So, we will continue our discussion on the
interaction of charged particles with electromagnetic
field in the quantum picture,
and in relativistic case. Essentially this
is what we call a quantum electrodynamics,
but we will come to the formal quantum electrodynamics
as a field. The quantum field theory at a
little stage, but this is basically to build
or give way to such a quantum field theory
how to think about that.
So, we will consider the discussion that we
have been doing in the past couple of lectures
forward.
So, we were discussing the gauge freedom associated
with the 4 vector potential picture of the
electromagnetic field. So, we said we could
express the electromagnetic potential A mu;
or express the electromagnetic field in terms
of the scalar potential and the vector 3 vector
potential putting them together as components
of a 4 vector, we can consider A mu, the 4
vector as the 4 vector potential representing
the electromagnetic field.
And we said, if we consider the measurable
quantities, electric and magnetic fields E
and B associated with this A mu, then we will
see that the A is not uniquely defined or
fixed. We can actually take A mu to A mu plus
derivative of some scalar function chi, which
will still give you the same electric and
magnetic fields. And this freedom to choose
any A mu related by this relation is basically
called gauge freedom and that is a kind of
an arbitrary defining the potential.
But it is taken as an advantage in quantum
electrodynamics. We will come to that a little
later when we discuss the gauge symmetry,
but at the moment this lets us fix the A mu,
so that doh mu A mu is equal to 0 that is
A mu can be made divergence less; 4 divergence
a is equal to 0 by appropriately choosing
the scalar function chi.
Now, this is what we saw yesterday and even
after fixing this there still we can actually
transform A mu to A mu plus doh mu lambda
the secondary A mu is the now A mu now mu
that satisfies the. So, let me call this A
mu prime to avoid any confusion. So, A mu
prime is the new field and A mu prime can
be transferred now to A mu prime plus some
doh mu lambda and then of course, this should
also give you the same electric and magnetic
field because it is similar to the earlier
transformation.
But we have to check whether the condition
the divergence condition of A prime mu is
also conserved if you take the divergence
of the new field A double prime you then we
will have divergence of A prime mu plus doh
mu of lambda. So, this first term is equal
to 0 and second term can be made 0, if we
choose a lambda appropriately. So, this is
equal to 0 for particular lambda.
So, first one transformation fixes the diver
a fixes A mu. So, that it is divergence less
and then we see that still there is a freedom
to choose A, the potential and we see that
there is always possibly this always possible
to choose A double prime by adding a derivative
term of a scalar function again to the A prime
as in the case of equation 2. This will not
change anything, it will not change this electric
and magnetic fields, the physical quantities,
neither will it change the condition that
the A prime is divergence less or it therefore,
A double prime is also divergence less, if
lambda satisfies the condition that doh mu
doh mu lambda is equal to 0.
So, the second transformation is not for any
scalar function, unlike the first transformation
where chi could have been any scalar function;
lambda is not any scalar function lambda is
the scalar function which satisfies the condition
doh mu doh mu is equal to 0. So, that is basically
the thing we will return back to this when
we discuss explicit expression for A mu later.
%%%%
Now, this actually says E is equal to minus
del phi minus doh A over doh t. and B equal
to curl of A, as satisfied by A mu or A prime
mu which is essentially equal to A mu plus
doh mu chi any scalar function chi, and A
double prime mu which is A prime mu plus doh
mu lambda where lambda is a scalar function.
So, that doh mu doh lambda is equal to 0.
And this gives doh mu A prime mu is equal
to 0 and also doh mu A double prime mu equal
to 0 divergence.
So, we have the same physical situation for
any of these fields and then in A double;
we will see that A double prime mu, we have
fixed the arbitrariness. So, that is unique
in the sense that I mean that fix that is
an in the sense that you cannot add any more
terms in the this thing, because lambda is
already lamb the choices of chi and lambda
will fix this thing for the condition for
A mu prime divergence less mu A prime is satisfied
here.
Let us look at the Maxwell’s equations with
all these conditions applied to the potential.
.
Maxwell’s equation written in terms of the
potential was doh nu doh mu A mu minus doh
mu doh nu A nu 4-divergence equal to j mu.
So, if we have a condition that gauge choice
is such that doh nu A nu is equal to 0 as
we discussed in earlier case; this condition
is called Lorenz gauge condition.
In that case the second term from the Maxwell’s
equation drops out and then we have doh nu
doh nu A mu equal to j mu. Or in a slightly
more compact way we can write it as box square
which is equivalent to doh nu doh nu or doh
mu doh mu where nu is summed over. And in
that case, I can write the box square A mu
equal to j mu slightly in need a way of writing
it meaning the same thing in any case.
Now, let us come to solutions of this equation.
This is the Maxwell’s equation with Lorenz
gauge condition applied. Maybe I should highlight
this. So, this is our Maxwell’s equation
with Lorenz condition. So, let us look at
it. So, we have del square A.
Mu equal to j mu equal to 0 for free electromagnetic
waves. Or I would say in the quantum picture,
free photons. We are not quantized it yet,
but let us say that we can do that and then
think about electromagnetic field as collection
of photons. In that case A mu represent the
photon. Now we are actually deviating a little
away from the earlier interpretation or earlier
picture of giving more physical meaning to
the electric field and the magnetic field,
and then considering the potential as an auxiliary
and function. That was the earlier picture
and then we said that we can actually have
different the potential is even which will
give you the same physical situation.
Now, in the quantum electrodynamics we will
deviate from this picture and then say that
photon, which is the quantum of the electromagnetic
field, is more meaningful physically and the
field associated with that is the potential;
is equal to the potential A mu, and therefore,
A mu is the one which represents the photon.
Associated to that, any photon we can actually
think about the electric field and the magnetic
field.
When you come down to see the picture of the
photon in the non-relativistic case we will
see the electric field and magnetic field
associated with photons with certain relations
between them. But A mu therefore, takes a
more physical meaning in quantum electrodynamics.
And arbitrariness can be fixed by choosing
the gauge conditions and then it somewhat
uniquely fixes the potential corresponding
to the photon or the wave function corresponding
to the photon or the field corresponding to
the function.
At this moment I will be a little vague in
defining what is wave function, what is the
field what is the potential, etcetera. Basically
I will interchange, I call A mu sometimes
as feel sometimes as the wave function corresponding
to the photons sometimes as the potential.
So, all these are the same thing, but looking
at from different pictures accordingly the
interpretation is slightly different, but
basically it is the same entity. So, for the
time being we will not in distinguish it clearly.
As we come to the clear representation of
the quantum field corresponding to the photon
we will, as an operator that is the correct
picture in quantum field theory or quantum
electrodynamics we will talk a little more
about what is the quantum field A mu. But
essentially that is the same as the potential
that we are considering in a relativistic
electron dynamics. So, this, the box square
is essentially called d’Alembertian. So,
d’Alembertian A mu, d’Alembertian is an
operator which acts on A mu here to give you
0 for free photons.
Solution well this is basically an easy guess.
If I take A mu to be some constant function
constant 4 vector as far as x is concerned;
d’Alembertian is an operator which acts
on the coordinates not on the momentum. So,
I can take this some constant function, constant
in x, function which is constant in x, but
could depend on momentum; a four vector and
explicit coordinate dependence in an oscillatory
exponential minus i q dot x, where q dot x
is for vector dot product q mu x mu where
mu is summed over.
Let me convince you that it is actually; indeed,
a solution. Or what are the conditions that
anything should satisfy for this to be a solution.
So, let me act on it by doh nu A mu is equal
to doh nu acting on epsilon mu q, e power
minus iq dot x which is equal to doh by doh
x nu acting on the same thing epsilon mu q,
e power minus i q 0 p minus q 1 x minus q
2 y minus q 3 z ok.
So, now let us take case by case: nu equal
to 0 which is doh by doh t of A mu which is
equal to doh by doh t of epsilon mu q, e power
minus i q 0 t; rest of it is 3 momentum dot
product with 3 coordinate x vector. So, there
is only one particular place where t appears
on the right hand side. It is exponential
minus i q 0 t. So, when I take the derivative
epsilon mu q is independent of t comes out,
E power minus i q 0 t, when the derivative
acts on its minus i q 0, E power minus i q
0 t and other things remain the same. minus
q dot x. So, this is essentially equal to
minus i q 0 A mu all right. So, now similarly
doh by doh x A mu is equal to doh by doh x
epsilon mu q exponential minus i q 0 t minus
q 1 x q 2 y minus q 3 z. So, that will give
you minus of minus plus i q 1 A. So, that
is one thing.
Now, what we want to have A doh nu doh nu
acting on mu which is equal to doh square
by doh t square acting on mu minus doh square
by doh x square acting on A mu minus doh square
by doh y square acting on A mu minus doh square
over doh z square acting on A mu. Same mu
and we had doh by doh t on A mu is equal to
minus iq 0 A mu.
So, doh square A mu by doh t square is equal
to minus iq 0 independent of t and doh by
doh t acting on A mu, which is another q 0
minus i q 0 A mu; which is equal to minus
q 0 square A mu. And doh square by doh x square
A mu will give you minus q 1 square A mu.
And doh square by doh y square A mu is minus
q 2 square A mu and doh square over doh z
square A mu is equal to minus q 3 square A
mu. Putting it together doh nu,
doh nu A mu is equal to minus q 0 square plus
q 1 square plus q 2 square plus q 3 square
which is equal to minus 4 vector square q
square A mu.
And we have Maxwell’s equation with Lorenz
gauge condition, del square, I mean the box
square A mu equal to 0. And that gives us
minus q square A mu equal to 0, we essentially
q square need to be equal to 0. And what is
q square q square is equal to q 0 square which
is essentially E square minus p square or
3 vector minus 3 vector square which is equal
to m square c square; and when we say q square
equal to 0 m square is equal to 0 q square
for vector square is equal to 0.
Now, as we said earlier when we consider the
A mu as representing the photon and we associated
f for momentum q with A mu like in this case
A mu is equal to epsilon mu q exponential
minus i q dot x is what we considered, right.
So, q is essentially then interpreted as the
momentum four momentum of the photon. And
Maxwell’s equation gave us for momentum
square is equal to 0 for the photon. Of course,
you can ask the question like this was in
the case of A mu satisfying the Lorenz condition.
What about if I consider A mu which is not
divergence less then still can I interpret
this as the photon with mass equal to 0.
So, this essentially gives you mass less photon.
I mean when I say mass less photon in the
sense it says that photon has no mass. But
then you remember this gauge condition is
not an additional condition that we can apply
we could have applied the gauge condition
and chosen any condition any kind of gauge
any conditions that satisfies the gauge transformation.
So, that electric field and magnetic field
remains the same from that point of view the
physics should not depend on whether I have
taken Lorenz gauge or not. Lorenz gauge condition
or not; which means the A mu that satisfies
the Maxwell’s equation, the way we had written
with the gauge condition or some A prime which
is not divergence less and do not satisfy
the Lorenz gauge condition, but is related
to this A mu through a gauge condition, gauge
transformation by adding A or differed by
A doh mu chi kind of a term then chi is a
scalar fit in that case they are physically
the same equivalent to each other and therefore,
all of those when we consider as photons if
in one case, it is mass less, then in all
other cases also it should be mass less.
So, this represents the same this thing and
then photon in any case whether there is a
Lorenz condition applied or not it is mass
less. Only that when you take it in this pattern
it is manifestly clear its manifested, there
it is very apparently there clearly there.
Now let me look at it a little more and then
see what is the interpretation that we can
give to epsilon.
So, we had we had the solution, A mu equal
to epsilon mu q E power minus E power minus
i q dot x for the photon field satisfying
the Maxwell’s equation.
Let me write this epsilon mu as epsilon 0
and 3 vector epsilon; E power minus i q x.
So, I have not explicitly written they q dependence
on the epsilon now. It is understood that
it depends on q. It is just for the sake of
clarity that we are not adding the bracket
with q there. What is E electric field is
equal to minus grad phi minus doh A by doh
t.
supposing we do not have any electrostatic
potential phi and electric field is only due
to the changing magnetic field, like in the
case of electromagnetic waves. So, for a free
photon we can imagine that there is nothing
else, but the changing electric field and
changing magnetic field; one inducing the
other. So, in that case this is equal to 0.
So, you have minus doh A by doh t as the electric
field and that is essentially epsilon exponential
doh by doh t acting on this thing will give
you minus i q 0 E power minus I q x. Or minus
i q naught epsilon which depends on q exponential
minus i q naught t exponential i q dot x.
So, this is basically the electric fields
with explicit dependence on the time and x
written.
We are familiar with this and the amplitude
is given by a complex amplitude here minus
i q naught epsilon 3 vector direction of the
electric field is the same as the direction
of the epsilon 3 vector. And therefore, we
can say that polarization direction of the
photon is that of the epsilon.
Now, let us see what happens to the polarization
vector in the case of Lorenz condition, let
us look at Lorenz condition which is doh mu
A mu equal to 0. A mu equal to epsilon mu
E power minus i q x will immediately give
you q mu or minus i q mu epsilon mu is equal
to 0 when 
doh mu a is acted on A mu. So, this says that,
Lorenz condition says not all the components
epsilon 0 1 2 3 are independent of each other.
But depends through the relation q mu epsilon
mu equal to 0 or q naught epsilon naught minus
q 1 minus q 1 epsilon one minus q 2 epsilon
2 minus q 3 epsilon 3 equal to 0. So, for
a fixed q epsilon; all the 4 components of
epsilon are not independent of each other.
So, there is a relation that they had to satisfy
therefore, there are only 3 independent. That
gives 3 independent components.
Further we can choose lambda. Let us choose
lambda equal to i a q sorry i a exponential
minus i q dot x what is doh mu lambda doh
mu lambda is equal to i a minus i q mu E power
minus i q dot x; and doh mu doh mu lambda
is equal to i a minus q square E power minus
iq dot x. But q square is equal to 0 says
that this is equal to 0.
So, a scalar function satisfying doh mu doh
mu lambda equal to 0 is i a exponential minus
i q dot x. Well this i a is kept like that
because of reason that will be clear just
now. So, this is 0 because q square is equal
to c. So, now, what does that tell you in
terms of the change?
In epsilon corresponding to a change a transformation
of A mu to mu plus doh mu lambda this is equal
to A mu plus a q. So, we had i a minus i q
mu exponential minus i q x.
So, if I write A mu as epsilon mu E power
minus i q dot x then plus i a times minus
i q mu is plus a q mu exponential minus iq
x, which is epsilon mu plus a q mu E power
minus i q x. If I denote this by some epsilon
prime E power minus iq x as A mu goes to a
prime mu equal to A mu plus doh mu lambda,
with this particular lambda, this gives epsilon
mu going to epsilon prime mu, which is equal
to epsilon mu plus a q mu right, fine.
So, I have epsilon prime mu equal to epsilon
mu plus a q mu. Now I can choose a, so that
epsilon prime 0 is equal to 0 all right. So,
this is why we had taken the lambda in that
particular way. So, I can choose this. So,
initially we had a condition on epsilon that
epsilon 0 q 0 epsilon q 1 epsilon 1 q 2 epsilon
2 q 3 epsilon 3 equal to 0. Now that epsilon
prime. So, this is applicable to epsilon prime
as well, because its the same this thing satisfying
same condition Lorenz condition this is basically
the Lorenz condition and changing it by lambda
which has doh mu doh mu is equal to doh mu
doh mu lambda equal to 0 will not change the
Lorenz condition. So, this actually tells
you that q 0 epsilon prime 0 minus.
Let me write it as q dot epsilon prime is
equal to 0. Since epsilon prime 0 is set to
0 by appropriately choosing appropriately
choosing a, we have q dot epsilon prime 3
vector is equal to 0. Epsilon 3 prime has
3 components, but then there is a condition
q dot epsilon 3 epsilon 3 vector prime is
equal to 0. So, essentially there is a relation
between these 3 components, and therefore,
that gives you 2 independent components.
.
Or we can say that photon has 2 degrees of
freedom. So, let us summarize what we have
been doing. We had the Maxwell’s equation
In terms of field tensor F mu is equal to
j nu. doh mu F mu nu equal to j nu and we
could write f mu nu in terms of vector potential
as doh mu A nu minus doh nu A mu that gave
the Maxwell’s equation as doh mu doh mu
A nu minus doh nu doh mu A mu equal to j nu.
Then we had the Lorenz gauge condition doh
mu A mu equal to 0; this Lorenz gauge condition
will not affect physics, because we can find
out any mu which satisfies the gauge condition
with the same electric and magnetic field
associated with it as the original case. And
that gives us Maxwell’s equation del square
sorry the box square A mu equal to j mu. Then
we considered the solutions A mu equal to
epsilon mu q exponential minus i q x for j
equal to 0 case; which is then interpreted
as representing the a free photon without
any charge density or current density in the
region isolated electromagnetic wave freely
propagating in the quantum vector equivalent
to a photon represented by A mu.
And that told us that we have q square equal
to 0; mass less photon and epsilon mu is essentially
the polarization vector in the 4 dimensional
case and especially in the 3 dimensional case
it is the familiar polarization associated
with electromagnetic field. And we saw that
it has 2 degrees of freedom or 2 independent
components maybe I should write it in that
fashion.
So, this with the Lorenz condition this has
2 independent components; that in turn tells
us that photon has 2 degrees of freedom. We
will come to the discussion of the interaction
of charged particle with such photons in the
coming discussion.
