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PROFESSOR: Today, we
continue with scattering.

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And we begin by reviewing
what were the main ideas that

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have already been explored.

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And here they are.

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I've summarized the main
results on the blackboard.

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And we begin with an
expansion of our solutions

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for the case of
central potentials.

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This psi effect
represents a solution away

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from the scattering center.

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The scattering
center is this region

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where there is a potential
that the particles fill.

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And this solution that
we've written on top

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is a solution written
for spherically symmetric

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potentials in the case
that the potential is 0.

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That is, that solution is
valid away from the potential.

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We constructed it with the
free particle solutions--

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the jl and the nl of kr--

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of the spherical Schrodinger
or radial Schrodinger equation.

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In particular, we
identified these objects

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called phase shifts
that we discussed

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in the process of scattering
with spherical waves.

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And the result is that the
way we introduce the phase

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shifts, the tangent
of the phase shift

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is determined by the ratio
of these two coefficients, Bl

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over Al.

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So there's a phase
shift for every value

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of l beginning from l equals 0.

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Then we have that given that
relation, the solution above,

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when you use asymptotic
expansions for jl and nl,

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takes that form with a sine
of kr minus l pi over 2

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plus a phase shift, which
is another way where

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you can read the phase
shift of your solution.

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We're going to focus today for
a little while in understanding

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the physics of the
phase shifts and how

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you find the phase shifts.

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But supposing you
find the phase shifts

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and that was the derivation that
was very non-trivial that we

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did last time, we find what
this scattering amplitude

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is in terms of the phase shift.

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So this was our main hard work
establishing that formula.

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Once you have that formula,
that f of k in preliminary work

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was shown to represent the
scattering solution far

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away from the source.

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Far away from the source
we have the incoming wave

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and the outgoing spherical
wave modulated by fk of theta.

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Now, in general, you can
have an fk of theta and phi,

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but then the phase shift method
is not quite suitable for that.

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For that, we will discuss
some other methods today

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that can allow us to do that.

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So once you have this f of k,
you know the wave far away.

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You know how it contributes
to the cross-section,

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and you know how
the phase shifts.

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Each phase shift contributes
to a total cross-section,

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which is given here.

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So these were our main results.

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So let's leave this
results here and discuss

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your general computation
of a phase shift.

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You're given a problem, maybe
like the problems in the p set,

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and you need to find
the phase shift.

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So where do you begin?

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There's nothing in
there that seems

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to say, OK, this
is the way you're

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going to find the phase shit.

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So let's do this general
computation of the phase shift.

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Computation of the phase shifts.

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So I think you all know--

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and there's clear intuition--
that if you want to figure out

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the phase shift,
at some point you

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have to look at this
region, r less than a,

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where the potential is.

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Because after all, the phase
shift depends on the potential.

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So that's unavoidable.

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You have to get dirty with
that potential and solve it.

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So what is the equation
that you have to solve?

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It's the radial
Schrodinger equation.

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So you have to solve--

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you set energy in the
Schrodinger equation equal h

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squared k squared
over 2m because we're

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using k from the
beginning to represent

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the energy of the state.

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So energy is going to be that.

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And the radial
Schrodinger equation

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is minus h squared
over 2m, d second dr

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squared plus the potential.

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Finally, the potential shows up.

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You have to solve that.

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Plus the potential centrifugal
barrier, h squared l times l

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plus 1 over 2m r squared.

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All that acting on
a solution u sub l--

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I will put the k as well of r--

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equal the energy,
which is h squared

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k squared over 2m ulk of r.

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So this step cannot be avoided.

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You have to solve it.

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You have to solve this equation.

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So you solve this
for r less than a.

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This for r less than a.

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You have some hints
of what you're

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going to encounter because
you know from solving

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radial equations that ul
behaves like r to the l plus 1

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as r goes to 0.

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That's a boundary condition
for the l-th wave.

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And well, with that, you're
supposed to solve for this ul.

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OK.

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So you solve for this ul.

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And so what?

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Where are the phase shifts?

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You're done with it.

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And you need to know where
are the phase shifts.

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Well, remember our
notation is that we have

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a radial function, l of k of r.

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In the radial equation you
know that the radial part

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of the solution of the
Schrodinger equation is--

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the full solution
is a radial part.

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So ylm.

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In this case, yl0.

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But this radial thing
is ulk of r over r.

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That is presumably something
you still remember,

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that the radial
solution was u over r.

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OK.

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So you've solved it.

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You've got this quantity.

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And you say, all right,
what do I do next?

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Well, what you
have to do next is

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to match to the solutions
that lie outside.

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There is these solutions that
hold for r greater than a,

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where the potential is 0.

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So those have to be
matched to this solution.

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So let me draw a little diagram.

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We have diagram for r.

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And here is a.

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In this region, the
solution is Rlk of r.

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That's the solution.

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And this solution here, this
is for the l-th partial wave.

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The solution outside is
the l-th partial wave

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that I've written there.

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So the solution here is some
Al Jl of kr plus Bl nl of kr.

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So this one you've
determined already.

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You did solve the
Schrodinger equation.

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We cannot help you.

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Now it depends on what
potential you have.

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You have to solve it each time.

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But you now have the
solution up to A.

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But you know the general
form of the solution

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for r greater than A, and
these two have to match here.

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There might be
cases in which there

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is a delta function
precisely here,

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like in the p set, in
which case you know

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that with a delta function,
the derivatives have

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to match in a funny way.

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There's a discontinuity
in the derivatives,

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and the wave functions
do have to match.

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So let's assume, in
general, that there's

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no delta function there.

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And let's do that case.

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If you have a delta function
you could do a similar case.

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And the way to do
this is actually

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to match the wave functions
and the derivatives.

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So matching at r equal a.

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So you must have Rlk of a is
equal to Al Jl of ka plus Bl

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nl of ka.

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The derivatives must also match.

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Let me use primes to
indicate derivatives

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with respect to the argument.

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Most people use that for primes.

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So the derivative of this
quantity Rlk a prime--

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so that means the
derivative of this function

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of r evaluated at a.

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And you differentiate
with respect to a in order

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to get the units not to change.

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You can multiply by an a.

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You have to multiply then by
an a on the right-hand side

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when I take the derivative.

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Now, when I take the derivative
here with respect to r,

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I take the derivative of J
with respect to the argument,

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and then the relative of the
argument with respect to r.

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That takes out the k out.

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Since I put an a, I get a
ka times Al J prime l at ka.

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The derivative of J with
respect to the argument

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evaluated at ka plus
Bl n prime l at ka.

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So here it is.

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I've matched the function
and the derivatives.

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So I'm giving you
an algorithm to do

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it any time and all times.

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At this point, the
right thing to do

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is to divide the
equations because that

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gets rid of a lot of things
that we are not interested in.

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We're not interested in
anything except the phase shift.

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That's all we want,
the phase shift.

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So how are we going to
get the phase shift?

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Well, the phase shift
comes from Bl over Al.

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So let's form the ratio here.

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So I'll form the ratio aRlk
prime of a over Rlk at a.

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And then I have ka.

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That's still there.

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And then let me divide--

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I'm going to divide
these two things,

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but then I can divide the
numerator and the denominator

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by Al.

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So here I'll have the ratio
of Jl prime ka plus Bl

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over Al n prime l of ka over--

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my equation came out
a little unbalanced,

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but it's not so bad.

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ka.

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Let's lower this.

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OK.

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Numerator.

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And here you get Jl of ka
plus Bl over Al nl of ka.

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But this is nothing but ka
and the ratio Jl prime at ka

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minus tan delta k-- no, delta l.

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That's what the
ratio Bl over Al is.

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n prime l of ka over Jl of
ka minus tan delta l eta.

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Not eta.

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It's n, actually. n of ka.

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That's it.

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Actually, we've
solved the problem.

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Why?

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Because this number is supposed
to be known because you've

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solved the equation.

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You did solve the equation,
the Schrodinger equation,

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from r equals 0 to r equal a.

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It's completely solved.

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So that number is known.

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The Bessel functions,
spherical Bessels, are known.

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And all you need to
find is tangent delta,

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and that equation determines it.

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Should I write the
formula for tangent delta?

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OK.

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Let's write.

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Let's write it.

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Yeah.

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OK.

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One minute to write it.

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It's just solving from here.

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Tan delta of l is
equal J prime l

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at ka minus Rl prime at a
over kRl of k of a Jl of ka

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over nl prime at ka minus
Rl prime of a over kRl of a,

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that same constant nl of ka.

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So that's solving
for tangent delta

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from this side
equal to this side.

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This shows that, in
principle, once you

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have the solution
from 0 to a, you

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have solved the problem
of the phase shift.

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You know the phase shift.

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It's a matter of just
calculating them.

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Might be a little messy.

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But if your calculation
is important

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and this is a problem
you really want to solve,

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you will be willing to look
at the spherical Bessel

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00:17:37,050 --> 00:17:42,020
functions, which, in fact,
are trigonometric functions

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00:17:42,020 --> 00:17:43,020
times powers.

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00:17:43,020 --> 00:17:49,080
They're not as difficult as
the ordinary Bessel functions.

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00:17:49,080 --> 00:17:52,910
So these are easy to work with.

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00:17:52,910 --> 00:17:55,650
And these things are, many
times, done numerically.

255
00:17:55,650 --> 00:17:59,700
For a given potential, you
calculate the first 10 phase

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00:17:59,700 --> 00:18:03,730
shifts, and you get a solution
that is very accurate.

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00:18:03,730 --> 00:18:10,740
So there's a lot of power
in this formulation.

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00:18:10,740 --> 00:18:12,180
So you have a solution.

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00:18:12,180 --> 00:18:14,770
You have the phase shifts.