PROFESSOR: Scattering, so where did we get to last time?
We did the following things.
This is a recap.
We wrote the solution, a scattering solution,
or I'll use r.
And this solution had a wave that
represented the incoming wave, and it had what
we call the scattered wave.
So there were two pieces to the solution.
We understood that setting up a scattering problem
with a central potential.
We had a wave that came in.
And then we had the spherical wave,
and that was an outgoing spherical wave.
I've added the subscript k to represent the wave
number or the k of the wave.
This is an energy island state we're calculating.
And therefore, the energy is h bar squared k squared over 2m.
And we're writing the solution with that energy.
That's an energy island state.
And this solution is, as written, not an exact solution
of the Schrodinger equation.
But it's approximately exact for r much greater than a,
where a is the range of your potential.
So this is only valid in those cases.
It's not valid near the scattering.
For r near zero, that's not true.
And it's not an exact solution.
This would have been an exact solution
of the wave of the time independent Schrodinger
equation.
And this without the f would have been an exact solution.
But with the f is an approximate solution.
But that's a solution that represents
the physics of scattering.
Then we also showed that the differential cross section
was in fact given by this function f.
So this function f is really what we're after.
And if we know f, we know the differential cross section,
which is something we measure experimentally.
We're scattering particles, and we detect them.
And the differential cross section
tells us about our ability and the number of particles
that each detector picks up.
Then we've restricted ourselves to the case
of central potentials.
And for those central potentials,
v of r with some function of just the scalar distance.
In such cases, the cross section would not
have a fine dependence.
You can imagine here is the object.
It's vertically symmetric, and you're shooting waves.
And therefore, the cross section and the amplitude
will be independent of the angle phi.
There's already a direction picked up by the incoming wave.
But otherwise, it's just that.
So we wrote solutions in those cases.
If we are going to solve, as we will
today, some of these problems, you
need to write complete solutions.
And spherical solutions are of this form.
Although, the relevant ones that we
will be using will have no m.
We'll be focusing in solutions that have m equals zero.
For us, m will be equal to zero.
But these are the general solutions.
So maybe I might as well constrain ourselves
to our case, already central potential.
So we will have solutions of this kind.
And these are solutions of the radial equation.
And we reviewed those and mentioned that psi of r
would be given by Al Jl of kr plus Bl
Nl of kr times Yl0 of omega.
This will be our solutions.
These are the spherical Bessel functions.
And those were solutions that are valid
as long as you are away from the scattering center.
So this is valid for r greater than a.
Indeed, you know that this could not be valid all the way
to the center of the part of where this scattering is
happening, because ada is divergent as r goes to zero.
So this is not the solution.
This is just the solution for r greater than a,
where a is the range of the potential.
Imagine a potential that totally becomes zero
after some distance a, then that is your solution for r
greater than a.
That is the most general solution
given the spherical symmetry of this equation.
Then the last thing we discussed was
that one part of our solution, e to the ikz,
could be written in that way, because it's a solution.
So it must admit an expansion of this form.
So it is like this.
You have a sum over all else with some funny coefficients,
including i to the l, Yl0 of omega.
I might as well no put theta, because there's
no phi dependence in Yl0.
Jl of kr.
So this is a pretty remarkable expression,
we commented last time, that represents your plane
waves as spherical waves.
Last, but not least, we have an expansion
that is useful for a large argument
of the spherical Bessel functions.
They both fall off, like one over the argument,
with sines and cosines.
And you see a constant shift there in this sines and cosines
of l pi over two.
And this is for x much greater than one, not exact either,
but approximate.
OK, so these are some of our ingredients are ready.
That's how far we got.
And now, we're going to try to get more information
and learn how to solve for fk.
We need to calculate fk.
If you have a given problem, you want the current section,
you need fk.
Now the thing we're going to do is a little intricate,
a lot of funny formulas, so let's try
to keep the ideas very clear about it.
So--