AIMS-pp-06.pdf

Quantum  Electrodynamics

Equations

Gauge Derivative

Field Strength

Electric Current

Summary so far:

One Dirac Eqn for each charged fermion

We can test QED by determining its predictions.

Bound States: Electromagnetism is the force that binds atoms together. By considering the Maxwell-Dirac equations for bound systems we can search for phenomena peculiar to QED (do later).

Scattering: As with the toy theory, we wish to compute

Fine-structure constant

q₁, q₂, q₃,…, q_j,

Feynman Rules for QED

photon

fermion

antifermion

Time 1 NOTATION

p

,

s

p

,

s

k

,

ε

p

,

s

′

p

,

s

′

′

p

,

s

′

′

p

,

s

′

′

k

,

ε

′

Free particles in the past  come from sources

Free particles in the future 

absorbed by detectors

Interactions in the middle

2 EXTERNAL LINES

scalars

=

1

incoming

outgoing

fermions

( )

incoming outgoing

anti

fermions

= u p

(

)

= v p,

( )

= v p

(

)

photons

µ p

( )

= εµ*'

p'

( )

=

1

•  The external lines are the wavefunctions of the incoming or

outgoing particles

•  Since these are free particles, we use the free particle wavefunctions

q

i

q

−

m

=

scalar

i

(

γ

⋅

q

+

m

)

q

−

m

=

fermion/ antifermion

µ

ν

−

i

q

g

=

photon

q

q

3 INTERNAL LINES

For each internal line, multiply by d 4

q

2π

( )

5 INTEGRATE OVER INTERNAL MOMENTA

4 CONSERVE MOMENTUM AT EACH VERTEX

6 VERTEX FACTOR

=

−

ig

γ

µ

µ

This is what characterizes QED.

To get all contributions for a given process draw all possible diagrams by joining up all internal vertex points to the external lines or to each other consistent with rule 6 that are topologically distinct. The sum of all

diagrams is − iM

7 TOPOLOGY

8 ANTISYMMETRIZATION

Include a minus sign between diagrams that differ

(a)  only in the interchange of two incoming (or outgoing)

fermions/anti-fermions of the same kind

(b) only in the interchange of an incoming fermion with an outgoing antifermion of the same kind (or vice versa)

These are the Feynman rules for QED

9 LOOPS

After integrating the result will have a factor

2π

( )

δ( )4 ′

p₁,+p′₂+ …+ p_m′ -p₁ − p₂- …- p_n

(

)

Cancel this out and what remains in - iM

10 CANCEL THE OVERALL DELTA FUNCTION

Example

e− + µ−

p

p

p

'

p

'

electron muon

q

p

p

p

'

p

'

=

_{+ …}

current propagator current

e

µ

e

µ

Rules give:

+

1

p p₂

2

' p 1

'

p

1

p p2

2

'

p

1 ' p

1 p 2 p 2 ' p 1 ' p 1

p p2

2 ' p 1 ' p Relative Sign? 1

p p2

2 ' p 1 ' p 1

p p2

2 ' p 1 ' p Switch incoming electron with outgoing positron 1

p p2

2 ' p 1 ' p

= So we

Rules give: p1

2 p

2

' p

1

'

p

2 p 1

p

1

'

p p'2

+

Back to: _e− + µ− → _e− + µ−

q

p

p

p

'

p

'

p

p

p

'

p

'

electron muon

=

_{+ …}

Spin-average & spin-sum:

How to compute this?

M2 =M†M= e4

p₁′ − p₁

(

)

′₎

(p₁′)γµu(i1)(p

1)

⎡

⎣ ⎤⎦ u(i2 ′₎

(p₂′ )γ_µu(i2)(p

2)

⎡

⎣ ⎤⎦

(

)

u(i1 ′₎

(p₁′)γνu(i1)(p

1)

⎡

⎣ ⎤⎦ u(i2 ′₎

(p₂′ )γ_νu(i2)(p

2)

⎡

⎣ ⎤⎦

(

)

Physically our machines don’t control spin:

Use completeness relations:

To demonstrate the Casimir Tricks:

Use Casimir trick here and here

a = γ µa_µ

In this case the Casimir trick is:

Result: This is a Lorentz-_{invariant scalar that}

can be put into a cross-section formula!

Lab frame:

p₁

p₂

p´₁

p´₂

Lµν

(

)

Final result:

Put this into the lab-frame cross-section & you’_{re done!}

Limits

Mott Formula

e

+

µ

→

e

+

µ

p₁

p₂

p'₂

p'₁

electron muon

q

=

_{+ …}

Summary:

p'₁

p₂

p'₂

spin- average

and spin- sum

p₁

Lab frame:

p₁ p₂

p´₁

p´₂

e− µ− _e−

Propagator

Features  of  QED  processes

Coupling Each vertex  α

p₁

q p'₁

p₂

p'₂

Matrix Element

at high energies

Dimensionality

at high energies area (length)2  2

(momentum)2  

2

electron-muon scattering

+

Electron-electron scattering

(Moller scattering)

+

Electron-positron scattering

(Bhaba scattering)

Elastic Processes

+

Pair creation

+

Pair annihilation

Inelastic Processes

Compton Scattering

+

Anomalous Magnetic Moment

µB=

e

2mc= 5.78838263×10

-5 _T-1

Bohr magneton

g

=

2

= _{+ +...}

= eφ − e

2m



σ •B _{= eφ α}

π dq q 0 ∞

∫

⎝⎜ ⎞⎠⎟σ • 

B

≈ eφ 1+ α π dq q 0 ∞

∫

2m 1+

α 2π ⎛

⎝⎜ ⎞⎠⎟σ•



B

renormalized charge anomalous magnetic

moment

m

agne

t

Let’s look at this in terms of QED diagrams:

To measure its magnetic moment, we need to allow the electron to interact with a magnet:

Electron

Muon

theory

experiment

the best agreement between theory and experiment that we have!

Lamb  Shift

Exact solution for H-atom

E = mrc

2

1+ ( )Zα

2

n−(j +1 / 2)+ (j +1 / 2)2 −( )Zα 2

⎡

⎣⎢ ⎤⎦⎥

2

n=principal quantum number

j = 1 2,

3 2,

5 2,…

= total angular momentum

E = m_rc2 1− 1 2

Zα

( )2

n2 +

3 8

Zα

( )4

n4 −

Zα

( )4

n3(2j+1)+

+

Running Coupling Constant

p₁

′

p₁ p′2

p₂

p₁

′

p₁ p′2

Fourier Transform

QED makes no sense for distances shorter than this

10-291_cm r

α =1 /137

-

--

-+ + ₊

+

+ +

+ +

The vacuum is a polarizable