Lecture 18: Mar 23 th
2012
Reading Griffiths chapter 7
HW: 7.15, 7.16, 7.18, 7.24, 7.25
1) Charge conjugation:
Given spin up electron
 1
= u
1
(p)e
-ipx
and positron
 1
=
 1
(p)e ipx
We would like to find an operator that takes us from the electron to the anti-particle positron.
Define

C
as i
 2 times the complex conjugate of transpose of

, i
 2
=
ç
ç
æ
ç
ç
1
-
1
-
1
1
ö
÷
÷
÷
Which will convert particles to antiparticles, since it flips u(p) and takes p

-p in the exponential

C
= i
 2  T
 1 elec
= u
1
(p)e
-ipx
,

C
=
 1 pos
=
 1
(p)e ipx
Note the transpose doesn’t do much to a vector except to shift it from column to row.
Next we define the ajoint operator. y
=
 +  0
The ajoint and other transposed vectors will be useful in defining bra-ket pairs when calculating probabilities. for the Dirac spinors u = u
+  0
, and u
=
 +  0 where
(
 k p k
- m)u = 0, (
 k p k
+ m)

= 0, u i u j
= 2m
 ij
, u i  j
= -2m
 ij
 u u =
 k p k
+ m,
 u
=
 k p k
– m u (
 k p k
- m) = 0, u
(
 k p k
+ m) = 0
These relationships will be useful for calculating matrix elements.

2) Consider Parity
Parity is something that two component objects can have intrinsically and can be conserved or violated in interactions. Therefore, it makes sense that we should check the parity of combinations of two particles with or without terms to cause interactions. y 
scalar, parity 1 y  5 
psuedoscalar, parity -1 y   
vector, parity -1 y  5   
psuedovector or axialvector, partity 1 y
σ
μν 
antisymmetric tensor, σ
μν
= (i/2)(γ
μ γ ν
- γ
μ γ ν
)
 5 = i
 0  1  2  3 =
æ
0 1
è 1 0
ö
ø
The matrices between the four vectors can be considered as a basis of the 16 possible ways to multiply the two four vectors together including scalar, vector and tensor products. Alternatively you can think of it as a basis of all the 4x4 matrices that could be made from the gamma matrices. Note that all such combinations can be expressed as one or two matrices.
When we start forming propagators for forces they will include several of these gamma matrix terms.
We can test these by forming a parity operator
L u m
=
æ
1
ç
ç
ç
-
1
-
1
÷
÷
ö
÷
, Reverses 3 space or momentum dimensions
-
1
÷
3) 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 space equation. p
 p  = 0, and wave equation -
¶  ¶


= 0
To factorize this we need to solve a much simpler problem. With general constants:

(
 k p k
)(
 l p l ) =
 k  l p k p l = 0, just using gamma to represent an arbitrary constant, not the necessarily the gamma matrix
Then we only need
 k
=
 l
,
 k  l
=
 lk
These conditions can be satisfied by I for the 0 th or -
 i
, for the 1, 2 and 3 components.
component and the 2D Pauli matrices
 i
,
(
  p  )

= (E +
 p )

= 0, (
  p  )

= (E -
 p )

= 0
E

= -
 p

, E

=
 p

Where for a massless fermion. E=| p | and
Then considering the helicity operator is
S z
× ˆ =
2 s × ˆ and S z
× ˆ c =
2 s × ˆ c = -
2 c
spin -½ and S z
× ˆ f =
2 s × ˆ f = -
2 f
spin ½
There are only two solutions each with definite helicity.
Usually these are put together in a four component form like the spin ½ mass fermions u(p) =
æ
è c f
ö
ø
To go further we have define the gamma matrices for u, (
  p  )u = 0
A valid representation is(different than for the massive fermions): g
0 =
é
ê
0 I
ë I 0
ù
û
, g i =
é
ë -
0 s i s
0 i
ù
û
,
 5
=
æ -
I
I
ö
ø
Note then if we have a V-A force, such as the weak force: ½(1-
 5 )
½
 
(1-
 5
)u(p) = g m ç
è
æ
I 0
0 0
ö
ø
÷
è
ç
ç
æ c f
ö
ø
÷
÷ = g m
æ
è
çç c
0
ö
ø
÷÷
The weak force only interacts with the left handed (negative helicity neutrino). Applying the charge conjugation procedure moves

into the top position of the two vector. Thus the right handed antineutrino also interacts with the V-A weak force. Also the two parts of the force have different parties intrinsically violating parity conservation!

4) The photon
Starting from Maxwell’s equations. Find the fields in terms of the scalar and vector potentials
, which means
, , which means
Including charges and currents
,
Thus we can express B and E in terms of the four vector potential A

(V, A ) and all of
Maxwels equations as.
¶ m
F mn = ¶ m
¶ m
A n ¶ n ¶ m
A m =
4 p c
J n where F if a 4x4 matrix with E and B terms.
A
J
¶ m n m
is the scalar potential, V, and the vector potential, A, expressed as a four vector.
is the scalar charge,

, the vector current, j, expressed as a four vector.
F mn
represents Maxwell’s equations expressed as single derivatives of E and B or double derivatives of the potentials.
J the the charge density and current and obeys the charge and current continuity equation:
J m =
( J , r
) ,
¶ m
J m =
0,
The single derivatives of E and B are given by a four current of charge density and the standard current and are double derivatives in terms of the scalar and vector potential
However since the physics of Maxwell’s equations is given by
F we can make the transformation A m ®
A m +¶ m l
( t , x
Consider a case where we use gauge freedom to set
)without changing
¶ m
A m mn = ¶ m
A
F n mn
¶ n
A m then
=
0 and there is no four current i.e. no charges.
Then
¶ m
F mn = ¶ m
¶ m
A n ¶ n ¶ m
A m = ¶ m
¶ m
A n = boxA m =
4 p c
J n =
0, box =

This is the Maxwell’s equation for a photon and we also recognize it as the Klein-Gordon equation for a massless particle box A

=0, ,
However, this time we must consider the more general case of four vector solutions. This will allow for particles with spin. Borrowing from previous solutions
A n
( x )
= ae
-
( i / ) x m p m e n
( p )
Plugging into the gauge condition gives p m e m
=
0
Which can be solved with: e
1
=
(0,1,0,0) , and e
1
=
(0,0,1,0) for a particle traveling in the z direction. This is appropriate for the photon which can have spin oriented along it’s direction of motion of opposite. Spin projection 1 or -1
Now consider the case with a charges so that J is not zero :
¶ m
¶ m
A n =
4 p c
J n
A is the potential caused by charges and currents. If we have an electron in this potential will feel a force proportional to E and B times the charge(and velocity) of the electron or in terms of A, e times derivatives of A. There will also be a A
2
term for the energy of the potential. Another way to think of this is that a force will cause a change in momentum F = dp/dt. Integrating to find the total change
D p
= ò
F
= e
ò
¶
A
µ eA
Therefore p
® p
+ eA . Quantizing i
¶ m ® i
¶ m + eA m
This is how you insert the effect of the potential in the wave equation
For simplicity let’s consider the effect on a massive scalar particles satisfying the Klein-
Gordan equation.
(
¶ m
¶ m + m
2 ie (
¶ m
A m +
A m
¶ m
)
e
2
A
2
) f =
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 (
¶ m
A m +
A m
¶ m
)
e
2
A
2
5) T: The transition rate from an initial to a final state

T
= i
f
* f
V f 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
* f
(
(
¶ m
A m +
A m
¶ m
)
) f i d
4 x integrating by parts
f
* f
¶ m
A m f i d
4 x
= -
( )
A m f i
and
T
= e
ò ( f
* f
i
-
f i
)
A m d
4 x then for Klein-Gordon equation solutions: f =
Ne
ip
× x
T
= ieN i
N f
ò ( p i
+ p f
) m e i ( p f
p i
)
× x
A m d
4 x
A the potential and represents the photon and eN i
N f
( p i
+ p f
) m e i ( p f
p i
)
× x
one particle
Consider Maxwell’s equation for A.
¶ m
¶ m
A n =
4 p
J n c
Considering that the potential is caused by the second particle a possible solution for J is
J n = eN i 2
N f 2
p i 2
+ p f 2
n e i ( p i 2
p f 2
)
× x
Plugging in the Klein-Gordon equation solution for A A n =
Ne
iq
× x e n
( q ) , q=p f2
-p i2
T
= i
ò
J
1 m
1 q
2
J m
2 d
4 x
The integrate out the exponentials over space which forces q = p f2
-p i2
= p i1
-p f1
T = -iN
( i 1
+ p f 1
) m
ç
æ
è g mn q
2
ö
ø
÷
( i 2
+ p f 2
) n
(2

)
4  4
(p i1
+ p i2
- p f1
- p f2
)
This is the form of the matrix element in EM theory.
Very similar to our toy theory except that there are the momentum terms. Recall that in the toy theory 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.