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PROFESSOR: Today we begin
with the harmonic oscillator.
00:00:29.900 --> 00:00:33.010
And before we get into
the harmonic oscillator,
00:00:33.010 --> 00:00:35.840
I want to touch on
a few concepts that
00:00:35.840 --> 00:00:40.690
have been mentioned in class
and just elaborate on them.
00:00:40.690 --> 00:00:45.550
It is the issue of nodes,
and how solutions look at,
00:00:45.550 --> 00:00:48.920
and why solutions have
more and more nodes, why
00:00:48.920 --> 00:00:52.360
the ground state has no nodes.
00:00:52.360 --> 00:00:55.220
This kind of stuff.
00:00:55.220 --> 00:01:02.600
So these are just a collection
of remarks and an argument
00:01:02.600 --> 00:01:07.830
for you to understand a
little more intuitively why
00:01:07.830 --> 00:01:11.620
these properties hold.
00:01:11.620 --> 00:01:15.430
So one first thing
I want to mention
00:01:15.430 --> 00:01:20.010
is, if you have a Schrodinger
equation for an energy
00:01:20.010 --> 00:01:21.020
eigenstate.
00:01:21.020 --> 00:01:28.040
Schrodinger equation for
an energy eigenstate.
00:01:34.180 --> 00:01:43.060
You have an equation of the
from minus h squared over 2m.
00:01:43.060 --> 00:01:53.230
d second dx squared psi of
x plus v of x psi of x equal
00:01:53.230 --> 00:01:57.360
e times psi of x.
00:01:57.360 --> 00:02:00.030
Now the issue of
this equation is
00:02:00.030 --> 00:02:05.820
that you're trying to solve for
two things at the same time.
00:02:05.820 --> 00:02:09.550
If you're looking at what
we call bound states, now
00:02:09.550 --> 00:02:11.570
what is a bound state?
00:02:11.570 --> 00:02:18.040
A bound state is something that
is not extended all that much.
00:02:18.040 --> 00:02:22.270
So a bound state will
be a wave function
00:02:22.270 --> 00:02:29.210
that goes to 0 as the absolute
value of x goes to infinity.
00:02:29.210 --> 00:02:34.930
So it's a probability function
that certainly doesn't
00:02:34.930 --> 00:02:36.980
extend all the way to infinity.
00:02:36.980 --> 00:02:39.090
It just collapses.
00:02:39.090 --> 00:02:41.000
It's normalizable.
00:02:41.000 --> 00:02:42.790
So these are bound
states, and you're
00:02:42.790 --> 00:02:45.560
looking for bound
states of this equation.
00:02:45.560 --> 00:02:49.680
And your difficulty is
that you don't know psi,
00:02:49.680 --> 00:02:52.160
and you don't know E either.
00:02:52.160 --> 00:02:57.700
So you have to solve a problem
in which, if you were thinking
00:02:57.700 --> 00:03:00.570
oh this is just a plain
differential equation,
00:03:00.570 --> 00:03:03.530
give me the value of E.
We know the potential,
00:03:03.530 --> 00:03:04.650
just calculate it.
00:03:04.650 --> 00:03:07.880
That's not the way it
works in quantum mechanics,
00:03:07.880 --> 00:03:11.450
because you need to have
normalizable solutions.
00:03:11.450 --> 00:03:15.380
So at the end of the day,
as will be very clear today,
00:03:15.380 --> 00:03:19.210
this E gets fixed.
00:03:19.210 --> 00:03:22.740
You cannot get
arbitrary values of E's.
00:03:22.740 --> 00:03:27.060
So I want to make a couple of
remarks about this equation.
00:03:27.060 --> 00:03:29.635
Is that there's this
thing that can't happen.
00:03:34.230 --> 00:03:48.300
Certainly, if v of x is a smooth
potential, then if you observer
00:03:48.300 --> 00:03:52.602
that the wave function
vanishes at some point,
00:03:52.602 --> 00:03:55.880
and the derivative
of the wave function
00:03:55.880 --> 00:03:59.600
vanishes at that same
point, these two things
00:03:59.600 --> 00:04:05.070
imply that psi of
x is identically 0.
00:04:05.070 --> 00:04:08.000
And therefore it
means that you really
00:04:08.000 --> 00:04:09.530
are not interested in that.
00:04:09.530 --> 00:04:12.430
That's not a solution of
the Schrodinger equation.
00:04:12.430 --> 00:04:16.530
Psi equals 0 is obviously solves
this, but it's not interesting.
00:04:16.530 --> 00:04:19.250
It doesn't represent
the particle.
00:04:19.250 --> 00:04:23.740
So what I claim here
is that, if it happens
00:04:23.740 --> 00:04:25.540
to be that you're
solving the Schrodinger
00:04:25.540 --> 00:04:28.035
problem with some
potential that is smooth,
00:04:28.035 --> 00:04:30.960
you can take derivatives of it.
00:04:30.960 --> 00:04:34.670
And then you encounter that
the wave function vanishes
00:04:34.670 --> 00:04:38.830
at some point, and its slope
vanishes at that same point.
00:04:38.830 --> 00:04:41.970
Then the wave function
vanishes completely.
00:04:41.970 --> 00:04:45.120
So you cannot have
a wave function,
00:04:45.120 --> 00:04:48.210
a psi of x that
does the following.
00:04:48.210 --> 00:04:51.510
Comes down here.
00:04:51.510 --> 00:04:54.850
It becomes an inflection
point and goes down.
00:04:54.850 --> 00:04:56.105
This is not allowed.
00:05:03.160 --> 00:05:07.610
If the wave function
vanishes at some point,
00:05:07.610 --> 00:05:11.940
then the wave function
is going to do this.
00:05:11.940 --> 00:05:16.450
It's going to hit at an angle,
because you cannot have that
00:05:16.450 --> 00:05:20.640
the wave function is 0 and needs
the derivative 0 at the same
00:05:20.640 --> 00:05:21.460
point.
00:05:21.460 --> 00:05:23.330
And the reason is simple.
00:05:23.330 --> 00:05:25.390
I'm not going to
prove it now here.
00:05:25.390 --> 00:05:29.930
It is that you have a second
order differential equation,
00:05:29.930 --> 00:05:32.200
and a second order
differential equation
00:05:32.200 --> 00:05:35.680
is completely determined
by knowing the function
00:05:35.680 --> 00:05:38.800
at the point and the the
derivative at the point.
00:05:38.800 --> 00:05:42.570
And if both are 0s, like
the most trivial kind
00:05:42.570 --> 00:05:46.610
of initial condition, the only
solution consistent with this
00:05:46.610 --> 00:05:49.610
is psi equals 0 everywhere.
00:05:49.610 --> 00:05:51.600
So this can't happen.
00:05:51.600 --> 00:05:54.460
And it's something good
for you to remember.
00:05:54.460 --> 00:05:57.040
If you have to do a
plot of a wave function,
00:05:57.040 --> 00:05:59.240
you should never have this.
00:05:59.240 --> 00:06:04.855
So this is what we call the
node in a wave function.
00:06:10.280 --> 00:06:13.780
It's a place where the
wave function vanishes,
00:06:13.780 --> 00:06:15.800
and the derivative
of the wave function
00:06:15.800 --> 00:06:17.360
better not vanish at that point.
00:06:21.720 --> 00:06:28.590
So this is one claim that
it's not hard to prove,
00:06:28.590 --> 00:06:30.830
but we just don't try to do it.
00:06:30.830 --> 00:06:35.270
And there's another claim that
I want you to be aware of.
00:06:35.270 --> 00:06:43.470
That for bound states
in one dimension,
00:06:43.470 --> 00:06:50.650
in the kind of thing that we're
doing now in one dimension,
00:06:50.650 --> 00:06:55.435
no degeneracy is possible.
00:07:01.410 --> 00:07:03.160
What do I mean by that?
00:07:03.160 --> 00:07:07.100
You will never find
two bound states
00:07:07.100 --> 00:07:09.640
of a potential that
are different that
00:07:09.640 --> 00:07:10.730
have the same energy.
00:07:10.730 --> 00:07:14.620
It's just absolutely impossible.
00:07:14.620 --> 00:07:19.076
It's a very wonderful
result, but also I'm
00:07:19.076 --> 00:07:20.200
not going to prove it here.
00:07:20.200 --> 00:07:24.070
Maybe it will be given as an
exercise later in the course.
00:07:24.070 --> 00:07:27.830
And it's discussed
in 805 as well.
00:07:27.830 --> 00:07:31.540
But that's another statement
that is very important.
00:07:31.540 --> 00:07:33.630
There's no degeneracy.
00:07:33.630 --> 00:07:38.610
Now you've looked at
this simple potential,
00:07:38.610 --> 00:07:45.910
the square well infinite one.
00:07:45.910 --> 00:07:47.430
And how does it look?
00:07:47.430 --> 00:07:53.670
You have x going from 0 to a.
00:07:53.670 --> 00:07:56.220
And the potential is 0.
00:07:56.220 --> 00:07:58.660
From 0 to a is
infinite otherwise.
00:07:58.660 --> 00:08:02.540
The particle is bound
to stay inside the two
00:08:02.540 --> 00:08:04.920
walls of this potential.
00:08:04.920 --> 00:08:08.830
So in here we've plotted the
potential as a function of x.
00:08:08.830 --> 00:08:11.270
Here is x.
00:08:11.270 --> 00:08:13.890
And the wave functions are
things that you know already.
00:08:13.890 --> 00:08:14.649
Yes?
00:08:14.649 --> 00:08:17.190
AUDIENCE: Is it true if you have
two wells next to each other
00:08:17.190 --> 00:08:20.299
that there's still no degeneracy
if it's an infinite barrier?
00:08:20.299 --> 00:08:22.090
PROFESSOR: If there's
two wells and there's
00:08:22.090 --> 00:08:24.370
an infinite barrier
between them,
00:08:24.370 --> 00:08:26.600
it's like having two universes.
00:08:26.600 --> 00:08:31.250
So it's not really a
one dimensional problem.
00:08:31.250 --> 00:08:33.870
If you have an infinite
barrier like two worlds
00:08:33.870 --> 00:08:35.250
that can talk to each other.
00:08:35.250 --> 00:08:37.750
So yes, then you
would have degeneracy.
00:08:37.750 --> 00:08:42.570
It's like saying you can have
here one atom of hydrogen
00:08:42.570 --> 00:08:46.200
or something in one energy
level, here another one.
00:08:46.200 --> 00:08:47.550
They don't talk to each other.
00:08:47.550 --> 00:08:49.520
They're degenerate states.
00:08:49.520 --> 00:08:53.375
But in general, we're talking
about normal potentials
00:08:53.375 --> 00:08:55.770
that are preferably smooth.
00:08:55.770 --> 00:08:59.170
Most of these things are true
even if they're not smooth.
00:08:59.170 --> 00:09:01.140
But it's a little more delicate.
00:09:01.140 --> 00:09:02.840
But certainly two
potentials that
00:09:02.840 --> 00:09:05.030
are separated by
an infinite barrier
00:09:05.030 --> 00:09:10.440
is not part of what we
really want to consider.
00:09:10.440 --> 00:09:16.870
OK so these wave functions
start with m equals 0, 1, 2,
00:09:16.870 --> 00:09:25.040
3, and psi ends of x are
square root of 2 over a sin n
00:09:25.040 --> 00:09:28.340
plus 1 pi x over a.
00:09:28.340 --> 00:09:30.700
Things that you've seen already.
00:09:30.700 --> 00:09:36.030
And En, the energies
are ever growing
00:09:36.030 --> 00:09:41.680
as a function of the integer
n that characterizes them.
00:09:41.680 --> 00:09:44.650
2ma squared.
00:09:44.650 --> 00:09:49.750
And the thing that you
notice is that psi 0
00:09:49.750 --> 00:09:52.790
has technically no nodes.
00:09:52.790 --> 00:09:55.120
That is to say these
wave functions have
00:09:55.120 --> 00:09:58.420
to varnish at the end, because
the potential becomes infinite,
00:09:58.420 --> 00:10:01.340
which means a particle
really can go through.
00:10:01.340 --> 00:10:03.370
The wave function
has to be continuous.
00:10:03.370 --> 00:10:06.570
There cannot be any wave
function to the left.
00:10:06.570 --> 00:10:08.730
So it has to vanish here.
00:10:08.730 --> 00:10:11.350
These things don't
count as nodes.
00:10:11.350 --> 00:10:16.020
It is like a bound state
has to vanish at infinity.
00:10:16.020 --> 00:10:18.470
And that's not what
we count the node.
00:10:18.470 --> 00:10:20.910
A node is somewhere
in the middle
00:10:20.910 --> 00:10:24.030
of the range of x where
the wave function vanishes.
00:10:24.030 --> 00:10:25.640
So this is the ground state.
00:10:25.640 --> 00:10:30.750
This is psi zero has no nodes.
00:10:30.750 --> 00:10:35.500
Psi one would be something
like that, has one node.
00:10:39.170 --> 00:10:42.900
And try the next ones.
00:10:42.900 --> 00:10:44.330
They have more and more nodes.
00:10:44.330 --> 00:10:47.015
So psi n has n nodes.
00:10:51.600 --> 00:10:58.480
And the interesting thing
is that this result actually
00:10:58.480 --> 00:11:02.550
is true for extremely
general potentials.
00:11:02.550 --> 00:11:05.200
You don't have to just
do the square well
00:11:05.200 --> 00:11:08.240
to see that the ground
state has no nodes.
00:11:08.240 --> 00:11:12.580
That first excited state was one
node, and so on and so forth.
00:11:12.580 --> 00:11:16.580
It's true in general.
00:11:16.580 --> 00:11:18.710
This is actually a
very nice result,
00:11:18.710 --> 00:11:21.590
but its difficult to prove.
00:11:21.590 --> 00:11:26.370
In fact, it's pretty
hard to prove.
00:11:26.370 --> 00:11:28.590
So there is a nice argument.
00:11:28.590 --> 00:11:32.540
Not 100% rigorous,
but thoroughly nice
00:11:32.540 --> 00:11:34.720
and really physical
that I'm going
00:11:34.720 --> 00:11:37.890
to present to you why
this result is true.
00:11:37.890 --> 00:11:40.030
So let's try to do that.
00:11:40.030 --> 00:11:41.430
So here is the general case.
00:11:41.430 --> 00:11:46.470
So I'm going to take
a smooth v of x.
00:11:46.470 --> 00:11:48.160
That will be of this kind.
00:11:48.160 --> 00:11:55.310
The potential, here is
x, and this potential
00:11:55.310 --> 00:12:01.210
is going to be like that.
00:12:01.210 --> 00:12:02.325
Smooth all over.
00:12:05.460 --> 00:12:09.440
And I will actually
have it that it actually
00:12:09.440 --> 00:12:12.630
goes to infinity as
x goes to infinity.
00:12:12.630 --> 00:12:15.290
Many of these things are
really not necessary,
00:12:15.290 --> 00:12:19.610
but it simplifies our life.
00:12:24.290 --> 00:12:33.272
OK, so here is a result
that it's known to be true.
00:12:33.272 --> 00:12:36.390
If this thing grows to
infinity, the potential never
00:12:36.390 --> 00:12:41.210
stops growing, you get
infinite number of bound states
00:12:41.210 --> 00:12:42.680
at fixed energies.
00:12:42.680 --> 00:12:45.380
One energy, two
energy, three energy,
00:12:45.380 --> 00:12:47.330
infinite number of bound states.
00:12:50.710 --> 00:12:53.125
Number of bound states.
00:12:57.020 --> 00:12:58.550
That's a fact.
00:12:58.550 --> 00:13:01.810
I will not try to prove it.
00:13:01.810 --> 00:13:04.000
We'll do it for the
harmonic oscillator.
00:13:04.000 --> 00:13:05.750
We'll see those infinite
number of states,
00:13:05.750 --> 00:13:09.110
but here we can't
prove it easily.
00:13:09.110 --> 00:13:11.120
Nevertheless, what
I want to argue
00:13:11.120 --> 00:13:15.510
for you is that these
states, as the first one,
00:13:15.510 --> 00:13:17.690
will have no nodes.
00:13:17.690 --> 00:13:20.490
The second state, the
first excited state,
00:13:20.490 --> 00:13:21.710
will have one node.
00:13:21.710 --> 00:13:24.290
Next will have two nodes,
three nodes, four nodes.
00:13:24.290 --> 00:13:29.050
We want to understand what
is this issue of the nodes.
00:13:29.050 --> 00:13:31.270
OK.
00:13:31.270 --> 00:13:34.000
You're not trying
to prove everything.
00:13:34.000 --> 00:13:36.360
But we're trying to
prove or understand
00:13:36.360 --> 00:13:39.120
something that is
very important.
00:13:39.120 --> 00:13:40.780
So how do we prove this?
00:13:40.780 --> 00:13:44.800
Or how do we understand that
the nodes-- so there will
00:13:44.800 --> 00:13:46.880
be an infinite number
of bound states,
00:13:46.880 --> 00:13:53.440
psi 0, psi 1, psi 2, up
to psi n, and it goes on.
00:13:53.440 --> 00:13:57.485
And psi n has n nodes.
00:14:01.410 --> 00:14:02.090
All right.
00:14:06.290 --> 00:14:11.030
So what I'm going to do, in
order to understand this,
00:14:11.030 --> 00:14:14.230
is I'm going to produce what we
will call screened potentials.
00:14:16.930 --> 00:14:19.620
Screened potentials.
00:14:23.370 --> 00:14:28.390
I'm going to select the
lowest point of the potential
00:14:28.390 --> 00:14:29.990
here for convenience.
00:14:29.990 --> 00:14:35.000
And I'm going to call it x
equals 0 is the lowest point.
00:14:40.770 --> 00:14:46.860
And the screen potential
will have a parameter a.
00:14:46.860 --> 00:14:52.280
It's a potential
which is equal to v
00:14:52.280 --> 00:14:57.470
of x if the absolute
value of x is less than a.
00:14:57.470 --> 00:15:04.580
And its infinity if the absolute
value of x is greater than a.
00:15:04.580 --> 00:15:07.810
So I come here,
and I want to see
00:15:07.810 --> 00:15:11.410
this is this
potential, v of x, what
00:15:11.410 --> 00:15:14.651
is the screened
potential for sum a?
00:15:14.651 --> 00:15:19.170
Well, I wanted colored
chalk, but I don't have it.
00:15:19.170 --> 00:15:23.190
I go mark a here minus a.
00:15:23.190 --> 00:15:26.600
Here are the points
between x and a.
00:15:26.600 --> 00:15:29.096
Absolute value of x less than a.
00:15:29.096 --> 00:15:32.290
Throughout this region
the screened potential
00:15:32.290 --> 00:15:35.540
is the potential that you have.
00:15:35.540 --> 00:15:39.160
Nevertheless, for the
rest, its infinite.
00:15:39.160 --> 00:15:42.820
So the screened
potential is this thing.
00:15:42.820 --> 00:15:45.750
Is infinite there, and
it's here this thing.
00:15:48.470 --> 00:15:52.410
So it's just some potential.
00:15:52.410 --> 00:15:55.150
You take it a screen
and you just see
00:15:55.150 --> 00:15:58.422
one part of the potential,
and let it go to infinity.
00:15:58.422 --> 00:15:59.630
So that's a screen potential.
00:16:02.190 --> 00:16:06.200
So now what I'm going
to do is that I'm
00:16:06.200 --> 00:16:08.550
going to try to
argue that you could
00:16:08.550 --> 00:16:13.590
try to find the bound state
of the screen potential.
00:16:13.590 --> 00:16:17.950
Unless you remove the
screen, you will find,
00:16:17.950 --> 00:16:20.830
as you let a go to
infinity, you will
00:16:20.830 --> 00:16:25.560
find the bound states of
the original potential.
00:16:25.560 --> 00:16:27.460
It's reasonable
that that's true,
00:16:27.460 --> 00:16:29.240
because as you
remove the screen,
00:16:29.240 --> 00:16:31.760
you're letting more of
the potential be exposed,
00:16:31.760 --> 00:16:33.730
and more of the
potential be exposed.
00:16:33.730 --> 00:16:37.440
And the wave functions
eventually die, so as the time
00:16:37.440 --> 00:16:40.570
that you're very far away,
you affect the wave functions
00:16:40.570 --> 00:16:41.425
less and less.
00:16:44.190 --> 00:16:45.680
So that's the argument.
00:16:45.680 --> 00:16:48.260
We're going to try
to argue that we're
00:16:48.260 --> 00:16:50.150
going to look at
the bound states
00:16:50.150 --> 00:16:54.550
of the screened potentials and
see what happened, whether they
00:16:54.550 --> 00:16:58.490
tell us about the bound
states the original potential.
00:16:58.490 --> 00:17:02.680
So for this, I'm going
to begin with a screen
00:17:02.680 --> 00:17:08.210
potential in which
a goes to 0 and say
00:17:08.210 --> 00:17:11.530
that a is equal to
epsilon, very small.
00:17:11.530 --> 00:17:13.839
So what potential so do I have?
00:17:13.839 --> 00:17:21.390
A very tiny potential here
from epsilon to minus epsilon.
00:17:21.390 --> 00:17:26.270
Now I chose the original point
down here to be the minimum.
00:17:26.270 --> 00:17:31.300
So actually, the bottom part of
the potential is really flat.
00:17:31.300 --> 00:17:35.240
And if you take
epsilon going to 0,
00:17:35.240 --> 00:17:38.220
well, the potential
might do this,
00:17:38.220 --> 00:17:42.390
but really at the bottom for
sufficiently small epsilon,
00:17:42.390 --> 00:17:48.500
this is an infinite square
well with psis to epsilon.
00:17:52.900 --> 00:17:57.520
I chose the minimum so that you
don't get something like this.
00:17:57.520 --> 00:17:59.310
If it would be a
point with a slope,
00:17:59.310 --> 00:18:00.710
this would be an ugly thing.
00:18:00.710 --> 00:18:02.460
So let's choose the minimum.
00:18:02.460 --> 00:18:05.770
And we have the screen
potential here, and that's it.
00:18:05.770 --> 00:18:07.070
Now look what we do.
00:18:07.070 --> 00:18:11.065
We say all right, here
there is a ground state.
00:18:13.680 --> 00:18:16.060
Very tiny.
00:18:16.060 --> 00:18:17.280
Goes like that.
00:18:17.280 --> 00:18:18.410
Vanishes here.
00:18:18.410 --> 00:18:21.090
Vanishes there.
00:18:21.090 --> 00:18:23.920
And has no nodes.
00:18:23.920 --> 00:18:24.720
Very tiny.
00:18:24.720 --> 00:18:28.190
You know the two 0s are
very close to each other.
00:18:28.190 --> 00:18:30.550
And now I'm going
to try to increase
00:18:30.550 --> 00:18:35.850
the value of the screen a.
00:18:35.850 --> 00:18:43.790
So suppose we've
increased the screen,
00:18:43.790 --> 00:18:49.040
and now the potential is here.
00:18:49.040 --> 00:18:53.050
And now we have a finite screen.
00:18:53.050 --> 00:18:55.060
Here is the potential.
00:18:55.060 --> 00:18:57.090
And I look at the wave function.
00:18:57.090 --> 00:18:58.520
How it looks.
00:18:58.520 --> 00:19:00.910
Here is psi 0.
00:19:00.910 --> 00:19:03.480
This ground state psi 0.
00:19:06.030 --> 00:19:13.410
Well, since this thing in here,
the potential becomes infinite,
00:19:13.410 --> 00:19:15.500
the wave function
still must vanish here
00:19:15.500 --> 00:19:18.150
and still must vanish here.
00:19:18.150 --> 00:19:23.830
Now just for your
imagination, think of this.
00:19:23.830 --> 00:19:28.900
At this stage, it still more
or less looks like this.
00:19:28.900 --> 00:19:30.830
Maybe.
00:19:30.830 --> 00:19:38.230
Now I'm going to ask, as I
increase, can I produce a node?
00:19:38.230 --> 00:19:40.550
And look what's going to happen.
00:19:40.550 --> 00:19:44.110
So suppose it might happen
that, as you increase,
00:19:44.110 --> 00:19:46.000
suddenly you produce a node.
00:19:46.000 --> 00:19:48.257
So here's what I'm saying here.
00:19:48.257 --> 00:19:49.340
I'm going to show it here.
00:19:49.340 --> 00:19:52.200
Suppose up to this
point, there is no node.
00:19:52.200 --> 00:19:56.000
But then when I double
it, when I increase it
00:19:56.000 --> 00:20:01.760
to twice the size, when I go
to screen potential like that,
00:20:01.760 --> 00:20:05.970
suddenly there is a
node in the middle.
00:20:05.970 --> 00:20:08.340
So if there is a
node in the middle,
00:20:08.340 --> 00:20:12.660
one thing that could have
happened is that you have this.
00:20:15.820 --> 00:20:19.920
And now look what must
have happened then.
00:20:19.920 --> 00:20:26.560
As I stretch this, this slope
must have been going down,
00:20:26.560 --> 00:20:31.080
and down, and down, until
it flips to the other side
00:20:31.080 --> 00:20:33.250
to produce a node here.
00:20:33.250 --> 00:20:34.960
It could have
happened on this side,
00:20:34.960 --> 00:20:36.810
but it's the same, so
the argument is just
00:20:36.810 --> 00:20:38.220
done with this side.
00:20:38.220 --> 00:20:41.780
To produce a node you
could have done somehow
00:20:41.780 --> 00:20:44.900
the slope here must
have changed sine.
00:20:44.900 --> 00:20:49.230
But for that to happen
continuously, at some point
00:20:49.230 --> 00:20:51.305
the this slope must have been 0.
00:20:54.660 --> 00:20:58.590
But you cannot have
a 0 and 0 slope.
00:20:58.590 --> 00:21:04.010
So this thing can't
flip, can't do this.
00:21:04.010 --> 00:21:05.810
Another thing that
could have happened
00:21:05.810 --> 00:21:09.060
is that when we
are here already,
00:21:09.060 --> 00:21:14.080
maybe the wave function
looks like that.
00:21:14.080 --> 00:21:17.120
It doesn't flip at the
edges, but produces something
00:21:17.120 --> 00:21:18.290
like that.
00:21:18.290 --> 00:21:21.590
But the only way this
can happen continuously,
00:21:21.590 --> 00:21:25.390
and this potential is
changing continuously,
00:21:25.390 --> 00:21:29.110
is for this thing at
some intermediate stage,
00:21:29.110 --> 00:21:34.310
as you keep stretching the
screen, this sort of starts
00:21:34.310 --> 00:21:37.560
to produce a depression here.
00:21:37.560 --> 00:21:42.000
And at some point, to get
here it has to do this.
00:21:42.000 --> 00:21:43.530
But it can't do this either.
00:21:43.530 --> 00:21:46.940
It cannot vanish and have
derivative like that.
00:21:46.940 --> 00:21:49.990
So actually, as you
stretch the screen,
00:21:49.990 --> 00:21:54.320
there's no way to
produce a node.
00:21:54.320 --> 00:21:57.360
That property forbids it.
00:21:57.360 --> 00:22:01.470
So by the time you go and
take the screen to infinity,
00:22:01.470 --> 00:22:04.255
this wave function has no nodes.
00:22:07.620 --> 00:22:12.100
So that proves it that the
ground state has no nodes.
00:22:12.100 --> 00:22:15.300
You could call this
a physicist proof,
00:22:15.300 --> 00:22:19.880
which means-- not in
the pejorative way.
00:22:19.880 --> 00:22:24.330
It means that it's
reasonable, it's intuitive,
00:22:24.330 --> 00:22:28.460
and a mathematician working
hard could make it rigorous.
00:22:32.300 --> 00:22:36.280
A bad physicist proof is
one that is a little sloppy
00:22:36.280 --> 00:22:39.990
and no mathematician could
fix it and make it work.
00:22:39.990 --> 00:22:44.370
So I think this is a good
physics proof in that sense.
00:22:44.370 --> 00:22:46.460
Probably you can
construct a real proof,
00:22:46.460 --> 00:22:50.410
or based on this, a
very precise proof.
00:22:50.410 --> 00:22:53.970
Now look at excited states.
00:22:53.970 --> 00:23:00.090
Suppose you take now here
this screen very little,
00:23:00.090 --> 00:23:05.375
and now consider the third
excited state, psi three.
00:23:08.250 --> 00:23:10.840
I'm sorry, we'll call this psi
2 because it has two nodes.
00:23:13.880 --> 00:23:15.565
Well, maybe I should do psi 1.
00:23:18.610 --> 00:23:19.710
Psi 1.
00:23:19.710 --> 00:23:21.670
One node.
00:23:21.670 --> 00:23:23.200
Same thing.
00:23:23.200 --> 00:23:27.750
As you increase
it, there's no way
00:23:27.750 --> 00:23:29.960
to create another
node continuously.
00:23:29.960 --> 00:23:32.620
Because again, you have
to flip at the edges,
00:23:32.620 --> 00:23:34.910
or you have to
depress in the middle.
00:23:34.910 --> 00:23:38.640
So this one will evolve
to a wave function that
00:23:38.640 --> 00:23:43.500
will have one node in
the whole big potential.
00:23:43.500 --> 00:23:49.680
Now stayed does that
state have more energy
00:23:49.680 --> 00:23:52.450
than the ground state?
00:23:52.450 --> 00:23:59.530
Well, it certainly begins with
a small screen with more energy,
00:23:59.530 --> 00:24:03.730
because in the square well
psi 1 has more energy.
00:24:03.730 --> 00:24:08.570
And that energy should
be clear that it's not
00:24:08.570 --> 00:24:12.440
going to go below the
energy of the ground state.
00:24:12.440 --> 00:24:13.640
Why?
00:24:13.640 --> 00:24:17.220
Because if it went below the
energy of the ground state
00:24:17.220 --> 00:24:21.500
slowly, at some point for
some value of the screen,
00:24:21.500 --> 00:24:25.160
it would have the same
energy as the ground state.
00:24:25.160 --> 00:24:29.960
But no degeneracy is possible
in one dimensional problems.
00:24:29.960 --> 00:24:31.750
So that can't happen.
00:24:31.750 --> 00:24:32.820
Cannot have that.
00:24:32.820 --> 00:24:35.980
So it will always
stay a little higher.
00:24:35.980 --> 00:24:39.810
And therefore with one node you
will be a little higher energy.
00:24:39.810 --> 00:24:43.390
With two nodes will
be higher and higher.
00:24:43.390 --> 00:24:44.570
And that's it.
00:24:44.570 --> 00:24:45.940
That's the argument.
00:24:45.940 --> 00:24:51.260
Now, we've argued by this
continuous deformation process
00:24:51.260 --> 00:24:54.760
that this potential not
only has these bound states,
00:24:54.760 --> 00:25:05.100
but this is n nodes and En is
greater than En prime for n
00:25:05.100 --> 00:25:08.890
greater than n prime.
00:25:08.890 --> 00:25:12.340
So the more nodes,
the more energy.
00:25:12.340 --> 00:25:15.920
Pretty nice result,
and that's really
00:25:15.920 --> 00:25:19.630
all I wanted to say
about this problem.
00:25:19.630 --> 00:25:21.210
Are there any questions?
00:25:21.210 --> 00:25:21.710
Any?
00:25:29.582 --> 00:25:31.070
OK.
00:25:31.070 --> 00:25:34.440
So what we do now is
the harmonic oscillator.
00:25:34.440 --> 00:25:39.570
That's going to keep us busy
for the rest of today's lecture.
00:25:39.570 --> 00:25:43.170
It's a very interesting problem.
00:25:43.170 --> 00:25:49.450
And it's a most famous quantum
mechanics problem in a sense,
00:25:49.450 --> 00:25:55.310
because it happens to be useful
in many, many applications.
00:25:55.310 --> 00:25:58.070
If you have any
potential-- so what
00:25:58.070 --> 00:26:02.010
is the characteristic of
the harmonic oscillator?
00:26:02.010 --> 00:26:03.245
Harmonic oscillator.
00:26:05.850 --> 00:26:08.720
Oscillator.
00:26:08.720 --> 00:26:15.580
Well, the energy operator is p
squared over 2m plus, we write,
00:26:15.580 --> 00:26:19.910
one half m omega
squared x squared
00:26:19.910 --> 00:26:23.180
where omega is this
omega that you always
00:26:23.180 --> 00:26:27.230
think of angular velocity,
or angular frequency.
00:26:27.230 --> 00:26:28.870
It's more like
angular frequency.
00:26:28.870 --> 00:26:32.670
Omega has units of 1 over time.
00:26:32.670 --> 00:26:36.880
It's actually put 2pi over
the period of an oscillation.
00:26:36.880 --> 00:26:40.090
And this you know from
classical mechanics.
00:26:40.090 --> 00:26:43.730
If you have a harmonic
oscillator of this form, yeah,
00:26:43.730 --> 00:26:47.620
it actually oscillates
with this frequency.
00:26:47.620 --> 00:26:49.680
And E is the energy
operator, and this
00:26:49.680 --> 00:26:53.020
is the energy of the oscillator.
00:26:53.020 --> 00:26:55.640
So what defines an oscillator?
00:26:55.640 --> 00:27:00.170
It's something in which the
potential energy, this term
00:27:00.170 --> 00:27:01.950
is v of x.
00:27:01.950 --> 00:27:04.880
v of x is quadratic in x.
00:27:04.880 --> 00:27:07.870
That is a harmonic oscillator.
00:27:07.870 --> 00:27:10.910
Then you arrange the
constants to make sense.
00:27:10.910 --> 00:27:13.260
This has units of
energy, because this
00:27:13.260 --> 00:27:16.700
has units of length squared.
00:27:16.700 --> 00:27:18.820
1 over time squared.
00:27:18.820 --> 00:27:21.210
Length over time
is velocity squared
00:27:21.210 --> 00:27:23.140
times mass is kinetic energy.
00:27:23.140 --> 00:27:26.680
So this term has
the units of energy.
00:27:26.680 --> 00:27:29.440
And you good with that.
00:27:29.440 --> 00:27:30.930
And why is this useful?
00:27:30.930 --> 00:27:39.190
Because actually in any sort
of arbitrary potential, rather
00:27:39.190 --> 00:27:42.030
general potential
at least, whenever
00:27:42.030 --> 00:27:47.190
you have a minimum where
the derivative vanishes,
00:27:47.190 --> 00:27:50.060
then the second derivative
need not vanish.
00:27:50.060 --> 00:27:53.790
Then it's a good
approximation to think
00:27:53.790 --> 00:27:57.810
of the potential at the minimum
as a quadratic potential.
00:27:57.810 --> 00:28:02.840
It fits the potential
nicely over a good region.
00:28:02.840 --> 00:28:06.190
And therefore when you have
two molecules with a bound
00:28:06.190 --> 00:28:09.250
or something oscillating,
there is a potential.
00:28:09.250 --> 00:28:11.780
It has a minimum at the
equilibrium position.
00:28:11.780 --> 00:28:15.060
And the oscillations
are governed
00:28:15.060 --> 00:28:18.110
by some harmonic oscillator.
00:28:18.110 --> 00:28:21.460
When you have photons
in space time traveling,
00:28:21.460 --> 00:28:23.810
there is a set of
harmonic oscillators
00:28:23.810 --> 00:28:27.420
that correspond to photons.
00:28:27.420 --> 00:28:28.830
Many, many applications.
00:28:28.830 --> 00:28:32.450
Endless amount of applications
for the harmonic oscillator.
00:28:32.450 --> 00:28:36.080
So we really want to
understand this system quantum
00:28:36.080 --> 00:28:37.570
mechanically.
00:28:37.570 --> 00:28:39.870
And what does that mean?
00:28:39.870 --> 00:28:43.790
Is that we really want
to calculate and solve
00:28:43.790 --> 00:28:45.790
the Schrodinger equation.
00:28:45.790 --> 00:28:49.570
This is our first step in
understanding the system.
00:28:49.570 --> 00:28:51.060
There's going to
be a lot of work
00:28:51.060 --> 00:28:54.320
to be done even once we have
the solutions of the Schrodinger
00:28:54.320 --> 00:28:54.940
equation.
00:28:54.940 --> 00:28:57.520
But the first thing
is to figure out
00:28:57.520 --> 00:29:01.530
what are the energy
eigenstates or the solutions
00:29:01.530 --> 00:29:04.500
of the Schrodinger
equation for this problem.
00:29:04.500 --> 00:29:08.750
So notice that here
in this problem
00:29:08.750 --> 00:29:12.910
there's an energy quantity.
00:29:16.160 --> 00:29:18.520
Remember, when you have
a harmontonian like that,
00:29:18.520 --> 00:29:22.000
and people say so what is
the ground state energy?
00:29:22.000 --> 00:29:25.390
Well, have to find the
ground state wave function.
00:29:25.390 --> 00:29:26.810
Have to do things.
00:29:26.810 --> 00:29:28.910
Give me an hour, I'll find it.
00:29:28.910 --> 00:29:29.900
And all that.
00:29:29.900 --> 00:29:33.170
But if you want an
approximate value,
00:29:33.170 --> 00:29:37.590
dimensional analysis will do it,
roughly what is it going to be.
00:29:37.590 --> 00:29:41.640
Well, with this constant how
do you produce an energy?
00:29:41.640 --> 00:29:44.730
Well, you remember
what Einstein did,
00:29:44.730 --> 00:29:51.020
and you know that h bar
omega has units of energy.
00:29:51.020 --> 00:29:55.320
So that's an energy associated
with Lagrangian energy
00:29:55.320 --> 00:29:57.150
like quantity.
00:29:57.150 --> 00:30:00.340
And we expect that
that energy is
00:30:00.340 --> 00:30:02.030
going to be the relevant energy.
00:30:02.030 --> 00:30:04.070
And in fact, we'll find
that the ground state
00:30:04.070 --> 00:30:06.010
energy is just one half of that.
00:30:08.870 --> 00:30:11.520
There's another quantity
that may be interesting.
00:30:11.520 --> 00:30:13.640
How about the length?
00:30:13.640 --> 00:30:16.205
How do you construct a
length from these quantities?
00:30:20.620 --> 00:30:26.070
Well, you can start
doing m omega h bar
00:30:26.070 --> 00:30:28.210
and put powers and struggle.
00:30:28.210 --> 00:30:31.100
I hate doing that.
00:30:31.100 --> 00:30:34.770
I always try to find some
way of doing it and avoiding
00:30:34.770 --> 00:30:35.780
that thing.
00:30:35.780 --> 00:30:42.030
So I know that
energies go like h
00:30:42.030 --> 00:30:47.240
over h squared over
m length squared.
00:30:47.240 --> 00:30:51.550
So I'm going to call
the length a quantity a.
00:30:51.550 --> 00:30:54.210
So ma squared.
00:30:54.210 --> 00:30:55.615
That has units of energy.
00:30:59.740 --> 00:31:01.917
And you should remember
that because energy
00:31:01.917 --> 00:31:07.790
is b squared over 2m,
and b by De Broglie
00:31:07.790 --> 00:31:10.910
is h bar over sub lamda.
00:31:10.910 --> 00:31:15.360
So h bar squared, lambda
squared, and m here,
00:31:15.360 --> 00:31:16.780
that's units of energy.
00:31:16.780 --> 00:31:18.570
So that's a length.
00:31:18.570 --> 00:31:20.970
On the other hand,
we have another way
00:31:20.970 --> 00:31:24.620
to construct an energy
is with this thing,
00:31:24.620 --> 00:31:27.190
m omega squared length squared.
00:31:27.190 --> 00:31:32.840
So that's also m omega
squared a squared.
00:31:32.840 --> 00:31:34.880
That's another energy.
00:31:34.880 --> 00:31:40.680
So from this equation I
find that a to the fourth
00:31:40.680 --> 00:31:46.160
is h squared over m
squared omega squared.
00:31:46.160 --> 00:31:49.390
And it's a little
complicated, so a squared
00:31:49.390 --> 00:31:52.495
is h bar over m omega.
00:31:55.360 --> 00:31:58.640
So that's a length.
00:31:58.640 --> 00:31:59.690
Length squared.
00:31:59.690 --> 00:32:01.480
I don't want to take
the square root.
00:32:01.480 --> 00:32:04.130
We can leave it
for a moment there.
00:32:04.130 --> 00:32:08.680
But that's important because
of that's a length scale.
00:32:08.680 --> 00:32:11.780
And if somebody would ask
you in the ground state,
00:32:11.780 --> 00:32:14.760
how far is this
particle oscillating,
00:32:14.760 --> 00:32:19.940
you would say probably
about a square root of this.
00:32:19.940 --> 00:32:27.710
Would be a natural answer
and probably about right.
00:32:27.710 --> 00:32:33.310
So OK, energy and
units is very important
00:32:33.310 --> 00:32:35.290
to begin your analysis.
00:32:35.290 --> 00:32:38.440
So what is the
Schrodinger equation?
00:32:38.440 --> 00:32:41.030
The Schrodinger
equation for this thing
00:32:41.030 --> 00:32:48.590
is going to be minus h
squared over 2m, d second psi,
00:32:48.590 --> 00:32:54.580
dx squared plus the potential,
one half m omega squared
00:32:54.580 --> 00:33:00.640
x squared psi is equal E psi.
00:33:00.640 --> 00:33:08.150
And the big problem is I don't
know psi and I don't know E.
00:33:08.150 --> 00:33:11.550
Now there's so many elegant
ways of solving the harmonic
00:33:11.550 --> 00:33:12.050
oscillator.
00:33:15.020 --> 00:33:18.222
You will see those next lecture.
00:33:18.222 --> 00:33:20.300
Allan Adams will be back here.
00:33:20.300 --> 00:33:25.850
But we all have to go
through once in your life
00:33:25.850 --> 00:33:30.670
through the direct, uninspired
method of solving it.
00:33:33.680 --> 00:33:38.390
Because most of the times
when you have a new problem,
00:33:38.390 --> 00:33:42.130
you will not come up with
a beautiful, elegant method
00:33:42.130 --> 00:33:44.830
to avoid solving the
differential equation.
00:33:44.830 --> 00:33:47.430
You will have to struggle with
the differential equation.
00:33:47.430 --> 00:33:51.220
So today we struggle with
the differential equation.
00:33:51.220 --> 00:33:53.080
We're going to just do it.
00:33:53.080 --> 00:33:56.600
And I'm going to do it slow
enough and in detail enough
00:33:56.600 --> 00:33:59.380
that I hope you
follow everything.
00:33:59.380 --> 00:34:01.280
I'll just keep a
couple of things,
00:34:01.280 --> 00:34:06.210
but it will be one line
computations that I will skip.
00:34:06.210 --> 00:34:10.949
So this equation is some sort
of fairly difficult thing.
00:34:10.949 --> 00:34:15.750
And it's complicated and
made fairly unpleasant
00:34:15.750 --> 00:34:18.960
by the presence of
all these constants.
00:34:18.960 --> 00:34:23.159
What kind of equation is that
with all these constants?
00:34:23.159 --> 00:34:26.719
They shouldn't be there,
all this constants, in fact.
00:34:26.719 --> 00:34:31.159
So this is the first step,
cleaning up the equation.
00:34:31.159 --> 00:34:32.510
We have to clean it up.
00:34:32.510 --> 00:34:33.010
Why?
00:34:33.010 --> 00:34:37.130
Because the nice
functions in life like y
00:34:37.130 --> 00:34:41.850
double prime is equal to
minus y have no units.
00:34:41.850 --> 00:34:44.909
The derivatives create no units.
00:34:44.909 --> 00:34:49.639
y has the same units of that,
and the solution is sine of x,
00:34:49.639 --> 00:34:52.400
where x must have no units,
because you cannot find
00:34:52.400 --> 00:34:55.960
the sine of one centimeter.
00:34:55.960 --> 00:34:59.400
So this thing, we should
have the same thing here.
00:34:59.400 --> 00:35:00.720
No units anywhere.
00:35:04.100 --> 00:35:05.830
So how can we do that?
00:35:05.830 --> 00:35:08.600
This is an absolutely
necessary first step.
00:35:08.600 --> 00:35:11.120
If you're going to be
carrying all these constants,
00:35:11.120 --> 00:35:13.610
you'll get nowhere.
00:35:13.610 --> 00:35:16.590
So we have to clean it up.
00:35:16.590 --> 00:35:18.520
So what I'm going
to try to see is
00:35:18.520 --> 00:35:22.260
that look, here is
psi, psi, and psi.
00:35:22.260 --> 00:35:26.750
So suppose I do the
following thing,
00:35:26.750 --> 00:35:30.050
that I will clean
up the right hand
00:35:30.050 --> 00:35:37.960
side by dividing by something
with units of energy.
00:35:37.960 --> 00:35:42.340
So I'm going to do
the following way.
00:35:42.340 --> 00:35:52.610
I'm going to divide all
by 1 over h bar omega.
00:35:52.610 --> 00:35:56.190
And this 2 I'm going
to multiply by 2.
00:35:56.190 --> 00:36:07.760
So multiply by 2
over h bar omega.
00:36:07.760 --> 00:36:11.190
So what do I achieve
with that first step?
00:36:11.190 --> 00:36:14.550
I achieve that
these 2s disappear.
00:36:14.550 --> 00:36:16.540
Well, that's not too bad.
00:36:16.540 --> 00:36:19.270
Not that great either, I think.
00:36:19.270 --> 00:36:23.050
But in the right hand side,
this has units of energy.
00:36:23.050 --> 00:36:26.490
And the right hand side will
not have units of energy.
00:36:26.490 --> 00:36:27.940
So what do we get here?
00:36:27.940 --> 00:36:29.510
So we get minus.
00:36:29.510 --> 00:36:39.280
The h becomes an h alone over--
the m disappears-- so m omega.
00:36:39.280 --> 00:36:44.000
The second psi the x squared.
00:36:44.000 --> 00:36:49.860
The 1/2 disappeared,
so m omega over h
00:36:49.860 --> 00:37:01.900
bar x squared psi equals
2 E over h bar omega psi.
00:37:01.900 --> 00:37:05.760
It looks actually
quite better already.
00:37:05.760 --> 00:37:07.070
Should agree with that.
00:37:07.070 --> 00:37:14.040
It looks a lot nicer Now
I can use a name for this.
00:37:14.040 --> 00:37:18.650
I want to call this
the dimensionless value
00:37:18.650 --> 00:37:19.750
of the energy.
00:37:19.750 --> 00:37:23.120
So a calligraphic e.
00:37:23.120 --> 00:37:25.510
It has no units.
00:37:25.510 --> 00:37:29.850
It's telling me if I find some
energy, that that energy really
00:37:29.850 --> 00:37:34.970
is this number, this pure
number is how many times bigger
00:37:34.970 --> 00:37:39.650
is e with respect
to h omega over 2.
00:37:39.650 --> 00:37:43.350
So I'll write this now as e psi.
00:37:43.350 --> 00:37:44.540
And look what I have.
00:37:44.540 --> 00:37:48.160
I have no units here.
00:37:48.160 --> 00:37:50.480
And I have a psi.
00:37:50.480 --> 00:37:52.140
And I have a psi.
00:37:52.140 --> 00:37:56.020
But things have
worked out already.
00:37:56.020 --> 00:38:00.000
Look, the same factor
here, h over m omega
00:38:00.000 --> 00:38:02.530
is upside down here.
00:38:02.530 --> 00:38:06.670
And this factor has
units of length squared.
00:38:06.670 --> 00:38:12.000
Length squared times d d
length squared has no units.
00:38:12.000 --> 00:38:14.320
And here's 1 over
length squared.
00:38:17.880 --> 00:38:20.620
1 over length squared
times length squared.
00:38:20.620 --> 00:38:22.660
So things have worked out.
00:38:22.660 --> 00:38:28.490
And we can now
simply say x is going
00:38:28.490 --> 00:38:32.560
to be equal to au,
a new variable.
00:38:32.560 --> 00:38:35.450
This is going to be your new
variable for your differential
00:38:35.450 --> 00:38:41.480
equation in which is this thing.
00:38:41.480 --> 00:38:49.570
And then this
differential equation
00:38:49.570 --> 00:38:53.025
really has cleaned
up perfectly well.
00:38:57.020 --> 00:38:59.450
So how does it look now?
00:38:59.450 --> 00:39:05.220
Well, it's all gone
actually, because if you
00:39:05.220 --> 00:39:14.050
have x equals au, d dx by
chain rule is 1 over a d du.
00:39:17.500 --> 00:39:19.440
And to derivatives
this with respect
00:39:19.440 --> 00:39:29.790
to x it's 1 over a squared
times the d second du squared.
00:39:29.790 --> 00:39:32.590
And this thing is a squared.
00:39:32.590 --> 00:39:36.470
So actually you
cancel this factor.
00:39:36.470 --> 00:39:39.360
And when I write x
equals to au, you
00:39:39.360 --> 00:39:41.205
get an a squared times this.
00:39:41.205 --> 00:39:43.950
And a squared times this is 1.
00:39:43.950 --> 00:39:46.190
So your differential
equations has
00:39:46.190 --> 00:39:53.230
become minus the
second psi du squared,
00:39:53.230 --> 00:39:56.740
where u is a dimensionless
quantity, because this has
00:39:56.740 --> 00:39:58.820
units of length, this
has units of length.
00:39:58.820 --> 00:40:01.640
No units here.
00:40:01.640 --> 00:40:02.720
You have no units.
00:40:02.720 --> 00:40:07.580
So minus d second
du squared plus u
00:40:07.580 --> 00:40:12.805
squared psi is equal to e psi.
00:40:18.320 --> 00:40:21.890
Much nicer.
00:40:21.890 --> 00:40:24.770
This is an equation
we can think about
00:40:24.770 --> 00:40:27.830
without being distracted
by this endless amount
00:40:27.830 --> 00:40:29.656
of little trivialities.
00:40:33.690 --> 00:40:38.760
But still we haven't
solved it, and how are we
00:40:38.760 --> 00:40:42.290
going to solve this equation?
00:40:42.290 --> 00:40:46.860
So let's again think
of what should happen.
00:40:46.860 --> 00:40:51.780
Somehow it should happen
that these e's get fixed.
00:40:51.780 --> 00:40:55.750
And there is some solution
just for some values of e's.
00:40:55.750 --> 00:41:00.090
It's not obvious at this stage
how that is going to happen.
00:41:00.090 --> 00:41:00.940
Yes?
00:41:00.940 --> 00:41:01.856
AUDIENCE: [INAUDIBLE].
00:41:06.290 --> 00:41:12.220
PROFESSOR: Here for example,
let me do this term.
00:41:12.220 --> 00:41:18.200
h bar over m omega is minus,
from that equation, a squared.
00:41:18.200 --> 00:41:26.640
But dx squared is 1 over
a squared d du squared.
00:41:26.640 --> 00:41:29.820
So a squared cancels.
00:41:29.820 --> 00:41:33.380
And here the x is equal
a squared times u,
00:41:33.380 --> 00:41:35.585
so again cancels.
00:41:39.060 --> 00:41:44.640
OK so what is the problem here?
00:41:44.640 --> 00:41:48.150
The problem is that most likely
what is going to go wrong
00:41:48.150 --> 00:41:51.760
is that this solution for
arbitrary values of e's is
00:41:51.760 --> 00:41:55.500
going to diverge at
infinity, and you're never
00:41:55.500 --> 00:41:57.810
going to be able
to normalize it.
00:41:57.810 --> 00:42:02.050
So let's try to understand
how the solution looks
00:42:02.050 --> 00:42:04.010
as we go to infinity.
00:42:04.010 --> 00:42:05.710
So this is the first
thing you should
00:42:05.710 --> 00:42:07.520
do with an equation like that.
00:42:07.520 --> 00:42:13.880
How does this solution
look as u goes to infinity?
00:42:13.880 --> 00:42:18.400
Now we may not be able to solve
it exactly in that case either,
00:42:18.400 --> 00:42:22.930
but we're going to gain
insight into what's happening.
00:42:22.930 --> 00:42:25.940
So here it is.
00:42:25.940 --> 00:42:30.950
When u goes to infinity,
this term, whatever psi is,
00:42:30.950 --> 00:42:33.190
this term is much
bigger than that,
00:42:33.190 --> 00:42:37.060
because we're presumably
working with some fixed energy
00:42:37.060 --> 00:42:38.810
that we still don't
know what it is,
00:42:38.810 --> 00:42:42.500
but it's a fixed number and,
for you, sufficiently large.
00:42:42.500 --> 00:42:44.430
This is going to dominate.
00:42:44.430 --> 00:42:46.620
So the equation
that we're trying
00:42:46.620 --> 00:42:53.840
to solve as u goes to infinity,
the equation sort of becomes
00:42:53.840 --> 00:42:58.240
psi double prime-- prime
is for two derivatives--
00:42:58.240 --> 00:43:02.330
is equal to u squared psi.
00:43:08.860 --> 00:43:13.270
OK, so how do we get
an idea what solves
00:43:13.270 --> 00:43:16.020
this is not all that obvious.
00:43:16.020 --> 00:43:19.720
It's certainly not a power of u,
because when you differentiate
00:43:19.720 --> 00:43:22.640
the power of u, you
lower the degree
00:43:22.640 --> 00:43:25.650
rather than increase the degree.
00:43:25.650 --> 00:43:29.200
So what function increases
degree as you differentiate?
00:43:29.200 --> 00:43:32.350
It's not the trivial function.
00:43:32.350 --> 00:43:34.420
Cannot be a polynomial.
00:43:34.420 --> 00:43:36.150
If it could be
even a polynomial,
00:43:36.150 --> 00:43:39.230
if you take two derivatives,
it kind cannot be equal to x
00:43:39.230 --> 00:43:40.440
squared times a polynomial.
00:43:40.440 --> 00:43:43.320
It's sort of upside down.
00:43:43.320 --> 00:43:46.390
So if you think about
it for a little while,
00:43:46.390 --> 00:43:48.180
you don't have an
exact solution,
00:43:48.180 --> 00:43:51.650
but you would imagine
that something like this
00:43:51.650 --> 00:43:55.780
would do it, an e
to the u squared.
00:43:55.780 --> 00:43:58.280
Because an e to
the u squared, when
00:43:58.280 --> 00:44:04.390
you differentiate with respect
to us, you produce a u down.
00:44:04.390 --> 00:44:05.830
When you one derivative.
00:44:05.830 --> 00:44:07.420
When you take
another derivative,
00:44:07.420 --> 00:44:09.650
well, it's more complicated,
but one term you
00:44:09.650 --> 00:44:11.980
will produce another u down.
00:44:11.980 --> 00:44:16.670
So that probably is quite good.
00:44:16.670 --> 00:44:18.290
So let's try that.
00:44:18.290 --> 00:44:21.310
Let's try to see if we
have something like that.
00:44:21.310 --> 00:44:22.890
So I will try something.
00:44:26.900 --> 00:44:33.290
I'll try psi equals 2.
00:44:37.420 --> 00:44:41.480
I'm going to try
the following thing.
00:44:41.480 --> 00:44:49.250
e to the alpha u squared over
2 where alpha is a number.
00:44:49.250 --> 00:44:50.590
I don't know how much it is.
00:44:50.590 --> 00:44:53.740
Alpha is some number.
00:44:53.740 --> 00:44:56.950
Now could try this
alone, but I actually
00:44:56.950 --> 00:45:01.860
want to emphasize
to you that if this
00:45:01.860 --> 00:45:05.090
is the behavior
near infinity, it
00:45:05.090 --> 00:45:07.430
won't make any difference
if you put here,
00:45:07.430 --> 00:45:12.800
for example, something
like u to the power k.
00:45:12.800 --> 00:45:17.250
It will also be
roughly a solution.
00:45:17.250 --> 00:45:20.660
So let's see that.
00:45:20.660 --> 00:45:22.475
So for that I have
to differentiate.
00:45:27.290 --> 00:45:31.460
And let's see what we get.
00:45:31.460 --> 00:45:37.850
So we're trying to see how the
function behaves far, far away.
00:45:37.850 --> 00:45:39.880
You might say well
look, probably
00:45:39.880 --> 00:45:41.850
that alpha should be negative.
00:45:41.850 --> 00:45:43.940
But let's see what
the equation tells us
00:45:43.940 --> 00:45:46.760
before we put anything in there.
00:45:46.760 --> 00:45:53.270
So if I do psi prime,
you would get what?
00:45:53.270 --> 00:45:54.900
You would get one
term that would
00:45:54.900 --> 00:46:02.170
be alpha u times this u to
the k into the alpha u squared
00:46:02.170 --> 00:46:02.760
over 2.
00:46:02.760 --> 00:46:06.580
I differentiated
the exponential.
00:46:06.580 --> 00:46:08.850
I differentiated
the exponential.
00:46:08.850 --> 00:46:10.640
And then you would
get a term where
00:46:10.640 --> 00:46:12.260
you differentiate the power.
00:46:12.260 --> 00:46:17.120
So you get ku to the k
minus 1 into the alpha u
00:46:17.120 --> 00:46:19.470
squared over 2.
00:46:19.470 --> 00:46:23.120
If you take a second
derivative, well, I
00:46:23.120 --> 00:46:25.620
can differentiate the
exponential again,
00:46:25.620 --> 00:46:28.840
so I will get alpha
u now squared,
00:46:28.840 --> 00:46:31.690
because each derivative
of this exponent
00:46:31.690 --> 00:46:35.120
produces a factor of alpha u.
00:46:35.120 --> 00:46:40.396
u to the k into the
alpha u squared over 2.
00:46:40.396 --> 00:46:50.240
And a couple more terms that
they all have less powers of u,
00:46:50.240 --> 00:46:56.020
because this term
has u to the k plus--
00:46:56.020 --> 00:46:58.270
already has u to the k plus 1.
00:46:58.270 --> 00:47:00.390
And this has u to the k minus 1.
00:47:00.390 --> 00:47:03.010
They differ by two powers of u.
00:47:03.010 --> 00:47:08.930
So for illustration,
please, if you want, do it.
00:47:08.930 --> 00:47:12.840
Three lines, you should skip
three lines in your notebook
00:47:12.840 --> 00:47:17.510
if you're taking notes
and get the following.
00:47:17.510 --> 00:47:21.080
No point in me doing
this algebra here.
00:47:21.080 --> 00:47:23.810
Alpha u squared over 2.
00:47:23.810 --> 00:47:28.550
Because actually it's
not all that important.
00:47:28.550 --> 00:47:37.840
Over alpha 1 over u squared
plus k minus 1 over alpha
00:47:37.840 --> 00:47:41.680
squared 1 over u to the fourth.
00:47:41.680 --> 00:47:42.840
That's all you get.
00:47:49.510 --> 00:47:56.510
Look, this is alpha
squared u squared
00:47:56.510 --> 00:48:01.260
times psi times these things.
00:48:01.260 --> 00:48:07.910
1 plus 2 k plus 1 over
alpha 1 over u squared.
00:48:07.910 --> 00:48:12.620
So when u goes to
infinity, your solution
00:48:12.620 --> 00:48:18.200
works, because these
thing's are negligible.
00:48:18.200 --> 00:48:21.400
So you get a number
times u squared.
00:48:21.400 --> 00:48:23.985
That is the equation you are
trying to solve up there.
00:48:27.020 --> 00:48:30.650
And therefore, you get
that the equation if alpha
00:48:30.650 --> 00:48:34.240
squared is equal to 1.
00:48:34.240 --> 00:48:40.050
And that means and really that
alpha can be plus minus 1.
00:48:40.050 --> 00:48:45.270
And roughly this
solution near infinity,
00:48:45.270 --> 00:48:48.990
probably there's two solutions.
00:48:51.820 --> 00:48:54.390
This is a second order
differential equation,
00:48:54.390 --> 00:48:57.910
so even near infinity there
should be two solutions.
00:48:57.910 --> 00:49:03.440
So we expect as u goes
to infinity psi of u
00:49:03.440 --> 00:49:09.290
will be some constant A
times u to the k times
00:49:09.290 --> 00:49:13.670
e to the minus u squared over 2.
00:49:13.670 --> 00:49:16.400
That's where alpha
equal minus 1.
00:49:16.400 --> 00:49:24.080
Plus Bu to the k into the
plus u squared over 2.
00:49:24.080 --> 00:49:26.340
And what is k?
00:49:26.340 --> 00:49:28.180
Well, we don't know what is k.
00:49:28.180 --> 00:49:30.700
It seems to work for all k.
00:49:30.700 --> 00:49:36.000
That may seem a little
confusing now, but don't worry.
00:49:36.000 --> 00:49:42.280
We'll see other things
happening here very soon.
00:49:42.280 --> 00:49:45.030
So look at what has happened.
00:49:45.030 --> 00:49:48.300
We've identified that most
likely your wave function
00:49:48.300 --> 00:49:50.920
is going to look like
this at infinity.
00:49:50.920 --> 00:49:54.540
So we're going to want to
this part not to be present.
00:49:54.540 --> 00:49:57.210
So presumably we're going
to want a solution that
00:49:57.210 --> 00:50:00.590
just has this, because
this is normalizable.
00:50:00.590 --> 00:50:05.930
The integral of any power times
a Gaussian is convergence.
00:50:05.930 --> 00:50:07.440
So this can be normalized.
00:50:07.440 --> 00:50:11.830
The Gaussian falls so
fast that any power can
00:50:11.830 --> 00:50:14.490
be integrated
against a Gaussian.
00:50:14.490 --> 00:50:17.660
Any power however big
doesn't grow big enough
00:50:17.660 --> 00:50:19.770
to compensate a Gaussian.
00:50:19.770 --> 00:50:21.655
It's impossible to
compensate a Gaussian.
00:50:25.210 --> 00:50:26.525
So we hope for this.
00:50:31.690 --> 00:50:34.700
But we want to
translate what we've
00:50:34.700 --> 00:50:38.630
learned into some
technical advantage
00:50:38.630 --> 00:50:40.430
in solving the
differential equation,
00:50:40.430 --> 00:50:44.960
because, after all, we wanted
be insight how it looks far way,
00:50:44.960 --> 00:50:48.100
but we wanted to solve
the differential equation.
00:50:48.100 --> 00:50:50.790
So how can we use
this insight we now
00:50:50.790 --> 00:50:57.560
have to simplify the solution
of the differential equation?
00:50:57.560 --> 00:51:03.000
The idea is to change
variables a little bit.
00:51:03.000 --> 00:51:14.660
So write psi of u to be
equal to h of u times e
00:51:14.660 --> 00:51:17.450
to the minus u squared over 2.
00:51:21.670 --> 00:51:23.780
Now you're going to say
wait, what are you doing?
00:51:23.780 --> 00:51:26.190
Are you making an
approximation now
00:51:26.190 --> 00:51:28.990
that this is what is
going to look far away?
00:51:28.990 --> 00:51:32.710
Or what are you putting there?
00:51:32.710 --> 00:51:35.050
I'm not making
any approximation.
00:51:35.050 --> 00:51:40.340
I'm just saying whatever
pis is, it can always
00:51:40.340 --> 00:51:42.100
be written in this way.
00:51:42.100 --> 00:51:42.600
Why?
00:51:42.600 --> 00:51:45.440
Because if you have
a psi of u, you
00:51:45.440 --> 00:51:51.840
can write it as psi of u
over e to the minus u squared
00:51:51.840 --> 00:51:55.940
over 2 times e minus
u squared over 2.
00:51:55.940 --> 00:51:59.210
Very trivially this
can always be done.
00:51:59.210 --> 00:52:02.640
As long as we say
that h is arbitrary,
00:52:02.640 --> 00:52:07.650
there's nothing,
no constraint here.
00:52:07.650 --> 00:52:10.040
I have not assume
anything, nothing.
00:52:10.040 --> 00:52:16.360
I'm just hoping that I have a
differential equation for psi.
00:52:16.360 --> 00:52:20.850
That because this is a very
clever factor, the differential
00:52:20.850 --> 00:52:24.500
equation for h will be simpler.
00:52:24.500 --> 00:52:30.010
Because part of the dependence
has been taken over.
00:52:30.010 --> 00:52:34.700
So maybe h, for example, could
be now a polynomial solution,
00:52:34.700 --> 00:52:38.520
because this product
has been taken care.
00:52:38.520 --> 00:52:43.580
So the hope is that by
writing this equation
00:52:43.580 --> 00:52:46.290
it will become an
equation for h of u,
00:52:46.290 --> 00:52:49.800
and that equation
will be simpler.
00:52:49.800 --> 00:52:52.070
So will it be simpler?
00:52:52.070 --> 00:52:56.340
Well, here again this
is not difficult.
00:52:56.340 --> 00:53:00.060
You're supposed to plug
into equation one--
00:53:00.060 --> 00:53:03.220
this is the equation
one-- plug into one.
00:53:11.900 --> 00:53:12.870
I won't do it.
00:53:12.870 --> 00:53:16.270
It's three lines of
algebra to plug into one
00:53:16.270 --> 00:53:18.580
and calculate the
equation for h of u.
00:53:18.580 --> 00:53:20.070
You should do it.
00:53:20.070 --> 00:53:24.185
It's the kind of thing that
one should do at least once.
00:53:26.830 --> 00:53:28.260
So please do it.
00:53:28.260 --> 00:53:30.890
It's three, four lines.
00:53:30.890 --> 00:53:32.520
It's not long.
00:53:32.520 --> 00:53:34.170
But I'll just write the answer.
00:53:37.710 --> 00:53:39.790
So by the time you
substitute, of course,
00:53:39.790 --> 00:53:41.980
the e to the minus
u squared over 2
00:53:41.980 --> 00:53:44.940
is going to cancel
from everywhere.
00:53:44.940 --> 00:53:46.210
It's very here.
00:53:46.210 --> 00:53:48.190
You just need to
take two derivatives,
00:53:48.190 --> 00:53:51.350
so it becomes a second
order differential equation.
00:53:51.350 --> 00:53:54.760
And indeed, it becomes
a tractable differential
00:53:54.760 --> 00:53:56.080
equation.
00:53:56.080 --> 00:54:08.070
The second h, du squared minus
2u dh du plus e minus 1 h
00:54:08.070 --> 00:54:08.880
equals 0.
00:54:15.740 --> 00:54:19.470
OK, that is our equation now.
00:54:19.470 --> 00:54:25.530
So now we face the problem
finally solving this equation.
00:54:25.530 --> 00:54:27.640
So before we start,
maybe there's
00:54:27.640 --> 00:54:31.441
some questions of what
we've done so far.
00:54:31.441 --> 00:54:31.940
Let's see.
00:54:31.940 --> 00:54:34.980
Any questions?
00:54:38.080 --> 00:54:38.970
Yes?
00:54:38.970 --> 00:54:42.130
AUDIENCE: Do you have
right there in the middle
00:54:42.130 --> 00:54:45.632
would be-- this equation
is linear, so can we just
00:54:45.632 --> 00:54:51.065
[INAUDIBLE] minus u squared over
2 and you stuck it to that u
00:54:51.065 --> 00:54:51.911
to the k.
00:54:51.911 --> 00:54:52.890
PROFESSOR: It's here?
00:54:52.890 --> 00:54:53.495
This thing?
00:54:53.495 --> 00:54:54.120
AUDIENCE: Yeah.
00:54:54.120 --> 00:54:57.900
Could you then just power series
what's going on at 0 with those
00:54:57.900 --> 00:55:01.120
u to the k terms [INAUDIBLE]?
00:55:01.120 --> 00:55:02.320
PROFESSOR: No.
00:55:02.320 --> 00:55:08.350
This is the behavior
as u goes to infinity.
00:55:08.350 --> 00:55:12.640
So I actually don't know
that the function near 0
00:55:12.640 --> 00:55:15.450
is going to behave
like u to the k.
00:55:15.450 --> 00:55:17.100
We really don't know.
00:55:17.100 --> 00:55:20.920
It suggest to you that
maybe the solution
00:55:20.920 --> 00:55:25.050
is going to be near 0 u to
the k times some polynomial
00:55:25.050 --> 00:55:26.060
or something like that.
00:55:26.060 --> 00:55:29.120
But it's not that, because
this analysis was just
00:55:29.120 --> 00:55:30.290
done at infinity.
00:55:30.290 --> 00:55:36.494
So we really have no information
still what's going on near 0.
00:55:36.494 --> 00:55:37.160
Other questions?
00:55:39.860 --> 00:55:40.958
Yes?
00:55:40.958 --> 00:55:45.369
AUDIENCE: So is k some arbitrary
number or is it an integer?
00:55:45.369 --> 00:55:47.660
PROFESSOR: At this moment,
actually, it doesn't matter.
00:55:47.660 --> 00:55:50.260
Is that right?
00:55:50.260 --> 00:55:51.320
Doesn't matter.
00:55:51.320 --> 00:55:56.420
The analysis that we did here
suggests it could be anything.
00:55:56.420 --> 00:56:00.640
That's why I just didn't
put it into h or u.
00:56:00.640 --> 00:56:03.520
I didn't put it because
would be strange to put here
00:56:03.520 --> 00:56:05.590
a u to the k.
00:56:05.590 --> 00:56:07.530
I wouldn't know
what to make of it.
00:56:07.530 --> 00:56:10.430
So at this moment,
the best thing to say
00:56:10.430 --> 00:56:14.050
is we don't know what it is,
and maybe we'll understand it.
00:56:14.050 --> 00:56:14.830
And we will.
00:56:14.830 --> 00:56:18.010
In a few seconds, we'll
sort of see what's going on.
00:56:20.570 --> 00:56:24.195
OK, so how does one
solve this equation?
00:56:27.900 --> 00:56:31.610
Well, it's not a
trivial equation, again.
00:56:31.610 --> 00:56:35.500
But it can be solved
by polynomials,
00:56:35.500 --> 00:56:36.670
and we'll see that.
00:56:36.670 --> 00:56:39.080
But the way we
solve this equation
00:56:39.080 --> 00:56:42.940
is by a power series expansion.
00:56:42.940 --> 00:56:45.670
Now you could do
it by hand first,
00:56:45.670 --> 00:56:48.370
and I did it when
I was preparing
00:56:48.370 --> 00:56:50.470
the lecture yesterday.
00:56:50.470 --> 00:56:57.060
I said I'm going to
just write h of u
00:56:57.060 --> 00:57:03.880
equal a constant a0 plus
a1u plus a2u squared
00:57:03.880 --> 00:57:07.360
plus a3u cubed.
00:57:07.360 --> 00:57:11.030
And I plugged it in here.
00:57:11.030 --> 00:57:13.220
And I just did the
first few terms
00:57:13.220 --> 00:57:15.490
and start to see what happened.
00:57:15.490 --> 00:57:17.850
And I found after
a little thinking
00:57:17.850 --> 00:57:24.410
that a2 is determined
by a0, and a3
00:57:24.410 --> 00:57:30.100
is determined by a1
once you substitute.
00:57:30.100 --> 00:57:34.090
It's not the obvious when you
look at this, but that happens.
00:57:34.090 --> 00:57:37.820
So when you face a
problem like that,
00:57:37.820 --> 00:57:41.060
don't go high power
to begin with.
00:57:41.060 --> 00:57:43.910
Just try a simple series
and see what happens.
00:57:43.910 --> 00:57:45.530
And you see a little pattern.
00:57:45.530 --> 00:57:48.670
And then you can do a more
sophisticated analysis.
00:57:48.670 --> 00:57:51.610
So what would be a more
sophisticated analysis?
00:57:51.610 --> 00:57:55.030
To write h of u
equal the sum from j
00:57:55.030 --> 00:58:00.262
equals 0 to infinity
aju to the j.
00:58:03.496 --> 00:58:06.370
Then if you take a
derivative, because we're
00:58:06.370 --> 00:58:11.780
going to need the
derivative, dh du
00:58:11.780 --> 00:58:18.860
would be the sum from
j equals 0 to infinity.
00:58:18.860 --> 00:58:24.500
j times aju to the j minus 1.
00:58:24.500 --> 00:58:27.360
You would say that
doesn't look very good
00:58:27.360 --> 00:58:31.000
because for j equals
0 you have 1 over u.
00:58:31.000 --> 00:58:31.990
That's crazy.
00:58:31.990 --> 00:58:35.650
But indeed for j equals 0,
the j here multiplies it
00:58:35.650 --> 00:58:37.090
and makes it 0.
00:58:37.090 --> 00:58:40.110
So this is OK.
00:58:40.110 --> 00:58:44.630
Now the term that we actually
need is minus 2u dh du.
00:58:44.630 --> 00:58:53.750
So here minus 2u dh du would
be equal to the sum from j
00:58:53.750 --> 00:59:02.745
equals 0 to infinity, and I
will have minus 2jaju to the j.
00:59:02.745 --> 00:59:07.990
The u makes this j minus 1 j,
and the constant went there.
00:59:07.990 --> 00:59:11.990
So here is so far h.
00:59:11.990 --> 00:59:14.250
Here is this other
term that we're
00:59:14.250 --> 00:59:18.582
going to need for the
differential equation.
00:59:18.582 --> 00:59:20.290
And then there's the
last term that we're
00:59:20.290 --> 00:59:22.125
going to need for the
differential equation,
00:59:22.125 --> 00:59:23.465
so I'm going to go here.
00:59:29.570 --> 00:59:32.110
So what do we get
for this last term.
00:59:32.110 --> 00:59:33.800
We'll have to take
a second derivative.
00:59:37.200 --> 00:59:40.290
So we'll take-- h
prime was there,
00:59:40.290 --> 00:59:46.790
so d second h du squared
will be the sum from j
00:59:46.790 --> 00:59:58.075
equals 0 of j times j minus
1 aju to the j minus 2.
01:00:03.980 --> 01:00:09.580
Now you have to rewrite this
in order to make it tractable.
01:00:09.580 --> 01:00:14.060
You want everything
to have u to the j's.
01:00:14.060 --> 01:00:17.960
You don't want actually to
have u to the j minus 2.
01:00:17.960 --> 01:00:20.130
So the first thing
that you notice
01:00:20.130 --> 01:00:28.690
is that this sum actually begins
with 2, because for 0 and 1
01:00:28.690 --> 01:00:29.370
it vanishes.
01:00:29.370 --> 01:00:37.600
So I can write j times j
minus 1 aj u to j minus 2.
01:00:37.600 --> 01:00:39.470
Like that.
01:00:39.470 --> 01:00:47.520
And then I can say let j
be equal to j prime plus 2.
01:00:52.270 --> 01:00:55.150
Look, j begins
with 2 in this sum.
01:00:55.150 --> 01:01:00.620
So if j is j prime plus 2,
j prime will begin with 0.
01:01:00.620 --> 01:01:06.490
So we've shifted the sum so it's
j prime equals 0 to infinity.
01:01:06.490 --> 01:01:09.850
And whenever I have a j I
must put j prime plus 2.
01:01:09.850 --> 01:01:13.210
So j prime plus 2.
01:01:13.210 --> 01:01:24.350
j prime plus 1 aj prime
plus 2 u to the j prime.
01:01:24.350 --> 01:01:28.670
Wherever I had j, I
put j prime plus 2.
01:01:28.670 --> 01:01:33.110
And finally you say j or
j prime is the same name,
01:01:33.110 --> 01:01:34.800
so let's call it j.
01:01:34.800 --> 01:01:36.830
j equals 0.
01:01:36.830 --> 01:01:38.610
j plus 2.
01:01:38.610 --> 01:01:41.480
j plus 1.
01:01:41.480 --> 01:01:43.997
aj plus 2 uj.
01:01:49.320 --> 01:01:52.730
So we got the series
expansion of everything,
01:01:52.730 --> 01:01:56.285
so we just plug into the
differential equation.
01:01:59.260 --> 01:02:01.170
So where is the
differential equation?
01:02:01.170 --> 01:02:03.150
It's here.
01:02:03.150 --> 01:02:04.990
So I'll plug it in.
01:02:04.990 --> 01:02:05.990
Let's see what we get.
01:02:08.590 --> 01:02:13.420
We'll get some from j
equals 0 to infinity.
01:02:13.420 --> 01:02:15.880
Let's see the second
derivative is here.
01:02:18.680 --> 01:02:35.510
j plus 2 times j plus 1 aj
plus 2 uj, so I'll put it here.
01:02:39.500 --> 01:02:41.830
So that's this second
derivative term.
01:02:41.830 --> 01:02:44.150
Now this one.
01:02:44.150 --> 01:02:44.870
It's easy.
01:02:44.870 --> 01:02:48.100
Minus 2j aj and the uj is there.
01:02:48.100 --> 01:02:49.390
So minus 2jaj.
01:02:55.050 --> 01:03:02.400
Last term is just e
minus 1, because it's
01:03:02.400 --> 01:03:05.880
the function this
times aj as well.
01:03:08.610 --> 01:03:10.890
That's h.
01:03:10.890 --> 01:03:13.870
And look, this whole
thing must be 0.
01:03:13.870 --> 01:03:16.970
So what you learn is that
this coefficient must
01:03:16.970 --> 01:03:19.920
be 0 for every value of j.
01:03:19.920 --> 01:03:23.710
Now it's possible to--
here is aj and aj,
01:03:23.710 --> 01:03:26.030
so it's actually
one single thing.
01:03:26.030 --> 01:03:27.720
Let me write it here.
01:03:27.720 --> 01:03:38.850
j plus 2 times j plus
1 aj plus 2 minus 2j
01:03:38.850 --> 01:03:46.870
plus 1 minus e aj uj.
01:03:46.870 --> 01:03:49.840
I think I got it right.
01:03:49.840 --> 01:03:51.130
Yes.
01:03:51.130 --> 01:03:52.300
And this is the same sum.
01:03:55.830 --> 01:04:01.370
And now, OK, it's a lot of
work, but we're getting there.
01:04:01.370 --> 01:04:03.330
This must be 0.
01:04:03.330 --> 01:04:08.230
So actually that solves for
aj plus 2 in terms of aj.
01:04:08.230 --> 01:04:11.760
What I had told you that you
can notice in two minutes
01:04:11.760 --> 01:04:12.880
if you try it a little.
01:04:12.880 --> 01:04:15.710
That a2 seems to be
determined by a0.
01:04:15.710 --> 01:04:18.740
And a3 seems to be
determined by a2.
01:04:18.740 --> 01:04:24.600
So this is saying
that aj plus 2 is
01:04:24.600 --> 01:04:36.380
given by 2j plus 1 minus e
over j plus 2 j plus 1 aj.
01:04:44.790 --> 01:04:47.460
A very nice recursive relation.
01:04:47.460 --> 01:04:51.780
So indeed, if you
put the value of a0,
01:04:51.780 --> 01:04:57.320
it will determine for you a2,
a4, a6, a8, all the even ones.
01:04:57.320 --> 01:05:04.070
If you put the value of a1, it
will determine for you a3, a5.
01:05:04.070 --> 01:05:10.070
So a solution is determined by
you telling me how much is a0,
01:05:10.070 --> 01:05:12.600
and telling me how much is a1.
01:05:12.600 --> 01:05:15.130
Two constants, two numbers.
01:05:15.130 --> 01:05:17.830
That's what you expect from
a second order differential
01:05:17.830 --> 01:05:18.660
equation.
01:05:18.660 --> 01:05:20.370
The value of the
function at the point,
01:05:20.370 --> 01:05:22.190
the derivative at a point.
01:05:22.190 --> 01:05:27.240
In fact, you are
looking at a0 and a1
01:05:27.240 --> 01:05:31.110
as the two constants that
will determine a solution.
01:05:31.110 --> 01:05:34.300
And this is the value of h at 0.
01:05:34.300 --> 01:05:37.590
This is the
derivative of h at 0.
01:05:37.590 --> 01:05:51.830
So we can now write
the following facts
01:05:51.830 --> 01:05:53.790
about the solution
that we have found.
01:05:57.720 --> 01:05:59.090
So what do we know?
01:05:59.090 --> 01:06:13.170
That solutions fixed
by giving a0 and a1.
01:06:15.680 --> 01:06:19.245
That correspond to the
value of the function at 0
01:06:19.245 --> 01:06:23.120
and the derivative
of the function at 0.
01:06:23.120 --> 01:06:25.350
And this gives one solution.
01:06:25.350 --> 01:06:29.780
Once you fix a0, you get a2, a4.
01:06:29.780 --> 01:06:33.330
And this is an even
solution, because it
01:06:33.330 --> 01:06:36.940
has only even powers.
01:06:36.940 --> 01:06:43.360
And then from a1, you fixed
a3, a5, all the other ones
01:06:43.360 --> 01:06:44.470
with an odd solution.
01:06:50.055 --> 01:06:50.555
OK.
01:06:53.680 --> 01:06:55.810
Well, we solve the
differential equation,
01:06:55.810 --> 01:06:58.690
which is really,
in a sense, bad,
01:06:58.690 --> 01:07:01.880
because we were expecting
that we can only solve it
01:07:01.880 --> 01:07:03.185
for some values of the energy.
01:07:05.690 --> 01:07:13.460
Moreover, you have a0,
you get a2, a4, a6, a8.
01:07:13.460 --> 01:07:18.220
This will go on forever
and not terminate.
01:07:18.220 --> 01:07:20.910
And then it will be an
infinite polynomial that
01:07:20.910 --> 01:07:24.160
grows up and doesn't
ever decline,
01:07:24.160 --> 01:07:28.070
which is sort of
contradictory with the idea
01:07:28.070 --> 01:07:31.710
that we had before
that near infinity
01:07:31.710 --> 01:07:37.160
the function was going to be
some power, some fixed power,
01:07:37.160 --> 01:07:39.410
times this exponential.
01:07:39.410 --> 01:07:44.950
So this is what we're looking
for, this h function now.
01:07:44.950 --> 01:07:47.000
It doesn't look
like a fixed power.
01:07:47.000 --> 01:07:50.240
It looks like it goes forever.
01:07:50.240 --> 01:07:53.790
So let's see what
happens eventually
01:07:53.790 --> 01:07:58.190
when the coefficient, the
value of the j index is large.
01:07:58.190 --> 01:08:02.460
For large j.
01:08:06.960 --> 01:08:13.500
aj plus 2 is roughly equal
to, for large a, whatever
01:08:13.500 --> 01:08:17.580
the energy is, sufficiently
large, the most important here
01:08:17.580 --> 01:08:20.565
is the 2j here, the j and the j.
01:08:20.565 --> 01:08:23.600
So you get 2 over j aj.
01:08:29.810 --> 01:08:37.649
So roughly for large j,
it behaves like that.
01:08:37.649 --> 01:08:41.300
And now you have to ask
yourself the question,
01:08:41.300 --> 01:08:45.140
if you have a power series
expansion whose coefficients
01:08:45.140 --> 01:08:51.020
behave like that, how
badly is it at infinity?
01:08:51.020 --> 01:08:52.130
How about is it?
01:08:55.380 --> 01:08:57.800
You know it's the
power series expansion
01:08:57.800 --> 01:09:01.770
because your h was all
these coefficients.
01:09:01.770 --> 01:09:03.670
And suppose they
behave like that.
01:09:03.670 --> 01:09:06.910
They grow in that way
or decay in this way,
01:09:06.910 --> 01:09:09.470
because they're decaying.
01:09:09.470 --> 01:09:11.770
Is this a solution
that's going to blow up?
01:09:11.770 --> 01:09:15.050
Or is it not going to blow up?
01:09:15.050 --> 01:09:19.569
And here comes an
important thing.
01:09:19.569 --> 01:09:22.380
This is pretty bad
behavior, actually.
01:09:22.380 --> 01:09:24.979
It's pretty awful behavior.
01:09:24.979 --> 01:09:28.439
So let's see that.
01:09:28.439 --> 01:09:29.290
That's pretty bad.
01:09:34.810 --> 01:09:35.950
How do we see that?
01:09:35.950 --> 01:09:39.210
Well you could do it
in different ways,
01:09:39.210 --> 01:09:41.189
depending on whether
you want to derive
01:09:41.189 --> 01:09:45.560
that this is a bad
behavior or guess it.
01:09:45.560 --> 01:09:48.580
I'm going to guess something.
01:09:48.580 --> 01:09:53.910
I'm going to look at how
does e to the u squared
01:09:53.910 --> 01:09:56.430
behave as a power series.
01:09:56.430 --> 01:09:59.030
Well, you know as a
power series exponential
01:09:59.030 --> 01:10:04.750
is 1 over n u squared to the n.
01:10:04.750 --> 01:10:06.080
Here's n factorial.
01:10:06.080 --> 01:10:07.715
n equals 0 to infinity.
01:10:10.950 --> 01:10:18.810
Now these two n's, u to the
2n, these are all even powers.
01:10:18.810 --> 01:10:21.720
So I'm going to
change letters here,
01:10:21.720 --> 01:10:28.400
and I'm going to work with j
from 0, 2, 4, over the evens.
01:10:28.400 --> 01:10:32.980
So I will write u to the j here.
01:10:32.980 --> 01:10:36.360
And that this correct, because
you produce u to the 0,
01:10:36.360 --> 01:10:39.510
u to the 2, u to the
fourth, these things.
01:10:39.510 --> 01:10:43.362
And j is really 2n,
so here you will
01:10:43.362 --> 01:10:48.560
have one over j
over 2 factorial.
01:10:48.560 --> 01:10:51.730
Now you might say, j over
2, isn't that a fraction?
01:10:51.730 --> 01:10:54.170
No, it's not a fraction,
because j is even.
01:10:54.170 --> 01:10:55.690
So this is a nice factorial.
01:10:58.600 --> 01:11:06.056
Now this is the
coefficient, cj u to the j.
01:11:06.056 --> 01:11:09.170
And let's see how this
coefficients vary.
01:11:09.170 --> 01:11:14.490
So this cj is 1 over
j over 2 factorial.
01:11:14.490 --> 01:11:20.800
What is cj plus 2 over cj?
01:11:20.800 --> 01:11:23.560
Which is the analogue
of this thing.
01:11:23.560 --> 01:11:30.940
Well, this would be 1 over
j plus 2 over 2 factorial.
01:11:30.940 --> 01:11:36.340
And here is up there,
so j over 2 factorial.
01:11:36.340 --> 01:11:40.310
Well, this has one more
factor in the denominator
01:11:40.310 --> 01:11:41.970
than the numerator.
01:11:41.970 --> 01:11:49.270
So this is roughly
one over j over 2
01:11:49.270 --> 01:11:52.105
plus 1, the last value of this.
01:11:52.105 --> 01:11:55.690
This integer is just
one bigger than that.
01:11:55.690 --> 01:11:58.800
Now if j is large,
this is roughly 1
01:11:58.800 --> 01:12:02.730
over j over 2,
which is 2 over j.
01:12:02.730 --> 01:12:06.470
Oh, exactly that stuff.
01:12:06.470 --> 01:12:09.145
So it's pretty bad.
01:12:09.145 --> 01:12:13.380
If this series
goes on forever, it
01:12:13.380 --> 01:12:18.080
will diverge like
e to the u squared.
01:12:18.080 --> 01:12:24.160
And your h will be like e to the
u squared with e to the minus
01:12:24.160 --> 01:12:27.780
u squared over 2 is going
to be like e to the plus.
01:12:27.780 --> 01:12:32.160
u squared over 2 is going
to go and behave this one.
01:12:32.160 --> 01:12:37.350
So it's going to do
exactly the wrong thing.
01:12:37.350 --> 01:12:43.280
If this series
doesn't terminate,
01:12:43.280 --> 01:12:46.290
we have not succeeded.
01:12:46.290 --> 01:12:51.200
But happily, the
series may terminate,
01:12:51.200 --> 01:12:54.420
because the j's are integers.
01:12:54.420 --> 01:12:58.670
So maybe for some energies
that are integers,
01:12:58.670 --> 01:13:01.720
it terminates, and
that's a solution.
01:13:01.720 --> 01:13:05.090
The only way to get a solution
is if the series terminates.
01:13:05.090 --> 01:13:08.360
The only way it can
terminate is that the e
01:13:08.360 --> 01:13:14.020
is some odd number over here.
01:13:14.020 --> 01:13:16.030
And that will solve the thing.
01:13:16.030 --> 01:13:19.200
So we actually need to do this.
01:13:19.200 --> 01:13:21.230
This shows the energy.
01:13:21.230 --> 01:13:24.500
You found why it's quantized.
01:13:24.500 --> 01:13:28.150
So let's do it then.
01:13:28.150 --> 01:13:31.550
We're really done
with this in a sense.
01:13:31.550 --> 01:13:36.460
This is the most important
point of the lecture,
01:13:36.460 --> 01:13:39.480
is that the series
must terminate,
01:13:39.480 --> 01:13:42.530
otherwise it will
blow up horrendously.
01:13:42.530 --> 01:13:44.550
If it terminates
as a polynomial,
01:13:44.550 --> 01:13:46.640
then everything is good.
01:13:46.640 --> 01:13:58.900
So to terminate you can choose
2j plus 1 minus e to be 0.
01:13:58.900 --> 01:14:05.010
This will make aj
plus 2 equal to 0.
01:14:07.970 --> 01:14:13.930
And your solution, your
h of u, will begin.
01:14:13.930 --> 01:14:17.800
aj will be the last
one that is non-zero,
01:14:17.800 --> 01:14:25.020
so it will be aj
times u to the j,
01:14:25.020 --> 01:14:33.500
and it will go down like aj
minus 2 u to the j minus 2.
01:14:33.500 --> 01:14:37.380
It will go down in steps of
2, because this recursion
01:14:37.380 --> 01:14:40.100
is always by steps of two.
01:14:40.100 --> 01:14:41.232
So that's it.
01:14:41.232 --> 01:14:44.350
That's going to be the solution
where these coefficients are
01:14:44.350 --> 01:14:47.160
going to be fixed by
the recursive relation,
01:14:47.160 --> 01:14:49.720
and we have this.
01:14:49.720 --> 01:14:56.180
Now most people
here call j equal n.
01:14:56.180 --> 01:14:58.080
So let's call it n.
01:14:58.080 --> 01:15:04.420
And then we have 2n
plus 1 minus e equals 0.
01:15:04.420 --> 01:15:12.580
And h of u would be an u to
the n plus all these things.
01:15:12.580 --> 01:15:14.120
That's the h.
01:15:14.120 --> 01:15:18.810
The full solution is h
times e to the minus u
01:15:18.810 --> 01:15:21.290
squared over 2 as we will see.
01:15:21.290 --> 01:15:25.250
But recall what e was.
01:15:25.250 --> 01:15:28.480
e here is 2n plus 1.
01:15:28.480 --> 01:15:36.620
But he was the true energy
divided by h omega over two.
01:15:36.620 --> 01:15:37.935
That was long ago.
01:15:41.310 --> 01:15:42.790
It's gone.
01:15:42.790 --> 01:15:46.150
Long gone.
01:15:46.150 --> 01:15:50.160
So what have you
found therefore?
01:15:50.160 --> 01:15:59.258
That the energy,
that' we'll call en,
01:15:59.258 --> 01:16:02.680
the energy of the
nth solution is
01:16:02.680 --> 01:16:07.770
going to be h omega
over 2 2n plus 1.
01:16:07.770 --> 01:16:13.890
So it's actually h omega, and
people write it n plus 1/2.
01:16:13.890 --> 01:16:15.570
Very famous result.
01:16:15.570 --> 01:16:20.440
The nth level of the harmonic
oscillator has this energy.
01:16:20.440 --> 01:16:25.060
And moreover, these
objects, people
01:16:25.060 --> 01:16:27.220
choose these-- you
see the constants are
01:16:27.220 --> 01:16:28.720
related by steps of two.
01:16:28.720 --> 01:16:32.770
So just like you could start
with a0, or a1 and go up,
01:16:32.770 --> 01:16:34.640
you can go down.
01:16:34.640 --> 01:16:38.990
People call these functions
Hermite functions.
01:16:38.990 --> 01:16:44.230
And they fix the notation so
that this an is 2 to the n.
01:16:44.230 --> 01:16:45.160
They like it.
01:16:45.160 --> 01:16:46.880
It's a nice normalization.
01:16:46.880 --> 01:16:52.410
So actually h of
n is what we call
01:16:52.410 --> 01:16:56.880
the Hermite function of u sub n.
01:16:56.880 --> 01:17:02.960
And it goes like 2 to the
n u to the n plus order u
01:17:02.960 --> 01:17:06.170
to the n minus 2 plus
n minus 4, and it
01:17:06.170 --> 01:17:07.855
goes on and on like that.
01:17:13.180 --> 01:17:18.300
OK, a couple things
and we're done.
01:17:18.300 --> 01:17:23.780
Just for reference,
the Hermite polynomial,
01:17:23.780 --> 01:17:25.950
if you're interested
in it, is the one
01:17:25.950 --> 01:17:28.230
that solves this equation.
01:17:28.230 --> 01:17:30.470
And the Hermite
sub n corresponds
01:17:30.470 --> 01:17:36.570
to e sub n, which is 2n plus 1.
01:17:36.570 --> 01:17:39.330
So the Hermite solution
from that the equation
01:17:39.330 --> 01:17:47.378
is that the Hermite polynomial
satisfies this minus 2u
01:17:47.378 --> 01:17:52.480
d Hn du plus 2n.
01:17:52.480 --> 01:17:55.050
Because en is 2n plus 1.
01:17:55.050 --> 01:17:59.240
So it's 2n Hn equals 0.
01:17:59.240 --> 01:18:02.130
That's the equation for
the Hermite polynomial,
01:18:02.130 --> 01:18:05.530
and interesting thing to know.
01:18:05.530 --> 01:18:08.100
Actually, if you want to
generate the efficiently
01:18:08.100 --> 01:18:11.490
the Hermite polynomials, there's
something called the generating
01:18:11.490 --> 01:18:13.450
function.
01:18:13.450 --> 01:18:17.730
e to the minus z
squared plus 2zu.
01:18:17.730 --> 01:18:23.440
If you expand it in
a power series of z,
01:18:23.440 --> 01:18:27.660
it actually gives you
n equals 0 to infinity.
01:18:27.660 --> 01:18:32.780
If it's a power series of z,
it will be some z to the n's.
01:18:32.780 --> 01:18:37.605
You can put a factor here
n, and here is Hn of u.
01:18:40.620 --> 01:18:43.740
So you can use your
mathematic program
01:18:43.740 --> 01:18:46.590
and expand this in powers of z.
01:18:46.590 --> 01:18:51.370
Collect the various powers of
u that appear with z to the n,
01:18:51.370 --> 01:18:56.260
and that's Hn It's the most
efficient way of generating Hn
01:18:56.260 --> 01:18:59.480
And moreover, if you want
to play in mathematics,
01:18:59.480 --> 01:19:03.840
you can show that
such definition of Hn
01:19:03.840 --> 01:19:05.950
satisfies this equation.
01:19:05.950 --> 01:19:08.380
So it produces the solution.
01:19:08.380 --> 01:19:10.810
So what have we found?
01:19:10.810 --> 01:19:13.910
Our end result is the following.
01:19:13.910 --> 01:19:17.080
Let me finish with that here.
01:19:17.080 --> 01:19:25.070
We had this potential, and the
first energy level is called E0
01:19:25.070 --> 01:19:27.620
and has energy h omega over 2.
01:19:27.620 --> 01:19:29.940
The next energy is E1.
01:19:29.940 --> 01:19:33.810
It has 3/2 h omega.
01:19:33.810 --> 01:19:39.830
Next one is E2 5/2 h omega.
01:19:39.830 --> 01:19:42.940
This polynomial is
nth degree polynomial.
01:19:42.940 --> 01:19:46.860
So it has n zeros,
therefore n nodes.
01:19:46.860 --> 01:19:50.470
So these wave functions will
have the right number of nodes.
01:19:50.470 --> 01:19:54.030
E0, the psi 0,
will have no nodes.
01:19:54.030 --> 01:19:59.300
When you have psi 0, the
Hn becomes a number for n
01:19:59.300 --> 01:20:00.510
equals zero.
01:20:00.510 --> 01:20:02.830
And the whole solution
is the exponential
01:20:02.830 --> 01:20:05.610
of u squared over 2.
01:20:05.610 --> 01:20:08.960
The whole solution, in
fact, is, as we wrote,
01:20:08.960 --> 01:20:15.150
psi n Hn of u e to the
minus u squared over 2.
01:20:15.150 --> 01:20:18.630
In plain English,
if you use an x,
01:20:18.630 --> 01:20:24.890
it will be Hn u with x over
that constant a we had.
01:20:24.890 --> 01:20:30.470
And you have minus x
squared over 2a squared.
01:20:30.470 --> 01:20:32.810
Those are your eigenfunctions.
01:20:32.810 --> 01:20:34.750
These are the solutions.
01:20:34.750 --> 01:20:39.020
Discrete spectrum, evenly
spaced, the nicest spectrum
01:20:39.020 --> 01:20:40.530
possible.
01:20:40.530 --> 01:20:42.730
All the nodes are there.
01:20:42.730 --> 01:20:46.620
You will solve this in a
more clever way next time.
01:20:46.620 --> 01:20:48.170
[APPLAUSE]