PROFESSOR: Simultaneous
eigenstates.
So let's begin with that.
We decided that we could
pick 1 l and l squared,
and they would commute.
And we could try
to find functions
that are eigenstates of both.
So if we have functions that
are eigenstates of those,
we'll try to expand in
terms of those functions.
And all this operator
will become a number
acting on those functions.
And that's why the
Laplacian simplifies,
and that's why we'll be able
to reduce the Schrodinger
equation to a radial equation.
This is the goal.
Schrodinger equation
has r theta and phi.
But theta and phi will deal
with all the angular dependents.
We'll find functions for
which that operator gives
a number acting on them.
And therefore, the whole
differential equation
will simplify.
So simultaneous eigenstates,
and given the simplicity of l z,
everybody chooses l z.
So we should find simultaneous
eigenstates of this two things.
And let's call them psi
l m of theta and phi.
Where l and m are numbers
that, at this moment,
are totally arbitrary, but
are related to the eigenvalues
of this equations.
So we wish that l
z acting on psi l m
is going to be a
number times psi l m.
That is to be an eigenstate.
The number must have the
right units, must be an H bar.
And then we'll use m.
We don't say what m is yet.
M. Where m belongs
to the real numbers.
Because the eigenvalues
of a Hermitian
operator are always real.
So this could be what we
would demand from l z.
From l squared on
psi l m, I can demand
that this be equal because
of units and h squared.
And then a number,
lambda psi l m.
Now this lambda-- do I know
anything about this lambda?
Well, I could argue that this
lambda has to be positive.
And the reason is
that this begins
as some sort of positive
operator, is L. Squared.
Now that intuition may
not be completely precise.
But if you followed it a little
more with an inner product.
Suppose we would have
an inner product,
and we can put psi l m here.
And l squared, psi l
m from this equation.
This would be equal to h squared
lambda, psi l m, psi I m.
An inner product if
you have it there.
And then if your wave functions
are suitably normalized,
this would be a 1.
But this thing is l x--
l x plus l y, l y plus l z, l z.
And l x, l x--
you could bring one l
x here, and you would
have l x, psi l m, l x psi l m.
Plus the same thing
for y and for z.
And each of these
things is positive.
Because when you have the
same wavefunction on the left
and on the right, you
integrate the norm squared.
It's positive.
This is positive.
This is positive.
So the sum must be positive,
and lambda must be positive.
So lambda must be positive.
This is our expectation.
And it's a reasonable
expectation.
And that's why, in
fact, anticipating
a little the answer, people
write this as l times
l plus 1 psi l m.
And where l is a real
number, at this moment.
And you say, well,
that's a little strange.
Why do you put it
as l times l plus 1.
What's the reason?
The reason is-- comes when
we look at the differential
equation.
But the reason you
don't get in trouble
by doing this is that as you
span all the real numbers,
the function l times
l plus 1 is like this.
l times l plus 1.
And therefore, whatever lambda
you have that is positive,
there is some l
for which l times
l plus 1 is a positive number.
So there's nothing wrong.
I'm trying to argue there's
nothing wrong with writing
that the eigenvalue is of
the form l times l plus 1.
Because we know the
eigenvalue's positive,
and therefore, whatever lambda
you give me that is positive,
I can always find, in fact, two
values of l, for which l times
l plus 1 is equal to lambda.
We can choose the positive one,
and that's what we will do.
So these are the equations
we want to deal with.
Are there questions in the
setting up of these equations?
This is the conceptual part.
Now begins a little bit of
play with the differential
equations.
And we'll have to do
a little bit of work.
But this is what the physical
intuition-- the commutators,
everything led us to believe.
That we should be able
to solve this much.
We should be able to find
functions that do all this.
All right, let's
do the first one.
So the first equation--
The first equation is--
let me call it equation 1 and 2.
The first equation is
h bar over i d d 5.
That's l z, psi l m,
equal h bar m psi l m.
So canceling the h bars,
you'll get dd phi of psi l m
is equal to i m, psi l m.
So psi l m is equal
to e to the i m phi
times some function of theta.
Arbitrary function
of theta this moment.
So this is my solution.
This is up psi l m
of theta and phi.
With the term in
the phi dependants,
and it's not that complicated.
So at this moment,
you say, well, I'm
going to use this
for wavefunctions.
I want them to behave normally.
So if somebody gives
me a value of phi,
I can tell them what
the wavefunction is.
And since phi increases by 2
pi and is periodic with 2 pi,
I may demand that psi l m
of theta, and 5 plus 2 pi
be the same as psi
l m theta and phi.
You could say, well, what if you
could put the minus sign there?
Well, you could try.
The attempt would
fail eventually.
There's nothing, obviously,
wrong with trying
to put the sine there.
But it doesn't work.
It would lead to rather
inconsistent things
soon enough.
So this condition here requires
that this function be periodic.
And therefore when
phi changes by 2 pi,
it should be a multiple of 2 pi.
So m belong to the integers.
So we found the
first quantization.
The eigenvalues of
l z are quantized.
They have to be integers.
That was easy enough.
Let's look at the
second equation.
That takes a bit more work.
So what is the second equation?
Well, it is most slightly
complicated differential
operator.
And let's see what it does.
So l squared.
Well, we had it there.
So it's minus h squared 1
over sine theta, dd theta,
sine theta, dd theta, plus
1 over sine squared theta,
d second d phi squared
psi l m equal h
squared l times
l plus 1 psi l m.
One thing we can do here is let
the dd phi squared act on this.
Because we know
what dd phi does.
Dd phi brings an i n factor,
because you know already
the phi dependents of psi l m.
So things we can do.
So we'll do the second d
5 squared gives minus--
gives you i m squared,
which is minus
m squared, multiplying
the same function.
You can cancel
the h bar squared.
Cancel h bar squared.
And multiply by minus
sine squared theta.
To clean up things.
So few things.
So here is what we have.
We have sine theta, dd theta.
This is the minus sine squared
that you are multiplying.
The h squared went away.
Sine theta, d p l m d theta.
Already I substituted
that psi was
into the i m phi times the p.
So I have that.
And maybe I should put
the parentheses here
to make it all look nicer.
Then I have in here
two more terms.
I'll bring the right-hand
side to the left.
It will end up with
l l plus 1, sine
squared theta, minus m
squared, p l m equals 0.
There we go.
That's our
differential equation.
It's a major, somewhat
complicated, differential
equation.
But it's a famous
one, because it
comes from [?
Laplatians. ?] You know,
people had to
study this equation
to do anything with Laplatians,
and so many problems.
So everything is
known about this.
And the first
thing that is known
is that theta really appears
as cosine theta everywhere.
And that makes sense.
You see, theta and cosine theta
is sort of the same thing,
even though it
doesn't look like it.
You need angles that
go from 0 to pi.
And that's nice.
But [? close ?] and
theta, in that interval
goes from 1 to minus 1.
So it's a good parameter.
People use 0 to 180
degrees of latitude.
But you could use from 1
to minus 1, the cosine.
That would be perfectly good.
So theta or cosine theta
is a different variable.
And this equation is simpler
for cosine theta as a variable.
So let me write that,
do that simplification.
So I have it here.
If x is closer in
theta, d d x is minus 1
over sine theta, d d theta.
Please check that.
And you can also show
that sine theta, d d theta
is equal to minus 1
minus x squared d d x.
The claim is that this
differential equation just
involves cosine theta.
And this operator you see in the
first term of the differential
equation, sine
theta, dd theta is
this, where x is cosine theta.
And then there is a
sine squared theta,
but sine squared theta is 1
minus cosine squared theta.
So this differential
equation becomes d d x--
well, should I write
the whole thing?
No.
I'll write the
simplified version.
It's not-- it's
only one slight--
m of the x plus l
times l plus 1 minus m
squared over 1 minus x
squared p l m of x equals 0.
The only thing
that you may wonder
is what happened to the 1
minus x squared that arises
from this first term.
Well, there's a 1
minus x squared here.
And we divided by all of it.
So it disappeared from the first
term, disappeared from here.
But the m squared ended up
divided by 1 minus x squared.
So this is our equation.
And so far, our
solutions are psi l m's.
Are going to be
some coefficients,
m l m's, into the i m phi
p l m of cosine theta.
Now I want to do a little
more before finishing today's
lecture.
So this equation is
somewhat complicated.
So the way physicists
analyze it is
by considering first the
case when m is equal to 0.
And when m is equal to 0,
the differential equation--
m equals 0 first.
The differential equation
becomes d d x 1 minus x
squared d p l 0.
But p l 0, people write as p l.
The x plus l times l
plus 1, p l equals 0.
So this we solve by
a serious solution.
So we write p l of x
equals some sort of a k--
sum over k, a k, x k.
And we substitute in there.
Now if you substituted it
and pick the coefficient
of x to the k, you get
a recursion relation,
like we did for the case
of the harmonic oscillator.
And this is a simple
recursion relation.
It reads k plus 1-- this
is a two-line exercise--
k plus 2, a k plus 2, plus
l times l plus 1, minus k
times k plus 1, a k.
So actually, this
recursive relation
can be put as a
[? ratio ?] form.
The [? ratio ?] form we're
accustomed, in which we
divide a k plus 2 by a k.
And that gives you
a k plus 2 over a k.
I'm sorry, all this
coefficient must be equal to 0.
And a k plus 2 over a k,
therefore is minus l times
l plus 1 minus k times k plus
1 over k plus 1 times k plus 2.
OK, good.
We're almost done.
So what has happened?
We had a general
equation for phi.
The first equation,
one, we solved.
The second became an [?
integrated ?] differential
equation.
We still don't know
how to solve it.
M must be an integer so far.
L we have no idea.
Nevertheless we now solve this
for the case m equal to 0,
and find this
recursive relation.
And this same
story that happened
for the harmonic
oscillator happens here.
If this recursion
doesn't terminate,
you get singular functions
that diverge at x equals 1
or minus 1.
And therefore this
must terminate.
Must terminate.
And if it terminates,
the only way
to achieve termination
on this series
is if l is an
integer equal to k.
So you can choose some case--
you choose l equals to k.
And then you get
that p l of x is
of the form of an x
to the l coefficient.
Because l is equal to k, and a
k is the last one that exists.
And now a l plus 2, k plus
2 would be equal to 0.
So you match this, the last
efficient is the value of l.
And the polynomial
is an elf polynomial,
up to some number at the end.
and you got a quantization.
L now can be any
positive integer or 0.
So l can be 0, 1, 2, 3, 4.
And it's the quantization of
the magnitude of the angular
momentum.
This is a little surprising.
L squared is an operator
that reflects the magnitude
of the angular momentum.
And suddenly, it is quantized.
The eigenvalues
of that operator,
where l times l plus 1, that
I had in some blackboard
must be quantized.
So what you get here are
the Legendre polynomials.
The p l's of x that satisfy
this differential equation are
legendre polynomials.
And next time, when we
return to this equation,
we'll find that m
cannot exceed l.
Otherwise you can't
solve this equation.
So we'll find the complete
set of constraints
on the eigenvalues
of the operator.
