WEBVTT
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That's a solution.
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It's an accomplishment
to have such a solution.
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If somebody gives you
a value of the energy,
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you can calculate what
is the phase shift,
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but we probably want
to do more with it.
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So you decide to plot
this on a computer.
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Again, there's lots of
variables going on here,
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so you would want to figure out
what are the right variables
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to plot this.
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And the right variables
suggest themselves.
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From k squared equal 2 me over
h squared, unit less constant
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are things like ka, k
prime a, and that's it.
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Well, so ka is a proxy
for the energies.
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OK, a squared is really 2me,
a squared over h bar squared.
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And so this we
could call anything.
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Well, let's call it u.
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On the other hand, k
prime squared then--
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if you have k prime a squared
that it's also unit free
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would be 2me a squared
over h squared plus 2mv0
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a squared over h squared.
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You probably recognize them.
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The first one is just u squared.
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I should call this
u squared, sorry.
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U squared, and this is
our friend z0 squared.
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It's that number that tells you
the main thing you always want
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to know about a square well.
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That ratio between the energy
v0 to the demand to the energy
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that you can build
with h bar m and a.
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So here we go.
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We have k prime a
given by this quantity,
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and therefore let me
manipulate this equation.
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Might as well do it.
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It probably easier to
consider just tan delta, which
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is the inverse of this.
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You would have 1
minus the inverse
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of this would be k prime a
over ka, put the a's always,
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so cot k prime a tan ka over
tan ka plus k prime aka cot k
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prime a.
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So in terms of
our variables, see
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k prime a is the
square root of this,
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so k prime a square root of
u squared plus z0 squared,
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and k prime a over ka,
you divide now by u.
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So it's square root of 1 plus
z0 squared over u squared.
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That's this quantity.
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So how big, how much space
do I need to write it?
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Probably, I should
write it here.
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1 minus square root of
1 plus z0 squared over u
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squared cot k prime a is the
square root of z0 squared
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plus u squared and tan of
k a, which is u over tan u
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plus square root
of 1 plus z0 over u
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squared cotangent of square root
of z0 squared plus u squared.
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OK, it's not terrible.
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That's tan delta.
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So if somebody gives
you a potential,
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you calculate what z0
is for this potential,
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you put z0 there, and you
plot as a function of u
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with Mathematica.
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And plotting as a
function of u is
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plotting as a function of ka.
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And that's perfectly
nice thing to do.
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And it can be done
with this expression.
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In this expression,
you can also see
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what goes on when u goes to 0.
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Not immediately, it takes
a little bit of thinking,
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but look at it.
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As u goes to 0,
well, these numbers
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are 1, that's perfectly OK.
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That seems to diverge, goes
like 1 over u, but u going to 0.
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This goes to 0.
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So the product goes to a number.
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So the whole-- the
numerator goes to a number,
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some finite number.
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On the other hand,
when u goes to 0,
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the denominator
will go to infinity,
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because while this term
goes to 0 the tan u,
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this number is finite.
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And here you have a 1/u.
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So the denominator
goes to infinity.
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And the numerator
remains finite.
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So as u goes to 0, tangent
of delta goes to zero.
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So you can choose delta
to be 0 for 0 energy.
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So as u goes to 0, you get
finite divided by infinity,
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and goes to zero.
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So tan delta goes to 0.
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And we can take delta of ka
equals 0, which is u to be 0.
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The phase shift
is 0 for 0 energy.
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Let me go here.
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So here is an example.
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z0 squared equal 3.4.
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That actually correspond
to 0.59pi for z0.
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z0 equal 0.59pi.
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You may wonder why
we do that, but let
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me tell you in a second.
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So here are a couple
of plots that occur.
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So here is u equals ka.
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And here's the phase
shift, delta of u.
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You have the tangent of
delta, but the phase shift
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can be calculated.
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And what you find
is that, yes, it
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starts at 0, as we mentioned.
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And then it starts
going down, but it
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stabilizes at minus pi,
which is a neat number.
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That's what the
phase shift does.
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The so-called scattering
amplitude, well you
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could say, when is this
scattering strongest?
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When you get an extra wave of
this propagating more strongly?
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So you must plot sine squared
delta and sine squared is
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highest for minus pi over 2.
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So this goes like this, up,
and decays as a function of u.
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Third thing, the delay, is 1/a.
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The delay is 1/a d delta
dk, as a function of u.
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So that, you can imagine,
that takes a bit of time,
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because you would have to
find the derivative of delta
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with respect to u, and do
all kinds of operations.
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Don't worry, you will have
a bit of exercises on this
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to do it yourselves.
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But here the delay turns
out to be negative.
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And this is unit-free.
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And here, comes to be
equal minus 4 for equals 0,
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and goes down to 0.
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So in this case, the
delay is negative.
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So the reflected
packet comes earlier
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than you would expected,
which is possible,
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because the reflected
packet is going slowly here.
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Finally, at this point, reaches
more kinetic energy, just--
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and then back.
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So that's the delay.
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And you can plot another thing.
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Actually it's kind of
interesting, is the quantity
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a, this coefficient here.
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That gives you an idea
of how big the wave
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function is in the well.
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How much does it
stick near the well?
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So it peaks to 1.
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And it actually goes
like this, and that's
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the behavior of this form.
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Basically, it does those things.
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So, so far so good.
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We got some information.
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And then you do a
little experiment,
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and try, for
example, z0 equals 5.
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And you have delta
as a function of u,
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and here is minus pi, minus 2pi.
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And actually, you find that it
just goes down, and approaches
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now minus 2pi.
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So actually, if you
increase this z0 a bit,
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it still goes to pi, a pi
excursion of the phase.
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But suddenly, at
some value, it jumps.
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And it now goes to 2pi.
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And if you do with a
larger value, at some point
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it goes to 3pi and 4pi.
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And it goes on like that.
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Well if z0 would have been
smaller, like half of this,
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the phase would go down
and would go back up,
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wouldn't go to minus pi.
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It does funny things.
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So what's really
happening is that there
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is a relation between
how much the phase moves,
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and how many bound states
this potential has.
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And you say, why in the world?
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This calculation had nothing
to do with bound states.
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Why would the phase shift
know about the bound states?
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Well actually, it does.
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And here is the thing.
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If you remember, you've
actually solved this problem
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in homework, the half
square well, in which you
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put an infinite wall here.
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And if you had the full square
well, from minus a to a,
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this problem has all
the old solutions
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of the full square well.
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All the old solutions exist.
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And if you remember
the plots that you
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would do in order
to find solutions,
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you have pi/2, pi, 3pi/2, 2pi.
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And here is the even solution.
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Here is the odd solution.
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I'll do it like that.
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Here is an even solution.
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Here is an odd solution.
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And I marked the odd
solutions, because we
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care about the odd ones, because
that's what this potential has.
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So z0 equals 0.59pi is
a little more than pi/2.
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So it corresponds
to one solution.
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So there is one bound
state for this z0.
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z0 equals 5 is about here.
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it's in between 3pi/2 and this.
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And there's two nodes,
two intersections.
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Therefore, two solutions
in the square well.
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And here we have that the phase
has an excursion of, not just
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pi for one, but 2pi.
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And if you did this
experiment for awhile,
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you would convince
yourself there's
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a magic relation between how
much the phase shift moves,
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and how many bound states
you have in this potential.
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This relation is called
Levinson's theorem.
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And that's what we're going
to prove in the last half
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an hour of this lecture.