PROFESSOR: Let's do a case that this mostly solvable
and illustrates all these things.
It's a very entertaining case.
It's called Landau-Zener transitions.
For these two people, Lev Landau, who you've
probably heard from Landau and Lifshitz.
He's the first person that tried to do this.
And Zener did it more carefully.
In fact, apparently found that Landau
made a factor of two error.
And the paper of Zener, it's actually quite nice,
and it's a very nice example that illustrates
the physics of this transition.
So we'll devote the rest of the lecture
to that Landau-Zener thing.
OK.
So it will give us a little bit into the spirit
of the adiabatic approximation in the language that Berry
used.
So Landau-Zener transitions.
OK, Zener and Landau were interested in molecules,
and some way of thinking of molecules
is to think of nuclei as fixed, separated by some distance R,
and then you assume they are fixed,
and they're separated by some distance R.
And then you calculate what is the electronic configuration.
So Zener imagined that you would have psi 1, one electronic
configuration.
It's a wave function that depends
on some x's for the electrons, but it represents the situation
where the two protons, say, for a simple molecule,
maybe they're more distances, but in particular, they
are separated by a distance, R. So that's
an electronic configuration.
Electronic configuration, protons a distance R away.
And suppose there's another configuration psi 2 of R.
It's another configuration, so two configurations.
Two different states.
Maybe in the first state, the electrons
are in some ground state.
In the second state, they're in some kind of excited state,
two different configurations.
Now, we could plot.
So we'll have [INAUDIBLE] here, distance R,
and here's the cloud of electrons.
We could plot a graph as a function
of the separation, what are the values of the energies.
And here is one possibility for the states.
And here's another one.
And that's the plot of the style that Zener drew in his paper.
And this represents E1 of R and this E2 of R.
That is the energy of the first state, the energy
of the second state as a function of R.
So we are having here--
oops-- two energy eigenstates.
So we have H of R. The Hamiltonian depends on the R.
And basically you're putting the two protons, the distance R,
and calculating the electrons, how they move.
Psi i of x R equals E i of R psi i of xR.
This is for i equal 1 and 2.
The case that the people were interested in
was the case where this molecule here, for example, in the state
2, for this value of R, there is a critical R 0,
where things, the levels get very close.
For some value of R, this molecule, for example,
could be a polar molecule.
A polar molecule is a permanent dipole moment.
It has plus charges and minus charges,
not evenly distributed.
So you get a dipole.
And maybe here, the molecule is non-polar.
And here, it's non-polar.
Here polar.
So if you would follow one of the energy eigenstates,
there's a critical value of R, where
the electronic configuration is such
that it goes from non-polar to polar
and in the other energy eigenstate,
it goes from polar to non-polar.
So the question is well, OK, what--
first of all, what does all this have
to do with instantaneous energy eigenstates and time
dependence?
Why are we thinking about this?
The issue is that sometimes, you can think of this molecules
as forming or being subjected to extra interactions in which you
will have a process or a reaction in which the radius
changes in time.
So it's possible under some configuration
that R becomes R of t.
And then, this Hamiltonian is a Hamiltonian
that depends on R of t.
This wave functions psi i are R of t E i's become R of t,
psi i's become x of R of t.
This is an important point.
It's simple, but important.
The most important points in physics are simple.
But you have to stop and recognize
that something slightly new is happening here.
If you have solved this equation for all values of R,
if you know those energy eigenstates
for all separations of the molecule,
you now have found instantaneous energy eigenstates
if it so happens that R is a function of t,
because if this is true for any value of this [INAUDIBLE] R,
well, then this is true for all times.
Because for any specific time, this is the R,
the same R is here, the same R is here,
and the same R is here.
And that equation holds for all R.
So if this can be solved for all R, this holds for all times.
And you have your instantaneous energy eigenstates.
You have found those instantaneous energy
eigenstates.
And therefore, the instantaneous energy eigenstate
are these ones.
And the instantaneous energies are this ones.
So many times in quantum mechanics, you do that.
You solve for the energy eigenstates for a whole range
of some parameters.
And then it so happens that those parameters
may change in time.
But then you have found the instantaneous energy
eigenstates for all times.
So in that picture, we have the following situation,
in which the energies now could be thought
if R is some function alpha of t,
then the same picture would basically hold true for time
here and the energies as a function of time,
because as time changes, R changes, and as R changes,
you already know how the figure looks.
So this is a figure of the energy
levels as a function of time.
And now the physical question is do we get a transition or not?
So the adiabatic theorem would say, OK,
you should state in your instantaneous energy
eigenstate, but we're going to get precisely to this situation
where these things could be so small, so little,
that there is a possibility of a non adiabatic transition,
in which you jump to the other one, because the gap is small.
So this goes to the real physics of the adiabatic theorem,
can we get an estimate or a calculation that tells us
how much probability you have of jumping the gap
and going to the other branch?
That's what we're going to try to do.
So for that, we'll do a particular example.
So let's do that.
It's an easy one to begin with.
I'll erase this.
So baby example, toy example.
So take a Hamiltonian, H of t, which
is going to be of this form, time dependent one,
but relatively simple.
Elements just along the diagonal.
OK, that's your Hamiltonian, two by two matrix,
elements on the diagonal, but just
simple things, the same thing.
So let's calculate the instantaneous energy
eigenstates.
OK, sounds like a task.
It's actually pretty simple.
The instantaneous energy eigenstates are 1, 0 and 0, 1.
They don't depend on time, because essentially
this Hamiltonian is just alpha t over 2, 1 minus 1, 0, 0 is--
it's a constant matrix times an overall factor.
The eigenstates of this matrix are 1, 0 and 0, 1.
And they are the eigenstates of this matrix for any time,
because the time goes in front.
The matrix doesn't quite change shape.
So these are the instantaneous energy eigenstates.
They are good forever.
To plot this, I will assume from now on that alpha is positive.
The energy of the first state is--
well, what do you get when you add with the Hamiltonian
on this state?
The matrix [INAUDIBLE] and this is just alpha t over 2.
And the energy of the second state
is going to be minus alpha t over 2.
We can plot those energies, and here
is the energy of the first state is alpha t over 2 with alpha
positive.
This is like this, thick output of here.
This is the state 1, 0 is here, 1, 0 is here,
alpha t over [? 6. ?] Here is the energy E1 of t.
The energy time dependent.
Here is time.
And here are energies.
This is E1.
And then we have the E2 is the other one that goes like this.
It's state 0, 1.
That's the state 2, 0, 1.
And the energy is E2 of t, which is minus alpha t over 2.
So it's negative for large positive time
and positive for the other one.
So these are your instantaneous energy eigenstates.
OK, and this is not quite what we wanted here.
We wanted things to avoid themselves.
But this is going to illustrate an important effect.
I claim, actually, that the true solutions of the Schrodinger
equation are in this case dressed up
versions of the instantaneous energy eigenstates.
So what I claim is the kind of-- you
do the adiabatic state corresponding to this,
the adiabatic state corresponding to that,
and those are exact solutions.
So there is no coupling between the states, 1 and 2.
So this is plausible.
So let's write those solutions.
I claim here is psi 1 of t I claim
is the exponential of minus i over h bar
integral up to t of E1 of t prime dt prime times
the state 1.
I claim this solves the Schrodinger equation.
i h bar dt of this psi should be equal to H psi.
Is it clear?
Yes, I think it's clear.
It solves it, because if you take
the time derivative of this thing, it multiplies by E1.
The i h bar cancels that factor of minus i over h bar.
The time derivatives brings out an E1 of t.
But this state, despite the phase when h [INAUDIBLE],,
it goes through the phase, hits the state 1,
and produces the E1 energy.
So this is solved by that equation,
and you can do the integral.
It looks OK.
It's exponential of minus i alpha t squared over 4 h bar 1,
and the state psi 2 of t is the same exponential with E2 with 2
here, and it's the exponential of plus i alpha
t squared over 4 h bar 2.
OK.
Let's appreciate the lesson again.
We got a very simple system, two levels, crossing--
they cross.
The energy levels cross.
That generally doesn't happen.
You have to have a very special Hamiltonian for the energy
levels to cross.
We found the instantaneous energy eigenstate,
and we found two exact solutions of this Schrodinger equation,
two perfect complete exact solutions of the Schrodinger
equation that represent the system doing just zoom,
like that, or doing like that.
Totally oblivious that there's a state they're crossing,
the Schrodinger equation doesn't couple them in this case.