H. A. Tanaka

Photons and Quantum Electrodynamics

The Photon• Apart from we need some other particle/object with definite Lorentz

transformation properties to make Lorentz invariants• What would we do with the “vector” term to get a Lorentz

scalar?• Recall the photon:

• Classically, we have Maxwell’s equations:

• Recall that we can re-express the Maxwell equations using potentials:

• these can in turn be combined to make a 4 vector:• Likewise for the “source” terms ρ and J:

��

⇥�µ⇥

⌅ · E = 4�⇥ ⌅ · B = 0

⌅⇤E +1cB = 0 ⌅⇤B� 1

cE =

4�

cJ

E = �⇥�

Aµ = (�,A)Jµ = (c�,J)

B = ⇥�A

Maxwell’s Equation in Lorentz Covariant Form• All four equations can be expressed as:

• The issue is that A is (far) from unique:• Consider:

• the last terms cancel, so the “new” Aµ is also a solution to Maxwell’s solution

• they are physically the same, so we can make some conventions:• “Lorentz gauge condition”:• “Coulomb gauge”

⇥µ⇥µA⇥ � ⇥⇥(⇥µAµ) =4�

cJ⇥

Aµ � Aµ + ⇥µ�

�µAµ = 0

A0 = 0

⇥µ⇥µA⇥ =4�

cJ⇥

⇥ · A = 0

⇥µ⇥µ(A� + ⇥��)� ⇥�(⇥µ(Aµ + ⇥µ�) =⇥µ⇥µA� � ⇥�(⇥µ(Aµ) + ⇥µ⇥µ⇥��� ⇥�⇥µ⇥µ�

Fµ⇥ = �µA⇥ � �⇥Aµ =�

⇧⇧⇤

0 �Ex �Ey �Ez

Ex 0 �Bz By

Ey Bz 0 �Bx

Ez �By Bx 0

⇥

⌃⌃⌅

Solutions to the Maxwell Equation in Free Space:

• “Free” means no sources (charges, currents): Jµ=0• Find solution as usual by ansatz:

• Now check:

• Conclusions:• Photon is massless• Polarization ε is transverse to photon direction:

• it has two degrees of freedom/polarizations

�µ�µA⇥ = 0

Aµ(x) = a e�ip·x�µ(p)

⇥µA⇥(x) = �ipµ a e�ip·x�⇥(p)

⇥µ⇥µA⇥(x) = (�i)2pµpµ a e�ip·x�⇥(p) = 0 p2 = m2c2 = 0

A0 = 0� �0 = 0

⇥µAµ = 0� pµ�µ(p) = 0

⇥ p · � = 0

Making a “scalar” object:

• In the end, these spaces must collapse:• In Lorentz space, this happens by contracting indices:• In spinor space, products of adjoint spinors with spinors (with gamma

matrices possibly in between): Г=(product of g matrices)• but some expressions have structure in both:

gµ⇥aµb⇥ = aµbµ

u1�v2

M = � g2e

(p1 + p2)2[u(3) �µ v(4)] [v(2) �µ u(1)]

sum over μ collapses the Lorentz structure

�⇥ = S�1�µS⇥x⇥

⇥xµ⇥

sum over μ collapses the Lorentz structure

Product of 4x4 matrices in spinor space

Contracted in spinor space, but not in Lorentz

Same here

Reminder of Dirac Spinors

• We can now construct the column vector u:

u1 = N

�

⇧⇧⇤

10

pzc/(E + mc2)(px + ipy)c/(E + mc2)

⇥

⌃⌃⌅ u2 = N

�

⇧⇧⇤

01

(px � ipy)c/(E + mc2)�pzc/(E + mc2)

⇥

⌃⌃⌅

Use “positive” energy solutions

Use “negative” energy solutions

positrons

electrons

�v2 ⌘ u3 = N

0

BB@

pz

c/(E +mc2)(p

x

+ ipy

)c/(E +mc2)10

1

CCA v1 ⌘ u4 = N

0

BB@

(px

� ipy

)c/(E +mc2)�p

z

c/(E +mc2)01

1

CCA

A second look at Dirac spinors

• “s” labels the spin states (two for electrons/positrons)• The exponential term sets the space/time = energy/momentum• Let’s look at the “spinor” part u,v which determines the “Dirac structure”:

• If we insert ψ into the Dirac equation, we get:

• If we take the adjoint of these equations, we get:

�(x) = ae�(i/�)p·x us(p) �(x) = ae(i/�)p·x vs(p)electrons positrons

(i��µ⇤µ �mc)⇥ = 0⇥ (�µpµ �mc)⇥ = 0⇥ (�µpµ �mc)us(p) = 0

(i��µ⇤µ �mc)⇥ = 0⇥ (��µpµ �mc)⇥ = 0⇥ (�µpµ + mc)vs(p) = 0

“momentum space Dirac equations”

us(�µpµ �mc) = 0 vs(�µpµ + mc) = 0

Orthogonality and Completeness of Spinors:

• From the explicit form of our u/v spinors, we can also show:

• We can also show:

uivj = viuj = 0uiuj = 2mc �ij

�

s=1,2

usus = (�µpµ + mc)�

s=1,2

vsvs = (�µpµ �mc)

vivj = �2mc �ij

�

⇧⇧⇤

a1

a2

a3

a4

⇥

⌃⌃⌅ (b1, b2, b3, b4) =

�

⇧⇧⇤

a1b1 a1b2 a1b3 a1b4

a2b1 a2b2 a2b3 a2b4

a3b1 a3b2 a3b3 a3b4

a4b1 a4b2 a4b3 a4b4

⇥

⌃⌃⌅

Photon Polarizations and Orthogonality:• We showed that the polarization 4-vector εμ with the Lorentz and Coloumb

gauge conditions must satisfy:

• We noted that this allows two degrees of freedom corresponding to transversely polarized electromagnetic fields. • We need to two orthogonal ε basis vectors to span the space

• For example, if the photon is moving in the z direction, we can choose:

• The polarization vectors satisfy orthogonality/completeness relations:

p · � = 0

�1µ =

�

⇧⇧⇤

0100

⇥

⌃⌃⌅

⇥i�µ ⇥µj = ��ij

�

s=1,2

⇥si ⇥

s�j = �ij � pipj

�2µ =

�

⇧⇧⇤

0010

⇥

⌃⌃⌅

The Feynman Rules: External Lines

• First right down the Feynman diagram(s) for the process and label the momentum flow• use p’s for external lines, q’s for internal (Griffiths convention).• Note that there are two flows:

• “particle/antiparticle”• momentum • These are separate

• Now the components of the expression• External Lines:

• Electrons: incoming outgoing• Positrons: incoming outgoing• Photons: incoming outgoing

p1

p2

p3

p4

q

us(p) us(p)

vs(p) vs(p)

�µ(p) ��µ(p)

The Feynman Rules: Vertices and Propagators:

• For each QED vertex:• where as before, momentum is “+” incoming, “-” outgoing from vertex

• Internal lines:• electron/positron propagator

• Photon propagator• indices match vertices/polarization

• Integral over momentum:

• Finally: cancel the overall delta function, what remains is -iM

p1

p2p3

p4

q

ige�µ

i(�µqµ + mc)q2 �m2c2

�igµ⇥

q2

d4q

(2�)4

(2⇥)4�4(k1 + k2 + k3)

Example:

• Order matters due to Dirac matrix structure (photon part doesn’t care)• Griffiths: go backward through the fermion lines:

• In the “final state”: • In the “initial state”:• Throw in the internal photon propagator:

u(3) ige�µ v(4)

v(2) ige�� u(1)

�igµ⇥

q2

1(2�)4

�d4q

(2⇥)4�4(q � p3 � p4)(2⇥)4�4(p1 + p2 � q)

[u(3) �µ v(4)] gµ⇥ [v(2) �⇥ u(1)]i(2⇥)4�4(p1 + p2 � p3 � p4)⇥g2

e

(p1 + p2)2

M = � g2e

(p1 + p2)2[u(3) �µ v(4)] [v(2) �µ u(1)]

p1

p2p3

p4

q

Example: e+ + e- → e+ + e-

p1

p2

p4

p3

q

e

e

e

e

p3

q

p4p1

p2

u(3) ige�⇢ v(4) v(2) ige�

� u(1)�ig⇢�

(p1 + p2)2

u(3) ige�µ u(1) v(2) ige�

⌫ v(4)�igµ⌫

(p1 � p3)2

(2⇡)4�4(p1 + p2 � p3 � p4)