Come to the interaction of charged particles.
In the earlier discussions, we had considered
interaction of interaction of charged particles
with some electromagnetic field represented
by say A mu. Charged particle, represented
by wave function phi. And we had written that
the equation Klein Gordon equation satisfied
by this charged particle then is dou mu dou
mu plus m square phi plus V phi is equal to
0 where m is the mass of the particle corresponding
to the wave function phi, and V is the potential
due to the electromagnetic field, which can
be written as V equal to minus i e dou mu
A mu plus A mu dou mu minus e square A mu
A mu.
Then we said the transition amplitude is minus
i integral j mu A mu d 4 x, where j mu is
the 4 vector current due to the particle minus
i e phi f star dou mu phi i minus dou mu phi
f star phii , where phi i denotes the wave
function of the particle before the interaction,
phi f denotes the wave function of the particle
after the interaction. e is the electric charge.
Pictorially this can be written as a current
j mu interacting with, and electromagnetic
field represented by A mu. In particular this
A mu could be due to another charged particle,
in that case we can associate A mu with this
mu charged particle, and let us say A mu is
the electric field due to that charged particle
and therefore, Maxwell’s equation will tell
you that box where A mu equal to j mu, where
this j mu is not the earlier current, but
this is the second current due to which the
first current, charged current, or the first
particle sees an electromagnetic field.
In this case we can if you write A mu as epsilon
mu function of q, e power minus i q x, then
box square A mu is essentially minus q square
A mu, but then now we also have del square
A mu equal to j 2 mu. Together, we then can
say that A mu is minus 1 over q square j mu.
If you observe carefully, if you are careful
with our earlier interpretations of q square
in the Maxwell’s equation the way we had
written; we had written q square equal to
we had q square equal to 0 for case which
represents a photo.
But now here if we take q square equal to
0 this will not work out, there will not be
any current q square equal to 0 works for
0 current, that is if you put j 2 mu equal
to 0, then it is 0 by 0 in the limit are you
will get something. We cannot, we do not take
that. So, what happens to the mass of the
photon, remember we are not having a free
electromagnetic field. So, it is not a free
photon which is propagating. Free photons
will satisfy q square equal to 0 or mass is
equal to 0.
But in the presence of matter or in when it
is interacting with other things, it can have
an effective mass either an effective positive
mass, or it can even have a negative, it can
be interpreted as taking away energy from
the system, or giving the energy to the system.
And remember also that we are working with
quantum mechanics, which allows us to violate
even the energy conservation, as long as it
happens for a short period of time.
Because delta E delta t is approximately equal
to h or it should be larger than or equal
to h for any transformation. So, you can actually
give some energy to some particular system,
in a very short time from a massive particle,
and then can remain mass less than the earlier
original actual mass of the particle, for
a short period of time as long as the uncertainty
principle allows you to do that.
So, actually it is going the other way around
it says that we will not really be able to
measure such deficits. So, this is one way
of, when we actually try to interpret it in
a in that way. And then the possible confusion
is that q square was taken to be 0 earlier
when we associated that with A and why is
it not 0 here, understood in this fashion
that when photon is interacting with matter,
or when we are considering the photon associated
with a charged particle, then we do not really
cannot take q square equal to 0 for that particle,
rather A mu there is then minus 1 over q square
j 2 mu, which is the current associated with
that particle.
Now, look at T f i; it is minus i j 1 mu A
mu d 4 x, which is equal to j 1 mu minus 1
over q square j 2 mu d 4 x that is there is
a j 1 mu interacting with a another current,
j 2 mu. Interaction is facilitated by transfer
of something say the photon..
And I can actually write the same T f i, as
we already known by raising the index on by
raising the index on j 2. So, j 2 nu I want
to write, but then I should have a g mu nu.
So, that everything is proper integrated over
for moment for position coordinates.
So, I will take it as integral j 1 mu, let
me take this g mu nu with this q square and
write it as g mu nu minus i; minus g mu nu
over q square j 2 mu. Now look at the picture
again. So, we have a current j 1 mu, interacting
with another current j 2 nu through exchange
of something, which is represented by g mu
nu over q square. So, these are 2 currents
j 1 and j 2, and the other 1 the factor is
basically propagation of the effect of j 2
as far as j 1 is concerned.
When j 1 looks at it, it sees the effect of
j 2 propagated to j 1, and the factor corresponding
to that coming in the transformation matrix
probability transition probability is minus
g mu nu over q square we call this the propagator
factor. So, we have in T f i a current interacting
with another current, a current j 1 interacting
with another current j 2 through a propagator,
then you sum over all the possible values
of x corresponding to this. And then that
will give you the transition amplitude.
I can actually say that. Let me do it in a
proper way at a particular point. Corresponding
to that I have A mu, and another point corresponding
to that I have an index nu. And this is j
mu, and this is j nu, and so the propagator
propagates from some point to another point,
and if the indices are nu, and mu the momentum
associated with that momentum transfer, which
means momentum that the propagator carries
is q, then the factor associated with that
is g mu nu over q square. This is one way
of diagrammatically representing it.
It gives us feeling that we understand clearly
what is happening there, but I would advise
that you should take it not as a real picture
or realistic case in that sense. This is only
an aid to get a feeling get an understanding
of what might be going on. But in quantum
picture you cannot actually trace the particle
perfectly as in a box; therefore, a current.
It is a very hazy picture in quantum case.
And then it is not possible to actually think
about a particular time a photon is emitted
by something, and absorbed by a particle.
But that does not say that what we are discussing
is a half-cooked idea. It incorporates this
uncertainty in it. So, when we actually look
at the transition probability it says the
probability for the transition to occur in
quantum mechanics. In a particular interaction
case it may or may not happen; one thing.
Second thing is that the expression itself
is mathematically correct up to whatever order
of perturbation that we are considering. It
is only that the way we want to see it perceiving,
picturize this, in our mind by actually looking
at a see if a particle is going in a current,
and a wiggled photon emitted by that which
is absorbed by it is interacting with another
particle, current etcetera is not completely
in that way.
There are a lot of issues with that and then
therefore, you should take it in that spirit.
So, I just mentioned this, because one should
not go away with a wrong picture of what is
happening there as such. So, it is only a
pictorial aid to let us write down the expression
for example, correct expressions for the transition
amplitude in this case. Although we call it
by either propagator the current etcetera
so, all these things are understood in that
sense.
Let us come to more proper understanding;
let us consider j mu 1, I can write it as
minus ie phi f star dou mu phi i minus dou
mu phi f star phi i. So, this is the current
due to a particle let us say this is the interaction
point. So, before that is phi i let me associate
momentum to that particle as P A, and at the
end after the interaction it has it is represented
by phi f, let us say the momentum is P B.
So, the current that we are talking about
is j1 mu is this. Now consider phi i equal
to some normalization exponential minus i
P A x. So, we have considered a plane wave
solution, plane wave solution the box normalization
or some appropriate normalization has to be
considered. We had to constraint the whole
thing in a finite volume in that sense. So,
N A is the normalization constant P A is the
momentum associated with this thing, and phi
f similarly is N B e power minus i P B x.
And dou mu phi i is equal to minus i P A mu
N A exponential minus i P A x, and dou mu
phi f star is equal to plus i P B plus, because
it is a phi star that I will consider not
phi N B e power minus not minus plus i P B
x.
.
Putting this together J 1 mu is equal to minus
ie dou mu phi star phi f star dou mu phi i.
So, dou mu phi is actually equal to minus
iP A phi i, and dou mu phi f star is plus
iP B phi f star. So, j mu is equal to phi
f star minus iP A phi i minus plus iP B phi
star phi i. This is equal to minus i minus
i it is minus e, then I have P A phi f star
phi i, second one I have minus i, and minus
i that is what is taken out.
So, it is plus P B phi f star phi i, which
is equal to minus e P A plus P B phi f star
phi i. Now this is equal to minus e P A plus
P B; N A N B e power minus i P A x e power
plus i P B x. Or I can write it as minus e
P A plus P B; N A N B e power minus i P A
minus P B x. So, that is the current that
I have j 2.
So, this was j 1 and j 2 nu. Let us consider
this to be some chi i initially with momentum
P C, and chi f with momentum P D. In that
case j 2 nu in his exactly similar fashion
as the j 1 mu; can be written as minus e P
C plus P D; N C N D e power minus i P C minus
P D x, where chi i P C is equal to N C e power
i P C x, and chi f P D the final state is
N D e power minus i P D x.
So, we have j 1 and j 2 putting together in
transition amplitude T f i, this gives us
minus i j 1 mu minus g mu nu over q square
j 2 nu d 4 x. This is our transition amplitude,
where q is essentially the momentum transfer.
So, that is let us say from the point of view
of the current 1, particle 1 it is P A minus
P B the for momentum terms, and in terms of
the wave function explicitly written, we have
integral minus e N A N B; P A plus P B e power
P B mu alright. So, in all these cases there
is mu which is missing. So, I should be similarly
for j 2 there is a nu. e power minus i P A
minus P B x times minus g mu nu over q square,
then the other current minus e N C N D; PC
plus P D nu e power minus i P C minus P D
x. Integrated over for volume d4x. This is
the transition amplitude written now, in terms
of the wave function explicitly written as
a function of the momentum and the position.
Now, let me do some simplification. Firstly,
there is a minus i N A N B N C N D integral
e P A plus P B mu minus g mu nu over q square
e P C plus P D nu. None of this depend on
the position. So, integration comes out of
the integration. Integral the exponential
minus i P A plus P C minus P B minus P D x
d4x.
What is integral e power minus i p x d x minus
infinity to plus infinity full right, this
is nothing but the Dirac delta function. Now
if I have more than 1 variable, I can write
e power say 3 dimensional case P dot x d 3
x, I will write it as exponential minus i
P 1 x plus P 2 y plus P 3 z dx dy dz. So,
each of these are independent of each other.
So, I can take it as integral e power minus
i P 1 x dx integral e power minus i P 2 y
dy integral e power minus i P 3 z dz. This
is equal to first one is a delta function
2 pi delta P 1, 2 pi delta P 2, 2 pi delta
P 3 together as a compact notation, I can
write it as 2 power 3 delta 3 vector P. Meaning
of the last line is actually explicit form
in the explicit form the line just before
that.
In a similar fashion look at what we have
in T fi last line integral e power minus i
P A plus P C minus P B plus P D. That can
be taken as some single entity q or some capital
q or whatever as a momentum combination of
something, and then total together is a for
moment times or dotted with x d4x.
So, that is essentially delta function minus
i P A plus P C minus P B minus P D x d 4 x,
where now x is the 4 vector x; x 0 x 1 x 2
x 3 or x 0 x y z all of this together.
So, we have 2 pi 4 delta 4 dimension P A plus
P C minus P B minus P D. This actually means
that 2 pi 4 1 delta function says P A 0 plus
P z 0, P B 0, P D 0; and another delta function
for P A 1 P C 1 P B 1 minus P D 1 into delta
P A 2 plus P C 2 minus P B 2 minus P D 2 into
delta P A 3 plus P C 3 minus P B 3 minus P
D3. With that, we can write T fi as minus
i N A N B N C 
N D;
Then you have e P A plus P B mu minus g mu
nu over q square e times P C plus P D nu,
and you have a 2 pi 4 delta 4 P A plus P C
minus P B minus P D.
So, we will write it as let me write it here,
T f i equal to minus i N A N B N C N D delta
4 P A plus P C minus P B minus P D into m,
where m is equal to e P A plus P B mu, minus
g mu nu over q square e P C plus P D.
Let us stare at it for some time. what is
this? This is the transition amplitude for
a particular interaction given by particle,
with momentum P A going to a particle, with
momentum P B, after interaction with a particle
another particle with momentum P C, and going
which will go to I am a particle with momentum
P D. Charges on each of this particle is e,
and the momentum transfer is q. This then
says that an N A, N B, N C, N D are the normalizations
associated with that, we will come to that.
Look at m. It is a Lorenz invariant quantity
as we said, this is the dot product between
P A plus P B and P C plus P D with a q square
which anyway is invariant under Lorenz transformation,
and e square which is again invariant under
this.
So, the first part e P A plus P B nu is associated
with the first current particle current, second
part is in the bracket g mu nu over q square
is associated with the propagator, and the
last part e P C plus P D nu is associated
with the current due to the second particle;
that is one part. Essentially that summarizes
all the dynamics in it. What it is, it will
give you the charge current interaction. Two
charged particles, the currents of two particles
are interacting with each other.
Now, the rest of it in T fi is; one is the
normalization factor it is a factor of minus
i also, apart from that there is a delta function
what does the delta function tell you, when
you actually put it in the cross section as
we will do in a moment, in the next class
as such. So, when we put it in the cross section
and try to find the number of particles which
are coming out etcetera, then we will have
to integrate it over the momentum, when we
do that immediately we realize, because of
the delta function that when you integrate
over all the momentum P A plus P C is equal
to P B plus P D that is because of the delta
function.
And what does that tell you. So, the delta
4 P A plus P C minus P B minus P D tells us
that P A plus P C is equal to P B plus P D,
what is P A it is essentially the initial
momentum of one particle, and P C is the initial
momentum of the other particle. So, if I write
it in components these are 4 momenta, I will
get P A 0 plus P C 0 is equal to P B 0 plus
P D 0 or essentially, when you consider the
4 momentum, this is the energy of the particle
A, plus energy of the particle C is equal
to energy of the particle B plus energy of
the particle D.
When I say particle D, I mean the corresponding
particle in the final set the second particle
in the final state. What is this? This is
basically that the sum of the initial energy
is equal to sum of the final energy; this
is the statement of energy conservation. Similarly,
when we look at the 3 momentum P A plus P
C is equal to P B plus P D, the sum of the
initial 3 momentum equal to sum of the final
3 momentum.
So, this is the linear momentum conservation.
So, the delta function makes sure that the
energy of the initial particle, sum of the
initial energies is equal to sum of the final
energies. Energy is conserved. And similarly
Lorenz the linear momentum is conserved. And
that only talks about the kinematics that
the energy momentum conservation that part.
The rest of it all the dynamics is in the
quantity m. And m is the is called the Lorenz
invariant transition amplitude. And we should
see that, note that when we consider these
charged particles initially, we have not considered
anything any information about this particle
other than its momentum and charge. We have
in fact represented the particle by a wave
function phi i equal to N A e power minus
i P A x. It has a momentum ok.
It has some charge associated with that. That
is understood, but supposing we consider it
as an electron, then what about the spin of
this particle? And underlying equation that
we considered was Maxwell’s equation. And
even before that the Klein Gordon equation.
And we will see that here with the Klein Gordon
equation satisfied by the particle phi, we
do not have any provision to discuss the spin
of the particle.
So, if we want to incorporate the spin of
the particle or the information related to
the spin of the particle, then we will have
to modify this, and find out slightly different
way of dealing with these particles. That
is one thing that we will have to understand
and then we will come to. The other thing
that we have not discussed is, what are these
normalization constants? And then how does
it, I mean what; at least what are the ways
to look at it. How does it come in the picture?
In the transition amplitude they apparently
appear, but then it should be independent
of this normalization as such the transition
amplitude how does that come about.
Actually, we will see that we can relate this
transitional amplitude to the cross section.
And the cross section which actually tells
you directly; which can be directly related
to the probability for finding a particle
scattered into a particular region etcetera,
will be independent of all these normalizations.
And we will see how that comes out in the
next class.
