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PROFESSOR: Scattering states are
energy eigenstates that cannot

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be normalized.

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And when you say this
cannot be normalized,

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so what's the use of them?

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They don't represent particles.

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Well, it's like
they e to the ipx

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over h bar, those
infinite plane waves.

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Each one by itself
cannot be normalized,

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but you can conserve wave
packets that are normalized.

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So the whole intuition that
you get with scattering states

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is based on the idea that
we're going to construct

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energy eigenstates.

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This time we cannot think of
them as states of a particle.

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Bound states, yes.

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We can think of them,
they're normalizable.

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But this energy eigenstates
and bounded scattering states

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are not states of one particle.

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So we definitely have to go
back and produce wave packets.

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But the intuition from
those energy eigenstates

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is very valuable.

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So scattering states.

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And we call them sometimes
scattering states

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because they look like
the process of scattering.

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This will be non-normalizable
energy eigenstates.

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And you've played a
little with some of them.

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And we'll now study
one case in detail.

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We'll try a couple of cases
between today and next lecture.

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So the step potential.

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And the step potential
is a potential

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that is 0 up to x equals 0.

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Here's the x-axis, and then
suddenly there's a step at v0.

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And here is the potential.

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But then the wave,
this is here, goes up.

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It's a step.

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And any energy eigenstates
here has to be bigger.

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The energy has to be
bigger than the lowest

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point of the potential.

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You know that, you kind
of have an energy that

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is like that less
because this would

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have decayed exponentially
for infinite distance.

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It just, all over it would
have to decay exponentially.

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It's impossible.

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So all the energy
states, eigenstates here,

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must have positive energy.

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So we have actually
qualitatively

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two possibilities.

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The energy may be less than
v0, might be greater than v0.

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It would look like you have to
solve the problem two times.

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Happily, we'll solve one,
then let the other happen

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by analytic continuation.

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So here is the energy.

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I'll take the energy
greater than v0.

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But whatever is the energy,
even if it's less than v0,

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the solution over
here is going to be

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an exponential or a cosine and
a sine, a non-decaying function,

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and therefore can't be
normalized because it's

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non-decaying forever and ever.

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So it cannot be normalized.

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So how do we write the solution
for the energy eigenstate?

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It's a psi of x.

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Well, I should write two
formulas: a formula for what's

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happening on the left side,
and the formula for what's

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happening on the right side.

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Now I have a choice
actually here.

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There's two ways of
visualizing this.

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I can visualize it as a wave
that is coming from the left,

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moving here.

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Or a wave that is
coming from the right.

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So let's visualize this
solution as a wave that's

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coming from the left.

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It will be a little easier.

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So I will write it.

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A e to the ikx.

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OK.

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Why is it coming from the left?

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Because if you put the energy--

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that I will not put it,
it's in stationary state,

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presumably this is a state
with some fixed energy.

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You will have a factor e to
the minus iEt over h bar.

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And when you see
kx minus Et, you

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know that that's a wave
that is moving to the right.

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So this A e to the ikx
is moving to the right.

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And then what will happen?

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Now it's a matter of
finding a solution

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of Schrodinger's equation.

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So you can try to find the
solution of Schrodinger

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equation, but you have to
write some answers for what's

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happening on the right.

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I will write an answer
here, that we'll

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put C e to the i k bar x.

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And another k.

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Well, we'll see now
what those k's are.

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I say the following.

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Here, the energy is
bigger than the potential

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so it has to be a wave.

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But here they energy is still
bigger than the potential

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so it also must be a wave.

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But a wave with
different kinetic energy,

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different momentum, therefore
different de Broglie

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wavelength and different k.

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But we know from Schrodinger's
equation what that should be.

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This wave is also
moving to the right,

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because probably if I have a
wave moving to the right here,

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it produces some transmitted
wave to the right.

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But then, you could try solving
the Schrodinger equation

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with this.

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It won't be enough
because physically you

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would expect the wave bouncing
back as well from here.

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So I will put a B e
to that minus ikx.

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That's a wave moving
towards the left

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with an unknown coefficient.

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And, now let's get
those constants.

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I'll finish in two minutes.

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What is k?

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Well if you have energy E,
you know that the energy is h

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squared k squared over 2 m.

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You can look at the
Schrodinger equation

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with 0 potential over there.

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And therefore, k squared
is also 2mE over h squared.

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It's a combination you've
been seeing quite a bit.

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The intuition for
k bar should be

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that k bar squared is 2m
times the kinetic energy,

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so it should be e minus
v0 over h squared.

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So these are k and k bar.

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And the wave function must
be continuous at x equals 0.

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That gives you A plus B equal
to C. At 0, all the exponentials

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vanish.

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And the derivative must be
continuous at x equals 0.

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And the derivative being
continuous because there's

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no delta function anywhere here.

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So you have ikA, that's the
derivative of the first term,

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minus ikB, the same k in
that region of course,

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is equal to i k
bar C. So from this

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you get A minus B is
equal to k bar over kC.

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Two equations and two unknowns.

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And that's OK, even though
there are three coefficients,

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because the way to think
of this is that you're

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sending in some wave
and you're going

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to get some reflection
and some transmission.

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So in some sense,
A is the input.

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You could want to
call it 1 or whatever.

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So what we're looking
for is what is B over A?

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And what is C over A?

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And these two equations,
it's a one line computation.

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I'll write the answer.

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B over A is k minus k
bar over k plus k bar.

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Do it for fun.

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And C over A is 2k
over k plus k bar.

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B gives you a sense of
how much is reflected.

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C, how much is transmitted.

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But this is the beginning.

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Because this is not
a particle coming in.

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So we'll have to
build the packet

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and send it in and see
how this relations tell

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you what's going to happen.

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So this is a nice story that
we will develop next time.