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PROFESSOR: Let's then try
to understand intuitively

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some issues with phase shift.

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And that, I think,
is pretty valuable.

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One shouldn't rush
through these things.

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I'm going to erase
this formula already.

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It's a little messy.

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It's not like you're
going to use it ever.

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You could use that
formula if you were

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to compute higher phase shifts.

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But at least this homework,
you don't have that.

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So let's try to
understand a little

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what phase shifts do for you.

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Phase shifts are useful when the
first few phase shifts dominate

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the cross section.

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If you have to do this
sum for all values of l,

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it can get very difficult.
So if the answer really

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necessitates the tail of l
going to infinity, it's hard.

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So in general, we
find that phase shifts

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are useful when the first few
dominate the cross section.

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So phase shifts are useful if
the first few dominate sigma.

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And that will tend to happen
when k a is much less than 1.

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And this will
happen in that case.

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That will be our argumentation.

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We will try to explain why
when k a is less than 1,

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this will happen.

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So what is k a less than 1?

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Well, it could be
that a is very small.

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So one possibility
is short range,

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which means a small
compared to other scales,

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and maybe there is a problem.

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Or the other possibility
is that low energy.

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That is, k is small, because
energy goes like k squared.

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If k is small, it's low energy.

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So these are the
possibilities that

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would allow us to
have a situation where

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only the first few
phase shifts contribute

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to the cross section.

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So let's see why
that is the case

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and what is the
intuition for this.

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This happens for--

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I think I want to add
here this happens when--

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so the claim above happens
when k a is less than 1.

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Let's have a picture of this.

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Consider an incident
particle on a potential.

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So here is a potential.

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Here is r equals 0 and
this is the potential.

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Maybe some region-- here's
an incident particle

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coming in with some momentum.

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So these ideas are
semi-classical ideas on how

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I'm looking at this particle
coming with some impact

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parameter. b is this distance
to the center, the maximum--

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the closest approach of the
particle to the potential.

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That's called the
impact parameter.

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And it has momentum p.

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So b is the impact parameter
and p is the momentum.

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And then, what is the angular
momentum of this particle

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with respect to the origin?

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Is r cross p.

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And so the angular
momentum l is equal to bp.

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OK.

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So now I'm going to try to
make sense of this equation

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and get an estimate,
considering our situation.

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So what is our situation?

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In our situation, when we
consider a partial wave

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expansion, we're
considering waves

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with different values of the
orbital angular momentum l.

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So when we consider
a little l, we're

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really saying that the angular
momentum is proportional to hl,

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where l is the quantum number.

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Number.

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The number that we're
fixing here each time

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we talk about the partial wave.

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Moreover, if we're asking
about the momentum,

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we know what is the
typical momentum

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of these waves-- it's hk.

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So if I substitute those
two pieces of information

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into the relation of the
angular momentum here--

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so l is equal to hl and
b multiplies h bar k,

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we get that the impact parameter
can be visualized as l over k.

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It's a somewhat
classical intuition.

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But it gives you an idea.

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As you're increasing l, you're
increasing the impact parameter

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intuitively.

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A partial wave
scattering process with l

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reflects the input parameter.

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If l is equal to 0, you're
hitting the object right on.

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And now, you can say, all
right, if b is greater than a,

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where a is the range
of the potential--

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range-- there should be
little or no scattering.

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So if b is greater than a,
is little or no scattering.

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That's intuitively clear from
the range of the potential.

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If the input parameter
is much bigger.

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So this means l greater than
k a, there is no scattering.

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And this physically-- this
is classical intrusion

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suggests that the contribution
to scattering from l's

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bigger than k a is negligible.

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There's no scattering.

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So this is saying that sigma--

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how should I say it?

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The delta l-- I'll just say
delta l's are small or very

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small--

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small-- for l greater than k a.

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The delta l, the phase
shifts, enter here,

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as sine squared of
the phase shifts.

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So they must be very small,
because they just pretty much

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don't contribute.

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This is intuition we
were claiming up there.

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If k a is rather small, very few
phase shifts will contribute.

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Because as soon as your
l is bigger than k a,

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you get nothing.

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If k a is much less than 1,
only l equals 0 will contribute.

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If k a is 1 or 2, it's--

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I think I would write here, if--

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we don't have to be very
small, but around 1, very few

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will contribute.

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Or essentially, you could say
that all the l less than k

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contribute.

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They contribute.

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So even if k a, say, was 5, you
could hope that the l's up to 5

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will be a big contribution.

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And after that, they
start to fall off.

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OK.

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So that's one
intuition into the l's.

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Second piece of intuition
comes from the solutions

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that you have of
the real equation.

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So it's interesting, actually.

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Let's look at our
partial waves directly.

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Second intuition for this fact--

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think of your solution
Jl of kr, Yl of theta.

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This is a solution.

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Represents a partial wave.

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It's the one that doesn't
diverge at the origin.

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So that makes some sense.

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It represents a free
wave, but still is

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giving us an intuition of
how this partial wave is

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supposed to behave.

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So I would like to
have an intuition

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of how big is this function
and where it is big.

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So this is a solution
of the radial equation.

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A radial equation
with no potential.

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That doesn't mean zero
effective potential,

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because there's always
the centrifugal barrier.

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So this is the solution.

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Solution of the v equals
0 radial equation, which

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has a potential,
which is v effective

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is h squared, l times l
plus 1 over 2mr squared.

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That's the effective potential.

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And it's a solution with energy
h squared, k squared, over 2m.

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So here is r.

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Here is the effective potential.

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And how does it look?

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Well, it diverges
and then falls off.

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And you have a solution
with energy equal h squared,

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k squared, over 2m.

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It's a solution of that
problem with that potential.

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So here is your intuition.

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This solution encounters
a barrier here.

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So it must be exponentially
suppressed in this distance.

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It will be
exponentially suppressed

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as you go below
this turning point.

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Therefore, this partial wave
will be 0 for some radius

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smaller to a certain
distance here.

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And that's back
of our intuition.

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If you have some impact
parameter this large,

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you have no support
over distances smaller

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than your impact parameter.

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There's no wave there.

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The wave stays away.

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So what we want to show is
that this wave stays away

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from the center r equals 0,
again, by a quantity related

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to this thing.

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So what do we get?

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Well, we can solve
for the turning point.

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So h squared l, l plus 1
over 2mr star squared--

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I can call this r star--

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is equal to h squared,
k squared, over 2m.

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Lots of things cancel.

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2m's, h squared.

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So you get kr star
squared is roughly equal

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to l times l plus 1.

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And here, I can
approximate and I can

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say that kr star is roughly l.

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That's the leading
order square root.

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And therefore, the
wave function vanishes.

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This Jl of kr must be
quite approximately

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0 for r less than r
star equal l over k,

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which is the impact parameter.

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So we know because this must
be exponentially suppressed,

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that the wave function
must be very much 0 here.

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00:15:46,240 --> 00:15:50,590
And that the radial
wave function is this.

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So you have derived
that the Jl of kr

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is essentially 0 for r less than
this r star, which is about l

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over k, that we identified
after impact parameter.

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So our ideas are consistent.

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This partial wave
doesn't reach the origin.

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It stays a bit the way.

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00:16:15,910 --> 00:16:17,740
How much?

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00:16:17,740 --> 00:16:19,540
The intuition is
that the distance

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00:16:19,540 --> 00:16:21,730
approximate to the
input parameter,

200
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which is l divided by k.

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OK.

202
00:16:27,100 --> 00:16:30,800
So that's another
piece of intuition.

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00:16:33,540 --> 00:16:37,110
There's one more
classical intuition

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00:16:37,110 --> 00:16:41,960
that they think
it's interesting.

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00:16:41,960 --> 00:16:45,770
It's very surprising,
actually, this third one,

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this semi-classical intuition
into what these things are.

207
00:16:50,430 --> 00:16:54,410
But they're important because
the subject of phase shift

208
00:16:54,410 --> 00:16:56,030
seems very technical.

209
00:16:56,030 --> 00:17:00,350
But the phase shifts
have important meaning

210
00:17:00,350 --> 00:17:05,510
and those partial
waves are interesting.

211
00:17:05,510 --> 00:17:12,784
So let's do one more way of
thinking about these matters.

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00:17:18,210 --> 00:17:20,609
OK.

213
00:17:20,609 --> 00:17:22,754
So I'd say, recall
this equation.

214
00:17:27,560 --> 00:17:34,190
Sigma is given by the total
cross section-- three--

215
00:17:34,190 --> 00:17:38,050
the total cross section
is given by this sum.

216
00:17:38,050 --> 00:17:43,640
So sigma is the
sum of sigma l's.

217
00:17:43,640 --> 00:17:45,280
And that's a nice notation.

218
00:17:45,280 --> 00:17:49,420
This is the stigma associated
with each partial wave.

219
00:17:49,420 --> 00:17:53,050
No, it's a good thinking.

220
00:17:53,050 --> 00:18:01,720
And the sigma l is equal
to 4 pi over k squared, 2l

221
00:18:01,720 --> 00:18:07,660
plus 1, sine squared of delta l.

222
00:18:14,080 --> 00:18:20,110
Now, just from this formula,
when somebody gives you a sigma

223
00:18:20,110 --> 00:18:24,220
l, you can know whether
they're making sense

224
00:18:24,220 --> 00:18:28,750
or not, because there
is a test, at least.

225
00:18:28,750 --> 00:18:36,210
Sigma l is always less than or
equal to 4 pi over k squared

226
00:18:36,210 --> 00:18:38,880
times 2l plus 1.

227
00:18:43,070 --> 00:18:45,880
And this just comes
because sine squared

228
00:18:45,880 --> 00:18:49,400
is less than or equal to 1.

229
00:18:49,400 --> 00:18:51,100
So it's a good test.

230
00:18:51,100 --> 00:18:54,910
People say, oh, I found the
partial cross section given

231
00:18:54,910 --> 00:18:56,930
by this for that partial wave.

232
00:18:56,930 --> 00:18:59,680
Well, it better satisfy
that, otherwise, you've

233
00:18:59,680 --> 00:19:02,170
done a wrong calculation.

234
00:19:02,170 --> 00:19:06,850
Some people call this
partial wave unitarity.

235
00:19:06,850 --> 00:19:10,240
Unitarity has to do with
conservation of probability

236
00:19:10,240 --> 00:19:14,050
and cross sections
have that thing.

237
00:19:14,050 --> 00:19:17,070
It's a probability to
scatter in some direction,

238
00:19:17,070 --> 00:19:18,350
in some measure.

239
00:19:18,350 --> 00:19:21,580
So this is called
partial wave unitarity.

240
00:19:24,790 --> 00:19:27,415
Wave unitarity.

241
00:19:31,030 --> 00:19:35,080
It's interesting
that these ideas

242
00:19:35,080 --> 00:19:38,380
have a connection in
classical physics, again, that

243
00:19:38,380 --> 00:19:45,610
allow you to think of,
actually, physically about

244
00:19:45,610 --> 00:19:49,690
these processes in a way
that illuminates what's

245
00:19:49,690 --> 00:19:51,910
happening quantum mechanically.

246
00:19:51,910 --> 00:19:54,610
So here is how you
can think of this.

247
00:19:54,610 --> 00:19:55,600
We have, again--

248
00:19:55,600 --> 00:19:57,240
I'll do a nice picture.

249
00:19:57,240 --> 00:19:59,320
I'll try at least.

250
00:19:59,320 --> 00:20:02,740
A source here and
there's a potential.

251
00:20:07,350 --> 00:20:16,070
And now I'm going to come
at some impact parameter.

252
00:20:16,070 --> 00:20:20,570
But let me think
of it as coming--

253
00:20:20,570 --> 00:20:26,990
here is the cylinder
of impact parameters.

254
00:20:26,990 --> 00:20:29,720
If you come anywhere on the
surface of this cylinder,

255
00:20:29,720 --> 00:20:33,160
you'll have input parameter b--

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00:20:33,160 --> 00:20:34,010
b is here.

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00:20:40,330 --> 00:20:42,775
Now, let's fit in the cylinder.

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00:20:47,200 --> 00:20:48,555
I'll fit in the cylinder.

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00:20:54,280 --> 00:21:02,220
To make it of thickness delta b.

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00:21:02,220 --> 00:21:03,820
Thickness.

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00:21:03,820 --> 00:21:09,320
So the thickness is delta b.

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00:21:09,320 --> 00:21:13,850
And you now consider
all these particles

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00:21:13,850 --> 00:21:19,760
that are going to come
in here and scatter.

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00:21:19,760 --> 00:21:23,120
Now, in classical
physics, you know

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00:21:23,120 --> 00:21:27,380
it's easy or reasonable
to track things,

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00:21:27,380 --> 00:21:29,540
because everything
is deterministic.

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00:21:29,540 --> 00:21:37,145
So I can imagine very well
that this shell just goes here.

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00:21:42,120 --> 00:21:50,800
And it becomes a shell
here of particles that

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00:21:50,800 --> 00:21:55,330
are going off at some angles.

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00:21:55,330 --> 00:21:57,360
So, like a trumpet.

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00:21:57,360 --> 00:22:02,230
This comes like that
and spreads out here.

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00:22:02,230 --> 00:22:05,860
And all these particles
get scattered.

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00:22:05,860 --> 00:22:07,870
Deterministically,
that's reasonable.

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00:22:07,870 --> 00:22:09,927
You have a charge
here, for example,

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00:22:09,927 --> 00:22:11,260
and these are charged particles.

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00:22:11,260 --> 00:22:13,890
They repel and they get
scattered like that.

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00:22:17,090 --> 00:22:26,600
So you could say,
OK, this contribution

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00:22:26,600 --> 00:22:31,850
is going to go into a
particular angle, the omega.

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00:22:31,850 --> 00:22:35,480
So I can think of
this as contributing

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00:22:35,480 --> 00:22:40,310
to the differential cross
section at an angle omega,

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00:22:40,310 --> 00:22:45,620
or some contribution to the
total cross section that

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00:22:45,620 --> 00:22:48,470
would come from
integrating and adding up

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00:22:48,470 --> 00:22:54,590
all these contributions
as you vary this thing.

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00:22:54,590 --> 00:22:58,460
So if I want to compute
how much this contributes

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00:22:58,460 --> 00:23:00,680
to the cross section--

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00:23:00,680 --> 00:23:04,430
estimate the classical
contribution.

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00:23:04,430 --> 00:23:18,760
Estimate the classical
contribution d sigma to sigma

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00:23:18,760 --> 00:23:23,470
from a single partial wave--

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00:23:23,470 --> 00:23:34,010
single partial wave-- if
I consider a partial wave

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00:23:34,010 --> 00:23:36,740
with some value of l.

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00:23:36,740 --> 00:23:40,370
Think of a partial wave
with a fixed value of l.

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00:23:40,370 --> 00:23:46,340
Then, I have fixed
the impact parameter.

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00:23:46,340 --> 00:23:52,970
If I change l by one
unit, the impact parameter

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00:23:52,970 --> 00:23:54,600
changes a little bit.

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00:23:54,600 --> 00:24:01,676
Think of this as
the b of a given l--

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00:24:01,676 --> 00:24:04,490
b of a given l.

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00:24:04,490 --> 00:24:08,870
And this delta b or db--

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00:24:08,870 --> 00:24:13,310
I'll call it delta
b with capital.

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00:24:13,310 --> 00:24:15,500
Delta b.

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00:24:15,500 --> 00:24:27,970
As they change in b
for delta l equal 1.

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00:24:27,970 --> 00:24:33,190
You see, each partial wave
corresponds to a different l.

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00:24:33,190 --> 00:24:35,530
Classically, there's
no quantization.

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00:24:35,530 --> 00:24:39,640
So I must think of
an impact parameter.

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00:24:39,640 --> 00:24:44,950
And the partial wave corresponds
to the thickness of this thing

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00:24:44,950 --> 00:24:48,820
when the thickness will
correspond to a change of b,

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00:24:48,820 --> 00:24:51,220
because l has changed one.

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00:24:51,220 --> 00:24:54,580
So you can think of
concentric cylinders.

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00:24:54,580 --> 00:24:59,290
The l equals 0 partial wave,
the l equal 1 partial wave,

309
00:24:59,290 --> 00:25:00,850
all of those are cylinders.

310
00:25:00,850 --> 00:25:04,250
And they all contribute areas.

311
00:25:04,250 --> 00:25:07,360
This is the area that
is going to get--

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00:25:07,360 --> 00:25:09,310
all the particles
are going to scatter.

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00:25:09,310 --> 00:25:13,210
So this little thin
cross-sectional area

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00:25:13,210 --> 00:25:17,200
here is the contribution
to the differential cross

315
00:25:17,200 --> 00:25:26,030
section from this partial wave
in the classical approximation.

316
00:25:26,030 --> 00:25:27,490
So let's do this.

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00:25:27,490 --> 00:25:34,670
So b of l is equal to l over k.

318
00:25:34,670 --> 00:25:40,490
And delta b is the change
in b when l changes by 1.

319
00:25:40,490 --> 00:25:45,950
So delta b is 1 over k.

320
00:25:45,950 --> 00:25:48,980
So what is the
differential cross section?

321
00:25:48,980 --> 00:25:52,900
The classical-- classical--

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00:25:52,900 --> 00:25:56,090
it's this area.

323
00:25:56,090 --> 00:25:58,550
This is the area
of the beam that

324
00:25:58,550 --> 00:26:03,860
is going to be scattered
in that partial wave.

325
00:26:03,860 --> 00:26:08,670
So it's going to
be 2 pi b delta b.

326
00:26:08,670 --> 00:26:15,350
And this is 2 pi l
over k times 1 over k.

327
00:26:15,350 --> 00:26:18,420
And that's not so bad.

328
00:26:18,420 --> 00:26:32,800
This is-- I'll write it as 1
over 4, 4 pi over k squared--

329
00:26:32,800 --> 00:26:35,190
the k squared came
out right-- times 2l.

330
00:26:40,600 --> 00:26:44,250
I wrote it like that because--

331
00:26:44,250 --> 00:26:50,650
here it is-- what the
quantum theory tells you

332
00:26:50,650 --> 00:26:54,820
the partial wave contribution
to the cross section is

333
00:26:54,820 --> 00:26:57,610
4 pi over k squared 2l plus 1.

334
00:26:57,610 --> 00:27:00,920
That's pretty much it.

335
00:27:00,920 --> 00:27:06,430
And the classical estimate
gives you one fourth of that.

336
00:27:06,430 --> 00:27:10,330
So pretty nice that
everything works out.

337
00:27:10,330 --> 00:27:12,850
And that there is
a correspondence.

338
00:27:12,850 --> 00:27:19,780
And there is a way to
estimate partial contributions

339
00:27:19,780 --> 00:27:23,470
to the cross section
from classical arguments.

340
00:27:26,230 --> 00:27:31,620
So that's our discussion of
the physics of phase shifts

341
00:27:31,620 --> 00:27:34,200
and the physics of
input parameters

342
00:27:34,200 --> 00:27:38,530
and l and your intuition
that you must have to them.