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PROFESSOR: So today
is going to be
00:00:25.750 --> 00:00:29.600
our last pass at bound states.
00:00:29.600 --> 00:00:32.750
So starting next week or
actually starting next lecture,
00:00:32.750 --> 00:00:34.389
we're going to
look at scattering.
00:00:34.389 --> 00:00:35.680
Scattering's going to be great.
00:00:35.680 --> 00:00:38.970
But we need to close
out bound states.
00:00:38.970 --> 00:00:45.600
So today's topic
is the finite well,
00:00:45.600 --> 00:00:46.695
finite the potential well.
00:00:52.892 --> 00:00:54.350
We've sort of
sketched this when we
00:00:54.350 --> 00:00:56.560
looked at qualitative structure
of wave functions of energy
00:00:56.560 --> 00:00:57.060
eigenstates.
00:01:00.280 --> 00:01:02.090
But we're going to
solve the system today.
00:01:06.830 --> 00:01:10.220
So good.
00:01:10.220 --> 00:01:14.185
So the system we're
interested in is going to be--
00:01:14.185 --> 00:01:15.560
The system we're
interested in is
00:01:15.560 --> 00:01:18.009
going to be a system
with a finite depth
00:01:18.009 --> 00:01:18.800
and a finite width.
00:01:18.800 --> 00:01:21.531
And I'll go into detail and
give you parameters in a bit.
00:01:21.531 --> 00:01:23.030
But first I want
to just think about
00:01:23.030 --> 00:01:26.020
how do we find energy
eigenfunctions of a potential
00:01:26.020 --> 00:01:31.470
of this form, v of x, which
is piecewise constant.
00:01:31.470 --> 00:01:34.370
So first off, is this is a
terribly realistic potential?
00:01:34.370 --> 00:01:36.360
Will you ever in the
real world find a system
00:01:36.360 --> 00:01:39.560
that has a potential which
is piecewise constant?
00:01:39.560 --> 00:01:40.210
Probably not.
00:01:40.210 --> 00:01:41.310
It's discontinuous.
00:01:41.310 --> 00:01:41.840
Right?
00:01:41.840 --> 00:01:43.210
So it's rather unphysical.
00:01:43.210 --> 00:01:45.130
But it's a very
useful toy model.
00:01:45.130 --> 00:01:47.730
So for example, if you take
a couple of capacitor plates,
00:01:47.730 --> 00:01:52.470
then you can induce a situation
where the electric field
00:01:52.470 --> 00:01:56.050
is nonzero in between
the capacitor plates
00:01:56.050 --> 00:01:59.621
and zero outside of
the capacitor plates.
00:01:59.621 --> 00:02:00.120
Right.
00:02:00.120 --> 00:02:04.209
So at a superficial level,
this looks discontinuous.
00:02:04.209 --> 00:02:05.750
It looks like the
electric field is--
00:02:05.750 --> 00:02:07.276
But actually, you know
that microscopically there
00:02:07.276 --> 00:02:08.859
are a bunch of
charges, and everything
00:02:08.859 --> 00:02:11.250
is nice and continuous
except for the behavior
00:02:11.250 --> 00:02:12.820
right at the charges.
00:02:12.820 --> 00:02:16.150
So but it's reasonable
to model this
00:02:16.150 --> 00:02:18.680
as a step function
for an electric field.
00:02:18.680 --> 00:02:20.490
So this is going to
be an idealization,
00:02:20.490 --> 00:02:22.660
but it's going to be a
very useful idealization,
00:02:22.660 --> 00:02:25.855
the constant potential.
00:02:25.855 --> 00:02:26.355
OK.
00:02:30.822 --> 00:02:31.780
So what's the equation?
00:02:31.780 --> 00:02:33.110
What are we trying to do?
00:02:33.110 --> 00:02:34.630
We want to find the energy
eigenstates for this
00:02:34.630 --> 00:02:36.190
because we want to study
the time evolution.
00:02:36.190 --> 00:02:38.220
And the easiest way to solve
the Schrodinger equation,
00:02:38.220 --> 00:02:40.345
the time evolution equation,
is to expand an energy
00:02:40.345 --> 00:02:41.210
eigenstates.
00:02:41.210 --> 00:02:46.220
So the equation we want to solve
is energy eigenvalue times phi
00:02:46.220 --> 00:02:51.790
e of x is equal to minus
h bar squared on 2m phi
00:02:51.790 --> 00:02:55.610
e phi prime plus v of x.
00:03:00.820 --> 00:03:03.900
And I'm going to put it in
the form we've been using.
00:03:03.900 --> 00:03:08.910
Phi prime prime e of x
is equal to 2m upon h bar
00:03:08.910 --> 00:03:13.980
squared v of x minus e.
00:03:13.980 --> 00:03:14.480
OK.
00:03:14.480 --> 00:03:15.862
So this is the
form that I'm going
00:03:15.862 --> 00:03:18.278
to use today to solve for the
energy eigenvalue equations.
00:03:18.278 --> 00:03:19.890
e is some constant.
00:03:19.890 --> 00:03:25.130
Do you expect the allowed
energies to be arbitrary?
00:03:25.130 --> 00:03:25.630
No.
00:03:25.630 --> 00:03:28.567
They should be discrete.
00:03:28.567 --> 00:03:29.150
Yeah, exactly.
00:03:29.150 --> 00:03:32.040
So we expect that there should
be discrete lowest energy
00:03:32.040 --> 00:03:35.730
state, some number
of bound states.
00:03:35.730 --> 00:03:37.850
And then, eventually,
if the energy
00:03:37.850 --> 00:03:41.220
is greater than the
potential everywhere,
00:03:41.220 --> 00:03:45.030
the energies will be continuous.
00:03:45.030 --> 00:03:47.035
Any energy will be allowed
above the potential.
00:03:47.035 --> 00:03:52.240
So we'll have a continuum of
states above the potential.
00:03:52.240 --> 00:03:54.650
And we'll have a discrete
set of bound states--
00:03:54.650 --> 00:03:57.000
Probably, it's reasonable
to expect some finite number
00:03:57.000 --> 00:03:58.416
of bound states
just by intuition.
00:04:01.240 --> 00:04:03.860
--from the infinite well.
00:04:03.860 --> 00:04:06.460
So we expect to have a finite
number of discrete energies
00:04:06.460 --> 00:04:09.890
and then a continuous set
of energies above zero.
00:04:09.890 --> 00:04:14.250
So if this is the asymptotic
value potential of zero.
00:04:14.250 --> 00:04:14.750
OK.
00:04:14.750 --> 00:04:16.566
And this is intuition
gained from our study
00:04:16.566 --> 00:04:18.649
of qualitative structure
of energy eigenfunctions.
00:04:18.649 --> 00:04:21.390
So we are going to talk
today about the bound states.
00:04:21.390 --> 00:04:24.980
And in recitation,
leaders should
00:04:24.980 --> 00:04:29.720
discuss the continuum
above zero energy.
00:04:29.720 --> 00:04:32.440
OK.
00:04:32.440 --> 00:04:35.340
So to solve for the actual
energy eigenfunctions
00:04:35.340 --> 00:04:37.750
and the energy eigenvalues,
what we need to do
00:04:37.750 --> 00:04:44.830
is we need to
solve this equation
00:04:44.830 --> 00:04:46.900
subject to some
boundary conditions.
00:04:46.900 --> 00:04:49.233
And the boundary conditions
we're going to want to solve
00:04:49.233 --> 00:04:51.240
are going to be finite.
00:04:51.240 --> 00:04:52.490
So it's normalizable infinity.
00:04:52.490 --> 00:04:55.640
The solution should
be vanishing far away.
00:04:55.640 --> 00:04:58.390
And the wave function
should be everywhere smooth.
00:04:58.390 --> 00:05:00.059
Well, at least it
should be continuous.
00:05:00.059 --> 00:05:02.350
So let's talk about what
exactly boundary conditions we
00:05:02.350 --> 00:05:02.974
want to impose.
00:05:05.770 --> 00:05:08.290
And so in particular,
we're going
00:05:08.290 --> 00:05:12.870
to want to solve for the energy
eigenfunctions in the regions
00:05:12.870 --> 00:05:16.150
where the potential is constant
and then patch together
00:05:16.150 --> 00:05:18.460
solutions at these boundaries.
00:05:18.460 --> 00:05:20.749
We know how to solve for
the energy eigenfunctions
00:05:20.749 --> 00:05:22.040
when the potential is constant.
00:05:22.040 --> 00:05:23.498
What are the energy
eigenfunctions?
00:05:26.501 --> 00:05:27.000
Yeah.
00:05:27.000 --> 00:05:29.420
Suppose I have a potential,
which is constant.
00:05:29.420 --> 00:05:30.564
v is equal to 0.
00:05:30.564 --> 00:05:31.980
What are the energy
eigenfunctions
00:05:31.980 --> 00:05:33.065
of this potential?
00:05:33.065 --> 00:05:34.220
AUDIENCE: [INAUDIBLE]
00:05:34.220 --> 00:05:36.031
PROFESSOR: Yeah, e to the ikx.
00:05:36.031 --> 00:05:36.530
Yeah.
00:05:36.530 --> 00:05:37.988
Now, what if I
happen to tell you--
00:05:37.988 --> 00:05:41.315
So we're at h bar
squared k squared upon 2m
00:05:41.315 --> 00:05:42.829
is equal to the energy.
00:05:42.829 --> 00:05:45.120
Suppose I happen to tell you
that here's the potential.
00:05:45.120 --> 00:05:47.730
And I want to find the
solution in this region
00:05:47.730 --> 00:05:50.397
where the energy is here
less than the potential.
00:05:50.397 --> 00:05:51.355
What are the solutions?
00:05:54.017 --> 00:05:55.850
That's a second order
differential equation.
00:05:55.850 --> 00:05:56.670
There should be two solutions.
00:05:56.670 --> 00:05:58.500
What are the solutions to that
differential equation when
00:05:58.500 --> 00:06:00.083
the energy is less
than the potential?
00:06:03.689 --> 00:06:04.980
AUDIENCE: Decaying and growing.
00:06:04.980 --> 00:06:06.855
PROFESSOR: Decaying and
growing exponentials.
00:06:06.855 --> 00:06:11.930
Exactly. e to the plus alpha
x and e to the minus alpha x.
00:06:11.930 --> 00:06:15.720
And the reason is
these are sinusoidal,
00:06:15.720 --> 00:06:18.320
and these have the
opposite concavity.
00:06:18.320 --> 00:06:21.551
They are growing and
dying exponentials.
00:06:21.551 --> 00:06:22.185
Cool?
00:06:22.185 --> 00:06:22.685
OK.
00:06:22.685 --> 00:06:24.163
So we've studied that.
00:06:24.163 --> 00:06:25.030
Yeah?
00:06:25.030 --> 00:06:27.412
AUDIENCE: Shouldn't
that be a phi e of x?
00:06:27.412 --> 00:06:28.120
PROFESSOR: Sorry.
00:06:28.120 --> 00:06:29.510
Oh, oh, yes, indeed.
00:06:29.510 --> 00:06:30.150
Sorry.
00:06:30.150 --> 00:06:31.470
Thank you.
00:06:31.470 --> 00:06:34.600
Phi e of x.
00:06:34.600 --> 00:06:36.600
It's early, and I'm still
working on the coffee.
00:06:40.310 --> 00:06:41.390
It won't take long.
00:06:46.120 --> 00:06:46.620
Good.
00:06:46.620 --> 00:06:49.520
So we know how to solve the
energy eigenvalue equation
00:06:49.520 --> 00:06:51.990
in all these regions where
the potential is constant.
00:06:51.990 --> 00:06:54.690
So our job is going to be to
find a solution where we patch
00:06:54.690 --> 00:06:56.482
them together at
these interfaces.
00:06:56.482 --> 00:06:57.440
We patch them together.
00:06:57.440 --> 00:07:00.520
And what condition
should we impose?
00:07:00.520 --> 00:07:02.250
So the basic condition
is going to be
00:07:02.250 --> 00:07:13.580
continuity of the wave
functions phi e of x.
00:07:13.580 --> 00:07:15.980
And so what are the
conditions that we need?
00:07:15.980 --> 00:07:17.380
Well, if v of x--
00:07:17.380 --> 00:07:19.210
Here's the way I like
to think about this.
00:07:19.210 --> 00:07:20.550
Suppose v of x is continuous.
00:07:26.040 --> 00:07:28.512
So if the potential
is continuous, then
00:07:28.512 --> 00:07:29.220
what can you say?
00:07:29.220 --> 00:07:33.450
You can say that phi prime prime
is a continuous function times
00:07:33.450 --> 00:07:35.635
phi of x.
00:07:35.635 --> 00:07:37.260
So very roughly, if
we look at a region
00:07:37.260 --> 00:07:39.860
where phi isn't varying
very much, so if we
00:07:39.860 --> 00:07:42.540
have a potential that's
varying in some way, then
00:07:42.540 --> 00:07:45.890
phi prime prime, in a
region where it's not
00:07:45.890 --> 00:07:50.310
varying much compared to its
value as a function of x,
00:07:50.310 --> 00:07:51.960
does something
smooth because it's
00:07:51.960 --> 00:07:55.050
varying with the potential.
00:07:55.050 --> 00:07:57.510
And so phi prime,
which is just going
00:07:57.510 --> 00:07:58.770
to be the integral of this--
00:07:58.770 --> 00:08:00.145
The integral of
a smooth function
00:08:00.145 --> 00:08:02.810
is again a smooth function.
00:08:02.810 --> 00:08:04.980
--and phi, the integral
of that function
00:08:04.980 --> 00:08:08.150
is also going to be a
nice, smooth function.
00:08:08.150 --> 00:08:08.710
OK?
00:08:08.710 --> 00:08:11.320
I've drawn it badly, but--
00:08:11.320 --> 00:08:14.430
So the key thing here is that
if the potential is continuous,
00:08:14.430 --> 00:08:17.360
then the energy eigenfunction
has a second derivative, which
00:08:17.360 --> 00:08:19.914
is also continuous.
00:08:19.914 --> 00:08:21.830
That means its first
derivative is continuous.
00:08:21.830 --> 00:08:23.830
So that means the function
itself is continuous.
00:08:27.720 --> 00:08:29.240
Everyone agree with this?
00:08:29.240 --> 00:08:31.262
Questions?
00:08:31.262 --> 00:08:33.220
So in regions where the
potential's continuous,
00:08:33.220 --> 00:08:35.344
the wave function and its
first two derivatives all
00:08:35.344 --> 00:08:36.816
have to be continuous.
00:08:36.816 --> 00:08:39.980
On the other hand, suppose
the potential has a step.
00:08:39.980 --> 00:08:42.209
v of x has a step discontinuity.
00:08:45.480 --> 00:08:46.030
OK.
00:08:46.030 --> 00:08:48.190
So the potential
does one of those.
00:08:50.399 --> 00:08:52.440
So what does that tell
you about phi prime prime?
00:08:56.416 --> 00:08:57.290
It's a function of x.
00:09:03.444 --> 00:09:05.360
So for example, let's
look at that first step.
00:09:05.360 --> 00:09:08.420
Suppose the potential is that
first step down by some amount.
00:09:08.420 --> 00:09:12.390
Then phi prime prime is going
to decrease precipitously
00:09:12.390 --> 00:09:13.745
at some point.
00:09:13.745 --> 00:09:15.370
And the actual amount
that it decreases
00:09:15.370 --> 00:09:17.300
depends on the value of
phi because the change
00:09:17.300 --> 00:09:21.770
in the potential time is
the actual value of phi.
00:09:21.770 --> 00:09:23.730
And if that's true of
phi prime prime what can
00:09:23.730 --> 00:09:24.629
you say of phi prime?
00:09:24.629 --> 00:09:25.670
So this is discontinuous.
00:09:30.900 --> 00:09:32.960
Phi prime of x,
however, is the integral
00:09:32.960 --> 00:09:35.627
of this discontinuous function.
00:09:35.627 --> 00:09:36.460
And what does it do?
00:09:36.460 --> 00:09:38.335
Well, it's linearly
increasing in this region
00:09:38.335 --> 00:09:40.340
because its derivative
is constant.
00:09:40.340 --> 00:09:44.410
And here, it's linearly
increasing less.
00:09:44.410 --> 00:09:49.590
So it's not differentiably
smooth, but it's continuous.
00:09:49.590 --> 00:09:52.060
And then let's look at the
actual function phi of x.
00:09:56.770 --> 00:09:57.270
OK.
00:09:57.270 --> 00:09:57.978
What is it doing?
00:09:57.978 --> 00:09:59.430
Well, it's quadratic.
00:09:59.430 --> 00:10:02.907
And then here it's quadratic
a little less afterwards.
00:10:02.907 --> 00:10:04.490
But that still
continuous because it's
00:10:04.490 --> 00:10:07.642
the integral of a
continuous function.
00:10:07.642 --> 00:10:09.150
Everyone we cool with that?
00:10:09.150 --> 00:10:12.150
So even when we have a step
discontinuity in our potential,
00:10:12.150 --> 00:10:14.470
we still have that
our derivative
00:10:14.470 --> 00:10:16.685
and the value of the
function are continuous.
00:10:19.240 --> 00:10:20.210
Yeah?
00:10:20.210 --> 00:10:26.340
However, imagine the potential
has this delta function.
00:10:26.340 --> 00:10:28.200
Let's just really push it.
00:10:28.200 --> 00:10:30.330
What happens if our potential
has a delta function
00:10:30.330 --> 00:10:31.616
singularity?
00:10:31.616 --> 00:10:32.740
Really badly discontinuous.
00:10:32.740 --> 00:10:39.424
Then what can you say about phi
prime prime as a function of x?
00:10:39.424 --> 00:10:40.840
So it has a delta
function, right?
00:10:40.840 --> 00:10:43.650
So the phi prime prime has to
look like something relatively
00:10:43.650 --> 00:10:48.580
slowly varying, and then
a step delta function.
00:10:48.580 --> 00:10:51.070
So what does that tell you
about the derivative of the wave
00:10:51.070 --> 00:10:52.799
function?
00:10:52.799 --> 00:10:53.840
It's got a step function.
00:10:53.840 --> 00:10:54.190
Exactly.
00:10:54.190 --> 00:10:56.689
Because it's the integral of
this, and the integral is 0, 1.
00:10:59.810 --> 00:11:00.310
Whoops.
00:11:00.310 --> 00:11:00.810
I missed.
00:11:06.510 --> 00:11:07.660
So this is a delta.
00:11:07.660 --> 00:11:09.796
This is a step.
00:11:09.796 --> 00:11:11.170
And then the wave
function itself
00:11:11.170 --> 00:11:14.615
is, well, it's the integral
of a step, so it's continuous.
00:11:18.840 --> 00:11:20.960
Sorry.
00:11:20.960 --> 00:11:22.430
It's certainly not
differentiable.
00:11:22.430 --> 00:11:23.763
Its derivative is discontinuous.
00:11:26.964 --> 00:11:28.880
Well, it's differentiable
but its derivative's
00:11:28.880 --> 00:11:29.671
not supposed to be.
00:11:29.671 --> 00:11:31.110
It is not continuous, indeed.
00:11:31.110 --> 00:11:33.750
So this is continuous.
00:11:33.750 --> 00:11:36.670
So we've learned something
nice, that unless our potential
00:11:36.670 --> 00:11:39.100
is so stupid as to have
delta functions, which
00:11:39.100 --> 00:11:41.690
sounds fairly unphysical--
00:11:41.690 --> 00:11:44.650
We'll come back to that
later in today's lecture.
00:11:44.650 --> 00:11:46.900
--unless our potential
has delta functions,
00:11:46.900 --> 00:11:50.473
the wave function and its first
derivative must be smooth.
00:11:50.473 --> 00:11:51.160
Yeah.
00:11:51.160 --> 00:11:53.590
This is just from the
energy eigenvalue equation.
00:11:53.590 --> 00:11:56.589
Now, we actually argued this
from the well definedness
00:11:56.589 --> 00:11:58.130
and the finiteness
of the expectation
00:11:58.130 --> 00:12:00.800
value of the momentum
earlier in the semester.
00:12:00.800 --> 00:12:02.992
But I wanted to give
you this argument for it
00:12:02.992 --> 00:12:04.700
because it's going to
play a useful role.
00:12:04.700 --> 00:12:08.040
And it also tells us that if
we do have a delta function
00:12:08.040 --> 00:12:10.510
singularity in the
potential, then the upshot
00:12:10.510 --> 00:12:13.234
is that the wave function
is going to be continuous,
00:12:13.234 --> 00:12:14.650
but its first
derivative will not.
00:12:14.650 --> 00:12:17.935
Its first derivative will
jump at the wave function.
00:12:17.935 --> 00:12:19.560
And that means that
the wave function--
00:12:19.560 --> 00:12:21.250
Let me draw this
slightly differently.
00:12:21.250 --> 00:12:23.790
--the wave function
will have a kink.
00:12:23.790 --> 00:12:26.080
Its derivative will
not be continuous.
00:12:26.080 --> 00:12:26.580
OK.
00:12:26.580 --> 00:12:28.330
So anywhere where we
have a delta function
00:12:28.330 --> 00:12:31.760
in the potential, we will have
a kink in the wave function
00:12:31.760 --> 00:12:35.630
where the first derivative
is discontinuous.
00:12:35.630 --> 00:12:36.470
Cool?
00:12:36.470 --> 00:12:37.410
Yeah.
00:12:37.410 --> 00:12:39.022
AUDIENCE: What does
that mean as far
00:12:39.022 --> 00:12:41.017
as the expectation
value of the momentum?
00:12:41.017 --> 00:12:42.850
PROFESSOR: Ah, that's
an excellent question.
00:12:42.850 --> 00:12:46.582
So what do you expect to
happen at such a point?
00:12:46.582 --> 00:12:48.470
AUDIENCE: Well, your
momentum blows up.
00:12:48.470 --> 00:12:49.511
PROFESSOR: Yeah, exactly.
00:12:49.511 --> 00:12:51.520
So we're going to
have some pathologies
00:12:51.520 --> 00:12:53.569
with expectation
values in the momentum.
00:12:53.569 --> 00:12:56.110
Let's come back to that when we
talk about the delta function
00:12:56.110 --> 00:12:57.360
potential, which should
be at the end of today.
00:12:57.360 --> 00:12:58.480
Hold that question in
the back of your head.
00:12:58.480 --> 00:12:59.580
It's a good question.
00:12:59.580 --> 00:13:02.430
Other questions?
00:13:02.430 --> 00:13:03.020
OK.
00:13:03.020 --> 00:13:07.426
So what we're going to do now
is we're going to use, you know,
00:13:07.426 --> 00:13:10.960
math to find the
energy eigenfunctions
00:13:10.960 --> 00:13:13.766
and eigenvalues for the
finite well potential, i.e.
00:13:13.766 --> 00:13:15.640
we're going to solve
this equation contingent
00:13:15.640 --> 00:13:17.330
on the boundary
conditions that the wave
00:13:17.330 --> 00:13:19.480
function and its derivative
are smooth everywhere.
00:13:19.480 --> 00:13:22.389
And in particular, they must
be smooth here and here.
00:13:22.389 --> 00:13:24.180
I mean they've got to
be smooth everywhere.
00:13:24.180 --> 00:13:25.790
We know what the
solution is inside here.
00:13:25.790 --> 00:13:27.331
We know what the
solution is in here.
00:13:27.331 --> 00:13:30.180
All we have to worry about is
what happens at the interface.
00:13:30.180 --> 00:13:32.830
And we're going to use
smoothness of the wave function
00:13:32.830 --> 00:13:35.510
and its derivative to
impose conditions that
00:13:35.510 --> 00:13:37.950
allow us to match
across that step.
00:13:37.950 --> 00:13:40.710
Everyone cool with that?
00:13:40.710 --> 00:13:42.540
OK.
00:13:42.540 --> 00:13:45.320
So let's do it.
00:13:45.320 --> 00:13:46.812
By the way, just
a quick side note.
00:13:46.812 --> 00:13:48.020
Let me give you a definition.
00:13:48.020 --> 00:13:49.090
I've used this
phrase many times,
00:13:49.090 --> 00:13:50.881
but I haven't given
you a definition of it.
00:13:50.881 --> 00:13:53.260
So I've used the
phrase, bound state.
00:13:53.260 --> 00:13:56.560
And its opposite is
called a scattering state.
00:13:56.560 --> 00:13:58.204
So here's what I mean
by a bound state.
00:13:58.204 --> 00:13:59.620
Intuitively, a
bound state, if you
00:13:59.620 --> 00:14:01.161
think about this
classically, imagine
00:14:01.161 --> 00:14:02.220
I have a potential well.
00:14:02.220 --> 00:14:04.910
And I have a marble,
and I let go from here.
00:14:04.910 --> 00:14:06.220
This marble is bound.
00:14:06.220 --> 00:14:06.720
Right?
00:14:06.720 --> 00:14:08.178
It's never getting
out of the well.
00:14:08.178 --> 00:14:09.073
It's stuck.
00:14:09.073 --> 00:14:10.040
Yeah?
00:14:10.040 --> 00:14:13.130
And so I would call
that a bound marble.
00:14:13.130 --> 00:14:17.180
On the other hand, a marble
that I give a huge kick to,
00:14:17.180 --> 00:14:19.240
big velocity so
that it can get out,
00:14:19.240 --> 00:14:21.400
well then it's not bound
to this potential well.
00:14:21.400 --> 00:14:23.910
So I'll call that a scattering
state just to give it a name.
00:14:23.910 --> 00:14:26.760
And next lecture we'll see why
we call it a scattering state.
00:14:26.760 --> 00:14:30.032
The important thing is
that bound configurations
00:14:30.032 --> 00:14:31.490
of a classical
potential are things
00:14:31.490 --> 00:14:33.870
basically in a well
that are stuck.
00:14:33.870 --> 00:14:34.370
OK.
00:14:34.370 --> 00:14:36.786
So the quantum version of that
statement is the following.
00:14:36.786 --> 00:14:39.142
Suppose I take this
potential, and I
00:14:39.142 --> 00:14:41.350
treat it quantum mechanically,
and I consider a state
00:14:41.350 --> 00:14:43.780
with total energy like this.
00:14:43.780 --> 00:14:45.760
Well, among other
things, the total energy
00:14:45.760 --> 00:14:49.290
is less than the value of
the potential asymptotically
00:14:49.290 --> 00:14:49.847
far away.
00:14:49.847 --> 00:14:51.430
So what is the form
of a wave function
00:14:51.430 --> 00:14:54.529
in this region with this energy?
00:14:54.529 --> 00:14:55.320
Exponential, right?
00:14:55.320 --> 00:14:57.770
It's e to the
minus some alpha x,
00:14:57.770 --> 00:15:01.440
where alpha squared is roughly
the difference, alpha x.
00:15:01.440 --> 00:15:05.009
And out here it's going to
be e to the plus alpha x.
00:15:05.009 --> 00:15:06.300
And that's for normalizability.
00:15:06.300 --> 00:15:09.540
We want to have a single
particle in this state.
00:15:09.540 --> 00:15:14.220
So what that tells us is that
the wave function falls off
00:15:14.220 --> 00:15:19.550
in these classically disallowed
regions exponentially.
00:15:19.550 --> 00:15:22.510
And so the probability
of measuring the particle
00:15:22.510 --> 00:15:27.360
at an arbitrarily far
position goes to zero.
00:15:27.360 --> 00:15:29.640
And it goes to zero
exponentially rapidly.
00:15:29.640 --> 00:15:30.140
Cool?
00:15:30.140 --> 00:15:34.030
So I will call a bound state, a
quantum bound state, an energy
00:15:34.030 --> 00:15:38.300
eigenstate, such that
the probability falls off
00:15:38.300 --> 00:15:42.252
exponentially as we go far
away from wherever we think
00:15:42.252 --> 00:15:44.460
is an interesting point,
like the bottom of the well.
00:15:44.460 --> 00:15:45.340
Cool?
00:15:45.340 --> 00:15:46.960
A bound state is
just a state which
00:15:46.960 --> 00:15:48.720
is exponentially localized.
00:15:48.720 --> 00:15:51.760
If you put it there,
it will stay there.
00:15:51.760 --> 00:15:52.680
Yeah.
00:15:52.680 --> 00:15:54.680
And it's important that
when I say a bound state
00:15:54.680 --> 00:15:57.810
I'm talking about
energy eigenstates.
00:15:57.810 --> 00:15:59.455
And the reason is this.
00:15:59.455 --> 00:16:01.320
Bound state equals
energy eigenstate.
00:16:04.890 --> 00:16:07.630
The reason is that,
consider by contrast
00:16:07.630 --> 00:16:10.610
a free particle,
so a free particle
00:16:10.610 --> 00:16:12.230
with constant potential.
00:16:12.230 --> 00:16:13.790
What are the wave functions?
00:16:13.790 --> 00:16:15.340
What are the energy
eigenfunctions?
00:16:15.340 --> 00:16:16.670
Well, they're plane waves.
00:16:16.670 --> 00:16:17.170
Right.
00:16:17.170 --> 00:16:19.860
So are these bound states?
00:16:19.860 --> 00:16:20.360
No.
00:16:20.360 --> 00:16:20.700
Good.
00:16:20.700 --> 00:16:21.000
OK.
00:16:21.000 --> 00:16:22.458
On the other hand,
I claimed that I
00:16:22.458 --> 00:16:27.180
can build a wave packet, a
perfectly reasonable wave
00:16:27.180 --> 00:16:30.600
packet, which is a Gaussian.
00:16:30.600 --> 00:16:31.780
OK.
00:16:31.780 --> 00:16:35.940
This is some wave
function, psi of x times 0.
00:16:35.940 --> 00:16:36.630
It's a Gaussian.
00:16:36.630 --> 00:16:37.630
It's nice and narrow.
00:16:37.630 --> 00:16:41.350
Is that a bound state?
00:16:41.350 --> 00:16:45.550
Well, it's localized
at this moment in time.
00:16:45.550 --> 00:16:47.880
But will it remain, and in
particular its probability
00:16:47.880 --> 00:16:49.470
distribution, which
is this thing norm
00:16:49.470 --> 00:16:51.980
squared is localized in space--
00:16:51.980 --> 00:16:52.480
Sorry.
00:16:52.480 --> 00:16:55.237
This was zero.
00:16:55.237 --> 00:16:56.320
--it's localized in space.
00:16:56.320 --> 00:16:57.270
The probability of
it being out here
00:16:57.270 --> 00:16:58.527
is not just
exponentially small, it's
00:16:58.527 --> 00:17:00.277
Gaussian so it's e to
the minus x squared.
00:17:00.277 --> 00:17:02.560
It's really not out here.
00:17:02.560 --> 00:17:04.329
But what happens if I let go?
00:17:04.329 --> 00:17:08.019
What happens if I look
at the system at time t?
00:17:08.019 --> 00:17:09.060
It's going to spread out.
00:17:09.060 --> 00:17:09.650
It's going to disperse.
00:17:09.650 --> 00:17:11.816
We're going to talk about
that in more detail later.
00:17:11.816 --> 00:17:13.339
So it's going to spread out.
00:17:13.339 --> 00:17:18.030
And eventually, it will get
out arbitrarily far away
00:17:18.030 --> 00:17:19.670
with whatever probably you like.
00:17:19.670 --> 00:17:23.538
So the probability distribution
is not time invariant.
00:17:23.538 --> 00:17:25.329
That is to say it's
not a stationary state.
00:17:25.329 --> 00:17:26.780
It's not an energy eigenstate.
00:17:26.780 --> 00:17:31.140
Saying something is bound
means that it never gets away.
00:17:31.140 --> 00:17:34.071
So bound states are
specifically energy eigenstates
00:17:34.071 --> 00:17:35.820
that are strictly
localized, that fall off
00:17:35.820 --> 00:17:38.870
at least exponentially as
we go away from the origin.
00:17:38.870 --> 00:17:39.780
Cool?
00:17:39.780 --> 00:17:43.110
It's just terminology, what
I mean by a bound state.
00:17:43.110 --> 00:17:44.760
Questions?
00:17:44.760 --> 00:17:45.280
OK.
00:17:45.280 --> 00:17:46.840
So let's talk about
the finite well.
00:17:50.680 --> 00:17:56.690
So I need to give you
definitions of the parameters.
00:17:56.690 --> 00:17:58.520
Let's draw this more precisely.
00:17:58.520 --> 00:17:59.850
Here's my well.
00:17:59.850 --> 00:18:02.920
Asymptotically, the
potential is zero.
00:18:02.920 --> 00:18:07.010
The potential depth, I'm
going to call minus v naught.
00:18:07.010 --> 00:18:07.630
OK.
00:18:07.630 --> 00:18:10.060
And I'm going to center
the well around zero.
00:18:10.060 --> 00:18:12.330
And I'll call the
sides minus l and l.
00:18:18.590 --> 00:18:21.020
And I want to find bound
states of this potential,
00:18:21.020 --> 00:18:22.880
just like we found bound states
of the harmonic oscillator,
00:18:22.880 --> 00:18:23.400
i.e.
00:18:23.400 --> 00:18:27.780
states with energy e,
which is less than zero.
00:18:27.780 --> 00:18:30.640
So these are going to give
us bound states because we're
00:18:30.640 --> 00:18:32.910
going to have exponential
fall offs far away.
00:18:37.830 --> 00:18:39.120
So a couple of things to note.
00:18:39.120 --> 00:18:40.494
The first is on,
I think, problem
00:18:40.494 --> 00:18:43.080
set three or four you
showed that if you
00:18:43.080 --> 00:18:45.540
have a potential,
which is symmetric,
00:18:45.540 --> 00:18:48.935
which is even, under x goes
to minus x, then every energy
00:18:48.935 --> 00:18:51.060
eigenfunction, or at least
every bound state energy
00:18:51.060 --> 00:18:53.890
eigenfunction, every energy
eigenfunction can be written
00:18:53.890 --> 00:19:02.450
as phi e symmetric or
phi e anti-symmetric.
00:19:02.450 --> 00:19:03.940
So it's either even or odd.
00:19:06.880 --> 00:19:10.122
It's either even or odd under
the exchange of x to minus x.
00:19:10.122 --> 00:19:11.580
So when our potential
is symmetric,
00:19:11.580 --> 00:19:13.590
the wave function or the
energy eigenfunctions
00:19:13.590 --> 00:19:16.410
are either symmetric
or anti-symmetric.
00:19:16.410 --> 00:19:17.580
OK.
00:19:17.580 --> 00:19:19.880
So we want to solve for
the actual eigenfunctions.
00:19:19.880 --> 00:19:21.745
So we want to solve
that equation.
00:19:25.930 --> 00:19:28.300
And we have this
nice simple fact
00:19:28.300 --> 00:19:30.440
that we know the
solutions in this region.
00:19:30.440 --> 00:19:31.820
We know the general
solution in this region.
00:19:31.820 --> 00:19:33.653
We know the general
solution in this region.
00:19:33.653 --> 00:19:38.189
So I'm going to call these
regions one, two, and three,
00:19:38.189 --> 00:19:39.230
just to give them a name.
00:19:39.230 --> 00:19:40.290
So in region one--
00:19:43.430 --> 00:19:45.290
That's actually sort of stupid.
00:19:45.290 --> 00:19:49.304
Let's call this inside,
left, and right.
00:19:49.304 --> 00:19:49.804
Good.
00:19:54.730 --> 00:19:55.230
OK.
00:19:55.230 --> 00:20:02.730
So let's look at this equation.
00:20:06.610 --> 00:20:08.170
We have two cases.
00:20:08.170 --> 00:20:13.220
If the energy is greater than
the potential in some region,
00:20:13.220 --> 00:20:17.020
then this is of the
form phi prime prime
00:20:17.020 --> 00:20:19.050
is equal to energy
greater than potential.
00:20:19.050 --> 00:20:20.635
This is a negative number.
00:20:24.050 --> 00:20:29.136
And so this is a
minus k squared phi.
00:20:29.136 --> 00:20:31.750
And we get exponentials.
00:20:31.750 --> 00:20:38.450
And if, on the other hand,
e is less than v of x, then
00:20:38.450 --> 00:20:45.530
phi prime prime is equal
to plus alpha squared phi.
00:20:45.530 --> 00:20:46.570
I should say oscillator.
00:20:54.240 --> 00:21:00.120
And in particular here, I want
the k squared is equal to 2m
00:21:00.120 --> 00:21:03.130
upon h bar squared.
00:21:03.130 --> 00:21:07.860
It's just the
coefficient v minus e.
00:21:07.860 --> 00:21:17.330
And alpha squared is equal to
2m over h bar squared e minus v.
00:21:17.330 --> 00:21:17.830
OK.
00:21:17.830 --> 00:21:19.510
So let's apply that here.
00:21:19.510 --> 00:21:22.294
So in this region, we're
going to get oscillations
00:21:22.294 --> 00:21:24.460
because we're in a classically
allowed region, where
00:21:24.460 --> 00:21:26.168
the energy is greater
than the potential.
00:21:26.168 --> 00:21:27.870
So we'll get
oscillatory solutions.
00:21:27.870 --> 00:21:31.390
And the salient value
of k inside, is k
00:21:31.390 --> 00:21:39.120
is equal to the square root
of 2m over h bar squared times
00:21:39.120 --> 00:21:40.940
v minus e.
00:21:40.940 --> 00:21:42.255
So that's minus v naught.
00:21:48.920 --> 00:21:51.560
What did I do?
00:21:51.560 --> 00:21:53.452
I did.
00:21:53.452 --> 00:21:54.670
It's e minus v. I thank you.
00:21:54.670 --> 00:21:55.170
Yes.
00:21:59.260 --> 00:22:03.170
And I want the other one to
be l so it'd be minus v. Good.
00:22:03.170 --> 00:22:03.950
Good.
00:22:03.950 --> 00:22:04.690
Excellent.
00:22:04.690 --> 00:22:06.326
So root 2 over h bar squared.
00:22:06.326 --> 00:22:07.950
And now we have e
minus v naught, which
00:22:07.950 --> 00:22:10.520
is the actual value of
e, which is negative.
00:22:10.520 --> 00:22:11.310
Right?
00:22:11.310 --> 00:22:15.960
But minus v naught
or plus v naught
00:22:15.960 --> 00:22:21.640
is positive and
greater in magnitude.
00:22:21.640 --> 00:22:23.350
So this is a nice
positive number,
00:22:23.350 --> 00:22:25.210
and k is the square root of it.
00:22:25.210 --> 00:22:27.520
This is controlling
how rapidly the wave
00:22:27.520 --> 00:22:30.720
function oscillates
in this region.
00:22:30.720 --> 00:22:33.370
Similarly, out here we
have alpha is equal to--
00:22:33.370 --> 00:22:35.020
Well, here the
potential is zero.
00:22:35.020 --> 00:22:36.310
So it's particularly easy.
00:22:36.310 --> 00:22:38.896
Alpha is equal to
the square root of 2m
00:22:38.896 --> 00:22:41.530
upon h bar squared of--
00:22:41.530 --> 00:22:45.850
Now z minus e is zero minus
e, which is a negative number.
00:22:45.850 --> 00:22:49.950
So we can just write
e absolute value.
00:22:55.770 --> 00:23:03.450
So we can write the general
solution of this eigenfunction,
00:23:03.450 --> 00:23:11.941
of this eigenvalue equation,
as phi e of x is equal to--
00:23:11.941 --> 00:23:13.690
Let's break it up into
inside and outside.
00:23:13.690 --> 00:23:19.340
Well, inside we know, since it's
constant with this value of k,
00:23:19.340 --> 00:23:23.055
we get superpositions of
oscillatory solutions.
00:23:23.055 --> 00:23:24.680
It' a second-order
difference equation.
00:23:24.680 --> 00:23:26.929
There are two solutions and
two integration constants.
00:23:26.929 --> 00:23:34.065
So first we have a cosine
of kx plus b sine kx.
00:23:37.172 --> 00:23:38.610
This is inside.
00:23:44.340 --> 00:23:52.520
And then on the left
we have a combination
00:23:52.520 --> 00:23:54.770
of exponentially growing and
exponentially decreasing.
00:23:54.770 --> 00:23:57.630
So the exponentially
growing is e to the alpha x
00:23:57.630 --> 00:24:02.180
plus de the minus alpha x.
00:24:02.180 --> 00:24:05.490
And on the right,
similarly by symmetry,
00:24:05.490 --> 00:24:07.088
we have some combination of e--
00:24:07.088 --> 00:24:09.046
But I don't want to call
it the energy, so I'll
00:24:09.046 --> 00:24:10.080
call it the curly e.
00:24:15.005 --> 00:24:22.240
--to the alpha x plus
fe to the minus alpha x.
00:24:22.240 --> 00:24:23.240
OK.
00:24:23.240 --> 00:24:24.630
So that's the general solution.
00:24:24.630 --> 00:24:27.770
We solve the problem as a
superposition of the two
00:24:27.770 --> 00:24:29.820
oscillatory solutions
or a superposition
00:24:29.820 --> 00:24:32.590
of the two exponentially
growing and damped solutions
00:24:32.590 --> 00:24:35.370
or exponentially
growing and collapsing
00:24:35.370 --> 00:24:37.090
functions on the left and right.
00:24:42.620 --> 00:24:44.400
Questions?
00:24:44.400 --> 00:24:46.530
So a couple of things
to note at this point.
00:24:49.500 --> 00:24:53.470
So the first is we have
boundary conditions to impose.
00:24:53.470 --> 00:24:56.730
We have boundary conditions
at these two interfaces.
00:24:56.730 --> 00:24:59.610
But we also have boundary
conditions off in infinity.
00:24:59.610 --> 00:25:02.470
What are the boundary
conditions at infinity?
00:25:02.470 --> 00:25:03.292
Yeah, exactly.
00:25:03.292 --> 00:25:04.000
It should vanish.
00:25:04.000 --> 00:25:05.708
So we want the system
to be normalizable.
00:25:11.610 --> 00:25:13.740
So normalizable is
going to say that phi
00:25:13.740 --> 00:25:18.320
goes to zero at minus infinity.
00:25:18.320 --> 00:25:25.930
Phi of x goes to minus infinity
goes to zero, which it equals.
00:25:25.930 --> 00:25:31.260
And phi of x goes to plus
infinity should also be zero.
00:25:31.260 --> 00:25:32.110
OK.
00:25:32.110 --> 00:25:33.484
And then we're
also going to have
00:25:33.484 --> 00:25:37.454
the conditions at
the left boundary,
00:25:37.454 --> 00:25:38.870
and we're going
to have conditions
00:25:38.870 --> 00:25:39.786
at the right boundary.
00:25:45.963 --> 00:25:50.382
[LAUGHTER]
00:25:50.382 --> 00:25:52.619
All right.
00:25:52.619 --> 00:25:54.910
So what are the boundary
condition at the left boundary
00:25:54.910 --> 00:25:56.840
condition?
00:25:56.840 --> 00:25:58.830
So first off, what are
the boundary conditions
00:25:58.830 --> 00:26:00.955
we want to impose at the
left and right boundaries?
00:26:00.955 --> 00:26:02.200
AUDIENCE: Continuous.
00:26:02.200 --> 00:26:03.360
PROFESSOR: Continuous,
and the derivative
00:26:03.360 --> 00:26:04.000
should be continuous.
00:26:04.000 --> 00:26:04.499
Exactly.
00:26:04.499 --> 00:26:08.060
So we have that phi is
continuous, and phi prime--
00:26:08.060 --> 00:26:09.110
Good god.
00:26:09.110 --> 00:26:10.960
--phi prime is continuous.
00:26:10.960 --> 00:26:14.775
Similarly, phi continuous,
phi prime continuous.
00:26:18.040 --> 00:26:20.690
OK.
00:26:20.690 --> 00:26:23.880
Do we have enough
boundary conditions
00:26:23.880 --> 00:26:27.360
to specify our function?
00:26:27.360 --> 00:26:28.910
So we have now for
our solution, we
00:26:28.910 --> 00:26:31.700
have six undetermined
coefficients.
00:26:31.700 --> 00:26:33.390
And we have six
boundary conditions.
00:26:33.390 --> 00:26:34.549
So that looks good.
00:26:34.549 --> 00:26:35.590
Are they all independent?
00:26:38.204 --> 00:26:38.870
Ponder that one.
00:26:43.300 --> 00:26:45.960
So in particular, let's
start with the normalizable.
00:26:45.960 --> 00:26:49.400
So in order for phi to go to
zero at minus infinity deep
00:26:49.400 --> 00:26:53.631
out on the left,
what should be true?
00:26:53.631 --> 00:26:54.715
Yeah, d goes to zero.
00:26:57.390 --> 00:27:00.200
Oops, equals zero.
00:27:00.200 --> 00:27:03.040
And on the right?
00:27:03.040 --> 00:27:05.800
Yeah, that curly
e equals 0, which
00:27:05.800 --> 00:27:08.860
is nice so I don't ever
have to write it again.
00:27:08.860 --> 00:27:09.690
So that's zero.
00:27:09.690 --> 00:27:10.440
And that's zero.
00:27:10.440 --> 00:27:10.940
OK.
00:27:10.940 --> 00:27:11.470
That's good.
00:27:16.280 --> 00:27:18.754
We can take advantage
though of something nice.
00:27:18.754 --> 00:27:20.170
We know that the
wave function has
00:27:20.170 --> 00:27:22.240
to be either symmetric
or anti-symmetric.
00:27:22.240 --> 00:27:22.920
Right?
00:27:22.920 --> 00:27:24.840
So we can exploit
that and say, look,
00:27:24.840 --> 00:27:28.420
the wave function is going to
be different from our boundary
00:27:28.420 --> 00:27:30.827
conditions, but
it's a true fact,
00:27:30.827 --> 00:27:32.160
and we can take advantage of it.
00:27:32.160 --> 00:27:33.590
We can use the
parody of the well.
00:27:36.610 --> 00:27:38.090
I can never--
00:27:38.090 --> 00:27:40.450
So we can use the
parody of the potential
00:27:40.450 --> 00:27:47.380
to say that the system is either
symmetric or anti-symmetric.
00:27:47.380 --> 00:27:51.669
And these are often
said as even or odd
00:27:51.669 --> 00:27:53.460
because the function
will be, as a function
00:27:53.460 --> 00:27:54.501
of x, either even or odd.
00:27:54.501 --> 00:27:57.150
You either pick up a
minus sign or a plus sign
00:27:57.150 --> 00:28:00.109
under taking x to minus x.
00:28:00.109 --> 00:28:01.650
So if the system is
even, what can we
00:28:01.650 --> 00:28:02.858
say about these coefficients?
00:28:08.110 --> 00:28:10.157
What must be true
of b, for example?
00:28:10.157 --> 00:28:10.990
AUDIENCE: It's zero.
00:28:10.990 --> 00:28:11.656
PROFESSOR: Yeah.
00:28:11.656 --> 00:28:13.130
So b equals 0.
00:28:13.130 --> 00:28:14.534
And what else?
00:28:14.534 --> 00:28:15.450
AUDIENCE: It equals f.
00:28:15.450 --> 00:28:16.220
PROFESSOR: This equals f.
00:28:16.220 --> 00:28:16.540
Yeah, good.
00:28:16.540 --> 00:28:17.040
OK.
00:28:17.040 --> 00:28:17.750
This equals f.
00:28:17.750 --> 00:28:19.310
And if the system
is anti-symmetric,
00:28:19.310 --> 00:28:21.440
then a equal to 0.
00:28:21.440 --> 00:28:24.845
And we see that c
is equal to minus f.
00:28:24.845 --> 00:28:25.345
Yeah?
00:28:28.269 --> 00:28:29.685
So that's a useful
simplification.
00:28:32.580 --> 00:28:35.852
So it's easy to see that we
could do this either way.
00:28:35.852 --> 00:28:37.810
We could do either
symmetric or anti-symmetric.
00:28:37.810 --> 00:28:39.476
I'm going to, for
simplicity in lecture,
00:28:39.476 --> 00:28:41.050
focus on the even case.
00:28:41.050 --> 00:28:45.010
b is equal to 0, and
c is equal to minus f.
00:28:45.010 --> 00:28:46.650
Sorry, c is equal to f.
00:28:46.650 --> 00:28:47.757
So plus c.
00:28:50.310 --> 00:28:55.670
So now we're
specifically working
00:28:55.670 --> 00:28:58.800
with the even solutions.
00:28:58.800 --> 00:29:01.179
And on your problem set,
you'll repeat this calculation
00:29:01.179 --> 00:29:02.095
for the odd functions.
00:29:05.420 --> 00:29:08.379
So we're going to focus
on the even solutions.
00:29:08.379 --> 00:29:09.920
And now what we have
to do is we have
00:29:09.920 --> 00:29:14.560
to impose the boundary
conditions for phi and phi
00:29:14.560 --> 00:29:16.680
prime.
00:29:16.680 --> 00:29:17.750
So that's easy enough.
00:29:23.010 --> 00:29:24.525
We have the function.
00:29:24.525 --> 00:29:28.995
All we have to do is impose
that the values are the same.
00:29:28.995 --> 00:29:31.495
So for example, let's focus on
the left boundary conditions.
00:29:36.250 --> 00:29:36.750
Sorry.
00:29:36.750 --> 00:29:38.166
Let's focus on the
right because I
00:29:38.166 --> 00:29:40.280
don't want to deal
with that minus sign.
00:29:40.280 --> 00:29:42.280
So let's focus on the
right boundary conditions.
00:29:42.280 --> 00:29:46.240
So this is x is equal to plus l.
00:29:46.240 --> 00:29:48.900
So when x is equal to
plus l, what must be true?
00:29:48.900 --> 00:29:51.150
Phi and phi prime
must be continuous.
00:29:51.150 --> 00:29:51.820
So what's phi?
00:29:51.820 --> 00:29:53.520
So phi is equal to--
00:29:56.540 --> 00:30:00.660
Well, from inside,
it's a cosine of kl.
00:30:05.310 --> 00:30:06.710
Yeah.
00:30:06.710 --> 00:30:09.690
And b is gone because we're only
looking at the even functions.
00:30:09.690 --> 00:30:15.060
On the right, however, it's
equal to c e to the minus alpha
00:30:15.060 --> 00:30:18.170
l because we're evaluating
at the right boundary.
00:30:18.170 --> 00:30:19.320
Yeah?
00:30:19.320 --> 00:30:19.820
OK.
00:30:19.820 --> 00:30:20.486
So this is cool.
00:30:20.486 --> 00:30:22.720
It allows us to determine
c in terms of a.
00:30:22.720 --> 00:30:25.100
And if we solve that
equation for c in terms of a,
00:30:25.100 --> 00:30:29.150
we'll get an
eigenfunction, phi even,
00:30:29.150 --> 00:30:32.760
with one overall
normalization coefficient, a.
00:30:32.760 --> 00:30:34.260
And then we can fix
that to whatever
00:30:34.260 --> 00:30:36.550
it has to be so that
everything integrates to one.
00:30:36.550 --> 00:30:37.630
Yeah?
00:30:37.630 --> 00:30:38.980
So that seems fine.
00:30:38.980 --> 00:30:42.076
It seems like we can
solve for c in terms of a.
00:30:42.076 --> 00:30:43.030
c is equal to--
00:30:43.030 --> 00:30:46.355
This is weight.
00:30:46.355 --> 00:30:47.030
Well, OK.
00:30:47.030 --> 00:30:53.890
So c is equal to a cosine
e to the plus alpha l.
00:30:53.890 --> 00:30:57.002
On the other hand, we also have
a condition on the derivative.
00:30:57.002 --> 00:30:58.460
And the condition
on the derivative
00:30:58.460 --> 00:31:02.246
is that phi prime is continuous.
00:31:02.246 --> 00:31:04.120
And the derivative of
this, well that's easy.
00:31:04.120 --> 00:31:05.790
It's the derivative of cosine.
00:31:05.790 --> 00:31:08.880
So this is going
to be minus sine.
00:31:08.880 --> 00:31:10.920
But we pull out a factor
of k, because we're
00:31:10.920 --> 00:31:12.420
taking derivative
with respect to x.
00:31:12.420 --> 00:31:14.470
Minus k is sine of kx.
00:31:14.470 --> 00:31:18.040
Evaluate it out, sine of kl, l.
00:31:18.040 --> 00:31:20.930
And this is going
to be equal to minus
00:31:20.930 --> 00:31:24.706
alpha c e to the minus alpha l.
00:31:24.706 --> 00:31:26.159
AUDIENCE: You forgot an a.
00:31:26.159 --> 00:31:26.950
PROFESSOR: Oh, yes.
00:31:26.950 --> 00:31:30.060
There should be an a.
00:31:30.060 --> 00:31:32.150
Thank you.
00:31:32.150 --> 00:31:32.789
OK.
00:31:32.789 --> 00:31:34.580
So but now we've got
a problem because this
00:31:34.580 --> 00:31:43.190
says that c is equal to minus
a times k over alpha sine
00:31:43.190 --> 00:31:47.918
of kl times e to the alpha l.
00:31:47.918 --> 00:31:51.670
And that's bad
because c can't be
00:31:51.670 --> 00:31:55.350
equal to two different
numbers at the same time.
00:31:55.350 --> 00:31:57.950
There's a certain monogamy
of mathematical equations.
00:31:57.950 --> 00:31:59.400
It just doesn't work.
00:31:59.400 --> 00:32:00.972
So how do we deal with this?
00:32:00.972 --> 00:32:03.055
Well, let's think about
what these equations would
00:32:03.055 --> 00:32:03.370
have meant.
00:32:03.370 --> 00:32:05.110
Forget this one for
the moment, and just
00:32:05.110 --> 00:32:06.610
focus on that first expression.
00:32:06.610 --> 00:32:08.240
I'm going to rewrite
this slightly.
00:32:08.240 --> 00:32:11.812
a cosine of kl.
00:32:11.812 --> 00:32:13.120
OK.
00:32:13.120 --> 00:32:14.610
So what does this
expression say?
00:32:14.610 --> 00:32:16.320
Well, it seems like
it's just saying
00:32:16.320 --> 00:32:20.100
if we fix c to be equal to this
value, for fixed value of kappa
00:32:20.100 --> 00:32:22.810
l and alpha, then
there's a solution.
00:32:22.810 --> 00:32:25.250
However, what is k?
00:32:25.250 --> 00:32:26.715
What are k and alpha.
00:32:26.715 --> 00:32:30.420
k and alpha are
functions of the energy.
00:32:30.420 --> 00:32:32.150
So it would seem from
this point of view,
00:32:32.150 --> 00:32:34.830
like for any value
of the energy,
00:32:34.830 --> 00:32:38.686
we get a solution
to this equation.
00:32:38.686 --> 00:32:40.150
Everyone see that?
00:32:40.150 --> 00:32:42.010
But we know that can't
possibly be right
00:32:42.010 --> 00:32:45.110
because we expect the
solutions to be discrete.
00:32:45.110 --> 00:32:47.410
We don't expect
any value of energy
00:32:47.410 --> 00:32:50.610
to lead to a solution of the
energy eigenvalue equation.
00:32:50.610 --> 00:32:52.610
There should be only
discrete set of energies.
00:32:52.610 --> 00:32:53.003
Yeah?
00:32:53.003 --> 00:32:54.669
AUDIENCE: Did you
pick up an extra minus
00:32:54.669 --> 00:32:56.951
sign in the expression for c?
00:32:56.951 --> 00:32:57.700
PROFESSOR: You do.
00:32:57.700 --> 00:32:58.200
Thank you.
00:32:58.200 --> 00:32:59.050
The sign's cancel.
00:32:59.050 --> 00:32:59.737
Yes, excellent.
00:32:59.737 --> 00:33:01.570
I've never written this
equation in my life.
00:33:01.570 --> 00:33:03.530
So thank you.
00:33:03.530 --> 00:33:04.590
Yes, extra minus sign.
00:33:04.590 --> 00:33:05.740
So what's going on here?
00:33:05.740 --> 00:33:08.510
Well, what we see is
that we've written down
00:33:08.510 --> 00:33:09.970
the general form
of the solution.
00:33:09.970 --> 00:33:12.428
Here were imposing that we've
already imposed the condition
00:33:12.428 --> 00:33:14.210
that we're normalizable
at infinity.
00:33:14.210 --> 00:33:16.250
Here, we're imposing
the continuity condition
00:33:16.250 --> 00:33:19.400
on the right.
00:33:19.400 --> 00:33:21.690
And if we impose just
the continuity condition
00:33:21.690 --> 00:33:25.800
for the wave function,
we can find a solution.
00:33:25.800 --> 00:33:28.420
Similarly, if we impose only
the continuity condition
00:33:28.420 --> 00:33:30.170
for the derivative,
we can find a solution
00:33:30.170 --> 00:33:31.860
for arbitrary values
of the energy.
00:33:31.860 --> 00:33:33.900
But in order to find
a solution where
00:33:33.900 --> 00:33:39.390
the wave function and its
derivative are both continuous,
00:33:39.390 --> 00:33:43.059
it can't be true that the energy
takes just any value because it
00:33:43.059 --> 00:33:45.100
would tell you that c
takes two different values.
00:33:45.100 --> 00:33:45.720
Right?
00:33:45.720 --> 00:33:47.260
So there's a
consistency condition.
00:33:47.260 --> 00:33:50.410
For what values of energy or
equivalently, for what values
00:33:50.410 --> 00:33:54.090
of k and alpha are these
two expressions equal
00:33:54.090 --> 00:33:56.850
to the same thing?
00:33:56.850 --> 00:33:58.214
Cool?
00:33:58.214 --> 00:33:59.630
So we can get that
by saying, look
00:33:59.630 --> 00:34:02.050
we want both of these
equations to be true.
00:34:02.050 --> 00:34:03.720
And this is easy.
00:34:03.720 --> 00:34:07.510
I can take this equation and
divide it by this equation.
00:34:07.510 --> 00:34:09.030
And I will lose
my coefficients c.
00:34:09.030 --> 00:34:10.280
I will lose my coefficients a.
00:34:10.280 --> 00:34:10.960
What do we get?
00:34:10.960 --> 00:34:12.739
On the right hand side, we get--
00:34:12.739 --> 00:34:14.929
And I'm going to put a
minus sign on everything.
00:34:14.929 --> 00:34:18.560
So minus, minus, minus.
00:34:18.560 --> 00:34:21.310
So if we take this equation and
we divide it by this equation,
00:34:21.310 --> 00:34:24.040
on the right hand
side, we get alpha,
00:34:24.040 --> 00:34:26.170
because the c
exponential drops off.
00:34:26.170 --> 00:34:27.929
And on this side, we lose the a.
00:34:27.929 --> 00:34:29.830
We get a k.
00:34:29.830 --> 00:34:32.170
And then we get sine over
cosine of kl, also known
00:34:32.170 --> 00:34:34.210
as tangent of kl.
00:34:37.280 --> 00:34:38.080
Here we have a kl.
00:34:38.080 --> 00:34:39.564
Here we have a k.
00:34:39.564 --> 00:34:40.980
These are all
dimensionful things.
00:34:40.980 --> 00:34:43.752
Let's multiply
everything by an l.
00:34:43.752 --> 00:34:45.210
And this is nice
and dimensionless.
00:34:45.210 --> 00:34:46.650
Both sides are dimensionless.
00:34:46.650 --> 00:34:47.960
So we get this condition.
00:34:47.960 --> 00:34:49.989
This is the
consistency condition,
00:34:49.989 --> 00:34:52.530
such that both the wave
function and its derivative
00:34:52.530 --> 00:34:56.230
can be continuous at
the right boundary.
00:34:56.230 --> 00:34:58.650
OK?
00:34:58.650 --> 00:35:01.390
And this is a pretty
nontrivial condition.
00:35:01.390 --> 00:35:03.790
It says, given a value
of k, you can always
00:35:03.790 --> 00:35:06.290
determine the value of alpha,
such as this equation as true.
00:35:06.290 --> 00:35:08.300
But remember that k and
alpha are both known
00:35:08.300 --> 00:35:09.450
functions of the energy.
00:35:09.450 --> 00:35:12.440
So this is really an equation, a
complicated, nonlinear equation
00:35:12.440 --> 00:35:13.220
for the energy.
00:35:17.020 --> 00:35:23.180
So this is equal to a horrible
expression, a condition, badly
00:35:23.180 --> 00:35:25.450
nonlinear, in fact,
transcendental condition
00:35:25.450 --> 00:35:26.130
on the energy.
00:35:29.846 --> 00:35:30.970
And where's it coming from?
00:35:30.970 --> 00:35:34.860
It's coming from normalizability
and continuity everywhere.
00:35:39.280 --> 00:35:41.020
And a useful thing
to check, and I
00:35:41.020 --> 00:35:43.105
invite you to do
this on your own,
00:35:43.105 --> 00:35:51.020
is to check that the boundary
conditions at the left wall
00:35:51.020 --> 00:35:52.305
give the same expression.
00:35:55.870 --> 00:35:57.581
Yeah.
00:35:57.581 --> 00:36:00.443
AUDIENCE: For our final
form of that equation,
00:36:00.443 --> 00:36:04.259
is there a reason that we prefer
to multiply both sides by l
00:36:04.259 --> 00:36:05.690
than divide both sides by k?
00:36:05.690 --> 00:36:06.400
PROFESSOR: Yeah.
00:36:06.400 --> 00:36:08.275
And it'll be little more
obvious in a second.
00:36:08.275 --> 00:36:10.920
But here's the reason.
00:36:10.920 --> 00:36:12.760
So let's divide through by l.
00:36:12.760 --> 00:36:15.439
This is the form that we got.
00:36:15.439 --> 00:36:16.230
What are the units?
00:36:16.230 --> 00:36:18.585
What are the dimensions
of this expression?
00:36:18.585 --> 00:36:23.140
k is a wave number, so it
has units of 1 upon length.
00:36:23.140 --> 00:36:23.640
Right?
00:36:23.640 --> 00:36:26.223
And that's good because that's
1 upon length times the length,
00:36:26.223 --> 00:36:29.370
and you'd better have something
dimensionless inside a tangent.
00:36:29.370 --> 00:36:31.632
But it seems there are two
things to say about this.
00:36:31.632 --> 00:36:34.090
The first is it seems like l
is playing an independent role
00:36:34.090 --> 00:36:36.200
from k in this equation.
00:36:36.200 --> 00:36:38.080
But this is dimensionless.
00:36:38.080 --> 00:36:40.407
These are both dimensionful
units of 1 over length.
00:36:40.407 --> 00:36:42.490
So we can make the entire
expression dimensionless
00:36:42.490 --> 00:36:46.680
and make it clear that k and l
don't have an independent life.
00:36:46.680 --> 00:36:49.020
The dimensionless quantity,
kl, times the tangent
00:36:49.020 --> 00:36:50.410
of that dimensionless
quantity is
00:36:50.410 --> 00:36:52.129
equal to this
dimensionless quantity.
00:36:52.129 --> 00:36:54.170
So the reason that this
is preferable is twofold.
00:36:54.170 --> 00:36:56.430
First off, it makes it
sort of obvious that k
00:36:56.430 --> 00:36:59.740
and l, you can't vary them
independently in this sense.
00:36:59.740 --> 00:37:04.190
But the second is that it makes
it nice and dimensionless.
00:37:04.190 --> 00:37:05.840
And you'll always,
whenever possible,
00:37:05.840 --> 00:37:07.548
want to put things in
dimensionless form.
00:37:10.434 --> 00:37:11.850
I mean it's just
multiplying by l.
00:37:11.850 --> 00:37:13.349
So it's obviously
not all that deep.
00:37:13.349 --> 00:37:16.960
But it's a convenient bit
of multiplication by l.
00:37:16.960 --> 00:37:19.360
Other questions?
00:37:19.360 --> 00:37:20.960
OK.
00:37:20.960 --> 00:37:23.395
So where are we?
00:37:23.395 --> 00:37:25.520
So I'd like to find the
solutions of this equation.
00:37:25.520 --> 00:37:26.680
So again, just to--
00:37:29.440 --> 00:37:37.920
Let me write this slightly
differently where k squared is
00:37:37.920 --> 00:37:44.320
equal to 2m upon h
bar squared v0 plus e.
00:37:44.320 --> 00:37:52.420
And alpha squared is equal
to 2m upon h bar squared e,
00:37:52.420 --> 00:37:55.140
the positive value of e.
00:37:55.140 --> 00:37:58.670
So this is a really complicated
expression as a function of e.
00:37:58.670 --> 00:38:00.980
So I'd like to solve for the
actual energy eigenvalues.
00:38:00.980 --> 00:38:05.420
I want to know what are the
energy eigenvalues of the bound
00:38:05.420 --> 00:38:08.654
states of the finite potential
well, as a function of l,
00:38:08.654 --> 00:38:09.154
for example.
00:38:11.820 --> 00:38:14.389
Sadly, I can't
solve this equation.
00:38:14.389 --> 00:38:15.680
It's a transcendental equation.
00:38:15.680 --> 00:38:18.570
It's a sort of canonically
hard problem to solve.
00:38:18.570 --> 00:38:21.180
You can't write down a closed
from expression for it.
00:38:21.180 --> 00:38:24.080
However, there are a bunch
of ways to easily solve it.
00:38:24.080 --> 00:38:27.760
One is take your
convenient nearby laptop.
00:38:27.760 --> 00:38:30.670
Open up Mathematica, and ask it
to numerically find solutions
00:38:30.670 --> 00:38:31.170
to this.
00:38:31.170 --> 00:38:32.075
And you can do this.
00:38:32.075 --> 00:38:32.950
It's a good exercise.
00:38:35.407 --> 00:38:37.490
I will encourage you to
do so on your problem set.
00:38:37.490 --> 00:38:39.450
And in fact, on
the problem set, it
00:38:39.450 --> 00:38:40.960
asks you to do a calculation.
00:38:40.960 --> 00:38:43.100
And it encourages you
do it using Mathematica.
00:38:43.100 --> 00:38:45.200
Let me rephrase the
statement in the problem set.
00:38:45.200 --> 00:38:48.900
It would be crazy for you to
try to do it only by hand.
00:38:48.900 --> 00:38:50.707
You should do it by
hand and on computer
00:38:50.707 --> 00:38:51.790
because they're both easy.
00:38:51.790 --> 00:38:53.560
And you can check
against each other.
00:38:53.560 --> 00:38:56.680
They make different
things obvious.
00:38:56.680 --> 00:38:59.455
This should be your default is
to also check on Mathematica.
00:38:59.455 --> 00:39:02.080
The second thing we can do is we
can get a qualitative solution
00:39:02.080 --> 00:39:03.890
of this equation
just graphically.
00:39:03.890 --> 00:39:06.280
And since this is such a useful
technique, not just here,
00:39:06.280 --> 00:39:08.122
but throughout physics
to graphically solve
00:39:08.122 --> 00:39:09.580
transcendental
equations, I'm going
00:39:09.580 --> 00:39:11.114
to walk through it a little bit.
00:39:11.114 --> 00:39:13.030
So this is going to be
the graphical solution.
00:39:17.336 --> 00:39:18.710
And we can extract,
it turns out,
00:39:18.710 --> 00:39:22.420
an awful lot of the
physics of these energy
00:39:22.420 --> 00:39:24.700
eigenstates and their
energies through
00:39:24.700 --> 00:39:27.464
this graphical technique.
00:39:27.464 --> 00:39:28.880
So the first thing
is I write this
00:39:28.880 --> 00:39:31.232
in nice, dimensionless form.
00:39:31.232 --> 00:39:33.440
And let me give those
dimensionless variables a name.
00:39:33.440 --> 00:39:37.720
Let me call kl is equal to
z, just define a parameter z.
00:39:37.720 --> 00:39:39.660
And alpha l is a parameter y.
00:39:42.530 --> 00:39:48.330
And I want to note that z
squared plus y squared is
00:39:48.330 --> 00:39:52.170
equal to a constant, which
is, I will call if you just
00:39:52.170 --> 00:39:54.620
plug these guys out, kl
squared plus al squared.
00:39:54.620 --> 00:39:55.320
That's easy.
00:39:55.320 --> 00:39:58.460
Kl squared is this
guy times l squared.
00:39:58.460 --> 00:40:01.070
Al squared is this
guy times l squared.
00:40:01.070 --> 00:40:03.680
And so the e and
the minus e cancel
00:40:03.680 --> 00:40:05.210
when we add them together.
00:40:05.210 --> 00:40:09.690
So we just get 2mv0 over h
bar squared times l squared.
00:40:09.690 --> 00:40:16.830
So 2m upon h bar
squared l squared v0.
00:40:16.830 --> 00:40:21.420
And I'm going to call
this r naught in something
00:40:21.420 --> 00:40:24.670
of a pathological
abusive notation.
00:40:24.670 --> 00:40:25.170
OK.
00:40:25.170 --> 00:40:26.253
So this is our expression.
00:40:27.827 --> 00:40:29.660
And I actually want to
call this r0 squared.
00:40:32.260 --> 00:40:32.760
I know.
00:40:32.760 --> 00:40:33.540
I know.
00:40:33.540 --> 00:40:34.120
It's awful.
00:40:34.120 --> 00:40:35.578
But the reason I
want to do this is
00:40:35.578 --> 00:40:37.670
that this is the
equation for a circle.
00:40:37.670 --> 00:40:38.620
Yeah?
00:40:38.620 --> 00:40:40.220
And a circle has a radius.
00:40:40.220 --> 00:40:42.680
The thing that goes
over here is r squared.
00:40:42.680 --> 00:40:43.180
OK.
00:40:43.180 --> 00:40:45.513
So at this point, you're
thinking like, come on, circle.
00:40:45.513 --> 00:40:46.229
So let's plot it.
00:40:46.229 --> 00:40:48.020
So how are we going to
solve this equation?
00:40:48.020 --> 00:40:49.260
Here's what I want to solve.
00:40:49.260 --> 00:40:55.340
I have now two equations
relating z and y.
00:40:55.340 --> 00:41:02.020
We have that from this equation
z tangent z is equal to y.
00:41:02.020 --> 00:41:04.890
And from this equation we
have that z squared plus y
00:41:04.890 --> 00:41:08.980
squared is a
constant r0 squared.
00:41:08.980 --> 00:41:11.892
Where r0 squared depends on
the potential and the width
00:41:11.892 --> 00:41:14.100
in a very specific way, on
the depth of the potential
00:41:14.100 --> 00:41:15.740
and the width in a
very specific way.
00:41:15.740 --> 00:41:16.580
So we want to find--
00:41:16.580 --> 00:41:17.079
Bless you.
00:41:17.079 --> 00:41:20.640
--simultaneously, we want to
find simultaneous solutions
00:41:20.640 --> 00:41:22.910
of these two equations.
00:41:22.910 --> 00:41:25.720
Yeah?
00:41:25.720 --> 00:41:27.770
So that's relatively easy.
00:41:27.770 --> 00:41:30.300
So here's y, and here's z.
00:41:34.790 --> 00:41:36.940
So this equation has solutions.
00:41:36.940 --> 00:41:38.424
Any time that y
plus z squared is
00:41:38.424 --> 00:41:40.840
equal to r0 squared, that means
any time we have a circle.
00:41:40.840 --> 00:41:45.350
So solutions for fixed
values of r0 lie on circles.
00:41:49.410 --> 00:41:52.853
Oh, I really should have
drawn this under here.
00:41:52.853 --> 00:41:53.353
Sorry.
00:41:57.650 --> 00:41:58.200
y and z.
00:42:05.760 --> 00:42:06.860
So those are the circles.
00:42:06.860 --> 00:42:10.300
Notice that I'm only focusing
on y and z, both positive.
00:42:10.300 --> 00:42:12.775
Why?
00:42:12.775 --> 00:42:14.930
Not yz, but W-H-Y .
00:42:14.930 --> 00:42:17.101
Why am I focusing
on the variables y
00:42:17.101 --> 00:42:17.975
and z being positive?
00:42:21.335 --> 00:42:23.710
Because we started out defining
them in terms of k and l,
00:42:23.710 --> 00:42:25.460
which were both positive,
and alpha and l,
00:42:25.460 --> 00:42:26.600
which were both positive.
00:42:26.600 --> 00:42:28.183
Can we find solutions
to this equation
00:42:28.183 --> 00:42:30.100
that have x and y negative?
00:42:30.100 --> 00:42:30.850
Sure.
00:42:30.850 --> 00:42:33.040
But they don't mean anything in
terms of our original problem.
00:42:33.040 --> 00:42:35.020
So to map onto solutions
of our original problem,
00:42:35.020 --> 00:42:37.200
we want to focus on the
positive values of y and z.
00:42:37.200 --> 00:42:38.260
Cool?
00:42:38.260 --> 00:42:39.240
OK.
00:42:39.240 --> 00:42:40.260
So that's this one.
00:42:40.260 --> 00:42:41.570
The solutions lie on circles.
00:42:41.570 --> 00:42:44.130
So given a value of y, you
can find a solution of z.
00:42:44.130 --> 00:42:46.877
But we want to also find a
solution of this equation.
00:42:46.877 --> 00:42:49.210
And this equation is a little
more entertaining to plot.
00:42:49.210 --> 00:42:50.130
Here's y.
00:42:50.130 --> 00:42:51.370
Here's z.
00:42:51.370 --> 00:42:53.370
So what does z tangent z do?
00:42:56.220 --> 00:42:57.257
Oh, shoot.
00:42:57.257 --> 00:42:58.340
I want to plot y vertical.
00:42:58.340 --> 00:43:00.930
Otherwise, it's going
to a giant pain.
00:43:00.930 --> 00:43:04.656
Happily, this plot
can be left identical.
00:43:04.656 --> 00:43:05.942
Let's plot y vertically.
00:43:05.942 --> 00:43:07.650
So the reason I want
to plot y vertically
00:43:07.650 --> 00:43:09.380
is that this is z tangent z.
00:43:09.380 --> 00:43:13.205
So first off, what does
tangent z look like?
00:43:13.205 --> 00:43:13.864
Yeah.
00:43:13.864 --> 00:43:14.530
This is awesome.
00:43:14.530 --> 00:43:16.020
Yeah, it looks like this.
00:43:16.020 --> 00:43:16.890
Yes, exactly.
00:43:16.890 --> 00:43:19.377
So tangent is sine over cosine.
00:43:19.377 --> 00:43:20.710
Sine is zero, and cosine is one.
00:43:20.710 --> 00:43:23.870
So it does this, as
you go to a value
00:43:23.870 --> 00:43:26.170
where the argument, let's
call the argument z.
00:43:26.170 --> 00:43:28.450
So if we just plot tangent--
00:43:28.450 --> 00:43:29.080
OK.
00:43:29.080 --> 00:43:33.510
So when z is equal to pi over
2, then the denominator cosine
00:43:33.510 --> 00:43:35.205
vanishes, and that diverges.
00:43:39.700 --> 00:43:41.750
Oops.
00:43:41.750 --> 00:43:42.250
OK.
00:43:42.250 --> 00:43:43.960
So here's pi over 2.
00:43:43.960 --> 00:43:45.620
Here's pi.
00:43:45.620 --> 00:43:46.230
Whoops.
00:43:46.230 --> 00:43:54.670
Pi, pi over 2, and here's
3pi over 2, and so on.
00:43:54.670 --> 00:43:56.840
Now, we're only interested
in the first quadrant.
00:43:56.840 --> 00:44:00.540
So I'm just ignore down here.
00:44:00.540 --> 00:44:01.300
OK.
00:44:01.300 --> 00:44:03.760
So this is pi over 2.
00:44:03.760 --> 00:44:07.760
This is pi, 3pi over 2.
00:44:07.760 --> 00:44:08.832
OK.
00:44:08.832 --> 00:44:10.540
But this is not what
we're interested in.
00:44:10.540 --> 00:44:13.100
We're not interested
in tangent of z.
00:44:13.100 --> 00:44:15.120
We're interested in z tangent z.
00:44:15.120 --> 00:44:16.760
And what does z
tangent z look like?
00:44:16.760 --> 00:44:18.250
Well, it's basically the same.
00:44:18.250 --> 00:44:18.750
Right?
00:44:18.750 --> 00:44:21.450
z tangent z, it has an
extra factor of zero here
00:44:21.450 --> 00:44:23.270
and remains extra
small at the beginning.
00:44:23.270 --> 00:44:25.020
But it still curves
off roughly like this.
00:44:25.020 --> 00:44:28.760
And z is just nice
and linear, nice
00:44:28.760 --> 00:44:29.940
and regular throughout this.
00:44:29.940 --> 00:44:31.481
So it doesn't change
the fact that we
00:44:31.481 --> 00:44:33.350
have a divergence at pi over 2.
00:44:33.350 --> 00:44:36.330
And it doesn't change the fact
that it vanishes again at pi
00:44:36.330 --> 00:44:37.500
and becomes positive again.
00:44:37.500 --> 00:44:39.480
It just changes the
shape of the curve.
00:44:39.480 --> 00:44:41.030
And in fact, the way it
changes the shape of the curve
00:44:41.030 --> 00:44:43.180
is this becomes a little
fatter around the bottom.
00:44:43.180 --> 00:44:44.890
It's just a little more round.
00:44:44.890 --> 00:44:46.829
And when we get out
to large values of z,
00:44:46.829 --> 00:44:48.620
it's going to have a
more pronounced effect
00:44:48.620 --> 00:44:50.920
because that slope
is, at every example
00:44:50.920 --> 00:44:52.920
where it crosses z, that
slope is getting larger
00:44:52.920 --> 00:44:56.542
because the coefficient
of z is getting larger.
00:44:56.542 --> 00:44:57.244
OK.
00:44:57.244 --> 00:44:59.160
So it's just going to
get more and more sharp.
00:44:59.160 --> 00:45:02.090
But anyway, with all
that said, here's 0.
00:45:02.090 --> 00:45:03.510
Here's pi.
00:45:03.510 --> 00:45:05.470
Here's pi over 2.
00:45:05.470 --> 00:45:06.910
Here's pi.
00:45:06.910 --> 00:45:09.590
Here's 3pi over 2.
00:45:09.590 --> 00:45:11.799
The second plot we want
to plot, y is z tangent z.
00:45:11.799 --> 00:45:12.840
We know how to plot this.
00:45:27.390 --> 00:45:28.720
Cool?
00:45:28.720 --> 00:45:31.140
And what we want to find
are simultaneous solutions
00:45:31.140 --> 00:45:34.570
of this, values of y and z, for
which this equation is solved
00:45:34.570 --> 00:45:38.410
and this equation is solved
for the same value of y and z.
00:45:38.410 --> 00:45:40.640
This is a graphical solution.
00:45:40.640 --> 00:45:43.080
So let's combine them together.
00:45:43.080 --> 00:45:45.610
And the combined
plots look like this.
00:45:45.610 --> 00:45:48.900
First we have pi over 2.
00:45:51.600 --> 00:45:52.930
So let's plot the tangents.
00:45:57.780 --> 00:46:00.557
And then we have these circles
for various values of r.
00:46:00.557 --> 00:46:02.390
So for a particular
value of r, for example,
00:46:02.390 --> 00:46:05.920
suppose this is the value of r.
00:46:05.920 --> 00:46:06.610
This is r0.
00:46:12.100 --> 00:46:15.250
So how many
solutions do we have?
00:46:15.250 --> 00:46:16.082
One.
00:46:16.082 --> 00:46:18.310
One set of common
points where at y and z
00:46:18.310 --> 00:46:19.185
solve both equations.
00:46:22.620 --> 00:46:25.962
So we immediately learn
something really lovely.
00:46:25.962 --> 00:46:27.670
What happens to the
radius of that circle
00:46:27.670 --> 00:46:28.753
as I make the well deeper?
00:46:32.294 --> 00:46:35.124
Yeah, as I make the well
deeper, that means v0 gets
00:46:35.124 --> 00:46:37.290
larger and larger magnitude,
the radius gets larger.
00:46:37.290 --> 00:46:38.081
So does the circle.
00:46:38.081 --> 00:46:40.730
So if I make the well deeper,
I make this the circle larger.
00:46:40.730 --> 00:46:42.345
Will I still have a solution?
00:46:42.345 --> 00:46:43.720
Yeah, I'll still
have a solution.
00:46:43.720 --> 00:46:45.235
But check this out.
00:46:45.235 --> 00:46:46.485
Now, I'll have a new solution.
00:46:50.570 --> 00:46:55.200
And you can even see the
critical value of the depth
00:46:55.200 --> 00:46:57.290
and the width of the well.
00:46:57.290 --> 00:47:01.030
In order to have exactly a
new bound state appearing,
00:47:01.030 --> 00:47:02.890
what must the value of r0 be?
00:47:02.890 --> 00:47:07.951
Well, it's got to be that value,
such that r0 squared is pi.
00:47:07.951 --> 00:47:08.450
Yeah?
00:47:10.792 --> 00:47:13.000
And similarly, let me ask
you the following question.
00:47:13.000 --> 00:47:15.000
As I make the well deeper
and deeper and deeper,
00:47:15.000 --> 00:47:17.780
holding the width, and make it
deeper and deeper and deeper,
00:47:17.780 --> 00:47:20.480
does the number of states
increase or decrease?
00:47:20.480 --> 00:47:21.025
It increases.
00:47:21.025 --> 00:47:22.400
If you make it
deeper and deeper,
00:47:22.400 --> 00:47:24.691
the radius of that circle is
getting bigger and bigger.
00:47:24.691 --> 00:47:30.010
There are more points where this
circle intersects this point.
00:47:30.010 --> 00:47:33.730
So here's another one.
00:47:33.730 --> 00:47:38.670
We've got one here, one here,
one here, three solutions.
00:47:38.670 --> 00:47:40.850
And the number of
solutions just goes.
00:47:40.850 --> 00:47:42.800
Every time we click
over a new point
00:47:42.800 --> 00:47:45.383
by increasing the radius of the
circle, we get a new solution.
00:47:45.383 --> 00:47:46.990
We get another bound state.
00:47:46.990 --> 00:47:49.730
But here's the thing that
I really want to focus on.
00:47:49.730 --> 00:47:51.390
Let's make the well
less and less deep.
00:47:51.390 --> 00:47:54.060
Let's make it shallower
and shallower.
00:47:54.060 --> 00:48:00.220
At what depth do we lose
that first bound state?
00:48:00.220 --> 00:48:01.131
We never do.
00:48:01.131 --> 00:48:01.630
Right?
00:48:01.630 --> 00:48:03.190
There is no circle
so small that it
00:48:03.190 --> 00:48:06.030
doesn't intersect this curve.
00:48:06.030 --> 00:48:09.590
In a 1D, finite well potential,
there is always at least one
00:48:09.590 --> 00:48:10.890
bound state.
00:48:10.890 --> 00:48:12.931
There are never
zero bound states.
00:48:12.931 --> 00:48:15.430
This will turn out not to be
true in three dimensions, which
00:48:15.430 --> 00:48:16.960
is kind of interesting.
00:48:16.960 --> 00:48:19.840
But it's true in one
dimension that we always
00:48:19.840 --> 00:48:21.970
have at least one bound state.
00:48:21.970 --> 00:48:23.800
And in fact, you
can decorate this.
00:48:23.800 --> 00:48:25.790
You can use this and
fancy it up a bit
00:48:25.790 --> 00:48:28.770
to argue that in
any potential in 1D,
00:48:28.770 --> 00:48:31.445
there's always at
least one bound state
00:48:31.445 --> 00:48:32.820
unless the potential
is constant,
00:48:32.820 --> 00:48:36.025
I mean any potential that varies
and goes to zero infinity.
00:48:38.640 --> 00:48:39.405
Yeah?
00:48:39.405 --> 00:48:41.030
And so we still don't
have any numbers.
00:48:41.030 --> 00:48:44.390
But we know an awful lot about
the qualitative structure
00:48:44.390 --> 00:48:49.476
of the set of energy eigenvalues
of the spectrum of the energy.
00:48:49.476 --> 00:48:49.975
Questions?
00:48:52.580 --> 00:48:53.710
Yeah?
00:48:53.710 --> 00:48:57.081
AUDIENCE: So what
happens if r is bigger
00:48:57.081 --> 00:49:03.622
than pi or y is bigger than
pi and you get two solutions?
00:49:03.622 --> 00:49:04.330
PROFESSOR: Great.
00:49:04.330 --> 00:49:06.720
So when we have two solutions,
what does that mean?
00:49:06.720 --> 00:49:11.270
Well, you've got to bound
states, two different energies.
00:49:11.270 --> 00:49:13.175
Right?
00:49:13.175 --> 00:49:14.050
It's a good question.
00:49:14.050 --> 00:49:17.890
Every solution here corresponds
to some particular value of y
00:49:17.890 --> 00:49:19.430
and some particular value of z.
00:49:19.430 --> 00:49:21.640
But those values
of y and z are just
00:49:21.640 --> 00:49:23.375
telling you what
k and alpha are.
00:49:26.050 --> 00:49:27.674
And so that's
determining the energy.
00:49:27.674 --> 00:49:29.090
So a different
value of k is going
00:49:29.090 --> 00:49:30.941
to give you a different
value of the energy.
00:49:30.941 --> 00:49:32.690
So we can just eyeball
this in particular.
00:49:32.690 --> 00:49:33.780
Let's look at alpha.
00:49:33.780 --> 00:49:35.020
Alpha is just e.
00:49:35.020 --> 00:49:36.090
Alpha squared is just e.
00:49:36.090 --> 00:49:37.716
Yes?
00:49:37.716 --> 00:49:38.840
So here's a quick question.
00:49:38.840 --> 00:49:40.920
If alpha is just e,
and alpha l is y--
00:49:40.920 --> 00:49:43.060
So this is our y value.
00:49:43.060 --> 00:49:46.350
y is roughly alpha,
the width, which
00:49:46.350 --> 00:49:49.450
means it's roughly the
energy square root.
00:49:49.450 --> 00:49:51.720
So this value,
the vertical value
00:49:51.720 --> 00:49:54.760
of each of these intersection
points on a given circle
00:49:54.760 --> 00:49:57.030
corresponds to the
square root of the energy
00:49:57.030 --> 00:49:59.700
times some coefficients.
00:49:59.700 --> 00:50:04.671
So which state has the
largest value of the energy?
00:50:04.671 --> 00:50:06.420
Absolute value, which
state is most deeply
00:50:06.420 --> 00:50:09.577
bound on this circle?
00:50:09.577 --> 00:50:10.410
Yeah, the first one.
00:50:10.410 --> 00:50:10.540
Right?
00:50:10.540 --> 00:50:11.880
Because it's got the
largest value of alpha.
00:50:11.880 --> 00:50:12.630
So this is nice.
00:50:12.630 --> 00:50:14.127
We see that the
first state always
00:50:14.127 --> 00:50:16.210
has a higher value of alpha
than the second state,
00:50:16.210 --> 00:50:18.420
which always has a higher value
of alpha than the third state.
00:50:18.420 --> 00:50:19.870
And every time we
add a new state,
00:50:19.870 --> 00:50:23.640
we make the depth of these
guys the binding energy
00:50:23.640 --> 00:50:25.820
of the already existing states.
00:50:25.820 --> 00:50:27.952
We make it just a
little bit deeper.
00:50:27.952 --> 00:50:29.700
We make them a little
more tightly bound.
00:50:29.700 --> 00:50:31.420
And only eventually
then do we get
00:50:31.420 --> 00:50:32.890
a new bound state appearing.
00:50:32.890 --> 00:50:34.764
And what's the energy
of that new bound state
00:50:34.764 --> 00:50:37.340
when it appears?
00:50:37.340 --> 00:50:38.300
Zero energy.
00:50:38.300 --> 00:50:41.200
It's appearing
just at threshold.
00:50:41.200 --> 00:50:42.010
OK.
00:50:42.010 --> 00:50:44.343
So we'll explore that in more
detail in the problem set.
00:50:44.343 --> 00:50:45.930
But for now, let me
leave it at that.
00:50:45.930 --> 00:50:46.430
Questions?
00:50:46.430 --> 00:50:47.096
Other questions?
00:50:47.096 --> 00:50:47.750
Yeah.
00:50:47.750 --> 00:50:49.708
AUDIENCE: You said that
this can be generalized
00:50:49.708 --> 00:50:53.250
to any nonconstant
function that you'd like.
00:50:53.250 --> 00:50:56.250
So there's always going to
be at least one bound state.
00:50:56.250 --> 00:50:58.250
What about, like
with delta function?
00:50:58.250 --> 00:50:58.790
PROFESSOR: Excellent question.
00:50:58.790 --> 00:50:59.610
What about the delta function?
00:50:59.610 --> 00:51:00.420
We're going to come back to
that in just a few minutes.
00:51:00.420 --> 00:51:01.503
It's a very good question.
00:51:01.503 --> 00:51:04.380
So the question is, look,
if any potential that
00:51:04.380 --> 00:51:07.077
goes to zero infinity
and wiggles inside,
00:51:07.077 --> 00:51:09.410
if any potential like that
in 1D has a bound state, what
00:51:09.410 --> 00:51:10.451
about the delta function?
00:51:10.451 --> 00:51:11.730
We briefly talked about that.
00:51:11.730 --> 00:51:14.146
So we're going to come back
to that in just a few minutes.
00:51:14.146 --> 00:51:15.361
But it's a pressing question.
00:51:15.361 --> 00:51:15.860
OK.
00:51:15.860 --> 00:51:16.526
Other questions?
00:51:22.550 --> 00:51:23.572
Yes.
00:51:23.572 --> 00:51:25.947
AUDIENCE: So the energy is
zero, but that's not possible.
00:51:28.015 --> 00:51:28.890
PROFESSOR: Thank you.
00:51:28.890 --> 00:51:29.070
OK.
00:51:29.070 --> 00:51:29.240
Good.
00:51:29.240 --> 00:51:31.040
So let me talk about that
in a little more detail.
00:51:31.040 --> 00:51:32.970
So I wasn't going to
go into this, but--
00:51:32.970 --> 00:51:36.130
So when new bound states appear,
so let's consider a point where
00:51:36.130 --> 00:51:39.950
our r0 is, let's say,
it's just the right value
00:51:39.950 --> 00:51:43.291
so that r0 is equal to pi.
00:51:43.291 --> 00:51:43.790
OK.
00:51:43.790 --> 00:51:48.400
And we see that we're just about
to develop a new bound state.
00:51:48.400 --> 00:51:50.870
So let's think about what
that bound state looks like.
00:51:50.870 --> 00:51:52.440
So this is the new bound state.
00:52:00.504 --> 00:52:02.170
And I'm going to put
this in parentheses
00:52:02.170 --> 00:52:04.545
because it's got bound state.
00:52:04.545 --> 00:52:05.545
And we say at threshold.
00:52:09.660 --> 00:52:11.100
OK.
00:52:11.100 --> 00:52:13.124
At threshold, i.e.
00:52:13.124 --> 00:52:21.199
at the energy is roughly
zero, and r0 is equal to pi.
00:52:21.199 --> 00:52:22.490
So this is really what we mean.
00:52:22.490 --> 00:52:24.030
This new state,
when r0 is pi and we
00:52:24.030 --> 00:52:26.600
have a solution on
that second branch.
00:52:26.600 --> 00:52:27.101
Cool?
00:52:27.101 --> 00:52:28.850
So what does this wave
function look like?
00:52:28.850 --> 00:52:30.350
What does it look
like when you have
00:52:30.350 --> 00:52:32.050
a wave function
that just appeared?
00:52:32.050 --> 00:52:33.252
It's just barely bound.
00:52:33.252 --> 00:52:35.210
Well, first off, what
does it mean to be bound?
00:52:35.210 --> 00:52:36.370
Let's just step back
and remember for now.
00:52:36.370 --> 00:52:37.870
What does it mean
to be a bound state?
00:52:37.870 --> 00:52:39.495
It means you're an
energy eigenfunction
00:52:39.495 --> 00:52:40.930
and you're localized.
00:52:40.930 --> 00:52:43.600
Your wave function
falls off at infinity.
00:52:43.600 --> 00:52:45.070
Now, if it falls
off at infinity,
00:52:45.070 --> 00:52:46.554
do these guys fall
off at infinity?
00:52:46.554 --> 00:52:48.095
These wave functions,
sure, they fall
00:52:48.095 --> 00:52:49.940
of with an exponential damping.
00:52:49.940 --> 00:52:51.898
And in particular, let's
look at the right hand
00:52:51.898 --> 00:52:52.690
side of the well.
00:52:52.690 --> 00:52:56.110
This new bound state is
appearing just at zero energy.
00:52:56.110 --> 00:52:58.130
So out here, what is
the wave function?
00:52:58.130 --> 00:52:59.860
It's e to the minus alpha x.
00:52:59.860 --> 00:53:02.560
But what's alpha?
00:53:02.560 --> 00:53:04.560
Zero, right?
00:53:04.560 --> 00:53:07.020
There it is, zero.
00:53:07.020 --> 00:53:09.572
So this is e to the minus alpha
x where alpha is equal to 0.
00:53:09.572 --> 00:53:10.280
This is constant.
00:53:13.040 --> 00:53:15.190
So what does the wave
function look like?
00:53:15.190 --> 00:53:17.540
Well, the wave function,
again over the same domain--
00:53:17.540 --> 00:53:19.360
Here's 0, l.
00:53:23.607 --> 00:53:24.690
And here's the value zero.
00:53:24.690 --> 00:53:28.184
We know that in this
domain it's oscillatory,
00:53:28.184 --> 00:53:29.600
and in this domain,
it's constant.
00:53:36.610 --> 00:53:40.515
And actually, since we know that
it's the first excited state,
00:53:40.515 --> 00:53:42.130
we know that it does this.
00:53:46.340 --> 00:53:51.990
So if we make the well
ever so slightly deeper,
00:53:51.990 --> 00:53:54.070
ever so slightly
deeper, which means
00:53:54.070 --> 00:53:57.320
making the radius of the
circle ever so slightly larger,
00:53:57.320 --> 00:54:00.450
we will get a nonzero value
for the alpha of this solution.
00:54:00.450 --> 00:54:00.950
Right?
00:54:00.950 --> 00:54:02.270
It'll be just tiny.
00:54:02.270 --> 00:54:03.637
But it'll be nonzero.
00:54:03.637 --> 00:54:05.720
So we make the well just
a little tiny bit deeper.
00:54:05.720 --> 00:54:06.428
We get something.
00:54:06.428 --> 00:54:08.130
OK, good.
00:54:08.130 --> 00:54:10.310
So what's going to happen
to the wave function?
00:54:10.310 --> 00:54:12.320
Well, instead of going flat.
00:54:12.320 --> 00:54:13.400
This is going to curve.
00:54:13.400 --> 00:54:14.649
It's got a little more energy.
00:54:14.649 --> 00:54:16.050
There's a little more curvature.
00:54:16.050 --> 00:54:18.950
So it curves just a
little tiny bit more.
00:54:18.950 --> 00:54:21.575
And then it matches onto a very
gradually decaying exponential.
00:54:24.710 --> 00:54:25.210
OK.
00:54:25.210 --> 00:54:27.920
So what's happening as we take
this bound state, the second
00:54:27.920 --> 00:54:30.510
bound state, and we make the
well a little more shallow?
00:54:30.510 --> 00:54:32.450
We make the well a
little more shallow.
00:54:32.450 --> 00:54:34.240
It's a little less
curved inside.
00:54:34.240 --> 00:54:37.490
And the evanescent tails,
the exponential tails
00:54:37.490 --> 00:54:40.210
become longer and longer
and longer and broader
00:54:40.210 --> 00:54:43.530
until they go off to infinity,
until they're infinitely wide.
00:54:43.530 --> 00:54:45.404
And is that
normalizable anymore?
00:54:45.404 --> 00:54:46.570
No, that's not normalizable.
00:54:46.570 --> 00:54:48.720
So the state really
isn't strictly localized
00:54:48.720 --> 00:54:49.350
at that point.
00:54:49.350 --> 00:54:50.990
It's not really a
normalizable state.
00:54:50.990 --> 00:54:54.490
And just when the state
ceases to be normalizable,
00:54:54.490 --> 00:54:56.627
it disappears.
00:54:56.627 --> 00:54:58.210
We make the well
just a little deeper,
00:54:58.210 --> 00:55:00.180
and there's no
state there at all.
00:55:00.180 --> 00:55:00.680
OK.
00:55:00.680 --> 00:55:02.179
So this tells you
a very nice thing.
00:55:02.179 --> 00:55:04.730
It's a good bit of intuition
that when states are appearing
00:55:04.730 --> 00:55:08.620
or disappearing, when states
are at threshold as you
00:55:08.620 --> 00:55:13.020
vary the depth, those threshold
bound states have exceedingly
00:55:13.020 --> 00:55:18.544
long evanescent tails, and
they're just barely bound.
00:55:18.544 --> 00:55:19.087
OK?
00:55:19.087 --> 00:55:21.420
This turns out to have all
sorts of useful consequences,
00:55:21.420 --> 00:55:22.290
but let me move on.
00:55:22.290 --> 00:55:23.800
Did that answer your question?
00:55:23.800 --> 00:55:24.300
Good.
00:55:24.300 --> 00:55:25.023
Yeah?
00:55:25.023 --> 00:55:27.861
AUDIENCE: So in this case,
the radius, as you say,
00:55:27.861 --> 00:55:29.753
is proportional to the length.
00:55:29.753 --> 00:55:32.118
Right?
00:55:32.118 --> 00:55:36.385
But also we have intuition
that as we increase
00:55:36.385 --> 00:55:37.884
the length of the
well, the energy's
00:55:37.884 --> 00:55:39.351
going to keep increasing--
00:55:39.351 --> 00:55:41.141
PROFESSOR: Fantastic.
00:55:41.141 --> 00:55:41.640
OK.
00:55:41.640 --> 00:55:43.160
So what's up with that?
00:55:43.160 --> 00:55:44.020
Right.
00:55:44.020 --> 00:55:45.020
So the question is this.
00:55:45.020 --> 00:55:47.210
We already have intuition that
is if we take a finite well
00:55:47.210 --> 00:55:48.760
and we make it a
little bit wider,
00:55:48.760 --> 00:55:51.050
the ground state
energy should decrease.
00:55:51.050 --> 00:55:53.550
The energy of the ground state
should get deeper and deeper,
00:55:53.550 --> 00:55:54.925
or the magnitude
should increase,
00:55:54.925 --> 00:55:55.950
another way to say it.
00:55:55.950 --> 00:55:56.450
Right?
00:55:56.450 --> 00:55:57.470
That was our intuition.
00:55:57.470 --> 00:55:58.790
So let's check if that's true.
00:55:58.790 --> 00:56:01.790
What happens if I
take the ground state
00:56:01.790 --> 00:56:03.350
some value of the
radius, and then I
00:56:03.350 --> 00:56:05.077
make the well a
little bit wider.
00:56:05.077 --> 00:56:06.910
Well, if I make the
well a little bit wider,
00:56:06.910 --> 00:56:09.080
what happens to r0, the
radius of the circle?
00:56:09.080 --> 00:56:12.020
Well, if I double the length
of the well, the width
00:56:12.020 --> 00:56:13.520
of the well, then
it will double r0,
00:56:13.520 --> 00:56:15.780
and it will double the
radius of the circle.
00:56:15.780 --> 00:56:18.142
So if I make it
wider, r0 gets bigger,
00:56:18.142 --> 00:56:19.350
and we go to a bigger circle.
00:56:19.350 --> 00:56:23.100
And what happened to the
energy of this state?
00:56:23.100 --> 00:56:26.330
Yeah, it got deeper
and deeper and deeper.
00:56:26.330 --> 00:56:28.210
And meanwhile, as
we make it wider,
00:56:28.210 --> 00:56:30.987
as we make the well wider, the
circle is getting bigger again.
00:56:30.987 --> 00:56:32.820
And we're going to get
more and more states.
00:56:32.820 --> 00:56:36.200
So as we make the well wider,
holding the depth fixed,
00:56:36.200 --> 00:56:37.390
we get more and more states.
00:56:37.390 --> 00:56:40.210
As we make the well deeper,
holding the width fixed,
00:56:40.210 --> 00:56:42.110
we get more and more states.
00:56:42.110 --> 00:56:43.650
And so how do you trade off?
00:56:43.650 --> 00:56:45.537
If I make it twice
as wide, how much--
00:56:45.537 --> 00:56:46.620
So here's a good question.
00:56:46.620 --> 00:56:51.140
Suppose I take a well,
and it has n states.
00:56:51.140 --> 00:56:53.720
Suppose I then make
it twice as wide.
00:56:53.720 --> 00:56:55.690
What must I do to the
energy so that it still
00:56:55.690 --> 00:56:59.181
has the same states
with the same energies?
00:56:59.181 --> 00:57:01.730
AUDIENCE: Divide it by 4.
00:57:01.730 --> 00:57:02.730
PROFESSOR: Yep, exactly.
00:57:02.730 --> 00:57:04.289
I've got to divide it by 4.
00:57:04.289 --> 00:57:05.830
Because I've doubled
the length, that
00:57:05.830 --> 00:57:07.600
means the radius
has gone up by four.
00:57:07.600 --> 00:57:09.200
But I get exactly
the same solutions
00:57:09.200 --> 00:57:10.700
if I just bring this out.
00:57:10.700 --> 00:57:12.570
Well, that's almost true.
00:57:12.570 --> 00:57:14.200
So if I put a
factor of 1/4 here,
00:57:14.200 --> 00:57:19.160
that's almost true, except for
the value of y is unchanged,
00:57:19.160 --> 00:57:22.391
but the alpha hasn't changed.
00:57:22.391 --> 00:57:22.890
Sorry.
00:57:22.890 --> 00:57:24.875
The alpha has changed
because there's an l.
00:57:24.875 --> 00:57:25.770
So y is fixed.
00:57:25.770 --> 00:57:29.274
But alpha's changed
because of the l.
00:57:29.274 --> 00:57:31.190
So the reason that it's
useful to write things
00:57:31.190 --> 00:57:33.740
in these dimensionless forms is
that you can see the play off
00:57:33.740 --> 00:57:36.550
of the various different
dimensionful parameters
00:57:36.550 --> 00:57:39.270
in changing the answer.
00:57:39.270 --> 00:57:41.210
Other questions.
00:57:41.210 --> 00:57:42.550
OK.
00:57:42.550 --> 00:57:43.745
So a couple of comments.
00:57:43.745 --> 00:57:45.620
So the first is let's
just check to make sure
00:57:45.620 --> 00:57:46.630
that this makes sense.
00:57:46.630 --> 00:57:49.262
We have already
solved this problem.
00:57:49.262 --> 00:57:50.720
We solved this
problem a while ago,
00:57:50.720 --> 00:57:52.950
but we solved it in
a particular limit.
00:57:52.950 --> 00:57:56.750
We solved it in the limit that
the potential one's arbitrarily
00:57:56.750 --> 00:57:58.804
deep.
00:57:58.804 --> 00:58:01.039
Right?
00:58:01.039 --> 00:58:03.080
So when the potential
one's are arbitrarily deep,
00:58:03.080 --> 00:58:05.390
holding the width fixed,
that was the infinite well.
00:58:05.390 --> 00:58:07.480
That was the first
problem we solved.
00:58:07.480 --> 00:58:11.100
So let's make sure that we
recover this in that limit.
00:58:11.100 --> 00:58:14.470
So what happens as we make
the well arbitrarily deep,
00:58:14.470 --> 00:58:18.030
holding the length fixed
or the width fixed.
00:58:18.030 --> 00:58:19.680
So if we make this
arbitrarily deep,
00:58:19.680 --> 00:58:21.270
v0 is getting arbitrarily large.
00:58:21.270 --> 00:58:22.460
That means r0 is getting--
00:58:22.460 --> 00:58:25.410
We've got a huge circle.
00:58:25.410 --> 00:58:27.520
So what do these
solutions look like when
00:58:27.520 --> 00:58:29.750
we have a huge circle.
00:58:29.750 --> 00:58:31.110
Let me not do that here.
00:58:33.696 --> 00:58:34.985
Let me do that here.
00:58:38.290 --> 00:58:43.072
So if we make the
potential nice and deep,
00:58:43.072 --> 00:58:44.905
Let's think about what
that plot looks like.
00:58:47.640 --> 00:58:51.100
So again, that first plot looks
identical with the tangents.
00:59:09.160 --> 00:59:11.370
So on and so forth.
00:59:11.370 --> 00:59:15.690
And what I want to do is
I want to dot, dot, dot.
00:59:15.690 --> 00:59:16.190
OK.
00:59:16.190 --> 00:59:17.280
So this is way up there.
00:59:22.580 --> 00:59:25.390
So these guys are basically
vertical lines at this point.
00:59:25.390 --> 00:59:28.770
So for very, very large values
of y, and in particular,
00:59:28.770 --> 00:59:31.470
for very large values that are
of order the gigantically deep
00:59:31.470 --> 00:59:36.819
radius r0, what does
the circle look like?
00:59:36.819 --> 00:59:39.360
So what does the second equation
look like, the second curve?
00:59:39.360 --> 00:59:40.970
Well, again it's circles.
00:59:40.970 --> 00:59:43.490
But now it's a gigantic circle.
00:59:43.490 --> 00:59:44.250
Yeah, exactly.
00:59:44.250 --> 00:59:46.370
If it's a gigantic circle,
it's basically flat.
00:59:51.907 --> 00:59:53.740
It's not exactly flat,
but it's almost flat.
00:59:53.740 --> 00:59:54.390
It's a circle.
01:00:02.620 --> 01:00:04.900
So what are the values?
01:00:04.900 --> 01:00:06.090
And here's the key question.
01:00:06.090 --> 01:00:07.400
What are the values of--
01:00:10.720 --> 01:00:11.750
Did do that right?
01:00:11.750 --> 01:00:12.250
Yeah, OK.
01:00:12.250 --> 01:00:12.750
Good.
01:00:12.750 --> 01:00:20.010
So what are the values of the
curve, where we get a solution?
01:00:20.010 --> 01:00:26.436
The values now of
z, are just exactly
01:00:26.436 --> 01:00:28.310
on these vertical lines,
on the separatrices.
01:00:28.310 --> 01:00:30.955
The value of z is at pi over 2.
01:00:34.410 --> 01:00:35.480
and 3pi over 2.
01:00:38.220 --> 01:00:43.281
And then another one at 5pi
over 2, and so on and so forth.
01:00:43.281 --> 01:00:43.780
Right?
01:00:43.780 --> 01:00:48.694
So what we find is that
the allowed values of z--
01:00:48.694 --> 01:00:49.690
Sorry.
01:00:49.690 --> 01:00:51.920
kl, yes, good.
01:00:51.920 --> 01:00:58.140
--allowed values of z are
equal to 2n plus 1 over pi.
01:00:58.140 --> 01:01:00.159
Whoops 2n plus 1
over 2 times pi.
01:01:00.159 --> 01:01:01.200
So let's just check that.
01:01:01.200 --> 01:01:02.505
So n is 0.
01:01:02.505 --> 01:01:04.630
That's 0 1/2 pi.
01:01:04.630 --> 01:01:05.710
n is 1.
01:01:05.710 --> 01:01:06.831
That's 3/2 pi.
01:01:06.831 --> 01:01:07.330
Good.
01:01:07.330 --> 01:01:09.420
So these are the
values of z, which
01:01:09.420 --> 01:01:18.580
says that kl is equal to 2n plus
1 upon 2 pi or k is equal to 2n
01:01:18.580 --> 01:01:22.520
plus 1 over 2l, which is
the width of this well
01:01:22.520 --> 01:01:26.820
because it's from
minus l to l pi.
01:01:26.820 --> 01:01:29.355
So is this the correct answer
for the infinite square well?
01:01:33.330 --> 01:01:35.860
Are these the
allowed values of k
01:01:35.860 --> 01:01:41.330
inside the well for the
infinite square well?
01:01:41.330 --> 01:01:42.850
Almost.
01:01:42.850 --> 01:01:45.720
Instead of 2n plus 1, it
should just be n plus 1.
01:01:45.720 --> 01:01:49.596
We seem to be missing about
half of the energy eigenvalues.
01:01:49.596 --> 01:01:51.096
AUDIENCE: That's
only the even ones.
01:01:51.096 --> 01:01:51.870
PROFESSOR: Yeah, thank you.
01:01:51.870 --> 01:01:52.820
This is only the even ones.
01:01:52.820 --> 01:01:54.030
We started out saying, oh, look.
01:01:54.030 --> 01:01:55.510
Let's look only
at the even ones.
01:01:55.510 --> 01:01:58.710
Where do you think the
odd ones are going to be?
01:01:58.710 --> 01:02:02.450
Ah, well the odd ones,
so this should be k even.
01:02:02.450 --> 01:02:03.380
So what about k odd?
01:02:03.380 --> 01:02:04.830
Well, we know the
answer already--
01:02:04.830 --> 01:02:05.440
Whoops.
01:02:05.440 --> 01:02:06.820
Odd, that's an odd spelling.
01:02:06.820 --> 01:02:12.000
--should be equal to 2n
over 2l plus 2 over 2l.
01:02:12.000 --> 01:02:12.500
Whoops.
01:02:12.500 --> 01:02:14.561
2 capital L, pi.
01:02:14.561 --> 01:02:15.060
OK.
01:02:15.060 --> 01:02:17.710
This is our guess
just from matching on
01:02:17.710 --> 01:02:18.902
to the infinite square well.
01:02:18.902 --> 01:02:19.860
So what does that mean?
01:02:19.860 --> 01:02:22.994
Well, that means it should
be this one and this one.
01:02:22.994 --> 01:02:25.160
So when you go through this
exercise on your problem
01:02:25.160 --> 01:02:28.215
set and you find the
solutions for the odd,
01:02:28.215 --> 01:02:30.740
and you repeat this analysis
for the odd ground states.
01:02:30.740 --> 01:02:31.710
What should you expect?
01:02:31.710 --> 01:02:33.400
Well, you should
expect to find this.
01:02:33.400 --> 01:02:37.327
And what do you
think the curves are
01:02:37.327 --> 01:02:39.410
that you're going to use
int he graphical solution
01:02:39.410 --> 01:02:42.804
to do your
transcendental equation?
01:02:42.804 --> 01:02:44.470
Yeah, it's really
tempting to say, look.
01:02:44.470 --> 01:02:46.700
It's just going to be
something shifted, like this.
01:02:46.700 --> 01:02:47.200
OK.
01:02:47.200 --> 01:02:53.850
So these are going to be the odd
question mark, question mark.
01:02:53.850 --> 01:02:55.330
OK.
01:02:55.330 --> 01:02:58.245
So you'll check whether that's
correct intuition or not
01:02:58.245 --> 01:03:00.274
on your problem set.
01:03:00.274 --> 01:03:01.230
OK.
01:03:01.230 --> 01:03:03.514
Questions?
01:03:03.514 --> 01:03:04.014
Yeah?
01:03:04.014 --> 01:03:07.316
AUDIENCE: What about the lower
end of these states, where
01:03:07.316 --> 01:03:08.774
it won't have gone
off high enough?
01:03:08.774 --> 01:03:09.730
Is that [INAUDIBLE]?
01:03:09.730 --> 01:03:10.605
PROFESSOR: Excellent.
01:03:10.605 --> 01:03:12.930
So that's a really
good question.
01:03:12.930 --> 01:03:15.600
How to say it?
01:03:15.600 --> 01:03:18.649
What we've done here to
make this an infinite well--
01:03:18.649 --> 01:03:20.690
So the question, let me
just repeat the question.
01:03:20.690 --> 01:03:24.400
The question is, look, what
about all the other states?
01:03:24.400 --> 01:03:26.110
OK, it's true that
r0 as gigantic.
01:03:26.110 --> 01:03:28.650
But eventually, we'll go
to a large enough z, where
01:03:28.650 --> 01:03:30.570
it's the circle coming
down over here too.
01:03:30.570 --> 01:03:31.600
So what's up with that?
01:03:31.600 --> 01:03:32.625
What are those states?
01:03:32.625 --> 01:03:33.250
Where are they?
01:03:33.250 --> 01:03:35.541
What do they mean in terms
if the infinite square well?
01:03:35.541 --> 01:03:39.798
Well, first off, what are
the energies of those states?
01:03:39.798 --> 01:03:40.652
AUDIENCE: Very low.
01:03:40.652 --> 01:03:42.360
PROFESSOR: They're
very low in magnitude,
01:03:42.360 --> 01:03:44.555
which means they're close
to what in absolute value?
01:03:44.555 --> 01:03:45.180
AUDIENCE: Zero.
01:03:45.180 --> 01:03:45.550
PROFESSOR: Zero.
01:03:45.550 --> 01:03:46.110
They're close to zero.
01:03:46.110 --> 01:03:47.819
So they're at the top
of the finite well.
01:03:47.819 --> 01:03:50.193
These are the states bound at
the top of the finite well.
01:03:50.193 --> 01:03:52.670
These are the states bound at
the bottom the finite well.
01:03:52.670 --> 01:03:55.824
But how many states are bound
at the top of the finite well
01:03:55.824 --> 01:03:58.240
when we take the limit that
the well goes infinitely deep?
01:04:01.630 --> 01:04:02.510
Yeah, none of them.
01:04:02.510 --> 01:04:02.760
Right?
01:04:02.760 --> 01:04:03.860
So when we make the
well infinitely deep,
01:04:03.860 --> 01:04:05.570
what we're saying
is, pay no attention
01:04:05.570 --> 01:04:07.050
to the top of the well.
01:04:07.050 --> 01:04:08.550
Look only at the
bottom of the well.
01:04:08.550 --> 01:04:11.607
And if it's really deep, it's
a pretty good approximation.
01:04:11.607 --> 01:04:12.940
So that's what we're doing here.
01:04:12.940 --> 01:04:13.731
We're saying, look.
01:04:13.731 --> 01:04:15.190
Pay no attention.
01:04:15.190 --> 01:04:16.960
There is no top of the well.
01:04:16.960 --> 01:04:17.960
There's just the bottom.
01:04:17.960 --> 01:04:19.030
And look at the
energy eigenvalues.
01:04:19.030 --> 01:04:19.880
Does that make sense?
01:04:19.880 --> 01:04:21.770
So what we're saying is if
you have a preposterously
01:04:21.770 --> 01:04:23.750
deep well, the energies
of a preposterously
01:04:23.750 --> 01:04:25.870
deep well should be
a good approximation
01:04:25.870 --> 01:04:29.350
to the low-lying energies
of an infinitely deep well.
01:04:29.350 --> 01:04:33.180
Because it's way up there,
what difference can it make?
01:04:33.180 --> 01:04:35.180
And that's what we're
seeing work out.
01:04:35.180 --> 01:04:37.420
Did that answer your question?
01:04:37.420 --> 01:04:38.350
AUDIENCE: Wait.
01:04:38.350 --> 01:04:39.696
I might have this backward.
01:04:39.696 --> 01:04:42.070
But when it says the high
energy instead of looking right
01:04:42.070 --> 01:04:43.275
at the top of the well--
01:04:43.275 --> 01:04:43.550
PROFESSOR: Yeah.
01:04:43.550 --> 01:04:43.950
OK, good.
01:04:43.950 --> 01:04:45.730
So this is an important
bit of intuition.
01:04:45.730 --> 01:04:49.590
So when we say this
is the energy zero,
01:04:49.590 --> 01:04:55.277
and the potential has
a minimum at minus v0,
01:04:55.277 --> 01:04:57.360
and we're measuring the
energies relative to zero,
01:04:57.360 --> 01:05:00.080
then the states at the
top of the potential well
01:05:00.080 --> 01:05:02.100
are the states with
energy close to zero.
01:05:02.100 --> 01:05:03.975
And the states at the
bottom of the potential
01:05:03.975 --> 01:05:07.550
well are those states with
the energy of order v0.
01:05:07.550 --> 01:05:08.050
cool?
01:05:08.050 --> 01:05:12.010
And what are the energies
of all these states?
01:05:12.010 --> 01:05:12.920
They're order of v0.
01:05:12.920 --> 01:05:13.672
Right?
01:05:13.672 --> 01:05:14.380
Because this is--
01:05:14.380 --> 01:05:15.480
Are they exactly v0?
01:05:18.080 --> 01:05:19.392
No, because this isn't linear.
01:05:19.392 --> 01:05:20.350
It's actually a circle.
01:05:20.350 --> 01:05:21.849
So there's going
to be a correction,
01:05:21.849 --> 01:05:23.720
and the correction is
going to be quadratic.
01:05:23.720 --> 01:05:25.636
If you work out that
correction, it's correct.
01:05:25.636 --> 01:05:28.410
The depth above the
bottom of the potential
01:05:28.410 --> 01:05:31.720
is correct for the energy of
the corresponding infinite well
01:05:31.720 --> 01:05:32.966
problem.
01:05:32.966 --> 01:05:35.920
I'll leave that to
you as an exercise.
01:05:35.920 --> 01:05:37.800
Other questions?
01:05:37.800 --> 01:05:39.210
OK.
01:05:39.210 --> 01:05:41.050
So there's another
limit of this system
01:05:41.050 --> 01:05:43.010
that's fun to think about.
01:05:43.010 --> 01:05:44.510
So this was the
infinite well limit.
01:05:47.110 --> 01:05:49.999
What I want to do is I
want to take advantage
01:05:49.999 --> 01:05:52.290
of the observation we made
a second ago that as we make
01:05:52.290 --> 01:05:54.680
the well deeper,
we get more states.
01:05:54.680 --> 01:05:58.720
As we make the well more
narrow, we get fewer states.
01:05:58.720 --> 01:06:05.300
To trade that off, consider
the following limit.
01:06:05.300 --> 01:06:14.829
I want to take a potential
well, which has a ground state.
01:06:14.829 --> 01:06:16.370
What does the ground
state look like?
01:06:16.370 --> 01:06:18.453
So the ground state wave
function is going to be--
01:06:18.453 --> 01:06:19.940
So here's zero.
01:06:19.940 --> 01:06:22.385
And here's
exponentially growing.
01:06:22.385 --> 01:06:23.800
Here's exponentially decreasing.
01:06:23.800 --> 01:06:26.425
And if it's a ground state, how
many nodes will it have inside?
01:06:28.010 --> 01:06:29.760
How many nodes will
the ground state have?
01:06:29.760 --> 01:06:30.634
AUDIENCE: Zero.
01:06:30.634 --> 01:06:31.300
PROFESSOR: Zero.
01:06:31.300 --> 01:06:31.820
Good.
01:06:31.820 --> 01:06:34.017
OK.
01:06:34.017 --> 01:06:36.100
So the ground state will
look something like this.
01:06:36.100 --> 01:06:39.340
We more conventionally
draw it like this.
01:06:39.340 --> 01:06:42.870
But just for fun, I'm going
to draw it in this fashion.
01:06:42.870 --> 01:06:45.330
In particular, it
has some slope here.
01:06:45.330 --> 01:06:46.700
And it has some slope here.
01:06:46.700 --> 01:06:47.280
Oh, shoot.
01:06:47.280 --> 01:06:48.800
Did I?
01:06:48.800 --> 01:06:50.028
Yes, I did.
01:06:50.028 --> 01:06:51.780
Dammit.
01:06:51.780 --> 01:06:55.430
I just erased the one thing
that I wanted to hold onto.
01:06:55.430 --> 01:06:55.930
OK.
01:06:55.930 --> 01:06:57.164
So there's my wave function.
01:06:57.164 --> 01:06:58.580
It has some
particular slope here.
01:06:58.580 --> 01:07:00.029
It has some
particular slope here.
01:07:00.029 --> 01:07:01.820
And this is the ground
state wave function,
01:07:01.820 --> 01:07:02.609
with some energy.
01:07:02.609 --> 01:07:03.150
I don't know.
01:07:03.150 --> 01:07:03.920
I'll call it this.
01:07:08.565 --> 01:07:10.190
Now, what I want to
do is we've already
01:07:10.190 --> 01:07:14.130
shown that as we make the well
more and more shallow and more
01:07:14.130 --> 01:07:17.849
and more narrow, the
energy of the ground state
01:07:17.849 --> 01:07:19.140
gets closer and closer to zero.
01:07:19.140 --> 01:07:20.870
But there remains
always a bound state.
01:07:20.870 --> 01:07:22.580
There is always at
least one bound state.
01:07:22.580 --> 01:07:24.230
We proved that.
01:07:24.230 --> 01:07:27.776
Proved, as a physicist would.
01:07:27.776 --> 01:07:28.650
So I want to do that.
01:07:28.650 --> 01:07:29.500
I want to take this seriously.
01:07:29.500 --> 01:07:31.166
But here's the limit
I want to consider.
01:07:31.166 --> 01:07:34.070
Consider the limit that we
make the potential v goes
01:07:34.070 --> 01:07:38.920
to infinity, v0, while
making l go to zero.
01:07:38.920 --> 01:07:41.010
So what I want to do is I
want to take this thing,
01:07:41.010 --> 01:07:44.464
and I want to make it
deeper and deeper, but more
01:07:44.464 --> 01:07:45.130
and more narrow.
01:07:52.320 --> 01:07:56.090
If I do this
repeatedly, eventually I
01:07:56.090 --> 01:07:58.860
will get a delta function.
01:07:58.860 --> 01:08:00.360
And I will get a
delta function if I
01:08:00.360 --> 01:08:02.820
hold the area of this guy fixed.
01:08:02.820 --> 01:08:03.320
Yeah?
01:08:03.320 --> 01:08:11.050
So if I do so, holding the
area under this plot fixed,
01:08:11.050 --> 01:08:13.876
I will get a delta function.
01:08:13.876 --> 01:08:14.900
Everyone cool with that?
01:08:18.347 --> 01:08:20.180
So let's think, though,
quickly about what's
01:08:20.180 --> 01:08:22.263
going to happen to the
ground state wave function.
01:08:27.240 --> 01:08:31.319
So as I make the potential,
let's take this wave function,
01:08:31.319 --> 01:08:34.979
and let's look at this
version of the potential.
01:08:34.979 --> 01:08:36.520
So as I make the
potential deeper,
01:08:36.520 --> 01:08:38.850
what happens to the rate
of the oscillation inside
01:08:38.850 --> 01:08:39.974
or to the curvature inside?
01:08:43.290 --> 01:08:44.000
It increases.
01:08:44.000 --> 01:08:44.665
Right?
01:08:44.665 --> 01:08:48.450
So the system is oscillating
more, it changes more rapidly,
01:08:48.450 --> 01:08:51.060
because phi double
dot or phi double
01:08:51.060 --> 01:08:53.655
prime is equal to v minus e phi.
01:08:53.655 --> 01:08:54.779
It oscillates more rapidly.
01:08:54.779 --> 01:08:56.749
So to make it deeper,
the system tends
01:08:56.749 --> 01:08:57.939
to oscillate more rapidly.
01:09:00.720 --> 01:09:04.290
However, as we make
it more narrow,
01:09:04.290 --> 01:09:08.389
the system doesn't have
as far to oscillate.
01:09:08.389 --> 01:09:09.930
So it oscillates
more rapidly, but it
01:09:09.930 --> 01:09:11.939
doesn't oscillate as far.
01:09:11.939 --> 01:09:13.284
So what's going to happen?
01:09:13.284 --> 01:09:16.410
Well, as we make it more
and more narrow and deeper
01:09:16.410 --> 01:09:21.450
and deeper, we again have
the wave function coming in.
01:09:21.450 --> 01:09:24.960
And now it oscillates
very rapidly.
01:09:24.960 --> 01:09:26.696
Let's do it again.
01:09:26.696 --> 01:09:31.630
The wave function comes in,
and it oscillates very rapidly.
01:09:31.630 --> 01:09:33.609
And the it evanescent tail out.
01:09:33.609 --> 01:09:39.450
And now as we have a delta
function, exponential damping,
01:09:39.450 --> 01:09:42.750
it oscillates extremely rapidly
over an arbitrarily short
01:09:42.750 --> 01:09:46.500
distance and gives
us the kink that we
01:09:46.500 --> 01:09:48.500
knew at the very beginning
we should expect when
01:09:48.500 --> 01:09:50.432
the potential is
a delta function.
01:09:50.432 --> 01:09:51.770
Right?
01:09:51.770 --> 01:09:54.069
From our qualitative
structure of the wave function
01:09:54.069 --> 01:09:55.330
at the very beginning
we saw that when
01:09:55.330 --> 01:09:56.788
we have a delta
function potential,
01:09:56.788 --> 01:09:59.620
we should see a kink
in the wave function.
01:09:59.620 --> 01:10:03.780
Because again, if we have phi
prime prime is delta function
01:10:03.780 --> 01:10:06.220
discontinuous, phi prime
is the integral of this.
01:10:06.220 --> 01:10:08.805
This is a step, and
phi is continuous.
01:10:13.330 --> 01:10:14.970
And so here we have
a step function.
01:10:14.970 --> 01:10:18.360
We get a discontinuity
in the second derivative.
01:10:18.360 --> 01:10:20.340
Here we have a delta
function in the potential,
01:10:20.340 --> 01:10:22.440
and we get a discontinuity
in the first derivative
01:10:22.440 --> 01:10:26.160
if we get a kink in
the wave function.
01:10:26.160 --> 01:10:27.914
Yeah?
01:10:27.914 --> 01:10:30.192
AUDIENCE: Would we
get a jump there?
01:10:30.192 --> 01:10:30.900
PROFESSOR: Sorry.
01:10:30.900 --> 01:10:31.910
Say again.
01:10:31.910 --> 01:10:32.631
For e1?
01:10:32.631 --> 01:10:33.256
AUDIENCE: Yeah.
01:10:33.256 --> 01:10:34.460
Would we get a jump?
01:10:34.460 --> 01:10:36.036
PROFESSOR: Very good question.
01:10:36.036 --> 01:10:37.410
So let me do this
more seriously.
01:10:37.410 --> 01:10:39.990
Let's do this more carefully.
01:10:39.990 --> 01:10:42.130
So the question is, for
the first excited state,
01:10:42.130 --> 01:10:43.727
do we get a jump?
01:10:43.727 --> 01:10:44.810
Do we get a discontinuity?
01:10:44.810 --> 01:10:46.680
What do we get for the
first excited state?
01:10:46.680 --> 01:10:47.225
Right?
01:10:47.225 --> 01:10:48.750
So let's talk about
that in detail.
01:10:48.750 --> 01:10:50.040
It's a very good question.
01:10:50.040 --> 01:10:56.132
Example v is equal to
minus v0 delta of x.
01:10:56.132 --> 01:10:58.090
Now, here I want to just
warn you of something.
01:10:58.090 --> 01:11:00.670
This is totally standard
notation for these problems,
01:11:00.670 --> 01:11:02.950
but you should be
careful about dimensions.
01:11:02.950 --> 01:11:06.020
What are the dimensions of
v0, the parameter of v0?
01:11:09.070 --> 01:11:10.650
It's tempting to say energy.
01:11:10.650 --> 01:11:11.377
That's an energy.
01:11:11.377 --> 01:11:12.085
That's an energy.
01:11:12.085 --> 01:11:12.890
But wait.
01:11:12.890 --> 01:11:14.823
What are the dimensions
of the delta function?
01:11:14.823 --> 01:11:15.610
AUDIENCE: Length.
01:11:15.610 --> 01:11:16.985
PROFESSOR: Whatever
length right?
01:11:16.985 --> 01:11:20.190
Because we know that
delta of alpha x
01:11:20.190 --> 01:11:23.500
is equal to 1 over
norm alpha delta of x.
01:11:27.290 --> 01:11:29.047
So if I write delta
of x, which is always
01:11:29.047 --> 01:11:31.380
a slightly ballsy thing to
do because this should really
01:11:31.380 --> 01:11:33.421
be dimensionless, but if
I write delta of x, then
01:11:33.421 --> 01:11:35.470
this has units of 1
over length, which
01:11:35.470 --> 01:11:39.270
means this must have units
of length times energy.
01:11:42.081 --> 01:11:42.580
OK.
01:11:42.580 --> 01:11:44.700
Just a little warning,
when you check your answers
01:11:44.700 --> 01:11:46.030
on a problem, you
always want to make sure
01:11:46.030 --> 01:11:47.820
that they're
dimensionally consistent.
01:11:47.820 --> 01:11:49.486
And so it will be
important to make sure
01:11:49.486 --> 01:11:52.810
that you use the
energy times the length
01:11:52.810 --> 01:11:55.330
for the dimensions
of that beast.
01:11:55.330 --> 01:11:57.910
So my question here is,
is there a bound state?
01:11:57.910 --> 01:11:59.580
So for this example,
for this potential,
01:11:59.580 --> 01:12:02.610
the delta function
potential bound state, which
01:12:02.610 --> 01:12:09.890
again is this guy, is
there a bound state?
01:12:12.460 --> 01:12:16.310
So again, we just ran
through the intuition
01:12:16.310 --> 01:12:19.410
where we made the potential
deep and deeper and deeper,
01:12:19.410 --> 01:12:26.860
v0 divided by epsilon
over width epsilon.
01:12:26.860 --> 01:12:33.710
So v0, in order for
this to be an energy
01:12:33.710 --> 01:12:35.689
has to be an energy
times a length
01:12:35.689 --> 01:12:37.480
because we're going to
divide it by length.
01:12:37.480 --> 01:12:39.313
So this is going to
give us a delta function
01:12:39.313 --> 01:12:40.290
potential in the limit.
01:12:40.290 --> 01:12:41.748
We have an intuition
that we should
01:12:41.748 --> 01:12:43.209
get a bound state with a kink.
01:12:43.209 --> 01:12:44.500
But let's check that intuition.
01:12:44.500 --> 01:12:47.300
We want to actually
solve this problem.
01:12:47.300 --> 01:12:49.170
So we'll do the same
thing we did before.
01:12:49.170 --> 01:12:55.390
We now write the general
solution in the places
01:12:55.390 --> 01:12:58.610
where the potential is
constant, which is on the left
01:12:58.610 --> 01:12:59.920
and on the right.
01:12:59.920 --> 01:13:02.460
And then we want to impose
appropriate boundary conditions
01:13:02.460 --> 01:13:05.562
at the interface and at
infinity, where these are going
01:13:05.562 --> 01:13:08.020
to be normalizable, and this
is whatever the right boundary
01:13:08.020 --> 01:13:08.590
conditions are.
01:13:08.590 --> 01:13:10.050
So we are going to have to
derive the appropriate boundary
01:13:10.050 --> 01:13:10.700
conditions.
01:13:10.700 --> 01:13:12.020
So let's just do that quickly.
01:13:12.020 --> 01:13:17.710
So phi with definite
e is equal to a--
01:13:17.710 --> 01:13:20.020
So in this region
in the left, it's
01:13:20.020 --> 01:13:22.270
either growing or
decreasing exponential.
01:13:22.270 --> 01:13:27.010
So ae to the alpha x plus
be to the minus alpha x.
01:13:27.010 --> 01:13:29.370
And this is x less than 0.
01:13:29.370 --> 01:13:34.130
And ce to the alpha x
plus de to the minus
01:13:34.130 --> 01:13:36.970
alpha x for x greater than 0.
01:13:36.970 --> 01:13:39.047
So first off, let's
hit normalizability.
01:13:39.047 --> 01:13:40.630
What must be true
for normalizability?
01:13:43.255 --> 01:13:45.740
Yeah, they'd better be
converging to zero here
01:13:45.740 --> 01:13:48.840
and converging to zero here,
which means that c had better
01:13:48.840 --> 01:13:52.381
vanish and b had better vanish.
01:13:52.381 --> 01:13:52.880
OK.
01:13:52.880 --> 01:13:56.260
So those guys are gone
from normalizability.
01:13:56.260 --> 01:13:59.170
And meanwhile, if
this is symmetric,
01:13:59.170 --> 01:14:01.790
what is going to be true
of the ground state?
01:14:01.790 --> 01:14:03.090
It's going to be symmetric.
01:14:03.090 --> 01:14:07.214
So a must be equal to d.
01:14:07.214 --> 01:14:08.490
Great.
01:14:08.490 --> 01:14:12.220
So a is now just some overall
normalization constant,
01:14:12.220 --> 01:14:13.720
which we can fix
from normalization.
01:14:13.720 --> 01:14:15.595
So it looks like this
should be the solution.
01:14:15.595 --> 01:14:17.060
We have an exponential.
01:14:17.060 --> 01:14:18.420
We have an exponential.
01:14:18.420 --> 01:14:23.080
But there's one more
boundary condition to fix.
01:14:23.080 --> 01:14:25.440
We have to satisfy
some matching.
01:14:25.440 --> 01:14:27.190
We have to satisfy the
boundary conditions
01:14:27.190 --> 01:14:28.106
at the delta function.
01:14:28.106 --> 01:14:30.310
So what are those?
01:14:30.310 --> 01:14:32.770
What are those
matching conditions?
01:14:32.770 --> 01:14:36.820
So we can get that from the
energy eigenvalue equation,
01:14:36.820 --> 01:14:39.850
which says that
phi prime prime is
01:14:39.850 --> 01:14:41.851
equal to h bar squared upon 2m.
01:14:41.851 --> 01:14:42.350
Sorry.
01:14:45.027 --> 01:14:46.110
Get your dimensions right.
01:14:46.110 --> 01:14:49.680
So it's 2m over h bar
squared v minus e.
01:14:49.680 --> 01:14:54.470
In this case, v is equal
to minus v0 delta function.
01:14:58.690 --> 01:14:59.910
That's very strange.
01:14:59.910 --> 01:15:05.392
So minus 2m over h bar squared
v0 delta of x minus e--
01:15:05.392 --> 01:15:06.350
I pulled out the minus.
01:15:06.350 --> 01:15:09.360
--so plus e phi.
01:15:12.090 --> 01:15:13.860
So this must be
true at every point.
01:15:13.860 --> 01:15:15.318
This of course, is
zero everywhere,
01:15:15.318 --> 01:15:20.460
except for at the origin.
01:15:20.460 --> 01:15:22.420
So what we want to do
is we want to turn this
01:15:22.420 --> 01:15:24.029
into a boundary condition.
01:15:24.029 --> 01:15:25.820
And we know what the
boundary condition is.
01:15:25.820 --> 01:15:28.604
If v is a delta function, that
means that phi prime prime
01:15:28.604 --> 01:15:30.270
is also a delta
function or proportional
01:15:30.270 --> 01:15:31.390
to a delta function.
01:15:31.390 --> 01:15:33.350
That means that phi
prime is a step function.
01:15:33.350 --> 01:15:34.308
And how did I get that?
01:15:34.308 --> 01:15:36.622
I got that by integrating
phi prime prime.
01:15:36.622 --> 01:15:38.955
You integrate a delta function,
you get a step function.
01:15:38.955 --> 01:15:39.960
Well, that's cool.
01:15:39.960 --> 01:15:42.700
How do we figure out what
step function discontinuity
01:15:42.700 --> 01:15:43.750
gives us?
01:15:43.750 --> 01:15:44.530
Let's integrate.
01:15:44.530 --> 01:15:46.488
Let's integrate right
across the delta function
01:15:46.488 --> 01:15:48.730
and figure out what
the discontinuity is.
01:15:48.730 --> 01:15:50.880
So let's take this
equation, integrate it
01:15:50.880 --> 01:15:52.649
from minus epsilon
to epsilon, where
01:15:52.649 --> 01:15:53.940
epsilon is a very small number.
01:15:56.780 --> 01:15:58.150
And that's epsilon to epsilon.
01:15:58.150 --> 01:16:00.025
So what is this going
to give us on the left?
01:16:00.025 --> 01:16:01.570
Well, integral of
a total derivative
01:16:01.570 --> 01:16:04.680
is just the value of the
thing at a value to the point.
01:16:04.680 --> 01:16:09.040
So this is going to be
phi prime at epsilon
01:16:09.040 --> 01:16:11.620
minus phi prime
and minus epsilon.
01:16:11.620 --> 01:16:12.460
What does that mean?
01:16:12.460 --> 01:16:15.230
The difference
between the derivative
01:16:15.230 --> 01:16:17.440
just after the origin and
just before the origin.
01:16:17.440 --> 01:16:20.350
This is the discontinuity
for very small epsilon.
01:16:20.350 --> 01:16:22.220
This is the discontinuity
of the derivative
01:16:22.220 --> 01:16:23.720
at the origin at
the delta function.
01:16:23.720 --> 01:16:26.053
And we already expected it
to have a step discontinuity.
01:16:26.053 --> 01:16:26.980
And there it is.
01:16:26.980 --> 01:16:28.242
And how big is it?
01:16:28.242 --> 01:16:29.700
Well on the right
hand side we have
01:16:29.700 --> 01:16:32.475
2m minus 2m upon h bar squared.
01:16:32.475 --> 01:16:33.850
And we're going
to get two terms.
01:16:33.850 --> 01:16:36.050
We get a term from
integrating the first term.
01:16:36.050 --> 01:16:39.700
But over this narrow
window, around,
01:16:39.700 --> 01:16:43.130
let's say epsilon was here, over
this narrow window we can treat
01:16:43.130 --> 01:16:47.160
the wave function as being
more or less constant.
01:16:47.160 --> 01:16:51.042
But in any case,
it's continuous.
01:16:51.042 --> 01:16:52.250
And this is a delta function.
01:16:52.250 --> 01:16:54.874
So we know what we get from the
integral of the delta function.
01:16:54.874 --> 01:16:58.870
We just get the value v0
phi at the delta function.
01:16:58.870 --> 01:17:01.660
Phi of the zero at the
delta functions, so phi 0.
01:17:01.660 --> 01:17:05.230
We get a second term, which
is plus the energy integrated
01:17:05.230 --> 01:17:06.491
against phi.
01:17:06.491 --> 01:17:07.490
The energy's a constant.
01:17:07.490 --> 01:17:09.150
And phi is continuous.
01:17:09.150 --> 01:17:11.350
So this, whatever else
you can say about it,
01:17:11.350 --> 01:17:12.995
is roughly the
constant value of phi
01:17:12.995 --> 01:17:15.220
at the origin times the
energy times the width, which
01:17:15.220 --> 01:17:17.070
is epsilon.
01:17:17.070 --> 01:17:20.466
So plus-order epsilon terms.
01:17:20.466 --> 01:17:21.984
Everyone cool with that?
01:17:21.984 --> 01:17:23.400
So now what I'm
going to do is I'm
01:17:23.400 --> 01:17:25.399
going to take the limit
as epsilon goes to zero.
01:17:28.830 --> 01:17:31.540
So I'm just going to take
that the discontinuity just
01:17:31.540 --> 01:17:32.040
across zero.
01:17:32.040 --> 01:17:37.590
So this is going to give
me, of this gives me
01:17:37.590 --> 01:17:41.240
the change in the
slope at the origin.
01:17:41.240 --> 01:17:42.090
OK.
01:17:42.090 --> 01:17:43.590
The derivative just
after the origin
01:17:43.590 --> 01:17:46.170
minus the derivative just
before the origin is equal to--
01:17:46.170 --> 01:17:48.490
These order epsilon
terms go away.
01:17:48.490 --> 01:17:54.800
--minus 2m upon h
bar squared v0 phi.
01:17:54.800 --> 01:17:59.060
So that's my continuity.
01:17:59.060 --> 01:18:00.680
That's the condition
for continuity
01:18:00.680 --> 01:18:03.410
of the derivative and
appropriate discontinuity
01:18:03.410 --> 01:18:07.540
of the first derivative
at the origin.
01:18:07.540 --> 01:18:10.510
And so this, when we
plug-in these values
01:18:10.510 --> 01:18:12.390
of this form for
the wave function,
01:18:12.390 --> 01:18:14.390
when we take a derivative,
all we're going to do
01:18:14.390 --> 01:18:16.919
is we're going to
pick up an alpha.
01:18:16.919 --> 01:18:18.460
And so when we work
all of this out--
01:18:18.460 --> 01:18:20.879
I'm not going to go
through the algebra.
01:18:20.879 --> 01:18:22.920
You're going to go through
it on the problem set.
01:18:22.920 --> 01:18:25.450
--when we take this condition,
when we impose this condition
01:18:25.450 --> 01:18:27.270
with this wave
function, it gives us
01:18:27.270 --> 01:18:28.870
a very specific value for alpha.
01:18:28.870 --> 01:18:33.560
This is only solvable
if alpha is equal to mv0
01:18:33.560 --> 01:18:34.640
upon h bar squared.
01:18:38.840 --> 01:18:39.830
Good.
01:18:39.830 --> 01:18:41.310
So let's just check the units.
01:18:41.310 --> 01:18:44.870
So this is momentum times
length, momentum times length.
01:18:44.870 --> 01:18:46.600
This is mass.
01:18:46.600 --> 01:18:48.940
This is an energy
times a length.
01:18:48.940 --> 01:18:52.770
So this has overall units
of, p squared over m,
01:18:52.770 --> 01:18:55.300
overall units of 1 upon
length, which is what we wan.
01:18:55.300 --> 01:18:56.900
So that's good.
01:18:56.900 --> 01:18:59.030
So we get alpha is
equal to mv0 upon h bar.
01:18:59.030 --> 01:19:00.890
And that gives us the
form of the potential.
01:19:00.890 --> 01:19:03.640
And it also tells us that the
energy, plugging this back in,
01:19:03.640 --> 01:19:08.530
is equal to minus h bar
squared alpha squared upon 2m,
01:19:08.530 --> 01:19:10.770
which we could then
plug-in the value of alpha
01:19:10.770 --> 01:19:12.110
and solve for v0.
01:19:12.110 --> 01:19:15.487
So what we found is that
there is a single bound state
01:19:15.487 --> 01:19:17.070
of the delta function
potential, which
01:19:17.070 --> 01:19:19.028
we could have gotten by
just taking this limit.
01:19:19.028 --> 01:19:20.960
It's a fun way to
rederive the same result.
01:19:20.960 --> 01:19:22.740
It's a nice check on
your understanding.
01:19:22.740 --> 01:19:25.370
So we find that there's
a single bound state
01:19:25.370 --> 01:19:27.110
of the delta function potential.
01:19:27.110 --> 01:19:29.980
Now, what about an
odd bound state?
01:19:29.980 --> 01:19:32.180
We assumed at an important
point that this was even.
01:19:32.180 --> 01:19:35.610
What if I assume
that it was odd?
01:19:35.610 --> 01:19:38.120
One node, what if we had
assumed that it was odd?
01:19:38.120 --> 01:19:40.020
What would be true
of the wave function?
01:19:40.020 --> 01:19:42.860
Well for odd, this would be
a, and this would be minus a.
01:19:42.860 --> 01:19:46.100
So the value of the wave
function at the origin is what?
01:19:46.100 --> 01:19:46.700
Zero.
01:19:46.700 --> 01:19:49.430
So that tells us the value
of the wave function is zero.
01:19:49.430 --> 01:19:51.780
What's the discontinuity?
01:19:51.780 --> 01:19:52.340
Zero.
01:19:52.340 --> 01:19:54.760
So it's as if
there's no potential
01:19:54.760 --> 01:19:57.480
because it has a zero right
at the delta function.
01:19:57.480 --> 01:19:59.430
Yeah?
01:19:59.430 --> 01:20:03.160
But that means that this wave
function, an odd wave function,
01:20:03.160 --> 01:20:06.980
doesn't notice the delta
function potential.
01:20:06.980 --> 01:20:09.430
So is there a bound state?
01:20:09.430 --> 01:20:11.100
No.
01:20:11.100 --> 01:20:13.630
So how many bound
states are there?
01:20:13.630 --> 01:20:17.180
Always exactly one for the
single isolated delta function.
01:20:17.180 --> 01:20:18.560
On your problem
set, you're going
01:20:18.560 --> 01:20:21.390
to use the result of the
single isolated delta function,
01:20:21.390 --> 01:20:24.520
and more broadly you're
going to derive the results
01:20:24.520 --> 01:20:26.050
for two delta functions.
01:20:26.050 --> 01:20:28.310
So you might say, why
two delta functions?
01:20:28.310 --> 01:20:30.470
And the answer is,
the two delta function
01:20:30.470 --> 01:20:33.080
problem, which
involves no math--
01:20:33.080 --> 01:20:33.580
Right?
01:20:33.580 --> 01:20:35.670
It's a totally straightforward,
simple problem.
01:20:35.670 --> 01:20:38.400
You can all do it right
now on a piece of paper.
01:20:38.400 --> 01:20:39.780
The two delta
function problem is
01:20:39.780 --> 01:20:42.370
going to turn out to
be an awesome model
01:20:42.370 --> 01:20:44.515
for the binding of atoms.
01:20:44.515 --> 01:20:47.810
And we're going to use it as
intuition on your problem set
01:20:47.810 --> 01:20:50.890
to explain how quantum
mechanical effects can lead
01:20:50.890 --> 01:20:54.390
to an attractive force
between two atoms.
01:20:54.390 --> 01:20:55.760
See you next time.
01:20:55.760 --> 01:20:57.310
[APPLAUSE]