We have been discussing quantization of a
scalar fields, although some of the results
that
we have derived is a much more general than
scalar field. So, it goes beyond scalar
fields, but mostly we were discussing a free
scalar fields as well as self interacting
scalar
fields and a bunch of scalar fields interacting
with each other. That is not what happens
in the real world, in real world there are
particles which have non zero spin, and then
they interact among each other, electromagnetic
field interacts with radiation field with
electron and so on. So, we need to consider
beyond scalar fields one of the classical
example of a quantum field which is not a
scalar field is the radiation field or
electromagnetic field, what we will discuss
now is we will discuss the quantization of
free radiation field.
.
So, let us consider the Quantization of Electromagnetic
Field, normally when you have a
bunch of scalar fields, let us say this scalar
fields are label by pi i then what do you
do
you. You consider the Lagrangian density and
then you derive the corresponding
.conjugate momentum, let us say you have a
free field then 
pi i phi i of x t pi j of y t. So,
you postulate this equal time commutation
relation to be i delta i j for this label
psi n j
and delta x minus y.
Now, we have the electromagnetic field which
I will denoted as a mu the Lagrangian
density, which governs the dynamics of this
field is given by L electromagnetic is a
minus 1 4th F mu nu F mu nu, where F mu nu
equal to del mu A mu minus del mu A mu.
The gauge field A mu is a vector field, which
amongst to saying that under Lorentz
transformation, this field A mu goes to lambda
nu mu A mu that is why it is a vector
field. And because it transforms nontrivially
under Lorentz transformation, it has non
zero spin and you can work out the details
and it turns out that, this represents a spin
one
field that is why we consider photons to represent
by this field.
In addition you have also a gauge transformation,
under which this field A mu goes to A
mu plus del mu lambda, where lambda is any
arbitrary function of space time. You can
notice that under this transformation this
field’s strength F mu nu remains invariant
correspondingly the electric as well as magnetic
field, remain invariant under this gauge
transformation. And hence your equations of
motion remain invariant under this gauge
transformation.
So, we will consider the quantization of this
electromagnetic field or the gauge field.
How do you do that, we will see what happens
when we try to do it in the usual way that
is to find the corresponding conjugate momentum
for this Lagrangian and then look at
the equal time commutation relations.
..
So, the conjugate momentum which I will denote
as pi mu or pi mu equal to del l over
del del 0 A mu right, there are four fields
A 1, A 2, A 3, A 4, four independent quantities.
So, they correspondingly you have four conjugate
momentum that I am labeling this
index mu here and I mu is given by this. From
the expression here, you can see that it is
just a minus 1 4th twice f alpha beta del
over del del 0 A mu f right; so this will
give me
minus f 0 mu.
So, therefore, we have pi mu the conjugate
momentum is given by this you might think
of considering this commutation relations
A mu pi mu A mu of x t pi mu x t. However,
it
is not, so trivial the reason is following,
if you consider pi 0, the conjugate variable
to the
field A 0, then because of anti-symmetry of
f mu nu pi 0 vanishes. Therefore, you have
a
difficulty in considering the commutation
relation here, because one of the conjugate
momentum actually vanishes so how to deal
with that.
There are many ways you can handle this problem
one of them is by exploiting the gauge
transformation here, because the theory is
invariant under this gauge transformation
you
can what you can do is, you can fix a gauge
and then you can quantize. Or you can start
with this Lagrangian you can add an additional
terms here and then you can quantize it in
different way.
So, we will what we will see is that, we will
quantize it into two different ways first
one
is we will first fix the gauge and then we
will quantize with theory and then what is
we
.will consider the covariant quantization,
which we will perform without fixing any
gauge. So, let us first do not worry about
the Lorentz covariance or gauge invariance
or
anything like that we will just do the gauge
fixing, so how do you do that. Let us write
the equations of motion corresponding to this
Lagrangian in a non covariant way.
.
So, what are the equations of the motion you
have del dot e equal to 0, because we are
considering the Maxwell’s equation in free
space and you have del cross E which is a
minus del B over del t del dot B equal to
0 right is this minus sign here correct. And
in
this notation the statement about gauge transformation
turns out to be, first of all this the
electric and magnetic fields are given by
B equal to del cross A and E equal to minus
grade phi minus del A over del t you can check
that from these two equations. This
implies that B has to be curl of a vector
field and now you can plug this forming in
this
one. And then you can see that it has to be
at this point or you can plug it here and
then
you can see that the electric field has to
here this form.
The gauge transformation says that the vector
potential A goes to A plus some gradient
of lambda whereas, the scalar potential phi
goes to phi plus del lambda over del t you
can
see that under this transformation I think
one of this has to be minus. If you make such
a
transformation, then the electric field as
well as the magnetic field remain invariant,
this
phi you can identify it with the 0th component
of the gauge field a mu, and the i th
components are identical to the i th components
of this vector potential here.
..
So, you can exploit this gauge transformation
to consider this gauge del dot A equal to
0,
this is known as the coulomb gauge. So, what
we will do is that, we will work with this
coulomb gauge, look at the equations of motion
you can consider, let us say this equation
here del dot E, E is minus grade phi minus
del A over del t. So, this is nothing but
minus
del square phi minus del over del t del dot
A, so in the gauge del dot A equal to 0, this
quantity here vanishes. Therefore, this equal
to 0 implies in the coulomb gauge, del
square phi equal to 0, if in addition you
impose the boundary condition that this scalar
potential vanishes at infinity, then it amongst
to saying that this vanishes throughout.
So, with suitable boundary condition, what
you have seen is that, we can just set these
and phi equal to 0, now you are left with
only the A field. Now, you can consider the
weather equations of motion appropriately
and then you can show that in the coulomb
gauge, this is what is the equation of motion
for the vector potential A. So, I leave it
as a
homework for you, you show that these equations
with the gauge del dot A equal to 0
and phi equal to 0 implies that del number
and acting on the vector potential A is equal
to 0.
So, this equation looks quite familiar to
you, this almost looks like the Klein Gordon
equation, except that the field A is now a
vector field, and also the mass term is no
longer there.
..
See what is the implication of that del square
A equal to 0 means that, for this field A
you will have a plane wave like solution,
which goes like e to the power minus i k dot
x
minus omega t, the amplitude here will be
sum vector epsilon of k. The field which
satisfies this condition del dot A equal to
0 is known as a transverse field, which it
so
because del dot A is simply given by, this
equal to 0 simply means k dot A equal to 0.
Therefore, the field is orthogonal to the
wave vector that is the reason you say that,
this is
a transverse field.
So, because the field is transverse, therefore
there are two independent polarizations
here, this epsilon here, this simply implies
that k dot epsilon of k equal to 0, this simply
implies that this epsilon k only have two
independent components, so it is a 3 vector.
But, it is not an arbitrary 3 vector, it is
a orthogonal to the direction of propagation
of the
wave, therefore it will have only two linearly
independent components. So, therefore, I
will denote, therefore this admits two linearly
independent solutions, I will denote them
to be epsilon 1 of k and epsilon 2 of k. I
can choose a linear combinations such that,
they
are orthogonal to each other, so epsilon r
of k dot epsilon s of k equal to delta r s.
..
So, the most general solution for the vector
field, this is summed over r integration d
cube k over 2 pi cube 2 k 0 and then epsilon
r of k a r k e to the power minus i k dot
x
minus omega t plus epsilon r of k a dagger
r of k e to the power minus i k dot x minus
omega t, this is the most general solution.
Now, do not have to worry about, the
conjugate momentum corresponding to the variable
A 0, because we have set this gauge
and the boundary condition. So, that A 0 equal
to 0 identically, so this field does not
appear at all.
So, you can now quantize it in the usual way,
you will find that I mean the analysis is
exactly identical to what we have discussed
for a free scalar field. You can show that,
this field a r of these creation at annihilations
operators a r, and a r daggers satisfy this
commutation relation a s dagger of k prime
commutator is equal to 2 pi cube 2 k 0 delta
r
s delta of k a minus k prime whereas, a r
k a s k prime equal to 0 equal to a r dagger
of k
a s dagger of k prime.
So, you can argue that, there exist a state
which we will call as the ground state, and
this
is annihilated by a r of k for all k and r
equal to 1, 2 then the n particles states
are created
by acting creation operators on this vacuum
here. So, the entire analysis will be exactly
identical to that of a real scalar field,
but the difficulty here is that in quantizing
this way,
the covariance of the theory is lost now,
this is no longer manifest Lorentz invariant.
.So, because you have already chosen a gauge
in which A 0 equal to 0, so Lorentz
invariance is lost in this process, what you
would like to see is, that you would like
to see
the quantization, where the Lorentz invariance
is manifest to do that what we will do is,
we will 
and then we will quantize.
.
So, this is your Lagrangian density minus
1 4th F mu nu F mu nu to this I have added
this term minus lambda over 4 del mu A mu
square, so of course, this is not a Maxwell's
Lagrangian, this is modified this term. Let
us see what is the equation of motion
corresponding to this Lagrangian density here,
where del L over del del alpha A beta,
from here you get minus F alpha beta. And
this one will give us minus lambda over 2
times what you will get eta alpha beta del
mu A mu.
Del alpha of del L over del del alpha A beta
equal to 0, so this simply means that you
have here del alpha F alpha beta minus lambda
over 2 del alpha it is A plus. So, let us
write it in the more simplified manner, this
is del alpha of F alpha beta is del alpha
A
beta minus del beta A alpha plus lambda over
2 eta alpha beta del alpha del mu 
a mu
equal to 0.
..
So, this simply means that you have, del square
acting on A beta minus 1 minus lambda
over 2 del beta of del mu A mu equal to 0,
I mean it is trivial from this that, if you
act
further on del beta, then you see that del
square of del mu Amu becomes 0. This also
gives del mu, there is a del mu A mu here,
del square del mu A mu, here again you get
a
del square del mu A mu and these two terms
add there is some common coefficient apart
from that this is 0.
So, this quantity del mu A mu, it acts like
A free field, free scalar field you can you
see
that you recover Maxwell's equation, if you
impose this gauge del mu A mu equal to 0,
if
del mu A mu equal to 0. Then this quantity
the equation of motion coincides with the
Maxwell's equations, so what we will do is
that, we will first quantize the
electromagnetic field here. And then later
we will see what do you mean by imposing
this condition del mu A mu equal to 0, what
does this mean, this is known as the Lorentz
gauge.
We can choose one specific value for this
parameter lambda here, which is lambda equal
to 1, lambda equal to 1 although it is not
a gauge it is known as the Feynman gauge,
so
we will work with Feynman gauge and then quantize
this theory. So, tomorrow we will
start with this lambda equal to 1 gauge and
then we will carry out the quantization and
we will impose this Lorentz gauge condition
and then we will see what it physically
means.
.
