Part III Quantum Field Theory · Quantum Field Theory Notes by Aﬁq Hatta 2 Classical Field Theory...

Part III Quantum Field Theory

Afiq Hatta

January 17, 2020

Contents1 Why do we care about Quantum Field Theory? 5

1.1 The Problem of Locality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Particle indistinguishability and Particle conservation . . . . . . . . . . . . . . . 51.3 Problems with QFT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4 Scaling in QFT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Classical Field Theory 62.1 Euler Lagrange equations for a point particle . . . . . . . . . . . . . . . . . . . . 62.2 Promotion of Point Particles to the Classical Field . . . . . . . . . . . . . . . . . 7

2.2.1 The Klein Gordon field . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.2.2 The Lagrangian for Electromagnetism . . . . . . . . . . . . . . . . . . . . 8

2.3 Lorentz invariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.4 Symmetries and Noether’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.4.1 The Energy-Momentum Tensor . . . . . . . . . . . . . . . . . . . . . . . . 122.5 Switching over to the Hamiltonian Formalism . . . . . . . . . . . . . . . . . . . . 132.6 Promoting discrete operators to fields . . . . . . . . . . . . . . . . . . . . . . . . 142.7 Reviewing the Harmonic Oscillator in One Dimension in Quantum Mechanics . . 152.8 Promoting fields to operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.8.1 Calculating the Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . 182.8.2 Infinity issues with the Hamiltonian . . . . . . . . . . . . . . . . . . . . . 19

2.9 Creating particle states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.9.1 Relativistic normalisation . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.10 Quantising the free complex scalar field . . . . . . . . . . . . . . . . . . . . . . . 242.11 Introducing time with the Heisenberg picture . . . . . . . . . . . . . . . . . . . . 26

3 Causality in QFT 283.1 Propagators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.2 The Feynman Propagator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4 Interacting fields and the approach with perturbation theory 334.1 Why do we need interacting fields and what are ’valid’ conditions for perturbation

expansions? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334.2 The interaction picture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

4.2.1 The Heisenberg and Schrodinger Picture . . . . . . . . . . . . . . . . . . . 35

1

Quantum Field Theory Notes by Afiq Hatta

4.2.2 The Interaction picture is a hybrid! . . . . . . . . . . . . . . . . . . . . . . 354.2.3 We can Fourier expand Interaction picture states! . . . . . . . . . . . . . 364.2.4 Operators defined from the Interaction picture obey the same commuta-

tion relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374.2.5 Introducing the time unitary operator . . . . . . . . . . . . . . . . . . . . 39

4.3 Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414.3.1 An example with Yukawa theory . . . . . . . . . . . . . . . . . . . . . . . 424.3.2 First order term in meson decay . . . . . . . . . . . . . . . . . . . . . . . 42

4.4 Wick’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444.5 Feynman Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444.6 Scattering revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 464.7 Amplitudes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 464.8 Computing the two point correlation function for the ground state of our per-

turbed Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 484.9 Wick’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 484.10 Applying Wick’s theorem for the interaction term . . . . . . . . . . . . . . . . . . 48

4.10.1 Symmetry factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504.11 Feynman Rules in φ4 theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4.11.1 Combinatoric factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 514.11.2 Deriving our Feynman rules . . . . . . . . . . . . . . . . . . . . . . . . . . 544.11.3 Vacuum Bubbles and Connected Diagrams . . . . . . . . . . . . . . . . . 54

4.12 Green’s function vacuum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554.12.1 LSZ Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

5 Cross sections and decay rates 585.0.1 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 585.0.2 Decay rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 595.0.3 2 → 2 scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 605.0.4 Decay rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

6 The Dirac Equation 626.1 What does the Dirac equation give us? . . . . . . . . . . . . . . . . . . . . . . . . 626.2 The Spinor Representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 636.3 Constructing a Lorentz Invariant action of ψ . . . . . . . . . . . . . . . . . . . . 676.4 The Dirac equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 686.5 Chiral Spinors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

6.5.1 The Weyl equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 696.6 Dirac Field Bilinears . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 706.7 Parity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 716.8 Symmetries and Currents of Spinors . . . . . . . . . . . . . . . . . . . . . . . . . 726.9 Lorentz transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 736.10 Internal vector symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 746.11 Axial Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 746.12 Plane-Wave Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 746.13 Helicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

6.13.1 Negative frequency solutions . . . . . . . . . . . . . . . . . . . . . . . . . 766.14 A canonical Basis for normalised Spinors . . . . . . . . . . . . . . . . . . . . . . . 76


6.15 Quantising the Dirac field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 786.16 The Dirac Hole Interpretation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 806.17 Fermi-Dirac Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

6.17.1 Going into the Heisenberg picture . . . . . . . . . . . . . . . . . . . . . . 806.17.2 The Feynman Propagator . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

6.18 Momentum Space Feynman Rules for Fermion Amplitudes . . . . . . . . . . . . . 846.19 Cross-Sections and Mod-squared expressions . . . . . . . . . . . . . . . . . . . . . 86

7 Quantum Electrodynamics 897.1 Quantisation of the Electromagnetic Field . . . . . . . . . . . . . . . . . . . . . . 91

7.1.1 Coulomb Gauge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 917.1.2 Lorentz Gauge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

7.2 Timelike γ pairs continued . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 957.2.1 γ Feynman propagator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

7.3 Interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 957.3.1 Coupling to Fermions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 967.3.2 Coupling to Scalars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

7.4 Warming up with spinors from the SO(3) rotation group . . . . . . . . . . . . . . 977.4.1 Rotations in Euclidean space . . . . . . . . . . . . . . . . . . . . . . . . . 977.4.2 Transformations with Unitary matrices . . . . . . . . . . . . . . . . . . . 98

7.5 Lorentz tranformations and spinors . . . . . . . . . . . . . . . . . . . . . . . . . . 997.6 Constructing the Chiral representation for the Dirac equation . . . . . . . . . . . 1017.7 Motivating Lorentz invariance of the Dirac equation . . . . . . . . . . . . . . . . 1017.8 Constructing boosts and rotations in block diagonal form . . . . . . . . . . . . . 1037.9 The Weyl equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1047.10 Constructing Plane-Wave solutions to the Weyl equation . . . . . . . . . . . . . . 105

8 Example sheet 1 1078.1 Question 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1078.2 Question 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1098.3 Question 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1108.4 Question 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1128.5 Question 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1138.6 Question 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

8.6.1 Gauge invariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1158.6.2 Constructing the energy-momentum tensor . . . . . . . . . . . . . . . . . 115

8.7 Question 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1178.8 Question 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

9 Example Sheet 2 1209.1 Question 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1209.2 Question 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1209.3 Question 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1209.4 Question 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1239.5 Question 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

10 Example Sheet 4 126


10.0.1 Part a . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12610.1 Question 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12910.2 Question 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13010.3 Question 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13210.4 Question 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134


1 Why do we care about Quantum Field Theory?

If you’re reading these set of notes, you’ve probably done a course at university on basic quantummechanics. However, there are several philosophical and observed issues in quantum mechanicsthat we’ve yet to address. I’ll go over them here.

1.1 The Problem of Locality

1.2 Particle indistinguishability and Particle conservation

In quantum mechanics, when we talk about multiparticle systems, we usually tensor producttheir state representation like |ψ〉 ⊗ |φ〉, and we usually don’t care about how these thingsinteract. In quantum mechanics, we view particle number as conserved. When we do thishowever, and attempt to make Schrodinger’s equation relativistic, we run into problems withregards to causality violation and unbounded energy. However, in QFT we have multiparticlestates, with creation and annihilation operators which remedy this situation.

1.3 Problems with QFT

In these notes, we’ll have a ’free theory’ where particles don’t interact, and we solve for statesanalytically. Later, we’ll also add on ’interaction terms’, where particles do interact and wehave to resort to perturbation techniques to study physical quantities of interest. Within thesenotes, we assume that these interaction terms are small. Physics where we try to understandhow to deal with larger interactions is still a big area of research (see lattice QCD where we tryto discretise space time and solve QFT numerically).

1.4 Scaling in QFT

Throughout QFT, we scale out the natural constants c, ~ to have c = ~ = 1. Because of this,we can form relationship between units of time, length and mass. c is a unit of speed, but sincethis is set to 1 we have that

[L][T ]−1 = 1 =⇒ [L] = [T ]

So we have that in this regime, which we call ’natural units’ length has units of time. Inimmediate consequence of this is that energy has the units of mass.

[E] = [M ][L]2[T ]−2 = [M ]

Our condition that the Planck constant goes to 1 implies that

[M ][L]2[T ]−1 = 1 =⇒ [M ] = [L]−1


2 Classical Field Theory

2.1 Euler Lagrange equations for a point particle

We’ll start off by reviewing some ideas in classical mechanics to motivate our approach toquantum field theory. In particular, the simple harmonic oscillator is a good place to start. Inclassical mechanics, we describe the motion of a particle x as a function of time, so x = x(t). Inhigh school, we learnt that Newton’s laws dictate that in one dimension, mass times accelerationis equal to the force on the particle. So, in the case of the simple harmonic oscillator,

mx = −kx

, where k is our spring constant. We then learnt as undergraduates that this is a more specificcase of the motion of a particle in a potential V , where in the case of a simple harmonic osccilator,V (x) = 1

2kx2. In this formalism, our equations of motion are governed by the equation

mx = −∇V

You can check that this recovers the above equation of motion.

Generalising one step further, we see that we can encode all of this infomation into a single usefulquantity, the Lagrangian, which is the quantity L = T − V , where T is a kinetic energy term.The Lagrangian should always be a scalar function, and in the case of our harmonic oscilator in3 dimensions it is

L =1

2mx2 − 1

2kx2

To derive the equations of motion, we want to apply the principle of least action. This principleis that the particle will move along a path which minimises the integral of the Lagrangian overtime. In other words, to find out the path of the particle between to points in time, say t1 andt2, we would like to minimise the action.

S =

∫ t2

t1

L(x, x)dt

A condition that we impose here is that the endpoints should be fixed, so x(t1) = x1, andx(t2) = x2. To minimise the action, we vary the curve a tiny bit by sending x → x+ δx. Thisalso induces a variation in x. We use the chain rule with respect to the variables x and x andto give

δS =

∫ t2

t1

δx · ∂L∂x

+ δx · ∂L∂x

But, we can perform integration by parts on the second term, by integrating the δx and differ-entiating in the integrand. This gives us a surface term which we can push to zero.

δS =[δx · ∂L

∂x

]t2t1+

∫ t2

t1

dt δx ·(∂L

∂x− d

dt

(∂L

∂x

))Now, since our variation was arbitrary, we have that the integrand needs to be zero, so we havethe Euler-Lagrange equation

d

dt

(∂L

∂x

)=∂L

∂x


2.2 Promotion of Point Particles to the Classical Field

Previously, we were formulating our problem in a point particle sense. We now alter the physicsslightly. Instead of a point particle, suppose we assign a value to every point in spacetime, andwork with that instead. This function φ = φ(x), we call a field. We write φ(x) synonymouslywith φ(x, t). In this regime, much like how we previously viewed i as a label for a spatialcoordinate in xi, we now have infinite degrees of freedom, where now the x term in φ(x, t)acts as a continuous, infinite label. In essense, fields are like coordinates but with infinite degreesof freedom. In fact, we don’t have to stop there.

We can attach components to this field, thus adding a few more degrees of freedom, and indexingthem like φa(x, t). So, if we have a indexing over a = 0, 1, 2, 3, then we’ve added an additionalfour degrees of freedom.

Our most familiar example of this are the electromagnetic fields (which yield 6 degrees of freedomin all). These are Ei(x, t) and Bi(x, t) where our index labels spatial directions. These sixfields are derived from the familiar electromagnetic vector potential Aµ = (φ,A), given by thefollowing

Ei = −∂Ai∂t

− ∂A0

∂xi

Bi = εijk∂Ak∂xj

Since we have a field over space now, our Lagrangian itself needs to be an integral over space.So, we formulate our problem in terms of something we call the Lagrangian density.

L =

∫d3xL(φ, ∂µφ)

where our density now is a function of both the field itself and also the first derivative (not sodifferent from the point particle case). And, like the point particle case, our action is once againthe integral over time so that

S =

∫dtL =

∫dt

∫d3xL =

∫d4x L

Our Euler Lagrange equations are derived in the same way, but in this case over the 4 space-timecoordinates, not just the time coordinate like we did earlier. Our change in action is

δS =

∫d4x δφ

∂L∂φ

+ δ∂µφ∂L

∂(∂µφ)

=

∫d4x δφ

∂L∂φ

− δφ∂µ∂L

∂(∂µφ)

And since our variation δS = 0, we have the Euler-Lagrange equations

∂µ

(∂L

∂(∂µφ)

)=∂L∂φ


2.2.1 The Klein Gordon field

We’ll now introduce one of the main equations for a free field theory, the Klein-Gordon field.The Klein-Gordon field looks roughly like a Lagrangian for a simple harmonic oscillator

L =1

2ηµν∂µφ∂νφ− 1

2m2φ2

ηµν denotes our Minkowski metric, which in the course of these notes we’ll take to be the’mostly negative’ metric ηµν = diag(+1,−1,−1,−1). In terms of time and spatial derivatives,our Lagrangian reads

L =1

2φ2 − 1

2(∇φ)2 − 1

2m2φ2

Which is reminiscent of our Lagrangian being of the form L = T − V , where T = 12 φ

2 is ourkinetic term and the rest is our potential term. We can compute our equations of motion fromthis. We have that

∂L∂φ

= m2φ,∂L

∂(∂µφ)= ∂µφ

where we’ve used the chain rule for differentiation both times. Substituting this into the Euler-Lagrange equations, we have the Klein-Gordon equation which reads

∂µ∂µφ−m2φ = 0

If we have an ansatz for a plane wave solution, that φ(x) = e−ip·x, then this yields the relataivisticdispersion relation that

pµpµ = m2

which we know is true.

2.2.2 The Lagrangian for Electromagnetism

Another important example of a Lagrangian we’ll be encountering is the Lagrangian associatedwith electromagnetism. This is given by

L = −1

2(∂µAν)(∂

µAν) +1

2(∂µA

µ)2

where Aµ is our familiar vector potential we described earlier. Notice that this Lagrangiandoesn’t depend on Aµ, so if we want to figure out the equations of motion we need to calculatethe term ∂L

∂(∂µAν). Using the identity that

∂(∂αAβ)

∂µAν= ηαµηβν

our use of the product rule gives us that

∂L∂(∂µAν)

= −∂µAν + ∂ρAρ∂(∂αA

α)

∂(∂µAν)

= −∂µAν + ∂ρAρη µα ηαν

= −∂µAν + ∂ρAρηµν


Our Euler Lagrange condition implies that

0 = ∂µ(−∂µAν) + ∂ν∂ρAρ = ∂µ(−∂µAν + ∂νAµ)

where in the second term, we relabelled the summed over indices from ρ → µ because they’redummy indices. za

2.3 Lorentz invariance

In QFT and special relativity, theories should be invariant under Lorentz transformations. Inthis section, we will consider active transformations, where we’re changing the direction of thefield under a Lorentz boost. Under a Lorentz boost, our scalar field φ changes like

φ(x) → φ′(x) = φ(x′) = φ(Λ−1x)

This is what we call an active transformation since our field is actually, physically shifted.We’re shifting our field by a Lorentz boost in our frame, with x → Λx which induces a changeφ(x) → φ′(x), but this change is equivalent to rotation our frame of reference in the oppositedirection first (I’ll show a diagram below). Our Lorentz boosts are defined by the propertythat

ΛµρηρτΛ ν

τ = ηµν

Lorentz boosts can simultaneously represent rotations in 3-space, as well as boosts in a particularaxis in which mixes time and space. For a rotation, our boost is represented as

Λµρ =

1 0 0 00 1 0 00 0 cos θ − sin θ0 0 − sin θ cos θ

and for a boost, Lorentz transformations are represented by

Λµρ =

γ −γv 0 0

−γv v 0 00 0 1 00 0 0 1

In terms of group theory, Lorentz transformations are representations of the Lie group SO(3, 1)on scalar fields. Let’s look at Lorentz invariant theories, which are theories where our physicalaction S is unchanged by Lorentz transformations. Consider the action

S =

∫d4x ∂µφ∂

µφ+ U(φ(x))

We’ve declared here that U(φ(x)) is some polynomial of φ(x). How does this object changeunder transformations of our field φ(x) → φ′(x)? Let’s do the polynomial term first. Under theactive transformation φ(x) → φ′(x) = φ(x′), we have that

U(x) = U(φ(x)) → U(φ′(x)) = U(φ(x′)) = U(x′)


This means that U(x) → U(x′). Now, to do the kinetic part of the Lagrangian. To do this, weobserve that our partial derivative of a field transforms as

∂µφ→ ∂µφ′(x) = ∂µφ(x

′) = (Λ−1)ρµ∂′ρφ(x

′), x′ = Λ−1x

Something important to note here is that we’re not transforming x simultaneously here aswell, since this is an active transform of the actual field. We’ve only expressed x diffferently toexpress our change in our field in terms of new coordinates, but the same function. Thus, ourKinetic term transforms as

1

2ηµν∂µφ∂νφ→ 1

2ηµν∂µφ

′∂νφ′

=1

2ηµν(Λ−1)σµ(Λ

−1)τν∂′ρ(x′)∂′τ (x

′)

=1

2ηρτ∂′ρφ(x

′)∂′τ (x′)

where we’ve used the transformation property of the Minkowski metric. Thus, every term in ourLagrangian density has replaced x with a modified x′ = Λ−1x. We thus have that our actionchanges like

S =

∫d4xL(x) → S′ =

∫d4xL(x′), x′ = Λ−1x

However, since the determinant (associated Jacobian) of our Lorentz boost is 1 (as Lorentzboosts are part of the special group SO(3, 1), we have that our measure doesn’t change:

d4x = det(Λ)d4x′ = d4x′ =⇒ S′ =

∫d4x′ L(x′) =⇒ S = S′

In the last step we have equality since we’re integrating over a dummy variable! Thus, Lagrna-gians of this form are Lorentz invariant.


2.4 Symmetries and Noether’s theorem

We’ll now take a look at the role of symmetries in this formalism. In physics, we can have multipletypes of symmetries, including Lorentz symmetries, internal symmetries, gauge symmetries andsupersymmetries. In terms of Lagrangians, we’ll now show that every continuous symmetryof a Lagrangian density gives rise to a conserved current jµ, which satisfies the conservationcondition

∂µjµ = 0

It’s easy to check that, from this quantity, we can construct a conserved charge given by

Q =

∫d3x j0

This is converved with respect to changes in time since

Q =

∫d3x

∂j0

dt= −

∫d3x∇ · j → 0

where the final term goes to zero, since the divergence theorem allows us to re-express this as asurface integral. We’re assuming that jµ(x) → 0 as xµ → 0. What do we mean by a symmetry?An infinitesimal transformastion of our scalar field

φ(x) → φ(x) + αδφ(x)

where α is small is a symmetry provided that the Lagrangian only changes by a total fourderivative:

L → L+ ∂µFµ

for some function Fµ. This is a symmetry because since our action integrates over our Lagrangiandensity, due to the divergence theorem this extra term vanishes, leaving our action S invariant.We can now derive Noether’s theorem. If we Taylor expand out our Lagrangian,

L(x) → L(x) + α∂L

∂φδφ+ α

∂L

∂(∂µφ)∂µ(δφ)

We can rewrite the above term as

L(x) + α∂µ

(∂L

∂(∂µφ)

)δφ+ αδφ

(∂L

∂φ

)− ∂µ

(∂L

∂(∂µφ)

)But, from the Euler Lagrange equations the final term vanishes, and comparing with our totalderivative term we find that a conserved current is

jµ =∂L

∂(∂µφ)δφ− Fµ

Let’s consider a Lagrangian with a complex scalar field ψ given by

ψ(x) =1√2(φ1(x) + φ2(x))


We treat ψ(x) and ψ(x)∗ as separate fields, and write a complex scalar lagrangian as

L = ∂µψ∂µψ∗ −m2ψ∗ψ − λ

2(ψ∗ψ)2

It’s clear that this Lagrangian is invariant under the phase symmetry

ψ → eiαψ, ψ∗ → e−iαψ∗

This induces the infinitesimal symmetry

δψ = iαψ, δpsi∗ = −iαψ∗

With Noether’s theorem, one can verify that the conserved current is

jµ = i(ψ∂µψ∗ − ψ∗∂µψ)

Physically, conserved charges could be anything from conserved electric charge, or conservedparticle number such as baryon or lepton number.

2.4.1 The Energy-Momentum Tensor

In this subsection, we’ll be constructing important conserved quantities from Noether’s theorem.These conserved quantities will arise specifically from translational symmetry in spacetime. Inclassical mechanics, energy conservation arose from time translational symmetry, and momentumconservation arose from spatial translational symmetry. In classical field theory, this is nodifferent. Our concept of conserved energy and momentum arise from translational symmetry inour Lagrangian, and we’ll combine them to form a single conserved tensor. Consider a translationtransformation in our spacetime coordinates

xν → xν + αεν

We expect that our symmetry here will give rise to a higher rank tensor because we have 4linearly independent directions arising from εν . Taylor expanding out, we have

φa(x) → φa(x)− αεν∂νφa(x)

And this induces a similar transformation on our Lagrangian, since L = L(φ(x)) = L(x), so wecan Taylor expand our expression in exactly the same way.

L(x) → L(x)− αεν∂νL(x) = L(x)− αεν∂µ(δνµL)

In the second expression, we’ve written things in a slightly weird yet more suggestive manner.Our rationale is as follows. Since εν has four degrees of freedom, we can write this out as fourseparate conserved currents as written in the bracket, where the distinct currents are indexedby ν. Hence, with Noether’s theorem, we can construct four different Noether currents usingthis form of δL.

Tµν = (jµ)ν = ∂νφa∂L∂µφa

− δµνL, ∂µTµν = 0


Different components of this tensor correspond to different quantities. We have

Total Energy E =

∫d3xT 00

Total momentum P i =

∫d3xT i0

We apply this to our Klein-Gordon field.

L =1

2∂µφ∂

µφ− 1

2m2φ2

This has an associated energy-momentum tensor, where raising the index on the delta functionyields the Minkowski

Tµν = ∂µφ∂νφ− ηνµL

Substituting our expression for conserved energy, and momentum

E =

∫d3xT 00 =

1

2

∫d3xφ2 + (∇φ)2 +m2φ2, P i =

∫φ∂iφ

If Tµν is non-symmetric, we can massage it into a form that’s symmetric which makes it easierto deal with by adding an arbitrary, antisymmetric ’gauge’ term

Tµν → Tµν + ∂ρΓρµν , Γ(ρµ)ν = 0

This doesn’t affect our conservation law, because we have

∂µ∂ρΓρµν = 0

2.5 Switching over to the Hamiltonian Formalism

In quantum mechanics, we’re already used to using the Hamiltonian H to investigate the energyspectrum and energy eigenstates of a system. Right now, we’ll build a way to switch overfrom our current Lagrangian understanding of a scalar field to a derivation of the associatedHamiltonian. To do this, we need to define what conjugate momenta, π is. Conjugate momentais defined as

π =∂L∂φ

where φ is the partial derivative of our field with respect to time. Perhaps some of our intuitionabout this is as follows. If we were given the Lagrangian L = 1

2x · x − V (x) in the discreteparticle case, then our conjugate momenta is given by

π =∂L∂x

which is in this case is x, the expression for our standard notion of momentum with unitmass.


In the discrete particle case, our Hamiltonian is related to our Lagrangian by a Legendre trans-form, where

H =∑i

π · x− L

Our sum over i denotes the sum of over particles in the system. However, in field theory, wemake this continuous by integrating over space instead, and then writing in the Lagrangiandensity in the integrand. Thus, our expression for the Hamiltonian in field theory is

H =

∫d3x π(x)φ(x)− L

In the case of the Lagrangian density, we can calculate our conjugate momenta

L =1

2φ2 − 1

2(∇φ)2 − V (φ), φ = φ

This gives ys an expression for our Hamiltonian density H, which is defined as the integrand ofH =

∫d3xH.

H =1

2(π2 + (∇φ)2) + V (φ)

Completely analogously to Hamilton’s equations in classical dynamics, we have Hamilton’s equa-tions defined in terms of partial field derivatives of our Hamiltonian density.

φ =∂H∂π

, π = −∂H∂φ

Our Hamiltonian formalism is not obviously a Lorentz invariant theory, but it is!

2.6 Promoting discrete operators to fields

We know from standard quantum mechanics how to find the spectrum of the quantum harmonicoscillator H for unit mass, given by

H =1

2p2 +

1

2mω2q2.

p, q represent our momentum and position operators respectively. Last year, we promoted gen-eralised coordinates involving position and momentum to operator values obeying commutationrelations. One recalls that we have our canonical commutation relations (which we can deriveby examining infinitesimal transformations)

[pa, pb] = [qa, qb] = 0, [qa, pb] = iδ b

a .

We can intuitively promote our scalar and momentum fields in the same way, where now afield is an operator valued function of space and time. For simplicity’s sake, let’s work in theSchrödinger picture where these operators don’t depend on time but just space (and we pushall our time dependence on states instead)

[φa(x), πb(y)] = iδ3(x− y)δ b

a

[φa(x), φb(y)] = [πa(x), π

b(y)] = 0.


Our final goal is to compute the spectrum from our Hamiltonian derived from quantum fields.However, this is a hard thing to do. In a free theory however, at any given point in space time,the field evolves independently on other points, and so one useful thing to do is find a basiswhere we can diagonalise this Hamiltonian. One thing we could do is to Fourier transform φ(x)into momentum space, so that (where we’ll put back in our time dependence)

φ(x, t) =

∫d3x

(2π)3eip·xφ(p, t)

Substituting this into the Klein-Gordon equation by differentiating from within the integrandgives us a simple harmonic oscillator with a momentum dependent frequency, we have theequation (

∂

∂t2+ (p2 +m2)

)φ(p, t) = 0

This is a simple harmonic oscillator vibrating at frequency

ωp =√p2 +m2

So, we have that for a field satisfying the free theory, at every point, we have an infinite super-position of simple harmonic oscillators!

2.7 Reviewing the Harmonic Oscillator in One Dimension in Quantum Me-chanics

Going back to the problem of the QM harmonic oscillator, our Lagrangian and Hamiltonian aregiven by

H =1

2p2 +

1

2mω2q2, L =

1

2q2 − 1

2ω2q2

Recall that we can reduce our description of the discrete Hamiltonian via creation and annihi-lation operators, with their important commutation relation

a =

(√ω

2q + i

√1

2ωp

), a† =

(√ω

2q − i

√1

2ωp

), [a, a†] = 1

We can invert these to give expressions for momentum and position as

q =1√2ω

(a+ a†) , p = − i√2ω

(a− a†)

The whole point of the exercise we’ve done above is to write our Hamiltonian in terms of a, a†,and hence make use of our commutation relations. We can write our Hamiltonian as follows,with the raising and lowering commutation relations

H = ω

(a†a+

1

2

), [H, a†] = ωa†, [H, a] = −ωa

This has the effect of raising and lowering energy eigenstates, which can be shown by applyingthe Hamiltonian to a raised or lowered state and then using our commutation relations whichwe derived earlier.

Ha† |E〉 = (E + ω)a† |E〉Ha |E〉 = (E − ω)a |E〉


Since we can step up or step down energy in this way, we’ve given rise to a ladder of energyvalues

. . . , E − 2ω,E − ω,E,E + ω,E + 2ω, . . .

There’s a caveat here because we can have infinite negative energy! Thus, lowering by a has tostop somewhere. This means that there’s a unique state |0〉such that

a |0〉 = 0 =⇒ H |0〉 = ω

2|0〉

We hence have a unique ground state with energy ω/2! We call this the zero point energy.Ignoring normalisation, we can step energy up n amount of times by applying raising operators,so that

|n〉 = (a†)n |0〉 , =⇒ H |n〉 = (n+1

2)ω |n〉

In physics however, we usually only care about energy differences, so this non-zero zero pointenergy is a bit annoying. We apply a procedure called normal ordering, where we reorder termsin an expression so that lowering terms are shoved to the right and raising terms are shoved tothe left. The has the notation : A :, so in the case of the Hamiltonian

: H :=1

2(: a†a : + : aa† :)ω = ωa†a

We now have the effect that our zero point energy is set to zero, with H |0〉 = 0. We get forfree that we can solve for this wavefunction. Since our momentum operator in position space isrepresented as p = −i ∂∂q , substituting this into a |0〉 = 0 gives us a differential equation for thestate |0〉 one we contract this in the position basis.

Rearranging, our expressions for the operators x, p are

x =1√2ω

(a+ a†

), p = −i

√ω

2

(a− a†

).

This gives rise to commutation relations, which we will show below.

2.8 Promoting fields to operators

From our definitions for position and momentum operators in the one dimensional case, we cangeneralise this notion to scalar and conjugate momentum fields. We do this by expanding overall modes, in what looks like a Fourier decomposition. Another motivation for the followingdefinitions is the fact that our operators should be hermitian, which is an extra reason for whywe’ve included terms for both operators.

φ(x) =

∫d3p

(2π)31√2ωp

(ape

ip·x + a†pe−ip·x

)Again using the form of momentum in 1 dimensional quantum mechanics to motivate definitions,we have a definition for conjugate momentum in terms of Fourier modes as well. This is givenby

π(x) =

∫d3p

(2π)3(−i)

√ωp

2

(ape

ip·x − a†pe−ip·x

)


This procedure is called second quantization, we’ve reconstructed the scalar and momentumfields completely in terms of and infinite amount of simple harmonic oscillators in momentumspace. Again, just as in quantum mechanics, we should impose the commutation relationsthat

[φ(x), φ(y)] = [π(x), π(y)] = 0, [φ(x), π(y)] = iδ3(x− y)

However, there’s an equivalence here which we’ll prove, which is that if we impose that ourcommutation relations in terms of ap and a†p as

[ap, aq] = [a†p, a†q] = 0, [ap, a

†q] = (2π)3δ3(p− q)

then this induces the commutation relations in terms of φ(x) and π(x), and vice versa.

Theorem. The commutation relations above are equivalent to each other.

Proof. We’ll prove first that the commutation relations on our scalar and momentum fields implythe commutation relations on the annihilation and creation operators. To do do this, we needsensible expansions of these. One can verify with our standard delta function identity that∫

d3xφ(x)e−ip·x =1√2ωp

ap +1√2ωp

a†−p∫d3xπ(x)e−ip·x = −i

√ωp

2ap + i

√ωp

2a†p

Solving this system gives us an expression for ap. Then, contracting the integral but with eip·x

instead gives us a†p. Hence, we have the expressions:

ap =

√ωp

2

∫d3xφ(x)e−ip·x + i

1√2ωp

∫d3xπ(x)e−ip·x

a†p =

√ωp

2

∫d3x d3xφ(x)eip·x − i

√1

2ωp

∫d3xπ(x)eip·x

Now we integrate these over two variables to get

[ap, a†q] = [

√ωp

2

∫d3xφ(x)e−ip·x + i

1√2ωp

∫d3xπ(x)e−ip·x

,

√ωq

2

∫d3y φ(y)eiy·q − i

√1

2ωq

∫d3y π(y)eiy·q]

We pull the commutators by linearity into the integral. Now, to make our lives easier, since thescalar fields commute with other scalar fields, and since the momentum fields commute with


other momentum fields, we have that the above expression

[ap, a†q] = − i

2

√ωp

ωq

∫d3xd3y e−ip·xeiq·y[φ(x), π(y)] +

i

2

√ωq

ωp

∫d3xd3y eiq·ye−ip·x[π(x), φ(y)]

= − i

2

√ωp

ωq

∫d3xd3y iδ(x− y)e−ip·xeiq·y +

i

2

√ωq

ωp

∫d3xd3y − iδ(x− y)eiq·ye−ip·x

=1

2

√ωp

ωq

∫d3x ei(q−p)·x +

1

2

√ωq

ωp

∫d3x e−i(p−q)·x

=1

2

√ωp

ωqδ(q− p)(2π)3 +

1

2

√ωq

ωpδ(p− q)(2π)3

= δ(p− q)(2π)3, since ωq = ωp when p = q

Similarly, in the opposite direction, we have that

[φ(x), π(y)] =

∫d3pd3q

(2π)3(−i)

√ωq

ωp

(−[ap, a

†q]e

ix·p−iq·y + [a†p, aq]e−ip·x+iq·y

)=

∫d3p

(2π)3(− i

2)(−eip·(x−y) − eip·(x−y)

)= iδ(x− y)

2.8.1 Calculating the Hamiltonian

Now that we’ve promoted our scalar and momentum fields as an infinite expansion of annihi-lation and creation operators, we can start computing objects of interest in terms of Fourierexpansions of these operators as well. We’ll start by calculating the most important quantity,our Hamiltonian. This is given by, as we calculated earlier,

H =

∫d3x

(1

2Π2 +

1

2(∇φ)2 + 1

2m2φ2

)We go term by term, and first calculate the most tricky term∫

d3x (∇φ)2 =∫d3xd3pd3q

(2π)3− 1

2√ωpωq

p · q(ape

ip·x − a†pe−ix·p

)(aqe

iq·x − a†qe−iq·x

)=

∫d3xd3pd3q

(2π)3− 1

2√ωpωq

p · q(apaqei(q+q)·x

+ a†pa†qe−i(p+q)·x − a†paqe

i(q−p)·x − q†apei(p−q)·x)

But recall that we can integrate over x first, and make use of the identity∫d3xeia·x = (2π)3δ(a)


This means that the above expression is equal to∫d3x (∇φ)2 =

∫d3p

(2π)3(−1)

p · p2ωp

(apa−p − a†pap − a†pap + a†pa

†−p

)=

∫d3p

(2π)3p2

2ωp

(a†pap + apa

†p

)Going into the second line, we’ve vanished the first and last terms since the function depends onjust the modulus of p, but taking the change of variables from p → p means that the expressionis equal to the negative of itself. It’s easy to show that∫

d3Π2 = 0,

∫d3x

1

2m2φ2 =

∫d3p

(2π)3m2

2ωp

(a†pap + apa

†p

)Hence our final expression for our full Hamiltonian is, using the equation for our dispersionrelation and commuting the annihilation and creation operators,

H =

∫d3p

(2π)3p2 +m2

2ωp

(a†pap + apa

†p

)=

∫d3p

ωp

(2π)3

(a†pap +

1

2(2π)3δ(0)

)

2.8.2 Infinity issues with the Hamiltonian

Let’s calculate the ground state of this Hamiltonian by letting it hit zero, to get H |0〉. Whenwe hit the annihilation operator in the first term, the term vanishes. Thus, we’re only left withthe second term. Since δ3(0) is constant, we naively pull this out of the integral to get

H |0〉 = 4π3δ(0)

∫d3pωp, ωp =

√|p|2 +m2

However, since we’re integrating over all momentum space, we get that ωp → ∞, and hence theintegral diverges (let’s ignore the problems associated with the delta function for now)! This iscalled a high-frequency, or ultraviolet, divergence. However, since we’re doing non-gravitationalphysics, all we care about are energy differences. So, to subtract of this infinity, we could’venaively wrote (as similar in the case of the one dimensional harmonic oscillator), that uponreordering of our indices we have that

H =

∫d3p

(2π)3ωpa

†pap =⇒ H |0〉 = 0

So with reordering, we have that zero point energy is just zero. One way to interpret this isthat in QFT, we have ’operator ambiguity’ in a sense that all operators are defined up to areordering. Now, we’re in a place to properly define normal ordering.

Definition. We denote a normal ordered string of operators, denoted

: φ(x1)φ(x2) . . . φ(xn) :

to be the same operator but with all annihilation operators pulled to the right.


Applying normal operators has the effect of removing unwanted infinities in our expression.It’s easy to check that

: H :=

∫d3p

(2π)3ωpa

†pap

Theorem. Our normal ordered Hamiltonian raises and lowers energy as in QM. With thisdefinition, our commutation relations with raising and lowering operators are now completelyanalogous to the case we saw in 1 dimensional QM.

[H, a†p] = ωpa†p, [H, ap] = −ωpap

Proof. We substitute in our definitions to find that

[H, a†p] =

∫d3q

(2π)3ωq[a

†qaq, a

†p]

=

∫d3q

(2π)3ωqa

†q[aq, a

†p]

=

∫d3q

(2π)3ωqa

†q(2π)

3δ(q− p)

= ωpa†p

The other case is completely similar!

2.9 Creating particle states

With this framework in mind, we can create excited states with a given energy ωp from our vac-uum by applying raising operators. Let’s denote |p′〉 = a†p′ |0〉. Then, applying our Hamiltonian,we have that

H∣∣p′⟩ = ∫ d3p

(2π)3ωpa

†papa

†p′ |0〉

=

∫d3p

(2π)3ωpa

†p[ap, a

†p′ ] |0〉

= ωp′a†p′ |0〉 = ωp′∣∣p′⟩

We thus interpret the raised state as a particle with mass m, energy ωp′ =√p′2 +m2 and

momentum p′, where the mass comes from the scalar term 12m

2φ2 in the Lagrangian. Fromnow on, we will relabel the energy ωp as Ep. From our energy momentum tensor, we could alsopromote our expression for momentum as an operator, to find that

P = −∫d3xπ(x)φ(x) =

∫d3p

(2π)3pa†pap

One can verify that this definition makes sense by acting on our momentum eigenstate via

P |p〉 = p |p〉


Similarly, constructing our angular momentum operator, J i as a cross product of our momentumand position operators, to find that for zero momentum eigenstates, J i |p = 0〉 = 0, in otherwords, the intrinsic angular momentum, or spin, in zero. Now, to create multi-particle states,we can apply a bunch of raising operators to the vacuum state, and interpret this as a systemof multiple particles. This is denoted as

|p1, . . .pn〉 = a†p1. . . a†pn

|0〉

Because creation operators commute, we have that particles are symmetric under exchange(|p,q〉 = |q,p〉), which means that this particle species are bosons.

Our full Hilbert space is then the space of all particles spanned by states of the form |0〉 , a†p |0〉 , a†pa†q |0〉 , . . .This is what we call Fock space. We can create another interesting operator on this space, calledthe number operator, defined as

N =

∫d3p

(2π)3a†pap

This has the property that it counts particle states. This is because since

N |p1 . . .pn〉 =∫

d3p

(2π)3a†papa

†p1a†p2

. . . a†pn|0〉

=

∫d3p

(2π)3a†p([ap, a

†p1] + a†p1

ap)a†p2. . . a†pn

[ket0

=

∫d3p

(2π)3a†p(2π)

3δ(p− p1)a†p2. . . a†pn

|0〉

+

∫d3p

(2π)3a†pa

†p1apa

†p2a†pn

|0〉

= |p1 . . .pn〉+∫

d3p

(2π)3a†pa

†p1apa

†p2. . . a†pn

|0〉

= N |p1 . . .pn〉

The last line is obtained by repeating the procedure until ap hits |0〉. To show that particlenumber is conserved in our Fock space, we need to show that this operator commutes with theHamiltonian. We simply write this out as

[H,N ] =

∫d3pd3q

(2π)6ωq[a

†qaq, a

†pap]

=

∫d3pd3q

(2π)6ωq

(a†q[aq, a

†pap] + [a†q, a

†pap]aq

)=

∫d3pd3q

(2π)6ωq

(a†q[aq, a

†p]ap + a†p[a

†q, ap]aq

)=

∫d3pd3q

(2π)6ωq(2π)

3δ(p− q)a†qap − (2π)3δ(p− q)a†paq

= 0

Hence we have that the particle operator commutes with the Hamiltonian, and thus the particlenumber is conserved.


Note momentum eigenstates aren’t localised. We can write a localised state via

|x〉 =∫

d3p

(2π)3e−ip·x |p〉

More generally, we describe a wave packet partially localised in both position and momentumspace. We can write

|ψ〉 =∫

d3p

(2π)3e−ip·xψ(p) |p〉 , ψ(p) ∝ e−p

2/2m2

Neither |x〉 nor |ψ〉 are eigenstates of H like in usual quantum mechanics.

2.9.1 Relativistic normalisation

From now on, let’s define our vacuum state to be normalised such that 〈0|0〉 = 1. Let’s do asimple thing, and just contract two momentum eigenstates, and see what we get;

〈p|q〉 = 〈0| apa†q |0〉 = 〈0| [apa†q] |0〉 = (2π)3δ(p− q)

But, this our delta function Lorentz invariant? If not, it would be in our interest to come upwith some expression that is Lorentz invariant, since manifest Lorentz invariance is easy (sincethings hold in all frames). We know definitely how four momentum transforms under a Lorentzboost, this is just given by

pµ → Λµνpν = p′µ

This induces a transformation on our actual eigenstates, where we map |p〉 → |p′〉. In our usualnotion of quantum theory, we would want |p〉 to be related to |p′〉 by a unitary transformation,because this forces the contraction to be Lorentz invariant. So we would want

|p〉 →∣∣p′⟩ = U(Λ) |p〉

(2π)3δ(p− q) = 〈p|q〉 → 〈p|U †(Λ)U(Λ) |p〉 =⟨p′∣∣q′⟩ = (2π)3δ(p′ − q′)

But we can’y assume that this holds in our quantum field theory case - we’ve made no studyinto how states change under relativistic transformations. So, the best we can do is to startfrom a different perspective, and use objects we know which are Lorentz invariant to constructa Lorentz invariant dot product.

Notice that our identity transformation

1 =

∫d3p

(2π)3|p〉〈p|

is Lorentz invariant, despite the individual states |p〉 not being Lorentz invariant. This meansthat or integration expression, or measure, is not Lorentz invariant. Thus, a good place tostart is to find a Lorentz invariant measure.

Theorem. We have that the measure∫d3p

2Ep, Ep =

√p2 +m2

is an invariant measure.


Proof. We construct this item from a chain of Lorentz invariant arguments. First, we assertthat the measure ∫

d4p

is Lorentz invariant. We know that this is true because when we transform 4-vectors d4p→ d4p′,the factor is changed by det(Λ). But since Lorentz transformations lie in the special orthogonalgroup SO(1, 3) this is just 1. In addition. we know that p2 = pµp

µ = p20 − p2 −m2 is a Lorentzinvariant quantity, hence putting this inside a delta function and sticking it inside the integrandalso gives us a Lorentz invariant quantity! So∫

d4p δ(p20 − p2 −m2

)is Lorentz invariant.

This all seems a bit rabbit-out-of-the-hat at the moment, but bear with us. Let’s separate ourthe time and space terms of our integral like this∫

d3p

∫dp0 δ

(p20 − p2 −m2

)=

∫d3p

∫dp0 δ (f(p0)) , where f(p0) = p20 − p2 −m2

Now, we appeal to an identity we learned in our undergraduate years which allows us to workwith delta functions of functions.

δ(f(x)) =∑xi

δ(x− xi)

|f ′(xi)|where xi are roots of f .

Now, our roots of f are atp±0 = ±

√p2 +m2

And, our derivative if just f ′(p0) = 2p0. Thus, our full function for our delta function is

δ(p20 − p2 −m2

)=δ(p0 −

√p2 +m2)

2√p2 +m2

+δ(p0 +

√p2 +m2)

2√p2 +m2

=δ(p0 − Ep)

2Ep+δ(p0 + Ep)

2Ep

Now, we don’t have to include both of these delta functions in our integral. It’s okay just tochoose p0 > 0, since we cannot Lorentz boost from positive time into negative time. Thus, ournotion of sign choice for p0 is also Lorentz invariant. So, selecting just the positive componentgives us ∫

d3p

∫dp0δ

(p20 − p2 −m2

)∣∣∣∣p0>0

=

∫d3p

∫dp0

δ(p0 − Ep)

2Ep=

∫d3p

2Ep

since the delta function absorbed into the∫p0 just goes to one.

With this fact, we motivate the concept of a relativistically normalised state; which we denoteas

|pµ〉 := |p〉 =√

2Ep |p〉This is manifestly Lorentz invariant because

I =

∫d3p

2Ep|p〉〈p|

is Lorentz invariant, and so is the measure. Hence the integrand must be Lorentz invariant.Thus, our normalised delta function looks like

〈p|q〉 = (2π)32√EpEqδ

3(p− q)


2.10 Quantising the free complex scalar field

Let’s explore what we get when we play around with a complex, free-like Lagrangian. We havethe Lagrangian attached with a mass term

L = ∂µψ∗∂µψ − µ2ψ∗ψ

This Lagrangian is associated with the following Euler-Lagrange equations

∂µ∂µψ + µ2ψ = 0

∂µ∂µψ∗ + µ2ψ∗ = 0

Now since the complex conjugate of our field ψ is not the same as ψ∗, when we promote ψ toa Schrodinger picture operator there’s no notion which suggests that the operator should beHermitian. So, we write out a normalised expansion that’s similar to what we had for a scalarfield but, add different operators inside. We have

ψ(x) =

∫d3p

(2π)31√2Ep

(bpeip·x + c†pe

−ip·x)

ψ(x)† =

∫d3p

(2π)31√2Ep

(b†pe−ip·x + cpe

ip·x)

In addition, we motivate the same expressions for our conjugate momenta. Our conjugatemomenta classically is found by differentiating the Lagrangian:

π =∂L∂ψ

= ψ∗

With the same Heuristic we applied to find conjugate momenta for our scalar field, we tentativelywrite that

π =

∫d3p

(2π)3i

√Ep

2(b†pe

−ip·x − cpe−ip·x)

π† =

∫d3p

(2π)3(−i)

√Ep

2(bpe

ip·x − c†peip·x)

Here, we’ve defined two operators b, c, which have the interpretation of creating and annihilatinga particle and antiparticle respectively. With this expansion one can check that the commutationrelations

[φ(x), π(y)] = iδ(x− y), [φ(x), π(y)†] = 0, all other relations derived from this

are equivalent to the commutation relations

[bp, b†q] = (2π)3δ(p− q), [cp, c

†q] = (2π)3δ(p− q), all other relations 0

We interpret the operators as particle and anti particle creation. These particles have spin 0, andthe same mass since we have a µ2 coupling term in there. Since a real scalar field obeys ψ∗ = ψ,


our interpretation of a particle created by a real scalar field is that it its’ own antiparticle. UsingNoether’s theorem, one can show that our conserved charge which is obtained by the symmetryof a phase rotation

ψ → eiθψ

gives rise to the conserved charge

Q = i

∫d3x ψ∗ψ − ψ∗ψ

Using the relation we derived earlier for conjugate momentum, this can be written as

Q =

∫d3xπψ − ψ†π†

Expanding this in terms of the creation and annihilation operators b, c, with normal orderingwe have that a conserved charge is

Q =

∫d3p

(2π)3(c†pcp − b†pbp) = Nc −Nb

This counts the number of of particles minus the number of anti-particles.


2.11 Introducing time with the Heisenberg picture

So far, in the Schrodinger picture, we aren’t entirely confident that our operators are manifestlyLorentz invariant. In the Schrodinger picture, operators don’t depend on time but states dosince the evolve according to the equation

id |p〉dt

= H |p〉 = Ep |p〉

This however, is entirely equivalent to placing time dependence on operators instead of states,and force operators to evolve according to

OH(t) = eiHtOse−iHt

where Os is our Schrodinger picture operator which coincides with our Heisenberg picture oper-ator at t = 0. At a fixed time, substituting this expressions now give us fixed time commutationrelations. Recall that

[UAU−1, UBU−1] = U [A,B]U−1

Application of this identity yields us the fixed time commutation relations

[φ(x, t), φ(y, t)] = [π(x, t), π(y, t)] = 0, [φ(x, t), π(y, t)] = iδ(x− y)

Taylor expanding out we have

OH(t+ δt) = 1 + δti[H,OH(t)] + . . .

so operators obey the Heisenberg equations of motion, that

idOHdt

= i[H,OH ]

We can use this to calculate our time derivatives of our operators, to get π and φ. We can thenrelate π = φ to obtain the Klein Gordon equation. For notation purposes, we write

ψ(x) := φ(x, t)

We can check, by computing the commutator [ap, e−iHt], and [a†p, e

−iHt], that

eiHtape−iHt = e−iEptap

eiHta†pe−iHt = eiEptap

Hence, we have

φ(x) =

∫d3p

(2π)31√2Ep

(ape−ip·x + a†pe

ip·x)

with this we can calculate φ amd π. (Insert section to check that this recovers the relativisticKlein Gordon equation). We promote φ(x), π(x) to adopt time dependence, so we write

φ(x, t) = φ(x)


and similiarly with π(x).

Our Hamiltonian which didn’t depend on time was

H =

∫d3x

(2π)3ape

ip·x + a†pe−ix·p

but to calculate the time dependent Hamiltonian we’d need to calculate

eiHtape−iHt, eiHtape

−iHt

The first operator is only non-zero when we contract it on both sides for the configuration

〈0| eiHtape−iHt |p〉 ,

and this gives us〈0| eiHtape−iHt |p〉 = e−iEpt

thus our conjugated operator is

eiHtape−iHt = ape

−iEpt

and similarly we have thateiHta†pe

−iHt = a†peiEpt

Since p = (Ep,p), altogether this gives our expression for our scalar field

φ(x) =

∫d3x

(2π)31√Ep


ip·x)


3 Causality in QFT

We still have that potentially, a hint of Lorentz variance because the commutation relations ofφ and π are equal time relations. If we shift to a differed time, this may not hold. Our previousfixed time commutation relations involved expressions of the form

[O2(~x, t),O1(~y, t)]

but, if we transform the coordinates with a Lorentz transformation, we would get this turninginto

[O1(~x′, t′),O2(~y

′, t′′)]

where t′, t′′ are not necessarily the same since ~x, ~y are different. What about arbitrary spacetime separations? The real requirement of causality is that all space like separated operatorscommute

[O(x1),O(x2)] = 0, ∀(x− y)2 < 0

The reason why commutation is synonymous with the fact that operators don’t affect one an-other, comes from the fact that they have a shared eigenbasis. Suppose we have A,B, hermitianoperators which commute. Then, we have that they share an eigenbasis which we label as |i〉,with the property that

A |i〉 = λi |i〉 , λi ∈ RB |i〉 = µi |i〉 , µi ∈ R

In particular, this means that if we expand a given quantum state in terms of the eigenbasis, weget

|ψ〉 =∑i

ai |i〉 =⇒ 〈ψ|BA |ψ〉 =∑i

aiλiµi

where in the last sum we used the orthogonality property of the eigenstates. However, noticethat this is the same by symmetry of calculating 〈ψ|AB |ψ〉. Hence, applying an operatorin the process of measuring the other one doesn’t change our outcomes. Thus, we have nocausality.

If we define ∆(x − y) = [φ(x), φ(y)], we’d like to check if this holds for space-time separa-tion.

[φ(x), φ(y)] =

∫d3p d3p′

(2π)6√4EpE′p

[a~p, a

†~p′ ]e

i(p·x−p′·y) + [a†~p, a~p′ ]ei(p·x−p′·y)

which is just

[φ(x), φ(y)] =

∫d3p

2Ep(2π)3

e−ip·(x−y) − eip·(x−y)

Well, what do we know about this function? We know that this is Lorentz invariant becausethe measure is, and so is the integrand. If x− y is space-like, it vanishes because we can take aLorentz transformation to y−x in the first term, giving 0. Another way to say this is that if wehave space-like separated events, then we don’t care about breaching causality in time, so wecan freely transform events in this way. We can’t however do this for time like events becausethat would mean time reversal. This integral doesn’t vanish, however, for timelike separations.


If we have time like separated events, we can Lorentz transform them such that they’re constantin space.

[φ(~x, 0), φ(~x, t)] =

∫d3p

(2π)32Ep(e−iEpt − eiEpt) 6= 0

We assert that this is non-zero. To show that this is non-zero, we can evaluate this expressionexplicitly. Let’s do the first term.∫

d3p

(2π)32Epe−iEpt ∼

∫dp

p2√p2 +m2

e−i√p2+m2t

=

∫ ∞m

dE√E2 −m2e−iEt

In the second line, we expressed the integral in polar form using p as a radial coordinate. Inthe second line, we used a change of variables and set E =

√p2 +m2. To evaluate the final

integral, we need to analytically continue it to C. Since we have a square root, this induces abranch cut at E = m. By the residue theorem, our whole contour evaluates to zero. However,the part on the x-axis is what we’d like to calculate. To do this, we calculate the arc, as well asthe contour which is show as the vertical line. Our integral along the arc is given by

CR =

∫dz√z2 −m2e−izt, z = m+Reiθ

If we do this integral in polars, for large R this integral goes like

CR ∼ i

∫dθeiθR

√R2e−imteRe

iθt

∼ e−imti

∫dθR2eiR cos θteR sin θt

Now, since we’re taking the contour in −π2 < θ < 0, and ignoring our oscillating eiR cos θt term,

this integral goes like∼ e−imti

∫dθe−θR2e−εRt

The negative epsilon comes from the fact that we’re taking a negative contour. Due to thenegative in the exponential, this term is bounded. So, CR ∼ e−imt. A similar analysis of thedownwards vertical contour shows that it grows in the same way. Hence, we have that∫

d3p

(2π)22Epe−Ept ∼ e−imt

Thus, our time-like separated integral goes like[φ(~x, 0), φ(~x, t)] ∼ e−imt − eimt

This is non-zero.

At equal times, we have

[φ(~x, t), φ(~y, t)] =

∫d3p

(2π)32Ep(ei~p·(~x−~y) − e−i~p·(~x−~y)) = 0

where in the second term we apply a Lorentz transformation put ~x− y → − ~y − x. We can do aLorentz transformation from within this integral because the whole integral was Lorentz invariantin the first place. Thus, the second term cancels, which agrees with equal time commutationrelations, since it goes to zero.


Re(E)Im(E)

m

Figure 1: Our contour integral which we use to calculate our timelike integrals

3.1 Propagators

If we prepare a particle at a point y, what’s the probability it ends up at x? Well, we couldidentify the creation of a particle at the point x, with φ(x). It’s instructive to view the similaritieswith QM. We have that 〈x|X |p〉 = eip·x in regular quantum mechanics, but we have

〈0|φ(x) |p〉 = e−ip·x

in QFT. Hence, we can somewhat identify φ(x) |0〉 as the position presentation of a createdparticle.

Our amplitude for a particle to propagate at x then travel to y is then

〈0|φ(x)φ(y) |0〉 =∫

d3pd3q

(2π)6√4EpEq

〈0| [a~p, a~p′ ] |0〉 e−i(p·xi−p′·y

which is=

∫d3p

(2π)32Epe−ip·(x−y) := D(x− y)

This quantity we call the propagator. For spacelike separations (x− y)2 < 0, we can show thatit decays as

D(x− y) ∼ e|~x−~y|

The quantum field leaks out of the light cone. But, we just saw that the space like separationscommute;

∆(x− y) = [φ(x), φ(y)] = D(y − x)−D(x− y) = 0, ∀(x− y)2 < 0

There’s no Lorentz invariant way of ordering the events and a particle can just as easily travelfrom x → y as it can from y → x. In a measurement, these amplitudes cancel. For a C scalarfield, we have that

[ψ(x), ψ(y)] = 0, outside the light cone

The amplitude to go from x→ y cancels the one for an antiparticle to go from y → x. For realfields, the particle is the antiparticle.


3.2 The Feynman Propagator

Our Feynman propagator is given by

∆F (x− y) = 〈0|Tφ(x)φ(y) |0〉 =

〈0|φ(x)φ(y)〉 x0 > y0

〈0|φ(y)φ(x) |0〉 x0 < y0

Our claim is that we can write this propagator in the compact form

∆F =

∫d4p

(2π)4i

p2 −m2e−ip·(x−y)

This is Lorentz invariant. Note that this is ill defined because for each p, the integral over p0has a pole where (p0)2 = ~p2 +m2. We need a prescription here, which means that we resolvethe ambiguity of the integral by choosing a specific contour. In our case, we need a contourthat recovers the exact expression above. We define the integration contour to be a straight linewhich gues under the residue at −Ep and above the residue at Ep. (draw diagram here). Noticethat

1

p2 −m2=

1

(p0)2 − E2p

=1

(p0 − Ep)(p0 + Ep)

The residue of the pole at p0 = ±Ep is ± 12Ep

. When x0 > y0, we close the contour in thelower half plane. When p0 → −i∞, e−ip0·(x0−y0) → e−∞ → 0. So, the lower contour doesn’tcontribute. Hence, we have

∆F (x− y) =

∫d3p

(2π)32Ep(−2πi)ie−iEp(x0−y0)+i~p·(~x−~y) =

∫d3p

(2π)32Epe−ip·(x−y)

This is D(x − y), which agrees which what we want. We had the minus sign due to clock-wise.

When x0 < y0, we close the contour in the upper half place to get

∆F (x− y) =

∫d3p

(2π)4(2πi)

(−2EpieiEp(x0−y0)+i~p·(~x−~y)

We flip the sign in the second term to get this as

=

∫d3p

(2π)31

2Epe−iEp(y0−x0)−i~p·(~y−~x) =

∫d3p

(2π)31

2Epe−ip·(y−x) = D(y − x)

Instead of this laborious process of specifying the contour, we can instead do the iε prescriptionwhere we set

∆F (x− y) =

∫d4p

(2π)4ie−ip·(x−y)

p2 −m2 + iε

This propagator ∆F is the Green’s function of the Klein Gordon operator. We should getthat

(∂2t −∇2 +m2)∆(x− y) =

∫d4p

(2π)4i

p2 −m2(−p2 −m2)e−ip·(x−y)


which is= −i

∫d4p

(2π)4e−p·(x−y) = −iδ(x− y)

It can be useful in some circumstances to pick other contours. For example, the retarded Green’sfunction (diagram of camel humps) over both poles. This is when

∆F (x− y) =

[φ(x), φ(y)] x0 > y0

0 y0 < x0

This is useful if we start with an initial field configuration and look at the evolution in thepresence of a source

∂µ∂µφ+m2φ = J(x)

The Feynman propagator ∆F is the most useful object in QFT.


4 Interacting fields and the approach with perturbation the-ory

4.1 Why do we need interacting fields and what are ’valid’ conditions forperturbation expansions?

This section follows David Tong’s notes, and some of Peskin and Schroesder

So far our Lagrangian L has only been quadratic in the field φ. We’ve referred to this as ourfree theory; it gives rise to the second order, linear PDE

∂µ∂µφ+m2φ = 0

which we can solve and quantise exactly. However, even though our analysis of this has consid-ered multi particle states, we have yet to consider theories in which there particles interact. Forexample, scattering scenarios.

Interaction terms can be written as higher order terms to our free Lagrangian;

L =1

2∂µφ∂

µφ− 1

2m2φ2 −

∑n=3

λnφn

n!

The λn coefficients are called coupling constants, because they couple our free theory withinteraction terms. For example, we could choose our constants to give rise to what we call φ4theory, with λ4 = λ and λn zero otherwise:

L =1

2∂µφ∂

µφ− 1

2m2φ2 − λ

4!φ4

However, if we have hopes of trying to solve our theory for expansions like this in a perturbativefashion, we then need to ensure that the λn terms are small. In other words, λn 1. But whatdoes this even mean? For something to be smaller than a number, it better first be dimensionless.So, let’s try figure out what the dimensions of our objects in question are.

Our action isS =

∫d4xL, [S] = 0

Our action should be dimensionless. We’ve denoted our mass dimension of a quantity Q as [Q],which is why we wrote [S] = 0 as the above. d4x is an integral, which means it has dimensions of4 in length terms. But, in natural units, we have that [M ] = [L]−1. So, we have that [d4x] = −4.This implies that our Lagrangian must have dimensions [L] = 4 so compensate.

So, what is [φ] ? To find this out, we look at our kinetic term ∂µφ∂µφ. Since we have two length

differentials which contribute a −2 to the dimension, we can then infer that [φ] = 1.

This means that since [L] = 4, our dimension of [λn] must satisfy [λn] + n = 4 =⇒

[λn] = 4− n

Let’s go case by case


• If we have n = 3, our dimension of λ3 is 1. This means, to force a dimensionless parameterwhat we actually require is that λ3/E 1, since E has the same dimensions as mass. Thismeans that a perturbation of this form is only valid at high energies! High energy scatteringis one case where we could apply this theory. This is called a relevant perturbation,because the perturbation is significant at low energies.

• In the case where d = 4, our mass dimensions [λ4] = 0, so we require that the perturbationis small when λ4 1. This, the boundary case, is called a marginal operator. We candeal with the infinities that come from this perturbation, so we call it a renormalisabletheory.

• The perturbation λnφn/n! are called irrelevant operators. This is because their associateddimensionless parameter is λEn−4, so the perturbation expansion is only valid at lowenergies. Their effect is only significant at high energies. These are non re normalisabletheories.

Let’s look at some of the properties of φ4 theory. This has Lagrangian

L =1

2∂µφ∂

µφ− 1

2m2φ2 − λφ4

4!, λ 1

We can already see from first glance that our Hamiltonian is going to contain a bunch of extraterms which are inherited from the last term. This means that our number operator

N =

∫d3p

(2π)3a†~pa~p

will satisfy [H,N ] 6= 0. This already tells us that particle number will not be conserved. We cansee this explicitly by expanding the last term, which will yield combinations of the form

· · ·∫a†~pa†~p′a†~p′′a†~p′′′

and other terms which create or destroy particles.

Another example is scalar Yukawa theory. THis is what we get when we lump together a a Cscalar field, a real scalar field, as well as an interaction term.

L = ∂µψ∂µψ − µ2ψψ∗ +

1

2∂µφ∂

µφ− m2

2φ2 − gψψ∗φ

This Lagrangian has been used for example in the interaction of mesons. Now, our g term hereis combined with three fields. This means that it’s like a λ3 term.

This means it’s a relevant operator. Hence, working perturbatively works for low energies. Ina relativistic setting, E > m. So, we can make perturbations small by taking g m. Wecan start to make preliminary statements about this Lagrangian by observing that its invariantunder the transformation

ψ → eiθψ, ψ∗ → e−iθψ∗, φ→ φ

This means that our Noether current for this symmetry exists. This Noether current Q is ourparticle minus anti particle number, and we have that [Q,H] = 0. Thus, with this Lagrangianthe number of particles minus the number of antiparticles stays constant.

We have no conservation however, for scalar particles generated by φ.


4.2 The interaction picture

This section combines Prof. B. Allanach’s lecture course, page 53 of QFT 1 Notes, Universityof Heidelberg, and the interaction section of Peskin and Schroesder

The whole point of this section is to reduce solutions in our full picture, and express themperturbatively in terms of free solutions. We may wish to compute a term like

〈Ω|Tφ(x)φ(y) |Ω〉

where we’re contracting by the ground states of the full theory, and using solutions of the fulltheory. Our aim is to do find out what this is in terms of free theory objects like |0〉!.

4.2.1 The Heisenberg and Schrodinger Picture

In quantum mechanics, our Schrodinger picture is when states evolve in time, and not operators.The obey the equation

id |ψ〉sdt

= H |ψ〉sOur operators, Os, are time independent. On the other hand, we can also choose to be inthe Heisenberg picture, where operators evolve instead. We define operators in the Heisenbergpicture to obey

OH = eiHtOse−iHt, |ψ〉H = eiHt |ψ〉s

This ensures consistency; one can verify that

〈ψ|H OH |ψ〉H = 〈ψ|S OH |ψ〉SWe see from the above that we can switch between our Schrodinger and Heisenberg pictures bymultiplying by an appropriate time evolution factor. Heisenberg operators obey the Heisenbergequation of motion

dOH

dt= i[H,OH ]

4.2.2 The Interaction picture is a hybrid!

In the interaction picture, however, is a hybrid of the Schrodinger and Heisenberg picture.We separate out the free and interaction parts of the Hamiltonian as

H = H0 +Hint

This split is arbitrary, but we choose H0 typically as the section which we can solve exactly.

Example 1. In φ4 theory, our Lagrangian and Hamiltonian terms associated with our interac-tion terms are

Lint = − λ

4!ψ4, Hint = −

∫d3xLint =

∫d3x

λ

4!φ4

Our Hamiltonian associated with the free term H0 is given by

H0 =

∫d3x

1

2π2 +

1

2(∇φ)2 + 1

2m2φ2


Our definition of our interaction picture is similar to the Heisenberg picture, exact that we areevolving states with the free Hamiltonian. This is in place of the full Hamiltonian. Theinteraction picture for a general operator is

OI = eiH0tOse−iH0t, =⇒ φI(x) = eiH0tφs(~x)e

−H0t

There’s a slightly more specific way to handle this however, which makes use of a generalreference time t0. To construct the interaction picture, choose some t0 in the Heisenberg picture,then evolve from t0 → t (philosophically, however, our Schrodinger picture is the same as ourHeisenberg picture at fixed time).

φI(t, ~x) = eiH0(t−t0)φH(t0, ~x)e−iH0(t−t0), πI(t, ~x) = eiH0(t−t0)πH(t0, ~x)e

−iH0(t−t0)

4.2.3 We can Fourier expand Interaction picture states!

Now, φI obeys the same commutation relations as φ did in the free theory: it obeys the samedynamics in the Heisenberg picture

φI = i[H0, φI ], πI = i[H0, πI ]

Theorem. We have that φI also obeys the Klein-Gordon equation.

(∂2 +m2)φI(x) = 0

Proof. Observe that for φ4 theory, our interaction term Hamiltonian is a function of just φ.Thus, we have that our commutation relations remain exactly the same for a operators at fixedtime. Thus, we have the relations for our conjugate momenta at fixed time, despite having anadded Lagrangian

Lkinetic =1

2∂µφ∂

µφ =⇒ π(t0, ~x) = φ(t0, ~x)

We use this fact to our advantage in finding what φI(t, ~x) is. Substituting our definitions, wecompute the Heisenberg equation motion

φI(t0, ~x) = i[H0, φI(x, )]

= i[H0, eiH0(t−t0)φ(t0, ~x)e

−iH0(t−t0)]

= ieiH0(t−t0)[H0, φ(t0, ~x)]e−iH0(t−t0)

= ieiH0(t−t0)(−iπ)(t0, x)e−iH0(t−t0)

= πI(t, x)

In the above, we used the commutation relation that [φ(~x), π(~y)] = iδ(~x − ~y) to calculate[H0, φ(t0, ~x)]. We can play exactly the same game with

πI(t, x0) = i[H0, πI(t, ~x)]

= ieiH0(t−t0)[H0, π(t0, ~x)]e−iH0(t−t0)

= ieiH0(t−t0)(−i∇2φ(t0, ~x) + im2φ(t0, ~x))ie−iH0(t−t0)

= ∇2φI(t, ~x)−m2φI(t, ~x)


Equating the above equation means that we deduce the Klein-Gordon equation for states in theinteraction picture

∂µ∂µφI(x) +m2φI(x) = 0

This means we could apply the exact same procedure to quantise this thing as we did for thefree case.

4.2.4 Operators defined from the Interaction picture obey the same commutationrelations

Since, in the interaction picture we evolve this state by conjugating it either side with eiH0t.This gives us the same result that we has for the free field. We have that

φI(x) =

∫d3p

(2π)31√2Ep

(a~pe−ip·x + a†~pe

ip·x)

There’s an important point to be made here. Philosophically, these are different annihilationand creation operators that in our free case, soreally, we should be writing,

φI(x) =

∫d3p

(2π)31√2Ep

(a∗~pe−ip·x + a∗†~p e

ip·x)

In this case, we denote a∗~p, a∗†~p as operators which are in the interaction picture, not necessarily

in the free picture.

Despite this however, we still have that as before, these annihilation and creation operatorssatisfy our commutation relation as we had in the free theory, where

[a~p, a†~p′ ] = (2π)3δ(~p− ~p′)

A natural question to ask then is how these interaction picture creation and operators actcommute with the free-field part of the Hamiltonian, as well as act on the ground state. If φ, π areoperators associated with our full theory, and H0 is our free Hamiltonian, then particular

H0 = eiH0(t−t0)H0e−iH0(t−t0) = eiH0(t−t0)

(∫d3x

1

2π2 +

1

2(∇φ)2 + 1

2m2φ2

)e−iH0(t−t0)

=

∫d3x

1

2π2I +

1

2(∇φI)2 +

1

2m2φ2I

The upshot of doing this is that we can transfer our previous analysis of raising and lower-ing operators to our interaction picture. In particular, we have that our interaction pictureannihilation and creation operators obey

[H0, a∗~p] = − ~Epa

∗~p, [H0, a

∗†~p ] = E~pa

∗†p

This means that we have a state |0〉 such that

H0 |0〉 = 0, a∗~p |0〉


where H0 is the free part of our full Lagrangian. Again, a philosophical point to be madeis that the free part of our interacting Lagrangian in of the same from as just in the case of apurely free Lagrangian. The whole point of the interacting picture was to introduce a new statewhich obeyed the Klein Gordon relation, hence having a Fourier mode expansion, and hencehad the same commutation relations as in the free case.

In addition, a~p annihilates the ground state of the free theory part in the Hamiltonian, soa~p |0〉 = 0.

An interesting question to would be whether the physical ground state of the free part of thefull Hamiltonian is the same as if we were to treat the full Hamiltonian by itself.

We have a diagram of our logical flow below

Get a Lagrangian L free + L interaction

Construct πsame as free case when L interaction doesn’t depend on ∂µφ

Construct H′ ’free’ Heisen. to show KG equationφI solves KG

KG =⇒ possible ap, a†p expansionderive commutation relations with H0

annihilation and ground state |0〉 definedH0 |0〉 = 0, ap |0〉 = 0


4.2.5 Introducing the time unitary operator

Let’s reparse things for the sake of generality for a bit. There’s nothing special in our choice oft = 0. We could have easily just fixed a time t0 as our reference point, and in our Heisenbergoperator our states evolve as

φ(x, t) = eiH(t−t0)φ(x, t0)e−iH(t−t0)

In the interaction picture, we set our perturbation constant λ = 0, and thus get the resultthat

φI(x, t) = eiH0(t−t0)φ(x, t0)e−iH0(t−t0)

Now, how do we switch the between the two pictures? Well, we invert our interaction picturethen multiply like so:

φ(x, t) = eiH(t−t0)e−iH0(t−t0)φI(x, t)eiH0(t−t0)e−iH(t−t0)

Now, if we define U(t, t0) = eiH0(t−t0)e−iH(t−t0) The above equation simply reads as

φ(x, t) = U(t, t0)−1φI(t, x)U(t, t0)

Our operator U is a unitary time evolution operators. It has the property that

U(t1, t2)U(t2, t3) = U(t1, t3), U(t, t) = 1

It also satisfies the equation that

i∂U(t, t0)

∂t= eiH0(t−t0)(H −H0)e

−iH(t−t0)

= eiH0(t−t0)Hinte−iH(t−t0)

= eiH0(t−t0)Hinte−iH0(t−t0)eiH0(t−t0)e−iH(t−t0)

= HI(t)U(t, t0)

In this case HI(t) is our interaction picture Hamiltonian with

HI(t) = eiH0(t−t0)Hinte−iH0(t−t0)

If HI were just a function, we could solve this by setting

U = exp

[−i∫ t

t0

HI(t′)dt′

]but this doesn’t work if there are ordering ambiguities. This is because we have that [HI(t

′),HI(t′′)] 6==

0. However, our differential equation for U(t, t0) implies that it satisfies the equation

U(t, t0) = 1− i

∫ t

t0

dt′HI(t′)U(t′, t0)

We substitute this back into itself to get the infinite series expansion

U(t, t0) = 1 + (−i)∫ t

t0

dt′HI(t′) + (−i)2

∫ t

t0

dt′∫ t′

t0

dt′′HI(t′)HI(t

′′) + . . .


We can already intuitively see that this satisfies the differential equation. Also, the boundarycondition that U(t0, t0) works out.

For the ranges of integration, the HI product is automatically time ordered! We’ll see why thisis the case - it has to do with clever relabelling and the diagram we’ll show below. Let’s lookspecifically at the second order term∫ t

t0

dt′∫ t′

t0


′′)

0 t0

t

t′′

t′

Figure 2: The integration variables add up to a square!

With this choice of dummy variable our integration variables are defined so that

t ≥ t′ ≥ t′′ > t0

This is shown in the figure 2. So, by the definition of time ordering, in this range, we havethat ∫ t

t0

dt′∫ t′

t0


′′) =

∫ t

t0

dt′∫ t′

t0

dt′′T (HI(t′)HI(t

′′))

However, integration variables are just dummies, so we can relabel them. If we relabel t′ → t′′,and t′′ → t′, then we have that∫ t

t0

dt′′∫ t′′

t0

dt′HI(t′′)HI(t

′) =

∫ t

t0

dt′∫ t′

t0


′′)

But in this case, our first expression has ranges defined by t ≥ t′′ ≥ t′ > t0. This is the redportion of our square in the diagram. This is also consistent with time ordering, so we can putour time ordering symbol in the integrand so that in this range, HI(t

′′)HI(t′) = T (HI(t

′)HI(t′′)).

Hence, we can write our second term as


∫ t

t0

dt′∫ t′

t0


′′) =1

2

(∫ t

t0

dt′′∫ t′′

t0

dt′T (HI(t′)HI(t

′′) +

∫ t

t0

dt′∫ t′

t0


′′))

)

So, we have the same term in the integrand, T (HI(t′)HI(t

′′)! But, the ranges combined on theright hand side of this expression is just the full square. Hence, we can just integrate over thewhole square to get that that∫ t

t0

dt′∫ t′

t0


′′) =1

2

∫ t

t0

dt′∫ t

t0


′′))

In full generality, we can expand this idea to more integrals and write∫ t

t0

dt1

∫ t1

t0

dt2· · ·∫ tn−1

t0

dtnHI(t1) . . .HI(tn) =1

n!

∫ t

t0

dt1· · ·∫ t

t0

T (HI(t1) . . . HI(tn))

Thus, we can write

U(t, t0) = 1 + (−i)∫ t

t0

dt′HI(t′) +

(−i)2

2

∫ t

t0

dt′∫ t

t0

dt′′THI(t

′)HI(t′′)

This is Dyson’s formula.

U(t, t0) = T exp−i∫ t

t0

dt′HI(t′)

Alternatively was have that, using the Lagrangian instead, that

U(t, t0) = T exp

i

∫ t

t0

d3xdt′LI(t′)

For scalar Yukawa theory, this would be

T exp

−ig

∫d4xψ∗ψφ

This is something of a formal result. We expand this to finite order to get some results forscattering amplitudes which we will derive below in the next section.

4.3 Scattering

The time evolution used in scattering theory is called the ’S-matrix’, where

S = limt→∞,t0→−∞

U(t, t0)

The initial state |i〉 and final state |φ〉 are some sense ’far away’ form each other and theinteraction. We assume that |i〉 , |f〉 behave like free particles; they’re eigenstates of H0 . THeamplitude is

limt→∞,t0→−∞

〈f |U(t, t0) |i〉 = 〈f |S |i〉

The latter expression is called the ’S-matrix’ matrix element.


4.3.1 An example with Yukawa theory

Going back to scalar Yukawa theory, the interaction Hamiltonian is

HI = gψ∗IψIφI

We model the creation of mesons with φ, and the creation of nucleons or anti nucleons with ψ.Concentrating on the creation and annihilation operators in the Fourier mode expansion,

φ ∼ a~p + a†~p

These things destroy and create mesons respectively. We have that

• ψ ∼ b~p + c†~p which destroys a nucleon and create an anti nucleon respectively.

• ψ∗ ∼ b†~p + c~p which creates a nucleon and destroys an anti nucleon.

With this interaction, we have the commutation relations

[a~p, a†~p′ ] = [b~p, b

†~p′ ] = [c~p, c

†~p′ ] = (2π)3δ(~p− ~p′)

All other commutation relations are zero. To our first order our interaction, expanding out ourinteraction Hamiltonian in g, we might have an integral term like∫

dp′dp′′dp′′′b†~pc†~p′a

How we we interpret this? Well, we could interpret this as a term which destroys a meson, thencreates a nucleon and anti-nucleon.

If we expand to second order in g, we find more involved terms which stem from our integral.For example, we could have the term ∫

(c†b†a)(bca†)

This is a two stage process. First, from our first bracket term we annihilate a nucleon and ananti-nucleon, and then create a scalar meson. Then, we annihilate that scalar meson an thencreate a nucleon and anti-nucleon. Thus, this is nucleon and anti-nucleon scattering, and isrepresented by the process

ψψ → φ→ ψψ

Thus, this term specifically contributes to nucleon and anti-nucleon scattering.

4.3.2 First order term in meson decay

In this part, we will be focusing on probability amplitudes that arise from certain processes.Let’s focus specifically on the case of our first order interaction with a meson φ decaying into anucleon and anti-nucleon

φ→ ψψ


Let’s give a momentum to our meson, ~p. Treating this as an initial state, we assume that wecan use the ground state of our free theory, |0〉, as a springboard to excite states from. Thus,we write down the momentum eigenstate for our initial state as

|i〉 =√2Epa

†~p |0〉

If we assign our final state |f〉 to be a nucleon and anti nucleon with momenta ~q1, ~q2, then wehave

|f〉 =√4Eq1Eq2b

†~q1c†~q2 |0〉

Our scattering amplitude is given by

〈f |S |i〉 = 〈0| bca† |0〉 − ig 〈f |∫d4xψ∗I (x)ψI(x)φI(x) |0〉+O(g2)

The zeroth order term is just zero, because a†, c commute. Once we commute the c past a†,it hits our vacuum state and sends it to zero. The form of the first order term deserved someexplanation. Since we’re taking the limit as t → ∞ and t0 → −∞, the form of our first orderterm is

〈f |∫ ∞−∞

dt

∫d3xψ∗I (x)ψI(x)φI(x) |i〉

we can compose the integrals together to get∫d4x. Let’s go slowly. We expand on the right

hand side our |i〉 term to get that

〈f |S |i〉 = −ig 〈f |∫d4xψ∗(x)ψ(x)

∫d3k

(2π)3√2Ek

(a~ka†~pe−ik·x + a†~k

a†~pe−k·x

)|0〉

=

∫d4x d3k 〈f |ψ∗(x)ψ(x) 1

(2π)3√2Ek

(a~ka†~pe−ik·x + a†~k

a†~pe−k·x

)|0〉

Now, since our a†~k commutes with b, c, b†, c†, we have that the second term trivially commutespast the ψ and ψ∗ term, and it the part goes to zero. As for the first term, we commute a~k pastand a†~p, to pick up a (2π)3δ(~k − ~p). Thus we replace

a~ka†~p |0〉 = [a~k, a

†~p] |0〉 = (2π)3δ(~p− ~k) |0〉

This means that our term above is

〈f |S |i〉 = −ig∫d4xd3k1d

3k2(2π)6

√4Eq1Eq2 〈0|

1√4Ek1Ek2

c~q2b~q1

(b†~k1eik1·x + c~k1e

−ik1·x)

(b~k2e

−ik2·x + c†~k2eik2·x

)e−ip·x |0〉

We can create a caricature of this product by ignoring the momenta and integrals. We get astring of operators that look like

∼ cbb†b+ cbb†c† + cbcb+ cbcc†

Now, all strings ending with b vanish since they operate on the |0〉 term to the right.


We’re left with the terms∼ cbb†c† + cbcc†

Now, in the second term, if we commute c past c†, we pick up a delta function. But this leavesa term of the form δcb, and the b still annihilates. So, our only surviving term is

∼ cbb†c† = c~q2b~q1b†~k1c†~k2

Commuting b past b† gives the following term in our integral!

(2π)3δ(~q1 − ~k1)c~q2c†~k2eix·(k1+k2−p)

Once again, commuting c past c† gives us another delta function to include. So the only termpreserved in our integral is

(2π)6δ(~q1 − ~k1)δ(~q2 − ~k2)eix·(k1+k2−p

Thus, performing our integrals over ~k1 and ~k2 gives us our only term at first order whichcontributes to our scattering.

〈f |S |i〉 = −ig 〈0|∫d4x ei(q1+q2−p)·p |0〉 = −igδ4(q1 + q2 − p)

But, this gives a final amplitude which is a delta function in 4 space. This means that for a nonzero scattering amplitude, we require our four momentum to be conserved!

4.4 Wick’s theorem

4.5 Feynman Diagrams

Feynman diagrams are a visual way to represent our complicated integrals we’ve defined above.We draw Feynman diagrams to represent the expansion of 〈f | (S − 1) |i〉 and learn to associatefunctions to them (functions of the four momentum). We draw an external line for each particlein |i〉 and |f〉, assigning a four dimensional 4-momentum to each. We add an arrow for C fieldsto show flow of charge.

The first step is to set up our initial and final states. We choose an in (out) going arrow in |i〉for a particle (antiparticle), and the opposite for final states 〈f |. These are ’external lines’ whichdon’t contribute an integral since we’re just exciting by an operator like b†~p |0〉.

We show what kind of diagrams you might have for different scattering cases in figure 3. Thenext step is to ’fill this in’. We have to be a bit careful here. This is is when we need to beaware of what our interaction term looks like. For the Yukawa potential, we have our interactionHamiltonian

HI =

∫d3xψ(x)∗ψ(x)φ(x)

this means that each set of connecting lines (what we call a vertex) in our Feynman diagramneeds to look like what we have in figure 4. We then fill up this diagram with vertices. The orderof our expansion in g corresponds to how many vertices we have, since each vertex corresponds to


Figure 3: A template diagram. On the left we have φ→ ψψ scattering and on the right we havea template for ψψ → ψψ scattering

Figure 4: This is what we must have at each point

Figure 5: We have some O(g2) diagrams for ψψ → ψψ scattering


an integral∫d4xHI(x) contributed in Dyson’s formula. Once we’ve joined together our vertices,

we then assign a momenta to each line. We have simple examples in the figure.

Our ’internal lines’ which are lines with vertices on both sides are dummy momenta which weintegrate over. On the other hand, we keep external lines with fixed momenta. Each diagramis in 1: 1 correspondence with terms in our expansion. To each diagram, we evaluate totalamplitude using Feynman rules for this problem.

1. Associate a momentum to each internal line, and keep the momenta associated with ex-ternal legs of the problem fixed.

2. Assign a factor of (−ig)(2π)4δ(∑ki) to each vertex, where

∑i ki is the sum of four mo-

menta flowing in to the vertex. If momenta flows out, this gives a negative momentumcontribution.

3. For each internal line with 4-momenta k, write factor∫d4k

(2π)4where D(k2) =

i

k2−m2−iε for φi

k2−µ2+iε for ψ

The total amplitude is the sum of all diagrams at the given order.

4.6 Scattering revisited

Now, to order O(g2)From our first diagram, we have that our contribution is given by the sumof the diagrams in figure 5. Applying our Feynman rules, we have that these diagrams give anintegral of

= (−ig)2∫

d4k

(2π)4i

k2 −m2 + iε(2π)8(δ(p1 − p′1 − k)δ(p2 + k − p′2)

+ δ(p1 − p′2 − k)δ(p2 + k − p′1))

Our final scattering momenta to second order is therefore

i(−ig)2

1

(p1 − p′1)2 −m2 + iε

+1

(p1 − p′2)2 −m2 + iε

(2π)4δ(p1 + p2 − p′1 − p′2)

Note that the meson doesn’t necessarily satisfy the relation that k2 = m2. If its doesn’t it’scalled an off shell or virtual particle. One might also think that the second diagram in figure 5is superfluous, but we need to count this as a distinct diagram since the ψ are indistinguishableunder exchange, and hence obey Bose-Einstein statistics for identical particles.

We can go to even higher order to the point where we introduce loops into our diagrams.

4.7 Amplitudes

In the above, we will always have a factor to impose momentum conservation between final andinitial states. Hence, this fact helps us simply things a bit. We will define the matrix element


M by〈f | (S − 1) |i〉 = iM(2π)4δ(

∑final state particles

pi −∑

initial state particlespi)

Our factor of i in our expression is done to match conventions with non-relativistic quantummechanics. Our delta function expression follows from translation invariance, and is common toall S-matrix elements we compute.

We now can define a new set of Feynman rules to compute iM, which makes things slightlyeasier.

1. We draw all possible diagrams with appropriate external legs, and impose 4 momentumconservation at each vertex with appropriate use of delta functions.

2. We assign a factor of (−ig) at each vertex to take into account the order of our expansion.

3. We integrate over closed loops in the diagram, since they give rise to dummy variableswhich we can integrate over. So, for closed loops, we do the integral

∫d4k(2π)4

.

We’ll now do an example with meson scattering.

For example, for meson-meson scattering φφ→ φφ, we need to draw diagrams. Our lowest orderdiagram is a bit tricky and our requirement that our vertex has to be a meson, nucleon andanti-nucleon trio means that there’ll be a loop. This is shown in figure 6.

Figure 6: Our lowest order scattering term for φφ→ φφ decay

In the loop, we build up our momentum labels by picking a side as k, then just working aroundwhilst using conservation of momentum. This is

M =

∫d4k

(2π)4(−ig)4 i4

(k2 − µ2 + iε)((k + p1)2 − µ2 + iε)

× 1

((k + p1 − p′1)2 − µ2 + iε)(k + p1)2 − µ2 + iε

The integrand goes as 1k8

, so we’re sure that this thing converges.


4.8 Computing the two point correlation function for the ground state ofour perturbed Hamiltonian

Let’s call |ω〉 our ground state for our Hamiltonian. A natural question to ask would be what〈ω|φ(x)φ(y) |ω〉 is in terms of the spectrum we already know for the free Hamiltonian H0.

One way to do this would be to take the ground state of the free Hamiltonian which we call|0〉 and expand it in terms of the energy eignestates of the full Hamiltonian. with the fullHamiltonian, the state evolves as

e−iHt |0〉 = e−iHt |ω〉〈ω|0〉+∑n≥1

e−iHt |n〉〈n|0〉

= e−iE0t |ω〉〈ω|0〉+∑n≥1

e−iEnt |n〉〈n|0〉

but by construction we’ve assigned these energy eignestates in terms of the magnitudes of theirenergy eigenvectors, so E0 < E1 . . . En < . . . . Thus, to make the terms in the sum disappear,we employ a clever trick. This trick is to basically make the e−iE0t term decay slower than thee−iEnt terms, by taking the limit of t to t → (1 − iε)∞, where ε is chosen sufficiently smallas to make the terms in the sum decay, but not the exponential term in front of the groundstate. Hence our final result is that, our ground state for the full hamiltonian |ω〉 can be writtenas

|ω〉 = limt→(1−iε)∞e−iHt (e−iE0t 〈ω|0〉

)−14.9 Wick’s theorem

4.10 Applying Wick’s theorem for the interaction term

Let’s apply Wick’s theorem to calculate the next term in the series of our quantity

〈0| T φ(x)φ(y) exp(−i∫dtHI(t)

)|0〉

. Let’s resume our discussion in the case of φ4 theory, where our interaction Hamiltonian is∫d3zφ4(z). Recall, our first term in the series expansion is just

〈0|φ(x)φ(y) |0〉 = ∆F (x− y)

and our expression in the series expansion is

〈0| − iλ

4!T φ(x)φ(y)

∫dt

∫d3zφ4(z) |0〉

We can write this a little more concisely by combining writing∫dt∫d3z =

∫d4z, so we’re

aware that it’s an integral over four space time dimensions. To make it clearer to understandthe nature of the contractions in this expression, we do something a bit weird and write out theterms in the φ4 term as φ(z)φ(z)φ(z)φ(z). We want to cookup a simplified expression for theterm

−i λ4!

〈0| T φ(x)φ(y)∫d4zφ(z)φ(z)φ(z)φ(z) |0〉


Since we have a time ordering operator there, the only non zero terms which surivive once weexpand this thing into a sum are the terms where every scalar φ in contracted with another. Theimportant thing to remember is that this includes the φ(z) terms incorporated into the integral.There are several contraction we could do here, but the most obvious one to start with is to firstcontract φ(x) and φ(y) together, and then pair up the terms in the integral as follows

− iλ4!φ(x)φ(y)

∫d4zφ(z)φ(z)φ(z)φ(z)

but this corresponds to, replacing the contractions with Feynman propagators,

− iλ4!∆F (x− y)

∫d4∆F (z − z)∆F (z − z)

Alternatively, we could have contracted the φ(z) terms differently in the integrand, and insteadcould have contracted the terms like so:

−i λ4!φ(x)φ(y)

∫d4zφ(z)φ(z)φ(z)φ(z)

This term here still corresponds to the integral

− iλ4!∆F (x− y)

∫d4z∆F (z − z)∆F (z − z)

We count that there are in total 3 different sets of contractions which yield this integral. So thecontribution that we have from contractions of this type is

−3iλ

4!∆F (x− y)

∫d4z∆F (z − z)∆F (z − z)

We represent this type of contraction diagramatically like before in the following diagram.

z x y

We also have another type of contraction that we can include. We contract φ(x) with one of theφ(z), and contract φ(y) with another one. This gives us 12 choices for our contraction. One ofthese may look like

− iλ4!φ(x)φ(y)

∫d4z φ(z)φ(z)φ(z)φ(z)

we count 12 of these, and represent this diagramatically in the figure adjacent. Our total first

x z y

order contribution is therefore

−3iλ

4!DF (x− y)

∫d4z DF (z − z)DF (z − z)− 12iλ

4!

∫d4z DF (x− z)DF (y − z)DF (z − z)

which exhausts all of the possible ways to contract terms in our sum above.


4.10.1 Symmetry factors

We now present a convinient way to find the coefficent of our contribution which a diagramcontributes. This is called the symmetry factor of the diagram. It is as follows. Given aparticular diagram,

1. Ascribe a symmetry factor of 2 for edges which loop to the same node.

2. Ascribe a symmetry factor of n for edges which can be interchanged

3. Ascribe a symmetry factor of 2 for vertices which are equivalent.

As an example, consider the diagram below.


4.11 Feynman Rules in φ4 theory

When we consider interaction terms which contain the same species of particle, things get a bitmore complicated, and we need to take into account different permutations we can get. Whenconsidering something like the Yukawa interaction, HI = g

∫d3xψψ∗φ, our interaction term

contains distinct particles. That means, for example, that our Feynman rules we derived earlierdoesn’t come with extraneous factors and we only have to add a prefactor (−ig) at each vertexpoint. However, for the case of an interacting theory like

HI =λ

4!φ4

we have to change our thinking a bit for the sake of making life easier. On might think that,similar to the above case, that a vertex drawn in the Feynman rules for φ4 theory might meanthat we add a factor of −iλ4! , however, taking into account the different permutations of the fourφ fields, it’s easier for us to just attach a −iλ for each vertex then divide by a small numberbased on the look of the diagram. Our final goal here will be to compute amplitudes whichcontain terms like

iM ∼ −iλ4!

∫d4x

⟨p′1, p

′2

∣∣ : φ(x)φ(x)φ(x)φ(x) |p1, p2〉4.11.1 Combinatoric factors

To motivate our discussion, we’ll start by considering interactions in position, and not momen-tum space. We’ll start by calculating a propagator-like object for a scattering system with mparticles, which is given by

〈0| T φ1 . . . φnS |0〉

In our perturbation expansion for the case of φ4 theory, we’ll have terms like

1

n!

(−iλ4!

)n ∫d4y1 . . . d

4yn 〈0|Tφ1 . . . φnφ

4(y1) . . . φ4(yn)

|0〉

To illustrate the idea of symmetry factors, we will consider the case of a first order expansion(n = 1), with four particles to consider. We expand our integral as

· · · = −iλ4

∫d4x 〈0| T

φ1 . . . φ4φ

4x

|0〉

Now, we apply Wick’s theorem to determine the non-vanishing components in the integrand.One possible configuration we could have is that we could contract each of the φi term s withone of the φx terms, which gives several terms up to permutation. Or we could contract two ofthe φi ’ s together and contract the rest with φx. Finally, we could have also just contractedthe φi’s amongst themselves, and t the φx ’s amongst themselves. In total, this sum then looks


like

. . . =−iλ4!

∫d4xφ1φ2φ3φ4φxφxφxφx

+ similar terms gained from contracting all φi with φx

+−iλ4!


+ similar contractions which only contract two φx

+−iλ4!


+ similar contractions where no φx is contracted with φi

Now, the upshot of this is that since contractions are C functions, Each of the terms in thesame ’category’ yield the same expression in terms of Feynman propagators. For the first term,we have a total of 24 permutations that yield the above. This is because we can contract φ1with any of the 4 φx ’s. Then, we can pair φ2 with the remaining 3 , and so on. So we have 4!permutations here. Each of these terms yields a factor of

−iλ4!

∫d4x∆F (x1 − x)∆F (x2 − x)∆F (x3 − x)∆F (x4 − x)

and since we have 24 of them, our total contribution is

−iλ∫d4x∆F (x1 − x)∆F (x2 − x)∆F (x3 − x)∆F (x4 − x)

Now, looking specifically at the terms where we have φ3 and φ4 contracted with φx, we have12 ways to configure our contractions. More so, we could have equally included terms whichcontracted φ1 and φ2 with φx instead, which again yields a term with 12 contractions (wehave 6 possible diagrams which give this configuration, each of them implcitly summing over12 possibilities). Finally, in the term where we contract x1 with x2, and x3 with x4, we have 3possible ways for the φ′xs to contract amongst themselves.

Thus, our total expansion in terms of Feynman propagators is given by

〈0|T φ1φ2φ3φ4φxφxφφx |0〉 = −iλ∫d4x∆F (x1 − x)∆F (x2 − x)∆F (x3 − x)∆F (x4 − x)

−iλ2

∫dx∆F (x1 − x2)∆F (x3 − x)∆F (x4 − x)∆F (x− x)

+ 5 similar permutations

− iλ

8

∫d4x∆F (x1 − x2)∆F (x3 − x4)∆F (x− x)∆F (x− x)

+ 2 similar permuations

Now, if we expand to higher order, to O(λ2), we might get a term like

1

2

(−iλ4!

)2 ∫d4xd4y 〈0|T

φ1φ2φ3φ4φ

4xφ

4y

|0〉


𝑥

𝑥1𝑥2

𝑥3 𝑥4

𝑥1 𝑥2

𝑥

𝑥3𝑥4

+ 5 similar terms which permute between

, , ,𝑥1 𝑥2 𝑥3 𝑥4

+

𝑥

𝑥1𝑥2

𝑥3 𝑥4

+ 2 similar terms which permute between

, , ,𝑥1 𝑥2 𝑥3 𝑥4

Figure 7: A diagrammatic expansion of our representation

Now, in this term, if we insist that x1 and x2 are contracted with φx and that φ3, φ4 arecontracted with φy, then we could get a contraction that looks like

φ1φ2φ3φ4φxφxφxφxφyφyφyφy

Now we ask how many terms of this type there are. We have a total of 12 possibilities from ourpossible contractions of φ4, φ3 with φy. And, we also have 12 times this from φ1, φ2 with φx.This means that our prefactor is going to be

1

2=

1

2

1

4!12× 12

Our associated diagram for this kind of object is shown in figure 8.

𝑥1𝑥

𝑥3 𝑦

𝑥2

𝑥4

𝑥1

𝑥3

𝑥2

𝑥4

𝑥 𝑦

𝑥1

𝑥 𝑥2

3 distinct diagrams

Figure 8: All diagrams of the above type with distinct contractions; each has a symmetry factorof 2

Our associated contribution to this scattering in terms of Feynman propagators is then

(−iλ)2

2

∫dx d4y∆F (x1 − x)∆F (x2 − x)∆F (x3 − y)∆F (x4 − y)(∆F (x− y))@


4.11.2 Deriving our Feynman rules

We can now use our analysis above to match a diagram with an integral expression. We concludethat the expression

〈0|Tφ1 . . . φm exp

(−iλ4!

∫d4xφ4x

)|0〉

is the sum off all possible diagrams with m ’external points’ which are just points with a singleline coming out of them, along with any number of internal vertex points (which are points withfour lines coming into them since we are dealing with φ4 theory. Our routine is to then, for eachdiagram that exists, have the Feynman rules where we

1. Attach a propagator ∆F (x − y) for any line connecting the points x and y, no matterwhether they represent external or internal points.

2. For every vertex (a point where four lines meet), this represents an integral over ourinteraction space so we ascribe a factor of (−iλ)

∫d4x.

3. We divide by the appropriate symmetry factor of the diagram.

Now, we have a nice way to write down amplitudes in φ4 theory in position space. The nextthing to do would then be to transfer this into some momentum rules. Recall our momentumexpansion of the Feynman propagator

DF (x− y) =

∫d4p

(2π)4i

p2 −m2 + iεe−ip·(x−y)

This integral has a nice interpretation. Since this represents a line going from y into x, weinterpret eip·y and e−ip·x as momentum factors we can attach at each vertex for momentumgoing out of y and into x. In addition, our i

p2−m2i ε

factor can be an expression attached to theline itself. At a vertex point, we’ll have some propagators coming in to meet at a point. Due tothe fact that we’re integrating over

∫d4x , a factor like this corresponds to∫

d4xe−ix·(p1+p2−p3−p4) = (2π)4δ4(p1 + p2 − p3 − p4)

for example. Hence, in momentum space, our Feynman rules look like

1. For each line, assign a momentum p and attach a factor of ip2−m2+iε

in the integral.

2. At each vertex, assign a factor of iλ, and also impose the conservation of momentum whichcomes from a delta function appearing.

3. Integrate over all other undetermined momenta∫

d4k(2π)4

4. Divide by the symmetry factor of the diagram

4.11.3 Vacuum Bubbles and Connected Diagrams

If we consider scattering both to and from our free theory vacuum state, we are describingthe term 〈0|S |0〉. This means however, that we have no external points. Our only validdiagrams are the ones where every point is an internal vertex (four lines going in). So, our total


1 + + + +

Second order diagrams

First order

Figure 9: All vacuum bubble diagrams contributing to the scattering

amplitude is given by the sum of all vacuum bubble diagrams shown in figure 9. Now, one canconvince themselves that we can write this sum as an exponential of just distinct vacuum bubblediagrams. This sum is shown diagrammatically in figure 10. In other words, we can write our

+ +

⟨0|𝑆|0⟩ = exp + ...

Figure 10: Combinatoric factors work out to give an exponential of distinct vacuum types

vacuum amplitude as

〈0|S |0〉 = exp (Distinct vacuum bubble types)

Now, from our previous discussion, we had that a term like

〈0|T φ1 . . . φnS |0〉 =∑

diagrams with m external points

but, amazingly, we have that this quantity factorizes into two parts, one factor being theexponential of distinct vacuum bubble types 〈0|S |0〉, and the other factor the sum of justconnected diagrams.

〈0|T φ1 . . . φmS |0〉 =(∑

connected diagrams)(〈0|S |0〉)

4.12 Green’s function vacuum

Let’s delve into the nuts and bolts of the vacuum state in our interacting theory. We denotevacuum state of our interacting theory as |Ω〉. This is a different state from our vacuum state


in our free theory, where we have that

H0 |0〉 = 0

However when we move to our interaction picture, we get that H = H0 +H int satisfies

H |Ω〉 = 0, 〈Ω| |Ω〉

We define ’Green’s functions’ as

G(n) = 〈Ω|T φH(x1) . . . φH(xn) |Ω〉

which can be interpreted as our n point correlator but this time, in our interaction picture.Notice that this is time dependent, and that our states are in the Heisenberg picture. Now,interaction picture vacuum states are difficult to deal with. The goal of this section is to be ableto write them as a function of our free theory vacuum state |0〉. Our claim is that

〈Ω|T φ1H . . . φmH |Ω〉 =〈0|T φ1I . . . φmIS |0〉

〈0|S |0〉

Let’s explain some of the notation here. Recall, that S is merely our time evolution operatorand since we are considering initial and final state scattering this is

S = limT→∞

U(T,−T ) = limT→∞

exp

(∫ T

−TdtHI(x)

)We denote states in the Heisenberg picture (on the left hand side), as φiH , and states in theinteraction picture as φiI . From the previous section, we showed that the numerator on theright has side is

〈0|T φ1I . . . φmIS |0〉 =∑

( connected diagrams with m external points ) 〈0|S |0〉

In other words, our Green’s function is the sum of connected diagrams with m externalpoints. To prove this, we take, without loss of generality, that x01 > x02 > · · · > x0m. This means,we can employ a trick to parition the integral and reorder things in the expression.

T(φ1Iφ2I . . . φmI exp

(−i∫ ∞−∞

dtHI(t)

))= T

φ1I(x

01)φ2I(x

02) . . . φmI(x

0m)

exp

(−i∫ x0m

−∞dtHI(x)− i

∫ x0m−1

x0m

dtHI(x)− · · · − i

∫ ∞x01

dtHI(t)

)The uposhot of writing this that we can now we apply to time ordering operator, which shufflesaround the terms in the exponential. Thus, the expression above reads

. . . = exp

(−i∫ ∞x01

dtHI(t)

)φ1I(x1) exp

(−i∫ x01

x02

dtHI(t)

). . .

... exp

(−i∫ x0n

x0n−1

dtHI(t)

)φIm exp

(−i∫ x0n

−∞dtHI(t)

)


4.12.1 LSZ Reduction

To describe scattering in interacting theory, with external states (eg |p1, p2〉, should be frominteracting theory. This means we exclude loops on external legs. THe above diagram is bannedsince it’s absorbed into the definition of an initial interacting state. See AQFT or look upamputated diagrams.

5 Cross sections and decay rates

In this section, we’ll look at calculating decay rates (the probability of particles undertaking acertain type of scattering due to the interaction potential).

5.0.1 Kinematics

Let’s first consider the case of 2 particles scattering into two particles. For example, this couldbe a nucleon anti-nucleon pair scattering into two mesons (ψψ → φφ.

To analyse this problem from a kinematic perspective, we analyse incoming and outgoing mo-menta. Suppose that pµ1 and pµ2 represent the 4 momenta of the two incoming particles, andthat qµ1 and qµ2 represent their outgoing momenta. Due to conservation of momentum, we havethat

pµ1 + pµ2 = qµ1 + qµ2

if we define the Mandelstam variables

s := (p1 + p2)2

t := (p1 − q1)2

u := (p1 − q2)2

The neat thing about these variables is that, when we add them up, we get the sum of thesquares of the rest masses of our constituent particles.

s+ t+ u = 3p21 + p22 + q21 + q22 + 2p1 · p2 − 2p1 · q1 − 2p1 · q2= 3p21 + p22 + q21 + q22 + 2p1 · p2 − 2p1 · (p1 + p2)

= 3p21 + p22 + q21 + q22 − 2p21

= p21 + p22 + q21 + q22

However, due to the relativistic dispersion relation, we have that p2 = m2, so we have the resultthat

s+ t+ u = m21 +m2

2 + (m′1)2 + (m′2)

2

We can write out scattering amplitude for ψψ → ψψ scattering we had earlier in terms of the


Mandelstam variables.

iM = (−ig)2

1

(p1 − q1)2 −m2+

1

(p1 − q2)2 −m2

= (−ig)2

1

t−m2+

1

u−m2

5.0.2 Decay rates

Recall that in our earlier section, we expressed the probability amplitude of scattering as aproduct of a Dirac delta function which imposed the conservation of momentum, and the factorM which we called the amplitude

〈φ|S − 1 |i〉 = iM(2π)4δ4(pi −∑i

qi)

Note that pi here is not an index, but a label for the total momentum of particles in the initialstate!

Just to refresh our memory, we expressed momentum conservation in terms of a delta functionwhich summed initial and final states in terms of of definite momenta. Now, we can ask ourselves,how do we convert this into a probability value that we can measure in experiment? Well, wetake the modulus of the amplitude, and divide by 〈f |f〉〈i|i〉 since these initial states necessarilyalready normalised. Our probability is given by

P =‖ 〈f | (S − 1) |i〉 ‖2

〈f |f〉〈i|i〉From this, we then define our differential cross section, which roughly is the incremental changein area we get when scattering particles through some interaction potential (see Peskin andSchroeder for a proof of this result).

dσ =(2π)4δ4 (p1 + p2 −

∑i qi) |M|2

FHere, F is our flux factor with

F = 4√

(p1 · p2)2 −m21m

22

To find our total cross section from i → f , we integrate over our final state momentum inthe usual Lorentz invariant way, that is, over integrals in momentum space with the measure∫ d3p

(2π)32Ep. Thus, integrating our differential cross section over each momentum variable gives

usσ =

∫1

Fdpf |M|2

Our measure is given by∫dpf = (2π)4

∫ ( n∏r=1

d3qr(2π)32Eqr

)δ4

(pi −

∑i

qi

)Hence, in the case of 2 → 2 scattering for example, we set here that n = 2, so we only have twointegrals.


5.0.3 2 → 2 scattering

We will now present an instructive example to illustrate calculating decay rates. Our aim forthis section will be to calculate the change in the cross section with respect to the Mandelstamvariable t. We write out this variable as

t = (p1 − q1)2 = m2

1 +m22 − 2Ep1Eq1 + 2~p1 · ~q1

We have thatdt

d cos θ= 2|~p1||~q1|

where cos θ is the angle between ~q1 and ~p1. It’s convinient for us to work in the center ofmass frame, the frame where our total momenum is 0. Our second Mandelstam variable iss = (p1 + p2)

2, which we can work out to be the centre of mass energy. This quantity is aconstant of the scattering. In the first section, we showed that our Lorentz invariant measurecould be written in terms of Lorentz invariant delta functions. For our purposes, we will returnto that formalism for q2 to aid computations. We set

d3q22Eq2

= d4q2δ(q22 − (m′2)

2)θ(q02)

where we’ve defined θ as the Heaviside step function. For the other measure d3q1, we rewritethe measure in terms of polar coordinates.

d3q12Eq1

=|~q1|2d|~q1|d(cos θ)dφ

2Eq1=

1

4|~p1|dEq1dφdt

With this in mind, we can write out the full Now, putting this together, we have an expres-sion

dσ

dt=

1

8φF|~p1|

∫dEq1 |M|2δ(s−m2

2 − (m′1)2 − 2~q1 · (~p1 + ~p2))

Now, if we boost back to the centre of mass fram, we have that

pµ1 = (√|~p1|2 −m2

1, ~p1)

pµ2 = (√|~p1|2 +m2

2,−~p1)

This means we compute th s mandelstam variable

s = (√

|~p1|2 +m21 +

√|~p2|2 +m2

2)2

This implies that we can rewrite

|~p1| = λ12 (S,m2

1,m22)/2

√s

where we have defined

λ(x, y, z) = x2 + y2 + z2 − 2xy − 2xz − 2yz

From this, we have that

F = 2√λ(s,m2

1,m22) =⇒ dσ

dt=

|M|2

16πλ(s,m21,m

22)


5.0.4 Decay rates

We defime a partial decay rate for |i〉 → |f〉, or partial width, as

Γf =1

2Ep

∫dpf |M|2

Note that this is not a Lorentz invariant quantity due to time dilation. Hence, our conventionis to quote this in terms of the res frame of the initial particle, where the energy Epi is equal tothe mass mi. We define the total decay with whcih we write as Γ. This is the sum of all possiblestates

Γ =∑f

Γf

The branching ratio for |i〉 → |f〉 is denoted as

BR(i→ f) =ΓfΓ

If we look at the units, we notice that the average lifetime isτ = 6.6× 10−25 × (1GeV )/Γ seconds

Summary

Mandelstam Variables

• Used in 2 → 2 scattering problems.

• Mandelstam variables are squares of sums of pertinent momentas = (p1 + p2)

2

t =(p1 − p′1

)2u =

(p1 − p′2

)2We can use conservation of momentum p1 + p2 = p′1 + p′2 to rewrite some of the abovevariables as well.

• We can work in the centre of mass frame so that ~p1 + ~p2 = ~p′1 + ~p′2 = 0. Hences = (p1 + p2)

2 = (E1 + E2)2

• In the centre of mass frame with identical particles,s = 4E2, t = 2p2 (1− cos θ) , u = −2p2 (1 + cos θ)

Cross sections

• The differential cross section formulae are written with resepct to both the t Mandelstamvariable as well as the solid angle.

dσ

dt=

|M |2

16πλ(s,m2

1,m22

) , dσ

dΩ=

|M |2

64π2s


6 The Dirac Equation

6.1 What does the Dirac equation give us?

The Dirac equation is important because it provides the framework for quantization of spin-12particles, like electrons and other fermions. Another startling prediction that the Dirac equationgives us is the existence of anti-matter. This equation is an amazing combination of the geometryof spinors in Minkowski space-time, and the wave-function formalisation of quantum mechanicsthat we all know and love. In this section, we’ll provide some motivation for constructing theDirac equation, solve it, then quantise this object to give rise to particles (fermions).

Right now we’ll try to analyse how fields with multiple components change under an underlyingLorentz transformation. Recall that when we have a Lorentz transformation in our coordinates,we induce a transform on our scalar field.

xµ → (x′)µ = Λµνxν , φ(x) → φ′(x) = φ(λ−1x)

Most particles have an intrinsic angular momentum-spin however, and so transform in a non-trivial way under boosts and rotations. For example a spin-1 vector field Aµ (for example, ourfamiliar electromagnetic vector potential) transforms under a Lorentz boost as

Aµ(x) → (A′)µ(x) = ΛµνAν(Λ−1x)

There’s no reason a priori why vector fields with multiple components need to transform withΛµν multiplying it out front. In general, we have that our field transforms as

φa(x) → Dab(Λ)φ

b(Λ−1x)

where our matrices D form a representation of our Lorentz group. Let’s explain why this is.We treat ψa as a vector which is acted on by a symmetry. However, for this symmetry toact on an object like a vector, it needs to be represented in the same space (for example,as a matrix). Up until now, we’ve been representing a Lorentz transformation group in it’sstandard form as a matrix, but this isn’t the only representation (recall the Lorentz group isn’tdefined in terms of matrices in the first place, but as members of the orthogonal group O(1, 3)). Representations obey the properties that the group structure is preserved. This means that,if D is a representation, then

D(Λ1)D(Λ2) = D(Λ1Λ2)

D(Λ−1) = D(Λ)−1

D(I) = 1

To find representations we look at the Lorentz algebra, considering infinitesimal transforma-tions

Λµν = δµν + εωµν +O(ε2)

Working to infinitesimal order, the find that the generators obey the relation

ΛµσΛνρ = ηµν


Substituting this into our relation we have that

(δµσ + εωµσ)(δνρ + εωνρ)η

σρ = ηµν +O(ε2)

This implies that our generator satisfies

ωµν = −ωνµ =⇒ ωµν is anti symmetric

This generator has 6 components, split up into 3 rotations and 3 boosts. Thus, we can introducea basis of six 4× 4 matrices

(Mρσ)µν = ηρµησν − ησµηρν

We can lower the index here to give

(Mρσ)µν = ηρµδσν − ησρδρν

For example, we have that

(M01)µν =

0 1 0 01 0 0 00 0 0 00 0 0 0

This specific generator generates a boost in the x1 direction. We can hence write our generatoras a linear sum of these basis matrices.

ωµν =1

2(ΩρσMρσ)µν

We call Mρσ the generators of our Lorentz group, and our corresponding Ωρσ the anti symmetricparameters of the Lorentz transformation. Our Lorentz algebra, one can calcuate, is

[Mρσ,Mτν ] = ηστMρν − ηρτMσν − ηρνMστ − ησνMρτ

What we’ve done here is that we’ve calculated the Lie bracket of two elements in our repre-sentation. This means that any other representation of the Lorentz group necessarily satisfiesthis relation, since representations of a Lie algebra needs to preserve the Lie bracket. Our cor-responding boost is exp

(12Ωρσ

)Mρσ. This is because when we are returning the infinitesimal

transformation back into its full form by exponentiating.

6.2 The Spinor Representation

Now, the aim of the game is to try and find other representations which satisfy the Lorentz alge-bra. To help us find representations that satisfy the Lorentz algebra, we start by defining some-thing called the Clifford algebra. This is an algebra whose elements satisfy the anti-commutationrelations

γµ, γν = 2ηµνI, γµ, γν = γµγν + γνγµ

where γµ are a set of four matrices. We have that our matrices in the algebra obey the proper-ties

γµγν = −γνγµ, µ 6= ν

(γ0)2 = I, (γi)2 = −I, i ∈ 1, 2, 3


Our simpliest representation which obeys these commutation relation are a set of 4×4 matrices,which we call the Chiral representation. These are given in block matrix form as

γ0 =

(0 I2I2 0

), γi =

(0 σi

−σi 0

)Here, σi are the Pauli matrices given by

σ1 =

(0 11 0

), σ2 =

(0 −ii 0

), σ3 =

(1 00 1

)These Pauli matrices obey the commutation and anti commutation relations

σi, σj= 2δijI2, [σj , σk] = 2iεjklσl

Given this representation, we can also generate a new representation by conjugating by a con-stant invertible matrix U , so UγµU−1 works also as a representation.

If we define

Sρσ :=1

4[γρ, γσ] =

0 ρ = σ12γ

ργσ ρ 6= σ=

1

2γργσ − 1

2ηρσI

we aim to show that this satisfies the Lorentz algebra, whilst being philosphically different tothe algebra we defined before. Now, expanding this whole thing out explicitly is difficult to do,so we prove a handful of useful relations to show this. We can

Claim. We have that[Sµν , γρ] = γµηνρ − γνηρµ

Proof. There’s some steps we have to follow and it’s not as straightforward as it seems from firstglance. The basic idea is to commute things out so that we only have one γ factor per term.

[Sµν , γρ] =1

2[γµγν − ηµν , γρ]

=1

2[γµγν , γρ]

=1

2(γµγνγρ − γργµγν)

Now, this is a bit tricky to manage. The key is to anti-commute things in the right order. Weanti-commute the last two matrices in the first term, and the first two matrices in the last term.The above is then equal to

. . . =1

2γµ(γν , γρ − γργν)− 1

2(γρ, γµ − γµγρ) γν

=1

2γµ (2ηνρ − γργν)− 1

2(2ηρµ − γµgγρ)γν

= γµηνρ − γνηρµ


Using this, we can then prove that Sρσ satisfies the Lorentz algebra! This means we would thenbe able to use this as a valid representation of the Lorentz group.

Claim.[Sρσ, Sτν ] = ηστSρν − ηρτSσν + ηρνSστ − ησνSρτ

Proof. We can expand out one side and appeal to the relation we proved earlier.

[Sρσ, Sτν ] =1

2[Sρσ, γτγν − ητµI]

=1

2[Sρσ, γτγν ]

=1

2[Sρσ, γτ ]γν +

1

2γτ [Sρσ, γν ]

=1

2(γρηστ − γσηρτ )γν +

1

2γτ (γρησν − γσηρν)

Now, to recover what we had earlier, we need to sub out our expressions with two gammas foran expression in S. To do this, we invert our relation for S so that

γµγν = 2Sµν + ηµνI

We then get that the above is equal to

. . . =1

2(2ρνηστ + ηστηρνI)

− 1

2(2Sσνηρτ + ησνηρτI)

+1

2(2Sτρησν + ηρτησνI)

− 1

2(2Sτσηρν + ηρνητσ)

= Sρνηστ − Sσνητρ + Sτρησν − Sτσηρν

The terms with two η terms cancel out. The object that we’re left with is precisely the Lorentzalgebra.

Thus, S provides a representation of the Lorentz group. We now define a Dirac spinor, ψα(x)whichis a four component vector that satisfies the following transformation law

ψα(x) → S[Λ]αβψβ(Λ−1x)

where we’ve defined Λ = exp(12ΩρσM

ρσ), and our spinor representation S[Λ] = exp

(12ΩρσS

ρσ).

We need to check that the Spinor representation is not equivalent to the usual vector represen-tation, in the sense that we need to check what we’ve written down gives something differentthan our standard Lorentz boost.

To check that what we have is different, we trial a particular Lorentz transformation (in thiscase, a rotation by 2π).


Example 2. (The Spinor representation is different) If we choose

Sij =1

4

[(0 σi

−σi 0

),

(0 σj

0 σk

)]= − iε

ijk

2

(σk 00 σk

)from σi algebra

If we write as a paramater Ωij = −εijkφk . Exponentiating, we get that

S[Λ] = exp

(1

2ΩρσS

ρσ

)=

(eiφ·σ/2 0

0 eiφ·σ/2

)Now, if we consider a rotation of 2π abou the x3 axis, we get that our corresponding represen-tation is

S[Λ] =

(eiπσ

3‘ 0

0 eiπσ3

)= −I4

In this notation, the exponential needs Thus, a rotation of 2π takes a spinor φσ(x) → −φσ(x),not like a vector, which goes like Λ = I as expected (in our fundamental representation).

Now, what happens with boosts of spinors

S0i =1

2

(−σi 00 σi

)If we write our boost parameter as Ω0i = χi, then our corresponding representation is

S[Λ] =

(e−χ·σ/2 0

0 eχ·σ/2

)Note that for rotations, S[Λ] is unitary, since S[Λ]†S[Λ] = I, but for boosts, it’s not.

Theorem. (No finite dimensional, unitary representations of the Lorentz group) There are nofinite dimensional unitary representations of the Lorentz group!

Proof. S[Λ] = exp[12ΩρσS

ρσ]

is only unitary if Sρσ are anti hermitian. Now, when we take theHermitian conjugate of a commutator, not only does each element pick up a hermitian conjugate,but the arguments in the commutator switch around. This means we have to pick up a minussign.

(Sρσ)† = −1

4[(γρ)†, (γσ)†]

can be anti-hermitian only if all γµ are hermitian or are all anti hermitian. However, this can’tbe arranged, and is impossible to do. This can’t be arranged, because

(γ0)2 = 1

This means in particular that γ0 has real eigenvalues, which means it can’t be hermitian sincehermitian matrices only have pure imaginary eigenvalues.

(γi)2 = 1 =⇒ γi can’t be hermitian

In general, there’s no way to pick γµ such that Sµν are anti-Hermitian.


6.3 Constructing a Lorentz Invariant action of ψ

Now, something of immediate interest is to see whether we can construct some sort of actionfrom this. To carry this out, we should try to construct Lorentz invariant quantities. We writeψ†(x) = (ψ∗)T (x). One question that we might have is to ask whether ψ†(x)ψ(x) is a Lorentzscalar. However, under a Lorentz transformation,

ψ†(x)ψ(x) → ψ†(Λ−1x)S[Λ]†S[Λ]ψ(Λ−1x)

This is not invariant in general since S[Λ] is not unitary. In the chiral representation, we havethat

(γ0)† = γ0, (γi)† = −(γi)

We can verify by checking for each index, that we can write out an identity to relate the hermitianconjugate of a gamma matrix to an expression which doesn’t involve conjugates as

(γµ)† = γ0γµγ0

Thus, we can find the Hermitian conjugate of S as

(Sµν)† = −1

4[(γµ)†, (γν)†] = −1

4[γ0γµγ0, γ0γνγ0] = −γ0Sµνγ0

We do this because, since γ0 = (γ0)−1, when we exponentiate this object we can pull out the γ0on both sides. Hence, our exponential gives

S[Λ]† = exp

(1

2Ωµν(S

µν)†)

= γ0S[Λ]−1γ0

Definition. (The Dirac adjoint) The Dirac adjoint of ψ is defined as

ψ(x) := ψ†(x)γ0

We define this object in the spirit of creating scalars, vectors and other objects which transformnicely under Lorentz transformations.

Claim. (Lorentz invariance of ψψ) Our claim is that ψψ is indeed a Lorentz scalar. To provethis observe that

ψψ = ψ†γ0φ→ ψ†(Λ−1x)S[Λ]†γ0S[Λ]ψ(Λ−1x)

Using the identity above, we get that this is the same.

Claim. (ψγµψ transforms as a Lorentz vector) Things that transform nicely under Lorentzboosts are called Lorentz vectors (or more familiarly, 4-vectors). This means that they transformmuch like how a position vector transforms under a Lorentz boost:

xµ → Λµνxν

We claim that the object ψγµψ transforms in this way to, where under a Lorentz transform wehave that

ψγµψ → Λµνψγνψ


Proof. Our vector transforms as follows (note that the γµ matrix doesn’t change since it’s aconstant object - only the spinors change.

ψγµψ → ψS[Λ]−1γµS[Λ]ψ

For this to be a Lorentz vector, we require that

S[Λ]−1γµS[Λ] = Λµνγν

To go from here, we need to write out the transformations in terms of the exponents of theirgenerators. In particular, we require that the coefficients of the generators Ωρσ are the same onboth sides - it’s just the matrices M and S that are different since we’re dealing with distinctrepresentations.

Λµν = exp

(1

2ΩρσMρσ

), S[Λ] = exp

(1

2ΩρσS

ρσ

)So, we need to prove that

(Mρσ)µνγν = −[Sρσ, γµ]

This can be shown to be true by observing that

(ηρµδσν − ησµδρν ) γν = ηρµγρ − γρησµ = −[Sρσ, γµ]

Now, the upshot of doing all this work is that we can now write down an action (which byright should be Lorentz invariant). We know that ψγµψ is a Lorentz four vector, which impliesthat ψγµ∂µψ is a Lorentz invariant quantity. We also already know that ψψ is also Lorentzinvariant on its own. Finally, we can construct a Lorentz invariant action from these objects,by setting

S =

∫d4xψ (γµ∂µ −m)ψ(x)

6.4 The Dirac equation

Varying ψ independently in the Euler Lagrange equations gives us the Dirac equation. Becausethere is no dependence on ψ, we have that our equation of motion is given by

∂L∂ψ

= 0

Specifically, this is the equation(iγµ∂µ −m)ψ = 0

This is first order, not second order like KG. To save some ink, we write objects which arecontracted with γµ with a slash. For example, we write Aµγµ = /A. In this notation, the Diracequation reads (

i/∂ −m)ψ = 0


Note that each component of ψ solves the KG equation.

(i/∂ +m)(i/∂ −m)ψ = 0

(γµγν∂µ∂ν +m2)ψ = 0

(1

2γµ, γν ∂µ∂ν +m2)ψ = 0

−(∂2 +m2)ψ = 0

6.5 Chiral Spinors

Since S[Λ] is block diagonal in our Chiral representation, we say that it’s reducible (since wecan write the representation as a direct sum of two different representations). It decomposesinto 2 irreducible representations. We write our spinor as

ψ =

(uLuR

)where uL, uR are 2 C component objects which we call Weyl, or chiral, spinors. These objectstransform identically under rotations, so that

uL,R → eφ·σ/2uL,R

but under boosts, they transform in the opposite fashion. This can be seen by just multiplyingthe matrix with the vector as ψ → S[Λ] rot ψ, which gives us

uL → e−χ·σ/2uL, uR → eχ·σ/2uR

Heuristically we write

uL ∼ (1

2, 0) uR ∼ (0,

1

2) ∈ (SU(2), SU(2))

we have that ψ is in (12 , 0)⊕ (0, 12).

6.5.1 The Weyl equation

If we take our Dirac LagrangianLD = ψ

(i/∂ −m

)ψ

which is equal to

· · · = iu†Lσµ∂µuL + iu†Rσ

µ∂µuR −m(u†LuR + u†RuL

)where we define σµ = (I, σ), σµ = (I,−σ). This is derived simply by writing out the componentsexplicitly, then multiplying by the required matrices. A massive fermion requires both uL anduR , but the massless limit m = 0 requires only uL or uR. The mass term is mixing up lefthanded and right handed spinors. In the case m = 0, we get the following EL equations

iσµ∂µuL = 0, iσµ∂µuR = 0


These are called Weyl’s equations. In classical particle mechanics, the number of degrees offreedom is given by half the dimension of phase space. in field theory, we need to do thingsdifferent and discuss the number of degrees of freedom per spacetime point. In field theory, wehave a scalar φ and Πφ = φ, so our real degrees of freedom is 1

2 × 2 = 1. However, for a spinorψα we have 8 degrees of freedom since it’s complex. For our conjugate momenta,

Πψ = iψi

which doesn’t add any degrees of freedom. This means that our real degrees of freedom amountsto 1

2 × 8 = 4. We have 2 for the spin up and spin down particle, and 2 for the spin up and spindown anti-particle.

6.6 Dirac Field Bilinears

Throughout this section, we’ve only been using a specific representation of S (which we dubbedthe chiral representation). However, note that in a different basis, S[Λ] is not necessarily blockdiagonal! We can transform to a different basis by applying the map

γµ → UγµU−1, ψ → Uψ

We useγ5 := iγ0γ1γ2γ3

to define Weyl spinors in a basis independent way. This object has very nice properties whichrelate agonistically with γµ for µ = 0, 1, 2, 3.

Claim. (Commutation relations for γ5) The γ5 matrix satisfies the relationsγµ, γ5

= 0,

(γ5)2

= I

Proof. To prove the first relation, it’s instructive to set µ = 0 just to see what’s going on. Wehave that the anti-commutator is

γµ, γ5= i(γ0γ0γ1γ2γ3 + γ0γ1γ2γ3γ0)

= i(γ1γ2γ3 − γ1γ0γ2γ3γ0)

= i(γ1γ2γ3 + γ1γ2γ0γ3γ0)

= i(γ1γ2γ3 − γ1γ2γ3(γ0)2) = 0

We’ve used liberally here the fact that

γµγν = −γνγµ when ν 6= µ

For the second relation, we use this fact as well. Note that when we pass a γj matrix past nmatrices γi where i never equals j, we pick up a factor of (−1)n. Thus, note that (γ5)2 is just

(γ5)2 = i2γ0γ1γ2γ3γ0γ1γ2γ3

= (−1)4γ1γ2γ3γ1γ2γ3

= (−1)3γ2γ3γ2γ3

= (−1)4I = I


Now, we can write out γ5 explicitly depending on the representation we choose. We check thatin our chiral representation,

γ5 =

(−I 00 I

)We define projection operators in the spirit of finding a way to extract the left-handed andright-handed spinors from the entire spinor. We define

PL =1

2(I4 − γ5), PR =

1

2(1 + γ5)

These obey the properties that come with what we usually know as a projection operator,namely, that projecting twice is the same as projecting once, and that PL and PR project onto’orthogonal’ spaces

P 2L = PL, P 2

R = PR, PLPR = 0

We then define ’left handed’ and ’right handed’ spinors ψL and ψR, which are the projectionspinors with these operators

ψL = PLψ, ψR = PRψ

for left handed and right handed spinors respectively. Under a Lorentz transformation,

ψ(x)γ5ψ(x) → ψ(Λ−1x)S[Λ]−1γ5S[Λ]ψ

Checking that [Sµν , γ5], we have that this is

· · · = ψ(Λ−1x)γ5ψ(Λ−1x)

This is called a pseudoscalar since it transforms under parity transformation. We can also definethe axial vector, ψγ5γµψ which transforms in the same way.

6.7 Parity

We’ll now explore how spinors transform under discrete transformations. We’ll look at a specificclass of transformations which flips either our time coordinate or our spatial coordinate. Thetransformations are called parity transformations. Ultimately, we’ll find that ψL and ψR arerelated by a parity trasnformation. In the Lorentz group O(1, 3), we’ve dealt with boosts androtations, which are a continuous group of transformations. However, there are other discreteLorentz transformations which are not part of the Lorentz group’s component connected to theidentity. These transformations are

Time Reversal T : x0 → −x0, xi → xi

The other one is parity.Parity P : x0 → x0, xi → −xi

Other transformations like x2 → −x2 are connected to the identity, because something like thiscan be written as a rotation. Under P , rotations don’t change sign, but boosts do.

uL,R → eφ·σ/2uL,R, uL,R → e∓χ·σ/2uL,R


This is equivalent to writing, in our new notations with ψL,R = 12(I ∓ γ5)ψ, that P exchanges

LH and RH spinors.P : ψL,R(~x, t) → ψR,L(−~x, t)

Since our parity transformation switches around the left and right handed components of thespinor, it is completely equivalent to applying the γ0 matrix to the spinor, (but also we have toremember to change the ~x sign)

P : ψ(~x, t) → γ0ψ(−~x, t)

How do our terms in our L density change? We look at each field one by one Firstly,

ψψ(~x, t) → ψψ(−~x, t)

Also,

ψγµψ(~x, t) →

µ = 0 ψγ0ψ(−~x, t)µ = i ψγ0γiγ0ψ(−~x, t) = −ψγiψ(−~x, t)

On the other hand, we have that under parity,

ψγ5γ(~x, t) → ψγ0γ5γ0ψ(−~x, t) = −ψγ5ψ

In addition,

ψγ5γµψ → ψγ0γ5γ0ψ =

µ = 0 −ψγ5γ0ψµ = i +ψγ5γiψ

To summarise, we write down the number of components as

ψψ1

ψγµψ4

ψSµν6

ψγ5ψ1

ψγ5γµψ4

(1)

this has a total of 16 components. Now we add extra terms to L involving γ5. This can breakP invariance. For example,

L = gWµψγµ(1− γ5)ψ

2

This represents a W boson vector field coupling only to LH ψ ’ s. If L treats ψL and ψR equallyit is called vector like. If they appear differently they’re called chiral.

6.8 Symmetries and Currents of Spinors

If we have a space time translation xµ → xµ − εµ, then ∆ψ = εµ∂µψ. This gives us the stressenergy tensor

Tµν = iψγµ∂νψ − ηµνL


We get a conserved current when equations of motion are obeyed, do we can impose them inTµν .

LD = ψ(i/∂ −m

)ψ = 0

This implies L = 0 in Tµν thusTµν = iψγµ∂νψ

S =

∫d4x

1

2ψ(i/∂→ −m

)ψ +

1

2ψ(−i/∂← −m)ψ

This is ∫d4x

1

2ψ(i/∂

↔ − 2m)ψ

where /ψ↔ = /∂→ − /∂

←. SoTµν =

i

2ψ(∂µ∂ν − ∂ν∂µ)ψ

6.9 Lorentz transformations

We know how a spinor is supposed to transform. We have that infinitesimally,

ψ → S[Λ]αβψβ(xµ − ωµνx

ν)

where we have that ω is generated as

ωµν =1

2Ωρσ(M

ρσ)µν

Since we defined earlier that(Mρσ)µν = ηρµδσν − ησµδρν

This implies that ωµν = Ωµν . Our infinitesimal change in our spinor is

δψα = iωµν[xν∂µψ

α − 1

2(Sρσ)αβψ

β

]Our corresponding conjugate change is

δψα = −ωµν[xν∂µψα +

1

2ψ(Sµν)

βα

]where the last term comes from ψβS[Λ]

−1. Our conserved current

(Jµ)ρσ = xρTµσ − xσTµρ − iψγµSρσψ

THe first term looks like a real scalar stress energy tensor. The new piece will give us spin 12

after we quantise the field. For example, looking at the last term (J0)ij = −iψγ0Sijψ. Usingthe representation in Pauli matrices, we get that the above is equal to

(J0)ij =1

2εijkψ†

(σk 00 σk

)ψ


6.10 Internal vector symmetry

We get that under the phase rotation

ψ → eiαψ =⇒ δψ = iσψ

This gives us a conserved quantitiy

jµν = ψγµψ a vector current

This gives us the conserved charge

Q =

∫d3xψγ0ψ =

∫d3xψ†ψ

in other words, an electric charge.

6.11 Axial Symmetry

For massless spinors in the m = 0 limit, we transform

ψα →(eiαγ

5) β

αψβ

This rotates left handed and right handed spinors in the opposite direction. This leads to theconserved axial vector current

jµA = ψγµγ5ψ

6.12 Plane-Wave Solutions

Now, let’s actually start writing down solutions to the Dirac equation. We want to solve (i/∂ −m)ψ = 0. Even though this is a first order differential equation and typically we wouldn’t thinkof using wave solutions, we try this anyway and make the ansatz

ψ = u~pe−ip·x

where u~p is a constant spinor depending on ~p . Substituting this in, using the chiral representa-tion of γµ, we have that

(γµpµ −m)u~p = 0 =

(−m pµσ

µ

pµσµ −m

)u~p = 0

Here, we are again using our notation that σµ = (1, σ), and σµ = (1,−σ).

Claim. (The Spinor plane-wave solution)

Our claim is that we have a solution given by

u~p =

(√p · σξ√p · σξ

)


for any two component spinor ξ such that ξ†ξ = 1. To prove this, we let u~p = (u1, u2), andsubstitute into the above such that

(p · σ)u2 = mu1, (p · σ)u1 = mu2

Either of these equations implies the other, since

(p · σ)(p · σ) = p20 − pipjσiσj

= p20 − pipj1

2

σi, σj

= pµp

µ = m2

Now, we try the ansatz u1 = (p ·σ)ξ for the two component spinor ξ′. Substituting this into theabove, we have that

u2 =1

m(p · σ)(p · σ)ξ = mξ′

Hence, any vector of the form u~p = A((p ·σ)ξ′,mξ′). To make this look more symmetric, chooseA = 1

m and ξ′ =√p · σξ with ξ const. Then, we have that

u~p =


)Examples. Let’s take p = 0, then we have that our solution looks like

u~p=0 =√m

(ξξ

)for any ξ

Under spatial rotations, we have that

ξ → eiσ·φ/2ξ

After quantisation, ξ describes the spin of the spinor. For example, we have that

ξ =

(10

)represents spin up along x3 axis

Let’s consider a particle boosted along the x3 direction. Now, the momentum is not going to bezero. Now what we have is that

u~p =

√E − p3

(10

)√E + p3

(10

)

which, in the m→ 0 or E → p3 limit, converges to

√2E

0010


Now, with instead ξ =

(01

), when we act on this we instead get that

u~p =

√E + p3

(01

)√E − p3

(01

)

which tends to

√2E

0100

6.13 Helicity

The Helicity operator projects angular momentum along the direction of motion. If we de-fine

h = ~p · ~s = 1

2pk

(σk 00 σk

)A massless spin up operator has h = 1

2 , and a massless spin down particle has h = −12 .

6.13.1 Negative frequency solutions

If we define ψ = v~peip·x. The Dirac equation then gives

v~p =

( √p · ση

−√p · ση

)where we have as well that η†η = 1 and η is a constant 2 -spinor.

6.14 A canonical Basis for normalised Spinors

In what follows, we’ll construct a basis for the two component spinors in a way such that somenice identities will hold when we’re doing calculations. It will be convenient for us to choose thebasis

ξ1 = η1 =

(10

), ξ2 = η2 =

(01

)This gives us the sweet property that these things are normalised, so

(ξr)†ξs = δrs, (ηr)†ηs = δrs

In this basis, we then get two independent solutions for both u(~p) and v(~p), which we will indexas us(~p) and vs(~p). We get these solutions by just plugging in ξs or ηs into the respective spinors- these linearly independent solutions are given by

us(~p) =


), vs(~p) =

( √p · ση

−√p · ση

)


The neat thing is that, now with our choice of basis for the two component spinors, we get thatus and vs are themselves normalised depending on how we contract them. We can contract themin two ways. The first thing that we can do is take the hermitian dot product, so something likeus†ur. Secondly, we could’ve also equally defined us = us†γ0, and we also check the normalisationusur. It turns out, both will be useful.

Let’s do the first contraction.

us†ur =(ξs†

√p · σ ξs†

√p · σ

)(√p · σξr√p · σξr

)= ξs†(p · σ)ξr + ξs†(p · σ)ξr

= 2p0ξs†ξr

= 2p0δrs

We can see clearly that this is not a necessarily Lorentz invariant contraction since p0 changesunder boosts. Now, lets do the other type of contraction which is indeed Lorentz invariant! Wehave that

us†ur =(ξs†

√p · σ ξs†

√p · σ

)(0 II 0

)(√p · σξr√p · σξr

)= 2ξs†

√(p · σ)(p · σ)

= 2mδrs

In the above we’ve used the relation we proved earlier that (p · σ)(p · σ) = m2.

Now, what about the ’negative frequency’ spinors vs(~p)? Well, it is easy to also check that thesespinors obey similar normalisation conditions, up to a minus sign.

vs†vr = 2p0δrs, vsvr = −2mδrs

Finally, something we will also have to use frequently in the coming section is the ’outer product’of the spinors. In this case, instead of doing a dot product, we sum over the spinors in the otherway round to create a matrix.

Claim. (Outer products of spinors) We have that the outer products of the two spinors us(~p)and vs(~p) satisfy ∑

s=1,2

us(~p)us(~p) = /p+m

And similarly, we have that our negative frequency solution acts somewhat in a conjugate fashion,with ∑

s=1,2

vs(~p)vs(~p) = /p−m

Proof. Proving this is just a simple matter of writing out the definitions, and using the factthat the outer product ξsξs† and ηsηs† is just the identity matrix. Similarly, we have that forvs(~p)vs(~p), we get that

vs(~p) =

( √p · σηs

−√p · σηs

), vs(~p) =

(−ηs†

√p · σ ηs†

√p · σ

)


Now, doing the outer product and summing over s gives∑s

vs(~p)vs(~p) =∑s

(−√

p · σηsηs†√p · σ √

p · σηsηs†√p · σ√p · σηsηs†

√p · σ −√

p · σηsηs†√p · σ

)=

(−mI p · σp · σ −mI

)

Now, with the observation that /p = γµpµ, which is

/p =

(0 p · σ

p · σ 0

)we are left with the result that

vs(~p)vs(~p) = /p−m

where the −m is implicitly multiplied by the identity matrix.

6.15 Quantising the Dirac field

In conclusion, we’ve found two plane wave solutions of the Dirac equation, which, for cleanliness,we will write as us(~p) and vs(~p). Recall, the s index let’s us choose between ξ1 or ξ2 in thedefinition us(~p), or η1 or η2 in the definition of vs(~p). When we take these solutions and writethem out in Fourier modes, we have that

ψ(~x) =

2∑i=1

∫d3p

(2π)3√2E~p

[bs~pu

s~pei~p·~x + cs†~p v

s~pe−i~p·~x

]ψ†(~x) =

2∑i=1

∫d3p

(2π)3√2E~p

[bs†~p u

s†~p e−~p·~x + cs~pv

s†~p e

i~p·~x]

Analogous to our previous case in nucleon anti-nucleon quantisation, we have that bs~p and bs†~prepresent the annihilation and creation operators for u(~p)s, and that cs~p, c

s†~p represent annihilation

and creation for vs(~p). In spinor quantisation, it turns out that we require anti commutationsrelations instead of commutation relations, with

A,B = AB +BA

We’ll go into more detail about why we need anti-commutation relations later, but for now justacknowledge that this is due to the fact that if we were to impose commutation relations, we’llcome across the problem of unbounded negative energy. The specific anti-commutation relationsare:

ψα(~x), ψβ(~y) = 0 =ψ†α(~x), ψ

†β(~y)

,ψα(~x), ψ

†β(~y)

= δαβδ

3(~x− ~y)

In the same vein as the case we had with scalar fields and nucleons, we construct a set ofcommutation relations on the creation and annihilation operators which are logically equivalentto our anti-commutation relations. We can prove that these are equivalent to

br~q, bs†~q

= (2π)3δ3(~p− ~q)δrs =

cr~p, c

s†~q


To show that this is true, we’ll do just one direction and check that our anti-commutationrelations hold true.

To show this, observe thatψ,ψ†

=∑r,s

∫d3pd3q

(2π)62√EpEq

[bs~p, b

r†~q

us~pu

r†~q e

i·(~p·~x−~q·~y +cs†~p , c

r~q

vs~pv

r†~q e−i(~p·~x−~q·~y)

]=∑r,s

∫d3pd3q

(2π)62√EpEq

[(2π)3δrsδ(~p− ~q)us~pu

r†~q e

i(~p·~x−~q·~y) + (2π)3δrsδ(~p− ~q)vs~pvr†~q e−i(~p·~x−~q·~y

]=∑s

∫d3p

(2π)32Ep

[us~pu

s~pγ

0ei~p·(~x−~y) + vs~pvr~pγ

0e−i~p·(~x−~y)]

=

∫d3p

(2π)32Ep

[(/p+m)γ0ei~p·(~x−~y) + (/p−m)γ0e−i~p·(~x−~y)

]=

∫d3p

(2π)3m

Epei~p·(~x−~y)

=

∫d3p

(2π)3Ep

[(γ0p0 − γipi +m)γ0ei~p·(~x− ~yu) + (γ0p0 + γipi −m)γ0ei~p·(~x−~y)

]= δ(~x− ~y)

Going into the second last line, we switched the 3 momentum variable ~p → −~p in the secondterm. Going into the last line, we also used the fact that (γ0)2 = 1.

We have that in the last line, since our measure is Lorentz invariant, we boosted into the restframe so that m cancels with Ep.

We do the Legendre transform to get our Hamiltonian density. It’s easy to see that our conjugatemomenta π is π = iψ†. This means that our Hamiltonian looks like

H = πψ − L= iψ†ψ − iψγ0∂0ψ − iψγi∂iψ +mψψ

= ψ(−iγi∂i +m)ψ

By the way, one needs to be careful that the index i here only sums over the spatial derivatives(this makes sense since the Hamiltonian shouldn’t be time independent anyway since it’s aconserved quantity). If we plug in ψ, ψ from the above, we use anti commutation relations andsome reuslts on inner products of spinors to get that

ur†~p us~p = vr†~p v

s~p = 2p0δ

rs, ur†~p vs~p = 0 = vr†~p u

s~p

This means that for our Hamiltonian, we fortunately get a positive definite quantity

H =

∫d3p

(2π)3Ep

2∑s=1

(bs†~p b

s~p + cs†~p c

s~p

)To derive this, the best way to go about doing this is to look at the operator

(−iγi∂i −m

)on

us(~p) and vs(~p). Due to Dirac’s equation, we eliminate these things in favour of γ0p0, and thenuse this to find what the above is. Then we substitute that back into our expression for H.


If we used commutation relations, we get a minus sign which means that we have unboundedlower energy and therefore unstable theory.

6.16 The Dirac Hole Interpretation

We can write the Dirac equation as

i∂ψ

∂t= (−iα · ∇+mβ)ψ, α = −γ0γ, β = γ0

The term in the brackets is considered as the 1-particle Hamiltonian H. This has positive andnegative energy solutions. There’s no consistent way of realising the Dirac equation to oneparticle states.

6.17 Fermi-Dirac Statistics

We expanded our spinor field with operators b and c, which had a spin index. These operatorsannihilate the vacuum

bs~p |0〉 = 0 = cs~p |0〉

Recall that these operators obey anti-commutation relations. We can check that these commutewith the Hamiltonian as follows

[H, br†~p ] = Epbr†~p ,

[H, br~p

]= −Epbr~p[

H, cr~p]= −E~pcr~p

[H, cr†~p

]= E~pc

r†~p

If we index this state, for example, as |~p1, r1〉 := br1†~p1 |0〉. Then the anti-commutation relationsgive us a sign change when we swap two things around. In particular, we have that

|~p1, r1; ~p2, r2〉 = − |~p2, r2; ~p1, r1〉

You can check this yourself.

6.17.1 Going into the Heisenberg picture

To study propagators in the theory, we make the operators ψ and ψ time dependent by movinginto the Heisenberg picture. Now, we have a time dependent operator ψ(x) satisfying ∂ψ

∂t =i [H,ψ] which is solved by the Heisenberg picture expansion

ψα(x) =2∑s=1

∫d3p

(2π)3√2Ep

(bs~pu

s~pαe−ip·x + cs†~p v

s~pαe

ip·x)

with an analogous expression for ψ†α. Honestly, the only thing we’re doing here is attachinga time dependence to the exponential previously. Now, we are in a position to define the


Feynman propagaator but in the case of these fermionic fields. We then define, in analogy with∆(x− y) = [φ(x), φ(y)], that for R scalars,

iSαβ(x− y) =ψα(x), ψβ(y)

where Sαβ is a four by four matrix. For brevity, we’ll now drop the α, β for brevity. Substitutingin our results from previously, we have that

iS(x− y) =∑r,s

∫d3pd3q

(2π)6√4EpEq

[bs~p, b

r†~q

e−i(p·x−q·y)us~pu

r~q +

cs†~p , c

r~q

vs~pv

r~qei(p·x−q·y)

]The anti commutators in this expression yield delta functions. So, this simplifies to the expres-sion

iSαβ(x− y) =

∫d3p

(2π)e2Ep

[∑s

us~pαus~pβe−ip·(x−y) +

∑s

vs~pαvs~pβeip·(x−y)

]Earlier we showed that the outer products of the spinors u and v summed over the spin indicesgive /p+m and /p−m. This thing then simplifies to

=(i/∂x +m

)D(x− y)−

(i/∂x +m

)D(y − x) =

(i/∂x +m

)(D(x− y)−D(y − x))

Note that for (x−y)2 < 0, we have that D(x−y)−D(y−x) = 0. We now haveψα(x), ψβ(y)

=

0∀(x − y)2 < 0. So what about causality? Our observables are bilinear in fermions. They docommute at space-like separations so the theory is causal.

Away from singularities, we have that

(i/px −m)iS(x− y) = 0

= (i/∂x −m)(i/∂ +m) [D(x− y)−D(y − x)]

= −(∂2x +m2) [D(x− y)−D(y − x)]

= 0 using pµpµ = m2

6.17.2 The Feynman Propagator

A similar calculation gives us us that the two point propagator between x and y are

〈0|ψα(x)ψβ(y) |0〉 =∫

d3p

(2π)32Ep(/p+m)αβe

−ip·(x−y)

〈0|ψβ(y)ψα(x) |0〉 =∫

d3p

(2π)32Ep(/p−m)αβe

ip·(x−y)

If we define SFαβ = 〈0|Tψα(x)ψβ(y) |0〉 as the object〈0|ψα(x)ψβ(y) |0〉 x0 > y0

−〈0|ψβ(y)ψα(x) |0〉 y0x0


The negative sign in the second term is required for Lorentz invariance when (x− y)2 < 0, sincethere exists no Lorentz invariant way to determine whether x0 < y0 or the other way around.In this case,

ψ(x), ψ(y)

= 0 and so T as defined is lorentz invariant. This is the same for

strings of fermionic operators in T - they anti commute. We see the same behaviour for normalordering, we get that

: ψ1ψ2 := − : ψ2ψ1 :

We can define our Feynman propagator as

ψ(x)ψ∗(y)

From this we get thatT (ψ(x), ψ(y) =: ψ(x)ψ(y)

Our propagator before can be written as a 4-momentum integral, which bears resemblance toour propagator for scalar fields.

SF (x− y) = i

∫d4p

(2π)4e−ip·(x−y)

p2 −m2 + iε(/p+m)

This propagator acts as a Green’s function for the Dirac equation. Writing /∂x to denote thederivative terms acting on x, we have that

(i/∂x −m)SF (x− y) = iδ4(x− y)

As an example, let’s look at Fermionic Yukawa theory. Our Lagrangian here looks like

L =1

2∂µφ∂

µφ+1

2µ2φ2 + ψ(i/∂ −m)ψ − λφψψ

We ask ourselves, is this theory normalisable? To answer this question, we need to figure out themass dimension of our coupling constant λ. A simple analysis of the mass dimension shows that[φ] = 1, [ψ] = 3

2 = [ψ′], which implies that [λ] = 0. In the language of statistical field theory, wecall this interaction term marginal. Thus, we have that our theory is renormalisable.

Example 3. (Nucleon to Nucleon scattering) Let’s revisit our problem of nucleon to nucleonscattering, ψψ → ψψ but treating are nucleons as fermions. Let’s label our initial state astwo nucleons with incoming momentum ~p and ~q, as well as spin indexes r and s, so that |i〉 =√

4EpEqbs†~p b

r†~q |0〉. Similarly, we label |f〉 as the state

√4Eq′Ep′b

s′†~p′ b

r′†~q′ |0〉.

This is shown in the following Feynman diagram

In computing the scattering amplitude, we’ll need to take the Hermitian conjugate of |f〉, whichis

〈f | =√4Ep′Eq′ 〈0| br

′

~q′bs′

~p′


We don’t pick up a minus sign here - we have to be careful. This is because we are just takingthe Hermitian conjugate. Let’s look at O(λ2) scattering terms from the fields in the interactionpicture. Our only non zero contribution to second order is given by Wick’s theorem

〈f | : ψ(x1)ψ(x1)ψ(x2)ψ(x2)φ(x1)φ(x2) : |i〉

Since in this configuration, the ψ ’s annihilate |i〉 and the ψ’s create 〈f |. We’ll now work oncalculating the scattering amplitude we get from this problem. The first step to do is to expandout ψ in terms of our creation and annihilation operators. This gives us that

: ψ(x1)ψ(x1)ψ(x2)ψ(x2) : bs†~p b

r†~q |0〉 = −

∫d3k1d

3k2(2π)6

[ψ(x1) · um(~k1)

] [ψ (x2) · un(~k2)

]=e−ik1·x1−ik2·x2√

4Ek1Ek2bm~k1bn~k2bs†~p b

r†~q |0〉

The terms in square brackets like[ψ (x1) · um

]illustrate how we’re summing over indices. Let’s

just be clear about what objects we’re dealing with here. For any value of m, un(~p) is afour component object, and so are the spinors ψ, so we’re taking the dot product here. Weadditionally need to sum over the values m and n, for the different spins. Note that we alsohave an extra minus sign in the beginning because we need to commute ψ(x1) past ψ(x2), sincewe need to move the b~k1 past the b† operator nested in ψ(x2).

The next mission is to commute the b operators past the b† operators. We do this by looking atanti commutation relations. We have that

bm~k1bn~k2bs†~p b

r†~q |0〉 =

(δmrδnsδ(~k2 − ~p)δ(~k1 − ~q)− δnrδmsδ

(~k1 − ~p

)δ(~k2 − ~p)

)|0〉

When we integrate over the delta functions, we have that the above expression

=−1

2√EpEq

([ψ(x1) · ur(~q)

] [ψ(x2) · us(~p)

]e−p·x2−iq·x1

−[ψ(x1) · us(~p)

] [ψ(x2) · ur(~q)

]e−ip·x1−iq·x2) |0〉

Now, we contract this with 〈f |. We’ll go slow and just do one term first. For the first term inour expression, we get, writing out 〈f | explicitly, the term

〈0| bs′~q′bs′

~p′[ψ(x1) · ur(~q)

] [ψ(x2) · us(~p)

]|0〉

Now, ignoring c operators again, we bring out the b† operators which come from ψ in theexpansion. This gives us a string of b operators on the left, which leaves the above expressionas

=eip

′·x1+iq′·x2

2√Ep′Eq′

[us

′(~p′) · ur(~q)

] [ur

′ (~q′)· us(~p′)

]− eip

′·x2+iq′·x1

2√Ep′Eq′

[ur

′ (~q′)· ur (~q)

] [us

′ (~p′)· us(~p)

]


By symmetry, the second term in our expression also gives the same contribution. Includingthe relativistic normalisation from the |i〉 and |f〉 states, our final expression for our scatteringamplitude is, including the expression for our contraction φ(x1)φ(x2), is given by

(−iλ)2∫d4x1d

4x2d4k

(2π)4ieik·(x1−x2)

k2 − µ2 + iε([us

′(~p′) · us(~p)

] [ur

′ (~q′)· ur(~q)

]eix1·(q

′−q)+ix2·(p′−p)

−[us

′ (~p′)· ur(~q)

] [ur

′(~q) · us(~p)

]eix1·(p

′−q)+ix2·(q′−p))

After integrating over x1 and x2, we can extract our scattering amplitude from the formula〈f |S − 1 |i〉 = iM(2π)4δ4(p + q − p′ − q′). Once we do this, our scattering amplitude is givenby

M = (iλ)2

[us

′(~p′) · us(~p)

] [ur

′(~q′) · ur(~q)

](p′ − p)2 − µ2 + iε

−

[us

′(~p′) · ur(~q)

] [ur

′(~q′) · us(~p)

](q′ − p)2 − µ2 + iε

6.18 Momentum Space Feynman Rules for Fermion Amplitudes

We are now in a good position to guess the Feynman rules for fermions. We can use our intuitionto guess these rules. Compared to momentum space Feynman rules for scalar fields, we needto attach a spin as well as a momentum for each incoming line. The factor we attach is aspinor.

• From our amplitude above, it’s clear that incoming particle with momentum ~p and a givenspin s contributes a us(~p) spinor to our amplitude. Similarly, outgoing particles contributewith spin r and momentum ~p contribute a ur(~p) spinor factor.

p

ur(p)

p

ur(~p)

• This is flipped when we consider anti-fermions. For an incoming particle, we associate theopposite frequency spinor v and for outgoing particles we associate the spinor v. For anti-fermions of momentum p and spin r, we associate the spinor vr(~p) for ingoing particles.For outgoing particles, we associate the spinor vr(~p).

p

vr(p) p

vr(~p)

• For vertices in our Feynman diagram, we attach a factor (−iλ)


• For each internal line, for scalars, we attach our standard propagator

~p

=i

p2 − µ2 + iε,

~p

=i(/p+m2

)p2 −m2 + iε

Importantly, we have that fermion lines need to move consistently throughout the diagram.This reflects fermion number conservation.

• Since we’re attaching spinors, which are vectors, to the external legs in the diagram, weneed a sensible way to turn these into a number which represents the probability. This issimple, all we do is contract the spinors which join at a vertex.

• As before, we need momentum conservation at each vertex. Propagators with undeter-mined momenta require an integral over their momentum.

• The difference with drawing diagrams for fermions versus scalar particles, is that we needto take into account fermion statistics when we switch around legs in fermion diagrams.

Example 4. (Nucleon scattering Fermion diagrams) At first order, our possible diagrams fornucleon to nucleon scattering are given below. We read of the associated amplitude from this

+

p, s

q, r q′, r′

p′, s′p, s

q, r

q′, r′

p′, s′

Figure 11: The Feynman diagram associated with nucleon to nucleon scattering. Notice thatwe have Fermi-Dirac statistics, so our switching of momentum leads to a minus sign picked upin the amplitude.

Feynman diagram to get

M = (−iλ)2[us

′(~p′) · us(~p)

] [ur

′(~q′) · ur(~q)

](p− p′)2 − µ2

−

[us

′(~p′) · ur(~q)

] [ur

′(~q′) · us(~p)

](p− q′)2 − µ2

Let’s justify this picture and the terms in it. In the first diagram, we have two fermions for thetop vertex - one going in and one going out with momenta ~p and ~p′ and these are contractedtogether. In addition, we have on the bottom vertex momenta ~q and ~q′, and we contract thespinors associated with these in a similar fashion.

Example 5. (Fermionic loops contribute a minus sign) It doesn’t matter so much for scalarfields, but we need to remember minus signs when dealing with diagrams. Fermionic loops


contribute a minus sign to our overall amplitude. We can see this in the following fermionic loop

Let’s label the left vertex x and the right vertex y. Our arrow pointing from x to y representsthe contraction between φ(x) and φ(y). The arrow pointing in the other direction represents acontraction from φ(y) and φ(x). However, Wick’s theorem gives us that our interaction term(without the contractions between φ term s) is proportional to

ψ∗α(x)ψα(x)ψ∗β(y)ψβ(y)

We have the following diagrams. The important thing is that operators need to remain the sameorder, unless we decide to anti-commute them and pick up a minus sign. Hence, for a closedfermionic loop which looks like : ψα(x)ψα(x)ψ

∗β(y)ψβ(y) : where we contract the middle two

fields, to contract the other two fields we need to pick up a minus sign (show here).

Question: what if L int = −λφψα(γ5)αβψβ ? This interaction preserves parity if φ is a pseudo-scalar, in other words

Pφ(~x, t) = −φ(−~x, t)This interaction looks like the below (diagram here) this contributes (−iλ)(γ5)αβ.

Example 6. (Nucleon to Meson Decay) Now we study the amplitudes for the problem ψψ → φφ.The key thing about this problem is to think about the statistics of the outgoing mesons. Sincethey’re Bosons, we don’t get the minus sign from switching the legs like last time. We assignthe momenta as shown in the figure.

6.19 Cross-Sections and Mod-squared expressions

We want to calculate amplitudes and mod-squared expressions like |M|2 in this formalisationwith fermions? We know how to do this in the case with scalar fields ( our scattering amplitudewill depend on the ingoing and outgoing momenta), but we’ve yet do discuss how to do thiswhen we’ve added different options for spin and so on. It turns out, probably for consistencypurposes, is that the way people do this is to also formulate our scattering amplitude so thatit also depends just on initial and final momenta, and not spin. The way we remove thedependence on spin is by averaging over the spins in the initial states, and summing over thespins in the final states. For example, if we consider the case of ψψ → ψψ decay, we need toinitialise a choice of spins r, s, s′, r′ for each initial and final fermion state.

|i〉 =√

4E~qE~pbr†~p b

s†~q |0〉

|f〉 =√

4E~q′E~p′bs′†~p′ b

r′†~q′ |0〉


Thus, we need find a scattering amplitude that’s a function of the momenta p, q, p′ and q′ butnot the spins?

In most experiments, and in the formalisation that follows, we make the big assumption thatthe spin states of incoming particles |i〉 are randomly distributed. Then, because the incomingspin states are random and so we average over them. For example, for ψψ, it would be 1

4

∑2r,s=1,

since we average over all the possible pairs of (r, s) with the assumption that the probability ofeach pair is14 .

Another reason we do this is that typically the spin states of the final state can’t be measured,so we sum over these.

M = B −A, in ψψ → ψψ

where B,A are the different terms. We wrote appropriate spin sums averages with a line ontop.

|M|2 = |A|2 + |B|2 −A†B −B†A

A =λ2[us~p′ · u

r~q

] [ur

′

~q′ · us~p

]µ2 − µ2 + iε

Note that we have (γ0)† = γ0.

|A|2 = |λ|4

4(µ2 −m2)2

∑r,s,s′,r′

us′

~p′αur~qαu

r~qβus

′

~p′βur

′

~q′βus~pγu

s~pδur

′

~q′δ

This is equal to|λ|4

4

(/p′ +m)αβ(/q +m)βα tr[(/q′ +m)(/p+m)

](µ2 −m2)2

Often, we are in the high energy limit, where we may wish to neglect particle masses. In thiscase, we then find that

|A|2 =|λ|4 tr

(/p′/q)tr(/q′/p)

4µ2

Similarly, we have that for |B|2, we get that

|B|2 = |λ|2

4t2tr(/q′/q)tr(/p′/p)

We also want −A†B −B†A = −2Re(A†B). We calculate this as

A†B =|λ|4

4ut

∑r,r′,s,s′

ur~qβus′

~p′βus~pαu

r′

~q′αur

′

~q′γur~qγu

s′

~p′δus~pδ

=|λ|2

4uttr(/p/p′/q/q′)

This gives us the Feynman rules for spin summed M2 diagrams. We have that complex conju-gation switches |i〉 and |f〉 in the diagram. Fermion lines are joined with identical momentum


on the LHS and RHS. Adter a spin sum, a closed fermion line in the |M|2 diagram is given bya trace over γ matrices, with appropriate γ5 ’ s etc in vertices at the correct posiition in thetrace. trace follows fermion arrows backwards.

IN nucleon to nucleon scattering, we have that

|M|2 = |λ|2

4

tr /q/q′ tr

(/p/p′)

t2+

tr(/q′/ptr

(/p′/q))

u2−

2R tr(/p/p′/q/q′

)ut

Using trace techniques from sheet three we can show that

|M|2 = |λ|4

4

[4q · q′4p · p′

t2+

4q′ · p4p′ · qu2

− 8

ut(p · q′p′ · q + p · p′q · q′ − p · qp′ · q′)

]It’s customary to express this in terms of s, t and u. We have that, in the m,µ → 0 limit, weget that

s = (p+ q)2 = (p′ + q′)2 =⇒ p · q = p′ · q′ = s

2

t = (p− p′)2 = (q − q′)2 =⇒ p · p′ = q · q′ = − t

2

u = (p− q′)2 = (q − p′)2 =⇒ p · q′ = q · p′ = −u2

This implies that|M|2 = |λ|4

1 + 1− 1

2ut

(u2 + t2 − s2

)Adding all the Mandelstam variables, so that

s+ t+ u =∑

m2i → 0, u = −s− t, =⇒ |M|2 = 3|λ|4

Going into the centre of mass frame, we have that

dσ

dt=

|M|2

16πλ(s, 0, 0).

dt

d cos θ| COM = 2|~p||~p′| = s

2

Writing out our solid angle Ω, we have that dΩ = d (cos θ) dφ. This means that

dσ

dΩ| COM =

|M|2

64π2s

We have identical particles in the final state. We integrate our final state angles over thehemisphere. This implies that our cross section is

σ =3|λ|4

32πs

This implies that σ > 0 even with negative quantum interference. This means that [λ] = 0, whichmeans that [σ] = [s]−1 = −2, which is an area. This means that everything is correct.


Summary

Trace technology

We use these trace relations to simplify cross-section calculations. We use anti-commutationrelations to prove this.

• Useful ones

tr (γµγν) = 4ηµν , tr(γαγβγγγδ

)= 4

(ηαβηγδ − ηαγηβδ + ηαδηβγ

)• The trace of an odd number of gamma matrices vanishes.

•

7 Quantum Electrodynamics

In this section, we’ll first look at the reason why Aµ has four components, but physically wehave only two distinct polarisation states for a photon. In electromagnetism, our Lagrangian isgiven by

L = −1

4FµνF

µν , where Fµν = ∂µAν − ∂νAµ

is the field strength tensor. As we shown in general relativity, this Lagrangian is motivated bythe fact that it’s pretty much the only thing we can write down in four dimensional space time,given a two form F . It comes from the action

S =

∫F ∧ F

Our Euler Lagrange equations read

∂µ

(∂L

∂(∂µAν)

)= 0 = ∂µF

µν

These are Maxwell’s equations in vacua. This can be written in terms of our fields as

~E = −∇φ− A

~B = ∇×A

Our sign convention is such that ∇ = (∂x, ∂y, ∂z), and Aµ =(φ, ~A

). From our definitions, we

have that ~E = (F01, F02, F03) =(−F 01,−F 02,−F 03

). This equality comes from the fact that

F0i = ∂0Ai − ∂iA0. In addition, we also have that for ~B = (B1, B2, B3) gives for example,B3 = ∂1A

2 − ∂2A1. Hence,

Fµν =

0 E1 E2 E3

−E1 0 −B3 B2

−E2 B3 0 −B1

−E3 −B2 B1 0


Our field strength tensor satisfies the Bianchi identity, where

∂λFµν + ∂µFνλ + ∂νFλµ = 0

When we set λ = 3, µ = 1, ν = 1, we have the equation ∇ · ~B = 0. Similarly, we have that forλ = 0, µ = i, ν = j then for i 6= j, we get that

B = −∇× ~E

Also,∂µF

µν = 0 =⇒ ∇ · ~E = 0, ~E = ∇× ~B

The massless vector field Aµ satisfies 4 real degrees for freedom, µ = 0, 1, 2, 3, but a photon γonly has 2 polarisation states. Note that A0 is not dynamical - there is no kinetic term in L.This means that in particular, our conjugate momentum π0 = 0, and solving for the equationsof motion we have that

∂L∂A0

= 0 6= an explicit function of t

This means we can also sub-out A0 in favour of the other degrees of freedom. Given Ai(~x, t) andAi(~x, t0), we have that A0 is fully determined. The constraint that ∇· ~E = 0 implies that

∇2A0 +∇ · ~A = 0

This has solutionA0(~x, t0) =

∫d3x′

∇ · ~A(~x′, t0)4π|~x′ − ~x|

where we have that

∇2

(1

| ~X − ~x′|

)= −4πδ3(~x− ~x′)

In addition there is a large symmetry group

Aµ(x) → Aµ(x) + ∂µλ(x)

where λ(x) is some function such that lim|~x|→∞ λ(x) → 0. It’s easy to show that Fµν is invariantunder this transformation. This is a new kind of symmetry, viewed as a redundancy in thedescription. Observe that

ηµν∂ρFρν = 0 =⇒ [ηµν∂ρ∂

ρ − ∂µpν ]Aν = 0

This operator, Oµν is not invertible since it annihilates any function of the form ∂νλ(x). There’sno way to uniquely determine the evolution of Aµ given Ai, Ai at t0. We can’t distinguishbetween Aµ and Aµ + ∂µλ which we identify as the same physical state. Below we have adiagram of gauge orbits in configuration space. We choose one representation from each gaugeorbit by cutting a slice. All states in on the orbit are related to by a gauge transformationand so are physically equivalent. It doesn’t matter which representative we choose from each,since they’re all physically equivalent. There are many possibilities, some of which make certaincalculations easier. Let’s go through some examples of gauges.


• The Lorentz gauge ∂µAµ = 0. We can always pick a representative that satisfies this.Start with a field A′µ such that ∂µA

′µ = f(x). Then we choose Aµ = A′µ + ∂µλ(x)where ∂µ∂

µλ(x) = −f(x), which we can always solve. This condition doesn’t pick aunique representative from the orbit, because we can make a further transformation with∂µ∂

µλ(x) which has non trivial solutions. The advantage of this is that the gauge conditionis Lorentz invariant.

• The coulomb Gauge or radiation gauge. Here we set ∇·A = 0, which exists by an argumentsimilar to the Lorentz gauge. This means that A0 = 0. The advantage is that it’s easyto see physical degrees of freedom. We have three comonents of ~A satisfying a physicalconstraint, leaving behind 2 physical degrees of freedom which are our polarisation states.

7.1 Quantisation of the Electromagnetic Field

We can quantise the field ay first looking at the conjugate momenta.

π0 =∂L∂A

= 0

πi =∂L∂Ai

= −Ai + ∂iA0 = F i0 = Ei

Now, when we put this into the Hamiltonian, we find that our expression is

H =

∫d3x

(πiAi − L

)=

∫d3x

1

2

(| ~E|2 + | ~B|2

)−A0 (∇ · E)

now, the element A0 acts as a Lagrange multiplier, which imposes the equation ∇ · ~E = 0(by the Euler Lagrange equations). Let’s recall the classical dynamics interpretation of this Toreview this interpretation, check out David Tong’s notes on Classical Dynamics. Our form ofthe Hamiltonian above is reminiscent of the Lagrangian of a system with constraints given byf i(~xa) = 0. The Lagrangian looks like

L′ = L+ λbfb(~xa)

Now, we identify A0 with the Lagrange multiplier here, and view ∇·E = 0 as the constraint. Nowthat it’s a constraint of a system, we can treat ~A just as the physical degrees of freedom.

7.1.1 Coulomb Gauge

In the Coulomb gauge, we’re working with the condition that ∇ · ~A = 0. From our discussionearlier, we learnt that we could substitute A0 in favour of ~A via the formula

A0(~x) =

∫d3x′

∇ ·(∂A∂t (~x)

)4π|~x− ~x′|

However, when we commute our time derivative operator with ∇, we simply get that the inte-grand is 0. Hence, we have that A0 = 0. Now, from our equation of motion, ∂µFµν = 0, we


have that

∂µ (∂νAµ − ∂µAν) = 0

∂µ∂νAµ − ∂µ∂

µAν = 0

∂ν(A0 −∇ ·A

)− ∂µ∂

µAν = 0

∂µ∂µAν = 0

In the last line, we used the Coulomb gauge where ∇ · ~A = 0 and also the fact that A0 was aconstant.

7.1.2 Lorentz Gauge

We’ll now work in the Lorentz gauge. The Lorentz gauge is what we get when we impose thecondition ∂µA

µ = 0, which is different from the Coulomb gauge because this is a divergence inspace-time.

This implies that the Maxwell’s equations which we started with ∂µFµν = 0, becomes the

relation∂µ∂

µAν = 0

We can view this from a different perspective, and construct a different Lagrangian where theLorentz gauge is imposed from the equations of motion. To do this, we modify the Lagrangianand add an extra term, so that it becomes

L = −1

4FµνF

µν − 1

2(∂µA

µ)2

Note that this Lagrangian is no longer gauge invariant under the gauge transformation of thefield Aµ → Aµ+∂µχ. But, the upshot of writing our Lagrangian in this form is that we get

∂µFµν + ∂ν (∂µA

µ) = 0 ⇐⇒ ∂µ∂µAν = 0

which is the condition we imposed before. We will now work with this new Lagrangian, only im-posing ∂µAµ = 0 later, at the operator level. In general, we could’ve written a Lagrangian

L = −1

4FµνF

µν − 1

2α(∂µA

µ)2

where this gives rise to a continuum of different theories. α = 1 is called the Feynman gauge,and α → 0 is called the Landau gauge. For illustration purposes, we’ll look at the Feynmangauge. Now, for α = 1, we have no gauge symmetry. But now, both A0 and the variables ~A aredynamical. We have that

π0 =∂L∂A0

= −∂µAµ

and we also have that∂i =

∂L∂Ai

= −Ai + ∂iA0

We impose the commutation relations

[Aµ(~x), Aν(~y)] = 0 = [πµ(~x), πν(~y)]


As well as a non-trivial commutation relation

[Aµ(~x), πν(~y)] = iδ3(~x− ~y)ηµν

We now expand our field Aµ(~x) in terms of this.

Aµ(~x) =

∫d3p

(2π)3√2|~p|

3∑λ=0

(ελµ(~p)a

λ~pei~p·~x + ελ∗µ (~p)aλ†~p e

−i~p·~x)

where ελ=0,1,2,3µ are four polairsation vectors. Similarly, we have that for our conjugate momen-

tum

πµ =

∫d3p|~p|

(2π)3√2i

3∑λ=0

(εµλ(~p)aλ~pe

i~p·~l − aλ∗~p e−i~p·~x (εµ∗(~p))λ

)In this formalisation we pick ε0µ to be timelike, and εiµ to be timelike. We choose normalisationελ = ε∗λ = ηλλ

′ We choose ε1µ and ε2µ to be transverse, in other words we have ε1 · p = ε2 · p = 0,and ε3µ to be our longitudinal polarisation. For example, we take γ to be travelling in an x3

direction. We set

pµ = |~p|

1001

, ε0µ =

1000

, ε1µ =

0100

, ε2µ =

0010

, ε3µ =

0001

The relations above are equivalent to the creation operators with[

aλ~p′ , aλ~q

]= 0 =

[aλ†~p , a

λ′†~q

]However, our non zero commutation relation is given by the following[

aλ~p , aλ†~q

]= −ηλλ′ (2π)3 δ3(~p− ~q)

We’re in a bit of trouble here since we have an minus sign in the commutation relation here! Thisis okay for λ, λ′ = 1, 2, 3 due to the fact that the Minkowski metric is negative in these indices,but is unusual for time-like photons, as we will see now. We have that aλ~p |0〉 = 0 as usual. Now,as usual we have |~p, λ〉 = aλ†~p |0〉 is fine for λ = 1, 2, 3 but actually, we have that

〈~p, λ = 0|~q, λ = 0〉 = 〈0| a0~pa0†~q |0〉 = −(2π)3δ3 (~p− ~q)

This is a negative norm state.! A Hilbert space with a negative norm state is problematic, butthe constraint equation comes to the rescue - so far we’ve yet to impose the equation ∂µAµ = 0,so we expect this to restrict our Hilbert space and resolve this problem. We’ll try to resolve theproblem by trying out new definitions of what physical states actually could be. Working in theHeisenberg picture, we have that

1. ∂µAµ = 0 which doesn’t work since π0 = −∂µAµ so commutation relations couldn’t beobeyed.


2. Impose conditions on Hilbert space. We split the state |φ〉. ∂µAµ |ψ〉 is too strong of a

condition. We split the integral up into two parts in the Heisenberg picture, into a partthat annihilates our vacuum state |0〉 and a part that doesn’t. We set

A+µ (x) =

∫d3p

(2π)3√2|~p|

3∑λ=0

ελµaλ~pe−ip·x

A−µ (x) =

∫d3p

(2π)3√2|~p|

ελ∗µ aλ†~p e

ip·x

Notice that in this formalisation, we have that A+µ |0〉 = 0, but unfortunately we have that

∂µA−µ |0〉 6= 0, so we can conclude that this is not the right way to classify physical states,since our vacuum state isn’t even a physical state itself.

3. The correct way to do this is to classify the physical states as states which satisfy thecondition

∂µA+µ (x) |ψ〉 = 0

This is the ’correct way’ to deliver the condition that ∂µAµ = 0 in the Lorentz gauge. Wehave that for physical states |ψ〉 and |ψ′〉, we have that 〈ψ′| ∂µAµ |ψ〉 = 0. In addition,we have that |0〉 is also counted as a physical state, so we don’t have the same problemswhich we encountered above. Now, by linearity, we have that the linear span of physicalstated constitute a Hilbert space themselves, which we denote as Hphys . So, this makessense.

Since our operators aλ~p and aλ†~p mutually commute, we can decompose the Fock space (the Hilbertspace which is generated by applying the operators aλ† to the vacuum state |0〉) into the directsum of two states. We decompose the state

|ψ〉 = |ψT 〉 |φ〉

where |ψT 〉 is the subspace of states generated by applying the creation operators a1,2†~p to |0〉,and |φ〉 is the space of states generated by applying a0,3† to |0〉. The states in the |ψT 〉 partrepresent transverse polarised states, and in the |φ〉 represent longitudinal polarised states.Taking p = (1, 0, 0, 1), we have that the operator takes the form

∂µA†µ = ∂µ∫

d3p

(2π)31√2|~p|

(3∑

λ=0

εrµar~p

)e−ipµx

µ

=

∫d3p

(2π)31√2|~p|

3∑λ=0

εrµar~p (−ip

µ) e−ipνxν

This, when we operate on it with |ψ〉, gives us the condition(a3~k

− a0~k

)|φ〉 = 0. This is significant

because it means that whenever a state contains a time-like photon of momentum ~p, generatedby a0†~p , then by the condition above (which is called the Gupta-Blauer condition), we have thatthe state necessarily contains a longitudinal photon, which is generated by a3†~p .

This is an important observation - it tells us that in the state |φ〉, our Hilbert space has a basis|φn〉 which denotes the number of pairs of a time-like photon and a longitudinal photon. Thus,


we have that|φ〉 =

∞∑n=0

cn |φn〉

The Gupta-Blauer condition gives us that the states obey an orthogonality relation

〈φm|φn〉 = δm0δn0

So states with transverse and longitudinal photon pairs have zero norm. This is problematic,but we can bunch physical states into the equivalence class if they differ by just transverse andlongitudinal pairs, and this solves the problem.

7.2 Timelike γ pairs continued

What’s the main point of this section? Why do we care about photon pairs? It seems to methat this whole section is just about being consistent with gauge choice. We have asserted thecondition (

a3~k − a0~k

)|φ〉 = 0

which implies the normalisation condition 〈φm|φn〉 = δm0δn0. If we set |φ1〉 = a0†~p1a3†~p2|0〉, and

|φ′1〉 = a0†~p3a3†~p4|0〉. If we contract these two things together, we have that⟨

φ′1∣∣φ1⟩ = 〈0| a3~p4a

0~p3|φ1〉

= 〈0| a3~p4a3~p3a0†~p1a

3†~p2|0〉

Now, I’m not sure how to prove this but the convention above asserts that this thing must bezero. This is because we have the condition that these states obey 〈φn|φm〉 = δ0nδ0m, and sincewe’re setting m,n = 1 in the above, the above should be zero.

7.2.1 γ Feynman propagator

Here we’ll go through the computation of the Feynman propagator, this time starting from theexpansion of the field Aµ(x). If we want to compute a propagator bewteen two points givenby

〈0|TAµ(x)Aν(y) |0〉 =∫

d4p

(2π)4−i

p2 + iε

[ηµν + (α− 1)

pµpνp2

]e−ip·(x−y)

for the general α gauge. The propagator takes the simple form ∼ − ηµνp2+iε

in the Feynman gaugewhen we take α = 1. This is also straightforward to prove on its own.

7.3 Interactions

If we want to explore interactions in the theory, we set

L = −1

4FµνF

µν − ejµAµ

e here is some dimensionless coupling constant, which we call the gauge coupling. In addition,jµ is some current which has some symmetry.


Claim. Interaction currents of this form are conserved. We show that jµ as actually conserved.That is, ∂µjµ = 0.

Proof. For the Euler-Lagrange equations, we have that

∂µFµν = ejν =⇒ ∂ν∂µF

µν = e∂νjν = 0

We have a zero since the expression on the left hand side contains Fµν , which is anti-symmetricin µν and the symmetric partial derivative terms ∂µ∂ν . This means that jν is a conservedcurrent.

7.3.1 Coupling to Fermions

Our Dirac part of our Lagrangian is now given by

LD = ψ(i/∂ −m

)ψ

which has an internal symmetry from taking ψ → e−iαψ, and ψ → eiαψ, where α ∈ R. Thisinternal symmetry induces a Noether current jµ = ψγµψ, which is conserved. This gives us ourfull Lagrangian for quantum electrodynamics.

LQED = −1

4FµνF

µν + ψ(i/∂ −m

)ψ − eψαγ

′µαβAµψβ

This yields the Feynman rule factor ie (γµ)αβ. Previously, when discussing free Maxwell theorywith L = −FµνFµν , gauge symmetry was one of the arguments we used to whittle down our 4possible polarisations to just 2 polarisations (we cancelled out unphysical polarisations). Do westill have gauge symmetry here under a transformation of the field Aµ as well as the Dirac spinorterms ψ? We will show that this indeed is still the case. We rewrite our Lagrangian as

LQED = −1

4FµνF

µν + ψ(i /D −m

)ψ

where we define Dµψ = ∂µψ + ieAµψ is called the covariant derivative. Under the gaugetransformation, L QED is gauge invariant under

Aµ(x) → Aµ(x) + ∂µλ(x)

ψ(x) → e−ieλ(x)ψ(x)

To prove this we have that

Dµψ = (∂µ + ieAµ)ψ → ∂µ(e−ieλψ)

= e−ieλDµψ

You can check this gauge invariance yourself!

We will show that observables don’t depend on timelike γ pairs, ie |φn〉. If we compute theHamiltonian, we get that

H =

∫d3p

(2π)3|~p|

(3∑i=1

ai†~p ai~p − a0†~p a

0~p

)


But our Gupta condition implies

〈ψ| a3†~p a3~p |ψ〉 = 〈ψ| a0†~p a

0~p |ψ〉

Gauge invariance of LQED implies that we remove the unphysical degrees of freedom in Aµ,leaving two polarisation states. e has the interpretation of electric charge, since EL are ∂µFµν =ejν . In electromagnetism, we interpret j0 as charge density. But as a quantum operstor, weset

Q = −e∫d3xψγ0ψ = −e

∫d3p

(2π)3

∑s

(bs†~p b

s~p − cs†~p c

s~p

)Which is −e (number of particles − number of anti particles)

7.3.2 Coupling to Scalars

We’ll now describe π±γ interactions. For a real scalar, there is no suitable current and we can;tcouple it to electromagentism. For a complex scalar φ, we use a covariant derivative

Dµφ = ∂µφ− ieqAµφ

where the q is just a real number, which was −1 for an electron. This is the charge for a scalarparticle. Under a gauge transformation, we have that φ(x) transforms as

φ(x) → eieqλ(x)φ(x)

So that Dµφ → eieqλDµφ. q is chage of φ in units of e. Our total coupled Lagrangian istherefore

L QED = −1

4FµνF

µν + (Dµφ)†Dµφ−m2φ†φ

This means our interaction lagrangian reads

Lintφ QED = ieq(φ†∂µφ− (∂µφ)† φ

)Aµ + e2q2AµA

µφ†φ

Our first Feynman rules are (insert pictures here) We can then derive a Noether current

jµ = ieq((Dµφ)† − φ†Dµφ

)is gauge invariant.ă A good model for electromagentic EM interactions of π± s and BSM withH± etc. To couple a U ( 1) guage field Aµ to any number of fields, φa (fermions and or bosons),replace space-time derviatives with covariant ones

∂µφa → Dµφ

a = ∂µφa − ieqaAµφ

a

7.4 Warming up with spinors from the SO(3) rotation group

Spinors are (complex) elements in a vector field which transform linearly when the underlyingcoordinate basis is rotated. In three dimensional Euclidean space, this transformation wouldtake the form of an element in SO(3), acting on three dimensional real vectors in R3. Our goalof this subsection is to build an equivalence between this and elements in SU(2), where thenthese elements act on 2 component complex vectors in C2.


7.4.1 Rotations in Euclidean space

Let’s revise quickly how to map rotations out in three dimensional space. We multiply a vectorv by a matrix to give the map v 7→ Rv. In the case of a rotation by an angle θ about the z-axis,our rotation matrix Rz takes the form

Rz =

(cos θ sin θ− sin θ cos θ

)In the terminology of Lie groups, this is an element of the Lie group SO(3), and has an associatedgenerator Jz, given by it’s derivative at θ = 0;

Jz =d

dθRz(θ)

∣∣∣∣θ=0

=

0 1 0−1 0 00 0 0

We can repeat this same procedure with rotations along the y and x axes, and differentiatewith respect to the parameters as well to obtain the generators Jx and Jy. One finds that theserotation generators obey the following commutation relations, or ’algebra’

[Ji, Jj ] = iεijkJk

We can then write out a full transformation, which is a rotation by an angle θ about the normalvector n, as

v 7→ ein·Jθv

where the exponent is simply the expected series sum of the matrices.

7.4.2 Transformations with Unitary matrices

We now explore the group SU(2), the group of 2 by 2 matrices which satisfy the relationU † = U−1, and detU = 1. One can show, that by comparing coefficients, that unitary matricescan be written in general of the form

U =

(a b

−b∗ a∗

)where we have the condition that |a|2 + |b|2 = 1. Recall that a and b are complex, so over thereals are specified by 2 parameters each. Our condition that |a|2 + |b|2 = 1 reduces our degreesof freedom by 1, so we have 3 degrees of freedom in total. We would like to see how unitarymatrices transform vectors of the form

ξ =

(ξ1ξ2

)We have that

ξ 7→ ξ′ = Uξ =

(aξ1 + bξ2

−b∗ξ1 + a∗ξ2

)=

(ξ′1ξ′2

)Now we ask the question of whether we can find other vectors derived from this which cantransform in the same way. Turns out, the vector (−ξ∗2 , ξ∗1) does as well. You can verify this


yourself! We employ the notation that Ryder uses here, using the sign ∼ to denote ’transformsas’, we’ve discovered that

(ξ1, ξ2) ∼ (−ξ∗2 , ξ∗1)

Or, setting

χ =

(0 −11 0

)we can alternatively write that ξ ∼ χξ∗. Now, this also means that, taking the hermitianconjugate of a vector, that ξ† transforms as (−ξ2, ξ1) which implies that ξξ† transforms as

H =

(ξ1ξ2

)(−ξ2, ξ1) =

(−ξ1ξ2 ξ21−ξ22 ξ1ξ2

)But we already know that since we have the transformation laws ξ 7→ Uξ and ξ† 7→ ξU †,

ξξ† 7→ Uξξ†U † =⇒ H 7→ UHU † = UHU−1

The main reason why we’ve invested so much into finding this matrix which transforms underU is that H is indeed traceless. So, we can write out a matrix which transforms under U in theway we want by giving ourselves a complex traceless matrix, which we’ll call h. Now, h can bewritten as

h =

(z x+ iy

x− iy −z

)= σ · x

where x is our original position vector we’ve rotated, and σ are the Pauli sigma matrices. So,we’ve taken a vector rotating in SO(3) and have reduced it down to induced rotations by SU(2).By contruction we’ve shown that h 7→ UhU−1, and so we’ve shown that unitary transformationsacting on spinors in SU(2) correspond to rotations acting on position vectors in SO(3). Wehave a specifc map from the spinors to the position vector, given by comparing elements of hand H

x =1

2(ξ1 + ξ2), y =

1

2i(ξ21 + ξ22), z = ξ1ξ2

7.5 Lorentz tranformations and spinors

To motivate the derivation of the Dirac equation, we need to first discuss in general, objectswhich transform sensibly with Lorentz transformations. If we have a coordinate system denotedby xµ, we Lorentz transform it with the map xµ → Λµνxν . If we had a scalar field ψ(x), thenthis passive transformation would induce the transformation

ψ(x) → ψ(Λ−1x)

Now, instead of a scalar field, we might wish to consider what happens to, say, a vector fieldwhen we induce a Lorentz transformation. In the case that our vector field transforms linearlyin response to Lorentz transformations, we have a spinor. More specifically, we have a complexvector field ψa(x) which obeys the transformation law

ψa(x) → Λabψb(x)


Suppose we had some eaution which a scalar field φ satisfies. For example, we have the Klein-Gordon equation (

∂µ∂µ −m2

)φ = 0

. If we were to perform a Lorentz boost which shifts our frame of reference, then it only makesphysical sense that the equation still holds, because nothing about the actual physical systemhas changed. Suppose that our Lorentz boost is written as

xµ → x′µ = Λµνxν

Then our corresponding scalar field should transform as

φ(x) → φ(Λ−1x)

We can chack that, under this transformation, the Klein-Gordon equation still holds.

We can generalise this kind of transformation to a field with multiple components which we indexas φa(x). As inspiration, we explore how some vector field V i in three dimensions transformsunder a rotation.

V i → V ′i = RijV j

and so we’d expect a Lorentz boost to have a scalar field transform like

φa(x) → φ′(x)a =M(Λ)baφb(Λ−1x)

In the above, we’re representing the Lorentz transformation as a matrix M(Λ). We call this arepresentation of our Lorentz transformation. Representations should obey the rule that

M(Λ)M(Λ′) =M(ΛΛ′)

or in other words the representation should respect the group structure of Lorentz transforma-tions. Proverbially, we are taking a methematical ’encoding’ of our transformation. How do wego about finding the representation M?

Again, we seek inspiration from rotation groups in 3 dimensions. A rotation in 3 dimensions isrepresented in real space by the matrices in SO(3), which is isomorphic to SU(2). This SU(2)has a basis, which are the Pauli matrices σi. Given a quantum state |ψ〉, how do we representits transformation?

Let |ψ〉 be a wavefunction representing a 12 spin particle. Then, we can represent rotations with

thefollowing transformation|ψ〉 →

∣∣ψ′⟩ = expin · J |ψ〉

In this case, Ji are the generators of this transformation. We say that the Lie group of rotationsin three dimensions are generated by a Lie algebra with commutation relations

Jk = iεijk[Ji, Jj ]

In three dimensions for our rotation group, this operator is given by

J = x× p = x× (−i∇)


Where we can index this angular momentum object as

Jνµ = i(xν∂µ − xµ∂nu)

This will precisely be our generalisation for our generatorfor our Lorentz group, except now theindices µ, ν are taken over µ, ν = 0, 1, 2, 3. This object obeys the Lorentz algebra.

[Jµν , Jρσ] = i(−gµρJνσ + gµσJνρ − gνσJµρ + gνρJµσ)

And in general, any expression for Jµν which satisfies this equation is a valid representationof our Lie algebra. We denote the components of Jµν by writing down the object with twoadditional indices αβ. Thus we denote our object by (Jµν)αbeta. A simple representation thatwe pull out of the hat is

(Jµν)αβ = i(δµαδ

νβ − δµβδ

να

)Why is this choice intuitive? Well, our Lorentz transformation is obtained by exponentiat-ing our generator that we have here. Thus, for an anti-symmetric 2nd rank tensor, our fulltransformation is given by

exp(iωµν2Jµν

)and infinitesimally applied to a 4-vector, our transformation is

V µ →(δµν −

i

2ωαβ(J

αβ)µν

)V µ

7.6 Constructing the Chiral representation for the Dirac equation

You can show that if we have a set of matrices γµµ which obey the identity

γµ, γν = −2gµν

then the objectSµν =

i

4[γµ, γν ]

obeys the Lorentz algebra which we derived above.

Whilst this choice of representation is not unique, we choose a canonical one called the Weylrepresentation which is given by

γ0 =

(0 II 0

)γi =

(0 σi

−σi 0

)I’ll touch on this a bit more later, but the important this is that this object represents boosts in3 different dimensions, given by S0i and rotations in 3 dimensions given by Sij . Thus, our fieldφ(x) transforms as, when we exponentiate Sij , as

φ(x) → φ′(x) = exp(iωµν2Sµν

)φ(Λ−1)

where we’ve done and inverse transform to represent our change in frame of reference, and alsodone a map on the field itself.


7.7 Motivating Lorentz invariance of the Dirac equation

In this section, we’ll present the Dirac equation and show that it’s Lorentz invariant. The wayto do this is slightly more convoluted than showing the Lorentz invariance of the Klein-Gordonequation, like we did in the previous section. It involves exploring how γµ transforms under boththe Sµν and Jµν representations of the Lorentz transformations, and how these representationsrelate to one another.

But first, we present the Dirac equation. A field ψ(x) satisfying the Dirac equation satisfies theequation

(iγµ∂µ −m)ψ(x) = 0

Note that, as opposed to the Klein-Gordon equation, we have a m as the constant term insteadof m2. Also notice the fact that we’re not taking two derivatives in this equation, we insteadcontract the derivative ∂µ with γµ. Now, under the Lorentz transformation where xµ → xνΛµν ,recall our field transforms as

ψ(x) 7→ Λ 12ψ(Λ−1x)

To show that the Dirac equation is invariant under this specific transformation, we require that(for reasons we will show later)

Λ−112

γµΛ 12= Λµνγ

ν

One can see that if this condition is satisfied, then, upon transforming the equation

(iγµ∂µ −m)ψ → (iγµ∂µ −m)Λ 12(φΛ−1x)

= Λ 12Λ−11

2

(iγµ(Λ−1) νµ ∂ν −m)Λ 1

2ψ(Λ−1x)

= Λ 12(iΛ−11

2

γµΛ 12(Λ−1) ν

µ ∂ν −m)ψ(Λ−1x)

= Λ 12(iγµ∂µ −m)ψ(Λ−1x)

= 0

Let’s review carefully what we’ve done here. In the first line, we’ve transformed ψ(x) appropri-ately according to our Lorentz transformation. In the second line, we’ve done two things. First,we multiply on the left by Λ 1

2Λ−11

2

since it’s just the identity. In addition, we also pull out afactor of Λ−1 due to the chain rule. Now, in the third line, we used the identity we’ve derivedabove to switch out our expression for Λ−11

2

γµΛ 12

with just Λγν . In doing this, the factors of Λhave cancelled each other out. The final expression is zero, which means that we’ve successfullyshown that the Dirac equation is Lorentz invariant.

Now, we’ll prove the expression above, by using the infinitesimal generators instead. Observethat the above condition is equivalent to

(1 +iωρσ2Sρσ)γµ(1− iωρσ

2Sρσ = (1− iωρσ

2(Jρσ) ν

µ γnu

So in turn, if we can show that[γµ, Sρσ] = γν(Jρσ)µν


we’ve proven the identity above. We do this as follows

(Jµν)αβ = i(δµαδ

νβ − δµβδ

να

)When we raise the index on a delta function, we get our metric back. So our expression is

(Jµν) βα = i

(δµαg

νβ − gµβδνα

)Thus

γα(Jµν) βα = i

(γµgνβ − γνgµβ

)As for the left hand side

[γβ, Sµν ] = [γβ,i

4[γµ, γν ]]

=i

4

(γβγµγν − γβγνγµ − γµγνγβ + γνγµγβ

)but we can use the identity that

γµ, γν = −gµν

and commute index pairs which are either (ν, β) or (µ, β), to givei

4

(−γµγβγν − 2gµβγν + γνγβγµ + 2gβνγµ + γµγβγν + 2gν betaγµ − γνγβγµ − 2γνgµβ

)but the terms above cancel to give the correct expression. Hence we’ve proven the identityabove, and have shown that Dirac’s equation is Lorentz invariant.

So we’ve done all this work - but let’s remind ourselves what it’s for. This commutation re-lation is important because it gives us an equation relating the generators Jµν and Sµν . Thiscommutation relation is equivalent to saying, with ωρσ as a parameter, that infinitesimally wehave (

1 +iωρσS

ρσ

2

)γµ(1− iωρσS

ρσ

2

)= (1− i(Jρσ)µν)γ

ν =⇒ Λ−112

γµΛ 12= Λµνγ

ν

And with this identity, we’ve shown that the Dirac equation is Lorentz invariant.

7.8 Constructing boosts and rotations in block diagonal form

This spinor representation is reducible, which means that we can write out our transformationsin block diagonal form. For example, we write out the components of Sµν explicitly in terms ofboosts and rotations, and discover that both the boost and rotations are parametrised by threeparameters ( hence we can encode them as vectors). In the boost case, we have that

S0i =i

2

(0 II 0

)(0 σ

−σi 0

)=i

2

(−σi 00 σi

)But, as a tranformation, contracting with ωµν , we’re only summing to get exp

(i2ω0iS

0i), and so

reparametrising ω0i = αi, we find that our transformation takes the form

Λ 12= exp

(iωµνS

µν

2

)=

(eα·

σ2 0

0 e−α·σ2

)


Bear in mind that since we’re summing over all the indices, we have an extra factor of 2. Now,notice that we have a switch in sign under boosts.

Now we explore what happens when we transform under rotations. Rotations are representedby Sij , where i, j = 0, 1, 2, 3. Let’s compute these matrix representations explicitly - this isalso good revision for remembering our Pauli Sigma matrix commutation relations. We havethat

Sij =i

4[γi, γj ]

=i

2γiγj

=i

2

(0 σi

−σi 0

)(0 σj

−σj 0

)=i

2

(−σiσj 0

0 −σiσj

)

Now, we make use of our Pauli-sigma relations to simplify this expression. One can easily verifythat

σiσj = iεijkσk =⇒ Sij =1

2εijk

(σk 00 σk

)Now, when we want to contract Sij with ωij we have that

ωijSij =

1

2εijkωij

(σk 00 σk

)But the cool thing abou this is that we can write βk = εijkωij , which gives us the final result fora rotatation transformation, which is

Λ 12=

(exp−

i2σ·β 0

0 exp−i2σ·β

)

Where we have a noticable difference with respect to boosts in that there’s no sign changebetween the block diagonal matrices. The fact that we’ve broken down this spinor representationof Lorentz transformations into two block diagonal matrices is super useful because now we canalso split up the spinor vector ψ into two parts, which we call the left and right handed parts ofour spinor.

ψ =

(ψLψR

)So infinitesimally, we have that

ψL 7→(1 + α · σ

2− iβ · σ

2

)ψL

ψR 7→(1− α · σ

2− iβ · σ

2

)ψR


7.9 The Weyl equations

Let’s go back to our Dirac equation, but this time let’s view it in the context of our spinor whichwe split up. We have that (

iγ0∂0 − γi∂i −m)ψ = 0

Now, we can write this in the form of a matrix equation, bearing in mind that m acts as a scalarmultiple of the identity. Thus, we have the matrix equation(

−m i(∂0 + σ · ∇)i(∂0 − σ · ∇) −m

)(ψLψR

)= 0

In this case, we’ve mixed the components ψR and ψL. However, we can separate out the mixingby exploring the massless case with m = 0, which reduces us to the equations

(∂0 + σ · ∇)ψR = 0

(∂0 − σ · ∇)ψL = 0

We can simplify this notation even further by ’extending’ out σ, and defining the objects

σµ = (1, σ), σµ = (1,−σ)

Which gives us a condensed form of the equations to read

σµ∂µψR = 0

σµ∂µψL = 0

7.10 Constructing Plane-Wave solutions to the Weyl equation

At it’s essence, the Dirac equation is a wave equation. In this section, we’ll explore solutionsof Dirac’s equation which have the form ψ(p) = e−ipxu(p), where p denotes momentum, andwe have the relativistic dispersion relation where pµpµ = m2. We can make our life easier byreducing the problem to that of the rest frame, where pµ = (Ep, 0). Our dispersion relation thentells us that pµ = (m, 0). Our Dirac equation in matrix form then reads(

−m i(∂0 + σ · ∇)i(∂0 − σ · ∇) −m

)e−ip·xu(p) =

(−m mm −m

)e−ip·xu(p) = 0

This implies that our general solution in the rest frame can be written as

u(p) =√m

(ξξ

)

Now, it’s a matter of boosting this solution to a non-rest frame. To do this, we need to do aLorentz boost on our 4-momentum vector, which for simplicity we’ll just consider a boost in thez-direction. We know from special relativity that a boost given by(

Ep3

)=

(1 + ν

(0 11 0

))(m0

)


Notice that we’re not using our spinor representation here; the spinor representation is not fortransforming 4-vectors; we’re just using our ordinary Lorentz boost representation. Exponenti-ating this gives us our full Lorentz boost j The associated spinor representation however for thisboost is

Λ 12= exp

(− i

2ωµνS

µν

)For a boost in the z-direction, our boost parameter looks is zero everywhere except in the zdirection, hence ω03 = −ω30 = µ, and contracting this object with Sµν (rembering to include anextra factor of two due to antisymmetry, gives our Lorentz boost

Λ 12= exp

(µ

2

(σi 00 −σi

))Now we’re in good shape to multiply our spinor by the Lorentz boost object.


8 Example sheet 1

Solved exercises and notes from example sheet 1.

8.1 Question 1

In this question, we’re asked to derive the momentum operator Pµ =∫d3xT 0µ. The first thing

we notice is that this object is derived from the momentum part of the energy-momentum tensor,and is simply integrating this over all of space. Hence, we expect it to be an operator on fields.The first thing we’ll do is derive the form of the energy momentum tensor.

The energy-momentum tensor is a conserved object which arises from translational symmetry.In Minkowski spacetime, translational symmetries in time correspond to the conservation ofenergy, and translational symmetries in space correspond to conserved momemtum. Hence, weconsider the translation

xµ → xµ + εµ

But, since this is a passive transformation of our reference frame, we have that our scalar fieldψ(x) transforms as

ψ(x) → ψ(x− ε) = ψ(x)− εµ∂µψ(x)

So our change in the scalar field δ(x) = εµ∂µψ(x). Similarly, our Lagrangian L in generaltransforms in the same way, with δL = εµ∂µL. This is indeed a symmetry of the Lagrangiansince δL = ∂µ(F

µ) where Fµ = εµL Thus, by Noether’s theorem we can write out conservedcurrent as

jµ = εµL− εν∂νψ

(∂L

∂(∂µψ)

)However, now we can ’factorise’ out the εµ since it’s arbitrary, and write our conserved currentas

Tµν = δµνL− ∂νψ

(∂L

∂(∂µψ)

)Now, substituting our expression for the Lagrangian (which is associated to the Klein-Gordonequation)

L =1

2∂αψ∂

αψ − 1

2m2ψ2

Our expression for our energy momentum tensor becomes

Tµν = ηµν(1

2∂αψ∂

αψ − 1

2m2ψ2

)− ∂νψ∂µψ

Let’s examine this term by term for∫d3xT 0µ. For starters, let’s take the ∂0ψ∂µψ term. First,

we Fourier expand this object to get

ψ(x) =

∫d3p

1

(2π)3√2Ep


ip·x)


If we differentiate this thing with ∂0 and ∂µ, we have that

∂0ψ =

∫d3p

p0

(2π)3√2Ep

(−ape−ip·x + a†pe

ip·x)

∂µψ =

∫d3p

pµ

(2π)3√2Ep


ip·x)

Now, amalgamating this all together, we have that the integral of this is∫d3xT 0µ =

∫d3x d3p d3q

p0qµ

(2π)6√2EpEq

(−apaqe−i(p+q)·x + a†paqe

i(p−q)·x + apa†qei(p−q)·x − a†pa

†qei(p+q)·x

)But here, we can use the identity that∫

d3x eix·(c = (2π)3δ(c)

In the spatial part to isolate out a delta function. In addition, we also use the fact that p0 = Ep

to cancel out with Ep in the denominator. Hence, we end up with the expression that the aboveis ∫

d3p

2(2π)3pµ(apa−p + a†pap + apa

†p + a†pa

†−p)

Now, the first and last terms disappear because under a change of variables p → −p, pµ is anodd function under the spatial integral, but apa−p and a†pa†−p are even functions, so these termsdisappear. Now, if we commute the third expression with our standard relation and remove ourdelta function since it’s an infinite term ( like we do with our Hamiltonian), we get the finalexpression.

One can show that the rest of the terms in the energy momentum tensor disappear underintegrating over

∫d3x.

Now, for the next part of the question, we’d like to show that in the Heisenberg picture

[Pµ, ψ(x)] = −i∂µψ(x)

This is achieved fairly easily by pulling the commutators to inside the integral and then usingour standard commutation relations.

[Pµ, ψ(x)] = [

∫d3p

(2π)3pµa†pap, ψ]

=

∫d3p

(2π)3[pµa†pap, ψ]

=

∫d3p

(2π)3pµ[a†pap, ψ]

Where, going into the third line, since pµ is not an operator, we’ve just pulled it out of thecommutator. By linearity of the integral, we can also pull the commutator inside the integral,


as we did going into the second line. Fourier expanding out, the above expression reads∫d3pd3q

(2π)6pµ[a†pap, aqe

−iq·x + a†qeiq·x] =

∫d3pd3q

(2π)6pµ(e−iq·x[a†p, aq]ap + eiq·xa†p[ap, a

†q])

=

∫d3p

(2π)3

(−pµape−ip·x + a†pp

µeip·x)

= −i∂µψ(x)

8.2 Question 2

In this question we’re trying to derive the Klein-Gordon equation, but via the Heisenberg picturefor operators. This is when operators evolve in time instead of states. In the Heisenberg picture,our equation for time evolution for an operator OH is

dOH

dt= i[H,OH ]

First, let’s look at computingφ(x) = i[H,φ(x)]

The Hamiltonian associated with our free field Lagrangian is

H =

∫d3y

1

2]π(y)2 +

1

2(∇φ(y))2 + 1

2m2φ(y)2

So, to compute the commutator required, we need to remind ourselves of the commutationrelations between the our scalar field φ(x) and our conjugate momentum field π(x). This ourentirely analogous to those we know from quantum mechanics:

[φ(x), φ(y)] = [π(x), π(y)] = 0

[φ(x), π(y)] = δ3(x− y)

I just wanted to note that here, when we write φ(x), this is condensed notation for φ(x, t), wherex. Now, when we bring the commutator inside the integral to calculate [H,φ(x)], all functionswhich are functions of φ(x) disappear under the commutator. The only expression we’re leftwith is

φ(x) =i

2

∫d3y [π(y)2, φ(x)]

However, we can apply the ’product rule’ for operator commutators which is

[AB,C] = A[B,C] + [A,C]B

so that the above expression reads as follows:


8.3 Question 3

Showing invariance

Our Lagrangian density is the Klein-Gordon field

L =1

2∂µφ∂

µφ− 1

2m2φ2

This question is about symmetries under rotations, transformations which infinitesimally takethe form, to first order in θ,

φa → φa + θεabcηbφc

Our Lagrangian changes as

L → L+ θεabcηb∂µφa∂µφc −m2εabcηbφaφc

The important thing to notice here is that the two terms added on go to zero, since εabc isantisymmetric in (a, c), yet ∂µφa∂µφc are symmetric in (a, c)! Antisymmetric objects contractedwith symmetric objects go to zero.

Constructing conserved quantities

Since L is invariant under our transformation our Noether current is reduced to

jµ = −δφa∂L

∂(∂µφa)∝ εabcηbφc∂

µφa

Now, to get a conserved quantity in integral form, we write out the space and time componentsseparately. Our condition that ∂µjµ = 0 implies that

∂j0

∂t−∇ · j = 0 =⇒ ∂

∂t

∫d3x j0 =

∫d3x∇ · j

Where we’ve simply integrated the equation and moved the spatial part to the left, and took thepartial derivative out of the integral. However, using divergence theorem, we have that∫

d3∇ · j =∫dS · j → 0

since currents should decay at infinity. Hence our conserved quantity is just∫d3x j0 =

∫d3xεabcηbφc∂

0φa = ηb

∫d3xεabcφcφa

However, there was freedom in our choice of εb this whole time, so we can ’strip’ out thiscomponent such that

∫d3x εabcφcφa is our conserved quantity, which is what the question asks

for up to relabelling of indices.


Verifying our conserved quantity with our field equations

It’s easy to check that the Euler-Lagrange equations yield the Klein-Gordon equation

(∂µ∂µ −m2)φ = 0

which with time and spatial derivatives gives

φa −∇2φa +m2φa = 0

Differetiating our conserved quantity with respect to time, and then applying our Klein-Gordonequation gives

Qa =

∫d3xεabcφbφc + εabcφbφc

=

∫d3x εabcφbφc

=

∫d3x εabc(m

2φb −∇2φb)φc

=

∫d3x εabcm

2φbφc + εabc(∇φb) · (∇φc)

= 0

Going into the second line, we have symmetric time derivatives in (b, c). Going into the penul-timate line, we’ve integrated by parts to shift one derivative in the Laplacian to the other side.The result is zero because both terms have φi expressions which are symmetric in (b, c).


8.4 Question 4

Minkowski metric is invariant under Lorentz transformations

We’re also exploring invariance under Lorentz transformations xµ → x′µΛµνxν . Our most basicquantity that we can make Lorentz invariant is simply our ’length’ of our 4-vector, which is ourMinkwoski metric contracted with our vectors

L = ηµνxµxν = ηµνx

′µx′ν

This implies that, expanding out the Lorentz transforms with slightly different summation in-dices, that

ηµνxµxν = ηρσΛ

ρµx

µΛρνxνxµ

However, since our choice for xµ was arbitrary, we must have that identically

ηµν = ηρσΛρµΛ

σν

Conditions on an infinitesimal Lorentz transformation

Our condition that the Minkowski metric is invariant under Lorentz transformations infinitesi-mally yields the condition that

ηρα = ηρα + α(δναω

νρηνµ + δµρω

ναηµν

)= ηρα + α

(ηµαω

µρ + ηρνω

µα

)= ηρα + α (ωρα + ωαρ)

Thus the tensor ω is antisymmetric here. The matrix which corresponds to rotations about thex3 direction is a basis element for antisymmetric 4 by 4 matrices, and is

ωνµ =

0 0 0 00 0 1 00 −1 0 00 0 0 0

Similary, for boosts, whilst ωµν is antisymmetric, ωµν isn’t since we’re contracting with aMinkwoski metric. Thus, a boost in the x direction is given by

ωνµ =

0 −1 0 0−1 0 0 00 0 0 00 0 0 0


8.5 Question 5

Our corresponding active transformation in the field it given by

ψ(xµ) → ψ(xµ − αωµνxν) = ψ(x)− αωµνx

ν∂µψ(x)

= ψ(x) + 0− αωµνxν∂µψ(x)

= ψ(x)− ωµνηµνψ(x)− αωµνx

ν∂µψ(x)

= ψ(x)− αωµν∂µ(xνψ(x))

where in the last equality we’re just performing a standard Taylor expansion. Since our La-grangian is a function of L(φ) = L(φ(x)) = L(x), our transformation on x induces exactly thesame transformation on L, and since ωµν is constant we can pull the derivative out:

L→ L− α∂µ(ωµνx

ν)L

Putting this together, this implies that our Noether current is

jµ = ωµνxνL− (ωρν∂ρx

νφ)∂L

∂(∂µφ)

But observe that this isjµ = ωρν

(δµρx

νL− ∂ρxν ∂L

∂(∂µφ)

)Our expression in the brackets is exactly the energy momentum tensor with one index contracteddown, with an extra factor of xν . Hence, our Noether current is

jµ = ωρνxνTµρ

Our conserved current is given by

Q =

∫d3x j0 = ωρν

∫d3xT 0ρxν

Since we shown earlier that ωµν is antisymmetric, if we consider indices only in the spatial part,we can write this thing as

ωij = εijknk =⇒ Q = nkεijk

∫d3xT 0ixj

Since nk is free, we can choose, relabelling indices, that

Qi = εijk

∫d3xT 0jxk =

1

2εijk

∫d3xT 0jxk − T 0kxj

where we’ve added the extra term due to antisymmetry in j, k. Similarly, we can do this withboosts by looking at ω0i = −ωi0 components, which gives our conserved current

Q = ω0i

∫d3xT 00xi − T 0ix0

Again, since ωi0 is free, the quantity

Qi =

∫d3xT 00xi − T 0ix0


is conserved. If we differentiate with respect to time, the LHS is zero and thus

d

dt

(∫d3xT 00xi

)=

d

dt

(∫d3xx0T 0i

)=

∫d3T 0i + t

d

dt

∫d3xT 0i

= const

This is because we already know that∫d3xT 0i is already a conserved quantity, so is constant.

The second term is zero since the derivative of a conserved quantity with respect to time is zero.(Also recall that x0 = t.


8.6 Question 6

8.6.1 Gauge invariance

To show that L is invariant under gauge transforms, is suffices to show that Fµν itself is invariant.The gauge transform we’re doing is Aµ → Aµ + ∂µξ, so Fµν transforms as

Fµν → ∂µ(Aν + ∂νξ)− ∂ν(Aµ + ∂µξ)

= ∂µAν + ∂µ∂νξ − ∂νAµ − ∂ν∂µξ

= ∂µAν − ∂νAµ

In the second line, we’ve applied symmetry of second partial derivatives.

8.6.2 Constructing the energy-momentum tensor

From our section earlier, we derived that our four conserved currents which satisfy ∂µTµν = 0,for a scalar field with multiple components φa, are given by

Tµν = Lδµν − ∂νφa

(∂L

∂(∂µφa)

)In EM field theory, this scalar field is simply replced with our four-vector potential Aρ! Also, weraise the indices so that the delta function in the first term becomes a Minkowski metric termηµν . Our expression for the energy-momentum tensor is EM theory is therefore

Tµν = Lηµν − ∂νAρ

(∂L

∂(∂µAρ)

)Now, we have a bit of a tricky term to deal with here. How do we compute the term L

∂(∂νAρ)?

We bear in mind that since we’re in phase space, each of our degrees of freedom mean that ∂µAνrepresents a separate variable for each index (so we have 16 variables here).Hence, we assert thefollowing about derivatives with these terms

∂(∂µAβ)

∂(∂νAα)= δνµδ

αβ

To assist us as well, we apply product rule and antisymmetry. We’ll now calculate this thing

∂L∂(∂νAρ)

= −1

4

∂(FµσFµσ)

∂(∂νAρ)

= −1

2Fµσ

∂Fµσ∂(∂νAρ)

= −Fµσ ∂(∂µAσ)∂(∂νAρ)

= −Fµσδνµδρσ= −F νρ


From this, we get something for free. Since our Lagrangian has no dependence on Aµ on it’sown, our Euler-Lagrange equations dictate that

∂µFµν = 0

We’ll use this fact later on. Now, it’s just a matter of substituting this in to the expression forenergy-momentum to get

Tµν = Fµρ∂νAρ −1

4ηµνFρσF

ρσ

Now, let’s add on our extra component −Fµρ∂ρAν . We check that this is conserved underdifferentiation:

∂µ(Fµρ∂ρA

ν) = (∂µFµν)∂ρA

ν + Fµρ∂µ∂ρAν = 0

the first term goes to zero as a result of the Euler-Lagrange equations. The second term goesto zero since we’re contracting the antisymmetric Fµρ with the symmetric ∂µ∂ρ. Hence, we’veshown that Ωµν is a conserved current!

To show symmetry, observe that

Ωµν = −1

4ηµνFσρF

σρ + Fµρ(∂νAρ − ∂ρAν) = −1

4ηµνFσρF

σρ + FµρF νρ

But this is clearly symmetric in indices µ, ν! Furthermore, since this tensor is composed entirelyfrom Fµν , it’s gauge invariant.

Finally, taking the trace of this object, we have that the trace of ηµν = 4, so

Ωνν = −FρσF ρσ + F νρFνρ = 0

Thus, this object is traceless! It’s a much more natural form of the energy momentum tensorthan what we had before.


8.7 Question 7

We want to derive the equations of motion for the Lagrangian

L = −1

4FµνF

µν +1

2m2CµC

µ

Instead of looking at the Euler-Lagrange equations, it’s slightly easier to do this by varying theaction directly instead. With the product rule, one can convince themselves pretty easily thatthe varied action is

δS =

∫−1

2FµνδF

µν +m2CµδCµ

=

∫−1

2Fµνδ (∂

µCν − ∂νCµ) +m2CµδCµ

=

∫−Fµνδ (∂µCν) +m2CµδC

µ

=

∫δCν

(∂µFµν +m2Cν

)

This implies that the equation of motion is

∂µFµν +m2Cν = 0

When m 6= 0, if we differentiate both sides by ∂ν , we have that

m2∂νCν = ∂ν∂µF

νµ = 0

Since, the second term as Fµν which is antisymmetric in µν, and ∂µ∂ν which is symmetric.Contracting this antisymmetric object with this symmetric one reduces the term to zero.

Our equation of motion dictates that, setting ν = 0,

∂µFµ0 = m2C0

The trick now is to expand this separately in terms of derivatives with respect to time and space.We have that

∂µ∂µC0 − ∂µ∂

0Cµ = m2C0

C0 − ∂i∂iC0 − ∂µC

µ = m2C0

C0 − ∂i∂iC0 − C0 + ∂iC

i = m2C0

Rearrangement gives our required result. Up to relabelling of indices, our Lagrangian is givenby

L = −1

2(∂µCν∂

νCν − ∂νCµ∂µCν)


To keep track of minus signs from contracting with up and down indices, we go slow and sumone index at a time. We find that

L = −1

2(∂0Cν∂

0Cν − ∂iCν∂iCν − ∂νC0∂

0∂ν + ∂νCi∂iCν

= −1

2(C0C

0 − CiCi − ∂iC0∂

iC0 + ∂iCj∂iCj − C0C

0 + ∂iC0Ci + Ci∂

iC0 − ∂jCi∂iCj)

Note that here, the C0C0 terms cancel. After some algebra, we get that

L =1

2(CiCi + ∂iC0∂

iC0 − (∂iCj)2 − 2Ci∂iC0 + ∂jCi∂

iCj) +1

2m2CµC

µ

Since there’s no dependence on C0, our conjugate momentum to C0 vanishes. Differentiating,we have that

πi =∂L∂Ci

= Ci − ∂iC0

Our Hamiltonian density is given by by

H = πiCi − L

Since we have an expression for πi, we substitute this expression in favour of Ci terms in boththe first term in the expression as well as the Lagrangian. After a significant amount of algebra,we have that

H =1

2(πiπ

i) +1

2(∂iCj)

2 +1

2∂jCi∂

iCj − 1

2m2CµCµ

This expression is positive definite!


8.8 Question 8

This question is a bit weird, because our discussion will include the effect of our measure d4xthat we have in our Lagrangian. Under the simultaneous scaling transformation, we’d like ouraction to transform, along with the transformation xµ → λxµ as∫

d4xL(x) →∫d4(x′)µ L′(x′)

= λ4∫d4xL′(x′)

Since we want our action to be invariant, the integrands here need be the same. Thus, ourcondition for invariance is

L(x) = λ4L′(x′) =⇒ L′(x′) = λ−4L(x)

More generally, for n+ 1 spacetime dimensions, we have that our condition changes to

L′(x′) = λ−(n+1)L(x)

Since our field transforms like φ→ λ−Dφ, and our derivative transforms like

∂′µ =∂

∂(x′)µ=

∂

λ∂xµ= λ−1∂µ

Our condition that the partial derivative terms are invariant imply that

λ−(n+1)∂µφ∂µφ = λ−2D−2∂µφ∂

µφ

For invariance, we thus require that in n+1 space time dimensions, we have the condition

2D + 2 = n+ 1, D =n+ 1

2− 1 =⇒ D = 1 for when n = 3

Now, for our mass term to be invariant, using the relation that we have above for the Lagrangian,we have the requirement that upon our transformation, we have

λ−(n+1) 1

2m2φ2 =

1

2m2λ−2Dφ2

which implies that D = n+12 , which contradicts the above assertion for the derivative terms to

be invariant. Hence, we have that m = 0 if we want a symmetry to appear. Similarly, comparingwith our φp term we have the condition that pD = n+1, which upon substituting for our valueof D, we yield the condition that

p = 2

(n+ 1

n− 1

)So in 4 spacetime dimensions, p = 4.

Constructing a Noether current

Working in 4 space time dimensions with D = 1, our field changes like φ→ λ−1φ = (1− log λ)φfor λ close to 1. In the argument above, we asserted that L → λ−4L = (1 − 4 log λ)L. Henceδφ = −φ log λδ.


9 Example Sheet 2

In general, save ink by writing exponents with either eix·(p−q) together, or either eix·(p+q) to-gether.

Save ink by not bothering to write the whole measure, and writing out just the operators we’reconcerned with. We can write out.

9.1 Question 1

Our stress energy tensor for µ = 1, 2, 3 is given by T 0µ = ∂µφφ = ηµν .

We can write out time evolution for ap explicitly by using the Heisenberg equation of mo-tion

dapdt

= i[H, ap] = iEpap

9.2 Question 4

Notes on this question.

We need to use the energy momentum tensor expansion. Tske the derivatives out.

Write out xj by differentiating in the very beginning, then only differentiate by parts afterthat.

Use liberally the argument involving odd and even.

9.3 Question 7

Be careful with counting! Especially with the double 8 diagram. In this question, we wish toverify the bubble diagram expansion for φ4 theory, for vacuum to vacuum scattering. Specifically,we’re interested in the expansion to λ2 order for 〈0|S |0〉, where, due to Dyson’s formula, wehave that

S = 1− iλ

4!

∫ ∞−∞

dtHI(x) +1

2

λ2

(4!)2

∫ ∞−∞

dt1

∫ ∞−∞

dt2T HI(t1, ~x)HI(t2, ~x)

Writing this in terms of our Hamiltonian density instead, we have that this simplifies downto

S = 1− iλ

4!

∫d4xHI(x) +

1

2

λ2

(4!)2

∫d4x1d

4x2T HI(x1)HI(x2)

where we have our interaction picture Hamiltonian given by HI(x) = φ4I(x). Now,let’s look atthe contribution to the first order in λ. This contribution is given by

−iλ4!

∫d4x 〈0| T φ(x)φ(x)φ(x)φ(x) |0〉


Now, due to Wick’s theorem, the integrand can be expanded out in terms of all possible con-tractions of the two fields. There are 3 ways to do this. This is because we could’ve contractedthe first field with 3 others, which leaves the last two fields to contract on their own. This meansthat our total contribution from first order from this field is

− iλ8

∫d4xD(x− x)D(x− x)

This is represented by a figure 8 diagram. Now, we aim to show that this matches up with thediagram shown in the exponential. To first order in λ, our exponential is given by

x

We know that from this diagram that, by assigning propagators to each line, that this is pro-portional to

∫d4xD(x − x)D(x − x). But, what about the symmetry factors? This diagram

has a symmetry factor of 8 since we assign a factor of two for being able to switch the ends ofthe bottom and top loops. In addition, we assign another factor of 2 since we can rotate thediagram 180 degrees.

Now, looking to second order in λ2, we’d like to find the amplitude given by the expression

λ2

2(4!)2

∫d4x dyT φxφxφxφxφyφyφyφy

Let’s examine the possible contractions that we get from this. We could have that the φx andφy terms are contracted internally amongst themselves. There are 3 ways to contract the φxamongst themselves and 3 ways to contract the φy amongst themselves, so we have 9 ways intotal. This means that our contribution is given by

1

2

λ2

82

∫d4xd4yD(x− x)D(x− x)D(y − y)D(y − y)

The factor of 12 comes from Dyson’s formula. As a diagram, this is represented by the dia-

gram

x y

Our symmetry factor for this diagram is 2×82, since, as we discussed before, we have a symmetryof 8 for each diagram. In addition, we could’ve swapped the diagrams as well. This agrees withthe above.


The next thing to do is that we could have contracted each of the φx terms with φy. Since wehave four choices to contract the first φx with, then 3, then 2, we have that our contribution tothis term is

λ24!

2(4!)2

∫d4x d4y D(x− y)D(x− y)D(x− y)D(x− y) =

λ2

2(4!)

∫d4x d4y D(x− y)D(x− y)D(x− y)D(x− y)

This corresponds to the diagramx

y

Finally, our last possible type of contraction is what happens when we contract just two of the

φx fields with two of the φy fields, and loop the rest. There are(42

)= 6 ways to choose which

to φx fields we contract with φy, and once we chose the φx fields, we have 12 = 4× 3 options towhich we can contract them. This means that the contribution gained from this is

−λ26× 4× 3

2(4!)2

∫d4x d4yD(x− x)D(y − x)D(y − x)D(y − y) =

− λ2

16

∫d4x d4yD(x− x)D(y − x)D(y − x)D(y − y)

This is associated with the diagram

x

y

We can switch around the two loops as well as the two propagators in the centre. Hence, we’veshowed that to second order in λ, the combinatoric and symmetry factors work out for eachdiagram when we exponentiate the bubble diagrams.


9.4 Question 8

For the Yukawa term, we need only attach p> ig to each vertex. We need to take into accountcombinatoric factors for each vertex if the same time of propagator is going into them.

To find the mass dimensions of g, h, k, l, we need to first find the mass dimensions of the fieldsφ and ψ. Our scalar action needs to be dimensionless, so∫

d4x1

2∂µφ∂

µφ

has mass dimension 0. This implies that [φ] = 1. Similarly, we find that [ψ] = 1. This meansthat the mass dimensions of g, k are [g] = [k] = 1. We have that [h] = 0, and [l] = −1.This theory contains both relevant, marginal and irrelevant operators. Thus, the theory isnon-renormalisable since the l dimension is irrelevant.

We have four interaction terms, which means that we get 4 distinct types of diagrams, eachwith their own set of Feynman rules. Let’s tackle the Yukawa interaction first. We knowthe momentum rules for this. Now, for the interaction h|ψ|4 = hψψ∗ψψ∗, we have that ourinteraction vertex is given by

Now, with this kind of diagram, our propagator comes from contracting just ψ,ψ∗ together,which gives us a propagator in momentum space which looks like

i

p2 − µ2 + iε

For the term kφ3, we have that our vertices look like

We need to also add in combinatoric factors, so we attach a 6 to this. Each of these objectscome with a propagator which comes from contracting two scalars together

i

p2 −m2 + iε

Finally, the non-trivial one is the propagator that comes from interactions with φ∂µψ∂µψ∗. In

this case, we have a different kind of propagator! Non-zero contributions come from contractionsof the form

φφψ∗∂µψ∂µψ∗ψ

Differentiating the contraction of ψ and ψ∗ pulls out a factor of iπµ. Thus, since we’ll alwayshave an even number of propagators involving the anti-nucleons, for each pair we attach a factorof −p · q

p2 −m2 + iε


9.5 Question 9

We want to find the decay width for the scattering process φ → ψψ. For studying the decaywidth up to order g2, we need only consider the trivial Feynman diagram with one vertex (sinceΓ ∼ |M|2.

Now, since this already imposes momentum conservation, our amplitude M is just ig. Thus,our decay width is given by the formula

Γ =1

2mg2

1

(2π)2

∫d3q1d

3q22Eq12Eq2

δ4(p1 − q1 − q2)

The trick here is to rewrite the invariant measure d3q2(2Eq2

as δ4((q22 − µ2

)(notice we’re using µ2

since this comes from the nucleon. Substituting this in, then integrating over q2, we get thatour decay width is

Γ =g2

(2π)22m

∫d3q12Eq1

δ4((p1 − q1)

2 − µ2)

Now, we’d like to find the set of q1 such that (p1− q1)2 = µ2. Our relativistic dispersion relationgives us that p21 = m2, and q21 = µ2. Also, in the convention of calculating decay widths, wework in the rest frame of the incoming particle (in this case, the meson). This means that wetake p1 = (m, 0).

p21 + q21 − 2p1 · q1 = µ2

m2 + µ2 − 2mEq1 = µ2

m2 − 2m√µ2 + ~q21 = 0

‖~q1‖ = ±√m2

4− µ2

Our decay width is the integral

Γ =g2

16π2m

∫d3q1Eq1

δ4(m2 − 2m

√µ2 + ‖~q1‖2

)Selecting the positive root in the delta function, using our standard identity for a delta functionof a function, this gives that the integral is

Γ =g2

16π2m

∫d3q1Eq1

δ (|~q1| − |~q∗1|)2m|~q∗1 |√µ2+|~q∗1 |2

=g2

16π2m

∫d|~q1|

4π|~q1|2δ(|~q1| − |~q∗1|)2mEq1

|~q∗1 |√µ2+|~q∗1 |2


In the last two lines we converted this integral into spherical coordinates. Integrating out thedelta function, this is just

Γ =g2(2π)

16π2m

1

m

|~q∗1||Eq∗1|Eq∗1 |

=g2

8πm2

√m2

4− µ2

=g2

16πm2

(1− 4µ2

m2

) 12

=g2

16πm

√1− (4µ2/m2)

This has dimensions of mass since g has mass dimension 1, thus decay make sense since this isa rate.


10 Example Sheet 4

Question 1

10.0.1 Part a

For ψψ → ψψ scattering, the only two possible diagrams to second order in λ are

~p, s

~q, r

~p′, s′

~q′, r′~p+ q

~p, s~p′, s′

~q, r ~q′, r′

~p− p′

Figure 12: E4q1

Due to Feynman rules, since we have two vertices in each diagram, we have a factor of (−iλ)2. Forthe first diagram, our ingoing nucleon spinor contributes u (~p)s, and the ingoing anti-nucleoncontributes v (~q)r. Since they meet at a vertex, we need to contract them to give a factorof u (~p)s · v (~q)r. Similarly, for the outgoing nucleon and anti-nucleon, we have a factor ofu (~p′)s

′· v (~q′)r

′. Our nucleon propagator in the centre contributes a factor of 1(

(p+q)2+µ2) . This

gives a contribution of

(−iλ)2[vr (~q) · us (~p)]

[us

′(~p′) · vr′ (~q′)

](p+ q)2 − µ2

For the second diagram, the appropriate contractions give

(−iλ)2[us

′(~p′) · us (~p)

] [vr (~q) · vr′ (~q′)

](p− p′)2 − µ2

But, due to fermi-dirac statistics, we need to add a minus sign relative to the first diagramsince we’re ’switching’ the out-going fermions. This gives the final contribution shown in thequestion.


p, s

q

p+ q

q′

p′, s′

p, s

q

p− q′

q′

p′, s′

+

Figure 13: ψφ→ ψφ

Part b

In the first diagram, we have an incoming nucleon with spinor u (~p)s and and outgoing nu-cleon with spinor u

(~p′)s′

. These spinors are connected with the nucleon propagator with fourmomentum p+ q. The nucleon propagator contributes the factor

γµ ((pµ + qµ) +m)

(p+ q)2 −m2

This is a matrix object. We need to contract the spinor objects on either side to give a contri-bution of

(−iλ)2 us′ (~p′) γµ ((p

µ + qµ) +m)us (~p)

(p+ q)2 −m2

In the second diagram, we have the same story but we’ve switched around the meson lines. Thismeans the 4 momentum of our nucleon propagator is different. We don’t add an extra minussign here since mesons are bosons. We get the contribution

(−iλ)2us

′(~p′) γµ

((pµ − q

′µ) +m)us (~p)

(p− q′)2 −m2

Adding these two amplitudes together gives our total contribution.

Question 3

According to Feynman rules, each vertex contributes a factor of −ieγµ. The ingoing and outgoingpolarisation vectors of the photons are contracted along these vertex indices, and the spinorindices are contracted across the propagator as before. This means that we get, for the diagramsbelow


εout(q)

p, s p+ q p′

εin(q)

εout(q)

p, s p− q′ p′

εin(q)

Figure 14: Nucleon and photon to nucleon and photon scattering

For the first diagram, we have the internal fermion propagator which contributes a matrix factorof

/p+ /q +m

(p+ q)2 −m2

From our vertices, we attach a factor of −ieγµ to each one, which we have to contract with thepolarisation vectors εout (~q′) and ε in (~q). Putting this on either side of the matrix, and thencontracting with our ingoing and outgoing spinors, means that we get our amplitude from thisdiagram as

iA1 = i (−ie)2 ur′(~p′)/εout

/p+ /q +m

(p+ q)2 −m2/εinu

s (~p)

The contribution from the second diagram is exactly the same, except that for the momentumassociated with the internal propagator, it’s now p− q′.

iA2 = i (−ie)2 ur′(~p′)/εout

/p− /q′ +m

(p− q′)2 −m2/εinu

s (~p)

The sum of these contributions give us our total amplitude for this scattering problem.

Now, what happens which we replace our polarisation vector ε in (~q) with the four vector q? For


our first amplitude iA1, schematically we have that

iA1 ∝(/p+ /q +m

)(p+ q)2 −m2

/qs (~p)

=/p/q + /q/q + /q/p

(p+ q)2 −m2us (~p)

=2p · q

(p+ q)2 −m2us (~p)

= 1

There are some important justifications to be made with the calculations above. Firstly, wehave that since q is a null vector,

/q/q = q2 = 0

Secondly we subbed out the m factor by using the on-shell spinor condition(/p−m

)us (~p) = 0

Finally, we use the fact that/p/q + /q/p = 2p · q

Similarly, for our second factor, we get a −1 from cancellations. So adding the two togethergives us a 0.

Now, to calculate the cross-section, we need to take the spin-sum amplitude. To do this, we notneed to only average over the incoming spins and sum over out-going spins, but we also need toaverage and sum over incoming and outgoing photons as well.

Our expression for our differential scattering with respect to the t Mandelstam variable is

dσ

dt=

|A|2

16πλ(s,m2

1,m22

)where λ (x, y, z) = x2 + y2 + z2 − 2xy − 2yz − 2zx. In this case, |A|2 is the spin and polarisedaverage. We decompose this into the separate amplitudes we calculated earlier, so that

|A|2 = |A1|2 + |A2|2 + 2Re (A∗1A2)

10.1 Question 4

To introduce scalars into our Lagrangian, we need to add in covariant derivatives to couple thephotons from Aµ to charged scalars φ. This acts as

Dµφ = ∂φ + ieAµφ

This enforces gauge invariance when we put it in the Lagrangian

L = DµφDµφ† = ∂µφ∂µφ

† + ieAµ(φ∂µφ

† − φ†∂µφ)+ e2AµAµφ†φ


p q


The first term is a kinetic term, and our two interaction terms give us two vertices for ourFeynman diagrams which we need to calculate Feynman rules for. We get this from Fouriertransforming.

The first diagram contributes a factor of ie(pµ + qµ) due to the derivative terms. The seconddiagram contributes a factor of 2ie2ηµν . We have two identical photons from the second diagram,so we need to add an additional combinatoric factor of 2.

10.2 Question 5

In QED, the only possible diagram which we can get is the one below

e+

e−

p

p′

µ+

p+ q

µ−

q

q′


As before, each vertex in this diagram contributes one factor of −ieγµ. We contract the incomingfermions e+ and e−. Our progagator is just a factor of iηµν

(p+q)2, since it’s a photon propagator,

and this contracts the two indices for the γµ matrices.


Question 6

Our amplitude from this scattering problem is

iA =ie2

q2(vs′(p′)γµus (p)

) (ur (k) γµvr′

(k′))

We calculate the spin sum-average of |A|2.

|A|2 = 1

4

∑r,r′,s,s′

[vs′(p′)γµus (p)

] [us (p) γ

νvs(p′)] [

ur (k) γµvr′(k′)] [

vr′(k′)γνur (k)

]In the massless case, we use the completeness relation for u and v this simplifies to

|A|2 = 1

4

e4

q4tr(/p′γµ/pγ

ν)tr(/kγµ/k

′γν)

where q2 = (p+ p′)2 = 4E2 Now here, we take the p, p′, k, k′ indices out and use the tracerelation

tr(γαγβγγγδ

)= 4

(ηαβηγδ − ηαγηβδ + ηαδηβγ

)After some simplification of the above, we arrive at the amplitude being

|A|2 = e4

2E4

((p · k)

(p′ · k′

)+(p′ · k

) (p · k′

))We recognise that we can use the Mandelstam variables to replace these momenta, and so weget that

|A|2 = 1

8

e4

E4

(t2 + u2

)However, since we’re scattering identical particles, we use the fact that, in this massless case,

t = −2p2 (1− cos θ) = −2E2 (1− cos θ) , u = 2p2 (1 + cos θ) = 2E2(1 + cos2 θ

)where θ is our scattering angle for this problem. This means that

|A|2 = e4(1 + cos2 θ

)We substitute this into our expression for our differential cross section

dσ

dΩ=

|A|2

64π2s=e4(1 + cos2 θ

)64π2s

Integrating along the whole solid angle, this gives

σ =e4

12πs=

4πα2

3s

The centre of mass frame is the frame where our total three momentum is zero. This meansthat if we have to incoming 4-momenta p1, p2 , say in a 2 → 2 scattering problem, then we havethat ~p1 = −~p2. As a result, we also have that

(p1 + p2)2 = (E1 + E2)

2

We use this to our advantage in simplifying aspects of the problem.


10.3 Question 1

In this question, we’re asked to derive the momentum operator Pµ =∫d3xT 0µ. The first thing

we notice is that this object is derived from the momentum part of the energy-momentum tensor,and is simply integrating this over all of space. Hence, we expect it to be an operator on fields.The first thing we’ll do is derive the form of the energy momentum tensor.

The energy-momentum tensor is a conserved object which arises from translational symmetry.In Minkowski spacetime, translational symmetries in time correspond to the conservation ofenergy, and translational symmetries in space correspond to conserved momemtum. Hence, weconsider the translation

xµ → xµ + εµ

But, since this is a passive transformation of our reference frame, we have that our scalar fieldψ(x) transforms as

ψ(x) → ψ(x− ε) = ψ(x)− εµ∂µψ(x)

So our change in the scalar field δ(x) = εµ∂µψ(x). Similarly, our Lagrangian L in generaltransforms in the same way, with δL = εµ∂µL. This is indeed a symmetry of the Lagrangiansince δL = ∂µ(F

µ) where Fµ = εµL Thus, by Noether’s theorem we can write out conservedcurrent as

jµ = εµL− εν∂νψ

(∂L

∂(∂µψ)

)However, now we can ’factorise’ out the εµ since it’s arbitrary, and write our conserved currentas

Tµν = δµνL− ∂νψ

(∂L

∂(∂µψ)

)Now, substituting our expression for the Lagrangian (which is associated to the Klein-Gordonequation)

L =1

2∂αψ∂

αψ − 1

2m2ψ2

Our expression for our energy momentum tensor becomes

Tµν = ηµν(1

2∂αψ∂

αψ − 1

2m2ψ2

)− ∂νψ∂µψ

Let’s examine this term by term for∫d3xT 0µ. For starters, let’s take the ∂0ψ∂µψ term. First,

we Fourier expand this object to get

ψ(x) =

∫d3p

1

(2π)3√2Ep


ip·x)

If we differentiate this thing with ∂0 and ∂µ, we have that

∂0ψ =

∫d3p

p0

(2π)3√2Ep


ip·x)

∂µψ =

∫d3p

pµ

(2π)3√2Ep


ip·x)


Now, amalgamating this all together, we have that the integral of this is∫d3xT 0µ =

∫d3x d3p d3q

p0qµ

(2π)6√2EpEq

(−apaqe−i(p+q)·x + a†paqe

i(p−q)·x + apa†qei(p−q)·x − a†pa

†qei(p+q)·x

)But here, we can use the identity that∫

d3x eix·(c = (2π)3δ(c)

In the spatial part to isolate out a delta function. In addition, we also use the fact that p0 = Ep

to cancel out with Ep in the denominator. Hence, we end up with the expression that the aboveis ∫

d3p

2(2π)3pµ(apa−p + a†pap + apa

†p + a†pa

†−p)

Now, the first and last terms disappear because under a change of variables p → −p, pµ is anodd function under the spatial integral, but apa−p and a†pa†−p are even functions, so these termsdisappear. Now, if we commute the third expression with our standard relation and remove ourdelta function since it’s an infinite term ( like we do with our Hamiltonian), we get the finalexpression.

One can show that the rest of the terms in the energy momentum tensor disappear underintegrating over

∫d3x.

Now, for the next part of the question, we’d like to show that in the Heisenberg picture

[Pµ, ψ(x)] = −i∂µψ(x)

This is achieved fairly easily by pulling the commutators to inside the integral and then usingour standard commutation relations.

[Pµ, ψ(x)] = [

∫d3p

(2π)3pµa†pap, ψ]

=

∫d3p

(2π)3[pµa†pap, ψ]

=

∫d3p

(2π)3pµ[a†pap, ψ]

Where, going into the third line, since pµ is not an operator, we’ve just pulled it out of thecommutator. By linearity of the integral, we can also pull the commutator inside the integral,as we did going into the second line. Fourier expanding out, the above expression reads∫

d3pd3q

(2π)6pµ[a†pap, aqe

−iq·x + a†qeiq·x] =

∫d3pd3q

(2π)6pµ(e−iq·x[a†p, aq]ap + eiq·xa†p[ap, a

†q])

=

∫d3p

(2π)3

(−pµape−ip·x + a†pp

µeip·x)

= −i∂µψ(x)


10.4 Question 2

In this question we’re trying to derive the Klein-Gordon equation, but via the Heisenberg picturefor operators. This is when operators evolve in time instead of states. In the Heisenberg picture,our equation for time evolution for an operator OH is

dOH

dt= i[H,OH ]

First, let’s look at computingφ(x) = i[H,φ(x)]

The Hamiltonian associated with our free field Lagrangian is

H =

∫d3y

1

2]π(y)2 +

1

2(∇φ(y))2 + 1

2m2φ(y)2

So, to compute the commutator required, we need to remind ourselves of the commutationrelations between the our scalar field φ(x) and our conjugate momentum field π(x). This ourentirely analogous to those we know from quantum mechanics:

[φ(x), φ(y)] = [π(x), π(y)] = 0

[φ(x), π(y)] = δ3(x− y)

I just wanted to note that here, when we write φ(x), this is condensed notation for φ(x, t), wherex. Now, when we bring the commutator inside the integral to calculate [H,φ(x)], all functionswhich are functions of φ(x) disappear under the commutator. The only expression we’re leftwith is

φ(x) =i

2

∫d3y [π(y)2, φ(x)]

However, we can apply the ’product rule’ for operator commutators which is

[AB,C] = A[B,C] + [A,C]B

so that the above expression reads as follows:

Part III Quantum Field Theory · Quantum Field Theory Notes by Aﬁq Hatta 2 Classical Field Theory...

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Transcript of Part III Quantum Field Theory · Quantum Field Theory Notes by Aﬁq Hatta 2 Classical Field Theory...