QFT Part 2 Notes

Ph444 Quantum Field Theory Winter 2012

Notes

March 7, 2013 Giordon Stark

Contents

1 January 8th, 2013 (Introduction to Fermions) 3

1.1 Fermions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2 Group Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.3 Labels Representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.4 Symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2 January 10th, 2013 17

2.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.2 Field Operators under the Lorentz Group . . . . . . . . . . . . . . . . . . . . . . . . 18

2.2.1 Motivation through Scalar Fields . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.3 Fermions, spin 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.4 (12 , 0) representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.5 (0, 12) representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.6 Lorentz Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.7 1-fermion ζα . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3 Jan. 29th, 2013 – Spin-1 Vectors 29

3.1 Photons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

3.2 Lorentz invariance, unitarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

3.2.1 Lorentz Boost . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4 February 5th, 2013 33

4.1 Renormalization of QED . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

4.1.1 Feynman Rules of Renormalization . . . . . . . . . . . . . . . . . . . . . . . . 36

4.1.2 Renormalization Conditions (on-shell scheme) . . . . . . . . . . . . . . . . . . 36

1


5.1 Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

5.1.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

5.1.2 Schwinger-Dyson Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

5.1.3 Renormalized QED . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

5.2 General 3-point Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

5.3 Yet Another (not complete) Proof for Ward Identity . . . . . . . . . . . . . . . . . . 42


6.1 Ward Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

6.2 Spontaneous Symmetry Breaking (SSB) . . . . . . . . . . . . . . . . . . . . . . . . . 49

7 February 28th, 2013 – Spontaneous Symmetry Breaking (contd) 52

7.1 SO(2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

8 March 5th, 2013 63

9 March 7th, 2013 68

9.1 Standard Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

10 March 5th, 2013 80

11 March 7th, 2013 85

11.1 Standard Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

2

List of Figures

1 Discrete symmetry D = 1 no SSB . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

3

1 JANUARY 8TH, 2013 (INTRODUCTION TO FERMIONS)QFT II

1 January 8th, 2013 (Introduction to Fermions)

I remember at the very beginning of the last quarter, I sort of draw you some schematic table, that

we can talk about particles, and the related issue in the following way. So you can classify them

by spin. Spin can be 0. What else it can be? 1/2. We are not talking about spin for the particles.

That is a very deep question. I don’t know what, I don’t know the answer. In principle it can

be as well but usually, maybe we will make a comment about that later. But even with this, it is

not very complete. What is wrong with this list? Every half integer, there is a spin. Why do we

usually stop with these two? The answer is that we only know how to write normal looking up to

spin 2. There is a reason for that, which we know we may not be able to get.

name scalar fermion spin 1 ?? graviton

spin 0 12 1 3

2 2

renormalization

EFT

quantization

s-matrix

symmetries

Particle data group, there is a particle data group. You Google it. Particle data. Used to be they

mail you a booklet every year. But these days they are all on-line. It is very fast. By the way, a

very good test whether you are any good as a particle physicist is to see how many pages of particle

data book you can understand not meaning just reading the English, but trying to understand the

properties.

There are a lot of confusing issues with quantum theory such as renormalization, EFT, quantization,

S-matrix, symmetries. But now in this quarter, we will mostly try to see what are the new issues

that arise in these categories for spin 1, you see that there are quite a bit of new things. We will also

have a chance to talk about normalization EFT again in the context of these series, just to give you

more examples of that. I think I also mentioned to you last quarter that historic Clee, quantum

4

1.1 Fermions 1 JANUARY 8TH, 2013 (INTRODUCTION TO FERMIONS)

– historically quantum field theory is confusing. Is it more confusing than necessary because

unfortunately the first example of quantum field theory, everyone knows what is the first example

of quantum field theory, dynamics, is these particles. All issues tangle together. Historically, our

predecessors have to understand everything all together. That is why it takes a long time. Why

are these things interesting? First of all, they are seen in nature. Most of the particles we know,

elementary particles in nature, spin 1/2, spin one. We might have one elementary particle in spin

0, not so long ago. But before that, everything we know are in these two categories. Even people

suspect there is no spin 0 elementary particles. But we can go through some of those reasons if we

have a chance later. But we still don’t know whether it is elementary particle, or not. Most of the

elementary are in these particles. They give you more differentiation of dynamics. We basically

are focusing on that. In this quarter we will basically do Fermian, as I said, and the spin 1. But

mostly, we will do this and that. And there is example, we will do this. We will see how much

those things already are taking up my time. But I do want to in the end, do a little bit of synthesis,

since I do want to talk a little bit about hicks mechanism, in the context of Q.E.D. Not our Q.E.D.

For the Q.E.D. in a different world. Sorry. Either for Q.E.D. in a different work or for Q.E.D. in

the BCS you conduct, inside of superconductor. That is sort of using all the knowledge, use all the

scalar, using the scalar field and using Fermian and Baxter. Also very, BCS series is also a beautiful

demonstration of effective field theory, by the way, if any of you actually read the journals. Let’s

begin with fermions.

1.1 Fermions

We’re not going to spend too much time because there is not too much deep about fermions and

there is quite a few things deep about scalars, about vectors.

It is very hard to keep track with consistent notation of everything deep. But concept is easy.

What we will do is not to talk about everything about Fermions, is to I’ll try to talk about some

basic concept about Fermions, then I will try to basically focus on one special case that we will

need later. That is the Fermions we need which is the probably simplest one. But before I do that,

if you want to ever become a grand master of Fermions, here is something you can, you must have.

There is a review article by this group of people, okay. What else do we want to know? This is

5

1.1 Fermions 1 JANUARY 8TH, 2013 (INTRODUCTION TO FERMIONS)

a physics report. It’s this big. You probably don’t want to read it from cover to cover. But you

want to read enough so that you can, this has every single formula you want, that you ever need

about Fermions and also, if you actually do calculation with Fermions, you know that there are 20

different conventions and so on.

H. Dreiner, H. Haber, S. Martin physical report 494 (2010) [pdf]

They have all of them, and teach you how to translate between them, and I don’t know whether

you have ever done latex, they have a macrofile that you can use your space time signature, so on,

relapse, the whole thing. It is a different convention, okay? But that is not the main point, why

this, that is one of the main point why this is useful. But the other thing, why this is useful is

these people are widely respected in the community for never make any mistakes, so everything,

anything I wrote on the board is inconsistent with them, they are right. I am wrong. In this

business, negative sign 1/2 is everything. There is nothing too deep about it. But you have to, if

you want to get the right number, you have to have, you have to get those things right.

What Fermion is, what distinguish Fermion and scalar is basically Fermions are representations,

different representation of Lorentz group. Why do we care about Lorentz group? Because it’s

symmetry. It’s a space/time symmetry. Why do we, why is that obvious? Why is Lorentz group

symmetry, used for symmetry? Lorentz group is certainly not symmetry of this classroom. You

are sitting there. I’m sitting here. We are certainly not in rotations. So it’s useful, if you think

about microscopic physics. When you go to a microscopically, we know for a fact that the theory

is, and we talk about the particles and that the scalars are the trivial representation of Lorentz

group. Fermions is the next representation basically. You will see that all these guys falls into

different representations of lower he wants group. It is interesting that nature use lower dimension

representations of loren t z group. As I said we have examples all of this, except of this. This one

we know how to write in theory but we haven’t observed. Nature has managed to use all of them.

That alone give you a strong motivation to suspect super symmetry is there. Now we are going to

talk more about representations and symmetry. Let me ask, nobody has never seen group theory

before, right? You heard of the word, group.

6

http://arxiv.org/pdf/0812.1594.pdf

1.2 Group Theory 1 JANUARY 8TH, 2013 (INTRODUCTION TO FERMIONS)

1.2 Group Theory

groups → symmetries → equivalent theories

So groups. I’m sure you can look up the definition of group. I’m not going to do that. Usually

we think groups are in physics, is a useful concept to handle symmetries. We know symmetry is a

powerful tool in physics. What symmetry says in physics is that there are different series that is

equivalent. Different theories are equivalent. This is the meaning of symmetry. If you think about

it, it is quite a powerful statement. Although, since we brought up in kindergarten getting used to

symmetry you think it is true but it is highly nontrivial. It means the, the difference is, in principle

I can write a theory facing this way and I can write a theory facing this way. The statement that

these two theories are equivalent is actually highly nontrivial statement. It means including all the

quantum corrections, calculate all the loop, no matter what you do, these two theories are the same.

Without doing that very complete calculation, you can already just, if there is a symmetry you can

say those two theories are the same. What is the symmetry, tells me that the theory, this way and

theory that way, is the same? What is the symmetry? That is the rotational symmetry. Typically

three spatial dimensions, for example, symmetries like SO(3), it’s called, everybody knows what

this means. Special orthogonal group. Rotations. How many parameters in this group? Three.

There is three rotation. For group there are elements, group elements.

Given group elements A,B,AB, we can have generators TA, TB with something like A = eiTAαA

(abstractly for SO(3)). In physics, we work with representations of a group. A representation is

something the group acts on. In terms of representations, group elements, it’s useful to think of them

as being matrices M(A) · · · . Every group corresponds to a matrix acting on some object.

For example, in SO(3) – we have a rotational matrix R for every group element. So

~r → R~r

where ~r is a 3-dimensional fundamental real representation. In quantum mechanics, we talk about

two components – spinors χs (2-dimensional complex representation) which acts like

ei~r2·~θχs

7


For representations, there are also two kinds of representations. Anybody knows what two kinds is?

You can classify them as so called reducible versus ir reducible representations. It’s very useful to

talk about irreducible. All kinds of other things like [inaudible] representations, that is not talking

about those things. Irreducible representations, sometimes called irraps. What is a irreducible

representation meaning? Irreducible representation means that let’s say I have a five members, it is

useful to think about all of this in terms of some finite representation. If I have certain five members

of this representation, and irreducible representation means I can act a full group on this irreducible

representation, it will go through all five of them. Nothing more, nothing less. If you take all the

elements of the group, acting on this representation, it cannot be a four-dimensional. It has to be

five dimensional. Let’s say, if I, so this is irreducible representation, for example. But I cannot

just take acts, for example. Acts. It does not form a representation of the rotational group. But,

and but on the other hand, this three-dimensional vector spin a representation. You don’t need

anything more. In principle, you can also a representation. But this is a reducible representation.

In quantum mechanics, we have, angular momentum. We learn angular momentum representations.

The one thing about irreducible representation is that they are all labeled by environs. A different

irreducible representation of a group, reduced, is labeled by environs. The value of those environs

or of that, of that group.

Generate all the reducible representations is a pain – let’s say SO(3). What are reducible repre-

sentations of SO(3)? In QM, you will learn angular momentum and spin: L,m along with the

eigenvalues

L2 = l(l + 1), l = 01

2, 1 · · ·

which one definition is to keep ~r2 invariant. So all irreducible for SO(3) is labeled by l(l + 1), the

different values correspond to different irreducible representations.

SO(1,3) Lorentz group keeps x0,2− ~x2 invariant and has 6 generators. Jµν , µ, ν = 0, 1, 2, 3 which is

8


anti-symmetric under µ↔ ν.

[Jµν , Jρσ] = i ([gνρJµσ − (µ↔ ν)]− [ρ↔ σ])

Li =1

2εijkJ jk, Ki = J0i, (i, j, k → 1, 2, 3)

(rotations) [Li, Lj ] = iεijkLk, (boosts) [Ki,Kj ] = −εijkLk

[Li,Kj ] = iεijkKk

It has six generators. Three of them are normal rotation. But there are also three others, what

are those three others? Boost. So a general definition, I can write a general definition of this.

It is orthogonal group. So mu, and then nu, these are generators, from 0 to 3. And this is ante

symmetric under interchange nu and mu. Okay? So there are six of them. Let’s just take J nu mu.

J sigma, so algebra is defined by the ante symmetric commutation relations. G ro, J mu sigma

minus mu .... okay. Ante semmette Rickize between these two. Next term is ante symmetrickize

between this set of indices and this set of indices, in the commute eighter, so the next set of terms,

the same thing here. But ro – sorry, ante symmetrickize between these two. These are, if you want

defines, the algebra of Lorentz group. If you want to write this in a slightly more familiar notation

... (pause). I can define Li, which is one-half and ki, which is just J0i. I haven’t done anything

else. I’m just calling, take part, take apart this, I should say iJk, is 1, 2, is 1, 2 and 3. There is

special directions. That is the only thing else. There is no J00 because it’s anti symmetric. I’m

just taking different elements of this nu mu matrix and calling it different names. Okay? It is likely

redefine a subset of them. This is just a change, I’m changing base of this operators. If you do

this, you observe ..... (pause). Okay? So this is the most familiar part. That is, this is the whole

point of writing it in this way. Okay? This is SO3 group. Epsilon iJk is anti symmetric answer,

which is also the structure constant of SO3. This is SO3, rotation. And this is saying that the

true boost, so these are the boosts so if you actually write the representation out, you can see that

this is boost. The two, the commutation, 2 boost does not commute. The equivalent to a location,

okay? That is just a saying that. And then moreover – sorry, yeah, strategically cause this to be

confusing, but this is upper – this is a generator. Rotation boost also does not commute. But you

can write it in terms of boosts. The whole point of writing this, this way, is just to understand

how the rotational group is embedded in the Lorentz group. This writing is obviously less coherent

9


to looking it as writing it, the physical meaning of this generators is better. I’m sure you are all

familiar with representation. Most [inaudible] representation is Lorentz group, four vectors. I’m

sure you can look it up in Jackson.

particles ↔ state

Now, what we care about are particles and fields, as a representations of Lorentz group. Okay? So

let’s get to that. Right off the bat, it’s already pretty confusing. You will see there is already the

first confusion is that as I, you have to distinguish between a particle state and the field, okay, in

this case. State is not a field. State is a quantum mechanical state. A field is an operator acting on

the space spanned by those states. Okay? This is one of the main confusing points. A state, you

can write it in, write in for example in some representation such as momentum representation, or

space or position representation, that becomes a wave function. Okay? However, a field operator

is different from that wave function, although sometimes they tend to use the same symbol. That

is why it is confusing. Just keep in mind all of these things. The two Fermion state does not anti

commute with each other. The field operator does anti commute. So for example, okay? So let’s

talk about particles first. Particles first. It means let’s talk about state first before we go on to

the Fermion field operator. Again having to be careful to distinguish these two things, okay? Now,

Lorentz transformation, we want to build representations of Lorentz group. Okay? That is the

goal. Lorentz transformation on state, okay, and this is a symmetry of our theory.

Now, for our theory, for our theory with symmetry, the symmetry transformation on the state, you

can also represent it as, with a matrix. Okay? That is a representation of the symmetry. So what

is the requirement of the matrix? Yeah, the matrix has to be unitary on a state. Why? Because

quantum mechanics, the inner product of two states has a very important interpretation as the

probability. So if you think something is a symmetry of the theory, it is better, first thing is better

preserve probability. We can not have non[inaudible] interpretations. Let’s say one state goes to

some U – sorry – for Lorentz transformation. Lambda, okay, not to be confused as the color we

talked about last quarter, because some people likes to use lambda. It’s on the state, is just state

goes to some state. Okay?

10


We want a Lorentz transformation on a state, so for a Lorentz transformation Λ:

|〉 → U(Λ)|〉

where U(Λ) is unitary by requirement.

Where this is a matrix, okay? This is a matrix which is unitary. So far I haven’t talked about

anything about Fermion. But for bosons, for scalars, we already derived what this matrix is.

It’s a pure phase. That is obviously unitary. But for Fermions, this can be nontrivial. But

whatever it is, it has to be unitary. Another way, this has to be a unit airy representation of –

unitary representation of Lorentz group. Unitary representation of Lorentz group. Off the bat

there is something already confusing because we know Lorentz group, I’m stating all this fact,

Lorentz group is noncompact. Meaning that you cannot, the parameter of Lorentz group does not

live in the compact space. Okay? This part, obviously, lives on the compact space, the angles,

rotational angles obviously lives in compact space. But this part doesn’t. It can have a, boost

parameter can be as big as you want it to be. This is noncompact. There is a theorem, we cannot

fight with the theorem, there is a theorem saying that for noncompact groups, there is no finite

dimensional unitary representations. Okay? So now there is a problem. So why does it make sense,

this make sense at all? Okay? This means that the representations, irreps(unitary) irreducible

representations are unitary, unitary representations which we know every particle state must be

in, is infinite dimensional, okay? Which you will see is indeed that is the case, they are infinite

dimensional.

Okay? I say that again, okay? I’m repeating myself. I say it again. This is on the state. Okay?

When we talk about field operators, that is another set of state. Okay? This is on the state.

Field operator, on the other hand, there is no requirement of field operator be unitary. Okay? In

fact, the field operators field, finite dimensional nonunitary representations of Lorentz group. The

rest of a couple lectures are not really that much about physics, but we have to get through this

to be able to do Fermions. Of course, from this you can already say that you have two Lorentz

transformation, lambda and lambda prime. They are obvious like group theory relations like this.

That is obviously, has to be the case. That is just representations theory from groups.

For two Lorentz transformations Λ,Λ′, we have U(Λ)U(Λ′) = U(ΛΛ′).

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1.3 Labels Representation 1 JANUARY 8TH, 2013 (INTRODUCTION TO FERMIONS)

1.3 Labels Representation

Okay. As we said, the different representations, different irreducible representations are labeled by

envariants – by invariants under the symmetry transformation. Okay? So, we wanted to, so let’s

begin by some state K which is labeled by the momentum. Not that K. Sorry. This is K. This is

momentum. I’m sorry about all this notation. But maybe it’s a good exercise, because the whole

thing about the, the whole difficult thing about the Fermions are conventions, so if you get through

all this, you ... okay? This is momentum, okay? What is the most obvious invariant of Lorentz

group?

(momentum) |k〉 k2 = kµkµ = m2 ← labels representation

First of all, we know that, so therefore each mass, and the particles with different mass don’t

transform into each other. Okay? The particle with certain mass after Lorentz transformation still

have the same mass. This means that the different representations with different mass obviously

does not transform into each other. So all the irreducible representations are labeled by their mass.

They all are just different mass corresponding to different representations. This is one of those

labels, labels representations. There can be additional labels, as you see, I didn’t call this irreducible

representation. There are additional labels we can assign to reduce, fully specify ir reducible

representation. How do we do that? I can tell you the result. But there is also a way systematic

way of doing it. Primarily devised by Eugene Wagner which is called induced representation or

commonly known as the “little group”.

start with n = (m, 0, 0, 0) – little group of n is symmetric which leaves n invariant.

The (0, 0, 0) part is a little group of SO(3). Denote elements of this by R. Any vector k is a boost

L(k) : n→ k. Then

L(k) R L−1(k) leaves k invariant

The last frame – also known as the rest frame for momentum. The little group, little group which is

a group of any vector, okay, any full vector, let’s say little group of N, is, you know, is the symmetry

12


which leaves N invariant, okay? Once the symmetry leaves this invariant, the little group of this is

SO3, so little group is SO3. So, writing this way is easy to see this. But a nontrivial statement is

that the, actually, the little group of any arbitrary vector is SO3. Any arbitrary vector with mass

M is SO3. Okay? Let’s try to see it. Let’s do it here. There is enough space. Any arbitrary vector

in this representation, K, of any vector, K, I can obtain by, from N, by doing a boost. Let’s say I

have a boost, lk which takes N to K.

Now, so this little group that keeps N invariant, I can just take any arbitrary element, without

SO3, is called R, R is an element of this little group. And you can see that, so obviously, R is SO3,

which leaves N invariant. But you can also see that the following – okay? So, this acting on K

means that this take you back to N, and R doesn’t do anything to N. R leaves N invariant. This

takes the N back to K. Is this okay? Is this okay? Okay. But you can show that this is also SO3.

All the elements look like this, this is also SO3. This is verified by doing whatever you want. Okay?

Commutation relations, whatever. So this tells me that the little group for any vector is always

SO3. Is this okay? Which is obvious. This is sort of obvious if you think about it. There is always

three notations that you can do to keep a fallback to invariant. Or said it the other way, you go to

the rest frame of the vector, rotate and then go back. So okay, now, let’s think about, this is going

to do something to me eventually I think. (chuckles). It’s okay. Now, let’s think about, let’s think

about how this is acting on vectors. Okay? State. Okay? Let’s call a state N. As I said every state

is labeled by the vector, for vector. So this is labeled by this. But I’m going to assign it another

indices, because obviously there’s additional things that transform under about the states. So this

is what little group acts on. This indices, this individual characteristics about the state that the

little group acts on. This will keep A invariant, and this is what, F is reserved for little group. And

we already argue that the little group is just SO3. We know what, the irreducible representation of

SO3. Maybe I’ve already erased it. This is labeled by the total angular momentum, and N. Okay?

Where L can be 0, 1/2, and so on. Okay? At least to start out this vector, we can write, we can –

it is pretty clear to me that in this vector space, what rotation really is, representation is. Okay?

It’s just that. I can write F or I can classify different offer according to L and M, if you think about

this additional quantum numbers, I can write this in terms of L and N.

where Mαβ(R) is a rotational matrix.

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state little group SO(3)

R ∈ SO(3) |n, α〉 L,m

U(R)|n, α〉 = Mαβ(R)|n, α〉

Okay? Therefore, a, for R, said it the other way, if R is SO3, UR on this state, okay, unitary

representation of this state is just a R for beta R, I can write it as a matrix. And this is just

nothing but the rotational matrix. I’m just writing this so that you don’t feel this is very mysterious.

This is the rotational matrix you know and love, okay? So spin 1/2 is just the E to the Is sigma

without theta we wrote earlier. Okay? Let’s ask ourselves, armed with all this information, let’s

ask ourselves what is U lambda? General transformation, on a general state vector. Okay? This

is the last ingredient we need to build irreducible representations. Okay? To do it in a very, how

shall I say, yeah, we will write it in a slightly different way. Okay? So this is the last sort of

nontrivial step. UL – I’ll write it. Then I’ll explain. Okay? So this is, looks very weird. But to

understand why this is the case, all you need to know is the K, lambda K, okay, obviously, this

Lorentz transformation is going to take K to lambda K. Okay?

U(Λ)|k, α〉 = U(L(Λk))U(L−1(Λk)ΛL(k))|n, α〉

boost rotation

with the Wigner rotation

W (Λ, k) = L−1(Λk)ΛL(k)

Λk = L(Λk)L−1(Λk)Λk

= L(Λk)L−1(Λk)ΛL(k)n

So you just have to understand, at least this will happen. Lambda K you can write it as L. That is

obvious. This cancels that and that. But I still think it’s not complicated enough. I want to make

a contact with the invariant factor. Now I’m going to call, separate this out, and that’s over there.

Now inside, remember that I told you that for two consecutive Lorentz transformation, this must

14


be true. This is any representation. Using these two, I get that. Is there any questions about that?

(pause). Let me write a little bit further, okay? But so far, whether I write F or not it doesn’t

even matter. Let’s understand a collection [inaudible] but otherwise there is no problem with this

one. But why do I want to make this thing more complicated than necessary? What you see is

that this is a rotation. Take any vector. This act on vector is just boosted to K. And transform

boost it back. It’s, therefore, the end result is a rotation. You can verify again, compute. This is

a rotation, and this is a boost, by definition because I use L to denote boost. Is it okay? So, this

means, why do I want to do this? When I, the first time I see this, I was totally confused. Why are

you bothering to do this? Let me write this as a rotation. Okay? Lambda and K, let me give you

the notation. Lambda K, and this is called a Wagner rotation, Mr. Vigner write it down first. So

this shows any unitary transformation. Unitary transformation of Lorentz group. This is not very

precise mathematical language. Any unitary transformation of Lorentz group can be obtained, can

be obtained from N, alpha by boost times rotation. Wigner rotation first, then do boost.

So any unitary transformation of Lorentz group can be obtained from n, α by a boost times a

rotation.

Okay? We already argued that the alpha here is, it can be classified so alpha is just irreps of

SO3. It is a label that, labeled by reduce representations of SO3. Suppose I’m already in some

ir reducible representation of SO3, labeled L and M. Labeled by L. Labeled by the total angular

momentum. Rotation is not going to change that. Boost is not going to change that either. Okay?

This means on the general Lorentz vector, any general Lorentz vector, if I’m already in the ir

reducible representation for this vector, I’m also in the ir reducible representation for any Lorentz,

general Lorentz vector. Boost, you do not change the mass. This basically tells you that there are

two labels of invariant labels of any reducible representation. Boost invariants is M, and rotation

invariants is L.

α irreps of SO(3) L. All irreps are m and L (0, 12 , ...). All members of that irrep ← |n, α〉 → boost

times rotation. Infinite dimensional, infinite number of k, k2 = m2.

Moreover, how to write all the state vectors in that ir reducible representation. So all state, yeah,

so let me say that again. So all irreps are abled by mass, and L. And all members of that irreducible

representation, can be obtained by starting from NLL and do rotation, do L times – sorry, do boost

15

1.4 Symmetries 1 JANUARY 8TH, 2013 (INTRODUCTION TO FERMIONS)

time rotation. In this sense, so you can obtain everybody in the irreducible representation, you can

induce everybody in that irreducible representation by doing a little group rotation, and a boost. So

this is the so-called induced representation method. For us, it is not necessarily absolutely needed

that we have to do this all the time, because irreducible representation is known. But I thought

it would be nice to understand this method at least once. It is very useful, and a nice method.

Okay? So, okay? So all the Lorentz irreducible representations will be NML and L again is a 0,

one-half and so on. This is obviously the Fermion and the 1 is vector and so on. What else? As I

promised, I said that so this is unitary representation. But it must be infinite dimensional. Okay?

Because Lorentz group is noncompact. How do I see it’s infinite dimensional? Representation for L

is finite dimensional. Okay? Infinite dimensional comes from there, so infinite dimensional comes

from, there is infinite number of K, of momentum K, sorry, momentum K which is set by K score

and N score. There is indeed infinite number of members in this representation. And that infinite

numbers are related to each other by exactly the noncompact part of the group which is the boost.

You can go to any K from N via boost. Is this okay? Okay. So, next time we will focus more on

Fermions. So this is a general representation of Lorentz group. Again, if you haven’t received the

E-mail, and want to be on the E-mail list, let me know. And send me an E-mail and I’ll add on to

it. I will see you on Thursday.

1.4 Symmetries

This is a little bit of a tangent. There are two kinds of symmetries. Kind of symmetry, this kind

of rotational symmetry is one kind. And as we will soon talk about generalization symmetry, and

but we also have concurring, translational, physics theory right here. It is the other end of the

universe is the same, maybe it’s the same. That is called concurrent symmetry. Translation is also.

But these are called space/time symmetries. So on, so forth. There are other kinds of symmetries.

Another kind of symmetry which we sometimes called a flavor symmetry. For those of you who

learn a little bit of standard model, this is not to be confused as the narrow sense, the actual flavors,

although that flavor is example of this kind of flavor symmetry. But flavor symmetry is some kind

of internal symmetry. An example with complex scalar being

φ→ eiθφ

16

1.4 Symmetries 1 JANUARY 8TH, 2013 (INTRODUCTION TO FERMIONS)

which rotates between the real and imaginary component of the field φ.

17

2 JANUARY 10TH, 2013

2 January 10th, 2013

Last time we started to talk about how to build a representation of a Lorentz group using the

“little group” method. We talked about, for a mass of particle, we can start with a single vector,

a special full vector – the little group is the one that keeps this vector invariant [SO(3)]

n = (m, 0, 0, 0)

Therefore, reducible representation of mass of particle has two things – one is its momentum which

satisfies k2 = m2. There is another label here which is irreducible representation corresponding to

the rotations of SO(3).

|k, α〉 (L,m = −L · · ·L)

This irreducible representation is obviously classified by the momentum, angular momentum, and

m (not the mass of the particle, but depends on L). That is a full representation of, so in other

words, the particles, particles in form representations of the Lorentz group, we can distinguish them

by their mass, by their spin and there are infinite number of members in irreducible representation

as they are infinite number of case that satisfy this. So it is infinite number of dimensional repre-

sentation which is a little bit different from the usual representations where we used to have the

representations of rotations and so on. Those are finite dimensional ones. This is, everything is

consistent, as we said that we need the unitary representations of Lorentz group, the states. The

particle states has to form unitary representations because we want to conserve property prob-

ability. On the other hand the Lorentz group is not compact because there is a boost and the

parameter space of boost is not compact. It is a 0 to infinity. Therefore, there is a theorem saying

that you cannot have a finite dimensional unitary representations of a noncompact group. But

this, everything is consistent, this is unitary but it’s infinite dimensional. This has to be, this needs

to be infinite dimensional because we know particles can come from infinite number of different

momentums. So everything is, is this okay with everyone? This is as group theory will be, okay?

So in this course, I think, it is useful at least to remember a little bit of that. The next thing, so

this is a mass of particle. This applies to any particle. It doesn’t, it is not even just Fermion or

anything you want. As we said, so far we know how to write nice looking field theories for these

18

2.1 Example 2 JANUARY 10TH, 2013

kind of spin par. For spin higher than 2, we can write down something but it is not very nice.

L = 0,1

2, 1,

3

2, 2

massless particle For massless particle, for massless particle there is also a special vector, momentum

vector of massless particle. But it’s not this one. It does not have a frame, this is the full vector for

vector particle and this one doesn’t. What is the special vector. Anybody know? Is what? That

is as special as it can be. You can put this anywhere. But let’s put there.

kµ = (k, 0, 0, k)

where the (0, 0, k) is SO(2) - also a rotational group. It’s rotation in two dimensions, not three

dimensions. How many generators are in SO2? Just one. Just one rotation. So this, the irreducible

representations of this guy, is also characterized by L. Okay? And the projections. Okay. The only

thing different is that there is only two projections here. (−L,L). Spin up, spin down.

2.1 Example

For a massive particle with spin 12 , we have possible values ±1

2 . For a massive particle with spin

1, we have values 1, 0,−1. For a massless particle with spin 12 we have values ±1

2 . For a massless

particle of spin 1, we have values ±1. (This last part underlies the Higgs Mechanism – the main

difference).

2.2 Field Operators under the Lorentz Group

How does a field operator transform? Under the Lorentz group. As I said, again, last time, its major

difference here, there are two. A is that the field operator does not need to be unitary. It is not a

state. It does not have a direct connection with a probability. It does not need to be, whatever unit

representation the field operators lives in, it does not have to be a unitary representation. These

is that we usually look for only finite dimensional representation of field operators. The reason is

the following. For particles, I said we can have infinite number of possibility, infinite number of

different momentums. We want to have infinite dimensional representation for particle. For field

operators, usually we only have finite number of fields to talk about. We have electrons. We have

19

2.2 Field Operators under the Lorentz Group 2 JANUARY 10TH, 2013

photon. We have infinite number of different protons. Different universe but usually we talk about

finite number of field. But again, you can see that everything is still consistent. I’m looking for a

finite number, finite dimensional representation of this group. And it is nonunitary. Again, it is

consistent with the fact that the Lorentz group is not compact.

For general field (do not need the α-label for a scalar)

Φα(x) M−1(Λ)Φα(x)M(Λ) = MαβΦβ(Λ−1x)

The field operator is a function of X. Usually the definition of Lorentz transformation is the fol-

lowing. Okay? Transformation, is realized by a matrix. It’s a matrix. Let’s try to motivate this

a little bit. Let’s first imagine, there is a scalar. Okay? There is nothing scalar so this does not

exist. There is at most .... for real scalar there is not even a phase. I must say this.

2.2.1 Motivation through Scalar Fields

So for scalar, there will be just fi X will go to Phi, okay, this is a particular way of

φ(x)→ φ(Λ−1x)

We are just imagining that if I imagine there is some field configuration, here, okay. I’m just taking

this to a Lorentz boost, I’m just taking it here. Therefore, the new Phi at the particular point will

be the older Phi as if you do inverse on X. I’m trying to explain this but you have to think about it

yourself. This does work. This is called the active way of thinking about symmetry transformation.

Transform the field. Another way of doing it is called passive way. You can try to think about

rotating the axis.

My way of thinking about the transformation is that transforming the field (shifting it) is effectively

equivalent to shifting the field operators instead for a specific transformation Λ.

So, let’s see what are the possible representations. To do this, we are looking for something else. We

are looking for a finite .... to do this, it’s useful to recognize the following fact. So again, remember

that we, during the last lecture we write the general formula for the Lorentz transformation on

20

2.2 Field Operators under the Lorentz Group 2 JANUARY 10TH, 2013

generators. Okay? As J, we write them as JN mu but we think it may be more useful to write them

down as the following, use another different base. Writing down them into, putting them into two

class of generators. IJK is from 1, 2, 3. This kind of Greek letter is also 0 to 3. This is a rotation.

This satisfied algebra is 03 and these are the boost. Because there is 1, 2, 3, sometimes we can just

use the normal 3, three dimensional vector notation by putting an arrow.

Li =1

2εijkJ jk ki = J0i

~J+ =1

2(~L+ i ~K) ~J− =

1

2(~L− i ~K)

[ ~J+, ~J−] = 0 [J i±, Jj±] = iεijkJk±

This is the thing that always bugs me about group theory and in general mass – there seems to be a

lot of tricks going on. There is. You have to be smart to be a mathematician. You have to somehow

stare at these things and notice all these facts. It is not like us, we start with .... ([inaudible] The

merit of doing this, is that you observe that this is zero. You separated the algebra, the algebra

into two parts that actually commute with each other. In other words, the full algebra is that sum

of sub algebra generated by J plus, another sum of sub algebra generated by J minus. Separately, J

plus/minus I, J plus/minus J, the algebra, sub algebra itself is SU2 or SO3. So the two sub algebra

here, so this way means that Lorentz group is – well, decomposed into the algebra, the algebra of

Lorentz group can be decomposed into a sum of two sub U2, SU2 sub algebras. Hmm? (pause).

K is not compact. The stuff generated by K is not compact. Whatever a symbol you put it here, I

don’t really care. But how you tell that this is what it means, okay, this is what it means.

Lorentz→ SU(2) SO(2)

So, therefore, all the irreducible representations, all the irreducible representations can be labeled

as a product of this plus and minus. All the irreducible representations are labeled by a quantum

number in plus and a quantum number in minus. That is all I wanted to say. The quantum number

in plus and minus, or in SU2, or SO3, same locally, are easy. Right? So everybody knows it’s the

Eigen value of 1/2, or Eigen value of, so the Eigen value of SU2, SA plus or minus is labeled by

J plus/minus. Again, this is where this Eigen value is 1 where N is again 1, 2.... okay? 0 as well.

21

2.3 Fermions, spin 12 2 JANUARY 10TH, 2013

That is trivial.

SU±(2) j± =n

2(n = 1, 2, · · · )

irreps (j+, j−) (2j+ + 1)(2j− + 1) dimensional Li = ji+ + ji−

All the representations can be denoted as J plus, J minus. Okay? That is a product of these

two numbers. As long as I specify the quantum number under plus and minus I specify the

representation. These are all finite dimensional representations. Okay? Again, it’s different. That

is not, I’m not looking for infinite dimensional representations anymore. Okay? So this, the

dimension is dimensional representations. Okay? (pause). The different, it is not here the term.

Let’s see how they transform. Very deep. (chuckles). Thank you. Okay. Now, of course, we know

that you can also have a better way of representing this. Well, let’s just do it.

(0, 0) : spin 0

(1

2, 0) : spin

1

2(fermion)

(0,1

2) : spin

1

2(fermion)

(1

2,1

2) : spin 1 (4 vector)

2.3 Fermions, spin 12

Okay. Now, let’s go to talk about Fermions. Let’s talk about these two representations. Field

operators in these two representations. (pause). 01, and 10, these are not irreducible. You have

to – yes. So, yeah. Okay. Let’s do Fermions. Firmiance – Fermions. We are almost through,

with Fermions. Okay, so spin 1/2. So these two representations, let’s just look at them. This

decomposition is useful, because I can immediately write down the transformation rules, because I

know how to write. I know how to write down the generations in arbitrary number of dimensions, so

that’s all known. But before we do that, let’s make another sort of general comment. In Fermions,

we are used to think about it, it’s a column vector, complex column vector, two-dimensional complex

column vector. This is what we usually think about. On the other hand, we know the location of

group is SO3. Whatever locally is SO3, it is rotation. There is a little bit of mystery here, because

22

2.3 Fermions, spin 12 2 JANUARY 10TH, 2013

SO3 we know it’s real. So why is more convenient to think about – of course, you can also say, it’s

SU2 why it’s more convenient to think about SU2 than SO3, because SO3 is not single value in spin

1/2 rotations. So as you learn, it is a great mystery why you rotate 360 degrees, get back maybe

one or not one. So SO3 is not single value. But what we will do is to lift SU3 into SU2, so SU2, so

the complex representation, fundamental representation of SU2, that is single value under the spin

1/2 rotation. Thinking about complex representation is always more convenient for Fermions. We

are going to do a very similar thing. In mathematical language, there is another fancy name, it’s

called universal covering group, use universal covering group. For SO3, we will still want to use a

complex, use this covering group to describe Fermions.

SO(1, 3)→ (1, ~σ) = σµ X = σµχµ

X is hermitian. Det(X) = χµχµ ← Lorentz invariant.

We don’t want to use literally as it is. This is a Lorentz group. What is that? So again, we can

try to identify it, by the following trick. So, again, so this we want to map it onto some complex,

complex, a group with complex parameters basically. So the trick, again, is to say, consider the

following quantity. Consider the following quantity which I call X, well, sorry. Obviously it has to

do something to do with the poly matrices, let me call it signal mu. This is the normal three signal

matrix and it does identity by two matrix. This form complete base of 2 by 2 is possible complete

2 by 2. (coughing). Matrices. Now, you form the following from it. You form the X, that is a

sigma mu, X mu, this is just some normal space time coordinate. Space/time format. So there are

some things, several things, X is a formation. Okay? Because these guys are permission. And the

determinant of X is just – okay? Can just work it out. Therefore, there is a Lorentzian variant –

Lorentz invariant. (oh, thanks, that’s a nice new word). X, whatever it is, it is a matrix. It’s a 2

by 2 matrix. Oh, X. Ki. How do you write X? Those are regular space time 4 vector. Whatever

it is, it is a 4-vector. Okay? This transform as SO3 too – SO13. Now, how do I transform this

2 by 2 matrix, that is satisfied these conditions. There is also a set of transformation on these

2 by 2 matrix that satisfy these kind of conditions, which preserve, which makes this Lorentz

value, preserve the determinants, and everything. So the transformation, that that’s consistent

with Lorentz transformation, has to satisfy any transformation on, like this, that satisfy the, be

23

2.4 (12 , 0) representation 2 JANUARY 10TH, 2013

consistent with Lorentz transformation, that defines a set of transformations.

X ′ = MXM−1 M → SL(2, C)

This is the universal cover we are looking for. An analogy in QM is where

x = ~σ · ~a

rotation: ei2~σ·~θxe−

i2~σ·~θ SU(2)

~a→ R(θ)~a

You say that, you say there is a vector, A. You do a rotation. Okay? But what you want to do

is to ask, what is this rotation, how do you translate into SU2? How do you translate into SU2?

What you do is, you form this combination, and ask how does this combination transform another

rotation. This transformation transform as, and this is SU2 rotation. If you remember, okay? You

can show that this, another way of saying, if you show this, you can show that this exercise in

quantum mechanics is that this, that after this, A goes to a rotation by theta of A. It is based on

vector. It is rotated by theta. Anybody knows what blog sphere is? Yeah, but so I’m just saying

that this is a common trick of mapping a real rotation into a universal cover group. In quantum

mechanics, the question is how to map SO3 into SU2. What you do is to form this combination,

and do the rotation on this combination. That is SU2. In particular, this is indeed a rotation

because this and that acting on X, in the end, is equivalent to just rotate this vector by theta. This

is the mapping between SU2 rotation and normal space time rotation. It’s very similar.

2.4 (12, 0) representation

Again this means that it’s 1/2 under J plus, is 0 under J minus. That is all it means. So the field

operator we call it C. Probably not the best. This is my way. If you have another way, use your

way. The official way is, I don’t know what is this. Whatever. Okay? The question, the problem

with this is, another one is zeta which looks like that but never mind. So because of this 1/2 0,

so there is the infinite, that just shows you how much less people try to put to make Fermion

rotation because this is 1/2 0. Because later on we will talk about 0 and 1/2 representation. So

in this representation of course these guys, these are two components. Two components, complex

in general. Okay? So we are labeling the components this by alpha. Alpha is 1 and 2. We only

24

2.4 (12 , 0) representation 2 JANUARY 10TH, 2013

are allowed to use alpha, not anything else. Alpha beta maybe. You will see where it is coming

from later. Under the Lorentz transformation, under Lorentz transformation goes to M, M is the

complex 2 by 2 we identified that way. We can identify in, well, you don’t even have to go through

the exercise. But if you do, well, that is the 2 by 2 matrix.

ζα →Mβαζβ

And we write alpha, beta. Okay. Notice these two are strategically located, and this one is

closer to and this one is further away from M. They have a lot of meanings, because this means

that this, this index is closer to that. In this case it is obvious. In some cases it is not very

obvious. This is not even very, not even very easy because usually beta will be just here. Anyway,

everything has a lot of meanings. You will see what. Okay. Alpha and beta are 1 and 2 obviously.

Immediately, the one thing you want to see for any transformation is what is the invariant tensor

in that transformation.

εαβ ε12 = −ε21 = 1

The form invariants, the point of doing upper and lower indices, easier to form invariants. Okay?

You identify, suppose I define something for epsilon alpha beta, where alpha, 12 is negative 21.

Answer is 1. Have to be careful with this matrix. Sometimes people use the other convention. That

is why convention is everything in this business. So

εαβM δαM

γβ = εδγ

This has alpha beta. It will also transform under this. Let’s make it also transform under that.

You can show that epsilon alpha beta M, okay, now alpha is on this side, beta is on this side.

Gamma on this side. This you can show is just a determinant of M, times another epsilon tensor

basically. But then the determinant is 1. Because S over there, SLC, you see it, and so this is –

okay? In group theory language, this means that this is a numeric tense or. This is like in Lorentz

transformation with Λµν , gµν .

ζα = εαβζβ ζα → ζβ(M−1)αβ ζη = ζαεαβηβ

= −ηβεαβζα

= ηβεβαζα

= ηζ

25

2.5 (0, 12) representation 2 JANUARY 10TH, 2013

First you can do this. There is negative sign because these are Fermion field operators. The anti

commute. Okay? The next thing I do is to flip this. This is – okay? Now if you see 2 components

Fermion field, you will flip it, the same sign. That is because there is two negative signs. We can

also just write down an expression of that. At least in infinite is malform. Just because we know

the generators in this, acting, how generator is acting on this representation, right? J minus is

trivial. J plus is just to spin 1/2 rotation basically. But with this generator. In this representation,

in this representation L is just normal Paulie – Pauli matrices. Okay? On the other hand, the K

is negative I. This I is crucial because K is [inaudible] M alpha beta, in this small transformation

is just this. Sorry. This is velocity. This is the velocity of the boost. Beta is V over C but we

take C as 1 usually. But this is beta. This is again the hermittian part. This is a noncompact

part. You can see that too just from this. If you wonder how a 1/2 spinor, 2 by 2 spinor transform,

it explicitly is like that, for this. Now you can transpose hermittian conjugate, that is something

else.

~L =~σ

2, ~K = −i~σ

2Mβα = 1− i~σ

2· ~θ − ~σ

2· ~β, ~β =

~v

c

2.5 (0, 12) representation

Now we talk about 0 1/2. Okay? Again, this is just taking the J minus, and switch the ro and

J minus and J plus. You say okay, that is trivial. That is trivial. But usually, there is, because

obviously, you can write down some relations between these two representations. That is why

sometimes it can get confusing. Okay? The fundamental, denoted by something called dagger

just to confuse you. Because it’s on this, it is Lorentz transformation on the right, if you want,

on this side. And 1/2, so I cannot use alpha anymore, because alpha I used to label the other

representation. Okay? So some genius come up with a representation called a dot representation.

Is it called alpha dot. That is not my fault. I have to tell you what the convention is. If I invented

weel it would be more confusing – Weyl. At this point there is no meaning. I’m not implying

this is a dagger of anything yet. That is just a notation. This is where I start. On the other

hand, obviously there is a root of this notation, where does this come from? You can show that

this representation and 1/2 0 presentation are conjugate of each. More specifically, which was

26

2.6 Lorentz Vectors 2 JANUARY 10TH, 2013

this,

ζ†α = (ζα)†

ζ†α → (M∗)βαζ†β

= ζ†β(M †)βα = ζ†M †

Now, it’s very easy to just derive the transformation laws of this, in terms of the transformation

laws on that. So, goes to M star alpha dot, beta dot. And this I can slightly rewrite it to make it

more hermitian conjugate, beta dot, alpha dot. See the subtle difference between these? It is all

very important. Or if you are lazy, write it like this. But you have to be very careful. You have to

understand what this means. Even this stuff sometimes there can be confusion. Some book decide

not to use this, they use bars. Also, you can in terms of this, in terms of this matrix, you can also

work out what is this guy. We formed earlier, we formed earlier how does this transform. Okay?

It just transforms as this close to M dagger on the other side. Therefore, you can figure out how

this transforms, how this guy transforms.

X = σµχµ → MXM †

σµαβ →Mα′α σ

µ

α′β′(M †)β

′

β

(ζη)† = η†ζ† = ζ†η†

As I said, I haven’t, so far we haven’t said anything really really deep. But there is all this

bookkeeping with Fermions you have to go through, because there is two kind of representations,

there is anti commuting itself, epsilon tensor and so on. Obviously, you also have sigma dagger.

This, okay. That is another invariant. You can define epsilon tensor with dot indices as well. This

is just dot. And this is ..... okay? These things are very useful, because so these are first set of

Lorentz invariants you can write down. Lorentz invariant is very important. You want to write

down LaGrangeians. This looks like a mass terms, because as one Fermion times another Fermion.

But we can also have other Lorentz invariants.

2.6 Lorentz Vectors

ζ†σµη σµ = (1,−~σ)

ζσµη† σµ(1, ~σ)

27

2.7 1-fermion ζα 2 JANUARY 10TH, 2013

The obviously, the right thing to do, I’m going to write down the right thing to do, is the following:

Now here is the first time we meet with bars, sigma mu bar is one. You can show this transform

is like a vector. This transform, these are Lorentz vectors. So is this one. They transform vector.

That is all I want to say. Okay. Very good. Now we are in a position to write down our first

LaGrangian.

2.7 1-fermion ζα

Okay? Let’s have one Fermion. Let’s just have that. Okay? Nothing else. This is Lorentz invariant

because that is a Lorentz vector.

L = iζ†σµ∂µζ −1

2m(ζζ + ζ†ζ†)

This is a mass of Fermion. I’ll write it down. Then I’ll make a few comments. You need this.

You need this, because this is the hermitian conjugate of that, first of all. And why isn’t Fermion

times Fermion not vanishing? Usually I would say Fermion times, because exclusion principle I

cannot have the same twice because this is different Fermions. There are two component in this,

two component in this. One component of that times second component of that. They are different

Fermions. That is the little bit, it is another confusing thing if you just write them like this. But

usually it’s much better. That is a minor, lump into this rotation. Okay, several things. Let’s

consider a limit, where mass M goes to 0.

limit where mass m→ 0 symmetry ζ → eiθF ζ “fermion #”Themasstermbreaksthissymmetry.Fermionmassbreaksthesymmetry.

Remember the difference between this and the scalar mass. What is scalar mass? What does scalar

mass do? Or Phi dagger Phi, whatever it is. This breaks no symmetry. If you met the symmetry

that this breaks, let me know. (chuckles) you met a symmetry that is also consistent with the

kinetic term that this breaks, okay, because you always have to have a kinetic term. Symmetry

is consistent with this, that this breaks. This is very important. This is a fundamental, this is

fundamental importance.

scalar: m2φ2 m2φ†φ (∂φ)2

28

2.7 1-fermion ζα 2 JANUARY 10TH, 2013

Second of all, let’s say do we know an example of this in nature? Something that we usually – the

answer is no. Okay? The Fermion, answer is no in nature, model is neutrino okay. We don’t know

whether it, it is consistent with this but we don’t know. Electrons allow this. So this set of, this

kind of Fermions are called, because it only has one Fermion, it is its own anti particle these are

called Majorana Fermions.

I’m doing the whole Fermions, if you notice, a slightly different from the older textbooks. Old

textbooks start with the same but I assure you Majoranna is much better – the two components

you use, the language is much better. The two components, this kind of two components language

and this is a generic thing called a Weyl Fermion (2-component). If you only have one Weyl Fermion

it’s Majorana, in four dimensions.

Consider 2 species (12 , 0) fermions ζi, i = 1, 2.

L =∑i=1,2

iζ†σµ∂µζ −1

2m(ζiζi + ζ†i ζ

†i )

limit m1 = m2 = m. SO(2) or U(1).

current: Jµ = i(ζ†1σµζ2 − ζ†2σ

µζ1)

Flavor symmetry, rotate to each other. There is SO2, or U1, symmetry, rotate these two into each

other. Cosine, sine theta, because again this is invariant. Only in this limit. In general, it’s not

true. If you follow the procedure, conserve current of this is J mu 1 sigma, 2, so minus, that is the

current of that. This is not a very – again, this is SO2 language. This is the SO2. In this case you

will see very explicitly why this SO2. But again, doing this kind of business, SO group is usually

not optimal. It is useful to think about U1 rather than SO2.

χ ≡ 1

2(ζ1 + iζ2), η =

1

2(ζ1 − iζ2)→ Jµ = χ†σµχ− η†σµη

where Jµ is U(1)Q charge current. This theory looks like QED with electron and positrons.

29

3 JAN. 29TH, 2013 – SPIN-1 VECTORS

3 Jan. 29th, 2013 – Spin-1 Vectors

Spin-1 vectors: photons γ, W and Z vectors, and gluons. These are called the force carriers. Why?

Can think about two electrons interacting via a photon intermediary. The force carrier refers to

the fact that you can draw this kind of diagram, saying that there are scalars instead of vectors

– using the Yukawa theory talked about earlier. There are other vectors, QED vectors, such as

ρ±, ρ0, a, φ.

In Effective Field Theory, the cutoff is much larger than the mass of the vectors. In QCD, the

cutoff is approximately the same as the vector, about 1 GeV. Also, last quarter, we said that for

theory that has a cutoff much bigger than the characteristic mass scale of the theory, there is a

pretty nice name to it – renormalizable – relevant operators are small, suppressed by some large

cutoff.

3.1 Photons

What kind of theory describes vectors? Gauge theory. Let’s discuss vectors and focus on photons for

now, E&M. One thing we learned is mγ = 0. This is spin 1 and for massless spin – representation

for Lorentz group, we see that a little group of SO(2) [irreducible representation] works here.

Identify a special vector k = (E, 0, 0, E). This is easiest because mass is zero. Why SO(2)? Two

polarizations (γ has 2 physical d.o.f). We can also see this from Maxwell’s equations - dF = 0 or

∂µFµν = 0.

Fµν = ∂µAν − ∂νAµ

What is the solution of the wavefunction of the photon? It is in terms of this vector potential with

some polarization vector

Aµ = εµeip·x

In momentum space (p-space)

pµFµν = pµ(pµεν − pνεµ) = p2εν − (p · ε)pν = 0

So there are two kinds of solutions.

30

3.2 Lorentz invariance, unitarity 3 JAN. 29TH, 2013 – SPIN-1 VECTORS

1. ε0 = pν(p·ε)p2 Fµν = 0 identically – “longitudinal” polarization

2. p2 = 0, (p · ε) = 0← “transverse”

We can see that there are two polarization vectors that satisfy this kind of condition. But we can

always add another term because p2 = 0

εµ = α+εµ+ + α−ε

µ− + αpµ

3.2 Lorentz invariance, unitarity

We always wanted to do things in the manifest Lorentz invariant way so we don’t have to check

every time. How do we describe a vector, writing down, writing it as a Lorentz invariant, as

possible. Obviously the way to do it is promote it to a full vector. This usually requires us write

things in terms of A mu. Not A1, A2 but Aµ. Not just two polarizations. This has 4 degrees of

freedom. But we immediately see there is a problem here – we know there are only two physical

degrees of freedom which means that we have a “redundant” description. There is something about

this theory that needs to tell me that I can remove this redundancy. That is the gauge equivalence,

gauge invariance. Redundant means there are also many equivalent descriptions of the same theory,

because there are two physical variables here.

We know Aµ ∝ εµeipx so how do we reduce it?

1. impose p · ε = 0 – take 4 d.o.f. → 3 d.o.f.

2. physical εµ± = 1√2

(0, 1,±i, 0) a quantization, but this is not manifestly Lorentz invariant.

Secretly, there is Lorentz invariant theory, so what is the mechanism? What is the secret of keeping

the theory Lorentz invariant?

3.2.1 Lorentz Boost

εµ → εµ + α(p)pµ ← (p · ε) = 0

We require the theory to be Lorentz invariant which means εµ and εµ+α(p)pµ is the same physical

state. It refers to the same particle, just view it in two different Lorentz frames. In a more familiar

31


way, the theory stays the same if

Aµ(x)→ Aµ(x) + ∂µα(x) ≡ Aα

This is known as the gauge transformation of the photon field. It is also known as a gauge invariance,

gauge equivalence,((((((((

(gauge symmetry, gauge redundancy. Speaking roughly, it means the true theory

using a path integral has to be identical∫D(A)S[A] identical≡

∫D(Aα)S[Aα]

Gauge symmetry is not a symmetry. It is a redundant theory. I think I sort of made this point

last quarter but let me make it again. It is very important to understand. We have two, we have

real symmetries, such as global symmetry. One global symmetry is rotational symmetry. I can just

rotate. Rotational symmetry means I have two different theories N. one theory I call this X, this

Y. In another theory I call this X, this Y. These two theories give you the same, if you calculate

the same physical quantity they give you the same result. That is a symmetry. In principle these

are two theories, and that’s, symmetry need not hold for any particular physical system. It is a

physical statement, whether I have two theories are different or not. On the other hand, we see

that this is not a symmetry. It means that this is just the same theory. I have infinite way of

describing the same theory. Talking about the same physical degree of freedom.

tl;dr – Gauge Redundancy is the best terminology.

We need to do gauge fixing. I implicitly fix the two gauges. This kind of requirement implies strong

constraints on physical observables such as S-matrix. We work with µ

M = εµMµ; µ invariant εµ → εµ + αpµ

→ pµMµ = 0 Ward identity (exact)

The Ward identity is exact, nothing about perturbation theory. It is a difference between four

different functions,M0,M1,M2,M3 for arbitrary external momentum. In QFT, exact results are

very rare. Later when we talk about QED, we will prove this identity within that context, but here

you see that it follows directly from gauge invariants.

32


Let’s see what else we can get. Consider a propagator of photons, from µ→ ν.

µk→

:::::::::::::ν = Dµν(k) =

i

k2 + iε

(Agµν +B

kµkνk2

)• unitarity of S-matrix → only 2 d.o.f. on initial & fixed state

• Consider a special process [include image with blocks]

M = JµDµνJν = Jµ

i

k2 + iε

(Agµν +B

kµkνk2

)Jν

Recall the Optical theorem

Now, we sum over the physical polarizations λ = ± and have two parts – the left part is εµλ

and the right part is εν∗λ . We haven’t done anything else yet, this is just unitarity. Remember

that the optical theorem relates an imaginary part of the amplitude to a physical amplitude

square with intermediate particle on shell. Therefore, because this is S-matrix element, you

sum only the physical polarizations.

Choose

kµ = E(1, 0, 0, 1), εµ± =1√2

(0, 1,±i, 0)

Define

kµ = E(1, 0, 0,−1)

33

4 FEBRUARY 5TH, 2013

so

→∑λ=±

εµλεν∗λ = −gµν +

kµkν + kν kµ

k · k

• Using the optical theorem → A = −1 (A is fixed)

• kµkν+kν kµk·k is not Lorentz invariant, does not contribute b/c of gauge invariance, b/c kµJ

µ = 0

(follows from Ward identity)

• B is not fixed (not physical) – will have to make a “gauge choice”

End result - propagator is

Dµν(k) =i

k2 + iε

(−gµν +B

kµkνk2

)• Summing over external state. S-matrix: Mλ = εµλMµ.∑

λ

|M|2 =∑λ=±

εµλεν∗λ MµM∗ν = −gµνMµM∗ν =((((

(((((

−|M0|2 + |M3|2 + |M1|2 + |M2|2

but from before, we saw that since the second part doesn’t contribute, we can replace∑λ=±

εµλεν∗λ → −gµν

and that we get something physical

|M1|2 + |M2|2 = |εµ+M+µ|2 + |εµ−M−µ|2

and something that needs to cancel for this statement to make sense. From Ward Identity

kµMµ = 0

and choose k = (E, 0, 0, E) so that

EM0 − EM3 = 0 → M0 =M3

4 February 5th, 2013

Okay. Everybody happy with QED so far? Today we will continue with QED. Last time we start

to calculate a scattering process. Let me say again, for the next couple lectures, we complete in

34


parallel with what we did in scalar field. We are going to go through the motions of calculate

some simple process and normalize QED. This is the basic stuff. The only new thing is now we are

dealing with Fermions and vectors. So the complication is more, so the computation is a little more

complicated. You have to deal with 4 by 4 matrices and so on. Otherwise, look at the big review

article by Steve Martin and so on, you would have done the two components. It does not actually

simplify a lot QED, but it does simplify a lot for later if you want to work on Susiea. (?) But for

now we have to deal with, let’s go through this fairly standard Q.E.D. stuff. By the way, what I’m

going to say is the small subset of chapter 5 in Peskin. There will be homework also doing similar

calculation, but you are strongly encouraged to go through the rest of chapter 5 in Peskin. It is all

fairly straightforward.

We started to calculate the following process last time

e+e− → µ+µ−

know, 100 and some – don’t have 500. Do we know why this is heavier than electron? Is there any

reason why this has to be very heavier? There is no reason. We don’t know why. This is called a

flavor problem. It has nothing to do with your food. But these are called flavors. Predecessors are

very good at giving names. These are called flavors. Later on you will learn colors. I think there

are people also, made models that particles, they are called – yes. Yes. Those names never caught

on. But mainly because that model doesn’t work, so it’s a technical model, doesn’t quite work.

The flavor, the physics of flavor used to be, there is old name that doesn’t quite use anymore, called

quantum flavor dynamics. In the physics, colors is called quantum chromo dynamics also known as

QCD. I’m trying to remember the word people use for the smell. Okay. Hopefully you don’t have

to deal with it.

The nice thing is we only have to deal with one diagram.

1

4

∑|M|2 =

1

4

∑ e4

q4

(ν(p′)γµu(p)u(p)γνν(p′)

) (u(k)γµν(k′)ν(k′)γνu(k)

)=

e4

4q4Tr[(p′ −me)γ

µ(p+me)γν]

Tr[(k +mµ)γµ(k

′ −mµ)γν]

where we replaced

u(p)u(p)→ p+me

ν(k′)ν(k′)→k′ −mµ k′ = k

′ργρ

35


So computing some traces

Trγµγν = 4gµν γµ, γν = 2gνµ · I4×4

γµγν + γνγµ = 2gµν

Tr(γµγν) + Tr(γνγµ) = 8gµν

We can also do more complicated traces by running through cyclical permutations (definition of

trace)

Tr(γµγνγργσ) = Tr(2gµνγργσ − γνγµγργσ)

= Tr(2gµνγργσ − γν2gµργσ + γνγρ2gµσ − γνγργσγµ)

= 4(gµνgρσ − gµρgνσ + gµσgνρ)

There are some other things you can say. For example, you can show that the trace of gamma mu,

gamma mu, gamma ro is 0. Or let’s begin with a simpler one. First of all, trace of gamma mu is 0.

Explicit verification, or there is nothing that is a number, okay? There is no constant that carries

a mu indices. There is nothing. As I said there is only, the only thing that is a number that carries

over indices is the G mu, mus, are the G mu, mus. Then you can show that the G mu G or gamma

mu gamma mu gamma ro is zero, mostly because we can permute this and that reduce down to

that.

Tr(γµ) = 0, Tr(γµγνγρ) = 0, Tr(odd # γ) = 0

Useful identities for γ-matrices

Back to the big-ass trace before, we’ll quote the result. So for example, the second trace

Tr[(k +mµ)γµ(k

′ −mµ)γν]

= 4[kµk

′ν + kνk

′µ − gµν

[k′k +m2

µ

]]

Tr[(p′ −me)γ

µ(p+me)γν]

= 4[p′µpν + p′νpµ − gµνp′p

]me → 0

I’m going to take the limit that the electron mass is zero. Just to simplify my equations. Muons

are much more heavier than electrons. In other words, in order for that process to go through, the

36

http://en.wikipedia.org/wiki/Gamma_matrices

4.1 Renormalization of QED 4 FEBRUARY 5TH, 2013

minimum energy, the center of mass energy of those two electrons has to be greater than 2 times

the mass of the muon. This means that |p|, |p′| me. Doing the math gives us

1

4

∑|M|2 =

8e4

q4

[(pk)(p′k′) + (pk′)(p′k) +m2

µ(pp′)]

4.1 Renormalization of QED

L = −1

4(F 0

µν)2 + ψ0(i∂ −m0)ψ0 − e0ψ0γµψ0A

0µ

ψ0 = ψz1/22 A0

µ = Aµz1/23

δ3 = z3 − 1, δ2 = z2 − 1, δ1 = z1 − 1, δm = z2m0 −m

L = −1

4(Fµν)2 + ψ(i∂ −m)ψ − eψγµψAµ −

δ3

4(Fµν)2 + ψ(iδ2∂ − δm)ψ − eδ1ψγ

µψAµ

4.1.1 Feynman Rules of Renormalization

And then there are some rules involved (including the Feynman gauge, photon propagator, full

propagator, fermion propagator, and so forth).

4.1.2 Renormalization Conditions (on-shell scheme)

∑(p = m) = 0 m is a pole

d

dp

∑(p)

∣∣∣∣p=m

= 0 canonical normalization

∏(q2 = 0) = 0 canonical normalization for Aµ

−ieΓµ(p′ − p = q = 0) = −ieγµ

37



Ward Identitiy QED (Ward-Takahasha Identity)

εµ εµ + α(p2)pµ εµMµ → pµMµ = 0

I remind you that the one identity, we mean that, so what we said is, it’s crucial to ensure that

this is the piece that ensures the series of Lorentzian variant, and, so sorry, it’s consistent to have

a gauge equivalence, and Lorentzian variants at the same time. So, you can take a different point

of view, as we said, at the beginning, if you just insist on, I’m going to talk about a quantity that’s

... if you want to start out with something, like this, and under Lorentz transformation, it goes

like this. It goes to like times some alpha, in general. Polarization vector has this kind of boost.

This can be a arbitrary function of P in principle, scatter function. And because S matrix element

is proportional to this, therefore, Lorentzian variance has to be consistent with this picture. I

have to require this is 0. Okay? This is the version of Lorentz, one identity. This is the gauge

transformation. That is the gauge transformation on the photon field. In this way of saying that,

the fact that the theory is the equivalent under this kind of, this set of gauge transformations,

is essential in guarantee the Lorentzian variants. It means that this must be zero. Okay? So,

this sounds like, this sounds very reasonable. This almost is a drirvation – derivation, sometimes

you can think of this as a dare ivation if you think gauge symmetry, this is really fundamental.

And that’s fine. That is probably the quickest way to see that this has to be true. If you buy

this, that the theory is gauge equivalents. But to be very strictly speaking, this is a statement of

just equivalence between the gauge transformation and the one identity. These two statements are

equivalent. But it is not really a proof of this identity itself. In particular, in the following sense,

we know that we haven’t shown that the theory, if you just compute these things, it actually satisfy

this property. Okay. You have just shown that these two statements are equivalent.

We have to fix the gauge, and we do this by introducing a term like this into the Lagrangian

−(∂ ·A)2

2ξ← LG.f.

Let me repeat that proof. It is interesting proof. You should go through it. Yeah. You mean

whether this term is, has a counter term or not? What is the question? You add this term from

the beginning. One of the things we will show is that this term is not renormalized. That is a

38

5.1 Proof 5 FEBRUARY 14TH, 2013

consequence that we will show. Okay? That is part of the result. After we are done, similarly you

will see, okay. That is what we do. We are going to start with LaGrangian gauge fixing term, and

show that that is the case. Okay? Without using explicit gauge symmetry, without using gauge

symmetry.

5.1 Proof

S-matrix

〈f |i〉 = iεµ∫

dx eik·x2x · · · (2xi +M2i )〈Ω|TAµ(x) · · · |Ω〉

5.1.1 Example

For example, if I have 5 masses interacting somehow, then I can expect

〈f |i〉 ∝ (p21 −m2

1)(p22 −m2

2) · · · 〈Ω|Tφ1φ2 · · · |Ω〉

with the large vacuum state component at the end

〈Ω|Tφ1φ2 · · · |Ω〉 ∝1

p21 −m2

1

1

p22 −m2

2

5.1.2 Schwinger-Dyson Equation

〈Ω|T δS

δφaφa1(x1) · · ·φan(xn)|Ω〉 = i

n∑i=1

〈Ω|Tφa1(x1) · · · δaaiδ(4)(x− xi) · · ·φaφan(xn)|Ω〉

which contains the action S with respect to any variation of any fields. (Note differentiation

follows the Feynman rules we’ve defined before along with the fourier transformation **** Ask

about functional differentiation)

δS

δφa= 0⇒ classical equation of motion

39

5.2 General 3-point Function 5 FEBRUARY 14TH, 2013

From above, combining everything, we have 2A = J+ δ.∫d4x d4xi e

ikxeikixiδ(x− xi)f(x, xi) =

∫d4x ei(k+ki)xf(x, xi = x)

= f(k + ki) to contribute to S-matrix, needs to have form

1

k2

1

k2i −m2

i

5.1.3 Renormalized QED

E.O.M. Z32Aµ = z1jµ where jµ = eψγµψ

〈f |i〉 = εµiz1

z3

∫d4x eikx · · · 〈Ω|jµ(x) · · · |Ω〉

= εµMµ ∂µMµ = 0 (Ward identity)

So

∂µ〈Ω|Tjµ(x)φ(x1) · · ·φ(xn)|Ω〉 = −i∑i

〈Ω|Tφ(x1) · · ·∆φδ(4)(x− xi) · · ·φ(xn)|Ω〉

with the conserved current symmetry δφ = ε∆φ. This is a statement of global symmetry and does

not depend on whether your gauge is fixed or not. Therefore, we can make the replacement

kµMµ = −∫

d4x eik·x · · · 〈Ω|T∂µjµ · · · |Ω〉 = 0

This is the Ward Identity.

5.2 General 3-point Function

insert a picture of a general three point function with jµ(x), p′, p, ψ(z), ψ(y) and µ with q

40


Gµ(q, p′, p) = iz1

∫d4x d4y d4z exp(ip′y − iqx− ipz)〈Ω|Tjµ(x)ψ(y)ψ(z)|Ω〉

= (2π)4δ(q + p− p′)×[S(p′)

(−ieΓµ(p′, p

)S(p)

]

S(p) =i

p−m−∑

(p)

qµGµ = −i(2π)4δ(q + p− p′)×

[S(p′) (−ieqµΓµ)S(p)

]= z1

∫dx dy dz e···∂xµ〈Ω|Tjµ(x)ψ(y)ψ(z)|Ω〉

But we can also see that this one is taking, again, the expression upstairs, it’s just so. Again, I

don’t write the exponential. TJ mu, okay? That is just that. I’m taking the derivative with respect

to X. Derivative with respect to X. Okay. Now we say, great. Why is that great? Because this is 0.

Okay? Unfortunately, no, that is not 0. Okay? We will have to be careful with this. Okay?

So, because now, I’m not talking about the S matrix anymore. Okay? I’m talking about, I’m talking

about actually a matrix element. Now you do have to worry about all those Delta functions. I’m

not doing S matrix anymore. It is not S matrix. It’s a greens function. So this is, this is again to

conserve the current of our U1 symmetry.

Jµ = z2eψγµψ = z2j

µ conserved current

Let’s explicitly write those Delta functions out. Okay? Okay, again I’m using Schwinger-Dyson

equation. I don’t know which board it’s on. I think I erased it. But that is just the Schwinger-

Dyson equation for conserve the current. Okay? I’m just writing it out. There are two Delta

functions corresponding to when X and Y gets closer, and, or X and Z gets closer.

U(1) symm.→ z2∂µ〈Ω|jµ(x)ψ(y)ψ(z)|Ω = −eδ4(x− y)〈Ω|Tψ(y)ψ(z)|Ω〉

+ eδ4(x− z)〈Ω|Tψ(y)ψ(z)|Ω〉

Okay? Again, so it’s these two point functions, obviously are connected to the propagators. These

are just the propagator, two propagator. So psiY sidebar Z, these are just the fourier transform of

41


the full propagator.

〈Ω|Tψ(y)ψ(z)|Ω〉 =

∫d4k

(2π)4eik(z−y)S(k)

This SK is just the, this propagator here. Okay? Now I’m going to put all these things together.

So I have this, okay? I have this. But I did not use this, provided in terms of the sum of two,

two-point functions. Two, two point functions are just a propagator. Let me put everything in,

and then write it more carefully. The end result for that is that I have this, E5P, C1 over C2.

Okay? So C1 I inherited from here, and C2, I inherit from here. This is important, C1 and C2,

normalization constants. Q, ESP minus ESP prime. Okay? These are, again these are the two,

two-point functions, which is just a full propagator.

qµGµ(q, p′, p) =z1

z2(2π)4δ4(q + p− p′)

[eS(p)− eS(p′)

]= qµ

[S(p′)(−ieΓµ)S(p)

]where the last line was derived from drawing diagrams and looking at the diagram. The reason

I wanted to do that, in the end there is a relation between this three-point function and those

two-point functions. So you see this is already quite nontrivial. Right? This means that ward

identity, or more specifically, the conservation of a global U1 symmetry, the global, exist in global

U1 symmetry. There is a nontrivial relation between a general three point function, three point

vertex function, and the full two point functions. Again, this is exact. Okay? Let me, when we

went through all this, not because we cannot calculate one loop to verify this but because this

is exactly wrong. The only thing that we relied on are just this existence of symmetry, and the

Schwinger-Dyson equations which are all exactly right.

These are 4 by 4 matrices. Let me multiply from this side S inverse from P and from this side

S inverse P. This side S inverse P and from this side S inverse P. Okay? Just to make it looks

nicer.

−iqµΓµ(p′ = p+ q, p) =z1

z2

(eS−1(p′)− eS−1(p)

)But before we even do that, let’s just say this is, behold, 1 over U, exactly Delta function. Let’s

see how to use it. As PQ goes to 0, so of course you say this is 0. Until you notice the other side

is also 0. Okay? If you say this means 0 equals 0 you haven’t graduated yet. This means that you

42

5.3 Yet Another (not complete) Proof for Ward Identity 5 FEBRUARY 14TH, 2013

need to expand both sides, to the second order, and see if there is, there is a nontrivial result, once

you obtain the result on both sides.

and then take limit as q → 0.

Γµ(q = 0) = γµ → z1 = z2

Yeah, you can divide both of them by Q. Either way you want to do it, there is a expansion you can

do, that is what I’m going to say. Okay? We know that origin of the table expansion, this is 0, this

is gamma mu. This is our renormalization symmetry. That is just the condition we choose.

Under this condition, after you do your tailor expansion which is trivial, when those things are

getting very close by it’s trivial, you will see that the nontrivial result you get is the factor in front

is what? Here you are using this. You are using that. You are using everything. Okay? This is

our proof that these two normalization constant is the same, as we demonstrated at the one loop

order already, Delta 1 is the same as Delta 2 but this is the exact proof.

5.3 Yet Another (not complete) Proof for Ward Identity

In fact, I’m not going to give you the full proof. But just show you how this thing will go, and in

the one intermediate step we will actually obtain another useful result. Okay? This is even more

like, even more direct in a sense you will see. We know that everything is specified by some Z

function. For QED partition function can be written down as like this. The passage, I haven’t told

you how to do past integral in Fermions and we are not going to use it here. Let’s just pretend we

can do it.

Z[J ] =

∫DADψDψ exp

[i

∫d4x

(Lgauge-inv −

(∂ ·A)2

2ξ+ J ·A+ · · ·

)]That is not the conserve current. We need to introduce os strom from Fermions. I’m not going

to use that so let me just dot dot dot. Okay? If you want to calculate correction functions with

Fermions [inaudible] only thing annoying is that the source term is familiar in your case. But we

haven’t talked about variables. We are not going to use it. Okay. That’s that. But that is as ab

initial yo as it can be, there is nothing more general than starting from that. Now let’s do gauge

transformation. Let’s transform. Just be done with it. Now let me write. Suppress all the indices.

43


Imagine these are some differential forms.

A = B − ∂α ψ = eiαψ′, DA = DB Dψ = Dψ′

We are changing variables from A → B and so on, these measures are not invariant. We’re doing

a phase shift.

Z[J ] =

∫DBDψ′Dψ′ exp

[i

∫Lgauge-inv(B,ψ′)− 1

2ξ(∂B −2α)2 + J(B − ∂α)

]From that I derive this DB, psi bar ... again, gauge invariant, that doesn’t change, because it’s

gauge invariant. I can just call the field new names. That is what I mean by gauge invariant. The

second term changes under this transformation. The second term changes, to alpha .. there is a

derivative acting on that. Becomes a box. That, and J, the third term changes too, by the way.

So J.E minus D alpha. Okay? So everything changes. Let’s do some manipulation. First I wanted

to, just to be confusing, I will relabel B as A. I just go back. I want to compare with the original

LaGrangian basically. It is just a name.

relabel B → A,ψ′ → ψ

Z[J ] =

∫DADψDψ exp

i

∫Lgauage-inv(A,ψ)− (∂A)2

2ξ+

(Jµ −

1

ξ∂µ2α

)Aµ −

1

2ξ(2α)2

× exp

(i

∫α∂µJ

)So all of this, I’m just coming from, I’m squaring this and recombining the form. Box alpha square,

so the whole thing, the whole thing times another exponential, I alpha, I alpha D mu J. Okay?

And I’m not asking you to follow every single step. But you see the gist of this. I’m just taking the

function and do a gauge, go from there. Not make any assumptions whatsoever. This is a constant.

It’s not a constant but it is a constant as far as past integral is concerned. We know any constant

in past integral doesn’t matter, I just get rid of. In principle I can write a overall constant in front.

But it’s cumbersome. Why don’t I just assume that’s one. Okay? So let’s see.

So, let’s see what this means to us. This means if you compare this with the original LaGrangian,

this actually means CJ is C, the same Z but only a shift is J. This looks like a shift on the source

term. Okay? Nothing, nothing fancy. I’m just shifting the source term.

Z[J ] = Z

[J − 1

ξ∂µ2α

]ei∫α∂µJµ

44


Okay? We know that this is the generating function of all correlation functions, W is the generating

function. That is why overall constant in that doesn’t matter, okay? This is the generating function,

or constant is this plus a constant but doesn’t depend on J, so it doesn’t matter.

Z[J ] = e−iW [J ]

Taking this derivative with respect to J, you get all the graing functions but this means that WJ

is just WJ1 over ... that is why I’m always keeping this J around

W [J ] = W

[J − 1

ξ∂2α

]−∫α∂µJ

µ

45



Anybody tried homework already? Easy? Trivial, right? It is only QED calculations. Later has

loops. But that is only QED loops. The first time is always hard, you have to do it at least

once. The second time, you begin to do actual, it will be harder. Then you discover people don’t

really do it. The software help you do it. You have to write software, because all the procedures

is pretty standard. The real difficult part is not to write down a loop, is to actually evaluate it.

The real process has hundreds of diagrams. You talk about QCD process, for example. You can

actually make a living doing that. Today we will wrap up ward identity and move on to the next

subject.

6.1 Ward Identities

Let me briefly recap what we did last time. I remind you, the last time, the goal is to start out with

a gauge fixed Lagrangian, and demonstrate ward identity. Last time, close to the end we embarked

on another different approach, trying to demonstrate explicitly ward identity is true.

We started out with the partition function – the grandaddy of everything – if we know it, we know

the whole theory. All the correlation functions can be obtained by differentiation with respect to

this current, for each field you have a current. Let’s just write it formally, gauge invariant (GI =

gauge invarient, GF = gauge fixing)

Z[J ] =

∫DADψDψ exp

[i

∫d4x (LGI(A) + LGF (A) + Jµ ·Aµ + · · · )

]The gauge fixing here is (as a function of A)

LGF = −(∂A)2

2ξ

Now we would like to start out with this lagrangian and do gauge transformation. Gauge transfor-

mation is doing this

Aµ = Bµ − ∂µα(x), ψ = eiαψ′

DA = DB DψDψ = Dψ′Dψ′

LGI(A) = LGI(B)relabel B → A→ LGI(A)

46

6.1 Ward Identities 6 FEBRUARY 19TH, 2013

This is just a shift, and as far as functional integral is concerned, a constant shift. The gauge

invariant part is obviously invariant. Let’s rewrite it in terms of the new variables

Z[J ] =

∫DADψDψ exp

[i

∫LGI(A)− 1

2

(∂ ·A)2

2ξ+

(Jµ −

1

ξ∂µ2α(x)

)Aµ −

1

2ξ(2α)2

]× ei

∫α∂µJµ

Let’s write the corrections. So always in this square bracket, J mu ... alpha X, A mu, okay, that

is one term. The other one is box alpha square. Okay? This one is a constant. It’s a constant so

it goes away. This means that this one is not, it does not have anything that will integrate over,

or integrate over A. It does not depend on J. It is not something that we care about in the end.

It is just a constant. So also known as doesn’t exist, okay? So we can see, the overall constant,

past integral doesn’t matter. The whole thing, so this also times yet another exponential I can –

A independent exponential looks like that. This one is not quite a constant because it depends on

J. It is a constant as far as the functional integral is concerned. But we want to keep it because it

depends on J. That’s it. That is the result. I haven’t made any assumptions. I just did a dumb

gauge transformation.

The new J looks like

Jµ → J ′µ = Jµ −1

ξ∂µ2α(x)

So now you can also write this, you recognize this is just the statement that Z[J ], the first part is

just old Lagrangian. The first part is the old Lagrangian. This one is, plus this term is just old

Z[J ].

Z[J ] = Z[J ′]ei∫α∂µJµ

And recall from before that

Z[J ] = eiW [J ]

which can be called the Ward identity - to write Z[J ] is a more explicit function. The usefulness

of W [J ] is the same as Z[J ] – where if we differentiate with respect to the J (currents), we obtain

correlation functions. But we can see that

W [J ] = W [J ′]−∫α∂µJ

µ

47


So let’s do this for very small α

W [J ] = W [J ]−∫

d4x1

ξ∂µ2α

δW [J ]

δJµ(x)+ · · ·+

∫d4xα∂µJ

µ(x)

So since alpha is arbitrary, this guy is arbitrary. Okay? The same thing you do with classical

mechanics variation of principles. You make a variation and the fact of the derivation the rest of

it is 0. So this kills that. You get our first identity, which is box minus plus mu J mu is 0. Okay?

Is this okay? Is everything okay? Let’s put like X back. This is all X. Everything is with respect

to X. Okay? You can call this ward identity too. You can call many things ward identity. Let’s

take another variation. Let’s take another functional derivative out of this equation. Okay? So

basically because this is identical is 0, let’s take another functional derivative of it. Okay? Of JY.

Just to be, so let’s be careful, let’s put labels on these things. Now, this gives you 1 over ... let

me be careful, box X, D square, Delta square WJ. Mu Y is the same. I’m moving this term to the

other side. Okay? Partial X mu, Delta X minus 1. This is still partial X. Mu becomes mu because

I’m taking derivative with Delta J mu. I’m just using Delta J mu X, Delta J mu Y, the Delta mu,

mu, Delta X minus Y basically. That is all I have done. The reason I wanted to do that is this

looks very much like a 2 point function already. This part is just a 2 point function. Remember

this is how we derived our propagators. Take the two functional derivative of generating functional

W.

δ

δJν(y)

(1

ξ∂µ2

δW [J ]

δJµ(x)− ∂µJµ(x)

)= 0

→ 1

ξ∂µx2x

δ2W [J ]

δJµ(x)δJν(y)= ∂xνδ(x− y)

This gives you a 2 point correlation function. Which is related to the full propagator. So in

particular, full propagator, two point function, is defined by just this. Again we did this with

scalar field last quarter without the mu and mu. But that is not, there is nothing new really with

mu. At J equal 0, this is our 2 point function. Okay? This is our 2 point function. Now, what does

this mean? I can take this definition, put it in here, and what does this mean? This means that 1

over XKC partial mu X mu box X, by the way, this all just means that I’m taking derivative with

respect to X. Mu, mu, X minus Y, D mu X Delta X minus Y. Okay? Or 1 over KCP mu P square

Delta mu, mu P is P mu.

δ2W [J ]

δJµ(x)δJν(y)

∣∣∣∣J=0

= ∆µν(x− y)

48


ξ∂µx2x∆µν(x− y) = ∂νxδ(x− y)

or

1

ξpµp2∆µν(p) = pν ***

So let’s see, I can get rid of this already. That equation means, so okay, let’s start out from another

point of view. Okay? What is the general form of mu new? Just general form. General form,

of mu and new, of Delta new mu. We know it has something to do with the momentum. The

momentum will enter, we know between mu and mu, the things that can give you mu and mu

are the momentum and G mu mu. Using that, you can just argue that the most general form of

whatever it is, is the following. It’s just this. Okay? That’s it. Okay?

∆µν =1

p2

[(gµν −

pµpνp2

)F (p2) +

pµpνp2

G(p2)

]derive this from an effective lagrangian

Let me call it, give it a symbol. Applying this to that gives you, let’s see P mu, P mu, cancels that,

so it gives you that P mu over KC. Sorry. Let me just .... (pause). Yes. Times GP square gives

P mu. Okay? I’m just using that identity on this one. This means P square, although in general

is a function, you think it can be a function of P square based on Lorentz invariants, but it’s not.

It is just a constant. Okay? Exact ward identity implication this term is just a constant which is

KC, period. Nothing – okay? All right. This means that for exactly, exact 2 point function, you

can always write this form. This is always KC. It is just the gauge fixing parameter. No matter

what you do. Okay? I mean independent of renormalization, and independent of anything. That

is why also we decompose it this way. This is exactly just the gauge fixing term basically, showing

up itself in propagators. Okay? We can also look at it a little bit carefully.

***→ pνξG(p2) = pν G(p2) = ξ

So this, you see that I can derive this from this effective LaGrangian. I can derive this from

an effective action, actually. But effective LaGrangian. How do I derive this propagator from

LaGrangian? It is coming from the LaGrangian of this form. P square plus P mu, P mu, A mu, P,

F minus 1P square and A mu. Okay? Starting out with this LaGrangian, you ask yourself what is

the propagator of A? You get that. Okay? Here, so now this is momentum space. You may not feel

comfortable with momentum space because we write it in LaGrangian position space. So but this,

the first term, the first term in position space is just, so the first term is just the following. F mu,

49

6.2 Spontaneous Symmetry Breaking (SSB) 6 FEBRUARY 19TH, 2013

mu, F nu mu, now it looks a little bit more familiar, right? Times some function, some function.

You feel very bad about this? (chuckles).

Lfull =1

2Aµ(p)(−gµνp2 + pµpν)Aν(p)

F−1(p2)

3+

1

2Aµ(p)pµpνAν(p)G−1(p2)

In position space, it would look something like∫d4x− 1

4Fµνfµν · (1 + f(2))← full quantum corrections

That is why I have this here. It’s all quantum corrections in it already. I’m not doing any – although

this looks familiar with your initial LaGrangian to begin with, but it is not. It has all the quantum

corrections already. But even at this LaGrangian level, the gauge fixing term is still this. Okay?

The conclusion is that, that is not renormalized. That is what this ward identity tells us. Even if

you write it exact operator, you invert it, you get your full LaGrangian, the gauge fixing term is

still this. The F nu mu square term on the other hand gets renormalized by this function. It’s a

highly nonintuitive statement. I want to emphasize that. It is not intuitive, because you are used

to thinking about the fact that once you open the Pandora box, everything goes. So you start to

calculate corrections, everything will correct, is corrected. It include the counter term of everything.

But we don’t ever include counter term for the gauge fixing term. That is why just because it’s

not renormalized, dictated by ward identity. In other words, another way of saying it is that the

gauge fixing term, something is renormalized, means that something is not physical. There is no,

the actual value of KC has no physical consequence. Otherwise it has to be renormalized. Okay?

That is another way of saying it. From hindsight, it is not that surprising. None of the physical

quantities you calculate depends on KC. Even if you want to set up a renormalizing condition,

based on physical quantities, to determine KC, there is no such a condition you can write down.

Okay? Because none of the physical quantity depends on KC.

6.2 Spontaneous Symmetry Breaking (SSB)

It is not a very good word, but that is the technical term. We will see why it’s not a very good

word. SSB. Basically, the idea of spontaneous symmetry breaking as opposed to explicit symmetry

breaking, so you have spontaneous symmetry and explicit symmetry breaking. Explicit means

there is no symmetry. But why do you call it a explicit symmetry breaking if there is no symmetry

50


to begin with? It is sometimes useful to think symmetry breaking that is small, okay, you have

approximate symmetry with a few small breaking effect. But that can be explicit breaking. So that

is explicit symmetry breaking. Spontaneous symmetry breaking is completely a totally different

beast. Historically there have been a lot of confusion about this again. But it is basically a

statement of symmetry of the dynamics, versus symmetry of the grand state. Okay? You will see

that even this is not a very good description. But let’s go on. This is the easiest way of getting

to this concept. Okay? Even this is not completely right. The most famous, everybody use that

example is a physicist inside a magnet, it is a small physicist inside a big magnet. Okay? Whatever,

let’s consider a magnet. Let’s consider a magnet. A magnet, a perfect place to use quantum field

theory. Not even Lorentz invariance. But it has many, a good model of magnet consisted of many

spins. Many, many spins, it can be dipoles, going up, going down, going up, going down. A lot of

dipoles inside. This is called a feral magnet, by the way. The Hamiltonian, to some leading order,

to some approximation, let’s do a simple model, okay? Let’s only consider the nearest neighbor

interaction, which is approximation. Is it a good approximation. Hamiltonian can be written down

as something like this. There is interacting spins basically. But it doesn’t even matter actually for

this argument, whether it’s nearest neighbor or not.

H = −λ∑i 6=j

~si · ~sj full 3D rotation symmetry, SO(3)

Let’s consider the grand state of this theory. Let’s consider the grand state, lowest energy level.

Ground state, the ground state means that obviously, you want to minimize H. Okay? There can

be many very complicated configurations. But the one that’s actually really minimize the local

ground state is everything is aligned. Everything is parallel. That is where the product is biggest

because there is a negative sign and this is a ground state. This is magnet. Here is a physicist

inside a magnet. A small physicist in the big magnet. What he will see is that all the spins are

aligned. This is the ground state. What he would conclude from that observation is that there is

no rotational symmetry. The quantum, the state does not have a rotational symmetry. Right? You

just pick out a direction. It could also pick up this direction. It could also pick up that direction.

That also has some deep meaning to it, by the way. But nevertheless, right now let’s all agree

that it will pick out a direction. It will point towards some direction. That is the ground state.

Okay? That ground state, consider there is only a 2D rotational symmetry around that axis, SO2.

This is the symmetry that preserve the ground state. That is example that ground state does not

51


have the same symmetry as the dynamics of the system. The dynamics of the system is this. I

think there is a theorem saying that the converse is actually true. If ground state is preserved some

symmetry, the dynamics must have a bigger symmetry, contains that symmetry. But not this way,

okay? Ground state can have much less symmetry than the dynamics of the system. Everybody

have heard analogy of you are sitting around a table, oh, okay. So I don’t know. So it’s, imagine

you are sitting around a table with your, I guess it’s – I don’t know, with your silverware. Now

imagine setup for that, is it unclear whether it’s on your left-hand side or on your right-hand side.

It is completely symmetrical, your plate in the middle, your silverware on the both sides. I don’t

know western food very well. So I may not do this right. But let’s just imagine there is such a

configuration. Now dynamically it’s completely ambiguous. You should do this. I’ll do that. Unless

you prefer your right hand. Let’s imagine that we have a, what is that word? There is a word you

can use. Yeah. Let’s imagine that everybody is that. Like that, okay. Now, dynamically there is

no preference. There is a complete refraction symmetry let’s say going that way and going that

way. It only up to one guy to make a mental decision, due to a small perturbation in his brain that

he wants to use this hand. The analogy is quite exact. Now you break the symmetry. Everybody,

okay, so now the configuration in the end, you end up with fork in your hand and start eating, is

that state breaks the dynamics of the system you begin with.

Consider real scalar field

L =1

2(∂φ)2 − V (φ) V (φ) =

1

2µ2φ2 +

λ

4φ4

φ→ −φ Z2 symmetry

ground state “vacuum”

and other random things that I’ll have to look in notes for at this point

52

7 FEBRUARY 28TH, 2013 – SPONTANEOUS SYMMETRY BREAKING (CONTD)

7 February 28th, 2013 – Spontaneous Symmetry Breaking (contd)

Talked about the Nambu-Goldstone theorem, with the Goldstone mΠ = 0 which can be seen by

symmetry or under the shift Π→ Π + vα.

Then we started to do a more slightly general analysis U(1) to see this breaking

φa → φa + α∆a(φ) linear function of fields

Symm: V (φa) = V (φa + α∆a(φ))→ ∆a(φ)∂V (φ)∂φa = 0

Vacc: ∂V∂φa

∣∣∣φa=φa0

= 0

→(∂∆a(φ)

∂φb

)(∂V (φ)

∂φa

)∣∣∣∣φab=φab0︸︷︷︸

=0

+ ∆a(φ)∂2V (φ)

∂φa∂φb

∣∣∣∣φab=φab0︸︷︷︸

=0

= 0

So from this I can already derive that – an equation, so I expand this out and take the derivatives

and derive the equation. So this actually – so let’s just write it down. This means delta A phi D

phi, phi A is zero. Identical is zero, I just explain. Because alpha is arbitrary, and – and I take

another derivative out of that. Okay? And set everything to its vacuum value. So the – this will

be – so I’m just repeating the last part of our lecture last time. So let’s take the derivative over

35 B, some other field strands, and D phi – D phi A, but I wanted to set this into this (inaudible)

vacuum value, okay? And there’s another term which you keep (maybe it was three phi B back

there). But taking the second derivative with phi AB equals phi AB zero, okay? And that’s zero.

Okay? So there are – there are in general two solutions, and – well, first of all let’s observe, this is

zero, because this is zero, because my condition that this is a vacuum, so this is – this is my ground

state. Now, then, therefore this has to be zero, and there are two cases where this can be zero. A

is – is delta A phi – well, error at the vacuum zero, okay? And this means the symmetry – this

means no SSB in phi A. There’s no spontaneous symmetry in that field, even you give it a value

it doesn’t break anything, doesn’t break symmetry. Or there’s no vac, either way. Okay? B – so

that’s a possibility. So that – but that’s just original paste, whatever my ground state is, doesn’t

break symmetry, okay?

• ∆a(φ = φ0) = 0 No SSB in φa

53

7 FEBRUARY 28TH, 2013 – SPONTANEOUS SYMMETRY BREAKING (CONTD)

• ∆a(φ = φ0) ∂2V(∂φa)(∂φb)

∣∣∣φab=φab0

= 0 ∆a(φ)→ Goldstone Mode

Now, for the case that ground state actually breaks some of the symmetry for those field values that

actually breaks the symmetry, you are – end up with – and let me write down this equation again.

Whatever – and D square – D phi A, D phi B. Okay? So then you ask, what’s the importance

of this one? You realize that for the potential, in general you can write them down as V phi

zero, where everything is plus of – you just Taylor expand, doing a Taylor expansion around its

ground state. So this is a constant, and the first (inaudible) terms all vanish, because again, my

condition. Then the secondary (inaudible) terms are like this. Okay? There’s small fluctuations.

It’s exactly that. Okay? They are higher order terms, but this term is exactly the mass matrix,

okay? This – now it’s a constant because I set all the fields into their vacuum value, so that this

is the mass matrix. Okay? Mass of the scaler field. It’s a matrix in general, just again, I said

like last time, where we’re used to the fact that if we have a coupled system, harmonic oscillators,

okay, so the frequency is – in the generic base is a matrix, okay? And they (inaudible)ize it, you

find eigen values, eigen (inaudible). So this is just like an eigenvalue equation and this tells me

that for everything – for every value that’s actually – for every field of variation that’s been broken

by giving fields (inaudible), there is this combination, which is – which is a serial eigenvalue state,

which is a ghost (inaudible) mode. So this is the ghostal mode (I don’t know what he’s saying

there. (maybe not serial).

V (φ) = V (φ0) +1

2δφaδφb

(∂2V

∂φa∂φb

)∣∣∣∣φab=φab0

+ · · ·

so in general in a spontaneous symmetry breaking (inaudible), you consider a co-sat, co-sat, so

what’s the symbol? So the original – the original global symmetry, you call it a G, global symmetry.

You call it a G. And after a symmetry breaking, there is still an unbroken subgroup that’s possible,

which we call H, okay? And then the goldstone mold lives in this co-sat. Okay? Which is a fancier

way for saying that for every broken generator can associate a goldstone mode. Okay? There’s a

slightly – there’s a much fancier formalism to talk about this called CCWZ (inaudible). Okay. But

we’ll see whether we have a chance to get to that or not. Probably – maybe not. Okay?

For an original global symmetry G, we have the Goldstone unbroken H – so the Goldstone number is

G/H. For example, if you have the Hamiltonian of a magnet H = −λ~S1 · ~S2 which has SO(3)/SO(2)

→ 2 Goldstones.

54

7.1 SO(2)7 FEBRUARY 28TH, 2013 – SPONTANEOUS SYMMETRY BREAKING (CONTD)

7.1 SO(2)

So let’s – let’s just try to see. So this may be a little bit more abstract. Let’s try to see this in this

U1 example. Let’s try to see how this works. Let’s try to write things this way. Okay. So let’s try

to do this for the U1 example. Let’s erase this co-sat stuff. So let – let’s – let’s not talk about the

U1 but instead talking about SO 2, okay? It’s just to – for – for a change. Okay? So we argue that

this is – equivalent talk about the U1 rotating by a complex face or just SO 2 (inaudible) in those

two. So phi complex field or – will be like this. I can equal it just – just think about the two real

fields, okay? In SO2 rotation, two real fields are rotated by this, okay.

Let’s start with φ = φ1 + iφ2.

φ′ =

cosα − sinα

sinα cosα

φ1

φ2

For small α

φ′ = φ+

−α

α

φ1

φ2

︸︷︷︸

∆(φ)

Now we know there’s a Goldstone and that that is actually telling me what the Goldstone mode

is. Okay? If I set – so that delta phi zero is where the Goldstone mode is, if I set the phi to its

vacuum value. Okay? So let’s see what this is.

vac: φ10 = v, φ20 = 0. So

∆(φ0) =

0

0

So first of all let’s recall that the last time I said that the existence of phase transition does depend

on dimensions. Okay, does depend on dimensions. So last time I already said it. In the case of

discrete symmetry breaking, okay, there is also a degenerative vacuum but there’s no Goldstone

mode, but even that case you can say that the D equals to 1, there is no excess speed, okay? In

one-dimensional, okay? Because one dimension that’s just quantum mechanics. And we argue in

quantum mechanics the ground state is not a symmetry breaking state. In quantum mechanics

there’s something like this, minus – the ground state is plus – this is the ground state period, in

55


quantum mechanics. But the one (inaudible) is quantum mechanics, as we said last, okay? So

discrete (inaudible) symmetry breaking only happens when the (inaudible) greater than 1. Okay?

So – so the icing model has a discrete symmetry breaking, for example.

Figure 1: Discrete symmetry D = 1 no SSB

Now, let’s talk about continuous. So we are worried about the fluctuation like that, okay? From

point to point is you can have a large fluctuation. So let’s again just take some simple example.

Let’s – I parameterize my field this way. I can parameterize my field this way. My worry is that

the value of my vacuum can fluctuate from place to place. My pi can be at different values, okay?

If that’s the case, then there’s – in the (inaudible) phase transition there’s no long-range order. I

can not talk about a single phase. Where I am pi is phi or where you are pi is 1, so I don’t know –

we are not in the same phase. What happens is really there is no meaningful way of talking about

phase transition. There is no other parameter in that case. Okay?

φ = veiΠ/v

Consider a space volume with size R (δΠ ∼ v on length scale R; ∂Π ∼ v/R)

S ∼∫

dDx (∂Π)2 ∼ RD v2

R2∼ v2RD−2

So the infinite wisdom of calculating action is that the – that’s actually what’s characterized the

contribution of a certain field configuration to the pass integral, it’s proportionate to the action.

Okay? So in particular is this. Okay? That’s your weight, okay? And by the way, this means

that, you know, when H (inaudible) R goes to zero, exactly you want to minimize the action, okay?

You want to minimize the action. That’s exactly where the classical limit is. I assume you’re all

familiar with that story. But I can plug this in ( I guess limit). This means HR V square, R, D

56


minus 2. Okay? So what – so what you want to – what you worry about is that any fluctuation

can destroy long-range order, okay? So long-range – spontaneous imagery is supposed to happen

in all the – in a big space time volume. Otherwise, you don’t even have a right to talk about it.

Long-range order. So I’m borrowing a lot of words from phase transition and (inaudible) physics

because they are already the same. So long-range means, you know, R is much, much bigger than

1 over V. Okay. Much, much bigger than 1 over V. So you see that this is suppressed. Okay? So

fluctuation – let me write it carefully – fluctuation suppressed for D greater than 2. Okay? So the

answer is that if you live in a space time dimension greater than 2, you don’t worry about those

things. Okay. It makes a lot of sense to say pi is just fixated, just fixed at the – where it is. You

don’t – the fluctuation in pi is highly suppressed at the large – at the large distance scales, and

therefore it doesn’t matter. Okay? All right.

e−S~ ∼ e−

v2

~ RD−2

long range order R 1

v

Fluctuation suppressed for D > 2. For D ≤ 2 – no SSB of continuous Goldstone symmetry.

Of course there is the converse of this story. What happens if D equals to – less than or equal to,

okay? No spontaneous imagery (inaudible), okay (I guess imagery). I can keep writing continuous

global symmetry. Because the fluctuation is big. It’s not suppressed. In fact, it’s enhanced, going

to allow larger distances. Okay? So because of this quantum fluctuation there’s no – and there

is a slightly fancier way of saying the same thing. Okay? A slightly fancier way of saying the

same thing is that a measure of the quantum fluctuation in the field is the expectation value of a

2 point function, okay? If you – if you measure – the expectation value of a one point function

is expectation value. The variance of the expression value is square, basically. That’s how much

things going around, and if you do this (expression not expectation I guess) in momentum space,

it’s like this. Okay? You see when D less than or equal to 2, there is infrared divergence. Okay?

Means that when – in dimension less than 2, the long wavelength mode is very, very important.

That’s another fancy way of saying there’s long-range fluctuations, that’s being very important,

okay? But it’s basically just this, okay? If you’re a little bit nervous about this, seeing the two-point

functions. Okay. But they are basically talking about the same thing.

〈ΠΠ〉 ∼∫

dDk

k2D ≤ 2 1R divergence

57


So here I – the expansion parameter is really H. Hbar here is very important because that actually

characterizes the size of the (inaudible) fluctuation, okay, you know, even in D – even in DD equals

1, if hbar zero, this is always suppressed, okay? If hbar goes to zero attention you’re shutting off

the quantum fluctuation, okay? So (quantum fluctuation). There is a completely analogous – there

is a – ah.

e−H/T D ≤ 2 thermal fluctuation – No SSB

Coleman’s theorem. Mermin-Wagner-Hohenberg.

Okay. Now let’s – let’s promote – so again, I remind you that we have said repeatedly that there is

a degeneracy of vacuum and there is Goldstone mode. Say those words in the quantum mechanical

way, okay? Let’s say those words.

Conserved current

∂µJµ = 0 Q =

∫d3xJ0(x)

dQ

dt= 0 [Q,H] = 0

Consider the ground state |ψ〉 and undergo a transformation |ψ′〉 = eiQα|ψ〉. Since [Q,H] = 0 then

|ψ〉, |ψ′〉 are degenerate in energy.

Okay, let me see. Remember, we always start with a conserved global symmetry. A conserved

global symmetry means there is a conserved current, meaning that equals zero modulo (inaudible)

terms. So for conserve current, I can define something, we’ll call it a conserved charge. Okay. And,

you know, the fact that the current is conserved means that the DQ/DP is zero. Okay? Okay. Now,

another way of saying this is a conserved current is that this is zero, okay? That’s just another way

of saying this is a conserved charge, okay. Now I consider – consider ground state sign, let’s just

call it the sign. And now let’s consider transformation of ground state. Okay? Which is E to the

IQ alpha (inaudible), okay? This is the generator of the symmetry, which is the charge, and that’s

a small, and let’s call this a side prime, but because QH is zero, so because QH is zero, zero, you

can show that the xi and the xi prime degenerate. In energy (zi or xi) so this is a statement you

have degeneracy of vacuum. And this is not a very good argument. It’s very easy to go through,

but it’s not a very – how should I say, it’s not – that’s the way you see it on textbook all the time,

I think, and it’s not very good in the sense that I have just argued, you know, this doesn’t even

58

http://en.wikipedia.org/wiki/Mermin%E2%80%93Wagner_theorem


depend on this is a ground state. Doesn’t even depend on spontaneous (inaudible), looks like. So

it’s not a very good argument.

Let’s talk about something manifestly symmetric – assume Q exists.

[Q,φ(x)] = −i∆φ(x)

Q|0〉 = 0

So now, let’s calculate this quantity

i〈0|[Q,φ(x)]|0〉 = 〈0|∆φ(x)|0〉 = 0

which we have in this vacuum. Therefore, we know that: 〈0|∆φ(x)|0〉 6= 0⇒ SSB where we

transform φ → φ + ε∆φ. So if there is a global symmetry, I can still construct a conservative

current. So let’s just write down the Ward identity

∂xµ〈0|TJµ(x)φ(y)|0〉 = −iδ(4)(x− y)〈0|∆φ|0〉

Consider

Gµ(x− y) = 〈0|TJµ(x)φ(y)|0〉

=

∫d4p

(2π)4e−ip·(x−y)Gµ(p)

Now combine these two statements and get

→ ∂µGµ(x− y) =

∫d4p

(2π)4(−ipµ)Gµ(p)e−ip·(x−y)

= 〈0|∆φ|0〉∫

d4p

(2π)4(−i)e−iP ·(x−y)

and note that 〈0|∆φ(x)|0〉 = 〈0|eiPx∆φ(0)e−iPx|0〉 = 〈0|∆φ(0)|0〉.

Okay. Now it’s time, let’s give you – this already tells you something about this. Right? So – for

example, this tells me – this tells me that the P mu – well, P mu, P (inaudible) mu, P, is just this

matrix element. Because the rest of them are the same. Okay.

And this leads to

→ pµGµ(p) = 〈0|∆φ|0〉 Gµ(p) = pµg(p2)

59


Now I can also say something about this. Okay, in general, because this carries mill indices, and

there’s only one thing that carries mill indices here which is P itself. You have to be able to write

this as a P mu times some scaler functional P (I guess it was P mu? Mill indices?) So now I

combine this. How do I combine this? – I can solve for GP square, so this means that the P square,

GP square is zero delta phi zero and GP square also known – well – okay, I won’t repeat the trivial

step, but you solve GP square plug (inaudible) back. This is the P mu. P square, okay? This is

that. Of course this is nothing if there’s no spontaneous symmetry breaking, is zero, but there is

something if spontaneous symmetry is (inaudible) down here, that we say spontaneous symmetry

means that – means that this, so spontaneous symmetry breaking means this is nonzero, okay? So

this is actually a legitimate greens function, not the trivial one. If there’s no axis B this is a trivial

one, this is zero.

p2g(p2) = 〈0|∆φ|0〉

Gµ(p) =pµ

p2〈0|∆φ|0〉︸︷︷︸

SSB: 6=0

pole at p2 = 0

Moreover, I claim this proves the Goldstone theorem. Why? Anybody knows why? So what’s the

meaning of I have a zero mass particle? It means that some greens function – there is a physical

pull at the mass equals zero (maybe he’s saying Lagrangian??)

So here is zero pull. Okay, this is a formal way of proof there is a massless particle, that you can

construct any green function you want. You identify the post, okay, or the physical post should

show up in the green function, in some greens functions. They don’t have to show up in all greens

function, any – all of they hope, bub at least if there is a mass – if there’s a physical particle with

mass, you should be able to come up with a greens function, where the Poe equal to the mass, okay?

And we came up with a green function and we find our Poe. That’s basically it, and it (inaudible)

depends on the fact that this is known zero (or nonzero). Is it okay? This is very formal. It doesn’t

depend on the fact that the – have to go around the Mexican hat potential and so on and so forth.

Okay? I don’t know whether you found this is very comforting, but I think that the shift symmetry

argument is pretty good already in the beginning, but this is just in case you wonder whether it’s

a theorem or not, this is a theorem.

60


〈0|Jµ(x)|p〉 = −ifpµeip·x

So, in fact, we can say a few things – one more thing. Let’s just say one more thing about base –

based on this set of equation we have our data (inaudible) now. Okay? Let’s say something about

the matrix element of J mu. Between that and the momentum P state. Okay? Let’s say something

– we can say one thing about this matrix element, turns up. Okay? That again, carries mu, right,

this momentum matrix (inaudible) carries mu, and the only thing that carries mu is here, okay?

And so this is (inaudible) P mu, and you can show that this under space/time boost, under space

time translation it translates a state with momentum P. So therefore it has to have something like

this. Okay? The argument I just said is a little bit quick, but it’s correct. So – and it is the most

general form you can write down. Of course we need a proportional constant, which we call F,

okay? It’s just some constant. And – and just to be devious, usually that’s the convention, okay?

But once you get used to this, this is the thing that everybody use, in (inaudible) theory and so

on, that’s the convention everybody uses, and you hate the other conventions. You hate the people

that don’t use your convention. Yeah. Okay?

But anyway, dealing with – being able to deal with, you know, different conventions is a test of

whether you have (inaudible) to become a theoretical physicist or not. You have to deal with

different conventions. Okay? So there’s nothing to complain.

So recall our – what the definition of G mu, and which is just a time order correlation function

between this and phi. So I can – so I can write it out and using – using translational variance and

insert a set of complete momentum states, okay, I can write the whole thing out. So here is the

result. It’s time ordered so it depends on which – whether X happen first or Y happen first. So there

is zeta function X zero minus Y zero. (zeta I guess) okay? So just – I’m (inaudible) I’m putting

P in between. Okay? So basically I’m inserting – inserting the following comment. I’m inserting

this quantity into this formula (quantity, not comment). So – and using my parameterizations, so

– I’m inserting things here. Okay? Using that formula. So that’s P mu – I’m just going to write

one of them. P phi zero – again use – use something very similar to what I did there with the

– with the space translation. Start – and the plus – another one was X zero Y zero interchange,

basically. Okay? And that one actually has – I’m just trying to remember that da, da, da, da,

61


WIP (inaudible). Okay. Okay?

Gµ(x− y) = 〈0|TJµ(x)φ(y)|0〉

= Θ(x0 − y0)

∫d3p

(2π)32Ep(−if)pµ〈p|φ(0)|0〉e−ip(x−y) + (x0 ↔ y0) · · · (+if) · · · eip(x−y)

= −if〈0|φ(0)|p〉(i∂

∂xµ

)∫d4p

(2π)4

i

p2 + iεe−ip(x−y)

=

∫d4p

(2π)4

pµ

p2f〈0|φ(0)|p〉eip(x−y)

So it takes a little bit of insight, but it doesn’t take very long insight to show – it’s similar –

similar to things that we go through and we derive the Cline Gordon propagator basically in the

first quarter to show that this is the following. Of course, I mean, these things all factor out, so

I’d write it out first, and the next step is a little bit you have to convince yourself a little bit, but

it’s – it’s the same that we did with – with our D 4 P (writing on board). It’s nothing but just

that. Okay? Again, this is a very – this is identical steps. We went through identical steps when

we derived (inaudible) and operator. You can go back to the easier because there are actually two

Poes you complete. Depends on which one is bigger, you each completed counter above or counter

below, exactly get these two terms. Okay?

And if you look at the work we’ve shown before, you can get that

f 〈0|φ(0)|p〉︸︷︷︸=1 (normalization)

= 〈0|∆φ|0〉

→ f = 〈0|∆φ|0〉

So, in fact, this is usually normalized to 1. You can check with our normalization. This matrix

element is 1. So in this case F is just a – okay? So this is the same. Let me just – let me just

say that in word so you know I’m not going through this scene just to have fun. So there is

a matrix element, okay, of the symmetry current, pitching vacuum and the warm particle state,

okay? That’s what this is (or one particle). Roughly speaking this is the probability amplitude

of getting a one particle state on a vacuum from the symmetry current itself. Okay? Statement

No. 1. And that matrix element modulo this general piece, that matrix element has a strength,

okay, which is (inaudible), okay? And the F – this is completely general but this F is exactly the

same as the strength of spontaneous symmetry breaking, okay? Remember, this – again, this is

62


the signal of strength of spontaneous symmetry breaking. So this setting in a more mundane word,

this is the (inaudible), okay? And the way that the – that symmetry – symmetry current acting

on vacuum is the probability of pulling out a particle by doing a symmetry transformation on the

vacuum is proportional to the strength of symmetry breaking. Okay? The probability of taking out

a pion from vacuum is proportional to F. Proportional to Vaf. So this is usually the VAF. Okay?

In some very different context this can be very different things, in spontaneous imagery breaking,

in another model this is called the hicks VAF. Okay, this is the miks VAF. And in QCD this is

called the pion K constant, but it’s the – it’s some VAF that breaks cierl – spontaneous symmetry

breaking, the global symmetry is the chi rale symmetry. This is the – doing this kind of formalism

you don’t refer to the hicks field. You don’t have to refer to the hicks field, and in QCD it’s not

clear whether there is a hicks field or not, but everything else (or fixed) everything I said still goes

through because it just relies – it doesn’t rely on the – my initial picture of Mexican hat potential.

Has to have a hicks field. But it doesn’t depend on that at all. That’s just – that’s just a cute

example, simplest example I can write down, but everything else goes through without relying on

the skater field, okay?

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8 MARCH 5TH, 2013

8 March 5th, 2013

Okay. So far, we have been talking about spontaneous [inaudible] just with the scalar field. Its

see so called spontaneous symmetry breaking of global symmetry. Although I think I’ve told you

also that gauge symmetry is not a symmetry. That is the only symmetry. Okay. That is always a

spontaneous, although you frequently heard a word of spontaneous symmetry, spontaneous breaking

of gauge symmetry, but that is a misnomer. So that is gauge symmetry is, I think I told you that

gauge symmetry is not really a symmetry. But we are going to use that word anyways, consistent, so

commonly used. Today we are going to talk about what happened if we gauge the global symmetry.

Remember still our canonical example is a complex scalar field.

V (φ) = µ2φ∗φ+ λ(φ∗φ)2

and when µ2 < 0 – you will have this Mexican hat potential. And there is a radial mode. Just to

remind you, we decided the distance away from the center, called v for vacuum expectation variable

of this field φ

〈φ〉 = v =|µ|√λ

1√2

m2h = 2λv2

This is what we have done so far with this scalar field. Now I want to put a photon into this system.

Okay? For the gauge field into this system, and so I will introduce a gauge field. A mu. Again, you

want gauge field. You want vector field let’s say, U1 gauge field. There is a gauge transformation.

I’m going to gauge transformation associated with this view 1.

Aµ → Aµ −1

g∂µα(x)

And I will also say φ has a charge +1, like an electric charge. So φ transforms φ → eiα(x)φ. The

Gauge transformation is just that, Aµ and φ with charge +1.

Kinetic Terms

1

2(∂µφ)2

Kinetic terms looks like in original theory, okay? Now this is not quite invariant under that. Okay?

Let’s call this gauge transformation. Okay. But as we already learn in QED the trick is write some

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8 MARCH 5TH, 2013

covariant derivatives. Whenever I say g, it’s gauge coupling; or if there is a charge, I use q.

Dµφ = (∂µ + igAµ)φ

Dµφ→

gauge trans eiα(x)Dµφ

hat is the benefits of including the – well, by putting them together basically. So therefore, it’s

clear to how to write a LaGrangian that is invariant under this gauge transformation which is this.

Okay? That is our LaGrangian

L = −1

4FµνF

µν + |Dµφ|2 − V (φ)

So, okay. Everything looks pretty innocent at this point. Our LaGrangian, let’s not worry about

the gauge fixing and anything like that, is just that. This is a kinetic term because I introduced the

A field. I have a kinetic term for this guy, so D mu ph iSquare over, then I’m getting to potential.

This is called a boolean hicks model. Everything has a name. It is very important. It is knowing the

name is more important than sometimes knowing the exactly the physics. Because you immediately

know what people are talking about, okay? So that is our LaGrangian. Now you have to go work

at it. Let’s try to do it. So to proceed, we will obviously start with this point, okay? I’ll expand my

field around this point, because I want it to be in the broken phase. So, let’s see, in broken phase,

what I will have, well, there are different parameterization, but the parameterization we have been

using is the following. By the way, there is an equal probability or in contrary parameterization

with square root and without square root too. Anyway, that is completely, it is very annoying. It

is annoying if you talk about standard model. You can talk about with the vacuum value of the

field, whether with or without square root. With square root 2 it is 100, it is the same as top mass.

With square root 2 it is 250 something DV. If you want to write a paper about why it’s the same

as top mass you have to be aware this is a convention dependent statement.

Broken Phase

φ(x) =1√2

(v + h(x))ei(π(x)v

)

I already have two degrees of freedom to parameterize. There is an angular variable that takes you

around the circle and there is this radial mode, which we call h. Now we have to work it out a

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8 MARCH 5TH, 2013

little bit on the board

Dµφ =eiπ(x)v

√2

[∂µh+ i∂µπ +

i

v(∂µπ)h+ ig(v + h)Aµ

]

|Dµφ|2 =

normal looking kinetic terms︷︸︸︷1

2(∂h)2 +

1

2(∂π)2 +

mass term and Higgs field︷︸︸︷1

2g2(v + h)2A2

+1

4v2(∂π)2h2 +

1√2

1

v(∂π)2h

+ g(∂π)(v + h)A+1√2

g

v(∂π)h(v + h)A︸︷︷︸

goes away by gauge transformation

Under Gauge transformation φ→ φ+ vα(x). So we “gauge away π” (π = 0; unitary gauge).

→ L =1

2(∂h)2 +

g2

2(v + h)2A2

=1

2(∂h)2 +

g2v2

2A2︸︷︷︸

=m2A2A2 “mass term”

+g2vhA2 +g2

2A2h2

And the g2v and g2

2 terms are fixed by Dφ. I want to in particular draw your attention to these

things. Okay? All these coefficient are fixed. All these things are fixed. Basically, by the form of

this, like this. All of these are completely fixed. Okay? So, you know, in other words, in order to see

whether, so, okay, let me first say a few – oh, in order to see whether, just in order to test whether

this scheme that we have drawn up so far, to give gauge a mass is correct or not, it is not enough

to just see that. Okay? It is far from enough to just see that, just see a Higgs boson as I would say,

I’ll comment later on. But it is very, you know, very important to also measure these couplings

because these couplings are precisely related to each other and related to the mass. That is why

it’s important even if, even as you discover some boson, it may not be the Higgs boson, you wanted

to measure these couplings. So you want to measure the coupling with a Higgs, with two As, and

you want to measure the coupling with the Higgs, two Higgs with two As. Okay? This is U1, but

you must have read a newspaper for SU2 version of this, for some reason that is more important.

And this is more or less measured already. This is almost impossible to measure. But that is the

prediction, even measuring this is already a big confirmation of how this works. You must know

what these things are called. This is called Higgs mechanism. I will clarify the terminology further

later. This is Higgs mechanism. But, this week there is a big meeting in morian, everybody know

where it is? I don’t. It is somewhere in Europe. It is a ski resort. Doing physics is great. You go

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8 MARCH 5TH, 2013

Phases Aµ φ

Coulomb µ2 > 0, no SSB mA=02 polarizations: λ=±1

charged2 massive scalars

µ2 < 00 mA 6=03 polarizations: λ=±1,0

π= would-be Goldstoneh= radial mode

Table 1: Phases

to ski resorts. And having meetings, and that is usually some place where big experimental result

gets announced. I was just told that the official name now for that meeting at least, of everybody

else is going to call it Higgs boson but for that meeting it is going to be called BEH boson (Brout,

Englert, Higgs).

And now we proceed to the dropping of the names.

So anyway, but those days, people, when you write a paper, you have to, it was even before my

time, so it was when you write a paper, you call up your friend say I’m going to send you a preprint.

Months later the preprint arrived and so on, so forth. These days, if you have anything in your

head, you put it on Facebook first.

Everything is uniquely normalized of course. But I’m just going to do the classic. Actually I’m

not going to normalize it. That is going to take another three weeks to do, introducing goldstone

which we haven’t even talked about. I’m just going to do the classical, mostly classical part of the

story. Okay? This is called the Higgs mechanism. Maybe you have been slightly wondering what

exactly is going on. This is nice. But when Peter Higgs likes to give a talk, I’ve seen one of his

talks, his talks, is my life as a boson. Have you seen that? You should go to see the talk. It is

interesting. He said when he first wrote that paper, he went to Princeton and Harvard and MIT to

give talks. Everybody assured him he is completely wrong, because the existence of goldstone after

symmetry breaking has been proved by some, I don’t know, called C style algebra or whatever,

some axiomatic quantum field theory. So let’s understand a little bit better, just in that vein. Of

course, axiomatic quantum field theory doesn’t understand in this. That is almost totally useless

endeavor. Okay. Let’s try to understand why exactly we get a mass. Or why do we get a mass?

This is a completely correct but you may want to, may want me to say it in a more deeper sounding

way to get convinced. Here is a slightly deeper way of saying it.

I have goldstone and I have radial mode. But this guy obviously becomes our H in the end. But this

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8 MARCH 5TH, 2013

guy has come here actually, become one of the polarizations. Usually I said it becomes longitudinal

polarization but that is slightly gauge dependent that statement. What these three polarization is

coming from is the taking a goldstone mode and taking this to form of three polarizations. You can

see this as I again, I want to, I can gain away the Pi. And it becomes part of A, basically. There is

no, there is not a invariant distinction between Pi and A basically. When you shift the Pi, A shift.

I’ll write it down. As you can see, any time you shift A to some, alpha Pi also shift. There is no,

with A around, there is no invariant distinction between Pi and A. They just become one of the

same state basically. This is the official statement of Higgs mechanism. The goldstone is again the

terminology is eaten by the gauge boson. Gauge boson become massive and the goldstone here is

not called goldstone anymore. This is called a would-be goldstone. If there is no A, it is its own

goldstone but if there is A, when there is A, this is not an independent observable physical state.

And these are also not independent observable physical state. They combine into one massive

vector basically. That is the official statement of Higgs mechanism. This is the Higgs boson. But

it is not, you see that it is actually just playing a peripheral role in this case. It is not completely

– we will argue that, why, it’s actually not necessary, Higgs boson here.

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9 MARCH 7TH, 2013

9 March 7th, 2013

Today we wrap up a few remaining issues on symmetry breaking with U1.

We already said that the gauge U1, in gauge symmetry, gauge symmetry breaking, the story is that

you are going from A mu, massless, with no symmetry breaking. This is massless. The polarization

is plus/minus 1. Another complex scalar field with two degree of freedom, okay. But then you

go through a phase where there is symmetry breaking, this becomes massive. The correlation is

plus/minus 1 and 0. On the other hand, we said that in, after spontaneous symmetry breaking,

the salear vector changes as well. It goes to give you a Goldstone, and another scalar. We said

this is not observable, directly observable, because it’s becoming part of the massive vector, with

three polarizations. This is called eaton. That’s basically I think where we are last time. Does this

picture bother anybody? We went through two ways of convincing ourselves, we just do the straight

LaGrangian way, and we did a more formal sounding way, talking about poles or propagators. Poles,

the second one is actually very close to actually prove a theorem. The first one is showing you

what is going on. For your understanding of standard model it’s important, particle physics. Let’s

talk about Fermions in connection with spontaneous symmetry breaking. Maybe let’s not do this.

Let’s do, let’s briefly remark on some more complicated spontaneous symmetry breaking pattern

first. How about I have a U1, still have a U1 but now I have multiple Higgs. Now I have multiple

field, phi1 with charge Q1 and phi2 with charge Q2 and so on. How about I have that? Let’s

say every one of them gets a V1. If any of these values is nonzero, U1 symmetry is spontaneously

broken. There is actually for each one of them, in general, for each one of them, there is a U1 global

symmetry as well. If I just do this, now, the question is, what happened to the theory, after I’m

breaking all those things. Okay? After I give all these things values. To be specific, let’s consider

two. The theory will be ... D mu phi1 is IGQ1A mu, 1 and so on. Okay? Write this down, this is

going to be homework. (chuckles). They just go do this, basically. Generally this has value, that

means there is Vphi1, phi2 as well, which I’m not writing down at this moment. But you know

what is going to happen. Roughly speaking you know what is going to happen. Roughly, you know.

What is going to happen is that there is going to be a – (coughing). Sorry. Of course the gauge

boson will become massive. What is going to happen, so you can do the whole thing and look at

the mass and so on. What is going to happen, A, you are going to have MA. The mass is, the

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9 MARCH 7TH, 2013

gauge boson is going to have a mass. Exercise to find out what the mass is, okay? It’s the sum of

the values and so on, so forth. But it is going to have a mass because the symmetry is broken. The

question is what happened to the rest of the spectrum? I start out here. These are two complex

scalar field. This is 4-degree of freedom.

U(1) : φ1 : Q1, φ2 : Q2 → 4 d.o.f.

〈φ1〉 = v1 〈φ2〉 = v2

so

|Dµφ1|2︸︷︷︸mA

+|Dµφ2|2 + V (φ1, φ2)Dµφ1 =︸︷︷︸A=±1,0

(∂µ + igQ1Aµ)φ1

You know one of them is gone. One of them which is whatever a Goldstone is going to be going

to the, is going to the mass, the massive A vector field. That massive vector field has to have one

Goldstone, has to eat one Goldstone to become massive because otherwise it doesn’t have three

polarizations, so it’s just wrong. But then the rest of the three, still remaining, you still have three.

The other three scalars. This theory has three extra scalars. Without going through the whole

exercise, the exercise itself is pretty simple, multiply things out and look at the mass and so on and

the couple. Now, what if I also do another exercise, suppose 2U1. 1U2. I have two U1 symmetry.

I have two gauge bosons. A1 and A2. Suppose I have that now, suppose I have a charge, a particle

that is charged under both. I can have that too, right? I can have a particle charge under 2U1

symmetries, and with Q1 and Q2. Suppose I have that. So this sometimes can be summarized in

the diagram like this. It is called U11, U12 and 5 I guess. Now, depends on what field are you in,

this diagram have different names. This diagram has different names. In some fields it’s called this.

This is actually called a minimum moose, longer lattice, quiver, but anyway, so I’m just preparing

you for something. These are the sides and this is called the link.

U(1)1)︸︷︷︸A1

×U(1)2︸︷︷︸A2

φ : Q1, Q2

and you can imagine that φ is the link between the two sites U(1)1, U(2)2.

Dµφ = (∂µ + iQ1g1Aµ1 + iQ2g2Aµ2)φ

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9 MARCH 7TH, 2013

and φ has 2 d.o.f. 〈φ〉 = v → would-be 1 Goldstone.

But it’s important to give them good names. So that they can be remembered. So anyway, string

theory, deeper in physics, this is called a quiver. Symmetric H theories. Anyway, suppose I have a

quiver, a moose like this. So what do I, what is the LaGrangian? It is actually very simple. Those

names are sometimes scary. But it is actually simple. D mu Phi is partial mu. Now you have to

include all the gauge field here. Because the Phi is a charged under two gauge fields. It has to be

locally gauge, has to be gauge under the transformation of the, there are two independent gauge

transformation. So the current derivative has to involve all of them. Q1 in general, there is a G1A

mu plus IQ2G2A mu 12, not using, choosing very illuminating notation, but this is what you do.

The LaGrangian is, again that. You can open this up. And do our usual thing. You can open this

up and look at the masses and so on, look at the mass of gauge bosons and so on. But you already

know what is going to happen in the end. Qualitatively, what is going to happen in the end, you

can work out the details, or you will work out the details, is that there are two degree of freedom

here. Okay? Only one of them shift under the gauge symmetry. After, again after SSB, so after

SSB, only one Goldstone, would be Goldstone, only one would be Goldstone. Okay? Therefore,

you know that what is going to happen is that in the end, there is going to be one unbroken, one

massless vector. Okay? It is going to be one massive vector, plus one massless vector. Because

there is only one Goldstone, it can only supply one of the vectors with its longitudinal mode. Just

by counting the degree of freedom, so, yeah. But this massive one is not any one of these. This

massless one is not any one of this. They are linear combinations. You have to work out the linear

combination. Basically, all of them are just linear combinations of A1 and A2. Okay? That you

have to, you open this up, square, there is a mass matrix, Again this is similar to what you do with

couple harmonic oscillator. You have to diagrammize the frequency mate I can to find the Eigen

mode. There are two Eigen mode. One of them is massive vector. You know there can only be

one because there is only one Goldstone to be eaten. Is this okay? Both of these two situations,

these two situations happen, this happens in standard model. A similar story happen here in the

standard model, similar story to that happens in EMSSM. Let’s also talk about the Carol Fermions.

Do I want to talk about Fermions here or later? Let’s talk about it here. Let’s remember what

the Fermion mass is. Fermion mass is something like that. But we know that, you still remember

Fermion part at the beginning of this quarter, we can write it as a, left-handed or right-handed

71

9 MARCH 7TH, 2013

component. Okay?

mψψ ψ =

(ψLψR

)ψ =

(ψ†R, ψ

†L

)= m(ψ†LψR + ψ†RψL)

So that, okay? That is that, okay? That is just that. Fermion mass is mass. This okay? I’m

multiplying them out. Okay? Lorentz invariance, this basically is fixed by Lorentz invariants.

Lorentz invariants basically is saying that left-handed and right-handed for massive Fermions are

not good quantum numbers, because they mix. That is basically what Lorentz invariants is telling

us for massive Fermions. On the other hand, there could be situations that those two charged under

different gauge symmetry. That would carry different charges under gauge symmetry. Okay?

Under U(1) Under U1, under U1, there can be two situations, okay? One is that Q left is the same

as Q right. The left hand Fermion and right-handed Fermion has the same charge under this U1.

So, ah, I should actually say something slightly different. Now, let’s look at the kinetic terms. Let’s

look at the kinetic terms. The kinetic terms looks like this, looks like this.

ψ†Lσµ(∂µ + iQLAµ)ψL + ψ†Rσ

µ(∂µ + iQRAµ)ψR

This is 2 by 2 poly matrices basically – Pauli. It doesn’t matter for our argument but these are

kinetic term. This is gauge invariant, kinetic terms. Gauge invariants does not tell me anything

about QL versus QR, okay, of the kinetic terms. There is no limit. I can just set this whatever I

want. There are two situations. Case one is QL equals QR. This is called a vector like, Fermions. It

doesn’t mean the Fermion becomes a vector. It just means that the left hand and right hand have

the same charge. This means that it is invariant. That just means that this term is invariant, if QL

is the same as QR, and the charge of this, rotation of this is opposite of the rotation of that.

i) QL = QR vector-like → mψψ invariant

ii) QL 6= QR chiral →mψψ not invariant

This is gauge invariant. Then there is another situation with QL not the same as QR. Okay? In

this case, this is called the kierl, the Fermion is called the curial. In that case, it is not invariant.

That cannot exist. A term like that cannot exi For kierl Fermion does not have mass basically,

that is the statement. Just because you can assign left-handed and right hand Fermions, different

charges. Let’s consider an extreme case.

72

9.1 Standard Model 9 MARCH 7TH, 2013

Consider QL = 1, QR = 0

yφ(ψ†LψR) + y∗φ†ψ†RψL)

〈φ〉 = v

→ yv︸︷︷︸mψ

(ψ†LψR) + h.c.

= mψψ†LψR + h.c.

QL is 1, QR is 0. That is one of those cases. There is nothing, you cannot write a mass term

basically. Okay? Look at this mass term, only QL carries charge. U1 rotation is certainly not

gaiblg – gauge invariant. So now you wonder is there any way to give those Fermion masses,

because after all we live in a world with Fermion masses, where Fermion has masses. Maybe you

say, okay, maybe the Fermion world is all vector like, but it is not. Unfortunately it’s not. We

will talk about it too. Suppose I give you a pair of Fermions, how do you give them mass? The

trick is again there is a spontaneous symmetry break. It is going through spontaneous symmetry

break.

Consider φ,Qφ = 1, 〈φ〉 = v.

φ = (v + h)→ yhψ†LψR + h.c. =mψ

vhψ†LψR

9.1 Standard Model

So, are you all, you have all seen standard model at some point? Particles and interactions? Not

see it but know about it. Standard model, so the story we just said about U1 is important for

understanding the most intriguing part of, not the most intriguing but at least the most, one of

the most interesting part of the standard model. Standard model are controlled by gauge similar

trees – symmetries.

gauge symmetry SU(3)C × SU(2)L × U(1)Y

SU(3)C strong interactions 8 gluons quarks are 3: u,d,c,s,t,b

SU(2)L × U(1)Y

SU(2)L doublet(ud

)L

which mirrors UR, dR singlet

73


doublet

scalar: H =

(h†

H0

)SSB=

(h†

v + h+ ia

)Yπ =

1

2

SU(2)L × U(1)Y

which gives 3 massive particles W±, Z and a massless photon. We have 3-Goldstones

yuH QL︸︷︷︸(uLdL)

uR + yDHcQLdR

actually which way is the left-handed. Let’s call this left-handed. Let’s call this right-handed. Okay.

They carry different charge, if you want, under SU2 gauge interaction in the most extreme way.

One of them couples to them, one of them don’t. Okay? But this too also interchange each other

under parity. Okay? So means that if you imagine you put a mirror here, this goes to that. Okay?

So if you put a mirror here, this just goes to that. Okay? Gauge interaction SU2L violate parity, so

this is parity violation. Is there anything wrong, violating parity? Nothing. It can be violated. It

is violated in nature. Nothing, nothing goes horribly wrong. For a long time we thought it would

be, parity is obviously a symmetry but it’s not. But again very similar to what I said here, very

similar to what I said here. You see the statement I just made is not completely consistent with

Lorentz invariants, of massive Fermions at least, because we know from experiment these things

gets, all have mass. By the way this is called opt because usually we write it on top, I don’t know,

top down, the term strange is a little bit, I don’t know why, maybe strange that they don’t think

it should be there, and charm is just try to make a good name. I don’t know. But this is not

consistent with Lorentz invariants because for massive Fermions, so there is a very intuitive way of

understanding why mass makes two polarizations, because for massive particle there is a rest frame.

Once you get into the rest room you lost a direction, there is no direction. In the rest room you can

pick any access of your polarization vector, and obviously left-handed can become right-handed.

Right-handed can become left-handed. It’s your choice. In the massless case it makes sense because

there is no rest room. You are always moving in that direction. You always use that direction as

your reference, polarization. For massless Fermions there is a direction. We also know that this

symmetry is broken. The good thing is, so at this point in order to accommodate massive Fermions

you need to do one of those two things. Either break the symmetry or break Lorentz symmetry.

Lorentz symmetry we are not breaking. We are not breaking this one. But we know this is broken.

74


The gauge bosons are massive. It’s broken by a doublet. It can be broken by a doublet, scalar

doublet. This is again a scalar. H is a – have to write it like this, H0. Just in a very suggestive

form, and this under spontaneous symmetry breaking, you can write this as an H plus, V plus H

plus IA. So that is how you do standard model doublet. And, okay. On the other hand, the H also

charges under this U1. So we haven’t talked about U1 yet. H also is charged under this U1. So YH

is 1/2, according to normalization, that is the usual normalization we use. So that’s that. So this H

when it gets verve here, first of all you may ask me why am I giving vev to this guy, why am I doing

this, why am I not giving to it that guy or whatever it is? (vev?) The answer is it doesn’t matter.

You can put it wherever you want, you can give vev to every component. It is similar to the U1

case, the valve can go any direction and the end result is the same. All those are vacuum, all those

vacuum are equivalent as we have said, along the circle. Similarly, you can take this to where SU2

rotation, give vev to anywhere you want, but all those vacuums are completely equivalent. Very

similar to the U1 case, okay? This breaks both SU2, cross U1Y. It breaks both of them. Okay?

On the other hand, there are only – well, the global symmetry breaking is SU2 down to nothing.

There are only three Goldstones which is this and this. It’s again very similar to the U1 case. It’s

not everything in the scalar, it can be a Goldstone. Okay? In this case, these and that can, gold

stones. These and that. This is a complex scalar so there are two degree of freedom. Between two

of them there is one of them. There are only three Goldstones and how many generators are here?

How many gauge bosons are here? SU3 has one, how many, SU2 has how many? Three. It’s SO(3)

basically, just three rotations. There are four generators. Three Goldstones. Okay? There is one

remain massless. Very similar to what I just erased, similar to what I just erased. Okay? So what

is going to happen is going to be three massive gauge bosons, and they all have names. They are W

plus/minus and the Z. And one massless. So because, again, which one is, what is the massless one

here? It’s the photon. That is the symmetry breaking. Again, but again, it’s not exactly this one

doesn’t become the photon, and this one is not exactly these three, and it is a linear combination.

In particular, Z and the photon are linear combination of the tau 3 here, and this U1 basically.

But let’s not work it out. It is going to take us too long. But so yeah, these are, I’m mentioning

this because this is continuous, it is a continuous story that we have talked about. You see there

is nothing to it. All you have to do is write down the current derivative and diegriz it, basically.

(?) With this H obviously we can write masses now for the Fermions. With this H I can start

75


to write YHQLUR, call this YU plus YD, H, conjugate, QLDR. Doesn’t quite matter what this

is. But you see these are the mass terms. Okay? This is, again, this is just my shorthand way

of writing down ULDL. These are the mass terms of the standard model quarks basically. They

have different Yukawa couplings in general. They don’t have the same. We all know that there are

more than one quarks. Right? In general, these are 3 by 3 matrices. They are three quarks, three

flavor of quarks. They are [inaudible] too. To understand that, that is called a flavor problem. We

don’t know really how to derive these Yukawa matrices from first principles, and it’s unlikely we

are going to know any time soon. Okay. That is basically the standard model. Everything what

I said has already been discovered, by the way. I’m not making any of this story up. Let’s take a

step back. Since we have been talking about Higgs mechanism and so on, let’s take a step back,

let’s try to understand Higgs mechanism slightly better, what does it really mean. For that let’s go

back to the U1 story. Let’s from a bottom up point of view, let’s take a bottom up point of view.

Let’s suppose I’m very dumb. I don’t know about – let’s just say I observe a massive gauge boson.

Okay?

−1

4FµνFµν +

m2A

2AµA

µ unitary gauge

I’ll write a mass term like that. Okay? You say wait, that is all wrong, right? It is not gauge

symmetry. We all know we need gauge symmetry, blah blah blah, but actually we don’t. First of

all we don’t. Let’s say we do, but the usual story tells you that now gauge symmetry forbid this

term. That is why I need to introduce, that is why I need to do this whole Higgs boson mechanism

that write a cowritten derivative, give you a mass, that part of the story is wrong. The reason is

that, the reason this is wrong is that this is a secret to the gauge symmetric.

in general gauge Okay? It’s not gauge symmetric when you fix to a unitary gauge which is not

such a big surprise. But suppose let’s work in the slightly more general gauge. In general, in a

general gauge I can write this as this following form. You see this is completely gauge invariant.

Remember in gauge transformation A mu goes to A mu D mu, alpha we already said there is no

need to realize Pi, Pi shift as of this. Under this gauge transformation this is completely invariant,

complete gauge invariant.

−1

4FµνFµν +

m2A

2

(Aµ + ∂µ

π

v

)2

76


with gauge transformation

Aµ → Aµ + ∂µα, π → π − vα

On the other hand we know these two are completely [inaudible] to each other, there is nothing

different. Using gauge transformation I can gauge only Pi it become this, become unitary gauge.

Therefore, this is actually gauge invariant. There is nothing wrong with that. Gauge invariance

does not prevent you to write down a mass term. It is completely fine. The only thing you have to

remember is now there are three degree of freedom. But that is the real physical statement, between

massive and massless gauge boaston is not whether it’s gauge symmetry or not. It’s whether there

is three degree of freedom or not. That what I’m trying to get at. This is a real physical statement

is that. Goldstone is everything. The eaten Goldstone captures the essence of the physics really.

This also tells you that the LaGrangian sometimes can be very deceiving. Although we are getting

really used to doing everything with LaGrangian, but LaGrangian sometimes can tell you, teach

you a very wrong lessons, because it can hide things. You can hide things with LaGrangian.

You can do horribly wrong things with LaGrangian. If something looks horribly wrong, it can be

completely fine. Okay? So for that, let’s just digress. Let me give you an example of something

that looks horribly wrong but is completely fine. Let’s actually take this example. Let’s take

this one. If I square it out, it goes like, I’m going to be using shorthand, but I think it’s pretty

understandable what these things are. It’s A square plus 2 Pi, 2D Pi, A plus, I’m actually setting,

by the way, I think I’m not being careful with gauge couplings, I’m setting gauge coupling to one

in this case.

−1

4F 2 − 1

2m2A

A2 + 2ππA+ (∂π)2︸︷︷︸−π2π

But general lesson is still there. This part of the story is a little bit schematic. That, okay. So this

is also known as a negative Pi box Pi after integrated by part. Now the next thing you observe

is that this is a quadratic in Pi. This is a quadratic in Pi. Nothing goes beyond that. When

something is a quadratic in a field, we don’t even need to do preservation theory. We can just do

exactly – it’s a gaussian, if it’s quadratic it’s gaussian. I can integrate this out basically. I can

perform D Pi exactly. Integrate out. In fact, I don’t need to do that. I can use the equation motion

77


of Pi to integrate up, because something quadratic, there is nothing wrong with it.

→ L =1

4F

(1 +

m2A

2

)F

Now I can integrate, I can substitute, integrate means I substitute everything I see a box Pi, I

substitute with DA. Okay? With everything, as every time we see Pi, I substitute so I make this

the following substitution. Probably having a heart attack with this one box. But that’s okay.

The thing we end up with is the following. Again, I’m not being careful with the factors. That

is the LaGrangian. So if I write it this LaGrangian to begin with you will kill me, because it’s

like, what is this? This is not local. But we know secretly it’s just that. That is completely fine.

Anyway, it’s not a very important for our course. But this is just an example of, LaGrangian can

fool you. If you are writing down this LaGrangian you think you are doing cutting edge research

in particle physics, you are not. You are just doing that. Okay? Okay. Now, this does bring up

an interesting point. But because now I completely trashed gauge symmetry, I told you early on

it is not symmetry and so on, but you see that I can make anything that doesn’t look like gauge

symmetric gauge symmetry. Anything is gauge symmetric. Just have to replace A, A, replace with

that and I’m done. The whole thing is gauge symmetry. Gauge symmetrical. Now you say now

we don’t have roots anymore. I can write down everything I want, anything I want to write. Now

I can write down anything I want to, so all hell breaks through. Again, I’m coming back to this

LaGrangian. I already told you that this term is fine. Okay?

L = −1

4F 2 +

m2A

2A2 + a(∂A)2 + bA4

Now you ask why stop here. I’m going to keep going. I’m going to add this one. Nothing wrong.

Because you do add something like this as a gauge fixing but this is not gauge fixing, I’m not even

doing that at this moment. This is a physical coupling I’m adding here. Okay? I can do this as well.

You don’t have dimensional terms but maybe those things are suppressed by some energy scalar.

But these are renormalizable looking. Nothing quite wrong. This is the case, where actually, we

are keeping track with the Pis are very important. You keep track Pis are wrong actually teach

you a lot, okay is this so let’s try to do that. You notice that the deep, my rule is to replace every

A with that.

∂A = ∂(A+ ∂π) = ∂A+ 2π

(∂A)2 = a[(∂A)2 + (2π)2 + · · ·

]78


Every single A with that, okay? I’m going to do that. Okay? I’m going to do that. Again

schematically, okay? Let’s imagine I redefine the Pis with V inside of Pi now, I’m just not going to

keep all these things around, but you can put all the factors in and the story is the same. Okay?

This one is DA plus box Pi. You say what’s wrong with that? You can put it back in. And so DA

square is A, DA square plus box Pi square plus ... okay? Now, but also A square contribute a term

like this. So now let’s put these two terms together, and A square, okay? You say what is wrong

with this story? It’s a little bit wrong.

→ −m2A

2π2π + a(2π)2 → 1

p2(p2 −m′2)

Because, well, let’s go do it. We know that this is completely fine. This is just a kinetic term for

the Goldstone. Goldstone needs to have kinetic terms. Anything that can legitimately be called

a Cal man field needs a kinetic term is fine but this is not fine with these two. With these two

now you can solve it. If I use this to write down my propagator, this is going to be a P square in

the propagator there is going to be a P to the force in the propagator. Therefore, the propagator

will have the form of P square, P square minus P prime square, something like that. That is my

propagator. There is a pole in P square equals 0, that is fine. That’s Goldstone. But there is

another pole. Okay. Once you have P to the fourth there is another pole. There is another pole

on the order of MA square over A. This is certainly unphysical. There is something wrong with

this theory, a theory just like that. Because you are introducing a spurious poles, this will be,

okay?

pole ∼m2A

a← unphysical a must be small

To make this theory healthy, A must be, be small so this pole is above your cutoff. You can put this

in. But there is a limit, how big this A is. This kind of story tells you that Pi teach us a lot. Keep

the Pis around, they really teach you a lot about what is allowed to have. Not gauge symmetry

itself. But you have the Pi’s around, tells you how healthy a theory is. Okay? So let’s do SU2

version of it, of Pi, SU2 version. Let’s do SU2 version. Imagine everything here SU2, let’s imagine

all these things are SU2s. Let’s define a few things. A will be sigma AA. This is SU2 generators.

Pauli matrices. The appropriate quantity now, we want to have is, another is the, instead of E to

the I Pi, we will have E to the I sigma A Pi A over V, which I will call E to the I Pi over V, but

79


the Pi is sigma A Pi A. This is the generators, SU2 generators.

A = σaAa U = eiσa πa

v = eiπv π = σaπa

Lπ = v2trace[|∂U |2

]+ trace

[(∂π)2 +

(∂π)2π2

v2+ · · ·

]

80

10 MARCH 5TH, 2013

10 March 5th, 2013

Okay. So far, we have been talking about spontaneous [inaudible] just with the scalar field. Its

see so called spontaneous symmetry breaking of global symmetry. Although I think I’ve told you

also that gauge symmetry is not a symmetry. That is the only symmetry. Okay. That is always a

spontaneous, although you frequently heard a word of spontaneous symmetry, spontaneous breaking

of gauge symmetry, but that is a misnomer. So that is gauge symmetry is, I think I told you that

gauge symmetry is not really a symmetry. But we are going to use that word anyways, consistent, so

commonly used. Today we are going to talk about what happened if we gauge the global symmetry.

Remember still our canonical example is a complex scalar field.

V (φ) = µ2φ∗φ+ λ(φ∗φ)2

and when µ2 < 0 – you will have this Mexican hat potential. And there is a radial mode. Just to

remind you, we decided the distance away from the center, called v for vacuum expectation variable

of this field φ

〈φ〉 = v =|µ|√λ

1√2

m2h = 2λv2

This is what we have done so far with this scalar field. Now I want to put a photon into this system.

Okay? For the gauge field into this system, and so I will introduce a gauge field. A mu. Again, you

want gauge field. You want vector field let’s say, U1 gauge field. There is a gauge transformation.

I’m going to gauge transformation associated with this view 1.

Aµ → Aµ −1

g∂µα(x)

And I will also say φ has a charge +1, like an electric charge. So φ transforms φ → eiα(x)φ. The

Gauge transformation is just that, Aµ and φ with charge +1.

Kinetic Terms

1

2(∂µφ)2

Kinetic terms looks like in original theory, okay? Now this is not quite invariant under that. Okay?

Let’s call this gauge transformation. Okay. But as we already learn in QED the trick is write some

81

10 MARCH 5TH, 2013

covariant derivatives. Whenever I say g, it’s gauge coupling; or if there is a charge, I use q.

Dµφ = (∂µ + igAµ)φ

Dµφ→

gauge trans eiα(x)Dµφ

hat is the benefits of including the – well, by putting them together basically. So therefore, it’s

clear to how to write a LaGrangian that is invariant under this gauge transformation which is this.

Okay? That is our LaGrangian

L = −1

4FµνF

µν + |Dµφ|2 − V (φ)

So, okay. Everything looks pretty innocent at this point. Our LaGrangian, let’s not worry about

the gauge fixing and anything like that, is just that. This is a kinetic term because I introduced the

A field. I have a kinetic term for this guy, so D mu ph iSquare over, then I’m getting to potential.

This is called a boolean hicks model. Everything has a name. It is very important. It is knowing the

name is more important than sometimes knowing the exactly the physics. Because you immediately

know what people are talking about, okay? So that is our LaGrangian. Now you have to go work

at it. Let’s try to do it. So to proceed, we will obviously start with this point, okay? I’ll expand my

field around this point, because I want it to be in the broken phase. So, let’s see, in broken phase,

what I will have, well, there are different parameterization, but the parameterization we have been

using is the following. By the way, there is an equal probability or in contrary parameterization

with square root and without square root too. Anyway, that is completely, it is very annoying. It

is annoying if you talk about standard model. You can talk about with the vacuum value of the

field, whether with or without square root. With square root 2 it is 100, it is the same as top mass.

With square root 2 it is 250 something DV. If you want to write a paper about why it’s the same

as top mass you have to be aware this is a convention dependent statement.

Broken Phase

φ(x) =1√2

(v + h(x))ei(π(x)v

)

I already have two degrees of freedom to parameterize. There is an angular variable that takes you

around the circle and there is this radial mode, which we call h. Now we have to work it out a

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10 MARCH 5TH, 2013

little bit on the board

Dµφ =eiπ(x)v

√2

[∂µh+ i∂µπ +

i

v(∂µπ)h+ ig(v + h)Aµ

]

|Dµφ|2 =

normal looking kinetic terms︷︸︸︷1

2(∂h)2 +

1

2(∂π)2 +

mass term and Higgs field︷︸︸︷1

2g2(v + h)2A2

+1

4v2(∂π)2h2 +

1√2

1

v(∂π)2h

+ g(∂π)(v + h)A+1√2

g

v(∂π)h(v + h)A︸︷︷︸

goes away by gauge transformation

Under Gauge transformation φ→ φ+ vα(x). So we “gauge away π” (π = 0; unitary gauge).

→ L =1

2(∂h)2 +

g2

2(v + h)2A2

=1

2(∂h)2 +

g2v2

2A2︸︷︷︸

=m2A2A2 “mass term”

+g2vhA2 +g2

2A2h2

And the g2v and g2

2 terms are fixed by Dφ. I want to in particular draw your attention to these

things. Okay? All these coefficient are fixed. All these things are fixed. Basically, by the form of

this, like this. All of these are completely fixed. Okay? So, you know, in other words, in order to see

whether, so, okay, let me first say a few – oh, in order to see whether, just in order to test whether

this scheme that we have drawn up so far, to give gauge a mass is correct or not, it is not enough

to just see that. Okay? It is far from enough to just see that, just see a Higgs boson as I would say,

I’ll comment later on. But it is very, you know, very important to also measure these couplings

because these couplings are precisely related to each other and related to the mass. That is why

it’s important even if, even as you discover some boson, it may not be the Higgs boson, you wanted

to measure these couplings. So you want to measure the coupling with a Higgs, with two As, and

you want to measure the coupling with the Higgs, two Higgs with two As. Okay? This is U1, but

you must have read a newspaper for SU2 version of this, for some reason that is more important.

And this is more or less measured already. This is almost impossible to measure. But that is the

prediction, even measuring this is already a big confirmation of how this works. You must know

what these things are called. This is called Higgs mechanism. I will clarify the terminology further

later. This is Higgs mechanism. But, this week there is a big meeting in morian, everybody know

where it is? I don’t. It is somewhere in Europe. It is a ski resort. Doing physics is great. You go

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10 MARCH 5TH, 2013

Phases Aµ φ

Coulomb µ2 > 0, no SSB mA=02 polarizations: λ=±1

charged2 massive scalars

µ2 < 00 mA 6=03 polarizations: λ=±1,0

π= would-be Goldstoneh= radial mode

Table 2: Phases

to ski resorts. And having meetings, and that is usually some place where big experimental result

gets announced. I was just told that the official name now for that meeting at least, of everybody

else is going to call it Higgs boson but for that meeting it is going to be called BEH boson (Brout,

Englert, Higgs).

And now we proceed to the dropping of the names.

So anyway, but those days, people, when you write a paper, you have to, it was even before my

time, so it was when you write a paper, you call up your friend say I’m going to send you a preprint.

Months later the preprint arrived and so on, so forth. These days, if you have anything in your

head, you put it on Facebook first.

Everything is uniquely normalized of course. But I’m just going to do the classic. Actually I’m

not going to normalize it. That is going to take another three weeks to do, introducing goldstone

which we haven’t even talked about. I’m just going to do the classical, mostly classical part of the

story. Okay? This is called the Higgs mechanism. Maybe you have been slightly wondering what

exactly is going on. This is nice. But when Peter Higgs likes to give a talk, I’ve seen one of his

talks, his talks, is my life as a boson. Have you seen that? You should go to see the talk. It is

interesting. He said when he first wrote that paper, he went to Princeton and Harvard and MIT to

give talks. Everybody assured him he is completely wrong, because the existence of goldstone after

symmetry breaking has been proved by some, I don’t know, called C style algebra or whatever,

some axiomatic quantum field theory. So let’s understand a little bit better, just in that vein. Of

course, axiomatic quantum field theory doesn’t understand in this. That is almost totally useless

endeavor. Okay. Let’s try to understand why exactly we get a mass. Or why do we get a mass?

This is a completely correct but you may want to, may want me to say it in a more deeper sounding

way to get convinced. Here is a slightly deeper way of saying it.

I have goldstone and I have radial mode. But this guy obviously becomes our H in the end. But this

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10 MARCH 5TH, 2013

guy has come here actually, become one of the polarizations. Usually I said it becomes longitudinal

polarization but that is slightly gauge dependent that statement. What these three polarization is

coming from is the taking a goldstone mode and taking this to form of three polarizations. You can

see this as I again, I want to, I can gain away the Pi. And it becomes part of A, basically. There is

no, there is not a invariant distinction between Pi and A basically. When you shift the Pi, A shift.

I’ll write it down. As you can see, any time you shift A to some, alpha Pi also shift. There is no,

with A around, there is no invariant distinction between Pi and A. They just become one of the

same state basically. This is the official statement of Higgs mechanism. The goldstone is again the

terminology is eaten by the gauge boson. Gauge boson become massive and the goldstone here is

not called goldstone anymore. This is called a would-be goldstone. If there is no A, it is its own

goldstone but if there is A, when there is A, this is not an independent observable physical state.

And these are also not independent observable physical state. They combine into one massive

vector basically. That is the official statement of Higgs mechanism. This is the Higgs boson. But

it is not, you see that it is actually just playing a peripheral role in this case. It is not completely

– we will argue that, why, it’s actually not necessary, Higgs boson here.

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11 MARCH 7TH, 2013

11 March 7th, 2013

Today we wrap up a few remaining issues on symmetry breaking with U1.

We already said that the gauge U1, in gauge symmetry, gauge symmetry breaking, the story is that

you are going from A mu, massless, with no symmetry breaking. This is massless. The polarization

is plus/minus 1. Another complex scalar field with two degree of freedom, okay. But then you

go through a phase where there is symmetry breaking, this becomes massive. The correlation is

plus/minus 1 and 0. On the other hand, we said that in, after spontaneous symmetry breaking,

the salear vector changes as well. It goes to give you a Goldstone, and another scalar. We said

this is not observable, directly observable, because it’s becoming part of the massive vector, with

three polarizations. This is called eaton. That’s basically I think where we are last time. Does this

picture bother anybody? We went through two ways of convincing ourselves, we just do the straight

LaGrangian way, and we did a more formal sounding way, talking about poles or propagators. Poles,

the second one is actually very close to actually prove a theorem. The first one is showing you

what is going on. For your understanding of standard model it’s important, particle physics. Let’s

talk about Fermions in connection with spontaneous symmetry breaking. Maybe let’s not do this.

Let’s do, let’s briefly remark on some more complicated spontaneous symmetry breaking pattern

first. How about I have a U1, still have a U1 but now I have multiple Higgs. Now I have multiple

field, phi1 with charge Q1 and phi2 with charge Q2 and so on. How about I have that? Let’s

say every one of them gets a V1. If any of these values is nonzero, U1 symmetry is spontaneously

broken. There is actually for each one of them, in general, for each one of them, there is a U1 global

symmetry as well. If I just do this, now, the question is, what happened to the theory, after I’m

breaking all those things. Okay? After I give all these things values. To be specific, let’s consider

two. The theory will be ... D mu phi1 is IGQ1A mu, 1 and so on. Okay? Write this down, this is

going to be homework. (chuckles). They just go do this, basically. Generally this has value, that

means there is Vphi1, phi2 as well, which I’m not writing down at this moment. But you know

what is going to happen. Roughly speaking you know what is going to happen. Roughly, you know.

What is going to happen is that there is going to be a – (coughing). Sorry. Of course the gauge

boson will become massive. What is going to happen, so you can do the whole thing and look at

the mass and so on. What is going to happen, A, you are going to have MA. The mass is, the

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11 MARCH 7TH, 2013

gauge boson is going to have a mass. Exercise to find out what the mass is, okay? It’s the sum of

the values and so on, so forth. But it is going to have a mass because the symmetry is broken. The

question is what happened to the rest of the spectrum? I start out here. These are two complex

scalar field. This is 4-degree of freedom.

U(1) : φ1 : Q1, φ2 : Q2 → 4 d.o.f.

〈φ1〉 = v1 〈φ2〉 = v2

so

|Dµφ1|2︸︷︷︸mA

+|Dµφ2|2 + V (φ1, φ2)Dµφ1 =︸︷︷︸A=±1,0

(∂µ + igQ1Aµ)φ1

You know one of them is gone. One of them which is whatever a Goldstone is going to be going

to the, is going to the mass, the massive A vector field. That massive vector field has to have one

Goldstone, has to eat one Goldstone to become massive because otherwise it doesn’t have three

polarizations, so it’s just wrong. But then the rest of the three, still remaining, you still have three.

The other three scalars. This theory has three extra scalars. Without going through the whole

exercise, the exercise itself is pretty simple, multiply things out and look at the mass and so on and

the couple. Now, what if I also do another exercise, suppose 2U1. 1U2. I have two U1 symmetry.

I have two gauge bosons. A1 and A2. Suppose I have that now, suppose I have a charge, a particle

that is charged under both. I can have that too, right? I can have a particle charge under 2U1

symmetries, and with Q1 and Q2. Suppose I have that. So this sometimes can be summarized in

the diagram like this. It is called U11, U12 and 5 I guess. Now, depends on what field are you in,

this diagram have different names. This diagram has different names. In some fields it’s called this.

This is actually called a minimum moose, longer lattice, quiver, but anyway, so I’m just preparing

you for something. These are the sides and this is called the link.

U(1)1)︸︷︷︸A1

×U(1)2︸︷︷︸A2

φ : Q1, Q2

and you can imagine that φ is the link between the two sites U(1)1, U(2)2.

Dµφ = (∂µ + iQ1g1Aµ1 + iQ2g2Aµ2)φ

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11 MARCH 7TH, 2013

and φ has 2 d.o.f. 〈φ〉 = v → would-be 1 Goldstone.

But it’s important to give them good names. So that they can be remembered. So anyway, string

theory, deeper in physics, this is called a quiver. Symmetric H theories. Anyway, suppose I have a

quiver, a moose like this. So what do I, what is the LaGrangian? It is actually very simple. Those

names are sometimes scary. But it is actually simple. D mu Phi is partial mu. Now you have to

include all the gauge field here. Because the Phi is a charged under two gauge fields. It has to be

locally gauge, has to be gauge under the transformation of the, there are two independent gauge

transformation. So the current derivative has to involve all of them. Q1 in general, there is a G1A

mu plus IQ2G2A mu 12, not using, choosing very illuminating notation, but this is what you do.

The LaGrangian is, again that. You can open this up. And do our usual thing. You can open this

up and look at the masses and so on, look at the mass of gauge bosons and so on. But you already

know what is going to happen in the end. Qualitatively, what is going to happen in the end, you

can work out the details, or you will work out the details, is that there are two degree of freedom

here. Okay? Only one of them shift under the gauge symmetry. After, again after SSB, so after

SSB, only one Goldstone, would be Goldstone, only one would be Goldstone. Okay? Therefore,

you know that what is going to happen is that in the end, there is going to be one unbroken, one

massless vector. Okay? It is going to be one massive vector, plus one massless vector. Because

there is only one Goldstone, it can only supply one of the vectors with its longitudinal mode. Just

by counting the degree of freedom, so, yeah. But this massive one is not any one of these. This

massless one is not any one of this. They are linear combinations. You have to work out the linear

combination. Basically, all of them are just linear combinations of A1 and A2. Okay? That you

have to, you open this up, square, there is a mass matrix, Again this is similar to what you do with

couple harmonic oscillator. You have to diagrammize the frequency mate I can to find the Eigen

mode. There are two Eigen mode. One of them is massive vector. You know there can only be

one because there is only one Goldstone to be eaten. Is this okay? Both of these two situations,

these two situations happen, this happens in standard model. A similar story happen here in the

standard model, similar story to that happens in EMSSM. Let’s also talk about the Carol Fermions.

Do I want to talk about Fermions here or later? Let’s talk about it here. Let’s remember what

the Fermion mass is. Fermion mass is something like that. But we know that, you still remember

Fermion part at the beginning of this quarter, we can write it as a, left-handed or right-handed

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11 MARCH 7TH, 2013

component. Okay?

mψψ ψ =

(ψLψR

)ψ =

(ψ†R, ψ

†L

)= m(ψ†LψR + ψ†RψL)

So that, okay? That is that, okay? That is just that. Fermion mass is mass. This okay? I’m

multiplying them out. Okay? Lorentz invariance, this basically is fixed by Lorentz invariants.

Lorentz invariants basically is saying that left-handed and right-handed for massive Fermions are

not good quantum numbers, because they mix. That is basically what Lorentz invariants is telling

us for massive Fermions. On the other hand, there could be situations that those two charged under

different gauge symmetry. That would carry different charges under gauge symmetry. Okay?

Under U(1) Under U1, under U1, there can be two situations, okay? One is that Q left is the same

as Q right. The left hand Fermion and right-handed Fermion has the same charge under this U1.

So, ah, I should actually say something slightly different. Now, let’s look at the kinetic terms. Let’s

look at the kinetic terms. The kinetic terms looks like this, looks like this.

ψ†Lσµ(∂µ + iQLAµ)ψL + ψ†Rσ

µ(∂µ + iQRAµ)ψR

This is 2 by 2 poly matrices basically – Pauli. It doesn’t matter for our argument but these are

kinetic term. This is gauge invariant, kinetic terms. Gauge invariants does not tell me anything

about QL versus QR, okay, of the kinetic terms. There is no limit. I can just set this whatever I

want. There are two situations. Case one is QL equals QR. This is called a vector like, Fermions. It

doesn’t mean the Fermion becomes a vector. It just means that the left hand and right hand have

the same charge. This means that it is invariant. That just means that this term is invariant, if QL

is the same as QR, and the charge of this, rotation of this is opposite of the rotation of that.

i) QL = QR vector-like → mψψ invariant

ii) QL 6= QR chiral →mψψ not invariant

This is gauge invariant. Then there is another situation with QL not the same as QR. Okay? In

this case, this is called the kierl, the Fermion is called the curial. In that case, it is not invariant.

That cannot exist. A term like that cannot exi For kierl Fermion does not have mass basically,

that is the statement. Just because you can assign left-handed and right hand Fermions, different

charges. Let’s consider an extreme case.

89


Consider QL = 1, QR = 0

yφ(ψ†LψR) + y∗φ†ψ†RψL)

〈φ〉 = v

→ yv︸︷︷︸mψ

(ψ†LψR) + h.c.

= mψψ†LψR + h.c.

QL is 1, QR is 0. That is one of those cases. There is nothing, you cannot write a mass term

basically. Okay? Look at this mass term, only QL carries charge. U1 rotation is certainly not

gaiblg – gauge invariant. So now you wonder is there any way to give those Fermion masses,

because after all we live in a world with Fermion masses, where Fermion has masses. Maybe you

say, okay, maybe the Fermion world is all vector like, but it is not. Unfortunately it’s not. We

will talk about it too. Suppose I give you a pair of Fermions, how do you give them mass? The

trick is again there is a spontaneous symmetry break. It is going through spontaneous symmetry

break.

Consider φ,Qφ = 1, 〈φ〉 = v.

φ = (v + h)→ yhψ†LψR + h.c. =mψ

vhψ†LψR

11.1 Standard Model

So, are you all, you have all seen standard model at some point? Particles and interactions? Not

see it but know about it. Standard model, so the story we just said about U1 is important for

understanding the most intriguing part of, not the most intriguing but at least the most, one of

the most interesting part of the standard model. Standard model are controlled by gauge similar

trees – symmetries.

gauge symmetry SU(3)C × SU(2)L × U(1)Y

SU(3)C strong interactions 8 gluons quarks are 3: u,d,c,s,t,b

SU(2)L × U(1)Y

SU(2)L doublet(ud

)L

which mirrors UR, dR singlet

90


doublet

scalar: H =

(h†

H0

)SSB=

(h†

v + h+ ia

)Yπ =

1

2

SU(2)L × U(1)Y

which gives 3 massive particles W±, Z and a massless photon. We have 3-Goldstones

yuH QL︸︷︷︸(uLdL)

uR + yDHcQLdR

actually which way is the left-handed. Let’s call this left-handed. Let’s call this right-handed. Okay.

They carry different charge, if you want, under SU2 gauge interaction in the most extreme way.

One of them couples to them, one of them don’t. Okay? But this too also interchange each other

under parity. Okay? So means that if you imagine you put a mirror here, this goes to that. Okay?

So if you put a mirror here, this just goes to that. Okay? Gauge interaction SU2L violate parity, so

this is parity violation. Is there anything wrong, violating parity? Nothing. It can be violated. It

is violated in nature. Nothing, nothing goes horribly wrong. For a long time we thought it would

be, parity is obviously a symmetry but it’s not. But again very similar to what I said here, very

similar to what I said here. You see the statement I just made is not completely consistent with

Lorentz invariants, of massive Fermions at least, because we know from experiment these things

gets, all have mass. By the way this is called opt because usually we write it on top, I don’t know,

top down, the term strange is a little bit, I don’t know why, maybe strange that they don’t think

it should be there, and charm is just try to make a good name. I don’t know. But this is not

consistent with Lorentz invariants because for massive Fermions, so there is a very intuitive way of

understanding why mass makes two polarizations, because for massive particle there is a rest frame.

Once you get into the rest room you lost a direction, there is no direction. In the rest room you can

pick any access of your polarization vector, and obviously left-handed can become right-handed.

Right-handed can become left-handed. It’s your choice. In the massless case it makes sense because

there is no rest room. You are always moving in that direction. You always use that direction as

your reference, polarization. For massless Fermions there is a direction. We also know that this

symmetry is broken. The good thing is, so at this point in order to accommodate massive Fermions

you need to do one of those two things. Either break the symmetry or break Lorentz symmetry.

Lorentz symmetry we are not breaking. We are not breaking this one. But we know this is broken.

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The gauge bosons are massive. It’s broken by a doublet. It can be broken by a doublet, scalar

doublet. This is again a scalar. H is a – have to write it like this, H0. Just in a very suggestive

form, and this under spontaneous symmetry breaking, you can write this as an H plus, V plus H

plus IA. So that is how you do standard model doublet. And, okay. On the other hand, the H also

charges under this U1. So we haven’t talked about U1 yet. H also is charged under this U1. So YH

is 1/2, according to normalization, that is the usual normalization we use. So that’s that. So this H

when it gets verve here, first of all you may ask me why am I giving vev to this guy, why am I doing

this, why am I not giving to it that guy or whatever it is? (vev?) The answer is it doesn’t matter.

You can put it wherever you want, you can give vev to every component. It is similar to the U1

case, the valve can go any direction and the end result is the same. All those are vacuum, all those

vacuum are equivalent as we have said, along the circle. Similarly, you can take this to where SU2

rotation, give vev to anywhere you want, but all those vacuums are completely equivalent. Very

similar to the U1 case, okay? This breaks both SU2, cross U1Y. It breaks both of them. Okay?

On the other hand, there are only – well, the global symmetry breaking is SU2 down to nothing.

There are only three Goldstones which is this and this. It’s again very similar to the U1 case. It’s

not everything in the scalar, it can be a Goldstone. Okay? In this case, these and that can, gold

stones. These and that. This is a complex scalar so there are two degree of freedom. Between two

of them there is one of them. There are only three Goldstones and how many generators are here?

How many gauge bosons are here? SU3 has one, how many, SU2 has how many? Three. It’s SO(3)

basically, just three rotations. There are four generators. Three Goldstones. Okay? There is one

remain massless. Very similar to what I just erased, similar to what I just erased. Okay? So what

is going to happen is going to be three massive gauge bosons, and they all have names. They are W

plus/minus and the Z. And one massless. So because, again, which one is, what is the massless one

here? It’s the photon. That is the symmetry breaking. Again, but again, it’s not exactly this one

doesn’t become the photon, and this one is not exactly these three, and it is a linear combination.

In particular, Z and the photon are linear combination of the tau 3 here, and this U1 basically.

But let’s not work it out. It is going to take us too long. But so yeah, these are, I’m mentioning

this because this is continuous, it is a continuous story that we have talked about. You see there

is nothing to it. All you have to do is write down the current derivative and diegriz it, basically.

(?) With this H obviously we can write masses now for the Fermions. With this H I can start

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to write YHQLUR, call this YU plus YD, H, conjugate, QLDR. Doesn’t quite matter what this

is. But you see these are the mass terms. Okay? This is, again, this is just my shorthand way

of writing down ULDL. These are the mass terms of the standard model quarks basically. They

have different Yukawa couplings in general. They don’t have the same. We all know that there are

more than one quarks. Right? In general, these are 3 by 3 matrices. They are three quarks, three

flavor of quarks. They are [inaudible] too. To understand that, that is called a flavor problem. We

don’t know really how to derive these Yukawa matrices from first principles, and it’s unlikely we

are going to know any time soon. Okay. That is basically the standard model. Everything what

I said has already been discovered, by the way. I’m not making any of this story up. Let’s take a

step back. Since we have been talking about Higgs mechanism and so on, let’s take a step back,

let’s try to understand Higgs mechanism slightly better, what does it really mean. For that let’s go

back to the U1 story. Let’s from a bottom up point of view, let’s take a bottom up point of view.

Let’s suppose I’m very dumb. I don’t know about – let’s just say I observe a massive gauge boson.

Okay?

−1

4FµνFµν +

m2A

2AµA

µ unitary gauge

I’ll write a mass term like that. Okay? You say wait, that is all wrong, right? It is not gauge

symmetry. We all know we need gauge symmetry, blah blah blah, but actually we don’t. First of

all we don’t. Let’s say we do, but the usual story tells you that now gauge symmetry forbid this

term. That is why I need to introduce, that is why I need to do this whole Higgs boson mechanism

that write a cowritten derivative, give you a mass, that part of the story is wrong. The reason is

that, the reason this is wrong is that this is a secret to the gauge symmetric.

in general gauge Okay? It’s not gauge symmetric when you fix to a unitary gauge which is not

such a big surprise. But suppose let’s work in the slightly more general gauge. In general, in a

general gauge I can write this as this following form. You see this is completely gauge invariant.

Remember in gauge transformation A mu goes to A mu D mu, alpha we already said there is no

need to realize Pi, Pi shift as of this. Under this gauge transformation this is completely invariant,

complete gauge invariant.

−1

4FµνFµν +

m2A

2

(Aµ + ∂µ

π

v

)2

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with gauge transformation

Aµ → Aµ + ∂µα, π → π − vα

On the other hand we know these two are completely [inaudible] to each other, there is nothing

different. Using gauge transformation I can gauge only Pi it become this, become unitary gauge.

Therefore, this is actually gauge invariant. There is nothing wrong with that. Gauge invariance

does not prevent you to write down a mass term. It is completely fine. The only thing you have to

remember is now there are three degree of freedom. But that is the real physical statement, between

massive and massless gauge boaston is not whether it’s gauge symmetry or not. It’s whether there

is three degree of freedom or not. That what I’m trying to get at. This is a real physical statement

is that. Goldstone is everything. The eaten Goldstone captures the essence of the physics really.

This also tells you that the LaGrangian sometimes can be very deceiving. Although we are getting

really used to doing everything with LaGrangian, but LaGrangian sometimes can tell you, teach

you a very wrong lessons, because it can hide things. You can hide things with LaGrangian.

You can do horribly wrong things with LaGrangian. If something looks horribly wrong, it can be

completely fine. Okay? So for that, let’s just digress. Let me give you an example of something

that looks horribly wrong but is completely fine. Let’s actually take this example. Let’s take

this one. If I square it out, it goes like, I’m going to be using shorthand, but I think it’s pretty

understandable what these things are. It’s A square plus 2 Pi, 2D Pi, A plus, I’m actually setting,

by the way, I think I’m not being careful with gauge couplings, I’m setting gauge coupling to one

in this case.

−1

4F 2 − 1

2m2A

A2 + 2ππA+ (∂π)2︸︷︷︸−π2π

But general lesson is still there. This part of the story is a little bit schematic. That, okay. So this

is also known as a negative Pi box Pi after integrated by part. Now the next thing you observe

is that this is a quadratic in Pi. This is a quadratic in Pi. Nothing goes beyond that. When

something is a quadratic in a field, we don’t even need to do preservation theory. We can just do

exactly – it’s a gaussian, if it’s quadratic it’s gaussian. I can integrate this out basically. I can

perform D Pi exactly. Integrate out. In fact, I don’t need to do that. I can use the equation motion

94


of Pi to integrate up, because something quadratic, there is nothing wrong with it.

→ L =1

4F

(1 +

m2A

2

)F

Now I can integrate, I can substitute, integrate means I substitute everything I see a box Pi, I

substitute with DA. Okay? With everything, as every time we see Pi, I substitute so I make this

the following substitution. Probably having a heart attack with this one box. But that’s okay.

The thing we end up with is the following. Again, I’m not being careful with the factors. That

is the LaGrangian. So if I write it this LaGrangian to begin with you will kill me, because it’s

like, what is this? This is not local. But we know secretly it’s just that. That is completely fine.

Anyway, it’s not a very important for our course. But this is just an example of, LaGrangian can

fool you. If you are writing down this LaGrangian you think you are doing cutting edge research

in particle physics, you are not. You are just doing that. Okay? Okay. Now, this does bring up

an interesting point. But because now I completely trashed gauge symmetry, I told you early on

it is not symmetry and so on, but you see that I can make anything that doesn’t look like gauge

symmetric gauge symmetry. Anything is gauge symmetric. Just have to replace A, A, replace with

that and I’m done. The whole thing is gauge symmetry. Gauge symmetrical. Now you say now

we don’t have roots anymore. I can write down everything I want, anything I want to write. Now

I can write down anything I want to, so all hell breaks through. Again, I’m coming back to this

LaGrangian. I already told you that this term is fine. Okay?

L = −1

4F 2 +

m2A

2A2 + a(∂A)2 + bA4

Now you ask why stop here. I’m going to keep going. I’m going to add this one. Nothing wrong.

Because you do add something like this as a gauge fixing but this is not gauge fixing, I’m not even

doing that at this moment. This is a physical coupling I’m adding here. Okay? I can do this as well.

You don’t have dimensional terms but maybe those things are suppressed by some energy scalar.

But these are renormalizable looking. Nothing quite wrong. This is the case, where actually, we

are keeping track with the Pis are very important. You keep track Pis are wrong actually teach

you a lot, okay is this so let’s try to do that. You notice that the deep, my rule is to replace every

A with that.

∂A = ∂(A+ ∂π) = ∂A+ 2π

(∂A)2 = a[(∂A)2 + (2π)2 + · · ·

]95


Every single A with that, okay? I’m going to do that. Okay? I’m going to do that. Again

schematically, okay? Let’s imagine I redefine the Pis with V inside of Pi now, I’m just not going to

keep all these things around, but you can put all the factors in and the story is the same. Okay?

This one is DA plus box Pi. You say what’s wrong with that? You can put it back in. And so DA

square is A, DA square plus box Pi square plus ... okay? Now, but also A square contribute a term

like this. So now let’s put these two terms together, and A square, okay? You say what is wrong

with this story? It’s a little bit wrong.

→ −m2A

2π2π + a(2π)2 → 1

p2(p2 −m′2)

Because, well, let’s go do it. We know that this is completely fine. This is just a kinetic term for

the Goldstone. Goldstone needs to have kinetic terms. Anything that can legitimately be called

a Cal man field needs a kinetic term is fine but this is not fine with these two. With these two

now you can solve it. If I use this to write down my propagator, this is going to be a P square in

the propagator there is going to be a P to the force in the propagator. Therefore, the propagator

will have the form of P square, P square minus P prime square, something like that. That is my

propagator. There is a pole in P square equals 0, that is fine. That’s Goldstone. But there is

another pole. Okay. Once you have P to the fourth there is another pole. There is another pole

on the order of MA square over A. This is certainly unphysical. There is something wrong with

this theory, a theory just like that. Because you are introducing a spurious poles, this will be,

okay?

pole ∼m2A

a← unphysical a must be small

To make this theory healthy, A must be, be small so this pole is above your cutoff. You can put this

in. But there is a limit, how big this A is. This kind of story tells you that Pi teach us a lot. Keep

the Pis around, they really teach you a lot about what is allowed to have. Not gauge symmetry

itself. But you have the Pi’s around, tells you how healthy a theory is. Okay? So let’s do SU2

version of it, of Pi, SU2 version. Let’s do SU2 version. Imagine everything here SU2, let’s imagine

all these things are SU2s. Let’s define a few things. A will be sigma AA. This is SU2 generators.

Pauli matrices. The appropriate quantity now, we want to have is, another is the, instead of E to

the I Pi, we will have E to the I sigma A Pi A over V, which I will call E to the I Pi over V, but

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the Pi is sigma A Pi A. This is the generators, SU2 generators.

A = σaAa U = eiσa πa

v = eiπv π = σaπa

Lπ = v2trace[|∂U |2

]+ trace

[(∂π)2 +

(∂π)2π2

v2+ · · ·

]

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QFT Part 2 Notes