PROFESSOR: OK.
Let's do another application.
I leave one thing.
So this is going to be-- but it's called Landau levels.
And they're pretty interesting.
So it's the problem of solving for the motion of an electrode
in a magnetic field.
So it's called Landau levels for the physicist Lev Landau
from Russia that discovered or did this calculation first.
So you have a mass m charge q, and the magnetic field
b in the z direction.
We will solve it.
And I think you will find it pretty interesting.
Well, you will be left with a little of an uneasy feeling,
I think, at the end of the lecture,
because the problem of gauging variance
is so dramatic that the physics will look a little strange
and a bit unrecognizable.
So let me remind you that in classical physics,
if you have a magnetic field, b, you
can have electrons that perform circular orbits,
and the main property of those orbits
is that they all run at the same frequency
called cyclotron frequency, qb over mc.
So this is just Newton's law, and the force being v over c
cross b should be a couple of lines
for you to remember that particle in the magnetic field
goes in circles with that angular frequency.
So OK, there's lots of comments we could make about this,
but let's assume we're going to solve
for the motion of this particle.
We're going to ignore spin.
Sometimes spin is important.
And when you have spin, you have a Zeeman effect,
spin in a magnetic field.
So that can change the energy levels by some quantities.
Let's ignore it for the moment.
It is sometimes relevant.
Sometimes it's not relevant, and you can easily take care of it.
So again, we're faced with having to represent
the magnetic field.
So you have the magnetic field [INAUDIBLE] of a,
and we'll take a solution in which a this time is minus
by 0 and 0.
Remember, that b is dx ay minus dy ax.
And therefore this works out.
It has an ax component, the derivative with respect to y
gives you the magnetic field b.
So what happens to your Hamiltonian?
Your Hamiltonian is now 1 over 2m px minus q over c times ax
plus 1 over 2m py squared plus 1 over 2m pz squared.
I will not consider z motion.
It's too simple.
It's not too interesting.
It's just motion, plane waves in the z direction.
It's not interesting.
All the physics is really happening in the plane.
It's the idea that we can constrain
ourselves to have orbits of electrons
that go like this in circles.
OK.
So our Hamiltonian is here.
Let's write it once again, 2m px plus
q by over c squared plus 1 over 2m py squared.
In order to solve a system like that,
it's convenient to see what is conserved, and px is conserved.
px commutes with the Hamiltonian,
because the Hamiltonian has no x dependence.
So h commutes with px.
It doesn't commute with py, because there is a y in there.
And, OK, so if it commutes with px,
we can hope for eigenstates, energy eigenstates that
are also eigenstates of px.
So I will write my wave function of x and y
as a wave function that depends on y times e to the i kx x.
And it's already a little strange.
We're looking for circular orbits maybe,
and the x dependence is really a little funny here.
It's almost like plane waves in the x direction.
OK, well, it's a fact.
It's true.
You could not do that if you would have
chosen a more symmetric gauge.
You see, if you had chosen a more symmetric gauge,
in which a doesn't have just an x component, but a y component.
The y component would have depended on x,
and then x would-- px and py, neither one
would have commuted with the Hamiltonian.
So things would have been different.
But this gauge is easy to solve equations.
It's called a Landau gauge.
And let's just explore what this says.
Trust that if you're solving the equations,
and you get a legal solution, it must mean the right thing.
So here, in this solution here, px is just h bar kx,
and though we can restrict the action of the Hamiltonian
to that subspace of momentum, px equal h kx,
we can look at the Hamiltonian, it
is like looking at the Hamiltonian for this wave
function, we already constrain ourselves
to have this kind of momentum.
So the Hamiltonian that acts on the y function
is the Hamiltonian constrained to the momentum kx,
and it's equal to 1 over 2 m py squared, the second term,
plus 1 over 2 m qb over cy plus h kx.
It's the q by over c plus the value of px, which is h kx,
and I should square it.
If you look at it carefully, you see
that this is a simple harmonic oscillator, Hamiltonian.
There's py squared, and there's a y where the origin shifted.
It's not just a y squared, but it's
some sort of y minus y0 at some point squared.
So let's write it like that.
So I have to make it look like a harmonic oscillator, py
squared plus 1 over 2.
For a harmonic oscillator, I should have an m here.
So I'm going to have an m squared in the denominator,
so m.
Let's get the qb mc squared out.
y, I got that out, minus--
I have to put all these things.
Minus h bar kx over--
times the c over qb.
Look at all that.
So we factor out qb over c.
With the m squared here, and the m--
this is the 1 over m there.
y, and I put two minus signs, because I always
like to write y minus some y0, which is the equilibrium
position of the oscillator.
Here, as you pull the c and the qb out,
you get all these things.
So this is a p squared y over 2n, plus 1 over m qb
over mc squared y minus y0 squared with y0 equal minus h
bar k xc over qb.
That's it.
That's the system.
One good thing happened already.
While this thing is a little funny or strange, I think,
at first sight, this harmonic oscillator
is resonating with an omega, which
is the cyclotron frequency.
So that's nice.
The harmonic oscillator is that.
And therefore, already, you know you're
going to get levels that are going to be separated by h bar
omega cyclotron.
So that classical cyclotron frequency
is going to separate the levels, and these levels
are called the Landau levels, the various Landau levels.
OK, so let's see what this means, or how it looks like.
For that, let's imagine a solution.
So how does a solution look?
So here is x and y.
And suppose I take a kx, which is negative.
This kx is any number you wish at this moment.
So kx negative means y0 positive.
So here is the y0 is here.
And the wave function is certainly the probability
densities independent of x.
So the wave function sort of has support all over x here.
And it represents an oscillator in the y direction.
In the y direction, you're oscillating,
and well, you're going to oscillate a little.
We know, in particular, is a length scale associated
with the oscillator.
So let's look at that length scale.
The length scale on an oscillator h bar over m omega,
in general.
But now, we have an expression for omega.
So h bar over m, and omega is qb over mc.
So this is square root of hc over qb.
And we're going to call this the magnetic length.
So this is for an oscillator, an arbitrary oscillator
takes these values.
So let's call the magnetic length hc over qb.
This magnetic length.
So that's a nice definition.
And that's roughly the width of the state.
Imagine you're in the ground state,
you oscillate from the harmonic oscillator, the typical length
scale, and in this case, is the magnetic length.
So in the ground state, this size is lb.
And that's how your orbit looks.
In particular, y0 has a simple interpretation here.
y0 is equal to minus kx times the square
of this magnetic length is a [? right ?] squared.
So it's minus kx times lb squared.
And the units work out because k has units of 1 over length,
and you have a length squared.
So that's y0.
So that's how the orbits look.
Now, you would say OK, so where are the circular orbits?
Well, they're not quite so visible here.
You have to do some work to find them.
And in particular, what is happening here
is a very strange thing.
I think-- well, I'll say one more thing.
If you have this as your harmonic oscillator,
your energies, that may depend of kx and ny,
those are the quantum numbers you have already.
Well, this is just a harmonic oscillator.
So it's h bar, the cyclotron frequency.
Some people call the Landau frequency.
Occupation number for the oscillator times 1/2,
the usual formula for a harmonic oscillator.
And here is the energy.
And well, you had the plane wave.
Why don't we have the p squared over 2m of a plane wave?
It's nowhere to be found.
That doesn't show and doesn't contribute to the energy.
The mathematics was shown for us.
So this kind of dependence doesn't do anything.
It's the absolute degeneracy, the kx.
So you could construct the superposition of states
here, using Fourier transforms in which you somehow
localize this in the x direction,
by superimposing degenerate energy eigenstate.
So this is the most important thing about this Landau levels.
The Landau levels are the different levels of ny,
so ny equals 0 is the lowest Landau level.
Then you go ny equal 1, ny equal 2, ny equal 3.
But each Landau level is infinitely degenerate
because you can put different values of kx,
and the energy doesn't change.