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PROFESSOR: Let's do a case
that this mostly solvable

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and illustrates
all these things.

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It's a very entertaining case.

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It's called Landau-Zener
transitions.

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For these two people,
Lev Landau, who you've

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probably heard from
Landau and Lifshitz.

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He's the first person
that tried to do this.

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And Zener did it more carefully.

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In fact, apparently
found that Landau

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made a factor of two error.

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And the paper of Zener,
it's actually quite nice,

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and it's a very nice
example that illustrates

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the physics of this transition.

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So we'll devote the
rest of the lecture

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to that Landau-Zener thing.

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OK.

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So it will give us a
little bit into the spirit

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of the adiabatic approximation
in the language that Berry

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used.

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So Landau-Zener transitions.

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OK, Zener and Landau were
interested in molecules,

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and some way of
thinking of molecules

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is to think of nuclei as fixed,
separated by some distance R,

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and then you assume
they are fixed,

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and they're separated
by some distance R.

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And then you calculate what is
the electronic configuration.

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So Zener imagined that you
would have psi 1, one electronic

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configuration.

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It's a wave function
that depends

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on some x's for the electrons,
but it represents the situation

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where the two protons,
say, for a simple molecule,

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maybe they're more distances,
but in particular, they

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are separated by a
distance, R. So that's

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an electronic configuration.

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Electronic configuration,
protons a distance R away.

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And suppose there's another
configuration psi 2 of R.

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It's another configuration,
so two configurations.

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Two different states.

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Maybe in the first
state, the electrons

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are in some ground state.

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In the second state, they're
in some kind of excited state,

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two different configurations.

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Now, we could plot.

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So we'll have [INAUDIBLE]
here, distance R,

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and here's the
cloud of electrons.

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We could plot a
graph as a function

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of the separation, what are
the values of the energies.

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And here is one
possibility for the states.

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And here's another one.

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And that's the plot of the style
that Zener drew in his paper.

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And this represents E1
of R and this E2 of R.

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That is the energy of the
first state, the energy

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of the second state
as a function of R.

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So we are having here--

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oops-- two energy eigenstates.

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So we have H of R. The
Hamiltonian depends on the R.

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And basically you're putting
the two protons, the distance R,

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and calculating the
electrons, how they move.

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Psi i of x R equals
E i of R psi i of xR.

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This is for i equal 1 and 2.

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The case that the people
were interested in

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was the case where this molecule
here, for example, in the state

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2, for this value of R,
there is a critical R 0,

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where things, the
levels get very close.

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For some value of R, this
molecule, for example,

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could be a polar molecule.

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A polar molecule is a
permanent dipole moment.

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It has plus charges
and minus charges,

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not evenly distributed.

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So you get a dipole.

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And maybe here, the
molecule is non-polar.

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And here, it's non-polar.

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Here polar.

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So if you would follow one
of the energy eigenstates,

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there's a critical
value of R, where

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the electronic
configuration is such

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that it goes from
non-polar to polar

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and in the other
energy eigenstate,

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it goes from polar to non-polar.

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So the question is
well, OK, what--

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first of all, what
does all this have

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to do with instantaneous
energy eigenstates and time

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dependence?

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Why are we thinking about this?

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The issue is that sometimes,
you can think of this molecules

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as forming or being subjected to
extra interactions in which you

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will have a process or a
reaction in which the radius

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changes in time.

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So it's possible under
some configuration

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that R becomes R of t.

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And then, this Hamiltonian
is a Hamiltonian

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that depends on R of t.

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This wave functions psi i are
R of t E i's become R of t,

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psi i's become x of R of t.

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This is an important point.

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It's simple, but important.

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The most important points
in physics are simple.

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But you have to
stop and recognize

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that something slightly
new is happening here.

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If you have solved this
equation for all values of R,

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if you know those
energy eigenstates

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for all separations
of the molecule,

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you now have found
instantaneous energy eigenstates

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if it so happens that
R is a function of t,

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because if this is true for any
value of this [INAUDIBLE] R,

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well, then this is
true for all times.

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Because for any specific
time, this is the R,

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the same R is here,
the same R is here,

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and the same R is here.

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And that equation
holds for all R.

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So if this can be solved for
all R, this holds for all times.

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And you have your instantaneous
energy eigenstates.

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You have found those
instantaneous energy

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eigenstates.

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And therefore, the
instantaneous energy eigenstate

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are these ones.

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And the instantaneous
energies are this ones.

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So many times in quantum
mechanics, you do that.

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You solve for the energy
eigenstates for a whole range

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of some parameters.

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And then it so happens
that those parameters

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may change in time.

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But then you have found
the instantaneous energy

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eigenstates for all times.

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So in that picture, we have
the following situation,

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in which the energies
now could be thought

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if R is some
function alpha of t,

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then the same picture would
basically hold true for time

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here and the energies
as a function of time,

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because as time changes, R
changes, and as R changes,

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you already know how
the figure looks.

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So this is a figure
of the energy

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levels as a function of time.

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And now the physical question is
do we get a transition or not?

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So the adiabatic
theorem would say, OK,

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you should state in your
instantaneous energy

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eigenstate, but we're going to
get precisely to this situation

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where these things could
be so small, so little,

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that there is a possibility
of a non adiabatic transition,

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in which you jump to the other
one, because the gap is small.

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So this goes to the real physics
of the adiabatic theorem,

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can we get an estimate or
a calculation that tells us

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how much probability you
have of jumping the gap

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and going to the other branch?

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That's what we're
going to try to do.

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So for that, we'll do
a particular example.

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So let's do that.

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It's an easy one to begin with.

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I'll erase this.

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So baby example, toy example.

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So take a Hamiltonian,
H of t, which

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is going to be of this
form, time dependent one,

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but relatively simple.

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Elements just
along the diagonal.

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OK, that's your Hamiltonian,
two by two matrix,

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elements on the
diagonal, but just

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simple things, the same thing.

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So let's calculate the
instantaneous energy

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eigenstates.

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OK, sounds like a task.

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It's actually pretty simple.

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The instantaneous energy
eigenstates are 1, 0 and 0, 1.

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They don't depend on
time, because essentially

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this Hamiltonian is just alpha
t over 2, 1 minus 1, 0, 0 is--

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it's a constant matrix
times an overall factor.

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The eigenstates of this
matrix are 1, 0 and 0, 1.

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And they are the eigenstates
of this matrix for any time,

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because the time goes in front.

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The matrix doesn't
quite change shape.

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So these are the instantaneous
energy eigenstates.

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They are good forever.

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To plot this, I will assume from
now on that alpha is positive.

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The energy of the
first state is--

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well, what do you get when
you add with the Hamiltonian

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on this state?

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The matrix [INAUDIBLE] and
this is just alpha t over 2.

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And the energy of
the second state

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is going to be minus
alpha t over 2.

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We can plot those
energies, and here

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is the energy of the first state
is alpha t over 2 with alpha

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positive.

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This is like this,
thick output of here.

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This is the state 1, 0
is here, 1, 0 is here,

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alpha t over [? 6. ?] Here
is the energy E1 of t.

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The energy time dependent.

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Here is time.

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And here are energies.

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This is E1.

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And then we have the E2 is the
other one that goes like this.

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It's state 0, 1.

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That's the state 2, 0, 1.

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And the energy is E2 of t,
which is minus alpha t over 2.

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So it's negative for
large positive time

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and positive for the other one.

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So these are your instantaneous
energy eigenstates.

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OK, and this is not quite
what we wanted here.

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We wanted things to
avoid themselves.

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But this is going to
illustrate an important effect.

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I claim, actually, that the true
solutions of the Schrodinger

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equation are in
this case dressed up

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versions of the instantaneous
energy eigenstates.

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So what I claim is
the kind of-- you

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do the adiabatic state
corresponding to this,

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the adiabatic state
corresponding to that,

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and those are exact solutions.

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So there is no coupling
between the states, 1 and 2.

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So this is plausible.

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00:16:32,780 --> 00:16:36,990
So let's write those solutions.

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00:16:36,990 --> 00:16:42,600
I claim here is
psi 1 of t I claim

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is the exponential
of minus i over h bar

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00:16:47,670 --> 00:16:57,640
integral up to t of E1
of t prime dt prime times

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00:16:57,640 --> 00:16:58,675
the state 1.

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00:17:01,690 --> 00:17:04,210
I claim this solves the
Schrodinger equation.

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00:17:04,210 --> 00:17:13,150
i h bar dt of this psi
should be equal to H psi.

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00:17:13,150 --> 00:17:14,980
Is it clear?

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00:17:14,980 --> 00:17:16,930
Yes, I think it's clear.

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00:17:16,930 --> 00:17:18,700
It solves it,
because if you take

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00:17:18,700 --> 00:17:29,330
the time derivative of this
thing, it multiplies by E1.

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00:17:29,330 --> 00:17:34,330
The i h bar cancels that
factor of minus i over h bar.

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00:17:34,330 --> 00:17:38,860
The time derivatives
brings out an E1 of t.

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00:17:38,860 --> 00:17:44,380
But this state, despite the
phase when h [INAUDIBLE],,

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it goes through the
phase, hits the state 1,

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and produces the E1 energy.

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So this is solved
by that equation,

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and you can do the integral.

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00:17:58,300 --> 00:17:59,530
It looks OK.

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00:17:59,530 --> 00:18:09,700
It's exponential of minus i
alpha t squared over 4 h bar 1,

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00:18:09,700 --> 00:18:21,940
and the state psi 2 of t is the
same exponential with E2 with 2

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00:18:21,940 --> 00:18:30,620
here, and it's the
exponential of plus i alpha

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00:18:30,620 --> 00:18:34,700
t squared over 4 h bar 2.

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OK.

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Let's appreciate
the lesson again.

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We got a very simple system,
two levels, crossing--

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they cross.

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The energy levels cross.

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That generally doesn't happen.

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You have to have a very special
Hamiltonian for the energy

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00:18:58,010 --> 00:18:59,570
levels to cross.

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We found the instantaneous
energy eigenstate,

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and we found two exact solutions
of this Schrodinger equation,

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00:19:10,320 --> 00:19:14,190
two perfect complete exact
solutions of the Schrodinger

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00:19:14,190 --> 00:19:17,970
equation that represent
the system doing just zoom,

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00:19:17,970 --> 00:19:22,050
like that, or doing like that.

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00:19:22,050 --> 00:19:26,740
Totally oblivious that there's
a state they're crossing,

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00:19:26,740 --> 00:19:31,100
the Schrodinger equation doesn't
couple them in this case.