combined

AQT Term 2

Lecturer: Arthur Lipstein

1 Introduction ..... 2
2 Path Integral in QM ..... 2
3 Path integral for free QFT ..... 5
4 Path integral for interacting QFT ..... 7
5 LSZ reduction ..... 12
6 Scattering Amplitudes and Feynman Rules ..... 14
7 Loops and Renormalisation ..... 17
8 Classical Maxwell Theory ..... 22
9 LSZ and Path Integral For Photons ..... 25
10 Dirac Equation ..... 28
11 Classical Solutions of the Dirac Equation ..... 32
12 Quantizing the Dirac field ..... 35
13 LSZ for spin half ..... 39
14 Fermionic Propagator ..... 41
15 Quantum Electrodynamics ..... 44
16 Scattering in QED ..... 47
17 Renormalisation of QED ..... 50

1 Introduction

References

Srednicki, QFT
Tong, lecture notes on QFT
Peskin and Schroeder, Intro to QFT
Kabat, Lecture notes on Particle Physics

Overview

Path integrals
Loops and Renormalization
Spin 1 (Maxwell)
Spin $1 / 2$ (Dirac)
QED

2 Path Integral in QM

Based on Srednicki Chap 6. In the first term you quantised scalar field theory by expanding the field in Fourier modes and promoting the coefficients to creation/annihilation operators. You also learned how to compute correlation functions using Wick's theorem in the interaction picture.

In this term, we will describe a more fundamental approach based on path integrds. This is useful for a number of reasons:

provides a non-perturbative definition of QFT
simplest way to quantise gauge theory
naturally generalises QM

Indeed, from the point of view of path integrals, QM can be thought of as 1d QFT (where

d

-dimensional QFT refers to a QFT living in

d

space-time dimensions). Let us therefore review path integrals in QM.

Consider the following Hamiltonian for a non-relativistic particle:

H (P, Q) = \frac{1}{2 m} P^{2} + V (Q)

where

[Q, P] = ℏ

(set

ℏ = 1

from now on). Let us compute the probability amplitude for the particle to start at position

q^{'}

at time

t^{'}

and end at position

q^{''}

at time

t^{''}

⟨ q^{''} | e^{- i H (t^{''} - t^{'})} | q^{'} ⟩

where

Q | q^{'} ⟩ = q^{'} | q^{'} ⟩

, etc. To do so, let's divide the time interval into

N + 1

pieces of duration

δ t = \frac{t^{''} - t^{'}}{N + 1}

Then introduce

N

complete sets af position eigenstates:

\begin{matrix} (1) & ⟨ q^{''} | e^{- i H (t^{''} - t^{'})} | q^{'} ⟩ = \int \prod_{j = 1}^{N} d q_{j} ⟨ q^{''} | e^{- i H δ t} | q_{N} ⟩ ⟨ q_{N} | e^{- i H δ t} | q_{N - 1} ⟩ \dots ⟨ q_{1} | e^{- i H δ t} | q^{'} ⟩ \end{matrix}

Note thati in the limit

δ t \to 0

e^{- i H δ t} = e^{- i δ t (P^{2} / 2 m)} e^{- i δ t V (Q)} + θ (δ t^{2})

where we used

\exp (A + B) = \exp (A) \exp (B) \exp (- \frac{1}{2} [A, B] \dots)

(Baker-Campbell-Hausdorf formula).
Now focus on one term in (1):

\begin{aligned} (2) & ⟨ q_{2} | e^{- i H δ t} | q_{1} ⟩ & = \int d p_{1} ⟨ q_{2} | e^{- i δ t (P^{2} / 2 m)} | p_{1} ⟩ ⟨ p_{1} | e^{- i δ t V (Q)} | q_{1} ⟩ \\ (3) & = \int d p_{1} e^{- i δ t (p_{1}^{2} / 2 m + V (q_{1}))} ⟨ q_{2} ∣ p_{1} ⟩ ⟨ p_{1} ∣ q_{1} ⟩ \\ (4) & = e^{- i δ t V (q_{1})} \int \frac{d p_{1}}{2 π} \exp [- i \frac{δ t}{2 m} p_{1}^{2} + i p_{1} (q_{2} - q_{1})] \\ (5) & = {(\frac{m}{2 π i δ t})}^{1 / 2} \exp [i δ t (\frac{m}{2} {(\frac{q_{2} - q_{1}}{δ t})}^{2} - V (q_{1}))] \end{aligned}

where we inserted a complete set of momentum eigenstates in the first line and performed the integral over

p_{1}

by completing the square and using

\begin{matrix} (6) & \int_{- \infty}^{\infty} d x e^{- a x^{2}} = \sqrt{π / a} \end{matrix}

Repeating this for all the factors in (1),

⟨ q^{''} | e^{- i H (t^{''} t^{'})} | q^{'} ⟩ = {(\frac{m}{2 π i δ t})}^{\frac{N + 1}{2}} \int \prod_{k = 1}^{N} d q_{k} \exp (\sum_{j = 0}^{N} i [\frac{m}{2} {(\frac{q_{j + 1} - q_{j}}{δ t})}^{2} - V (q_{j})] δ t)

In the continuum limit

N \to \infty

, this gives

⟨ q^{''} | e^{- i H (t^{''} t^{'})} | q^{'} ⟩ = \int D q \exp [i \int_{t^{'}}^{t^{''}} d t L (\dot{q} (t), q (t))]

where

L = \frac{1}{2} m {\dot{q}}^{2} - V (q)

is the Lagrangian and

D q = {(\frac{m}{2 π i s t})}^{\frac{N + 1}{2}} \prod_{k = 1}^{N} d q_{k}

Summary: the amplitude can be computed by integrating over al paths with fixed endpoints, weighted by

e^{i S}

, where

S

is the action of the path!

Now let us consider inserting the operator

Q

time

t_{1}

⟨ q^{''} | e^{- i H (t^{''} - t_{1})} Q e^{- i H (t_{1} - t^{'})} | q^{'} ⟩ = ⟨ q^{''}, t^{''} | Q (t_{1}) | q^{'}, t^{'} ⟩

where

| q^{'}, t^{'} ⟩ = e^{i H t^{'}} | q^{'} ⟩

and

Q (t_{1}) = e^{i H t_{1}} Q e^{- i H t_{1}}

in the Heisenberg picture. Using similar manipulations as before, this will result in a path integral with an insertion of

q (t_{1})

(the location of the particle at time

t_{1}

for each given path):

⟨ q^{''}, t^{'} | Q (t_{1}) | q^{'}, t^{'} ⟩ = \int D q q (t_{1}) e^{i S}

Reversing this logic shows that

\int D q q (t_{1}) q (t_{2}) e^{i S} = ⟨ q^{''}, t^{''} | T Q (t_{1}) Q (t_{2}) | q^{'}, t^{'} ⟩

where

T

is "time-ordering symbol": operators at later times placed to left of operators at earlier times.

Amplitudes with an arbitrary number of time-ordered insertions can then be generated from the following object:

{⟨ q^{''}, t^{''} ∣ q^{'}, t^{'} ⟩}_{f} = \int D q \exp (i \int_{t^{'}}^{t^{''}} d t [L (q; q) + f (t) q (t)])

using "functional derivatives":

\frac{δ}{δ f (t_{2})} f (t_{1}) = δ (t_{1} - t_{2})

In particular,

\begin{aligned} ⟨ q^{''}, t^{''} & | T Q (t_{1}) \dots Q (t_{n}) | q^{'}, t^{'} ⟩ \\ = \int D q e^{i S} q (t_{1}) \dots q (t_{n}) \\ = {\frac{1}{i} \frac{δ}{δ f (t_{1})} \dots \frac{1}{i} \frac{δ}{δ f (t_{n})} ⟨ q^{''}, t^{'} ∣ q^{'}, t^{'} ⟩ |}_{f = 0} \end{aligned}

Finally, note that if we take

t^{''} \to \infty, t^{'} \to - \infty

, this will generically project the external states onto the ground state. To see this, note that

\begin{aligned} | q^{'}, t^{'} ⟩ & = e^{i H t^{'}} | q^{'} ⟩ \\ = lim_{ϵ \to 0^{+}} e^{i (1 - i ϵ) H t^{'}} | q^{'} ⟩ \\ = lim_{ϵ \to 0^{+}} \sum_{n = 0}^{\infty} e^{i (1 - i ϵ) H t^{'}} | n ⟩ ⟨ n ∣ q^{'} ⟩ \\ = lim_{ϵ \to 0^{+}} \sum_{n = 0}^{\infty} e^{i E_{n} t^{'}} e^{i E_{n} t^{'}} ⟨ n ∣ q^{'} ⟩ | n ⟩ \end{aligned}

Hence, if

t^{'} \to - \infty

, the ground state will dominate (assuming

⟨ n ∣ q^{'} ⟩ \neq 0

).
Hence the following object generates correlators with initial and final states given by the ground state:

Z (f) = \int D q \exp [i \int_{- \infty}^{\infty} d t (L + f q)]

In particular, an

n

-point correlator is given by

⟨ 0 | T Q (t_{1}) \dots Q (t_{n}) | 0 ⟩ = {\frac{\frac{1}{i} \frac{δ}{δ f (t_{1})} \dots \frac{1}{i} \frac{δ}{δ f (t_{n})} Z (f)}{Z (f)} |}_{f = 0}

The above two formae naturally generalise to QFT.

3 Path integral for free QFT

This is based on Srednicki Chapter 8. There is a simple generalisation from QM to QFT:

\begin{array}{ll} Q (t) \to \hat{ϕ} (t, \vec{x}) & ( field operator ) \\ q (t) \to ϕ (t, \vec{x}) & ( field configuration ) \\ f (t) \to J (t, \vec{x}) & ( source ) \end{array}

Now the path integral is over all feld configurations on spacetime (which we take to be 4 d ):

Z (J) = \int D ϕ \exp i \int d^{4} x [L (\partial_{μ} ϕ, ϕ) + J (x) ϕ (x)]

where

x

is shorthand for

x^{μ} = (t, \vec{x}), L

is the Lagrangian,

D ϕ \propto \prod_{x} d ϕ (x)

.
Let us first consider free theory:

L_{0} = - \frac{1}{2} \partial^{μ} ϕ \partial_{μ} ϕ - \frac{1}{2} m^{2} ϕ^{2}

In this case, we denote the path integral as

Z_{0} (J)

. To proceed, it is convenient to work in momentum space:

ϕ (x) = \int \frac{d^{4} k}{(2 π)^{4}} e^{i k \cdot x} ϕ (k) ⟺ ϕ (k) = \int d^{4} x e^{- i k \cdot x} ϕ (x)

where

k \cdot x = - k^{0} t + \vec{k} \cdot \vec{x}

. One then finds that

\begin{aligned} S_{0} = \frac{1}{2} \int d^{4} x [- \partial_{μ} ϕ (x) \partial^{n} ϕ (x) - m^{2} ϕ (x)^{2} + 2 J (x) ϕ (x)] \\ = \frac{1}{2} \int d^{4} x \frac{d^{4} k}{(2 π)^{4}} \frac{d^{4} k^{'}}{(2 π)^{4}} \int - (- k \cdot k^{'}) ϕ (k) ϕ (k^{'}) - m^{2} ϕ (k) ϕ (k) + 2 J (k) ϕ (k^{'})] e^{i x \cdot (k + k^{'})} \\ = \frac{1}{2} \int \frac{d^{4} k}{(2 π)^{4}} \frac{d^{4} k^{'}}{(2 π)^{4}} (2 π)^{4} δ (k + k^{'}) [- k^{'} \cdot k^{'} ϕ (k) ϕ (k^{'}) - m^{2} ϕ (k) ϕ (k^{'}) + 2 J (k) ϕ (k^{'})] \\ = \frac{1}{2} \int \frac{d^{4 k}}{(2 π)^{4}} [- ϕ (k) (k^{2} + m^{2}) ϕ (- k) + J (k) ϕ (- k) + J (- k) ϕ (k)] \end{aligned}

Now change path integration variables:

\tilde{ϕ} (k) = ϕ (k) - \frac{J (k)}{k^{2} + m^{2}}, D \tilde{ϕ} = D ϕ

We then obtain a path integral with action

\to S_{0} = \frac{1}{2} \int \frac{d^{4} k}{(2 π)^{4}} [\frac{J (k) J (- k)}{k^{2} + m^{2}} - \tilde{ϕ} (k) (\tilde{k} + m^{2}) \tilde{ϕ} (- k)]

The Integral over

\tilde{ϕ}

just gives overall normalisation

Z_{0} (0)

, so

\frac{Z (J)}{Z (0)} = \exp (\frac{i}{2} \int \frac{d^{4} k}{(2 π)^{4}} \frac{J (k) J (- k)}{k^{2} + m^{2}})

Plugging in

J (k) = \int d^{4} x e^{- i k - x} J (x)

\frac{Z (J)}{Z (0)} = \exp [\frac{1}{2} \int d^{4} x d^{4} x^{'} J (x) Δ (x - x^{'}) J (x^{'})]

where

Δ (x - x^{'}) = \int \frac{d^{4} k}{(2 π)^{4}} \frac{e^{i k \cdot (x - x^{'})}}{k^{2} + m^{2}}

is the propagator.
Note that this is Green's function of Klein-Gordon eqaution:

(- \partial_{x}^{2} + m^{2}) Δ (x - x^{'}) = δ^{4} (x - x^{'})

Actually

Δ (x - x^{'})

is ill-defined because there is a pole in the denominator when

k^{2} = - m^{2}

. Need to specify a prescription for avoiding this pole:

Δ (x - x^{'}) = \int \frac{d^{4} k}{(2 π)^{4}} \frac{e^{i k (x - x^{'})}}{k^{2} + m^{2} - i ϵ}

where

ϵ > 0

. This is known as the Feynman propagator. Other prescriptions are possible, but this gives rise to time ordered propagation, so this is the one we will use (since we are interested in time-ordered path integrals). To see this, perform

k^{0}

integral:

Δ (x - x^{'}) = - \int \frac{d^{3} k}{(2 π)^{3}} e^{i \vec{k} \cdot (\vec{x} - {\vec{x}}^{'})} \int \frac{d k^{0}}{2 π} \frac{e^{- i k^{0} \cdot (t - t^{'})}}{{(k^{0})}^{2} - (ω_{k}^{2} - i ϵ)}, ω_{k} = \sqrt{{\vec{k}}^{2} + m^{2}}

The denominator can be written as

(k^{0} - (ω_{k} - i ϵ)) (k^{0} + ω_{k} - i ϵ)

where we have expnded in

ϵ

and rescaled it by positive numbers. If

t > t^{'}

, then close in lower hall plane so that numentor decays as

| k^{0} | \to \infty

. This picks up
the residue at

k^{0} = ω_{k} - i ϵ

. Similarly if

t < t^{'}

, then close the contour in upper hall plane and pick up residue at

k^{0} = - ω_{k} + i ϵ

. In the end, we obtain

Δ (x - x^{'}) = i θ (t - t^{'}) \int \tilde{d k} e^{i k \cdot (x - x)} + i θ (t^{'} - t) \int \tilde{d k} e^{- i k \cdot (x - x^{'})}

where

\tilde{d k} = \frac{d^{3} k}{(2 π)^{3} 2 ω_{k}}

and

k^{μ} = (ω_{k}, \vec{k})

. Hence, propagation is indeed time ordered.

Now let's compute time-ordered correlators:

⟨ 0 | T ϕ (x_{1}) \dots ϕ (x_{n}) | 0 ⟩ = {\frac{\frac{1}{i} \frac{δ}{δ J (x_{1})} \dots \frac{1}{i} \frac{δ}{δ J (x_{n})} Z_{0} (J)}{Z_{0})} |}_{J = 0}

As an example, consider the 2-point:

\begin{aligned} ⟨ 0 | T ϕ (x_{1}) ϕ (x_{2}) | 0 ⟩ & = {\frac{1}{i} \frac{δ}{δ J (x_{1})} \frac{1}{i} \frac{δ}{δ J (x_{2})} \exp [\frac{i}{2} \int d^{4} x d^{4} x J (x) Δ (x - x^{'}) J (x^{'})] |}_{J = 0} \\ = {\frac{1}{i} \frac{δ}{δ J (x_{1})} (\int d^{4} x^{'} Δ (x_{2} - x^{'}) J (x^{'})) \exp [\dots] |}_{J = 0} \\ = {(\frac{1}{i} Δ (x_{2} - x_{1}) + terms with J) \exp [\dots] |}_{J = 0} \\ = \frac{1}{i} Δ (x_{2} - x_{1}) \equiv \overset{\leftarrow}{ϕ (x_{1}) ϕ} (x_{2}) \end{aligned}

where the brakcet is a Wick contraction. More generally odd-point correlators will vanish since there will always be a left-over

J

in the prefator after taking an odd number of functional derivatives, and even-point correlators will be given by sum over all Wick contractions. This is Wick's theorem, which you proved last term using oscillators. As another example, consider the 4-point correlator:

\begin{aligned} ⟨ 0 | T ϕ (x_{1}) ϕ (x_{2}) ϕ (x_{3}) ϕ (x_{4}) | 0 ⟩ = \tilde{ϕ (x_{1}) ϕ} (x_{2}) ϕ (x_{3}) ϕ (x_{4}) \\ + \overset{ϕ (x_{1}) ϕ (x_{2}) ϕ}{} (x_{3}) ϕ (x_{4}) \\ + \emptyset (x_{1}) ϕ (x_{2}) ϕ (x_{3}) ϕ (x_{4}) \end{aligned}

4 Path integral for interacting QFT

This will be based on Srednicki Chapter 9. Let us now consider an interacting QFT:

L = \underset{L_{0}}{\underset{⏟}{- \frac{1}{2} \partial_{μ} ϕ \partial^{μ} ϕ - \frac{1}{2} m^{2} ϕ^{2}}} \underset{L_{1}}{\underset{⏟}{- \frac{1}{4!} g ϕ^{4}}}

We would like to evaluate the path integral

\begin{aligned} Z (J) & = \int D ϕ \exp [i \int d^{4} x (L_{0} + L_{1} + J ϕ)] \\ = \exp [i \int d^{4} x L_{1} (\frac{1}{i} \frac{δ}{δ J (x)})] Z_{0} (J) \end{aligned}

where

Z_{0} (J) = Z_{0} (0) \exp (\frac{i}{2} \int d^{4} x d^{4} x^{'} J (x) Δ (x - x^{'}) J (x^{'})) .

Taylor expanding exponentials:

\begin{aligned} (7) & Z (J) \propto & \sum_{V = 0}^{\infty} \frac{1}{V!} {[- \frac{i g}{4!} \int d^{4} x {(\frac{1}{i} \frac{δ}{δ J (x)})}^{4}]}^{V} \\ (8) & \times \sum_{P = 0}^{\infty} \frac{1}{P!} {[\frac{i}{2} \int d^{y} y^{4} z J (y) Δ (y - z) J (z)]}^{P} \end{aligned}

where we will not worry about normalisation for now.
We can visualise the terms in this sum using Feynman diagrams:

A term in the sum with a particular value of

V

and

P

corresponds to a diagram with

V

vertices and

P

propagators. The number of surviving sources after taking all the functional derivatives is

E = 2 P - 4 V

Diagrams with

E = 0

are known as vacuum bubbles:

E = 0, V = 1 E = 0, V = 2

There are no diagrams with an odd number of external sources. Here are some examples for

E = 2

E = 2, V = 0

E = 2, V = 1

And some examples for

E = 4

+

E = 4, V = 1

E = 4, V = 1

ง

E = 4, V = 0

Hence, the sum in (8) can be expressed as a sum over Feynman diagrams. The examples given above are connected diagrams, but the sum also contains disconnected diagrams, which can be expressed as products of connected ones:

E = 4, V = 0

E = 4, V = 1

We will soon see that

Z (J)

can be expressed in terms of the exponential of the sum of connected diagrams, so let's focus on those for now.

Question: What is the coefficient of a given connected diagram appearing in the sum in (8)? First note that

Z (J) \propto \sum_{V = 0}^{\infty} \frac{1}{V!} {[- \frac{i g}{4!} \int d^{4} x {(\frac{1}{i} \frac{δ}{δ J (x)})}^{4}]}^{V} \exp (\frac{1}{2} \cdot .)

Now look at some examples:

\begin{aligned} V = 0, P = 1 : \frac{1}{2} \\ V = 0, P = 2 : \frac{- i g}{4!} \int d^{4} x {(\frac{1}{i} \frac{δ}{δ J (x)})}^{4} \frac{1}{2} {(\frac{1}{2} \dots)}^{2} \\ = (- i g) \frac{1}{4!} {(\frac{1}{2})}^{3} \int d^{4} x {(\frac{1}{i} \frac{δ}{δ J (x)})}^{3} 4 (\sqrt{x}) + 2 {(\frac{1}{x})}^{2}) \\ = (- i g) \frac{1}{4!} {(\frac{1}{2})}^{3} 4 \int d^{4} x {(\frac{1}{i} \frac{δ}{δ J (x)})}^{2} (2) \\ = (- i g) \frac{1}{4!} {(\frac{1}{2})}^{3} 4 \int d^{4} x \frac{1}{i} \frac{δ}{δ J (x)} (2) \\ = \frac{1}{8} \end{aligned}

In general, the ceofficient of a connected diagram corresponds to

1 / S

, where

S

is the symmetry factor of the diagram. The two-point diagram

has symmetry factor

S = 2

because it is invariant under exchange of the two sources. Moreover, the vacuum diagram

has a symmetry factor 2 coming from the reflection symmetry of each bubble and another factor of 2 coming the freedom to exchange the two bubbles, giving a total symmetry factor of

S = 2^{3} = 8

Using similar reasoning you can read of the following symmetry factors:

s = 2^{2} = 4

S = 4! = 24

S = 2^{4} = 16

S = 2 \times 3! = 12

If in doubt, you can always compute

S^{- 1}

by brute force by computing functional derivatives.

Let us return to the sum over all diaghams in

Z (J)

. Let

C_{I}

stand for a particular connected diagram

I

, including its inverse symmetry factor. A general diagram

D

can then be expressed as

D = \frac{1}{S_{D}} Π_{I} {(C_{I})}^{n_{I}}

where

I / S_{D}

is an additional symmetry factor and the product is over distinct connected diagrams . If there are

n_{I}

copies of diagram

I

, then there is a symmerry factor of

n_{I}

! associated with exchanging all the copies. Hence

S_{D} = Π_{I} n_{I}!

Now we have

\begin{aligned} Z (J) & \propto \sum_{{n_{I}}} D \\ = \sum_{{n_{I}}} \prod_{I} \frac{1}{n_{I}!} {(C_{I})}^{n_{I}} \\ = \prod_{I} \sum_{n_{I} = 0}^{\infty} \frac{1}{n_{I}!} {(C_{I})}^{n_{I}} \\ = \prod_{I} \exp (C_{I}) \\ = \exp (\sum_{I} C_{I}) \end{aligned}

where the sum over

{n_{I}}

denotes the sum over sets of diagrams with

n_{I}

copies of each distinct connected diagram

I

In summary,

Z (J)

is proportional to the sum of connected diagrams. It follows that

Z (0)

is proportional to the sum of connected vacuum diagrams (those with no sources). Hence,

\frac{Z (J)}{Z (0)} = \exp [i W (J)]

where

i W (J)

is the sum of connected non-vacuum diagrams including inverse symmetry factors. In practice,

i W (J)

is the object of interest for computing scattering amplitudes. Before we explain how to compute scattering amplitudes using Feynman diagrams, let's review how to extract them from correlators.

5 LSZ reduction

This is based on Srednick Chapter 5). Let us review how to extract scattering amplitudes from correlation functions. First recall that

ϕ (x) = \int \tilde{d k} (a (\vec{k}) e^{i k \cdot x} + a^{†} (\vec{k}) e^{- i k \cdot x})

where

k \cdot x = - ω_{k} t + \vec{k} \cdot \vec{x}, ω_{k} = \sqrt{{\vec{k}}^{2} + m^{2}}

, and

\tilde{d k} = \frac{d^{3} k}{(2 π)^{3} 2 ω_{k}}

. It follows that

\begin{aligned} \int d^{3} x e^{- i k \cdot x} ϕ (x) & = \int {\tilde{d k}}^{'} d^{3} x [e^{i t (ω_{k} - ω_{k^{'}})} e^{i ({\vec{k}}^{'} - \vec{k}) \cdot \vec{x}} a ({\vec{k}}^{'}) + e^{i t (ω_{k} + ω_{k^{'}})} e^{- i ({\vec{k}}^{'} + \vec{k}) \cdot \vec{x}} a^{†} ({\vec{k}}^{'})] \\ = \int {\tilde{d k}}^{'} [(2 π)^{3} δ^{3} (\vec{k} - {\vec{k}}^{'}) e^{i t (ω_{k} - ω_{k^{'}})} a ({\vec{k}}^{'}) + (2 π)^{3} δ^{3} (\vec{k} + {\vec{k}}^{'}) e^{i t (ω_{k} + ω_{k^{'}})} a^{†} ({\vec{k}}^{'})] \\ = \frac{1}{2 ω_{k}} (a (\vec{k}) + e^{2 i ω_{k} t} a^{†} (- \vec{k})) \end{aligned}

Similarly,

\begin{aligned} \int d^{3} x e^{- i k \cdot x} \partial_{t} ϕ (x) = \frac{1}{2 ω_{k}} ((- i ω_{k}) a (\vec{k}) + (i ω_{k}) e^{2 i ω_{k} t} a^{†} (- \vec{k})) \\ = - \frac{i}{2} a (\vec{k}) + \frac{i}{2} e^{2 i ω_{k} t} a^{†} (- \vec{k}) \end{aligned}

Hence,

\begin{aligned} a (\vec{k}) & = \int d^{3} x e^{- i k x} (i \partial_{t} ϕ (x) + ω_{k} ϕ (x)) \\ = i \int d^{3} x e^{- i k \cdot x} {\overset{\leftrightarrow}{\partial}}_{t} ϕ (x), f {\overset{\leftrightarrow}{\partial}}_{μ} g = f \partial_{μ} g - \partial_{μ} f g \\ \to a^{†} (\vec{k}) = - i \int d^{3} x e^{i k \cdot x} {\overset{\leftrightarrow}{\partial}}_{t} ϕ (x) \end{aligned}

In the interacting theory,

(a, a^{†})

become time-dependent. Assume that particles are non-interacting at

t = \pm \infty

. Suppose we have two incoming and two outgoing particles:
Incoming state:

| i ⟩ = lim_{t \to - \infty} a^{†} ({\vec{k}}_{1}, t) a^{†} ({\vec{k}}_{2}, t) | 0 ⟩

Final state:

| f ⟩ = lim_{t \to + \infty} a^{†} ({\vec{k}}_{3}, t) a^{†} ({\vec{k}}_{4}, t) | 0 ⟩

The scattering amplitude is then given by:

\begin{aligned} ⟨ f ∣ i ⟩ & = ⟨ 0 | a ({\vec{k}}_{4}, \infty) a ({\vec{k}}_{3}, \infty) a^{†} ({\vec{k}}_{2}, - \infty) a^{†} ({\vec{k}}_{1}, - \infty) | 0 ⟩ \\ = ⟨ 0 | T a ({\vec{k}}_{4}, \infty) \dots a^{†} ({\vec{k}}_{1}, - \infty) | 0 ⟩ \end{aligned}

where we trivially inserted

T

. Now compute

\begin{aligned} a^{†} (\vec{k}, \infty) - a^{†} (\vec{k}, - \infty) = \int_{- \infty}^{\infty} d t \partial_{t} a^{†} (\vec{k}, t) \\ = - i \int d^{4} x \partial_{t} (e^{i k \cdot x} {\overset{\leftrightarrow}{\partial}}_{t} ϕ (x)) \\ = - i \int d^{4} x e^{i k \cdot x} (\partial_{t}^{2} + w_{k}^{2}) ϕ (x) \\ = - i \int d^{4} x e^{i k \cdot x} (\partial_{t}^{2} - {\overset{\leftarrow}{\nabla}}^{2} + m^{2}) ϕ (x) \\ = - i \int d^{4} x c^{i k \cdot x} (- \partial^{2} + m^{2}) ϕ (x), \partial^{2} = - \partial_{t}^{2} + {\vec{\nabla}}^{2} \\ ∴ a^{†} (\vec{k}, - \infty) = a^{†} (\vec{k}, + \infty) + i \int d^{4} x e^{i k \cdot x} (- \partial^{2} + m^{2}) ϕ (x) \\ a (\vec{k}, + \infty) = a (\vec{k}, - \infty) + i \int d^{4} x e^{- i k \cdot x} (- \partial^{2} + m^{2}) ϕ (x) \end{aligned}

If we now plug these formulae into scattering amplitude, we see that time ordering moves

a^{†} (k, + \infty)

to the left annihilating

⟨ 0 |

and

a (\vec{k}, - \infty)

to the right
annihilating

| 0 ⟩

. Hence we obtain

\begin{aligned} ⟨ f ∣ i ⟩ = (i)^{4} \int & d^{4} x_{1} e^{i k_{1} \cdot x_{1}} (- \partial_{1}^{2} + m^{2}) d^{4} x_{2} e^{i k_{1} \cdot x_{1}} (- \partial_{2}^{2} + m^{2}) \\ d^{4} x_{3} e^{- i k_{1} \cdot x_{1}} (- \partial_{3}^{2} + m^{2}) d^{4} x_{4} e^{- i k_{4} \cdot x_{4}} (- \partial_{4}^{2} + m^{2}) ⟨ 0 | T ϕ (x_{1}) \dots ϕ (x_{4}) | 0 ⟩ \end{aligned}

This is the LSZ reduction formula. Recalling that

(- \partial_{x}^{2} + m^{2}) Δ (x - y) = δ^{4} (x - y)

the diff ops trivialise or "amputate" the external props of the Feynman diagrams which contribute to the amplitude. Moreover, the integrals Fourier transform the corrdator to momentum space, where external momenta are on-shell, i.e.

k^{2} = - m^{2}

Summary: Compute scattering amplitudes by Fourier transforming correlator to momentum space, amputating external legs, and putting them on-shell. In practice, one can also allow the legs to be off-shell (tor example when renormalising). We will refer to such quantities as amputated correlators.

6 Scattering Amplitudes and Feynman Rules

This is based on Srednicki Chapter 10 and Peskin section 4.6. Let us apply the formalism developed so far to compute some scattering amplitudes. The first step is to compute a correlation function by taking functional derivatives of the generating functional. In general, this will give connected diagrams as well as disconnected diagrams (i.e. products of connected ones), but we only want to consider connected diagrams. To isolate the connected diagrams recall that

\frac{Z (J)}{Z (0)} = \exp (i W (J))

where

i W (J)

is the sum of al connected diagrams with external sources (no vacuum bubbles). Hence, we will focus on connected correlators:

⟨ 0 | T ϕ (x_{1}) \dots ϕ (x_{n}) | 0 ⟩_{c} = {\frac{1}{i} \frac{δ}{δ J (x_{1})} \dots \frac{1}{i} \frac{δ}{δ J (x_{n})} i W (J) |}_{J = 0}

We then apply the LSZ reduction formula to extract the amplitude.
As an example, let's compute a 3 -point scattering amplitude in

ϕ^{3}

theory:

L = - \frac{1}{2} \partial^{μ} ϕ \partial_{μ} ϕ - \frac{1}{2} m^{2} ϕ^{2} - \frac{1}{3!} \partial ϕ^{3}

At 3 points, the following diagram contributes to

i W (J)

at leading order in order in

g

We then have

\begin{aligned} ⟨ 0 | T ϕ (x_{1}) ϕ (x_{2}) ϕ (x_{3}) | 0 ⟩_{c} = {\frac{1}{i} \frac{1}{δ J (x_{1})} \frac{1}{i} \frac{δ}{δ J (x_{2})} \frac{1}{i} \frac{δ}{δ J (x_{3})} i W (J) |}_{J = 0} \\ = 3! \frac{1}{3!} (- i g) \int d^{4} y {(\frac{1}{i})}^{3} Δ (x_{1} - y) Δ (x_{2} - y) Δ (x_{3} - y) \\ = \frac{- i g}{i^{3}} \int d^{4} y Δ (x_{1} - y) Δ (x_{2} - y) Δ (x_{3} - y) \end{aligned}

where the factor of 3 ! arises from the functional derivatives.
Now suppose that legs 1 and 2 are incoming and leg 3 is outgoing. LSZ then gives

⟨ 3 ∣ 1, 2 ⟩ =

\begin{aligned} i^{3} \int d^{4} x_{1} d^{4} x_{2} d^{4} x_{3} e^{i (k_{1} \cdot x_{1} + k_{2} \cdot x_{2} - k_{3} \cdot x_{3})} \\ \times (- \partial_{1}^{2} + m^{2}) (- \partial_{2}^{2} + m^{2}) (- \partial_{3}^{2} + m^{2}) ⟨ 0 | T ϕ (x_{1}) ϕ (x_{2}) ϕ (x_{3}) | 0 ⟩_{c} \\ = - i g \int d^{4} x_{1} d^{4} x_{2} d^{4} x_{3} d^{4} y δ^{4} (x_{1} - y) δ^{4} (x_{2} - y) δ^{4} (x_{3} - y) e^{i (k_{1} \cdot x_{1} + k_{2} \cdot x_{2} - k_{3} \cdot x_{3})} \\ = - i g \int d^{4} y e^{i (k_{1} + k_{2} - k_{3}) \cdot y} \\ = - i g (2 π)^{4} δ^{4} (k_{1} + k_{2} - k_{3}) \end{aligned}

In general we will always get a delta function imposing momentum conservation so we will drop it:

\begin{aligned} ⟨ f ∣ i ⟩ = (2 π)^{4} g^{4} (k_{in} - k_{out}) i M \\ \to i M_{3} = - i g \end{aligned}

Note that this can be obtained from the following Feynman rule:

Now consider a 4-point amplitude. First we must compute

⟨ 0 | T ϕ (x_{1}) \dots ϕ (x_{4}) | 0 ⟩_{c} = {\frac{1}{i} \frac{δ}{δ J (x_{1})} \dots \frac{1}{i} \frac{δ}{δ J (x_{4})} i W (J) |}_{J = 0}

At leading order, this will come from acting with functional derivatives on the following connected diagram with four external sources:

i W (J) = \frac{1}{8}

There are 4 ! ways to act with the

4 δ / δ J

on the 4 sources. These 24 diagrams can be collected into 3 groups of 8 identical diagrams.

This factor of 8 cancels the symmetry factor. We then obtain

\begin{aligned} ⟨ 0 | T ϕ (x_{1}) \dots ϕ (x_{4}) | 0 ⟩_{c} = (- i g)^{2} {(\frac{1}{i})}^{5} \int d^{4} y d^{4} z Δ (y - z) \\ Δ (x_{1} - y) Δ (x_{2} - y) Δ (x_{3} - z) Δ (x_{4} - z) \\ Δ (x_{1} - y) Δ (x_{2} - z) Δ (x_{3} - z) Δ (x_{4} - y) \\ Δ (x_{1} - y) Δ (x_{2} - z) Δ (x_{3} - y) Δ (x_{4} - z) \end{aligned}

Take legs 1,2 incoming and 3,4 outgoing and apply LSZ:

\begin{aligned} ⟨ f ∣ i ⟩ = & i^{4} \frac{1}{i^{5}} (- i g)^{2} \int d^{4} y d^{4} z Δ (y - z) \\ [\exp (i (k_{1} + k_{2}) \cdot y - i (k_{3} + k_{4}) \cdot z) \\ + \exp (i (k_{1} - k_{4}) \cdot y + i (k_{2} - k_{3}) \cdot z) \\ + \exp (i (k_{1} - k_{3}) \cdot y + i (k_{2} - k_{4}) \cdot z)] \end{aligned}

Now plug in

Δ (y - z) = \int \frac{d^{4} k}{(2)^{4}} \frac{e^{i k \cdot (y - z)}}{k^{2} + m^{2}}

. Then we obtain

\begin{aligned} ⟨ f ∣ i ⟩ = (- i g)^{2} (2 π)^{4} δ^{4} (k_{1} + k_{2} - k_{3} - k_{4}) [\frac{- i}{{(k_{1} + k_{2})}^{2} + m^{2}} + \frac{- i}{{(k_{1} - k_{4})}^{2} + m^{2}} + \frac{- i}{{(k_{1} - k_{3})}^{2} + m^{2}}] \\ \to i M_{4} = (- i g)^{2} [\frac{- i}{{(k_{1} + k_{2})}^{2} + m^{2}} + \frac{- i}{{(k_{1} - k_{4})}^{2} + m^{2}} + \frac{- i}{{(k_{1} - k_{3})}^{2} + m^{2}}] \end{aligned}

This could have been directly calculated from the following Feynman diagrams in momentum space:

where arrows indicate momentum flow, we associate

- i g

to each vertex, and use the following Feynman rule for internal lines:

For external lines, we simply have a factor of 1 because of LSZ. We also impose momentum conservation at every vertex.

In general, we can compute any scattering amplitude by drawing all connected Fexnman diagrams contributing up to a given order in the coupling and evaluating each diagram using the rules above. In practice we can deduce the Feynman rules more directly using Wick's theorem in the interaction picture:

where we expanded

e^{- i H_{int}}

O (g)

and ignore spacetime location of the fields. We also introduced the rule

\hat{ϕ} | \bar{p} ⟩ = 1

which follows from LSZ. This corresponds to an external line of a Feynman diagram. Contractions of

ϕ

's give internal lines. We first contracted

⟨ 3 |

with one of the three

ϕ

's giving a factor of 3 , then contracted

| 2 ⟩

with one of the 2 ramaining

ϕ

's giving a factor of 2 , and then contracted

| 1 ⟩

with the last remaing

ϕ

. For derivative interactions, take

\partial_{μ} \to \pm i k_{μ}

when acting on an incoming/outgoing leg with momertun

k_{μ}

. Contractions giving internal lines are replaced by the momentum space propagator given above.

7 Loops and Renormalisation

This is based on Srednicki chapters 14, 29 and Kabat section 7.2.1. So far we have only talked about tree-level Feynman diagrams. Now let's discuss looplevel diagrams, ie. diagrams with closed loops. Tree-level diagrams encode classical physics, while loop-level diagrams encode quantum corrections (this
can be seen by restoring

ℏ

).
To compute loop-level diagrams, we can use the Feynman rues described before, along with two more rules:

for each closed loop, there will be an unfixed internal momentum $l^{μ}$ flowing through the loop. Integrate each loop momatum with measure $d^{4} l / (2 π)^{4}$ .
Divide by the symmetry factor associated with exchanging only internal lines.

Let's illustrate how this works at 1-loop 4-point in

ϕ^{4}

theory:

L = - \frac{1}{2} \partial_{μ} ϕ \partial^{u} ϕ - \frac{1}{2} m^{2} ϕ^{2} - \frac{1}{4!} g ϕ^{4}

Take legs 1,2 incoming and legs 3,4 outgoing:

We will refer to the the three diagrams on the right-hand-side as the s , t , and u channels, respectively. Let's compute the first diagram using Wick contractions:

i M_{(s)}^{1 - loop} = ⟨ 3, 4 | \frac{1}{2} (\frac{- i g}{4!} ϕ_{x}^{4}) (\frac{- i g}{4!} ϕ_{y}^{4}) | 1, 2 ⟩

where we've expanded

\exp (- i H_{int})

O (g^{2})

and the subscripts

x, y

simply remind us that the vertices are at different locations. Performing the Wick contractions gives

where we first contracted leg 1 with one the four

ϕ_{y}

's, giving a factor of 4 as well as a factor of 2 because we could have also contracted it onto one of the

ϕ_{x}

's. Then we contracted leg 2 with one of the three remaining

ϕ_{y}

's giving a factor of 3 (if we contracted leg 2 onto a different vertex than leg 1 , this would give the second or third diagram). Then we contracted leg 4 with one of the four

ϕ_{x}

's giving a factor of 4 , and leg 3 with one of the remaining

3 ϕ_{x}

's giving a factor of 3 . Finally, we contracted a

ϕ_{x}

with one of the two remaining

ϕ_{y}

's giving
a factor of 2 , and contracted the remaining

ϕ_{x}

with the remaing

ϕ_{y}

. The two internal contractions give two internal propagators which form a closed loop.

Now assign loop momenta to the diagram:

We then find that

M_{(s)}^{1 - loop} = \frac{1}{2} (- i g)^{2} \int \frac{d^{4} l}{(2 π)^{4}} \frac{- i}{l^{2} + m^{2} - i ϵ} \frac{- i}{{(l + k_{1} + k_{2})}^{2} + m^{2} - i ϵ}

where the factor of

\frac{1}{2}

is a symmetry factor associated with exchanging internal lines. Note that the integral is divergent. This can be seen by taking the

l \to \infty

limit of the integrand. In general if a loop integral goes like

\int d^{4} l l^{D - 4}

and

D \geq 0

, then the "superficial degree of divergence" is

D

. For example, if

D = 2

then the integral will be quadratically divergent. In the present example

D = 0

, which corresponds to a logarithmic divergence. These divergences must be regulated. A standard regulator is dimensional regularisation where we take spacetime dimension

d

to be

d = 4 - ϵ

. In this course, we will use a cut-off on the loop momentum for pedagogical reasons. Let's see how this works.

First introduce "Feynman parameters":

\frac{1}{A B} = \int_{0}^{1} d x \frac{1}{(x A + (1 - x) B)^{2}}

This can be used to write the loop integral as follows:

\begin{aligned} \int \frac{d^{4} l}{(l^{2} + m^{2}) ({(l + k_{1} + k_{2})}^{2} + m^{2})} & = \int d^{4} l \int_{0}^{1} d x {[x ({(l + k_{1} + k_{2})}^{2} + m^{2}) + (1 - x) (l^{2} + m^{2})]}^{- 2} \\ = \int_{0}^{1} d x \int d^{4} q {(q^{2} + M^{2})}^{- 2} \end{aligned}

where

q = l + x (k_{1} + k_{2})

and

M^{2} = x (1 - x) {(k_{1} + k_{2})}^{2} + m^{2}

. Next, Wick rotate to Euclidean signature. Let

q^{μ} = (i {\bar{q}}^{0}, {\bar{q}}^{1}, {\bar{q}}^{2}, {\bar{q}}^{3}) \to q^{2} = {({\bar{q}}^{0})}^{2} + (\bar{q})^{2} + {({\bar{q}}^{2})}^{2} +

{({\bar{q}}^{3})}^{2}

. The the loop integral can be written as

i \int_{0}^{1} d x \int d^{4} \bar{q} {({\bar{q}}^{2} + M^{2})}^{- 2}

We can then foliate

R^{4}

with 3 -spheres and write

d^{4} \bar{q} = 2 π^{2} {\bar{q}}^{3} d \bar{q}

, where

2 π^{2}

is the area of a unit 3 -sphere and

\bar{q}

is the radius. In summary, we find that

\int \frac{d^{4} l}{(l^{2} + m^{2}) ({(l + k_{1} + k_{2})}^{2} + m^{2})} = 2 π^{2} i \int_{0}^{1} d x \int_{0}^{Λ} d \bar{q} {\bar{q}}^{3} {({\bar{q}}^{2} + M^{2})}^{- 2}

where we have introduced a cutoff

Λ

. The integral over

\bar{q}

is given by

\int_{0}^{Λ} d \bar{q} {\bar{q}}^{3} {({\bar{q}}^{2} + M^{2})}^{- 2} = \frac{1}{2} [\ln (\frac{Λ^{2} + M^{2}}{M^{2}}) - \frac{Λ^{2}}{Λ^{2} + M^{2}}] \sim \frac{1}{2} (\ln (\frac{Λ^{2}}{M^{2}}) - 1)

, for

Λ^{2} ≫ M^{2}

Including the tree-level contribution, the total 4-point amplitude is depicted below

and is given by

i M_{4} = - i g + \frac{i ℏ}{32 π^{2}} \int_{0}^{1} d x [\ln (\frac{Λ^{2}}{M^{2} (s)}) + \ln (\frac{Λ^{2}}{M^{2} (t)}) + \ln (\frac{Λ^{2}}{M^{2} (u)})]

where we have restored

ℏ

, and

\begin{aligned} M^{2} (s) = x (1 - x) s + m^{2} \\ s = {(k_{1} + k_{2})}^{2}, t = {(k_{1} - k_{4})}^{2}, u = {(k_{1} - k_{3})}^{2} \end{aligned}

where

s, t, u

are known as Mandelstam variables.
From the point of view of the path integral, we can think of

Λ

as a cutoff above which we integrate out all the higher energy modes of the field (you will explore this in the HW). Hence, we should think of this as describing an effective field theory of modes with energy below the cut-off. For low energy observers, the amplitude must be independent of the cutoff. If we allow

g

to depend on

Λ

, we then have

\begin{aligned} 0 & = \frac{d}{d \ln Λ} (i M_{4}) \\ = - i \frac{d g}{d \ln Λ} + \frac{d}{d \ln Λ} (\frac{i ℏ g^{2}}{16 π^{2}} 3 \ln Λ + \dots) + O (ℏ^{2}) \\ = - i \frac{d g}{d \ln Λ} + \frac{3 i ℏ g^{2}}{16 π^{2}} + O (ℏ^{2}) \end{aligned}

where we neglect

ℏ \underset{O (ℏ)}{\underset{⏟}{d g / d \ln Λ}} = O (ℏ^{2})

. From this we compute the beta function

β (g) \equiv \frac{d g}{d \ln Λ} = \frac{3 g^{2}}{16 π^{2}}

where we set

ℏ = 1

.
The solution to this differential equation is given by

\frac{1}{g (Λ)} = \frac{1}{g (μ)} - \frac{3}{16 π^{2}} \ln (\frac{Λ}{μ})

where

1 / g (μ)

is an integration constant,

g (Λ)

is the bare coupling,

g (μ)

is the renormalised coupling (measured by an experiment), and

μ

is the renormalisation scale (in practice, this is the energy scale of the experiment). This formula tells us how the coupling

g (Λ)

has to change in order to compensate for a change in

Λ

. For each value of

g (μ)

, there is a family of EFT's related by "renormalisation group flow" whose physical consequences are the same:

Note that

g (Λ)

diverges at

Λ_{max} = μ \exp (16 π^{2} / 3 g (μ))

. This is known as a "Landau pole". Some new physics has to kick in before the scale

Λ_{max}

. Equivalently, it we hold

Λ

and

g (Λ)

fixed, then we see that renormalised coupling

g (μ)

must change as we vary renormalisation scale

μ

. This is referred to as a "running coupling." At high energy, the theory becomes strongly coupled so perturbation theory not reliable.

Let us express the amplitude in terms of

g (μ)

. First note that

\begin{aligned} g (Λ) & = {(\frac{1}{g (μ)} - \frac{3 ℏ}{16 π^{2}} \ln (\frac{Λ}{μ}))}^{- 1} \\ = g (μ) {(1 - \frac{3 ℏ g (μ)}{16 π^{2}} \ln (\frac{Λ}{μ}))}^{- 1} \\ = g (μ) + \frac{3 ℏ g (μ)^{2}}{16 π^{2}} \ln (\frac{Λ}{μ}) + O (ℏ^{2}) \end{aligned}

Plugging this back into amplitude and dropping terms of

O (ℏ^{2})

then gives

i M_{4} = - i g (μ) + \frac{i ℏ g (μ)^{2}}{32 π^{2}} \int_{0}^{1} d x [\ln (\frac{μ^{2}}{M^{2} (s)}) + \ln (\frac{μ^{2}}{M^{2} (t)}) + \ln (\frac{μ^{2}}{M^{2} (μ)}) - 3]

Hence, the

Λ

-dependence cancels out and the amplitude is explicitly finite. It has been "renormalised."

Let us now consider more general

β

function:

\frac{d g}{d \ln λ} = β (g)

Suppose that

β (g) = - c g^{2}

, where

c > 0

. Then

\frac{1}{g (Λ)} = \frac{1}{g (μ)} + c \ln (\frac{Λ}{μ})

We plot the solution below:

Hence, coupling goes to zero at high energies and becomes strong at low energies. This is known as "asymptotic freedom."

Now suppose that

β (g_{*}) = 0

for some particular value

g_{*}

. Then

g_{*}

is a "fixed point" because if the coupling starts at

g_{*}

, it will stay there as we vary

Λ

since

{\frac{d g}{d Λ} |}_{g = g_{*}} = {\frac{1}{Λ} \frac{d g}{d \ln Λ} |}_{g = g_{*}} = \frac{1}{Λ} β (g_{*}) = 0

β^{'} (g_{*}) < 0

, then

g

is driven to

g_{*}

Λ

increases, correspponding to "ultraviolet (UV) fixed point." If

β^{'} (g_{*}) > 0

, then

g

is driven away from

g_{*}

Λ

increases, corresponding to an "infrared (IR) fixed point" (see HW). Such fixed points can describe phase transitions and quantum gravity via AdS/CFT.

8 Classical Maxwell Theory

This is based on Srednicki Chapters 54, 55. Let us recall Maxwell's equations:

\begin{aligned} \vec{\nabla} \cdot \vec{E} = ρ, \vec{\nabla} \times \vec{B} - \partial_{t} \vec{E} = \vec{J} \\ \vec{\nabla} \cdot \vec{B} = 0, \vec{\nabla} \times \vec{E} + \partial_{t} \vec{B} = 0 \end{aligned}

where

ρ

is the charge density and

\vec{J}

is the current density. Note that

\vec{\nabla} \cdot \vec{B} = 0 \to \vec{B} = \vec{\nabla} \times \vec{A}

where

\vec{A}

is some vector potential. Plugging this into the other homogeneous equation gives

\begin{aligned} 0 & = \vec{\nabla} \times \vec{E} + \partial_{t} \vec{B} \\ = \vec{\nabla} \times (\vec{E} + \partial_{t} \vec{A}) \to \vec{E} + \partial_{t} \vec{A} = - \vec{\nabla} ϕ \to \vec{E} = - \partial_{t} \vec{A} - \vec{\nabla} ϕ \end{aligned}

where

ϕ

is some scalar potential.
The scalar and vector potential can be combined into a 4 -vector called a gauge field:

A^{μ} (x) = (ϕ (x), \vec{A} (x))

Maxwell's equations can be expressed in a Lorentz-covariant way in terms of

A^{μ}

. First define the field strength:

F^{μ ν} = \partial^{μ} A^{ν} - \partial^{ν} A^{μ}

Note that

F^{μ ν} = - F^{ν μ}

. Using the above definitions, we find that

F^{t i} = E^{i}, F^{i j} = ϵ^{i j k} B_{k}

where

ϵ^{i j k} = - ϵ^{j k} = - ϵ^{i k j}, ϵ^{123} = 1

and indices are raised and lowered using

η_{μ ν} = diag {- 1, 1, 1, 1}

. Now Maxwell's equations take the following form:

\begin{aligned} \partial_{ν} F^{μ ν} = J^{μ}, J^{μ} = (ρ, \vec{J}) \\ \partial^{μ} F^{ρ λ} + cyclic = 0 \end{aligned}

Note that the first equation implies current conservation:

\partial_{μ} J^{μ} = \partial_{μ} \partial_{ν} F^{μ ν} = 0

Gauge symmetry:

Maxwell's equations are invariant under

A_{μ} \to A_{μ} - \partial_{μ} Γ

(gauge transformations). Indeed, under this transformation

F_{μ ν} \to \partial_{μ} (A_{ν} - \partial_{μ} Γ) - \partial_{ν} (A_{μ} - \partial_{μ} Γ) = F_{μ ν} \underset{0}{\underset{⏟}{- \partial_{μ} \partial_{ν} Γ + \partial_{ν} \partial_{μ} Γ}} = F_{μ ν}

Action:

S = \int d^{4} x (- \frac{1}{4} F_{μ ν} F^{μ v} + J^{μ} A_{μ})

Let's check gauge invariance. Under

δ A_{μ} = - \partial_{μ} Γ

\begin{aligned} δ S & = \int d^{4} x (- \frac{1}{4} 2 F_{μ ν} δ F^{μ ν} + J^{μ} δ A_{μ}) \\ = - \int d^{4} x J_{μ} \partial_{μ} Γ & = 0 \end{aligned}

where we used integration by parts followed by current conservation.

Equations of motion:

Now consider a general variation

δ A_{μ}

\begin{aligned} δ S & = \int d^{4} x (- \frac{1}{4} 2 F_{μ v} δ F^{μ v} + J_{ν} δ A^{ν}) \\ = \int d^{4} x (- \partial^{μ} δ A^{v} F_{μ v} + J_{ν} δ A^{ν}) \\ = \int d^{4} x \int A^{ν} (\partial^{μ} F_{μ ν} + J_{ν}) \to \partial^{μ} F_{μ ν} + J_{ν} = 0 \end{aligned}

Gauge-fixing:

Using

δ A_{μ} = - \partial_{μ} Γ

, we can impose Lorenz gauge:

\partial_{μ} A^{μ} = 0

. Setting

J^{μ} = 0

, the EOM reduce to

\begin{aligned} 0 = \partial_{μ} F^{μ ν} & = \partial_{μ} (\partial^{μ} A^{ν} - \partial^{ν} A^{μ}) \\ = \partial^{2} A^{ν} - \partial^{ν} \underset{0}{\underset{⏟}{\partial \cdot A}} \\ = \partial^{2} A^{v} \end{aligned}

These are just the EOM of 4 massless scalars! On the other hand, there are fewer that 4 degrees of freedom due to gauge-invariace. To see this, let's Fourier transform to momentum space:

A_{μ} (x) = \int \frac{d^{4} k}{(2 π)^{4}} e^{i k \cdot x} A_{μ} (k)

After doing so, the EOM reduce to

k^{2} A (k) = 0 \to k^{0} = \pm ω_{k}, ω_{k} = | \vec{k} |

Note that there is a residual gauge symmetry which preserves the gauge condition:

A_{μ} (x) \to A_{μ} (x) - \partial_{μ} ω (x)

where

\partial^{2} ω (x) = 0

. Fourier transforming to momentum space:

\begin{aligned} ω (x) = \int \frac{d^{4} k}{(2 π)^{4}} δ (k^{2}) e^{i k \cdot x} ω (k) \\ A_{μ} (x) = \int \frac{d^{4} k}{(2 π)^{4}} δ (k^{2}) e^{i k \cdot x} ϵ_{μ} (k) \end{aligned}

where

δ (k^{2})

enforces the equations of motion and

ϵ_{μ}

is a polarisation vector. After doing so, the gauge condition corresponds to

k \cdot ϵ (k) = 0

and the residual gauge transformation corresponds to

ϵ_{μ} (k) \to ϵ_{μ} (k) + i ω (k) k_{μ}

. Hence

ϵ_{μ}

and

ϵ_{μ} + i α k_{μ}

are physically equivalent, where

α

is arbitrary and

k^{2} = 0

Since

ϵ

satisfies

ϵ \cdot k = 0

and has a residual gauge symmetry, it has two physical degrees of freedom,

ϵ_{\pm}

. To see this more explicitly, choose a frame where

k^{μ} = (ω_{k}, 0, 0, ω_{k})

Then

ϵ \cdot k = 0 \to ϵ^{μ} = (0, b, c, 0) + a (1, 0, 0, 1)

. Using the residual gauge symmetry, we may then take

ϵ^{μ} \to ϵ^{μ} - \frac{a}{ω_{k}} k^{μ} = (0, b, c, 0)

The two independent solutions are then given by

ϵ_{\pm}^{μ} = \frac{1}{\sqrt{2}} (0, 1, \mp i, 0)

, where we have fixed the normalisation. These correspond to positive and negative helicity. In summary, the polarisation vectors satisfy

ϵ_{\pm} (k) \cdot k = ϵ_{\pm} (k) \cdot ϵ_{\pm} (k) = 0, ϵ_{+}^{μ} (k) = ϵ_{-}^{μ} (k)^{*}, ϵ_{+} (k) \cdot ϵ_{-} (k) = 1

where we assume that

k^{2} = 0

. These properties were derived in a particular frame, but they are Lorentz-invarlient so hold in all frames.

Hence, we find the following solution to the EOM:

A^{μ} (x) = \int \frac{d^{3} k}{(2 π)^{3}} \int d k^{0} δ (k^{2}) θ (k^{0}) \sum_{h = \pm} [a_{h} (\vec{k}) ϵ_{- h}^{μ} (\vec{k}) e^{i k \cdot x} + a_{h}^{*} (\vec{k}) ϵ_{h}^{μ} (\vec{k}) e^{- i k \cdot x}]

where

k^{μ} = (ω_{k}, \vec{k})

. Note that

A^{μ} = {(A^{μ})}^{*}

since

ϵ_{h}^{*} = ϵ_{- h}

. Using

\int d x δ (f (x)) = \sum_{i} | 1 / f^{'} (x_{i}) |, f (x_{i}) = 0

we then perform the

k^{0}

integral to obtain

\int \tilde{d k} \sum_{h = \pm} [ϵ_{- h}^{μ} (\vec{k}) a_{h} (\vec{k}) e^{i k \cdot x} + ϵ_{h}^{μ} (\vec{k}) a_{h}^{*} (\vec{k}) e^{- i k \cdot x}]

Note that this is essatially the mode expansion for two scalar fields dressed with polarisations. In particular, we can quantise by promoting the Fourier coefficients to creation and annihilation operators, as we did for the scalar field.

9 LSZ and Path Integral For Photons

This is based on Srednicki Chapters

56, 57, 62

. We previously found that the general solution to Maxwell theory in the absence of sources is given by

A^{μ} (x) = \int \tilde{d k} \sum_{h = \pm} [ϵ_{- h}^{μ} (\vec{k}) a_{h} (\vec{k}) e^{i \dot{k} \cdot x} + ϵ_{h}^{μ} (+ \vec{k}) a_{h}^{*} (\vec{k}) e^{- i k \cdot x}]

We can quantise as we did for scalar fields. Taking

a^{*} \to a^{†}

\begin{aligned} [a_{h} (\vec{k}), a_{h^{'}} ({\vec{k}}^{'})] = [a_{h}^{†} (\vec{k}), a_{h^{'}}^{†} ({\vec{k}}^{'})] = 0 \\ [a_{h} (\vec{k}), a_{h^{'}}^{†} ({\vec{k}}^{'})] = (2 π)^{3} 2 ω_{k} δ^{3} (\vec{k} - {\vec{k}}^{'}) δ_{h h^{'}} \end{aligned}

where

a_{h}^{†} (\vec{k}) / a_{h} (\vec{k})

creates/annihilates a photon with momentum

\vec{k}

and polarisation

ϵ_{h}^{μ}

. Moreover, we can obtain a formula for creation and annihilation operators in exactly the same way we did for scalar fields:

\begin{aligned} a_{h}^{†} (\vec{k}) = - i ϵ_{- h}^{μ} (\vec{k}) \int d^{3} x e^{i k \cdot x} {\overset{\leftrightarrow}{\partial}}_{t} A^{μ} (x) \\ a_{h} (\vec{k}) = + i ϵ_{h} (\vec{k}) \int d^{3} x e^{- i k \cdot x} {\overset{\leftrightarrow}{\partial}}_{t} A^{μ} (x) \end{aligned}

where we used

ϵ_{h^{'}} (\vec{k}) \cdot ϵ_{- h} (\vec{k}) = δ_{h h^{'}}

Following the same steps used to derive LSZ for scalars, we can make the following replacements when computing photon amplitudes:

\begin{aligned} a_{h}^{†} (\vec{k}, - \infty) & \to i ϵ_{- h}^{μ} ({\overset{`}{k}}^{'}) \int d^{4} x e^{i k \cdot x} (- \partial^{2}) A_{μ} (x) \\ a_{h} (ℏ_{k}, + \infty) & \to i ϵ_{h}^{μ} (ℏ_{k}) \int d^{4} x e^{- i k \cdot x} (- \partial^{2}) A_{μ} (x) \end{aligned}

where the

a^{†}

creates incoming photons and the

a

annihilates outgoing photons.
Summary: To compute photon amplitudes, first compute time-ordered correlators of

A_{μ} (x)

, Fourier transform to momentum space, amputate external propagators, and dress with polarisation vectors. In practice we will not distinguish between

ϵ_{+}^{μ}

and

ϵ_{-}^{μ}

and simply dress the amputated correlator with

ϵ^{μ} (\vec{k})

. Due to gauge symmetry, the amplitude must be invariant if we shift an external polarisation by its momentum:

ϵ_{μ} (k) \to ϵ_{μ} (k) + α k μ

It follows that if we write

i M = i ϵ_{μ} (k) M^{μ} (k) \to k^{μ} M_{μ} (k) = 0

i.e. if we replace a polarisation with its momentum, then the amplitude vanishes. This is known as the "Ward identity."

Now we must understand how to compute time-ordered corrclators of

A_{μ} (x)

. This is easiest to do using the path integral formalism. We have

Z_{0} (J) = \int D A \exp [i \int d^{4} x (- \frac{1}{4} F_{μ ν} F^{μ ν} + J^{μ} A_{μ})]

Let us denote the argument of the exponential

i S_{0}

. Fourier transforming to momentum space:

A_{μ} (k) = \int d^{4} x e^{- i k \cdot x} A_{μ} (x), A_{μ} (x) = \int \frac{d^{4} k}{(2 π)^{4}} e^{i k \cdot x} A_{μ} (k)

we find that

\begin{aligned} i S_{0} = i \int d^{4} x \frac{d^{4} k}{(2 π)^{y}} \frac{d^{4} k^{'}}{(2 π)^{4}} e^{i (k + k^{'}) \cdot x} \\ [- \frac{1}{4} (i k_{μ} A_{ν} (k) - i k_{ν} A_{μ} (k)) (i k^{' μ} A^{ν} (k^{'}) - i k^{'} ν A^{μ} (k^{'})) \\ + \frac{1}{2} (J^{μ} (k) A_{μ} (k^{'}) + J^{μ} (k^{'}) A_{μ} (k))] \end{aligned}

The

x

-integral gives

(2 π)^{4} δ^{4} (k + k^{'})

. Performing

k^{'}

integral then gives

i S_{0} = \frac{i}{2} \int \frac{d^{η} k}{(2 π)^{4}} [- A_{μ} (k) (k^{2} η^{μ ν} - k^{μ} k^{ν}) A_{ν} (- k) + J^{μ} (k) A_{μ} (- k) + J^{μ} (- k) A_{μ} (k)]

The next step is to complete the square but the

4 \times 4

matrix

k^{2} η^{μ ν} - k^{μ} k^{ν}

is not invertible. To see this, consider the matrix

P^{μ ν} (k) = η^{μ ν} - k^{μ} k^{ν} / k^{2}

This is a projection matrix:

p^{μ ν} (k) P_{ν}^{λ} (k) = p^{μ λ} (k)

Hence, the only allowed eigenvalues of

P

are zero or one. There is at least one zero eigenvalue because

P^{μ ν} (k) k_{μ} = 0

On the other hand, the sum of the eigenvalues is given by the trace

η_{μ ν} P^{μ ν} (k) = 3 .

Hence the remaining three eigenvalues are all 1. From this, we see that the component of

A_{μ} (k)

along

k_{μ}

does not contribute to the quadratic term in the action. Moreover, it does not contribute to the linear term since

\partial_{μ} J^{μ} (x) =

0 \to k^{μ} J_{μ} (k) = 0

. Thus the path integral over this component just gives an

\infty

constant which we can discard.

The non-trivial part of the path integral is over the components of

A_{μ} (k)

in the space orthogonal to

k_{μ}

. The matrix

P^{μ ν} (k)

projects 4 -vectors into this subspace and in this subspace it is the identity matrix. Therefore with in this subspace the inverse of

k^{2} P^{μ ν} (k)

1 / k^{2} P_{μ ν} (k)

and we can complete the square in the action to give

Z_{0} (J) \propto \exp [\frac{i}{2} \int \frac{d^{4} k}{(2 π)^{4}} J_{μ} (k) \frac{P^{μ ν} (k)}{k^{2} - i ϵ} J_{ν} (k)]

where we have restored the Feynman

i ϵ

prescription. Since

k^{μ} J_{μ} (k) = 0

, we are thee to shift

P^{μ v} (k) \to P^{μ ν} (k) + ξ \frac{k^{μ} k^{v}}{k^{2}} .

Then the propagator in momentum space becomes

\overset{m}{m} = \frac{- i}{k^{2} - i \in} (η_{μ v} - (1 - ξ) \frac{k_{μ} k_{ν}}{k^{2}})

where

ξ = 1

is called "Feynman gauge" and

ξ = 0

is called "Lorenz gauge" or "Landau gauge." This form of the propagator can be derived by adding a "gauge-fixing" term to the Lagrangian:

L = - \frac{1}{4} F_{μ ν} F^{μ v} - \frac{1}{2 ξ} {(\partial^{μ} A_{μ})}^{2}

After adding this term and going to momentum space, the

4 \times 4

matrix in the qaudratic part of the action becomes invertable and the inverse is the propagator given above.

In this way, we can compute photon amplitudes by drawing Feynman diagrams. We can also compute them from Wick's theorem in the interaction picture replacing

{\hat{A}}^{μ} A^{ν}

with the propagator and

\hat{A_{μ} | k, ϵ ⟩} = ϵ_{μ} (k)

which follows from LSZ.

10 Dirac Equation

This is based on Tong sections 4.1-4.3, 4.6.

Representations of Lorentz group:

Under a Lorentz transformation

x^{μ} \to x^{μ} = Λ^{μ}_{ν} x^{ν}

we have seen that

\begin{array}{ll} ϕ (x) \to ϕ (Λ^{- 1} x) & (spin-0) \\ A_{μ} (x) \to Λ^{μ}_{v} A_{μ} (Λ^{- 1} x) & (spin-) \end{array}

In general, we can write

Λ = \exp (\frac{1}{2} Ω_{ρ σ} M^{ρ σ})

where

Ω_{ρ σ} = - Ω_{σ ρ}

are six numbers that pananeterise the Lorentz transformation ( 3 rotations, 3 boosts) and

M^{ρ σ}

are generators, whose representation
depends on the spin of the particle. For spin-1, the generator of rotations in the

x^{1} - x^{2}

plane is

{(M^{12})}_{ν}^{μ} = (\begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & - 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array})

The generator of boosts along

x^{1}

direction is:

{(M^{01})}_{ν}^{μ} = (\begin{array}{llll} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array})

In general the spin-1 generators can be written as follows:

{(M^{ρ σ})}_{ν}^{μ} = η^{σ μ} δ_{ν}^{ρ} - η^{ρ μ} δ_{ν}^{σ}

These obey the algebra

[M^{μ ν}, M^{ρ σ}] = (η^{μ σ} μ^{ν ρ} - μ \leftrightarrow ν) - ρ \leftrightarrow σ

The genentors in any representation must obey this algebra.
For spin-s, a Lorentz transformation becomes trivial for a rotation of

2 π / s

. For spin- 1 , consider a rotation by

θ

in the

x^{1} - x^{2}

plane:

Ω_{12} = θ \to Λ (θ) = \exp (Ω_{12} M^{12}) = (\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & \cos θ & - \sin θ & 0 \\ 0 & \sin θ & \cos θ & 0 \\ 0 & 0 & 0 & 1 \end{array})

Hence

Λ (2 π) = 1

, as expected.

Spin-1/2

Next we will look at spin-

1 / 2

, which decribes electrons. In this case, we expect that a spatial rotation by

4 π

will be trivial but not by

2 π

, i.e.

Λ (4 π) = 1

but

Λ (2 π) \neq 1

. Since

Λ (4 π) = (Λ (2 π))^{2} \to Λ (2 π) = - 1

. Let us postulate the following generators and verify that they imply this property:

S^{ρ σ} = \frac{1}{4} [γ^{ρ}, γ^{σ}] γ^{μ} = (\begin{array}{cc} 0 & σ^{μ} \\ {\bar{σ}}^{μ} & 0 \end{array}) "Dirac matrices"

where

σ^{μ} = (1, \vec{σ}), {\bar{σ}}^{μ} = (1, - \vec{σ}), σ^{1} = (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}), σ^{2} = (\begin{array}{cc} 0 & - i \\ i & 0 \end{array}), σ^{3} = (\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array})

The Dirac matrices obey a Clifford algebra:

{γ^{μ}, γ^{ν}} = - 2 η^{μ ν}, {A, B} = A B + B A

Using this algebra, you can check that

S^{ρ σ}

indeed obey the Lorentz algebra (HW)

Using these generators, the Lorentz transformations are given by

S [Λ] = \exp (\frac{1}{2} Ω_{ρ σ} S^{ρ σ})

This is known as the "spinor representation." Let's see what rotations and boosts look like. For a rotation by

θ

x^{1} - x^{2}

plane,

\begin{matrix} S^{12} = \frac{1}{4} [γ^{1}, γ^{2}] = - \frac{i}{2} (\begin{array}{cc} σ^{3} & 0 \\ 0 & σ^{3} \end{array}) \\ S [Λ] = \exp (S^{12} θ) = (\begin{array}{cc} \exp (- i θ σ^{3} / 2) & 0 \\ 0 & \exp (- i θ σ^{3} / 2) \end{array}) \end{matrix}

For

θ = 2 π, S [Λ] = - 1

as expected. Boosts are generated by

S^{0 i} = \frac{1}{4} [γ^{0}, γ^{i}] = \frac{1}{2} (\begin{array}{cc} - σ^{i} & 0 \\ 0 & σ^{i} \end{array})

. A boost along the

x^{3}

direction with rapidity

η

is then given by

S [Λ] = (\begin{array}{cl} \exp (- η σ^{3} / 2) & 0 \\ 0 & \exp (η σ^{3} / 2) \end{array})

Constructing an Action:

The object which transforms under

S [Λ]

is calleda "Dirac spinor"

ψ^{α} (x) \to S [Λ]^{α}_{β} ψ^{β} (Λ^{- 1} x)

where

α = 1, 2, 3, 4

. To construct a Lorentz-invariant action, we need some identities:

\begin{aligned} {(γ^{μ})}^{†} = γ^{0} γ^{μ} γ^{0} & \to {(S^{μ ν})}^{†} = - γ^{0} S^{μ ν} γ^{0} \\ \to S [Λ]^{†} = γ^{0} S [Λ]^{- 1} γ^{0} \end{aligned}

You will show this on the HW. A natural guess for a Lorentz invariant is

ψ^{†} ψ

but under Lorentz transformation we get

ψ^{†} ψ \to ψ^{†} S [Λ]^{†} S [Λ] ψ = ψ^{†} γ^{0} S [Λ]^{- 1} γ^{0} S [Λ] ψ

so this is not invariant.
To get an invariant define

\bar{ψ} (x) = ψ^{†} (x) γ^{0}

Then

\bar{ψ} ψ

is a Lorentz scalar:

\begin{aligned} \bar{ψ} ψ & \to (ψ^{†} S [Λ]^{†} γ^{0}) S [Λ] ψ \\ = (ψ^{†} γ^{0} S [Λ]^{- 1} γ^{0} γ^{0}) S [Λ] ψ \\ = (\bar{ψ} S [Λ]^{- 1}) S [Λ] ψ = \bar{ψ} ψ \end{aligned}

Moreover

\bar{ψ} γ^{μ} ψ

is a Lorentz vector:

\bar{ψ} γ^{μ} ψ \to \bar{ψ} S [Λ]^{- 1} γ^{μ} S [Λ] ψ = Λ_{ν}^{μ} \bar{ψ} γ^{ν} ψ

as you will show in the HW.
Hence, the following Lagrangian is Lorentz-invariant:

L = \bar{ψ} (x) (γ^{μ} \partial_{μ} - m) ψ (x)

We will denote

γ^{μ} \partial_{μ} = ∂̸

. The choice of relative coefficients will become clear when we show the EOM

\to

the Klein-Gordon equation.

Equations of motion:

Varying with respect to

\bar{ψ}

\begin{aligned} 0 & = δ S = \int d^{4} x δ \bar{ψ} (i ∂̸ - m) ψ \\ \to (i ⊅ - m) ψ = 0 \end{aligned}

This is the "Dirac equation." Note that the Dirac equation implies the KleinGordon equation:

0 = (i ∂̸ \partial + m) (i ∂̸ - m) ψ = (- ∂̸ ∂̸ \partial - m^{2}) ψ = (\partial^{2} - m^{2}) ψ

where we noted that

γ^{μ} γ^{ν} \partial_{μ} \partial_{ν} = \frac{1}{2} {γ^{μ}, γ^{ν}} \partial_{μ} \partial_{ν} = - \partial^{2}

Hamiltonian:

For a scalar field the Hamiltonian density is

H = \frac{\partial L}{\partial (\partial_{0} ϕ)} \partial_{0} ϕ - L

. Similarly, for a Dirac field

\begin{aligned} H & = \frac{\partial L}{\partial (\partial_{0} ψ)} \partial_{0} ψ - L \\ = (i \bar{ψ} γ^{0}) (\partial_{0} ψ) - \bar{ψ} (i ∂̸ - m) ψ \\ = \bar{ψ} (- i γ^{i} \partial_{i} + m) ψ \to H = \int d^{3} x \bar{ψ} (- i γ^{i} \partial_{i} + m) ψ \end{aligned}

where

i = 1, 2, 3

Electric charge:

Note that

L

is invariant under

ψ \to e^{- i α} ψ, \bar{ψ} \to e^{i α} \bar{ψ}

where

α

is a constant. This is a "

U (1)

global symmetry." For scalars the Noether current of a global symmetry is

j^{μ} = \frac{\partial L}{\partial (\partial_{μ} ϕ)} δ ϕ

where

δ ϕ

is an infinitesimal symmetry variation. For a Dirac field, the Noether current associated with the above symmetry is

j^{μ} = \frac{\partial L}{\partial (\partial_{μ} ψ)} δ ψ = (i \bar{ψ} γ^{μ}) (- i α ψ) \sim \bar{ψ} γ^{μ} ψ

where we dropped

α

, and recalled that infinitesimally

e^{- i α} ψ = (1 - i α + \dots) ψ \to δ ψ = - i α ψ

Let us check that the current is conserved:

\partial_{μ} j^{μ} = (\partial_{μ} \bar{ψ}) γ^{μ} ψ + \bar{ψ} ∂̸ ψ = i m \bar{ψ} ψ - i m \bar{ψ} ψ = 0 .

where we noted that

\begin{aligned} ∂̸ ψ = - i m ψ & \to (∂̸ ψ)^{†} = i m ψ^{†} \\ \to i m \bar{ψ} = (∂̸ ψ)^{†} γ^{0} = \partial_{μ} ψ^{†} {(γ^{μ})}^{†} γ^{0} = \partial_{μ} \bar{ψ} γ^{μ} \end{aligned}

The Noether charge corresponding to the current in (10) then given by

Q = \int d^{3} x j^{0} = \int d^{3} x \bar{ψ} γ^{0} ψ

11 Classical Solutions of the Dirac Equation

This is based on Srednicki pg 238-239 and Tong section 4.7. Recall that the Dirac equation implies Klein-Gordon equation. Hence, it admits plane-wave solutions:

ψ (x) = u (\vec{p}) e^{i p \cdot x} + v (\vec{p}) e^{- i p \cdot x}

where

p^{0} = ω_{p} = \sqrt{{\vec{p}}^{2} + m^{2}}, u (\vec{p})

and

v (\vec{p})

are 4 component Dirac spinors. Plugging this into Dirac equation gives

(- i ∂̸ + m) ψ = 0 \to (p̸ + m) u (\vec{p}) = (- p̸ + m) v (\vec{p}) = 0

Each of these equations has two linearly independent solutions that we will call

u_{s} (\vec{p})

and

v_{s} (\vec{p})

, where

s = 1, 2

. The general solution is then

ψ (x) = \sum_{s = 1, 2} \int \tilde{d p} [b_{s} (\vec{p}) u_{s} (\vec{p}) e^{i p \cdot x} + d_{s}^{†} (\vec{p}) v_{s} (\vec{p}) e^{- i p \cdot x}]

where

\tilde{d p} = \frac{d^{3} p}{(2 π)^{3} 2 ω_{p}}

Let us now work out

u (\vec{p})

from its EOM:

\begin{aligned} 0 = (p̸ + m) u (\vec{p}) = (\begin{array}{cc} m & p \cdot σ \\ p \cdot \bar{σ} & m \end{array}) (\binom{u_{L} (\vec{p})}{u_{R} (\vec{p})}) \\ σ^{μ} = (1, \vec{σ}), {\bar{σ}}^{μ} = (1, - \vec{σ}) \to p \cdot σ = - p^{0} + \vec{p} \cdot \vec{σ}, p \cdot \bar{σ} = - p^{0} - \vec{p} \cdot \vec{σ} \end{aligned}

u_{L / R}

are called "left/right-handed chiral spinors." We now have

(p \cdot σ) u_{R} = - m u_{L}, (p \cdot \bar{σ}) u_{L} = - m u_{R}

Note that each equation implies the other:

u_{L} = - \frac{1}{m} (p \cdot σ) u_{R} \to (p \cdot \bar{σ}) u_{L} = - \frac{1}{m} \underset{+ m^{2}}{\underset{⏟}{(p \cdot σ) (p \cdot \bar{σ})}} u_{R} = - m u_{R}

where we noted that

\begin{aligned} (p \cdot σ) (p \cdot \bar{σ}) = (- p^{0} + \vec{p} \cdot \vec{σ}) (- p^{0} - \vec{p} \cdot \vec{σ}) \\ = {(p^{0})}^{2} - (\vec{p} \cdot \vec{σ})^{2} = + {(p^{0})}^{2} - {\vec{p}}^{2} = - p^{2} = + m^{2} \end{aligned}

where

p_{i} p_{j} σ^{i} σ^{j} = \frac{1}{2} {σ^{i}, σ^{j}} p_{i} p_{j} = δ^{i j} p_{i} p_{j} = {\vec{p}}^{2}

.
Now let us make the following ansatz:

u_{L} = A (p \cdot σ) ξ^{'}

for some 2component spinor

ξ^{'}

. The second equation in (11) implies

\begin{aligned} u_{R} & = - \frac{1}{m} (p \cdot \bar{σ}) u_{L} \\ = - \frac{A}{m} (p \cdot \bar{σ}) (p \cdot σ) ξ^{'} = - m A ξ^{'} \\ \to u (\vec{p}) & = A (\binom{p \cdot σ ξ^{'}}{- m ξ^{'}}) \end{aligned}

To make the solution more symmetric, choose

A = - 1 / m, ξ^{'} = \sqrt{- p \cdot \bar{σ}} ξ

, where square root of a matrix is defined by going to a basis where it is diaganal and taking square root of its eigenvalues. Note that in rest frame,

p^{μ} = (m, \vec{0})

p \cdot \bar{σ} = - m 1 \to \sqrt{- p \cdot \bar{σ}} = \sqrt{m} 1 .

Then we have

u (\vec{p}) = - \frac{1}{m} (\begin{array}{cc} p \cdot σ \sqrt{- p \cdot \bar{σ}} & ξ \\ - m \sqrt{- p \cdot \bar{σ}} & ξ \end{array}) = (\begin{array}{cc} \sqrt{- p \cdot σ} & ξ \\ \sqrt{- p \cdot \bar{σ}} & ξ \end{array})

Similarly, we find that

v (\vec{p}) = (\begin{array}{cc} \sqrt{- p \cdot σ} & η \\ - \sqrt{- p \cdot \bar{σ}} & η \end{array})

where

η

is some 2-componnent spinor. Indeed,

\begin{aligned} (- p̸ + m) v (\vec{p}) = (\begin{array}{cc} m & - p \cdot σ \\ - p \cdot \bar{σ} & m \end{array}) (\binom{\sqrt{- p \cdot σ} η}{- \sqrt{- p \cdot \bar{σ}} η}) \\ = (\binom{(m \sqrt{- p \cdot σ} + p \cdot σ \sqrt{- p \cdot \bar{σ}}) η}{(- p \cdot \bar{σ} \sqrt{- p \cdot σ} - m \sqrt{- p \cdot \bar{σ}}) η}) = (\binom{0}{0}) \end{aligned}

where we noted that

p \cdot σ \sqrt{- p \cdot \bar{σ}} = \sqrt{- p \cdot σ} \sqrt{(- p \cdot σ) (- p \cdot \bar{σ})} = - m \sqrt{- p \cdot σ}

.
It's convenient to introduce a basis

ξ_{s}

and

η_{s}, s = 1, 2

such that

ξ_{r}^{†} ξ_{s} = δ_{r s}, η_{r}^{†} η_{s} = δ_{r s}

Then the independent solutions are given by

u_{s} (\vec{p}) = (\begin{array}{cc} \sqrt{- p \cdot σ} & ξ_{s} \\ \sqrt{- p \cdot \bar{σ}} & ξ_{s} \end{array}), v_{s} (\vec{p}) = (\begin{array}{cc} \sqrt{- p \cdot σ} & η_{s} \\ - \sqrt{- p \cdot \bar{σ}} & η_{s} \end{array})

The classical solutions obey various useful identities (which you will prove in HW ):

\begin{aligned} u_{s^{'}}^{†} (\vec{p}) u_{s} (\vec{p}) = v_{s}^{†} (\vec{p}) v_{s} (\vec{p}) = 2 ω_{p} δ_{s s^{'}} \\ {\bar{u}}_{s^{'}} (\vec{p}) u_{s} (\vec{p}) = - {\bar{v}}_{s^{'}} (\vec{p}) v_{s} (\vec{p}) = 2 m δ_{s s^{'}} \\ u_{s^{'}}^{†} (\vec{p}) v_{s} (- \vec{p}) = v_{s^{'}}^{†} (\vec{p}) u_{s} (- \vec{p}) = 0 \\ {\bar{u}}_{s^{'}} (\vec{p}) v_{s} (\vec{p}) = {\bar{v}}_{s^{'}} (\vec{p}) u_{s} (\vec{p}) = 0 \\ \sum_{s} u_{s}^{α} (\vec{p}) {\bar{u}}_{s}^{β} (\vec{p}) = (- p̸ + m)^{α β} \\ \sum_{s} v_{s}^{α} (\vec{p}) {\bar{v}}_{s}^{β} (\vec{p}) = (- p̸ - m)^{α β} \end{aligned}

Examples:

rest frame:

\begin{aligned} p^{μ} & = (m, \vec{0}) \to - p \cdot σ = - p \cdot \bar{σ} = m \to \\ u_{s} (\vec{0}) & = \sqrt{m} (\binom{ξ_{s}}{ξ_{s}}), v_{s} (\vec{0}) = \sqrt{m} (\binom{η_{s}}{- η_{s}}) \end{aligned}

for $p^{μ} = (p^{0}, 0, 0, p^{3}), p_{3} > 0$ , and $ξ = η = (\binom{1}{0})$ we have

\begin{aligned} - p \cdot σ = (\begin{array}{cc} + p^{0} - p^{3} & 0 \\ 0 & + p^{0} + p^{3} \end{array}), - p \cdot \bar{σ} = (\begin{array}{cc} + p^{0} + p^{3} & 0 \\ 0 & t p^{0} - p^{3} \end{array}) \\ \to \sqrt{- p \cdot σ} ξ = \sqrt{- p \cdot σ} η = \sqrt{p^{0} - p^{3}} (\binom{1}{0}) \\ \sqrt{- p \cdot \bar{σ}} ξ = \sqrt{- p \cdot \bar{σ}} η = \sqrt{p^{0} + p^{3}} (\binom{1}{0}) \\ \to u (\vec{p}) = (\binom{\sqrt{p^{0} - p^{3}} (\binom{1}{0})}{\sqrt{p^{0} + p^{3}} (\binom{1}{0})}), v (\vec{p}) = (\binom{\sqrt{p^{0} - p^{3}} (\binom{1}{0})}{- \sqrt{p^{0} + p^{3}} (\binom{1}{0})}) \end{aligned}

In the ultrarelativistic (or massless) limit

p^{3} = p^{0}

this reduces to

lim_{m \to 0} u (\vec{p}) = \sqrt{2 p^{0}} (\begin{array}{l} 0 \\ 0 \\ 1 \\ 0 \end{array}), lim_{m \to 0} v (\vec{p}) = - \sqrt{2 p^{0}} (\begin{array}{l} 0 \\ 0 \\ 1 \\ 0 \end{array})

Similarly, for $p^{μ} = (p^{0}, 0, 0, p^{3}), p^{3} > 0$ , and $ξ = η = (\binom{0}{1})$ we obtain

u (\vec{p}) = (\binom{\sqrt{p^{0} + p^{3}} (\binom{0}{1})}{\sqrt{p^{0} - p^{3}} (\binom{0}{1})}), v (\vec{p}) = (\binom{\sqrt{p^{0} + p^{3}} (\binom{0}{1})}{- \sqrt{p^{0} - p^{3}} (\binom{0}{1})})

and in the massless limit this reduces to

lim_{m \to 0} u (\vec{p}) = \sqrt{2 p^{0}} (\begin{array}{l} 0 \\ 1 \\ 0 \\ 0 \end{array}), lim_{m \to 0} v (\vec{p}) = \sqrt{2 p^{0}} (\begin{array}{l} 0 \\ 1 \\ 0 \\ 0 \end{array})

Hence in massless limit, Dirac spinors become chiral spinors.

12 Quantizing the Dirac field

This is based on Srednicki Chapter 38, Tong section 5.1-5.3. For the Dirac Lagrangian

L = i \bar{ψ} ∂̸ ψ - m \bar{ψ} ψ

we have the following canonical momentum:

Π_{ψ} = \frac{\partial L}{\partial \dot{ψ}} = i \bar{ψ} γ^{0} = i ψ^{†}

and the following canonical (anti)commutation relations:

\begin{aligned} {ψ_{α} (t, \vec{x}), ψ_{β} (t, \vec{y})} = 0 \\ {ψ_{α} (t, \vec{x}), {(Π_{ψ})}_{β} (t, \vec{y})} = i δ_{α β} δ^{3} (\vec{x} - \vec{y}) \to {ψ_{α} (t, \vec{x}), ψ_{β}^{†} (t, \vec{y})} = δ_{α β} δ^{3} (\vec{x} - \vec{y}) \end{aligned}

Note that we need anticommutators. Using commutators would lead to various inconsistencies such as negative-norm states or a Hamiltonian which is unbounded from below, as you will see in the HW. In general, integer-spin fields must be commuting and half-integer spin fields must be anti-commiting This known as "spin-statistics theorem".

Now recall the expansion:

ψ (x) = \sum_{s = 1, 2} \int \tilde{d p} [b_{s} (\vec{p}) u_{s} (\vec{p}) e^{i p \cdot x} + d_{s}^{†} (\vec{p}) v_{s} (\vec{p}) e^{- i p \cdot x}]

Let us use this to express

b, d

in terms of

ψ, ψ^{†}

and then deduce the algebra of

b, d

. We have that

\begin{aligned} \int d^{3} x e^{- i p \cdot x} ψ (x) & = \sum_{s} \int {\tilde{d p}}^{'} \int d^{3} x [b_{s} ({\vec{p}}^{'}) u_{s} ({\vec{p}}^{'}) e^{i (p^{'} - p) \cdot x} + d_{s}^{†} ({\vec{p}}^{'}) v_{s} ({\vec{p}}^{'}) e^{- i (p^{'} + p) \cdot x}] \\ = \sum_{s} \int {\tilde{d p}}^{'} [b_{s} ({\vec{p}}^{'}) u_{s} ({\vec{p}}^{'}) (2 π)^{3} δ^{3} ({\vec{p}}^{'} - \vec{p}) e^{+ i (ω_{p} - ω_{p^{'}}) t} \\ + d_{s}^{†} ({\vec{p}}^{'}) v_{s} ({\vec{p}}^{'}) (2 π)^{3} δ^{3} ({\vec{p}}^{'} + {\vec{p}}^{'}) e^{i (ω_{p^{'}} + ω_{p}) t}] \\ = \sum_{s} \frac{1}{2 ω_{p}} [b_{s} (\vec{p}) u_{s} (\vec{p}) + e^{2 i ω_{p} t} d_{s}^{†} (- \vec{p}) v_{s} (- \vec{p})] \end{aligned}

From this we see that

\begin{aligned} \int d^{3} x e^{- i p \cdot x} u_{s}^{†} (\vec{p}) ψ (x) & = \sum_{s^{'}} \frac{1}{2 ω_{p}} [b_{s}^{'} (\vec{p}) u_{s}^{†} (\vec{p}) u_{s^{'}} (\vec{p}) + e^{2 i ω_{p} t} d_{s}^{+} (- \vec{p}) u_{s}^{†} (\vec{p}) v_{s^{'}} (- \vec{p})] \\ = b_{s} (\vec{p}) \end{aligned}

Similarly, computing

\int d^{3} x e^{i p \cdot x} ψ (x)

, multiplying by

v_{s}^{†} (\vec{p})

, and using

v_{s}^{†} (\vec{p}) v_{s^{'}} (\vec{p}) =

2 ω_{p} δ_{s s^{'}}, v_{s}^{†} (\vec{p}) u_{s^{'}} (- \vec{p}) = 0

, we find

d_{s}^{†} (\vec{p}) = \int d^{3} x e^{i p \cdot x} v_{s}^{†} (\vec{p}) ψ (x)

Taking complex conjugates:

\begin{aligned} b_{s}^{†} (\vec{p}) = \int d^{3} x e^{i p \cdot x} ψ^{†} (x) u_{s} (\vec{p}) \\ d_{s} (\vec{p}) = \int d^{3} x e^{- i p \cdot x} ψ^{†} (x) v_{s} (\vec{p}) \end{aligned}

From this, we immediately see that

{b_{s} (\vec{p}), b_{s^{'}} ({\vec{p}}^{'})} = {d_{s} (\vec{p}), d_{s^{'}} ({\vec{p}}^{'})} = {b_{s} (\vec{p}), d_{s^{'}}^{†} ({\vec{p}}^{'})} = 0

since they all involve

{ψ, ψ} = 0

. Similarly, the hermitian conjugates also vanish. Moreover

\begin{aligned} {b_{s} (\vec{p}), d_{s^{'}} ({\vec{p}}^{'})} & = \int d^{3} x d^{3} y e^{- i p \cdot x - i p^{'} \cdot y} u_{s}^{†} (\vec{p})^{α} v_{s^{'}} {({\vec{p}}^{'})}^{β} {ψ_{α} (x), ψ_{β}^{†} (y) \\ = \int d^{3} x e^{- i (p + p^{'}) \cdot x} u_{s}^{†} (\vec{p}) v_{s^{'}} (- \vec{p}) \\ = (2 π)^{3} δ^{3} (\vec{p} + {\vec{p}}^{'}) u_{s}^{†} (\vec{p}) v_{s^{'}} (- \vec{p}) = 0 \end{aligned}

The hermitian conjugate must therefore vanish as well.
Now let's look at the following anticommutator:

\begin{aligned} {b_{s} (\vec{p}), b_{s^{'}}^{†} ({\vec{p}}^{'})} & = \int d^{3} x d^{3} y e^{- i p \cdot x + i p^{'} \cdot y} u_{s}^{† α} (\vec{p}) u_{s^{'}}^{β} ({\vec{p}}^{'}) {ψ_{α} (x), ψ_{β}^{†} (y)} \\ = \int d^{3} x e^{- i (p - p^{'}) \cdot x} u_{s}^{†} (\vec{p}) u_{s^{'}} ({\vec{p}}^{'}) \\ = 2 ω_{p} δ_{s, s^{'}} (2 π)^{3} δ^{3} (\vec{p} - {\vec{p}}^{'}) \end{aligned}

Similarly, we find

\begin{aligned} {d_{s}^{†} (\vec{p}), d_{s^{'}} ({\vec{p}}^{'})} & = \int d^{3} x d^{3} y e^{i p \cdot x - i p^{'} \cdot y} v_{s}^{† α} (\vec{p}) v_{s^{'}}^{β} (\vec{p}) {ψ_{α} (x), ψ_{β}^{†} (y)} \\ = \int d^{3} x e^{i (p - p^{'}) \cdot x} v_{s}^{†} (\vec{p}) v_{s^{'}} ({\vec{p}}^{'}) \\ = 2 ω_{p} δ_{s s^{'}} (2 π)^{3} δ^{3} (\vec{p} - {\vec{p}}^{'}) \end{aligned}

In summary:

{b_{s} (\vec{p}), b_{s^{'}}^{†} ({\vec{p}}^{'})} = {d_{s} (\vec{p}), d_{s^{'}}^{†} ({\vec{p}}^{'})} = (2 π)^{3} 2 ω_{p} δ^{3} (\vec{p} - \vec{p}) δ_{s s^{'}}

, and all other anticommutation relations vanish.

Hamiltonian:

Recall that

H = \int d^{3} x \bar{ψ} (- i γ^{i} \partial_{i} + m) ψ

Noting that

\begin{aligned} (- i γ^{i} \partial_{i} + m) ψ = \sum_{s} \int \tilde{d p} & [b_{s} (\vec{p}) (γ^{i} p_{i} + m) u_{s} (\vec{p}) e^{i p \cdot x} \\ + d_{s}^{†} (\vec{p}) (- γ^{i} p_{i} + m) v_{s} (\vec{p}) e^{- i p \cdot x}] \end{aligned}

and using

(p̸ + m) u (p) = (- p̸ + m) v (p) = 0

, we find that

\begin{aligned} H & = \sum_{s, s^{'}} \int \tilde{d p} {\tilde{d p}}^{'} d^{3} x (b_{s^{'}}^{†} ({\vec{p}}^{'}) {\bar{u}}_{s^{'}} ({\vec{p}}^{'}) e^{- i p^{'} \cdot x} + d_{s^{'}} ({\vec{p}}^{'}) {\bar{v}}_{s^{'}} ({\vec{p}}^{'}) e^{i p^{'} \cdot x}) \\ ω_{p} γ^{0} (b_{s} (\vec{p}) u_{s} (\vec{p}) e^{i p \cdot x} - d_{s}^{†} (\vec{p}) v_{s} ({\vec{p}}^{'}) e^{- i p \cdot x}) \\ - \sum_{s, s^{'}} \int \tilde{d p} \tilde{d p} d^{3} x ω_{p} [{\vec{p}}^{'}) b_{s^{'}}^{†} (\vec{p}) d_{s}^{†} (\vec{p}) v_{s^{'}}^{†} ({\vec{p}}^{'}) v_{s} (\vec{p}) e^{+ i (p^{'} - p) \cdot x}) u_{s} (\vec{p}) e^{- i (p^{'} - p) \cdot x} \\ - b_{s^{'}}^{†} ({\vec{p}}^{'}) d_{s}^{†} (\vec{p}) u_{s^{'}}^{†} ({\vec{p}}^{'}) v_{s} (\vec{p}) e^{- i (p + p^{'}) \cdot x} \\ + d_{s^{'}} ({\vec{p}}^{'}) b_{s} (\vec{p}) v_{s^{'}}^{†} ({\vec{p}}^{'}) u_{s} (\vec{p}) e^{i (p + p^{'}) \cdot x}] \\ = \sum_{s, s^{'}} \int \tilde{d p} \frac{1}{2} (b_{s^{'}}^{†} (\vec{p}) b_{s} (\vec{p}) u_{s}^{†} (\vec{p}) u_{s} (\vec{p}) - d_{s^{'}} (\vec{p}) d_{s}^{†} (\vec{p}) v_{s}^{†} (\vec{p}) v_{s} (\vec{p}) \\ - b_{s^{'}}^{†} (- \vec{p}) d_{s}^{†} (\vec{p}) \underset{0}{\underset{⏟}{u_{s^{'}}^{†} (- \vec{p}) v_{s} (\vec{p})}} + d_{s} (- \vec{p}) b_{s} (\vec{p}) \underset{s_{s^{'}}^{†} (\vec{p}) \cdot u_{s} (\vec{p}))}{\underset{⏟}{}} \\ = \sum_{s} \int \tilde{d p} ω_{p} [b_{s}^{†} (\vec{p}) b_{s} (\vec{p}) - d_{s} (\vec{p}) d_{s}^{†} (\vec{p})] \\ = \sum_{s} \int \tilde{d p} ω_{p} [b_{s}^{†} (\vec{p}) b_{s} (\vec{p}) - d_{s}^{†} (\vec{p}) d_{s} (\vec{p}) + {d_{s} (\vec{p}), d_{s}^{†} (\vec{p})} \\ = \sum_{s} \int \tilde{d p} ω_{p} [b_{s}^{†} (\vec{p}) b_{s} (\vec{p}) + d_{s}^{†} (\vec{p}) d_{s} (\vec{p})] - 4 E_{0} V \end{aligned}

where we noted that

{d_{s} (\vec{p}), d_{s}^{†} (\vec{p})} = 2 ω_{p} (2 π)^{3} δ^{3} (\vec{0}) = 2 ω_{p} V

, where

V

is the volume of space, and

E_{0} = \frac{1}{2} \int \frac{d^{3} p}{(2 π)^{3}} ω_{p}

is the vacuum energy. Hence, the Hamiltonion counts the number of b-type particles plus the number of d-type particles weighted by their energy.

ground state: $b_{s} (\vec{p}) | 0 ⟩ = d_{s} (\vec{p}) | 0 ⟩ = 0$ .
single-particle states: $b_{s}^{†} (\vec{p}) | 0 ⟩, d_{s}^{†} (\vec{p}) | 0 ⟩$
two-particle states:

\begin{aligned} | {\vec{p}}_{1}, s_{1}; {\vec{p}}_{2}, s_{2} ⟩ & = b_{s_{1}}^{†} ({\vec{p}}_{1}) b_{s_{2}}^{†} ({\vec{p}}_{2}) | 0 ⟩ \\ = - b_{s_{2}}^{†} ({\vec{p}}_{2}) b_{s_{1}}^{†} ({\vec{p}}_{1}) | 0 ⟩ = - | {\vec{p}}_{2}, s_{2}; {\vec{p}}_{1}, s_{1} ⟩ \end{aligned}

The antisymmetry of the states under particle exchange is konwn as "FermiDirac statistics" and implies that

\to | \vec{p}, s, \vec{p}, s ⟩ = 0

, which is known as "Pauli exclusion principle," i.e. two fermions cannot occupy the same state. In contrast, integer-spin fields satisfy bosonic (commuting) statistics, so multiple bosons can occupy the same state. As mentioned above, quantising spin-

\frac{1}{2}

fields using commuting statistics would lead to inconsistencies. Hence, Dirac theory implies Pauli exclusion principle!

Electric charge:

Let us now right the electric charge in terms of creation and annihilation operators.

\begin{aligned} Q = \int d^{3} x ψ^{†} ψ \\ = \sum_{s, s^{'}} \int \tilde{d p} \tilde{d p^{'}} d^{3} x (b_{s^{'}}^{†} ({\vec{p}}^{'}) u_{s^{'}}^{†} ({\vec{p}}^{'}) e^{- i p^{'} \cdot x} + d_{s^{'}} ({\vec{p}}^{'}) v_{s^{'}}^{†} ({\vec{p}}^{'}) e^{i p^{'} \cdot x}) \\ (b_{s} (\vec{p}) u_{s} (\vec{p}) e^{i p \cdot x} + d_{s}^{†} (\vec{p}) v_{s} (\vec{p}) e^{- i p \cdot x}) \\ = \sum_{s} \int \tilde{d p} (b_{s}^{†} (\vec{p}) b_{s} (\vec{p}) + d_{s} (\vec{p}) d_{s}^{†} (\vec{p})) \\ = \sum_{s} \int \tilde{d p} (b_{s}^{†} (\vec{p}) b_{s} (\vec{p}) - d_{s}^{†} (\vec{p}) d_{s} (\vec{p})) + const, \end{aligned}

where we performed manipluations similar to the ones used to derive (12). In summary, the electric charge is equal to the number of b-type particles minus the number of d-type particles. We can think of b-type particles as electrons and d-type particles as positrons.

13 LSZ for spin half

This is based on Srednicki Chapter 41 and Peskin pg 118-119. Recall that

ψ (x) = \sum_{s} \int d_{p} [b_{s} (\vec{p}) u_{s} (\vec{p}) e^{i p \cdot x} + d_{s}^{†} (\vec{p}) v_{s} (\vec{p}) e^{- i p \cdot x}]

where

b

annihilates

b

-type particles (ie. electrons) and

d^{†}

aretes

d

-type particles (ie. positrons). Hence,

ψ

annihilates particles and creates antiparticles. Similarly,

\bar{ψ}

annihilates antiparticles and creates particles. We can use this fact to compute scattering amplitudes of Dirac particles. For example, the

e^{-} e^{-} \to e^{-} e^{-}

amplitude is given by

⟨ f ∣ i ⟩ = ⟨ 0 | T b_{s_{4}} ({\vec{p}}_{4}, + \infty) b_{s_{3}} ({\vec{p}}_{3}, + \infty) b_{s_{2}}^{†} ({\vec{p}}_{2}, - \infty) b_{s_{1}}^{†} ({\vec{p}}_{1}, - \infty) | 0 ⟩

where

T

denotes time ordering. Since the operators ate anti-commuting, any odd permutation gives a minus sign. We expect this to be related to the following time-ordered corrolator:

⟨ 0 | T ψ (x_{4}) ψ (x_{3}) \bar{ψ} (x_{2}) \bar{ψ} (x_{1}) | 0 ⟩

where

\bar{Ψ}

's create the incoming electrons and

Ψ

's anmithate the two outgoing electrons. The precise relation to the scatting amplitude is given by the LSZ formula.

We previously showed that

b_{s}^{†} (\vec{p}) = \int d^{3} x e^{i p \cdot x} \bar{ψ} (x) γ^{0} u_{s} (\vec{p})

In an interacting theory,

b_{s}^{†} (\vec{p})

can be time-dependent:

\begin{aligned} b_{s}^{†} (\vec{p}, - \infty) - b_{s}^{†} (\vec{p}, + \infty) = - \int_{- \infty}^{\infty} d t \partial_{0} b^{†} (\vec{p}, t) \\ = - \int d^{4} x \partial_{0} (e^{i p \cdot x} \bar{ψ} (x) γ^{0} u_{s} (\vec{p})) \\ = - \int d^{4} x \bar{ψ} (x) (γ^{0} \overset{\leftarrow}{\partial} \underset{⏟}{- i γ^{0} p^{0}}) u_{s} (\vec{p}) e^{i p \cdot x} \\ - i γ^{i} p_{i} - i m, using (p̸ + m) u_{s} (\vec{p}) = 0 \\ = - \int d^{4} x \bar{Ψ} (x) (γ^{0} \overset{\leftarrow}{\partial} - γ^{i} {\vec{\partial}}_{i} - i m) u_{s} (\vec{p}) e^{i p \cdot x} \\ = i \int d^{4} x \bar{ψ} (x) (i {\overset{\leftarrow}{\partial}}_{μ} γ^{μ} + m) u_{s} (\vec{p}) e^{i p \cdot x} \end{aligned}

Note that in free theory

\bar{ψ} (i {\overset{\leftarrow}{\partial}}_{μ} γ^{μ} + m) = 0

by the EOM, but in an interacting theory this does not vanish so

b^{†}

indeed changes over time. Hence,

b_{s}^{†} (\vec{p}, - \infty) = b_{s}^{†} (\vec{p}, + \infty) + i \int d^{4} x \bar{ψ} (x) (i {\overset{\leftarrow}{\partial}}_{μ} γ^{μ} + m) u_{s} (\vec{p}) e^{i p \cdot x}

On the other hand, when we plug this into scattering amplitude, time-ordering moves

b^{†} (\vec{p}, + \infty)

to the left and it annihilates

⟨ 0 |

. Hence, we can make the replacement

b_{s}^{†} (\vec{p}, - \infty) \to i \int d^{4} x \bar{ψ} (x) (i {\overset{\leftarrow}{\partial}}_{μ} γ^{μ} + m) u_{s} (\vec{p}) e^{i p \cdot x}

Similarly we find

\begin{aligned} b_{s} (\vec{p}, + \infty) & \to i \int d^{4} x e^{- i p \cdot x} {\bar{u}}_{s} (\vec{p}) (- i ∂̸ + m) ψ (x) \\ d_{s}^{†} (\vec{p}, - \infty) & \to - i \int d^{4} x e^{i p \cdot x} {\bar{v}}_{s} (\vec{p}) (- i ∂̸ + m) ψ (x) \\ d_{s} (\vec{p}, + \infty) & \to - i \int d^{4} x \bar{ψ} (x) (i \overset{\leftarrow}{\partial}_{μ} γ^{μ} + m) v_{s} (\vec{p}) e^{- i p \cdot x} \end{aligned}

Hence, to obtain a scattering amplitude, Fourier transform a time-ordered correlator to momentum space and act with the Dirac operator, amputating external propagators. Incoming particles/outgoing antipartides are created/annihilated by

\bar{ψ}

, while incoming antiparticles/outgoing particles are created/annihilated by

ψ

, as expected. Coming back to the example of

e^{-} e^{-} \to

e^{-} e^{-}

scattering, LSZ gives

\begin{aligned} ⟨ f ∣ i ⟩ = (i)^{4} \int d^{4} x_{1} d^{4} x_{2} d^{4} x_{3} d^{4} x_{4} \\ e^{- i p_{4} \cdot x_{4}} {[{\bar{u}}_{s_{4}} ({\vec{p}}_{4}) (- i {∂̸}_{4} + m)]}_{α_{4}} e^{- i p_{3} \cdot x_{3}} {[{\bar{u}}_{s_{3}} ({\vec{p}}_{3}) (- i {∂̸}_{3} + m)]}_{α_{3}} \\ ⟨ 0 | T ψ_{α_{4}} (x_{4}) ψ_{α_{3}} (x_{3}) {\bar{ψ}}_{α_{2}} (x_{2}) {\bar{ψ}}_{α_{1}} (x_{1}) | 0 ⟩ \\ {[(i {\overset{\leftarrow}{\partial}}_{x_{2}^{μ}}^{μ} γ^{μ} + m) u_{s_{2}} ({\vec{p}}_{2})]}_{α_{2}} e^{i p_{2} \cdot x_{2}} {[(i {\overset{\leftarrow}{\partial}}_{x_{1}^{μ}} γ^{μ} + m) u_{s_{1}} ({\vec{p}}_{1})]}_{α_{1}} e^{i p_{1} \cdot x_{1}} \end{aligned}

In terms of momentum space Feynman rules, we have

where arrows on lines indicate charge plow. Note that

In general we move fields around such that contractions are untangled and pick up factors of

(- 1)

in the process. When doing so, keep the ordering of labels in

⟨ \dots |

and

| \dots ⟩

. In terms of Feynman diagrams, this means that diagrams related by an odd permutation of external legs will have a relative

(- 1)

14 Fermionic Propagator

This is based on Srednicki Chapter 42. Recall that

\begin{aligned} ψ (x) = \sum_{s} \int \tilde{d p} [b_{s} (\vec{p}) u_{s} (\vec{p}) e^{i p \cdot x} + d_{s}^{†} (\vec{p}) v_{s} (\vec{p}) e^{- i p \cdot x}] \\ \bar{ψ} (x) = \sum_{s} \int \tilde{d p} [b_{s}^{†} (\vec{p}) {\bar{u}}_{s} (\vec{p}) e^{- i p \cdot x} + d_{s} (\vec{p}) {\bar{v}}_{s} (\vec{p}) e^{i p \cdot x}] \end{aligned}

Let's compute

\begin{aligned} ⟨ 0 | T ψ_{α} (x) {\bar{ψ}}_{β} (y) | 0 ⟩ & = θ (x^{0} - y^{0}) ⟨ 0 | ψ_{α} (x) {\bar{ψ}}_{β} (y) | 0 ⟩ \\ - θ (y^{0} - x^{0}) ⟨ 0 | {\bar{ψ}}_{β} (y) ψ_{α} (x) | 0 ⟩ \\ ↑ \\ Fermi statistics \end{aligned}

First compute

⟨ 0 | ψ \bar{ψ} | 0 ⟩

. Note that the

d^{†}

ψ

and

d

\bar{ψ}

can be dropped since they annihilate the vacuum on left and right, respectively. We are then left with

⟨ 0 | ψ (x) \bar{ψ} (y) | 0 ⟩ = \sum_{s, s^{'}} \int \tilde{d p} {\tilde{d p}}^{'} e^{i p \cdot x - i p^{'} \cdot y} u_{s} (\vec{p}) {\bar{u}}_{s^{'}} ({\vec{p}}^{'}) ⟨ 0 | b_{s} (\vec{p}) b_{s^{'}}^{†} ({\vec{p}}^{'}) | 0 ⟩ .

where spinor indices are implicit. Using

b_{s} (\vec{p}) b_{s^{'}}^{†} ({\vec{p}}^{'}) = - b_{s^{'}}^{†} ({\vec{p}}^{'}) b_{s} (\vec{p}) + {b_{s} (\vec{p}), b_{s^{'}}^{†} ({\vec{p}}^{'})} = - b_{s^{'}}^{†} ({\vec{p}}^{'}) b_{s} (\vec{p}) + (2 π)^{3} 2 ω_{p} δ^{3} (\vec{p} - {\vec{p}}^{'}) δ_{s, s^{'}}

We obtain

⟨ 0 | ψ (x) \bar{ψ} (y) | 0 ⟩ = \int \tilde{d p} e^{i p \cdot (x - y)} \sum_{s} u_{s} (\vec{p}) {\bar{u}}_{s} (\hat{p}) = \int \tilde{d p} e^{i p \cdot (x - y)} (- p̸ + m)

Similarly, in

⟨ 0 | \bar{ψ} ψ | 0 ⟩

car neglect

b^{†}

\bar{ψ}

and

b

ψ

since they annihilate vacuum and we obtain:

\begin{aligned} ⟨ 0 | \bar{ψ} (y) ψ (x) | 0 ⟩ & = \sum_{s, s^{'}} \int \tilde{d p} {\tilde{d p}}^{'} e^{- i p \cdot x + i p^{'} \cdot y} v_{s} (\vec{p}) v_{s^{'}} ({\vec{p}}^{'}) ⟨ 0 | d_{s^{'}} ({\vec{p}}^{'}) d_{s}^{†} (\vec{p}) | 0 ⟩ \\ = \int \tilde{d p} e^{- i p \cdot (x - y)} (- p̸ - m) \end{aligned}

Hence, we obtain

\begin{aligned} ⟨ 0 | T ψ (x) \bar{ψ} (y) | 0 ⟩ = & θ (x^{0} - y^{0}) \int \tilde{d p} e^{i p \cdot (x - y)} (- p̸ + m) \\ + θ (y^{0} - x^{0}) \int \tilde{d p} e^{- i p \cdot (x - y)} (p̸ + m) \end{aligned}

Recall that

\int \frac{d^{4} p}{(2 π)^{4}} \frac{e^{i p \cdot (x - y)} f (p)}{p^{2} + m^{2} - i ϵ} = i θ (x^{0} - y^{0}) \int \tilde{d p} e^{i p \cdot (x - y)} f (p) + i θ (y^{0} - x^{0}) \int \tilde{d p} e^{- i p \cdot (x - y)} f (- p)

where

f (p)

is a polynomial in

p^{μ}

and

p^{0} = ω_{p}

on the right-hand-side. This can be derived by contour integration in complex

p^{0}

plane. Puting everything together we find

⟨ 0 | T ψ (x) \bar{ψ} (y) | 0 ⟩ = \frac{1}{i} \int \frac{d^{4} p}{(2 π)^{4}} e^{i p \cdot (x - y)} \frac{(- p̸ + m)}{p^{2} + m^{2} - i ϵ}

we define this to be the Wick contraction of the fields:

⟨ 0 | T ψ (x) \bar{ψ} (y) | 0 ⟩ \equiv ψ (x) \frac{◻}{ψ} (y)

Let

S (x - y) = i ⟨ 0 | T ψ (x) \bar{ψ} (y) | 0 ⟩

. This is the Green's function of the Dirac operator:

\begin{matrix} (- i \partial_{x} + m) S (x - y) = \int \frac{d^{4} p}{(2 π)^{4}} \frac{(p̸ + m) (- p̸ + m)}{p^{2} + m^{2}} e^{i p \cdot (x - y)} \\ = \int \frac{d^{4} p}{(2 π)^{4}} \frac{- {p̸}^{2} + m^{2}}{p^{2} + m^{2}} e^{i p \cdot (x - y)} = δ^{4} (x - y) \end{matrix}

where we noted that

- {p̸}^{2} = - p_{μ} p_{ν} γ^{μ} γ^{ν} = - p_{μ} p_{\underset{―}{ν}} \frac{1}{2} {γ^{μ}, γ^{ν}} = p^{2}

. Moreover it is easy to see that

⟨ 0 | T ψ (x) ψ (y) | 0 ⟩ = ⟨ 0 | T ψ (x) ψ (y) | 0 ⟩ = 0

since there is no way pair up a

b

with a

b^{†}

or a

d

with a

d^{†}

. Hence the only nontrivial Wick contraction is between

ψ

and

\bar{ψ}

, similar to a complex scalar field.

We may now evaluate time-ordered correlators of Dirac fields using Wick contractions:

Hence,

⟨ 0 | T ψ_{α} (x_{1}) {\bar{ψ}}_{β} (x_{2}) ψ_{γ} (x_{3}) {\bar{ψ}}_{δ} (x_{4}) | 0 ⟩ = {(\frac{1}{i})}^{2} [S {(x_{1} - x_{2})}_{α β} S {(x_{3} - x_{4})}_{γ δ} - S {(x_{1} - x_{4})}_{α δ} S {(x_{3} - x_{2})}_{γ β}]

In momentum space, we have the following Feynman rule for the Dirac propgator:

which arises from the Wick contraction of

ψ

with

\bar{ψ}

. The arrow on the line indices flow of charge. If momentum flows in the opposite direction of the charge
flow, take

p \to - p

in the Feynman rule. As we will see later, the direction of the arrows on internal lines are determined by the arrows on external lines (which come from LSZ) and charge conservation at the interaction vertices. One can then simply take the momentum on an internal line to point in the same direction as the charge flow and use the above Feynman rule.

15 Quantum Electrodynamics

This is based on Srednicki chapter 58 and Peskin pg 120. Recall Maxwell theory:

L = - \frac{1}{4} F_{μ ν} F^{μ ν} + J_{μ} A^{μ}

and the Dirac theory:

L = i \bar{ψ} p̸ ψ - m \bar{ψ} ψ

And recall the Noether current

J_{μ} = e \bar{ψ} γ^{μ} ψ

where

e

is the charge of electron. A simple guess for how to couple electrons/positrons to photons is to identify Noether current with the source in Maxwell theory:

L_{QED} = - \frac{1}{4} F_{μ ν} F^{μ v} + i \bar{ψ} ∂̸ ψ - m \bar{ψ} ψ + e \bar{ψ} γ^{μ} ψ A_{μ}

Is this gauge-invariant? Recall that under a gauge transformation

A_{μ} \to A_{μ} - \partial_{μ} Γ

ψ

must transform somehow in order to cancel the gauge transformation of the last term. The following definition does the job:

ψ (x) \to e^{- i e Γ (x)} ψ (x)

This can be obtained by promoting the

U (1)

global symmetry to a local one, where the

U (1)

transformation can vary from point to point in spacetime. This known as "gauging" the global symmetry. After gauging, the fermionic mass term remains invariant, but the derivative term does not.

Define the covariant derivative:

D_{μ} ψ = \partial_{μ} ψ - i e A_{μ} ψ

Under a gauge transformation we have:

D_{μ} ψ \to \partial_{μ} (e^{- i e Γ} ψ) - i e (A_{μ} - \partial_{μ} Γ) (e^{- i e Γ} ψ) = e^{- i e P} D_{μ} ψ

Hence,

\bar{ψ} D ψ \to (\bar{ψ} e^{i e Γ}) (e^{- i e Γ} D ψ) = \bar{ψ} D̸ ψ

so this is gauge-invariant. Moreover,

\begin{aligned} [D_{μ}, D_{ν}] & ψ = \\ = & (\partial_{μ} - i e A_{μ}) (\partial_{ν} ψ - i e A_{ν} ψ) - μ \leftrightarrow ν \\ - i e A_{μ} \partial_{ν} ψ - e^{2} A_{μ} A_{ν} ψ) - μ \leftrightarrow ν \\ = & - i e F_{μ ν} ψ \end{aligned}

More abstractly, we can write

F_{μ ν} = \frac{i}{e} [D_{μ}, D_{ν}]

and the auge transformation of

D_{μ} ψ

can be written as

D_{μ} ψ \to (e^{- i e Γ} D_{μ} e^{i e Γ}) (e^{- i e Γ} ψ)

D_{μ} \to e^{- i e Γ} D_{μ} e^{i e Γ}

. This provides another perspective on gauge -invariance of

F_{μ ν}

since under a gauge transformation we have

\begin{aligned} F_{μ ν} = \frac{i}{e} [D_{μ}, D_{ν}] & \to \frac{i}{e} [e^{- i e Γ} D_{μ} e^{i e Γ}, e^{- i e Γ} D_{ν} e^{i e Γ}] \\ = \frac{i}{e} e^{- i e Γ} [D_{μ} D_{ν}] e^{i e Γ} \\ = e^{- i e Γ} F_{μ ν} e^{i e Γ} \\ = F_{μ ν} \end{aligned}

Hence, a gange-invariant Lagrangian is given by

L = - \frac{1}{4} F_{μ ν} F^{μ ν} + i \bar{ψ} D ψ - m \bar{ψ} ψ

which agrees with our initial guess.

Feynman rules:

Photons are represented by wavy lines and fermions are represented by solid lines with arrows. We already derived the momentum space Feynman vales for external legs (from LSZ) and internal propagators. Now we just need to derive the interaction vertex coming from

L_{int} = e \bar{ψ} γ^{μ} ψ A_{μ}

Consider the tree-level amplitude

e^{-} \to e^{-} γ

\begin{aligned} p_{p_{2}} ↑ \hat{\sim} λ_{p_{3}}^{\overset{―}{u_{s}} (p_{2})} = ⟨ e^{-} (p_{2}) γ (p_{3}) | + i e^{\frac{7}{ψ} γ^{μ} ψ A_{μ} | e^{-} (p_{1}) ⟩} \\ = tie \in ({\vec{p}}_{3}) {\bar{u}}_{s_{2}} ({\vec{p}}_{2}) γ^{μ} u_{s_{1}} (\vec{p}) \end{aligned}

Stripping off the external polarisation and spinors, the Feynman rule for the interaction vertex is given by

Note that one arrow always points towards vertex and one always points away from vertex due to charge conservation.

Summary:

Fermions described by solid lines with arrows, photons by wavy lines.
For incoming $e^{-}$ , arrow and $p^{μ}$ point toward vertex. For outgoing $e^{-}$ , arrow and $p^{μ}$ point away from vertex. For incoming $e^{+}$ , arrow points away from vertex and $p^{μ}$ points toward vertex. For outgoing $e^{+}$ , arrow points toward and vertex and $p^{μ}$ points away from vertex.
Arrows on internal solid lines are determined by charge conservation at vertices. It is conventional to choose $p^{μ}$ to point along internal arrows. Then conserve $p^{μ}$ at each vertex.
For each internal photon, use
For each internal fermion, use
Spinor indices are contracted in opposite direction of arrows along fermion lines starting with a $\bar{u}$ or $\bar{v}$ (corresponding to an outgoing $e^{-}$ or an incoming $e^{+}$ ) and ending with $u$ or $v$ (corresponding to an in coming $e^{-}$ or outgoing $e^{+}$ ).
Dress external photons with polarisations $ϵ_{μ}$
The overall sign of a diagram is determined by drawing all fermian lines pointing in the same direction with a fixed ordering of the incoming arrows. If the ordering of the outgoing arrows is an odd permutation of some given ordering, then that diagram gets a ( -1 ).

For example, at tree-level

e^{-} e^{-} \to e^{-} e^{-}

is described by

so the second diagram comes with a ( -1 ).

Fermion loops:

To compute Feynman diagrams with fermion loops, we need two more rules:

Take the trace of the product of $γ$ matrices from the vertices and fermionic propagator numerators that arise when going backwards along fermion lines.
Multiply by $(- 1)$ for each fermion loop.

Example:

where we leave out the

γ

matrices coming from the interaction vertices and the minus sign comes from anticommuting the

ψ

on the right all the way to the left.

16 Scattering in QED

This is based on Tong sections 6.6, 3.5.2, and Peskin pg 121,125, 126. Let us consider a few examples of tree-level scattering amplitudes.

$e^{-} e^{-} \to e^{-} e^{-}$

where the minus sign in the second term arises from the exchange of legs 3 and 4 , and we use shorthand: $u_{i} = u_{s_{1}} ({\vec{p}}_{i})$ .
$e^{-} e^{+} \to γ γ$

= (i e)^{2} [{\bar{v}}_{2} γ_{μ} ϵ_{3}^{μ} \frac{i ({p̸}_{1} - {p̸}_{4} - m)}{{(p_{1} - p_{4})}^{2} + m^{2}} γ_{ν} ϵ_{4}^{ν} u_{1} + {\bar{v}}_{2} γ_{μ} ϵ_{4}^{μ} \frac{i ({p̸}_{1} - {p̸}_{3} - m)}{{(p_{1} - p_{3})}^{2} + m^{2}} γ_{ν} ϵ_{3}^{ν} u_{1}]

In this case there is no relative minus sign since photons are bosonic

$- e^{-} e^{+} \to e^{-} e^{+}$ (Bhabha scattering)

= (i e)^{2} [{\bar{u}}_{4} γ^{μ} u_{1} \frac{- i η_{μ ν}}{{(p_{1} - p_{4})}^{2}} {\bar{v}}_{2} γ^{ν} v_{3} - {\bar{v}}_{2} γ^{μ} u_{1} \frac{- i η_{μ ν}}{{(p_{1} + p_{2})}^{2}} {\bar{u}}_{4} γ^{ν} v_{3}]

The relative minus sign can be understood by drawing all fermion lines in the same direction with a fixed order of the incoming lines:

After doing so, we see that the first diagram is related to second one by exchanging legs 2 and 4.

$e^{-} γ \to e^{-} γ$ (Compton scattering)

You will look at this in the HW.

Coulomb potential:

We will now show how the Coulomb potential arises from charged partides exchanging photons. To see this, let's take the non-relativistic limit of the following diagram which contributes to

e^{-} e^{-} \to e^{-} e^{-}

Recall that in this limit.

u_{s} (\vec{0}) = \sqrt{m} (\binom{ξ_{s}}{ξ_{s}}) .

From this we find

{\bar{u}}_{s} (\vec{0}) γ^{μ} u_{s^{'}} (\vec{0}) = m ξ_{s}^{†} (σ^{μ} + {\bar{σ}}^{μ}) ξ_{s^{'}} = {\begin{cases} 2 m δ_{s, s^{'}} & μ = 0 \\ 0 & μ \neq 0 \end{cases}

To compute the denominator, note that in this limit

p^{μ} = (m, \vec{p}) \to {(p_{1} - p_{4})}^{2} = {({\vec{p}}_{1} - {\vec{p}}_{4})}^{2}

Hence we obtain

\frac{i M}{(2 m)^{2}} = \frac{- i e^{2} δ_{s_{1}, s_{4}} δ_{s_{3}, s_{2}}}{{(\vec{p_{1}} - \vec{p_{4}})}^{2}} \equiv A_{4} δ_{s_{1}, s_{4}} δ_{s_{3}, s_{2}}

In non-relativistic QM, the amplitude for a particle to scatter from

| {\vec{p}}_{1} ⟩

| {\vec{p}}_{4} ⟩

due to a potential

V (\vec{x})

(in the Born approximation) is the Fourier transform of the potential:

A_{4} ({\vec{p}}_{4} - {\vec{p}}_{p}) = - i \int d^{3} x V (\vec{x}) e^{- i ({\vec{p}}_{4} - {\vec{p}}_{1}) \cdot \vec{x}}

Hence,

\begin{aligned} \int d^{3} x V (\vec{x}) e^{- i ({\vec{p}}_{4} - {\vec{p}}_{1}) \cdot \vec{x}} = \frac{e^{2}}{{({\vec{p}}_{4} - {\vec{p}}_{1})}^{2}} \\ \to V (\vec{x}) = e^{2} \int \frac{d^{3} p}{(2 π)^{3}} \frac{1}{{\vec{p}}^{2}} e^{i \vec{p} \cdot \vec{x}} = \frac{e^{2}}{4 π | \vec{x} |} \end{aligned}

which is the Coulomb potential. For details on how to compute integral on LHS, see Tong pg 67. Alternatively, it easy to see that both sides satisfy the same differential equation:

◻_{x} V (\vec{x}) = - δ^{3} (\vec{x})

17 Renormalisation of QED

This is based on Kabat section 7.2.2 and Appendix C. The Coulomb potential receives quantum corrections from diagrams of the form

This has observable effects, for example the "Lamb shift" of the energy levels of

Hydrogen. Heuristically the virtual electron and positron in the loop screen the electron. This is known as "vacuum polarisation." This screening effect causes the electron charge to decrease at long distances, or low energies. Mathematically, this can be derived from renormalisation.

Consider QED with a cutoff

Λ

on Euclidean momentum:

L = - \frac{1}{4} Z_{A} F_{μ ν} F^{μ ν} + i \bar{ψ} D ψ - m \bar{ψ} ψ, D_{μ} = \partial_{μ} - i e A_{μ}

where

Z_{A}

encodes the normalisation of the gauge field

A_{μ}

. As we will see shortly, the normalisation will run with the scale

Λ

. This is known as "wavefunction renormalisation". Now imagine lowering the cutoff to

Λ^{'} = Λ - δ Λ

and write down a new theory

L^{'} = - \frac{1}{4} Z_{A}^{'} F_{μ ν} F^{μ ν} + i \bar{ψ} D ψ - m \bar{ψ} ψ

where Euclidean momentum is restricted to

| {\vec{k}}_{E} | < Λ^{'}

. The two theories will agree provided we relate

Z_{A}

and

Z_{A}^{'}

appropriately. For

δ Λ / Λ

small, write

Z_{A}^{'} = Z_{A} - \frac{d Z_{A}}{d Λ} δ Λ

Plugging this into primed Lagrangian, we see that it has an extra term compared to the unprimed Lagrangian:

L^{'} = L + \frac{1}{4} F_{μ ν} F^{μ ν} \frac{d Z_{A}}{d Λ} δ Λ

The extra term can be thought of as a new 2-point interaction vertex with the following Feynman rule:

Note that Fourier transforming

F_{μ ν} F^{μ ν}

to momentum space gives

2 (η_{μ ν} k^{2} - k_{μ} k_{ν})

, as we've seen before. An addifiond factor of 2 comes from the 2 possible ways of Wick contracting the two external states onto the vertex.

On the other hand in the unprimed theory, the photon propagator receives corrections from vacuum polarisation diagram given at the beginning of the chapter where the Euclidean loop momentum is restricted to

Λ^{'} < | ℓ_{E} | < Λ

and we neglect higher-loop corrections. In the HW, you will show that after amputating external legs this diagram evaluates to

= - 4 i e^{2} (η_{μ ν} k^{2} - k_{μ} k_{ν}) \int_{0}^{1} d x \int \frac{d^{4} q}{(2 π)^{4}} \frac{2 x (1 - x)}{{(q^{2} + k^{2} x (1 - x) + m^{2})}^{2}} + \dots

where

q

is a Euclidean loop momentum with

Λ^{'} < | \vec{q} | < Λ

, and ... denotes terms that break gauge invariance. Such terms are an artifact of working with a cutoff and don't arise if we use a regulator that preserves gauge invariance like dimensional regularisation, so we ignore them.

Hence, for the two theories to agree, we need

\frac{d Z_{A}}{d Λ} δ Λ = - 4 e^{2} \int_{0}^{1} d x \int \frac{d^{4} q}{(2 π)^{4}} \frac{2 x (1 - x)}{{(q^{2} - k^{2} x (1 - x) + m^{2})}^{2}}

Assuming that

k^{2} ≪ Λ^{2}

and

m^{2} ≪ Λ^{2}

, we can set

k^{2} = m^{2} = 0

\begin{aligned} \frac{d Z_{A}}{d Λ} δ Λ & = - 4 e^{2} \int_{0}^{1} d x 2 x (1 - x) \int_{Λ^{'}}^{Λ} \frac{2 π^{2} q^{3} d q}{(2 π)^{4}} \frac{1}{q^{2}} \\ \to \frac{d Z_{A}}{d Λ} & = - \frac{e^{2}}{6 π^{2} Λ} \end{aligned}

Hence normalisation of Maxwell kinetic term depends on cutoff! If we rescale

A_{μ} \to 1 / \sqrt{Z_{A}} A_{μ}

to get a canonical kinetic term for the Maxwell field, then the interaction term becomes

L_{int} = e \bar{ψ} γ^{μ} ψ A_{μ} \to e_{phys} \bar{ψ} γ^{μ} ψ A_{μ}

where

e_{phys} = \frac{e}{\sqrt{Z_{A}}}

. Hence, we find that

β (e_{phys}^{2}) = \frac{d}{d \ln Λ} e_{phys}^{2} = \frac{d}{d \ln Λ} (\frac{e^{2}}{Z_{A}}) = - \frac{e^{2}}{Z_{A}^{2}} \frac{d Z_{A}}{d \ln Λ} = \frac{e_{phys}^{4}}{6 π^{2}} > 0

This differential equation is then solved by

\frac{1}{e_{phys}^{2} (Λ)} = \frac{1}{e_{phys}^{2} (μ)} - \frac{1}{6 π^{2}} \ln (\frac{Λ}{μ})

where

μ

is the renormalisation scale. Hence

e_{phys}^{2}

increases as

Λ

increases, and there is a Landau pole at

Λ_{max} = μ \exp (6 π^{2} / e_{phys}^{2} (μ))

, like in the

ϕ^{4}

theory.

AQT Term 2

Lecturer: Arthur Lipstein

Contents

1 Introduction

References

Overview

2 Path Integral in QM

3 Path integral for free QFT

4 Path integral for interacting QFT

5 LSZ reduction

6 Scattering Amplitudes and Feynman Rules

7 Loops and Renormalisation

8 Classical Maxwell Theory

Gauge symmetry:

Action:

Equations of motion:

Gauge-fixing:

9 LSZ and Path Integral For Photons

10 Dirac Equation

Representations of Lorentz group:

Spin-1/2

Constructing an Action:

Equations of motion:

Hamiltonian:

Electric charge:

11 Classical Solutions of the Dirac Equation

Examples:

12 Quantizing the Dirac field

Hamiltonian:

Electric charge:

13 LSZ for spin half

14 Fermionic Propagator

15 Quantum Electrodynamics

Feynman rules:

Summary:

Fermion loops:

16 Scattering in QED

Coulomb potential:

17 Renormalisation of QED