WEBVTT
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BARTON ZWIEBACH:
The next thing I
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want to talk about
for a few minutes is
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about the node theorem.
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Theorem.
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And it's something
we've seen before.
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We've heard that if you have
a one-dimensional potential
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and you have bound states,
the ground state has no nodes.
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The first excited
state has 1 node.
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Second, 2, 3, 4.
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All I want to do is give
you a little intuition
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as to why this happens.
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So this will be an argument
that is not mathematically
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very rigorous, but
it's fairly intuitive
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and it captures the
physics of the problem.
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So it begins by making
two observations.
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So in the node theorem, if
you have psi 1, psi 2, psi 3,
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all energy--
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energy-- eigenstates of a
one-dimensional potential--
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bound states.
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Bound states.
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With energy E1 less than
E2, less than E3 and E4,
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psi n has n minus 1 nodes.
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Those are points where
the wave function
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vanishes inside the range of x.
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So for this square
well, you've proven this
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by calculating all the
energy eigenstates.
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The first state is
the ground state.
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It has no nodes.
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The next state is the
first excited state.
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It has one node.
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And you can write
all of them, and we
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saw that each one has one
more node than the next.
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Now I want to argue that in
an arbitrary potential that
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has bound states,
this is also true.
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So why would that be true
for an arbitrary potential?
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The argument we're going to
make is based on continuity.
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Suppose you have a
potential like this--
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V of x-- and I want to argue
that this potential will
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have bound states and will
have no node, 1 node, 2 nodes,
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3 nodes.
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How could I argue that?
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Well, I would do the following.
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Here is the argument.
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Identify the minimum here.
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Call this x equals 0.
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Oh, I want to say one more thing
and remind you of another fact
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that I'm going to use.
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So this is the first
thing, that the square well
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realizes this theorem,
and the second
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is that psi of x0 being
equal to psi prime at x0
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being equal to 0
is not possible.
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The wave function and its
derivative cannot vanish
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at the same point.
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Please see the notes about this.
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There is an explanation
in last lecture's notes.
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It is fact that for a second
order differential equation,
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psi and psi prime tell you
how to start the solution,
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and if both psi and psi
prime are equal to 0,
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the general solution of
the differential equation
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is always 0 everywhere.
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So this kind of thing doesn't
happen to a wave function--
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the point where it's 0
and the derivative is 0.
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That never happens.
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This happens-- 0 wave
function with the derivative.
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But this, no.
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Never happens.
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So those two facts.
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And now let's do the following.
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Let's invent a new potential.
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Not this potential,
but a new one
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that I'll mark the point minus
a here and the point a here
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and invent a new potential
that is infinite here,
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infinite there, and has
this part I'll write there.
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So this will be called
the screened potential.
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Screened potential.
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Va of x in which Va
of x is equal to V
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of x for x less than a, and it's
infinity for x greater than a.
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So that's a potential in
which you turn your potential
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into an infinite square
well whose bottom follows
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the potential.
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It's not flat.
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And now, we intuitively argue
that as I take a to infinity,
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the bound states of
the screened potentials
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become the bound states of
your original potential.
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Because when the screen is
very, very, very far away,
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up to infinity, you've got all
your potential, and by the time
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you have bound states
that are decaying,
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so the screen is not going
to do much at infinity.
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And anyway, you can move
it even further away.
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If you move it one light year
away or two light years away,
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shouldn't matter.
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So the idea is that
the bound states--
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bound states-- of Va of
x as a goes to infinity
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are the bound states of V of x.
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And moreover, as
you slowly increase
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the width of the screen,
the bound states evolve,
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but they evolve continuously.
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At no point a bound state
blows up and reappears
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or does something like that.
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It just goes continuously.
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These are physically
reasonable, but a mathematician
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would demand a
better explanation.
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But that's OK.
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We'll stick to this.
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So let's continue there.
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So here is the
idea, simply stated.
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If a is going to 0, if
the width of the screen
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is extremely narrow, you're
sitting at the bottom
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of the potential at x equals 0.
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And the screened
potential is basically
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a very, very narrow
thing, and here, there's
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the bottom of the potential.
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And for sufficiently small a--
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since you picked the bottom
of the potential there--
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it's basically flat.
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And then I can use the states
of the infinite square well
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potential.
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As a goes to 0, yes,
you have a ground state
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with no nodes, a first
excited state with one node,
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and all the states have
the right number of nodes
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because they are the states
of the infinite square well,
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however narrow it is.
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So the only thing we have to now
show is that if you have a wave
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function-- say, let's begin
with one with no nodes--
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as you increase the
width of the screen,
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you cannot get more nodes.
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It's impossible to change the
number of nodes continuously.
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So here it is.
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I'm going to do
a little diagram.
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So for example, let's
assume the screen
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is this big at this
moment, that you have
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some ground state like this.
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You've been growing
this, and then
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as the screen grows bigger,
you somehow have maybe a node.
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Could this have happened?
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As you increase this
screen, you get a node.
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Now I made it on this point.
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I didn't intend to do that, so
let me do it again somewhere.
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Do you get a node?
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Well, here was the
original screen,
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and here the derivative
psi prime is negative.
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Psi prime is negative.
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On the other hand, psi prime
here is already positive.
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So as you grew this screen,
this PSI prime that was here
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must have turned from
negative to positive,
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the way it looks here.
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But for that, there
must have been a point
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somewhere here when it was
horizontal if it's continuous.
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And therefore,
there must have been
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some point at which psi
and psi prime were both 0
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at the endpoint x equals a,
whatever the value of a was,
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because psi prime here is
positive, and here is negative.
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So at some point it
was 0, but since it's
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at the point where you have
the infinite square well,
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psi is also 0.
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And you would have both
psi and psi prime equal 0,
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which is impossible.
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So basically, you can't quite
flip this and produce a node
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because you would have to flip
here, and you can't do it.
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One could try to make a very
precise, rigorous argument,
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but if you have
another possibility
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that you might think, well, you
have this wave function maybe.
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And then suddenly it starts
doing this, and at some stage,
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it's going to try to do this.
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But before it does
that, at some point,
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it will have to be just
like this and cross,
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but at this point, psi
and psi prime would be 0.
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So you can intuitively
convince yourself
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that this thing doesn't
allow you to produce a node.
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So if you start with
whatever wave function
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that has no nodes, as
you increase the screen,
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you just can't produce a node.
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So the ground state of
the whole big potential
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will have no nodes.
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And if you start with the
first excited state that
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has one node, as you
increase the screen,
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you still keep one node.
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So the next state of
the full potential
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will have one node as well.
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And that way, you argue
that all your bound states
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of the complete
potential will just
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have the right number of
nodes, which is 0, 1, 2, 3, 4.
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And it all came, essentially,
from the infinite square
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well and continued.