III The Standard Model (Full)

Part III — The Standard Model

Based on lectures by C. E. Thomas

Notes taken by Dexter Chua

Lent 2017

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.

The Standard Mo del of particle physics is, by far, the most successful application

of quantum field theory (QFT). At the time of writing, it accurately describes all

exp erimental measurements involving strong, weak, and electromagnetic interactions.

The course aims to demonstrate how this model, a QFT with gauge group

(3)

(2)

U(1) and fermion fields for the leptons and quarks, is realised in nature. It is

intended to complement the more general Advanced QFT course.

We begin by defining the Standard Model in terms of its local (gauge) and global

symmetries and its elementary particle content (spin-half leptons and quarks, and

spin-one gauge bosons). The parity

, charge-conjugation

and time-reversal

transformation properties of the theory are investigated. These need not be symmetries

manifest in nature; e.g. only left-handed particles feel the weak force and so it violates

parity symmetry. We show how

violation becomes possible when there are three

generations of particles and describe its consequences.

Ideas of spontaneous symmetry breaking are applied to discuss the Higgs Mechanism

and why the weakness of the weak force is due to the spontaneous breaking of the

(2)

U(1) gauge symmetry. Recent measurements of what appear to be Higgs boson

decays will be presented.

We show how to obtain cross sections and decay rates from the matrix element squared

of a process. These can be computed for various scattering and decay processes in

the electroweak sector using perturbation theory because the couplings are small. We

touch upon the topic of neutrino masses and oscillations, an important window to

physics beyond the Standard Model.

The strong interaction is described by quantum chromodynamics (QCD), the non-

abelian gauge theory of the (unbroken)

(3) gauge symmetry. At low energies quarks

are confined and form bound states called hadrons. The coupling constant decreases

as the energy scale increases, to the point where perturbation theory can be used. As

an example we consider electron- positron annihilation to final state hadrons at high

energies. Time permitting, we will discuss nonperturbative approaches to QCD. For

example, the framework of effective field theories can be used to make progress in the

limits of very small and very large quark masses.

Both very high-energy experiments and very precise experiments are currently striving

to observe effects that cannot be described by the Standard Model alone. If time

permits, we comment on how the Standard Model is treated as an effective field theory

to accommo date (so far hypothetical) effects beyond the Standard Model.

Pre-requisites

It is necessary to have attended the Quantum Field Theory and the Symmetries, Fields

and Particles courses, or to be familiar with the material covered in them. It would

b e advantageous to attend the Advanced QFT course during the same term as this

course, or to study renormalisation and non-abelian gauge fixing.

Contents

0 Introduction

1 Overview

2 Chiral and gauge symmetries

2.1 Chiral symmetry

2.2 Gauge symmetry

3 Discrete symmetries

3.1 Symmetry operators

3.2 Parity

3.3 Charge conjugation

3.4 Time reversal

3.5 S-matrix

3.6 CPT theorem

3.7 Baryogenesis

4 Spontaneous symmetry breaking

4.1 Discrete symmetry

4.2 Continuous symmetry

4.3 General case

4.4 Goldstone’s theorem

4.5 The Higgs mechanism

4.6 Non-abelian gauge theories

5 Electroweak theory

5.1 Electroweak gauge theory

5.2 Coupling to leptons

5.3 Quarks

5.4 Neutrino oscillation and mass

5.5 Summary of electroweak theory

6 Weak decays

6.1 Effective Lagrangians

6.2 Decay rates and cross sections

6.3 Muon decay

6.4 Pion decay

6.5 K

mixing

7 Quantum chromodynamics (QCD)

7.1 QCD Lagrangian

7.2 Renormalization

7.3 e

−

→ hadrons

7.4 Deep inelastic scattering

0 Introduction

In the Michaelmas Quantum Field Theory course, we studied the general theory

of quantum field theory. At the end of the course, we had a glimpse of quantum

electrodynamics (QED), which was an example of a quantum field theory that

described electromagnetic interactions. QED was a massively successful theory

that explained experimental phenomena involving electromagnetism to a very

high accuracy.

But of course, that is far from being a “complete” description of the universe.

It does not tell us what “matter” is actually made up of, and also all sorts of other

interactions that exist in the universe. For example, the atomic nucleus contains

positively-charged protons and neutral neutrons. From a purely electromagnetic

point of view, they should immediately blow apart. So there must be some force

that holds the particles together.

Similarly, in experiments, we observe that certain particles undergo decay.

For example, muons may decay into electrons and neutrinos. QED doesn’t

explain these phenomena as well.

As time progressed, physicists managed to put together all sorts of experi-

mental data, and came up with the Standard Model. This is the best description

of the universe we have, but it is lacking in many aspects. Most spectacularly,

it does not explain gravity at all. There are also some details not yet fully

sorted out, such as the nature of neutrinos. In this course, our objective is to

understand the standard model.

Perhaps to the disappointment of many readers, it will take us a while before

we manage to get to the actual standard model. Instead, during the first half of

the course, we are going to discuss some general theory regarding symmetries.

These are crucial to the development of the theory, as symmetry concerns impose

a lot of restriction on what our theories can be. More importantly, the “forces”

in the standard model can be (almost) completely described by the gauge group

we have, which is

(3)

× SU

(2)

U(1). Making sense of these ideas is crucial

to understanding how the standard model works.

With the machinery in place, the actual description of the Standard Model is

actually pretty short. After describing the Standard Model, we will do various

computations with it, and make some predictions. Such predictions were of course

important — it allowed us to verify that our theory is correct! Experimental

data also helps us determine the values of the constants that appear in the

theory, and our computations will help indicate how this is possible.

Historically, the standard model was of course discovered experimentally,

and a lot of the constructions and choices are motivated only by experimental

reasons. Since we are theorists, we are not going to motivate our choices by

experiments, but we will just write them down.

1 Overview

We begin with a quick overview of the things that exist in the standard model.

This description will mostly be words, and the actual theory will have to come

quite some time later.

– Forces are mediated by spin 1 gauge bosons. These include

◦

The electromagnetic field (EM ), which is mediated by the photon.

This is described by quantum electrodynamics (QED);

◦

The weak interaction, which is mediated by the

and

bosons;

and

◦

The strong interaction, which is mediated by gluons

. This is

described by quantum chromodynamics (QCD).

While the electromagnetic field and weak interaction seem very different,

we will see that at high energies, they merge together, and can be described

by a single gauge group.

–

Matter is described by spin

fermions. These are described by Dirac

spinors. Roughly, we can classify them into 4 “types”, and each type comes

in 3 generations, which will be denoted G1, G2, G3 in the following table,

which lists which forces they interact with, alongside with their charge.

Type G1 G2 G3 Charge EM Weak Strong

Charged leptons e µ τ −1 3 3 7

Neutrinos ν

0 7 3 7

Positive quarks u c t +

3 3 3

Negative quarks d s b −

3 3 3

The first two types are known as leptons, while the latter two are known

as quarks. We do not know why there are three generations. Note that a

particle interacts with the electromagnetic field iff it has non-zero charge,

since the charge by definition measures the strength of the interaction with

the electromagnetic field.

–

There is the Higgs boson, which has spin 0. This is responsible for giving

mass to the

, Z

bosons and fermions. This was just discovered in 2012

in CERN, and subsequently more properties have been discovered, e.g. its

spin.

As one would expect from the name, the gauge bosons are manifestations of

local gauge symmetries. The gauge group in the Standard Model is

SU(3)

× SU(2)

× U(1)

We can talk a bit more about each component. The subscripts indicate what

things the group is responsible for. The subscript on

(3)

means “colour”,

and this is responsible for the strong force. We will not talk about this much,

because it is complicated.

The remaining

(2)

U(1)

bit collectively gives the electroweak inter-

action. This is a unified description of electromagnetism and the weak force.

These are a bit funny. The

(2)

is a chiral interaction. It only couples to

left-handed particles, hence the

. The U(1)

is something we haven’t heard of

(probably), namely the hypercharge, which is conventionally denoted

. Note

that while electromagnetism also has a U(1) gauge group, it is different from

this U(1) we see.

Types of symmetry

One key principle guiding our study of the standard model is symmetry. Sym-

metries can manifest themselves in a number of ways.

(i)

We can have an intact symmetry, or exact symmetry. In other words,

this is an actual symmetry. For example, U(1)

and

(3)

are exact

symmetries in the standard model.

(ii)

Symmetries can be broken by an anomaly. This is a symmetry that exists

in the classical theory, but goes away when we quantize. Examples include

global axial symmetry for massless spinor fields in the standard model.

(iii)

Symmetry is explicitly broken by some terms in the Lagrangian. This is

not a symmetry, but if those annoying terms are small (intentionally left

vague), then we have an approximate symmetry, and it may also be useful

to consider these.

For example, in the standard model, the up and down quarks are very

close in mass, but not exactly the same. This gives to the (global) isospin

symmetry.

(iv)

The symmetry is respected by the Lagrangian

, but not by the vacuum.

This is a “hidden symmetry”.

(a)

We can have a spontaneously broken symmetry: we have a vacuum

expectation value for one or more scalar fields, e.g. the breaking of

SU(2)

× U(1)

into U(1)

(b)

Even without scalar fields, we can get dynamical symmetry breaking

from quantum effects. An example of this in the standard model is

the SU(2)

× SU(2)

global symmetry in the strong interaction.

One can argue that (i) is the only case where we actually have a symmetry, but

the others are useful to consider as well, and we will study them.

2 Chiral and gauge symmetries

We begin with the discussion of chiral and gauge symmetries. These concepts

should all be familiar from the Michaelmas QFT course. But it is beneficial

to do a bit of review here. While doing so, we will set straight our sign and

notational conventions, and we will also highlight the important parts in the

theory.

As always, we will work in natural units

= 1, and the sign convention

is (+, −, −, −).

2.1 Chiral symmetry

Chiral symmetry is something that manifests itself when we have spinors. Since

all matter fields are spinors, this is clearly important.

The notion of chirality is something that exists classically, and we shall begin

by working classically. A Dirac spinor is, in particular, a (4-component) spinor

field. As usual, the Dirac matrices

are 4

4 matrices satisfying the Clifford

algebra relations

{γ

, γ

} = 2g

µν

where g

µν

is the Minkowski metric. For any operator A

, we write

A = γ

Then a Dirac fermion is defined to be a spinor field satisfying the Dirac equation

∂ − m)ψ = 0.

We define

= +iγ

which satisfies

(γ

)

= I, {γ

, γ

} = 0.

One can do a lot of stuff without choosing a particular basis/representation for

the

-matrices, and the physics we get out must be the same regardless of which

representation we choose, but sometimes it is convenient to pick some particular

representation to work with. We’ll generally use the chiral representation (or

Weyl representation), with



0 1

1 0



, γ



0 σ

−σ



, γ



−1 0

0 1



where the σ

∈ Mat

Chirality

is a particularly interesting matrix to consider. We saw that it satisfies

(

)

= 1. So we know that

is diagonalizable, and the eigenvalues of

are

either +1 or −1.

Definition

(Chirality)

A Dirac fermion

is right-handed if

, and

left-handed if γ

ψ = −ψ.

A left- or right-handed fermion is said to have definite chirality.

In general, a fermion need not have definite chirality. However, as we know

from linear algebra, the spinor space is a direct sum of the eigenspaces of

. So

given any spinor ψ, we can write it as

ψ = ψ

+ ψ

where ψ

is left-handed, and ψ

is right-handed.

It is not hard to find these ψ

and ψ

. We define the projection operators

(1 + γ

), P

(1 − γ

It is a direct computation to show that

= −P

, γ

= P

, (P

R,L

)

= P

R,L

, P

+ P

= I.

Thus, we can write

ψ = (P

+ P

)ψ = (P

ψ) + (P

ψ),

and thus we have

Notation.

= P

ψ, ψ

= P

ψ.

Moreover, we notice that

= P

= 0.

This implies

Lemma. If ψ

is left-handed and φ

is right-handed, then

= 0.

Proof. We only do the left-handed case.

= ψ

†

= (P

)

†

)

= ψ

†

= ψ

†

= 0,

using the fact that {γ

, γ

} = 0.

The projection operators look very simple in the chiral representation. They

simply look like



I 0

0 0



, P



0 0

0 I



Now we have produced these

and

, but what are they? In particular, are

they themselves Dirac spinors? We notice that the anti-commutation relations

give

∂γ

ψ = −γ

∂ψ.

Thus,

does not commute with the Dirac operator (

∂ −m

), and one can check

that in general,

and

do not satisfy the Dirac equation. But there is one

exception. If the spinor is massless, so

= 0, then the Dirac equation becomes

∂ψ = 0.

If this holds, then we also have

∂γ

−γ

∂ψ

= 0. In other words,

is still

a Dirac spinor, and hence

Proposition. If ψ is a massless Dirac spinor, then so are ψ

and ψ

More generally, if the mass is non-zero, then the Lagrangian is given by

L =

ψ(i

∂ − m)ψ =

∂ψ

− m(

In general, it “makes sense” to treat ψ

and ψ

as separate objects only if the

spinor field is massless. This is crucial. As mentioned in the overview, the weak

interaction only couples to left-handed fermions. For this to “make sense”, the

fermions must be massless! But the electrons and fermions we know and love

are not massless. Thus, the masses of the electron and fermion terms cannot

have come from a direct mass term in the Lagrangian. They must obtain it via

some other mechanism, namely the Higgs mechanism.

To see an example of how the mass makes such a huge difference, by staring

at the Lagrangian, we notice that if the fermion is massless, then we have a

U(1)

global symmetry — under an element (

, α

)

∈

U(1)

the fermion transforms as





7→



iα



The adjoint field transforms as





7→



−iα



and we see that the Lagrangian is invariant. However, if we had a massive

particle, then we would have the cross terms in the Lagrangian. The only way

for the Lagrangian to remain invariant is if

, and the symmetry has

reduced to a single U(1) symmetry.

Quantization of Dirac field

Another important difference the mass makes is the notion of helicity. This is

a quantum phenomenon, so we need to describe the quantum theory of Dirac

fields. When we quantize the Dirac field, we can decompose it as

ψ =

s,p



(p)u

(p)e

−ip·x

+ d

s†

(p)v

(p)e

−ip·x



We explain these things term by term:

– s is the spin, and takes values s = ±

– The summation over all p is actually an integral

(2π)

(2E

)

– b

†

and

†

operators that create positive and negative frequency particles

respectively. We use relativistic normalization, where the states

|pi = b

†

(p) |0i

satisfy

hp|qi = (2π)

(3)

(p − q).

–

The

(

) and

(

) form a basis of the solution space to the (classical)

Dirac equation, so that

(p)e

−ip·x

, v

(p)e

ip·x

are solutions for any

and

. In the chiral representation, we can write

them as

(p) =



√

p · σξ

√

p · ¯σξ



, v

(p) =



√

p · ση

−

√

p · ¯ση



where as usual

= (I, σ

), ¯σ

= (I, −σ

and {ξ

} and {η

} are bases for R

We can define a quantum operator corresponding to the chirality, known as

helicity.

Definition

(Helicity)

We define the helicity to be the projection of the angular

momentum onto the direction of the linear momentum:

h = J ·

p = S ·

where

J = −ir × ∇ + S

is the total angular momentum, and S is the spin operator given by

ijk



0 σ



The main claim about helicity is that for a massless spinor, it reduces to the

chirality, in the following sense:

Proposition. If we have a massless spinor u, then

hu(p) =

u(p).

Proof. Note that if we have a massless particle, then we have

pu = 0,

since quantumly,

is just given by differentiation. We write this out explicitly

to see

= (γ

− γ · p)u = 0.

Multiplying it by γ

gives

u(p) = γ

u(p).

Again since the particle is massless, we know

)

− p · p = 0.

p = p/p

. Also, by direct computation, we find that

= 2S

So it follows that

u(p) = 2hu(p).

In particular, we have

L,R

= ∓

L,R

So u

L,R

has helicity ∓

This is another crucial observation. Helicity is the spin in the direction of the

momentum, and spin is what has to be conserved. On the other hand, chirality is

what determines whether it interacts with the weak force. For massless particles,

these to notions coincide. Thus, the spin of the particles participating in weak

interactions is constrained by the fact that the weak force only couples to left-

handed particles. Consequently, spin conservation will forbid certain interactions

from happening.

However, once our particles have mass, then helicity is different from spin.

In general, helicity is still quite closely related to spin, at least when the mass is

small. So the interactions that were previously forbidden are now rather unlikely

to happen, but are actually possible.

2.2 Gauge symmetry

Another important aspect of the Standard Model is the notion of a gauge

symmetry. Classically, the Dirac equation has the gauge symmetry

ψ(x) 7→ e

iα

ψ(x)

for any constant

, i.e. this transformation leaves all observable physics un-

changed. However, if we allow

to vary with

, then unsurprisingly, the kinetic

term in the Dirac Lagrangian is no longer invariant. In particular, it transforms

ψi

∂ψ 7→

ψi

∂ψ −

ψγ

ψ∂

α(x).

To fix this problem, we introduce a gauge covariant derivative D

that transforms

ψ(x) 7→ e

iα(x)

ψ(x).

Then we find that

ψi

Dψ transforms as

ψi

Dψ 7→

ψi

Dψ.

So if we replace every

∂

with D

in the Lagrangian, then we obtain a gauge

invariant theory.

To do this, we introduce a gauge field A

(x), and then define

ψ(x) = (∂

+ igA

)ψ(x).

We then assert that under a gauge transformation

(

), the gauge field

transforms as

7→ A

−

∂

α(x).

It is then a routine exercise to check that D

transforms as claimed.

If we want to think of

(

) as some physical field, then it should have a

kinetic term. The canonical choice is

= −

µν

where

µν

= ∂

− ∂

, D

We call this a U(1) gauge theory, because

iα

is an element of U(1). Officially,

is an element of the Lie algebra

(1), but it is isomorphic to

, so we did

not bother to make this distinction.

What we have here is a rather simple overview of how gauge theory works.

In reality, we find the weak field couples only with left-handed fields, and we

have to modify the construction accordingly.

Moreover, once we step out of the world of electromagnetism, we have to

work with a more complicated gauge group. In particular, the gauge group will

be non-abelian. Thus, the Lie algebra

has a non-trivial bracket, and it turns

out the right general formulation should include some brackets in

µν

and the

transformation rule for A

. We will leave these for a later time.

3 Discrete symmetries

We are familiar with the fact that physics is invariant under Lorentz transfor-

mations and translations. These were relatively easy to understand, because

they are “continuous symmetries”. It is possible to “deform” any such transfor-

mation continuously (and even smoothly) to the identity transformation, and

thus to understand these transformations, it often suffices to understand them

“infinitesimally”.

There is also a “trivial” reason why these are easy to understand — the

“types” of fields we have, namely vector fields, scalar fields etc. are defined by

how they transform under change of coordinates. Consequently, by definition,

we know how vector fields transform under Lorentz transformations.

In this chapter, we are going to study discrete symmetries. These cannot be

understood by such means, and we need to do a bit more work to understand

them. It is important to note that these discrete “symmetries” aren’t actually

symmetries of the universe. Physics is not invariant under these transformations.

However, it is still important to understand them, and in particular understand

how they fail to be symmetries.

We can briefly summarize the three discrete symmetries we are interested in

as follows:

– Parity (P ): (t, x) 7→ (t, −x)

– Time-reversal (T ): (t, x) 7→ (−t, x)

–

Charge conjugation (C ): This sends particles to anti-particles and vice

versa.

Of course, we can also perform combinations of these. For example, CP corre-

sponds to first applying the parity transformation, and then applying charge

conjugation. It turns out none of these are symmetries of the universe. Even

worse, any combination of two transformations is not a symmetry. We will

discuss these violations later on as we develop our theory.

Fortunately, the combination of all three, namely CPT, is a symmetry of a

universe. This is not (just) an experimental observation. It is possible to prove

(in some sense) that any (sensible) quantum field theory must be invariant under

CPT, and this is known as the CPT theorem.

Nevertheless, for the purposes of this chapter, we will assume that C, P, T

are indeed symmetries, and try to derive some consequences assuming this were

the case.

The above description of P and T tells us how the universe transforms

under the transformations, and the description of the charge conjugation is just

some vague words. The goal of this chapter is to figure out what exactly these

transformations do to our fields, and, ultimately, what they do to the

-matrix.

Before we begin, it is convenient to rephrase the definition of P and T as

follows. A general Poincar´e transformation can be written as a map

7→ x

0µ

= Λ

+ a

A proper Lorentz transform has

det

Λ = +1. The transforms given by parity and

time reversal are improper transformations, and are given by

Definition (Parity transform). The parity transform is given by

= P







1 0 0 0

0 −1 0 0

0 0 −1 0

0 0 0 −1







Definition

(Time reversal transform)

The time reversal transform is given by







−1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1







3.1 Symmetry operators

How do these transformations of the universe relate to our quantum-mechanical

theories? In quantum mechanics, we have a state space

, and the fields are

operators

(

t, x

) :

H → H

for each (

t, x

)

∈ R

1,3

. Unfortunately, we do not have

a general theory of how we can turn a classical theory into a quantum one, so

we need to make some (hopefully natural) assumptions about how this process

behaves.

The first assumption is absolutely crucial to get ourselves going.

Assumption.

Any classical transformation of the universe (e.g. P or T) gives

rise to some function f : H → H.

In general

need not be a linear map. However, it turns out if the transfor-

mation is in fact a symmetry of the theory, then Wigner’s theorem forces a lot

of structure on f .

Roughly, Wigner’s theorem says the following — if a function

H → H

a symmetry, then it is linear and unitary, or anti-linear and anti-unitary.

Definition

(Linear and anti-linear map)

Let

be a Hilbert space. A function

f : H → H is linear if

f(αΦ + βΨ) = αf (Φ) + βf(Ψ)

for all α, β ∈ C and Φ, Ψ ∈ H. A map is anti-linear if

f(αΦ + βΨ) = α

∗

f(Φ) + β

∗

f(Ψ).

Definition

(Unitary and anti-unitary map)

Let

be a Hilbert space, and

f : H → H a linear map. Then f is unitary if

hfΦ, fΨi = hΦ, Ψi

for all Φ, Ψ ∈ H.

If f : H → H is anti-linear, then it is anti-unitary if

hfΦ, fΨi = hΦ, Ψi

∗

But to state the theorem precisely, we need to be a bit more careful. In

quantum mechanics, we consider two (normalized) states to be equivalent if they

differ by a phase. Thus, we need the following extra assumption on our function

Assumption.

In the previous assumption, we further assume that if Φ

∈ H

differ by a phase, then fΦ and f Ψ differ by a phase.

Now what does it mean for something to “preserve physics”? In quantum

mechanics, the physical properties are obtained by taking inner products. So

we want to require f to preserve inner products. But this is not quite what we

want, because states are only defined up to a phase. So we should only care

about inner products defined up to a phase. Note that this in particular implies

f preserves normalization.

Thus, we can state Wigner’s theorem as follows:

Theorem

(Wigner’s theorem)

Let

be a Hilbert space, and

H → H

be a

bijection such that

– If Φ, Ψ ∈ H differ by a phase, then fΦ and f Ψ differ by a phase.

– For any Φ, Ψ ∈ H, we have

|hfΦ, fΨi| = |hΦ, Ψi|.

Then there exists a map

H → H

that is either linear and unitary, or

anti-linear and anti-unitary, such that for all Φ

∈ H

, we have that

Φ and

differ by a phase.

For all practical purposes, this means we can assume the transformations C,

P and T are given by unitary or anti-unitary operators.

We will write

and

for the (anti-)unitary transformations induced on

the Hilbert space by the C, P, T transformations. We also write

(Λ

, a

) for the

induced transformation from the (not necessarily proper) Poincar´e transformation

7→ x

0µ

= Λ

+ a

We want to understand which are unitary and which are anti-unitary. We will

assume that these W (Λ, a) “compose properly”, i.e. that

W (Λ

, a

)W (Λ

, a

) = W (Λ

, Λ

+ a

Moreover, we assume that for infinitesimal transformations

= δ

+ ω

, a

= ε

where ω and ε are small parameters, we can expand

W = W (Λ, a) = W (I + ω, ε) = 1 +

µν

+ iε

where

µν

are the operators generating rotations and boosts, and

are the

operators generating translations. In particular, P

is the Hamiltonian.

Of course, we cannot write parity and time reversal in this form, because

they are discrete symmetries, but we can look at what happens when we combine

these transformations with infinitesimal ones.

By assumption, we have

P = W (P, 0),

T = W (T, 0).

Then from the composition rule, we expect

P W

−1

= W (PΛP

−1

, Pa)

T W

−1

= W (TΛT

−1

, Ta).

Inserting expansions for

in terms of

and

on both sides, and comparing

coefficients of ε

, we find

P iH

−1

= iH

T iH

−1

= −iH.

So iH and

P commute, but iH and

T anti-commute.

To proceed further, we need to make the following assumption, which is a

natural one to make if we believe P and T are symmetries.

Assumption.

The transformations

and

send an energy eigenstate of energy

E to an energy eigenstate of energy E.

From this, it is easy to figure out whether the maps

and

should be

unitary or anti-unitary.

Indeed, consider any normalized energy eigenstate Ψ with energy

E 6

= 0.

Then by definition, we have

hΨ, iHΨi = iE.

Then since

P Ψ is also an energy eigenstate of energy E, we know

iE = h

P Ψ, iH

P Ψi = h

P Ψ,

P iHΨi.

In other words, we have

P Ψ,

P iHΨi = hΨ, iHΨi = iE.

P must be unitary.

On the other hand, we have

iE = h

T Ψ, iH

T Ψi = −h

T Ψ,

T iHΨi.

In other words, we obtain

T Ψ,

T iHΨi = −hΨ, iHΨi = hΨ, iHΨi

∗

− iE.

Thus, it follows that

must be anti-unitary. It is a fact that

is linear and

unitary.

Note that these derivations rely crucially on the fact that we know the

operators must either by unitary or anti-unitary, and this allows us to just check

one inner product to determine which is the case, rather than all.

So far, we have been discussing how these operators act on the elements of

the state space. However, we are ultimately interested in understanding how

the fields transform under these transformations. From linear algebra, we know

that an endomorphism

H → H

on a vector space canonically induces a

transformation on the space of operators, by sending

W φW

−1

. Thus, what

we want to figure out is how W φW

−1

relates to φ.

3.2 Parity

Our objective is to figure out how

P = W (P, 0)

acts on our different quantum fields. For convenience, we will write

7→ x

= (x

, −x)

7→ p

= (p

, −p).

As before, we will need to make some assumptions about how

behaves.

Suppose our field has creation operator a

†

(p). Then we might expect

|pi 7→ η

∗

for some complex phase η

∗

. We can alternatively write this as

P a

†

(p) |0i = η

∗

†

) |0i.

We assume the vacuum is parity-invariant, i.e.

P |0i

|0i

. So we can write

this as

P a

†

(p)

−1

|0i = η

∗

†

) |0i.

Thus, it is natural to make the following assumption:

Assumption.

Let

†

(

) be the creation operator of any field. Then

†

(

)

transforms as

P a

†

(p)

−1

= η

∗

†

)

for some η

∗

. Taking conjugates, since

P is unitary, this implies

P a(p)

−1

= ηa(p

We will also need to assume the following:

Assumption.

Let

be any field. Then

P φ

(

)

−1

is a multiple of

(

), where

“a multiple” can be multiplication by a linear map in the case where

has more

than 1 component.

Scalar fields

Consider a complex scalar field

φ(x) =



a(p)e

−ip·x

+ c(p)

†

+ip·x



where

(

) is an annihilation operator for a particle and

(

)

†

is a creation

operator for the anti-particles. Then we have

P φ(x)

−1



P a(p)

−1

−ip·x

P c

†

(p)

−1

+ip·x





∗

a(p

−ip·x

+ η

∗

†

+ip·x



Since we are integrating over all p, we can relabel p

↔ p, and then get



a(p)e

−ip

·x

+ η

∗

†

(p)e

+ip

·x



We now note that x · p

= x

· p by inspection. So we have



a(p)e

−ip·x

+ η

∗

†

(p)e

ip·x



By assumption, this is proportional to

(

). So we must have

∗

≡ η

Then we just get

P φ(x)

−1

= η

φ(x

Definition

(Intrinsic parity)

The intrinsic parity of a field

is the number

∈ C such that

P φ(x)

−1

= η

φ(x

For real scalar fields, we have

, and so

∗

So η

= ±1.

Definition

(Scalar and pseudoscalar fields)

A real scalar field is called a

scalar field (confusingly) if the intrinsic parity is

−

1. Otherwise, it is called a

pseudoscalar field.

Note that under our assumptions, we have

by the composition rule

of the

(Λ

, a

). Hence, it follows that we always have

1. However, in

more sophisticated treatments of the theory, the composition rule need not hold.

The above analysis still holds, but for a complex scalar field, we need not have

= ±1.

Vector fields

Similarly, if we have a vector field V

(x), then we can write

(x) =

p,λ

(λ, p)a

(p)e

−ip·x

+ E

µ∗

(λ, p)c

†λ

(p)e

ip·x

where E

(λ, p) are some polarization vectors.

Using similar computations, we find

P V

−1

p,λ



(λ, p

(p)e

−ip·x

+ E

µ∗

(λ, p

†λ

(p)e

+ip·x

∗



This time, we have to deal with the

(

λ, p

) term. Using explicit expressions

for E

, we have

(λ, p

) = −P

(λ, p).

So we find that

P V

)

−1

= −η

where for the same reasons as before, we have

= η

∗

Definition

(Vector and axial vector fields)

Vector fields are vector fields with

= −1. Otherwise, they are axial vector fields.

Dirac fields

We finally move on to the case of Dirac fields, which is the most complicated.

Fortunately, it is still not too bad.

As before, we obtain

P ψ(x)

−1

p,s



(p)u(p

−p·x

+ η

∗

s†

(p)v

+ip·x



We use that

) = γ

(p), v

) = −γ

(p),

which we can verify using Lorentz boosts. Then we find

P ψ(x)

−1

= γ

p,s



(p)u(p)e

−p·x

− η

∗

s†

(p)v

(p)e

+ip·x



So again, we require that

= −η

∗

We can see this minus sign as saying particles and anti-particles have opposite

intrinsic parity.

Unexcitingly, we end up with

P ψ(x)

−1

= η

ψ(x

Similarly, we have

ψ(x)

−1

= η

∗

ψ(x

)γ

Since γ

anti-commutes with γ

, it follows that we have

P ψ

−1

= η

So the parity operator exchanges left-handed and right-handed fermions.

Fermion bilinears

We can now determine how various fermions bilinears transform. For example,

we have

ψ(x)ψ(x) 7→

ψ(x

)ψ(x

So it transforms as a scalar. On the other hand, we have

ψ(x)γ

ψ(x) 7→ −

ψ(x

)γ

ψ(x

and so this transforms as a pseudoscalar. We also have

ψ(x)γ

ψ(x) 7→ P

ψ(x

)γ

ψ(x

and so this transforms as a vector. Finally, we have

ψ(x)γ

ψ(x) 7→ −P

ψ(x

)γ

ψ(x

So this transforms as an axial vector.

3.3 Charge conjugation

We will do similar manipulations, and figure out how different fields transform

under charge conjugation. Unlike parity, this is not a spacetime symmetry. It

transforms particles to anti-particles, and vice versa. This is a unitary operator

C, and we make the following assumption:

Assumption.

is an annihilation operator of a particle, and

is the annihi-

lation operator of the anti-particle, then we have

Ca(p)

−1

= ηc(p)

for some η.

As before, this is motivated by the requirement that

C |pi

∗

|¯pi

. We will

also assume the following:

Assumption.

Let

be any field. Then

Cφ

(

)

−1

is a multiple of the conjugate

, where “a multiple” can be multiplication by a linear map in the case where

has more than 1 component. Here the interpretation of “the conjugate” depends

on the kind of field:

–

is a bosonic field, then the conjugate of

∗

= (

†

)

. Of course, if

this is a scalar field, then the conjugate is just φ

†

– If φ is a spinor field, then the conjugate of φ is

Scalar and vector fields

Scalar and vector fields behave in a manner very similar to that of parity

operators. We simply have

Cφ

−1

= η

†

Cφ

†

−1

= η

∗

for some

. In the case of a real field, we have

†

. So we must have

which is known as the intrinsic

-parity of the field. For complex fields, we can

introduce a global phase change of the field so as to set η

= 1.

This has some physical significance. For example, the photon field transforms

(x)

−1

= −A

(x).

Experimentally, we see that

only decays to 2 photons, but not 1 or 3. Therefore,

assuming that c-parity is conserved, we infer that

= (−1)

= +1.

Dirac fields

Dirac fields are more complicated. As before, we can compute

Cψ(x)

−1

= η

p,s



(p)u

(p)e

−ip·x

+ b

s†

(p)v

(p)e

+ip·x



We compare this with

(x) =

p,s



s†

(p)¯u

(p)e

+ip·x

+ d

(p)¯v

(p)e

−ip·x



We thus see that we have to relate ¯u

and v

somehow; and ¯v

with u

Recall that we constructed

and

in terms of elements

, η

∈ R

. At

that time, we did not specify what

and

are, or how they are related, so we

cannot expect

and

to have any relation at all. So we need to make a specific

choice of ξ and η. We will choose them so that

= iσ

s∗

Now consider the matrix

C = −iγ



iσ

0 −iσ



The reason we care about this is the following:

Proposition.

(p) = C ¯u

, u

(p) = C¯v

(p).

From this, we infer that

Proposition.

(x) ≡

Cψ(x)

−1

= η

(x)

(x) ≡

ψ(x)

−1

= η

∗

(x)C = −η

∗

(x)C

−1

It is convenient to note the following additional properties of the matrix C:

Proposition.

(Cγ

)

= Cγ

C = −C

= −C

†

= −C

−1

(γ

)

= −Cγ

−1

, (γ

)

= +Cγ

−1

Note that if ψ(x) satisfies the Dirac equation, then so does ψ

(x).

Apart from Dirac fermions, there are also things called Majorana fermions.

They have

(

) =

(

). This means that the particle is its own antiparticle,

and for these, we have

(x) = ψ(x).

These fermions have to be neutral. A natural question to ask is — do these

exist? The only type of neutral fermions in the standard model is neutrinos, and

we don’t really know what neutrinos are. In particular, it is possible that they

are Majorana fermions. Experimentally, if they are indeed Majorana fermions,

then we would be able to observe neutrino-less double

decay, but current

experiments can neither observe nor rule out this possibility.

Fermion bilinears

We can look at how fermion bilinears change. For example,

ψ(x)γ

ψ(x).

Then we have

(x)

−1

Cψ

−1

= −η

∗

−1

We now notice that the

∗

and

cancel each other. Also, since this is a scalar,

we can take the transpose of the whole thing, but we will pick up a minus sign

because fermions anti-commute

ψ(C

−1

ψC

µT

−1

)

= −

ψγ

= −j

(x).

Therefore A

(x)j

(x) is invariant under

C. Similarly,

ψγ

ψ 7→ +

ψγ

ψ.

3.4 Time reversal

Finally, we get to time reversal. This is more messy. Under T, we have

= (x

, x

) 7→ x

= (−x

, x

Momentum transforms in the opposite way, with

= (p

, p

) 7→ p

= (p

, −p

Theories that are invariant under

look the same when we run them backwards.

For example, Newton’s laws are time reversal invariant. We will also see that

the electromagnetic and strong interactions are time reversal invariant, but the

weak interaction is not.

Here our theory begins to get more messy.

Assumption. For any field φ, we have

T φ(x)

−1

∝ φ(x

Assumption. For bosonic fields with creator a

†

, we have

T a(p)

−1

= ηa(p

)

for some η. For Dirac fields with creator b

s†

, we have

T b

(p)

−1

= η(−1)

1/2−s

−s

)

for some η.

Why that complicated rule for Dirac fields? We first justify why we want

to swap spins when we perform time reversal. This is just the observation that

when we reverse time, we spin in the opposite direction. The factor of (

−

−s

tells us spinors of different spins transform differently.

Boson field

Again, bosonic fields are easy. The creation and annihilation operators transform

T a(p)

−1

= η

a(p

)

T c

†

(p)

−1

= η

†

Note that the relative phases are fixed using the same argument as for

and

. When we do the derivations, it is very important to keep in mind that

anti-linear.

Then we have

T φ(x)

−1

= η

φ(x

Dirac fields

As mentioned, for Dirac fields, the annihilation and creation operators transform

T b

(p)

−1

= η

(−1)

1/2−s

−s

)

T d

s†

(p)

−1

= η

(−1)

1/2−s

−s†

)

It can be shown that

(−1)

−s

−s∗

) = −Bu

(p)

(−1)

−s

−s∗

) = −Bv

(p),

where

B = γ

C =



iσ

0 iσ



in the chiral representation.

Then, we have

T ψ(x)

−1

= η

p,s

(−1)

−s



−s

s∗

(p)e

+ip·x

+ d

−s†

s∗

(p)e

−ip·x



Doing the standard manipulations, we find that

T ψ(x)

−1

= η

p,s

(−1)

−s+1



(p)u

−s∗

−ip·x

+ d

s†

(p)v

−s∗

+ip·x



= +η

p,s



(p)BU

(p)e

−ip·x

+ d

s†

(p)BV

(p)e

+ip·x



= η

Bψ(x

Similarly,

ψ(x)

−1

= η

∗

ψ(x

−1

Fermion bilinears

Similarly, we have

ψ(x)ψ(x) 7→

ψ(x

)ψ(x

)

ψ(x)γ

ψ(x) 7→ −T

ψ(x

)γ

ψ(x

)

This uses the fact that

−1

µ∗

B = −T

So we see that in the second case, the 0 component, i.e. charge density, is

unchanged, while the spacial component, i.e. current density, gets a negative

sign. This makes physical sense.

3.5 S-matrix

We consider how S-matrices transform. Recall that S is defined by

, p

, ···|S |k

, k

, ···i =

out

, p

, ···|k

, k

, ···i

= lim

T →∞

, p

, ···|e

−iH2T

, k

, ···i

We can write

S = T exp



−i

∞

−∞

V (t) dt



where T denotes the time-ordered integral, and

V (t) = −

x L

(x),

and L

(x) is the interaction part of the Lagrangian.

Example. In QED, we have

= −e

ψ(x)γ

(x)ψ(x).

We can draw a table of how things transform:

P C T

(x) L

) L

(x) L

)

V (t) V (t) V (t) V (−t)

S S S ??

A bit more care is to figure out how the

matrix transforms when we reverse

time, as we have a time-ordering in the integral. To figure out, we explicitly

write out the time-ordered integral. We have

S =

∞

n=0

(−i)

∞

−∞

···

n−1

−∞

V (t

)V (t

) ···V (t

Then we have

T S

−1

(+i)

∞

−∞

···

n−1

−∞

V (−t

)V (−t

) ···V (−t

)

We now put τ

= −t

n+1−2

, and then we can write this as

(+i)

∞

−∞

−τ

−∞

···

−τ

−∞

V (τ

)V (τ

n−1

) ···V (τ

)

(+i)

∞

−∞

dτ

∞

dτ

n−1

···

∞

dτ

V (τ

)V (τ

n−1

) ···V (τ

We now notice that

∞

−∞

dτ

∞

dτ

n−1

∞

−∞

dτ

n−1

−∞

dτ

We can see this by drawing the picture

n−1

= τ

So we find that

(+i)

∞

−∞

dτ

−∞

dτ

···

n−1

−∞

dτ

V (τ

)V (τ

n−1

) ···V (τ

We can then see that this is equal to S

†

What does this tell us? Consider states |ηi and |ξi with

|η

i =

T |ηi

|ξ

i =

T |ξi

The Dirac bra-ket notation isn’t very useful when we have anti-linear operators.

So we will write inner products explicitly. We have

(η

, Sξ

) = (

T η, S

T ξ)

= (

T η, S

†

T ξ)

= (

T η,

T S

†

, ξ)

= (η, S

†

ξ)

∗

= (ξ, Sη)

where we used the fact that

T is anti-unitary. So the conclusion is

hη

|S |ξ

i = hξ|S |ηi.

So if

T L

(x)

−1

= L

then the

-matrix elements are equal for time-reversed processes, where the

initial and final states are swapped.

3.6 CPT theorem

Theorem

(CPT theorem)

Any Lorentz invariant Lagrangian with a Hermitian

Hamiltonian should be invariant under the product of P, C and T.

We will not prove this. For details, one can read Streater and Wightman,

PCT, spin and statistics and all that (1989).

All observations we make suggest that CPT is respected in nature. This

means a particle propagating forwards in time cannot be distinguished from the

antiparticle propagating backwards in time.

3.7 Baryogenesis

In the universe, we observe that there is much more matter than anti-matter.

Baryogenesis is the generation of this asymmetry in the universe. Sakarhov came

up with three conditions that are necessary for this to happen.

(i)

Baryon number violation (or leptogenesis, i.e. lepton number asymmetry),

i.e. some process X → Y + B that generates an excess baryon.

(ii)

Non-equilibrium. Otherwise, the rate of

B → X

is the same as the

rate of X → Y + B.

(iii)

C and CP violation. We need C violation, or else the rate of

X → Y

and the rate of

X →

would be the same, and the effects cancel out

each other.

We similarly need CP violation, or else the rate of

X → nq

and the rate

X → nq

, where

L,R

are left and right handed quarks, is equal to the

rate of ¯x → n¯q

and ¯x → n¯q

, and this will wash out our excess.

4 Spontaneous symmetry breaking

In this chapter, we are going to look at spontaneous symmetry breaking. The

general setting is as follows — our Lagrangian

enjoys some symmetry. However,

the potential has multiple minima. Usually, in perturbation theory, we imagine

we are sitting near an equilibrium point of the system (the “vacuum”), and then

look at what happens to small perturbations near the equilibrium.

When there are multiple minima, we have to arbitrarily pick a minimum

to be our vacuum, and then do perturbation around it. In many cases, this

choice of minimum is not invariant under our symmetries. Thus, even though

the theory itself is symmetric, the symmetry is lost once we pick a vacuum. It

turns out interesting things happen when this happens.

4.1 Discrete symmetry

We begin with a toy example, namely that of a discrete symmetry. Consider a

real scalar field φ(x) with a symmetric potential V (φ), so that

V (−φ) = V (φ).

This gives a discrete Z/2Z symmetry φ ↔ −φ.

We will consider the case of a φ

theory, with Lagrangian

L =

∂

φ∂

φ −



−



for some λ.

We want the potential to

→ ∞

φ → ∞

, so we necessarily require

λ >

However, since the

term dominates for large

, we are now free to pick the

sign of m

, and still get a sensible theory.

Usually, this theory has m

> 0, and thus V (φ) has a minimum at φ = 0:

S(φ)

However, we could imagine a scenario where

0, where we have “imaginary

mass”. In this case, the potential looks like

V (φ)

To understand this potential better, we complete the square, and write it as

V (φ) =

(φ

− v

)

+ constant,

where

v =

−

We see that now

= 0 becomes a local maximum, and there are two (global)

minima at

±v

. In particular,

has acquired a non-zero vacuum expectation

value (VEV ).

We shall wlog consider small excitations around φ = v. Let’s write

φ(x) = v + cf(x).

Then we can write the Lagrangian as

L =

∂

f∂

f − λ



+ +vf



plus some constants. Therefore f is a scalar field with mass

= 2v

λ.

This

is not invariant under

f → −f

. The symmetry of the original Lagrangian

has been broken by the VEV of φ.

4.2 Continuous symmetry

We can consider a slight generalization of the above scenario. Consider an

N-component real scalar field φ = (φ

, φ

, ··· , φ

)

, with Lagrangian

L =

(∂

φ) · (∂

φ) − V (φ),

where

V (φ) =

As before, we will require that λ > 0.

This is a theory that is invariant under global O(

) transforms of

. Again,

0, then

= 0 is a global minimum, and we do not have spontaneous

symmetry breaking. Thus, we consider the case m

< 0.

In this case, we can write

V (φ) =

(φ

− v

)

+ constant,

where

v = −

> 0.

So any

with

gives a global minimum of the system, and so we have a

continuum of vacua. We call this a Sombrero potential (or wine bottle potential).

V (φ)

Without loss of generality, we may choose the vacuum to be

= (0, 0, ··· , 0, v)

We can then consider steady small fluctuations about this:

φ(x) = (π

(x), π

(x), ··· , π

N−1

(x), v + σ(x))

We now have a π field with N − 1 components, plus a 1-component σ field.

We rewrite the Lagrangian in terms of the π and σ fields as

L =

(∂

π) ·(∂

π) +

(∂

σ) ·(∂

σ) −V (π, σ),

where

V (π, σ) =

+ λv(σ

+ π

)σ +

(σ

+ π

)

Again, we have dropped a constant term.

In this Lagrangian, we see that the σ field has mass

√

2λv

but the

fields are massless. We can understand this as follows — the

field

corresponds to the radial direction in the potential, and this is locally quadratic,

and hence has a mass term. However, the

fields correspond to the excitations

in the azimuthal directions, which are flat.

4.3 General case

What we just saw is a completely general phenomenon. Suppose we have an

-dimensional field

= (

, ··· , φ

), and suppose our theory has a Lie group

of symmetries

. We will assume that the action of

preserves the potential

and kinetic terms individually, and not just the Lagrangian as a whole, so that

G will send a vacuum to another vacuum.

Suppose there is more than one choice of vacuum. We write

= {φ

: V (φ

) = V

min

}

for the set of all vacua. Now we pick a favorite vacuum

∈

, and we want

to look at the elements of G that fix this vacuum. We write

H = stab(φ

) = {h ∈ G : hφ

= φ

This is the invariant subgroup, or stabilizer subgroup of

, and is the symmetry

we are left with after the spontaneous symmetry breaking.

We will make some further simplifying assumptions. We will suppose

acts

transitively on Φ

, i.e. given any two vacua

, φ

, we can find some

g ∈ G

such

that

= gφ

This ensures that any two vacua are “the same”, so which

we pick doesn’t really

matter. Indeed, given two such vacua and

relating them, it is a straightforward

computation to check that

stab(φ

) = g stab(φ

−1

So any two choices of vacuum will result in conjugate, and in particular isomor-

phic, subgroups H.

It is a curious, and also unimportant, observation that given a choice of

vacuum φ

, we have a canonical bijection

G/H Φ

g gφ

∼

So we can identify G/H and Φ

. This is known as the orbit-stabilizer theorem.

We now try to understand what happens when we try to do perturbation

theory around our choice of vacuum. At φ = φ

+ δφ, we can as usual write

V (φ

+ δφ) − V (φ

) =

δφ

∂

∂φ

δφ

+ O(δφ

This quadratic ∂

V term is now acting as a mass. We call it the mass matrix

∂

∂φ

Note that we are being sloppy with where our indices go, because (

)

is ugly.

It doesn’t really matter much, since

takes values in

(or

), which is a

Euclidean space.

In our previous example, we saw that there were certain modes that went

massless. We now want to see if the same happens. To do so, we pick a basis

{

} of h (the Lie algebra of H), and then extend it to a basis {t

, θ

} of g.

We consider a variation

δφ = εθ

around φ

Note that when we write

, we mean the resulting infinitesimal change of

when

acts on field. For a general Lie algebra element, this may be zero.

However, we know that

does not belong to

, so it does not fix

. Thus

(assuming sensible non-degeneracy conditions) this implies that

is non-zero.

At this point, the uncareful reader might be tempted to say that since

is a

symmetry of V , we must have V (φ

+ δφ) = V (φ

), and thus

(θ

φ)

(θ

φ)

= 0,

and so we have found zero eigenvectors of

. But this is wrong. For example,

if we take

M =



1 0

0 −1



, θ

φ =





then our previous equation holds, but

is not a zero eigenvector. We need to

be a bit more careful.

Instead, we note that by definition of

-invariance, for any field value

and

a corresponding variation

φ 7→ φ + εθ

φ,

we must have

V (φ + δφ) = V (φ),

since this is what it means for G to be a symmetry.

Expanding to first order, this says

V (φ + δφ) − V (φ) = ε(θ

φ)

∂V

∂φ

= 0.

Now the trick is to take a further derivative of this. We obtain

0 =

∂

∂φ



(θ

φ)

∂V

∂φ





∂

∂φ

(θ

φ)



∂V

∂φ

+ (θ

φ)

Now at a vacuum φ = φ

, the first term vanishes. So we must have

(θ

)

= 0.

By assumption,

= 0. So we have found a zero vector of

. Recall that

we had

dim G − dim H

of these

in total. So we have found

dim G − dim H

zero eigenvectors in total.

We can do a small sanity check. What if we replaced

with

? The same

derivations would still follow through, and we deduce that

)

= 0.

But since

is an unbroken symmetry, we actually just have

= 0, and this

really doesn’t say anything interesting.

What about the other eigenvalues? They could be zero for some other reason,

and there is no reason to believe one way or the other. However, generally, in

the scenarios we meet, they are usually massive. So after spontaneous symmetry

breaking, we end up with

dim G−dim H

massless modes, and

N−

(

dim G−dim H

)

massive ones.

This is Goldstone’s theorem in the classical case.

Example.

In our previous

(

) model, we had

(

), and

(

N −

1).

These have dimensions

dim G =

N(N − 1)

, dim H =

(N − 1)(N − 2)

Then we have

∼

N−1

the (N − 1)-dimensional sphere. So we expect to have

N − 1

(N − (N − 2)) = N − 1

massless modes, which are the

fields we found. We also found

N −

(

N −

1) = 1

massive mode, namely the σ field.

Example.

In the first discrete case, we had

and

{e}

. These are

(rather degenerate) 0-dimensional Lie groups. So we expect 0

−

0 = 0 massless

modes, and 1 − 0 = 1 massive modes, which is what we found.

4.4 Goldstone’s theorem

We now consider the quantum version of Goldstone’s theorem. We hope to get

something similar.

Again, suppose our Lagrangian has a symmetry group

, which is sponta-

neously broken into

H ≤ G

after picking a preferred vacuum

|0i

. Again, we take

a Lie algebra basis

, θ

}

, where

}

is a basis for

. Note that we will

assume that

runs from 1

, ··· , dim G

, and we use the first

dim H

labels to refer

to the t

, and the remaining to label the θ

, so that the actual basis is

, t

, ··· , t

dim H

, θ

dim H+1

, ··· , θ

dim G

For aesthetic reasons, this time we put the indices on the generators below.

Now recall, say from AQFT, that these symmetries of the Lagrangian give

rise to conserved currents j

(x). These in turn give us conserved charges

x j

(x).

Moreover, under the action of the

and

, the corresponding change in

given by

δφ = −[Q

, φ].

The fact that the θ

break the symmetry of |0i tells us that, for a > dim H,

h0|[Q

, φ(0)] |0i 6= 0.

Note that the choice of evaluating

at 0 is completely arbitrary, but we have

to pick somewhere, and 0 is easy to write. Writing out the definition of

, we

know that

x h0|[j

(x), φ(0)] |0i 6= 0.

More generally, since the label

is arbitrary by Lorentz invariance, we are really

looking at the relation

h0|[j

(x), φ(0)] |0i 6= 0,

and writing it in this more Lorentz covariant way makes it easier to work with.

The goal is to deduce the existence of massless states from this non-vanishing.

For convenience, we will write this expression as

= h0|[j

(x), φ(0)] |0i.

We treat the two terms in the commutator separately. The first term is

h0|j

(x)φ(0) |0i =

h0|j

(x) |nihn|φ(0) |0i. (∗)

We write p

for the momentum of |ni. We now note the operator equation

(x) = e

iˆp·x

(0)e

−iˆp·x

So we know

h0|j

(x) |ni = h0|j

(0) |nie

−ip

·x

We can use this to do some Fourier transform magic. By direct verification, we

can see that (∗) is equivalent to

(2π)

(k)e

−ik·x

where

iρ

(k) = (2π)

(k − p

) h0|j

(0) |nihn|φ(0) |0i.

Similarly, we define

i˜ρ

(k) = (2π)

(k − p

) h0|φ(0) |nihn|j

(0) |0i.

Then we can write

= i

(2π)



(k)e

−ik·x

− ˜ρ

(k)e

+ik·x



This is called the K¨allen-Lehmann spectral representation.

We now claim that

(

) and

˜ρ

(

) can be written in a particularly nice

form. This involves a few observations (when I say

, I actually mean

and

˜ρ

– ρ

depends Lorentz covariantly in

. So it “must” be a scalar multiple of

– ρ

(k) must vanish when k

> 0, since k

≤ 0 is non-physical.

–

By Lorentz invariance, the magnitude of

(

) can depend only on the

length (squared) k

of k.

Under these assumptions, we can write can write ρ

and ˜ρ

(k) = k

Θ(k

)ρ

˜ρ

(k) = k

Θ(k

)˜ρ

where Θ is the Heaviside theta function

Θ(x) =

(

1 x > 0

0 x ≤ 0

So we can write

= i

(2π)

Θ(k

)(ρ

−ik·x

− ˜ρ

+ik·x

We now use the (nasty?) trick of hiding the

by replacing it with a derivative:

= −∂

(2π)

Θ(k

)



−ik·x

+ ˜ρ

ik·x



Now we might find the expression inside the integral a bit familiar. Recall that

the propagator is given by

D(x, σ) = h0|φ(x)φ(0) |0i =

(2π)

Θ(k

)δ(k

− σ)e

−ik·x

where σ is the square of the mass of the field. Using the really silly fact that

) =

dσ ρ

(σ)δ(k

− σ),

we find that

= −∂

dσ (ρ

(σ)D(x, σ) + ˜ρ

(σ)D(−x, σ)).

Now we have to recall more properties of

. For spacelike

, we have

(

x, σ

) =

(

−x, σ

). Therefore, requiring

to vanish for spacelike

by causality, we see

that we must have

(σ) = −˜ρ

(σ).

Therefore we can write

= −∂

dσ ρ

(σ)i∆(x, σ), (†)

where

i∆(x, σ) = D(x, σ) − D(−x, σ) =

(2π)

δ(k

− σ)ε(k

−ik·x

and

ε(k

) =

(

+1 k

> 0

−1 k

< 0

This ∆ is again a different sort of propagator.

We now use current conservation, which tells us

∂

= 0.

So we must have

∂

= −∂

dσ ρ

(σ)i∆(x, σ) = 0.

On the other hand, by inspection, we see that ∆ satisfies the Klein-Gordon

equation

(∂

+ σ)∆ = 0.

So in (†), we can replace −∂

∆ with σ∆. So we find that

∂

dσ σρ

(σ)i∆(x, σ) = 0.

This is true for all

. In particular, it is true for timelike

where ∆ is non-zero.

So we must have

σρ

(σ) = 0.

But we also know that

(

) cannot be identically zero, because

is not. So

the only possible explanation is that

(σ) = N

δ(σ),

where N

is a dimensionful non-zero constant.

Now we retrieve our definitions of ρ

. Recall that they are defined by

iρ

(k) = (2π)

(k − p

) h0|j

(0) |nihn|φ(0) |0i

(k) = k

Θ(k

)ρ

So the fact that

(

) =

(

) implies that there must be some states, which

we shall call |B(p)i, of momentum p, such that p

= 0, and

h0|j

(0) |B(p)i 6= 0

hB(p)|φ(0) |0i 6= 0.

The condition that

= 0 tell us these particles are massless, which are those

massless modes we were looking for! These are the Goldstone bosons.

We can write the values of the above as

h0|j

(0) |B(p)i = iF

hB(p)|φ(0) |0i = Z

whose form we deduced by Lorentz in/covariance. By dimensional analysis, we

know F

is a dimension −1 constant, and Z

is a dimensionless constant.

From these formulas, we note that since

(0)

|0i

is rotationally invariant, we

deduce that |B(p)i also is. So we deduce that these are in fact spin 0 particles.

Finally, we end with some computations that relate these different numbers

we obtained. We will leave the computational details for the reader, and just

outline what we should do. Using our formula for ρ, we find that

h0|[j

(x), φ(0)] |0i = −∂

δ(σ)i∆(x, σ) dσ = −iN

∂

∆(x, 0).

Integrating over space, we find

h0|[Q

, φ(0)] |0i = −iN

x ∂

∆(x, 0) = iN

Then we have

h0|t

φ(0) |0i = i h0|[Q

, φ(0)] |0i = i · iN

= −N

So this number

sort-of measures how much the symmetry is broken, and we

once again see that it is the breaking of symmetry that gives rise to a non-zero

, hence the massless bosons.

We also recall that we had

Θ(k

δ(k

) =

2|p|

(k − p) h0|j

(0) |B(p)ihB(p)|φ(0) |0i.

On the RHS, we can plug in the names we gave for the matrix elements, and on

the left, we can write it as an integral

2|p|

(k − p)ik

2|p|

(k − p)ip

So we must have

We can repeat this process for each independent symmetry-breaking

∈ g \ h

and obtain a Goldstone boson this way. So all together, at least superficially, we

find

n = dim G − dim H

many Goldstone bosons.

It is important to figure out the assumptions we used to derive the result.

We mentioned “Lorentz invariance” many times in the proof, so we certainly

need to assume our theory is Lorentz invariant. Another perhaps subtle point is

that we also needed states to have a positive definite norm. It turns out in the

case of gauge theory, these do not necessarily hold, and we need to work with

them separately.

4.5 The Higgs mechanism

Recall that we had a few conditions for our previous theorem to hold. In the case

of gauge theories, these typically don’t hold. For example, in QED, imposing

a Lorentz invariant gauge condition (e.g. the Lorentz gauge) gives us negative

norm states. On the other hand, if we fix the gauge condition so that we don’t

have negative norm states, this breaks Lorentz invariance. What happens in this

case is known as the Higgs mechanism.

Let’s consider the case of scalar electrodynamics. This involves two fields:

– A complex scalar field φ(x) ∈ C.

– A 4-vector field A(x) ∈ R

1,3

As usual the components of

(

) are denoted

(

). From this we define the

electromagnetic field tensor

µν

= ∂

− ∂

and we have a covariant derivative

= ∂

+ iqA

As usual, the Lagrangian is given by

L = −

µν

+ (D

φ)

∗

φ) − V (|φ|

)

for some potential V .

A U(1) gauge transformation is specified by some

(

)

∈ R

. Then the fields

transform as

φ(x) 7→ e

iα(x)

φ(x)

(x) 7→ A

(x) −

∂

α(x).

We will consider a φ

theory, so that the potential is

V (|φ|

) = µ

|φ|

+ λ|φ|

As usual, we require

λ >

0, and if

0, then this is boring with a unique

vacuum at φ = 0. In this case, A

is massless and φ is massive.

If instead µ

< 0, then this is the interesting case. We have a minima at

|φ

= −

2λ

≡

Without loss of generality, we expand around a real φ

, and write

φ(x) =

√

iθ(x)/v

(v + η(x)),

where η, θ are real fields.

Now we notice that

isn’t a “genuine” field (despite being real). Our theory

is invariant under gauge transformations. Thus, by picking

α(x) = −

θ(x),

we can get rid of the θ(x) term, and be left with

φ(x) =

√

(v + η(x)).

This is called the unitary gauge. Of course, once we have made this choice, we

no longer have the gauge freedom, but on the other hand everything else going

on becomes much clearer.

In this gauge, the Lagrangian can be written as

L =

(∂

η∂

η + 2µ

) −

µν

+ L

int

where L

int

is the interaction piece that involves more than two fields.

We can now just read off what is going on in here.

– The η field is massive with mass

= −2µ

= 2λv

> 0.

– The photon now gains a mass

= q

–

What would be the Goldstone boson, namely the

field, has been “eaten”

to become the longitudinal polarization of

and

has gained a degree

of freedom (or rather, the gauge non-degree-of-freedom became a genuine

degree of freedom).

One can check that the interaction piece becomes

int

+ qm

η −

− m

So we have interactions that look like

This η field is the “Higgs boson” in our toy theory.

4.6 Non-abelian gauge theories

We’ll not actually talk about spontaneous symmetry in non-abelian gauge theories.

That is the story of the next chapter, where we start studying electroweak theory.

Instead, we’ll just briefly set up the general framework of non-abelian gauge

theory.

In general, we have a gauge group

, which is a compact Lie group with

Lie algebra

. We also have a representation of

, which we will assume

is unitary, i.e. each element of

is represented by a unitary matrix. Our field

ψ(x) ∈ C

takes values in this representation.

A gauge transformation is specified by giving a

(

)

∈ G

for each point

the universe, and then the field transforms as

ψ(x) 7→ g(x)ψ(x).

Alternatively, we can represent this transformation infinitesimally, by producing

some t(x) ∈ g, and then the transformation is specified by

ψ(x) 7→ exp(it(x))ψ(x).

Associated to our gauge theory is a gauge field

(

)

∈ g

(i.e. we have an element

for each

and

), which transforms under an infinitesimal transformation

t(x) ∈ g as

(x) 7→ −∂

t(x) + [t, A

The gauge covariant derivative is again give by

= ∂

+ igA

where all fields are, of course, implicitly evaluated at

. As before, we can define

µν

= ∂

− ∂

− g[A

, A

] ∈ g,

Alternatively, we can write this as

, D

] = igF

µν

We will later on work in coordinates. We can pick a basis of

, say

, ··· , t

Then we can write an arbitrary element of

(

)

. Then in this basis, we

write the components of the gauge field as

, and similarly the field strength is

denoted F

µν

. We can define the structure constants f

abc

, t

] = f

abc

Using this notation, the gauge part of the Lagrangian L is

= −

µν

aµν

= −

Tr(F

µν

We now move on to look at some actual theories.

5 Electroweak theory

We can now start discussing the standard model. In this chapter, we will account

for everything in the theory apart from the strong force. In particular, we will

talk about the gauge (electroweak) and Higgs part of the theory, as well as the

matter content.

The description of the theory will be entirely in the classical world. It is only

when we do computations that we quantize the theory and compute

-matrices.

5.1 Electroweak gauge theory

We start by understanding the gauge and Higgs part of the theory. As mentioned,

the gauge group is

G = SU(2)

× U(1)

We will see that this is broken by the Higgs mechanism.

It is convenient to pick a basis for su(2), which we will denote by

Note that these are not genuinely elements of

(2). We need to multiply these

to actually get members of

(2) (thus, they form a complex basis for the

complexification of

(2)). As we will later see, these act on fields as

iτ

instead

of e

. Under this basis, the structure constants are f

abc

= ε

abc

We start by describing how our gauge group acts on the Higgs field.

Definition

(Higgs field)

The Higgs field

is a complex scalar field with

two components,

(

)

∈ C

. The

(2) action is given by the fundamental

representation on C

, and the hypercharge is Y =

Explicitly, an (infinitesimal) gauge transformation can be represented by

elements

(

)

, β

(

)

∈ R

, corresponding to the elements

(

)

∈ su

(2) and

β(x) ∈ u(1)

∼

R. Then the Higgs field transform as

φ(x) 7→ e

iα

(x)τ

β(x)

φ(x),

where the

factor of β(x) comes from the hypercharge being

Note that when we say

is a scalar field, we mean it transforms trivially via

Lorentz transformations. It still takes values in the vector space C

The gauge fields corresponding to

(2) and U(1) are denoted

and

where again

runs through

= 1

3. The covariant derivative associated to

these gauge fields is

= ∂

+ igW

for some coupling constants g and g

The part of the Lagrangian relating the gauge and Higgs field is

gauge,φ

= −

Tr(F

µν

W,µν

) −

µν

B,µν

+ (D

φ)

†

φ) − µ

|φ|

− λ|φ|

where the field strengths of W and B respectively are given by

W,a

µν

= ∂

− ∂

− gε

abc

µν

= ∂

− ∂

In the case

0, the Higgs field acquires a VEV, and we wlog shall choose

the vacuum to be

√





where

= −λv

< 0.

As in the case of U(1) symmetry breaking, the gauge term gives us new things

with mass. After spending hours working out the computations, we find that

φ)

†

φ) contains mass terms



)

+ g

)

+ (−gW

+ g



This suggests we define the following new fields:

√

∓ iW

)







cos θ

−sin θ

sin θ

cos θ





where we pick θ

such that

cos θ

+ g

, sin θ

+ g

This θ

is called the Weinberg angle. Then the mass term above becomes



((W

)

+ (W

−

)

) +

+ g

)



Thus, our particles have masses

+ g

= 0,

where

is the

particle, which is the photon. Thus, our original

(2)

U(1)

breaks down to a U(1)

symmetry. In terms of the Weinberg angle,

= m

cos θ

Thus, through the Higgs mechanism, we find that the

and

bosons gained

mass, but the photon does not. This agrees with what we find experimentally.

In practice, we find that the masses are

≈ 80 GeV

≈ 91 GeV

< 10

−18

GeV.

Also, the Higgs boson gets mass, as what we saw previously. It is given by

−2µ

√

2λv

Note that the Higgs mass depends on the constant

, which we haven’t seen

anywhere else so far. So we can’t tell

from what we know about

and

Thus, until the Higgs was discovered recently, we didn’t know about the mass of

the Higgs boson. We now know that

≈ 125 GeV.

Note that we didn’t write out all terms in the Lagrangian, but as we did before,

we are going to get W

, Z-Higgs and Higgs-Higgs interactions.

One might find it a bit pointless to define the

and

−

fields as we did,

as the kinetic terms looked just as good with

and

. The advantage of

this new definition is that

is now the complex conjugate

−

, so we can

instead view the

bosons as given by a single complex scalar field, and when

we quantize, W

will be the anti-particle of W

−

5.2 Coupling to leptons

We now look at how gauge bosons couple to matter. In general, for a particle

with hypercharge Y , the covariant derivative is

= ∂

+ igW

+ ig

Y B

In terms of the W

and Z bosons, we can write this as

= ∂

√

+ W

−

) +

igZ

cos θ

(cos

− sin

Y )

+ ig sin θ

(τ

+ Y ),

where

= τ

± iτ

By analogy, we interpret the final term

ig sin θ

(

) is the usual coupling

with the photon. So we can identify the (magnitude of the) electron charge as

e = g sin θ

while the U(1)

charge matrix is

Q = U(1)

= τ

+ Y.

If we want to replace all occurrences of the hypercharge

with

, then we need

to note that

cos

− sin

Y = τ

− sin

We now introduce the electron field. The electron field is given by a spinor field

e(x). We will decompose it as left and right components

e(x) = e

(x) + e

(x).

There is also a neutrino field

(

). We will assume that neutrinos are massless,

and there are only left-handed neutrinos. We know this is not entirely true,

because of neutrinos oscillation, but the mass is very tiny, and this is a very

good approximation.

To come up with an actual electroweak theory, we need to specify a represen-

tation of SU(2) × U(1). It is convenient to group our particles by handedness:

R(x) = e

(x), L(x) =



(x)



Here

(

) is a single spinor, while

(

) consists of a pair, or (

(2)) doublet of

spinors.

We first look at

(

). Experimentally, we find that

only couples to

left-handed leptons. So

(

) will have the trivial representation of

(2). We

also know that electrons have charge

−

1. Since

acts trivially on

, we

must have

Q = Y = −1

for the right-handed leptons.

The left-handed particles are more complicated. We will assert that

has

the “fundamental” representation of SU(2), by which we mean given a matrix

g =



a b

−

b ¯a



∈ SU(2),

it acts on L as

gL =



a b

−

b ¯a







aν

+ be

−

bν

+ ¯ae



We know the electron has charge

−

1 and the neutrino has charge 0. So we need

Q =



0 0

0 −1



Since Q = τ

+ Y , we must have Y = −

Using this notation, the gauge part of the lepton Lagrangian can be written

concisely as

lepton

DL +

DR.

Note that

means we take the transpose of the matrix

, viewed as a matrix

with two components, and then take the conjugate of each spinor individually,

so that

L =



¯ν

(x) ¯e

(x)



How about the mass? Recall that it makes sense to deal with left and right-

handed fermions separately only if the fermion is massless. A mass term

(¯e

+ ¯e

)

would be very bad, because our SU(2) action mixes e

with ν

But we do know that the electron has mass. The answer is that the mass

is again granted by the spontaneous symmetry breaking of the Higgs boson.

Working in unitary gauge, we can write the Higgs boson as

φ(x) =

√



v + h(x)



with v, h(x) ∈ R.

The lepton-Higgs interactions is given by

lept,φ

= −

√

2λ

(

LφR +

Rφ

†

L),

where

is the Yukawa coupling. It is helpful to make it more explicit what we

mean by this expression. We have

LφR =



¯ν

(x) ¯e

(x)





√



v + h(x)



(x) =

√

(v + h(x))¯e

(x)e

(x).

Similarly, we can write out the second term and obtain

lept,φ

= −λ

(v + h)(¯e

+ ¯e

)

= −m

¯ee − λ

h¯ee,

where

= λ

We see that while the interaction term is a priori gauge invariant, once we have

spontaneously broken the symmetry, we have magically obtained a mass term

. The second term in the Lagrangian is the Higgs-fermion coupling. We see

that this is proportional to m

. So massive things couple more strongly.

We now return to the fermion-gauge boson interactions. We can write it as

EM,int

lept

√

+ J

µ†

−

) + eJ

2 cos θ

where

= −¯eγ

= ¯ν

(1 − γ



¯ν

(1 − γ

)ν

− ¯eγ

(1 − γ

− 4 sin



Note that

has a negative sign because the electron has negative charge.

These currents are known as the EM current, charge weak current and neutral

weak current respectively.

There is one thing we haven’t mentioned so far. We only talked about

electrons and neutrinos, but the standard model has three generations of leptons.

There are the muon (

) and tau (

), and corresponding left-handed neutrinos.

With these in mind, we introduce





, L





, L





= e

, R

= µ

, R

= τ

These couple and interact in exactly the same way as the electrons, but have

heavier mass. It is straightforward to add these into the

EW,int

lept

term. What is

more interesting is the Higgs interaction, which is given by

lept,φ

= −

√



φR

+ (λ

†

)

†



where

and

run over the different generations. What’s new now is that we

have the matrices λ ∈ M

(C). These are not predicted by the standard model.

This

is just a general matrix, and there is no reason to expect it to be

diagonal. However, in some sense, it is always diagonalizable. The key insight

is that we contract the two indices

with two different kinds of things. It is a

general linear algebra fact that for any matrix

at all, we can find some unitary

matrices U and S such that

λ = U ΛS

†

where Λ is a diagonal matrix with real entries. We can then transform our fields

7→ U

7→ S

and this diagonalizes L

lept,φ

It is also clear that this leaves

lept

invariant, especially if we look at the

expression before symmetry breaking. So the mass eigenstates are the same as

the “weak eigenstates”. This tells us that after diagonalizing, there is no mixing

within the different generations.

This is important. It is not possible for quarks (or if we have neutrino mass),

and mixing does occur in these cases.

5.3 Quarks

We now move on to study quarks. There are 6 flavours of quarks, coming in

three generations:

Charge First generation Second generation Third generation

Up (u) Charm (c) Top (t)

−

Down (d) Strange (s) Bottom (b)

each of which is a spinor.

The right handed fields have trivial

(2) representations, and are thus

(2)

singlets. We write them as





which have Y = Q = +

, and





which have Y = Q = −

The left-handed fields are in SU(2) doublets

















and these all have Y =

. Here i = 1, 2, 3 labels generations.

The electroweak part of the Lagrangian is again straightforward, given by

quark

+ ¯u

It takes a bit more work to couple these things with

. We want to do so in a

gauge invariant way. To match up the

(2) part, we need a term that looks

φ,

as both

and

have the fundamental representation. Now to have an

invariant U(1) theory, the hypercharges have to add to zero, as under a U(1)

gauge transformation

(

), the term transforms as

β(x)

. We see that the

term

φd

works well. However, coupling

with

is more problematic.

has

hypercharge

−

and

has hypercharge +

. So we need another term of

hypercharge −

. To do so, we introduce the charge-conjugated φ, defined by

(φ

)

= ε

αβ

∗β

One can check that this transforms with the fundamental

(2) representation,

and has

−

. Inserting generic coupling coefficients

u,d

, we write the

Lagrangian as

quark,φ

= −

√

2(λ

φd

+ λ

+ h.c.).

Here h.c. means “hermitian conjugate”.

Let’s try to diagonalize this in the same way we diagonalized leptons. We

again work in unitary gauge, so that

φ(x) =

√



v + h(x)



Then the above Lagrangian can be written as

quark,φ

= −(λ

(v + h)d

+ λ

¯u

(v + h)u

+ h.c.).

We now write

= U

†

= U

†

where Λ

u,d

are diagonal and

u,d

, U

u,d

are unitary. We can then transform the

field in a similar way:

7→ U

, d

7→ U

, u

7→ S

, d

7→ S

and then it is a routine check that this diagonalizes

quark,φ

. In particular, the

mass term looks like

−

+ m

¯u

+ h.c.),

where

= vΛ

How does this affect our electroweak interactions? The

¯u

and

are staying fine, but the different components of

“differently”. In particular,

the

piece given by

is not invariant. That piece, can be explicitly

written out as

√

±µ

where

µ+

= ¯u

Under the basis transformation, this becomes

¯u

†

)

This is not going to be diagonal in general. This leads to inter-generational

quark couplings. In other words, we have discovered that the mass eigenstates

are (in general) not equal to the weak eigenstates.

The mixing is dictated by the Cabibbo–Kabyashi–Maskawa matrix (CKM

matrix )

CKM

= U

†

Explicitly, we can write

CKM









where the subscript indicate which two things it mixes. So far, these matrices

are not predicted by any theory, and is manually plugged into the model.

However, the entries aren’t completely arbitrary. Being the product of two

unitary matrices, we know

CKM

is a unitary matrix. Moreover, the entries are

only uniquely defined up to some choice of phases, and this further cuts down

the number of degrees of freedom.

If we only had two generations, then we get what is known as Cabibbo

mixing. A general 2

2 unitary matrix has 4 real parameters — one angle and

three phases. However, redefining each of the 4 quark fields with a global U(1)

transformation, we can remove three relative phases (if we change all of them by

the same phase, then nothing happens). We can then write this matrix as

V =



cos θ

sin θ

−sin θ

cos θ



where

is the Cabibbo angle. Experimentally, we have

sin θ

≈

22. It turns

out the reality of this implies that CP is conserved, which is left as an exercise

on the example sheet.

In this case, we explicitly have

µ†

= cos θ

¯u

+ sin θ

¯u

− sin θ

¯c

+ cos θ

¯c

If the angle is 0, then we have no mixing between the generations.

With three generations, there are nine parameters. We can think of this as

3 (Euler) angles and 6 phases. As before, we can re-define some of the quark

fields to get rid of five relative phases. We can then write

CKM

in terms of

three angles and 1 phase. In general, this

CKM

is not real, and this gives us

CP violation. Of course, if it happens that this phase is real, then we do not

have CP violation. Unfortunately, experimentally, we figured this is not the case.

By the CPT theorem, we thus deduce that T violation happens as well.

5.4 Neutrino oscillation and mass

Since around 2000, we know that the mass eigenstates and weak eigenstates for

neutrinos are not equivalent, as neutrinos were found to change from one flavour

to another. This implies there is some mixing between different generations of

leptons. The analogous mixing matrix is the Pontecorov–Maki–Nakagawa–Sakata

matrix , written U

P M NS

As of today, we do not really understand what neutrinos actually are, as

neutrinos don’t interact much, and we don’t have enough experimental data. If

neutrinos are Dirac fermions, then they behave just like quarks, and we expect

CP violation.

However, there is another possibility. Since neutrinos do not have a charge,

it is possible that they are their own anti-particles. In other words, they are

Majorana fermions. It turns out this implies that we cannot get rid of that many

phases, and we are left with 3 angles and 3 phases. Again, we get CP violation

in general.

We consider these cases briefly in turn.

Dirac fermions

If they are Dirac fermions, then we must also get some right-handed neutrinos,

which we write as

= ν

= (ν

, ν

µR

, ν

τR

Then we modify the Dirac Lagrangian to say

lept,φ

= −

√

2(λ

φR

+ λ

+ h.c.).

This is exactly like for quarks. As in quarks, we obtain a mass term.

−

(¯ν

+ ¯ν

Majorana neutrinos

If neutrinos are their own anti-particles, then, in the language we had at the

beginning of the course, we have

(p) = b

(p).

Then ν(x) = ν

(x) + ν

(x) must satisfy

(x) = C ¯ν

= ν(x).

Then we see that we must have

(x) = ν

(x),

and vice versa. So the right-handed neutrino field is not independent of the

left-handed field. In this case, the mass term would look like

−

(¯ν

+ ¯ν

As in the case of leptons, postulating a mass term directly like this breaks gauge

invariance. Again, we solve this problem by coupling with the Higgs field. It

takes some work to find a working gauge coupling, and it turns out it the simplest

thing that works is

L,φ

= −

)C(φ

) + h.c..

This is weird, because it is a dimension 5 operator. This dimension 5 operator is

non-renormalizable. This is actually okay, as along as we think of the standard

model as an effective field theory, describing physics at some low energy scale.

5.5 Summary of electroweak theory

We can do a quick summary of the electroweak theory. We start with the picture

before spontaneous symmetry breaking.

The gauge group of this theory is

(2)

U(1)

, with gauge fields

∈

(2) and

∈ u

(1). The coupling of U(1) with the particles is specified by a

hypercharge

, and the

(2) couplings will always be trivial or fundamental.

The covariant derivative is given by

= ∂

+ igW

Y B

where

are the canonical generators of

(2). The field strengths are given by

W,a

µν

= ∂

− ∂

− gε

abc

µν

= ∂

− ∂

The theory contains the scalar Higgs field

φ ∈ C

, which has hypercharge

and the fundamental

(2) representation. We also have three generations of

matter, given by

Type G1 G2 G2 Q Y

Positive quarks u c t +

Negative quarks d s b −

−

Leptons (e, µ, τ) e µ τ −1 −

−1

Leptons (neutrinos) ν

0 −

???

Here G1, G2, G3 are the three generations,

is the charge,

is the hypercharge

of the left-handed version, and

is the hypercharge of the right-handed version.

Each of these matter fields is a spinor field.

From now on, we describe the theory in the case of a massless neutrino,

because we don’t really know what the neutrinos are. We group the matter fields

L =

 

 

 

R = ( e

)

= ( u

)

= ( d

)

 

 

 

The Lagrangian has several components:

– The kinetic term of the gauge field is given by

gauge

= −

Tr(F

µν

W,µν

) −

µν

B,µν

– The Higgs field couples with the gauge fields by

gauge,φ

= (D

φ)

†

φ) − µ

|φ|

− λ|φ|

After spontaneous symmetry breaking, this gives rise to massive

, Z

and Higgs bosons. This also gives us

, Z

-Higgs interactions, as well as

gives us Higgs-Higgs interactions.

– The leptons interact with the Higgs field by

lept,φ

= −

√

2(λ

φR

+ h.c.).

This gives us lepton masses and lepton-Higgs interactions. Of course, this

piece has to be modified after we figure out what neutrinos actually are.

– The gauge coupling of the leptons induce

lept

DL +

DR,

with an implicit sum over all generations. This gives us lepton interactions

with

, Z, γ

. Once we introduce neutrino masses, this is described by

the PMNS matrix, and gives us neutrino oscillations and (possibly) CP

violation.

– Higgs-quark interactions are given by

quark,φ

= −

√

2(λ

φd

+ λ

+ h.c.).

which gives rise to quark masses.

– Finally, the gauge coupling of the quarks is given by

quark

+ ¯u

which gives us quark interactions with

, Z, γ

. The interactions are

described by the CKM matrix. This gives us quark flavour and CP

violation.

We now have all of the standard model that involves the electroweak part

and the matter. Apart from QCD, which we will describe quite a bit later, we’ve

got everything in the standard model.

6 Weak decays

We have mostly been describing (the electroweak part of) the standard model

rather theoretically. Can we actually use it to make predictions? This is what

we will do in this chapter. We will work out decay rates for certain processes,

and see how they compare with experiments. One particular point of interest in

our computations is to figure out how CP violation manifest itself in decay rates.

6.1 Effective Lagrangians

We’ll only consider processes where energies and momentum are much less than

, m

. In this case, we can use an effective field theory. We will discuss

more formally what effective field theories are, but we first see how it works in

practice.

In our case, what we’ll get is the Fermi weak Lagrangian. This Lagrangian

in fact predates the Standard Model, and it was only later on that we discovered

the Fermi weak Lagrangian is an effective Lagrangian of what we now know of

as electroweak theory.

Recall that the weak interaction part of the Lagrangian is

= −

√

+ J

µ†

−

) −

2 cos θ

Our general goal is to compute the S-matrix

S = T exp



x L

(x)



As before,

denotes time-ordering. The strategy is to Taylor expand this in

Ultimately, we will be interested in computing

hf|S |ii

for some initial and final states

|ii

and

|fi

. Since we are at the low energy

regime, we will only attempt to compute these quantities when

|ii

and

|fi

do not

contain

bosons. This allows us to drop terms in the Taylor expansion

having free W

or Z components.

We can explicitly Taylor expand this, keeping the previous sentence in mind.

How can we possibly get rid of the

and

terms in the Taylor expansion?

If we think about Wick’s theorem, we know that when we take the time-ordered

product of several operators, we sum over all possible contractions of the fields,

and contraction practically means we replace two operators by the Feynman

propagator of that field.

Thus, if we want to end up with no

term, we need to contract all

the

and

fields together. This in particular requires an even number of

and

terms. So we know that there is no

(

) term left, and the first

non-trivial term is O(g

). We write the propagators as

µν

(x − x

) = hT W

−

(x)W

µν

(x − x

) = hT Z

(x)Z

)i.

Thus, the first interesting term is

one. For initial and final states

|ii

and

|fi

we have

hf|S |ii = hf|T



1 −

x d



µ†

(x)D

µν

(x − x

)

cos

µ†

µν

(x − x

)



+ O(g

)



|ii

As always, we work in momentum space. We define the Fourier transformed

propagator

Z,W

µν

(p) by

Z,W

µν

(x − y) =

(2π)

−ip·(x−y)

Z,W

µν

(p),

and we will later compute to find that

µν

Z,W

µν

(p) =

− m

Z,W

+ iε

−g

µν

Z,W

Here

µν

is the metric of the Minkowski space. We will put aside the computation

of the propagator for the moment, and discuss consequences.

At low energies, e.g. the case of quarks and leptons (except for the top quark),

the momentum scales involved are much less than

Z,W

. In this case, we can

approximate the propagators by ignoring all the terms involving

. So we have

Z,W

µν

(p) ≈

µν

Z,W

Plugging this into the Fourier transform, we have

Z,W

µν

(x − y) ≈

µν

Z,W

(x − y).

What we see is that we can describe this interaction by a contact interaction,

i.e. a four-fermion interaction. Note that if we did not make the approximation

p ≈

0, then our propagator will not have the

(4)

(

x − y

), hence the effective

action is non-local.

Thus, the second term in the S-matrix expansion becomes

−



µ†

(x)J

(x) +

cos

µ†

(x)J

nµ

(x)



We want to define the effective Lagrangian

eff

to be the Lagrangian not involving

, Z such that for “low energy states”, we have

hf|S |ii = hf|T exp



x L

eff



|ii

= hf|T



1 + i

x L

eff

+ ···



|ii

Based on our previous computations, we find that up to tree level, we can write

eff

(x) ≡ −

√



µ†

(x) + ρJ

µ†

nµ

(x)



where, again up to tree level,

√

, ρ =

cos

Recall that when we first studied electroweak theory, we found a relation

cos θ

. So, up to tree level, we have

= 1. When we look at higher levels,

we get quantum corrections, and we can write

ρ = 1 + ∆ρ.

This value is sensitive to physics “beyond the Standard Model”, as the other

stuff can contribute to the loops. Experimentally, we find

∆ρ ≈ 0.008.

We can now do our usual computations, but with the effective Lagrangian rather

than the usual Lagrangian. This is the Fermi theory of weak interaction. This

predates the idea of the standard model and the weak interaction. The

in some sense indicates that Fermi theory breaks down at energy scales near

, as we would expect.

It is interesting to note that the mass dimension of

−

2. This is to

compensate for the dimension 6 operator

µ†

. This means our theory is

non-renormalizable. This is, of course, not a problem, because we do not think

this is a theory that should be valid up to arbitrarily high energy scales. To

derive this Lagrangian, we’ve assumed our energy scales are  m

, m

Computation of propagators

We previously just wrote down the values of the

and

-propagators. We

will now explicitly do the computations of the

propagator. The computation

for

is similar. We will gloss over subtleties involving non-abelian gauge

theory here, ignoring problems involving ghosts etc. We’ll work in the so-called

-gauge.

In this case, we explicitly write the free Z-Lagrangian as

Free

= −

(∂

− ∂

)(∂

− ∂

) +

To find the propagator, we introduce an external current

coupled to

. So

the new Lagrangian is

L = L

free

+ j

(x)Z

(x).

Through some routine computations, we see that the Euler–Lagrange equations

give us

∂

− ∂

∂ · Z + m

= −j

. (∗)

We need to solve this. We take the ∂

of this, which gives

∂

∂ · Z − ∂

∂ · Z + m

∂ · Z = −∂ · j.

So we obtain

∂ · Z = −∂ · j.

Putting this back into (∗) and rearranging, we get

(∂

+ m

= −



µν

∂



We can write the solution as an integral over this current. We write the solution

(x) = i

y D

µν

(x − y)j

(y), (†)

and further write

µν

(x − y) =

(2π)

−ip·(x−y)

µν

(p).

Then by applying (∂

+ m

) to (†), we find that we must have

µν

(p) =

−i

− m

+ iω



µν

−



6.2 Decay rates and cross sections

We now have an effective Lagrangian. What can we do with it? In this section,

we will consider two kinds of experiments:

–

We leave a particle alone, and see how frequently it decays, and into what.

–

We crash a lot of particles into each other, and count the number of

scattering events produced.

The relevant quantities are decay rates and cross sections respectively. We will

look at both of these in turn, but we will mostly be interested in computing

decay rates only.

Decay rate

Definition

(Decay rate)

Let

be a particle. The decay rate Γ

is rate of

decay of

in its rest frame. In other words, if we have a sample of

, then this

is the number of decays of

per unit time divided by the number of

present.

The lifetime of X is

τ ≡

We can write

X→f

where Γ

X→f

is the partial decay rate to the final state f

Note that the

is usually a complicated mixture of genuine sums and

integrals.

Often, we are only interested in how frequently it decays into, a particular

particle, instead of just the total decay ray. Thus, for each final state

, we

want to compute Γ

X→f

, and then sum over all final states we are interested in.

As before, we can compute the S-matrix elements, defined by

hf|S |ii.

We will take

. As before, recall that there is a term 1 of

that corresponds

to nothing happening, and we are not interested in that. If

|fi 6

|ii

, then this

term does not contribute.

It turns out the interesting quantity to extract out of the

-matrix is given

by the invariant amplitude:

Definition (Invariant amplitude). We define the invariant amplitude M by

hf|S − 1 |ii = (2π)

(4)

− p

)iM

When actually computing this, we make use of the following convenient fact:

Proposition. Up to tree order, and a phase, we have

= hf|L(0) |ii,

where L is the Lagrangian. We usually omit the (0).

Proof sketch. Up to tree order, we have

hf|S − 1 |ii = i

x hf |L(x) |ii

We write

in momentum space. Then the only

-dependence in the

hf|L

(

)

|ii

factor is a factor of

i(p

−p

)·x

. Integrating over

introduces a factor of

(2π)

(4)

− p

). Thus, we must have had, up to tree order,

hf|L(x) |ii = M

i(p

−p

)·x

So evaluating this at x = 0 gives the desired result.

How does this quantity enter the picture? If we were to do this naively, we

would expect the probability of a transition i → f is

P (i → f) =

|hf|S − 1 |ii|

hf|fihi|ii

It is not hard to see that we will very soon have a lot of

functions appearing

in this expression, and these are in general bad. As we saw in QFT, these

functions came from the fact that the universe is infinite. So what we do is that

we work with a finite spacetime, and then later take appropriate limits.

We suppose the universe has volume

, and we also work over a finite

temporal extent T . Then we replace

(2π)

(4)

(0) 7→ V T, (2π)

(3)

(0) 7→ V.

Recall that with infinite volume, we found the normalization of our states to be

hi|ii = (2π)

(3)

(0).

In finite volume, we can replace this with

hi|ii = 2p

Similarly, we have

hf|fi =

(2p

V ),

where r runs through all final state labels. In the S-matrix, we have

|hf|S − 1 |ii| =



(2π)

(4)

− p

)



We don’t really have a δ(0) here, but we note that for any x, we have

(δ(x))

= δ(x)δ(0).

The trick here is to replace only one of the

(0) with

V T

, and leave the other

one alone. Of course, we don’t really have a

(

−p

) function in finite volume,

but when we later take the limit, it will become a δ function again.

Noting that p

= m

since we are in the rest frame, we find

P (i → f) =

(2π)

(4)

−

V T





We see that two of our V ’s cancel, but there are a lot left.

The next thing to realize is that it is absurd to ask what is the probability of

decaying into exactly

. Instead, we have a range of final states. We will abuse

notation and use

to denote both the set of all final states we are interested in,

and members of this set. We claim that the right expression is

i→f

P (i → f)



(2π)



The obvious question to ask is why we are integrating against

(2π)

, and

not, say, simply d

. The answer is that both options are not quite right. Since

we have finite volume, as we are familiar from introductory QM, the momentum

eigenstates should be discretized. Thus, what we really want to do is to do an

honest sum over all possible values of p

But of course, we don’t like sums, and since we are going to take the limit

V → ∞

anyway, we replace the sum with an integral, and take into account of

the density of the momentum eigenstates, which is exactly

(2π)

We now introduce a measure

dρ

= (2π)

(4)

−

(2π)

and then we can concisely write

i→f

dρ

Note that when we actually do computations, we need to manually pick what

range of momenta we want to integrate over.

Cross sections

We now quickly look at another way we can do experiments. We imagine we set

two beams running towards each other with velocities

and

respectively.

We let the particle densities be ρ

and ρ

The idea is to count the number of scattering events,

. We will compute

this relative to the incident flux

F = |v

− v

|ρ

which is the number of incoming particles per unit area per unit time.

Definition (Cross section). The cross section is defined by

n = F σ.

Given these, the total number of scattering events is

N = nρ

V = F σρ

V = |v

− v

|ρ

V σ.

This is now a more symmetric-looking expression.

Note that in this case, we genuinely have a finite volume, because we have

to pack all our particles together in order to make them collide. Since we are

boring, we suppose we actually only have one particle of each type. Then we

have

= ρ

In this case, we have

σ =

− v

We can do computations similar to what we did last time. This time there are a

few differences. Last time, the initial state only had one particle, but now we

have two. Thus, if we go back and look at our computations, we see that we will

gain an extra factor of

in the frequency of interactions. Also, since we are

no longer in rest frame, we replace the masses of the particles with the energies.

Then the number of interactions is

N =

(2E

)(2E

dρ

We are often interested in knowing the number of interactions sending us to each

particular momentum (range) individually, instead of just knowing about how

many particles we get. For example, we might be interested in which directions

the final particles are moving in. So we are interested in

dσ =

− v

dN =

− v

|4E

dρ

Experimentalists will find it useful to know that cross sections are usually

measured in units called “barns”. This is defined by

1 barn = 10

−28

6.3 Muon decay

We now look at our first decay process, the muon decay:

−

→ e¯ν

This is in fact the only decay channel of the muon.

−

¯ν

We will make the simplifying assumption that neutrinos are massless.

The relevant bit of L

eff

−

√

α†

where the weak current is

= ¯ν

(1 − γ

)e + ¯ν

(1 − γ

)µ + ¯ν

(1 − γ

)τ.

We see it is the interaction of the first two terms that will render this decay

possible.

To make sure our weak field approximation is valid, we need to make sure

we live in sufficiently low energy scales. The most massive particle involved is

= 105.658 371 5(35) MeV.

On the other hand, the mass of the weak boson is

= 80.385(15) GeV,

which is much bigger. So the weak field approximation should be valid.

We can now compute

M =



−

(k)¯ν

(q)ν

)



eff



−

(p)



Note that we left out all the spin indices to avoid overwhelming notation. We

see that the first term in

is relevant to the electron bit, while the second is

relevant to the muon bit. We can then write this as

M = −

√



−

(k)¯ν

(q)



¯eγ

(1 − γ

)ν

|0ihν

)| ¯ν

(1 − γ

)µ



−

(p)



= −

√

¯u

(k)γ

(1 − γ

(q)¯u

)γ

(1 − γ

(p).

Before we plunge through more computations, we look at what we are interested

in and what we are not.

At this point, we are not interested in the final state spins. Therefore, we

want to sum over the final state spins. We also don’t know the initial spin of

−

So we average over the initial states. For reasons that will become clear later,

we will write the desired amplitude as

spins

|M|

spins

∗

spins



¯u

(k)γ

(1 − γ

(q)¯u

)γ

(1 − γ

(p)





¯u

(p)γ

(1 − γ

)¯v

(q)γ

(1 − γ

(k)



spins



¯u

(k)γ

(1 − γ

(q)¯v

(q)γ

(1 − γ

(k)





¯u

)γ

(1 − γ

(p)¯u

(p)γ

(1 − γ

)



We write this as

αβ

2αβ

where

αβ

spins

¯u

(k)γ

(1 − γ

(q)¯v

(q)γ

(1 − γ

(k)

2αβ

spins

¯u

)γ

(1 − γ

(p)¯u

(p)γ

(1 − γ

To actually compute this, we recall the spinor identities

spins

u(p)¯u(p) =

p + m,

spins

v(p)¯v(p) =

p − m

In our expression for, say,

, the

¯v

is already in the right form to apply

these identities, but

¯u

and

are not. Here we do a slightly sneaky thing. We

notice that for each fixed

α, β

, the quantity

αβ

is a scalar. So we trivially have

αβ

(

αβ

). We now use the cyclicity of trace, which says

(

) =

(

This applies even if

and

are not square, by the same proof. Then noting

further that the trace is linear, we get

αβ

= Tr(S

αβ

)

spins

¯u

(k)γ

(1 − γ

(q)¯v

(q)γ

(1 − γ

)

spins

(k)¯u

(k)γ

(1 − γ

(q)¯v

(q)γ

(1 − γ

)

= Tr

(

k + m

)γ

(1 − γ

)

qγ

(1 − γ

)

Similarly, we find

2αβ

= Tr

(1 − γ

)(

p + m

)γ

(1 − γ

)

To evaluate these traces, we use trace identities

Tr(γ

···γ

) = 0 if n is odd

Tr(γ

) = 4(g

µν

ρσ

− g

µρ

νσ

+ g

µσ

νρ

)

Tr(γ

) = 4iε

µνρσ

This gives the rather scary expressions

αβ

= 8



+ k

− (k · q)g

αβ

− iε

αβµρ



2αβ

= 8



+ q

− (q

· p)g

αβ

− iε

αβµρ

0µ



but once we actually contract these two objects, we get the really pleasant result

|M|

= 64G

(p · q)(k · q

It is instructive to study this expression in particular cases. Consider the case

where

−

and

go out along the +

direction, and

¯ν

along

−z

. Then we have

k · q

+ k

− k

As m

→ 0, we have → 0.

This is indeed something we should expect even without doing computations.

We know weak interaction couples to left-handed particles and right-handed

anti-particles. If

= 0, then we saw that helicity are chirality the same. Thus,

the spin of ¯ν

must be opposite to that of ν

and e

−

¯ν

⇐

So they all point in the same direction, and the total spin would be

. But the

initial total angular momentum is just the spin

. So this violates conservation

of angular momentum.

But if

= 0, then the left-handed and right-handed components of the

electron are coupled, and helicity and chirality do not coincide. So it is possible

that we obtain a right-handed electron instead, and this gives conserved angular

momentum. We call this helicity suppression, and we will see many more

examples of this later on.

It is important to note that here we are only analyzing decays where the

final momenta point in these particular directions. If m

= 0, we can still have

decays in other directions.

There is another interesting thing we can consider. In this same set up,

= 0, but neutrinos are massless, then we can only possibly decay to

left-handed neutrinos. So the only possible assignment of spins is this:

¯ν

⇐

⇒

Under parity transform, the momenta reverse, but spins don’t. If we transform

under parity, we would expect to see the same behaviour.

¯ν

⇐

⇒

But (at least in the limit of massless neutrinos) this isn’t allowed, because weak

interactions don’t couple to right-handed neutrinos. So we know that weak

decays violate P.

We now now return to finish up our computations. The decay rate is given

Γ =

(2π)

× (2π)

(4)

(p − k − q − q

)

|M|

Using our expression for |M |, we find

Γ =

8π

k d

q d

(4)

(p − k − q − q

)(p · q)(k · q

To evaluate this integral, there is a useful trick.

For convenience, we write Q = p − k, and we consider

µν

(Q) =

(4)

(Q − q − q

By Lorentz symmetry arguments, this must be of the form

µν

(Q) = a(Q

+ b(Q

µν

where a, b : R → R are some scalar functions.

Now consider

µν

(4)

(Q − q − q

)q · q

= (a + 4b)Q

But we also know that

(q + q

)

= q

+ q

+ 2q · q

= 2q · q

because neutrinos are massless. On the other hand, by momentum conservation,

we know

q + q

= Q.

So we know

q · q

So we find that

a + 4b =

, (1)

where

I =

(4)

(Q − q − q

We can consider something else. We have

µν

= a(Q

+ b(Q

(4)

(Q − q − q

)(q · Q)(q

· Q).

Using the masslessness of neutrinos and momentum conservation again, we find

that

(q · Q)(q

· Q) = (q · q

)(q · q

So we find

a + b =

. (2)

It remains to evaluate

, and to do so, we can just evaluate it in the frame where

Q = (σ, 0) for some σ. Now note that since q

= q

= 0, we must have

= |q|.

So we have

I =

|q|

δ(σ − |q| − |q

i=1

δ(q

− q

)

|q|

δ(σ − 2|q|)

= 4π

∞

d|q| δ(σ − 2|q|)

= 2π.

So we find that

Γ =

(2π)



2p · (p − k)k · (p − k) + (p · k)(p − k)



Recall that we are working in the rest frame of µ. So we know that

p · k = m

E, p ·p = m

, k · k = m

where E = k

. Note that we have

≈ 0.0048  1.

So to make our lives easier, it is reasonable to assume

= 0. In this case,

|k| = E, and then

Γ =

(2π)



E − 2(m

− 2(m

+ m



(2π)

k (3m

− 4E)

4πG

(2π)

dE E

(3m

− 4E)

We now need to figure out what we want to integrate over. When

is at rest,

then

min

= 0. The maximum energy is obtained when

, ¯ν

are in the same

direction and opposite to e

−

. In this case, we have

E + (E

¯ν

+ E

) = m

By energy conservation, we also have

E − (E

¯ν

+ E

) = 0.

So we find

max

Thus, we can put in our limits into the integral, and find that

Γ =

192π

As we mentioned at the beginning of the chapter, this is the only decay channel

off the muon. From experiments, we can measure the lifetime of the muon as

= 2.1870 × 10

−6

This tells us that

= 1.164 × 10

−5

GeV

Of course, this isn’t exactly right, because we ignored all loop corrections (and

approximated

= 0). But this is reasonably good, because those effects are

very small, at an order of 10

−6

as large. Of course, if we want to do more

accurate and possibly beyond standard model physics, we need to do better than

this.

Experimentally,

is consistent with what we find in the

τ → e¯ν

and

µ → e¯ν

decays. This is some good evidence for lepton universality, i.e. they

have different masses, but they couple in the same way.

6.4 Pion decay

We are now going to look at a slightly different example. We are going to study

decays of pions, which are made up of a pair of quark and anti-quark. We will

in particular look at the decay

−

(¯ud) → e

−

¯ν

−

¯ν

This is actually quite hard to do from first principles, because

−

is made up of

quarks, and quark dynamics is dictated by QCD, which we don’t know about

yet. However, these quarks are not free to move around, and are strongly bound

together inside the pion. So the trick is to hide all the things that happen in the

QCD side in a single constant

, without trying to figure out, from our theory,

what F

actually is.

The relevant currents are

lept

= ¯ν

(1 − γ

had

= ¯uγ

(1 − γ

)(V

d + V

s + v

b) ≡ V

had

− A

had

where the V

had

contains the γ

bit, while A

had

contains the γ

bit.

Then the amplitude we want to compute is

M =



−

(k)¯ν

(q)



eff



−

(p)



= −

√



−

(k)¯ν

(q)



†

α,lept

|0ih0|J

had



−

(p)



= −

√



−

(k)¯ν

(q)



¯eγ

(1 − γ

)ν

|0ih0|J

had



−

(p)



√

¯u

(k)γ

(1 − γ

(q) h0|V

had

− A

had



−

(p)



We now note that

had

does not contribute. This requires knowing something

about QCD and

−

. It is known that QCD is a P-invariant theory, and experi-

mentally, we find that

−

has spin 0 and odd parity. In other words, under P, it

transforms as a pseudoscalar. Thus, the expression

h0| ¯uγ



−

(p)



transforms as an axial vector. But since the only physical variable involved is

, the only quantities we can construct are multiples of

, which are vectors.

So this must vanish. By a similar chain of arguments, the remaining part of the

QCD part must be of the form

h0| ¯uγ



−

(p)



= i

√

for some constant F

. Then we have

M = iG

¯u

(k)

p(1 − γ

(q).

To simplify this, we use momentum conservation

, and some spinor

identities

¯u

(k)

k = ¯u

(k)m

(q) = 0.

Then we find that

M = iG

¯u

(k)(1 −γ

(q).

Doing a manipulation similar to last time’s, and noting that

(1 − γ

)γ

(1 + γ

) = 2(1 − γ

)γ

Tr(

q) = 4k · q

Tr(γ

q) = 0

we find

spins e,¯ν

|M|

spins

[¯u

(k)(1 −γ

(q)¯v

(q)(1 + γ

(k)]

= 2|G

(

k + m

)(1 − γ

)

= 8|G

(k · q).

This again shows helicity suppression. The spin-0

−

decays to the positive he-

licity

¯ν

, and hence decays to a positive helicity electron by helicity conservation.

−

⇒

¯ν

⇐

But if

= 0, then this has right-handed chirality, and so this decay is forbidden.

We can now compute an actual decay rate. We note that since we are working

in the rest frame of

−

, we have

= 0; and since the neutrino is massless,

we have q

= |q| = |k|. Finally, writing E = k

for the energy of e

−

, we obtain

π→e¯ν

(2π)

(4)

(p − k − q)

8|G

(k · q)

E|k|

δ(m

− E − |k|)(E|k|+ |k|

)

To simplify this further, we use the property

δ(f(k)) =

δ(k − k

)

where k

runs over all roots of f. In our case, we have a unique root

− m

and the corresponding derivative is

)| = 1 +

Then we get

π→e¯ν

4π

4πk

E + k

1 + k

δ(k − k

)

4π



1 −



Note that if we set

→

0, then this vanishes. This time, it is the whole decay

rate, and not just some particular decay channel.

Let’s try to match this with experiment. Instead of looking at the actual

lifetime, we compare it with some other possible decay rates. The expression for

π→µ¯ν

is exactly the same, with m

replaced with m

. So the ratio

π→e¯ν

π→µ¯ν



− m



≈ 1.28 × 10

−4

Here all the decay constants cancel out. So we can actually compare this with

experiment just by knowing the electron and muon masses. When we actually

do experiments, we find 1

230(4)

−4

. This is pretty good agreement, but

not agreement to within experimental error. Of course, this is not unexpected,

because we didn’t include the quantum loop effects in our calculations.

Another thing we can see is that the ratio is very small, on the order of 10

−4

This we can understand from helicity suppression, because m

 m

Note that we were able to get away without knowing how QCD works!

6.5 K

mixing

We now move on to consider our final example of weak decays, and look at

mixing. We will only do this rather qualitatively, and look at the effects

of CP violation.

Kaons contain a strange quark/antiquark. There are four “flavour eigenstates”

given by

(¯sd),

(

ds), K

(¯su), K

−

(¯us).

These are the lightest kaons, and they have spin

= 0 and parity

−ve

. We

can concisely write these information as J

= 0

−

. These are pseudoscalars.

We want to understand how these things transform under CP. Parity trans-

formation doesn’t change the particle contents, and charge conjugation swaps

particles and anti-particles. Thus, we would expect CP to send K

, and

vice versa. For kaons at rest, we can pick the relative phases such that





= −









= −





So the CP eigenstates are just





√

(





∓





Then we have





= ±





Let’s consider the two possible decays

→ π

and

→ π

−

. This

requires converting one of the strange quarks into an up or down quark.

¯s ¯u

−

¯s ¯u

From the conservation of angular momentum, the total angular momentum of

ππ is zero. Since they are also spinless, the orbital angular momentum L = 0.

When we apply CP to the final states, we note that the relative phases of

and π

−

(or π

and π

) cancel out each other. So we have



−



= (−1)



−





−



where the relative phases of π

and π

−

cancel out. Similarly, we have









Therefore ππ is always a CP eigenstate with eigenvalue +1.

What does this tell us about the possible decays? We know that CP is

conserved by the strong and electromagnetic interactions. If it were conserved

by the weak interaction as well, then there is a restriction on what can happen.

We know that

→ ππ

is allowed, because both sides have CP eigenvalue +1, but

−

→ ππ

is not. So

is “short-lived”, and

−

is “long-lived”. Of course,

−

will still

decay, but must do so via more elaborate channels, e.g. K

−

→ πππ.

Does this agree with experiments? When we actually look at Kaons, we find

two neutral Kaons, which we shall call

and

. As the subscripts suggest,

has a “short” lifetime of

τ ≈ 9 × 10

−11

, while

has a “long” lifetime of

τ ≈ 5 × 10

−8

But does this actually CP is not violated? Not necessarily. For us to be

correct, we want to make sure

never decays to

ππ

. We define the quantities

+−

|hπ

−





|hπ

−

|H |K

, η







|hπ

|H |K

Experimentally, when we measure these things, we have

≈ η

≈ 2.2 × 10

−3

6= 0.

So K

does decay into ππ.

If we think about what is going on here, there are two ways CP can be

violated:

– Direct CP violation of s → u due to a phase in V

CKM

– Indirect CP violation due to K

→

or vice-versa, then decaying.

Of course, ultimately, the “indirect violation” is still due to phases in the CKM

matrix, but the second is more “higher level”.

It turns out in this particular process, it is the indirect CP violation that is

mainly responsible, and the dominant contributions are “box diagrams”, where

the change in strangeness ∆S = 2.

u, c, t

¯s

¯u, ¯c,

W W

¯s

Given our experimental results, we know that

and

aren’t quite

and

−

themselves, but have some corrections. We can write them as





1 + |ε

(





+ ε



−



) ≈









1 + |ε

(



−



+ ε





) ≈



−



where

, ε

∈ C

are some small complex numbers. This way, very occasionally,

can decay as K

We assume that we just have two state mixing, and ignore details of the

strong interaction. Then as time progresses, we can write have

(t)i = a

(t)





+ b

(t)





(t)i = a

(t)





+ b

(t)





for some (complex) functions

, b

, a

, b

. Recall that Schr¨odinger’s equation

says

|ψ(t)i = H |ψ(t)i.

Thus, we can write





= R





where

R =







 









 







and

is the next-to-leading order weak Hamiltonian. Because Kaons decay

in finite time, we know

is not Hermitian. By general linear algebra, we can

always write it in the form

R = M −

Γ,

where

and Γ are Hermitian. We call

the mass matrix , and Γ the decay

matrix .

We are not going to actually compute

, but we are going to use the known

symmetries to put some constraint on

. We will consider the action of Θ =

The CPT theorem says observables should be invariant under conjugation by Θ.

So if

is Hermitian, then Θ

−1

. Now our

is not actually Hermitian,

but as above, we can write it as

= A + iB,

where A and B are Hermitian. Now noting that Θ is anti-unitary, we have

ΘH

−1

= A − iB = H

0†

In the rest frame of a particle

, we know









, and similarly for

. So we have





= −





, Θ





= −





Since we are going to involve time reversal, we stop using bra-ket notation for

the moment. We have

= (K

, H

) = (Θ

−1

ΘK

, H

−1

ΘK

) = (

, H

0†

)

∗

= (H

)

∗

= (

, H

) = R

Now if

was a good symmetry (i.e.

is good as well), then a similar

computation shows that

= R

We can show that we in fact have

= ε

= ε =

√

−

√

So if CP is conserved, then R

= R

, and therefore ε

= ε

= ε = 0.

Thus, we see that if we want to have mixing, then we must have

, ε

= 0.

So we need R

6= R

. In other words, we must have CP violation!

One can also show that

+−

= ε + ε

, η

= ε − 2ε

where

measures the direct source of CP violation. By looking at these two

modes of decay, we can figure out the values of

and

. Experimentally, we find

|ε| = (2.228 ± 0.011) × 10

−3

and



= (1.66 ± 0.23) × 10

−3

As claimed, it is the indirect CP violation that is dominant, and the direct one

is suppressed by a factor of 10

−3

Other decays can be used to probe

L,S

. For example, we can look at

semi-leptonic decays. We can check that

→ π

−

is possible, while

→ π

−

¯ν

is not.

has the opposite phenomenon. To show these, we just have to try to

write down diagrams for these events.

Now if CP is conserved, we’d expect the decay rates

Γ(K

L,S

→ π

−

) = Γ(K

L,S

→ π

−

¯ν

since we expect K

L,S

to consist of the same amount of K

and

We define

Γ(K

→ π

−

) − Γ(K

→ π

−

¯ν

)

Γ(K

→ π

−

) + Γ(K

→ π

−

¯ν

)

If this is non-zero, then we have evidence for CP violation. Experimentally, we

find

= (3.32 ± 0.06) × 10

−3

≈ 2 Re(ε).

This is small, but certainly significantly non-zero.

7 Quantum chromodynamics (QCD)

In the early days of particle physics, we didn’t really know what we were doing.

So we just smashed particles into each other and see what happened. Our initial

particle accelerators weren’t very good, so we mostly observed some low energy

particles.

Of course, we found electrons, but the more interesting discoveries were in

the hadrons. We had protons and neutrons,

and

, as well as pions

, π

and

. We found that

and

behaved rather similarly, with similar interaction

properties and masses. On the other hand, the three pions behaved like each

other as well. Of course, they had different charges, so this is not a genuine

symmetry.

Nevertheless, we decided to assign numbers to these things, called isospin.

We say

and

have isospin

, while the pions have isospin

= 1. The

idea was that if a particle has spin

, then it has two independent spin states;

If it has spin 1, then it has three independent spin states. This, we view

n, p

as different “spin states” of the same object, and similarly

, π

are the three

“spin states” of the same object. Of course, isospin has nothing to do with actual

spin.

As in the case of spin, we have spin projections

. So for example,

has

and

has

−

. Similarly,

, π

and

have

= +1

, −

1 respectively.

Mathematically, we can think of these particles as living in representations of

(2). Each “group”

{n, p}

{π

, π

, π}

corresponded to a representation

(2), and the isospin labelled the representation they belonged to. The

eigenvectors corresponded to the individual particle states, and the isospin

projection I

referred to this eigenvalue.

That might have seemed like a stretch to invoke representation theory. We

then built better particle accelerators, and found more particles. These new

particles were quite strange, so we assigned a number called strangeness to

measure how strange they are. Four of these particles behaved quite like the

pions, and we called them Kaons. Physicists then got bored and plotted out

these particles according to isospin and strangeness:

−

Remarkably, the diagonal lines join together particles of the same charge!

Something must be going on here. It turns out if we include these “strange”

particles into the picture, then instead of a representation of

(2), we now have

representations of su(3). Indeed, this just looks like a weight diagram of su(3).

Ultimately, we figured that things are made out of quarks. We now know

that there are 6 quarks, but that’s too many for us to handle. The last three

quarks are very heavy. They weren’t very good at forming hadrons, and their

large mass means the particles they form no longer “look alike”. So we only

focus on the first three.

At first, we only discovered things made up of up quarks and down quarks.

We can think of these quarks as living in the fundamental representation

su(2), with

u =





, d =





These are eigenvectors of the Cartan generator

, with weights +

and

−

(using

the “physicist’s” way of numbering). The idea is that physics is approximately

invariant under the action of

(2) that mixes

and

. Thus, different hadrons

made out of

and

might look alike. Nowadays, we know that the QCD part

of the Lagrangian is exactly invariant under the

(2) action, while the other

parts are not.

The anti-quarks lived in the anti-fundamental representation (which is also

the fundamental representation). A meson is made of two quarks. So they live

in the tensor product

⊗ V

= V

⊕ V

The

was the pions we found previously. Similarly, the protons and neutrons

consist of three quarks, and live in

⊗ V

= (V

⊕ V

) ⊗ V

= V

⊕ V

One of the V

’s contains the protons and neutrons.

The “strange” hadrons contain what is known as the strange quark,

. This

is significantly more massive than the

and

quarks, but are not too far off,

so we still get a reasonable approximate symmetry. This time, we have three

quarks, and they fall into an su(3) representation,

u =









, d =









, s =









This is the fundamental

representation, while the anti-quarks live in the

anti-fundamental

3. These decompose as

3 ⊗

3 = 1 ⊕

3 ⊗ 3 ⊗ 3 = 1 ⊕ 8 ⊕ 8 ⊕ 10.

The quantum numbers correspond to the weights of the eigenvectors, and hence

when we plot the particles according to quantum numbers, they fall in such a

nice lattice.

There are a few mysteries left to solve. Experimentally, we found a baryon

∆

uuu

with spin

. The wavefunction appears to be symmetric, but this

would violate Fermi statistics. This caused theorists to go and scratch their

heads again and see what they can do to understand this. Also, we couldn’t

explain why we only had particles coming from

3 ⊗

and

3 ⊗3 ⊗3

, and nothing

else.

The resolution is that we need an extra quantum number. This quantum

number is called colour . This resolved the problem of Fermi statistics, but also,

we postulated that any bound state must have no “net colour”. Effectively,

this meant we needed to have groups of three quarks or three anti-quarks, or

a quark-antiquark pair. This leads to the idea of confinement. This principle

predicted the Ω

−

baryon sss with spin

, and was subsequently observed.

Nowadays, we understand this is all due to a

(3) gauge symmetry, which

is not the

(3) we encountered just now. This is what we are going to study

in this chapter.

7.1 QCD Lagrangian

The modern description of the strong interaction of quarks is quantum chromo-

dynamics, QCD. This is a gauge theory with a

(3)

gauge group. The strong

force is mediated by gauge bosons known as gluons. This gauge symmetry is

exact, and the gluons are massless.

In QCD, each flavour of quark comes in three “copies” of different colour. It

is conventional to call these colours red, green and blue, even though they have

nothing to do with actual colours. For a flavour

, we can write these as

red

green

and q

blue

. We can put these into an triplet:





red

green

blue





Then QCD says this has an

(3) gauge symmetry, where the triplet transforms

under the fundamental representation. Since this symmetry is exact, quarks

of all three colours behave exactly the same, and except when we are actually

doing QCD, it doesn’t matter that there are three flavours.

We do this for each quark individually, and then the QCD Lagrangian is

given by

QCD

= −

aµν

µν

¯q

D − m

where, as usual,

= ∂

+ igA

µν

= ∂

− ∂

− gf

abc

Here T

for a = 1, ··· , 8 are generators of su(3), and, as usual, satisfies

, T

] = if

abc

One possible choice of generators is

where the

are the Gell-Mann matrices. The fact that we have 8 independent

generators means we have 8 gluons.

One thing that is very different about QCD is that it has interactions between

gauge bosons. If we expand the Lagrangian, and think about the tree level

interactions that take place, we naturally have interactions that look like

but we also have three and four-gluon interactions

Mathematically, this is due to the non-abelian nature of

(2), and physically,

we can think of this as saying the gluon themselves have colour charge.

7.2 Renormalization

We now spend some time talking about renormalization of QCD. We didn’t

talk about renormalization when we did electroweak theory, because the effect

is much less pronounced in that case. Renormalization is treated much more

thoroughly in the Advanced Quantum Field Theory course, as well as Statistical

Field Theory. Thus, we will just briefly mention the key ideas and results for

QCD.

QCD has a coupling constant, which we shall call

. The idea of renormal-

ization is that this coupling constant should depend on the energy scale

are working with. We write this as

(

). However, this dependence on

is not

arbitrary. The physics we obtain should not depend on the renormalization point

we chose. This imposes some restrictions on how

(

) depends on

, and this

allows us to define and compute the quantity.

β(g(µ)) = µ

dµ

g(µ).

The β-function for non-abelian gauge theories typically looks like

β(g) = −

16π

+ O(g

)

for some

. For an

(

) gauge theory coupled to fermions

{f}

(both left- and

right-handed), up to one-loop order,we have

N −

where

is the Dynkin index of the representations of the fermion

. For the

fundamental representation, which is all we are going to care about, we have

In our model of QCD, we have 6 quarks. So

= 11 − 4 = 7.

So we find that the β-function is always negative!

This isn’t actually quite it. The number of “active” quarks depends on the

energy scale. At energies

 m

top

≈ 173 GeV

, then the top quark is no longer

active, and

= 5. At energies

∼ 100 MeV

, we are left with three quarks, and

then

= 3. Matching the

functions between these regimes requires a bit of

care, and we will not go into that. But in any case, the

function is always

negative.

Often, we are not interested in the constant

itself, but the strong coupling

4π

It is an easy application of the chain rule to find that, to lowest order,

dα

dµ

= −

2π

We now integrate this equation, and see what we get. We have

(µ)

(µ

)

dα

= −

2π

dµ

So we find

(µ) =

2π

log(µ/µ

) +

2π

(µ

)

There is an energy scale where

diverges, which we shall call Λ

QCD

. This is

given by

log Λ

QCD

= log µ

−

2π

(µ

)

In terms of Λ

QCD

, we can write α

(µ) as

(µ) =

2π

log(µ/Λ

QCD

)

Note that in the way we defined it,

is positive. So we see that

decreases

with increasing

. This is called asymptotic freedom. Thus, the divergence occurs

at low

. This is rather different from, say, QED, which is the other way round.

Another important point to get out is that we haven’t included any mass

term yet, and so we do not have a natural “energy scale” given by the masses.

Thus,

QCD

is scale invariant, but quantization has led to a characteristic scale

QCD

. This is called dimensional transmutation.

What is this scale? This depends on what regularization and renormalization

scheme we are using, and tends to be

QCD

∼ 200-500MeV.

We can think of this as approximately the scale of the border between perturbative

and non-perturbative physics. Note that non-perturbative means we are in low

energies, because that is when the coupling is strong!

Of course, we have to be careful, because these results were obtained by only

looking up to one-loop, and so we cannot expect it to make sense at low energy

scales.

7.3 e

−

→ hadrons

Doing QCD computations is very hard

. Partly, this is due to the problem of

confinement. Confinement in QCD means it is impossible to observe free quarks.

When we collide quarks together, we can potentially produce single quarks or

anti-quarks. Then because of confinement, jets of quarks, anti-quarks and gluons

would be produced and combine to form colour-singlet states. This process is

known as hadronization.

Confinement and hadronization are not very well understood, and these

happen in the non-perturbative regime of QCD. We will not attempt to try to

understand it. Thus, to do computations, we will first ignore hadronization,

which admittedly isn’t a very good idea. We then try to parametrize the

hadronization part, and then see if we can go anywhere.

Practically, our experiments often happen at very high energy scales. At

these energy scales,

is small, and we can expect perturbation theory to work.

We now begin by ignoring hadronization, and try to compute the amplitudes

for the interaction

−

→ q ¯q.

The leading process is

−

¯q

We let

q = k

+ k

= p

+ p

and

be the quark charge. Repeating computations as before, and neglecting

fermion masses, we find that

M = (−ie)

Q¯u

)γ

)

−ig

µν

¯v

)γ

We average over initial spins and sum over final states. Then we have

spins

|M|

Tr(

) Tr(

)



· k

)(p

· k

) + (p

· k

)(p

· k

)



= e

(1 + cos

θ),

where

is the angle between

and

, and we are working in the COM frame

of the e

and e

−

Now what we actually want to do is to work out the cross section. We have

dσ =

− v

(2π)

(4)

(q − k

− k

) ×

spins

|M|

We first take care of the

− v

factor. Since

= 0 in our approximation,

they travel at the speed of light, and we have

− v

= 2. Also, we note that

= k

≡ |k| due to working in the center of mass frame.

Using these, and plugging in our expression for |M|, we have

dσ =

(2π)

|k|

(4)

(q − k

− k

)(1 + cos

θ).

This is all, officially, inside an integral, and if we are only interested in what

directions things fly out, we can integrate over the momentum part. We let dΩ

denote the solid angle, and then

= |k|

d|k| dΩ.

Then, integrating out some delta functions, we obtain

dσ

dΩ

d|k|

8π

− |k|

(1 + cos

θ)

(1 + cos

θ),

where, as usual

α =

4π

We can integrate over all solid angle to obtain

σ(e

−

→ q ¯q) =

4πα

We can compare this to

−

→ µ

−

, which is exactly the same, except we

put Q = 1 because that’s the charge of a muon.

But this isn’t it. We want to include the effects of hadronization. Thus, we

want to consider the decay of

−

into any possible hadronic final state. For

any final state X, the invariant amplitude is given by

hX|J

|0i¯v

)γ

where

¯q

and Q

is the quark charge. Then the total cross section is

σ(e

−

→ hadrons) =

spins

(2π)

(4)

(q − p

)|M

We can’t compute perturbatively the hadronic bit, because it is non-perturbative

physics. So we are going to parameterize it in some way. We introduce a hadronic

spectral density

µν

(q) = (2π)

X,p

(4)

(q − p

) h0|J

|XihX|J

|0i.

We now do some dodgy maths. By current conservation, we have

µν

= 0.

Also, we know

has positive energy. Then Lorentz covariance forces

to take

the form

µν

(q) = (−g

µν

+ q

)Θ(q

)ρ

We can then plug this into the cross-section formula, and doing annoying

computations, we find

σ =

(2π)e

4(p

1µ

2ν

− p

· p

µν

+ p

1ν

2µ

)(−g

µν

+ q

)ρ

)

16π

Of course, we can’t compute this

directly. However, if we are lazy, we can

consider only quark-antiquark final states

. It turns out this is a reasonably

good approximation. Then we obtain something similar to what we had at the

beginning of the section. We will be less lazy and include the quark masses this

time. Then we have

µν

) = N

(2π)

(4)

(q − k

− k

)

× Tr[(

+ m

)γ

(

− m

)γ

where

is the number of colours and

is the quark

mass. We consider

the quantity

µν

(4)

(q − k

− k



We can argue that we can write

µν

= A(q

+ B(q

µν

We contract this with

µν

and

(separately) to obtain equations for

A, B

We also use

= (k

+ k

)

= 2m

+ 2k

· k

We then find that

) =

12π

Θ(q

− 4m

)

1 −

1/2

+ 2m

That’s it! We can now plug this into the equation we had for the cross-section.

It’s still rather messy. If all

→

0, then this simplifies very nicely, and we find

that

) =

12π

Then after some hard work, we find that

−

→ hadrons) = N

4πα

where LO denotes “leading order”.

An experimentally interesting quantity is the following ratio:

R =

σ(e

−

→ hadrons)

σ(e

−

→ µ

−

)

Then we find that

= N











when u, d, s are active

when u, d, s, c are active

when u, d, s, c, b are active

In particular, we expect “jumps” as we go between the quark masses. Of course,

it is not going to be a sharp jump, but some continuous transition.

We’ve been working with tree level diagrams so far. The one-loop diagrams

are UV finite but have IR divergences, where the loop momenta

→

0. The

diagrams include

−

¯q

−

¯q

−

¯q

However, it turns out the IR divergence is cancelled by tree level

−

→ q ¯qg

such as

−

¯q

7.4 Deep inelastic scattering

In this chapter, we are going to take an electron, accelerate it to really high

speeds, and then smash it into a proton.

If we do this at low energies, then the proton appears pointlike. This is

Rutherford and Mott scattering we know and love from A-levels Physics. If we

increase the energy a bit, then the wavelength of the electron decreases, and

now the scattering would be sensitive to charge distributions within the proton.

But this is still elastic scattering. After the interactions, the proton remains a

proton and the electron remains an electron.

What we are interested in is inelastic scattering. At very high energies, what

tends to happen is that the proton breaks up into a lot of hadrons

. We can

depict this interaction as follows:

−

This led to the idea that hadrons are made up of partons. When we first

studied this, we thought these partons are weakly interacting, but nowadays, we

know this is due to asymptotic freedom.

Let’s try to understand this scattering. The final state

can be very

complicated in general, and we have no interest in this part. We are mostly

interested in the difference in momentum,

q = p − p

as well as the scattering angle

. We will denote the mass of the initial hadron

H by M , and we shall treat the electron as being massless.

It is conventional to define

≡ −q

= 2p · p

= 2EE

(1 − cos θ) ≥ 0,

where

and

, since we assumed electrons are massless. We also let

ν = p

· q.

It is an easy manipulation to show that p

= (p

+ q)

≥ M

. This implies

≤ 2ν.

For simplicity, we are going to consider the scattering in the rest frame of the

hadron. In this case, we simply have

ν = M(E − E

)

We can again compute the amplitude, which is confusingly also called M:

M = (ie)

¯u

)γ

(p)



−

µν

hX|J

|H(p



Then we can write down the differential cross-section

dσ =

4ME|v

− v

(2π)

X,p

(2π)

(4)

(q + p

− p

)

spins

|M|

Note that in the rest frame of the hadron, we simply have |v

− v

| = 1.

We can’t actually compute this non-perturbatively. So we again have to

parametrize this. We can write

spins

|M|

µν

hH(p

)|J

|XihX|J

|H(p

)i,

where

µν

= Tr



pγ

) = 4(p

− g

µν

p · p

+ p



We define another tensor

µν

4π

(2π)

(4)

(p + p

− o

) hH|J

|XihX|J

|Hi.

Note that this

should also include the sum over the initial state spins. Then

we have

dσ

8ME(2π)

4π

µν

We now use our constraints on

µν

such as Lorentz covariance, current conser-

vation and parity, and argue that W

µν

can be written in the form

µν



−g

µν



(ν, Q

)



−

· q



−

· q



× W

(ν, Q

Now, using

µν

= q

µν

= 0,

we have

µν

= 4(−2p · p

+ 4p · p

+ 4(2p · p

· p

− p

p · p

)

= 4Q

+ 2M

(4EE

− Q

We now want to examine what happens as we take the energy

E → ∞

. In

this case, for a generic collision, we have

→ ∞

, which necessarily implies

ν → ∞

. To understand how this behaves, it is helpful to introduce dimensionless

quantities

x =

2ν

, y =

· p

known as the Bjorken

and the inelasticity respectively. We can interpret

as the fractional energy loss of the electron. Then it is not difficult to see that

≤ x, y ≤

1. So these are indeed bounded quantities. In the rest frame of the

hadron, we further have

y =

E − E

This allows us to write L

µν

≈ 8EM



xyW

(1 − y)

νW



where we dropped the −2M

term, which is of lower order.

To understand the cross section, we need to simplify d

. We integrate out

the angular φ coordinate to obtain

7→ 2πE

d(cos θ) dE

We also note that by definition of Q, x, y, we have

dx = d



2ν



= −2EE

d cos θ + (···) dE

dy = −

Since (dE

)

= 0, the d

part becomes

7→ πE

dy = 2πE

ν dx dy.

Then we have

dσ

dx dy

8(2π)

2πν · 8EM



xyW

(1 − y)

νW



8πα



+ (1 − y)F



where

≡ W

, F

≡ νW

By varying

and

in experiments, we can figure out the values of

and

Moreover, we expect if we do other sorts of experiments that also involve hadrons,

then the same

and

will appear. So if we do other sorts of experiments and

measure the same

and

, we can increase our confidence that our theory is

correct.

Without doing more experiments, can we try to figure out something about

and

? We make a simplifying assumption that the electron interacts with

only a single constituent of the hadron:

−

k + q

We further suppose that the EM interaction is unaffected by strong inter-

actions. This is known as factorization. This leading order model we have

constructed is known as the parton model . This was the model used before we

believed in QCD. Nowadays, since we do have QCD, we know these “partons” are

actually quarks, and we can use QCD to make some more accurate predictions.

We let f range over all partons. Then we can break up the sum

(2π)

k Θ(k

)δ(k

)

spins

where we put δ(k

) because we assume that partons are massless.

To save time (and avoid unpleasantness), we are not going to go through the

details of the calculations. The result is that

µν

k Tr



µν

H,f

, k) +

µν

H,f

, k)



where

µν

(

k +

q)γ

δ((k + q)

)

H,f

, k)

βα

(4)

− k − p

) hH(p

)| ¯q

f,α

ihX

f,β

|H(p

)i,

where α, β are spinor indices.

In the deep inelastic scattering limit, putting everything together, we find

(x, Q

) ∼

(x) + ¯q

(x)]

(x, Q

) ∼ 2xF

for some functions

(

¯q

(

). These functions are known as the parton

distribution functions (PDF ’s). They are roughly the distribution of partons

with the longitudinal momentum function.

The very simple relation between

and

is called the Callan–Gross

relation, which suggests the partons are spin

. This relation between

and

is certainly something we can test in experiments, and indeed they happen.

We also see the Bjorken scaling phenomenon —

and

are independent of

. This boils down to the fact that we are scattering with point-like particles.