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PROFESSOR: Here is x.
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And here is a.
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Various copies of the x-axis.
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For the ground state, what
is the lowest energy state?
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It's not zero energy,
because n begins with 1.
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So it's this.
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The lowest energy
state is a sine.
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So the wave function
looks like this.
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This corresponds to sine.
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1.
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Or n equals 1.
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The next one corresponds to n
equals 2, and begins as a sine.
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And it just goes up like this.
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You add half a wave each time.
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Remember, we quantize
k a with n pi.
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So each time that you increase
n, you're adding pi to k a.
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So the phase, you see,
you have sine of kx.
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So you have sine that
goes from 0 up to k a.
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And k a is equal to n pi.
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So you go from 0 to pi, from 0
to 2 pi, then from 0 to 3 pi.
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So it would be one up,
one down, like that.
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And I could do one more.
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This one would be with four
cycles, two, three, four.
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So this is psi 1, psi
2, psi 3, and psi 4.
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Wave functions do
more and more things.
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So what can we learn
from this wave function?
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There are several things
that we need to understand.
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So one important thing is
that the wave function,
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these are all normalizable
wave functions.
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The ground state-- this
is the ground state--
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has no nodes.
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A node, in a wave function,
is called the point where
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the wave function vanishes.
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But it's not the endpoints
or the points at infinity,
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if you could have a range
that goes up to infinity.
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It's an interior
point that vanishes.
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And the ground
state has no nodes.
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So a node, node, so zero
of the wave function,
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not at the end of the domain.
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And of the domain.
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Because if we
included that, I would
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have to say that the ground
states has two nodes already,
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you'll see, around 0.
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But the 0 at the
end of the domain
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should not be counted as a node.
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Nodes are the zeros inside.
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And look.
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This has no nodes, and
it's a general fact
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about states of potentials.
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The next excited
state has one node.
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It's here.
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The next has two nodes, and
then the next is three nodes.
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So the number of
nodes of the wave
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function increases in potential.
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You have more and
more wave functions
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with higher and higher excited
states, and the number of nodes
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increases one by one
on each solution.
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That's actually
a theorem that is
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valid for general potentials
that have bound states.
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Bound states are states
that are normalizable.
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So the decay at infinity.
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You see, a state that
is not normalizable,
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like a plane, where it
is not a bound state.
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It exists all over.
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And it's a general
theorem that this phase,
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this one-dimensional
potentials, whenever
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you have bound states,
the number of nodes
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increases with the
energy of the eigenstate.
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We will see a lot
of evidence for this
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as we move along the course,
and a little bit of a proof.
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Not a very rigorous proof.
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The other thing
I want to comment
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on this thing that is
extremely important
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is the issue of symmetry.
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This potential for simplicity,
to write everything nicely,
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was written from 0 to a.
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So all the wave functions
are sine of n pi x over a.
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But in some ways, it perhaps
would have been better
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to put the 0 here.
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And you say, why?
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What difference does it make?
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Well, you have a 0 at the middle
of the interval, the potential
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and the domain of
the wave function
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are symmetric with respect
to x going to minus x.
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So actually, when you look at
the wave functions thinking
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you can rethink this as an
infinite box from a over 2
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to minus a over 2, and the
solutions, you just copy them,
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and you see now, this line
that I drew in the middle,
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the ground state is symmetric.
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The next state is anti-symmetric
with respect to the midpoint.
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The next state is now symmetric.
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And the following
one, anti-symmetric.
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So this is also a true fact.
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If you have bound states
of a symmetric potential--
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I will prove this one,
probably on Wednesday.
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A symmetric potential
is a potential for which
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V of minus x is V of x.
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Bound states of a
symmetric potential
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are either odd or even.
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This is not a completely
simple thing to prove.
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We will prove it, but you
need, in fact, another result.
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It will be in the homework.
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Not this week's homework,
but next week's homework.
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In fact, homework that is due
this week is due on Friday.
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So the bound states of
a symmetric potential,
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a potential that satisfies
this, are either odd or even.
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And that's exactly
what you see here.
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That's not a coincidence.
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It's a true fact.
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The number of nodes increase.
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And the other fact that is very
important, of bound states,
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of one-dimensional potentials--
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supremely important fact.
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No degeneracies.
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If you have a bound
state of a potential that
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is either localized like
this or goes to infinity,
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there are no degenerate
energy eigenstates.
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Each energy eigenstate here,
there was no degeneracy.
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Now, that is violated by
our particle in a circle.
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The particle in
the circle did have
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degenerate energy eigenstates.
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But as you will see, when you
have a particle in a circle,
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you cannot prove that theorem.
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This theorem is valid for
particles in infinitely--
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not in a circle.
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For x's that go from minus
infinity to infinity or x's
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with vanishing conditions
at some hard walls.
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In those cases, it's true.
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So, look, you're
seeing at this moment
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the beginning of very
important general results,
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of very fundamental
general results that
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allow you to understand the
structure of the wave function
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in general.
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We're illustrating it here, but
they are very much, truly now,
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general potential.
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So what are they?
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For one-dimensional potential
unbound states, no degeneracy,
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number of nodes
increasing one by one.
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If the potential is
symmetric, the wave functions
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are either even or odd.