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Physics 535 lecture notes: - 17 Nov 1 st
, 2007
HW: 7.4, 7.7,7.8
1) Solutions to the Dirac equation. Spin ½ particles and anti-particles. The solutions are simultaneous Eigenstates of the Hamiltonian, helicity, and total angular momentum, parity (in pairs) and charge conjugation (squared).
Solutions to Maxwell’s equation in free space lead to the Klein-Gordon equation for a massless particle. Spin 1 massless particles that are simultaneous Eigenstates of the
Hamiltonian, helicity, and total angular momentum, parity (odd) and charge conjugation
(the photon is its own anti-particle).
2) The Neutrino
A simpler solution to the Dirac equation is to consider mass 0 spin ½ fermions. The known nearly mass less spin ½ particle is the neutrino.
Consider the original energy momentum equation. p
 p  = 0, and wave equation -
  


= 0
We need to identify what general constants will factor the Klein-Gordon equation to get a
Dirac equation.
 
(
 k p k
)(
 l p l
) =
 k  l p k p l
= 0, just using gamma to represent an arbitrary constant, not the previous gamma matrix
Then we only need
 k =
 l ,
 k  l =
 lk
These conditions can be satisfied by I for the 0 matrices
 i
, or -
 i th
component and the simpler 2D Pauli
, for the 1, 2 and 3 components. We have to actually revamp all of our gamma matrices to agree with this definition. Since now
 i
=



 
We will have a 4 vector solution for two sets of possible particles, i

 i



and







(
  p  )

= (E +
 p )

= 0, (
  p  )

= (E -
 p )

= 0  for a massless fermion. E=| p |, dividing out an E


1
   ˆ   
0 ,

1
   ˆ   
0 considering the helicity operator:

S z
 ˆ 
2
  ˆ
S z
 ˆ  
2
  ˆ   
2

spin -½ and

S z
 ˆ  
2
  ˆ   
2

spin ½






Note then if we have a V-A force, such as the weak force: ½(1-
 5
)
In this representation
 5
=
 
I
I


½(1 5
)u(p) =

I 0
 0











0


The weak force only interacts with the left handed (negative helicity neutrino)
Note that these are all just various solutions of the same wave equation. Dirac type solutions for spin ½ particles and Klein-Gordon type solutions for integer spin particles
3) Photon propagator and the scattering cross section.
Starting from Maxwell’s equations


F
  

 
A
    

A
  

 
A
 
4

J
 c
The single derivatives of E and B are given by a four current of charge density and the standard current.
 
E

4

,
 
B

1


E t

4

J c c and are equal to double derivatives of a scalar and vector potential
E  V

1 c

A
 t
, B
  
A
We solved this for the photon to get a Klein-Gordon wave equation by considering the case where there was no four current and taking advantage of gauge freedom to set


A
 
0 .

Not eliminating the current:


 
A
 
4

J
 c
Now A is the potential caused by charges and currents. If we have an electron in this or in terms of A, e times appropriate derivatives of A. The standard way to represent this in the a wave equation is to made the substitution i
   i
   eA

.
For simplicity let’s consider a scalar massive particles satisfying the Klein-Gordan equation.












(


 
(


 
 m
2
)
 
0

((

 m
2  ie (


A


 ieA


A

 
)(
   ieA

)
 m
2
)
 
0
)
 e
2
A
2
)
 
0
Note that this gives the expected form of e times derivatives of A. Now if we treat this in perturbation theory then we can say we have a base wave equation and electromagnetic perturbation potential, V, that can change initial states to final states.
V
  ie (


A
 
A

 
)
 e
2
A
2
The transition rate from an initial to a final state is (we will square T for the probability):
T
  i
  f
*
V
 i d
4 x
Where the A 2 part of V will not contribute since it’s a constant and can not convert initial to final states.
T
  e
 
* f

(


A
 
A

 
)

 i d
4 x integrating by parts
 
* f


A
  i d
4 x
 
  
A
  i
and
T
  e
 

* f
 
   i

A
 d
4 x
 then for Klein-Gordon equation solutions:  
Ne
 ip
 x
T
  ieN i
N f
  p i
 p f

 e i ( p i
 p f
)
 x
A

 d
4 x
However, now we need a propagator solution for the Maxwell’s equation for A.


 
A
 
4 c

J

Note that previously from the continuity equation we identified terms such as
J

 
* f
 

 
 i
as the current of a charged particle.
This suggests a solution, since the potential is caused by other particle in the scattering event.


 
A
 
4
 c
J

, J
  eN i 2
N f 2
 p i 2
 p f 2
  e i ( p i 2
 p f 2
)
 x








Plugging in the Klein-Gordon equation solution for A A



 
A
 
4
 c
J
   q
2
A
 
4
 c
J
 
A
 

1
 q
2
4
 c
J

T
  i

J
1

1 q
2
J

2 d
4 x

Ne
 iq
 x
 
( q ) , q=p f2
-p i2
The integration over space will force q = p f2
-p i2
= p i1
-p f1
T = -iNM(2

)
4  4
(p i1
+ p i2
- p f1
- p f2
) i 1
 p f 1
 


 ig
 q
2


 i 1
 p f 1
 
Very similar to our toy theory except that there are the momentum terms. Recall that before we found that for units to work out the coupling constants had to have units of momentum. Now that unit of momentum is explicitly in the form and we have an identical propagator to the one we used for the toy theory for a massless propagator.
4) Adjoint and the electron propegator.
Also define the adjoint, which will be used to complete bra-ket pairs
Definition came from our study of charge conjugation

T
=
 +  0
=

*
 0

C
= i
 2  + = C
 0  + = C

T for the Dirac spinors u = u
+  0
, and

=
 +  0 where

(
 k p k

+ m)u = 0, (
 k p k

- m)

, u (
 k p k
+ m) = 0,

(
 k p k
- m) u i u j
= 2m
 ij
,
 i  j
= 2m
 ij
 u u =
 k p k
+ m,
 
=
 k p k
– m


This suggest a form for the electron propagator which will involve an electron current
 u u =
 k p + m and from a similar derivation to the one above giving a q
2
– m
2
in the denominator. Now we have all the necessary tools to start writing down the Feynman rules.
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