III Quantum Field Theory (Full)

Part III — Quantum Field Theory

Based on lectures by B. Allanach

Notes taken by Dexter Chua

Michaelmas 2016

These notes are not endorsed by the lecturers, and I have modified them (often

significantly) after lectures. They are nowhere near accurate representations of what

was actually lectured, and in particular, all errors are almost surely mine.

Quantum Field Theory is the language in which modern particle physics is formulated.

It represents the marriage of quantum mechanics with special relativity and provides the

mathematical framework in which to describe the interactions of elementary particles.

This first Quantum Field Theory course introduces the basic types of fields which play

an important role in high energy physics: scalar, spinor (Dirac), and vector (gauge)

fields. The relativistic invariance and symmetry properties of these fields are discussed

using Lagrangian language and Noether’s theorem.

The quantisation of the basic non-interacting free fields is firstly developed using the

Hamiltonian and canonical methods in terms of operators which create and annihilate

particles and anti-particles. The associated Fock space of quantum physical states is

explained together with ideas about how particles propagate in spacetime and their

statistics. How these fields interact with a classical electromagnetic field is described.

Interactions are described using perturbative theory and Feynman diagrams. This

is first illustrated for theories with a purely scalar field interaction, and then for a

couplings between scalar fields and fermions. Finally Quantum Electrodynamics, the

theory of interacting photons, electrons and positrons, is introduced and elementary

scattering processes are computed.

Pre-requisites

You will need to be comfortable with the Lagrangian and Hamiltonian formulations of

classical mechanics and with special relativity. You will also need to have taken an

advanced course on quantum mechanics.

Contents

0 Introduction

1 Classical field theory

1.1 Classical fields

1.2 Lorentz invariance

1.3 Symmetries and Noether’s theorem for field theories

1.4 Hamiltonian mechanics

2 Free field theory

2.1 Review of simple harmonic oscillator

2.2 The quantum field

2.3 Real scalar fields

2.4 Complex scalar fields

2.5 The Heisenberg picture

2.6 Propagators

3 Interacting fields

3.1 Interaction Lagrangians

3.2 Interaction picture

3.3 Wick’s theorem

3.4 Feynman diagrams

3.5 Amplitudes

3.6 Correlation functions and vacuum bubbles

4 Spinors

4.1 The Lorentz group and the Lorentz algebra

4.2 The Clifford algebra and the spin representation

4.3 Properties of the spin representation

4.4 The Dirac equation

4.5 Chiral/Weyl spinors and γ

4.6 Parity operator

4.7 Solutions to Dirac’s equation

4.8 Symmetries and currents

5 Quantizing the Dirac field

5.1 Fermion quantization

5.2 Yukawa theory

5.3 Feynman rules

6 Quantum electrodynamics

6.1 Classical electrodynamics

6.2 Quantization of the electromagnetic field

6.3 Coupling to matter in classical field theory

6.4 Quantization of interactions

6.5 Computations and diagrams

0 Introduction

The idea of quantum mechanics is that photons and electrons behave similarly.

We can make a photon interfere with itself in double-slit experiments, and

similarly an electron can interfere with itself. However, as we know, lights are

ripples in an electromagnetic field. So photons should arise from the quantization

of the electromagnetic field. If electrons are like photons, should we then have

an electron field? The answer is yes!

Quantum field theory is a quantization of a classical field. Recall that in

quantum mechanics, we promote degrees of freedom to operators. Basic degrees

of freedom of a quantum field theory are operator-valued functions of spacetime.

Since there are infinitely many points in spacetime, there is an infinite number

of degrees of freedom. This infinity will come back and bite us as we try to

develop quantum field theory.

Quantum field theory describes creation and annihilation of particles. The

interactions are governed by several basic principles — locality, symmetry and

renormalization group flow. What the renormalization group flow describes is

the decoupling of low and high energy processes.

Why quantum field theory?

It appears that all particles of the same type are indistinguishable, e.g. all

electrons are the same. It is difficult to justify why this is the case if each particle

is considered individually, but if we view all electrons as excitations of the same

field, this is (almost) automatic.

Secondly, if we want to combine special relativity and quantum mechanics,

then the number of particles is not conserved. Indeed, consider a particle trapped

in a box of size

. By the Heisenberg uncertainty principle, we have ∆

p & ~/L

We choose a particle with small rest mass so that m  E. Then we have

∆E = ∆p ·c &

When ∆

E &

, then we can pop a particle-antiparticle pair out of the

vacuum. So when

L .

2mc

, we can’t say for sure that there is only one particle.

We say

(

) is the “compton wavelength” — the minimum distance

at which it makes sense to localize a particle. This is also the scale at which

quantum effects kick in.

This argument is somewhat circular, since we just assumed that if we have

enough energy, then particle-antiparticle pairs would just pop out of existence.

This is in fact something we can prove in quantum field theory.

To reconcile quantum mechanics and special relativity, we can try to write a

relativistic version of Schr¨odinger’s equation for a single particle, but something

goes wrong. Either the energy is unbounded from below, or we end up with some

causality violation. This is bad. These are all fixed by quantum field theory by

the introduction of particle creation and annihilation.

What is quantum field theory good for?

Quantum field theory is used in (non-relativistic) condensed matter systems.

It describes simple phenomena such as phonons, superconductivity, and the

fractional quantum hall effect.

Quantum field theory is also used in high energy physics. The standard model

of particle physics consists of electromagnetism (quantum electrodynamics),

quantum chromodynamics and the weak forces. The standard model is tested

to very high precision by experiments, sometimes up to 1 part in 10

. So it is

good. While there are many attempts to go beyond the standard model, e.g.

Grand Unified Theories, they are mostly also quantum field theories.

In cosmology, quantum field theory is used to explain the density pertur-

bations. In quantum gravity, string theory is also primarily a quantum field

theory in some aspects. It is even used in pure mathematics, with applications

in topology and geometry.

History of quantum field theory

In the 1930’s, the basics of quantum field theory were laid down by Jordan,

Pauli, Heisenberg, Dirac, Weisskopf etc. They encountered all sorts of infinities,

which scared them. Back then, these sorts of infinities seemed impossible to

work with.

Fortunately, in the 1940’s, renormalization and quantum electrodynamics

were invented by Tomonaga, Schwinger, Feynman, Dyson, which managed to

deal with the infinities. It was a sloppy process, and there was no understanding

of why we can subtract infinities and get a sensible finite result. Yet, they

managed to make experimental predictions, which were subsequently verified by

actual experiments.

In the 1960’s, quantum field theory fell out of favour as new particles such

as mesons and baryons were discovered. But in the 1970’s, it had a golden

age when the renormalization group was developed by Kadanoff and Wilson,

which was really when the infinities became understood. At the same time, the

standard model was invented, and a connection between quantum field theory

and geometry was developed.

Units and scales

We are going to do a lot of computations in the course, which are reasonably

long. We do not want to have loads of

and

’s all over the place when we do

the calculations. So we pick convenient units so that they all vanish.

Nature presents us with three fundamental dimensionful constants that are

relevant to us:

(i) The speed of light c with dimensions LT

−1

;

(ii) Planck’s constant ~ with dimensions L

−1

;

(iii) The gravitational constant G with dimensions L

−1

−2

We see that these dimensions are independent. So we define units such that

= 1. So we can express everything in terms of a mass, or an energy, as we

now have

. For example, instead of

(

), we just write

−1

We will work with electron volts

. To convert back to the conventional SI

units, we must insert the relevant powers of

and

. For example, for a mass of

= 10

eV, we have λ

= 2 × 10

−12

After getting rid of all factors of

and

, if a quantity

has a mass dimension

d, we write [X] = d. For example, we have [G] = −2, since we have

G =

where M

∼ 10

GeV is the Planck scale.

1 Classical field theory

Before we get into quantum nonsense, we will first start by understanding

classical fields.

1.1 Classical fields

Definition

(Field)

A field

is a physical quantity defined at every point of

spacetime (x, t). We write the value of the field at (x, t) as φ(x, t).

The “physical quantities” can be real or complex scalars, but later we will

see that we might have to consider even more complicated stuff.

In classical point mechanics, we have a finite number of generalized coordi-

nates

(

). In field theory, we are interested in the dynamics of

(

x, t

), where

and

are both labels. Here the

labels the different fields we can have, and

(

x, t

) labels a spacetime coordinate. Note that here position has been relegated

from a dynamical variable (i.e. one of the q

) to a mere label.

Example.

The electric field

(

x, t

) and magnetic field

(

x, t

), for

= 1

are examples of fields. These six fields can in fact be derived from 4 fields

(x, t), for µ = 0, 1, 2, 3, where

∂A

∂t

−

∂A

∂x

, B

ijk

∂A

∂x

Often, we write

= (φ, A).

Just like in classical dynamics, we would specify the dynamics of a field

through a Lagrangian. For particles, the Lagrangian specifies a number at each

time

. Since a field is a quantity at each point in space, we now have Lagrangian

densities, which gives a number at each point in spacetime (t, x).

Definition

(Lagrangian density)

Given a field

(

x, t

), a Lagrangian density is

a function L(φ, ∂

φ) of φ and its derivative.

Note that the Lagrangian density treats space and time symmetrically. How-

ever, if we already have a favorite time axis, we can look at the “total” Lagrangian

at each time, and obtain what is known as the Lagrangian.

Definition

(Lagrangian)

Given a Lagrangian density, the Lagrangian is defined

L =

x L(φ, ∂

φ).

For most of the course, we will only care about the Lagrangian density, and

call it the Lagrangian.

Definition

(Action)

Given a Lagrangian and a time interval [

, t

], the action

is defined by

S =

dt L(t) =

x L.

In general, we would want the units to satisfy [

] = 0. Since we have

x] = −4, we must have [L] = 4.

The equations of motion are, as usual, given by the principle of least action.

Definition

(Principle of least action)

The equation of motion of a Lagrangian

system is given by the principle of least action — we vary the field slightly,

keeping values at the boundary fixed, and require the first-order change

δS

= 0.

For a small perturbation φ

7→ φ

+ δφ

, we can compute

δS =



∂L

∂φ

δφ

∂L

∂(∂

)

δ(∂

)





∂L

∂φ

− ∂



∂L

∂(∂

)



δφ

+ ∂



∂L

∂(∂

)

δφ



We see that the last term vanishes for any term that decays at spatial infinity,

and obeys δφ

(x, t

) = δφ

(x, t

) = 0.

Requiring δS = 0 means that we need

Proposition

(Euler-Lagrange equations)

The equations of motion for a field

are given by the Euler-Lagrange equations:

∂



∂L

∂(∂

)



−

∂L

∂φ

= 0.

We can begin by considering the simplest field one can imagine of — the

Klein–Gordon field. We will later see that this is a “free field”, and “particles”

don’t interact with each other.

Example.

The Klein–Gordon equation for a real scalar field

(

x, t

) is given by

the Lagrangian

L =

∂

φ∂

φ −

−

(∇φ)

−

We can view this Lagrangian as saying L = T − V , where

T =

is the kinetic energy, and

V =

(∇φ)

is the potential energy.

To find the Euler-Lagrange equation, we can compute

∂L

∂(∂

φ)

= ∂

φ = (

φ, −∇φ)

and

∂L

∂φ

= −m

φ.

So the Euler-Lagrange equation says

∂

φ + m

φ = 0.

More explicitly, this says

φ − ∇

φ + m

φ = 0.

We could generalize this and add more terms to the Lagrangian. An obvious

generalization would be

L =

∂

φ∂

φ − V (φ),

where

(

) is an arbitrary potential function. Then we similarly obtain the

equation

∂

φ +

∂V

∂φ

= 0.

Example. Maxwell’s equations in vacuum are given by the Lagrangian

L = −

(∂

)(∂

) +

(∂

)

To find out the Euler-Lagrange equations, we need to figure out the derivatives

with respect to each component of the A field. We obtain

∂L

∂(∂

)

= −∂

+ (∂

)η

µν

So we obtain

∂



∂L

∂(∂

)



= −∂

∂

+ ∂

(∂

) = −∂

(∂

− ∂

We write

µν

= ∂

− ∂

So we are left with

∂



∂L

∂(∂

)



= −∂

µν

It is an exercise to check that these Euler-Lagrange equations reproduce

∂

= 0,

= ε

ijk

∂

Using our F

µν

, we can rewrite the Lagrangian as

L = −

µν

How did we come up with these Lagrangians? In general, it is guided by two

principles — one is symmetry, and the other is renormalizability. We will discuss

symmetries shortly, and renormalizaility would be done in the III Advanced

Quantum Field Theory course.

In all of these examples, the Lagrangian is local. In other words, the terms

don’t couple φ(x, t) to φ(y, t) if x 6= y.

Example. An example of a non-local Lagrangian would be

y φ(x)φ(x − y)

A priori, there is no reason for this, but it happens that nature seems to be

local. So we shall only consider local Lagrangians.

Note that locality does not mean that the Lagrangian at a point only depends

on the value at the point. Indeed, it also depends on the derivatives at

. So we

can view this as saying the value of

only depends on the value of

at an

infinitesimal neighbourhood of x (formally, the jet at x).

1.2 Lorentz invariance

If we wish to construct relativistic field theories such that

and

are on an

equal footing, the Lagrangian should be invariant under Lorentz transformations

7→ x

0µ

= Λ

, where

στ

= η

µν

and η

µν

is the Minkowski metric given by

µν







1 0 0 0

0 −1 0 0

0 0 −1 0

0 0 0 −1







Example. The transformation







1 0 0 0

0 1 0 0

0 0 cos θ −sin θ

0 0 sin θ cos θ







describes a rotation by an angle around the x-axis.

Example. The transformation







γ −γv 0 0

−γv γ 0 0

0 0 1 0

0 0 0 1







describes a boost by v along the x-axis.

The Lorentz transformations form a Lie group under matrix multiplication —

see III Symmetries, Field and Particles.

The Lorentz transformations have a representation on the fields. For a scalar

field, this is given by

φ(x) 7→ φ

(x) = φ(Λ

−1

x),

where the indices are suppressed. This is an active transformation — say

is the point at which, say, the field is a maximum. Then after applying the

Lorentz transformation, the position of the new maximum is Λ

. The field

itself actually moved.

Alternatively, we can use passive transformations, where we just relabel the

points. In this case, we have

φ(x) 7→ φ(Λ(x)).

However, this doesn’t really matter, since if Λ is a Lorentz transformation, then

so is Λ

−1

. So being invariant under active transformations is the same as being

invariant under passive transformations.

A Lorentz invariant theory should have equations of motion such that if

(

)

is a solution, then so is

(Λ

−1

). This can be achieved by requiring that the

action S is invariant under Lorentz transformations.

Example. In the Klein–Gordon field, we have

L =

∂

φ∂

φ −

The Lorentz transformation is given by

φ(x) 7→ φ

(x) = φ(Λ

−1

x) = φ(y),

where

= (Λ

−1

)

We then check that

∂

φ(x) 7→

∂

∂x

(φ(Λ

−1

x))

∂

∂x

(φ(y))

∂y

∂x

∂

∂y

(φ(y))

= (Λ

−1

)

(∂

φ)(y).

Since Λ

−1

is a Lorentz transformation, we have

∂

φ∂

φ = ∂

∂

In general, as long as we write everything in terms of tensors, we get Lorentz

invariant theories.

Symmetries play an important role in QFT. Different kinds of symmetries

include Lorentz symmetries, gauge symmetries, global symmetries and super-

symmetries (SUSY).

Example.

Protons and neutrons are made up of quarks. Each type of quark

comes in three flavors, which are called red, blue and green (these are arbitrary

names). If we swap around red and blue everywhere in the universe, then the

laws don’t change. This is known as a global symmetry.

However, in light of relativity, swapping around red and blue everywhere in

the universe might be a bit absurd, since the universe is so big. What if we only

do it locally? If we make the change differently at different points, the equations

don’t a priori remain invariant, unless we introduce a gauge boson. More of this

will be explored in the AQFT course.

1.3 Symmetries and Noether’s theorem for field theories

As in the case of classical dynamics, we get a Noether’s theorem that tells us

symmetries of the Lagrangian give us conserved quantities. However, we have

to be careful here. If we want to treat space and time equally, saying that a

quantity “does not change in time” is bad. Instead, what we have is a conserved

current, which is a 4-vector. Then given any choice of spacetime frame, we can

integrate this conserved current over all space at each time (with respect to the

frame), and this quantity will be time-invariant.

Theorem

(Noether’s theorem)

Every continuous symmetry of

gives rise to

a conserved current j

(x) such that the equation of motion implies that

∂

= 0.

More explicitly, this gives

∂

+ ∇ · j = 0.

A conserved current gives rise to a conserved ch arge

Q =

since

= −

∇ · j d

= 0,

assuming that j

→ 0 as |x| → ∞.

Proof.

Consider making an arbitrary transformation of the field

7→ φ

δφ

We then have

δL =

∂L

∂φ

δφ

∂L

∂(∂

)

δ(∂

)



∂L

∂φ

− ∂

∂L

∂(∂

)



δφ

+ ∂



∂L

∂(∂

)

δφ



When the equations of motion are satisfied, we know the first term always

vanishes. So we are left with

δL = ∂



∂L

∂(∂

)

δφ



If the specific transformation

δφ

we are considering is a symmetry, then

δL

= 0 (this is the definition of a symmetry). In this case, we can define a

conserved current by

∂L

∂(∂

)

and by the equations above, this is actually conserved.

We can have a slight generalization where we relax the condition for a

symmetry and still get a conserved current. We say that

is a symmetry if

δL

∂

(

) for some

(

), i.e. a total derivative. Replaying the calculations,

we get

∂L

∂(∂

)

− F

Example

(Space-time invariance)

Recall that in classical dynamics, spatial

invariance implies the conservation of momentum, and invariance wrt to time

translation implies the conservation of energy. We’ll see something similar in

field theory. Consider x

7→ x

− ε

. Then we obtain

(x) 7→ φ

(x) + ε

∂

(x).

A Lagrangian that has no explicit x

dependence transforms as

L(x) 7→ L(x) + ε

∂

L(x),

giving rise to 4 currents — one for each ν = 0, 1, 2, 3. We have

)

∂L

∂(∂

)

∂

− δ

This particular current is called

, the energy-momentum tensor. This satisfies

∂

= 0,

We obtain conserved quantities, namely the energy

E =

x T

and the total momentum

x T

Example. Consider the Klein–Gordon field, with

L =

∂

φ∂

φ −

We then obtain

µν

= ∂

φ∂

φ − η

µν

So we have

E =



(∇φ)



The momentum is given by

φ∂

φ.

In this example,

µν

comes out symmetric in

and

. In general, it would

not be, but we can always massage it into a symmetric form by adding

µν

= T

µν

+ ∂

ρµν

with Γ

ρµν

a tensor antisymmetric in ρ and µ. Then we have

∂

ρµν

= 0.

So this σ

µν

is also invariant.

A symmetric energy-momentum tensor of this form is actually useful, and is

found on the RHS of Einstein’s field equation.

Example (Internal symmetries). Consider a complex scalar field

ψ(x) =

√

(φ

(x) + iφ

(x)),

where φ

and φ

are real scalar fields. We put

L = ∂

∗

∂

ψ − V (ψ

∗

ψ),

where

V (ψ

∗

ψ) = m

∗

ψ +

(ψ

∗

ψ)

+ ···

is some potential term.

To find the equations of motion, if we do all the complex analysis required,

we will figure that we will obtain the same equations as the real case if we treat

ψ and ψ

∗

as independent variables. In this case, we obtain

∂

ψ + m

ψ + λ(ψ

∗

ψ)ψ + ··· = 0

and its complex conjugate. The

has a symmetry given by

ψ 7→ e

iα

. Infinites-

imally, we have δψ = iαψ, and δψ

∗

= −iαψ

∗

This gives a current

= i(∂

∗

)ψ − i(∂

ψ)ψ

∗

We will later see that associated charges of this type have an interpretation of

electric charge (or particle number, e.g. baryon number or lepton number).

Note that this symmetry is an abelian symmetry, since it is a symmetry under

the action of the abelian group U(1). There is a generalization to a non-abelian

case.

Example

(Non-abelian internal symmetries)

Suppose we have a theory with

many fields, with the Lagrangian given by

L =

a=1

∂

−

a=1

− g

a=1

This theory is invariant under the bigger symmetry group

(

). If we

view the fields as components of complex fields, then we have a symmetry under

2) or even

(

2). For example, the symmetry group

(3) gives the

8-fold way.

Example.

There is a nice trick to determine the conserved current when our

infinitesimal transformation is given by

δφ

αφ

for some real constant

Consider the case where we have an arbitrary perturbation

(

). In this

case,

δL

is no longer invariant, but we know that whatever formula we manage

to come up with, it has to vanish when

is constant. Assuming that we only

have first-order derivatives, the change in Lagrangian must be of the form

δL = (∂

α(x))h

(φ).

for some h

. We claim that h

is the conserved current. Indeed, we have

δS =

x δL = −

x α(x)∂

using integration by parts. We know that if the equations of motion are satisfied,

then this vanishes for any

(

) as long as it vanishes at infinity (or the boundary).

So we must have

∂

= 0.

1.4 Hamiltonian mechanics

We can now talk about the Hamiltonian formulation. This can be done for field

theories as well. We define

Definition

(Conjugate momentum)

Given a Lagrangian system for a field

we define the conjugate momentum by

π(x) =

∂L

∂

This is not to be confused with the total momentum P

Definition (Hamiltonian density). The Hamiltonian density is given by

H = π(x)

φ(x) − L(x),

where we replace all occurrences of

φ(x) by expressing it in terms of π(x).

Example. Suppose we have a field Lagrangian of the form

L =

−

(∇φ)

− V (φ).

Then we can compute that

π =

φ.

So we can easily find

H =

(∇φ)

+ V (φ).

Definition (Hamiltonian). The Hamiltonian of a Hamiltonian system is

H =

x H.

This agrees with the field energy we computed using Noether’s theorem.

Definition (Hamilton’s equations). Hamilton’s equations are

φ =

∂H

∂π

, ˙π = −

∂H

∂φ

These give us the equations of motion of φ.

There is an obvious problem with this, that the Hamiltonian formulation is

not manifestly Lorentz invariant. However, we know it actually is because we

derived it as an equivalent formulation of a Lorentz invariant theory. So we are

safe, so far. We will later have to be more careful if we want to quantize the

theories from the Hamiltonian formalism.

2 Free field theory

So far, we’ve just been talking about classical field theory. We now want to

quantize this, and actually do quantum field theory.

2.1 Review of simple harmonic oscillator

Michael Peskin once famously said “Physics is the subset of human experience

that can be reduced to coupled harmonic oscillators”. Thus, to understand

quantum mechanics, it is important to understand how the quantum harmonic

oscillator works.

Classically, the simple harmonic oscillator is given by the Hamiltonian

H =

where

is the momentum and

is the position. To obtain the corresponding

quantum system, canonical quantization tells us that we should promote the

and

into complex “operators”

ˆp, ˆq

, and use the same formula for the Hamiltonian,

namely we now have

H =

ˆp

ˆq

In the classical system, the quantities

and

used to satisfy the Poisson brackets

{q, p} = 1.

After promoting to operators, they satisfy the commutation relation

[ˆq, ˆp] = i.

We will soon stop writing the hats because we are lazy.

There are a few things to take note of.

(i)

We said

and

are “operators”, but did not say what they actually

operate on! Instead, what we are going to do is to analyze these operators

formally, and after some careful analysis, we show that there is a space the

operators naturally act on, and then take that as our state space. This is,

in general, how we are going to do quantum field theory (except we tend

to replace the word “careful” with “sloppy”).

During the analysis, we will suppose there are indeed some states our

operators act on, and then try to figure out what properties the states

must have.

(ii)

The process of canonical quantization depends not only on the classical

system itself, but how we decide the present our system. There is no

immediate reason why if we pick different coordinates for our classical

system, the resulting quantum system would be equivalent. In particular,

before quantization, all the terms commute, but after quantization that is

no longer true. So how we decide to order the terms in the Hamiltonian

matters.

Later, we will come up with the notion of normal ordering. From then

onwards, we can have a (slightly) more consistent way of quantizing

operators.

After we’ve done this, the time evolution of states is governed by the

Schr¨odinger equation:

|ψi = H |ψi.

In practice, instead of trying to solve this thing, we want to find eigenstates

|Ei

such that

H |Ei = E |Ei.

If such states are found, then defining

|ψi = e

−iEt

|Ei

would give a nice, stable solution to the Schr¨odinger equation.

The trick is to notice that in the classical case, we can factorize the Hamilto-

nian as

H = ω



q +

√

2ω



q +

−i

√

2ω



Now

is a product of two terms that are complex conjugates to each other,

which, in operator terms, means they are adjoints. So we have the benefit

that we only have to deal with a single complex object

√

2ω

(and its

conjugate), rather than two unrelated real objects. Also, products are nicer than

sums. (if this justification for why this is a good idea doesn’t sound convincing,

just suppose that we had the idea of doing this via divine inspiration, and it

turns out to work well)

We now do the same factorization in the quantum case. We would not

expect the result to be exactly the above, since that working relies on

and

commuting. However, we can still try and define the operators

a =

√

2ω

p +

q, a

†

−i

√

2ω

p +

These are known as creation and annihilation operators for reasons that will

become clear soon.

We can invert these to find

q =

√

2ω

(a + a

†

), p = −i

(a − a

†

We can substitute these equations into the commutator relation to obtain

[a, a

†

] = 1.

Putting them into the Hamiltonian, we obtain

H =

ω(aa

†

+ a

†

a) = ω



†

a +

[a, a

†

]



= ω



†

a +



We can now compute

[H, a

†

] = ωa

†

, [H, a] = −ωa.

These ensure that a, a

†

take us between energy eigenstates — if

H |Ei = E |Ei,

then

†

|Ei = (a

†

H + [H, a

†

]) |Ei = (E + ω)a

†

|Ei.

Similarly, we have

Ha |Ei = (E − ω)a |Ei.

So assuming we have some energy eigenstate

|Ei

, these operators give us loads

more with eigenvalues

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

If the energy is bounded below, then there must be a ground state

|0i

satisfying

a |0i

= 0. Then the other “excited states” would come from repeated applications

of a

†

, labelled by

|ni = (a

†

)

|0i,

with

H |ni =



n +



ω |ni.

Note that we were lazy and ignored normalization, so we have hn|ni 6= 1.

One important feature is that the ground state energy is non-zero. Indeed,

we have

H |0i = ω



†

a +



|0i =

|0i.

Notice that we managed to figure out what the eigenvalues of

must be, without

having a particular state space (assuming the state space is non-trivial). Now we

know what is the appropriate space to work in. The right space is the Hilbert

space generated by the orthonormal basis

{|0i, |1i, |2i, ···}.

2.2 The quantum field

We are now going to use canonical quantization to promote our classical fields

to quantum fields. We will first deal with the case of a real scalar field.

Definition

(Real scalar quantum field)

A (real, scalar) quantum field is an

operator-valued function of space

, with conjugate momentum

, satisfying the

commutation relations

[φ(x), φ(y)] = 0 = [π(x), π(y)]

and

[φ(x), π(y)] = iδ

(x − y).

In case where we have many fields labelled by

a ∈ I

, the commutation relations

are

[φ

(x), φ

(y)] = 0 = [π

(x), π

(y)]

and

[φ

(x), π

(y)] = iδ

(x − y)δ

The evolution of states is again given by Schr¨odinger equation.

Definition (Schr¨odinger equation). The Schr¨odinger equation says

|ψi = H |ψi.

However, we will, as before, usually not care and just look for eigenvalues of

As in the case of the harmonic oscillator, our plan is to rewrite the field in

terms of creation and annihilation operators. Note that in quantum mechanics, it

is always possible to write the position and momentum in terms of some creation

and annihilation operators for any system. It’s just that if the system is not

a simple harmonic oscillator, these operators do not necessarily have the nice

properties we want them to have. So we are just going to express everything in

creation and annihilation operators nevertheless, and see what happens.

2.3 Real scalar fields

We start with simple problems. We look at free theories, where the Lagrangian

is quadratic in the field, so that the equation of motion is linear. We will see that

the whole field then decomposes into many independent harmonic oscillators.

Before we jump into the quantum case, we first look at what happens in a

classical free field.

Example.

The simplest free theory is the classic Klein–Gordon theory for a

real scalar field φ(x, t). The equations of motion are

∂

φ + m

φ = 0.

To see why this is free, we take the Fourier transform so that

φ(x, t) =

(2π)

ip·x

φ(p, t).

We substitute this into the Klein–Gordon equation to obtain



∂

∂t

+ (p

+ m

)



φ(p, t) = 0.

This is just the usual equation for a simple harmonic oscillator for each

independently, with frequency

+ m

. So the solution to the classical

Klein–Gordon equation is a superposition of simple harmonic oscillators, each

vibrating at a different frequency (and a different amplitude).

For completeness, we will note that the Hamiltonian density for this field is

given by

H =

(π

+ (∇φ)

+ m

So to quantize the Klein–Gordon field, we just have to quantize this infinite

number of harmonic oscillators!

We are going to do this in two steps. First, we write our quantum fields

(

)

and π(x) in terms of their Fourier transforms

φ(x) =

(2π)

ip·x

φ(p)

π(x) =

(2π)

ip·x

˜π(p)

Confusingly,

represents both the conjugate momentum and the mathematical

constant, but it should be clear from the context.

If we believe in our classical analogy, then the operators

(

) and

˜π

(

) should

represent the position and momentum of quantum harmonic oscillators. So we

further write them as

φ(x) =

(2π)

2ω



ip·x

+ a

†

−ip·x



π(x) =

(2π)

(−i)



ip·x

− a

†

−ip·x



where we have

= p

+ m

Note that despite what we said above, we are multiplying

†

−ip·x

, and not

ip·x

. This is so that φ(x) will be manifestly a real quantity.

We are now going to find the commutation relations for the

and

†

Throughout the upcoming computations, we will frequently make use of the

following result:

Proposition. We have

(2π)

−ip·x

= δ

(x).

Proposition. The canonical commutation relations of φ, π, namely

[φ(x), φ(y)] = 0

[π(x), π(y)] = 0

[φ(x), π(y)] = iδ

(x − y)

are equivalent to

, a

] = 0

†

, a

†

] = 0

, a

†

] = (2π)

(p − q).

Proof.

We will only prove one small part of the equivalence, as the others are

similar tedious and boring computations, and you are not going to read it anyway.

We will use the commutation relations for the

to obtain the commutation

relations for φ and π. We can compute

[φ(x), π(y)]

p d

(2π)

(−i)



−[a

, a

†

ip·x−iq·y

+ [a

†

, a

−ip·x+iq·y



p d

(2π)

(−i)

(2π)



−δ

(p − q)e

ip·x−iq·y

− δ

(q − p)e

−ip·x+iq·y



(−i)

(2π)



−e

−ip·(x−y)

− e

ip·(y−x)



= iδ

(x − y).

Note that to prove the inverse direction, we have to invert the relation between

φ(x), π(x) and a

, a

†

and express a

and a

†

in terms of φ and π by using

x φ(x) e

ip·x

2ω



−p

+ a

†



x π(x) e

ip·x

= (−i)



−p

− a

†



So our creation and annihilation operators do satisfy commutation relations

similar to the case of a simple harmonic oscillator.

The next thing to do is to express

in terms of

and

†

. Before we plunge

into the horrendous calculations that you are probably going to skip, it is a

good idea to stop and think what we are going to expect. If we have enough

faith, then we should expect that we are going to end up with infinitely many

decoupled harmonic oscillators. In other words, we would have

H =

(2π)

(a harmonic oscillator of frequency ω

But if this were true, then we would have a serious problem. Recall that each

harmonic oscillator has a non-zero ground state energy, and this ground state

energy increases with frequency. Now that we have infinitely many harmonic

oscillators of arbitrarily high frequency, the ground state energy would add up to

infinity! What’s worse is that when we derived the ground state energy for the

harmonic oscillator, the energy is

[

a, a

†

]. Now our [

, a

†

] is (2

)

(

p − q

i.e. the value is infinite. So our ground state energy is an infinite sum of infinities!

This is so bad.

These problems indeed will occur. We will discuss how we are going to avoid

them later on, after we make ourselves actually compute the Hamiltonian.

As in the classical case, we have

H =

x (π

+ (∇φ)

+ m

For the sake of sanity, we will evaluate this piece by piece. We have

x π

= −

x d

p d

(2π)

√

ip·x

− a

†

−ip·x

)(a

iq·x

− a

†

−iq·x

)

= −

x d

p d

(2π)

√

i(p+q)·x

− a

†

i(q−p)·x

− a

†

i(p−q)·x

+ a

†

−i(p+q)·x

)

= −

p d

(2π)

√

(p + q) − a

†

(p − q)

− a

†

(p − q) + a

†

(p + q))

(2π)

((a

†

+ a

†

) − (a

−p

+ a

†

−p

)).

That was tedious. We similarly compute

x (∇φ)

x d

p d

(2π)

√

(ipa

ip·x

− ipa

†

−ip·x

)

(iqa

iq·x

− iqa

†

−iq·x

)

(2π)

2ω

((a

†

+ a

†

) + (a

−p

+ a

†

−p

))

x m

(2π)

2ω

((a

†

+ a

†

) + (a

−p

+ a

†

−p

)).

Putting all these together, we have

H =

x (π

+ (∇φ)

+ m

)

(2π)



−ω



−p

+ a

†

−p

)





†

+ a

†

)



Now the first term vanishes, since we have ω

= p

+ m

. So we are left with

(2π)

(ω

+ p

+ m

)(a

†

+ a

†

)

(2π)

†

+ a

†

)

(2π)



†

, a

†

]



(2π)



†

(2π)

(0)



Note that if we cover the

(2π)

, then the final three lines are exactly what we

got for a simple harmonic oscillator of frequency ω

+ m

Following the simple harmonic oscillator, we postulate that we have a vacuum

state |0i such that

|0i = 0

for all p.

When

acts on this, the

†

terms all vanish. So the energy of this ground

state comes from the second term only, and we have

H |0i =

(2π)

(0) |0i = ∞|0i.

since all ω

are non-negative.

Quantum field theory is always full of these infinities! But they tell us

something important. Often, they tell us that we are asking a stupid question.

Let’s take a minute to explore this infinity and see what it means. This is

bad, since what we want to do at the end is to compute some actual probabilities

in real life, and infinities aren’t exactly easy to calculate with.

Let’s analyze the infinities one by one. The first thing we want to tackle is

the δ

(0). Recall that our δ

can be thought of as

(2π)

(p) =

x e

ip·x

When we evaluate this at

, we are then integrating the number 1 over all

space, and thus the result is infinite! You might say, duh, space is so big. If

there is some energy everywhere, then of course the total energy is infinite. This

problem is known as infrared divergence. So the idea is to look at the energy

density, i.e. the energy per unit volume. Heuristically, since the (2

)

(

) is

just measuring the “volume” of the universe, we would get rid of it by simply

throwing away the factor of (2π)

(0).

If we want to make this a bit more sophisticated, we would enclose our

universe in a box, and thus we can replace the (2

)

(

) with the volume

This trick is known as an infrared cutoff . Then we have

(2π)

We can then safely take the limit as

V → ∞

, and forget about this

(

) problem.

This is still infinite, since

gets unbounded as

p → ∞

. In other words, the

ground state energies for each simple harmonic oscillator add up to infinity.

These are high frequency divergences at short distances. These are called

ultraviolet divergences. Fortunately, our quantum field theorists are humble

beings and believe their theories are wrong! This will be a recurring theme —

we will only assume that our theories are low-energy approximations of the real

world. While this might seem pessimistic, it is practically a sensible thing to do

— our experiments can only access low-level properties of the universe, so what

we really see is the low-energy approximation of the real theory.

Under this assumption, we would want to cut off the integral at high mo-

mentum in some way. In other words, we just arbitrarily put a bound on the

integral over

, instead of integrating over all possible

. While the cut-off point

is arbitrary, it doesn’t really matter. In (non-gravitational) physics, we only care

about energy differences, and picking a different cut-off point would just add a

constant energy to everything. Alternatively, we can also do something slightly

more sophisticated to avoid this arbitrariness, as we will see in the example of

the Casimir effect soon.

Even more straightforwardly, if we just care about energy differences, we can

just forget about the infinite term, and write

H =

(2π)

†

Then we have

H |0i = 0.

While all these infinity cancelling sound like bonkers, the theory actually fits

experimental data very well. So we have to live with it.

The difference between this

and the previous one with infinities is an

ordering ambiguity in going from the classical to the quantum theory. Recall

that we did the quantization by replacing the terms in the classical Hamiltonian

with operators. However, terms in the classical Hamiltonian are commutative,

but not in the quantum theory. So if we write the classical Hamiltonian in a

different way, we get a different quantized theory. Indeed, if we initially wrote

H =

(ωq − ip)(ωq + ip),

for the classical Hamiltonian for a single harmonic operator, and then did the

quantization, we would have obtained

H = ωa

†

It is convenient to have the following notion:

Definition (Normal order). Given a string of operators

) ···φ

the normal order is what you obtain when you put all the annihilation operators

to the right of (i.e. acting before) all the creation operators. This is written as

:φ

) ···φ

): .

So, for example,

:H: =

(2π)

†

In the future, we will assume that when we quantize our theory, we do so in a

way that the resulting operators are in normal order.

Applications — The Casimir effect

Notice that we happily set

= 0, claiming that only energy differences are mea-

sured. But there exists a situation where differences in the vacuum fluctuations

themselves can be measured, when we have two separated regions of vacuum. In

this example, we will also demonstrate how the infrared and ultraviolet cutoffs

can be achieved. We are going to enclose the world in a box of length

, and

then do the ultraviolet cutoff in a way parametrized by a constant

. We then

do some computations under these cutoffs, derive an answer, and then take the

limit as

L → ∞

and

a →

0. If we asked the right question, then the answer will

tend towards a finite limit as we take these limits.

To regulate the infrared divergences, we put the universe in a box again. We

make the

direction periodic, with a large period

. So this is a one-dimensional

box. We impose periodic boundary conditions

φ(x, y, z) = φ(x + L, y, z).

We are now going to put two reflecting plates in the box at some distance

d  L

apart. The plates impose φ(x) = 0 on the plates.

The presence of the plates means that the momentum of the field inside them is

quantized

p =



πn

, p



for

n ∈ Z

. For a massless scalar field, the energy per unit area between the

plates is

E(d) =

∞

n=1

(2π)



πn



+ p

The energy outside the plates is then E(L − d). The total energy is then

E = E(d) + E(L − d).

This energy (at least naively) depends on

. So there is a force between the

plates! This is the Casimir effect, predicted in 1945, and observed in 1958. In

the lab, this was done with the EM field, and the plates impose the boundary

conditions.

Note that as before, we had to neglect modes with

|p|

too high. More precisely,

we pick some distance scale

a  d

, and ignore modes where

|p|  a

−1

. This is

known as the ultraviolet cut-off. This is reasonable, since for high momentum

modulus, we would break through the plates. Then we have

E(d) = A

(2π)

|p|e

−a|p|

Note that as we set a → 0, we get the previous expression.

Since we are scared by this integral, we consider the special case where we

live in the 1 + 1 dimensional world. Then this becomes

1+1

(d) =

∞

n=1

−anπ/d

= −

−anπ/d

= −

1 − e

−aπ/d

(1 − e

−aπ/d

)

2πa

−

24d

+ O(a

Our total energy is

E = E(d) + E(L − d) =

2πa

−



L − d



+ O(a

a →

0, this is still infinite, but the infinite term does not depend on

. The

force itself is just

∂E

∂d

2π

24d

+ O





+ O(a

which is finite as

a →

0 and

L → ∞

. So as we remove both the infrared and UV

cutoffs, we still get a sensible finite force.

In 3 + 1 dimensions, if we were to do the more complicated integral, it turns

out we get

∂E

∂d

480d

The actual Casimir effect for electromagnetic fields is actually double this due

to the two polarization states of the photon.

Recovering particles

We called the operators

and

†

the creation and annihilation operators.

Let us verify that they actually create and annihilate things! Recall that the

Hamiltonian is given by

H =

(2π)

†

Then we can compute

[H, a

†

] =

(2π)

†

, a

†

]

(2π)

†

(2π)

(p − q)

= ω

†

Similarly, we obtain

[H, a

] = −ω

which means (like SHM) we can construct energy eigenstates by acting with

†

We let

|pi = a

†

|0i.

Then we have

H |pi = ω

|pi,

where the eigenvalue is

= p

+ m

But from special relativity, we also know that the energy of a particle of mass

and momentum p is given by

= p

+ m

So we interpret

|pi

as the momentum eigenstate of a particle of mass

and

momentum

. And we identify

with the mass of the quantized particle. From

now on, we will write E

instead of ω

Let’s check this interpretation. After normal ordering, we have

P =

π(x)∇φ(x) d

x =

(2π)

†

So we have

P |pi =

(2π)

†

|0i)

(2π)

†

+ (2π)

(p − q)) |0i

= pa

†

|0i

= p |pi.

So the state has total momentum p.

What about multi-particle states? We just have to act with more

†

’s. We

have the n-particle state

, p

, ··· , p

i = a

†

···a

†

|0i.

Note that

|p, qi

|q, pi

for any

p, q

. So any two parts are symmetric under

interchange, i.e. they are bosons.

Now we can tell what our state space is. It is given by the span of particles

of the form

|0i, a

†

|0i, a

†

|0i, ···

This is known as the Fock space. As in the case of SHM, there is also an operator

which counts the number of particles. It is given by

N =

(2π)

†

So we have

N |p

, ··· , p

i = n |p

, ··· , p

It is easy to compute that

[N, H] = 0

So the particle number is conserved in the free theory. This is not true in general!

Usually, when particles are allowed to interact, the interactions may create or

destroy particles. It is only because we are in a free theory that we have particle

number conservation.

Although we are calling these states “particles”, they aren’t localized —

they’re momentum eigenstates. Theoretically, we can create a localized state via

a Fourier transform:

|xi =

(2π)

−ip·x

|pi.

More generally, we can create a wave-packet, and insert ψ(p) to get

|ψi =

(2π)

−ip·x

ψ(p) |pi.

Then this wave-packet can be both partially localized in space and in momentum.

For example, we can take ψ to be the Gaussian

ψ(p) = e

−p

/2m

Note now that neither

|xi

nor

|ψi

are

-eigenstates, just like in non-relativistic

quantum mechanics.

Relativistic normalization

As we did our quantum field theory, we have completely forgotten about making

our theory Lorentz invariant. Now let’s try to recover some. We had a vacuum

state

|0i

, which we can reasonably assume to be Lorentz invariant. We then

produced some 1-particle states

|pi

†

|0i

. This is certainly not a “scalar”

quantity, so it would probably transform as we change coordinates.

However, we can still impose compatibility with relativity, by requiring that

its norm is Lorentz invariant. We have chosen |0i such that

h0|0i = 1.

We can compute

hp|qi = h0|a

†

|0i = h0|a

†

+ (2π)

(p − q) |0i = (2π)

(p − q).

There is absolutely no reason to believe that this is a Lorentz invariant quantity,

as p and q are 3-vectors. Indeed, it isn’t.

As the resulting quantity is a scalar, we might hope that we can come up

with new states

|pi = A

|pi

so that

hp|qi = (2π)

∗

(p − q)

is a Lorentz invariant quantity. Note that we write non-bold characters for

“relativistic” things, and bold characters for “non-relativistic” things. Thus, we

think of the p in |pi as the 4-vector p, and the p in |pi as the 3-vector p.

The trick to figuring out the right normalization is by looking at an object

we know is Lorentz invariant, and factoring it into small pieces. If we know that

all but one of the pieces are Lorentz invariant, then the remaining piece must be

Lorentz invariant as well.

Our first example would be the following:

Proposition. The expression

is Lorentz-invariant, where

= p

+ m

for some fixed m.

Proof. We know

p certainly is Lorentz invariant, and

= p

− p

is also a Lorentz-invariant quantity. So for any m, the expression

p δ(p

− p

− m

)

is also Lorentz invariant. Writing

= p

+ m

integrating over p

in the integral gives

and this is Lorentz invariant.

Thus, we have

Proposition. The expression

(p − q)

is Lorentz invariant.

Proof. We have

· (2E

(p − q)) = 1.

Since the RHS is Lorentz invariant, and the measure is Lorentz invariant, we

know 2E

(p − q) must be Lorentz invariant.

From this, we learn that the correctly normalized states are

|pi =

†

|0i.

These new states satisfy

hp|qi = (2π)

(2E

)δ

(p − q),

and is Lorentz invariant.

We can now define relativistically normalized creation operators by

†

(p) =

†

Then we can write our field as

φ(x) =

(2π)

(a(p)e

ip·x

+ a

†

(p)e

−ip·x

2.4 Complex scalar fields

What happens when we want to talk about a complex scalar field? A classical

free complex scalar field would have Lagrangian

L = ∂

∗

∂

ψ − µ

∗

ψ.

Again, we want to write the quantized

as an integral of annihilation and

creation operators indexed by momentum. In the case of a real scalar field, we

had the expression

φ(x) =

(2π)



ip·x

+ a

†

−ip·x



We needed the “coefficients” of

ip·x

and those of

−ip·x

to be

and its

conjugate

†

so that the resulting

would be real. Now we know that our

is a complex quantity, so there is no reason to assert that the coefficients are

conjugates of each other. Thus, we take the more general decomposition

ψ(x) =

(2π)

ip·x

+ c

†

−ip·x

)

†

(x) =

(2π)

†

−ip·x

+ c

ip·x

Then the conjugate momentum is

π(x) =

(2π)

†

−ip·x

− c

ip·x

)

†

(x) =

(2π)

(−i)

ip·x

− c

†

−ip·x

The commutator relations are

[ψ(x), π(y)] = [ψ

†

(x), π

†

(y)] = iδ

(x − y),

with all other commutators zero.

Similar tedious computations show that

, b

†

] = [c

, c

†

] = (2π)

(p − q),

with all other commutators zero.

As before, we can find that the number operators

(2π)

†

, N

(2π)

†

are conserved.

But in the classical case, we had an extra conserved charge

Q = i

x (

∗

ψ − ψ

∗

ψ) = i

x (πψ − ψ

∗

Again by tedious computations, we can replace the

and

with their operator

analogues, expand everything in terms of

and

, throw away the pieces of

infinity by requiring normal ordering, and then obtain

Q =

(2π)

†

− b

†

) = N

− N

Fortunately, after quantization, this quantity is still conserved, i.e. we have

[Q, H] = 0.

This is not a big deal, since

and

are separately conserved. However, in

the interacting theory, we will find that

and

are not separately conserved,

but Q still is.

We can think of

and

particles as particle and anti-particle, and

computes

the number of particles minus the number of antiparticles. Looking back, in the

case of a real scalar field, we essentially had a system with

. So the particle

is equal to its anti-particle.

2.5 The Heisenberg picture

We have tried to make our theory Lorentz invariant, but it’s currently in a terrible

state. Indeed, the time evolution happens in the states, and the operators depend

on states only. Fortunately, from IID Principles of Quantum Mechanics, we

know that there is a way to encode the time evolution in the operators instead,

via the Heisenberg picture.

Recall that the equation of motion is given by the Schr¨odinger equation

|ψ(t)i = H |ψ(t)i.

We can write the solution formally as

|ψ(t)i = e

−iHt

|ψ(0)i.

Thus we can think of

−iHt

as the “time evolution operator” that sends

|ψ(0)i

forward in time by t.

Now if we are given an initial state

|ψ(0)i

, and we want to know what an

operator

does to it at time

, we can first apply

−iHt

to let

|ψ(0)i

evolve to

time

, then apply the operator

, then pull the result back to time 0. So in

total, we obtain the operator

(t) = e

iHt

−iHt

To evaluate this expression, it is often convenient to note the following result:

Proposition. Let A and B be operators. Then

−A

= B + [A, B] +

[A, [A, B]] +

[A, [A, [A, B]]] + ··· .

In particular, if [A, B] = cB for some constant c, then we have

−A

= e

Proof. For λ a real variable, note that

dλ

λA

−λA

) = lim

ε→0

(λ+ε)A

−(λ+ε)A

− e

λA

−λA

= lim

ε→0

λA

εA

−εA

− B

−λA

= lim

ε→0

λA

(1 + εA)B(1 − εA) − B + o(ε)

−λA

= lim

ε→0

λA

(ε(AB −BA) + o(ε))

−λA

= e

λA

[A, B]e

−λA

So by induction, we have

dλ

λA

−λA

) = e

λA

[A, [A, ···[A, B] ···]]e

−λA

Evaluating these at λ = 0, we obtain a power series representation

λA

−λA

= B + λ[A, B] +

[A, [A, B]] + ··· .

Putting λ = 1 then gives the desired result.

In the Heisenberg picture, one can readily directly verify that the commutation

relations for our operators

(

x, t

) and

(

x, t

) now become equal 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).

For an arbitrary operator O

, we can compute

iHt

−iHt

)

= iHe

iHt

−iHt

+ e

iHt

(−iHe

−iHt

)

= i[H, O

where we used the fact that the Hamiltonian commutes with any function of

itself. This gives the time evolution for the operators.

Recall that in quantum mechanics, when we transformed into the Heisenberg

picture, a miracle occurred, and the equations of motion of the operators became

the usual classical equations of motion. This will happen again. For

(

we have

φ(x, t) = i[H, φ(x, t)]

= i

[π

(y, t) + (∇

φ(y, t))

+ m

(y, t), φ(x, t)]

= i

[π

(y, t), φ(x, t)]

= i



π(y, t)[π(y, t), φ(x, t)] + [π(y, t), φ(x, t)]π(y, t)



= i

y π(y, t)(−iδ

(x − y))

= π(x, t).

Similarly, we have

˙π(x, t) = i[H, π(x, t)]



[(∇

φ(y, t))

, π(x, t)] + m

[φ

(y, t), π(x, t)]



Now notice that ∇

does not interact with π(x, t). So this becomes



−(∇

φ(y, t) ·∇

(x − y)) − m

φ(y, t)δ

(x − y)



Integrating the first term by parts, we obtain



(∇

φ(y, t))δ

(x − y) − m

φ(y, t)δ

(x − y)



= ∇

φ(x, t) − m

φ(x, t).

Finally, noting that ˙π =

φ and rearranging, we obtain

∂

φ(x, t) + m

φ(x, t) = 0.

This is just the usual Klein–Gordon equation!

It is interesting to note that this final equation of motion is Lorentz invariant,

but our original

dφ

= i[H, φ]

is not. Indeed,

dφ

itself singles out time, and to define

we also need to single

out a preferred time direction. However, this statement tells us essentially that

generates time evolution, which is true in any frame, for the appropriate

definition of H and t in that frame.

What happens to the creation and annihilation operators? We use our

previous magic formula to find that

iHt

−iHt

= e

−iE

iHt

†

−iHt

= e

†

These extra factors of

±iE

might seem really annoying to carry around, but

they are not! They merge perfectly into the rest of our formulas relativistically.

Indeed, we have

φ(x) ≡ φ(x, t) =

(2π)



−iE

ip·x

+ a

†

−ip·x



(2π)



−ip·x

+ a

†

ip·x



where now

p · x

is the inner product of the 4-vectors

and

! If we use our

relativistically normalized

†

(p) =

†

instead, then the equation becomes

φ(x) =

(2π)



a(p)e

−ip·x

+ a(p)

†

ip·x



which is made up of Lorentz invariant quantities only!

Unfortunately, one should note that we are not yet completely Lorentz-

invariant, as our commutation relations are equal time commutation relations.

However, this is the best we can do, at least in this course.

Causality

Since we are doing relativistic things, it is important to ask ourselves if our

theory is causal. If two points

x, y

in spacetime are space-like separated, then a

measurement of the field at

should not affect the measurement of the field at

. So measuring

(

) after

(

) should give the same thing as measuring

(

)

after φ(x). So we must have [φ(x), φ(y)] = 0.

Definition

(Causal theory)

A theory is causal if for any space-like separated

points x, y, and any two fields φ, ψ, we have

[φ(x), ψ(y)] = 0.

Does our theory satisfy causality? For convenience, we will write

∆(x − y) = [φ(x), φ(y)].

We now use the familiar trick that our

is Lorentz-invariant, so we can pick

a convenient frame to do the computations. Suppose

and

are space-like

separated. Then there is some frame where they are of the form

x = (x, t), y = (y, t)

for the same t. Then our equal time commutation relations tell us that

∆(x − y) = [φ(x), φ(y)] = iδ

(x − y) = 0.

So our theory is indeed causal!

What happens for time-like separation? We can derive the more general

formula

∆(x − y) = [φ(x), φ(y)]

p d

(2π)



, a

i(−p·x−q·y)

+ [a

, a

†

i(−p·x+q·y)

+[a

†

, a

i(p·x−q·y)

+ [a

†

, a

†

i(p·x+q·y)



p d

(2π)

(p − q)



i(−p·x+q·y)

− e

i(p·x−q·y)



(2π)

ip·(y−x)

− e

−ip·(y−x)

This doesn’t vanish for time-like separation, which we may wlog be of the form

x = (x, 0), y = (x, t), since we have

[φ(x, 0), φ(x, t)] ∝ e

−imt

− e

imt

2.6 Propagators

We now introduce the very important idea of a propagator. Suppose we are

living in the vacuum. We want to know the probability of a particle appearing

and then subsequently being destroyed at

. To quantify this, we introduce

the propagator.

Definition

(Propagator)

The propagator of a real scalar field

is defined to

D(x − y) = h0|φ(x)φ(y) |0i.

When we study interactions later, we will see that the propagator indeed

tells us the probability of a particle “propagating” from x to y.

It is not difficult to compute the value of the propagator. We have

h0|φ(x)φ(y) |0i =

(2π)

h0|a

†

|0ie

−ip·x+ip

·y

with all other terms in

(

)

(

) vanishing because they annihilate the vacuum.

We now use the fact that

h0|a

†

|0i = h0|[a

, a

†

] |0i = i(2π)

(p − p

since a

†

|0i = 0. So we have

Proposition.

D(x − y) =

(2π)

−ip·(x−y)

The propagator relates to the ∆(

x −y

) we previously defined by the following

simple relationship:

Proposition. We have

∆(x − y) = D(x − y) − D(y − x).

Proof.

∆(x − y) = [φ(x), φ(y)] = h0|[φ(x), φ(y)] |0i = D(x − y) − D(y − x),

where the second equality follows as [

(

)

, φ

(

)] is just an ordinary function.

For a space-like separation (x −y)

< 0, one can show that it decays as

D(x − y) ∼ e

−m(|x−y|)

This is non-zero! So the field “leaks” out of the light cone a bit. However,

since there is no Lorentz-invariant way to order the events, if a particle can

travel in a spacelike direction

x → y

, it can just as easily travel in the other

direction. So we know would expect

(

x −y

) =

(

y −x

), and in a measurement,

both amplitudes cancel. This indeed agrees with our previous computation that

∆(x − y) = 0.

Feynman propagator

As we will later see, it turns out that what actually is useful is not the above

propagator, but the Feynman propagator:

Definition (Feynman propagator). The Feynman propagator is

∆

(x − y) = h0|T φ(x)φ(y) |0i =

(

h0|φ(x)φ(y) |0i x

> y

h0|φ(y)φ(x) |0i y

> x

Note that it seems like this is not a Lorentz-invariant quantity, because we

are directly comparing

and

. However, this is actually not a problem, since

when

and

are space-like separated, then

(

) and

(

) commute, and the

two quantities agree.

Proposition. We have

∆

(x − y) =

(2π)

− m

−ip·(x−y)

This expression is a priori ill-defined since for each

, the integrand over

has

a pole whenever (

)

. So we need a prescription for avoiding this.

We replace this with a complex contour integral with contour given by

−E

Proof.

To compare with our previous computations of

(

x −y

), we evaluate the

integral for each p. Writing

− m

)

− E

)(p

+ E

)

we see that the residue of the pole at p

= ±E

is ±

When

> y

, we close the contour in the lower plane

→ −i∞

, so

−p

−t

)

→ e

−∞

= 0. Then

picks up the residue at

. So the

Feynman propagator is

∆

(x − y) =

(2π)

(−2πi)

−iE

−y

)+ip·(x−y)

(2π

)

−ip·(x−y)

= D(x − y)

When x

< y

, we close the contour in the upper-half plane. Then we have

∆

(x − y) =

(2π)

2πi

−2E

−y

)+ip·(x−y)

(2π)

−iE

−x

)−ip·(y−x)

We again use the trick of flipping the sign of p to obtain

(2π)

−ip·(y−x)

= D(y − x).

Usually, instead of specifying the contour, we write

∆

(x − y) =

(2π)

−ip·(x−y)

− m

+ iε

where

is taken to be small, or infinitesimal. Then the poles are now located at

−E

So integrating along the real axis would give us the contour we had above. This

is known about the “iε-prescription”.

The propagator is in fact the Green’s function of the Klein–Gordon operator:

(∂

− ∇

+ m

)∆

(x − y) =

(2π)

− m

(−p

+ m

−ip·(x−y)

= −i

(2π)

−ip·(x−y)

= −iδ

(x − y).

Other propagators

For some purposes, it’s useful to pick other contours, e.g. the retarded Green’s

function defined as follows:

−E

This can be given in terms of operators by

Definition

(Retarded Green’s function)

The retarded Green’s function is given

∆

(x − y) =

(

[φ(x), φ(y)] x

> y

0 y

> x

This is useful if we have some initial field configuration, and we want to see

how it evolves in the presence of some source. This solves the “inhomogeneous

Klein–Gordon equation”, i.e. an equation of the form

∂

φ(x) + m

φ(x) = J(x),

where J(x) is some background function.

One also defines the advanced Green’s function which vanishes when

< x

instead. This is useful if we know the end-point of a field configuration and want

to know where it came from.

However, in general, the Feynman propagator is the most useful in quantum

field theory, and we will only have to use this.

3 Interacting fields

Free theories are special — we can determine the exact spectrum, but noth-

ing interacts. These facts are related. Free theories have at most quadratic

terms in the Lagrangian. So the equations of motion are linear. So we get

exact quantization, and we get multi-particle states with no interactions by

superposition.

3.1 Interaction Lagrangians

How do we introduce field interactions? The short answer is to throw more

things into the Lagrangian to act as the “potential”.

We start with the case where there is only one real scalar field, and the field

interacts with itself. Then the general form of the Lagrangian can be given by

L =

∂

φ∂

φ −

−

∞

n=3

These λ

are known as the coupling constants.

It is almost impossible to work with such a Lagrangian directly, so we want to

apply perturbation theory to it. To do so, we need to make sure our interactions

are “small”. The obvious thing to require would be that



1. However, this

makes sense only if the

are dimensionless. So let’s try to figure out what the

dimensions are.

Recall that we have

[S] = 0.

Since we know S =

xL, and [d

x] = −4, we have

[L] = 4.

We also know that

[∂

] = 1.

So the ∂

φ∂

φ term tells us that

[φ] = 1.

From these, we deduce that we must have

[λ

] = 4 − n.

isn’t always dimensionless. What we can do is to compare it with some

energy scale

. For example, if we are doing collisions in the LHC, then the

energy of the particles would be our energy scale. If we have picked such an

energy scale, then looking at

n−4

would give us a dimensionless parameter

since [E] = 1.

We can separate this into three cases:

(i) n

= 3: here

n−4

−1

decreases with

. So

−1

would be small at

high energies, and large at low energies.

Such perturbations are called relevant perturbations at low energies. In

a relativistic theory, we always have

E > m

. So we can always make the

perturbation small by picking

 m

(at least when we are studying

general theory; we cannot go and change the coupling constant in nature

so that our maths work out better).

(ii) n

= 4: Here the dimensionless parameter is just

itself. So this is small

whenever λ

 1. This is known as marginal perturbation.

(iii) n >

4: This has dimensionless parameter

n−4

, which is an increasing

function of E. So this is small at low energies, and high at large energies.

These operators are called irrelevant perturbations.

While we call them “irrelevant”, they are indeed relevant, as it is typically difficult

to avoid high energies in quantum field theory. Indeed, we have seen that we

can have arbitrarily large vacuum fluctuations. In the Advanced Quantum Field

Theory course, we will see that these theories are “non-renormalizable”.

In this course, we will consider only weakly coupled field theories — one that

can truly be considered as small perturbations of the free theory at all energies.

We will quickly describe some interaction Lagrangians that we will study in

more detail later on.

Example (φ

theory). Consider the φ

theory

L =

∂

φ∂

φ −

−

where

λ 

1. We can already guess the effects of the final term by noting that

here we have

[H, N] 6= 0.

So particle number is not conserved. Expanding the last term in the Lagrangian

in terms of

, a

†

, we get terms involving things like

†

, and all other combinations you can think of. These will create or

destroy particles.

We also have a Lagrangian that involves two fields — a real scalar field and

a complex scalar field.

Example

(Scalar Yukawa theory)

In the early days, we found things called

pions that seemed to mediate nuclear reactions. At that time, people did not

know that things are made up of quarks. So they modelled these pions in terms

of a scalar field, and we have

L = ∂

∗

∂

ψ +

∂

φ∂

φ − M

∗

ψ −

− gψ

∗

ψφ.

Here we have

g  m, M

. Here we still have [

Q, H

] = 0, as all terms are at

most quadratic in the

’s. So the number of

-particles minus the number of

-antiparticles is still conserved. However, there is no particle conservation for

φ.

A wary — the potential

V = M

∗

ψ +

+ gψ

∗

ψφ

has a stable local minimum when the fields are zero, but it is unbounded below

for large −gφ. So we can’t push the theory too far.

3.2 Interaction picture

How do we actually study interacting fields? We can imagine that the Hamil-

tonian is split into two parts, one of which is the “free field” part

, and the

other part is the “interaction” part

int

. For example, in the

theory, we can

split it up as

H =



(π

+ (∇φ)

+ m

)

| {z }

|{z}

int



The idea is that the interaction picture mixes the Schr¨odinger picture and the

Heisenberg picture, where the simple time evolutions remain in the operators,

whereas the complicated interactions live in the states. This is a very useful

trick if we want to deal with small interactions.

To see how this actually works, we revisit what the Schr¨odinger picture and

the Heisenberg picture are.

In the Schr¨odinger picture, we have a state

|ψ(t)i

that evolves in time, and

operators

that are fixed. In anticipation of what we are going to do next, we

are going to say that the operators are a function of time. It’s just that they are

constant functions:

(t) = 0.

The states then evolve according to the Schr¨odinger equation

|ψ(t)i

= H |ψ(t)i

Since

is constant in time, this allows us to write the solution to this equation

|ψ(t)i

= e

−iHt

|ψ(0)i

Here

−iHt

is the time evolution operator, which is a unitary operator. More

generally, we have

|ψ(t)i

= e

−iH(t−t

)

|ψ(0)i

= U

(t, t

) |ψ(t

where we define

(t, t

) = e

−iH(t−t

)

To obtain information about the system at time

, we look at what happens

when we apply O

(t) |ψ(t)i

In the Heisenberg picture, we define a new set of operators, relating to the

Schr¨odinger picture by

(t) = e

iHt

(t)e

−iHt

|ψ(t)i

= e

iHt

|ψ(t)i

In this picture, the states

|ψ(t)i

are actually constant in time, and are just

equal to |ψ(0)i

all the time. The equations of motion are then

= i[H, O

]

|ψ(t)i

= 0.

To know about the system at time t, we apply

(t) |ψ(t)i

= e

iHt

(t) |ψ(t)i

In this picture, the time evolution operator is just the identity map

(t, t

) = id .

In the interaction picture, we do a mixture of both. We only shift the “free”

part of the Hamiltonian to the operators. We define

(t) = e

(t)e

−iH

|ψ(t)i

= e

|ψ(t)i

In this picture, both the operators and the states change in time!

Why is this a good idea? When we studied the Heisenberg picture of the free

theory, we have already figured out what conjugating by

does to operators,

and it was really simple! In fact, it made things look nicer. On the other hand,

time evolution of the state is now just generated by the interaction part of the

Hamiltonian, rather than the whole thing, and this is much simpler (or at least

shorter) to deal with:

Proposition. In the interaction picture, the equations of motion are

= i[H

, O

]

|ψ(t)i

= H

|ψ(t)i

where H

is defined by

= (H

int

)

= e

int

)

−iH

Proof.

The first part we’ve essentially done before, but we will write it out again

for completeness.

−iH

) = lim

ε→0

(t+ε)

−iH

(t+ε)

− e

−iH

= lim

ε→0

−iH

− O

−iH

= lim

ε→0

(1 + iH

ε)O

(1 − iH

ε) − O

+ o(ε)

−iH

= lim

ε→0

iε(H

− O

) + o(ε)

−iH

= e

i[H

, O

−iH

= i[H

, O

For the second part, we start with Schr¨odinger’s equation

d |ψ(t)i

= H

|ψ(t)i

Putting in our definitions, we have

−iH

|ψ(t)i

) = (H

+ H

int

)

−iH

|ψ(t)i

Using the product rule and chain rule, we obtain

−iH

|ψ(t)i

+ ie

−iH

|ψ(t)i

= (H

+ H

int

)

−iH

|ψ(t)i

Rearranging gives us

d |ψ(t)i

= e

int

)

−iH

|ψ(t)i

To obtain information about the state at time

, we look at what happens

when we apply

(t) |ψ(t)i

= e

(t) |ψ(t)i

Now what is the time evolution operator

(

t, t

)? In the case of the Schr¨odinger

picture, we had the equation of motion

|ψ(t)i

= H |ψ(t)i

So we could immediately write down

|ψ(t)i = e

−iH(t−t

)

|ψ(t

)i.

In this case, we cannot do that, because

is now a function of time. The next

best guess would be

(t, t

) = exp



−i

) dt



For this to be right, it has to satisfy

(t, t

) = H

(t)U

(t, t

We can try to compute the derivative of this. If we take the series expansion of

this, then the second-order term would be

(−i)



) dt



For the equation of motion to hold, the derivative should evaluate to

times

the first order term, i.e.

−iH



(−i)

) dt



We can compute its derivative to be

(−i)

(t)



) dt



(−i)



) dt



(t).

This is not equal to what we wanted, because

(

) and

(

) do not commute

in general.

Yet, this is a good guess, and to get the real answer, we just need to make a

slight tweak:

Proposition

(Dyson’s formula)

The solution to the Schr¨odinger equation in

the interaction picture is given by

U(t, t

) = T exp



−i

)dt



where

stands for time ordering: operators evaluated at earlier times appear

to the right of operators evaluated at leter times when we write out the power

series. More explicitly,

T {O

)} =

(

) t

> t

) t

> t

We do not specify what happens when

, but it doesn’t matter in our case

since the operators are then equal.

Thus, we have

U(t, t

) = 1 − i

) +

(−i)



)

+ ··· .

We now notice that

) =

since we are integrating over all

≤ t

on both sides. Swapping

and

shows that the two second-order terms are indeed the same, so we have

U(t, t

) = 1 − i

) + (−i)

) + ··· .

Proof.

This from the last presentation above, using the fundamental theorem of

calculus.

In theory, this time-ordered exponential is very difficult to compute. However,

we have, at the beginning, made the assumption that the interactions

are

small. So often we can just consider the first one or two terms.

Note that when we evaluate the time integral of the Hamiltonian, a convenient

thing happens. Recall that our Hamiltonian is the integral of some operators

over all space. Now that we are also integrating it over time, what we are doing

is essentially integrating over spacetime — a rather relativistic thing to do!

Some computations

Now let’s try to do some computations. By doing so, we will very soon realize

that our theory is very annoying to work with, and then we will later come up

with something more pleasant.

We consider the scalar Yukawa theory. This has one complex scalar field and

one real scalar field, with an interaction Hamiltonian of

= gψ

†

ψφ.

Here the

ψ, φ

are the “Heseinberg” versions where we conjugated by

, so we

have

ip·x

rather than

ip·x

in the integral. We will pretend that the

particles

are mesons, and that the ψ-particles are nucleons.

We now want to ask ourselves the following question — suppose we start

with a meson, and we let it evolve over time. After infinite time, what is the

probability that we end up with a ψ

ψ pair, where

ψ is the anti-ψ particle?

We denote our initial state and “target” final states by

|ii =

†

|0i

|fi =

†

|0i.

What we now want to find is the quantity

hf|U (∞, −∞) |ii.

Definition (S-matrix). The S-matrix is defined as

S = U(∞, −∞).

Here

stands for “scattering”. So we often just write our desired quantity

hf|S |ii

instead. When we evaluate this thing, we will end up with some

numerical expression, which will very obviously not be a probability (it contains

a delta function). We will come back to interpreting this quantity later. For

the moment, we will just try to come up with ways to compute it in a sensible

manner.

Now let’s look at that expression term by term in the Taylor expansion of

The first term is the identity 1, and we have a contribution of

hf|1 |ii = hf|ii = 0

since distinct eigenstates are orthogonal. In the next term, we are integrating

over

†

ψφ

. If we expand this out, we will get loads of terms, each of which is a

product of some

b, c, a

(or their conjugates). But the only term that contributes

would be the term

†

, so that we exactly destroy the

†

particle and

create the

and

particles. The other terms will end up transforming

|ii

something that is not

|fi

(or perhaps it would kill

|ii

completely), and then the

inner product with |fi will give 0.

Before we go on to discuss what the higher order terms do, we first actually

compute the contribution of the first order term.

We want to compute

hf|S |ii = hf|1 − i

dt H

(t) + ···|ii

We know the

contributes nothing, and we drop the higher order terms to get

≈ −i hf|

dt H

(t) |ii

= −ig hf|

x ψ

†

(x)ψ(x)φ(x) |ii

= −ig

x hf |ψ

†

(x)ψ(x)φ(x) |ii.

We now note that the

|ii

interacts with the

terms only, and the

hf|

interacts

with the

terms only, so we can evaluate

(

)

|ii

and

†

(

)

(

)

|fi

separately.

Moreover, any leftover a

†

, b

†

, c

†

terms in the expressions will end up killing the

other object, as the a

†

’s commute with the b

†

, c

†

, and h0|a

†

= (a |0i)

†

= 0.

We can expand φ in terms of the creation and annihilation operators a and

†

to get

φ(x) |ii =

(2π)

−iq·x

+ a

†

iq·x

†

|0i

We notice that nothing happens to the

†

terms and they will end up becoming

0 by killing h0|. Swapping a

and a

†

and inserting the commutator, we obtain

(2π)

([a

, a

†

] + a

†

−iq·x

|0i

(2π)

(p − q)e

−iq·x

|0i

= e

−ip·x

|0i.

We will use this to describe the creation and destruction of mesons.

We can similarly look at

†

(x)ψ(x) |fi

(2π)

†

·x

+ c

−ik

·x

)(b

−ik

·x

+ c

†

·x

) |f i

As before, the only interesting term for us is when we have both

and

present to “kill” b

†

in |f i. So we look at

(2π)

†

−i(k

)·x

|0i

(2π)

− k

)δ

− k

−i(k

)·x

|0i

= e

−i(q

)·x

|0i.

So we have

hf|S |ii ≈ −ig

x (e

−i(q

)·x

|0i)

†

−ip·x

|0i)

= −ig

x e

i(q

−p)·x

= −ig(2π)

+ q

− p).

If we look carefully at what we have done, what allowed us to go from the initial

state

|ii

†

|0i

to the final

state

|fi

†

|0i

was

the presence of the term

†

in the

matrix (what we appeared to have

used in the calculation was

instead of their conjugates, but that is because

we were working with the conjugate ψ

†

ψ |fi rather than hf |ψ

†

ψ).

If we expand the

-matrix to, say, second order, we will, for example, find

a term that looks like (

†

)(

cba

†

). This corresponds to destroying a

pair

to produce a

, then destroying that to recover another

pair, which is an

interaction of the form

3.3 Wick’s theorem

Now the above process was still sort-of fine, because we were considering first-

order terms only. If we considered higher-order terms, we would have complicated

expressions involving a mess of creation and annihilation operators, and we would

have to commute them past each other to act on the states.

Our objective is thus to write a time-ordered product

T φ

(

)

(

)

···φ

(

)

as a sum of normal-ordered things, since it is easy to read off the action of a

normal-ordered operator on a collection of states. Indeed, we know

hf|:O: |ii

is non-zero if the creation and annihilation operators

:O:

match up exactly with

those used to create |f i and |ii from the vacuum.

We start with a definition:

Definition (Contraction). The contraction of two fields φ, ψ is defined to be

φψ = T (φψ) − :φψ: .

More generally, if we have a string of operators, we can contract just some of

them:

···φ(x

) ···φ(x

) ··· ,

by replacing the two fields with the contraction.

In general, the contraction will be a c-function, i.e. a number. So where we

decide to put the result of the contraction wouldn’t matter.

We now compute the contraction of our favorite scalar fields:

Proposition. Let φ be a real scalar field. Then

φ(x

)φ(x

) = ∆

− x

Proof. We write φ = φ

+ φ

−

, where

(x) =

(2π)

−ip·x

, φ

−

(x) =

(2π)

†

ip·x

Then we get normal order if all the φ

−

appear before the φ

When x

> y

, we have

T φ(x)φ(y) = φ(x)φ(y)

= (φ

(x) + φ

−

(x))(φ

(y) + φ

−

(y))

= φ

(x)φ

(y) + φ

−

(x)φ

(y) + [φ

(x), φ

−

(y)]

+ φ

−

(y)φ

(x) + φ

−

(x)φ

−

(y).

So we have

T φ(x)φ(y) = :φ(x)φ(y): + D(x − y).

By symmetry, for y

> x

, we have

T φ(x)φ(y) = :φ(x)φ(y): + D(y − x).

Recalling the definition of the Feynman propagator, we have

T φ(x)φ(y) = :φ(x)φ(y): + ∆

(x − y).

Similarly, we have

Proposition. For a complex scalar field, we have

ψ(x)ψ

†

(y) = ∆

(x − y) = ψ

†

(y)ψ(x),

whereas

ψ(x)ψ(y) = 0 = ψ

†

(x)ψ

†

(y).

It turns out these computations are all we need to figure out how to turn a

time-ordered product into a sum of normal-ordered products.

Theorem

(Wick’s theorem)

For any collection of fields

(

)

, φ

), ···, we have

T (φ

···φ

) = :φ

···φ

: + all possible contractions

Example. We have

T (φ

) = :φ

: + φ

:φ

: + φ

:φ

: + ··· + φ

+ ···

Proof sketch. We will provide a rough proof sketch.

By definition this is true for

= 2. Suppose this is true for

···φ

, and

now add φ

with

> x

for all k ∈ {2, ··· , n}.

Then we have

T (φ

···φ

) = (φ

+ φ

−

)(:φ

···φ

: + other contractions).

The

−

term stays where it is, as that already gives normal ordering, and the

term has to make its way past the

−

operators. So we can write the RHS

as a normal-ordered product. Each time we move past the

−

, we pick up a

contraction φ

Example.

We now consider the more complicated problem of nucleon scattering:

ψψ → ψψ.

So we are considering interactions of the form

something happens

Then we have initial and final states

|ii =

†

|0i

|fi =

†

|0i

We look at the order

term in

hf|

(

S − 1

)

|ii

. We remove that

as we are not

interested in the case with no scattering, and we look at order

because there

is no order g term.

The second order term is given by

(−ig)



†

)ψ(x

)φ(x

)ψ

†

)ψ(x

)φ(x

)



Now we can use Wick’s theorem to write the time-ordered product as a sum of

normal-ordered products. The annihilation and creation operators don’t end up

killing the vacuum only if we annihilate two

and then construct two new

This only happens in the term

:ψ

†

)ψ(x

)ψ

†

)ψ(x

): φ(x

)φ(x

We know the contraction is given by the Feynman propagator ∆

(

− x

). So

we now want to compute

, p

|:ψ

†

)ψ(x

)ψ

†

)ψ(x

): |p

, p

The only relevant term in the normal-ordered product is

ZZZZ

(2π)

16E

†

·x

+ik

·x

−ik

·x

−ik

·x

Now letting this act on hp

, p

| and |p

, p

i would then give

·(p

−p

)+ix

·(p

−p

)

+ e

−p

)+ix

−p

)

, (∗)

plus what we get by swapping

and

by a routine calculation (swap the

-operators in the

with those that create

|ii, |fi

, insert the delta functions as

commutators, then integrate).

What we really want is to integrate the Hamiltonian. So we want to integrate

the result of the above computation with respect to all spacetime to obtain

(−ig)

(∗) ∆

− x

)

(−ig)

(∗)

(2π)

ik·(x

−x

)

− m

+ iε

= (−ig)

(2π)

i(2π)

− m

+ iε



− p

+ k)δ

− p

− k) + δ

− p

+ k)δ

− p

− k)



= (−ig)

(2π)



− p

)

− m

− p

)

− m



+ p

− p

What we see again is a δ-function that enforces momentum conservation.

3.4 Feynman diagrams

This is better, but still rather tedious. How can we make this better? In

the previous example, we could imagine that the term that contributed to the

integral above represented the interaction where the two

-particles “exchanged”

-particle. In this process, we destroy the old

-particles and construct new

ones with new momenta, and the

-particle is created and then swiftly destroyed.

The fact that the

-particles were just “intermediate” corresponds to the fact

that we included their contraction in the computation.

We can pictorially represent the interaction as follows:

The magical insight is that every term given by Wick’s theorem can be interpreted

as a diagram of this sort. Moreover, the term contributes to the process if and

only if it has the right “incoming” and “outgoing” particles. So we can figure

out what terms contribute by drawing the right diagrams.

Moreover, not only can we count the diagrams like this. More importantly,

we can read out how much each term contributes from the diagram directly!

This simplifies the computation a lot.

We will not provide a proof that Feynman diagrams do indeed work, as it

would be purely technical and would also involve the difficult work of what it

actually means to be a diagram. However, based on the computations we’ve

done above, one should be confident that at least something of this sort would

work, maybe up to some sign errors.

We begin by specifying what diagrams are “allowed”, and then specify how

we assign numbers to diagrams.

Given an initial state and final state, the possible Feynman diagrams are

specified as follows:

Definition

(Feynman diagram)

In the scalar Yukawa theory, given an initial

state |ii and final state |f i, a Feynman diagram consists of:

–

An external line for all particles in the initial and final states. A dashed

line is used for a φ-particle, and solid lines are used for ψ/

ψ-particles.

–

Each

-particle comes with an arrow. An initial

-particle has an incoming

arrow, and a final

-particle has an outgoing arrow. The reverse is done

for

ψ-particles.

–

We join the lines together with more lines and vertices so that the only loose

ends are the initial and final states. The possible vertices correspond to

the interaction terms in the Lagrangian. For example, the only interaction

term in the Lagrangian here is

†

ψφ

, so the only possible vertex is one

that joins a φ line with two ψ lines that point in opposite directions:

Each such vertex represents an interaction, and the fact that the arrows

match up in all possible interactions ensures that charge is conserved.

–

Assign a directed momentum

to each line, i.e. an arrow into or out of

the diagram to each line.

The initial and final particles already have momentum specified in the

initial and final state, and the internal lines are given “dummy” momenta

(which we will later integrate over).

Note that there are infinitely many possible diagrams! However, when we lay

down the Feynman rules later, we will see that the more vertices the diagram has,

the less it contributes to the sum. In fact, the

-vertices diagrams correspond

to the nth order term in the expansion of the S-matrix. So most of the time it

suffices to consider “simple” diagrams.

Example

(Nucleon scattering)

If we look at

ψ → ψ

, the simplest

diagram is

On the other hand, we can also swap the two particles to obtain a diagram of

the form.

These are the ones that correspond to second-order terms.

There are also more complicated ones such as

This is a 1-loop diagram. We can also have a 2-loop diagram:

If we ignore the loops, we say we are looking at the tree level.

To each such diagram, we associate a number using the Feynman rules.

Definition

(Feynman rules)

To each Feynman diagram in the interaction, we

write down a term like this:

(i) To each vertex in the Feynman diagram, we write a factor of

(−ig)(2π)

where the sum goes through all lines going into the vertex (and put a

negative sign for those going out).

(ii)

For each internal line with momentum

, we integrate the product of all

factors above by

(2π)

D(k

where

D(k

) =

− m

+ iε

for φ

D(k

) =

− µ

+ iε

for ψ

Example.

We consider the case where

is small. Then only the simple diagrams

with very few vertices matter.

As promised by the Feynman rules, the two diagrams give us

(−ig)

(2π)

(δ

− p

− k)δ

+ k −p

) + δ

− p

− k)δ

+ k −p

)).

Now integrating gives us the formula

(−ig)

(2π)

i(2π)

− m

+ iε

(δ

− p

− k)δ

+ k −p

) + δ

− p

− k)δ

+ k −p

)).

Doing the integral gives us

(−ig)

(2π)



− p

)

− m

− p

)

− m



+ p

− p

which is what we found before.

There is a nice physical interpretation of the diagrams. We can interpret

the first diagram as saying that the nucleons exchange a meson of momentum

− p

. This meson doesn’t necessarily satisfy the relativistic

dispersion relation

(note that

is the 4-momentum). When this

happens, we say it is off-shell, or that it is a virtual meson. Heuristically, it

can’t live long enough for its energy to be measured accurately. In contrast, the

external legs are on-shell, so they satisfy p

= µ

3.5 Amplitudes

Now the important question is — what do we do with that quantity we computed?

We first define a more useful quantity known as the amplitude:

Definition

(Amplitude)

The amplitude

f,i

of a scattering process from

|ii

|fi is defined by

hf|S − 1 |ii = iA

f,i

(2π)

− p

where

is the sum of final state 4-momenta, and

is the sum of initial

state 4-momenta. The factor of

sticking out is by convention, to match with

non-relativistic quantum mechanics.

We can now modify our Feynman rules accordingly to compute A

f,i

–

Draw all possible diagrams with the appropriate external legs and impose

4-momentum conservation at each vertex.

– Write a factor (−ig) for each vertex.

– For each internal line of momentum p and mass m, put in a factor of

− m

+ iε

– Integrate over all undetermined 4-momentum k flowing in each loop.

We again do some examples of computation.

Example

(Nucleon-antinucleon scattering)

Consider another tree-level process

ψ +

ψ → φ + φ.

We have a diagram of the form

and a similar one with the two φ particles crossing.

The two diagrams then say

A = (−ig)



− p

)

− µ

− p

)

− µ



Note that we dropped the

iε

terms in the denominator because the denominator

never vanishes, and we don’t need to integrate over anything because all momenta

in the diagram are determined uniquely by momentum conservation.

Example (Meson scattering). Consider

φ + φ → φ + φ.

This is a bit more tricky. There is no tree-level diagram we can draw. The best

we can get is a box diagram like this:

k − p

k + p

− p

k + p

We then integrate through all possible k.

This particular graph we’ve written down gives

iA = (−ig)

(2π)

− µ

+ iε)((k + p

)

− µ

+ iε)((k + p

− p

)

− µ

+ iε)((k −p

)

− µ

+ iε)

For large

, this asymptotically tends to

, which fortunately means the

integral converges. However, this is usually not the case. For example, we

might have

, which diverges, or even

. This is when we need to do

renormalization.

Example (Feynman rules for φ

theory). Suppose our Lagrangian has a term

We then have single-vertex interactions

Any such diagram contributes a factor of (

−iλ

). Note that there is no

. This

is because if we expand the term

−iλ

, p

|:φ(x

)φ(x

): |p

, p

there are 4! ways of pairing the

(

) with the

, p

, so each possible interaction

is counted 4! times, and cancels that 4! factor.

For some diagrams, there are extra combinatoric factors, known as symmetry

factors, that one must take into account.

These amplitudes can be further related to experimentally measurable quan-

tities. We will not go into details here, but in The Standard Model course, we

will see that if we have a single initial state

and a collection of final states

{f}

then the decay rate of i to {f } is given by

i→f

f,i

dρ

where

dρ

= (2π)

− p

)

(2π)

and r runs over all momenta in the final states.

3.6 Correlation functions and vacuum bubbles

The

-matrix elements are good and physical, because they correspond to

probabilities of certain events happening. However, we are not always interested

in these quantities. For example, we might want to ask the conductivity of some

quantum-field-theoretic object. To do these computations, quantities known as

correlation functions are useful. However, we are not going to use these objects

in this course, so we are just going to state, rather than justify many of the

results, and you are free to skip this chapter.

Before that, we need to study the vacuum. Previously, we have been working

with the vacuum of the free theory |0i. This satisfies the boring relation

|0i = 0.

However, when we introduce an interaction term, this is no longer the vacuum.

Instead we have an interacting vacuum |Ωi, satisfying

H |Ωi = 0.

As before, we normalize the vacuum so that

hΩ|Ωi = 1.

Concretely, this interaction vacuum can be obtained by starting with a free

vacuum and then letting it evolve for infinite time.

Lemma. The free vacuum and interacting vacuum are related via

|Ωi =

hΩ|0i

(t, −∞) |0i =

hΩ|0i

(t, −∞) |0i.

Similarly, we have

hΩ| =

hΩ|0i

h0|U (∞, t).

Proof.

Note that we have the last equality because

(

t, −∞

) and

(

t, −∞

)

differs by factors of e

−iHt

which acts as the identity on |0i.

Consider an arbitrary state |Ψi. We want to show that

hΨ|U (t, −∞) |0i = hΨ|ΩihΩ|0i.

The trick is to consider a complete set of eigenstates

{

Ω

, |xi}

for

with energies

. Then we have

U(t, t

) |xi = e

−t)

|xi.

Also, we know that

1 = |ΩihΩ| +

dx |xihx|.

So we have

hΨ|U (t, −∞) |0i = hΨ|U (t, −∞)



|ΩihΩ| +

dx |xihx|



|0i

= hΨ|ΩihΩ|0i + lim

→−∞

dx e

−t)

hΨ|xihx|0i.

We now claim that the second term vanishes in the limit. As in most of the

things in the course, we do not have a rigorous justification for this, but it is not

too far-stretched to say so since for any well-behaved (genuine) function

(

the Riemann-Lebesgue lemma tells us that for any fixed a, b ∈ R, we have

lim

µ→∞

f(x)e

iµx

dx = 0.

If this were to hold, then

hΨ|U (t, −∞) |0i = hΨ|ΩihΩ|0i.

So the result follows. The other direction follows similarly.

Now given that we have this vacuum, we can define the correlation function.

Definition

(Correlation/Green’s function)

The correlation or Green’s function

is defined as

(n)

, ··· , x

) = hΩ|T φ

) ···φ

) |Ωi,

where φ

denotes the operators in the Heisenberg picture.

How can we compute these things? We claim the following:

Proposition.

(n)

, ··· , x

) =

h0|T φ

) ···φ

)S |0i

h0|S |0i

Proof.

We can wlog consider the specific example

> t

> ··· > t

. We start

by working on the denominator. We have

h0|S |0i = h0|U(∞, t)U(t, −∞) |0i = h0|ΩihΩ|ΩihΩ|0i.

For the numerator, we note that S can be written as

S = U

(∞, t

, t

) ···U

, −∞).

So after time-ordering, we know the numerator of the right hand side is

h0|U

(∞, t

)φ

, t

)φ

) ···φ

(n)U

, −∞) |0i

h0|U

(∞, t

)φ

, t

)φ

) ···φ

(n)U

, −∞) |0i

= h0|U

(∞, t

)φ

) ···φ

, −∞) |0i

= h0|ΩihΩ|T φ

) ···φ

) |ΩihΩ|0i.

So the result follows.

Now what does this quantity tell us? It turns out these have some really nice

physical interpretation. Let’s look at the terms

h0|T φ

(

)

···φ

(

)

S |0i

and

h0|S |0i individually and see what they tell us.

For simplicity, we will work with the

theory, so that we only have a single

field, and we will, without risk of confusion, draw all diagrams with solid lines.

Looking at

h0|S |0i

, we are looking at all transitions from

|0i

. The

Feynman diagrams corresponding to these would look like

1 vertex 2 vertex

These are known as vacuum bubbles. Then

h0|S |0i

is the sum of the amplitudes

of all these vacuum bubbles.

While this sounds complicated, a miracle occurs. It happens that the different

combinatoric factors piece together nicely so that we have

h0|S |0i = exp(all distinct (connected) vacuum bubbles).

Similarly, magic tells us that

h0|T φ

) ···φ

)S |0i =



connected diagrams

with n loose ends



h0|S |0i.

So what

(n)

(

, ··· , x

) really tells us is the sum of connected diagrams modulo

these silly vacuum bubbles.

Example.

The diagrams that correspond to

(4)

(

, ··· , x

) include things

that look like

Note that we define “connected” to mean every line is connected to some of the

end points in some way, rather than everything being connected to everything.

We can come up with analogous Feynman rules to figure out the contribution

of all of these terms.

There is also another way we can think about these Green’s function. Consider

the theory with a source J(x) added, so that

H = H

+ H

int

− J(x)φ(x).

This

is a fixed background function, called a source in analogy with electro-

magnetism.

Consider the interaction picture, but now we choose (

int

) to be the

“free” part, and −Jφ as the “interaction” piece.

Now the “vacuum” we use is not

|0i

but

|Ωi

, since this is the “free vacuum”

for our theory. We define

W [J] = hΩ|U

(−∞, ∞) |Ωi.

This is then a functional in J. We can compute

W [J] = hΩ|U

(−∞, ∞) |Ωi

= hΩ|T exp



x − J(x)φ

(x)



|Ωi

= 1 +

∞

n=1

(−1)

···d

J(x

) ···J(x

(n)

, ··· , x

Thus by general variational calculus, we know

(n)

, ··· , x

) = (−1)

W [J]

δJ(x

) ···δJ(x

)



J=0

Thus we call W [J] a generating function for the function G

(n)

, ··· , x

4 Spinors

In this chapter, we return to the classical world where things make sense. We

are going to come up with the notion of a spinor, which is a special type of field

that, when quantized, gives us particles with “intrinsic spin”.

4.1 The Lorentz group and the Lorentz algebra

So far, we have only been working with scalar fields. These are pretty boring,

since when we change coordinates, the values of the field remain unchanged. One

can certainly imagine more exciting fields, like the electromagnetic potential.

Ignoring issues of choice of gauge, we can imagine the electromagnetic potential

as a 4-vector

at each point in space-time. When we change coordinates by a

Lorentz transformation Λ, the field transforms by

(x) 7→ Λ

(Λ

−1

x).

Note that we had to put a Λ

−1

inside

because the names of the points have

changed. Λ

−1

in the new coordinate system labels the same point as

in the

old coordinate system.

In general, we can consider vector-valued fields that transform when we

change coordinates. If we were to construct such a field, then given any Lorentz

transformation Λ, we need to produce a corresponding transformation D(Λ) of

the field. If our field

takes values in a vector space

(usually

), then this

D(Λ) should be a linear map V → V . The transformation rule is then

x 7→ Λx,

φ 7→ D(Λ)φ.

We want to be sure this assignment of

(Λ) behaves sensibly. For example, if we

apply the Lorentz transformation Λ =

, i.e. we do nothing, then

(

) should

not do anything as well. So we must have

D(1) = 1.

Now if we do Λ

and then Λ

, then our field will transform first by

(Λ

), then

(Λ

). For the universe to make sense, this had better be equal to

(Λ

So we require

D(Λ

)D(Λ

) = D(Λ

)

for any Λ

. Mathematically, we call this a representation of the Lorentz

group.

Definition

(Lorentz group)

The Lorentz group, denoted O(1

3), is the group

of all Lorentz transformations. Explicitly, it is given by

O(1, 3) = {Λ ∈ M

4×4

: Λ

ηΛ = η},

where

η = η

µν







1 0 0 0

0 −1 0 0

0 0 −1 0

0 0 0 −1







is the Minkowski metric. Alternatively, O(1

3) is the group of all matrices Λ

such that

hΛx, Λyi = hx, yi,

for all

x, y ∈ R

1+3

, where

hx, yi

denotes the inner product given by the Minkowski

metric

hx, yi = x

ηy.

Definition

(Representation of the Lorentz group)

A representation of the

Lorentz group is a vector space

and a linear map

(Λ) :

V → V

for each

Λ ∈ O(1, 3) such that

D(1) = 1, D(Λ

)D(Λ

) = D(Λ

)

for any Λ

, Λ

∈ O(1, 3).

The space V is called the representation space.

So to construct interesting fields, we need to find interesting representations!

There is one representation that sends each Λ to the matrix Λ acting on

1+3

but that is really boring.

To find more representations, we will consider “infinitesimal” Lorentz trans-

formations. We will find representations of these infinitesimal Lorentz trans-

formations, and then attempt to patch them up to get a representation of the

Lorentz group. It turns out that we will fail to patch them up, but instead we

will end up with something more interesting!

We write our Lorentz transformation Λ as

= δ

+ εω

+ O(ε

The definition of a Lorentz transformation requires

σρ

= η

µν

Putting it in, we have

(δ

+ εω

)(δ

+ εω

)η

ρσ

+ O(ε

) = η

µν

So we find that we need

µν

+ ω

νµ

= 0.

is antisymmetric. Thus we figured that an infinitesimal Lorentz transfor-

mation is an antisymmetric matrix. This is known as the Lie algebra of the

Lorentz group.

Definition (Lorentz algebra). The Lorentz algebra is

o(1, 3) = {ω ∈ M

4×4

: ω

µν

+ ω

νµ

= 0}.

It is important that when we say that ω is antisymmetric, we mean exactly

µν

νµ

= 0. Usually, when we write out a matrix, we write out the entries of

instead, and the matrix we see will not in general be antisymmetric, as we

will see in the examples below.

A 4

4 anti-symmetric matrix has 6 free components. These 6 components

in fact correspond to three infinitesimal rotations and three infinitesimal boosts.

We introduce a basis given by the confusing expression

ρσ

)

= η

ρµ

− η

σµ

Here the

and

count which basis vector (i.e. matrix) we are talking about,

and

µ, ν

are the rows and columns. Usually, we will just refer to the matrix as

ρσ

and not mention the indices µ, ν to avoid confusion.

Each basis element will have exactly one pair of non-zero entries. For example,

we have

)







0 1 0 0

1 0 0 0

0 0 0 0







, (M

)







0 0 0 0

0 0 −1 0

0 1 0 0

0 0 0 0







These generate a boost in the

direction and a rotation in the

plane

respectively, as we will later see.

By definition of a basis, we can write all matrices

in the Lorentz algebra

as a linear combination of these:

ω =

Ω

ρσ

Note that here

ρσ

and

σρ

will be the negative of each other, so

ρσ

}

doesn’t

really form a basis. By symmetry, we will often choose Ω so that Ω

ρσ

−

Ω

σρ

as well.

Now we can talk about representations of the Lorentz algebra. In the case of

representations of the Lorentz group, we required that representations respect

multiplication. In the case of the Lorentz algebra, the right thing to ask to be

preserved is the commutator (cf. III Symmetries, Fields and Particles).

Definition

(Representation of Lorentz algebra)

A representation of the Lorentz

algebra is a collection of matrices that satisfy the same commutation relations as

the Lorentz algebra.

Formally, this is given by a vector space V and a linear map R(ω) : V → V

for each ω ∈ o(1, 3) such that

R(aω + bω

) = aR(ω) + bR(ω

), R([ω, ω

]) = [R(ω), R(ω

)]

for all ω, ω

∈ o(1, 3) and a, b ∈ R.

It turns out finding representations of the Lorentz algebra isn’t that bad. We

first note the following magic formula about the basis vectors of the Lorentz

algebra:

Proposition.

ρσ

, M

τν

] = η

στ

ρν

− η

ρτ

σν

+ η

ρν

στ

− η

σν

ρτ

This can be proven, painfully, by writing out the matrices.

It turns out this is the only thing we need to satisfy:

Fact.

Given any vector space

and collection of linear maps

ρσ

V → V

they give a representation of the Lorentz algebra if and only if

ρσ

, L

τν

] = η

στ

ρν

− η

ρτ

σν

+ η

ρν

στ

− η

σν

ρτ

Suppose we have already found a representation of the Lorentz algebra. How

can we find a representation of the Lorentz group? We will need to make use of

the following fact:

Fact.

Let Λ be a Lorentz transformation that preserves orientation and does not

reverse time. Then we can write Λ =

exp

(

) for some

ω ∈ o

3). In coordinates,

we can find some Ω

ρσ

such that

Λ = exp



Ω

ρσ



. (∗)

Thus if we have found some representation

(

ρσ

), we can try to define a

representation of the Lorentz group by

R(Λ) = exp(R(ω)) = exp



Ω

ρσ

R(M

ρσ

)



. (†)

Now we see two potential problems with our approach. The first is that we can

only lift a representation of the Lorentz algebra to the parity and time-preserving

Lorentz transformations, and not all of them. Even worse, it might not even be

well-defined for these nice Lorentz transformations. We just know that we can

find some Ω

ρσ

such that (

∗

) holds. In general, there can be many such Ω

ρσ

, and

they might not give the same answer when we evaluate (†)!

Definition

(Restricted Lorentz group)

The restricted Lorentz group consists

of the elements in the Lorentz group that preserve orientation and direction of

time.

Example. Recall that we had

)







0 0 0 0

0 0 −1 0

0 1 0 0

0 0 0 0







We pick

Ω

= −Ω

= −φ

with all other entries zero. Then we have

Ω

ρσ







0 0 0 0

0 0 φ

0 −φ

0 0

0 0 0 0







So we know

Λ = exp



Ω

ρσ









1 0 0 0

0 cos φ

sin φ

0 −sin φ

cos φ

0 0 0 1







which is a rotation in the

plane by

. It is clear that

is not uniquely

defined by Λ. Indeed,

+ 2

nπ

for any

would give the same element of the

Lorentz group.

More generally, given any rotation parameters

= (

, φ

), we obtain a

rotation by setting

Ω

= −ε

ijk

4.2 The Clifford algebra and the spin representation

It turns out there is a useful way of finding a representation of the Lorentz

algebra which gives rise to nice properties of the representation. This is via the

Clifford algebra.

We will make use of the following notation:

Notation (Anticommutator). We write

{A, B} = AB + BA

for the anticommutator of two matrices/linear maps.

Definition

(Clifford algebra)

The Clifford algebra is the algebra generated by

, γ

subject to the relations

{γ

, γ

} = γ

+ γ

= 2η

µν

More explicitly, we have

= −γ

for µ 6= ν.

and

(γ

)

= 1, (γ

)

= −1.

A representation of the Clifford algebra is then a collection of matrices (or linear

maps) satisfying the relations listed above.

Note that if we have a representation of the Clifford algebra, we will also

denote the matrices of the representation by γ

, instead of, say, R(γ

Example

(Chiral representation)

The simplest representation (in the sense

that it is easy to write out, rather than in any mathematical sense) is the

4-dimensional representation called the chiral representation. In block matrix

form, this is given by



0 1

1 0



, γ



0 σ

−σ



where the σ

are the Pauli matrices given by



0 1

1 0



, σ



0 −i

i 0



, σ



1 0

0 −1



which satisfy

{σ

, σ

} = 2δ

Of course, this is not the only representation. Given any 4

4 matrix

, we

can construct a new representation of the Clifford algebra by transforming by

7→ Uγ

−1

It turns out any 4-dimensional representation of the Clifford algebra comes from

applying some similarity transformation to the chiral representation, but we will

not prove that.

We now claim that every representation of the Clifford algebra gives rise to

a representation of the Lorentz algebra.

Proposition.

Suppose

is a representation of the Clifford algebra. Then the

matrices given by

ρσ

[γ

, γ

] =

(

0 ρ = σ

ρ 6= σ

−

ρσ

define a representation of the Lorentz algebra.

This is known as the spin representation.

We will need the following technical result to prove the proposition:

Lemma.

µν

, γ

] = γ

νρ

− γ

ρµ

Proof.

µν

, γ

] =



−

µν

, γ



−

= γ

(η

νρ

− γ

) − (η

µρ

− γ

)γ

= γ

νρ

− γ

ρµ

Now we can prove our claim.

Proof of proposition. We have to show that

µν

, S

ρσ

] = η

νρ

µσ

− η

µρ

νσ

+ η

µσ

νρ

− η

νσ

µρ

The proof involves, again, writing everything out. Using the fact that

ρσ

commutes with everything, we know

µν

, S

ρσ

] =

µν

, γ

]



µν

, γ

]γ

+ γ

µν

, γ

]





νρ

− γ

µρ

+ γ

νσ

− γ

µσ



Then the result follows form the fact that

= 2S

µσ

+ η

µσ

So we now have a representation of the Lorentz algebra. Does this give us a

representation of the (restricted) Lorentz group? Given any Λ

∈ SO

3), we

can write

Λ = exp



Ω

ρσ



We now try to define

S[Λ] = exp



Ω

ρσ



Is this well-defined?

We can try using an example we have previously computed. Recall that given

any rotation parameter φ = (φ

, φ

), we can pick

Ω

= −ε

ijk

to obtain a rotation denoted by

. What does this give when we exponentiate

with S

ρσ

? We use the chiral representation, so that



0 σ

−σ





0 σ

−σ



= −

ijk



0 σ



Then we obtain

Proposition. Let φ = (φ

, φ

), and define

Ω

= −ε

ijk

Then in the chiral representation of S, writing σ = (σ

, σ

), we have

S[Λ] = exp



Ω

ρσ





iφ·σ/2

0 e

iφ·σ/2



In particular, we can pick

= (0

). Then the corresponding Lorentz

transformation is the identity, but

S[1] =

iπσ

0 e

iπσ

= −1 6= 1.

So this does not give a well-defined representation of the Lorentz group, because

different ways of representing the element

in the Lorentz group will give

different values of

[

]. Before we continue on to discuss this, we take note of

the corresponding formula for boosts.

Note that we have



0 1

1 0



0 σ

−σ





−σ

0 σ



Then the corresponding result is

Proposition. Write χ = (χ

, χ

). Then if

Ω

= −Ω

= −χ

then

Λ = exp



Ω

ρσ



is the boost in the χ direction, and

S[Λ] = exp



Ω

ρσ





χ·σ/2

0 e

−χ·σ/2



Now what is this

[Λ] we have constructed? To begin with, writing

[Λ]

is a very bad notation, because

[Λ] doesn’t only depend on what Λ is, but

also additional information on how we “got to” Λ, i.e. the values of Ω

ρσ

. So

going from

by rotating by 2

is different from going from

by doing

nothing. However, physicists like to be confusing and this will be the notation

used in the course.

Yet,

[Λ] is indeed a representation of something. Each point of this “some-

thing” consists of an element of the Lorentz group, and also information on

how we got there (formally, a (homotopy class of) paths from the identity to Λ).

Mathematicians have come up with a fancy name of this, called the universal

cover , and it can be naturally given the structure of a Lie group. It turns out

in this case, this universal cover is a double cover, which means that for each

Lorentz transformation Λ, there are only two non-equivalent ways of getting to

Λ.

Note that the previous statement depends crucially on what we mean by

ways of getting to Λ being “equivalent”. We previously saw that for a rotation,

we can always add 2

nπ

to the rotation angle to get a different Ω

ρσ

. However,

it is a fact that whenever we add 4

to the angle, we will always get back the

same

[Λ] for any representation

of the Lorentz algebra. So the only two

non-equivalent ways are the original one, and the original one plus 2π.

(The actual reason is backwards. We know for geometric reasons that

adding 4

will give an equivalent path, and thus it must be the case that any

representation must not change when we add 4

. Trying to supply further details

and justification for what is going on would bring us deep into the beautiful

world of algebraic topology)

We give a name to this group.

Definition

(Spin group)

The spin group

Spin

3) is the universal cover of

3). This comes with a canonical surjection to

3), sending “Λ and

a path to Λ” to Λ.

As mentioned, physicists like to be confusing, and we will in the future keep

talking about “representation of the Lorentz group”, when we actually mean the

representation of the spin group.

So we have the following maps of (Lie) groups:

Spin(1, 3) SO

(1, 3) O(1, 3) .

This diagram pretty much summarizes the obstructions to lifting a representation

of the Lorentz algebra to the Lorentz group. What a representation of the Lorentz

algebra gives us is actually a representation of the group

Spin

3), and we want

to turn this into a representation of O(1, 3).

If we have a representation

of O(1

3), then we can easily produce a

representation of

Spin

3) — given an element

M ∈ Spin

3), we need to

produce some matrix. We simply apply the above map to obtain an element of

O(1, 3), and then the representation D gives us the matrix we wanted.

However, going the other way round is harder. If we have a representation of

Spin

3), then for that to give a representation of

3), we need to make

sure that different ways of getting to a Λ don’t give different matrices in the

representation. If this is true, then we have found ourselves a representation

3). After that, we then need to decide what we want to do with the

elements of O(1, 3) not in SO

(1, 3), and this again involves some work.

4.3 Properties of the spin representation

We have produced a representation of the Lorentz group, which acts on some

vector space V

∼

. Its elements are known as Dirac spinors.

Definition

(Dirac spinor)

A Dirac spinor is a vector in the representation

space of the spin representation. It may also refer to such a vector for each point

in space.

Our ultimate goal is to construct an action that involves a spinor. So we

would want to figure out a way to get a number out of a spinor.

In the case of a 4-vector, we had these things called covectors that lived in the

“dual space”. A covector

can eat up a vector

and spurt out a number

(

Often, we write the covector as

and the vector as

, and then

(

) =

When written out like a matrix, a covector is represented by a “row vector”.

Under a Lorentz transformation, these objects transform as

λ 7→ λΛ

−1

v 7→ Λv

(What do we mean by

λ 7→ λ

−1

? If we think of

as a row vector, then this is

just matrix multiplication. However, we can think of it without picking a basis

as follows —

−1

is a covector, so it is determined by what it does to a vector

v. We then define (λΛ

−1

)(v) = λ(Λ

−1

v))

Then the result λv transforms as

λv 7→ λΛ

−1

Λv = λv.

So the resulting number does not change under Lorentz transformations. (Math-

ematically, this says a covector lives in the dual representation space of the

Lorentz group)

Moreover, given a vector, we can turn it into a covector in a canonical way,

by taking the transpose and then inserting some funny negative signs in the

space components.

We want to do the same for spinors. This terminology may or may not be

standard:

Definition

(Cospinor)

A cospinor is an element in the dual space to space of

spinors, i.e. a cospinor

is a linear map that takes in a spinor

as an argument

and returns a number

Xψ

. A cospinor can be represented as a “row vector” and

transforms under Λ as

X 7→ XS[Λ]

−1

This is a definition we can always make. The hard part is to produce some

actual cospinors. To figure out how we can do so, it is instructive to figure out

what S[Λ]

−1

is!

We begin with some computations using the γ

matrices.

Proposition. We have

= (γ

)

†

Proof. This is true by checking all possible µ.

Proposition.

S[Λ]

−1

= γ

S[Λ]

†

where

[Λ]

†

denotes the Hermitian conjugate as a matrix (under the usual basis).

Proof. We note that

µν

)

†

[(γ

)

†

, (γ

)

†

] = −γ



[γ

, γ

]



= −γ

µν

So we have

S[Λ]

†

= exp



Ω

µν

)

†



= exp



−

Ω

µν



= γ

S[Λ]

−1

using the fact that (

)

and

exp

(

−A

) = (

exp A

)

−1

. Multiplying both sides

on both sides by γ

gives the desired formula.

We now come to our acclaimed result:

Proposition. If ψ is a Dirac spinor, then

ψ = ψ

†

is a cospinor.

Proof.

ψ transforms as

ψ 7→ ψ

†

S[Λ]

†

= ψ

†

(γ

S[Λ]

†

) =

ψS[Λ]

−1

Definition

(Dirac adjoint)

For any Dirac spinor

, its Dirac adjoint is given

ψ = ψ

†

Thus we immediately get

Corollary.

For any spinor

, the quantity

ψψ

is a scalar, i.e. it doesn’t transform

under a Lorentz transformation.

The next thing we want to do is to construct 4-vectors out of spinors. While

the spinors do have 4 components, they aren’t really related to 4-vectors, and

transform differently. However, we do have that thing called

, and the indexing

should suggest that

transforms like a 4-vector. Of course, it is a collection

of matrices and is not actually a 4-vector, just like

∂

behaves like a 4-vector

but isn’t. But it behaves sufficiently like a 4-vector and we can combine it with

other things to get actual 4-vectors.

Proposition. We have

S[Λ]

−1

S[Λ] = Λ

Proof. We work infinitesimally. So this reduces to



1 −

Ω

ρσ





1 +

Ω

ρσ





1 +

Ω

ρσ



This becomes

ρσ

, γ

] = −(M

ρσ

)

But we can use the explicit formula for M to compute

−(M

ρσ

)

= (η

σµ

− η

ρµ

)γ

= γ

σµ

− γ

ρµ

and we have previously shown this is equal to [S

ρσ

, γ

Corollary.

The object

ψγ

is a Lorentz vector, and

ψγ

transforms as a

Lorentz tensor.

ψγ

, the symmetric part is a Lorentz scalar and is proportional to

µν

ψψ

, and the anti-symmetric part transforms as an antisymmetric Lorentz

tensor, and is proportional to

ψS

µν

ψ.

4.4 The Dirac equation

Armed with these objects, we can now construct a Lorentz-invariant action. We

will, as before, not provide justification for why we choose this action, but as we

progress we will see some nice properties of it:

Definition (Dirac Lagrangian). The Dirac Lagrangian is given by

L =

ψ(iγ

∂

− m)ψ.

From this we can get the Dirac equation, which is the equation of motion

obtained by varying

and

independently. Varying

, we obtain the equation

Definition (Dirac equation). The Dirac equation is

(iγ

∂

− m)ψ = 0.

Note that this is first order in derivatives! This is different from the Klein–

Gordon equation. This is only made possible by the existence of the

matrices.

If we wanted to write down a first-order equation for a scalar field, there is

nothing to contract ∂

with.

We are often going to meet vectors contracted with

. So we invent a

notation for it:

Notation (Slash notation). We write

≡

Then the Dirac equation says

∂ − m)ψ = 0.

Note that the

here means

, for

the identity matrix. Whenever we have a

matrix equation and a number appears, that is what we mean.

Note that the

matrices are not diagonal. So they mix up different

components of the Dirac spinor. However, magically, it turns out that each

individual component satisfies the Klein–Gordon equation! We know

(iγ

∂

− m)ψ = 0.

We now act on the left by another matrix to obtain

(iγ

∂

+ m)(iγ

∂

− m)ψ = −(γ

∂

+ m

)ψ = 0.

But using the fact that ∂

∂

commutes, we know that (after some relabelling)

∂

{γ

, γ

}∂

∂

= ∂

∂

So this tells us

−(∂

∂

+ m

)ψ = 0.

Now nothing mixes up different indices, and we know that each component of

satisfies the Klein–Gordon equation.

In some sense, the Dirac equation is the “square root” of the Klein–Gordon

equation.

4.5 Chiral/Weyl spinors and γ

Recall that if we picked the chiral representation of the Clifford algebra, the

corresponding representation of the spin group is

S[Λ] =











χ·σ

0 e

−

χ·σ

for boosts

φ·σ

0 e

φ·σ

for rotations

It is pretty clear from the presentation that this is actually just two independent

representations put together, i.e. the representation is reducible. We can then

write our spinor ψ as

ψ =



−



where

and

−

are 2-complex-component objects. These objects are called

Weyl spinors or chiral spinors.

Definition

(Weyl/chiral spinor)

A left (right)-handed chiral spinor is a 2-

component complex vector

and

−

respectively that transform under the

action of the Lorentz/spin group as follows:

Under a rotation with rotation parameters φ, both of them transform as

7→ e

iφ·σ/2

Under a boost χ, they transform as

7→ e

±χ·σ/2

So these are two two-dimensional representations of the spin group.

We have thus discovered

Proposition.

A Dirac spinor is the direct sum of a left-handed chiral spinor

and a right-handed one.

In group theory language,

is the (0

) representation of the Lorentz

group, and U

−

is the (

, 0) representation, and ψ is in (

, 0) ⊕ (0,

As an element of the representation space, the left-handed part and right-

handed part are indeed completely independent. However, we know that the

evolution of spinors is governed by the Dirac equation. So a natural question to

ask is if the Weyl spinors are coupled in the Dirac Lagrangian.

Decomposing the Lagrangian in terms of our Weyl spinors, we have

L =

ψ(i

∂ − m)ψ



†

−





0 1

1 0





0 ∂

+ σ

∂

− σ

∂



− m



1 0

0 1



−



= iU

†

−

∂

−

+ iU

†

¯σ

∂

− m(U

†

−

+ U

†

−

where

= (1, σ), ¯σ

= (1, −σ).

So the left and right-handed fermions are coupled if and only if the particle is

massive. If the particle is massless, then we have two particles satisfying the

Weyl equation:

Definition (Weyl equation). The Weyl equation is

i¯σ

∂

= 0.

This is all good, but we produced these Weyl spinors by noticing that in our

particular chiral basis, the matrices

[Λ] looked good. Can we produce a more

intrinsic definition of these Weyl spinors that do not depend on a particular

representation of the Dirac spinors?

The solution is to introduce the magic quantity γ

Definition (γ

= −iγ

Proposition. We have

{γ

, γ

} = 0, (γ

)

= 1

for all γ

, and

µν

, γ

] = 0.

Since (γ

)

= 1, we can define projection operators

(1 ± γ

Example. In the chiral representation, we have



1 0

0 −1



Then we have



1 0

0 0



, P

−



0 0

0 1



We can prove, in general, that these are indeed projections:

Proposition.

= P

, P

−

= P

−

= 0.

Proof. We have

(1 ± γ

)

(1 + (γ

)

± 2γ

) =

(1 ± γ

and

−

(1 + γ

)(1 − γ

) =

(1 − (γ

)

) = 0.

We can think of these

as projection operators to two orthogonal subspaces

of the vector space

of spinors. We claim that these are indeed representations

of the spin group. We define

= {P

ψ : ψ ∈ V }.

We claim that

[Λ] maps

to itself. To show this, we only have to compute

infinitesimally, i.e. that

µν

maps

to itself. But it follows immediately from

the fact that S

µν

commutes with γ

that

µν

ψ = P

µν

ψ.

We can then define the chiral spinors as

= P

ψ.

It is clear from our previous computation of the

in the chiral basis that these

agree with what we’ve defined before.

4.6 Parity operator

So far, we’ve considered only continuous transformations, i.e. transformations

continuously connected to the identity. These are those that preserve direction

of time and orientation. However, there are two discrete symmetries in the full

Lorentz group — time reversal and parity:

T : (t, x) 7→ (−t, x)

P : (t, x) 7→ (t, −x)

Since we defined the spin representation via exponentiating up infinitesimal

transformations, it doesn’t tell us what we are supposed to do for these discrete

symmetries.

However, we do have some clues. Recall that we figured that the

trans-

formed like 4-vectors under continuous Lorentz transformations. So we can

postulate that

also transforms like a 4-vector under these discrete symmetries.

We will only do it for the parity transformation, since they behave interestingly

for spinors. We will suppose that our parity operator acts on the γ

P : γ

7→ γ

7→ −γ

Because of the Clifford algebra relations, we can write this as

P : γ

7→ γ

So we see that

is actually conjugating by

(note that (

)

−1

), and this

is something we can generalize to everything. Since all the interesting matrices

are generated by multiplying and adding the

together, all matrices transform

via conjugation by γ

. So it is reasonable to assume that P is γ

Axiom. The parity operator P acts on the spinors as γ

So in particular, we have

Proposition.

P : ψ 7→ γ

ψ, P :

ψ 7→

ψγ

Proposition. We have

P : γ

7→ −γ

Proof.

The

, γ

each pick up a negative sign, while

does not change.

Now something interesting happens. Since

switches the sign of

, it

exchanges P

and P

−

. So we have

Proposition. We have

P : P

7→ P

∓

In particular, we have

P ψ

= ψ

∓

still acts as right-multiplication-by-

−1

on the cospinors, we know

that scalar quantities etc are still preserved when we act by

. However, if we

construct something with

, then funny things happen, because

gains a sign

when we transform by P . For example,

P :

ψγ

ψ 7→ −

ψγ

ψ.

Note that here it is important that we view

as a fixed matrix that does

not transform, and

only acts on

and

. Otherwise, the quantity does not

change. If we make

act on everything, then (almost) by definition the resulting

quantity would remain unchanged under any transformation. Alternatively, we

can think of P as acting on γ

and leaving other things fixed.

Definition

(Pseudoscalar)

A pseudoscalar is a number that does not change

under Lorentz boosts and rotations, but changes sign under a parity operator.

Similarly, we can look at what happens when we apply

ψγ

. This

becomes

ψγ

ψ 7→

ψγ

ψ =

(

−

ψγ

ψ µ = 0

ψγ

ψ µ 6= 0

This is known as an axial vector.

Definition

(Axial vector)

An axial vector is a quantity that transforms as

vectors under rotations and boosts, but gain an additional sign when transforming

under parity.

Type Example

Scalar

ψψ

Vector

ψγ

Tensor

ψS

µν

Pseudoscalar

ψγ

Axial vector

ψγ

We can now add extra terms to

that use

. These terms will typically break

the parity invariance of the theory. Of course, it doesn’t always break parity

invariance, since we can multiply two pseudoscalars together to get a scalar.

It turns out nature does use

, and they do break parity. The classic example

is a

-boson, which is a vector field, which couples only to left-handed fermions.

The Lagrangian is given by

L = ··· +

ψγ

(1 − γ

)ψ,

where (1 − γ

) acts as the left-handed projection.

A theory which puts

on an equal footing is known as vector-like. Otherwise,

it is known as chiral.

4.7 Solutions to Dirac’s equation

Degrees of freedom

If we just look at the definition of a Dirac spinor, it has 4 complex components,

so 8 real degrees of freedom. However, the equation of motion imposes some

restriction on how a Dirac spinor can behave. We can compute the conjugate

momentum of the Dirac spinor to find

∂L

∂

= iψ

†

So the phase space is parameterized by

and

†

, as opposed to

and

in the

usual case. But

†

is completely determined by

! So the phase space really

just has 8 real degrees of freedom, as opposed to 16, and thus the number of

degrees of freedom of

itself is just 4 (the number of degrees of freedom is in

general half the number of degrees of freedom of the phase space). We will see

this phenomenon in the solutions we come up with below.

Plane wave solutions

We want to solve the Dirac equation

∂ − m)ψ = 0.

We start with the simplest ansatz (“guess”)

ψ = u

−ip·x

where

is a constant 4-component spinor which, as the notation suggests,

depends on momentum. Putting this into the equation, we get

(γ

− m)u

= 0.

We write out the LHS to get



−m p

¯σ

−m



= 0.

Proposition. We have a solution



√

p · σξ

√

p · ¯σξ



for any 2-component spinor ξ normalized such that ξ

†

ξ = 1.

Proof. We write





Then our equation gives

(p · σ)u

= mu

(p · ¯σ)u

= mu

Either of these equations can be derived from the other, since

(p · σ)(p · ¯σ) = p

− p

= p

− p

= p

= m

We try the ansatz

= (p · σ)ξ

for a spinor ξ

. Then our equation above gives

(p · ¯σ)(p · σ)ξ

= mξ

So any vector of the form

= A



(p · σ)ξ

mξ



is a solution, for A a constant. To make this look more symmetric, we choose

A =

, ξ

√

p · ¯σξ,

with ξ another spinor. Then we have

(p · σ)

√

p · ¯σξ =

√

p · σξ.

This gives our claimed solution.

Note that solving the Dirac equation has reduced the number of dimensions

from 4 to 2, as promised.

Example. A stationary, massive particle of mass m and p = 0 has

√





for any 2-component spinor

. Under a spacial rotation, we have a transformation

ξ 7→ e

iφ·σ/2

ξ,

which rotates ξ.

Now let’s pick

= (0

, φ

), so that

φ · σ

is a multiple of

. We then pick

ξ =





This is an eigenvector of

with positive eigenvalue. So it is invariant under

rotations about the

axis (up to a phase). We thus say this particle has spin

up along the

axis, and this will indeed give rise to quantum-mechanical spin

as we go further on.

Now suppose our particle is moving in the x

direction, so that

= (E, 0, 0, p

Then the solution to the Dirac equation becomes







√

p · σ





√

p · ¯σ

















E − p





E + p











In the massless limit, we have E → p

. So this becomes

√













For a spin down field, i.e.

ξ =





we have







E + p





E − p











→

√













Helicity operator

Definition

(Helicity operator)

The helicity operator is a projection of angular

momentum along the direction of motion

h =

p · J =

ˆp



0 σ



From the above expressions, we know that a massless spin up particle has

h = +

, while a massless spin down particle has helicity −

Negative frequency solutions

We now try a different ansatz. We try

ψ = v

ip·x

These are known as negative frequency solutions. The

solutions we’ve found

are known as positive frequency solutions.

The Dirac solution then becomes



√

p · ση

−

√

p · ¯ση



for some 2-component η with η

†

η = 1.

This is exactly the same as

, but with a relative minus sign between

and v

A basis

It is useful to introduce a basis given by

= ξ





, η

= ξ





We then define



√

p · σξ

√

p · ¯σξ



, v



√

p · ση

−

√

p · ¯ση



These then form a basis for the solution space of the Dirac equation.

4.8 Symmetries and currents

We now look at what Noether’s theorem tells us about spinor fields. We will

only briefly state the corresponding results because the calculations are fairly

routine. The only interesting one to note is that the Lagrangian is given by

L =

ψ(i

∂ − m)ψ,

while the equations of motion say

∂ − m)ψ = 0.

So whenever the equations of motion are satisfied, the Lagrangian vanishes. This

is going to simplify quite a few of our results.

Translation symmetry

As usual, spacetime translation is a symmetry, and the infinitesimal transforma-

tion is given by

7→ x

− ε

ψ 7→ ψ + ε

∂

ψ.

So we have

δψ = ε

∂

ψ, δ

ψ = ε

∂

ψ.

Then we find a conserved current

µν

= i

ψγ

∂

ψ − η

µν

L = i

ψγ

∂

ψ.

In particular, we have a conserved energy of

E =

x i

ψγ

ψ(−iγ

∂

+ m)ψ,

where we used the equation of motion in the last line.

Similarly, we have a conserved total momentum

x T

x iψ

†

∂

ψ.

Lorentz transformations

We can also consider Lorentz transformations. This gives a transformation

ψ 7→ S[Λ]ψ(x

− ω

Taylor expanding, we get

δψ = −ω

∂

ψ +

ρσ

)ψ

= −ω

µν



∂

ψ −

µν

)ψ



Similarly, we have

ψ = −ω

µν



∂

ψ +

ψ(S

µν

)



The change is sign is because

transforms as

[Λ]

−1

, and taking the inverse of

an exponential gives us a negative sign.

So we can write this as

= −ω

ρσ



ψγ

∂

ψ − S

ρσ

ψ)



+ ω

= −ω

ρσ



ψγ

∂

ψ − S

ρσ

ψ)



So if we allow ourselves to pick different

ρσ

, we would have one conserved

current for each choice:

)

ρσ

= x

µρ

− x

µσ

− i

ψγ

ρσ

ψ.

In the case of a scalar field, we only have the first two terms. We will later see

that the extra term will give us spin

after quantization. For example, we have

)

= −i

ψS

ψ =

ijk

†



0 σ



ψ.

This is our angular momentum operator.

Internal vector symmetry

We also have an internal vector symmetry

ψ 7→ e

iα

ψ.

So the infinitesimal transformation is

δψ = iαψ.

We then obtain

ψγ

ψ.

We can indeed check that

∂

= (∂

ψ)γ

ψ +

ψγ

(∂

ψ) = im

ψψ − im

ψψ = 0.

The conserved quantity is then

Q =

x j

ψγ

ψ =

x ψ

†

ψ.

We will see that this is electric charge/particle number.

Axial symmetry

When

= 0, we have an extra internal symmetry obtained by rotating left-

handed and right-handed fermions in opposite signs. These symmetries are called

chiral. We consider the transformations

ψ 7→ e

iαγ

ψ,

ψ 7→

ψe

iαγ

This gives us an axial current

ψγ

ψ.

This is an axial vector. We claim that this is conserved only when

= 0. We

have

∂

= ∂

ψγ

ψ +

ψγ

∂

= 2im

ψγ

ψ.

This time the two terms add rather than subtract. So this vanishes iff m = 0.

It turns out that this is an anomalous symmetry, in the sense that the

classical axial symmetry does not survive quantization.

5 Quantizing the Dirac field

5.1 Fermion quantization

Quantization

We now start to quantize our spinor field! As it will turn out, this is complicated.

Recall that for a complex scalar field, classically a solution of momentum

can be written as

−ip·x

+ c

∗

ip·x

for some constants

, c

. Here the first term is the positive frequency solution,

and the second term is the negative frequency solution. To quantize this field,

we then promoted the b

, c

to operators, and we had

φ(x) =

(2π)

ip·x

+ c

†

−ip·x

Similarly, classically, every positive frequency solution to Dirac’s equation of

momentum p can be written as

+ b

−ip·x

for some b

, and similarly a negative frequency solution can be written as

+ c

ip·x

for some

. So when we quantize our field, we could promote the

and

operators and obtain

ψ(x) =

s=1

(2π)



ip·x

+ c

s†

−ip·x



†

(x) =

s=1

(2π)



s†

−ip·x

+ c

s†

ip·x



In these expressions,

and

are operators, and

and

are the elements of

the spinor space we have previously found, which are concrete classical objects.

We can compute the conjugate momentum to be

π =

∂L

∂

= i

ψγ

= iψ

†

The Hamiltonian density is then given by

H = π

ψ − L

= iψ

†

ψ − i

ψγ

ψ − i

ψγ

∂

ψ + m

ψψ

ψ(−iγ

∂

+ m)ψ.

What are the appropriate commutation relations to impose on the

, or, equiva-

lently, b

and c

s†

? The naive approach would be to impose

[ψ

(x), ψ

(y)] = [ψ

†

(x), ψ

†

(y)] = 0

and

[ψ

(x), ψ

†

(y)] = δ

αβ

(x − y).

These are in turn equivalent to requiring

, b

s†

] = (2π)

(p − q),

, c

s†

] = −(2π)

(p − q),

and all others vanishing.

If we do this, we can do the computations and find that (after normal

ordering) we obtain a Hamiltonian of

H =

(2π)

s†

− c

s†

Note the minus sign before

s†

. This is a disaster! While we can define a

vacuum

|0i

such that

|0i

= 0, this isn’t really a vacuum, because we

can keep applying c

†

to get lower and lower energies. This is totally bad.

What went wrong? The answer comes from actually looking at particles in

the Real World

. In the case of scalar fields, the commutation relation

†

, a

†

] = 0.

tells us that

†

|0i = |p, qi = |q, pi.

This means the particles satisfy Bose statistics , namely swapping two particles

gives the same state.

However, we know that fermions actually satisfy Fermi statistics. Swapping

two particles gives us a negative sign. So what we really need is

= −b

In other words, instead of setting the commutator zero, we want to set the

anticommutator to zero:

, b

} = 0.

In general, we are going to replace all commutators with anti-commutators. It

turns out this is what fixes our theory. We require that

Axiom. The spinor field operators satisfy

{ψ

(x), ψ

(y)} = {ψ

†

(x), ψ

†

(y)} = 0,

and

{ψ

(x), ψ

†

(y)} = δ

αβ

(x − y).

Proposition. The anti-commutation relations above are equivalent to

, c

s†

} = {b

, b

s†

} = (2π)

(p − q),

and all other anti-commutators vanishing.

Note that from a computational point of view, these anti-commutation

relations are horrible. When we manipulate our operators, we will keep on

introducing negative signs to our expressions, and as we all know, keeping track

of signs is the hardest thing in mathematics. Indeed, these negative signs will

come and haunt us all the time, and we have to insert funny negative signs

everywhere.

If we assume these anti-commutation relations with the same Hamiltonian

as before, then we would find

Proposition.

H =

(2π)



s†

+ c

s†



We now define the vacuum as in the bosonic case, where it is annihilated by

the b and c operators:

|0i = c

|0i = 0.

Although the

’s and

’s satisfy the anti-commutation relations, the Hamiltonian

satisfies commutator relations

[H, b

r†

] = Hb

r†

− b

r†

(2π)

s†

+ c

s†

†

− b

r†

(2π)

s†

+ c

s†

)

= E

r†

Similarly, we have

[H, b

] = −E

and the corresponding relations for c

and c

r†

Heisenberg picture

As before, we would like to put our Dirac field in the Heisenberg picture. We

have a spinor operator at each point x

ψ(x) = ψ(x, t).

The time evolution of the field is given by

∂ψ

∂t

= i[H, ψ],

which is solved by

ψ(x) =

s=1

(2π)



−ip·x

+ c

s†

ip·x



†

(x) =

s=1

(2π)



s†

ip·x

+ c

s†

−ip·x



We now look at the anti-commutators of the fields in the Heisenberg picture.

We define

Definition.

αβ

(x − y) = {ψ

(x),

(y)}.

Dropping the indices, we write

iS(x − y) = {ψ(x),

ψ(y)}.

If we drop the indices, then we have to remember that

is a 4

4 matrix with

indices α, β.

Substituting ψ and

ψ in using the expansion, we obtain

iS(x − y) =

s,r

p d

(2π)



, b

r†

¯u

−i(p·x−q·y)

+ {c

s†

, c

¯v

i(p·x−q·y)



(2π)



¯u

−p·(x−y)

+ v

¯v

ip·(x−y)



(2π)

(

p + m)e

−ip·(x−y)

+ (

p − m)e

ip·(x−y)

Recall that we had a scalar propagator

D(x − y) =

(2π)

−ip·(x−y)

So we can write the above quantity in terms of the D(x − y) by

iS(x − y) = (i

∂

+ m)(D(x − y) − D(y − x)).

Some comments about this: firstly, for spacelike separations where

(x − y)

< 0,

we had D(x − y) − D(y − x) = 0.

For bosonic fields, we made a big deal of this, since it ensured that

[

(

)

, φ

(

)] = 0. We then interpreted this as saying the theory was causal.

For spinor fields, we have anti-commutation relations rather than commutation

relations. What does this say about causality? The best we can say is that all

our observables are bilinear (or rather quadratic) in fermions, e.g.

H =

x ψ

†

(−iγ

∂

+ m)ψ.

So these observables will commute for spacelike separations.

Propagators

The next thing we need to figure out is the Feynman propagator. By a similar

computation to that above, we can determine the vacuum expectation value

h0|ψ(x)

ψ(y) |0i =

(2π)

(

p + m)e

−ip·(x−y)

h0|

ψ(y)ψ(x) |0i =

(2π)

(

p − m)e

ip·(x−y)

We define the Feynman propagator by

Definition

(Feynman propagator)

The Feynman propagator of a spinor field

is the time-ordered product

(x − y) = h0|T ψ

(x)

(y) |0i =

(

h0|ψ

(x)

(y) |0i x

> y

−h0|

(y)ψ

(x) |0i y

> x

Note that we have a funny negative sign! This is necessary for Lorentz

invariance — when (

x − y

)

0, then there is no invariant way to determine

whether one time is bigger than the other. So we need the expressions for the two

cases to agree. In the case of a boson, we had

h0|φ

(

)

(

)

|0i

h0|φ

(

)

(

)

|0i

but here we have anti-commutation relations, so we have

ψ(x)

ψ(y) = −

ψ(y)ψ(x).

So we need to insert the negative sign to ensure time-ordered product as defined

is Lorentz invariant.

For normal ordered products, we have the same behaviour. Fermionic opera-

tors anti-commute, so

:ψ

: = −:ψ

: .

As before, the Feynman propagator appears in Wick’s theorem as the contraction:

Proposition.

ψ(x)

ψ(y) = T(ψ(x)

ψ(y)) − :ψ(x)

ψ(y): = S

(x − y).

When we apply Wick’s theorem, we need to remember the minus sign. For

example,

:ψ

: = −:ψ

: = −ψ

:ψ

: .

Again, S

can be expressed as a 4-momentum integral

(x − y) = i

(2π)

−ip·(x−y)

p + m

− m

+ iε

As in the case of a real scalar field, it is a Green’s function of Dirac’s equation:

∂

− m)S

(x − y) = iδ

(x − y).

5.2 Yukawa theory

The interactions between a Dirac fermion and a real scalar field are governed by

the Yukawa interaction. The Lagrangian is given by

L =

∂

φ∂

φ +

ψ(iγ

∂

− m)ψ − λφ

ψψ.

where

is the mass of the scalar and

is the mass of the Lagrangian. This is

the full version of the Yukawa theory. Note that the kinetic term implies that

[ψ] = [

ψ] =

Since [φ] = 1 and [L] = 4, we know that

[λ] = 0.

So this is a dimensionless coupling constant, which is good.

Note that we could have written the interaction term of the Lagrangian as

int

= −λφ

ψγ

ψ.

We then get a pseudoscalar Yuakwa theory.

We again do painful computations directly to get a feel of how things work,

and then state the Feynman rules for the theory.

Example. Consider a fermion scattering ψψ → ψψ.

We have initial and final states

|ii =

s†

r†

|0i

|fi =

†

|0i.

Here we have to be careful about ordering the creation operators, because they

anti-commute, not commute. We then have

hf| =

h0|b

We can then look at the O(λ

) term in hf |(S − 1) |ii. We have

(−iλ)



ψ(x

)ψ(x

)φ(x

)

ψ(x

)ψ(x

)φ(x

)



The contribution to scattering comes from the contraction

ψ(x

)ψ(x

)

ψ(x

)ψ(x

): φ(x

)φ(x

The two

’s annihilate the initial state, whereas the two

create the final state.

This is just like the bosonic case, but we have to be careful with minus signs

and spinor indices.

Putting in |ii and ignoring c operators as they don’t contribute, we have

ψ(x

)ψ(x

)

ψ(x

)ψ(x

): b

s†

†

|0i

= −

(2π)

[

ψ(x

][

ψ(x

−ik

·x

−ik

·x

†

r†

|0i

where the square brackets show contraction of spinor indices

= −



[

ψ(x

][

ψ(x

−iq·x

−ip·x

−[

ψ(x

][

ψ(x

−ip·x−iq·x



|0i.

The negative sign, which arose from anti-commuting the

’s, is crucial. We put

in the left hand side to get

h0|b

[

ψ(x

][

ψ(x

]



[¯u

][¯u

·x

+iq

·x

− [¯u

][¯u

·x

+iq

·x



Putting all of this together, including the initial relativistic normalization of the

initial state and the propagator, we have

hf|S − 1 |ii

= (−iλ)

(2π)

ik·(x

−x

)

− µ

+ iε



[¯u

· u

][¯u

· u

·(q

−q)+ ix

·(p

−p)

− [¯u

][u

·(p

−q)+ ix

·(q

−p)



= i(−iλ)

k(2π)

− µ

+ iε



[¯u

· u

][u

· u

]δ

− q + k)δ

− p + k)

−[¯u

· u

][¯u

· u

]δ

− q + k)δ

− p + k)



So we get

hf|S − 1 |ii = iA(2π)

(p + q − p

− q

where

A = (−iλ)

[¯u

· u

][¯u

· u

]

− p)

− µ

+ iε

−

[¯u

· u

][¯u

· u

]

− p)

− µ

+ iε

5.3 Feynman rules

As always, the Feynman rules are much better:

(i)

An incoming fermion is given a spinor

, and an outgoing fermion is given

a ¯u

¯u

(ii)

For an incoming anti-fermion, we put in a

¯v

, and for an outgoing anti-

fermion we put a v

¯v

(iii) For each vertex we get a factor of (−iλ).

(iv) Each internal scalar line gets a factor of

− µ

+ iε

and each internal fermion we get

p + m)

− m

+ iε

(v)

The arrows on fermion lines must flow consistently, ensuring fermion

conservation.

(vi)

We impose energy-momentum conservation at each vertex, and if we have

a loop, we integrate over all possible momentum.

(vii) We add an extra minus sign for a loop of fermions.

Note that the Feynman propagator is a 4

4 matrix. The indices are contracted

with each vertex, either with further propagators or external spinors.

We look at computations using Feynman rules.

Example (Nucleon scattering). For nucleon scattering, we have diagrams

p, s

q, r

, s

, r

p − p

p, s

q, r

, s

, r

p − q

In the second case, the fermions are swapped, so we get a relative minus sign.

So by the Feynman rules, this contributes an amplitude of

A = (−iλ)

[¯u

· u

][¯u

· u

]

− p)

− µ

+ iε

−

[¯u

· u

][¯u

· u

]

− p)

− µ

+ iε

Example. We look at nucleon to meson scattering

ψ → φφ.

We have two diagrams

p, s

q, r

p, s

q, r

This time, we flipped two bosons, so we are not gaining a negative sign. Then

the Feynman rules give us

A = (−iλ)



¯v

(

p −

+ m)u

(p − p

)

− m

+ iε

¯v

(

p −

+ m)u

(p − q

)

− m

+ iε



Example. We can also do nucleon anti-nucleon scattering

ψ → ψ

ψ.

As before, we have two contributions

p, s

q, r

, s

, r

p − p

p, s

q, r

, s

, r

This time we have an amplitude of

A = (−iλ)

−[¯u

· u

][¯v

· v

]

(p − p

)

− µ

+ iε

[ ¯v

· u

][¯u

· v

]

(p + q)

− µ

+ iε

We have these funny signs that we have to make sure are right. We have an

initial state

|ii =

s†

r†

|0i ≡ |p, s; q, ri

|fi =

†

|0i ≡ |p

, s

; q

, r

To check the signs, we work through the computations, but we can ignore all

factors because we only need the final sign. Then we have

ψ ∼ b + c

†

ψ ∼ b

†

+ c

So we can check

hf|:

ψ(x

)ψ(x

)

ψ(x

)ψ(x

): b

s†

r†

|0i

∼ hf|[¯v

ψ(x

)][

ψ(x

s†

r†

|0i

∼ hf|[¯v

ψ(x

)][

ψ(x

] |0i

∼ h0|c

m†

†

[¯v

· v

][¯u

· u

] |0i

∼ −[¯v

· v

][¯u

· u

where we got the final sign by anti-commuting

m†

and

to make the

’s and

b’s get together.

We can follow a similar contraction to get a positive sign for the second

diagram.

6 Quantum electrodynamics

Finally, we get to the last part of the course, where we try to quantize electro-

magnetism. Again, this will not be so straightforward. This time, we will have

to face the fact that the electromagnetic potential

is not uniquely defined,

but can have many gauge transformations. The right way to encode this infor-

mation when we quantize the theory is not immediate, and will require some

experimentation.

6.1 Classical electrodynamics

We begin by reviewing what we know about classical electrodynamics. Classically,

the electromagnetic field is specified by an electromagnetic potential

, from

which we derive the electromagnetic field strength tensor

µν

= ∂

− ∂

We can then find the dynamics of a free electromagnetic field by

L = −

µν

The Euler-Lagrange equations give

∂

µν

= 0.

It happens that F satisfies the mystical Bianchi identity

∂

µν

+ ∂

νλ

+ ∂

λµ

= 0,

but we are not going to use this.

We can express these things in terms of the electromagnetic field. We write

A = (φ, A). Then we define

E = −∇φ −

A, B = ∇ ∧ A.

Then F

µν

can be written as

µν







0 E

−E

0 −B

−E

0 −B

−E

−B







Writing out our previous equations, we find that

∇ · E = 0

∇ · B = 0

B = −∇ ∧ E

E = ∇ ∧ B.

We notice something wrong. Photons are excitations of the electromagnetic

field. However, the photon only has two polarization states, i.e. two degrees of

freedom, while this vector field A

has four. How can we resolve this?

There are two parts to the resolution. Firstly, note that the

field is not

dynamical, as it has no kinetic term in the Lagrangian (the

∂

terms cancel

out by symmetry). Thus, if we’re given

and

at some initial time

, then

is fully determined by ∇ · E = 0, since it says

∇ ·

A + ∇

= 0.

This has a solution

(x) =

∇ ·

A(x

)

4π(x − x

)

is not independent, and we’ve got down to three degrees of freedom.

What’s the remaining one?

The next thing to note is that the theory is invariant under the transformation

(x) 7→ A

(x) + ∂

λ(x),

Indeed, the only “observable” thing is

µν

, and we can see that it transforms as

µν

7→ ∂

+ ∂

λ) − ∂

+ ∂

λ) = F

µν

It looks like that we now have an infinite number of symmetries.

This is a different kind of symmetry. Previously, we had symmetries acting

at all points in the universe in the same way, say

ψ 7→ e

iα

for some

α ∈ R

This gave rise to conservation laws by Noether’s theorem.

Now do we get infinitely many symmetries from Noether’s theorem? The

answer is no. The local symmetries we’re considering now have a different

interpretation. Rather than taking a physical state to another physical state,

they are really a redundancy in our description. These are known as local or

gauge symmetries. Seeing this, we might be tempted to try to formulate the

theory only in terms of gauge invariant objects like

and

, but it turns out

this doesn’t work. To describe nature, it appears that we have to introduce

quantities that we cannot measure.

Now to work with our theory, it is common to fix a gauge. The idea is that

we specify some additional rules we want

to satisfy, and hopefully this will

make sure there is a unique

we can pick for each equivalence class. Picking

the right gauge for the right problem can help us a lot. This is somewhat like

picking a coordinate system. For example, in a system with spherical symmetry,

using spherical polar coordinates will help us a lot.

There are two gauges we would be interested in.

Definition (Lorenz gauge). The Lorenz gauge is specified by

∂

= 0

Note that this is Lorenz, not Lorentz!

To make sure this is a valid gauge, we need to make sure that each electro-

magnetic potential has a representation satisfying this condition.

Suppose we start with

such that

∂

0µ

. If we introduce a gauge

transformation

= A

+ ∂

λ,

then we have

∂

= ∂

∂

λ + f.

So we need to find a λ such that

∂

λ = −f.

By general PDE theory, such a λ exists. So we are safe.

However, it turns out this requirement does not pick a unique representation

in the gauge orbit. We are free to make a further gauge transformation by with

∂

λ = 0,

which has non-trivial solutions (e.g. λ(x) = x

Another important gauge is the Coulomb gauge:

Definition (Coulomb gauge). The Coulomb gauge requires

∇ · A = 0.

Of course, this is not a Lorentz-invariant condition.

Similar to the previous computations, we know that this is a good gauge.

Looking at the integral we’ve found for A

previously, namely

∇ ·

A(x

)

4π|x − x

we find that

= 0 all the time. Note that this happens only because we do

not have matter.

Here it is easy to see the physical degrees of freedom — the three components

satisfy a single constraint

∇ · A

= 0, leaving behind two physical degrees

of freedom, which gives the desired two polarization states.

6.2 Quantization of the electromagnetic field

We now try to quantize the field, using the Lorenz gauge. The particles we

create in this theory would be photons, namely quanta of light. Things will go

wrong really soon.

We first try to compute the conjugate momentum

of the vector field

We have

∂L

∂

= 0.

This is slightly worrying, because we would later want to impose the commutation

relation

(x), π

(y)] = iδ

(x − y),

but this is clearly not possible if π

vanishes identically!

We need to try something else. Note that under the Lorenz gauge

∂

= 0,

the equations of motion tell us

∂

= 0.

The trick is to construct a Lagrangian where this is actually the equation of

motion, and then later impose ∂

= 0 after quantization.

This is not too hard. We can pick the Lagrangian as

L = −

µν

−

(∂

)

We can work out the equations of motion of this, and we get

∂

µν

+ ∂

(∂

) = 0.

Writing out the definition of

µν

, one of the terms cancels with the second bit,

and we get

∂

= 0.

We are now going to work with this Lagrangian, and only later impose

∂

= 0

at the operator level.

More generally, we can use a Lagrangian

L = −

µν

−

2α

(∂

)

where

is a fixed constant. Confusingly, the choice of

is also known as a

gauge. We picked

= 1, which is called the Feynman gauge. If we take

α →

we obtain the Landau gauge.

This new theory has no gauge symmetry, and both

and

are dynamical

fields. We can find the conjugate momenta as

∂L

∂

= −∂

∂L

∂

= ∂

−

We now apply the usual commutation relations

(x), A

(y)] = [π

(x), π

(y)] = 0

(x), π

(y)] = iδ

(x − y)δ

Equivalently, the last commutation relation is

(x), π

(y)] = iδ

(x − y)η

µν

As before, in the Heisenberg picture, we get equal time commutation relations

(x, t),

(y, t)] = iη

µν

(x − y).

As before, we can write our operators in terms of creation and annihilation

operators:

(x) =

(2π)

2|p|

λ=0

(λ)

(p)[a

ip·x

+ a

λ†

−ip·x

(x) =

(2π)

|p|

λ=0

(λ)

(p))

ip·x

− a

λ†

−ip·x

where for each

, the vectors

(0)

(

)

, E

(1)

(

)

, E

(2)

(

)

, E

(3)

(

)

}

form a basis of

3,1

. The exact choice doesn’t really matter much, but we will suppose we pick

a basis with the following properties:

(i) E

(0)

(p) will be a timelike vector, while the others are spacelike.

(ii)

Viewing

as the direction of motion, we will pick

(3)

(

) to be a longi-

tudinal polarization, while

(1),(2)

(

) are transverse to the direction of

motion, i.e. we require

(1),(2)

(p) · p

= 0.

(iii) The normalization of the vectors is given by

(λ)

(λ

)

= η

λλ

We can explicitly write down a choice of such basis vectors. When

p ∝

(1, 0, 0, 1), then we choose

(0)

(p) =













, E

(1)

(p) =













, E

(2)

(p) =













, E

(3)

(p) =













Now for a general

, we pick a basis where

p ∝

1), which is always possible,

and then we define

(

) as above in that basis. This gives us a Lorentz-invariant

choice of basis vectors satisfying the desired properties.

One can do the tedious computations required to find the commutation

relations for the creation and annihilation operators:

Theorem.

, a

] = [a

λ†

, a

†

] = 0

and

, a

†

] = −η

λλ

(2π)

(p − q).

Notice the strange negative sign!

We again define a vacuum |0i such that

|0i = 0

for λ = 0, 1, 2, 3, and we can create one-particle states

|p, λi = a

λ†

|0i.

This makes sense for

= 1

3, but for

= 0, we have states with negative

norm:

hp, 0|q, 0i ≡ h0|a

0†

|0i

= −(2π)

(p − q).

A Hilbert space with a negative norm means that we have negative probabilities.

This doesn’t make sense.

Here is where the Lorenz gauge comes in. By imposing

∂

= 0, we are

going to get rid of bad things. But how do we implement this constraint? We

can try implementing it in a number of ways. We will start off in the obvious

way, which turns out to be too strong, and we will keep on weakening it until it

works.

If we just asked for

∂

= 0, for

the operator, then this doesn’t work,

because

= −∂

and if this vanishes, then the commutation conditions cannot possibly be obeyed.

Instead, we can try to impose this on the Hilbert space rather than on the

operators. After all, that’s where the trouble lies. Maybe we can try to split the

Hilbert space up into good states and bad states, and then just look at the good

states only?

How do we define the good, physical states? Maybe we can impose

∂

|ψi = 0

on all physical (“good”) states, but it turns out even this condition is a bit too

strong, as the vacuum will not be physical! To see this, we decompose

(x) = A

(x) + A

−

(x),

where

has the annihilation operators and

−

has the creation operators.

Explicitly, we have

(x) =

(2π)

2|p|

(λ)

−ip·x

−

(x) =

(2π)

2|p|

(λ)

λ†

ip·x

where summation over λ is implicit. Then we have

∂

|0i = 0,

but we have

∂

−

|0i 6= 0.

So not even the vacuum is physical! This is very bad.

Our final attempt at weakening this is to ask the physical states to satisfy

∂

(x) |ψi = 0.

This ensures that

hψ|∂

|ψi = 0,

∂

will kill the right hand side and

∂

−

will kill the left. So

∂

has

vanishing matrix elements between physical states.

This is known as the Gupta-Bleuler condition. The linearity of this condition

ensures that the physical states span a vector space H

phys

What does

phys

look like? To understand this better, we write out what

∂

−

is. We have

∂

−

(2π)

2|p|

(λ)

(−ip

−ip·x

(2π)

2|p|

i(a

− a

ip·x

using the properties of our particular choice of the

. Thus the condition is

equivalently

− a

) |ψi = 0.

This means that it is okay to have a timelike or longitudinal photons, as long as

they come in pairs of the same momentum!

By restricting to these physical states, we have gotten rid of the negative

norm states. However, we still have the problem of certain non-zero states having

zero norm. Consider a state

|φi = a

0†

3†

|0i.

This is an allowed state, since it has exactly one timelike and one longitudinal

photon, each of momentum

. This has zero norm, as the norm contributions

from the

part cancel from those from the

part. However, this state is

non-zero! We wouldn’t want this to happen if we want a positive definite inner

product.

The solution is to declare that if two states differ only in the longitudinal

and timelike photons, then we consider them physically equivalent, or gauge

equivalent! In other words, we are quotienting the state space out by these

zero-norm states.

Of course, if we want to do this, we need to go through everything we do and

make sure that our physical observables do not change when we add or remove

these silly zero-norm states. Fortunately, this is indeed the case, and we are

happy.

Before we move on, we note the value of the Feynman propagator:

Theorem.

The Feynman propagator for the electromagnetic field, under a

general gauge α, is

h0|T A

(x)A

(y) |0i =

(2π)

−i

+ iε



µν

+ (α − 1)



−ip·(x−y)

6.3 Coupling to matter in classical field theory

We now move on to couple our EM field with matter. We first do it in the clear

and sensible universe of classical field theory. We will tackle two cases — in the

first case, we do coupling with fermions. In the second, we couple with a mere

complex scalar field. It should be clear one how one can generalize these further

to other fields.

Suppose the resulting coupled Lagrangian looked like

L = −

µν

− A

plus some kinetic and self-interaction terms for the field we are coupling with.

Then the equations of motion give us

∂

µν

= j

Since F

µν

is anti-symmetric, we know we must have

0 = ∂

∂

µν

= ∂

So we know

is a conserved current. So to couple the EM field to matter, we

need to find some conserved current.

Coupling with fermions

Suppose we had a spinor field ψ with Lagrangian

L =

ψ(i

∂ − m)ψ.

This has an internal symmetry

ψ 7→ e

−iαψ

ψ 7→ e

iα

ψ,

and this gives rise to a conserved current

ψγ

ψ.

So let’s try

L = −

µν

ψ(i

∂ − m)ψ −e

ψγ

ψ,

where e is some coupling factor.

Have we lost gauge invariance now that we have an extra term? If we just

substituted

for

∂

, then obviously the Lagrangian changes. However,

something deep happens. When we couple the two terms, we don’t just add a

factor into the Lagrangian. We have secretly introduced a new gauge symmetry

to the fermion field, and now when we take gauge transformations, we have to

transform both fields at the same time.

To see this better, we rewrite the Lagrangian as

L = −

µν

ψ(i

D − m)ψ,

where D is the covariant derivative given by

ψ = ∂

ψ + ieA

ψ.

We now claim that L is invariant under the simultaneous transformations

7→ A

+ ∂

λ(x)

ψ 7→ e

−ieλ(x)

ψ.

To check that this is indeed invariant, we only have to check that

Dψ

term.

We look at how D

ψ transforms. We have

ψ = ∂

ψ + ieA

7→ ∂

−ieλ(x)

ψ) + ie(A

+ ∂

λ(x))e

−ieλ(x)

= e

−ieλ(x)

ψ.

So we have

Dψ 7→

Dψ,

i.e. this is gauge invariant.

So as before, we can use the gauge freedom to remove unphysical states after

quantization. The coupling constant

has the interpretation of electric charge,

as we can see from the equation of motion

∂

µν

= ej

In electromagnetism, j

is the charge density, but after quantization, we have

Q = e

ψγ

(2π)

s†

− c

s†

)

= e × (number of electrons − number of anti-electrons),

with an implicit sum over the spin

. So this is the total charge of the electrons,

where anti-electrons have the opposite charge!

For QED, we usually write e in terms of the fine structure constant

α =

4π

≈

137

for an electron.

Coupling with complex scalars

Now let’s try to couple with a scalar fields. For a real scalar field, there is no

suitable conserved current to couple

to. For a complex scalar field

, we can

use the current coming from ϕ → e

iα

ϕ, namely

= (∂

ϕ)

∗

ϕ − ϕ

∗

∂

ϕ.

We try the obvious thing with an interaction term

int

= −i[(∂

ϕ)

∗

ϕ − ϕ

∗

∂

ϕ]A

This doesn’t work. The problem is that we have introduced a new term

the Lagrangian. If we compute the conserved current due to

ϕ 7→ e

iα

under

the new Lagrangian, it turns out we will obtain a different

, which means in

this theory, we are not really coupling to the conserved current, and we find

ourselves going in circles.

To solve this problem, we can try random things and eventually come up

with the extra term we need to add to the system to make it consistent, and

then be perpetually puzzled about why that worked. Alternatively, we can also

learn from what we did in the previous example. What we did, at the end, was

to invent a new covariant derivative:

ϕ = ∂

ϕ + ieA

ϕ,

and then replace all occurrences of

∂

with D

. Under a simultaneous gauge

transformation

7→ A

+ ∂

λ(x)

ϕ 7→ e

−ieλ(x)

ϕ,

the covariant derivative transforms as

ϕ 7→ e

iλ(x)

ϕ,

as before. So we can construct a gauge invariant Lagrangian by

L = −

µν

+ (D

ϕ)

†

ϕ) − m

∗

ϕ.

The current is the same thing as we’ve had before, except we have

= i(D

ϕ)

†

ϕ − ϕ

†

ϕ.

In general, for any field

taking values in a complex vector space, we have a

U(1) gauge symmetry

φ 7→ e

iλ(x)

φ.

Then we can couple with the EM field by replacing

∂

φ 7→ D

φ = ∂

φ + ieλ(x)A

φ.

This process of just taking the old Lagrangian and then replacing partial deriva-

tives with covariant derivatives is known as minimal coupling. More details on

how this works can be found in the Symmetries, Fields and Particles course.

6.4 Quantization of interactions

Let’s work out some quantum amplitudes for light interacting with electrons.

We again have the Lagrangian

L = −

µν

ψ(i

D − m)ψ,

where

= ∂

+ ieA

For a change, we work in the Coulomb gauge

∇ · A

= 0. So the equation of

motion for A

∇

= e

ψγ

ψ = ej

This equation has a solution

(x) = e

, t)

4π|x − x

In the Coulomb gauge, we can rewrite the Maxwell part (i.e.

µν

part) of

the Lagrangian (not density!) as

− B

)



(

A + ∇A

)

−





(∇A

)

−



where the gauge condition means that the cross-term vanishes by integration by

parts.

Integrating by parts and substituting our A

, we get that



(x)j

)

4π|x − x



−



This is weird, because we now have a non-local term in the Lagrangian. This

term arises as an artifact of working in Coulomb gauge. This doesn’t appear in

Lorenz gauge.

Let’s now compute the Hamiltonian. We will use capital Pi (

) for the

conjugate momentum of the electromagnetic potential

, and lower case pi (

)

for the conjugate momentum of the spinor field. We have

Π =

∂L

∂

∂L

∂

= iψ

†

Putting all these into the Hamiltonian, we get

H =



ψ(−iγ

∂

+ m)ψ − ej ·A

(x)j

)

4π|x − x



where

j =

ψγψ, j

ψγ

ψ.

After doing the tedious computations, one finds that the Feynman rules are as

follows:

(i) The photons are denoted as squiggly lines:

Each line comes with an index

that tells us which component of

are talking about. Each internal line gives us a factor of

+ iε



−

|p|



while for each external line,

(i)

we simply write down a polarization vector

(i)

corresponding to the

polarization of the particle.

(ii) The ej · A term gives us an interaction of the form

This contributes a factor of

−ieγ

, where

is the index of the squiggly line.

(iii)

The non-local term of the Lagrangian gives us instantaneous non-local

interactions, denoted by dashed lines:

The contribution of this factor, in position space, is given by

i(eγ

)

δ(x

− y

)

4π|x − y|

Whoa. What is this last term saying? Recall that when we previously derived

our Feynman rules, we first obtained some terms from Wick’s theorem that

contained things like

ip·x

, and then integrated over

to get rid of these terms,

so that the resulting formula only included momentum things. What we’ve

shown here in the last rule is what we have before integrating over

and

. We

now want to try to rewrite things so that we can get something nicer.

We note that this piece comes from the

term. So one possible strategy

is to make it into a

piece of the photon propagator. So we treat this as

a photon indexed by

= 0. We then rewrite the above rules and replace all

indices i with µ and let it range over 0, 1, 2, 3.

To do so, we note that

δ(x

− y

)

4π|x − y|

(2π)

ip·(x−y)

|p|

We now define the γ-propagator

µν

(p) =











+iε



µν

−

|p|



µ, ν 6= 0

|p|

µ = ν = 0

0 otherwise

We now have the following Feynman rules:

(i) The photons are denoted as squiggly lines:

Each internal line comes with an index

ranging over 0

3 that tells

us which component of

we are talking about, and give us a factor of

µν

, while for each external line,

(µ)

we simply write down a polarization vector

(µ)

corresponding to the

polarization of the particle.

(ii) The ej · A term gives us an interaction of the form

This contributes a factor of

−ieγ

, where

is the index of the squiggly line.

Now the final thing to deal with is the annoying

µν

formula. We claim that

it can always be replaced by

µν

(p) = −i

µν

i.e. in all contractions involving

µν

we care about, contracting with

µν

gives

the same result as contracting with

−i

µν

. This can be proved in full generality

using momentum conservation, but we will only look at some particular cases.

Example. Consider the process

−

→ e

−

We look at one particular diagram

p, s

q, r

, s

, r

We have two vertices, which contribute to a term of

[¯u(p

)γ

u(p)]D

µν

(k)[¯u(q

)γ

u(q)],

where

k = p − p

= −q + q

We show that in this case, we can replace D

µν

(k) by

µν

(k) =

−iη

µν

The proof follows from current conservation. Recall that u(p) satisfies

(

p − m)u(p) = 0.

We define the spinor combinations

= ¯u(p

)γ

u(p)

= ¯u(q

)γ

u(q)

What we have is that

= ¯u(p

)(

−

p)u(p) = ¯u(p

)(m − m)u(p) = 0.

Similarly, k

is also zero. So our Feynman diagram is given by

µν

= i



α · β

−

(α · k)(β · k)

|k|



But we know that α

= β

= 0. So this is equal to



α · β

−

|k|



= i



α · β

−

|k|

− k

)α



= i



α · β

−

|k|



= −i

α · β

= α



−iη

µν



What really is giving us this simple form is current conservation.

In general, in Lorenz gauge, we have

µν

= −



µν

+ (α − 1)



and the second term cancels in all physical processes, for similar reasons.

Charged scalars

We quickly go through the Feynman rules for charged complex scalar fields. We

will not use these for anything. The Lagrangian coming from minimal coupling

L = (D

ψ)

†

ψ −

µν

We can expand the first term to get

ψ)

†

ψ = ∂

†

∂

ψ − ieA

(ψ

†

∂

ψ − ψ∂

†

) + e

†

ψ.

The Feynman rules for these are:

(i) The first interaction term gives us a possible vertex

p q

This contributes a factor of −ie(p + q)

(ii) The A

†

ψ term gives diagrams of the form

p q

This contributes a factor of 2ie

µν

6.5 Computations and diagrams

We now do some examples. Here we will not explain where the positive/negative

signs come from, since it involves some tedious work involving going through

Wick’s theorem, if you are not smart enough to just “see” them.

Example. Consider the process

−

→ e

−

We again can consider the two diagrams

p, s

q, r

, s

, r

p, s

q, r

, s

, r

We will do the first diagram slowly. The top vertex gives us a factor of

−ie[¯u

The bottom vertex gives us

−ie[¯u

The middle squiggly line gives us

−

iη

µν

− p)

So putting all these together, the first diagram gives us a term of

−i(−ie)

[¯u

][¯u

]

− p)

Similarly, the second diagram gives

−i(−ie)

[¯u

][¯u

]

(p − q)

Example. Consider the process

−

→ γγ.

We have diagrams of the form

p, s

q, r

, p

, q

This diagram gives us

i(−ie)



[¯v

(

p −

+ m)γ

]

(p − p

)

− m



Usually, since the mass of an electron is so tiny relative to our energy scales, we

simply ignore it.

Example

(Bhabha scattering)

This is definitely not named after an elephant.

We want to consider the scattering process

−

→ e

−

with diagrams

p, s

q, r

, s

, r

p, s

q, r

, s

, r

These contribute

−i(−ie)

−

[¯u

][¯v

]

(p − p

)

[¯v

][¯u

]

(p + q)

Example (Compton scattering). Consider scattering of the form

γe

−

→ γe

−

We have diagrams

¯u

(p)

)

p + q

¯u

(p)

)

q − p

Example. Consider the

γγ → γγ

scattering process. We have a loop diagram

Naively, we might expect the contribution to be proportional to

This integral itself diverges, but it turns out that if we do this explicitly, the

gauge symmetry causes things to cancel out and renders the diagram finite.

Example

(Muon scattering)

We could also put muons into the picture. These

behave like electrons, but are actually different things. We can consider scattering

of the form

−

→ e

−

This has a diagram

−

We don’t get the diagram obtained by swapping two particles because electrons

are not muons.

We can also get interactions of the form

−

→ µ

−