1
00:00:00,500 --> 00:00:02,603
PROFESSOR: Simultaneous
eigenstates.

2
00:00:15,680 --> 00:00:18,710
So let's begin with that.

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00:00:27,160 --> 00:00:35,100
We decided that we could
pick 1 l and l squared,

4
00:00:35,100 --> 00:00:37,680
and they would commute.

5
00:00:37,680 --> 00:00:40,920
And we could try
to find functions

6
00:00:40,920 --> 00:00:44,850
that are eigenstates of both.

7
00:00:44,850 --> 00:00:48,720
So if we have functions that
are eigenstates of those,

8
00:00:48,720 --> 00:00:51,780
we'll try to expand in
terms of those functions.

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And all this operator
will become a number

10
00:00:54,990 --> 00:00:56,740
acting on those functions.

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And that's why the
Laplacian simplifies,

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and that's why we'll be able
to reduce the Schrodinger

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equation to a radial equation.

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This is the goal.

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Schrodinger equation
has r theta and phi.

16
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But theta and phi will deal
with all the angular dependents.

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We'll find functions for
which that operator gives

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a number acting on them.

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And therefore, the whole
differential equation

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00:01:26,250 --> 00:01:28,830
will simplify.

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So simultaneous eigenstates,
and given the simplicity of l z,

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everybody chooses l z.

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So we should find simultaneous
eigenstates of this two things.

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And let's call them psi
l m of theta and phi.

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Where l and m are numbers
that, at this moment,

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are totally arbitrary, but
are related to the eigenvalues

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of this equations.

28
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So we wish that l
z acting on psi l m

29
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is going to be a
number times psi l m.

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That is to be an eigenstate.

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The number must have the
right units, must be an H bar.

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And then we'll use m.

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We don't say what m is yet.

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M. Where m belongs
to the real numbers.

35
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Because the eigenvalues
of a Hermitian

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operator are always real.

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So this could be what we
would demand from l z.

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From l squared on
psi l m, I can demand

39
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that this be equal because
of units and h squared.

40
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And then a number,
lambda psi l m.

41
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Now this lambda-- do I know
anything about this lambda?

42
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Well, I could argue that this
lambda has to be positive.

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And the reason is
that this begins

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as some sort of positive
operator, is L. Squared.

45
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Now that intuition may
not be completely precise.

46
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But if you followed it a little
more with an inner product.

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Suppose we would have
an inner product,

48
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and we can put psi l m here.

49
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And l squared, psi l
m from this equation.

50
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This would be equal to h squared
lambda, psi l m, psi I m.

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An inner product if
you have it there.

52
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And then if your wave functions
are suitably normalized,

53
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this would be a 1.

54
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But this thing is l x--

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l x plus l y, l y plus l z, l z.

56
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And l x, l x--

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you could bring one l
x here, and you would

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have l x, psi l m, l x psi l m.

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Plus the same thing
for y and for z.

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And each of these
things is positive.

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Because when you have the
same wavefunction on the left

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and on the right, you
integrate the norm squared.

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It's positive.

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This is positive.

65
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This is positive.

66
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So the sum must be positive,
and lambda must be positive.

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So lambda must be positive.

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This is our expectation.

69
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And it's a reasonable
expectation.

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And that's why, in
fact, anticipating

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a little the answer, people
write this as l times

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l plus 1 psi l m.

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And where l is a real
number, at this moment.

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And you say, well,
that's a little strange.

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Why do you put it
as l times l plus 1.

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What's the reason?

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The reason is-- comes when
we look at the differential

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equation.

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But the reason you
don't get in trouble

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by doing this is that as you
span all the real numbers,

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the function l times
l plus 1 is like this.

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l times l plus 1.

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And therefore, whatever lambda
you have that is positive,

84
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there is some l
for which l times

85
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l plus 1 is a positive number.

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So there's nothing wrong.

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I'm trying to argue there's
nothing wrong with writing

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that the eigenvalue is of
the form l times l plus 1.

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Because we know the
eigenvalue's positive,

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and therefore, whatever lambda
you give me that is positive,

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I can always find, in fact, two
values of l, for which l times

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l plus 1 is equal to lambda.

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We can choose the positive one,
and that's what we will do.

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So these are the equations
we want to deal with.

95
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Are there questions in the
setting up of these equations?

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This is the conceptual part.

97
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Now begins a little bit of
play with the differential

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equations.

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And we'll have to do
a little bit of work.

100
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But this is what the physical
intuition-- the commutators,

101
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everything led us to believe.

102
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That we should be able
to solve this much.

103
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We should be able to find
functions that do all this.

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All right, let's
do the first one.

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So the first equation--

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The first equation is--

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let me call it equation 1 and 2.

108
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The first equation is
h bar over i d d 5.

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That's l z, psi l m,
equal h bar m psi l m.

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So canceling the h bars,
you'll get dd phi of psi l m

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is equal to i m, psi l m.

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So psi l m is equal
to e to the i m phi

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times some function of theta.

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Arbitrary function
of theta this moment.

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So this is my solution.

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This is up psi l m
of theta and phi.

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With the term in
the phi dependants,

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and it's not that complicated.

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So at this moment,
you say, well, I'm

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going to use this
for wavefunctions.

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I want them to behave normally.

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So if somebody gives
me a value of phi,

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I can tell them what
the wavefunction is.

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And since phi increases by 2
pi and is periodic with 2 pi,

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I may demand that psi l m
of theta, and 5 plus 2 pi

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be the same as psi
l m theta and phi.

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You could say, well, what if you
could put the minus sign there?

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Well, you could try.

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The attempt would
fail eventually.

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There's nothing, obviously,
wrong with trying

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to put the sine there.

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But it doesn't work.

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It would lead to rather
inconsistent things

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soon enough.

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So this condition here requires
that this function be periodic.

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And therefore when
phi changes by 2 pi,

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it should be a multiple of 2 pi.

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So m belong to the integers.

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So we found the
first quantization.

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The eigenvalues of
l z are quantized.

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They have to be integers.

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That was easy enough.

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Let's look at the
second equation.

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That takes a bit more work.

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So what is the second equation?

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Well, it is most slightly
complicated differential

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operator.

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And let's see what it does.

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So l squared.

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Well, we had it there.

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So it's minus h squared 1
over sine theta, dd theta,

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sine theta, dd theta, plus
1 over sine squared theta,

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d second d phi squared
psi l m equal h

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squared l times
l plus 1 psi l m.

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One thing we can do here is let
the dd phi squared act on this.

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Because we know
what dd phi does.

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Dd phi brings an i n factor,
because you know already

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the phi dependents of psi l m.

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So things we can do.

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So we'll do the second d
5 squared gives minus--

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gives you i m squared,
which is minus

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m squared, multiplying
the same function.

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You can cancel
the h bar squared.

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Cancel h bar squared.

165
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And multiply by minus
sine squared theta.

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To clean up things.

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So few things.

168
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So here is what we have.

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We have sine theta, dd theta.

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This is the minus sine squared
that you are multiplying.

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The h squared went away.

172
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Sine theta, d p l m d theta.

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Already I substituted
that psi was

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into the i m phi times the p.

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So I have that.

176
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And maybe I should put
the parentheses here

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to make it all look nicer.

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Then I have in here
two more terms.

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I'll bring the right-hand
side to the left.

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It will end up with
l l plus 1, sine

181
00:13:52,630 --> 00:14:00,115
squared theta, minus m
squared, p l m equals 0.

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00:14:08,490 --> 00:14:11,130
There we go.

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00:14:11,130 --> 00:14:14,480
That's our
differential equation.

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00:14:14,480 --> 00:14:18,720
It's a major, somewhat
complicated, differential

185
00:14:18,720 --> 00:14:20,968
equation.

186
00:14:20,968 --> 00:14:23,010
But it's a famous
one, because it

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00:14:23,010 --> 00:14:25,020
comes from [?
Laplatians. ?] You know,

188
00:14:25,020 --> 00:14:28,680
people had to
study this equation

189
00:14:28,680 --> 00:14:32,490
to do anything with Laplatians,
and so many problems.

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00:14:32,490 --> 00:14:34,930
So everything is
known about this.

191
00:14:34,930 --> 00:14:37,560
And the first
thing that is known

192
00:14:37,560 --> 00:14:46,890
is that theta really appears
as cosine theta everywhere.

193
00:14:46,890 --> 00:14:48,500
And that makes sense.

194
00:14:48,500 --> 00:14:52,580
You see, theta and cosine theta
is sort of the same thing,

195
00:14:52,580 --> 00:14:54,080
even though it
doesn't look like it.

196
00:14:54,080 --> 00:14:58,770
You need angles that
go from 0 to pi.

197
00:14:58,770 --> 00:14:59,690
And that's nice.

198
00:14:59,690 --> 00:15:02,210
But [? close ?] and
theta, in that interval

199
00:15:02,210 --> 00:15:06,630
goes from 1 to minus 1.

200
00:15:06,630 --> 00:15:08,390
So it's a good parameter.

201
00:15:08,390 --> 00:15:13,670
People use 0 to 180
degrees of latitude.

202
00:15:13,670 --> 00:15:18,220
But you could use from 1
to minus 1, the cosine.

203
00:15:18,220 --> 00:15:19,830
That would be perfectly good.

204
00:15:19,830 --> 00:15:24,410
So theta or cosine theta
is a different variable.

205
00:15:24,410 --> 00:15:30,200
And this equation is simpler
for cosine theta as a variable.

206
00:15:30,200 --> 00:15:33,710
So let me write that,
do that simplification.

207
00:15:38,560 --> 00:15:44,110
So I have it here.

208
00:15:44,110 --> 00:15:55,530
If x is closer in
theta, d d x is minus 1

209
00:15:55,530 --> 00:16:00,190
over sine theta, d d theta.

210
00:16:00,190 --> 00:16:03,040
Please check that.

211
00:16:03,040 --> 00:16:10,540
And you can also show
that sine theta, d d theta

212
00:16:10,540 --> 00:16:17,770
is equal to minus 1
minus x squared d d x.

213
00:16:17,770 --> 00:16:20,590
The claim is that this
differential equation just

214
00:16:20,590 --> 00:16:22,030
involves cosine theta.

215
00:16:22,030 --> 00:16:26,050
And this operator you see in the
first term of the differential

216
00:16:26,050 --> 00:16:28,690
equation, sine
theta, dd theta is

217
00:16:28,690 --> 00:16:32,350
this, where x is cosine theta.

218
00:16:32,350 --> 00:16:34,330
And then there is a
sine squared theta,

219
00:16:34,330 --> 00:16:38,540
but sine squared theta is 1
minus cosine squared theta.

220
00:16:38,540 --> 00:16:46,060
So this differential
equation becomes d d x--

221
00:16:46,060 --> 00:16:54,550
well, should I write
the whole thing?

222
00:16:54,550 --> 00:16:55,960
No.

223
00:16:55,960 --> 00:16:58,960
I'll write the
simplified version.

224
00:16:58,960 --> 00:17:02,688
It's not-- it's
only one slight--

225
00:17:02,688 --> 00:17:11,790
m of the x plus l
times l plus 1 minus m

226
00:17:11,790 --> 00:17:21,230
squared over 1 minus x
squared p l m of x equals 0.

227
00:17:26,349 --> 00:17:28,170
The only thing
that you may wonder

228
00:17:28,170 --> 00:17:33,270
is what happened to the 1
minus x squared that arises

229
00:17:33,270 --> 00:17:36,220
from this first term.

230
00:17:36,220 --> 00:17:40,110
Well, there's a 1
minus x squared here.

231
00:17:40,110 --> 00:17:42,150
And we divided by all of it.

232
00:17:42,150 --> 00:17:45,970
So it disappeared from the first
term, disappeared from here.

233
00:17:45,970 --> 00:17:52,030
But the m squared ended up
divided by 1 minus x squared.

234
00:17:52,030 --> 00:17:55,985
So this is our equation.

235
00:17:58,670 --> 00:18:04,120
And so far, our
solutions are psi l m's.

236
00:18:04,120 --> 00:18:06,010
Are going to be
some coefficients,

237
00:18:06,010 --> 00:18:15,370
m l m's, into the i m phi
p l m of cosine theta.

238
00:18:20,000 --> 00:18:24,021
Now I want to do a little
more before finishing today's

239
00:18:24,021 --> 00:18:24,520
lecture.

240
00:18:28,050 --> 00:18:31,380
So this equation is
somewhat complicated.

241
00:18:31,380 --> 00:18:36,080
So the way physicists
analyze it is

242
00:18:36,080 --> 00:18:40,310
by considering first the
case when m is equal to 0.

243
00:18:40,310 --> 00:18:47,930
And when m is equal to 0,
the differential equation--

244
00:18:47,930 --> 00:18:49,830
m equals 0 first.

245
00:18:53,060 --> 00:19:00,050
The differential equation
becomes d d x 1 minus x

246
00:19:00,050 --> 00:19:04,420
squared d p l 0.

247
00:19:04,420 --> 00:19:08,090
But p l 0, people write as p l.

248
00:19:08,090 --> 00:19:17,290
The x plus l times l
plus 1, p l equals 0.

249
00:19:20,800 --> 00:19:27,160
So this we solve by
a serious solution.

250
00:19:27,160 --> 00:19:32,620
So we write p l of x
equals some sort of a k--

251
00:19:32,620 --> 00:19:39,070
sum over k, a k, x k.

252
00:19:39,070 --> 00:19:40,890
And we substitute in there.

253
00:19:44,740 --> 00:19:52,950
Now if you substituted it
and pick the coefficient

254
00:19:52,950 --> 00:19:58,560
of x to the k, you get
a recursion relation,

255
00:19:58,560 --> 00:20:02,370
like we did for the case
of the harmonic oscillator.

256
00:20:02,370 --> 00:20:04,890
And this is a simple
recursion relation.

257
00:20:04,890 --> 00:20:09,930
It reads k plus 1-- this
is a two-line exercise--

258
00:20:09,930 --> 00:20:19,500
k plus 2, a k plus 2, plus
l times l plus 1, minus k

259
00:20:19,500 --> 00:20:23,360
times k plus 1, a k.

260
00:20:28,070 --> 00:20:37,340
So actually, this
recursive relation

261
00:20:37,340 --> 00:20:40,640
can be put as a
[? ratio ?] form.

262
00:20:40,640 --> 00:20:44,540
The [? ratio ?] form we're
accustomed, in which we

263
00:20:44,540 --> 00:20:48,500
divide a k plus 2 by a k.

264
00:20:51,260 --> 00:20:56,750
And that gives you
a k plus 2 over a k.

265
00:20:59,810 --> 00:21:03,555
I'm sorry, all this
coefficient must be equal to 0.

266
00:21:08,640 --> 00:21:15,360
And a k plus 2 over a k,
therefore is minus l times

267
00:21:15,360 --> 00:21:27,790
l plus 1 minus k times k plus
1 over k plus 1 times k plus 2.

268
00:21:27,790 --> 00:21:29,430
OK, good.

269
00:21:29,430 --> 00:21:32,490
We're almost done.

270
00:21:32,490 --> 00:21:33,900
So what has happened?

271
00:21:33,900 --> 00:21:37,410
We had a general
equation for phi.

272
00:21:37,410 --> 00:21:39,840
The first equation,
one, we solved.

273
00:21:39,840 --> 00:21:43,500
The second became an [?
integrated ?] differential

274
00:21:43,500 --> 00:21:44,670
equation.

275
00:21:44,670 --> 00:21:46,470
We still don't know
how to solve it.

276
00:21:46,470 --> 00:21:49,410
M must be an integer so far.

277
00:21:49,410 --> 00:21:51,960
L we have no idea.

278
00:21:51,960 --> 00:21:57,250
Nevertheless we now solve this
for the case m equal to 0,

279
00:21:57,250 --> 00:22:00,210
and find this
recursive relation.

280
00:22:00,210 --> 00:22:02,520
And this same
story that happened

281
00:22:02,520 --> 00:22:05,580
for the harmonic
oscillator happens here.

282
00:22:05,580 --> 00:22:08,490
If this recursion
doesn't terminate,

283
00:22:08,490 --> 00:22:16,770
you get singular functions
that diverge at x equals 1

284
00:22:16,770 --> 00:22:19,196
or minus 1.

285
00:22:19,196 --> 00:22:24,180
And therefore this
must terminate.

286
00:22:24,180 --> 00:22:26,580
Must terminate.

287
00:22:26,580 --> 00:22:28,980
And if it terminates,
the only way

288
00:22:28,980 --> 00:22:31,020
to achieve termination
on this series

289
00:22:31,020 --> 00:22:36,880
is if l is an
integer equal to k.

290
00:22:36,880 --> 00:22:43,530
So you can choose some case--
you choose l equals to k.

291
00:22:43,530 --> 00:22:49,620
And then you get
that p l of x is

292
00:22:49,620 --> 00:22:54,750
of the form of an x
to the l coefficient.

293
00:22:54,750 --> 00:23:00,320
Because l is equal to k, and a
k is the last one that exists.

294
00:23:00,320 --> 00:23:05,880
And now a l plus 2, k plus
2 would be equal to 0.

295
00:23:05,880 --> 00:23:12,950
So you match this, the last
efficient is the value of l.

296
00:23:12,950 --> 00:23:16,440
And the polynomial
is an elf polynomial,

297
00:23:16,440 --> 00:23:20,190
up to some number at the end.

298
00:23:20,190 --> 00:23:23,410
and you got a quantization.

299
00:23:23,410 --> 00:23:28,130
L now can be any
positive integer or 0.

300
00:23:28,130 --> 00:23:33,060
So l can be 0, 1, 2, 3, 4.

301
00:23:33,060 --> 00:23:38,490
And it's the quantization of
the magnitude of the angular

302
00:23:38,490 --> 00:23:39,940
momentum.

303
00:23:39,940 --> 00:23:42,210
This is a little surprising.

304
00:23:42,210 --> 00:23:48,540
L squared is an operator
that reflects the magnitude

305
00:23:48,540 --> 00:23:50,810
of the angular momentum.

306
00:23:50,810 --> 00:23:54,000
And suddenly, it is quantized.

307
00:23:54,000 --> 00:23:57,270
The eigenvalues
of that operator,

308
00:23:57,270 --> 00:24:03,570
where l times l plus 1, that
I had in some blackboard

309
00:24:03,570 --> 00:24:04,580
must be quantized.

310
00:24:04,580 --> 00:24:08,430
So what you get here are
the Legendre polynomials.

311
00:24:08,430 --> 00:24:13,860
The p l's of x that satisfy
this differential equation are

312
00:24:13,860 --> 00:24:15,130
legendre polynomials.

313
00:24:18,400 --> 00:24:22,030
And next time, when we
return to this equation,

314
00:24:22,030 --> 00:24:26,740
we'll find that m
cannot exceed l.

315
00:24:26,740 --> 00:24:28,950
Otherwise you can't
solve this equation.

316
00:24:28,950 --> 00:24:33,010
So we'll find the complete
set of constraints

317
00:24:33,010 --> 00:24:35,940
on the eigenvalues
of the operator.