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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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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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The psi incident is coming
from the left into the barrier,
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and that's OK.
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But then what happens with
psi incident as t is positive?
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As t is positive, psi
incident, well, it's
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just another integral.
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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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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.
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It's a little wiggle.
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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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there will be something.
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But for positive t, since
you only look at negative x,
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you don't satisfy
stationarity, so you're not
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going to get much.
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So that's interesting.
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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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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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d dk this time would be
kx with a different sign,
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minus kx minus E t over h
bar at k naught equals 0.
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For the reflected wave, the
phase is really the same.
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Yeah, this factor is a
little more complicated,
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00:17:32,325 --> 00:17:34,720
but it doesn't have
any phase in it.
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It's real, so [INAUDIBLE].
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So I just change a
sign, so this time I'm
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going to get the change of sign.
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x is equal to minus h
bar k naught over m t.
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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.
267
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.