We will look at the scattering cross section
between 2 particles like electron, or electron
and positron. or any charged particle, which
could be described using Dirac equation.
So, let us go back to our earlier discussion
on the scattering cross section, where we
had said that the transition rate per unit
volume can be written in terms of the transition
amplitude like; W f i; let me denote this
transition rate equal to T f i square over,
whatever is the time taken for this thing
per unit volume ok.
So, T f i as we know is the transition amplitude
for phi i to go to phi f, when it undergoes
an interaction; sorry, to phi f . When in
undergoes an interaction in time interval
T and in a volume spatial volume V ok. So,
usually we consider that there is no interaction
before this with outside this time interval.
One way to see it is that consider the Rutherford,
kind of scattering; the particle, which is
coming from in the beam coming towards this
one is free actually until it actually it
comes close to the wall; foil or whatever
the nucleus, that it will interact with. And
there is a short time interval where the interaction
happens, and then it flies off again asymptotically
as a free particle .
So, that is the kind of picture we have, now
we had in the earlier lectures described T
f i as 2 pi power 4 delta P A plus P C minus
P B minus P D .
So, let me first consider the T f i itself
is equal to this into whatever is the normalization
constant NA NB NC ND and m, invariant amplitude
. This is for wave functions like N e power
i P x . So, there is a particle; initial particle
with momentum say phi A with momentum P A
coming in after interaction goes out as phi
B with momentum P B and it is interacting
actually with another particle phi C P C,
which goes to phi D with momentum P D .
So, this case each of the normalization we
consider as N A, N B, NC, N D. Invariant amplitude,
we had written down earlier . In the case
of wave functions which obey the Klein Gordon
equation as well as the case of wave functions
which obey the Dirac equations .
What we want in the transition amplitude;
transition rate is T f i square probability
square amplitudes probability amplitude square
gives the probability. So, this gives me 2
pi power 4; and let us write the delta function
2 of them are there. So, let me write it first
us D P A plus P C the initial momentum P B
minus P D final momentum . And duplicate it
that is another P A plus P C minus P B minus
P D when is square it and you have N A N B
N C N D square; in variant amplitude square.
N A, NB; we will come to the normalization
later. First let us look at 2 pi power 4 delta
4 P A plus P C minus P B minus P D square
ok. This I can split into 2 pi into delta
E A plus E C minus E B minus E D into 2 pi
delta 3 3 momentum P A plus P C minus 
P B minus P D 2 pi power 3 .
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So, when I have two such delta 4’s, I have
1 term or 1 part 2 pi and delta E A plus E
C minus E B minus E D into the same thing
2 pi delta E A plus E C minus E P minus E
d. So, let me write . So, the first one as
it is; E A plus E C minus E B minus E D. Second
one in the integral form minus e power minus
i E A plus E C minus E B minus E D t integrate
over t d t. And let me take the interval as
minus T by 2 to T by 2, where essentially
the integration in interaction happens .
So, this now look at this the integral inside
this has an exponential E A plus E C minus
E B minus E D t. Owing to the other delta
function this can be said to equal to 0 . So,
essentially we have 2 pi delta E A plus E
C minus E B minus E D integral d t minus d
by 2 plus d by 2 .
So, that will give me 2 pi, one delta function
as it is and factor T whatever is the time
interval in a similar way we get 2 pi power
4. Not to pi power 4, we have taken the time
for energy part away only the 3 momentum now.
P A plus P C minus P B minus P D square of
it; let me denote it as the square of it;
equal to one of the delta functions as it
is P A minus sorry P A plus P C minus P B
minus P D . And then the conjugate variable
which is basically the space coordinate. And,
now we have a 3 dimensional thing therefore,
it is the spatial volume relevant to that
that is coming into picture, when we are integrating
over the other or when we consider the other
delta function as an in the integral form
and then integrate it over that this V .
So, T f i square is 2 pi now I can write it
as 2 pi 4 P A plus P C minus P B minus P D
in 4 momentum and it is delta 4. And you have
a T and V the time interval and the volume
involved in. And you have an N A N B N C N
D square and invariant mass square; invariant
amplitude squares. Now W f i, the rate is
T f i over T V per unit volume rate per unit
volume is now 2 pi the T V cancels with this
T V. So, what you have is 2 pi power 4 delta
P A plus P C minus P B minus P D, the 4 momenta,
delta 4; T V cancels with this; and you have
an N A N B N C N D square invariant amplitude
square.
Now, when we take phi to be 1 over root V
in the case of 3 dimensional box normalization,
we take this minus i P x . And that will give
you as per our earlier this thing rho is equal
to 2 E N square, if you remember or please
look back in the case of Klein Gordon equation
we can actually obtain this rho equal to 2
E N square as our density. Probability density
or if you multiply it by E, then it will be
charge density. So, number of particles essentially
that it will correspond to. So, this is 2
E over V for this . So, this actually says
that there are 2 E particles in V volume.
So, the density is 2 E over whatever the volume
V ok.
So, that is way one way to interpret this.
With all this; that put in W f i is equal
to 2 pi power 4 delta 4 P A plus P C minus
P B minus P D; one over V square, because
each of this. So, this says that N is equal
to one over root V .
So, we have not one over and this in to yes
1 over V square, which is N A N B N C N D.
And you have the invariant amplitude square
. We need; want to relate this to the number
of the cross section ok. So, the cross section
of this scattering process is equal to; or
cross section of the interaction is rate per
unit volume into number of final states ok;
per particle divided by; so, the cross section
into the cross section into initial flux into
target density ok .
So, the cross section times initial flux times
the target density is essentially going to
give you the W f i the rate into the number
of final states, whenever we just going into
available final states per particle . So,
this is what basically the cross section is
we had discussed this earlier.
Now number of final states is equal to, as
we said earlier the phase space volume V plus
d 3 p; when we consider a volume element between
P and P plus d P then it is V d 3 P over h
cube 2 pi h cross cube. and since we are taking
h cross to be equal to 1 this is 2 pi 3 for
h cross is equal to one in our units .
So, number of final set we said in volume
V, we have essentially 2 E number of particles
that, but that is what the normalization told
us. So, 2 E particles in V volume will give
you final state available for 1 particle is
equal to V d 3 P over 2 pi 3 2 E for 2 E it
is this much and then for one particle it
is one over 2 pi of this .
And, initial flux 
equal to number of particles per unit volume
into the velocity of a particle; right? Essentially
look at what is happening; there is a beam
which is going in . So, what is the initial
flux in a cross sectional area. Number density
and in a unit area, cross section. If you
take the number density for unit area of cross
section; the volume per unit time is going
to be this unit area into v t corresponding
to that. And when t is equal to 1 it is just
v.
So, it is cross section times weak the area
times v t is the volume. So, these times for
unit area it is equal to 1 into v into 1 . So,
that is the volume. So, 1 into v into 1 is
v essentially . Since we are considering initial
beam as the a particle beam and let me denote
it by v A.
So, this is the magnitude of the velocity
v A and if the other particle is a target
particle, which is at rest in the lab frame
target density is to be considered that is
going 2 E C other particles.
So, essentially what we have is initially
you have particles of type E A and E A and
C . So, same way you have V. Now um if we
consider colliding beams, then this v is actually
the relative v. So, there are 2 particles
then one is going with a velocity v A the
other is going with A velocity v C .
So, in the rest frame of C type of particle
one beam the other we will have velocity v
minus v A minus v C right . So, in this case
this is for fixed target . So, for colliding
beams we have initial flux equal to 2 E A
over V is the number of particles, but then
this is going to be v the magnitude of this
relative velocity v A minus v C. Target density
then gives 2 E C over V, because it is in
the rest frame of that particle that we are
considering the initial flux that is why the
velocity is the relative velocity .
So, that will give you the cross section equal
to W f i into V d 3 P B over 2 pi 3 2 E B,
V d 3 P D over 2 pi 3 2 E D. Initial flux
times target density is v A minus v C 2 E
A times 2 E C divided by V in each case. So,
we have a V power 4 W f i over V A minus V
C magnitude of that, 2 E A 2 E C and the number
of phase space d 3 P B over 2 pi 3 2 E B,
d 3 P D over 2 pi 3 2 E D .
