We are discussing the Dirac equation. Now
for interacting particles, what should we
do? Interacting particles:
So, the interaction with an electromagnetic
field represented by A mu. So, this we had
already discuss, will change P to P mu plus
charge of the particle times A mu, but charge
of the particle is minus e electron like particle.
And this gives gamma mu P mu minus m psi for
free particle, that is changed to gamma mu
P mu plus e gamma mu A mu minus m psi; sorry
is equal to 0; equal to 0 for interacting
particles. This involves the kinetic energy
term, the total energy, the free particle
term plus the interacting potential term in
the operator.
So, we have to first; will first isolate the
potential; exactly what it is and then use
the perturbation method to understand the
transition amplitude and eventually the cross
section. What did we do just to remind us
we are interested in and isolating the potential
of interaction.
For that let us first look at what we did
in the case of Schrödinger equation ok. What
we have in the case of Schrödinger equation?
It is minus h cross square over 2 m del square
psi plus V psi equal to E psi. I have not
put h cross equal to 1 in this for clarity,
but in the other equations I have not put
this h cross square. I will just take them
to be equal to 1.
So, Dirac equation with interaction term is
gamma mu P mu plus gamma mu e A mu minus m
psi equal to 0. We will expand the first term
gamma 0 e minus gamma dot P, then we have
plus e gamma mu A mu minus m 
psi equal to 0. So, I will take e to one side
just to compare it with the Schrödinger equation;
multiply it by gamma 0 that will give you
E psi equal to take the rest of it to the
right hand side; I will have a gamma 0 gamma
dot P psi, minus e gamma 0 gamma mu A mu psi,
plus m gamma 0 psi. This is the Dirac equation
with interaction term.
Comparing with the Schrödinger equation,
first term as the kinetic energy term, second
term is the potential. So, V, it should be
positive the potential. So, V therefore, V
is equal to minus e gamma 0 gamma mu A mu.
So, that is the potential coming in the Dirac
equation.
Now, using perturbation theory we have the
transition amplitude T f i equal to minus
i integral psi f dagger, v psi i d 4 x. That
is equal to minus i integral psi f dagger,
minus e gamma 0 gamma mu A mu psi i d 4 x.
I will take the gamma 0 with psi f dagger
write it as psi bar. So, I have minus i integral
minus e psi bar gamma mu psi A mu. So, psi
f psi i d 4 x. This is nothing but the j f
i mu A mu d 4 x.
So, that the transition amplitude is exactly
in the same form as we had earlier with this
j mu, now defined in this fashion. Again,
like in the earlier case we consider interaction
between 2 charged particles, ok.
So, the A mu is caused by some other charged
particle. In that case I have T f i is equal
to minus i integral j mu of 1 particle interacting
with the A mu, which is caused by another
particle, but again we can take j 1 mu j 2
mu and del square A mu is equal to j 2 mu.
This is just to remind you; and that gives
you A mu equal to minus one over q square
j 2 mu. This is what we had earlier.
So, we have here minus one over q square j
2 mu. Or I could write T f i as minus i two
contravariant currents j 1 mu minus I; there
is no i minus g mu nu over q square j 2 nu
d 4 x.
Now, let us look at each currents; j 1 mu
I can think about a particular situation I
have an incoming particle with momentum P
A outgoing, but after interaction goes as
P B. In that case minus e psi A; sorry this
is the final particle. So, psi B bar gamma
mu psi A is what I have. So, when I have psi
A written as u A e power minus i P A x similarly
for psi B, the bar conjugate will actually
take u bar and exponential plus i P A x.
So, psi A bar is going to be u A bar e power
plus i P A x. Keeping that in mind, we have
minus e u B bar gamma mu u A and exponential
minus i P A minus P B x. Similarly for the
second current; the second current let us
consider P C and P D as the initial and final
momenta. In that case I have minus e u d bar
gamma nu u B not u B u C e power minus i P
C minus P D x. Putting together I have T f
i.
T f i equal to minus i integral minus e u
B bar gamma mu u A e power minus i P A minus
P B x. This is j 1. And j 2 is minus e u D
bar gamma nu u C e power minus i P C minus
P D x. And you have the propagator g mu nu
over q square, where q square is now the momentum
transfer, which is P A minus P B, which is
also equal to P C minus P D.
This is equal to minus i minus e u B bar gamma
mu u A minus g mu nu over q square. So, let
me write it slightly up; q is equal to P A
minus P B. Then again that from the other
current u D bar gamma nu u C, into integral
e power minus i P A plus P C two of the initial
momenta minus P B minus P D the final momenta,
x, d 4 x. And this integral is nothing, but
a delta function.
So, we can write T f i s minus i 2 pi power
4 delta 4 P A plus P C some of the initial
momenta minus P B minus P D minus some of
the final momenta, into an invariant amplitude
m is equal to minus e u B bar gamma mu u A,
propagator factor g mu nu over P A minus P
B square ok. And minus e u D bar gamma nu
u C. We will look at the cross section; how
we can write the cross section in the next
class.
