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PROFESSOR: In order to learn
more about this subject,

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we must do the wave packets.

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So this is the place
where you really

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connect this need solution
of Schrodinger's equation,

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the energy eigenstates,
to a physical problem.

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So we'll do our wave packets.

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So we've been dealing
with packets for a while,

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so I think it's not going to
be that difficult. We've also

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been talking about
stationary phase

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and you've practiced that,
so you have the math ready.

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We should not have
a great difficulty.

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So let's new wave packets with--

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I'm going to use A
equals 1 in the solution.

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Now, I've erased every
solution, and we'll

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work with E greater than
v naught to begin with.

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The reason I want to work
with E greater than v naught

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is because there is
a transmitted wave.

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So that's kind of nice.

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So what am I going to do?

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I'm going to write it
this following way.

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So here is a solution
with A equals to 1.

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e to the i kx plus k
minus k bar over k plus k

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bar e to the minus i kx.

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And this was for x less than 0.

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And indeed, when there
was an A, there was a B,

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if A is equal to 1, remember we
solve for the ratio of B to A

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and it was this number
so I put it there.

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I'm just writing it
a little differently,

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but this is a solution.

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And for x greater
than 0, the solution

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is C, which was 2k over k
plus k bar e to the i k bar x.

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Now we have to
superimpose things.

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But I will do it very slowly.

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First, is this a solution of
the full Schrodinger equation?

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No.

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Is this a solution a
time-independent Schrodinger

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equation?

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What do I need to
make it the solution

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of the full
Schrodinger equation?

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I need e to the
minus i Et over h.

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And I need it here as well.

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And this is a psi
now of x and t.

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There's two options for x,
less than 0 and x less than 0,

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and those are solutions.

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So far so good.

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Now I'm going to
multiply each solution.

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So this is a solution of the
full Schrodinger equation,

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not just the time-independent
one, all of it.

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It has two expressions
because there's

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a little discontinuity in
the middle, but as a whole,

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it is a solution.

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That is the solution.

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Some mathematicians would
put a theta function here,

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theta of minus x and add
this with a theta of x so

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that this one exists for
less than x, x less than 0,

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and this would exist
for less greater than 0.

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I would not do that.

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I will write cases, but
the philosophy is the same.

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So let's multiply
by a number, f of k.

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Still a solution.

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k is fixed, so this
is just the number.

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Now superposition.

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That will still
be a solution if I

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do the same superposition in
the two formulas, integral dk

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and integral dk.

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That's still a solution
of a Schrodinger equation.

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Now I want to ask
you what limits

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I should use for that integral.

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And if anybody has
an opinion on that,

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it might naively be minus
infinity to infinity,

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and that might be good,
but maybe it's not so good.

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It's not so good.

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Why is not so good, minus
infinity to infinity?

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Would I have to do from
minus infinity on my force.

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Does anybody force me?

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No.

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You're superimposing solutions.

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For different values of
k, you're superimposing.

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You had a solution,
and another solution

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for another, another solution.

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What goes wrong if I go from
minus infinity to infinity?

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Yes.

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AUDIENCE: [INAUDIBLE] the wave
packet's going to be in one

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direction, so it [? shouldn't
?] be [? applied. ?]

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PROFESSOR: That's right.

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You see here, this
wave packet is

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going to be going in the
positive x direction,

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positive direction, as
long as k is positive.

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It's just that the direction is
determined by the relative sign

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within those quantities.

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E is positive in this
case. k is positive.

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This moves to the right.

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If I start putting things
where k is negative,

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I'm going to start producing
things and move to the left

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and to the right in
a terrible confusion.

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So yes, it should go 0 to
infinity, 0 to infinity.

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And f of k, what is it?

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Well, in our usual
picture, k f of k

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is some function that is
peaked around some k naught.

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And this whole thing
is psi of x and t,

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the full solution
of a wave packet.

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So now you see how the
A, B, and C coefficients

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enter into the construction
of a wave packet.

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I look back at the
textbook in which I

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learned quantum mechanics,
and it's a book by Schiff.

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It's a very good book.

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It's an old book.

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I think was probably
written in the '60s.

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And it goes through some
discussion of wave packets

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and then presents a jewel,
says with a supercomputer,

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we've been able to evaluate
numerically these things,

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something you can do now with
three seconds in your laptop,

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and it was the only
way to do this.

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So you produce an f of k.

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You fix their energy and
send in a wave packet

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and see what happens.

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You can do numerical
experiments with wave packets

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and see how the packet gets
distorted at the obstacle

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and how it eventually
bounces back or reflects,

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so it's very nice.

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So there is our solution.

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Now we're going to say
a few things about it.

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I want to split it a little bit.

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So lets go here.

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So how do we split it?

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I say the solution
is this whole thing,

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so let's call the
incident wave that

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is going to be defined
for x less than 0

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and t, this is x less than 0.

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And the incident
wave packet is dk 0

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to infinity f of k e to i kx e
to the minus i et over h bar.

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And this is just defined
for x less than 0,

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and that's so important
that write it here.

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For x less than 0, you have
an incident wave packet.

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And then you also have
a reflected wave packet,

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x less than 0 t is
the second part dk

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f of k k minus k bar over k
plus k bar e to the minus i kx

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e to the minus i et over h bar.

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It's also for x
less than 0, and we

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have a psi transmitted for
x greater than 0 and t,

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and that would be 0 to infinity
dk f of k 2k over k plus k bar

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e to the i k bar x e to
the minus i et over h bar.

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Lots of writing, but
that's important.

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And notice given
our definitions,

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the total psi of x and t is
equal to psi incident plus psi

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reflected for x less than 0,
and the total wave function of x

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and t is equal to
psi transmitted

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for x greater than 0.

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Lots of equations.

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I'll give you a second to copy
them if you are copying them.

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So now comes if we really
want to understand this,

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we have to push it
a little further.

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And perhaps in exercises we
will do some numerics to play

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with this thing as well.

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So I want to do stationary
phase approximation here.

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Otherwise, we don't
see what these packets,

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how they're moving.

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So you have some practice
already with this.

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You're supposed to have a
phase whose derivative is 0,

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and it's very, very
slowly at that place where

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there could be a contribution.

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Now every integral
has the f of k.

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So that still dominates
everything, of course.

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You see, if f of
k is very narrow,

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you pretty much could
evaluate these functions

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at the value of k naught
and get a rather accurate

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interpretation of the answer.

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The main difficulty would
be to do the leftover

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part of the integral.

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But again, here we
can identify phases.

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We're going to take f of k to
be localized and to be real.

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So there is no phase
associated with it,

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and there is no
phases associated

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with these quantities either,
so the phases are up there.

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So let's take, for
example, psi incident.

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What is this stationary
phase condition?

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Would be that the
derivative with respect

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to k that we are
integrating of the phase,

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kx minus Et over h bar must
be evaluated at k naught

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and must be equal to 0.

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So that's our stationary
phase approximation

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for the top interval.

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Now remember that E is equal
to h squared k squared over 2m.

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So what does this give you?

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That the peak of the
pulse of the wave packet

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is localized at the place where
the following condition holds.

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x minus de dk, with an h bar
will give an h k t over m

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evaluated at k naught equals 0.

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So this will be x equals
h bar k naught over m t.

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That's where the incident
wave is propagating.

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Now, look at that incident wave.

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What does it do
for negative time?

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As time is infinite and
negative, x is negative,

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and it's far away.

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Yes, the packet is very much
to the left of the barrier

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at time equals minus infinity.

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And that's consistent
because psi incident is only

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defined for x less than 0.

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It's only defined there.

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So as long as t is negative,
yes, the center of the packet

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is moving in.

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I'll maybe draw it here.

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The center of the packet is
moving in from minus infinity

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00:15:05,690 --> 00:15:10,450
into the wall, and
that is the picture.

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The packet is here, and
it's moving like that,

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and that's t negative.

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00:15:21,726 --> 00:15:30,220
The psi incident is coming
from the left into the barrier,

208
00:15:30,220 --> 00:15:31,305
and that's OK.

209
00:15:34,060 --> 00:15:43,080
But then what happens with
psi incident as t is positive?

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00:15:43,080 --> 00:15:46,800
As t is positive, psi
incident, well, it's

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00:15:46,800 --> 00:15:48,030
just another integral.

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00:15:48,030 --> 00:15:50,400
You might do it and
see what you get,

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00:15:50,400 --> 00:15:53,490
but we can see what
we will get, roughly.

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00:15:53,490 --> 00:16:00,090
When t is positive, the answer
would be you get something

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00:16:00,090 --> 00:16:02,370
if you have positive x.

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00:16:02,370 --> 00:16:06,240
But psi incident is
only for negative x.

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00:16:06,240 --> 00:16:09,540
So for negative x, you cannot
satisfy the stationarity

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00:16:09,540 --> 00:16:17,410
condition, and therefore, for
negative x and positive time,

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00:16:17,410 --> 00:16:21,930
t positive, psi
incident is nothing.

220
00:16:21,930 --> 00:16:25,090
It's a little wiggle.

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00:16:25,090 --> 00:16:26,860
There's probably
something, a little bit--

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00:16:26,860 --> 00:16:29,280
look at it with Mathematica--

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00:16:29,280 --> 00:16:30,280
there will be something.

224
00:16:30,280 --> 00:16:36,730
But for positive t, since
you only look at negative x,

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00:16:36,730 --> 00:16:39,280
you don't satisfy
stationarity, so you're not

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00:16:39,280 --> 00:16:40,120
going to get much.

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00:16:40,120 --> 00:16:41,950
So that's interesting.

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00:16:41,950 --> 00:16:45,010
Somehow automatically
psi incident just

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exists for negative time.

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For time near 0 is very
interesting because somehow

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00:16:54,324 --> 00:16:55,990
stationary [INAUDIBLE],
when you assess,

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00:16:55,990 --> 00:16:58,150
you still get
something, but you're

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going to see what the packet
does as it hits the thing.

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00:17:02,140 --> 00:17:05,550
Let's do the second
one of psi reflected.

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00:17:10,670 --> 00:17:17,440
d dk this time would be
kx with a different sign,

236
00:17:17,440 --> 00:17:24,339
minus kx minus E t over h
bar at k naught equals 0.

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00:17:24,339 --> 00:17:30,090
For the reflected wave, the
phase is really the same.

238
00:17:30,090 --> 00:17:32,325
Yeah, this factor is a
little more complicated,

239
00:17:32,325 --> 00:17:34,720
but it doesn't have
any phase in it.

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00:17:34,720 --> 00:17:37,090
It's real, so [INAUDIBLE].

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00:17:37,090 --> 00:17:40,150
So I just change a
sign, so this time I'm

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00:17:40,150 --> 00:17:41,790
going to get the change of sign.

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00:17:41,790 --> 00:17:48,820
x is equal to minus h
bar k naught over m t.

244
00:17:48,820 --> 00:17:55,170
And this says that for t
positive, you get things.

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00:17:55,170 --> 00:17:59,860
And in fact, as t is positive
your are at x negative.

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00:17:59,860 --> 00:18:04,170
And remember psi reflected is
only defined for x negative,

247
00:18:04,170 --> 00:18:06,390
so you can satisfy
stationary, and you're

248
00:18:06,390 --> 00:18:08,040
going to get something.

249
00:18:08,040 --> 00:18:11,170
So for t positive,
you're going to get

250
00:18:11,170 --> 00:18:14,790
as t increases, a thing that
goes more and more to the left

251
00:18:14,790 --> 00:18:17,130
as you would expect.

252
00:18:17,130 --> 00:18:23,340
So you will get psi
reflected going to the left.

253
00:18:26,070 --> 00:18:30,645
I will leave for you to
do the psi transmitted.

254
00:18:33,490 --> 00:18:37,550
It's a little different
because you have now k bar,

255
00:18:37,550 --> 00:18:39,910
and you have to take
the derivative of k bar

256
00:18:39,910 --> 00:18:41,350
with respect to k.

257
00:18:41,350 --> 00:18:46,720
It's going to be a little
more interesting example.

258
00:18:46,720 --> 00:18:55,360
But the answer is that this
one moves as x equals h bar k

259
00:18:55,360 --> 00:19:00,027
bar over m t.

260
00:19:00,027 --> 00:19:05,700
k bar is really the
momentum on the right,

261
00:19:05,700 --> 00:19:17,690
and since psi transmitted
exists only for positive x,

262
00:19:17,690 --> 00:19:21,290
this relation can be
satisfied for positive t.

263
00:19:21,290 --> 00:19:33,370
For positive t, there
will be a psi transmitted.

264
00:19:33,370 --> 00:19:41,940
The psi transmitted certainly
exists for negative t,

265
00:19:41,940 --> 00:19:47,910
but for negative t, stationarity
would want x to be negative,

266
00:19:47,910 --> 00:19:50,810
but that's not defined.

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00:19:50,810 --> 00:20:00,220
So for negative t on the right,
yes, psi transmitted maybe

268
00:20:00,220 --> 00:20:03,160
it's a little bit of
something especially for times

269
00:20:03,160 --> 00:20:05,090
that are not too negative.

270
00:20:05,090 --> 00:20:09,880
But the picture is that
stationary phase tells you

271
00:20:09,880 --> 00:20:11,920
that these packets,
psi incident,

272
00:20:11,920 --> 00:20:15,370
pretty much exist just
for negative t and psi

273
00:20:15,370 --> 00:20:19,090
reflected and psi transmitted
exist for positive t.

274
00:20:19,090 --> 00:20:21,460
And these are
consequences of the fact

275
00:20:21,460 --> 00:20:25,910
that psi incident and psi
reflected exist for negative x.

276
00:20:25,910 --> 00:20:29,350
The other exists for
positive x, and that coupled

277
00:20:29,350 --> 00:20:35,020
with stationarity produces
the physical picture that you

278
00:20:35,020 --> 00:20:38,770
expect intuitively, that
the incident wave is just

279
00:20:38,770 --> 00:20:40,940
something, part of
the solution that

280
00:20:40,940 --> 00:20:43,730
exists just at the beginning.

281
00:20:43,730 --> 00:20:46,740
And somehow it whistles away.

282
00:20:46,740 --> 00:20:51,940
Some of it becomes transmitted,
some of it becomes reflected.