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PROFESSOR: Scattering, so
where did we get to last time?

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We did the following things.

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This is a recap.

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We wrote the solution,
a scattering solution,

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or I'll use r.

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And this solution
had a wave that

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represented the incoming
wave, and it had what

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we call the scattered wave.

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So there were two
pieces to the solution.

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We understood that setting
up a scattering problem

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with a central potential.

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We had a wave that came in.

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And then we had
the spherical wave,

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and that was an
outgoing spherical wave.

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I've added the subscript
k to represent the wave

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number or the k of the wave.

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This is an energy island
state we're calculating.

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And therefore, the energy is h
bar squared k squared over 2m.

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And we're writing the
solution with that energy.

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That's an energy island state.

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And this solution is, as
written, not an exact solution

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of the Schrodinger equation.

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But it's approximately exact
for r much greater than a,

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where a is the range
of your potential.

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So this is only
valid in those cases.

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It's not valid near
the scattering.

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For r near zero,
that's not true.

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And it's not an exact solution.

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This would have been
an exact solution

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of the wave of the time
independent Schrodinger

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equation.

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And this without the f would
have been an exact solution.

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But with the f is an
approximate solution.

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But that's a solution
that represents

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the physics of scattering.

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Then we also showed that the
differential cross section

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was in fact given
by this function f.

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So this function f is
really what we're after.

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And if we know f, we know the
differential cross section,

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which is something we
measure experimentally.

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We're scattering particles,
and we detect them.

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And the differential
cross section

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tells us about our ability
and the number of particles

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that each detector picks up.

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Then we've restricted
ourselves to the case

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of central potentials.

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And for those
central potentials,

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v of r with some function
of just the scalar distance.

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In such cases, the
cross section would not

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have a fine dependence.

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You can imagine
here is the object.

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It's vertically symmetric,
and you're shooting waves.

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And therefore, the cross
section and the amplitude

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will be independent
of the angle phi.

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There's already a direction
picked up by the incoming wave.

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But otherwise, it's just that.

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So we wrote solutions
in those cases.

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If we are going to
solve, as we will

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today, some of
these problems, you

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need to write
complete solutions.

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And spherical solutions
are of this form.

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Although, the
relevant ones that we

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will be using will have no m.

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We'll be focusing in solutions
that have m equals zero.

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For us, m will be equal to zero.

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But these are the
general solutions.

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So maybe I might as
well constrain ourselves

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to our case, already
central potential.

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So we will have
solutions of this kind.

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And these are solutions
of the radial equation.

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And we reviewed those and
mentioned that psi of r

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would be given by
Al Jl of kr plus Bl

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Nl of kr times Yl0 of omega.

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This will be our solutions.

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These are the spherical
Bessel functions.

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And those were
solutions that are valid

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as long as you are away
from the scattering center.

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So this is valid for
r greater than a.

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Indeed, you know that this
could not be valid all the way

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to the center of the part
of where this scattering is

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happening, because ada is
divergent as r goes to zero.

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So this is not the solution.

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This is just the solution
for r greater than a,

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where a is the range
of the potential.

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Imagine a potential that
totally becomes zero

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after some distance a, then
that is your solution for r

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greater than a.

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That is the most
general solution

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given the spherical
symmetry of this equation.

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Then the last thing
we discussed was

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that one part of our
solution, e to the ikz,

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could be written in that
way, because it's a solution.

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So it must admit an
expansion of this form.

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So it is like this.

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You have a sum over all else
with some funny coefficients,

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including i to the
l, Yl0 of omega.

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I might as well no put
theta, because there's

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no phi dependence in Yl0.

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Jl of kr.

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So this is a pretty
remarkable expression,

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we commented last time,
that represents your plane

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waves as spherical waves.

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Last, but not least,
we have an expansion

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that is useful for
a large argument

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of the spherical
Bessel functions.

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They both fall off, like
one over the argument,

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with sines and cosines.

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And you see a constant shift
there in this sines and cosines

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of l pi over two.

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And this is for x much greater
than one, not exact either,

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but approximate.

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OK, so these are some of
our ingredients are ready.

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That's how far we got.

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And now, we're going to
try to get more information

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and learn how to solve for fk.

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We need to calculate fk.

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If you have a given problem,
you want the current section,

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you need fk.

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Now the thing we're going
to do is a little intricate,

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a lot of funny
formulas, so let's try

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to keep the ideas
very clear about it.

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So--