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PROFESSOR: I'm going to
write this e to the ikz

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somewhat differently so that
you appreciate more what it is.

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So e to the ikz, I'll write it
as square root of 4 pi over k.

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You say, where does
that k come from?

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We'll see in a second.

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Sum over l, square root of
2l plus 1, i to the l, yl0,

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1 over 2i.

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Basically what I'm going
to do is I'm expanding

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this jl function for large x.

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So I'm going to
take this equation

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and I'm going to expand it so
that the argument is large,

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which means r is large.

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I want to describe for
you those waves that

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are really going on here.

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So since the argument is
kr and you have a kr here,

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that's the origin of the k
that I pulled out in here.

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And then you have e to the i
kr minus l pi over 2 over r,

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minus e to the minus i kr
minus l pi over 2 over r.

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And this is valid for
r much bigger than a.

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It's an approximate thing,
because we approximated

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the Bessel function.

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This is exact.

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This equation is
exact for all r.

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But that equation, now, is not.

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That equation is
showing you that you

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have here an outgoing wave
because is e to the i kr times

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e to the minus i et,
the energy terms time.

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So that's an outgoing wave.

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This is an ongoing wave.

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So this is ingoing
and this is outgoing.

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So I'm going to say
a couple of things

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about this that are going
to play some role later.

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It's a series of comments.

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Because this is a particular
expansion and due to the fact

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that this is a
general expansion,

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you could say that this
is an approximate solution

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of the Schrodinger
equation far away.

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But in fact, each
term, each value of l

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is an approximate solution.

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This is not an
approximate solution

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because we added over
many ls and somehow they

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helped to create an
approximate solution.

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This is an approximate
solution because each l

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is an approximate solution,
because to get a solution,

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you could have said
l is equal to 50,

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and that's all that
I'm going to use,

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and that's an exact solution.

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And therefore when
I look far away

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it will look like
that with l equal 50,

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and that would be an
approximate solution that can

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be extended to a full solution.

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So each l term here is
actually independent.

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If you tell me a wave
looks like that far away, I

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would say good,
yes, that's possible

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and that comes from a solution.

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You're getting an
approximate solution.

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And you could make it exact
by turning the exponentials

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into Bessel functions.

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Even more, we will show--

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we'll discuss it
a little later--

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I could get a solution that
is approximate by considering

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either this wave or that wave.

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Suppose we just have
the outgoing wave.

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The fact that the
Schrodinger equation

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works for this one
approximately doesn't

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require the ingoing wave.

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The Schrodinger equation doesn't
mix, really, the ingoing waves

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and the outgoing waves.

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It keeps them separate.

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So this is an
approximate solution

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and this is an approximate
solution as well.

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Every term here, independently,
every l and every ingoing

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and every outgoing wave is
a good approximate solution

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of the Schrodinger equation.

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Now this is important to
emphasize because this

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is called partial wave.

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So the term partial wave,
waves, refers to the various l's

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that work independently.

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Now to motivate what
we're going to do,

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I'm going to go back to
the case of one dimension

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so that you need the
map to orient yourself

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in the argument we're
going to do now.

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So in order to
understand partial waves

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and how they work, let's think
about the one-dimensional case

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for a minute.

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I used to do that in 804.

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I don't know if that has been
done by other instructors,

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but we've done in
804 that stuff.

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So let me tell you about it
and how it works in that case.

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So this is an aside
1D case of scattering.

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So in scattering
in one dimension,

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one usually puts a hard
wall at x equals 0.

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There is some potential
up to x equals a,

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and then there's nothing.

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It's a finite range potential.

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So even in the case of d equals
1, you look for a solution

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when there is no potential.

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This is the v of x.

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This is the x-axis.

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If v is equal to 0, you
would have a solution.

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Phi of x would be a
solution, energy eigenstate,

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and would be sine of kx.

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This is the analog of our e
to the ikz, that we call it,

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phi of r.

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Here you have a phi of x,
which is the sine of kx,

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and it's your energy eigenstate.

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That's the solution
if v is equal to 0.

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If v is equal to 0, the wave
must vanish at the hard wall,

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at x equals 0.

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So that's your solution.

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On the other hand--

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or, well, we can also write it
as 1 over 2i e to the ikx minus

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e to the minus ikx.

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And this wave is
the incoming wave,

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and this is the outgoing wave.

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Very analogous to
above, because this x

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is defined to be positive
and plays the role of radius.

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In this problem, incoming
means going down in x.

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Incoming in spherical dimensions
means going down in r.

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So we're doing something
very analogous.

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That's why the study of
scattering in one dimension

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is a good preparation for
the study of scattering

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in three dimensions.

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So you have an incoming
wave and an outgoing wave,

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also because, again, there's an
e to the minus iet over h bar.

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So here it is.

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What we try to do is to
write a scattering solution

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corresponding to
the same physics--

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so the scattering solution, psi
of x, is the full solution--

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is going to have the
same incoming wave

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because that's the physics that
we're trying to understand.

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We put the same incoming wave
and we try to see what happens.

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That's what do we do
when we do scattering.

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So if we want to compare
our solution to the case

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when v is equal to 0, we
put the same incoming wave.

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But then solving the
scattering problem

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means finding the outgoing wave,
which now may be different.

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Because, sadly, when
you have a potential,

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you have something in, that you
have control what you send in,

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but you have no
control what comes out.

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That's solving the
scattering problem.

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So the thing that
you have here must

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be an outgoing wave, e to
the ikx, with the same energy

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because energy is conserved.

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We're talking about an
energy eigenstate anyway,

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so it must have the
same energy, and it

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must have the same probability
flux as the incoming wave.

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So the magnitude of this wave
and the magnitude of this wave

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must be the same.

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It almost seems that you'd have
to write the same thing that I

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wrote here, but no.

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I can add one more
little factor.

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It's conventionally
written as 2i delta of k.

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A phase shift, delta of k,
an extra phase, and that's

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the claim.

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That's the whole thing.

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You have to solve
what's going on here.

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But outside-- and this is only
valid for x greater than a--

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outside, where the
potential is zero,

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you must have the
solution is of this form.

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Here, this is for 0
potential everywhere.

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So this was valid everywhere,
but in particular is also

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valid for x greater
than a, and here we

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compare these two things.

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So this is the general
solution for scattering here.

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And now we can also
add another definition.

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We can say that
psi of x is going

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to be equal to phi of x
plus the scattered wave.

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Our intuition is
that this wave here

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has the same incoming component
as the reference wave,

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but nevertheless, it has the
different outgoing component

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and that's the scattered wave.

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It is exactly analogous
to what we did here,

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in which we have the total
wave being a reference

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incoming wave.

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Well, actually, incoming
and outgoing wave.

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The e to the ikz has
incoming and outgoing wave

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and needs a solution when
the potential is equal to 0,

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just like this one, this phi.

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So psi is equal to 5
plus scattering wave.

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Here it is the scattering wave.

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It's purely outgoing, because
the incoming part is already

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taken care of.

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So completely analogous thing.

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So let's solve for the
scattered wave here.

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The scatter wave.

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So in this equation,
we have 1 over 2i,

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e to the ikx plus 2i delta
k, minus e to the minus ikx

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is equal to 1 over 21, e to the
ikx minus e to the minus ikx

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plus psi s of x.

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Well, these two terms cancel.

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It is the fact that they are
ingoing waves are the same.

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And then I can
subtract this other two

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to find psi s is
equal to 1 over 2i.

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I can factor the e to the ikx.

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That tells me it's
an outgoing wave.

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And then I have e to
the 2i delta k minus 1.

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That's a term there.

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I think I got everything there.

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It's written--
I'll write it here,

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there's a little bit of space--

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psi s as e to the ikx, e to
the i delta k, sine of delta k.

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00:16:07,420 --> 00:16:15,190
I factor out 1 times i
delta k, so this becomes

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e to the i delta k minus e to
the minus i delta k, the form

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00:16:20,740 --> 00:16:26,410
together with a 2i in
front, a sign of delta k.

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00:16:26,410 --> 00:16:32,920
So this is your
outgoing scattered wave.

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00:16:32,920 --> 00:16:37,030
That's the shape of the
outgoing scattered wave

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in terms of the phase shift.

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00:16:42,530 --> 00:16:48,610
OK, so now we have to redo
this in the slightly more

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00:16:48,610 --> 00:16:52,900
complicated case of three
dimensions and get it right.

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00:16:52,900 --> 00:16:54,250
So that is our task.

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00:16:54,250 --> 00:16:58,040
That's what we have to
figure out how to do.