, 207 Feynman’s x, 183 Goldstone’s theorem, 207 Feynman-Stückelberg interpretation, 21 Grassman (anticommuting) numbers, 41 field, 28 gravity, 147, 176 Dirac, 29, 32, 33, 40–42 group, 8, 150 electromagnetic, 26, 29, 34 GRV parton distribution functions, 186 scalar, 28, 31 hadron-hadron scattering, 182–186, 194–198 vector, 29, 32, 34 hadronization, 183 Weyl, 32 half life, 122 field theory, 29 Hamilton’s principle, 29 fine structure constant, 69, 81 Hamiltonian four-fermion interaction, 127, 145, 147 free Dirac field, 42 four-vector, 4–7 free scalar field, 35, 37–39 free field theory, 43 general classical system, 35 fundamental representation, 152 interaction, 43, 44, 67 gµν (metric tensor), 5 single Dirac particle, 12, 18 γ matrices, 13 single scalar particle, 10

γ5 matrix, 20, 242 hard scattering of partons, 182 helicity, 19, 20 Lagrangian, 29, 31 helicity conservation, 92, 142, 231, 232 Lagrangian density, 31 helicity ﬂip, 142 LAMPF, 125 helicity suppression, 142 Landau gauge, 73, 75, 163 Higgs boson, 49, 66, 117–119, 214–217 Large Hadron Collider (LHC), 182, 189, 193, Higgs ﬁeld 194, 238, 239 generic, 203, 204, 207, 208 left-handed fermions Standard Model, 212–217 and weak interactions, 126, 128, 139, 157, Higgs mechanism, 203, 204, 207, 208 208, 209, 217 hole (absence of an electron), 20 Dirac, 20

projection matrix (PL), 20, 22 identity element of group, 150 Weyl, 24, 25 i² factor in propagators, 62 left-handed polarization of photon, 72 ignorance, parameterization of, 140, 173 lepton number conservation, 124, 125 IN state, 44, 45 Levi-Civita (²µνρσ) tensor, 9 index of representations, 152 Lie algebra, 152 SU(2), 156 Lie group, 151 SU(3), 158 linear collider, 86, 240 adjoint, 154 local (gauge) symmetry, 150, 199 inﬁnite momentum frame, 183 breaking, 202–205, 207 infrared slavery, 179 longitudinal polarization of massive vector bo- integrated luminosity, 49 son, 204 interaction Hamiltonian, 43, 44, 67 longitudinal rapidity, 195 interaction Lagrangian, 43 loop corrections to muon decay, 134 internal line, 60 loops in Feynman diagrams, 63, 64, 67, 76, 146, inverse metric tensor (gµν), 5 165, 171–176, 220 inverse muon decay, 136 fermion minus sign, 76 inverse of group element, 150 Lorentz transformations, 8 irreducible representation, 153 Lorentz-invariant phase space isospin global symmetry, 157 n-body, 52, 114 Jacobi identity, 153 2-body, 53, 54, 108, 114 Jacobian, 196, 197 3-body, 120, 121 jet, 83, 169, 183 lowering operator, 36–38 J/ψ particle, 85 luminosity, 49

Klein-Gordon equation, 11, 16, 32 Mandelstam variables (s,t,u), 61, 62, 93, 229, Klein-Nishina formula for Compton scattering, 231, 232 109 Mandelstam variables, partonic (ˆs, t,ˆ uˆ), 182, 190 mass diagonalization, 221 photon polarization, 71, 72 mass eigenstate ﬁelds, 210, 221–223, 236 photon propagator, 72–75 massive vector boson, 148, 165, 204–208, 210 picobarns (pb), 48

propagator, 149 pion (π±) decay, 140–145 matrix element, 47–49 pion decay constant (fπ), 140, 141 reduced, 50 polarization vector, 71 Maxwell’s equations, 26, 27, 34 and gauge transformations, 102 mean lifetime, 113 massive vector boson, 148 metric tensor (gµν), 5 polarized beams, 86 Mexican hat potential, 200 positron, 12, 21 Moller scattering, 100 potential energy, 29, 30, 199 momentum conservation, 6, 50, 75, 76, 225 projection matrices for helicity (PL, PR), 20, 242 momentum fraction (Feynman’s x), 183 propagator MRS parton distribution functions, 186 charged massive vector boson, 147, 149 MS renormalization scheme, 211 Dirac fermion, 67, 75 muon decay, 124, 127–132 generic, 66 + muon production in e−e collisions, 77–82 gluon, 168 Higgs boson, 205 Nambu-Goldstone boson, 201, 202, 207 massive vector boson, 147, 149 nanobarns (nb), 48 photon, 72–74 natural units, 241 scalar, complex (π±), 143 neutron decay, 122 scalar, real, 62 non-Abelian gauge invariance, 150, 151, 160– propagator 163 W ± bosons, 147 non-renormalizable theories, 146, 147, 166, 175 Yang-Mills vector ﬁelds, 163 nuclear weak decays, 122, 123 proper interval, 5 on-shell, 6 proper Lorentz transformations, 7, 8 OUT state, 45 pseudo-scalar fermion bilinear, 126 pseudo-scalar Higgs boson, 233 parity, 7, 126 parity violation, 7, 126–128 QCD scale (ΛQCD), 178 partial width, 114 quadratic Casimir invariant, 153 parton distribution function (PDF), 183–189 Quantum Chromo-Dynamics (QCD), 83, 158, parton model, 182 167, 168 partonic subprocess, 181, 182 Quantum Electro-Dynamics (QED), 69–71 Pauli exclusion principle, 20, 41 Feynman rules, 71–76 Pauli matrices (σ1,2,3), 12, 242 quark-quark scattering, 169–171 photon ﬁeld (Aµ), 29, 69–73 raising operator, 36–38 rapidity, 5, 15, 17, 195 singlet representation, 151, 154 longitudinal, 195 slash notation, 13 reduced matrix element, 50 speed of light (c), 4, 241 reducible representation, 153 spin operator, 18 regularization, 39, 63, 146, 172, 179 spin-sum identities, 23, 243 renormalizable theories, 146, 147 spinor, 11 renormalization, 39, 63, 134, 135 Dirac, 12 renormalization group, 176, 177 Weyl, 24 renormalization scale, 174, 176 spontaneous symmetry breaking, 199 renormalized (running) coupling, 173–176 global, 199–202 electroweak, 211 local (gauge), 202–205 renormalized (running) mass, 175, 176 Standard Model, 208–217 representation of group, 151 structure constants of non-Abelian group, 152 representations SU(2) Lie algebra, 156, 157

of SU(2), 156, 157 SU(2)L (weak isospin), 157, 208, 209 of SU(3), 158–160 SU(3) Lie algebra, 158–160 of SU(N), 160 SU(N) Lie algebra, 157, 160 of U(1), 151 subprocess, partonic, 181, 182 of non-Abelian group, 151–156 SuperKamiokande, 125

Rhadrons, 84, 85 symmetry factor (of Feynman diagram), 63, 64, right-handed antifermions and weak interactions, 76 126, 127, 137 t-channel, 61, 62, 93 right-handed fermions tensor, 9 Dirac, 20 tensor product of representations, 154 Weyl, 24, 25 Tevatron, 48, 49, 117, 118, 182, 189, 191, 193, right-handed polarization of photon, 72 237–240 right-handed projection matrix (PR), 22 Thomson scattering, 110 rotations, 8 three-body phase space, 119–121 Rutherford scattering, 97 time reversal (T), 7, 133 scalar ﬁeld, 28, 31 trace trick, 79 complex, 143 traces of γ matrices, 14, 242, 243 scalar function, 8 transverse momentum (pT ), 192, 195 s-channel, 60–62 transverse polarization, 71, 204 Schrodinger equation, 10 tree-level diagrams, 63 Schrodinger picture of quantum mechanics, 43, triangle function (λ), 115 44 two-body phase space, 108 sea partons, 184, 185, 188, 189 U(1) as a Lie group, 151 U(1) for electromagnetism, 151, 167

U(1)Y (weak hypercharge), 208, 209, 211–213 u-channel, 61, 62 unitarity of CKM matrix, 224 unitarity of quantum mechanical time evolution, 145, 147, 166 unitary gauge, 205, 206 Standard Model, 214–216 units, 11, 241 vacuum expectation value (VEV), 200 antiquark-quark, 207 Standard Model, 213 Standard Model numerical, 215 vacuum state, 28, 37, 40 valence quarks, 140, 183–189, 193 volume element, 9

Ward identity, 102, 103 weak hypercharge [U(1)Y ], 208, 209, 211–213 weak isospin [SU(2)L], 157, 208, 209 weak mixing (Weinberg) angle, 210 weak-interaction charged current, 139, 148 Weyl (chiral) representation of Dirac matrices, 13, 242 Weyl equation, 24 Weyl fermion, 24 Weyl spinor, 24 would-be Nambu-Goldstone boson, 204, 205, 207 Standard Model, 214 x, Feynman (momentum fraction), 183 ξ (gauge-ﬁxing parameter), 73

Yang-Mills theories, 149, 160–165 Yukawa coupling, 66, 116–118, 218–220, 222