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PROFESSOR: Levinson's theorem,
in terms of derivations,
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that we do in this
course, this is probably
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the most subtle derivation
of the semester.
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It's not difficult, but
it's kind of interesting
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and a little subtle.
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And it's curious, because
it relates to things that
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seem to be fairly unrelated.
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But the key thing
that one has to do
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is you have to use something.
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How all of the
sudden are you going
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to relate phase shifts
to bound states?
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The one thing you
have to imagine
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is that if you have
a potential and you
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have states of a potential,
if the potential changes,
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the states change.
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But here comes
something very nice.
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They never appear or disappear.
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And this is something probably
you haven't thought about this
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all that much.
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Because you had, for example,
a square, finite square well.
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If you made this more deep,
you've got more bound states.
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If you made it less deep,
the bound states disappear.
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What does it mean the
bound states disappear?
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Nothing, really can disappear.
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What really is happening,
if you have the bounds,
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the square well.
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You have a couple of
bound states, say.
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But then you have an infinity
of scatterings states.
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And as you make this
potential less shallow,
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the last bound state
is approaching here.
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And at one state,
it changes identity
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and becomes a scatterings
state, but it never gets lost.
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And how, when you make
this deeper and deeper
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you get more state, is
the scattering state
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suddenly borrowing, lending
you a state that goes down?
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The states never get
lost or disappear.
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And you will, say, how
could you demonstrate that?
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That sounds like
science fiction,
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because, well, there's
infinitely many states here.
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How do you know it borrowed one?
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Well, you can do it by putting
it in a very large box.
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And then the states
here are going
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to be finitely
countable and discrete.
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And then you can
track and see indeed
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how the states become bound.
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So you'd never lose
or gain states.
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And that's a very, very powerful
statement about quantum states
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in a system.
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So this is what
we're going to need
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to prove Levinson's theorem.
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So Levinson theorem theorem--
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so it relates a number of
bound states of the potential
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to the excursion
of the phase shift.
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So let's state it completely.
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It relates the number N of
bound states of the potential
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to the excursion, excursion of
the phase from E equals 0 to E
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equals infinity.
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So in other words, it says
that N is 1 over pi delta of 0
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minus delta of
infinity, a number
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of bound states
of your potential
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is predicted by the behavior of
the phase shift of scattering.
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So how do we do this?
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This is what we want to prove.
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So consider, again, our
potential of range R and 0
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here and here is x.
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And I want you to be
able to count states,
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but discovering states are
a continuous set of states.
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So in order to
count states, we're
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going to put a
wall here, as well,
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at some big distance L
much bigger than A than R.
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And we're going, therefore--
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now the states are
going to be quantized.
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They're never going be
quite scatterings states.
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They're going to look
like scatterings states.
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But they're precisely
in the way that they
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vanished at this point.
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Now you would say, OK, that's
already a little dangerous
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to be.
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Oh, you've changed the
problem a little, yes.
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But we're going
to do the analysis
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and see if the result depends
on L. If it doesn't depend on L
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and L is very large, we'll
take the limit this L goes
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to infinity.
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And we claim, we have answer.
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So we argue that L is
a regulator, regulator
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to avoid a continuum,
continuum of states
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to avoid that
continuum of states.
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All right, so let's
begin to count.
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To count this thing,
we start with the case
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where there is no
potential again.
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And why is that?
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Because we're going
to try to compare
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the situation with no
potential to the situation
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with potential.
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So imagine let V identical
is 0, no potential
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and consider positive
energy states.
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These are the only
states that exist.
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There are no bound states,
because the potential is 0.
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Well, the solutions
were found before, we
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mentioned that these are what we
call phi effects or sine of kx.
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But now we require that
phi of L is equal to 0,
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because we do have
the wall there.
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And therefore, si of kl must
be 0 and kl must be n pi and n
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is 1, 2 to infinity.
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You know we manage with the
wall to discretize this state,
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because the whole world is now
an infinite, a very big box,
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not infinite, but very big.
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So you've discrete the state.
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The separations are
microelectron volts,
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but they are discrete.
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You can count them.
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And then with this
state over here,
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we think of counting them.
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And you say, well, I
can count them with n.
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So if I imagine the
k line from 0 to 50,
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the other states are
over, all the values of k.
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And they could say,
well, I even want
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to see how many states there
are in a little element dk.
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And for the that you
would have that dk taken
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a differential here
is dn times pi.
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So of the number of states
that there are in dk, dn--
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let me right here--
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dn equals l over pi dk is
the number of positive energy
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states in dk.
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In the range dk, in
the range of momentum,
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dk that little interval, there
are that many positive energy
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eigenstates.
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So far, so good.
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So now let's consider
the real case.
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Repeat for the case
there is some potential.
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Well, you would say, well,
I don't know how to count.
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I have to solve the problem
of when the potential makes
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a difference.
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But no, you do
know how to count.
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So repeat for V
different from 0.
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This time we have a
solution for x greater
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than R. We know the solution.
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This is our universal
solution with the phase.
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So there you have the si
effect is e to i delta sine
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kx plus delta of k.
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That's a solution.
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Yes, you have the
solution always.
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You just don't
know what delta is.
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But you don't know what delta--
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you don't need to know what
delta is to prove the theorem,
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You just need to
know it's there.
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So here it goes.
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And this time the wall will
also do the same thing.
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We'll demand the si of x
vanishes for x equal L.
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So this time we get that kx--
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no kL plus delta of k is
equal to some other number n
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prime times pi multiple of pi.
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This phase-- this total phase
has to be a multiple of pi.
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And what is n prime?
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I don't know what is n prime?
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It is some integers.
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I don't know whether it starts
from 1, 2, 3 or from 100
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or whatever.
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The only thing we
care is that, again,
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taking a little
differential, you
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have dk times L
plus d delta, dk,
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times dk is equal to
the dn prime times pi.
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We take an infinitesimal version
of this equation, which again
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tells me how many positive--
all these states are
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positive energy states.
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They have k.
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So all these are
positive energy states.
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So from this equation, we get
that dn prime is equal to L
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over pi dk plus 1 over pi
d delta dk times dk, which
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gives me if I know the dk,
again, the number of states
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that you have in
that range of k,
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You see it's like
momentum is now quantized.
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So for any little
range of momentum,
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you can tell how many
states there are.
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And here it is how many states
there are, positive energy
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states, positive.
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This is the number of positive
energy states in dk with V
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different from 0, here
is with V equal 0.