WEBVTT

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PROFESSOR: Simultaneous
eigenstates.

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So let's begin with that.

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We decided that we could
pick 1 l and l squared,

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and they would commute.

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And we could try
to find functions

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that are eigenstates of both.

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So if we have functions that
are eigenstates of those,

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we'll try to expand in
terms of those functions.

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And all this operator
will become a number

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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.

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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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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.

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So we wish that l
z acting on psi l m

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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.

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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

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that this be equal because
of units and h squared.

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And then a number,
lambda psi l m.

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

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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.

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Now that intuition may
not be completely precise.

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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,

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and we can put psi l m here.

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And l squared, psi l
m from this equation.

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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.

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And then if your wave functions
are suitably normalized,

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this would be a 1.

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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.

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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.

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This is positive.

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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.

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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,

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there is some l
for which l times

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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.

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Are there questions in the
setting up of these equations?

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This is the conceptual part.

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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.

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

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everything led us to believe.

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That we should be able
to solve this much.

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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.

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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.

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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.

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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.

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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.

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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

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squared theta, minus m
squared, p l m equals 0.

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There we go.

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That's our
differential equation.

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It's a major, somewhat
complicated, differential

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equation.

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But it's a famous
one, because it

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comes from [?
Laplatians. ?] You know,

00:14:25.020 --> 00:14:28.680
people had to
study this equation

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to do anything with Laplatians,
and so many problems.

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So everything is
known about this.

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And the first
thing that is known

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is that theta really appears
as cosine theta everywhere.

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And that makes sense.

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You see, theta and cosine theta
is sort of the same thing,

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even though it
doesn't look like it.

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You need angles that
go from 0 to pi.

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And that's nice.

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But [? close ?] and
theta, in that interval

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goes from 1 to minus 1.

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So it's a good parameter.

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People use 0 to 180
degrees of latitude.

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But you could use from 1
to minus 1, the cosine.

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That would be perfectly good.

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So theta or cosine theta
is a different variable.

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And this equation is simpler
for cosine theta as a variable.

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So let me write that,
do that simplification.

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So I have it here.

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If x is closer in
theta, d d x is minus 1

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over sine theta, d d theta.

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Please check that.

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And you can also show
that sine theta, d d theta

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is equal to minus 1
minus x squared d d x.

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The claim is that this
differential equation just

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involves cosine theta.

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And this operator you see in the
first term of the differential

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equation, sine
theta, dd theta is

00:16:28.690 --> 00:16:32.350
this, where x is cosine theta.

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And then there is a
sine squared theta,

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but sine squared theta is 1
minus cosine squared theta.

00:16:38.540 --> 00:16:46.060
So this differential
equation becomes d d x--

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well, should I write
the whole thing?

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No.

00:16:55.960 --> 00:16:58.960
I'll write the
simplified version.

00:16:58.960 --> 00:17:02.688
It's not-- it's
only one slight--

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m of the x plus l
times l plus 1 minus m

00:17:11.790 --> 00:17:21.230
squared over 1 minus x
squared p l m of x equals 0.

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The only thing
that you may wonder

00:17:28.170 --> 00:17:33.270
is what happened to the 1
minus x squared that arises

00:17:33.270 --> 00:17:36.220
from this first term.

00:17:36.220 --> 00:17:40.110
Well, there's a 1
minus x squared here.

00:17:40.110 --> 00:17:42.150
And we divided by all of it.

00:17:42.150 --> 00:17:45.970
So it disappeared from the first
term, disappeared from here.

00:17:45.970 --> 00:17:52.030
But the m squared ended up
divided by 1 minus x squared.

00:17:52.030 --> 00:17:55.985
So this is our equation.

00:17:58.670 --> 00:18:04.120
And so far, our
solutions are psi l m's.

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Are going to be
some coefficients,

00:18:06.010 --> 00:18:15.370
m l m's, into the i m phi
p l m of cosine theta.

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Now I want to do a little
more before finishing today's

00:18:24.021 --> 00:18:24.520
lecture.

00:18:28.050 --> 00:18:31.380
So this equation is
somewhat complicated.

00:18:31.380 --> 00:18:36.080
So the way physicists
analyze it is

00:18:36.080 --> 00:18:40.310
by considering first the
case when m is equal to 0.

00:18:40.310 --> 00:18:47.930
And when m is equal to 0,
the differential equation--

00:18:47.930 --> 00:18:49.830
m equals 0 first.

00:18:53.060 --> 00:19:00.050
The differential equation
becomes d d x 1 minus x

00:19:00.050 --> 00:19:04.420
squared d p l 0.

00:19:04.420 --> 00:19:08.090
But p l 0, people write as p l.

00:19:08.090 --> 00:19:17.290
The x plus l times l
plus 1, p l equals 0.

00:19:20.800 --> 00:19:27.160
So this we solve by
a serious solution.

00:19:27.160 --> 00:19:32.620
So we write p l of x
equals some sort of a k--

00:19:32.620 --> 00:19:39.070
sum over k, a k, x k.

00:19:39.070 --> 00:19:40.890
And we substitute in there.

00:19:44.740 --> 00:19:52.950
Now if you substituted it
and pick the coefficient

00:19:52.950 --> 00:19:58.560
of x to the k, you get
a recursion relation,

00:19:58.560 --> 00:20:02.370
like we did for the case
of the harmonic oscillator.

00:20:02.370 --> 00:20:04.890
And this is a simple
recursion relation.

00:20:04.890 --> 00:20:09.930
It reads k plus 1-- this
is a two-line exercise--

00:20:09.930 --> 00:20:19.500
k plus 2, a k plus 2, plus
l times l plus 1, minus k

00:20:19.500 --> 00:20:23.360
times k plus 1, a k.

00:20:28.070 --> 00:20:37.340
So actually, this
recursive relation

00:20:37.340 --> 00:20:40.640
can be put as a
[? ratio ?] form.

00:20:40.640 --> 00:20:44.540
The [? ratio ?] form we're
accustomed, in which we

00:20:44.540 --> 00:20:48.500
divide a k plus 2 by a k.

00:20:51.260 --> 00:20:56.750
And that gives you
a k plus 2 over a k.

00:20:59.810 --> 00:21:03.555
I'm sorry, all this
coefficient must be equal to 0.

00:21:08.640 --> 00:21:15.360
And a k plus 2 over a k,
therefore is minus l times

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.

00:21:27.790 --> 00:21:29.430
OK, good.

00:21:29.430 --> 00:21:32.490
We're almost done.

00:21:32.490 --> 00:21:33.900
So what has happened?

00:21:33.900 --> 00:21:37.410
We had a general
equation for phi.

00:21:37.410 --> 00:21:39.840
The first equation,
one, we solved.

00:21:39.840 --> 00:21:43.500
The second became an [?
integrated ?] differential

00:21:43.500 --> 00:21:44.670
equation.

00:21:44.670 --> 00:21:46.470
We still don't know
how to solve it.

00:21:46.470 --> 00:21:49.410
M must be an integer so far.

00:21:49.410 --> 00:21:51.960
L we have no idea.

00:21:51.960 --> 00:21:57.250
Nevertheless we now solve this
for the case m equal to 0,

00:21:57.250 --> 00:22:00.210
and find this
recursive relation.

00:22:00.210 --> 00:22:02.520
And this same
story that happened

00:22:02.520 --> 00:22:05.580
for the harmonic
oscillator happens here.

00:22:05.580 --> 00:22:08.490
If this recursion
doesn't terminate,

00:22:08.490 --> 00:22:16.770
you get singular functions
that diverge at x equals 1

00:22:16.770 --> 00:22:19.196
or minus 1.

00:22:19.196 --> 00:22:24.180
And therefore this
must terminate.

00:22:24.180 --> 00:22:26.580
Must terminate.

00:22:26.580 --> 00:22:28.980
And if it terminates,
the only way

00:22:28.980 --> 00:22:31.020
to achieve termination
on this series

00:22:31.020 --> 00:22:36.880
is if l is an
integer equal to k.

00:22:36.880 --> 00:22:43.530
So you can choose some case--
you choose l equals to k.

00:22:43.530 --> 00:22:49.620
And then you get
that p l of x is

00:22:49.620 --> 00:22:54.750
of the form of an x
to the l coefficient.

00:22:54.750 --> 00:23:00.320
Because l is equal to k, and a
k is the last one that exists.

00:23:00.320 --> 00:23:05.880
And now a l plus 2, k plus
2 would be equal to 0.

00:23:05.880 --> 00:23:12.950
So you match this, the last
efficient is the value of l.

00:23:12.950 --> 00:23:16.440
And the polynomial
is an elf polynomial,

00:23:16.440 --> 00:23:20.190
up to some number at the end.

00:23:20.190 --> 00:23:23.410
and you got a quantization.

00:23:23.410 --> 00:23:28.130
L now can be any
positive integer or 0.

00:23:28.130 --> 00:23:33.060
So l can be 0, 1, 2, 3, 4.

00:23:33.060 --> 00:23:38.490
And it's the quantization of
the magnitude of the angular

00:23:38.490 --> 00:23:39.940
momentum.

00:23:39.940 --> 00:23:42.210
This is a little surprising.

00:23:42.210 --> 00:23:48.540
L squared is an operator
that reflects the magnitude

00:23:48.540 --> 00:23:50.810
of the angular momentum.

00:23:50.810 --> 00:23:54.000
And suddenly, it is quantized.

00:23:54.000 --> 00:23:57.270
The eigenvalues
of that operator,

00:23:57.270 --> 00:24:03.570
where l times l plus 1, that
I had in some blackboard

00:24:03.570 --> 00:24:04.580
must be quantized.

00:24:04.580 --> 00:24:08.430
So what you get here are
the Legendre polynomials.

00:24:08.430 --> 00:24:13.860
The p l's of x that satisfy
this differential equation are

00:24:13.860 --> 00:24:15.130
legendre polynomials.

00:24:18.400 --> 00:24:22.030
And next time, when we
return to this equation,

00:24:22.030 --> 00:24:26.740
we'll find that m
cannot exceed l.

00:24:26.740 --> 00:24:28.950
Otherwise you can't
solve this equation.

00:24:28.950 --> 00:24:33.010
So we'll find the complete
set of constraints

00:24:33.010 --> 00:24:35.940
on the eigenvalues
of the operator.