physics 253a - quantum field theory i - github pages · 2018-12-12 · physics 253a notes 4 1...

Physics 253a - Quantum Field Theory I

Taught by SchwartzNotes by Dongryul Kim

Fall 2018

¡+instructor+¿ ¡+meetingtimes+¿ ¡+textbook+¿ ¡+enrolled+¿ ¡+grading+¿¡+courseassistants+¿

Contents

1 September 4, 2018 41.1 Quantum theory of radiation . . . . . . . . . . . . . . . . . . . . 4

2 September 6, 2018 62.1 Special relativity . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 Quantum mechanics . . . . . . . . . . . . . . . . . . . . . . . . . 7

3 September 11, 2018 93.1 Operators on the Fock space . . . . . . . . . . . . . . . . . . . . . 93.2 Classical field theory . . . . . . . . . . . . . . . . . . . . . . . . . 103.3 Noether’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 11

4 September 13, 2018 124.1 Coulomb’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124.2 Green’s functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

5 September 18, 2018 155.1 Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155.2 Two-to-two scattering . . . . . . . . . . . . . . . . . . . . . . . . 16

6 September 20, 2018 186.1 LSZ reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186.2 Feynman propagators . . . . . . . . . . . . . . . . . . . . . . . . 19

7 September 25, 2018 217.1 Schwinger–Dyson equations . . . . . . . . . . . . . . . . . . . . . 217.2 Feynman diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . 23

1 Last Update: November 29, 2018

8 September 27, 2018 258.1 Feynman diagrams in momentum space . . . . . . . . . . . . . . 258.2 Hamiltonian derivation . . . . . . . . . . . . . . . . . . . . . . . . 268.3 Matrix element for the two-to-two scattering . . . . . . . . . . . 27

9 October 2, 2018 299.1 Writing down the Lagrangian . . . . . . . . . . . . . . . . . . . . 29

10 October 4, 2018 3210.1 Representations of the Poincare group . . . . . . . . . . . . . . . 3210.2 Induced representations . . . . . . . . . . . . . . . . . . . . . . . 33

11 October 9, 2018 3511.1 Scalar quantum electrodynamics . . . . . . . . . . . . . . . . . . 3511.2 Photon propagator . . . . . . . . . . . . . . . . . . . . . . . . . . 3611.3 Feynman rules for scalar QED . . . . . . . . . . . . . . . . . . . 37

12 October 11, 2018 3912.1 Gauge invariance and the Ward identity for scalar QED . . . . . 3912.2 Lorentz invariance and soft photons . . . . . . . . . . . . . . . . 4012.3 Spinors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

13 October 16, 2018 4313.1 Representations of the Lorentz group . . . . . . . . . . . . . . . . 4313.2 Dirac spinors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

14 October 18, 2018 4614.1 The Dirac equation . . . . . . . . . . . . . . . . . . . . . . . . . . 4614.2 Quantum electrodynamics . . . . . . . . . . . . . . . . . . . . . . 48

15 October 23, 2018 4915.1 Identical particles . . . . . . . . . . . . . . . . . . . . . . . . . . . 4915.2 Propagator for QED . . . . . . . . . . . . . . . . . . . . . . . . . 51

16 October 25, 2018 5216.1 Feynman rules for quantum electrodynamics . . . . . . . . . . . . 52

17 October 30, 2018 5517.1 Path integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5517.2 Generating functionals . . . . . . . . . . . . . . . . . . . . . . . . 57

18 November 1, 2018 5918.1 Feynman rules from the path integral . . . . . . . . . . . . . . . 5918.2 Gauge invariance from path integrals . . . . . . . . . . . . . . . . 61

19 November 6, 2018 6219.1 Renormalization . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

2

20 November 8, 2018 6520.1 Examples of renormalization—φ4 . . . . . . . . . . . . . . . . . . 6520.2 Examples of renormalization—vacuum polarization I . . . . . . . 66

21 November 13, 2018 6821.1 Examples of renormalization—vacuum polarization II . . . . . . 6821.2 Dimensional regularization . . . . . . . . . . . . . . . . . . . . . . 70

22 November 15, 2018 7222.1 Examples of renormalization—anomalous magnetic moment . . . 72

23 November 20, 2018 7523.1 Renormalizability of QED . . . . . . . . . . . . . . . . . . . . . . 75

24 November 27, 2018 7824.1 1PI and amputation . . . . . . . . . . . . . . . . . . . . . . . . . 7824.2 Renormalized perturbation theory . . . . . . . . . . . . . . . . . 79

25 November 29, 2018 8125.1 Renormalizability of QED . . . . . . . . . . . . . . . . . . . . . . 8125.2 Nonrenormalizable theories . . . . . . . . . . . . . . . . . . . . . 82

3

Physics 253a Notes 4

1 September 4, 2018

You need at least 10 hours a week to take this course. This course will get moredifficult as we go into renormalization. Then it will get easier once we pass thisand get to applications.

We will start with special relativity and quantum mechanics, put them to-gether and see what happens. We won’t start with the axioms, because theyare just statements that sound reasonable but cannot be tested.

1.1 Quantum theory of radiation

When you turn on the lights, the number of particles increase. How does thishappen? Max Planck in the 1900s observed that discrete energy can explainblackbody radiation. Einstein in 1916 explained spontaneous/stimulated emis-sion, and Paul Dirac in 1927 invented quantum electrodynamics, the microscopictheory of radiation.

We have a box of size L, poke a hole and heat it up. Then light comesout. We know that the wave numbers associated with the box are ~k = 2π

L ~n,

and ω = |~k|c. This is classical prediction. Then the number of modes ≤ n isproportional to n3, and the classical equipartition theorem predicts that eachmode has the same energy. Sow we would have

dI(ω) ∼ ω2dω.

This is called the ultraviolet catastrophe. But experimentally, we have expo-nential decay.

Planck said that energy E is quantized, so that En = hωn. Here, ωn = 2πL n

where n = |~n|. Then each mode gets excited an integer number of times, Etotn

is an integer times En. The probability of Etotn ∼ e−βEn . Then

〈En〉 =

∑∞j=0(~jωn)e−j~ωnβ∑∞

j=0 e−j~ωnβ

=~ωn

e~ωnβ − 1.

Then the total energy up to ω is

E(ω) =

∫ ω

0

d3n~ωn

e~ωnβ − 1= ~

∫ 1

−1

d cos θ

∫ 2π

0

dφ

∫ Lω/2π

0

n2dnωn

e~ωnβ − 1

= ~L3

(2π)34π

∫ ω

0

ω3

e~ωβ−1.

So we get Planck’s formula

I(ω) =K

2π2

ω3

e~ωβ − 1× 2.

The point here is that each mode gets excited an integer number of times.This is called second quantization. This really is just quantization, because


the first quantization refers to ~k = 2πL ~n, which is just classically solving wave

equations with boundary conditions.Let us now look at a number of atoms, either in the ground state or the

excited state with energy difference E2 − E1 = ~ω. Let n1, n2 be the numberof atoms with energy E1, E2. Also assume that there is a bath of photons of

frequency ω, with intensity I(ω) and number nω = π2

ω3 I(ω). If we look at theprobability of atoms getting excited or emitting, we get

dn2 = −An2 −BI(ω)n2 +B′I(ω)n1.

Here, the first term is spontaneous emission, the second is stimulated emission,and the third is stimulated absorption. It’s not obvious that the second termshould exist, but it turns out to be nonzero. In equilibrium, we have

I(ω)(B′n1 −Bn2) = An2.

So we get

I(ω) =A

B′ n1

n2−B

=A

B′eβ~ω −B

because n1 = e−βE1 and n2 = e−βE2 .Matching with Planck’s formula, we get the relations

B = B′, A =~π2ω3B,

called Einstein’s equations. The number B can be calculated by quantum me-chanics. So we can calculate A using this relation and quantum mechanics.

This is what got to Dirac. It’s great that we can compute the coefficientof spontaneous emission, but it will be good to calculate this without usingthermal systems, just from fundamental laws. The second quantization reallylooks like the simple harmonic oscillator. So we are going to identify

|n〉 = n photon state = nth excited state of the oscillator.

Consider a† the creation operator and a the annihilation operator so that[a, a†] = 1 and N = a†a is the number operator with N |n〉 = n|n〉. We cancompute

a†|n〉 =√n+ 1|n+ 1〉, a|n〉 =

√n|n− 1〉.

This turns out to be a powerful tool.Now Fermi’s golden rules says that the transition rate is Γ ∼ |M |2δ(Ef−Ei).

If we use this, we get at the end,

|M2→1|2 = |M0|2(nω + 1), |M1→2|2nω|M0|2.

So this algebra of creation and annihilation operation gives us the relation be-tween spontaneous emission and stimulated absorbtion. Then more algebragives

dn2 = −|M0|2(

1 +π2

~ωI(ω)

)n2 +

π2

~ω3I(ω)n1.


2 September 6, 2018

Today we are going to start the systematic development of the field. We letc = 1 and ~ = 1.

2.1 Special relativity

There are rotations on the plane,(xy

)→(

cos θ sin θ− sin θ cos θ

)(xy

), xi → Rijxj .

We can also rotate row vectors as

xi → xi(RTij),

and the rotations satisfy RTij · 1jkRkl = 1il. This is because rotations should

preserve xixi = x2 + y2. In 3 dimensions, we have x2 + y2 + z2, and in 4dimensions, we have t2 − x2 − y2 − z2. So Lorentz transformations satisfy

ΛT gΛ = g, gµν =

1−1

−1−1

.

Examples include

Λθz =

1

cos θz sin θz− sin θz cos θz

1

, Λβx =

coshβx sinhβxsinhβx coshβx

11

.

Four momentum is defined as

pµ = (E, px, py, pz),

and it satisfies p2 = pµpµ = E2 − ~p2 = m2. Usually, ~x or xi denotes a 3-dimensional vector, and x or xµ denotes a 4-dimensional vector.

Tensors transform asTµν → Λµ

αΛνβTαβ .

We define the d’Alembertian as

= ∂2µ = gµν∂µ∂ν = ∂2

t − ~∇2 = ∂2t −∆.

We say that a vector is timelike if V 2 > 0, and spacelike if V 2 < 0, andlightlike if V 2 = 0.

The proper orthochronous Lorentz group has det Λ = 1 and Λ00 > 0.There are four components of the Lorentz group, and this is the connectedcomponent at the identity. The Poincare group are Lorentz transformationsplus translations.


2.2 Quantum mechanics

Remember we had normal modes in a box last time. These frequencies arequantized classically. Then Planck said that the energy should be associatedto the frequency E = j~ω. Einstein was the one who interepreted these asparticles, which we call photons, and Dirac developed this microscopic theoryof H = H0a

† +H0a.Let us review the simple harmonic oscillator. We have a ball with a spring

on it, and its equation of motion is

md2x

dt2+ kx = 0.

You can solve this, and you get

x(t) = cos(√ k

mt).

The classical Hamiltonian is given by

H =1

2

p2

m+

1

2mω2x2.

Then we quantize this using [x, p] = i~, and define

a =

√mω

2~

(x+

ip

mω

), a† = · · · , [a, a†], H = ~ω(N + 1

2 ), N = a†a.

We found last time that

N |n〉 = n|n〉, a†|n〉 =√n+ 1|n〉, a|n〉 =

√n|n− 1〉.

Then in the Heisenberg picture,

a(t) = e−iωta(0).

Now what can the equation of motion for the scalar field be? It should beLorentz invariant, so the simplest possible equation is

φ = 0 = (∂2t − ~∇2)φ = 0.

Take the Fourier transform, and let

φ(~x, t) =

∫d3p

(2π)3[ap(t)e

i~p·~x + a∗p(t)e−i~p·~x]

Then the equation becomes

(∂2t + ~p2)ap(t) = 0.

Now each component is just a classical simple harmonic oscillator. So we canquantize each separately, and then put them back together.


Electromagnetic saves are oscillators,

Fµν =

0 Ex Ey Ez−Ex 0 −Bz By−Ey Bz 0 −Bx−Ez −By Bx 0

.

This concisely encodes Maxwell’s equations

∂µFνρ + ∂νFρµ + ∂ρFµν = 0, ∂µFµν = 0

in empty space. It’s also helpful to write

Fµν = ∂µAν − ∂νAµ.

This vector potential Aµ is more useful for field theory, because there are only4 components, and also because it is invariant under the transformation

Aµ → Aµ + ∂µα(x),

called gauge invariance.We can choose ∂µAµ = 0, and this is called Lorentz gauge. When you do

that, Maxwell’s equations become

0 = ∂µFµν = Aν .

So then we can make Aν(x, t) into a set of harmonic oscillators. We write

Aν(x, t) =

∫d3p

(2π)3(Apν(t)ei~p·~x +Ap∗ν (t)e−i~p·~x), (∂2

t + ~p2)A~pν = 0.

Then the free electromagnetic field is equivalent to an infinite number of simpleharmonic oscillators, labeled by 3 vectors ~p with frequencies ωp = |~p|.

Now we quantize as in quantum mechanics. Then

H0 =

∫d3p

(2π)3ωp(a

†pap + 1

2 ).

The relations between these creation and annihilation operators are

[ak, a†p] = (2π)3δ3(~p− ~k), ap|0〉 = 0, a†p|0〉 =

1√2ωp|p〉.

What we have done is that we have constructed the Hilbert space

F =⊕p

Hp,

called the Fock space.


3 September 11, 2018

Last time we reviewed the simple harmonic oscillator. To quantize this theory,we defined H = ω(a†a + 1

2 ). For fields, we classically had Aµ(x) = 0 or( +m2)φ(x) = 0. We do the Fourier transform, and we get something like

A(x, t) =

∫d3p

(2π)3[ap(t)e

i~p·~x + a∗p(t)e−i~p·~x].

Then the equation becomes [∂2t + ω2

p]ap(t) = 0 and ωp =√~p2 +m2. Then we

quantize and get

H =

∫d3p

(2π)3[ωp(a

†pap +

1

2].

3.1 Operators on the Fock space

The Fock space is then

F =⊕p

Hp =⊕n

Hn

where p is the momentum and n is the number of particles. The creation andannihilation operators then behave as

[ak, a†p] = (2π)3δ3(~p− ~k).

We normalizeap|0〉 = 0, |p〉 =

√2ωpa

†p|0〉.

Then we get

〈p|k〉 =√

2ωp√

2ωk〈0|apa†k|0〉 = 2ωp(2π)3δ3(~p− ~k).

We also have

1 =

∫d3p

(2π)3

1

2ωp|p〉〈p|.

Then you can check |k〉 = 1|k〉.Also, we define

A(x) =

∫d3p

(2π)3

1√2ωp

[apei~p·~x + a†pe

−i~p·~x].

This is like a creation operator in position space. Indeed, we compute

〈p|A(x)|0〉 =

∫d3kδ3(p− k)〈0|0〉e−i~k·~x = e−~p~x.

But A(x)A(y)|0〉 is not just particles at x and y.In quantum field theory, we work with the Heisenberg picture, so we define

a†p(t) = eiωpta†p(0). Then

φ(x, t) =

∫d3p

(2π)3

1√2ωp

[ap(0)ei~p·~x−iωpt + a†p(0)eiωpt−i~p·~x].

Here, you can interpret the exponent as pµxµ, because pµ = (ωp, ~p).


3.2 Classical field theory

The main object is the Hamiltonian

H(p, x) = energy = K + V.

This is not Lorentz invariant, and generates time translation. On the otherhand, the Lagrantian

L[x, x] = K − V

is not a conserved quantity, but it is Lorentz-invariant and the dynamics isdetermined by minimizing the action S =

∫dεL.

For fields, we are going to have

L[φ, φ, ~∇φ] = L[φ, ∂µφ], H[φ, π, ~∇φ].

We still talk about kinetic terms

K = things like1

2φφ,

1

4F 2µν ,

1

2m2φ2, φ∂µA

µ,

and interactions

V = things like Aφ3, eψAψ, e(∂µφ)φ∗Aµ.

Example 3.1. Consider

L =1

2(∂µφ)2 − V (φ) =

1

2φ2 − 1

2(~∇φ)2 − V (φ).

To minimize the action, we perturb the field a little bit and look at thedifference. Then

δS =

∫d4x[∂L∂φ

δφ+∂L

∂(∂µφ)δ(∂µφ)

]=

∫d4x[∂L

∂φ− ∂µ

∂L∂(∂µφ)

]+ ∂µ

[ ∂L∂(∂µφ)

δφ].

Here, we assume φ(∞) = 0, so we get

∂L∂φ

= ∂µ∂L

∂(∂µφ).

This is called the Euler–Lagrangian equations.

Example 3.2. In the above example, we get

−V ′(φ) = ∂µ[∂µφ] = φ.


3.3 Noether’s theorem

Suppose L is invariant under some specific continuous variation. For instance,take

L = ∂µφ∗∂µφ−m2φφ∗

which is invariant under φ→ eiαφ. Then

0 =δLδα

=∑n

[ ∂L∂φn

− ∂µ∂L

∂(∂µφn)+ ∂µ

[ ∂L∂(∂µφn)

δφn∂α

.

So if the Euler–Lagrange equations are satisfied, the first term is zero so

∂µJµ = 0, Jµ =

∑n

∂L∂(∂µφn)

δφnδα

.

Then if we define Q =∫d3xJ0, we have ∂tQ = 0. This is the statement and

proof of Noether’s theorem.Let’s think about what we get for φ 7→ eiαφ. We have

Jµ =∂L

∂(∂µφ)iφ+

∂L∂(∂µφ∗)

(−iφ∗) = iφ∂µφ∗ − iφ∗∂µφ.

We can check

∂µJµ = i∂µφ∂µφ

∗ + iφφ∗ − i∂µφ∗∂µφ− iφ∗φ = iφφ∗ − iφ∗φ.

This is zero because at the equations of motion, we have φ = m2φ.



Noether’s theorem says that if and action has a continuous symmetry, then thereexists a current Jµ with ∂νJ

µ = 0 when the equations of motion are satisfied.In this case,

Q =

∫d3xJ0

satisfies ∂tQ = 0.Consider translation invariance. When we look at a translate of L, we get

∂µ(gµνL) = ∂νL =[ ∂L∂φn

− ∂µ∂L

∂(∂µφn)

]δφn∂ξν

+ ∂µ

[ ∂L∂(∂µφ)

δφ

∂ξν

].

Because the first term vanishes at equation on motion. So we have

∂µTµν

where

Tµν =∂L

∂(∂µφn)∂νφn − gµνL.

This is called the energy-momentum tensor. Here, we note that

E = T00 =∑ ∂L

∂φnφn − L = πφ− L = H

is just the energy. So energy E =∫d3xT 00 is conserved over time.

4.1 Coulomb’s law

We are going to introduce an external current

L = −1

4F 2µν − JµAµ.

(When I say current, I don’t mean Noether current here.) Because Fµν =∂µAν − ∂νAµ, we have

L = −1

4(∂µAν − ∂νAµ)2 − JµAµ

= −1

2∂µAν∂µAν +

1

2∂µAν∂ν ‘Aµ − JµAµ.

Then ∂L/∂Aν = −Jν and ∂L/∂∂µAν = −∂µAν + ∂νAµ = −Fµν . Then theEuler–Lagrange equation is

∂µFµν = Jν ,

which is Maxwell’s equations. If we go to Lorentz gauge, we get

Aν = Jν .


We are going solve this by inverting the d’Alembertian . Here, note thatwe have Fourier transform

f(x) =

∫dk

2πf(k)eikx, δ(x) =

∫ ∞−∞

dk

2πeikx.

Then the inverse is

f(k) =

∫dxf(x)e−ikx.

We can compute

f(x) =

∫dkkf(x)eikx =

∫d4k(−k2)f(k)eikx.

So corresponds to −k2 in Fourier space.We want to solve the equation when there is a point charge, when J0 = δ3(x)

and ~J = 0. Then

A0(x) =e

δ3(x) = − e

∆δ3(x) =

∫d3k

(2π)3

e

~k2ei~k~x

=e

i4π2

∫ ∞0

dkeikr − e−ikr

ikr=

e

4πr.

This is the Coulomb potential.

4.2 Green’s functions

Let’s look at a complicated example,

L = −1

2hh+

1

3λh3 + hJ.

This is a toy example for gravity, because gravitons interact with each other.Then the Euler–Lagrange equation is

h− λh2 − J = 0.

We now work perturbatively in λ. For λ = 0, we know

h0 =1

J.

If λ 6= 0, we can write h = h0 + h1, where h1 = O(λ). If we plug in into theoriginal equation, we get h1 = λh2

0. So we can write

h1 =λ

( 1

J)2

.

So we get

h =1

J + λ

( 1

)( 1

J)( 1

J)

+ · · · .


We can interpret each of these in terms of Feynman diagrams. Think of each Jas a source, 1

as a propagation or a branch coming our from a source, and λas an interaction between these branches. Then this is something like the Sunemitting a graviton, emitting another graviton, and they interact and becomeone. There are other diagrams we can draw but are not represented in thesolution, and these are purely quantum mechanical effects that we will discuss.What we are doing now is classical.

If we look at the solution for xA(x) = J(x) again, we have

A(x) = −∫d4yΠ(x, y)J(y), Π(x, y) =

∫d4k

(2π)4eik(x−y) 1

k2.

Then you can check that xΠ(x, y) = −δ(x − y). We call this a propagatoror the Green’s function. (We have 1

= −Π.)Let us do what we did this above in this context. Then

h(x) =

∫d4yδ4(x− y)h(y) = −

∫d4π(x, y)yh(y) = −

∫d4yΠ(x, y)J(y).

So this is the propagator of the potential from the source. We can do the samething on the next order. We have

h(x) = −∫d4yΠ(x, y)J(y)+λ

∫d4w

∫d4y

∫d4zΠ(x,w)Π(w, y)Π(w, z)J(y)J(z).

Then these have good physical interpretation. In quantum field theory, therewill also be interactions in loops and so on.



This week and next week will be a bit dry. Why do we talk about cross sectionsin scattering? Scattering is a universal way of probing something that we can’tsee. We are skipping Chapter 4, which is old-fashioned perturbation theory.

5.1 Scattering

In quantum mechanics, we calculate amplitudes, 〈f |i〉, and probabilities, |〈f |i〉|2.In field theory, we calculate the same objects.

Let us consider the situation where two particles collide, and two or moreparticles come out. In the Schrodinger picture, we want to calculate

〈f ; t =∞|i; t = −∞〉.

In this Heisenberg picture, we are trying to measure 〈f |S|i〉. We are interestedin this matrix S.

Classically, if we throw a beam of particles on a large particle,s we canconsider the cross-section area as

σ =#particles scattered

time× number density of beam× velocity of beam.

We may think this as N = Lσ, where L is the luminosity.What we want to do now is to talk about quantum mechanics. Here, σ is just

a cross-section. Particles have a probability of scattering: P = Nscatter/Nincident.We are going to let Nincident = 1, so we are throwing one particles at a time.Then the flux is

Flux =|~v|V

=|~v1 − ~v2|

V.

Now our formula for σ is

dσ =V

T

1

|~v1 − ~v2|dP, dP =

|〈f |S|i〉|2

〈f |f〉〈i|i〉dΠ.

The last factor dΠ is the density of states. On a line of size L, momenta arepn = 2π

L n and so dp = 2πL dn. So we have

dΠ =∏j

V

(2π)3d3pj .

The initial and final states are given by

|i〉 = |p1〉|p2〉, |f〉 = |p3〉 · · · |pn〉.

Because we are working in a box, we consider |p|p〉 = 2Epδ3(0) = 2EpV . Then

|i|i〉 = 2E12E2V2, |f |f〉 =

n∏j=3

(2Ej)V.


Then we have

dσ =V

T

|〈f |S|i〉|2

|~v1 − ~v2|∏j(2Ei)2E12E2V n

∏n

V

(2π)3d3pi.

We writeS = 1 + iT,

where T = (2π)4δ4(p1 + p2− p3− · · ·− pn)M , because momentum is conserved.Then

|〈f |S|i〉|2f 6=i = (2π)8δ4(∑p)δ4(0)|M |2.

If we plug this in, we get

dσ =1

(2E1)(2E2)|~v1 − ~v2|× |M2|

∏j

d3pj(2π)3

1

2Ej(2π)4δ4(

∑p).

This second term is also called the Lorentz-invariant phase space, dΠLIPS. Sowe can write the decay wrate as

dΓ =1

2E1|M |2dΠLIPS.

There is no flux factor, and no 1/2E2.

5.2 Two-to-two scattering

Let us look at the example of a 2→ 2 scattering. Let us call the four particlesp1, p2, p3, p4. In the center of mass frame, we have

|~p1| = |~p2| = pi, |~p3| = |~p4| = pf .

Energy conservation is E1 + E2 = E3 + E4 = ECM. Now we look at

dΠLIPS = (2π)4δ4(p1 + p2 − p3 − p4)d3p3

(2π)3

1

2E3

d3p4

(2π)3

1

2E4.

But this has a lot of redundancies, so we can express in terms on the direction.If we integrate over ~p4, we get

dΠLIPS =1

4(2π)2

dp3

E3E4δ(E3 + E4 − ECM)

=dΩ

16π2

∫p2fdpf

1

E3E4δ(√m2

3 + p2f +

√m2

4 + p2f − ECM)

=dΩ

16π2

∫ ∞m3+m4−ECM

dxδ(x)pfECM

=dΩ

16π2

pEECM

θ(ECM −m3 −m4).

So we get

dσ

dΩ=

1

2E12E2|~v1 − ~v2|1

16π2

pfECM

|M |2 =1

64π2E2CM

pfpi|M |2.


in the center of mass frame. (This dΩ is the spherical angle dφd cos θ, so thatd3p3 = p2

3dp3dΩ.)Let us look at the non-relativistic limit. Consider the Born approximation

dσ

dΩ=m2e

4π2|V (k)|2.

Here, V (k) is the Fourier transformation

V (k) =

∫d3xe−i

~k~xV (x) =e2

~k2

in the Coulomb potential. So we have

dσ

dΩ=m2e

4π2

( e2

~k2

)2

.

Let us how see this agrees with what we have done so far.The free theory for the proton and the electron is

L = −1

4F 2µν − φ∗e( +m2

e)φe − φ∗p( +m2p)φp

− ieAµ(φ∗e∂µφe − φe∂µφ∗e) + ieAµ(φ∗p∂µφp − φp∂µφ∗p).

If we take the non-relativistic limit, we have pµ = (E, ~p) = (√m2 + ~p2, ~p) ≈

(m, 0), and so ∂tφ ≈ imφ. So we redefine φ → eimetφ so that the phases don’trotate. If we do that, the Lagrangian becomes

L = −1

4F 2µν + φe~∇2φe + 2emeφ

∗eφeA0 + φp~∇2φp − 2empφ

∗pφpA0.

The matrix M is going to be

M =(2eme)(−2emp)

~k2

because the coefficient of φ∗eφeA0 is 2eme and this is the interaction between the

electron, electron, photon, and likewise for (−2emp), and 1/|~k2| is the Green’sfunction. Then

dσ

dΩ=

1

64π2m2p

|M |2 =1

64π2m2p

16e4m2em

2p

k4=

1

4π2

m2ee

4

|~k2|2.

This is the same formula we had for the Born approximation.



We also have to need this other technical theorem. We recall that light satisfiesAµ = 0, and so (∂2

t + |~k|2)Aµ = 0. We had our operator

φ(x) =

∫d3p

(2π)3

1√2ωp

(a†pei~p·~x + ape

−i~p·~x).

Then i∂tap = −[H,Ap] = ωpap, so we can define even for difference time

φ(x, t) =

∫d3p

(2π)3

1√2ωp

(a†~peipx + a~pe

−ipx).

6.1 LSZ reduction

We also talked about cross sections,

dσ =1

(2E1)(2E2)|~v1 − ~v2||M |2dΠLIPS,

where S = 1 + iT and T = (2π)4δ4(∑p)M . So again, our initial state is

|i〉 = |p1p2〉 =√

2ω1

√2ω2a

†p1

(−∞)a†p2(−∞)|Ω−∞〉

We are using |Ω〉 because the vacuum is time-dependent. Similarly, in the finalstate, we can write

|f〉 = |p3 · · · pn〉 =√

2ω3 · · ·√

2ωna†p3

(+∞) · · · a†pn(+∞)|Ω+∞〉.

Now the matrix element between the two things is

〈f |S|i〉 =√

2ω1 · · ·√

2ωn〈Ω∞|ap3(∞) · · · apn(∞)a†p1

(−∞)a†p2(−∞)|Ω−∞〉.

So how do we create |p〉 from φ(x)? WE have

〈p|φ(x)|0〉 = eipx, φ(x)|0〉 =

∫d3p

1

2ωpei~p·~x|p〉 =

∫d4pδ(p2−m2)θ(p0)ei~p~x|p〉.

But we have

−2πiδ(p2 −m2) = limε→0

[ 1

p2 +m2 + iε− 1

p2 −m2 − iε

].

So we roughly have ∫e−ipx( +m2)φ(x)|0〉 = |p〉.

The precise expression is

i

∫d4xeipx( +m2)φ(x, t) =

√2ωp(ap(∞)− ap(−∞)).


Let me try to derive this. If we do spatial integration by parts, we get

i

∫d4xeipx( +m2)φ(x, t) = i

∫d4xeipx(∂2

t + ω2p)φ(x, t)

=

∫dt∂t

[eiωpt

∫d3xe−i~p~x(i∂t + ωp)φ(x, t)

]=

∫dt∂t

[eiωpt

√2ωpape

−iωpt]

=√

2ωp[ap(∞)− ap(−∞)].

Here, we are assuming that at t = ±∞, the field behaves like a free field, so wecan compute the spatial integral simply. Now if we take the complex conjugateof both sides, we get

−i∫d4xe−ipx( +m2)φ(x, t) =

√2ωp(a

†p(∞)− a†p(−∞)).

Now can compute 〈f |i〉. Here, we introduce a time-ordering operation Twhich just puts things in the correct time order. Then we have

〈Ω|ap3(∞) · · · apn(∞)a†p1

(−∞)a†p2(−∞)|Ω〉

= 〈Ω|T[ap3(∞)− ap3

(−∞)] · · · [apn(∞)− apn(−∞)]

[a†p1(∞)− a†p1

(−∞)][a†p2(∞)− a†p2

(−∞)]|Ω〉,

because all other terms become 0. So we get

〈f |i〉 =

[i

∫d4x1e

ip1x1( +m21)

]· · ·[−i∫d4xne

−ipnxn]

× 〈Ω|Tφ(x1) · · ·φ(xn)|Ω〉.

This is called the LSZ reduction formula.

6.2 Feynman propagators

The simplest example is

DF (x, y) = 〈0|Tφ0(x)φ0(y)|0〉 = limε→0

∫d4k

(2π)4

i

k2 −m2 + iεeik(x−y).

This is called the Feynman propagator, and satisfies (x + m2)DF (x, y) =∫d4keik(x−y) = δ4(x − y). So this is the factor you put in when you want to

talk about propagation between to interactions.The first thing we do is to calculate without time ordering. We have

〈0|φ(x1)φ(x2)|0〉 =

∫d3k1

(2π)3

∫d3k2

(2π)3

1√2ω1

√2ω2〈0|ak1a

†k2|0〉ei(k2x2−k1x1)

=

∫d3k

(2π)3

1

2ωkeik(x1−x2).


If we do have time-ordering, we get

〈0|Tφ(x1)φ(x2)|0〉 = 〈0|φ(x1)φ(x2)|0〉θ(t1 − t2) + 〈0|φ(x2)φ(x1)|0〉θ(t2 − t1)

=

∫d3k

(2π)3

1

2ωk[eik(x2−x1)θ(τ) + eik(x1−x2)θ(−τ)]

=

∫d3k

(2π)3

1

2ωk[ei~k(~x1−~x2)e−iωkτθ(τ) + e−i

~k(~x1−~x2)eiωkτθ(−τ)]

=

∫d3k

(2π)3

1

2ωke−i

~k(~x1−~x2)[e−iωkτθ(τ) + eiωkτθ(−τ)].

Here, we have

θ(τ) = limε→0

∫ ∞−∞

dωeiωτ

ω + iε

1

2πi.

So we get

〈0|Tφ0(x)φ0(y)|0〉 =

∫d3k

(2π)3

dω

2π

i

ω2 − ~p2 −m2 + iεeik(x−y)

=

∫d4k

i

k2 −m2 + iεeik(x−y).

This ε we are going to think of as uncertainty of energy. Feynman’s ingeniousidea is that you can add the two possible time orderings in one propagator, sothat we don’t have to think of the two cases separately.



Last time we showed that

〈P3 · · ·Pn|P1P2〉 =

[i

∫d4xe−ip1x1(1 +m2

1)

]· · ·[

i

∫d4xne

+ipnxn(n +m2n)

]〈Ω|Tφ(x1) · · ·φ(xn)|Ω〉.

Then for free field, we got the Feynman propagator as

DF (x, y) = 〈0|Tφ0(x)φ0(y)|0〉 = limε→0

∫d4k

(2π)4

i

k2 −m2eik(x−y).

Then(x +m2)DF (x, y) = −iδ4(x− y).

We will check this later.

7.1 Schwinger–Dyson equations

We want to specify the dynamics and the commutations relations. We can usethings like [x, p] = i~ and [θ,H] = i∂tθ, but then we need to worry about theseparation of space and time.

Assume that φ satisfies the equations of motion as an operator,

L = −1

2φ( +m2)φ+ Lint[φ].

Then the Euler–Lagrange equations become

( +m2)φ = L′int[φ].

The commutation relations are

[φ(~x, t), ∂tφ(~y, t)] = i~δ(~x− ~y), [φ(~x, t), φ(~y, t)] = 0.

These are equal-time commutations relations. The second relation should bethought of as, the two points (~x, t) and (~y, t) are causally unrelated.

We can check in the free theory that these hold. Here, we have

φ0(x, t) =

∫d3p

(2π)3

1√2ωp

(ape−ipx + a†pe

ipx), [ap, a†q] = (2π)3δ3(~p− ~q).

Our equations of motion are

( +m2)φ0(x, t) =

∫d3p

(2π)3(−p2 +m2)

1√2ωp

(· · · ) = 0


because p2 = m2. For the commutation relation, we first compute

∂tφ(~y, t) = −i∫

d3q

(2π)3

√ωq2

(aqe−iqy − a†qeiqy).

Then

[φ(~x, t), ∂tφ(~y, t)] = −i∫

d3p

(2π)3

d3q

(2π)3

√ωq4ωp

([a†p, aq]eipx−iqy − [ap, a

†q]e−ipx+iqy)

= − i2

∫d3p

(2π)3(−eip(x−y) − e−ip(x−y)) = iδ3(x− y).

So this is satisfied in the free theory, and we’re assuming this holds also in theinteracting theory. This is not so surprising, because we are looking at one timeslice.

What we are going to do now, is to look at the time ordering operator usingthese relations. We can compute

∂t〈Ω|Tφ(x), φ(x′)|Ω〉 = ∂t[〈φ(x)φ(x′)〉θ(t− t′) + 〈φ(x′)φ(x)〉θ(t′ − t)]= 〈T∂tφ(x)φ(x′)〉+ 〈φ(x)φ(x′)〉δ(t− t′)− 〈φ(x′)φ(x)〉δ(t− t′)= 〈T∂)tφ(x)φ(x′)〉

because φ(x) and φ(x′) commute at t = t′. Then the second derivative is

∂2t 〈Tφ(x)φ(x′)〉 = ∂t〈T∂tφ(x)φ(x′)〉

= 〈T∂2t φ(x)φ(x′)〉+ 〈[∂tφ(x), φ(x′)]〉δ(t− t′)

= 〈T∂2t φ(x)φ(x′)〉 − i~δ4(x− x′).

So we finally get

(x +m2)〈Tφ(x)φ(x′)〉 = 〈T( +m2)φ(x)φ(x′)〉 − i~δ4(x− x′).

This is(x +m2)DF (x, y) = −i~δ4(x− y).

Now we can assume that we have more than terms. Then we can show that

x〈Tφ(x)φ(y1) · · ·φ(yn)〉 = 〈Txφ(x)φ(y1) · · ·φ(yn)〉

− i~∑j

δ4(x− yj)〈Tφ(y1) · · ·φ(yj−1)φ(yj+1) · · ·φ(yn)〉.

This is called the Schwinger–Dyson equations. Historically, this was howSchwinger showed that the perturbative way and Feynman diagram way of doingquantum field theory are equivalent.


7.2 Feynman diagrams

Write δxi = δ4(x−xi) and Dij = Dji = DF (xi, xj) and Dxi = DF (x, xi). ThenxDx1 = −iδx1 and we have

〈φ1φ2〉 =

∫d4xδx1〈φxφ2〉 = i

∫d4x(xDx1)〈φxφ2〉 = i

∫d4xDx1〈φxφ2〉.

For instance in the free theory, x〈φxφy〉 = −iδxy so

〈φ1φ2〉 = i

∫d4xDx1(−iδx2) = D12.

If we have four terms,

〈φ1φ2φ3φ4〉 = i

∫d4xDx1x〈φxφ2φ3φ4〉

=

∫d4xDx1[δx2〈φ3φ4〉+ δx3〈φ2φ4〉+ δx4〈φ2φ3〉]

= D12D34 +D13D24 +D14D23.

We can represent these terms as diagrams that connect dots between 1, 2, 3, 4.

Let us now look at a theory with interactions, with the simplest possibleinteraction

L = −1

2φφ+

g

3!φ3, φ =

g

2φ2.

In this case,

〈φ1φ2〉 = i

∫d4D1xx〈φxφ2〉

= i

∫d4xD1x[〈 g2φ

2xφ2〉 − eδx2]

= D12 −g

2

∫d4xd4yD1xDy2y〈φ2

xφy〉

= D12 −g2

4

∫d4xd4yD1xD2y〈φ2

xφ2y〉+ ig

∫d4xD1xD2y〈φx〉.

And we can go on, by removing one x and putting in two x, until we get theorder of g we want. For 〈φ2

xφ2y〉, we can just assume the free field, because we

already have g, and so

〈φ2xφ

2y〉 = 〈φxφxφyφy〉 = 2D2

xy +DxxDyy +O(g),

〈φx〉 = i

∫d4yDyxy〈φy〉 =

ig

2

∫d4yDxyDyy.


So we finally get

〈φ1φ2〉 = D12−g2

∫d4xd4y

(1

2D1xD

2xyDy2+

1

4D1xDxxDyyDy2+

1

2D1xD2xDxyDyy

).

You can draw the diagrams for each term as well.We can generalize this process to position space Feynman rules.

(1) Points are x for each external position.

(2) Draw a line from each point.

(3) A line can either connect to another line or split due to an interaction.

(4) A vertex is proportional to the coefficient of L′int[φ]× i.(5) At a given order of perturbation theory, sum all possible diagrams and

integrate over internal positions.

But what are the numerical factors like 1/2 or so on? We conventionally nor-malize n! for each φn, like

L =g

2!6!φ2

1φ62.

If two line connect to each other, we get a symmetry factor. If the diagramhas a symmetry, then we need to divide by the number of symmetries. (Youactually rarely need symmetry factors.)



We had the formula for the cross section

dσ

dΩ=

1

64π2E2CM

|M|2, S = 1 + iδ4(p)M,

and then we looked at

〈f |S|i〉 =

∫d4x1(1 +m2)eipx1 · · · 〈Ω|Tφ(x1) · · ·φ(xn)|Ω〉.

Then we were able to interpret

〈Ω|Tφ(x1) · · ·φ(xn)|Ω〉

as a sum of Feynman diagrams. We are now going to try and do this in momen-tum space. This way, we don’t have to take the Fourier transform and thingsbecome more simpler.

8.1 Feynman diagrams in momentum space

Let’s take a example of a g6φ

3 interaction and the following diagram.

Then the corresponding term is

T =(ig)2

2

∫d4x

∫d4yDF (x, x)DF (x, y)2DF (y, x2).

By Fourier transforming, we get

T = −g2

2

∫d4xd4yd4p1 · · · d4p4e

ip1(x1−x)eip2(y−x2)eip3(x−y)eip4(x−y)

· i

p21 + iε

· · · i

p24 + iε

.

If we do the x-integral, we get momentum conservation δ4(p3 +p4−p1) and if wedo the y-integral, we get δ4(p2−p3−p4). Then the p3-integral sets p3 = p1−p4

and so δ4(p2−p3−p4) becomes δ4(p2−p1). So if we set k = p4 and p3 = p1−k,we get

T = −g2

2

∫d4k

(2π)4

d4p1

(2π)4

d4p2

(2π)4δ4(p1 − p2)(2π)4eip1x1e−ip2x2

i

p21 + iε

i

p22 + iε

i

(p1 − k)2 + iε

i

k2 + iε.


LSZ says that

〈pf |S|pi〉 =

[−i∫d4x1e

−ipix1p2i

][−i∫d4x2e

+ipfx2p2f

]T.

The integrals over x1 and x2 give δ4(p1− pi) and δ4(p2− pf ), and this becomes

〈pf |S|pi〉 = (2π)4δ4(pi − pf )

[−g

2

2

∫d2k

(2π)4

i

(pi − k)2 + iε

i

k2 + iε

].

So here are the momentum space Feynman rules:

(1) Internal lines get ip2−m2+iε .

(2) Vertices come from Lint.

(3) External lines do not get propagators.

(4) 4-momentum is conserved at each vertex.

(5) Integrate over undetermined momentum.

8.2 Hamiltonian derivation

Recall that in the Lagrangian formalism, we have

[φ(x, t), φ(x, t)] = i~δ3(~x− ~y), ( +m2)φ = L′int[φ].

In the Hamiltonian picture, we have

[φ(x, t), π(x, t)] = i~δ3(~x− ~y), i∂tφ = [H,φ].

Our fields are

φ(~x) =

∫d3p

1√2ωp

(apei~p~x + a†pe

−i~p~x).

In the interaction picture, fields have interaction,

H = H0 +Hint, φ0(x, t) = eiH0tφ(~x)e−iH0t =

∫d3p(ape

ipx + a†pe−ipx).

We now write

φ(x, t) = eiHtφ(x, 0)e−Ht = U†(t, 0)φ0(x, t)U(x, 0).

Here, U can be formally written as

U(t, 0) = eiH0te−Ht ≈ e−iV .

To be precise, we have

U21 = U(t2, T1) = eiH0(t2−t1)e−iH(t2−t1) = Texp(−i∫ t2t1dtVI(t)),


where VI is the potential for the interacting theory. For instance, φ(x, t2) =U21φ(x, t1)U12. So we get

〈Ω|Tφ(x1) · · ·φ(xn)|Ω〉 =〈0|TU∞0U01φ0(x1)U01U02φ(x2)U20 · · ·U0−∞|0〉

〈0|U∞−∞|0〉

=〈0|Tφ0(x1) · · ·φ0(xn)U−∞∞|0〉

〈0|U−∞∞|0〉,

because U commute with each other. So we have

〈Ω|Tφ1φ2|Ω〉 =

⟨0

∣∣∣∣Tφ0(x1)φ0(x2)− ig∫d4yφ0(y)3φ0(x1)φ0(x2)

+(−ig)2

2

∫d4x

∫d4yφ0(x)3φ0(y)3φ0(x1)φ0(x2)

∣∣∣∣0⟩.You can write out this and see that this is really the same thing as the Feynmandiagrams.

8.3 Matrix element for the two-to-two scattering

Let us take our favorite example

L = −1

2φ( +m2)φ+

g

3!φ3.

Thendσ

dΩ=

1

64π2E2|M|2.

If we look at

we get

iM = igi

k2 −m2 + iε=

−ig2

(p1 + p2)2 −m2 + iε, Mt =

g2

s−m2 + iε, s = (p1+p2)2.

But there are other diagrams. We have


with

Mt =−g2

t−m2 + iε, t = (p1 − p3)2,

and there is

which is

Mu =−g2

t−m2 + iε, u = (p1 − p4)2,

Then we can add them up and get

Mtot = −g2( 1

s−m2+

1

t−m2+

1

u−m2

), s+ t+ u =

4∑i=1

m2i .

Electron positron e−e+ → µ−µ+ only has an s-channel, Rutherford scatteringonly has a t channel, electron scattering has t and u channels, etc.

The Lagrangian sometimes have derivative couplings. For instance,

L = λφ1(∂µφ2)(∂µφ3).

Our field are

φ =

∫d3p

1√2ωp

(a†peipx + ape

−ipx),

so we get out a factor of ipµ if a particle is created, and −ipµ if a particle isannihilated. This means that the momentum leaves or enters the vertex. If welook at φ1φ2 → φ1φ2, and the diagram

φ3

φ1

φ2

φ1

φ2

corresponds to

iM = (iλ)2 (−ipµ2 )(ikµ)(−ikµ)(−ipµ4 )

k2 −m2 + iε.


9 October 2, 2018

The course up to this point can be summarized as the momentum space Feyn-man rules. The example we’ve been studying is scalar field theory

L = −1

2φ( +m2)φ+

g

3!φ3.

Then you draw all the Feynman diagrams and internal lines get factors ofi

p2−m2+iε . Then we get factors ofig

from the tree level, and then

(ig)

∫d4k

(2π)4

i

k2 −m2 + iε

i

(k − p1)2 −m2 + iε

i

(k − p2)2 −m2 + iε

from the next level with one loop. If you have a derivative term, you put in ipfor annihilation and −ip for creation.

9.1 Writing down the Lagrangian

But how do we write down the Lagrangian for the theory? We start withsymmetries of the theory. There are

• translation invariance φ(x)→ φ(x+ a),

• Lorentz invariance xµ → Λµνxν , including rotations and boosts,

• unitarity, that is, conservation of probabilities, 〈φ|φ′〉 = 〈ψ|e−iHteiHt|φ′〉,preserved by time-translation and other symmetries,

• internal symmetries, e.g., phase rotation φ→ eiαφ.

Lorentz invariance and translation invariance are together called Poincare in-variance.

Symmetries mix states in the Hilbert space. For instance, e− has two states

|↑〉, Jz =1

2, |↓〉, Jz = −1

2.

If we rotate our apparatus, we should also be rotating the spins. Another exam-ple is polarization of light. We go as far as defining a particle as a (minimal) setof states that mix under Poincare transformations. More mathematically, par-ticles transform as an irreducible unitary representations of the Poincare group.This is good because we have reduced the physics to a mathematical problem.

Definition 9.1. A group G = gi with a rule gigj = gk. The conditions are

1. there exists a 1 with 1g = g1 = g for all g,

2. there is an inverse g−1i with g−1 · g = 1,

3. it is associative, g1(g2g3) = (g1g2)g3.


Examples are rotations which are 3× 3 matrices with RTij = R−1ij .

Definition 9.2. A representation is an embedding of G into a set of linearoperators acting on a vector space.

An example is the trivial representation gi 7→ 1. Another representation ofthe 3× 3 rotations is the 4-dimensional representation given by just

A 7→

0

A 00

0 0 0 1

.

Definition 9.3. An irreducible representation is a representation where nosubspace of vector space is preserved under G.

But there is a problem between Poincare invariance and unitarity. If Pij isa transformation, then I should have

〈ψ|ψ′〉 = 〈ψ|P †P |ψ〉, P †P = 1.

In general, this is easy to do, but if we have Lorentz invariance, this is not easy.Let’s take a 4-dimensional space for instance, and write

|ψ〉 = c0|V0〉+ c1|V1〉+ c2|V2〉+ c3|V3〉.

Then〈ψ|ψ〉 = |c0|2 + |c1|2 + |c2|2 + |c3|2 ≥ 0.

But this is not Lorentz invariant. So we really want 〈Vµ|Vν〉 = gµν . But thenthis can’t be interpreted as probability. If you study representations of thePoincare group, you find the following.

Proposition 9.4. There is no finite-dimensional unitary representations of theLorentz group of the Poincare group.

However, there exist infinite-dimensional unitary representations of the Poincaregroup. Wigner studied this, and it turns out that there are two classes of rep-resentations, classified by mass m and spin J where

• m > 0 and there are 2J + 1 states,

• m = 0 and there are 2 states.

So how do we interpret this physically? We want E > 0 in a classical theory.This can be computed by the energy-momentum tensor

E = T00 =∑n

∂L∂φn

φn − L.


1. In the spin 0 case, there is one degree of freedom, with J = 0 and m ≥ 0one state. Then

L =1

2(∂µφ)(∂µφ)− 1

2m2φ2, ( +m2)φ = 0.

Then

E =∂L∂φ− L =

1

2[φ2 + (~∇φ)2 +m2φ2] ≥ 0.

This is satisfies as long as m2 ≥ 0.

2. In massive spin 1, we expect 2J + 1 = 3 degrees of freedom. Minimally,we can embed this in Aµ(x), which is 4 degrees of freedom, but this splitsinto 4 = 3 + 1. Then we can write down

L = −1

2(∂µAν)(∂νAµ) +

1

2m2A2

µ, ( +m2)Aµ = 0.

Then the energy density is

E =1

2[(∂t ~A)2 + (∇iAj)2 +m2 ~A]2 − 1

2[(∂tA0)2 + (~∇A0)2 +m2A2

0].

This holds generally for any four scalar fields, but if we further imposethe condition that the Lorentz group representation corresponding to Aµis the standard representation, we can write more things like

L =a

2AµAµ +

b

2Aµ∂µ∂νAν +

1

2m2A2

µ.

Then the equations of motion becomes

aAµ + b∂µ∂νAν +m2Aµ = 0, ((a+ b) +m2)(∂µAµ) = 0.

If a = −b, then we will get ∂µAµ = 0. What this is doing is projecting outthe 1-dimensional trivial representation. If we choose a = 1 and b = −1,then we get

L = −1

4F 2µν +

1

2m2A2

µ, Fµν = ∂µAν − ∂νAµ

with energy density

E =1

2( ~E2 + ~B2) +

1

2m2(A2

0 + ~A2)−A0∂t(∂µAµ)−A0( +m2)A0 + ∂i[A0F0i],

=1

2( ~E2 + ~B2) +

1

2m2(A2

0 + ~A2) + ∂i[A0F0i] ≥ 0

with the equations of motion ( + m2)Aµ = 0 and m2∂νAµ = 0. Thisis called the Proca Lagrangian. Note that the positive energy densitycondition forced our Lagrangian to be this.


10 October 4, 2018

There was this theorem due do Wigner.

Theorem 10.1 (Wigner). A particle is associated to an irreducible unitaryrepresentation of the Poincare group.

There are these things Aµ, Tµν that are finite-dimensional, irreducible, andnon-unitary. Then there are fields

Aµ(x), Tµν(x)

that are infinite-dimensional, reducible, non-unitary representations. Mathe-matically, unitary irreducible representations correspond to two invariants mand J . These are

• J > 0 and m > 0: 2J + 1 polarizations,

• J > 0 and m = 0: 2 polarizations,

• J = 0 : 1 polarization,

where J = 0, 12 , 1,

32 , . . .. We want to describe the physics. What Wigner did

was describe the Hilbert space, with vectors εi(p) = |εi; p〉 for i = 1, . . . , n.

10.1 Representations of the Poincare group

So we want to first construct the representations εiµ(p), and then construct aLagrangian so that the extra degree of freedom Aµ does not get produced. Let’sreview what we did last time. We connected unitarity with E > 0. For a scalarfield, we can write

L =1

2(∂µφ)2 − 1

2m2φ2,

and then we got

E =1

2[φ2 + (~∇φ)2 +m2φ2] ≥ 0

if m2 ≥ 0. For a massive spin 1 particle, we wrote

L = a(∂µAµ)2 + b(∂µAν)2 +m2A2µ,

and this constraint E ≥ 0 was satisfied for a = −b and m2/a > 0. Then theconstraint was ∂µAµ = 0 and ( + m2)Aµ = 0. So the representation Aµ(x)splits into a spin 1 and a spin 0 representation.

So how do we quantize this theory? We can write

Aµ(x) =

3∑j=1

∫d3p

(2π)3

1√2ωp

(εjµ(p)eipxaj†p +εjµ(p)e−ipxajp), [ajp, ak†q ] = (2π)3δ3(p−k)δjk.

Then we can write

|εj ; p〉 = aj†p |0〉, 〈0|Aµ(x)|εi; p〉εiµ(p)eipx.


But we haven’t talked about what the representation looks like. Suppose wehave pµ = (m, 0, 0, 0), where ∂µAµ = 0 looks like pµε

jµ = 0. Then if we have

εµ1 = (0, 1, 0, 0), εµ2 = (0, 0, 1, 0), εµ3 = (0, 0, 0, 1),

how do we boost it to pµ = (E, 0, 0, pz) with p2z + m2 = E2? This uses the

method of induced representations.

10.2 Induced representations

How do we construct unitary representations of the Poincare group?

1. Find a subgroup that stabilizes pµ. Here, we can take pµ = (m, 0, 0, 0)and the little group is SO(3).

2. Then construct finite-dimensional irreducible representations of the littlegroup. For SO(3), we have 0, 1/2, 1, . . .. In our case, we have J = 1/2 andε1µ = (0, 1, 0, 0 and so on.

3. Any g in the Lorentz group can be written as g = b · r where b is theboosts and r is the rotation, where b is in the coset SO(1, 3)/ SO(3).

4. Once we can write this, we find a basis εiµ(b · pµ).

5. Now we define the representation by

g · ε = (r · ε)(b · pµ).

What we really want to get is a photon. But here, we have to deal with themassless case. So how do we want to even construct this theory? Let’s just seewhat happens. Let’s take the massive theory and just take the limit. We expecttwo polarizations. Then natural thing to try is

L = −1

4F 2µν .

This does two things: we lose the ∂µAµ = 0 constraint. If we just look at theequations of motion, what we get is

Aµ + ∂µAνAν = 0, E =1

2( ~E2 + ~B2) ≥ 0.

So somehow, despite these problems, remarkably things come out naturally.What happened is that because

Fµν = ∂µAν − ∂νAµ,

this is invariant under Aµ(x) → Aµ(x) + ∂µα(x) for any α(x). This is calledgauge invariance, and this is some redundancy in the embedding of the physicsof εµ into Aµ(x).


Because of this redundancy, I can impose additional condition and choosewhat A to use. This is called gauge choice. We are going to choose Aµ so that∂iAi = 0. We can do this because the gauge transformation is given by

∂iAi → ∂iAi + ~∇2α; A0 → A0 + ∂t,

and then you can always solve this Laplace equation. Moreover, we can evenset A0 = 0. At the end, we have

A0 = 0, ∂iAi = 0, ∂µAµ = 0.

This is called the Coulomb gauge.So let’s take p = (E, 0, 0, E) some photon. The two constraints on the

polarization vectors are ε0 = 0 and pµεµ = 0. Then the two polarization thatare allowed are

ε1 = (0, 1, 0, 0), ε2 = (0, 0, 1, 0).

In this case, we have this decomposition

Aµ = 1⊕ 0⊕ 0

of dimension 4 = 2 + 1 + 1.

Example 10.2. Let me take pµ = (E, 0, 0, E), so that

εµ1 = (0, 1, 0, 0), εµ2 = (0, 0, 1, 0).

If we take

Λ =

3/2 1 0 −1/21 1 0 10 0 1 0

1/2 1 0 1/2

,

we get that Λ is a Lorentz transformation in the little group. You can compute

Λε1 = (1, 1, 0, 1) = ε1 +1

Epµ.

This means that this acts on ε1 to give ε1.


11 October 9, 2018

In a massless spin 1 theory, we can quantize the theory by looking at

Aµ(x) =

2∑j=1

∫d3p

(2π)3

1√2ωp

(aj†p ε∗µ(p)eipx + ajpεµ(p)e−ipx).

Here, we have|ε1(p)〉 = a1†

p |0〉, |ε2(p)〉 = a2†p |0〉.

Under Lorentz transformations, this transforms as εiµ → εiµ + pµ. So this isLorentz-invariant only if pµM

µ = 0. This is called the Ward identity.

11.1 Scalar quantum electrodynamics

The Lagrangian

L = −1

4F 2µν , Fµν = ∂µAν − ∂νAµ

is invariant under Aµ → Aµ + 1e∂µα, and this is exactly εµ → εµ + pµ under the

Fourier transformation.If we wanted an interaction with a scalar field, we can write

L = −1

4F 2µν +Aµφ∂µφ.

This is not Lorentz-invariant, and you can check this by

Aµφ∂µφ+1

e(∂µα)φ∂µφ.

So what we will do is to remove redundancy using this. This only possibly workswith more than one field. Let us try φ1 and φ2, with

φ = φ1 + iφ2 → eiαφ.

If we define a covariant derivative, it transforms as

Dµφ = [∂µ + ieAµ]φ→ e−iαDµφ.

So for scalar QED, we can write down the Lagrangian

LSQED = −1

4F 2µν + |Dµφ|2 −m2|φ|2

= −1

4F 2µν + ieAµ(φ∂µφ

∗ − φ∗∂µφ) + e2A2µφφ

∗ −m2φφ∗.

If you recognize the term

Jµ = φ∂µφ∗ − φ∗∂µφ,


this is the Noether current associated to phase rotation. So it has ∂µJµ = 0

on the equation of motion. It’s not a coincidence that Aµ couples to a Noethercurrent; the current is something that can be measured.

masslessspin 1

⇒ gauge invariantLagrangian

⇒ globalsymmetry

⇒ conservedcharge

.

The equations of motion is given by

( +m2)φ = i(−eAµ)∂µφ+ i∂µ(−eAµφ) + (−eAµ)2φ,

( +m2)φ∗ = i(eAµ)∂µφ∗ + i∂µ(eAµφ

∗) + (eAµ)2φ∗.

So another consequence of a massless spin 1 particle, is that there is an an-tiparticle associated to a particle. For instance, π− and π+ are spin 0 particlesthat couple to the photon.

11.2 Photon propagator

We want to develop the Feynman rules for scalar QED. So we need the propa-gator for the photon. Recall that for a scalar, we had

DF = 〈0|Tφ(x)φ(y)|0〉 =

∫d4peip(x−y) i

p2 −m2 + iε.

Then we had ( + m2)DF = −iδ4(x − y). Classically, we can think of this asinverting

( +m2)φ = J, φ(x) = Π(x, y)J(y),

which is the equations of motion to the classical Lagrangian.In the photon case, we have

L = −1

4F 2µν +AµJ

µ, ∂µFµν = Jν .

If we try to Fourier transform and invert this, we get

(k2gµν − kµkν)Aµ = Jν .

But then this matrix is not invertible, which we should have expected becausethere is gauge-invariance. One choice is to choose a gauge and then substi-tute into the Lagrangian. But this is annoying, so we are going to deform theLagrangian and write

L = −1

4F 2µν −

1

2ξ(∂µAµ)2 + JA.

(This deformation is very special, and it’s okay for reasons that are not obvious.)Then if we invert this, we get

iΠµν = −igµν − (1− ξ)kµkνk2

k2 + iε.


This you can check this explicitly, that

−[k2gµν(1− 1ξ )kµkν ]Πνα = gµα.

Generically, propagators are like

Π ∼∫

d4k

(2π)4eik(x−y)

∑s|s〉〈s|

k2 −m2 + iε

because we want the propagator to preserve the spin states.Now we are almost done. Let us quantize the theory

φ1 =

∫d3p

(2π)3(ap,1e

−ipx + a†p,1eipx),

φ2 =

∫d3p

(2π)3(ap,2e

−ipx + a†p,2eipx).

Then if we write ap = ap,1 + iap,2 and bp = ap,1 − iap,2, then we get

φ =

∫d3p

(2π)3(ape

−ipx + b†peipx),

φ∗ =

∫d3p

(2π)3(bpe

−ipx + a†peipx).

So φ is creating an antiparticle and annihilating a particle, and φ∗ is creating aparticle and annihilating an antiparticle.

11.3 Feynman rules for scalar QED

Now we compute

〈0|Tφ(x)∗φ(y)|0〉 =

∫d4p

(2π)4

i

p2 −m2 + iεeip(x−y).

There are interaction terms

−ieAµ(φ∗∂µφ− φ∂µφ∗) + e2AµAµφφ∗.

So let’s see. The term φ creates π− with ipµ, and φ∗ creates π+ with iπµ. Or,φ annihilates π+ with −ipµ and φ∗ annihilates π− with −ipµ.

If you analyze all the cases, you are going to get Figure 1. But if you lookat this, you see that you can package all of this compactly. We can invert thearrows for π+, and then only the interactions we are allowed to have are arrowscontinuous thought the vertices. Then all interaction terms are just

−ie(pµ1 + pµ2 ).

This is called the Feynman–Stueckelberg interpretation. You can interpretas a photon decaying to e−e+ as a electron bouncing off a photon.


−, p1

−, p2

= −ie(p1µ + p2

µ) +, p1

+, p2

= ie(p1µ + p2

µ)

−, p1

+, p2

= ie(−p1µ + p2

µ)

−, p1

+, p2

= ie(−p1µ + p2

µ)

Figure 1: Feynman diagrams for scalar QED: time from left to right

Dirac had this interpretation of thinking of antiparticles as holes in thisDirac sea of negative energy. Dirac didn’t like the Klein–Gordan equation

( +m2)φ = 0

having negative energy states, Ep = ±√~p2 +m2. The mathematics is just the

same, by looking at creation operators as just dagger of annihilation operators,but this language is totally unnecessary. Also, there is no physical justificationfor this interpretation.

The other interaction is

and this has contribution (i2e2gµν).


12 October 11, 2018

So we have our first theory of scalar QED. There is the Lagrangian

L = −1

4F 2µν − φ( +m2)φ∗ − ieAµφ∗∂µφ+ ieAµφ∂µφ

∗ + e2A2µφ∗φ.

with all these Feynman rules. We should be able to check gauge invariance andthe Ward identity.

12.1 Gauge invariance and the Ward identity for scalarQED

Let us consider the Mollar scattering e−e− → e−e− in scalar QED.

The t-channel contribution is

iMt = (−ie)(pµ1 + pµ2 )−i[gµν + (1− ξ)k

µkν

k2

k2(−ie)(pν2 + pν4).

But thenkµ(pµ1 + pµ3 ) = (pµ1 − p

µ3 )(pµ1 + pµ3 ) = m2 −m2 = 0

and similarly kν(pν2 + pν4) = 0. So we just have

Mt = e2 s− ut

.

Likewise, we have

Mu = e2 s− tu

,dσ

dΩ=

1

64π2s(Mt +Mu)2.

Here, the fact that the ξ term vanishes shows that the theory is gauge invariant.Let’s now check the Ward identities, using the process π+π− → γγ. Here,

gives the contribution

iMt = (−ie)2 (2pµ1 − pµ3 )ε∗µ3 (pν4 − 2pν2)εν4

(p1 − p3)2 −m2= Mµνε

3∗µ ε

4∗ν .


Now checking the Ward identity is checking if this is zero if εµ∗3 = pµ3 . But thenwe get

Mwt = e2[pν4 − 2pν2 ]εν4 .

This is nonzero because there are other diagrams. If we look at the u-channel,we get

Mwu = e2[p4 − 2pν3 ]εν4 , M2

4 = 2e2p3ε4,

where M4 is the contribution of the 4-particle interaction.

12.2 Lorentz invariance and soft photons

Suppose we have a diagram, and there is some p1 going in, with contributionM0(pi). Here, we can modify the diagram by just adding a photon. The Mmatrix for this is

Mi = (−ie) i[2p1 − q]ε(p− q)2 −m2

M0.

If we use p2 = m2 and q2 = 0 and εq = 0, then we can approximate this as

Mi = −e pεpqM0(pi + q) ≈ −e pε

pqM0(pi)Qi.

Likewise, we can add one photon to the diagram at the outgoing edge. Thenwe get a contribution of

Mi = +epε

pqM0Qi.

So adding a photon in some edge gives a contribution of

M = eM0

[ ∑incoming

Qipiε

piq−

∑outgoing

Qipiε

piq

].

Under the Lorentz transformation, we get εµ → εµ + Λqµ. So for M to beLorentz-invariant, we must have

eM0

[ ∑incoming

Qi −∑

outgoing

Qi

]= 0.

So we get conservation of charge from this.This is really universal. Even if the interaction term is arbitrary, say

−ieΓµ(p, q)εµ = −ie(Fjpµ +Gjqµ)εµ, Fj(p

2, q2, pq) = Fj

( pqm2

),

we can just look at the q → 0 limit and get

−ieFj( pqm2

)p→ −ieFj(0)p.


So we can just look at an arbitrary theory and define Fj(0) = Qj as the charge.So we can think of charge as the interaction with low-energy photons for longdistances. In this case, we can do the same thing and get conservation∑

in

Fj(0) =∑out

Fj(0).

This gets more interesting. Consider a massless spin-2 particles, and em-bed into εµν(p) polarization tensors. These are transverse and traceless andsymmetric:

gµνεµν = 0, qµεµν = 0, gµν = gνµ.

But because it is massless, it transforms in the same weird way

εµν → εµν + Λνqµ + ΛµqνΛqµν .

If we look at the soft limit, we some contribution

εµνΓµν = εµνpµpνF

(p · qm2

).

Then if we o the same thing,

M =M0

[ ∑incoming

F (0)piµp

iν

(pi · q)−

∑outgoing

pµi pνi

(pi · q)

]εµν .

The Ward identity gives

0 =M0Λν

[∑in

Fi(0)pνi −∑out

Fj(0)pνj

].

We already have momentum conservation, and this lets us solve for p1 as afunction of the other pj . So this is consistent only when Fi(0) = GN for someuniversal constant GN . This is saying that gravity is universal, and coupleswith every particle with the same coupling constant. This is the only way tohave a consistent theory of a spin-2 particle, and this particle must be unique.What about spin-3? If you do the same thing, we should get∑

in

γjpνj pµj =

∑out

γjpνj pµj .

If we only look at the (0, 0)-component, we get∑γjE

2 =∑γjE

2. This is im-possible, and so there are no consistent interacting theories of massless particleswith J > 2.

12.3 Spinors

You’ve seen spinors before. In non-relativistic quantum mechanics, there is this|ψ〉 = |↑〉 and |↓〉. The quantum mechanics is governed by the Schrodinger–Pauliequation

i∂tψ = Hψ =

[(p2

2m+ V (r)− µB ~B · ~L

)(1 00 0

)−(

Bz Bx − iByBx + iBy −Bz

)]ψ.


Here, we can write

σ1 =

(0 11 0

), σ2 =

(0 −ii 0

), σ3 =

(1 00 −1

), [σi, σj ] = 2iεijkσk

and then that matrix is just ~σ · ~B.Here, ~B is a 3-vector, and ~σ · ψ also transforms like a 3-vector. Also, ∂iσiψ

is rotation invariant, and then we can guess and check that

∂tψ − ∂iσiψ = 0

is Lorentz-invariant. In fact, if we define σµ = (1, ~σ), then

σµ∂µψ = 0

is the Dirac equation for m = 0. Maybe we could guess that σµ∂µψ = mψ isthe massive case, but this is wrong. So enough guessing.

This follows naturally from representations of the Lorentz group. There arethese

ΛRz = R(θz) =

1

cos θz sin θz− sin θz cos θz

1

, Bβx =

coshβx sinhβxsinhβx coshβx

11

.

Now we can look at the infinitesimal generators and extract the Lie algebra.Then we get

Rz = R(θz) =

0

θz−θz

0

, βx =

βx

βx0

0

.

Then we get∆V0 = βiVi, ∆Vi = βiV0 − εijkθjVk.

If we write the i times the rotation generators as Ji and the i times the boostgenerators as Ki, hen we have

[Ji, Jj ] = iεijkJk, [Ji,Kj ] = iεijkKk, [Ki,Kj ] = −iεijkJk.


13 October 16, 2018

We were talking about representations of the Lorentz group. The Lorentz groupis defined as

L = ΛF : (ΛF )†gΛF = g.

Once we have these matrices, there is a group operation, so we get a groupO(1, 3). There is the obvious 4-vector representation

Vµ → ΛµνVν .

To find a representation, we could instead look at the representations of the Liealgebra, which is generated by iJ1, iJ2, iJ3 and iK1, iK2, iK3.

13.1 Representations of the Lorentz group

The Lie algebra structure is given by

[Ji, Jj ] = iεijkJk, [Ji,Ki] = iεijkKk, [Ki,Kj ] = −iεijkJk,

which is called o(1, 3). How do we find representations of this? We observe thatthere is a convenient linear combination,

J+i =

1

2(Ji + iKi), J−i =

1

2(Ji − iKi).

Then the relations become

[J+i , J

+j ] = iεijkJ

+k , [J−i , J

−j ] = iεijkJ

−k , [J+

i , J−j ] = 0.

This means that we really have o(1, 3) = su(2)× su(2).But we know the finite-dimensional representations of su(2): they are just

(2j+1)-dimensional vector spaces. So irreducible representations of so(1, 3) arejust indexed by (A,B), which is a (2A+ 1)(2B + 1)-dimensional representationobtained by taking the tensor (box) product.

Now given any representation of so(1, 3), we will also get a representations

of so(3) by looking at ~J = ~J+ + ~J−. Now we can think about which spinrepresentations appear in each (A,B). Here is a table of this for small values:

(A,B) dim spin of so(3)-representations(0, 0) 1 0( 1

2 , 0) 2 12

(0, 12 ) 2 1

2( 1

2 ,12 ) 4 1, 0

(1, 0) 3 1

Table 1: Representations of su(2)× su(2) restricted to so(3)


13.2 Dirac spinors

So what are these representations? If we look at ( 12 , 0) and (0, 1

2 ), these differentrepresentations with spin 1

2 . These are called right-handed spinors ψL

J+ =σi2, J− = 0, ~J = J+ + J− =

~σ

2, ~K = i(J− − J+) = −i~σ

2,

and left-handed spinors ψR

J+ = 0, J− =σi2, ~J =

~σ

2, ~K = +i

~σ

2.

The representations then can be written as

δψL =1

2(iθi − βi)σiψL, δψ†L =

1

2(−iθi − βi)ψ†Lσi,

δψR =1

2(iθi + βi)σiψR, δψ†R =

1

2(−iθi + βi)ψ

†Rσi.

Now what would the Lagrangian be? We could try and write down

L = ψ†LψL,

but thenδ(ψ†LψL) = −βiψ†Lσiψi 6= 0.

This is somewhat expected, because we’re using the Hermitian norm on some-thing that transforms under Lorentz transformations. It turns out that theexpression

L = m[ψ†LψR + ψ†RψL]

is Lorentz invariant.This is the mass term. What about the kinetic term? It turns out that the

LagrangianL = ψ†LψR

is invariant if ψL is a spinor or if ψL is two scalars. Let us define

σµ = (1, ~σ), σµ = (1,−~σ).

Then the Lagrangian

L = iψ†Rσµ∂µψR + iψ†Lσ

µ∂µψL −m(ψ†LψR + ψ†RψL)

turns to be Lorentz invariant.To simplify notation, let us define

ψ =

(ψLψR

), ψ =

(ψ†R ψ†L

)= ψ†

(1 00 1

), ψψ = ψ†LψR + ψ†RψL.


This is called the Dirac spinor. If we moreover define

γµ4×4 =

(0 σµ

−σµ 0

),

then we can write the Dirac Lagrangian in the following compact form:

LDirac = iψγµ∂µψ +mψψ.

This representation really is different from just the vector representation. Ifwe look at J3 with the Dirac representation, we get

JDirac3 =

1

2

1−1

1−1

, Jvector3 = i

0

−11

0

.

When we look at the eigenvalues, the first one has 12 ,

12 ,−

12 ,−

12 , and the second

one has −1, 0, 0, 1. So we see that the first one has two spin 12 representations,

while the second one has one spin 0 and one spin 1.What happens if we rotate by 2π? We get

Λ3 = exp[iθJDe ] =

−1

−1−1

−1

.

So what we really get is that if we rotate by 2π, then we get the −1 of itself. Thisis not really a representations of the Lorentz group, because 1 is not mappedto 1. Instead, we have constructed a projective representation, where groupmultiplication holds up to a phase. Projective representations of a group G areequivalent to representations of the universal cover of G = SO(1, 3), which is inthis case SL(2,C).

Here is another interesting consequence. If we rotate by π, we get the matrix

Λ3(θ = π) =

i−i

i−i

.

So if we have |↑↑〉 and rotate around by π, we get −(|↑↑〉), so it is indistinguish-able from itself. So this is somehow expressing the Pauli exclusion principle.


14 October 18, 2018

Last time we constructed the Dirac spinors. We looked at irreducible finite-dimensional representations of the Lorentz group, characterized by two half-integers, and then put ψL and ψR together. They transform as

ψL → exp[ iθj

2σj +

βj2σj

]ψL, ψR → exp

[ iθj2σj −

βj2σj

]ψR.

14.1 The Dirac equation

Then we wanted the Lagrangian, and we got

L = iψ†Rσµ∂µψR + iψ†Lσ

µ∂µψL −mψ†LψR −mψ†RψL,

where σµ = (1, ~σ) and σµ = (1,−~σ). Then we defined ψ = ( ψLψR ) and γµ =

( 0 σµ

σµ 0 ) and then wroteL = iψγµ∂µψ −mψψ.

The equations of motion is then given by

(iγµ∂µ −m)ψ = 0.

We can think of these matrices γµ as algebraic objects, independent of theirrepresentations. They satisfy the anti-commutation relations

γµ, γν = 2gµν ,i

4[γµ, γν ] = Sµν =

1

2σµν .

There are different choices of γ, for instance, the Majorana representations

γ0 =

(0 σ2

σ2 0

), γ1 =

(iσ3

iσ3

), γ2 =

(0 −σ2

σ2 0

), γ3 =

(−iσ1

−iσ1

)that is purely imaginary.

Let’s look at the Dirac equation and multiply it by (iγµ∂µ + m). Then wehave

0 = (iγµ∂µ +m)(iγν∂ν −m)ψ = (−γµγν∂µ∂ν −m2)ψ

= (− 12γµ, γν −m

2)ψ = −( +m2)ψ.

So anything satisfying the Dirac equations also satisfy the Klein–Gordan equa-tions.

If we want photon interactions, we can write

Dµψ = [∂µ + ieAµ]ψ, LQED = iψγµDµψ −mψψ,

so that there is a cubic interaction term, and the equations of motion become

(iγµ∂µ − eγµAµ −m)ψ = 0.


If we do the same trick of multiplying (iγµ∂µ − eγµAµ +m) on the left, weget

0 =

[−e

2Fµν

i

2[γµ, γν ] +D2

µ −m2

]ψ.

First, this tells us that this is not just the Klein–Gordan equation. If we try tofigure out what σµν is, you can calculate that

σ0i = − i2

(σi

σi

), σij = εijk

(σk

σk

).

Then we get[(∂µ − imAµ)2 +m2 − e

(( ~B + i ~E)~σ 0

0 ( ~B − i ~E)~σ

)]ψ = 0.

You might remember doing this in quantum mechanics. The electron hassome spin, and this interacts with the magnetic field. But we did not knowhow strong this is going to be. But if we incorporate Lorentz invariant, thisexactly fixes the strength of the interaction. In fact, this prediction is that theelectron’s magnetic dipole moment is given by

µB =e

2m.

The measured values is about 1.002µB , so this shows that the spinor businessis really going on in nature.

This theory is symmetric under interchanging L and R, and so we say thatit is not chiral. But nature is actually chiral because of weak interactions, asdiscovered in the 50s.

Spin is a vector, and a spin operator ~J is an appropiate representation ofSO(1, 3). For spinors, we have

~J =~σ

2= ~S.

So ψ has spin s along the ~v axis if ~v · ~Sψ = sψ. Helicity is defined as theprojection of the spin on the direction of motion. So we have

~p · ~S|~p|s

= hψ,

and this is ±1. This is a useful concept in the massless limit. In this case, weget look at

iγµ∂µψL = 0, iγµ∂µψR = 0,

so we get (E− ~p ·~σ)ψR = 0 and (E+ ~p ·~σ)ψL = 0. Then ψR has helicity always+1 and ψL has helicity always −1.


14.2 Quantum electrodynamics

Now we want to quantize the theory. Remember how this was done. We had

φ(x) =

∫d3p

(2π)3

1√2ωp

(ape−ipx + b†pe

ipx),

φ∗(x) =

∫d3p

(2π)3

1√2ωp

(bpe−ipx + a†pe

ipx).

Then we can just do the same thing and write

ψ(x) =

2∑s=1

∫d3p

(2π)3

1√2ωp

(aspuspe−ipx + bs†p v

speipx),

ψ(x) =

2∑s=1

∫d3p

(2π)3

1√2ωp

(bspvpe−ipx + a†pu

speipx).

Here, usp and vsp are the wavefunctions for the spin, for the particle and theanti-particle.

To get the spin, we need to solve(−m pµσµpµσµ −m

)us(p) = 0,

(−m −pµσµ−pµσµ −m

)vs(p) = 0.

If we solve this, the solutions are

u1(p) =

(√pσ( 1

0 )√pσ( 1

0 )

), u2(p) =

( √pσ( 0

1 )−√pσ( 0

1 )

).

If we choose p = (E, 0, 0, pE), then we can show that these are√E − pE

0√E − pE

0

,

0√

E + pE0√

E + pE

.

But most of the time we will only care about the sum of these, the average overthe spins. Then we get

usus′ = u†sγ0us′ = 2mδss′ , vsvs′ = −2mδss′ .

So we can define the inner product

〈us|us′〉 = usus′ = 2mδss′ ,

and also the outer product

2∑s=1

us(p)us(p) = γµpµ +m.

We will calculate this later, but it will turn out that the propagator is given by

〈ψψ〉 =i(γµ∂µ +m)

p2 −m2 + iε.


15 October 23, 2018

We derived the Lagrangian of QED, (we write /X for γµXµ)

LQED = ψ(i /D −m)ψ, (i/∂ − e /A−m)ψ = 0.

In the non-relativistic limit, we get

i∂tψ = Hψ =

(~p2

2m+

2e

m~B · ~σ + 0

e

m~E · ~σ

)ψ,

so we expect 2e/m to be the electron’s dipole moment.To quantize the theory, we had to introduce polarizations usp and vsp, and

they satisfied

/pu = mu, −/pv = mv.

So we solved this and got

u1p =

√p · σ0√p · σ0

, u2p =

0√p · σ0√p · σ

, v1p =

√p · σ0

−√p · σ0

, v2p =

0√p · σ0

−√p · σ

.

In the non-relativistic limit, these are the four polarizations of the particle andthe antiparticle.

15.1 Identical particles

Let us look at creation and annihilation operators for every particle. There are

a†~pi,si,ni , si, ni,

where si is the spin of the particle and ni are the “internal numerators” of theparticles (so there are many particles with the same state). States created withthe same ni are called identical particles. So when we fix n, we get

|ψ12〉 = |s1p1ns2p2n〉 =√

2ω1

√2ω2a

†p1s1a

†p2s2 |0〉.

If we act in a different order, we will get

|ψ21〉 =√

2ω1

√2ω2a

†p2s2a

†p1s1 |0〉.

We should have that |ψ12〉 is the same physical state as |ψ21〉. So we only havea difference in phase, so

|ψ12〉 = α|ψ21〉.

What α can this be? This should be a representation of the Lorentz group, andthere are no one-dimensional representations except for the trivial representa-tion. But α might depend on the path in |ψ12〉 → |ψ21〉.


Example 15.1. In 2-dimensions, if we have two particles at A and B, thereare Z-many topological ways to start with A,B and end with A,B. Thisis saying that the fundamental group of SO(2) is Z, so we could get |ψ12〉 7→α|ψ12〉 7→ α2|ψ12〉 7→ · · · .

But in 3-dimensions, we have too much room, so there is only one way ofinterchanging A and B. So we have

|ψ12〉 = α|ψ21〉,

with α = ±1. If α = 1, then the particle is called a boson and we get

[a†p1s1 , a†p2s2 ] = 0.

On the other hand, if α = −1 then we call it a fermion and we have

a†p1s1 , a†p2s2 = 0.

If α is not ±1 (which may happen only in 2-dimensions), then the particle iscalled an anyon. We haven’t seen how this connects to the spin- 1

2 particle, butthis at least shows that we can have a consistent Lorentz-invariant theory withα 6= −1.

If we had a Jsz , then we would have

Jsz =

−s

−s+ 1. . .

s

.

So if we rotate about the z-axis by angle π, then we would have

|AB〉 = (eiπJz )2|BA〉 = e2πis|BA〉,

where s is the spin. This shows that if s is an integer, this is a boson, and if sis a half-integer, it is a fermion. For a fermion, we have 0 = a†p, a†p = 2(a†p)

2,so there are no two fermions in the same state.

This spin statistics is related to stability. If you choose the right statisticsand compute the energy density, you get

E =∑s

∫d3qωq(a

†qaq + b†qbq)

which is positive. But if you choose the wrong statistics, you get something like

E =∑s

∫d3qωq(a

†qaq − b†qbq)

which implies that the universe is not stable. Also, to get causality [φ(x), φ(y)] =0 or φ(x), φ(y) = 0, we need to choose the correct statistics.


15.2 Propagator for QED

This is also important in getting a Lorentz-invariant S matrix. Recall that

Tφ(x)φ(y) = φ(x)φ(y)θ(x0 − y0) + φ(y)φ(x)θ(y0 − x0)

for scalar fields. If we try to do this for fermions, then we have to define

Tψ(x)χ(y) = ψ(x)χ(y)θ(x0 − y0)− χ(y)ψ(x)θ(y0 − x0)

because otherwise Tψ(x)χ(y) 6= −Tχ(y)ψ(x). If we use this definition andtry to calculate the time-ordered product, we get

〈0|Tψ(0)ψ(x)|0〉 =

∫d4p

(2π)4

i(/p+m)

p2 −m2 + iεeipx,

which is Lorentz-invariant. If we use the wrong statistics, we would have gottensomething that is not Lorentz-invariant.

We can also see what the propagator is, directly from the Dirac equation.Recall that the free Dirac equation is

(i/∂ −m)ψ = 0, (/p−m)u = 0.

Because /p2 = γµγνpµpν = p2, we can write

〈0|Tψ(x)ψ(0)|0〉 =i

/p−m/p+m

/p+m=i(/p+m)

p2 −m2,

which is what we had.


16 October 25, 2018

Let us now write down the Feynman rules for quantum electrodynamics.

16.1 Feynman rules for quantum electrodynamics

We have the photon propagator

i

p2 + iε

[−gµν + (1− ξ)p

µpν

p2

]and the electron propagator

i(/p+m)

p2 −m2 + iε,

and interactions −iεγµ. Then for photons coming out from a vertex, there is aε∗µ, for a photon going in, there is εµ, and for an outgoing u-spinor, u(p), for anincoming u-spinor, u(p), and for an outgoing v-spinor, v, and for an outgoingv-spinor, v(p).

So if we have a diagram like

e− e−

becomes

ε∗µ4u(p3)γµ(/q +m)γνu(p1)

q2 −m2 + iεεν2 × (−ie)2i.

But electrons are fermions, so there are extra minus signs coming from

• each fermion loop,

• each time you swap two external fermions.

To see why this is the case, consider e−e− → e−e− in the free theory. Then tocompute

〈0|Tψ(x3)ψ(x1)ψ(x4)ψ(x2)|0〉.Then when we try to move things around, we need to put minus signs. Forinstance,

DF (1, 3)DF (2, 4) = 〈0|a3a†1|0〉〈0|a4a

†2|0〉,

DF (1, 4)DF (2, 3) = 〈0|a3a†20〉〈0|a4a

†1|0〉.

have a relative minus sign. There is always a minus sign for loops. When I havea loop, I am doing something like writing a1a

†5a5a

†3 and then identify a3 and a1.

Then we havea1a†5a5a

†1 = −a†1a1a

†5a5

and so we need a minus sign.


Example 16.1. Let us write down the matrix element for the fermion loop.

We have

iM = (−ie)2(−1)

∫d4k

(2π)4εµ1 ε∗ν2

i(/p+ /k +m)

(p+ k)2 −m2 + iεγµαβ

[i(/k +m)

k2 −m2 + iε

]γνβα

= e2εµ1 ε∗ν2

∫d4k

(2π)4

Tr[(/p+ /k +m)γµ(/k +m)γν ]

((k + p)2 −m2 + iε)(k2 −m2 + iε).

We are not going to compute this integral, but maybe we want to know thatthis trace is. The nice property we have is that traces are cyclic. If we writeγ5 = iγ0γ1γ2γ3 then γ2

5 = 1 and γ5γµ = −γµγ5. Moreover, γµ, γν = 2gµν andso we can do stuff like

Tr[γµ] = Tr[γ5γ5γµ] = Tr[γ5γ

µγ5] = −Tr[γ5γ5γµ] = −Tr[γµ].

So this is zero. Similarly you can calculate these things without choosing rep-resentations.

Example 16.2. If you calculate e+e− → µ+µ−, then we can approximateme = 0 because muon is heavier, and then

dσ

dΩ=

α2

16E2

√1−

m2µ

E2

(1 +

m2µ

E2+

(1−

m2µ

E

)cos2 θ

).

In the ultrarelativistic E mµ limit, we get α2

16E2 (1 + cos2 θ).

Example 16.3. Let us look at Rutherford scattering e−p+ → e−p+. Here, wehave

dσ

dΩ=

α2

4v2p2 sin4 θ2

(1− v2 sin2 θ

2

)in the p+ rest frame, where mp me. This is called the Mott formula.

Example 16.4. There is also Compton scattering γe− → γe−, and the crosssection is given by

dσ

d cos θ

π2α

m2e

(ω′ω

)2[ω′ω

+ω

ω′− sin2 θ

], ω′ =

ω

1 + ωme

(1− cos θ).

This is called the Klein–Nishna formula. If me ω, we get

dσ

d cos θ=πα2

m2(1 + cos2 θ),

which is Thompson scattering. The Klein–Nishna formula was discovered in1929, so this was the first achievements in quantum electrodynamics.


Let’s just wee how we write down the tree level diagram for e+e− → µ+µ−.

Then we have

iM = (−ie)2 u(p3)γµv(p2)v(p4)v(p2)γνu(p1)

k2i

(−gµν + (1− ε)k

µkν

k2

).

Here, we see that the gauge term vanishes because /k = /p3+ /p4

. Then we cansee that

M =e2

s[v(p2)γµu(p1)][u(p3)γµv(p4)].

If we take the conjugate transpose, we get

M† =e2

s[v(p4)γµu(p3)][u(p1)γµv(p2)].

So to get the spin sums, we should add them∑spins

MM† =∑spins

[v(p4)γνu(p3)][u(p3)γµv(p4)][u(p1)γνv(p2)][v(p2)γµu(p1)]

= Tr[(/p3+m)γµ(/p4

−m)γν ] Tr[· · · ]

and then use the trace formulas.


17 October 30, 2018

Today we are going to look at path integrals. They are a beautiful we to thinkabout quantum field theory, or quantum mechanics in general.

17.1 Path integrals

Consider the double slit experiment. Here, you add the two wavefunctions thatpasses through the two holes. But suppose there are three hole, or many slitswith many holes. Then you add up all the possible paths. So we have somethinglike

〈f |i〉 =

∫ x(tf )=xf

x(ti)=xi

Dx(t)eiS[x(t)].

In quantum field theory, we get

〈Ω|Tφ(x1) · · ·φ(xn)|Ω〉 =

∫DφeiS[φ]φ(x1) · · ·φ(xn)∫

DφeiS[φ].

This is actually a really remarkable formula, because the right hand side is clas-sical. First we are going to derived this formula, in two ways, non-perturbativelyusing the Hamiltonian and also perturbatively using Feynman rules.

Let me first think of the one-dimensional example

Z0 =

∫ ∞−∞

dφe−12aφ

2

=

√2π

a.

Then

〈φφ〉 ∼ 1

Z0

∫ ∞−∞

dφe−12aφ

2

φ2 =1

a.

So in the free theory, we are going to get

1

Z0

∫dφe−

12φ(+m2)φ =

1

+m2

the propagator back. If we do this for φ4, we get something like

1

Z0

∫dφe−

12φ(+m2)φφ4 =

3

( +m2)2,

which is just the three Feynman diagrams.First, let us get this mathematics out of the way. We have∫ ∞

−∞dpe−

12ap

2+Jp =

√2π

aeJ2

2a .

For many pi, we can look at a general quadratic form ap2 → piAijpj , and then∫dp1 · · · dpne−

12 ~p·A~p+ ~J·~p =

√(2π)n

detAe

12~JA−1 ~J


Here, we have

H[φ, π] =

∫d3x[ 1

2 π2 + V (φ) + 1

2 (~∇φ)2].

If we do this, we get

〈Φj+1|e−iH(tj)δt|Φj〉 =

∫d3Π〈Φ|Π〉〈Π|e−iHjδt|Φj〉

= N exp

[iδt

∫d3xL[φj , φj ]δt

].

At the end, we get

〈0; tf |0; ti〉 =

∫DΦ(x, t)eiS[Φ].

Here, DΦ is choosing a configuration of the field for each t. This is the denom-inator of the path integral. You can see from this that the dominating termis the when S is slowly moving, which is when S is extremized. In the clas-sical limit ~ → 0, we have really the normalization i

~S[φ], and we recover theEuler–Lagrange equations.

Let us now look at the time-ordered product. Note that

f(xj , tj) =

∫DΦ(x, t)eiS[Φ]Φ(xj , tj) = 〈0|φ(tj , xj)|0〉,

because I can just stick in φ(xj , tj) inside thie expansion we had, and then weare picking up the eigenvalue Φ(xj , tj) of |Φj〉〈Φj |. If we have two fields, weneed to order the two fields, so we really have∫

DΦeiS[Φ]Φ(x1, t1)Φ(x2, t2) = 〈0|Tφ(x1, t1), φ(x2, t2)|0〉.

This is why we get the correlation functions in the form we wanted.

17.2 Generating functionals

Consider

Z[J ] =

∫DφeiS[φ]+i

∫d4xφ(x)J(x).

This is called the generating functional. One useful thing is Z[0], which isthe denominator of the path integral. If I take the derivative with respect to J ,this is

1

Z[0](−i) dZ

dJ(x1)=

∫DφeiS[φ]+i

∫Jφφ(x1).

Then if I take the derivative n times, with respect to J(xj), we get

1

Z[0](−i)n dnZ

dJ(x1) · · · dJ(xn)

∣∣∣J=0

= 〈0|Tφ(x1) · · ·φ(xn)|0〉.


So if we know everything about the generating functionals J , we can obtaineverything about the theory.

Let us try to calculate this in the free theory. Here, we have L = − 12φ( +

m2)φ as we all know. Then

Z[J ] =

∫Dφei

∫d4x[− 1

2φ(+m2)φ+Jφ]

= Ne12J

−i+m2 J = Nei

∫d4xd4yJ(x)Π(x−y)J(y).

What this means is that

〈0|Tφ(x)φ(y)|0〉 =1

Z[0]

∂2Z

∂J(x)∂J(y)

∣∣∣J=0

= iΠ(x− y),

where ( +m2)Π(x− y) = −δ(x− y).


18 November 1, 2018

Today we are going to finish the deriving the Feynman rules. In quantummechanics, the path integral was another way of calculating 〈xf |e−iHt|xi〉 =∫xDxeiS[x]. In quantum field theory, we have

〈φf |e−iHt|φi〉 =

∫ φ(tf )=φf

φ(ti)=φi

DφeiS[φ].

Here, |φf 〉 are the eigenvectors, φ(x)|φf 〉 = φf (x)|φf 〉. Then we had the masterformula

〈Ω|Tφ(x1) · · ·φ(xn)|Ω〉 =

∫DφeiS[φ]φ(x1) · · ·φ(xn)∫

DφeiS[φ].

Then we had the generating functional

Z[J ] =

∫Dφei

∫d4x[L+Jφ].

Then we were able to recover everything by taking

(−i)n ∂nZ[J ]

∂J(x1) · · · ∂J(xn)

∣∣∣∣J=0

= 〈Ω|Tφ(x1) · · ·φ(xn)|Ω〉.

18.1 Feynman rules from the path integral

In the free theory, this is given by

Z[J ] =

∫Dφei

∫d4x[− 1

2φ(+m2)φ+Jφ]

= N exp

[i

∫d4xd4yJ(x)Π(x, y)J(y)

].

For example, a 4-point function in free theory is given by

〈0|Tφ(x1) · · ·φ(xn)|0〉 =1

Z

d4Z

dJ1dJ2dJ3dJ4e−

12

∫dxdyJxDxyJy |J=0

=1

Z

d4

dJ1dJ2dJ3(−JzDz4)Z|J=0

=1

Z

d2

dJ1dJ2(−D34 + JzDz3JwDw4)Z

= D34D12 +D23D14 +D13D24.

So that’s for the free theory. Let us now look at interactions, for instance,

L = −1

2φ( +m2)φ+

g

3!φ3.


Then we have

Z[J ] =

∫Dφ exp

[i

∫d4x(− 1

2φ( +m2)φ+ Jφ)

](

1 + i

∫d4x

g

3!φ3 −

∫d4xd4y

(g

3!

)2

φ(x)3φ(y)3 + · · ·).

But then we have

〈Ω|Tφ(x1)φ(x2)|Ω〉 = 〈0|Tφ0(x1)φ0(x2)|0〉

+g

3!

∫d4x〈0|Tφ0(x1)φ0(x2)φ0(x)3|0〉 − · · · .

So you get the Feynman rules.In the propagator, we didn’t get an iε. This should come from the constraint

on fields. We want to evaluate the time evolution operator

∞〈Ω|S|Ω〉−∞ =

∫ φ(∞)=φ0

φ(−∞)=φ0

DφeiS =

∫DφeiS+∞〈Ω|φ(t = +∞)〉〈φ(t = −∞)|Ω〉−∞.

Remember how we quantized the theory. Here, the ground state or vacuum foreach ~p is given by

φground(x) = 〈x|0〉 = e−12x

2mω.

In quantum field theory, the ground state is a wave functional Ψ0[φ], given by

Ψ0[φ] = exp

[−∫d3pφ∗pφpωp

]= exp

[−∫d3pd3xd3yei~p(~x−~y)φ(x)φ(y)ωp

].

To evaluate this, we can use the following slick trick due to Weinberg:

〈Ω|φ∞〉〈φ−∞|Ω〉 = limε→0

exp

[−ε∫ ∞−∞

dεd3pd3xd3yφ(x, t)φ(y, t)eip1(x−y)ωpe−ε|t|

]= exp

[−∫d3pd3xd3yωpe

ip(x−y)[φ(x,∞)φ(y,∞) + φ(x,−∞)φ(y,−∞)]

]= Ψ0[φ(x,−∞)]Ψ∗0[φ(y,−∞)]

= limε→0

exp

[−∫d4xd3yd3pei~p(~x−~y)φ(x, t)φ(y, t)εωp

]= limε→0

exp

[−∫d4xεφ(x)φ(x)

].

So we really have∫Dφ exp[−

∫d4x[− 1

2φ[ + m2 − iε]φ] and so we get Π =( +m2 − iε)−1 when we calculate the propagator. I don’t think this is math-ematically rigorous, but this should be morally correct.


18.2 Gauge invariance from path integrals

Remember we took the deformation

LQED = −1

4F 2µν + iψ( /D −m)ψ − 1

2ξ(∂µAµ)2.

Let me start by defining a function

f(ξ) =

∫Dπe−i

∫d4x 1

2ξ (π)2

.

Now I am going to change variables π → π − 1∂µAµ. This is what you would

choose to put your field in Lorentz gauge. If I look at scalar QED, define

F [O] =

∫DφDφ∗DAei

∫d4x[− 1

4F2µν+Dµφ

2]O 1

f(ξ)

∫Dπe−

i2ξ

∫d4x(π−∂µAµ)2

,

so that 〈Ω|TO|Ω〉 = F [O]/F [1]. If we make a change of variables Aµ →Aµ + ∂µπ and φ→ eiπφ and φ∗ → e−iπφ∗, then we get

F [O] =

∫Dπ

f(ξ)

∫DADφDφ∗ exp

[i

∫−1

4F 2µν + (Dµφ)2 − 1

2ξ(∂µAµ)2

].

So we see that this cancels out when we take the ratio.For fermions, we know that spinors have to anticommute, ψ1(x)ψ2(y) =

−ψ2(y)ψ1(x). Then the path integral for fermions is

Z[η, η] =

∫DψDψ exp

[i

∫d4xiψ /∂ψ −mψψ + ψη + ηψ

].

So we want a things that all anticommute. These are called Grassmann num-bers, Y = θi over C such that θiθj = −θjθi. So these are like differentialforms, but it is not a useful analogy.


19 November 6, 2018

Now we want to start renormalization, which is the heart of quantum fieldtheory.

19.1 Renormalization

If you try to look at the photon propagator, you can look at the ones withone fermion loop and ones with two fermion loops or so on. Then you try toadd all these loop corrections, and you get infinity. You can do this with otherthings like mass correction of the electron, or splitting of the energy levels ofthe hydrogen atom. This was so bad that Dirac tried to get rid of quantumfield theory, even though he invented it. The idea is that we can be very carefulabout calculating things that are only physical. Maybe the energy potential isinfinite, but the force or energy difference is finite.

If we consider

then we the integral∫d4p

i

p2 + iε

i(k + 2p)µ(2p+ k)µ

(2p+ k)2 + iε≈∫p3dp

p2

p4∼∫ Λ

pdp ∼ Λ2

if we cut off at Λ. This is a slight complicated, so we are not going to start withthis.

A simpler example is the vacuum energy, which has to do with the zeromodes of the harmonic oscillators. Then we have

vacuum energy = 〈0|H|0〉 =

∫d3k

(2π)3

ωk2

= 4π

∫k2dk

k

2∼ Λ4,

which is quartically divergent.So let us look at the what we are doing. We can first consider this box of

size r, and restrict things to this box. Then we get

ωn =π

rn, E(r) =

∞∑n=1

ωn2

=∞, F (r) = −dEdr

= − π

2r2

∞∑n=1

=∞.

This didn’t help much. This first missing the things outside the box, but it isstill infinite after putting in these terms. So we take a cut off Λ so that onlylow frequency ω < πΛ are quantized. Then we get

E(r) =π

4rbΛrc(bΛrc+ 1).

So define x = Λa− bΛac and if we add the energy for a and L− a we get

Etot(a) =π

4

(Λ2L− x(1− x)

a− x(1− x)

L− a

).


Now if we take the limit L a, and then look at F , we get

F = − d

daEtot = − π

24a2= − π~c

24a2.

If we did the same exercise in three dimensions, we would get

F (a) = − π~c240a4

A = −1.3× 10−27N ·m2 1

a4.

This is the Casmir effect, and was experimentally measured in 2002 withconstant (−1.22± 0.16)× 10−27.

This seems like a pretty arbitrary way to cut off and compute the number, solet us look at a different regularization, called the heat kernel regularization.Here, we assume a damping

E(r) =1

2

∑n

ωne−ωn/πΛ =

e1rΛ

(e1rΛ − 1)2

≈ π

2Λ2r − π

24r+ · · · , ωn =

nπ

r.

Now if we look at the difference, we get

Etot = E(a) + E(L− a) =π2

2Λ2L− π

24

(1

a+

1

L− a

), F = − π

24a2.

You could do a Gaussian regulator

E(r) =∑ ωn

2e−(ωnπn )2

, Λ→∞,

or a ζ-function regulator

E(r) =1

2

∑ωn

(ωnM

)−s, s→ 0,

and all of these will give the same prediction for Casmir effect.But you should be skeptical of this, because what if you get something else

by using some other cut off? Here is a regulartor-independent calculation, givenby Casmir himself in 1948. Let us consider

E(r) =π

2

∑n

n

rf

(n

rΛ

),

where f is something like e−x or θ(|x| − 1) or e−x2

or something that doesn’taffect small x but kills off large x. Then we have

E(L−a) =π

2(L−a)Λ2

∞∑n=1

n

(L− a)2Λ2f

(n

(L− a)Λ

)=π

2(L−a)Λ2

∫xdxf(x)


because we can write x = n(L−a)Λ and assume that the spacing is getting close.

Then we can write

Etot(0) = E(a) + E(L− a) = ρL+π

2a

[∑n

nf

(n

aΛ

)−∫ndnf

(n

aΛ

)].

This can be written as a Euler–MacLaurin series

N∑n=1

F (n)−∫ N

0

F (n)dn =F (0) + F (N)

2+F ′(N)− F ′(0)

12+ · · · .

If you plug in F (n) = nf( naΛ ) then we see that only the term that survives is

Etot = ρL− π

24af(0) = ρL− π

24a.

This is universal, in the sense that you can see this effect as long as the box canhold photons.


20 November 8, 2018

Let me start by summarizing the principles of renormalization. The basic ideais that

• long-distance / infrared(IR) / low-energy physics is independent of short-distance / ultraviolet(UV) / high-energy physics.

• observables are finite.

• infinites such as UV divergences appear at intermediate stages of the cal-culation.

• a regulator like Λ or ε is introduced to cut off UV divergence, and thelimit as Λ→∞ and ε→ 0 exists for the observable.

20.1 Examples of renormalization—φ4

Let’s take the φ4 theory,

L = −1

2φφ− λ

4!φ4.

If we look at φφ→ φφ, the leading order is just iM = −iλ. Then the next termis

iM2 = (−iλ)2

∫d4k

(2π)4

i

( 12p− k)2( 1

2p+ k)2∼∫k3 dk

(k2 − s4 )2∼ log Λ,

where s = p2 and p = p1 + p2. There is this trick due to Feynman, which is tolook at

∂M2(s)

∂s=λ2

2

∫d4k

(2π)4

− 12

( s4 − k2)3= − λ2

32π

1

s.

So we can integrate this and write

M2(s) = − λ2

32π2log s+ const = − λ2

32π2log

s

Λ2.

Even if we can’t make sense out of M(s), we can still make sense out of theirrelative difference

M(s1)−M(s2) = − λ2

32π2log

s1

s2.

But M should be a measurable quantity, because we can scatter pions andget a number. What I measure is the sum of all the Fyenman diagrams, so Ican measure for s0 and then write this as

M(s0) = −λR.

This λR is the renormalized coupling, and it is defined at a reference scale s0 toall orders in perturbation theory. Now if we try to expand λ out in λR, we willsee that

λ = λR −λ2R

32π2log

s0

Λ2+ · · · .


If we take this λ and try to plug this into s, we get

M(s) = −λR +λ2R

32π2log

s0

s.

20.2 Examples of renormalization—vacuum polarization I

Now let us look at the one-fermion loop given by

(−ie)2

∫d4k

(2π)4

i

(p− k)2 −m2 + iε

i

k2 −m2 + iεTr[γµ(/p− /k −m)γν(/k −m)].

Here, we can write

iΠµν = −ie2(p2gµν − pµpν)Π(p2),

Π(p2) =1

2π2

∫ 1

0

dxx(1− x) logΛ2

m2 − p2x(1− x).

Using this, let us try to compute the interaction between the proton and theelectron in the hydrogen atom,

Then if we add them together, we get

−igµν

p2(1− e2Π2(p2) +O(e4)), V (p2) = e2 1− e2Π2(p2)

p2.

This again is going to be infinite. So we now consider e as a measurement,though V (r) or at some r. Then we define the renormalized coupling as

e2p = p2

0V (p20)

exactly at some p20. In that case, can solve perturbatively, e2

R = p20V (p2

0) =e2 − e4Π2(p2

0) and soe2 = e2

R + e4RΠ2(p0)2 + · · · .

Now we see that

p2V (p2) = e2R−e4

RΠ2(p2)+e4RΠ2(p2

0) = e2R

e4R

2π

∫dxx(1−x) log

m2 − p2x(1− x)

m2 − p20x(1− x)

.

Now if we choose p0 = 0, we see that e2R/4π = 1

137 = α. Then we cancompute things like

V (p2) = −e2R

p2

[1 +

e2R

2π2

∫ 1

0

dxx(1− x) log

(1− p2

m2x(1− x)

)+ · · ·

],


and so if we take the Fourier transform, we get

V (r) = − e2

4πr

(1 +

e2R

6π

∫ 1

0

dxe−2mrx 2x2 + 1

2x4

√x2 − 1 + · · ·

).

If we numerically integrate, we see that

∆E2s1/2= −27MHz, ∆E2p1/2

= 0.

You can also calculate other loops in Feynman diagrams, and then we see that∆E between the two orbitals are 1024MHz. This is in excellent agreement withmeasurements.

Consider the scale p2 m2 and then take the reference scale p20 = −m2. In

this case, we have ∫dxx(1− x) log

(−p2

m2

)=

1

6log

(− p2

m2

)so we have

V (p2) =e2R

p2

(1 +

e2R

12π2log−p2

m2

)=e2R

p2(1 + 0.0077 log

Q2

m2).

This is really telling us that perturbative QED is breaking down when Q is veryvery large, for instance, when 10270eV. If we add all these diagrams involvingone loop and two loops and so on, we see that we have

Π =e2R

p2

(1 +

e2R

12π2log

p2

m2+

(e2R

12π2log

p2

m2

)2

+ · · ·)

=1

p2

(e2R

1− e2R12π2 log −p

2

m2

).

So when we approach Q ≈ 10270eV, also called the Landau pole, we see thatthe energy really is going to infinity. So QED is breaking down at this crazyenergy level. One way to understand this is that there is some screening forvacuum, given by all these diagrams which are like virtual electron-positronpairs. In smaller scales, this effect becomes small so the force becomes stronger.


21 November 13, 2018

Last time we looked at the 1-loop contribution in QED. We didn’t compute it,but today we are going to do this. We had

V (p2) =e2

p2

e4

p2Π2(~p2)→ · · · ,

and we renormalized this to e2R = V (p2

0)p0. Then we were able to predict

p2V (p2) = e2R +

e4R

2π

∫ 1

0

dxx(1− x) logm2 − p2x(1− x)

m2 − p20x(1− x)

which is finite. If you Fourier transform, we get an exponentially decayingcorrection term in the Coulomb potential.

21.1 Examples of renormalization—vacuum polarizationII

Let’s actually compute

Here, we can write this as

−(−ie)2

∫d4k

(2π)4

i

(p− k)2 −m2 + iε

i

k2 −m2 + iεTr[γµ(/k − /p+m)γν(/k +m)]

= −4e2

∫d4k

(2π)4

[−pµkν − kµpν + 2kµkν + gµν(−k2 + pk +m2)]

((p− k)2 −m2 + iε)(k2 −m2 + iε)

= ∆1

(p2

m2

)p2gµν + ∆2

(p2

m2

)pµpν .

Here, by the Ward identity we should get that ∆1 = −∆2. So we can onlycompute ∆1. Then we can drop the first two terms in the numerator, becausethey only contribute to ∆2.

First we are going to regulate the integral, combine denominators, completethe squares to make everything in terms of k2, drop the odd terms, and theWick rotate k0 → ik0 to integrate. To combine the denominators, we use thetrick

1

AB=

∫ 1

0

dx

[A+ (B −A)x]2.

Then for A = (p− k)2 +m2 + iε and B = k2 −m2 + iε, we get

A+ (B −A)x = [k − p(1− x)]2 + p2x(1− x)−m2 + iε.


Now our integral is,∫d4k

(2π)4

∫ 1

0

dxf(k, p)

[[k − p(1− x)]2 + p2x(1− x)−m2 + iε]2

=

∫f(k + p(1− x), p)

[k2 + p2x(1− x)−m2 + iε]2.

Now if we drop all terms that are odd in k, we are left with

iM = −4e2

∫d4k

(2π)4

∫ 1

0

dx2kµkν − gµν [k2 − x(1− x)p2 −m2]

[k2 + p2x(1− x)−m2 + iε]2.

So we should know how to integrate something like∫d4k

(2π)4

1 or k2 or kµkν

(k2 −∆ + iε)n.

Here, we note that the denominator looks like

k2 −∆ + iε = [k0 − (

√~k2 + ∆− iε)][k0 − (−

√~k2 + ∆ + iε)],

and as function of k0, has zeros at a (negative plus small imaginary) and a(positive minus small imaginary). This tells us that if we do a contour along afigure-eight figure, we can change the integral along the real axis to an integralalong the imaginary axis. So when we Wick rotate this becomes

i

∫d4k

(2π)4

1

−k2E −∆)n

= i

∫dΩ4

(2π)4

∫ ∞0

k3EdkE(−1)n

1

(k2E + ∆)n

.

For instance, for n = 3 this integral is −i/32π2∆.For n = 2, it is infinite, and we should expect this because we have to

regulate. Define

I2(∆) =

∫d4k

(2π)4

1

(k2 −∆ + iε)2=

i

8π2

∫ ∞0

k3 dk

(k2 + ∆)2.

Then the derivative is

I ′2(∆) = −2I3(∆) =i

16π2∆, I2(∆) = − i

16π2log

∆

Λ2.

But this only gives a feel for the answer, and we need to be careful because wehave lots of different integrals and be consistent about cutoffs.

We are going to do dimensional regularization, but let me first talk aboutPauli–Villars. Here, we introduce a fictitious ghost particles of mass Λ. Wesay this PV ghost has all the same interactions but with a minus sign. In thiscase,

I(∆) =

∫d4k

(2π)4

[1

(k2 −∆ + iε)2− 1

(k2 − Λ2 + iε)2

]= − i

16π2log

∆

Λ2.

Here we get a convergent integral because the 1/k4 cancels and then the leadingterm is 1/k6. Here, this is a systematic change of the theory. But people don’tuse this anymore.


21.2 Dimensional regularization

The regulator that is used universally is dimensional regularization. Here,the key is that the integral is continuous in less than 4 dimensions. So wecalculate it for d dimensions and analytically continue to d = 4 − ε. We notethat ∫

ddk

(k2 + ∆)2= Ωd

kd−1dk

(k2 + ∆)2,

and this is finite if d < 4.We first note that we have

L = −1

4(∂µAν − ∂νAµ)2 + ψ(iγµ∂µ −m)ψ − eψγµψA.

If we do dimensional analysis, we see that

[A] =d− 2

2, [ψ] =

d− 1

2, [m] = 1, [e] =

4− d2

.

But it is good to have the interaction constants dimensional. So we introducea scale µ of [µ] = 1 and then write

L = −1

4(∂µAν − ∂νAµ)2 + ψ(iγµ∂µ −m)ψ − eψγµψAµ

4−d2 .

Also note that gµνg − µν = tr(g2) = d.Now we can do the integrals. Write∫

ddkkµkνf(k2) = gµνX.

If we contract with gµν , then we get∫ddkk2f(k2) = dX. So we have∫

ddkkµkνf(k2) = gµν1

d

∫k2f(k2).

Also, we have

Ωd =

∫dΩd =

2πd/2

Γ(d/2).

This Γ(x) is the analytic continuation of Γ(x) = (x − 1)! and has poles atx = 0,−1,−2, . . .. At ε, we have Γ(ε) = 1

ε − γE . So if we do the integral, wehave

e2

∫ddk

(k2 −∆ + iε)2= µ4−d i

(4π)d/21

∆2−d/2 Γ

(4− d

2

).

Near d = 4, we will get

i

16π2

[2

ε+ log

µ24πe−γE

∆+O(ε)

].


The final answer is going to be

Π2(p2) =1

2π2

∫ 1

0

dxx(1− x)

[2

ε+ log

µ2

m2 − p2x(1− x)

], µ2 = µ24πe−γE .

If we take the difference Π2(p20) − Π2(p2), both ε and µ2 drop out and we get

the right answer.



We looked at the Pauli–Villars ghost particle, which is physical but not veryuseful, and also the dimensional regularization. The main thing here is to keepthe e dimensionless, and so we introduce a dimensionful µ. At the end, we got

Π2(p2) =1

2π2

∫ 1

0

dxx(1− x)

[2

ε+ log

4πµ2e−γE

m2 − p2x(1− x)+O(ε)

].

Only the difference is meaningful, so we look at

V (p2) =e2

p2− e

4

p2Π2(p2), e2

R = V (p20)p2

0 = e2−e4Π2(p20), e2 = e2

R+e4RΠ2(p2

0).

Then we get

p2V (p2) = e2R + e4

RΠ2(p20)− e4

RΠ2(p2)

= e2R +

e4R

2π

∫ 1

0

dxx(1− x) logm2 − p2x(1− x)

m2 − p20x(1− x)

.

22.1 Examples of renormalization—anomalous magneticmoment

This was done around 1940, and it got people really excited because there wasdata that we can compare theory to. Remember we had this non-relativisticHamiltonian

Hscalar =~p2

2m+V (r)e+

e

2m~B ·~L, Hspinor =

~p2

2m+V (r)e+

e

2m~B ·~L+

e

2mg ~B · ~S.

Here, only g = 2 agrees with Lorentz-invariance, which is the Dirac equation.But is really g = 2? The challenge here is to disentangle the corrections to eand the corrections to g.

Remember that we had the Dirac equation (i /D −m)ψ = 0, and multipliedthis with (i /D +m) to get

(D2µ + e

2Fµνσµν −m2)ψ = 0.

For a scalar, we would only have (D2µ −m2)φ = 0.

How do we interpret this in QED? For the Feynman diagrams in the tree-level, we have, for a scalar,

= −ie(qµ1 + qµ2 )

and for a spinor, we have


= −ieu(q2)γµu(q1).

Here, the Gordon identity gives

uγµu =1

2mu(qµ1 + qµ2 )u− i

2muσµνu(qν1 − qν2 ),

and so we can write the M-matrix as

−ie[qµ1 + qµ2

2muu+

i

2muσµνupν

]εµ.

The first term can be considered as what we have for the scalar particle, becauseusus′ = 2mδss′ and we are looking at the limit q1− q2 → 0. The second term iswhat predicts g = 2.

So suppose we computed the loop correction and got

= −iau(q2)γµu(q1) +e

2mbuσµνupν .

This is making the shift

e→ e+ a, g → g + 2b.

So suppose we calculate this out and get

εµu(q2)Γµu(q1), Γµ = f1γµ + f2p

µ + f3qµ1 + f4q

µ2 .

Then using p = q1 − q2, we can first eliminate f2, and then the Ward identitiestell us that f1/p + f3pq1 + f4pq2 = 0. So we can eliminate f4 and have twoindependent terms. So now we can use the Gordon identity to choose the twoterms as

Γµ = −ie[F1

(p2

m2e

)γµ +

iσµν

2mepνF2

(p2

m2e

)].

This F1 can be interpreted as the scale-dependent electric charge, and F2 givesus the correction to the electron’s magnetic dipole moment. In the tree level,F1 = 1 and F2 = 0 and so g = 2. Then after we calculate the correction, wewill have

g = 2 + 2F2(0).

If we look at the 1-loop correction, there are four diagrams, and the onlydiagram where we can get some σµν between the u is only the following diagram.

= (−ie)3

∫d4k

(2π)4

−igνα

(k − q1)2uγν

i(/p+ /k +m)

(p+ k)2 −m2γµi(/k +m)

k2 −m2γαu


Now first we combine the denominators and complete the square. At the end,we get

iM =

∫d4k

∫dxdydzδ(1− x− y − z) Nµ

(k2 −∆)3, ∆ = −xyp2 + (1− z)2m2,

Nµ = [− 12k

2 + (1− x)(1− y)p2 + (1− 4z + z2)m2]uγµu

+ imz(1− z)pν uσµνu+mz(1− z)(x− y)pµuu.

Actually, the last term in Nµ vanishes after integration, so we can use

Nµ = [− 12k

2 + (1− x)(1− y)p2 + (1− 4z + z2)m2]uγµu+ imz(1− z)pν uσµνu.

This really is as expected.We are only interested in the F2 term, so we don’t really have to care about

the other term. Here, the growth is like 1/k6, so there is no divergence as well.So we get

F2 =2m

e(4ie3m)

∫dxdydzδ(1− x− y − z)

∫d4k

(2π)4

z(1− z)(k2 −∆ + iε)2

· · · .

We can Wick rotate, and then we get

F2(p2) =8e2

32π2m2

∫d3xδ(x+ y + z − 1)

z(1− z)(1− z)2m− xyp2

.

Because g is defined as F2(0), we can integrate

F2(0) =8e2

32π2

∫ 1

0

dz

∫ 1

0

dy

∫ 1

0

dxδ(1− x− y − z) z

1− z

=α

π

∫ 1

0

dz

∫ 1−z

0

dyz

1− z=α

π

∫ 1

0

z =α

2π.

Thereforeg = 2 +

α

π≈ 2.00232.



Today we are going to do mass renormalization. The main renormalization wetalked about was the fermion loop. We defined αR so that αR = limr→∞ rV (r),and then we got

V (r) =αRr

[1 +

αR(mr)3/2

e−mr + · · ·].

Here the renormalization condition was a specification of a parameter in thetheory through some observable. So we can define a renomalizatble theoryas a theory of which a finite number of renormalization conditions are neededto define the theory completely. A theory can be renormalized, even if it is notrenormalizable.

23.1 Renormalizability of QED

We will show that QED is renormalizable. Last time we looked at

= uσµνupµ + uγµu

We saw that σµν was finite, but this other term uγµu is going to be UV-divergent. Similarly, the electron self-energy is UV-divergent.

But there can be other many different graphs we can draw. So we wantto consider showing that all Green’s functions or S-matrix elements can berenormalized in an increasing number of fields/extenral momenta. For 0-pointfunctions, we want

〈Ω|Ω〉 = 1.

But then we draw all diagrams with no external legs, and this in fact is stronglydivergent. But this doesn’t really affect the physics, so we can just set thisto ΛR and really, set this to whatever we want. This is important in generalrelativity because the measure

√g involves the constant.

Now we look at 1-point functions. Actually, all 1-point functions vanish inQED. We could have something like

which is called a tadpole graph, but this is zero because if we have

〈Ω|ψ|Ω〉 6= 0 or 〈Ω|ψ|Ω〉 6= 0

this violates Lorentz invariance. On the other hand, if we have a scalar field,for instance like

L = cφ− 1

2φ(−m2)φ,


this can have nonzero contribution for 1-point functions. We can think of asthere being a quadratic potential V (φ), and so if we shift φ so that φ = 0 isthe minimal energy state, there won’t be such a problem. So these tadpolesindicate instability of the theory. For instance, e−e− = φ2− interaction in somematerials can be attractive. Here, c = 〈0|φ|0〉 6= 0, and then the ground stateshould be charged. If you do this, you see that photon gets a mass, and thenwe get screening of magnetic fields in type II superconductors. This is the BCStheory of superconductivity.

Anyways, let’s now talk about 2-point functions. There is 〈AµAν〉 we talkedabout, and we also have 〈ψψ〉 given by

given by

iΣ2(/p) = (−ie)2

∫d4k

(2π)4γµ

i(/k +m)γµ)

k2 −m2 + iε

−i(p− k)2 + iε

=α

4π

∫ 1

0

dx[(2− ε)x/p− (4− ε)m]

[2

ε+ log

µ2

(1− x)(m2 − p2x)

]=α

π

(/p− 4m

2ε+ finite

).

So now we would like to remove UV divergence by redefining m. If we takem0 as the bare mass, and define m0 = mR + mRδm up to order e2 for mR therenormalized mass, we get

〈ψψ〉 =i

/p−m0+

i

/p−m0[iΣ2(/p)]

i

/p−m0+ · · ·

=i

/p−mR+

i

/p−mR[iΣ2(/p)− iδmmR]

i

/p−mR+ · · · .

To make this finite, we choose δm = −απ42ε .

Here is how you can think about this. We have the bare Lagrangian

L = −1

4F 2µν + iψ0 /∂ψ0 −m0ψ0ψ0 + ie0ψ0 /Aψ0,

and we renormalized this to

ψ0 =√Z2ψR, A0 =

√Z3AR,

for Z2 = 1 + δ2 and Z3 = 1 + δ3. Then the renormalized Lagrangian can bewritten as

L = Z3(− 14 (∂µA

µR−∂νA

νR)2)+iZ2ψR /∂ψR−mR(1+δm)Z2ψRψ+ieR(1+δe)Z2

√Z3ψR /ARψR.


So coming back to the calculation, we can now write

〈ψRψR〉 =1

Z2〈ψ0ψ0〉 =

1

1 + δ2

i

/p−mR − δmmR

=i

/R−mR

+i

/p−mR[i(δ2/p− (δ2 + δm)mR + Σ(/p))]

i

/p−mR+O(e4).

At the end we will have

δ2 =α

2π

1

ε, δm = −3α

2π

1

ε+ finite.

How do we compute this renormalization condition? Recall that the masswas defined from the representation of the Poincare group. We should have〈ψRψR〉 = (i/p + m)/(p2 −m2) has a pole at p2 = m2. It seems that there aretoo many poles, but actually we are adding over many different corrections andthis does shift the pole. So you can get

〈ψRψR〉 =i

/p−mR + ΣR(/p), ΣR(/p) = Σ2(/p) + δ2/p− (δm + δ2)mR +O(e4).

Then the renormalization condition is that there is a pole at /p = mR withresidue i,

〈ψRψR〉 =i

/p−mR+ regular.

This means that ΣR(mR) = 0 and Σ′R(mR) = 0.We can solve this for δ2 and δm. Because we have this explicit form

ΣR(/p) = Σ2(/p) + δ2/p− (δm + δ2)mR,

evaluating this at mR gives

δm =α

2π

(−3

ε− 3

2log

µ2

m2R

− 2

), δ2 = − α

2π

(1

ε+

1

2log

µ2

mR+ 2 +

m2γ

m2R

),

where m2γ is the photon mass in the IR regulator. This is called the on-shell

renormalization. An alternative is minimal subtraction. Here, we drop allfinite contributions to δ2 and δm in dimensional regularization.



We wanted to show that all Green’s functions can be made UV-finite throughrenormalization. We encountered infinities and we redefined the mass m0 =mR(1 + δm) and also ψ0 = 1

Z2ψR where Z2 = 1 + δ2. The finite parts of δ2

and δm are not unique, so we need a convention for defining. This is called thesubtraction scheme or the renormalization scheme. This is the prescriptionfor fixing the finite parts of δs. We used the on-shell scheme, so that mR = mpole

is the location of the pole in the propagator. Once we did that we got

δm =e2

8π

(−3

ε− 4− 3

2log

µ2

m2p

).

The other scheme we used was minimal subtraction, and this means that thefinite part is zero. This is simpler but not physical. Then we get the MS-massmMSR . There is also that MS-bar where the finite part is log e−γE4π. In this

scheme we have

δm =e2

8π2

(−3

ε

).

The differences between them are just finite corrections. Then

m0 = mR(1 + δMSm ) = mp(1 + δOS

m ),

and so they are related by

mR = mp

(1− α

4π

(4 + 3 log

µ2

m2p

)).

For example, the top quark has

mp = 174GeV, mMS = 163GeV.

24.1 1PI and amputation

The other concept we introduced was one-particle irreducibility or just 1PI.A 1PI graph cannot be split into two disconnected graphs by cutting a singleinternal line. The good thing about this is that we have

=1

1− (1PI).

If we can show that all 1PI graphs are finite, then all graphs are finite. So thisreduces the program of showing renormalizability to only 1PI graphs.

In S-matrix elements, external momenta are on-shell. But then, we reallyare including contributions of graphs where the external legs have some bubbles.So the LSZ formula gives

〈p3 · · · pn|p1p2〉 =

∫d4x1 · · · d4xne

ip1x1 · · · e−ipnxn(/p1−mp) · · · (/pn −mp)

× 〈Ω|TψR(x1) · · ·ψR(xn)|Ω〉.


So this is just the sum of the amputated diagrams, diagrams that don’t containbubbles on external legs. Here, we are actually having

ψOSR =

√ZMS

2

ZOS2

ψMSR

and so we really have

〈p3 · · · pn|p1p2〉 = (Amputated)×(√

ZR2ZOS

2

)n.

This is the renormalized version of the LSZ reduction formula, and it is cleanerin some sense.

24.2 Renormalized perturbation theory

We have the bare Lagrangian

L = −1

4(∂µA

0ν − ∂νA0

µ)2 + ψ0(i/∂ − e0 /A0 −m0)ψ0.

Now we renormalize ψ0 =√Z2ψ

R and A0µ =

√Z3A

Rµ with e0 = ZeeR and

m0 = ZmmR. Then

L = −1

4Z3(∂µA

Rν −∂νARµ )2 +iZ2ψR /∂ψR−Z2ZmmRψψ−eRZeZ2

√Z3ψR /A

RψR.

Now we define

ZeZ2

√Z3 = Z1 = 1 + δ1, Z2 = 1 + δ2, Z3 = 1 + δ3, Zm = 1 + δm.

We have that δ are formally of order at least 2 in eR even if they are infinite.So we can do perturbation theory in eR and then write

L = −1

4F 2µν + iψ /∂ψ −mRψψ − eRψ /Aψ

− 1

4δ3F

2µν + iδ2ψ

2 /∂ψ − (δm + δ2 + δmδ2)mRψψ − eRδ1ψ /Aψ

where all these fields are now renormalized. Then we can read off the Feynmanrules for these graphs, and for instance,

= i(/pδ2 − (δm + δ2 + δ2δm)mR).

We can think of this as these loops contracted to a point. We now want tochoose δ so that we cancel out all the infinite parts.

Now let us expand the electron self energy for instance. The we get

i

/p−mR+

i

/p−mRΣ2(/p)

i

/p−mR+

i

/p−mR(i/pδ2 − (δm + δ2)mR)

i

/p−mR.


Similarly, we see that the photon self energy is

−ie2R(p2gµν − pµpν)Π2(p2)− i(p2gµν − pµpν)δ3,

Π2 =1

2π

∫ 1

0

dxx(1− x)

[2

ε+ log

µ2

m2R − p2x(1− x)

].

To see the on-shell renormalization condition, we need to look at the sum of allthe loop corrections and get

igµν

p2(1 + Π(p2)), Π(p2) = e2

RΠ2(p2) + δ3.

This has to look like igµν/p2, so we get

δOS3 = − e2

R

6π2

1

ε− e2

R

12π2log

µ2

m2R

.

So we have dealt with all graph with two external legs. Now let us go tothree legs. We have

G3 = 〈Ω|TψAµψ|Ω〉.

Here, we can look at the 1PI graphs and write

Γµ(p2) = F1(p2)γµ + iσµν

2mpνF2(p2),

F1(p2) = 1 + e2R(· · · ), F2(p2) = 0 +

α

π+O(p2).

The loop graph contributing to this is just the graph we used to correct g − 2,and then we see that we get

F1(p2) = 1− 2ie2R

∫d3xδ(1− x− y − z)

∫d4k

(2π)4

k2 − 2(1− x)(1− y)p2 + · · ·[k2 −m2

R(1− x)2 · · · ].

As p→ 0, we should get the regular QED interaction, so we should have the δ1correction to be cancel out F1, that is,

δ1 = −F (2)1 (0) = − α

2π

(1

ε+

1

2log

µ2

m2R

+ 2 + logm2γ

m2R

).

In fact, this is equal to δ2 and so Z1 = Z2. The reason for this is essentiallycharge conservation.



Today we will prove that QED is renormalizable. We introduced these renor-malizations for the photon self-interaction, the electron self-interaction, and theinteraction term. Then we had

L = −1

4Z3F

2µν + iZ2ψ /∂ψ − eRZ1ψ /Aψ −mRZ2Zmψψ.

Then we had renormalization conditions that say that things look like tree-levelin long distances. This fixes the 1-loop terms.

The reason we have δ1 = δ2 is because of the conservation of charge. Thisimplies that Ze = 1/

√Z3. This is saying that the correction to the interaction

term is equal to the correction to the photon propagator term. So in somesense, the photon interaction is universal. This is because if we a symmetryabout charge, for instance, if we have a quark in our theory and there is thissymmetry

ψ → eiαψ, ψq → e−23 iαψq,

then L is invariant under this symmetry, and this is true for all renormalizations.So even if the interaction terms get renormalized, the symmetries stay the same.

25.1 Renormalizability of QED

So far, we found 3 UV-divergent graphs in QED, and then removed those byZ1, Z2, Zm, Z3. But are there more 1-loop UV-divergences? For 2-point and3-point diagrams, we don’t have anything more. You can worry about threephoton legs, but this is not possible. For 4-point diagrams, we have

∫d4k

1

k2

1

/k

1

k2

1

/k∼ 1

Λ2

and this is UV finite. Similarly, we can compute a similar thing for the diagramwith two external photons and two external legs. Finally, for four externalphotons, we get zero because the divergent term should look like

c log Λεµενεαεβ(gµνgαβ + gµαgνβgµβgνα).

Then the Ward identities show that this is only possible if c = 0. This diagramis still going to be nonzero, but it is finite because the divergent part is zero.More external lines are less UV-divergent, so we don’t have to worry about this.This shows that QED is renormalizable in 1-loop.

What about 2-loops? There can be a 1PI two-loop electron self-interaction,but then we can renormalize this using the terms that are higher-order in eR.You can also realize that adding more loops don’t make things more divergent,because you can do dimensional analysis and see that adding more loops doesn’t


change the order of divergence. So we need to use the same parameters and justcorrect the parameters in higher order.

The formal proof can be found in Weinberg, and the proof is due to BPHZ.To prove renormalizability of the Standard Model, you need take more care, butit is a nice thing to have because this says that you can do a finite number ofmeasurements and then start calculating.

Here is the dimensional analysis part. We can superficially calculate thedegree of divergence D, which makes M∼ ΛD. Indeed, we have

[M] = 4− nb − 32nf = D,

where nb is the number of external bosons and nf is the number of externalfermions. So there is only a finite number of (nb, nf ) such that the superficialdegree of divergence D is nonnegative. This really is all there is to the proof ofrenormalizability of QED.

25.2 Nonrenormalizable theories

QED is special because the only coupling constant eR is dimensionless. But ifthere are other scales in the problem, for instance,

L = ψ /∂ψ +Gψψψψ,

then we have [G] = −2. So how does this screw up our power counting? If wehave a ψψ → ψψ with two Fermi interactions, we get

G2

∫d4k

1

/k/k∼ G2Λ2, [M] = −2.

Moreover, if I have any diagram, I can start adding in edges and then we getUV-divergent graphs. So any small mass dimension −1 interaction screws upthe renormalizability of the theory.

A graph with nj factors of a coupling of dimension ∆j has dimension

D = 4− nb − 32nf −

∑j

nj∆j .

If some coupling has ∆j < 0 then there are an infinite number of UV divergentamplitudes. We call a theory

• renormalizable if we have ∆j ≥ 0 for all j,

• non-renormalizable if we have ∆j < 0 for all j,

• super-renormalizable if we have ∆j > 0 for all j.

As we will see, some non-renormalizable theories can be renormalized, andsuper-renormalizable theories are horrible.

Let’s take a scalar theory

L = −1

2φφ+Gφ2(∂µφ)(∂µφ), [G] = −2


for instance. At the tree level, we just have a interaction iGpiµpjµ and everything

is fine. At 1-loop, we can have things like

G2(c1Λ4 + c2Λ2p2 + c3p4 log Λ + c4Λ4 log p2 + finite).

So we can first renormalize G using GRZG, ZG = 1 + δG. Then we can setδG = c2Λ2G to cancel the c2 term. But then there are three more terms. Forinstance, to cancel out p4 log Λ, we really need something with 4 derivatives. Sowe write

L = −1

2φφ+GRZGφ

2φ2 + κRδRφ22φ+ λRzλφ

4

and set something like

δκ = −G2

κRc3 log Λ, δλ = − 1

λG2Λ4G.

So we get rid of c1, c2, c3. It turns out that c4 actually never arises.

Theorem 25.1. The UV-divergences are always polynomials in external mo-menta.

The Weinberg proves this is to differentiate. We can differentiate with re-spect to external momenta until the integral is convergent, and then integrate.We can look at something like

I(p) =

∫dk

k

k + p∼ Λ

and then integrate this. So we get

I ′(p) ∼ log Λ, I ′′(p) =1

p.

Now if we try to integrate this, we get

I ′(p) = log p+ const = logp

Λ,

I(p) = p logp

Λ− p+ c1Λ = p log p− p log Λ− p+ c1Λ.

So any divergences are only polynomial in p.It’s really important that we don’t get terms like log that are not local.

In non-renormalizable theories,

(1) we must add every local term to the Lagrangian that are consistent withsymmetries,

(2) then we need an infinite number of renormalization conditions to definethe theory,

(3) but the theory is still predictive because the new terms are suppressed atlow energy.

Index

antiparticle, 36anyon, 50

boson, 50

Casmir effect, 63chiral theory, 47Coulomb gauge, 34

d’Alembertian, 6dimensional regularization, 70Dirac Lagrangian, 45Dirac spinor, 45

energy-momentum tensor, 12Euler–Lagrangian equations, 10

fermion, 50Feynman propagator, 19Feynman rule, 24Feynman rules, 26Feynman-–Stueckelberg

interpretation, 37Fock space, 8

gauge invariance, 33Gaussian regulator, 63generating functional, 57Grassmann numbers, 61Green’s function, 14

heat kernel regularization, 63helicity, 47

Klein–Nishna formula, 53

Landau pole, 67Lie algebra, 42lightlike, 6

Lorentz gauge, 8Lorentz transformation, 6LSZ reduction formula, 19

Mott formula, 53

Noether’s theorem, 11

on-shell renormalization, 77one-particle irreducibility, 78orthochronous, 6

particle, 29Pauli–Villars ghost, 69Poincare group, 6Proca Lagrangian, 31projective representation, 45propagator, 14

renormalizability, 75, 82renormalization scheme, 78representation, 30

irreducible, 30

Schwinger–Dyson equations, 22second quantization, 4spacelike, 6spin, 47spinor, 44subtraction scheme, 78

time-ordering operation, 19timelike, 6

Ward identity, 35Wick rotation, 69

ζ-function regulator, 63

84

physics 253a - quantum field theory i - github pages · 2018-12-12 · physics 253a notes 4 1...

Documents