Quantum Electrodynamics

Wei Wang and Zhi-Peng Xing

Wei Wang and Zhi-Peng Xing (SJTU) QED 1 / 35

Contents

2 e+e− → µ+µ−

3 Helicity Amplitudes

4 ep→ ep

5 Compton Scattering

6 e+e− → 2γ

Quantum Electrodynamics

Lagrangian density for QED ,

L = ψ(x)γµ(i∂µ − eAµ)ψ(x)−mψ(x)ψ(x)− 14FµνF

Equation of motion are

(iγµ∂µ −m)ψ(x) = eAµγµψ(x) non-linear coupled equations

∂νFµν = eψ(x)γµψ(x)

QuantizationWriteL = L0 + Lint

L0 = ψ(x)(iγµ∂µ −m)ψ(x)− 14FµνF

Lint = −eψ(x)γµψ(x)Aµ

whereL0,free field Lagrangian, Lint is interaction part.Conjugate momenta for fermion

∂L∂(∂0ψα)

= iψ†α(x)

For em fields choose the gauge

~∇ · ~A = 0

Conjugate mometa

πi = ∂L∂(∂0Ai)

= −F 0i = Ei

From equation of motion

∂νF0ν = eψ†ψ ⇒ −∇2A0 = eψ†ψ

A0is not an independent field ,

A0 = e∫d3x′ ψ

†(x′,t)ψ(x′,t)

4π|~x′−~x|= e

∫d3x′ ρ(x

′,t)

|~x′−~x|

Commutation relations:

[ψα(~x, t), ψ†β(~x′, t)] = δαβδ3(~x− ~x′), [ψα(~x, t), ψβ(~x′, t)] = ... = 0

[Ai(~x, t), Aj(~x′, t)] = iδtrij (~x− ~x′)

δtrij (~x− ~y) =∫

d3k(2π)3

ei~k·(~x−~y)(δij − kikj

Commutators involving A0:

[A0(~x, t), ψα(~x′, t)] = e∫

d3x′′

4π|~x−~x′′| [ψ†(~x′′, t)ψ(~x′′, t), ψα(~x′, t)] =

− e4π

ψα(~x′,t)|~x−~x′|

Hamiltonian density H = ∂L∂(∂0ψα)

ψα + ∂L∂(∂0Ak)

Ak − L= ψ†(−i~α · ~∇+ βm)ψ + 1

2( ~E2 − ~B2) + ~E · ~∇A0 + eψγµψAµ

H =∫d3xH =

∫d3xψ†[~α · (−i~∇− e ~A) + βm]ψ + 1

2( ~E2 − ~B2)

A0does not appear in the interaction,But if we write

~E = ~El + ~Et where ~El = −~∇A0 , ~Et = ∂ ~A∂t

∫d3x( ~E2 − ~B2) = 1

∫d3x~E2

l +∫d3x( ~E2

t − ~B2)

longitudinal part is

∫d3x~E2

l = e4pi

∫d3xd3y ρ(~x,t)ρ(~y,t)|~x−~y| Coulomb interaction

Without classical solutions, can not do mode expansion to get creationand annihilation operators We can only do perturbation theory.Wei Wang and Zhi-Peng Xing (SJTU) QED 5 / 35

Recall that the free field part ~A0 satisfy massless Klein-Gordon equation

~A(0) = 0

The solution is

~A(0)(~x, t) =

∫d3k√

2ω(2π)3

~ε(k, λ)[a(k, λ)eikx + a†(k, λ)eikx] w = k0 = |~k|

~ε(k, λ), λ = 1, 2 with ~k · ~ε(k, λ) = 0

Standard choice

~ε(k, λ) · ~ε(k, λ′) = δλλ′ , ~ε(−k, 1) = −~ε(k, 1), ~ε(−k, 2) = ~ε(−k, 2)

It is convienent to write the mode expansion as,

Aµ(~x, t) =

∫d3k

2ω(2π)3

εµ(k, λ)[a(k, λ)e−ikx + a†(k, λ)eikx]

εµ(k, λ) = (0,~ε(k, λ))

Photon PropagatorFeynman propagatpr for photon is

iDµν(x, x′) = 〈0|T (Aµ(x)Aν(x′))|0〉= θ(t− t′)〈0|Aµ(x)Aν(x′)|0〉+ θ(t− t′)〈0|Aν(x′)Aµ(x)|0〉

Using mode expansion,

iDµν(x, x′) =

(2π)4e−ik(x

′−x)

k2 + iε

2∑λ

= 1εν(k, λ)εµ(k, λ)

polarization vectorsεµ(k, λ), λ = 1, 2 are perpendicular to each other. Add2 more unit vectors to form a complete set

ηµ = (1, 0, 0, 0), kµ =kµ − (k · η)ηµ√

(k · η)2 − k2completeness relation is then,

2∑λ=1

εν(k, λ)εµ(k, λ) = −gµν − ηµην − kµkν

= −gµν −kµkν

(k · η)2 − k2 +(k · η)(kµην + ηµkν)

(k · η)2 − k2 − k2ηµην(k · η)2 − k2

If we define propagator in momentum space as

Dµν(x, x′) =

(2π)4e−ik·(x

′−x)Dµν(k)

Dµν(k) =1

k2 + iε[−gµν −

kµkν(k · η)2 − k2 +

(k · η)(kµην + ηµkν)

(k · η)2 − k2 − k2ηµην(k · η)2 − k2 ]

terms proportional tokµ will not contribute to physical processes and thelast term is of the formδµ0 δν0will be cancelled by the Coulom interaction..

Feynman rule in QEDThe interaction Hamiltonian is ,

Hint = e

∫d3xψγµψAµ

The Feynman propagators, vertices and external wave functions are givenbelow.

e+e− → µ+µ−: Total Cross Section

The momenta for this reaction is used as

e+(p′) + e−(p)→ µ+(k′) + µ−(k)

and the Feynman diagram is given as

e−(p)

e+(p′)

µ−(k)

µ+(k′)

Use Feynman rule to write the matrix element as

iM = v(p′, s′)(−ieγµ)u(p, s)−igµνq2

u(k′, r′)(−ieγν)v(k, r)

q2v(p′, s′)γµu(p, s)u(k′, r′)γµv(k, r),

where q = p+ p′.Wei Wang and Zhi-Peng Xing (SJTU) QED 10 / 35

e+e− → µ+µ−

Note that electron vertex have property,

qµv(p′)γµu(p) = (p+ p′)µv(p′)γµu(p) = v(p′)(p/+ p/′)u(p) = 0.

This shows the term proportional to photon momentum qµ will notcontribute in the physical processes. For cross section, we needM∗ whichcontains factor(vγµu)∗

(vγµu)∗ = u†(γµ)†(γ0)†v = u†γ0γ

More generally,

(vΓu)∗ = uΓv, Γ = γ0Γ†γ0.

e+e− → µ+µ−

It is easy to see

γµ = γµ

γµγ5 = −γµγ56 a 6 b · ·· 6 p =6 p · ·· 6 b 6 a

unpolarized cross section which requires the spin sum,∑s

uα(p, s)uβ(p, s) = (6 p+m)αβ∑s

vα(p, s)vβ(p, s) = (6 p−m)αβ

A typical calculation is,∑s,s′

vα(p′, s′)(γµ)αβuβ(p, s)uρ(p, s)(γν)ρσvσ(p, s)

=∑s,s′

vα(p′, s′)(γµ)αβ(p/+m)βρ(γν)ρσvσ(p, s)

= (γµ)αβ(p/+m)βρ(γν)ρσ(p/′ −m)σα

= Tr[γµ(p/+m)γν(p/′ −m)]

trace of product of γ matrices:

Tr[γµ] = 0

Tr[γµγν ] = 4gµν

Tr[γµγνγαγβ] = 4(gµνgαβ − gµαgνβ + gµβgνα),

Tr[a/1a/2 · · · a/n] = 0, n odd

With these tools∑spin

|M|2 =e4

q4Tr[(p/′ −me)γ

µ(p/+me)γν ]Tr[(k/′ +mµ)γµ(k/+mµ)γν ]

e+e− → µ+µ−: cross section

Tr[(p/′ −me)γµ(p/+me)γ

ν ] = Tr[p/′γµp/γν ]−m2Tr[γµγν ]

= 4[p′µpν − gµν(p · p′) + pµp′ν ]− 4m2egµν ,

Tr[(k/′ +mµ)γµ(k/+mµ)γν ] = Tr[k/′γµk/γν ]−m2Tr[γµγ

= 4[k′µkν − gµν(k · k′) + kµk′ν ]− 4m2µg

for energies mµ.

∑spin

|M|2 = 8e4

q4[(p · k)(p′ · k′) + (p′ · k)(p · k′)]

In center of mass,

pµ = (E, 0, 0, E), p′µ = (E, 0, 0,−E)

kµ = (E,~k), k′µ = (E,−~k), with~k · z = E cos θ

If we set mµ = 0, E = |~k|and

q2 = (p+ p′)2 = 4E2, p · k = p′ · k′ = E2(1− cos θ)p′ · k = p · k′ = E2(1 + cos θ)Wei Wang and Zhi-Peng Xing (SJTU) QED 14 / 35

e+e− → µ+µ−: cross section

We choose the kinematics as follows:

∑spin

|M |2 = 88e4

16E4[E4(1− cos θ)2 + E4(1 + cos θ)2] = e4(1 + cos2 θ).

Note that under the parity θ → π − θ. this matrix element conserves theparity The cross section is

dσ =1

2E(2π)4δ4(p+ p′ − k − k′)1

∑spin′

|M|2 d3k

(2π)32ω

d3k′

(2π)32ω′Wei Wang and Zhi-Peng Xing (SJTU) QED 15 / 35

use the δ-function to carry out integrations . introduce the quantityρ,called the phase space, given by

∫(2π)4δ4(p+ p′ − k − k′) d3k

(2π)32ω

d3k′

(2π)32ω′

∫δ(2E − ω − ω′) d

4ωω′=

∫δ(E − ω)

k2dkdΩ

The flux factor is

√(p1 · p2)2 −m2

1m22 =

E22E2 = 2

The differential crossection is then

dσ =1

∑spin

|M|2) dΩ

16E2(1 + cos2 θ)

whereα = e2

4π is the fine structure constant.The total cross section is

σ(e+e− → µ+µ−) =α2π

3swith s = (p1 + p2)

2 = 4E2

e+e− →hadrons

One of the interesting procesess ine+e− collider is the reactione+e− →hadrons

According to QCD, theory of strong interaction, this processes will gothrough

e+e− → qqand then qq trun into hadrons. Since coupling of γ to qq differs from thecoupling toµ+µ− only in their charges cross section forqq as

σ(e+e− → qq) = 3(Q2q)

4α2π

3s= 3(Q2

q)σ(e+e− → µ+µ−),

Qq is the electric charge of quark q. The factor of 3 because each quarkhas 3 colors. Then

R =σ(e+e− → hadrons)

σ(e+e− → µ+µ−)= 3(

Q2i ),

This has used the quark-hadron duality.Wei Wang and Zhi-Peng Xing (SJTU) QED 17 / 35

e+e− →hadrons

Summation is over quarks which are allowed by the avaliable energies. e.g., for energy below the the charm quark only u,d, and s quarks should beincluded,

σ(e+e− → µ+µ−)= 3[(

3)2 + (

3)2] = 2

which is not far from the reality.

e+e− →hadrons

Summation is over quarks which are allowed by the avaliable energies. e.g., for energy below the the charm quark only u,d, and s quarks should beincluded,

σ(e+e− → µ+µ−)= 3[(

3)2 + (

3)2] = 2

which is not far from the reality.

Helicity Amplitudes

In the following, we will give the helicity amplitudes.

One can parametrize the momentum as

The spinor on the direction (θ, φ) is given as

The polarization vector corresponding to the momentum (θ, φ) isgiven as:

e−µ− → e−µ− and ep→ ep

e(k)+p(p)→e(k’)+p(p’)

e−(k)

e(k′)

p(p′)

Proton has strong interaction. First consider proton has no stronginteraction and include strong interaction later. The lowest ordercontribution is ,

M(e+ p→ e+ p) = u(p′, s′)(−ieγµ)u(p, s)(−igµνq2

)u(k′, r′)(−ieγν)u(k, r)

q2u(p′, s′)γµu(p, a)u(k′, r′)(−ieγν)u(k, r)

where q = k- k’ . For unploarized cross section, sum over the spins ,

∑spin

|M(e+ p→ e+ p)|2 =e4

q4Tr[(6 p′ +M)γµ(6 p+M)γν ]Tr[(6 k′ +Me)γµ(6 k +Me)γ

ν ]Wei Wang and Zhi-Peng Xing (SJTU) QED 21 / 35

Again neglectme. Compute the traces

Tr[6 k′γµ 6 kγν ] = 4[k′µkν − gµν(k · k′) + kµk′ν ]Tr[(6 p′ +M)γµ(6 p+M)γν ] = 4[p′µpν − gµν(p · p′) + pµp′ν ] + 4M2gµν

∑spin

|M(e+ p→ e+ p)|2 =e4

q48[(p · k)(p′ · k′) + (p′ · k)(p · k′)]− 8M2(k · k′)

More useful to use the laboratoy frame

pµ = (M, 0, 0, 0), kµ = (E,~k), k′µ = (E′,~k′)

p · k = ME, p · k′ = ME′, k · k′ = EE′(1− cos θ)p′ ·k′ = (p+k−k′)·k′ = p·k′+k ·k′, p′ ·k = (p+k−k′)·k = p·k−k ·k′

q2 = (k − k′)2 = −2k · k′ = −2EE′(1− cos θ)

Differential cross section is

dσ =1

2k0(2π)4δ4(p+ k − p′ − k′)1

∑spin′

|M |2 d3p′

(2π)32p′0

d3k′

(2π)32k′0

The phase space is

∫(2π)4δ4(p+ k − p′ − k′) d3p′

(2π)32p′0

d3k′

(2π)32k′0(1)

∫δ(p0 + k0 − p′0 − k′0)

d3k′

2p′02k′0

p′0 =

√M2 + (~p+ ~k − ~k′)2 =

√M2 + (~k − ~k′)2

Use the momenta in lab frame,

∫δ(M + E − p′0 − E′)

k′2dk′dΩ

2p′02E′

∫δ(M + E − p′0 − E′)

E′dE′dΩ

p′0Let

x = −E + p′0 + E′

dx = dE′(1 +dp′0dE′

) = dE′(p′0 + E′ − E cos θ

p′0)

∫δ(x−M)

dωE′dx

(p′0 + E′ − E cos θ)=

4π2dΩE′

M + E(1− cos θ)

From the argument of the δ− function we get therelation,M = x = −E + p′0 + E′

Solve for E’ ,

E′ =ME

E(1− cos θ) +M=

1 + (2EM ) sin2 θ2

The phase space is then

ρ =dΩ

4π2ME

(M + E(1− cos θ))2=

4π2E′2

The flux factor is

MEp · k = 1

The differential cross section is then

dσ =1

2k0(2π)4δ4(p+ k − p′ − k′)1

∑spin′

|M |2 d3p′

(2π)32p′0

d3k′

(2π)32k′0

4π2E′2

∑spin′

|M |2 = (E′

16π2M2

q48[(p · k)(p′ · k′) + (p′ · k)(p · k′)]− 8M2(k · k′)

It is staightforward to get

[(p · k)(p′ · k′) + (p′ · k)(p · k′)]−M2(k · k′)= [(p · k)((p+ k − k′) · k′) + (p · k′)(p+ k − k′) · k −M2(k · k′)]

= 2EE′M2 +M2EE′(1− cos θ)(− q2

2M2− 1)

= 2EE′M2[cos2θ

2− q2

2M2sin2 θ

dΩ= (

E)2α2

(4EE′ sin2 θ2)

2EE′M2[cos2θ

2− q2

2M2sin2 θ

sin4 θ2

[cos2θ

2− q2

2M2sin2 θ

sin4 θ2

[cos2 θ2 −q2

2M2 sin2 θ2 ]

[1 + (2EM ) sin2 θ2 ]

Now include the strong interaction. Use the fact that theγPP interactionis local to parametrize theγPPmatrix element as

〈p′|Jµ|p〉 = u(p′, s′)[γµF1(q2) +

iσµνqν

2MF2(q

2)]u(p, s) with q = p− p′ (2)

Lorentz covariance and current conservation have been used. Anotheruseful relation is the Gordon decomposition

u(p′)γµu(p) = u(p′) = u(p′)[(p+ p′)µ

2m+iσµν(p′ − p)ν

2m]u(p)

F1(q2), charge form factor

F2(q2), magnetic form factor .

Note thatF1(q2) = 1andF2(q

2) = 0 correspond to point particle.The charge form factor satifies the conditionF1(0) = 1.Form

Q|p〉 = |p〉

we get

〈p′|Q|p〉 = 〈p′|p〉 = 2E(2π)3δ3(~p− ~p′)Wei Wang and Zhi-Peng Xing (SJTU) QED 27 / 35

On the other hand from Eq(2) we see that

〈p′|Q|p〉 =

∫d3x〈p′|J0(x)|p〉 =

∫d3〈p′|J0(0)|p′〉ei(p′−p)·x

= 2π)3δ3(~p− ~p′)u(p′, s′)γ0u(p, s)F1(0)

= 2E(2π)3δ3(~p− ~p′)F1(0)

compare two equations ⇒ F1(0) = 1. To gain more insight, write Q interms of charge density

∫d3xρ(x) =

∫d3xJ0(x)

〈p′|J0(x)|p〉 = eiq·x〈p′|J0(0)|p〉 = eiq·xF1(q2)u(p′, s′)γ0u(p, s)

F1(q2)is the Fourier transform of charge density distribution i.e.

F1(q2) ∼

∫d3xρ(x)e−i~q·~x

ExpandF1(q2)in powers ofq2,

F1(q2) = F1(0) + q2F ′1(0) + · · ·

F1(0) is total charge andF ′1(0)is related to the charge radius.Calulate cross section as before,

[cos2 θ2( 11−q2/4M2 )[G2

E − (q2/4M2)G2M ]− q2

2M2 sin2 θ2G

sin4 θ2 [1 + (2EM ) sin2 θ

GE = F1 + q2

GM = F1 + F2

Experimentally,GEandGM have the form,

GE(q2) ≈ GM (q2)

χp≈ 1

(1− q2/0.7Gev2)2

whereχp = 2.79 magnetic moment of the proton. If proton were pointlike, we would haveGE(q2) = GM (q2) = 1Dependence ofq2 in Eq(3)⇒ proton has a structure. For largeq2 theelastic cross section falls off rapidly asGE ≈ GM ∼ q−4.

Compton Scattering

γ(k) + e(p)→ γ(k′) + e(p′)

Two diagrams contribute,

The amplitude is given by

M(γe→ γe) = u(p′)(−ieγµ)ε′µ(k′)i

6 p− 6 k −m(−ieγν)εν(k)u(p)

+u(p′)(−ieγµ)εµ(k)i

6 p− 6 k′ −m(−ieγν)ε′ν(k′)u(p)

Put theγ− matrices in the numerator,

M = −ie2ε′µεν [u(p′)γµ6 p+ 6 k +m

2p · k γνu(p) + u(p′)γµ6 p+ 6 k′ +m

2p · k′ γνu(p)]Wei Wang and Zhi-Peng Xing (SJTU) QED 30 / 35

Using the relations,

( 6 p+m)γνu(p) = 2pνu(p)

we get

M = −ie2u(p′)[6 ε′ 6 k 6 ε+ 2(p · ε) 6 ε′

2p · k +− 6 ε 6 k′ 6 ε′ + 2(p · ε) 6 ε′

−2p · k′ ]u(p)

The photon polarizations are,

εµ = (0,~ε), with ~ε · ~k = 0, ε′µ = (0,~ε′), with ~ε′ · ~k′ = 0,

Lab frame ,pµ = (m, 0, 0, 0),⇒ (p · ε) = (p · ε′) = 0and

M = −ie2u(p′)[6 ε′ 6 k 6 ε2p · k +

6 ε 6 k′ 6 ε′2p · k′ ]u(p)

Summing over spin of the electron

∑spin

|M |2 = e4T r( 6 p′ +m)[6 ε′ 6 k 6 ε2p · k +

6 ε 6 k′ 6 ε′2p · k′ ](6 p+m)[

6 ε′ 6 k 6 ε2p · k +

6 ε 6 k′ 6 ε′2p · k′ ]

The cross section is given by

dσ =1

2k0(2π)4δ4(p+ k − p′ − k′)1

∑spin′

|M |2 d3p′

(2π)32p′0

d3k′

(2π)32k′0

phase space

∫(2π)4δ4(p+ k − p′ − k′) d3p′

(2π)32p′0

d3k′

(2π)32k′0is exactly the same as the case for ep scattering and the result is

ρ =dΩ

4π2ω′2

It is straightforward to compute the trace with result,

4m2(ω′

ω)2[ω′

ω′+ 4(ε · ε′)2 − 2]

This is Klein-Nishima relation. In the limit ω → 0,

dΩ=α2

m2(ε · ε′)2

here αm is classical electron radius.

For unpolarized cross section, sum over polarization of photon,∑λλ′

[ε(k, λ) · ε′(k′, λ′)]2 =∑λλ′

[~ε(k, λ) · ~ε′(k′, λ′)]2

Since~ε(k, 1),~ε(k, 2)anf~k form basis in 3-dimension, completeness relationis ∑

εi(k, λ)εj(k, λ) = δij − kikj

∑λλ′

[ε(k, λ) · ε′(k′, λ′)]2 = (δij − kikj)(δij − k′ik′j) = 1 + cos2 θ

wherek · k′ = cos θ . The cross section is

2m2(ω′

ω)2[ω′

ω′− sin2 θ]

The total cross section,

σ =πα2

−1dz 1

[1 + ωm(1− z)]3 +

[1 + ωm(1− z)] −

1− z2[1 + ω

m(1− z)]2

At low energies,ω → 0,we

σ =8πα2

and at high energies

σ =πα2

ωm[ln

e+e− → 2γ and cross symmetry

Quantum Electrodynamics - SJTU

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