Announcements

10/22

• Today: Read 7D – 7F• Wednesday: 7.3, 7.5, 7.6• Friday: 7.7, 9, 10, 13

On Web•Exam 1•Solutions to Exam 1

Questions from the Reading Quiz“The q vector is a little unclear to me. I think I understand how it relates to epsilon, but I'm not sure where it comes from or what values it can have.”

i iEtx e k x

• Scalar field:

• Dirac Field:

• Vector (gauge) field:

ik xe , four-momentumk E k

, i iEtx u s e p xp

2 2 2 2k E M k

, four-momentump E p2 2 2 2p E m p

, i iEtA x e q xq , four-momentumq E q2 2 2 0q E q

A simple problem

Suppose a photon is heading in along the x3 axis. Find an interesting way to designate the two polarizations in Coulomb gauge.

0 0 0q *, , q q

,0,0,q E E

0,?,?,0

,1 0,1,0,0

,2 0,0,1,0

q

q

• The boring solution:

• More interesting:

12

12

,1 0,1,1,0

,2 0,1, 1,0

q

q

• Most interesting:

12

12

,1 0,1, ,0

, 2 0,1, ,0

q i

q i

Feynman Rules:

,u p s

,v p s

,v p s

,u p s

,q s * ,q s

External LinesPropagators:

p 2 2

i p m

p m

q

2

ig

q

The Vertex

ieQ

Announcements

10/24

• Today: 7.3, 7.5, 7.6• Friday: 7.7, 9, 10, 13• Monday: Read 8A-8D

Laser Tag money is due

Pair Creation

Calculate the cross-section for e-e+ ff*, where f is a fermion with charge Q

1 2p p

1e p

2e p

3f p

4f p

2 1 3 4 2

1 2

igi iev u ieQu v

p p

M 2 2

2 1 3 4

ie Qv u u v

s

2

*

1 2 4 3

ie Qi u v v u

s

M

4 2

2

2 1 1 2 3 4 4 32

e Qi v u u v u v v u

s

M

4 2

2

2 1 3 42spins

1Tr Tr .

4 4 e e

e Qi p m p m p m p m

s

M

Pair Creation (2)

4 22

2 1 3 42spins

1Tr Tr .

4 4

e Qi p p p p

s

M

• To make any particle that isn’t an electron, must have E > 200 me

• Let’s also assume m is negligible.

2 1 2 1 1 2 1 2Tr 4p p p p p p p p g

4 2

2

2 1 1 2 1 2 3 4 4 3 3 42spins

1 4

4

e Qi p p p p p p g p p p p p p g

s

M

4 2

2 3 1 4 2 4 1 3 2 1 3 42

42 2 1 1 1 1 4

e Qp p p p p p p p p p p p

s

4 2

2

2 3 1 4 2 4 1 32spins

1 8

4

e Qi p p p p p p p p

s M

Pair Creation (3)

1 3

2 4

,0,0, , , sin cos , sin sin , cos ,

,0,0, , , sin cos , sin sin , cos .

p E E p E E E E

p E E p E E E E

• Need the dot products

2 21 3 2 4 1 4 2 31 cos , 1 cos .p p p p E p p p p E

4 2

2

2 3 1 4 2 4 1 32spins

1 8

4

e Qi p p p p p p p p

s M

4 2

2 2 242

spins

1 81 cos 1 cos

4

e Qi E

s M

4 2 42

2

161 cos

e Q E

s

2 4 2 2

spins

11 cos

4i e Q M

Pair Creation (4)

28

D

E

• Work to the cross-section

2 4 2 2

spins

11 cos

4i e Q M

2142 2

spins

1

8 16 cm

pi d

E E M

4 22

2 21 cos

64

e Qd

E

2

4

e

2 2

22

1 cos16

Qd

E

2 2

23

Q

E

Electron-Positron ScatteringCalculate the cross-section for e-e+ e-e+

2 1 3 4 2 4 3 12 2

1 2 1 3

ig igi iev u ieu v iev v ieu u

p p p p

M

2e p

1e p 3e p

4e p1 2p p

1e p

2e p

3e p

4e p

1 3p p

• Relative minus sign• Treat electrons as massless• Note this diverges at = 0• Total cross-section is infinity• Must be coming from the second diagram

22 2

2

3 cos

16 1 cos

d

d E

Electron-Positron Scattering (2) 2

1 3p p

1 3,0,0, , , sin cos , sin sin , cos ,p E p p E p p p

• Expression vanishes in forward direction• Could it be due to assuming zero mass?

1 3 0 as 0p p

• Cross-section is genuinely infinite• Large probability of scattering by a very small angle• Experimentally, very small angle scattering not measurable

Electron-Positron Annihilation

Calculate the cross-section for e-e+

2 * *2 22 2

i p q m i q p mi ie v u v u

p q m q p m

M

2 4 * *

2 2 2 2

p q q p p q q pi e v uu v

p q p q p q p q

M

• Treat electron as massless

Electron-Positron Annihilation (2)2 4 * *

2 2 2 2

p q q p p q q pi e v uu v

p q p q p q p q

M

*

*

g

g

2 4

spins

1Tr

4 2 2 2 2

p q q p p q q pi e g g p p

p q p q p q p q

M

2 4

spins

1Tr

4 2 2 2 2

p q q p p q q pi e p p

p q p q p q p q

M

Electron-Positron Annihilation (3)2 2

2 2

1 cos

4 1 cos

d

d E

• Identical particles – factor of two• Infinite at = 0 and • Actually, these two angles are the same

• Infinity is logarithmic and due to approximations

4 2 2 2 2 2 22 2 2 2

22 2 2 2 2 2 22 2 2 2

coscos

4 cos coscos

m E p m E pd E p

d Ep E p E E pE E p

• Final integral is finite

Questions from the Reading Quiz“Why is it that we determine if a Feynman amplitude is gauge invariant by finding that it will equal zero by replacing epsilon-mu with q-mu? I may just be getting lost in the math, but I don't follow where this comes from.”

Gauge Invariance

• The Feynman amplitude must be gauge invariant• Recall: we have choice about how to write the EM field for a photon

2 * *2 22 2

i p q m i q p mi ie v u v u

p q m q p m

M

iq xA e A A A Let iq xce

iq xA icq e icq

• Feynman amplitude should remain unchanged

Gauge Invariance (2)

• The Feynman amplitude must be gauge invariant

2 * *2 22 2

i p q m i q p mi ie v u v u

p q m q p m

M

icq

2 * * *2 22 2

i p q m i q p mie ic q v u v u

p q m q p m

2 *2 22 2

0i p q m i q p m

ie q v u v up q m q p m

• To check gauge invariance, replace any polarization vector by its corresponding momentum

Announcements

10/24

• Today: 7.7, 9, 10• Monday: Read 8A-8D• Wednesday: Read 8E-8F, 7.10, 7.12 (me = 0), 7.13

Problem 7.13 uses identity from problem 2.2c

Questions from the Reading Quiz“I am a little confused about the loops in QED section. So if loop diagrams contribute extra factors to g, couldn't we add an infinite number of loops to a diagram so that g is always increasing?”

Questions from the Reading Quiz“I think what confused me the most was the idea of charged scalars. I'm not sure I understand the purpose seeing as they're not in the standard model. This seems to just be covering all interactions but are they really necessary?”