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ROBERT FIELD: Today is
the first of two lectures
00:00:24.870 --> 00:00:26.790
on the rigid rotor.
00:00:26.790 --> 00:00:30.300
And the rigid rotor is
really our first glimpse
00:00:30.300 --> 00:00:32.340
of central force problems.
00:00:32.340 --> 00:00:35.940
And so although the rigid rotor
is an exactly solved problem,
00:00:35.940 --> 00:00:38.820
it's also a problem that we--
00:00:38.820 --> 00:00:42.160
it involves something
that is universal.
00:00:42.160 --> 00:00:46.500
If we have a spherical
object, then the angular part
00:00:46.500 --> 00:00:50.220
of the Hamiltonian for
that spherical object
00:00:50.220 --> 00:00:52.810
is solved by the rigid rotor.
00:00:52.810 --> 00:00:57.280
And so if we understand
the rigid rotor
00:00:57.280 --> 00:01:01.380
and know how to
draw pictures, we're
00:01:01.380 --> 00:01:04.650
going to be able to
understand the universal part
00:01:04.650 --> 00:01:07.410
of all central force problems.
00:01:07.410 --> 00:01:09.240
So it's not just a curiosity.
00:01:09.240 --> 00:01:11.910
It's a really major thing.
00:01:11.910 --> 00:01:12.420
OK.
00:01:12.420 --> 00:01:16.170
And I'm going to do a lot
of unconventional stuff
00:01:16.170 --> 00:01:18.360
in this lecture and
the next lecture,
00:01:18.360 --> 00:01:20.500
as far as the rigid
rotor is concerned.
00:01:20.500 --> 00:01:24.750
Because it's covered
to death, but with lots
00:01:24.750 --> 00:01:28.140
of equations in all textbooks.
00:01:28.140 --> 00:01:34.260
And there is very little
effort to give you
00:01:34.260 --> 00:01:36.480
independent insights.
00:01:36.480 --> 00:01:40.740
And so my discussion
of the vector model
00:01:40.740 --> 00:01:43.290
is something that I
consider really special.
00:01:43.290 --> 00:01:45.900
And what I want you
to be able to do
00:01:45.900 --> 00:01:49.290
is to draw cartoons
that capture essentially
00:01:49.290 --> 00:01:51.660
all of the important points.
00:01:51.660 --> 00:01:55.305
And that you won't find
anywhere except in my notes.
00:01:58.650 --> 00:02:01.080
But I think it's
really important
00:02:01.080 --> 00:02:04.390
that you have your own point
of view on these problems.
00:02:04.390 --> 00:02:10.229
So let me just do a quick
review, with the exam in mind,
00:02:10.229 --> 00:02:14.520
of the perturbation
theory stuff.
00:02:14.520 --> 00:02:17.050
So we had a harmonic oscillator.
00:02:17.050 --> 00:02:22.080
And it's perturbed by
some anharmonicity terms.
00:02:22.080 --> 00:02:28.770
So the potential is going
to 1/2 kQ squared plus bQ
00:02:28.770 --> 00:02:35.040
cubed plus cQ to the fourth.
00:02:35.040 --> 00:02:36.240
And we could go on.
00:02:36.240 --> 00:02:37.410
But this is enough.
00:02:37.410 --> 00:02:41.010
Because it gives a glimpse
of all of the important stuff
00:02:41.010 --> 00:02:44.310
that emerges qualitatively
as well as quantitatively
00:02:44.310 --> 00:02:47.460
from perturbation theory.
00:02:47.460 --> 00:02:53.130
So this is part
of the H0 problem.
00:02:53.130 --> 00:02:56.960
And this is H1.
00:02:56.960 --> 00:03:02.390
And so you know now that doing
second order perturbation
00:03:02.390 --> 00:03:05.060
theory is tedious and ugly.
00:03:05.060 --> 00:03:09.560
And it's easy to get
overwhelmed by the equations.
00:03:09.560 --> 00:03:12.020
But what I want to
do is to just review
00:03:12.020 --> 00:03:16.950
how you get the crucial
structure of the problem.
00:03:16.950 --> 00:03:23.590
So the Q cubed' term--
00:03:23.590 --> 00:03:27.680
there are no diagonal
elements of the Q cubed term.
00:03:27.680 --> 00:03:31.790
And so we're always going to be
talking about the second order
00:03:31.790 --> 00:03:33.650
corrections to the energy.
00:03:33.650 --> 00:03:38.000
The selection rules are delta
v equals plus and minus 3
00:03:38.000 --> 00:03:42.522
and plus n minus 1.
00:03:42.522 --> 00:03:47.200
And so what I told you is, if
you're going to do the algebra,
00:03:47.200 --> 00:03:50.010
which you probably
don't want to ever do,
00:03:50.010 --> 00:03:52.980
you're going to want to
combine the delta v of plus 3
00:03:52.980 --> 00:03:56.010
and minus 3 terms of
the perturbation sum.
00:03:56.010 --> 00:03:58.650
Because the algebra simplifies.
00:03:58.650 --> 00:04:01.290
They have similar expressions.
00:04:01.290 --> 00:04:09.810
And you have a denominator,
which is 3 Hc omega and minus 3
00:04:09.810 --> 00:04:10.590
Hc omega.
00:04:10.590 --> 00:04:11.440
That factors out.
00:04:11.440 --> 00:04:15.150
And you just have a plus 3 and
minus 3 in the denominator.
00:04:15.150 --> 00:04:19.110
And similarly here,
so when you do this,
00:04:19.110 --> 00:04:21.300
you lose the highest order term.
00:04:21.300 --> 00:04:22.830
What's the highest order term?
00:04:22.830 --> 00:04:29.160
It's v plus 1/2 to
the third power.
00:04:29.160 --> 00:04:33.510
Because we have delta v of 3.
00:04:33.510 --> 00:04:35.940
We have Q cubed.
00:04:35.940 --> 00:04:39.150
And the matrix
elements of Q cubed
00:04:39.150 --> 00:04:43.230
are a plus a-dagger cubed.
00:04:43.230 --> 00:04:45.440
And the selection
rule for them--
00:04:45.440 --> 00:04:49.650
or the matrix-- must go as
the square root of the power
00:04:49.650 --> 00:04:51.900
of the a's.
00:04:51.900 --> 00:04:54.540
And so all of the
matrix elements
00:04:54.540 --> 00:04:59.640
have the leading term v
plus 1/2 to the 3/2 power.
00:04:59.640 --> 00:05:02.470
But we're squaring.
00:05:02.470 --> 00:05:06.590
And we lose the highest order
because of the subtraction.
00:05:06.590 --> 00:05:11.495
So this one gives rise to
terms in v plus 1/2 squared.
00:05:14.080 --> 00:05:22.100
Q4 has selection rules
delta v plus or minus 4,
00:05:22.100 --> 00:05:24.820
plus or minus 2, and 0.
00:05:24.820 --> 00:05:25.770
This is important.
00:05:25.770 --> 00:05:28.950
Because this one says,
we have a contribution
00:05:28.950 --> 00:05:30.840
from the first order--
00:05:30.840 --> 00:05:33.490
in the first order of energy.
00:05:33.490 --> 00:05:35.560
And so in the first
order of energy,
00:05:35.560 --> 00:05:41.745
we're going to have v plus
1/2 to the fourth power--
00:05:46.140 --> 00:05:49.620
I'm sorry-- to the 4/2 power.
00:05:53.460 --> 00:05:56.022
Because we only have
the matrix element,
00:05:56.022 --> 00:05:57.480
we don't square
the matrix element.
00:06:01.630 --> 00:06:08.320
So this gives rise to a
term in v plus 1/2 squared.
00:06:08.320 --> 00:06:11.980
But it's sensitive
to the sign of the Q
00:06:11.980 --> 00:06:14.470
to the fourth term
in the Hamiltonian.
00:06:14.470 --> 00:06:18.180
It's the only thing that's
sensitive to the sine.
00:06:18.180 --> 00:06:29.740
We also have terms that are v
plus 1/2 to the 4/2 times 2.
00:06:29.740 --> 00:06:39.500
So we can have matrix
elements to the four--
00:06:39.500 --> 00:06:43.670
because it's a fourth
power term, we get 4/2.
00:06:43.670 --> 00:06:46.220
And because it's squared,
we multiply by 2.
00:06:46.220 --> 00:06:47.780
And then we lose
the highest order.
00:06:47.780 --> 00:06:54.010
So we get, from this, v
plus 1/2 to the third power.
00:06:54.010 --> 00:06:58.870
So what we're hoping to get
from the perturbation theory
00:06:58.870 --> 00:07:04.060
is the highest order terms
in the energy expression.
00:07:04.060 --> 00:07:09.730
And so Q to the fourth gives a
signed v plus 1/2 squared term.
00:07:09.730 --> 00:07:14.950
And it gives an unsigned always
positive v plus 1/2 cubed term.
00:07:14.950 --> 00:07:21.790
And this one gives v
plus 1/2 squared term.
00:07:24.380 --> 00:07:26.800
OK.
00:07:26.800 --> 00:07:28.390
So this would tell
you, when you're
00:07:28.390 --> 00:07:32.290
organizing your work, what to
expect and how to organize it.
00:07:32.290 --> 00:07:35.020
And I'm really not
expecting you to do
00:07:35.020 --> 00:07:39.890
very much with this level of
complexity on an exam problem.
00:07:39.890 --> 00:07:42.940
But on homework,
all bets are off.
00:07:42.940 --> 00:07:44.680
OK.
00:07:44.680 --> 00:07:48.010
So that's all I want to do as
far as the review is concerned.
00:07:50.890 --> 00:07:54.010
So the rigid rotor--
00:07:54.010 --> 00:07:57.070
we have a molecule.
00:08:11.060 --> 00:08:14.900
So we have a bond
of length r sub 0.
00:08:14.900 --> 00:08:16.680
And we have two masses.
00:08:16.680 --> 00:08:18.850
And if the masses aren't
equal, the center of mass
00:08:18.850 --> 00:08:21.789
isn't in the middle.
00:08:21.789 --> 00:08:23.330
The center of mass--
you need to know
00:08:23.330 --> 00:08:25.670
how to calculate where
the center of mass is.
00:08:25.670 --> 00:08:28.700
There are all sorts
of simple algebra--
00:08:28.700 --> 00:08:30.800
but what is
happening is this guy
00:08:30.800 --> 00:08:34.914
is rotating without stretching
about the center of mass.
00:08:43.549 --> 00:08:47.870
Now, what we want to do is
think about this problem
00:08:47.870 --> 00:08:48.860
as if it were--
00:08:53.710 --> 00:08:55.850
here's our 0.
00:08:55.850 --> 00:08:58.740
And this is the reduced mass.
00:08:58.740 --> 00:09:04.740
This is a motion of a
fictitious particle of mass mu.
00:09:04.740 --> 00:09:16.360
Mu is m1 m2 over m1 plus m2
on the surface of a sphere.
00:09:16.360 --> 00:09:18.220
They're all the same problem.
00:09:18.220 --> 00:09:20.680
But the question is,
how do we interpret
00:09:20.680 --> 00:09:23.500
what we get from the
solution to this problem?
00:09:27.120 --> 00:09:32.210
So what we care about
is a description of,
00:09:32.210 --> 00:09:34.910
where is the molecular axis?
00:09:34.910 --> 00:09:37.190
The molecule is rotating.
00:09:37.190 --> 00:09:41.090
And we're solving the
Schrodinger equation.
00:09:41.090 --> 00:09:46.760
And we get things like the
expectation value of J squared
00:09:46.760 --> 00:09:52.390
and Jz and maybe some
other things where
00:09:52.390 --> 00:09:55.360
these are the angular momenta.
00:09:55.360 --> 00:10:04.270
And how do the eigenfunctions
for these states--
00:10:04.270 --> 00:10:05.620
we have a state.
00:10:05.620 --> 00:10:12.470
And we have quantum numbers
Jm and their probability
00:10:12.470 --> 00:10:14.480
amplitudes and theta and phi.
00:10:14.480 --> 00:10:16.085
That's a whole lot
of stuff to digest.
00:10:19.270 --> 00:10:21.580
We want to go as
quickly as we can
00:10:21.580 --> 00:10:27.650
to how this is related to the
thing we really care about,
00:10:27.650 --> 00:10:29.080
which is, where
is the bond axis?
00:10:33.100 --> 00:10:34.970
So the molecule is rotating.
00:10:34.970 --> 00:10:40.220
And so the bond axis is
moving in laboratory frame.
00:10:40.220 --> 00:10:43.450
And we want to be able
to take, from that,
00:10:43.450 --> 00:10:46.525
the minimal amount
of information
00:10:46.525 --> 00:10:49.730
that we memorize or remember--
00:10:49.730 --> 00:10:50.990
I don't like "memorize."
00:10:50.990 --> 00:10:53.810
Because "memorize" doesn't
involve understanding.
00:10:53.810 --> 00:10:56.060
But "remembering" does.
00:10:56.060 --> 00:11:01.120
We want to be able to,
at a drop of a hat,
00:11:01.120 --> 00:11:05.240
be able to say, yes, if
we pick these two quantum
00:11:05.240 --> 00:11:08.180
numbers, which are related to
the eigenvalues of these two
00:11:08.180 --> 00:11:12.320
operators, we can describe
the spatial distribution
00:11:12.320 --> 00:11:13.325
of the molecular axis.
00:11:17.020 --> 00:11:18.520
There is an extra complication.
00:11:25.700 --> 00:11:29.100
So we live in the laboratory.
00:11:29.100 --> 00:11:31.520
And we have a coordinate
system that we
00:11:31.520 --> 00:11:34.850
care about, the
lab-fixed coordinates.
00:11:34.850 --> 00:11:36.860
And they're always
going to be represented
00:11:36.860 --> 00:11:39.020
with capital letters.
00:11:39.020 --> 00:11:42.350
And there's also the body frame.
00:11:42.350 --> 00:11:45.150
And things are in
the body frame.
00:11:45.150 --> 00:11:47.690
That's what the
molecule cares about.
00:11:47.690 --> 00:11:50.180
And how do things
in the body frame
00:11:50.180 --> 00:11:51.830
relate to the laboratory frame?
00:11:51.830 --> 00:11:54.890
It's not trivial.
00:11:54.890 --> 00:11:58.970
And that's where the real
effort at understanding comes.
00:11:58.970 --> 00:12:04.150
And it's kind of trivial
for a diatomic molecule.
00:12:04.150 --> 00:12:10.860
But it's far from trivial when
you have a nonlinear molecule,
00:12:10.860 --> 00:12:15.760
a molecule with many atoms and
properties of each of the atoms
00:12:15.760 --> 00:12:19.840
that somehow combine to give
things that you observe.
00:12:19.840 --> 00:12:25.990
And so these two coordinate
systems are very different.
00:12:25.990 --> 00:12:30.010
And actually, the solutions
of the rigid rotor
00:12:30.010 --> 00:12:37.030
Hamiltonian gives you the
relationship of the body
00:12:37.030 --> 00:12:40.540
coordinates to the
laboratory coordinates.
00:12:40.540 --> 00:12:42.920
And that's kind of important.
00:12:42.920 --> 00:12:47.070
So we'll see how this
develops as I go on.
00:12:47.070 --> 00:12:48.970
And I feel quite
passionate about this.
00:12:48.970 --> 00:12:54.820
Because one of the things that
I do as an experimentalist is
00:12:54.820 --> 00:12:58.000
I observe fully
resolved spectra.
00:12:58.000 --> 00:13:01.000
And the big-- the most
information-rich feature
00:13:01.000 --> 00:13:04.460
of the spectrum is the
rotational spectrum.
00:13:04.460 --> 00:13:09.930
And so it contains a lot
of really good stuff.
00:13:09.930 --> 00:13:12.320
OK.
00:13:12.320 --> 00:13:20.630
So first of all, the Hamiltonian
is just the kinetic energy.
00:13:20.630 --> 00:13:23.160
Because it's a rigid
rotor, it's free.
00:13:23.160 --> 00:13:24.642
There's no potential.
00:13:27.740 --> 00:13:30.110
But that's the last
nice thing about it.
00:13:30.110 --> 00:13:36.980
Because we have to go from
Cartesian to spherical polar
00:13:36.980 --> 00:13:38.390
coordinates.
00:13:38.390 --> 00:13:40.850
And there are a lot
of unfamiliar things
00:13:40.850 --> 00:13:44.740
in this kinetic
energy expression.
00:13:44.740 --> 00:13:47.099
And when I write it down,
you're not going to like it.
00:13:47.099 --> 00:13:48.890
Unless you're a
mathematician, and you say,
00:13:48.890 --> 00:13:52.240
oh, yeah, I want to be able to
solve these kinds of equations.
00:13:52.240 --> 00:13:56.320
Because the Schrodinger
equation is,
00:13:56.320 --> 00:14:01.820
considering the simplicity
of the problem, terrible.
00:14:01.820 --> 00:14:04.370
And except for mathematicians
who just love it
00:14:04.370 --> 00:14:05.420
because they know--
00:14:05.420 --> 00:14:06.620
oh, I know that equation.
00:14:06.620 --> 00:14:09.572
I know how to write down
everything that I care about.
00:14:09.572 --> 00:14:11.030
But we care about
different things.
00:14:14.400 --> 00:14:21.250
So we want to know the energy
levels of the rigid rotor.
00:14:21.250 --> 00:14:25.255
And we also want pictures.
00:14:28.530 --> 00:14:33.670
And remember, a picture is a
reduced form of all the details
00:14:33.670 --> 00:14:35.990
that you have at your disposal.
00:14:35.990 --> 00:14:38.140
And you have to
really sweat to make
00:14:38.140 --> 00:14:41.980
sure you understand every detail
of the pictures and the stuff
00:14:41.980 --> 00:14:44.560
that you've averaged
over, or you've concealed,
00:14:44.560 --> 00:14:47.800
because you don't need it, OK?
00:14:51.631 --> 00:14:52.130
All right.
00:14:52.130 --> 00:14:55.900
So I'm going to be
generating a lot of pictures.
00:14:55.900 --> 00:14:59.400
And one of the things we
want to understand is--
00:14:59.400 --> 00:15:04.460
so we have this vector, J.
The J is the angular momentum
00:15:04.460 --> 00:15:07.340
of the molecule.
00:15:07.340 --> 00:15:09.760
And it's a vector as
opposed to a scalar.
00:15:09.760 --> 00:15:12.850
That means it has
three components.
00:15:12.850 --> 00:15:16.140
So for this part of the problem
that I've just hidden over
00:15:16.140 --> 00:15:19.230
here, that's a one
dimensional problem.
00:15:19.230 --> 00:15:22.740
We have a momentum and
a conjugate coordinate
00:15:22.740 --> 00:15:25.230
and it's 1D.
00:15:25.230 --> 00:15:26.040
This is 3D.
00:15:26.040 --> 00:15:28.260
And there's all sorts of
subtle stuff that goes on.
00:15:30.950 --> 00:15:31.670
OK.
00:15:31.670 --> 00:15:37.760
So we would like to understand
how this vector moves.
00:15:37.760 --> 00:15:39.680
Well, there is a question.
00:15:39.680 --> 00:15:40.994
I said moves.
00:15:40.994 --> 00:15:43.160
We're talking about the
time independent Schrodinger
00:15:43.160 --> 00:15:44.690
equation.
00:15:44.690 --> 00:15:45.560
Nothing moves.
00:15:49.750 --> 00:15:54.280
We can use our concept of
motion from classical mechanics
00:15:54.280 --> 00:15:57.490
to describe certain
features of the average
00:15:57.490 --> 00:16:00.470
of some quantum
mechanical system.
00:16:00.470 --> 00:16:02.180
And so there is
kind of a motion.
00:16:02.180 --> 00:16:04.401
But we'll see what that is.
00:16:04.401 --> 00:16:04.900
OK.
00:16:04.900 --> 00:16:07.000
We have some operators
that we like.
00:16:20.580 --> 00:16:21.360
OK.
00:16:21.360 --> 00:16:24.900
Total angular momentum,
projection of the angular
00:16:24.900 --> 00:16:28.350
momentum on a
laboratory z-axis--
00:16:28.350 --> 00:16:30.990
it's kind of hard to draw
capital Z and small z
00:16:30.990 --> 00:16:32.610
if you haven't got
anybody nearby.
00:16:32.610 --> 00:16:36.390
But anyway, this is
a capital Z. And this
00:16:36.390 --> 00:16:40.230
is the thing that's analogous
to the creation and annihilation
00:16:40.230 --> 00:16:40.830
operators.
00:16:43.350 --> 00:16:44.970
We need them.
00:16:44.970 --> 00:16:46.980
And it gives us
a lot of insight.
00:16:46.980 --> 00:16:49.500
And that will be the
subject of Monday's lecture.
00:16:52.610 --> 00:16:57.000
And so we're going to have
solutions, which are described
00:16:57.000 --> 00:16:59.370
by J and m quantum numbers.
00:16:59.370 --> 00:17:06.530
And we want to understand
what these things look like.
00:17:06.530 --> 00:17:07.885
Where are the nodal surfaces?
00:17:10.650 --> 00:17:14.030
The nodal surfaces give
us basically everything
00:17:14.030 --> 00:17:15.199
we want to know.
00:17:18.349 --> 00:17:21.280
There is the direct correlation
to where the nodes are
00:17:21.280 --> 00:17:26.230
and how many of them there are
with these quantum numbers.
00:17:26.230 --> 00:17:27.790
That's one of the
important things
00:17:27.790 --> 00:17:28.920
you have to come away with.
00:17:34.461 --> 00:17:34.960
OK.
00:17:34.960 --> 00:17:38.830
In my next lecture
on angular momentum,
00:17:38.830 --> 00:17:42.010
I'm going to introduce
a commutation rule,
00:17:42.010 --> 00:17:55.330
Ji Jj equals I H bar sum
over K epsilon I Jk Jk.
00:17:55.330 --> 00:17:56.990
Isn't that a strange
looking thing?
00:17:56.990 --> 00:18:01.252
Because I haven't told
you what this epsilon is.
00:18:01.252 --> 00:18:02.460
There's lots of names for it.
00:18:02.460 --> 00:18:03.000
It won't help.
00:18:03.000 --> 00:18:04.833
Because I'm not going
to talk about it here.
00:18:04.833 --> 00:18:08.430
But this is actually the
fundamental definition
00:18:08.430 --> 00:18:10.970
of an angular momentum.
00:18:10.970 --> 00:18:12.770
And from this
commutator rule, we
00:18:12.770 --> 00:18:18.140
can generate all the matrix
elements of angular momentum.
00:18:18.140 --> 00:18:23.170
And there's another family
of commutation rules
00:18:23.170 --> 00:18:25.880
where we have a component
of angular momentum
00:18:25.880 --> 00:18:29.710
and a component of what we
call a spherical tensor.
00:18:29.710 --> 00:18:33.310
Any operator can be classified
according to spherical tensor
00:18:33.310 --> 00:18:34.570
rank.
00:18:34.570 --> 00:18:37.300
And that then determines
all the matrix elements
00:18:37.300 --> 00:18:40.070
of that operator.
00:18:40.070 --> 00:18:42.860
And so you can imagine that
this is a really powerful way
00:18:42.860 --> 00:18:44.510
of approaching stuff.
00:18:44.510 --> 00:18:51.470
And what I said before about the
commutation rule of x and P sub
00:18:51.470 --> 00:18:54.800
x is that many people regard
that as the foundation
00:18:54.800 --> 00:18:56.697
of quantum mechanics.
00:18:56.697 --> 00:18:58.280
And this is just an
extension of that.
00:19:01.160 --> 00:19:03.580
So there is a way of getting
all quantum mechanics
00:19:03.580 --> 00:19:07.092
from a few well-chosen
commutation rules.
00:19:07.092 --> 00:19:08.050
And that's really neat.
00:19:11.251 --> 00:19:11.750
OK.
00:19:11.750 --> 00:19:12.875
Let's get down to business.
00:19:15.920 --> 00:19:16.920
So I already drew this.
00:19:16.920 --> 00:19:18.930
But I'll draw it again.
00:19:18.930 --> 00:19:21.790
Because I'm going
to drop a companion.
00:19:21.790 --> 00:19:22.290
OK.
00:19:22.290 --> 00:19:23.730
So we have here--
00:19:35.560 --> 00:19:36.060
OK.
00:19:36.060 --> 00:19:37.800
So we have an angular momentum.
00:19:37.800 --> 00:19:40.260
And we have the bond axis.
00:19:40.260 --> 00:19:46.270
And this angular momentum is
perpendicular to the bond axis.
00:19:46.270 --> 00:19:49.150
So if we know something about
how the angular momentum is
00:19:49.150 --> 00:19:52.570
distributed in space, we know
something about how the bond
00:19:52.570 --> 00:19:54.460
axis is distributed in space.
00:19:54.460 --> 00:19:57.800
But it's not trivial.
00:19:57.800 --> 00:20:00.250
But once you've learned how
to make those connections,
00:20:00.250 --> 00:20:02.031
it's fine.
00:20:02.031 --> 00:20:02.530
OK.
00:20:02.530 --> 00:20:03.730
So now, let's draw--
00:20:12.250 --> 00:20:16.250
So we have a right-handed
coordinate system.
00:20:16.250 --> 00:20:20.110
And we have, say,
the vector J. Now,
00:20:20.110 --> 00:20:22.000
this is the laboratory frame.
00:20:22.000 --> 00:20:34.150
We have J. And we
have a molecule, which
00:20:34.150 --> 00:20:40.490
is perpendicular
to J. And we have
00:20:40.490 --> 00:20:45.950
the projection of J on the
body axis, on the z-axis,
00:20:45.950 --> 00:20:49.470
and that's M.
00:20:49.470 --> 00:20:49.970
OK.
00:20:49.970 --> 00:20:53.900
So when we talked about
the quantum number
00:20:53.900 --> 00:20:58.310
J, which is the length of
this vector, and m, which
00:20:58.310 --> 00:21:00.860
is the projection of
that vector on this axis,
00:21:00.860 --> 00:21:04.790
that's beginning to be how
we understand how this works.
00:21:04.790 --> 00:21:06.620
Because this is
perpendicular to that.
00:21:09.480 --> 00:21:15.135
Now, here we have
one of the sins.
00:21:25.150 --> 00:21:28.330
So we begin by saying,
well, this is a picture.
00:21:28.330 --> 00:21:36.780
And J precesses, or
moves, on a cone around z.
00:21:36.780 --> 00:21:39.840
How do we have motion?
00:21:39.840 --> 00:21:43.990
This is a time independent
Schrodinger equation.
00:21:43.990 --> 00:21:49.630
It's a way of saying, well,
it doesn't matter where J is.
00:21:49.630 --> 00:21:52.660
It's more or less
equally distributed
00:21:52.660 --> 00:21:57.170
in probability on this cone.
00:21:57.170 --> 00:21:58.734
But not an amplitude.
00:22:01.460 --> 00:22:03.130
Remember, when we
have something that's
00:22:03.130 --> 00:22:06.700
moving in the time independent
Schrodinger equation,
00:22:06.700 --> 00:22:09.520
we get oscillations.
00:22:09.520 --> 00:22:11.830
But we need a complex
function in order
00:22:11.830 --> 00:22:13.750
to have those oscillations.
00:22:13.750 --> 00:22:18.430
If we're going to take psi star
psi, there is no complex part.
00:22:18.430 --> 00:22:22.900
And you do have
uniform amplitude of J.
00:22:22.900 --> 00:22:25.720
And so it's sensible to
say, well, it got that way
00:22:25.720 --> 00:22:27.910
because J precesses.
00:22:27.910 --> 00:22:30.450
Maybe.
00:22:30.450 --> 00:22:32.970
There is something
if you say, well,
00:22:32.970 --> 00:22:37.020
maybe if I created a
system at t equal 0
00:22:37.020 --> 00:22:42.040
where J was at a
particular position,
00:22:42.040 --> 00:22:45.380
and then I let
that thing evolve--
00:22:45.380 --> 00:22:46.670
this is not an eigenstate.
00:22:46.670 --> 00:22:50.330
It would be a complicated
superposition.
00:22:50.330 --> 00:22:52.610
That thing would precess.
00:22:52.610 --> 00:22:55.955
And you would observe what
we call polarization quantum
00:22:55.955 --> 00:22:56.780
beats.
00:22:56.780 --> 00:22:59.660
Now, that's getting
way ahead of things.
00:22:59.660 --> 00:23:07.360
But it is helpful to think
about J precessing on this cone.
00:23:07.360 --> 00:23:12.840
Because that gives you a
sense that the length of J is
00:23:12.840 --> 00:23:14.140
conserved.
00:23:14.140 --> 00:23:17.800
And the orientation
about the z-axis is not.
00:23:17.800 --> 00:23:21.610
And it's more or less
uniform probability, but not
00:23:21.610 --> 00:23:23.560
uniform probability amplitude.
00:23:23.560 --> 00:23:27.340
But you don't need that
for a lot of the things.
00:23:27.340 --> 00:23:29.040
If you're looking at
the wave function,
00:23:29.040 --> 00:23:31.030
yeah, there's going to
be some oscillation.
00:23:31.030 --> 00:23:32.740
The real part and
the imaginary part
00:23:32.740 --> 00:23:36.890
will oscillate in such a
way that the probability
00:23:36.890 --> 00:23:38.920
is constant about that.
00:23:38.920 --> 00:23:40.810
I finally realized
that this morning.
00:23:40.810 --> 00:23:43.540
So it's not as much
of a lie as you think.
00:23:43.540 --> 00:23:45.160
OK.
00:23:45.160 --> 00:23:49.750
So this is the picture.
00:23:49.750 --> 00:23:53.950
And this is what it looks
like in the laboratory.
00:23:53.950 --> 00:24:02.650
And now, I'm going
to prepare you for--
00:24:02.650 --> 00:24:06.040
so suppose you had a molecule.
00:24:06.040 --> 00:24:14.670
And you have a little person
standing on the molecule frame.
00:24:14.670 --> 00:24:17.875
And now, you have somebody out
here observing as the molecule
00:24:17.875 --> 00:24:18.375
rotates.
00:24:22.982 --> 00:24:26.790
Well, and here's J. All
information that you're
00:24:26.790 --> 00:24:30.800
allowed to know from
outside the molecule
00:24:30.800 --> 00:24:33.620
comes from the
projection of whatever
00:24:33.620 --> 00:24:41.650
is in the molecule
frame on J. And that's
00:24:41.650 --> 00:24:42.950
what the vector model does.
00:24:42.950 --> 00:24:44.920
It tells you how
to take stuff you
00:24:44.920 --> 00:24:48.160
know about the individual
atoms and project it
00:24:48.160 --> 00:24:51.974
on the thing that communicates
with the outside world.
00:24:51.974 --> 00:24:53.390
There are a couple
of other things
00:24:53.390 --> 00:24:54.765
that I'm going to
say about this.
00:24:54.765 --> 00:24:57.190
But now, let's get
down to the business
00:24:57.190 --> 00:25:00.430
of actually doing a
little bit of the solution
00:25:00.430 --> 00:25:01.718
of the Schrodinger equation.
00:25:14.180 --> 00:25:18.170
So for our free
rotor, the potential
00:25:18.170 --> 00:25:25.220
is r equals r0 theta phi.
00:25:25.220 --> 00:25:29.630
And it's equal to
0 if r equals r0.
00:25:29.630 --> 00:25:32.060
And it's equal to
infinity if it's not.
00:25:32.060 --> 00:25:35.810
So this is very much like a
particle in an infinite box.
00:25:35.810 --> 00:25:40.810
In fact, you can do a
really cheap solution
00:25:40.810 --> 00:25:44.740
if you say, well, let's
consider a particle
00:25:44.740 --> 00:25:48.020
in an infinite circular box.
00:25:48.020 --> 00:25:51.620
Well, you can solve this problem
just using the de Broglie
00:25:51.620 --> 00:25:53.380
wavelength.
00:25:53.380 --> 00:25:57.250
And you get that the energy
levels for this circular box
00:25:57.250 --> 00:26:01.800
problem go as the
quantum number squared.
00:26:01.800 --> 00:26:04.870
And that's almost exactly
like the solutions
00:26:04.870 --> 00:26:09.980
to the free rotor.
00:26:09.980 --> 00:26:14.390
The difference is the
energies for this go
00:26:14.390 --> 00:26:16.040
as the quantum number squared.
00:26:16.040 --> 00:26:18.530
And the energies for this
go as the quantum number
00:26:18.530 --> 00:26:20.476
times the quantum number plus 1.
00:26:20.476 --> 00:26:21.350
It's almost the same.
00:26:24.250 --> 00:26:25.870
OK.
00:26:25.870 --> 00:26:31.930
So because the potential is
constant as long as r is fixed,
00:26:31.930 --> 00:26:35.800
the Hamiltonian is just theta--
00:26:42.540 --> 00:26:43.040
OK.
00:26:43.040 --> 00:26:47.520
And so we have to understand
this kinetic energy.
00:26:54.080 --> 00:26:57.960
For 1D problems,
the kinetic energy
00:26:57.960 --> 00:27:04.618
is P squared over 2 mu,
the linear momentum.
00:27:04.618 --> 00:27:08.260
I should put a hat on this.
00:27:08.260 --> 00:27:12.430
Well, we have motion
in three dimensions,
00:27:12.430 --> 00:27:14.150
or two dimensions,
theta and phi.
00:27:16.810 --> 00:27:21.860
And so we want to do something.
00:27:21.860 --> 00:27:24.880
We want to generate a kinetic
energy Hamiltonian, which
00:27:24.880 --> 00:27:28.480
is analogous to this, some
kind of momentum squared
00:27:28.480 --> 00:27:30.520
over some kind of a mass.
00:27:30.520 --> 00:27:32.050
And so we can go by analogy.
00:27:32.050 --> 00:27:34.310
Or we can go back to
classical mechanics.
00:27:34.310 --> 00:27:40.190
We know that the orbital
angular momentum is r cross P.
00:27:40.190 --> 00:27:42.340
So these are two vectors.
00:27:42.340 --> 00:27:44.320
Cross product is a vector.
00:27:44.320 --> 00:27:45.550
We know all that stuff.
00:27:45.550 --> 00:27:48.280
We know how to write
the cross product
00:27:48.280 --> 00:27:51.017
in terms of a matrix
involving unit vectors
00:27:51.017 --> 00:27:51.850
and stuff like that.
00:27:51.850 --> 00:27:55.390
You've done that before.
00:27:55.390 --> 00:27:56.580
OK.
00:27:56.580 --> 00:28:09.020
So now, we have an
angular motion, omega.
00:28:09.020 --> 00:28:17.210
And we would like to know what
the velocity of the mass points
00:28:17.210 --> 00:28:19.130
are on a sphere.
00:28:19.130 --> 00:28:26.750
Or if we think of this as just
a particle of mass mu rotating,
00:28:26.750 --> 00:28:28.550
then we want to
know its velocity.
00:28:28.550 --> 00:28:31.670
And to get from the
angular velocity
00:28:31.670 --> 00:28:36.800
to the linear velocity, the
linear velocity is r omega.
00:28:44.790 --> 00:28:49.810
So we can say, all
right, knowing--
00:28:49.810 --> 00:28:55.180
well, for a vector cross
product, if the two things--
00:28:55.180 --> 00:28:57.700
the r and the P--
00:28:57.700 --> 00:29:07.010
are always orthogonal, well,
then we just do r times P.
00:29:07.010 --> 00:29:11.970
And in fact, for motion
on this sphere, here's r.
00:29:11.970 --> 00:29:15.750
And the motion is
always orthogonal to r.
00:29:15.750 --> 00:29:24.900
And so we can write that the
angular momentum is m1 r1
00:29:24.900 --> 00:29:33.560
squared omega plus m2
r2 squared times omega.
00:29:33.560 --> 00:29:36.980
Or it's just I omega.
00:29:43.250 --> 00:29:44.840
OK.
00:29:44.840 --> 00:29:52.970
And so we can write the
kinetic energy term just
00:29:52.970 --> 00:29:58.840
goes as L squared
over 2I where I
00:29:58.840 --> 00:30:05.530
is the sum of the individual
masses M sub r I squared.
00:30:09.190 --> 00:30:10.490
OK.
00:30:10.490 --> 00:30:15.900
So we have an angular
momentum operator,
00:30:15.900 --> 00:30:18.560
which is the guts of
the kinetic energy.
00:30:22.340 --> 00:30:24.230
OK.
00:30:24.230 --> 00:30:28.060
Now, we're thinking Cartesian.
00:30:28.060 --> 00:30:30.680
And this is a spherical problem.
00:30:30.680 --> 00:30:33.070
And so we want to
go from Cartesian
00:30:33.070 --> 00:30:37.338
coordinates to spherical
polar coordinates.
00:30:37.338 --> 00:30:40.290
And that's a
non-trivial problem.
00:30:40.290 --> 00:30:43.620
And it's also an
extremely boring problem.
00:30:43.620 --> 00:30:45.300
And what you get
is also something
00:30:45.300 --> 00:30:48.600
that's not terribly rewarding,
except it's the differential
00:30:48.600 --> 00:30:50.170
equation you have solve.
00:30:50.170 --> 00:30:54.990
So T, when you go to
spherical polar coordinates,
00:30:54.990 --> 00:31:06.150
is minus h squared over
2I 1 over sine theta
00:31:06.150 --> 00:31:10.380
partial with respect
to theta times
00:31:10.380 --> 00:31:19.200
sine theta partial with respect
to theta plus 1 over sine
00:31:19.200 --> 00:31:25.990
squared theta second
partial with respect to phi.
00:31:25.990 --> 00:31:30.360
So that's the kinetic
energy operator.
00:31:30.360 --> 00:31:35.760
So first of all, you say,
what am I going to do?
00:31:35.760 --> 00:31:39.410
And the first thing you say
is, separate the variables.
00:31:39.410 --> 00:31:41.080
So you do that.
00:31:41.080 --> 00:31:43.960
And you do the standard trick
for separating variables.
00:31:43.960 --> 00:31:46.630
And you get two differential
equations-- the theta
00:31:46.630 --> 00:31:48.250
equation and the phi equation.
00:31:51.230 --> 00:31:51.950
I'm sorry?
00:31:51.950 --> 00:31:53.140
AUDIENCE: H bar!
00:31:53.140 --> 00:31:56.380
ROBERT FIELD: When I wrote it,
I said, why is that not H bar?
00:31:56.380 --> 00:31:58.790
OK.
00:31:58.790 --> 00:32:05.370
All right, so when you
separate variables,
00:32:05.370 --> 00:32:06.820
we have this result--
00:32:06.820 --> 00:32:10.450
1 over phi of phi.
00:32:10.450 --> 00:32:14.430
This is the phi part
of the equation.
00:32:22.185 --> 00:32:23.740
Whoops, yeah, that's right.
00:32:30.470 --> 00:32:33.330
m squared is the
separation constant.
00:32:33.330 --> 00:32:37.280
So we arrange to have a
differential equation that
00:32:37.280 --> 00:32:43.130
depends only on phi and another
that depends only on theta.
00:32:43.130 --> 00:32:45.360
And they're equal to each other.
00:32:45.360 --> 00:32:48.660
And so we call the thing that
they're equal to a constant.
00:32:48.660 --> 00:32:50.090
And we call it m squared.
00:32:50.090 --> 00:32:53.370
We like that because m can
be positive or negative.
00:32:53.370 --> 00:32:57.200
And the sign of m
determines whether you
00:32:57.200 --> 00:33:01.820
have an oscillating function
or an exponential function.
00:33:01.820 --> 00:33:05.490
You're all familiar with that.
00:33:05.490 --> 00:33:07.320
And so this is the phi equation.
00:33:09.880 --> 00:33:12.340
And this one, you
can solve easily.
00:33:12.340 --> 00:33:14.530
You know the solution to this.
00:33:14.530 --> 00:33:17.500
You could, with a
little bit of thought,
00:33:17.500 --> 00:33:23.450
write down the solution and
have normalized functions.
00:33:23.450 --> 00:33:24.400
So I'll just do that.
00:33:29.580 --> 00:33:36.290
So the phi part of
the solution is--
00:33:42.210 --> 00:33:46.870
it's not the order in
which I wrote my notes--
00:33:46.870 --> 00:33:52.050
1 over the square root of 2
pi times e to the i m phi.
00:33:52.050 --> 00:33:56.560
And m is equal to 0 plus
and minus 1 plus and minus
00:33:56.560 --> 00:33:58.450
2 et cetera.
00:33:58.450 --> 00:34:02.490
Quantization comes
from imposition
00:34:02.490 --> 00:34:06.410
of what we call periodic
boundary conditions.
00:34:06.410 --> 00:34:11.590
The wave function has to
be the same for phi and phi
00:34:11.590 --> 00:34:14.600
plus 2 pi and phi plus 4 pi.
00:34:14.600 --> 00:34:17.590
And so that gives quantization.
00:34:17.590 --> 00:34:18.820
So this is the phi part.
00:34:18.820 --> 00:34:19.530
That's simple.
00:34:19.530 --> 00:34:20.139
That's simple.
00:34:20.139 --> 00:34:20.764
It's wonderful.
00:34:20.764 --> 00:34:23.949
We understand that perfectly.
00:34:23.949 --> 00:34:26.110
And one of the nice things
about the vector model
00:34:26.110 --> 00:34:28.480
is mostly that's what
you're focused on.
00:34:37.060 --> 00:34:41.550
So immediately, we can
draw some pictures.
00:34:41.550 --> 00:34:47.610
And we can look at the
nodes in the xy plane.
00:34:51.080 --> 00:35:00.760
So m equals 0,
there are no nodes.
00:35:03.930 --> 00:35:12.910
m equals 1, first of all,
let's draw where phi is 0.
00:35:12.910 --> 00:35:14.200
So this is phi.
00:35:14.200 --> 00:35:19.969
And so there can be a
nodal plane like that.
00:35:19.969 --> 00:35:21.760
And there could be a
nodal plane like that.
00:35:29.450 --> 00:35:34.320
And already, you know this
looks like an S orbital.
00:35:34.320 --> 00:35:37.810
And this looks like a Px
orbital and a Py orbital.
00:35:40.830 --> 00:35:45.660
And then we could also have--
00:35:45.660 --> 00:35:56.150
yeah, m equals 2, we could
have this, and this, and that--
00:35:56.150 --> 00:35:59.550
no node, one node, two nodes.
00:35:59.550 --> 00:36:00.650
I'm sorry, yeah.
00:36:05.040 --> 00:36:06.630
What am I doing here?
00:36:06.630 --> 00:36:08.160
This is one node.
00:36:12.890 --> 00:36:16.790
I have to be a little
bit more careful here.
00:36:16.790 --> 00:36:20.687
So what I wrote down
is not true here.
00:36:20.687 --> 00:36:22.020
There are going to be two nodes.
00:36:22.020 --> 00:36:23.853
So they're probably
like this and like that.
00:36:27.000 --> 00:36:32.550
But it's clear that what I
have in my handwritten notes--
00:36:32.550 --> 00:36:34.640
I'm not sure what the
printed notes say.
00:36:34.640 --> 00:36:36.290
But there are going
to be two nodes.
00:36:36.290 --> 00:36:40.850
And so immediately, you
know something about--
00:36:40.850 --> 00:36:42.560
if you draw a reduced
picture and you
00:36:42.560 --> 00:36:47.990
show the nodal structure
in the xy plane,
00:36:47.990 --> 00:36:54.500
you can tell what the projection
of the angular momentum quantum
00:36:54.500 --> 00:36:58.049
number is, which is
really important.
00:37:11.530 --> 00:37:13.900
There's the other
differential equation.
00:37:13.900 --> 00:37:17.145
And that's the theta
differential equation.
00:37:20.460 --> 00:37:25.880
And that is complicated.
00:37:25.880 --> 00:37:30.170
It is an exactly solved
differential equation.
00:37:30.170 --> 00:37:32.160
But one of the
things it gives you,
00:37:32.160 --> 00:37:35.150
which is easily memorized--
remembered, I'm sorry--
00:37:35.150 --> 00:37:39.170
is that the energy levels have--
00:37:46.260 --> 00:37:49.780
where this is the
overlying momentum.
00:37:49.780 --> 00:37:51.370
So you can solve this.
00:37:51.370 --> 00:37:53.950
And you can get
the eigenenergies.
00:37:53.950 --> 00:37:58.690
And the eigenenergies
do not depend on m.
00:37:58.690 --> 00:38:01.519
They only depend on
the L quantum number.
00:38:01.519 --> 00:38:02.727
And they have this nice form.
00:38:06.000 --> 00:38:06.540
OK.
00:38:06.540 --> 00:38:09.690
Now, in my notes and
in all the textbooks,
00:38:09.690 --> 00:38:15.750
there are these horrendously
beautiful detailed expressions
00:38:15.750 --> 00:38:20.790
of the solution, the
mathematical form
00:38:20.790 --> 00:38:24.240
to the solution of
the theta equation.
00:38:24.240 --> 00:38:26.910
And it's the Legendre equation.
00:38:26.910 --> 00:38:29.070
And there's Legendre
polynomials.
00:38:29.070 --> 00:38:31.500
And remember, with the harmonic
oscillator, I told you,
00:38:31.500 --> 00:38:34.180
you don't ever want to
look at these things.
00:38:34.180 --> 00:38:35.560
The computer will look at them.
00:38:35.560 --> 00:38:37.185
And if you have
integrals, the computer
00:38:37.185 --> 00:38:38.790
will know how to
do those integrals.
00:38:38.790 --> 00:38:40.530
Because you told it how.
00:38:40.530 --> 00:38:42.630
And you don't have to
keep that in your head.
00:38:42.630 --> 00:38:46.870
You've got better things to do
with your limited attention.
00:38:46.870 --> 00:38:49.390
So there is a solution.
00:38:49.390 --> 00:38:51.810
And it has a form,
which I have decided
00:38:51.810 --> 00:38:55.050
that I won't talk about.
00:38:55.050 --> 00:38:59.280
Because you get much more
insight into real problems
00:38:59.280 --> 00:39:01.530
from the vector model.
00:39:01.530 --> 00:39:07.360
Now, the vector
model is really only
00:39:07.360 --> 00:39:12.150
marginally useful for
a diatomic molecule
00:39:12.150 --> 00:39:17.250
in an electronic state, which
you don't know about yet, which
00:39:17.250 --> 00:39:22.100
is called sigma sigma plus,
where there is no angular
00:39:22.100 --> 00:39:25.120
momentum associated
with the electron
00:39:25.120 --> 00:39:27.160
and when there is no
angular momentum associated
00:39:27.160 --> 00:39:30.760
with the nuclei.
00:39:30.760 --> 00:39:34.680
So this is just
practice for real life
00:39:34.680 --> 00:39:39.780
when you have to understand
other things about what is
00:39:39.780 --> 00:39:42.310
living in the molecular frame.
00:39:42.310 --> 00:39:48.450
But this is hard enough to
understand completely that it's
00:39:48.450 --> 00:39:50.340
a worthwhile investment.
00:39:50.340 --> 00:39:52.680
And you will be doing
most of the understanding
00:39:52.680 --> 00:39:55.710
of more complicated
problems on your own
00:39:55.710 --> 00:39:58.500
if you ever do anything
with diatomic molecules
00:39:58.500 --> 00:40:00.520
or polyatomic molecules.
00:40:00.520 --> 00:40:03.040
And so, OK.
00:40:08.300 --> 00:40:09.910
So what is the vector model?
00:40:16.970 --> 00:40:20.540
Well, the things we want to
know about the vector model
00:40:20.540 --> 00:40:23.870
is, how long is J?
00:40:23.870 --> 00:40:28.640
So if we have a state with
J and m quantum numbers,
00:40:28.640 --> 00:40:32.320
how long is the vector
associated with J?
00:40:32.320 --> 00:40:36.190
And what is the angle--
00:40:36.190 --> 00:40:42.910
what is the projection of
J on the laboratory z-axis?
00:40:42.910 --> 00:40:45.670
So we want to know
the length of J.
00:40:45.670 --> 00:40:55.200
And we want to know the
projection of J on z.
00:40:55.200 --> 00:40:56.280
OK.
00:40:56.280 --> 00:41:03.260
And we want to know the angle
between the z-axis and J.
00:41:03.260 --> 00:41:04.700
That's the beginning
of a picture.
00:41:07.830 --> 00:41:12.700
And again, we stick
with this idea that J--
00:41:12.700 --> 00:41:14.750
so here's the z-axis.
00:41:14.750 --> 00:41:19.850
And J precesses about z.
00:41:23.970 --> 00:41:27.860
And here is an angle
we want to know.
00:41:27.860 --> 00:41:31.950
This theta is a different
theta from the theta in--
00:41:31.950 --> 00:41:35.860
in fact, I should call it alpha.
00:41:35.860 --> 00:41:36.360
OK.
00:41:40.000 --> 00:41:42.370
So some things we know--
00:41:42.370 --> 00:41:49.570
we know that the eigenvalues
of the rigid rotor equation
00:41:49.570 --> 00:41:51.430
are this.
00:41:51.430 --> 00:42:00.010
And we know that Jz
operator operating on y Lm
00:42:00.010 --> 00:42:07.584
gives H bar m y Lm.
00:42:11.900 --> 00:42:21.590
And J squared operating on y Lm
gives H bar squared J J plus 1.
00:42:25.070 --> 00:42:27.180
So what do we do with
these two things?
00:42:27.180 --> 00:42:31.640
Well, one is to say, all
right, the length of J
00:42:31.640 --> 00:42:34.640
is going to be the square root--
00:42:34.640 --> 00:42:37.700
times y Lm.
00:42:37.700 --> 00:42:43.790
So the length of J is
going to be the square root
00:42:43.790 --> 00:42:47.390
of H bar squared J J plus 1.
00:42:51.980 --> 00:42:58.109
So that becomes H bar and
approximately J plus 1/2.
00:42:58.109 --> 00:42:59.650
Because the square
root of J J plus 1
00:42:59.650 --> 00:43:01.390
is almost exactly J plus 1/2.
00:43:06.810 --> 00:43:09.180
When you get really
high J, it's exactly.
00:43:09.180 --> 00:43:12.420
At low J, It's a little
bit less than exact.
00:43:12.420 --> 00:43:15.470
So we know the
length of J is this.
00:43:15.470 --> 00:43:19.670
And we know the length
of m is H bar m.
00:43:19.670 --> 00:43:21.790
The length of J z is H bar m.
00:43:34.930 --> 00:43:43.280
So now, we know that
this is H bar m.
00:43:43.280 --> 00:43:48.120
And this is H bar
times J plus 1/2.
00:43:51.750 --> 00:43:57.300
And so this is alpha.
00:43:57.300 --> 00:44:03.210
So we know how to calculate
the cosine of alpha.
00:44:03.210 --> 00:44:07.230
So the cosine of
alpha is equal to H
00:44:07.230 --> 00:44:20.150
bar m over H bar J
plus 1/2, or m over J--
00:44:20.150 --> 00:44:24.170
oh, right, plus
1/2, really small.
00:44:24.170 --> 00:44:27.320
So the angle, the
cosine of the angle
00:44:27.320 --> 00:44:35.180
that j makes with the
z-axis is m over J.
00:44:35.180 --> 00:44:37.820
And all of a sudden,
now, we have lengths
00:44:37.820 --> 00:44:40.250
of things and angles of things.
00:44:40.250 --> 00:44:41.810
And we can do all
sorts of stuff.
00:44:41.810 --> 00:44:46.730
Remember, what we care
about is, if this is J,
00:44:46.730 --> 00:44:50.210
the body axis is like that.
00:44:50.210 --> 00:44:52.190
So we now know
where the body axis
00:44:52.190 --> 00:44:55.460
is relative to the laboratory.
00:44:55.460 --> 00:45:00.320
So we have the body axis
is perpendicular to J.
00:45:00.320 --> 00:45:03.670
And so we can then calculate
where everything is.
00:45:12.560 --> 00:45:15.870
All right, so special cases--
00:45:15.870 --> 00:45:25.250
if m is equal to
plus or minus J, then
00:45:25.250 --> 00:45:27.590
what does that mean
for this picture?
00:45:27.590 --> 00:45:30.210
Well, we have the z-axis.
00:45:30.210 --> 00:45:37.150
And we have J. And J is almost
exactly along the z-axis.
00:45:37.150 --> 00:45:39.795
It would be exactly along if
there wasn't this plus 1/2.
00:45:47.100 --> 00:45:51.690
And so what that
means is the bond
00:45:51.690 --> 00:45:56.770
axis is almost exactly
rotating in the xy plane.
00:45:59.930 --> 00:46:03.590
So if we choose m
equals plus or minus J,
00:46:03.590 --> 00:46:07.420
the molecule is rotating
in the xy plane.
00:46:07.420 --> 00:46:10.160
That's kind of nice to know.
00:46:10.160 --> 00:46:15.010
The other easy
case is m equals 0.
00:46:15.010 --> 00:46:21.150
If m equals 0, J is
perpendicular to the z-axis.
00:46:21.150 --> 00:46:28.720
And so the body axis
is in the yz plane.
00:46:28.720 --> 00:46:31.700
Or if we're coming around
here, it's in the xz plane.
00:46:31.700 --> 00:46:36.330
So we have xz yz.
00:46:36.330 --> 00:46:39.010
And together, what
it says is the body
00:46:39.010 --> 00:46:44.690
axis is more along z
than along anything else.
00:46:44.690 --> 00:46:46.760
That's a lot of insight.
00:46:46.760 --> 00:46:51.310
And so we can go to our friendly
extreme cases, special cases.
00:46:51.310 --> 00:46:54.806
And we could say,
where's the body axis?
00:46:54.806 --> 00:46:56.180
In the laboratory,
which is where
00:46:56.180 --> 00:46:57.470
we're making the observations.
00:47:00.410 --> 00:47:03.580
That should make you happy.
00:47:03.580 --> 00:47:06.680
Because there's
no equations here.
00:47:06.680 --> 00:47:10.940
There's just a lot of pictures
that you can understand
00:47:10.940 --> 00:47:12.815
and develop your intuition.
00:47:18.040 --> 00:47:21.331
Now, I want a blackboard.
00:47:21.331 --> 00:47:21.830
OK.
00:47:21.830 --> 00:47:25.370
Suppose you have a
laboratory in which you
00:47:25.370 --> 00:47:26.570
can make a molecular beam.
00:47:30.932 --> 00:47:32.140
And that's easy enough to do.
00:47:32.140 --> 00:47:34.360
You squirt molecules
out of a pinhole.
00:47:34.360 --> 00:47:38.350
And you have some
kind of an aperture.
00:47:38.350 --> 00:47:44.770
And so you have a directed flow
of molecules in one direction.
00:47:44.770 --> 00:47:46.570
And you could also
say, all right,
00:47:46.570 --> 00:47:48.760
I'm going to do
something with my laser.
00:47:48.760 --> 00:47:50.950
And I'm going to make
the molecules, which
00:47:50.950 --> 00:47:58.390
are like a helicopter,
in other words,
00:47:58.390 --> 00:48:04.420
the molecular axis is
rotating in a circle
00:48:04.420 --> 00:48:08.620
about the propagation axis
of the beam, so a helicopter.
00:48:12.390 --> 00:48:14.620
And we can also do this.
00:48:14.620 --> 00:48:21.210
And so these are two ways
that you can prepare molecules
00:48:21.210 --> 00:48:21.780
in this beam.
00:48:24.630 --> 00:48:28.710
And you could say, oh, I've
got a whole bunch of molecules
00:48:28.710 --> 00:48:30.120
that are not in the beam.
00:48:30.120 --> 00:48:32.910
And these molecules
are going to collide
00:48:32.910 --> 00:48:39.015
with the molecules outside
the beam and get scattered.
00:48:39.015 --> 00:48:43.650
Well, which one is going
to be scattered more?
00:48:43.650 --> 00:48:47.480
The one that sweeps out a
big volume or the one that
00:48:47.480 --> 00:48:54.640
sweeps out something only
essentially one dimension?
00:48:54.640 --> 00:48:55.265
That's insight.
00:48:57.824 --> 00:48:59.240
There's lots of
other things, too.
00:48:59.240 --> 00:49:01.280
Now, let's do something
which is actually
00:49:01.280 --> 00:49:02.420
slightly exam relevant.
00:49:06.480 --> 00:49:08.670
Suppose we have a
diatomic molecule
00:49:08.670 --> 00:49:14.450
which is positive on one end
and negative on the other, OK?
00:49:14.450 --> 00:49:17.500
And we're going to
apply an electric field.
00:49:17.500 --> 00:49:20.640
And if the electric field
is in this direction
00:49:20.640 --> 00:49:23.890
and the molecule is doing this,
the molecule doesn't care.
00:49:27.380 --> 00:49:30.130
But if the electric field
is in that direction
00:49:30.130 --> 00:49:34.450
and the molecule is doing
this, the energy levels--
00:49:34.450 --> 00:49:38.142
the energy goes up and down
as the molecule rotates.
00:49:38.142 --> 00:49:39.850
Well, how is that?
00:49:39.850 --> 00:49:42.580
The molecule was
free to precess.
00:49:42.580 --> 00:49:45.860
And we knew the amplitude
along this axis.
00:49:45.860 --> 00:49:50.230
But if there's some angle
dependence, what we're doing
00:49:50.230 --> 00:49:55.870
is we're mixing
in a different J.
00:49:55.870 --> 00:50:00.750
And so that's telling you that,
if you apply an electric field,
00:50:00.750 --> 00:50:03.500
there will be a Stark effect.
00:50:03.500 --> 00:50:07.490
And it will affect the m
equals 0 levels differently
00:50:07.490 --> 00:50:10.440
from the m equals J levels.
00:50:10.440 --> 00:50:15.480
Now, I'm not going to tell you
which one is more affected.
00:50:15.480 --> 00:50:18.330
But it's clear that
one is hardly affected.
00:50:18.330 --> 00:50:21.780
And the other is
profoundly affected.
00:50:21.780 --> 00:50:24.870
And that gives rise to a
splitting of the energy
00:50:24.870 --> 00:50:27.790
levels in an
electric field, which
00:50:27.790 --> 00:50:31.540
depends on J at m in a way
that you would recognize.
00:50:35.410 --> 00:50:36.390
Time to quit.
00:50:36.390 --> 00:50:39.300
I did what I was
hoping I could do.
00:50:39.300 --> 00:50:42.720
Now, there is some stuff
at the end of the notes
00:50:42.720 --> 00:50:45.620
which is standard textbook
material about polar plots
00:50:45.620 --> 00:50:47.670
of the wave functions.
00:50:47.670 --> 00:50:49.620
These polar plots are useful.
00:50:49.620 --> 00:50:52.860
But most of the time, when
you first encounter them,
00:50:52.860 --> 00:50:56.910
you have no idea what they
mean, except that they
00:50:56.910 --> 00:51:00.109
express where the nodes are.
00:51:00.109 --> 00:51:01.650
Now, you want to go
back and you want
00:51:01.650 --> 00:51:05.280
to actually think about
what these polar plots mean
00:51:05.280 --> 00:51:10.500
and whether you can use
them as an auxiliary
00:51:10.500 --> 00:51:12.720
to this vector picture.
00:51:12.720 --> 00:51:13.220
OK.
00:51:13.220 --> 00:51:17.615
So I will then lecture,
and it's exam relevant,
00:51:17.615 --> 00:51:21.890
on commutation rules on Monday.
00:51:21.890 --> 00:51:27.280
And then have a good weekend.