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PROFESSOR: So that's
part of this story.

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Let's try to understand
it a little better still.

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So what does the
state look like?

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Well the state looks like
e to the ikxx times a state

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of the oscillator and y.

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OK.

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So what is happening here?

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We could re-solve this problem
using a different gauge.

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And you will.

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You will solve it at
least once or twice.

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Again, this problem
using a different gauge.

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And the solutions are going to
be looking a little different

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but, of course, you're going
to find the same Landau

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levels and the same
infinite degeneracy.

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The wave functions will
look sometimes a little

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more intuitive.

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So in this case,
one calculation that

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is interesting to
try to understand

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the physics of this degeneracy
is to work roughly a little

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heuristically on a
finite size sample.

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Imagine a material but
it's now finite size.

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So let me remark to you what
are the degrees of freedom you

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have here.

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Suppose you solve this
Schrodinger equation

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in a different gauge, a
symmetric gauge in which there

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is an ay and an ax.

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Solutions then are going
to look a little more

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like circular orbits.

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There's a little
more mathematics

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involved in solving
it but they're

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going to look a little nicer.

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But how are they
related to this one?

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Well anyone with
a circular orbit

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must be related to this
solution by first forming

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a superposition of
those solutions,

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maybe localizing it or doing
something and then doing

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a gauge transformation.

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So in order to compare your
solutions in different gauges,

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you have to dig into gun, you
have an infinite degeneracy,

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and you have gauge
transformation.

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So to see what state
here corresponds

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to a particular
state here, it may

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be the gauge transformation
of a particular superposition

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in this side.

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So it's, in general,
not all that easy to do.

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OK.

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So let's take a count
state in a finite sample.

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So same picture, but now
the material is here.

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And we'll put Lx and Ly here.

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So finite size in the
Lx, finite size in Ly.

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So given our intuition
with quantization,

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this suggests that we impose
periodic boundary conditions

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in x and try to
quantize the kx here.

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In general, if you're
imposing thinking

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of very large boxes,
which is the case here,

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it doesn't matter
much whether you

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impose periodic or vanishing
boundary conditions or anything

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essentially at large number of
states it makes no difference.

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So we quantize in x.

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So we want e to the ikx
times x to be periodic

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under x goes to x plus Lx.

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I'm almost done with sine.

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So kxLx will have to be
equal to a multiple with Nx.

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Since we know that Y0 is
equal to minus kxlb squared,

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we should take Nx negative so
that you're within the sample.

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You must be in y
positive and therefore kx

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should be negative,
Nx should be negative.

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And now I have a way to count
because I can take Nx negative

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up to some value minus Nx bar.

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And when Nx grows,
kx grows and y grows.

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So I can take the
last Nx that I can

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use is the one in which the
orbit is still in the sample up

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to the value Y0.

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So this number is
really the degeneracy

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because this is how many
values of Nx I can have,

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from minus Nx up to 0,
are the number of values

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that are consistent with a
state still in this sample.

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So Y0 equal to Ly should
be equal to minus kx times

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lb squared.

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So it's minus 2 pi Nx bar
over Lx times lb squared.

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And this gives you Nx.

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We can solve for Nx there.

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Nx is LyLx over lb
squared over 1 over 2 pi.

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So Nx is the degeneracy.

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This is the degeneracy.

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And it's equal to the
area divided by lb

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squared, which is h bar c
over qB times 1 over 2 pi.

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So it's equal to area times b
divided by 2 pi h bar c over q.

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So we're back to
the kind of thing

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we were saying before in
which the degeneracy is

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equal to the flux divided
by the flux quantum

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that we figured
out earlier today.

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So this is how much states
you can put on the sample.

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So you're given a
magnetic field, a Tesla

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and you have some area.

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You find the fi,
you divide by fi0,

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and that's the number
of degenerate states

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of each Landau level.

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So in particular, given that
we have that number that fi0

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is equal to about 2 times
10 to the minus 7 Gauss then

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centimeters squared.

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If you have a sample
of 1 centimeter squared

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and you put one Gauss, the
value of the flux over fi0

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would be 1 Gauss centimeter
squared over 2 times 10

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to the minus 7 same units.

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So it's about 5 million states.

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That's just to give you an idea
of how big the numbers are.

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That's the degeneracy.

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So this is a classic problem.

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Very important in
condensed matter physics.

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Is a first step in trying
to understand quantum Hall

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effect on many things.

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And it's important to solve it
and think it in several ways.

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And I think you
will be doing that

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in homework and recitation.