We have considered the S matrix for QED and
then we locked into all the terms in the
first order as well as second order in e,
and then we looked at various physical process
and then we discussed which term in the S
matrix will contribute to what physical
process and so on. We also have seen that,
that can be more than one term, which can
contribute to a given physical process and
each of these terms can actually infect the
represented in terms of this local Feynman
diagrams.
.
So, for example, suppose you consider a process
like Compton's scattering let us say e
plus plus gamma e minus plus gamma going to
e plus gamma then there are two terms in
the S matrix, which gives non zero contribution
to this process. It is second order and
these two terms are represented by the following
to Feynman diagram. So, you have an
incoming electron and you have a incoming
photon and then you have an outgoing
electron and you have an outgoing photon at
X 1 or you too can also have an electron e
minus, which emits a photon at X 2.
.And there is an Fermion propagator between
X 1 and X 2, and incoming photon is
observed at this specific point x 1, finally
there is an outgoing electron. So, these are
the
two possibilities and correspondingly there
are two terms in the S matrix, so it gives
non
zero contribution for this physical process.
And these are the two Feynman diagrams for
the two terms in the S matrix, what you need
to do is, we actually need to consider the
S
matrix, the relevant terms in the s matrix
for this process and then we have to compute
the amplitude for this process.
So, you have some initial state given by these
incoming electron and photon, and you
have outgoing state represented by some outgoing
electron and outgoing photon. And
you know what are the terms in the S matrix
per second order, you need to compute this
amplitude in order to know, what is the cross
x or anything you want to do in the
quantum field theory, you need to evaluate
this amplitude. What we will do now is, we
will formulate a set of rules to get this
amplitude so that, you do not have to evaluate
again and again every time, you want to compute
this amplitude.
So, we will get a simplified form for this
amplitude from what is known as the Feynman
rules. We will derive the Feynman rules for
QED and then you can write down this
amplitude using Feynman rules, instead of
working out it is and every time whenever
you need it. So, this is what we will be doing
in a today’s lecture, let us first summarize
the use full expressions that will be needed
in this lecture.
.
.We will require the mode expansion for the
photon field, which is give by d cube k over
2 pi cube 2 k 0, sum over the polarization
lambda is going to 1 to 3 a lambda k epsilon
lambda of k e to the power minus i k dot x
plus a lambda dagger of k epsilon mu lambda
star of k e to the power i k dot x. Then if
the Dirac psi of x which is given by d cube
p
over 2 pi cube m over p 0 is the normalization
I have chosen. And sum over this spins c s
p u s p e to the power minus i p dot x then
d s dagger of 9 v s p e to the power i p dot
x.
Correspondingly, we have psi bar of x, this
is given by d cube p over 2 phi cube m over
p
0 sum over s, c s dagger p u s bar of p e
to the power i p dot x plus d s p v s bar
of p e to
the power minus i p dot x. And these operator
is a’s and c’s and d’s, they obey the
following computation and anti computation
relations, a lambda of k, a lambda prime
dagger of k, prime commutator is given by
minus eta lambda lambda prime 2 pi cube
two k 0 delta k minus k prime.
Whereas, c r k c s dagger k prime anti commutator
is given by delta r s 2 pi cube k 0 over
n delta k minus k prime, which is also same
as the commutator of d s d r k, d s dagger
k
prime and all other commutators and anti commutators
relating to the Feynman and
prosodic operators from this. So, these are
the non vanishing commutation and anti
commutation relations. So, what we will now
do is, we will lock at the first order term
in
the S matrix that I will call as S.
.
.I have denoted as S 1, which is i e times
d 4 X normal under that are psi bar of X A
slash
of X psi of X. And we look at a physical process,
for which this may, this might give a
non zero amplitude although will see later
that, there is no physical process, to which
this
one gives a non zero contribution and then
we will evaluate the amplitude for this
process in detailed. So, let us consider this
process of e minus going to e minus plus
gamma, this is one possibility, because you
can have this term.
For example, can contain one term, it is a
psi bar minus of X gamma mu a mu minus of
x
psi plus of X, one of the terms in this normal
under ring given by this and this psi plus
represents any lesser than electron at X.
Whereas, m mu minus m psi bar minus
represents emission of or creation of m emitton
at it is phase time point X. If this is X
then an electron is an elected here and an
outgoing photon and an outgoing electron are
created at x. So, this process this is the
term in S matrix, so it can contribute to
this
process.
So, we will consider this process and then
we will evaluate, so S 1 1 I will call, i
e times
to this d 4 X, so we will take this e minus
incoming electron with some momentum p and
some polarization s, S 1 1 e minus e prime
s prime gamma of k prime lambda prime. So,
the point is, suppose you have this incoming
electron with momentum p and the
polarization of s, going to some outgoing
electron of momentum p prime polarization
s
prime and some outgoing photon of momentum
k prime and polarization lambda prime.
Then, what is the amplitude for this process,
that is what is the question that we would
like to ask. So, naturally, we need to evaluate,
what happens when this psi plus acts on
these states and a mu minus acts on this states
and so on. So, this is what we need to
know.
..
So, to do that, let us look at first the electron
state, let us say this represents the state
with
a momentum p and spin s, therefore this state
for electron is given by c s dagger of the
acting on the vacuum. Now, we would like to
know, what happens when I consider psi
plus of X acting on this tact. So, what is
psi plus of X, you can look at this expression
here, psi plus of X is given by this term,
so 
d cube p over 2 pi cube and let us use the
integration variable to be k, m over k 0 sum
over s c s k.
Let us denote this dummy variable by r, c
r k u r k e to the power minus i k dot x,
this
one acts on this state, which is given by
c s dagger of p acting on vacuum. So, what
do
we get, I can rewrite this term as d cube
k over 2 pi cube m over k 0, sum over r or
this
time e to the minus i k dot x then you have
sum over r u r k c r k c s dagger of p acting
on
the ground state. I just rearrange the terms
and this expression.
..
Now, what I will do is, I will take this term
here, c r k c s dagger of p, I can rewrite
it as
the anti commutator of c r k c s dagger p
minus c s dagger p c r k. Now, when this whole
thing acts on the ground state, this simply
means that, this and this acting on the ground
state. But, as you can see, the second term
acting on ground state, there is an annihilation
operator. So, when c r k acts on 0 on the
vacuum, it gives you zero contribution, so
only
thing that is left is the anti commutator
of c r k c s dagger p.
But, the anti commutator we know what it is,
this is nothing delta r s, upto some
normalization this is delta r s and delta
k minus p. So, you can just substitute that,
we can
say that, this is equal to the anti commutator
of these two operators acting on the ground
state. And we will put the known value for
the anti commutator which is nothing d cube
k over 2 pi cube m over k 0 e to the minus
i k dot x 
sum over r u r k then 2 pi cube k 0
over m delta r s delta k minus p, this acting
on the ground state. So, the sum over r, I
can
evaluate and also I can evaluate the integration
over d cube k, that will give me simply
this 2 phi cube and k 0 over m, they will
cancel each other. So, what is left is, u
s p e to
the power minus i p dot x acting on the ground
state.
..
So therefore, what is saw is that, from these
result, psi plus of X acting on e minus which
nothing psi plus X c s dagger p on the ground
state is equal to u s p e to the power minus
i p dot X acting on the ground state. Similarly,
you can evaluate what happens, when m u
plus acts on a one photon state.
.
So, let us say that, you have a photon gamma
of polarization lambda and momentum k
and you want to know, what do you get when
you consider a mu X acting on that. So,
this is nothing a mu plus X acting on, so
a lambda dagger of k acting on the ground
state.
.We can substitute the value for a mu plus,
which is given by these, so this is nothing
integration, let us say this is a momentum
p d cube k over 2 pi cube 2 k 0 sum over
lambda a.
Let us say you are considering some over lambda
prime a lambda prime k epsilon mu
lambda prime k e to the power minus i k dot
x acting on a lambda dagger k 0. So, let us
rearrange the terms, so this is integration
d cube k over 2 pi cube 2 k 0 e to the power
minus i k dot x and then sum over lambda prime
epsilon mu lambda prime k a lambda
prime k a lambda dagger p. Again you can write
this term as the commutator of a lambda
prime k and a lambda prime dagger minus a
term with ((Refer Time: 21:10)) when acting
on the ground state.
Therefore, I will just substitute the commutator
of a lambda prime and a lambda dagger
in this place. And we know what is the commutator
is, this is nothing 2 pi cube 2 k 0
delta k minus p times minus eta lambda lambda
prime. So, you can now carry out the k
integration, when you carry out the k integration,
of course this 2 pi cube 2 k will cancel
and what you are left with is minus eta lambda
lambda prime epsilon mu lambda prime p
acting on the ground state, where there is
a sum over lambda prime is transfer from 0
to
3.
So, this time says the upper e to the power
i p dot x, now suppose you are considering
a
transfer photon here then for transfer photon,
you can just consider this is to be minus
delta lambda lambda prime. And what you will
get is, this whole thing is equal e to the
power minus i p dot x epsilon mu lambda p
times the ground state. So, if you have a
photon with a transfer polarization then this
is what you are going to get.
..
So, let me summarize, at here this is what
you will get for the electron, and for photon.
You will get a mu plus X gamma is equal to
e to the power minus i p dot x epsilon mu
lambda p and the ground state. So, we will
now use these two results and their
conjugates to evaluate the amplitude for this
process, e minus going to e minus plus
gamma.
.
So, let us do that, so what we need is, we
have an outgoing electron of momentum p
prime and a polarization s and you have an
outgoing photon of momentum, let us say k
.prime and polarization lambda, this is what
we have. So, we have already seen, which
term in the S matrix, that may give a non
zero contribution, let us see minus p prime
s
prime gamma k prime lambda and i e times integration
d 4 X psi bar minus of X gamma
mu a mu minus of X psi plus of X acting on
e minus p s.
.
So, this part I can directly substitute it
here and this one as you can see, psi bar
minus
acting on these will give me, from this term
itself I can conclude that, if I have an
electron state e minus of e prime X prime,
if psi bar minus of x on it from right then
this
will give me 0 times u s bar of p e to the
power i p dot x, p prime and s prime.
..
Similarly, I have, if a mu of X, a mu plus
X acts on this state then what I will get
here is i
e integration d 4 X then these are numbers
as far as the ground state is associated,
because 
these are the operator from the ground state,
but these numbers simply
multiplied them. So, I can just tack these
things out here, as far as this state is concerned,
these are just some multiplicative coefficients.
So, this is just 0 0, I will normalized the
ground state to be 1 then all that is left
with it is u bar s prime of p prime e to the
power i
p prime dot x.
And then gamma mu then when m u minus acts
on this, what I have is e to the power i k
prime dot x and epsilon mu lambda k prime
and then this one is just u s p e to the power
minus i p dot x. Epsilon star, so let us consider
the photon to be all transverse and rear.
So, here write this in principles sort of
a star, but I will not, I will assume from
now on
that, all the incoming and outgoing photons
are represented by real polarization as a
transverse.
So, what I have is i e d 4 X e to the power
i p prime plus k prime minus p dot x and then
you have u s prime bar p prime gamma mu epsilon
mu lambda k prime u s of p, this
quantity I will call this to be. So, i e times
this quantity, I will represented it by m,
which
I will cal as the Feynman amplitude. So, what
I get is 2 pi 4 delta, the four dimensional
delta function p prime plus k prime minus
p times m, where m s prime s where m means,
i e times u bar p prime epsilon slash lambda
of k prime u p.
.So, this is what I get for this amplitude,
but now we will see, because this delta function
is there, whether it can actually be satisfied
for this process. So, this is what is imposed
by the delta function.
.
It p prime plus k prime, so it actually be
equal to p, therefore p 0 will have to be
equal to
p 0 prime plus k 0 prime or since p 0 is the
energy of the electron, it is nothing mod
p
square plus m square, should be equal to square
root of mod p prime square plus m
square plus mod k 0 is more simply mod k prime,
because photon is massless. What
about the i th component of the momentum,
p prime plus k prime is equals to p, now
since the incoming electron is a massive particle,
you can go the rest frame of the
incoming electron and then you can do this
calculations here.
So, in the rest frame of the incoming electron,
p equal to 0, therefore in the rest frame,
p
prime is equal to minus k prime. Now, what
is this equation gives me in the rest frame,
this gives me, so the RHS is simply equal
to mod p prime square plus m square plus mod
p prime. This quantity is as you can see,
this must be greater than or equal to m and
the
equality will be saturated, this inequality
is saturated only when p prime is equal to
0. So
therefore, this will be satisfied only when
p prime and k prime is equal to 0 or in other
words, when there is no outgoing photon at
all.
..
For k prime non zero, this will not be satisfied,
therefore this delta function, because of
delta function, this amplitude will always
0 for any outgoing real photon. So, you can
see, you can try to consider what are possibilities
in the S matrix, and you can see that,
this first attempt does not give any contribution
at all, because you will always get delta
function like that and then you will see that,
it will never be satisfied at all. So, there
is
no physical process, to which the first order
term in the S matrix gives any contribution.
Now, what we will do is, we will consider
as simple example at the second order. Let
us
say, let us consider the example of Compton's
scattering.
.
.So, in the case of Compton's scattering,
we have a e minus plus gamma going to e minus
plus gamma and we saw that, at the second
order, there are two terms which can
contribute. So, S 2 e minus plus gamma going
to e minus plus gamma as two terms,
which I will call as S a plus S b, where S
a is equal to minus e square d 4 X 1 d 4 X
2 psi
bar minus X 1 gamma alpha i S F X 1 minus
X 2 gamma beta psi plus of x 2.
And S b is minus e square d 4 X 1 d 4 X 2
then psi bar minus X 1 gamma alpha i S F of
X 1 minus X 2 gamma beta psi plus X 2 times
A beta minus of X 2 A alpha plus X 1,
this is what we will get. Strictly significance,
because there is a normal ordering, I should
have written it here, but it does not matter,
because this A minus commutes with psi plus,
same thing here. And there are two diagrams,
which actually contributes to these two
processes, one of them is given by there is
an incoming photon and there is an outgoing
photon at X 2 X 1, this is gamma, this is
e minus, e minus.
So, another one is, there is an outgoing photon
at X 2 e minus and there is the incoming
photon is observed at X 1, these are the two
processes, a physical processes. So, what
we
need to do is, we need to consider the instate,
which is it is one electron and one photon
and the out state or final state f again,
there is one electron and one photon. We need
to
consider the spin and the polarization momentum
of the photon as well as the electron in
initial and final state.
And we need to consider, we need to find what
is the amplitude for this process, what is
f
S a i and what is f S b i. And these two matrix
elements we need to consider and we need
to take this sum of these two matrix elements
to compute the amplitude for this process
at lowest order. So, let us evaluate first
this quantity here, we already know, what
we get
when psi plus X 1, the initial state and so
on. So, let us do that and see, what we get
for
the amplitude.
..
So, I will write f S a i, this is minus e
square d 4 X 1 d 4 X 2 and then 
so this initial state
here i and again has some electron of momentum
p and polarization I will call this to be
s
and it is a photon of momentum k and polarization
lambda, this I can take is e minus p s
and gamma k lambda and this state, this psi
plus will act on this one. Whereas, the A
plus
will act on this state here.
.
So therefore, what I can do is, I can just
consider the inner product e minus p prime,
let
us suppress the spin index for the moment,
we will keep track of this spin and this is
later
.on, psi bar minus X 1 gamma alpha i S F X
1 minus X 2 for the Fermi on propagator
then gamma beta psi plus of X 2. This acts
on the incoming electron and it is momentum
p and spin s, this times the photon of momentum
k prime A alpha minus X 1 A beta plus
X 2 acting on the photon state of momentum
k.
So, what is this, this we will already know,
we have evaluated this piece, we have a
evaluated this piece, we have evaluated this
one and as well as this one. So, all these
thing we already know, so we can put all these
things together and then we will see what
we get. So, let us do that and finally, what
I will do is that, this Fermion propagator
I can
write it in terms of it is four year component.
So, this quantity is minus e square d 4 X
1
d 4 X 2 and then this will give me, as we
have already seen u bar of p prime e to the
power i p prime dot x one.
And then this one gamma alpha, this one in
terms of it is four year component is given
by 1 over 2 pi, fourth the integration d 4
q i S F q times e to the power minus i q dot
x 2
minus x 1. You will see very quickly, why
I wanted to write it this way and then I have
gamma beta and then psi plus acting on e minus
will give me u p e to the power minus i
p dot x 2. Then finally, I have this which
is nothing epsilon alpha k prime and I am
suppressing this spin as well as the polarization
indexes, e to the power i k prime dot x 1
and epsilon beta of k e to the power minus
i k dot x 2, that is all I have.
So, let us rearrange all these terms and then
see what do we get, what I will do is, I will
collect the coefficients of e to the power,
I will put all these exponential terms together,
I
will write it in terms of two terms. One collecting
in the coefficients of x 1 and then I
will carry out the X 1 integration here. Then
the second exponential, I will also again
I
will carry out these this second integration,
so this is nothing minus e square d 4 X 1
d 4
X 2 and then d 4 q over 2 pi fourth.
And finally, e to the power i p prime minus
q plus k prime dot x 1 e to the power minus
i
t minus q plus k dot x 2 and u bar of p prime
epsilon slash of k prime i S F q epsilon
slash k u of p. So, as you can see, you have
e to the power i p prime plus dot x 1 e to
the
power i k prime dot x 1 and then e to the
power, so let us write it as plus and then
e to the
power minus i q dot x 1 here. Whereas, in
the second term, I have if I just look at
the x
2’s, I have e to the power i q dot x 2,
here i minus i p dot x 2 and minus i k dot
x 2, that
when I put together I get here.
.And then I took out the q integration outside
and then remember this S F is an operator
which is inserted between gamma alpha and
the gamma beta. So, I do not have the
freedom of moving it anywhere else, I have
to do that, but this epsilon alphas are
numbers, so I can write, I can bring this
epsilon alpha here, that will give me epsilon
slash k prime. So, you have this u bar p,
this is a number, I just took it outside,
but this
row is already there.
So, u bar p and then gamma alpha, epsilon
alpha will give me epsilon slash k prime then
I have S F q which is here and finally, I
have the gain this gamma beta, I can take
this
number epsilon beta k, that will give me epsilon
slash k and finally, I have u p.
Remember, i do not have the freedom to move
these things anywhere other than where
they are, of course I can write them terms
of their components and finally, I can move
the components around, because they are just
numbers, but these are not numbers, so
they are there.
Now, I can carry out the X 1 and X 2 integration
and also I can carry out the q
integration. First let us carry out the X
1 and X 2 integration, that will give me the
two
delta functions. Here I get delta p prime
minus q plus k prime, here I will get delta
p
minus q plus k and when I integrate out over
q, I will get delta p prime plus k prime
minus p plus k. So, this is what I get and
then there are this factor of 2 pi to the
power
fourth is here.
But, when you carry out two X integrations,
you will get two factors of 2 pi to the power
fourth in the numerator. One of them will
cancel with this one and the remaining is
given
by 2 pi to the power fourth delta p plus k
minus p prime minus k prime, times this
quantity, minus e square u bar p prime epsilon
slash k prime i S F of, q is nothing p plus
k epsilon slash of k u p, so this is what
is the Feynman amplitudes. So, what I did
is, I
evaluated this and then I wrote in the form
of a delta function, times sum matrix here.
I will do this similar thing for S b and then
finally, to get the full amplitude for this
process, I will add them up. But, again I
will give you a set of rules so that, just
by
looking at the Feynman diagrams, you can evaluate
this matrix element here, instead of
going through all these algebras, which we
will do in the next lecture.
.
