WEBVTT
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PROFESSOR: In order to learn
more about this subject,
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we must do the wave packets.
00:00:06.060 --> 00:00:09.060
So this is the place
where you really
00:00:09.060 --> 00:00:14.520
connect this need solution
of Schrodinger's equation,
00:00:14.520 --> 00:00:19.440
the energy eigenstates,
to a physical problem.
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So we'll do our wave packets.
00:00:27.170 --> 00:00:31.600
So we've been dealing
with packets for a while,
00:00:31.600 --> 00:00:35.110
so I think it's not going to
be that difficult. We've also
00:00:35.110 --> 00:00:37.170
been talking about
stationary phase
00:00:37.170 --> 00:00:41.620
and you've practiced that,
so you have the math ready.
00:00:41.620 --> 00:00:43.630
We should not have
a great difficulty.
00:00:43.630 --> 00:00:53.700
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
00:01:00.360 --> 00:01:04.300
work with E greater than
v naught to begin with.
00:01:06.900 --> 00:01:09.600
The reason I want to work
with E greater than v naught
00:01:09.600 --> 00:01:11.535
is because there is
a transmitted wave.
00:01:14.610 --> 00:01:16.230
So that's kind of nice.
00:01:16.230 --> 00:01:22.300
So what am I going to do?
00:01:22.300 --> 00:01:24.920
I'm going to write it
this following way.
00:01:24.920 --> 00:01:28.921
So here is a solution
with A equals to 1.
00:01:28.921 --> 00:01:40.590
e to the i kx plus k
minus k bar over k plus k
00:01:40.590 --> 00:01:45.300
bar e to the minus i kx.
00:01:48.270 --> 00:01:50.850
And this was for x less than 0.
00:01:53.940 --> 00:01:58.140
And indeed, when there
was an A, there was a B,
00:01:58.140 --> 00:02:02.495
if A is equal to 1, remember we
solve for the ratio of B to A
00:02:02.495 --> 00:02:06.060
and it was this number
so I put it there.
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I'm just writing it
a little differently,
00:02:08.440 --> 00:02:11.845
but this is a solution.
00:02:11.845 --> 00:02:18.860
And for x greater
than 0, the solution
00:02:18.860 --> 00:02:28.680
is C, which was 2k over k
plus k bar e to the i k bar x.
00:02:37.060 --> 00:02:41.070
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
00:02:52.610 --> 00:02:53.810
equation?
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What do I need to
make it the solution
00:02:56.420 --> 00:02:58.090
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.
00:03:10.470 --> 00:03:16.340
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,
00:03:19.490 --> 00:03:20.710
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?
00:06:34.410 --> 00:06:40.640
Well, in our usual
picture, k f of k
00:06:40.640 --> 00:06:44.580
is some function that is
peaked around some k naught.
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And this whole thing
is psi of x and t,
00:06:58.150 --> 00:07:01.480
the full solution
of a wave packet.
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So now you see how the
A, B, and C coefficients
00:07:06.160 --> 00:07:09.640
enter into the construction
of a wave packet.
00:07:09.640 --> 00:07:13.450
I look back at the
textbook in which I
00:07:13.450 --> 00:07:19.130
learned quantum mechanics,
and it's a book by Schiff.
00:07:19.130 --> 00:07:20.695
It's a very good book.
00:07:20.695 --> 00:07:22.660
It's an old book.
00:07:22.660 --> 00:07:26.710
I think was probably
written in the '60s.
00:07:26.710 --> 00:07:31.420
And it goes through some
discussion of wave packets
00:07:31.420 --> 00:07:36.180
and then presents a jewel,
says with a supercomputer,
00:07:36.180 --> 00:07:40.510
we've been able to evaluate
numerically these things,
00:07:40.510 --> 00:07:45.790
something you can do now with
three seconds in your laptop,
00:07:45.790 --> 00:07:49.250
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.
00:08:09.310 --> 00:08:11.450
So there is our solution.
00:08:11.450 --> 00:08:15.100
Now we're going to say
a few things about it.
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I want to split it a little bit.
00:08:18.250 --> 00:08:22.720
So lets go here.
00:08:32.240 --> 00:08:35.570
So how do we split it?
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I say the solution
is this whole thing,
00:08:41.580 --> 00:08:52.540
so let's call the
incident wave that
00:08:52.540 --> 00:08:56.390
is going to be defined
for x less than 0
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and t, this is x less than 0.
00:09:00.400 --> 00:09:05.190
And the incident
wave packet is dk 0
00:09:05.190 --> 00:09:17.170
to infinity f of k e to i kx e
to the minus i et over h bar.
00:09:17.170 --> 00:09:20.800
And this is just defined
for x less than 0,
00:09:20.800 --> 00:09:23.250
and that's so important
that write it here.
00:09:26.900 --> 00:09:29.940
For x less than 0, you have
an incident wave packet.
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And then you also have
a reflected wave packet,
00:09:36.960 --> 00:09:43.030
x less than 0 t is
the second part dk
00:09:43.030 --> 00:09:54.780
f of k k minus k bar over k
plus k bar e to the minus i kx
00:09:54.780 --> 00:09:59.210
e to the minus i et over h bar.
00:09:59.210 --> 00:10:02.370
It's also for x
less than 0, and we
00:10:02.370 --> 00:10:11.700
have a psi transmitted for
x greater than 0 and t,
00:10:11.700 --> 00:10:22.012
and that would be 0 to infinity
dk f of k 2k over k plus k bar
00:10:22.012 --> 00:10:30.480
e to the i k bar x e to
the minus i et over h bar.
00:10:30.480 --> 00:10:36.420
Lots of writing, but
that's important.
00:10:36.420 --> 00:10:39.280
And notice given
our definitions,
00:10:39.280 --> 00:10:48.960
the total psi of x and t is
equal to psi incident plus psi
00:10:48.960 --> 00:10:56.180
reflected for x less than 0,
and the total wave function of x
00:10:56.180 --> 00:11:01.160
and t is equal to
psi transmitted
00:11:01.160 --> 00:11:02.800
for x greater than 0.
00:11:07.400 --> 00:11:08.450
Lots of equations.
00:11:08.450 --> 00:11:15.293
I'll give you a second to copy
them if you are copying them.
00:11:21.710 --> 00:11:28.490
So now comes if we really
want to understand this,
00:11:28.490 --> 00:11:31.730
we have to push it
a little further.
00:11:31.730 --> 00:11:36.380
And perhaps in exercises we
will do some numerics to play
00:11:36.380 --> 00:11:39.220
with this thing as well.
00:11:39.220 --> 00:11:47.460
So I want to do stationary
phase approximation here.
00:11:47.460 --> 00:11:49.920
Otherwise, we don't
see what these packets,
00:11:49.920 --> 00:11:51.540
how they're moving.
00:11:51.540 --> 00:11:55.210
So you have some practice
already with this.
00:11:55.210 --> 00:12:00.120
You're supposed to have a
phase whose derivative is 0,
00:12:00.120 --> 00:12:03.840
and it's very, very
slowly at that place where
00:12:03.840 --> 00:12:05.600
there could be a contribution.
00:12:05.600 --> 00:12:09.330
Now every integral
has the f of k.
00:12:09.330 --> 00:12:13.560
So that still dominates
everything, of course.
00:12:13.560 --> 00:12:20.230
You see, if f of
k is very narrow,
00:12:20.230 --> 00:12:23.550
you pretty much could
evaluate these functions
00:12:23.550 --> 00:12:32.070
at the value of k naught
and get a rather accurate
00:12:32.070 --> 00:12:35.140
interpretation of the answer.
00:12:35.140 --> 00:12:38.280
The main difficulty would
be to do the leftover
00:12:38.280 --> 00:12:39.480
part of the integral.
00:12:39.480 --> 00:12:45.200
But again, here we
can identify phases.
00:12:45.200 --> 00:12:51.010
We're going to take f of k to
be localized and to be real.
00:12:51.010 --> 00:12:54.530
So there is no phase
associated with it,
00:12:54.530 --> 00:12:56.600
and there is no
phases associated
00:12:56.600 --> 00:12:59.380
with these quantities either,
so the phases are up there.
00:12:59.380 --> 00:13:03.320
So let's take, for
example, psi incident.
00:13:03.320 --> 00:13:05.900
What is this stationary
phase condition?
00:13:05.900 --> 00:13:08.660
Would be that the
derivative with respect
00:13:08.660 --> 00:13:12.290
to k that we are
integrating of the phase,
00:13:12.290 --> 00:13:20.490
kx minus Et over h bar must
be evaluated at k naught
00:13:20.490 --> 00:13:23.220
and must be equal to 0.
00:13:23.220 --> 00:13:25.980
So that's our stationary
phase approximation
00:13:25.980 --> 00:13:27.390
for the top interval.
00:13:27.390 --> 00:13:34.870
Now remember that E is equal
to h squared k squared over 2m.
00:13:34.870 --> 00:13:37.780
So what does this give you?
00:13:37.780 --> 00:13:43.470
That the peak of the
pulse of the wave packet
00:13:43.470 --> 00:13:47.490
is localized at the place where
the following condition holds.
00:13:47.490 --> 00:13:58.310
x minus de dk, with an h bar
will give an h k t over m
00:13:58.310 --> 00:14:01.650
evaluated at k naught equals 0.
00:14:01.650 --> 00:14:12.670
So this will be x equals
h bar k naught over m t.
00:14:12.670 --> 00:14:18.960
That's where the incident
wave is propagating.
00:14:18.960 --> 00:14:23.670
Now, look at that incident wave.
00:14:23.670 --> 00:14:27.370
What does it do
for negative time?
00:14:27.370 --> 00:14:33.210
As time is infinite and
negative, x is negative,
00:14:33.210 --> 00:14:34.270
and it's far away.
00:14:34.270 --> 00:14:37.930
Yes, the packet is very much
to the left of the barrier
00:14:37.930 --> 00:14:40.470
at time equals minus infinity.
00:14:40.470 --> 00:14:44.640
And that's consistent
because psi incident is only
00:14:44.640 --> 00:14:49.070
defined for x less than 0.
00:14:49.070 --> 00:14:50.910
It's only defined there.
00:14:50.910 --> 00:14:55.630
So as long as t is negative,
yes, the center of the packet
00:14:55.630 --> 00:14:57.910
is moving in.
00:14:57.910 --> 00:15:00.255
I'll maybe draw it here.
00:15:00.255 --> 00:15:05.690
The center of the packet is
moving in from minus infinity
00:15:05.690 --> 00:15:10.450
into the wall, and
that is the picture.
00:15:16.123 --> 00:15:19.000
The packet is here, and
it's moving like that,
00:15:19.000 --> 00:15:21.726
and that's t negative.
00:15:21.726 --> 00:15:30.220
The psi incident is coming
from the left into the barrier,
00:15:30.220 --> 00:15:31.305
and that's OK.
00:15:34.060 --> 00:15:43.080
But then what happens with
psi incident as t is positive?
00:15:43.080 --> 00:15:46.800
As t is positive, psi
incident, well, it's
00:15:46.800 --> 00:15:48.030
just another integral.
00:15:48.030 --> 00:15:50.400
You might do it and
see what you get,
00:15:50.400 --> 00:15:53.490
but we can see what
we will get, roughly.
00:15:53.490 --> 00:16:00.090
When t is positive, the answer
would be you get something
00:16:00.090 --> 00:16:02.370
if you have positive x.
00:16:02.370 --> 00:16:06.240
But psi incident is
only for negative x.
00:16:06.240 --> 00:16:09.540
So for negative x, you cannot
satisfy the stationarity
00:16:09.540 --> 00:16:17.410
condition, and therefore, for
negative x and positive time,
00:16:17.410 --> 00:16:21.930
t positive, psi
incident is nothing.
00:16:21.930 --> 00:16:25.090
It's a little wiggle.
00:16:25.090 --> 00:16:26.860
There's probably
something, a little bit--
00:16:26.860 --> 00:16:29.280
look at it with Mathematica--
00:16:29.280 --> 00:16:30.280
there will be something.
00:16:30.280 --> 00:16:36.730
But for positive t, since
you only look at negative x,
00:16:36.730 --> 00:16:39.280
you don't satisfy
stationarity, so you're not
00:16:39.280 --> 00:16:40.120
going to get much.
00:16:40.120 --> 00:16:41.950
So that's interesting.
00:16:41.950 --> 00:16:45.010
Somehow automatically
psi incident just
00:16:45.010 --> 00:16:49.120
exists for negative time.
00:16:49.120 --> 00:16:54.324
For time near 0 is very
interesting because somehow
00:16:54.324 --> 00:16:55.990
stationary [INAUDIBLE],
when you assess,
00:16:55.990 --> 00:16:58.150
you still get
something, but you're
00:16:58.150 --> 00:17:02.140
going to see what the packet
does as it hits the thing.
00:17:02.140 --> 00:17:05.550
Let's do the second
one of psi reflected.
00:17:10.670 --> 00:17:17.440
d dk this time would be
kx with a different sign,
00:17:17.440 --> 00:17:24.339
minus kx minus E t over h
bar at k naught equals 0.
00:17:24.339 --> 00:17:30.090
For the reflected wave, the
phase is really the same.
00:17:30.090 --> 00:17:32.325
Yeah, this factor is a
little more complicated,
00:17:32.325 --> 00:17:34.720
but it doesn't have
any phase in it.
00:17:34.720 --> 00:17:37.090
It's real, so [INAUDIBLE].
00:17:37.090 --> 00:17:40.150
So I just change a
sign, so this time I'm
00:17:40.150 --> 00:17:41.790
going to get the change of sign.
00:17:41.790 --> 00:17:48.820
x is equal to minus h
bar k naught over m t.
00:17:48.820 --> 00:17:55.170
And this says that for t
positive, you get things.
00:17:55.170 --> 00:17:59.860
And in fact, as t is positive
your are at x negative.
00:17:59.860 --> 00:18:04.170
And remember psi reflected is
only defined for x negative,
00:18:04.170 --> 00:18:06.390
so you can satisfy
stationary, and you're
00:18:06.390 --> 00:18:08.040
going to get something.
00:18:08.040 --> 00:18:11.170
So for t positive,
you're going to get
00:18:11.170 --> 00:18:14.790
as t increases, a thing that
goes more and more to the left
00:18:14.790 --> 00:18:17.130
as you would expect.
00:18:17.130 --> 00:18:23.340
So you will get psi
reflected going to the left.
00:18:26.070 --> 00:18:30.645
I will leave for you to
do the psi transmitted.
00:18:33.490 --> 00:18:37.550
It's a little different
because you have now k bar,
00:18:37.550 --> 00:18:39.910
and you have to take
the derivative of k bar
00:18:39.910 --> 00:18:41.350
with respect to k.
00:18:41.350 --> 00:18:46.720
It's going to be a little
more interesting example.
00:18:46.720 --> 00:18:55.360
But the answer is that this
one moves as x equals h bar k
00:18:55.360 --> 00:19:00.027
bar over m t.
00:19:00.027 --> 00:19:05.700
k bar is really the
momentum on the right,
00:19:05.700 --> 00:19:17.690
and since psi transmitted
exists only for positive x,
00:19:17.690 --> 00:19:21.290
this relation can be
satisfied for positive t.
00:19:21.290 --> 00:19:33.370
For positive t, there
will be a psi transmitted.
00:19:33.370 --> 00:19:41.940
The psi transmitted certainly
exists for negative t,
00:19:41.940 --> 00:19:47.910
but for negative t, stationarity
would want x to be negative,
00:19:47.910 --> 00:19:50.810
but that's not defined.
00:19:50.810 --> 00:20:00.220
So for negative t on the right,
yes, psi transmitted maybe
00:20:00.220 --> 00:20:03.160
it's a little bit of
something especially for times
00:20:03.160 --> 00:20:05.090
that are not too negative.
00:20:05.090 --> 00:20:09.880
But the picture is that
stationary phase tells you
00:20:09.880 --> 00:20:11.920
that these packets,
psi incident,
00:20:11.920 --> 00:20:15.370
pretty much exist just
for negative t and psi
00:20:15.370 --> 00:20:19.090
reflected and psi transmitted
exist for positive t.
00:20:19.090 --> 00:20:21.460
And these are
consequences of the fact
00:20:21.460 --> 00:20:25.910
that psi incident and psi
reflected exist for negative x.
00:20:25.910 --> 00:20:29.350
The other exists for
positive x, and that coupled
00:20:29.350 --> 00:20:35.020
with stationarity produces
the physical picture that you
00:20:35.020 --> 00:20:38.770
expect intuitively, that
the incident wave is just
00:20:38.770 --> 00:20:40.940
something, part of
the solution that
00:20:40.940 --> 00:20:43.730
exists just at the beginning.
00:20:43.730 --> 00:20:46.740
And somehow it whistles away.
00:20:46.740 --> 00:20:51.940
Some of it becomes transmitted,
some of it becomes reflected.