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PROFESSOR: What
are we going to do?

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We're going to explore only
the first Born approximation.

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And the first Born
approximation corresponds

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to just taking this part.

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So this would be the
first Born approximation.

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It corresponds to what
we were doing here.

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What did we do here?

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Well, we're simplifying the
second term, the integral term,

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by using what the Green's
function looked like.

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We simplified this term.

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So all what we did
here was valuable,

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except that there's
one little difference.

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We-- in that Born
approximation, we

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replaced the psi that
appears inside the integral

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by the incident psi,
which is the psi in here.

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So we will do that
now to simplify

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this quantity in the so-called
first Born approximation.

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When can we use it?

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So, very good.

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So when is the
Born approximation

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a good approximation?

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Well, we are throwing
away terms, in general.

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When we're putting an expansion
of this form, we're saying, OK,

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we can set the wave function
equal to the free part

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plus the interacting part.

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So we have the free part,
and it gave us this quantity.

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And the interacting part
gave us the second one.

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So the Born
approximation is good

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when sort of the
free Hamiltonian

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dominates over the perturbation.

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So if a scattering center
has a finite-energy bump,

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and you're sending things
with very high energy,

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the Born approximation
should be very good.

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It's a high-energy
approximation,

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in which you are
basically saying

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that inside the
integral, you can

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replace the plane-incident
wave, because that dominates.

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That's not the whole solution.

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The whole solution then
becomes the plane integral wave

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plus the scattered wave.

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But the plane wave dominates
over the scattering process.

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So it should be
valid in high energy.

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It's better and better in
high-energy approximation.

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So we have it here.

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And let's, therefore,
clean it up.

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So if we call this equation
"equation A," we say,

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back to A. The first Born
approximation, back to A.

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The first Born approximation
gives us psi of r equals

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e to the i k i r.

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Now we put all the-- and
this is all the arrows.

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And we have here minus 1 over
4 pi, integral d cubed r prime,

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e to the minus i k n dot
r prime, u of r prime,

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and e to the i k i dot r prime.

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All this multiplied by
e to the i k r over r.

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So let's put a few vectors here.

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That's it.

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OK.

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So it's basically
this same thing here,

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but now replacing the
incident wave here.

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That's the so-called
first Born approximation.

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But now this is really good.

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We can compare this with what we
usually called f theta of phi.

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The expression of our brackets
is the scattering amplitude

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f theta phi.

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So here we have an answer.

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f at wave number k of theta and
phi is equal to this integral.

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Let's write it out.

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Minus 1 over 4 pi,
integral d cubed r prime.

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And now we will combine
the exponentials.

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Happily, the two exponentials
depend on r prime.

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So it's a difference
of exponents,

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and we will call it
e to the minus i,

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capital-K vector dot
r prime, u of r prime,

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where this capital-K
vector is equal--

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I kept the sign--

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k n minus-- this time we'll
enter with a minus in that k--

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k i or k s, the scattering k,
minus the incident i vector.

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Remember, we defined
there, on that blackboard,

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the scattered momentum
as k times the direction

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of observation,
that unit vector.

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So in combining these two
exponentials into a single one,

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we have this
capital-K vector that

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is a pretty important vector.

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And now, this is a nice formula.

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It kind of tells you story
that there are not many ways

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to generate things
that are interesting.

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Here it says that f k,
the scattering amplitude,

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as a function of theta and phi,
is nothing else than a Fourier

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transform of the potential
evaluated at what we would

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call the transfer momentum.

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So the scattering and bridges
are doing Fourier transforms

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of the potential.

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Pretty nice.

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Pretty pictorial way
of thinking about it.

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Fourier transforms
are functions of--

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I think when people
look at this formula,

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there's a little uneasiness,
because the angles don't

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show up on the right side.

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You have theta and
phi on the left,

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but I don't see a theta,
nor a phi, on the right.

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So I think that has to
be always clarified.

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So for that, if you want to
use, really, theta and phi,

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I think most people will assume
that k incident is indeed

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in the z-direction.

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z-direction.

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So here is k incident.

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And it has some length.

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k scattered has the same length.

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It's made by the same wave
number k without any index,

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but multiplied by
the unit vector n.

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As opposed to k incident,
that is the same k multiplied

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by the unit z vector.

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So the scattered vector is
the vector in the direction

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that you're looking at.

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So this is k s.

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It's over here.

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And therefore this vector is the
one that has the phi and theta

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directions.

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That is that vector.

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And the vector k is k
scattered minus k initial.

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So the vector k is
the transfer vector--

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is the vector that takes
you from k initial to k

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s, is the vector that
must be added to k initial

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to give you the
scattered vector.

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So this vector,
capital vector K--

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it's a little cluttered here.

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Let me put the z in here.

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That vector is over
there, and that vector

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is a complicated vector, not so
easy to express in terms of k i

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and k s, because it
has a component down.

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But it has an angle phi as well.

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But one thing you can say about
this vector is its magnitude

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is easily calculable, because
there is a triangle here

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that we drew that has
k incident and k s.

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And here is k.

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So this has length
k, this has length k,

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and this has length capital K.
The triangle with angle theta.

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So if you drop a
vertical line, you

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see that k is twice
this little piece, which

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is little k sine theta over 2.

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So that's one way that
formula on the right-hand side

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has the information of theta.

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It also has the
information of phi,

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because you also need phi
to determine the vector k.

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So this is an
approximation, but look

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how powerful approximations
are, in general, in physics.

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This approximation
is an approximation

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for the scattering amplitude.

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It-- first, it's a very nice
physical interpretation,

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in terms of a Fourier
transform of potential.

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Second, it gives
you answers even

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in the case where the potential
is not spherically symmetric.

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You remember, when potentials
were spherically symmetric,

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the scattering amplitude
didn't depend on phi,

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and we could use partial waves.

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And that's a nice way
of solving things.

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But here, at the expense
of not being exact,

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we have been able to calculate
a scattering amplitude

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when the potential is not
spherically symmetric.

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So you manage to go very
far with approximations.

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You don't get the
exact things, but you

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can go into results that
are a lot more powerful.

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So this is an explanation
of this formula.

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And we can use this formula.

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In fact, we'll do
a little example.

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And these are the
things that you still

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have to do a bit in
the homework as well.

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Many of them are in Griffiths.

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And indeed,
technically speaking,

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what you need to finish
for tomorrow hinges a bit

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in what I'm saying.

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But the final formulas are
well written in Griffiths

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explicitly, and in fact, half of
the problems are solved there.

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So it shouldn't be so difficult.