So, we are discussing the covariant quantization,
the Lagrangian density is given by f mu
nu f mu nu. We have modified this Lagrangian
density by A d, which is proportional to
del dot a del mu a mu whole square. This is
the Lagrangian density we will we consider
and then we will quantize this. Of course,
this is not as the same as the Maxwell say
Lagrangian density, but because of the presence
of this additional term, but we will make
sure that the additional term does not have
any physical effect adding this additional
term
does not have any physical effect.
So let us so the Feynman gets in this notation
correspond to lambda equal to one.
Although, you can consider this for any general
lambda its we will fix a lambda to be a
specific value. We will do quantize you can
consider lambda to be any arbitrary number.
Then you can quantize it nothing will change.
So, yes I could include yesterdays notation
it will be lambda equal to half, which is
the gauge will we will be working the equation
of motion is a given by del alpha A beta del
alpha equal to 0.
.So, let us compute this quantity del l over
del alpha A beta this one will give me A minus
f alpha beta. This one will give me A minus
eta alpha beta times lambda times del mu a
mu. Now, you know why I chosen this vector
two here, because I want to do this to
cancel, here that is the reason I am choosing
to.
So, the equation of motion is just del alpha
of f alpha beta minus lambda theta alpha beta
del alpha del mu A mu equal to 0 plus. So,
let us work out what we get del alpha del
alpha a beta minus del beta A alpha plus lambda
times, del beta del mu A mu equal to 0.
This one is nothing but del square acting
on a beta minus del beta acting on del alpha
A
alpha plus lambda del beta acting on del mu
A mu equal to 0. Because, mu is a dummy
index here I can change it to alpha.
Then what I see is the equation of motion
is basically del square a beta minus 1 minus
lambda times del beta del alpha A alpha equal
to 0. This is the equation of motion you
can see that. This equation here implies del
square acting on del alpha A alpha equal to
0.
So, del alpha A alpha acts like a free field
by suitable boundary condition, by choosing
suitable boundary condition.
.
You can set del alpha A alpha to be 0 classically,
which is equivalent Lawrence sketch.
So, if you choose this Lawrence gauge that
del alpha a alpha equal to 0 throughout this
space time. Then this is equivalent to saying
that you actually restrict yourselves to the
.Maxwell Lagrangian, but this is all classically,
quantum mechanically it is A not
substitute forward.
We will see what the this equation here simply
implies that the conjugate momentum to
the field A beta, which I will denote by pi
beta is given by minus f 0 beta minus eta
0
beta times del mu A mu with a factor of lambda.
So, this is the conjugate momentum to
the field A beta you see that pi 0 specially
is no longer zero. So, as such you they not
have any difficulty in considering the equal
time commutation relations, which are given
by A mu of x t pi nu of Y t is basically i
delta mu nu delta x minus y.
.
So, as an operator equation you can consider
this equal time commutation relation, but
you can see that pi 0 from this equation is
basically f 0 is 0. So, pi 0 is simply minus
lambda times del mu A mu. So, you cannot set
del mu A mu equal to 0 as an operator
equation quantum mechanically. It will be
inconsistent to just set this to be 0, because
if
you set this to be 0. Then the equal time
commutation relation you will have for sum
this
terms you will have the left hand side 0,
but the right hand side is not 0.
So, this is going to pose a problem that we
will see how to solve this. Therefore, as
an
operator equation we cannot impose the condition
that del mu A mu equal to 0. So, we
will see how actually we will what is the
physical condition that we have to impose.
So,
that we restrict ourselves to the Maxwell’s
theory. That is what we will discuss in a
.moment, but before this we will just a carry
on with A what we have learnt. So, then we
will start quantizing.
So, the equation of motion is del square A
beta minus one minus lambda del alpha A
alpha equal to 0. As I say this although we
can consider this lambda to be any arbitrary
parameter. It will simplify much if you if
we choose this parameter lambda to be 1,
because for when we restrict ourselves to
the ourselves to the choice lambda equal to
1.
Then the equation of motion becomes del square
a beta equal to 0 and we know how to
handle this equation. So, we will we will
solve this equation.
This restriction of this parameter lambda
equal to one is known as the Feynman gauge.
So, we work in the Feynman gauge for, which
the equation of motion is simply del
square acting on a beta equal to 0. So, this
almost looks like again the claim Jordan mass
less claim Jordan equation for each of these
fields A beta. So, we know how to solve this
equation exactly, there will be four linearly
independent solution. Now, because we are
no longer doing any kind of gauge fixing here,
therefore here A beta will go like some
epsilon beta of k e to the power minus i k
dot x.
So, let us lets parameterize this four linearly
independent polarization to be epsilon
lambda mu of k lambda varies from 0, 1, 2,
3; just like in the previous case. When we
have done when we had performed the gauge
fixing, we had A 2 linearly independent
polarization. We had two transverse polarizations
and those we denoted by epsilon r of k
with r equal to 1 and 2. Here, there will
be four linearly independent polarizations
two of
them are transverse polarizations. The remaining
two are known as the longitudinal
polarization, as well as the scalar polarization.
So, epsilon 0 mu of k is known as the
scalar polarization and epsilon three mu of
k is the longitudinal polarization. The most
general solution for the field A can be written
as A mu of x t is again in terms of the four
real components.
..
You will have sum over lambda as where is
0 to 3 and integration over d cube k over
2 pi
cube to k 0 times epsilon lambda mu of k A
lambda k e to the power minus i k dot x plus
epsilon lambda star mu of k A lambda dagger
of k e to the power i k dot x. Here, k 0 in
this integration is given by mode k. Because,
this is a mass less field this is the most
general solution.
Now, we are considering this a mu to be a
quantum field, therefore these operators,
now
we can look at the canonical commutation relations.
Then we can see what to the mean
in terms of these operators A and A dagger.
So, what in this gauge lambda equal to 1 the
conjugate momentum pi beta is given by minus
f 0 beta minus eta 0 beta del mu A mu.
You can see that this implies A dot mu of
x t A nu of y t the commutator of A dot and
A
is given by i eta mu nu delta of x minus y.
Whereas, A dot mu of x t A nu of y t A mu
dot of y t is equal to 0, similarly A mu of
x t a nu of y t equal to 0.
So, this is the set of equal time commutation
relations, we can substitute this here. Then
we can see what do we get in for the creation
for the operators A and A dagger before to
do that we will make a choice for the polarizations.
First of all we will assume that the
epsilons are real.
..
So, epsilon 
mu lambda of k is real and we will choose
these polarizations such that.
Suppose, n hat is unit vector along the time
direction, then we will make our choice such
that epsilon epsilon 0 mu is equal to n hat
mu. Then we will choose epsilon three mu of
k
to be the vector, which lies in the t k x
0 plane or in the n mu hat k mu. This A plane
and
this is one linearly independent vector in
this plane spin by n hat and k mu. You choose
epsilon three to be the unit vector, which
is orthogonal to n hat mu, which lies in this
plane. So, that is the choice for epsilon
3, epsilon 1 and epsilon 2 are two linearly
independent vectors, which are orthogonal
to each other, which are orthogonal to this
plane
I mean let say your time is along x 0 direction
and let us assume that k is along x 3
direction. Then these four linearly independent
vectors are simply epsilon 0 is equal to 1
0 0 0 epsilon 1 is a 0 1 0 0 epsilon 2 to
be 0 0 1 0 and epsilon 3 to be 0 0 0 1. I
mean not
exactly k mu, but you just look at A. I mean
you can this is just one choice that I have
given you need not restrict yourselves to
four linearly independent.
..
In general they with satisfy this equation
that is A given by sum over lambda epsilon
lambda mu of k epsilon lambda nu of k star.
You do not have to choose them to be real
and epsilon alpha lambda k epsilon beta lambda
k eta alpha beta. This is eta mu nu and
epsilon lambda mu epsilon lambda prime mu
eta mu nu is eta lambda lambda prime with
A star.
So, with this choice for the polarizations
we can look at the equal time commutation
relations. You can simply substitute the mode
expansions in these equations. Then you
can show the I mean this is almost identical
to the free scalar field case, except for
the
presence of these vectors here the polarization
vectors. Now, we are making this
particular choice for the polarizations. Then
you can substitute them and you can show
that the operators a and a daggers satisfy
the following relation, a lambda of k a lambda
prime dagger of k prime is given by minus
2 pi cube 2 k 0 eta lambda lambda prime delta
of k minus k prime.
..
Whereas, a lambda k a lambda prime k prime
equal to 0 and show is a lambda dagger of
k a lambda prime dagger of k prime. This is
what we get for the commutation relations
for the a s and a daggers. So, you can again
interpret these a s to be the annihilation
operators a daggers to be the creation operators.
You can define their vacuum to be the
one, which is annihilated by a lambda k for
lambda equal to 0, 1, 2, 3. For all k 
then the
one particle states two particle states etcetera
will be obtained from the vacuum by acting
the creation operators submit.
So, you might think that everything is perfectly
all right you have now the whole
spectrum. Then you can consider the Hamiltonian
and then what are the n and j Eigen
values and so on, but it is not so simple.
The reason is the following you look at the
one
particle states, which correspond to the let
say scalar polarization. So, folks space of
one
prime states are usually of this form. That
you just consider the cube k over 2 pi cube
to
k 0 sum distribution in the momentum space
f of k a lambda dagger of k acting on the
vacuum. So, a typical one particle state,
which has a finite spread in the momentum
space looks like this. You consider such a
state for the scalar polarization. So, you
just
set a lambda equal to.
So, you just consider the case when lambda
equal to 0 and then look at the norm for this
one particle state. So, what do you get for
the norm of this one particle state. So, this
will
be the d cube k prime over 2 pi cube 2 k 0
prime. Then you have f star of k prime f of
k
.then a lambda I am considering the one particle
state corresponding to this scalar
polarization. So, a lambda acting on k prime
a 0 dagger k this is what is the norm for
the
one particle state. Now, I can use this commutation
relation to flip this a and a dagger
here, but when I do that what I get is the
following. So, a 0 k prime a 0 dagger of k
is
basically the commutator of a 0 k prime a
0 dagger of k and plus a 0 dagger of k a 0
of k
prime.
.
Now, I will substitute this in the expression
for the norm of this one particle state
corresponding to the scalar polarization.
When I substitute that the contribution from
this
term vanishes, because I am considering this
inside this. So, therefore you have a
vacuum here and there is a ground state acting
on this is an annihilation operator. It will
just when acting on the ground state it will
give you 0. So, this term vanishes therefore,
what we get for the norm here is d cube k
over 2 pi cube 2 k 0 d cube k prime over 2
pi
cube 2 k 0 prime. This the commutator here,
but what is the commutator? If you look at
the commutator its minus 2 pi cube 2 k 0 eta
lambda lambda prime.
So, for lambda equal to 0 and lambda prime
equal to 0, it will give you minus 2 pi cube
2
k 0. Because, of this minus sign here you
have a minus 2 pi cube 2 k 0 times delta k
minus k prime, this is all you get. It has
zero, 0 is one, therefore this is what we
have.
Now, you can perform the k prime integration
and then there is f of k f star of k prime.
If
.you perform the k prime integration what
you get is norm of the ground state. There
is a
minus sign integration d cube k over 2 pi
cube 2 k 0 mode f of k square.
So, what you saw here is that the norm of
this one particle states are negative, therefore
when you quantize when you consider this theory.
Then you quantize it you see that
there are negative norm states present in
the Hilbert space. Therefore, you this is
of
course, a problem you need to get it off all
this negative norms state. So, need to consider
how to consider, how to impose a physical
condition, which will remove all the negative
norms states from the Hilbert space. Remember,
we had avoided one of the issues earlier,
which was the condition setting the condition
del mu a mu equal to 0.
.
If we set this as an operator equation, then
we run into inconsistency. In the sense that
the canonical commutation relations that we
have adapted cannot be satisfied, if we
consider this as an operator equation left
hand side becomes zero, but its right hand
side
none 0 and so on. So, we need to worry how
to impose this condition also in order to
recover the Maxwell’s theory. What we will
see is that we will see the states in the
Hilbert space are physical. If they satisfy
this condition that suppose psi is A state
for
which del mu A mu the expectation value of
this operator in this state is 0.
..
Then this is a physical state. So, this is
the physical state condition that that we
are going
to impose. Although, we cannot set del mu
a mu equal to 0 as an operator equation. We
can consider the expectation value of this
operator in any arbitrary state. Then we will
require that this expectation value is always
0. So, we are saying that state is a physical
state if it satisfies these conditions that
expectation value of del mu A mu equal to
0.
Otherwise, it is not a physical state. Then
we will see what is the implication of this
physical state condition?
We will see that this physical state condition
not only will give us the Maxwell’s theory.
It will also remove all the unwanted negative
norms states that appear in this spectrum
here. So, the physical Hilbert space will
not content of any of these negative norm
states,
that is what we will see in a moment. So,
this condition here is equivalent to imposing
this that to del mu A mu plus acting on this
state psi equal to 0. You can see that this
specially implies this one.
So, let us look at this condition here A mu
plus contents the positive frequency part
of A
the mode expansion. Therefore, A mu plus of
x according to our definition is d cube k
over 2 pi cube 2, k 0 sum over lambda equal
to 0 to 3 epsilon lambda mu of k a lambda,
of k e to the power minus i k dot x.
..
So, what do we get when I impose this condition
here? When I impose the condition that
this quantity acting on some state gives you
0, so to do that lets consider a basis in
the
Hilbert space ,which is given by this form
a basis of states, if I can choose a basis
of
states, which are of this form psi t phi.
.
Do you understand what is this psi t correspond
to the states, which I raise due to the
transverse polarization, whereas this phi
here correspond to the longitudinal and scalar
polarization. So, this is a scalar or this
correspond to transverse polarization. So,
this now
.what we get is we will see what happens,
when A mu plus acts on a state like this,
because this corresponds to this scalar polarization.
Therefore, this state phi here will be
of this form it will be obtained by a 0 k
1 a 0 dagger k 1 a 0 dagger k 2. There will
be a
number of a 0 operators.
.
Then there will be a number of a 3 dagger
k 1 prime a 3 dagger of k n prime etcetera
acting on the ground state here. So, a general
n particle state corresponding to this will
have n such operators here. Whereas, this
psi t here is just given by again this state
psi t
will be of this form, where you have transverse
polarization a 1 dagger of k 1 and a 1
dagger of k r number of such operators. Then
you will have also a 2 dagger of k 1 a
dagger of k n acting on the translate here.
Now, what happens is that, now what we will
see what a mu plus of x acting on this state
psi transverse and phi equal to 0 gets itself,
this is there is this del mu a mu acting on
this.
..
So, what do you get from here as you can see
this d cube k over 2 pi cube 2 k 0. This
derivative acting on this will give me sum
over lambda equal to 0 to 3 k mu epsilon mu
lambda of a k with a minus i 
a lambda k e to the power minus i k dot x
phi equal to 0.
Now, because of this appearance of this k
mu here the transverse polarization will simply
go away, because for lambda equal to 1 and
2 this is simply 0. Although, lambda is
summed over all the values lambda equal to
1 and 2, correspond to transverse
polarization.
So, this simply kills the lambda equal to
1 and 2. Therefore, what is left here is a
d cube
k over 2 pi cube 2 k 0. There is a minus i
and then a 0 of k minus a 3 k times e to the
power minus i k dot x, is this clear? This
lambda equal to 1 and 2 just goes away and
k
dot epsilon just gives you this minus sign
here, k is just a null vector and 
k you can
choose to be something like 1 0 0 1.
..
Then this will simply imply if according to
our choice epsilon 0 is just a 1 0 0 0.
Whereas, epsilon 3 0 0 0 1 and then this factor
will simply gave you this, because this
involves only the scalar and longitudinal
polarization. It’s irrelevant to consider
this
transverses part here.
.
So, these simply imply that 
a 0 k minus a 3 k acting on some n particles
state gives you
0. So, this is what is equivalent to imposing
this condition del mu a mu plus on this state
.psi equal to 0. Let say consider the number
operator corresponding to this scalar and
longitudinal polarization.
.
.
You can work out the expression for the number
operator you can show that the number
operator is simply d cube k over 2 pi cube
2 k 0 a 0 k a 0 dagger minus a 3. I think
there
is a 3 here, there is a 3 here minus a 0 k
a 0 dagger of k a 0. This is the number operator
corresponding to the longitudinal and scalar
polarization. You can look at some n
particles state phi n any arbitrary state
phi here can be express as sum coefficient
c n
times phi n summed over n, where phi n here
is n n particles state corresponding to this
scalar and longitudinal polarization. So,
there will be a 0 daggers and a 3 daggers
and
there will be n such objects acting on the
vacuum. That is what I am denoting by phi
n
and a phi in general is summed over all these
things.
..
So, now if you look at this quantity here
you can show that 
this 2 k 0 and a 3 dagger k a
three of k minus a 0 dagger k a 0 k 
phi n here, but now this phi n satisfies this
condition,
that a 0 acting on phi n is equal to a 3 acting
on phi n. So, I can used that relation when
I
used that this will become phi n d cube k
over 2 pi cube 2 k 0. This is a 3 dagger k
minus
a 0 dagger of k a 0 of k acting on phi n.
Suppose, phi n is not equal to phi 0 suppose
phi
n of course, in general it will have c 0 times
ground state plus c times one particles state
and so on.
.
.In general phi n will have something, so
suppose n is not 0 if n is not 0 then this
is not
ground state. So, in that case what you can
see is that this quantity here is simply equal
to
0, because you can consider this acting on
this is simply this is always equal to 0,
this
quantity here is always equal to 0. This is,
because if n is ground state then it of course,
it
annihilates ground state even if n is not
equal to ground state you can consider this.
Then
this when it acts on this one, from the right
again you can used the conjugate of this
equation here. This conjugate shows that this
is equal to 0. So, what does it mean? It is
simply means that 
if n is the number operator.
.
Then n acting on phi n is just n times phi
n right, but this calculation shows that this
is
equal to 0. Therefore, phi n, which is n times
phi n phi n, which is equal to 0.
..
Basically, implies that phi n phi n equal
to 0, when n is not equal to 0. In other words
if n
equal to 0 this is just a ground state, which
is unique norm, therefore this quantity is
delta
m 0. Therefore, if you consider an arbitrary
state corresponding to scalar and longitudinal
polarization, which satisfy this condition
here, then this state can either be ground
state
or it will have a 0 norm.
So, these are all null states expect for the
ground state all the n particle states
corresponding to this scalar and longitudinal
polarization, which satisfy the physical state
condition are all null state. Then you can
see that if you consider any physical observable
and then if you compute the expectation value.
For example, let say you consider the
Hamiltonian of the theory.
Then you compute the expectation value of
the Hamiltonian in the physical Hilbert
space, then these states these null states
completely decoupled. They just drop out from
the expectation values that is what you can
see when you do in an explicit computation.
So, although these states are there they do
not contribute to any physical measurable
quantity. We have said the physical Hilbert
space is formed by states of this type psi
equal to psi t and phi.
..
These involve the transverse polarization
whereas, these involve the scalar polarization.
However, not all these states in the in the
full Hilbert space satisfy the physical state
condition, which is given by del dot a plus
acting on psi is equal to 0. For example,
we
have considered the state the one particle
state, which is given by 1 is equal to integration
d cube k over 2 pi cube 2 omega f of k times
a 0 dagger k, acting on the vacuum. Then
we have seen that this state has negative
norm the norm of the state is less than 0.
So, this state for example does not satisfy
the physical state condition del dot a plus
acting on psi equal to 0 any state of this
type, which is psi t. If you consider any
state of
this type, which is actually given by the
creation, which is actually generated by acting
creation operators for the transverse polarizations,
which is a dagger 2 k, so on a dagger
1 k prime and so on like this.
So, states of this type do satisfy the physical
state condition. So, they are they are in
the
physical Hilbert space there are other states.
For example, states like this they do not
satisfy the physical state condition. So,
they do not belong to the physical Hilbert
space.
However, in addition to these states there
are a bunch of states, which are also generated
by this scalar. So, which are of this type
a 0 dagger k acting on this or a 3 dagger
k acting
on this or a bunch of operators, like a 0
dagger k a 3 dagger k, so on acting on the
state.
..
So, phi 0 I will call them phi n these are
the n particle states, which have 0 norms.
So,
phi n phi n norm of these states are 0. There
are a bunch of states, which are generated
by
acting these operators on the vacuum with
0 norm, this will be there in the physical
Hilbert space however, because these states
have 0 norm. Therefore, you consider any
observable quantity expectation values of
any operator. These states do not contribute
at
all. It is only these states, which are generated
by acting the transverse on the by acting
the creation operators corresponding to the
transverse polarizations. They only generate
the physical Hilbert space.
.
