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PROFESSOR: All right.
Welcome, everyone.
Hi.
So today we're going
to pick up where
we left off last time in our
study of angular momentum
and rotations in
quantum mechanics.
Before I get
started, let me open
up for questions, pragmatic
and physics related.
Yeah?
AUDIENCE: When we were solving
for the 3D harmonic oscillator
we solved for the energy
eigenfunction that
was a product of phi
x, phi y, and phi z.
We made an assumption
that phi e was
equal to phi sub n
of x plus that sum.
How did you get
from [INAUDIBLE]??
PROFESSOR: Good.
So what we had was that
we had that the energy--
we wanted to find the energy
of the 3D harmonic oscillator.
And we wanted to find the energy
eigenfunctions and eigenvalues.
And they way we did
this was by saying,
look, the energy of the 3D
harmonic oscillator, which
I can think of as a
function of x and px
and y and py and z and
pz, has this nice form.
We could write it as
the energy operator
purely in terms of x,
p squared x upon 2m
plus m omega squared
upon 2x squared.
Plus-- so this is a single
1D harmonic oscillator energy
operator in the x direction.
Plus E 1D in the y
direction, plus a harmonic
oscillator energy 1D
in the z direction.
So that was the
first observation.
And then we said that given that
this splits in this fashion,
I'm going to write my
energy eigenfunction, phi
of x, y, and z in separated
form as a product.
Phi x of x, phi y of
y, and phi z of z.
And we used this and deduced
that in order for this
to be an eigenfunction of
the 3D harmonic oscillator
it must be true that
5x of x was itself
an eigenfunction of the
x harmonic oscillator
equals with some energy
epsilon, which I will
call epsilon sub x, phi sub x.
And ditto for y and z.
And if this is true, if phi sub
x is an eigenfunction of the 1D
harmonic oscillator, then
this is an eigenfunction
of the 3D harmonic oscillator.
But we know what
the eigenfunctions
are of the harmonic oscillator.
The eigenfunctions
of the 1D harmonic
oscillator we gave a name.
We call them pi sub n.
So what we then said
was, look, if this
is an eigenfunction of
the 1D harmonic oscillator
in the x direction, then
it's labeled by an n.
And if this one is
in the y direction,
it's labeled by an l.
And if this one is in the z
direction it's labeled by a z.
But on top of that we know more.
We know what these
energy eigenvalues are.
The energy eigenvalues
corresponding to these guys
are if this is
phi n, this is En.
I can simply write this as En
for the 1D harmonic oscillator.
And taking this, using the fact
that they're 1D eigenfunctions
and plugging it into
the energy eigenvalue
equation for the 3D
harmonic oscillator
tells us that the energy
eigenvalues for the 3D case
are of the form E1d
x plus E1d in the y
plus E1d in the z, which is
equal to, since these were
the same frequency, h bar
omega times n plus l plus m,
from each of them
a 1/2, so plus 3/2.
Cool?
That answer your question?
Excellent.
Other questions?
Yeah?
AUDIENCE: Energy
and angular momentum
have to be related
somehow, right?
PROFESSOR: Yeah.
AUDIENCE: I mean, of course.
Because it's both.
PROFESSOR: Indeed.
AUDIENCE: [INAUDIBLE].
The thing is, we have a ladder.
Is there limits on them?
PROFESSOR: Very good question.
So this is where we're
going to pick up.
Let me rephrase this.
This is really two questions.
Question number one.
Look, there should be a
relationship between angular
momentum and energy.
But we're just talking
about angular momentum.
Why?
Second question, look,
we've got a ladder.
But is the ladder infinite?
So let me come back to
the second question.
That's going to be the
beginning of the lecture.
On the first question,
yes angular momentum
is going to play a role when
we calculate the energy.
But two quick things to note.
First off, consider
a system which
is spherically symmetric,
rotationally invariant.
That means that the energy
doesn't depend on a rotation.
If I rotate the system I
haven't changed the energy.
So if the system is
rotationally invariant,
that's going to imply some
constraints on the energy
eigenvalues and how they
depend on the angular momentum,
as we discussed last time.
Let me say that
slightly differently.
When we talk about the free
particle, 1D free particle--
we've talked about
this one to death.
Take the 1D free particle.
We can write the
energy eigenfunctions
as momentum eigenfunctions,
because the momentum commutes
with the energy.
And so the way the
eigenfunctions of the energy
operate are indeed e to
iKX there are plane waves,
they're eigenfunctions of the
momentum operator as well.
Similarly, when we
talk about a 3D system
it's going to be useful in
talking about the energy
eigenvalues to know a basis of
eigenfunctions of the angular
momentum operator.
Knowing the angular
momentum operator
is going to allow us to
write energy eigenfunctions
in a natural way
and a simply way,
in the same way that knowing
the momentum operator allowed
us to write energy
eigenfunctions
in a simple way in the 1D case.
That make sense?
We're going to have a glorified
version of the Fourier theorum
where instead of
something over e to iKX,
we're going to have something
over angular momentum
eigenstates.
And those are called
the spherical harmonics.
And they are the analog
of Fourier expansion
for this year.
But you're right.
We're going to have
to understand how
that interacts with the energy.
And that'll be the topic
of the next lecture.
We're going to finish up
angular momentum today.
Other questions?
OK, so from last time, these
are the commutation relations
which we partially
derived in lecture
and which you will be
driving on your problem set.
It's a really good exercise.
Commit these to memory.
They're your friends.
Key thing here to keep in mind.
h bar has units of angular
momentum, so this makes sense.
Angular momentum, angular
momentum, angular momentum.
So when you see an h
bar in this setting,
its job, in some sense, is to
make everything dimensionally
sensible.
So the important things here are
that lx and ly do not commute.
They commute to lz.
Can you have a state with
definite angular momentum
in the x direction and definite
angular momentum in the y
direction simultaneously?
No, because of this
commutation relation.
It would have to vanish.
This is, say, the x component
of the angular momentum.
Can you have a state with
definite angular momentum
in the x direction and total
angular momentum all squared?
Yes.
OK, great.
That's going to be
important for us.
So in order to construct
the eigenfunctions
it turns out to be
useful to construct
these so-called raising
and lowering operators,
which are Lx plus iLy.
They have a couple
of nice properties.
The first is, since these are
built up out of Lx and Ly,
both of which commute
with L squared,
the L plus minuses
commute with L squared.
So these guys commute.
So if we have an
eigenfunction of L squared,
acting with L plus does
not change its eigenvalue.
Similarly, L plus
commuting with Lz
gives us h bar L plus or minus,
with a plus or minus out front.
This is just like the raising
and lowering operators
for the harmonic oscillator.
But instead of the energy we
have the angular momentum.
So this is going to tell us
that the angular momentum
eigenvalues, the
eigenvalues of Lz,
are shifted by
plus or minus h bar
when we raise or
lower with L plus or L
minus, just like the
energy was shifted--
for the 1d harmonic oscillator
the energy was shifted,
plus h bar omega a
dagger, was shifted
by h bar omega when
we acted with a plus
on an energy eigenstate.
Same thing.
Questions on the commutators
before we get going?
In some sense, we're
going to just to just take
advantage of these
commutation relations
and explore their
consequences today.
So our goal is going to be
to build the eigenfunctions
and eigenvalues of the
angular momentum operators,
and in particular of the most
angular momentum operators
we [INAUDIBLE] complete set
of commuting observables,
L squared and Lz.
You might complain,
look, why Lz?
Whoops.
I don't mean a commutator.
I mean the set.
You might say, why Lz?
Why not Lx?
And if you call this
the z direction then
I will simply choose
a new basis where
this is called the x direction.
So it makes no
difference whatsoever.
It's just a name.
The reason we're
going to choose Lz
is because that coordinate
system plays nicely
with spherical coordinates,
just the conventional choice
of spherical coordinates where
theta equals 0 is the up axis.
But there's nothing
deep about that.
We could have
taken any of these.
OK.
So this is our goal.
So let's get started.
So first, because of these
commutation relations
and in particular
this one, we know
that we can find common
eigenfunctions of L squared
and Lz.
Let us call those common
eigenfunctions by a name,
Y sub lm such that--
so let these guys
be the common eigenfunctions
of L squared and Lz.
I.e., L squared Ylm is equal
to-- well first off, units.
This has units of
angular momentum squared,
h bar squared.
So that got rid of the units.
And we want our eigenvalue, lm.
And because I know
the answer, I'm
going to give--
instead of calling this
a random dimensionless number,
which would be the eigenvalue,
I'm going to call it a very
specific thing, l, l plus 1.
This is a slightly
grotesque thing to do,
but it will make the
algebra much easier.
So similarly, so that's
what the little l is.
Little l is labeling the
eigenvalue of L squared
and the actual value
of that eigenvalue.
I'm just calling h
bar squared ll plus 1.
That doesn't tell you
anything interesting.
This was already a
real positive number.
So this could have been any
real positive number as well,
by tuning L.
Similarly, Lz Ylm-- I want
this to be an eigenfunction--
this has units of
angular momentum.
So I'll put an h bar.
Now we have a dimensionless
coefficient Ylm.
And I'll simply call that m.
So if you will, these
are the definitions
of the symbols m and little l.
And Ylm are just the names I'm
giving to the angular momentum
eigenfunctions.
Cool?
So I haven't actually
done anything.
I've just told you that
these are the eigenfunctions.
What we want to know is what
properties do they have,
and what are the actual allowed
values of the eigenvalues.
So the two key things to note
are first that L plus and minus
leave the eigenvalue
of L squared alone.
So they leave l alone.
So this is the statement
that L squared on L plus Ylm
is equal to L plus
on L squared Ylm
is equal to h bar
squared ll plus 1,
the eigenvalue of L squared
acting on Ylm, L plus Ylm.
And so L plus Ylm is just as
much an eigenfunction of L
squared as Ylm was itself,
with the same eigenvalue.
Two.
So that came from--
where are we--- this
commutation relation.
OK, so similarly, L plus
minus raise or lower m by one.
And the way to see that is to
do exactly the same computation,
Lz on L plus, for example,
Ylm is equal to L plus--
now we can write
the commutator--
Lz, L plus, plus L plus Lz Ylm.
But the commutator of Lz
with L plus we already have,
is plus h bar L plus.
So this is equal
to h bar L plus.
And from this term, L plus
Lz of Ylm, Lz acting on Ylm
gives us h bar m, plus h bar
m L plus Ylm is equal to,
pulling this out, this is h bar
times m plus 1 times L plus.
h bar m plus 1 L plus Ylm.
So L plus has raised
the eigenvalue m by one.
This state, what we get by
acting on Ylm with the raising
operator is a thing
with m greater by one.
And that came from the
commutation relation.
And if we had done the
same thing with minus,
if you go through the minus
signs, it just gives us this.
What does that tell us?
What this tells us is we
get a ladder of states.
Let's look at what
they look like.
Each ladder, for a given value
of L, if you raise with L plus
and lower with L minus,
you don't change L
but you do change m.
So we get ladders that are
labeled by L. So for example,
if I have some value L1,
this is going to give me
some state labeled by m.
Let me put the m to the side.
I can raise it to get
m plus 1 by L plus.
And I can lower it to
get m minus 1 by L minus.
So I got a tower.
And if we have another value,
a different value of L,
I'll call it L2, we
got another tower.
You get m, m plus 1, m plus
2, dot, dot, dot, minus
1, dot, dot, dot.
So we have separated towers with
different values of L squared.
And within each tower we can
raise and lower by L plus,
skipping by one.
OK, questions?
So that was basically the
end of the last lecture,
said slightly differently.
Now here's the question.
So this is the question
that a student asked right
at the beginning.
Is this tower infinite?
Or does it end?
So I pose to you the question.
Is the tower infinite,
or does it end?
And why?
AUDIENCE: [INAUDIBLE] direction.
PROFESSOR: OK, so it's tempting
to say it's infinite in one
direction, because--?
AUDIENCE: There are no bounds
to the angular momentum.
PROFESSOR: OK, so
it's because there
are no bounds to the angular
momentum one can have.
That's tempting.
AUDIENCE: But at
the same time, when
you act a raising or
lowering operator on L,
the eigenvalue of L
squared remains the same.
So then you can't
raise the z [INAUDIBLE]
of the angular momentum
above the actual momentum.
PROFESSOR: Thank you.
Exactly.
So here's the statement.
Let me restate that.
That's exactly right.
So look, L squared is the
eigenvalue of the total angular
momentum.
Roughly speaking, it's giving
you precisely in the state Ylm
it tells you the expected value
of the total angular momentum,
L squared.
That's some number.
Now, if you act with L
plus you keep increasing
the expected value of Lz.
But if you keep increasing
it and keep increasing it
and keep increasing
it, Lz will eventually
get much larger than the
square root of L squared.
That probably isn't true.
That sounds wrong.
That was the statement.
Excellent.
Exactly right.
So let's make that precise.
So is the tower infinite?
No.
It's probably not, for
precisely that reason.
So let's make that precise.
So here's the way
we're going to do it.
This is a useful
trick in general.
This will outlive
angular momentum
and be a useful trick throughout
quantum mechanics for you.
I used it in a paper once.
So here's the nice observation.
Suppose it's true
that the tower ends.
Just like for the raising
and lowering operators
for the harmonic oscillator
in one dimension,
that tells us that in
order for the power
to end that state must
be 0 once we raise it.
The last state, so Yl, and I'll
call this m plus, must be 0.
There must be a max.
Oh, sorry.
Let me actually, before
I walk through exactly
this statement--
So let me make,
first, this, the no,
slightly more
obvious and precise.
So let's turn that argument
into a precise statement.
L squared is equal to,
just from the definition,
Lx squared plus Ly
squared plus Lz squared.
Now let's take the
expectation value
in this state Ylm of both
sides of this equation.
So on the left-hand side we
get h bar squared ll plus 1.
And on the right-hand side we
get the expectation value of Lx
squared plus the expectation
value of Ly squared
plus the expectation
value of Lz squared.
But we know that the
expectation value of Lz squared
is h bar squared, m squared.
But the expectation of
Lx squared and Ly squared
are strictly positive.
Because this can
be written as a sum
over all possible
eigenvalues of Lx squared,
which is the square of
the possible eigenvalues
of Lx times the
probability distribution.
That's a sum of positive,
strictly positive [INAUDIBLE]..
These are positive [INAUDIBLE].
So ll plus 1 is
equal to positive
plus positive plus h
bar squared m squared.
In particular that
tells you that it's
greater than or equal to
h bar squared m squared.
Maybe these are 0.
So the least it can be is--
so the most m
squared can possibly
be is the square
root of L plus 1.
Or the most m can be.
So m is bounded.
There must be a maximum m,
and there must be a minimum m.
Because this is squared.
The sign doesn't matter.
Everyone cool with that?
OK, so let's turn this into,
now, a precise argument.
What are the values
of m plus and m minus?
What is the top of the
value of each tower?
Probably it's going to
depend on total L, right?
So it's going to depend on each
value of L. Let's check that.
So here's the nice trick.
Suppose we really do
have a maximum m plus.
That means that if I try to
raise the state Ylm plus,
I should get the state
0, which ends the tower.
So suppose this is true.
In that case, in particular
here's the nice trick,
L plus Ylm, if we
take its magnitude,
if we take the magnitude of
this state, the state is 0.
It's the zero function.
So what's its magnitude?
Zero.
You might not think that's all
that impressive an observation.
OK, but note what this is.
We know how to work with this.
And in particular
this is equal to--
I should have done
this, dot, dot, dot--
I'm now going to use
the Hermitian adjoint
and pull this over to
the right-hand side.
The adjoint of L
plus is L minus.
So this gives us Ylm,
L minus, L plus Ylm.
This also doesn't look like
much of an improvement,
until you notice from the
definition of L plus and
L minus that what's
L minus L plus?
Well, L minus L plus, we're
going to get an Lx squared.
So Ylm, we get an Lx
squared plus an Ly squared.
I'm going to get my sign
right, if I'm not careful.
Plus-- well, let's just do it.
So we have L minus L plus.
So we're going to get a
Lx iLy minus i, iLyLx.
So plus i commutator
of Lx with Ly.
Good.
So that's progress, Ylm.
But it's still not progress,
because the natural operators
with which to act on the
Ylm's are L squared and Lz.
So can we put this in the
form L squared and Lz?
Sure.
This is L squared
minus Lz squared.
Ylm L squared minus Lz squared.
And this is i h bar Lz
times i is minus h bar Lz.
So minus h bar Lz, Ylm.
And this must be equal to 0.
But this is equal to--
from L squared we get h
bar squared, ll plus 1.
From the Lz squared we get
minus m squared h bar squared.
And from here we get
minus h bar, h bar m,
so minus m each bar squared.
Notice that the
units worked out.
And all of this was
multiplying Ylm, Ylm.
But Ylm, Ylm, if it's a properly
normalized eigenstate, which
is what we were assuming at
the beginning, is just 1.
So we get this times 1 is 0.
Aha.
And notice that in all
of this, this was m plus.
We were assuming,
we were working here
with the assumption that L plus
annihilated this top state.
And so grouping this together,
this says therefore h bar
squared times-- pulling out
a common h bar squared from
of all this--
times l, l plus 1 minus m plus,
m plus, plus 1 is equal to 0.
And this tells us that
m plus is equal to l.
And if you want to be
strict, put a plus sign.
Everyone cool with that?
This is a very useful trick.
If you know something
is 0 as function,
you know its norm is 0.
And now you can use things
like Hermitian adjoints.
Very, very useful.
OK questions about that?
Yeah.
AUDIENCE: [INAUDIBLE]
PROFESSOR: Excellent.
Where this came from
is I was just literally
taking L plus and L minus,
taking the definitions,
and plugging them in.
So if I have L minus
L plus, L minus
is, just to write
it out explicitly,
L minus is equal to Lx.
Because the [INAUDIBLE]
are observables,
and so they're Hermitians,
so they're self-adjoint,
the minus iLy.
So L minus L plus
gives me an LxLx.
It gives me an LxiLy.
It gives you a minus iLyLx.
And it gives me a minus ii,
which is plus 1 Ly squared.
Cool?
Excellent.
I'm in a state of
serious desperation here.
Good.
Other questions?
Yeah.
AUDIENCE: [INAUDIBLE] How do
you get from that to that?
PROFESSOR: This to here?
OK, good.
Sorry, I jumped a step here.
So here what I
said is, look, I've
got this nice set of
operators acting on Ylm.
I know how each of these
operators acts on Ylm.
Lz gives me an h bar m,
minus some h bar squared.
Lz squared, L squared.
And then overall that was a just
some number times Ylm, which
I can pull out of
the inner product.
Cool?
Yeah?
AUDIENCE: [INAUDIBLE]
PROFESSOR: Good.
This subscript plus
meant that, look,
there was a maximum value of m.
So m squared had to be
less than or equal to this.
There's a maximum value
and a minimum value.
Maybe they're different.
I don't know.
I'm just going to be
open-minded about that.
Others?
Yeah?
AUDIENCE: Will that mean,
then, that you can't ever
have all the angular
momentum [INAUDIBLE]??
PROFESSOR: Yeah,
awesome observation.
Exactly.
We were going to get
there in a little bit.
I like that.
That's exactly right.
Let me go through a couple
more steps, and then I'll
come back to that observation.
And I promise I will say so.
So if we go through
exactly a similar argument,
let's see what happens if
we did L minus on Ylm minus,
just to walk through the logic.
So if this were-- you lower
the lowest one, you get 0.
Then we'd get the same
story, L minus L minus.
When we take the adjoint
we'd get L plus, L minus.
And L plus L minus,
what changes?
The only thing that's
going to change
is that you get this
commutator the other direction.
And there are minus
signs in various places.
The upshot of which is that
m minus is equal to minus l.
So this is quite parsimonious.
It's symmetric.
If you take z to
minus z, if you switch
the sign of the angular momentum
you get the same thing back.
That's satisfying, perhaps.
But it's way more than that.
This tells us a lot about
the possible eigenvalues,
in the following way.
Look back at our towers.
In our towers we have that
the angular momentum is
raised and lowered by L plus.
So the Lz angular momentum is
raised and lowered by L plus.
l remains the same.
But there's a maximum
state now, which is plus l.
And there's a minimum
stay in here, l1.
And here there's a
minimum state, minus l1.
Similarly for the
other tower, there's
a minimum state minus l2
and a maximum state, l2.
So how big each tower is depends
on the total angular momentum.
That kind of makes sense, right?
If you've got more
angular momentum
and you can only step
Lz by one, you've
got more room to move
with a large value of l
than with a small value of l.
OK.
So what does that tell
us about the values of m?
Notice that m spans the
values from its minimal value,
minus l to l in integer
steps, in unit steps.
So if you think about m
is l, then m is l minus 1.
And if I keep lowering I get
down to m is minus l plus 1.
And then m is equal to minus l.
So the difference in the Lz
eigenvalue between these guys,
the difference is 2l.
But the number of
unit steps in here
is one fewer, because one,
dot, dot, dot 2l minus 1.
So this is some
integer, which is
the number of states minus 1.
So if these are n states, and
I'll call this N sub l states,
then this difference twice
l is N sub l minus 1,
because it's unit steps.
Cool?
So for example, if there
were two states, so N is 2,
2l is equal to--
well, that's 1,
which is 2 minus 1,
the number of states
minus 1, also known as 1.
So l is 1/2.
And more generally
we find that l
must be, in order
for this process
to make sense, in
order for the m plus
and the m minus to
match up, we need
that l is of the form an
integer N sub l minus 1 upon 2.
So this tells us that
where Nl is an integer--
Nl is just the number
of states in this tower,
and it's a strictly
positive integer.
If you have zero
states in the tower,
then that's not
very interesting.
So this tells us that l is
an integer a half integer,
but nothing else.
Cool?
In particular, if
it's a half integer,
that means 1/2, 3/2, 5/2, if
it's an integer or it can be 0.
There's nothing wrong
with it being 0.
But then 1 on.
So that means we can plot our
system in the following way.
If we have l equals 0, how
many states do we have?
If little l is 0,
how many states are?
What's the largest value of Lz?
What's the largest allowed
value of Lz or of m
if little l is equal to 0?
AUDIENCE: 0.
PROFESSOR: 0.
Because it goes from
plus 0 to minus 0.
So that's pretty much 0.
So we have a single state with
m equals 0 and l equals 0.
If l is equal to-- what's
the next possible value?--
1/2, then m can be
either 1/2 or we lower it
by 1, which is minus 1/2.
So there are two
states, m equals
1/2 and m equals minus 1/2.
So there's one power.
It's a very short tower.
This is the shortest
possible tower.
Then we also have the state l
equals 1, which has m equals 0.
It has m equals 1.
And it has m equals minus 1.
And that's it.
And so on and so forth.
Four states for l equals 3/2,
with states 3/2, 1/2, this is m
equals 1/2, m equals
minus 1/2, and minus 3/2.
The values of m span from minus
l to l with integer steps.
And this is possible for
all values of l which are
of the form all integer
or half integer l's.
So for every different
value of the total angular
momentum, the total
amount of angular momentum
you have, you have
a tower of states
labeled by Lz's eigenvalue
in this fashion.
Questions?
Does anyone notice anything
troubling about these?
Something physically
a little discomfiting
about any of these?
What does it mean to say
I'm in the state l equals 0,
m equals 0?
What is that telling you?
Zero angular momentum.
What's the expectation
value of l squared?
Zero.
The expectation value of Lz?
And by rotational invariance,
the expected value of Lx or Ly?
That thing is not rotating.
There's no angular momentum.
So the no angular momentum
state is one with l equals 0,
m equals 0.
So this is not spinning.
What about this guy?
L equals 1, m equals 0.
Does this thing carry
angular momentum?
Yeah, absolutely.
So it just doesn't carry
any angular momentum
in the z direction.
But its total angular
momentum on average,
its expected total
angular momentum
is ll plus 1, which is
2, times h bar squared.
So if you measure Lx and Ly
what do you expect to get?
Well, if you measure Lx
squared or Ly squared
you probably expect to
get something non-zero.
We'll come back to
that in just a second.
And what about the state
l equals 1, m equals 1?
Your angular
momentum-- you've got
as much angular momentum
in the z direction as you
possibly can.
So that definitely
carries angular momentum.
So there's a state
that has no angular
momentum in the z direction.
And there's a state
that has some,
and there's a state
that has less.
That make sense.
Yeah?
AUDIENCE: If m equals
1 and l equals 1,
that means that the angular
momentum for the z direction
is the total angular momentum?
PROFESSOR: Is it?
AUDIENCE: And then
Lx and Ly was 0?
PROFESSOR: OK, that's
an excellent question.
Let me answer that
now, and then I'll
come back to the point
I wanted to make.
Hold on a second.
Let me just answer
that question.
I'll work here.
So here's the crucial thing.
Even in this state, so you
were asking about the state l
equals 1, m equals 1.
And the question that was asked,
a very good question, is look,
does that mean all the angular
momentum is in the z direction,
and Lx and Ly are 0?
But let me just ask
this more broadly.
Suppose we have a state
with angular momentum l,
and m is equal to l.
The most angular
momentum you can possibly
have in the z direction,
same question.
Is all the angular momentum
in the z direction?
And what I want to
emphasize you is no,
that's absolutely not the case.
So two arguments for that.
The first is, suppose it's true
that Lx and Ly are identically
0.
Can that satisfy the
uncertainty relation
due to those commutators?
No.
There must be
uncertainty in Lx and Ly,
because Lz has a non-zero
expectation value.
So it can't be that Lx
squared and Ly squared
have zero expectation value.
But let's be more
precise about this.
The expectation value of L
squared is easy to calculate.
It's h bar squared l, plus 1.
Because we're in the state Ylm.
This is the state
Y l sub l, or ll.
The expectation of Lz squared
is equal to h bar squared,
m squared.
And m squared now,
m is equal to l.
So h bar squared, l squared.
Aha.
So the expected
value of l square
is not the same as
Lx squared, but this
is equal to the expected value
of Lx squared plus Ly squared
plus Lz squared.
Therefore the
expected value of Lx
squared plus the expected
value of Ly squared
is equal to the difference
between this and this.
We just subtract this off h bar
squared ll plus 1 minus h bar
squared, l squared,
h bar squared l.
And by symmetry you
don't expect the symmetry
to be broken between Lx and Ly.
You can actually
do the calculation
and not just be glib about it.
But both arguments give
you the correct answer.
The expectation
value of Lx squared
is equal to 1/2 h bar squared l.
And ditto for y, in this state.
So notice two things.
First off is we make the total
angular momentum little l
large.
The amount by which we fail
to have all the angular
momentum in the z direction
is getting larger and larger.
We're increasing the crappiness
of putting all the angular
momentum in the z direction.
However, as a ratio of the total
angular momentum divided by L
squared, so this
divided by L squared,
and this is h bar
squared, l, l plus 1,
and in particular this
is l squared plus l,
so if we took the ratio,
the rational mismatch
is getting smaller and smaller.
And that's good.
Because as we go to very
large angular momentums
where things should
start getting classical
in some sense, we should get
back the familiar intuition
that you can put all the angular
momentum in the z direction.
Yeah?
AUDIENCE: Why are we imposing
the [INAUDIBLE] condition
that the expectation values of
Lx and Ly should be identical?
PROFESSOR: Excellent.
That's why I was saying
this is a glib argument.
You don't have to impose--
that needs to be done a
little bit more delicately.
But we can just
directly compute this.
And you do so on
your problem set.
Yeah?
AUDIENCE: Why
doesn't the existence
of the l equals 0
eigenfunction where the angular
momentum the L squared
is definitely 0
violate the
uncertainty principle?
PROFESSOR: Awesome.
On your problem set you're
going to answer that,
but let me give you
a quick preview.
This is such a great response.
Just Let me give
you a quick preview.
So from that Lx--
OK, this is a
really fun question.
Let me go into it
in some detail.
Wow, I'm going fast today.
Am I going way too fast?
No?
A little?
Little too fast?
OK, ask me more questions
to slow me down.
I'm excited.
I didn't get much
sleep last night.
One of the great joys
of being a physicist
is working with
other physicists.
So yesterday one of
my very good friends
and a collaborator I really
delight in talking with
came to visit.
And we had a late night dinner.
And this led to me late at
night, not doing my work,
but reading papers about what
our conversation was about.
And only then at the very end,
when I was just about to die
did I write the response to
the reviewer on the paper
that I was supposed to be
doing by last night's deadline.
So I'm kind of tired, but
I'm in a really good mood.
It's a really good job.
Now I've totally lost
my train of thought.
What was the question again?
AUDIENCE: The existence
of the l equals
0 state where the total angular
momentum is definitely--
PROFESSOR: Excellent,
and the uncertainty.
So the question is, why don't
we violate the uncertainty when
we know that L
equals-- where am I;
I just covered it-- when we know
that L equals 0 and m equals 0,
doesn't that destroy
our uncertainty?
Because we know that the
angular momentum Lz is 0.
Angular momentum for Lx is 0.
Angular momentum for Ly is 0.
All of them vanish.
Doesn't that violate the
uncertainty principle?
So let's remind ourselves what
the form of that uncertainty
relation is.
The form of the uncertainty
relation following from Lx Ly
is i h bar Lz.
Recall the general statement.
The uncertainty in A times
the uncertainty in B, squared,
squared, is equal to--
let me just write
it as h bar upon 2--
sorry, 1/2.
The absolute value of
the expectation value
of the commutator,
A with B. Good lord.
Dimensionally, does this work?
Yes.
OK, good.
Because units of A, units of
B. Unites of A, units of B.
Triumph.
And these are going to be
quantum-mechanically small,
because commutators have h bars.
And commutators have h bars why?
Not because God hates us.
Why do commutators have h bars?
What happens classically?
In classical mechanics,
do things commute?
Yes.
Why are the h bars
and commutators
physical observables?
Because we exist.
Because there's a
classical limit.
OK so this is going to make
quantum-mechanically small,
so we expect the
uncertainty relation to also
be quantum-mechanically small.
Just important intuition.
So let's look at the specific
example of Lx, Ly, and Lz.
So the uncertainty in
Lx, in any particular--
remember that this is
defined as the uncertainty
in a particular state psi,
in a particular state psi.
And this expectation value is
taken in a particular state
psi, that same state.
So the uncertainty of
Lx, in some state Ylm,
times the uncertainty of
Ly in that same state Ylm
should be greater
than or equal to 1/2
the absolute value of
the expectation value
of the commutator of Lx and Ly.
But the commutator of
Lx and Ly is i h bar Lz.
And i, when we pull it
through this absolute value,
is going to give me just 1. h
bar is going to give me h bar.
So h bar upon 2, expectation
value of Lz, absolute value.
Yeah?
Can Lx and Ly have
zero uncertainty?
When?
Expectation value of Lz is 0.
So that sounds good.
It sounds like if Lz has
expectation value of 0,
then we can have Lx
and Ly, definite.
But that's bad.
Really?
Really, can we do that?
Why not?
AUDIENCE: [INAUDIBLE]
[LAUGHTER]
PROFESSOR: Bless you, my son.
Can we have Lx and Ly both
take definite values, just
because Lz?
Why?
What else do we have to satisfy?
What other uncertainty
relations must we satisfy?
There are two more.
And I invite you to go look
at what those two more are
and deduce that this is
only possible if Lx squared,
Ly squared, and Lz squared all
have zero expectation value.
In fact, I think it's just
a great question that I
think it's on your problem set.
So thank you for that question.
It's a really good question.
There was another
question in here.
Yeah?
AUDIENCE: Something that
you said earlier [INAUDIBLE]
half integer.
So were you deriving
this by counting
the number of equations
and somehow asserting--
Why does--?
PROFESSOR: Great.
So the question
is, wait, really?
Why is L and integer
a half integer.
That was a little too quick.
Is that roughly the
right statement?
AUDIENCE: [INAUDIBLE].
PROFESSOR: Good.
Excellent.
Let me go through the logic.
So the logic goes like this.
I know that the Ylm's--
if the Ylm's are eigenfunctions
of L squared and Lz,
and I've constructed this
tower of them using the raising
and lowering operators,
we've already
shown that the largest
possible value of m is l
and the least possible
value is minus l.
And these states
must be separated
by integer steps in m.
OK, good.
So pick a value of l,
a particular tower.
And let the number of states
in that tower be N sub l.
So there's N sub l of them.
Great.
And what's the distance
between these guys?
We haven't assumed
l is an integer yet.
We haven't assumed that.
So this N sub l is an integer.
Because it's the
number of states.
And the number of
states can't be a pi.
Now let's count that--
that so how many
states are there?
There are Nl.
But if I count one,
two, three, four,
the total angular
momentum down here is 2l.
I had some pithy way of
giving this a fancy name.
But I can't remember
what it was.
So the length of
the tower in units
of h bar, the height
of this tower is 2l.
But the number of steps I
took in here was Nl minus 1.
And that number of
steps is times 1.
So we get that 2l is N minus 1.
There's nothing fancy here.
I'm just saying if L is 0
we go from here to here.
There's just one element.
So number states is 1, L is 0.
So it's that same logic just
repeated for every value of L.
Other questions?
Coming back to this,
something on this board
should cause you some
serious physical discomfort.
We've talked about the l
equals 0 m equals 0 state.
This is a state which has no
angular momentum whatsoever,
in any direction at all.
We've talked about the l
equals 1 m equals 0 state.
It carries no angular
momentum in the z direction,
but it presumably has non-zero
expectation value for L
squared x and a Ly squared.
These guys are also fine.
What about these guys?
AUDIENCE: [INAUDIBLE].
PROFESSOR: Yes!
That's disconcerting.
Do I have to have angular
momentum in the z direction?
AUDIENCE: [INAUDIBLE]
[LAUGHTER]
PROFESSOR: That should
go on a shirt somewhere.
Let me ask the question
more precisely,
or in a way that's a little
less threatening to me.
Do you have to have angular
momentum in the z direction?
I'm sorry, what's your name?
Does David need to have angular
momentum in the z direction?
AUDIENCE: Yes.
PROFESSOR: Does this chalk
need to have angular dimension
in the--
well, the chalk's--
OK, classically no.
It can have some total
angular momentum, which is 0,
and it can be rotating
not in the z direction.
It can be rotating
in the zx plane.
If it's rotating
in the zx plane,
it's got total angular
momentum L squared as non-zero.
But its angular momentum
in the z direction is 0.
Its axis is exactly
along the y direction.
And so it's got no
angular moment-- that's
perfectly possible classically.
And that's perfectly possible
when L is an integer.
Similarly when L is 2.
This is the particular
tower that I love the most.
2, 1, 0, minus 1, minus 2.
The reason I love
this the most is
that it's related to gravity,
which is pretty awesome.
That's a whole other story.
I really shouldn't
have said that.
That's only going
to confuse you.
So for any integer, it's
possible to have no angular
momentum in the z direction.
That means it's
possible to rotate
around the x-axis or the y-axis,
orthogonal to the z-axis.
That make sense?
But for these half integer guys,
you are inescapably spinning.
There is no such thing as
a state of this guy that
carries no angular momentum.
Anything well described
by these quantum states
is perpetually
rotating or spinning.
It carries angular
momentum, we say precisely.
Perpetually carries angular
momentum in the z direction.
Any time you measure it, it
carries an angular momentum,
either plus a half integer
or minus a half integer.
But never, ever zero.
AUDIENCE: But in the
classical limit where
you have very large L, the
m equals one half state,
or the m equals
minus a half state is
going to get arbitrarily
small compared to the angular
momentum.
So isn't it just
like where we said,
well, OK, you can never
have your angular momentum
only in the z direction,
but we don't care?
Because in the classical
limit it gets arbitrarily
close to there.
PROFESSOR: See, one of the
nice things about writing
lectures like this is that you
get to leave little landlines.
So this is exactly one of those.
Thank you for asking
this question.
Let me rephrase that question.
Look, we all took
high school chemistry.
We all know about spin.
The nuclei have spin.
They have some angular momentum.
But if you build
up a lot of them,
you build up a piece
of chalk, look,
as we said before,
while it's true
that there's some
mismatch in the angular
momentum in the z direction
for some large L state,
it's not only angular momentum.
Some is in Lz, Lx,
and Ly as well.
It's preposterously small
for a macroscopic object
where L is macroscopic.
It's the angular
momentum in Planck units,
in units of the Planck constant.
10 to the 26th--
something huge.
Why would we even notice?
But here's the real statement.
The statement isn't
just that this
is true of macroscopic objects.
But imagine you take
a small particle.
Imagine you take a single atom.
We're deep in the quantum
mechanical regime.
We're not in the
classical regime.
We take that single atom.
And if it carries
angular momentum
and it's described by
L equals 1/2 state,
that atom will never, ever, ever
be measured to have its angular
momentum in the z direction,
or indeed any direction, be 0.
You will never,
in any direction,
measure its angular
momentum to be 0.
That atom perpetually
carries angular momentum.
And that is weird.
OK, maybe you don't
find it weird.
This is good.
You've grown up in an era
when that's not a weird thing.
But I find this
deeply disconcerting.
And you might say, look, we
never actually measure an atom.
But we do.
We do all the time.
Because we measure things
like the spectre of light,
as we'll study when we study
atoms in a week, a couple
weeks because of the exam--
sorry guys, there
has to be an exam--
as we will find
when we study atoms
in more detail,
or indeed, at all,
we'll be sensitive
to the angular
momentum of the
constituents of the atom.
And we'll see different spectra.
So it's an observable property
when you shine light on gases.
This is something we
can really observe.
So what this suggests
is one of two things.
Either these are just
crazy and ridiculous
and we should ignore
them, or there's
something interesting
and intrinsically
quantum-mechanical about
them that's not so familiar.
And the answer is
going to turn out
to be the second,
the latter of those.
The ladder?
That was not intentional.
Maybe it was subconscious.
OK, so I want to think
about some more consequences
of the structure of the
Ylm's and the eigenvalues,
in particular of
this tower structure.
I want to understand some more.
What other physics can we
extract from this story?
First, a very
useful thing is just
to get a picture of
these guys in your head.
Let's draw the angular
momentum eigenfunctions.
So what does that mean?
Well first, when we talked about
the eigenfunctions of momentum,
linear momentum
in one dimension,
we immediately went
to the wave function.
We talked about
how the amplitude
to beat a particular
spot varied in space.
And the amplitude was
just e to the iKX.
So the amplitude
was an oscillating--
the phase rotated.
And the absolute value of
the probability density
was completely constant.
Everybody cool with that?
That was the 1D plane wave.
So the variables there, we had
an angular momentum eigenstate.
And that's a function
of the position.
Or, sorry, a linear
momentum eigenstate.
And that's a function
of the position.
Angular momentum
eigenstates are going
to be functions of
angular position.
So I want to know what these
wave functions look like, not
just the eigenvalues.
But I want to know what is
the wave function associated
to eigenvalues
little l and little m
look like, y sub lm of
the angles theta and phi.
What do these guys look like?
How do we get them?
This is going to be your goal
for the next two minutes.
So the first thing
to notice is that we
know what the form of the
eigenvalues and eigenfunctions
are.
If we act with Lz
we get h bar m back.
If we act with L squared we get
h bar squared ll plus 1 back.
But we also have other
expressions for Lz and L
squared.
In particular-- I wrote
them down last time
in spherical coordinates.
So I'm going to working in the
spherical coordinates where
the declination from the
vertical is an angle theta,
and the angle around the
equator is an angle phi.
And theta equals 0 is going
to be up in the z direction.
It's just a choice
of coordinates.
There's nothing deep.
There's nothing even shallow.
It's just definitions.
So we're going to work in
the spherical coordinates.
And in spherical
coordinates we observe
that this angular
momentum, just following
the definition from
r cross p, takes
a particularly simple form.
That's a typo.
h bar upon i, dd phi.
And instead of
writing L squared I'm
going to write L plus
minus, because it's shorter
and also because it's going
to turn out to be more useful.
So this takes the form h
bar, e to the plus minus i
phi, d theta, plus or minus
cotangent of theta, d phi.
So these are the
expressions for Lz
and L plus minus in
spherical coordinates,
in these spherical coordinates.
I want to know how Ylm
depends on theta and phi.
And it's clearly
going to be easier
to ask about the
Lz eigenequation.
So let's look at that.
Lz on Ylm gives me h--
Oh yeah?
AUDIENCE: Is it h or h bar?
PROFESSOR: Oh, Jesus.
When I write letters
by hand, which
is basically when I write to
my mom, all my h's are crossed.
I can't help it.
So this is like the inverse.
It's been a long time
since I made that mistake.
It's usually the other.
So Lz acting on Ylm gives
me h bar m, acting on Ylm.
But that's what I get
when I act with Lz,
so let's just write out
the differential equation .
So h bar m Ylm where m is an
integer, or a half integer,
depending on whether l is
an integer or half integer.
h bar m Ylm is equal to h
bar upin i, d phi of Ylm.
Using the awesome power of
division and multiplication
I will divide both
sides by h bar.
And I will multiply
both sides by i.
And we now have the equation
for the eigenfunctions of Lz,
which we actually
worked on last time.
And we can solve
this very simply.
This says that, remember, Ylm
is a function of theta and phi.
Here we're only looking
at the phi dependence,
because that's all that
showed up in this equation.
So this tells us that the
eigenfunctions Ylm are
of the form of theta and phi,
are of the form e to the im
phi times some
remaining dependence
on theta, which I'll
write as p of theta.
And that p could
depend on l and m.
Cool?
Already, before we even
ask about that dependence
on theta, the p dependence, we
learn something pretty awesome.
Look at this wave function.
Whatever else we know,
its dependence on phi
is e to the im phi.
Now, remember what phi is.
Phi is the angle
around the equator.
So it goes from 0 to 2 pi.
And when it comes back to
2 pi it's the same point.
Phi equals 0 and
phi equals pi are
two names for the same point.
Yes?
But that should worry you.
Because note that as
a consequence of this,
Ylm of theta 0 is equal
to, well, whatever it is.
Sorry, theta of 2 pi is equal
to e to the im 2 pi times Plm
to theta.
But this is equal to--
oh, now we're in trouble.
If m is an integer, this is
equal to e to the i integer 2
pi 1.
Yes, OK.
You're supposed to
cheer at that point.
It's like the coolest
identity in the world.
So e to the i 2 pi, that's one.
So if m is an integer
then this is just
Plm of theta, which is also
what we get by putting in phi
equals 0.
So this is equal to
Ylm of theta, comma, 0.
But if m is a half integer
then e to the i half
integer times 2 pi is minus 1.
And so that gives us
minus Ylm of theta and 0,
if m is a half integer.
So let me say that again,
in the same words, actually.
But let me just say it again
with different emphasis.
What this tells us
is that Ylm at 0,
as function of the coordinates
theta and phi, Ylm--
so let's take m as an integer.
Ylm at theta and 0 is equal
to Ylm at theta and 0.
That's good.
But if Y is a half integer,
then Ylm at theta and 2 pi,
which is the same point
as Ylm at theta and 0,
is equal to minus
Ylm at theta and 0.
That's less good.
What must be true of
Ylm, of theta and 0?
0.
And was there anything
special about the point 0?
No.
I could have just
taken any point
and rotated it around by pi.
So this tells us that
Ylm of theta and phi
is identically equal to
0 if m is a half integer.
Huh.
That's bad.
Because what's the
probability density
of being found at any
particular angular position?
0.
Can you normalize that state?
No.
That is not a state that
describes a particle.
That is a state that describes
the absence of a particle.
That is not what we want.
So these states cannot describe
these values of l and m,
which seem like perfectly
reasonable values of l and m,
perfectly reasonable
eigenvalues of L squared and lm.
They cannot be used to label
wave functions of physical
states corresponding to
wave functions on a sphere.
You can't do it.
Because if you try, you find
that those wave functions
identically vanish.
OK?
So these cannot be used.
These do not describe wave
functions of a particle
in quantum mechanics.
They cannot be used.
Those values, those towers
cannot be used to describe
particles moving in
three dimensions.
Questions about that?
This is a slightly
subtle argument.
Yeah?
AUDIENCE: You said
something earlier about
how atoms would never have
any zero angular momentum.
And so the ones that have
no zero angular momentum,
we just said they're
not possible.
So [INAUDIBLE]?
PROFESSOR: Excellent question.
I said it slightly diff--
so the question is,
look, earlier you were
saying, yeah, yeah, yeah.
There are atoms in the world and
they have half integer angular
momentum.
And you can shine
a light on them
and you can tell and stuff.
But you just said
these can't exist.
So how can those two
things both be true?
Thank you for this question.
It's a very good question.
I actually said a
slightly different thing.
What I said was, states where
angular momentum lm are half
integers cannot be
described by a wave function
of the coordinates.
We're going to need some
different description.
And in particular, we're going
to need a different description
that does what?
Well, as we take phi from 0 to
2 pi, as we rotate the system,
we're going to pick
up a minus sign.
So in order to describe
an object with lm
being a half integer, we
can't use the wave function.
We need something that is
allowed to be doubly valued.
And in particular, we need
something that behaves nicely.
When you rotate
around by 2 pi, we
need to come back to a
minus sign, not itself.
So at some object
that's not a function,
it's called a spinner.
So we'll talk about it later.
We need some object
that does that.
So there's this
classic demonstration
at this point, which is
supposed to be done in a quantum
mechanics class.
So at this point
the lecturist is
obliged to do the
following thing.
You say, blah, blah, blah,
if things rotate by 2 pi
they have to come
back to themselves.
And then you do this.
I'm going to rotate my
hand like a record by 2 pi.
And it will not
come back to itself.
OK?
And it quite uncomfortably
has not come back to itself.
But I can do a further
rotation to show you
that it's a minus sign.
I can do a further
the rotation by 2 pi
and have it come back to itself.
And I kept the axis
vertical, yeah?
OK, so at this point you're all
supposed to go like, oh, yes,
uh-huh, mmm.
So now that we've got that
out of the way, I have an arm.
So the story is a little
more complicated than that.
This is actually a
fair demonstration,
but it's a slightly
subtle story.
If you want to understand it,
ask your recitation instructor
or come to my office hours.
AUDIENCE: What about USB sticks?
PROFESSOR: USB sticks?
AUDIENCE: You insert them here
and they don't go and insert
the other way.
[LAUGHTER]
PROFESSOR: What
about USB sticks?
You insert them this
way, they don't work.
You insert them this
way, they don't work.
But if you do it
again, then they do.
[LAUGHTER]
[APPLAUSE]
PROFESSOR: OK.
That's pretty good.
So for the moment, as long as
we want to describe our system
with a wave function
of position,
which means we're thinking
about where will we find it
as a function of angle, we
cannot use the half integer l
or m.
So if we can't use the
half interger l or m, fine.
We'll just throw them
out for the moment
and we'll use the
integer l and m.
And let's keep going.
What we need to
determine now is the P.
We've determined
the phi dependence,
but we need the
theta dependence.
We can get the theta
dependence in a sneaky fashion.
Remember the harmonic
oscillator in 1D.
When we wanted to find the
ground state wave function,
we could either solve the energy
eigenvalue equation, which
is a second order differential
equation and kind of horrible,
or we could solve the
ground state equation that
said that the ground state is
annihilated by the annihilation
operator, which is a first
order difference equation
and much easier to solve.
Yeah?
Let's do the same thing.
We have an
annihilation condition.
If we have the top state, L
plus on Y ll is equal to 0.
But L plus is a first order
differential operator.
This is going to be easier.
So we need to find a
solution to this equation.
And do I want to
go through this?
Yeah, why not.
OK, so I want this
to be equal to 0.
But L plus Yll is
equal to-- well,
it's h bar, e to the plus i phi,
d theta, plus cotangent theta,
d phi on Yll.
And Yll is e to the i l
phi times Plm of theta.
Cool?
So dd phi on e to the il phi,
there's no phi dependence here.
It's just going to
give us a vector of il.
Did I get the--
yeah.
So that's going to give us
a plus il and no dd phi.
And this e to the il
phi we can pull out.
But this has to be equal to 0.
So this says that 0 is equal to
d theta plus il cotangent theta
P lm of theta.
And this is actually
much better than it
seems for the following reason.
dd theta-- find the dd theta.
Cotangent of theta
is cosine over sine.
That's what you get if you
take the derivative of lots
of sine functions--
sine to the l, say.
Take a derivative.
You lose a power of sine and
you pick up a power of cosine.
So multiplying by cotangent
gets rid of a power of sign
and gives you a power of cosine,
which is a derivative of sine.
So Plm, noticing the l--
and I screwed up an i somewhere.
I think I wanted an i up here.
Let's see.
i.
Sorry.
Yes, I want an i cotangent.
OK, that's much better.
So i cotangent-- good, good.
And then the i squareds
give me a minus l.
And this tells us that Plm,
so if this is sine to the l,
then d theta gives us an l sine
to the l minus 1 cosine, which
is what I get by
taking sine to the l
and multiplying by
cosine, dividing by sine,
and multplying by l.
So this gives me Pll.
This is for the
particular state ll.
We're looking at the top state
and we're annhilating it.
So Pll is equal to
some coefficient,
so I'll just say proportional
to sine to the l of theta.
So this tells us that Yll is
equal to some normalization,
sub ll, just some number,
times from the phi dependence
e to the il phi, and from
the theta dependence,
sine to the l of theta.
So this is the form.
Sorry, go ahead.
AUDIENCE: What's
the symbol there?
PROFESSOR: Oh, this
twisted horrible thing?
It's proportional to.
But it was a long night.
So now we have the wave function
explicitly as a function phi
and as a function of theta
completely understood
for the top state in any tower.
This is for any L. The
top state in any tower
is e to the il phi.
Does that make sense?
Well Lz is h bar upon id phi.
So that gives us
h bar as m as l.
So that's good.
That's the top state.
And from the sine
theta we just checked.
We constructed that
this indeed has the--
well, if you then check you
will find that the L squared
eigenvalue, which you'll
do on the problem set,
the l squared eigenvalue
is h bar squared ll plus 1.
AUDIENCE: Quick question.
The expression we have for
the L plus minus operators,
how did we construct the
expressions for Lx and Ly?
PROFESSOR: Good.
It's much easier than you think.
So Lx is equal to--
L is r cross b, right?
So this is going to
be yPz minus zPy.
And this is equal to h
bar upon i, ydz minus zdy.
But you know what y is
in spherical coordinates.
And you know what
derivitive with respect to z
is in spherical coordinates.
Because you know what z is,
and you know the chain rule.
So taking this and just
plugging in explicit expression
for the change of variables
to spherical coordinates
takes care of it.
Yeah?
AUDIENCE: What does the
superscript of the sine
indicate?
Is that sine to the power of l?
PROFESSOR: Sorry.
This is bad notation.
It's not bad notation, it's
just not familiar notation.
It's notation that is used
throughout theoretical physics.
It means this, sine
theta to the l.
The lth power of sine.
For typesetting reasons
we often put the power
before the argument.
Yeah, no, it's a
very good question.
Thank you for asking,
because it was unclear.
I appreciate that.
Other questions?
AUDIENCE: [INAUDIBLE]
L hat [INAUDIBLE]??
PROFESSOR: How did we come
up with the L hat plus minus?
That was from this.
So we know what the components
of the angular momentum
are in Cartesian coordinates.
And you know how,
because it's coordinates,
to change variables from
Cartesian to spherical.
So you just plug this in for Lx.
But L plus is Lx plus
ioy, and so you just
take these guys
in spherical form
and add them together
with the relative i.
And that gives you
that expression.
AUDIENCE: Maybe I missed
this, but can you just
explain the distinction
between Y sub ll and Y sub lm?
PROFESSOR: Yeah, absolutely.
So Y sub ll, it means Y sub
lm where m is equal to l.
AUDIENCE: Oh, OK.
It was just the
generic [INAUDIBLE]..
PROFESSOR: It's a generic
eigenfunction of the angular
momentum, with the angular
momentum in the z direction
being equal to the angular
momentum in the total angular
momentum.
At least for these numbers.
OK?
Cool.
Good, so now, if we know this,
how do we just as a side note--
suppose we know-- well,
suppose we know this?
We know this.
We know what the top state
in the tower looks like.
How do I get the next
state down in the tower?
how do I get Yl l minus 1?
AUDIENCE: Lower it.
PROFESSOR: Lower it.
Exactly.
So this is easy, L minus on Yll.
And we have to be careful
about normalization.
So again, it's proportional to.
But this is easy.
We don't have to solve
any difference equations.
We just have to
take derivatives.
So it's just like
the raising operator
for a harmonic oscillator.
We can raise and
lower along the tower
and get the right wave function.
To give you some examples--
yeah, let's do that here.
Let me just quickly
give you a few examples
of the first few
spherical harmonics.
Sorry, I should give
these guys a name.
These functions, Ylm
of theta and phi,
they're called the
spherical harmonics.
They're called these because
they solve the Laplacian
equation on the sphere, which
is just the eigenvalue equation.
L squared on them is equal to
a constant times those things
back.
Just to tabulate a couple of
examples for you concretely,
consider the l equals 0 states.
What are the allowed values
of m for little l equals 0?
0.
So Y0,0 is the only state.
And if you properly normalize
it, it's 1 over root 4 pi.
OK, good.
what about l equals 1?
Then we have Y1, 0 and
we have Y1 minus 1.
And we have Y1,1.
So these guys take a
particularly simple form.
Root 3-- I'm not even going to
worry about the coefficient.
They're in the notes.
You can look them up anywhere.
So first off let's
look at Y1, 1.
So non-linear today.
So Y1, 1, it's going to
be some normalization.
And what is the form?
It's just e to the il phi
sine theta to the l. l is 1.
So this is some constant times
e to the i phi, sine theta.
Who 1, 0?
Well, it's got angular momentum
0 in the z direction, in Lz.
So that means how
does it depend on phi?
It doesn't.
And you can easily
see that, because when
we lower we get an e
to the i minus i phi.
Anyway, so this gives
us a constant times
no e to the i phi,
no phi dependents,
and cosine of theta.
And if you get a
cosine of theta,
the d theta and
the cotangent d phi
will give you the same thing.
And Y1 minus 1 is again
a constant, times e
to the minus i phi.
So it's got m equals minus
1 and sine theta again.
Notice a pleasing
parsimony here.
The theta dependence is the
same for plus m and minus m.
So what about Y2, 2?
Some constant e to the i 2 phi,
sine squared theta, dot, dot,
dot, Y0, 0.
And here it's interesting.
Here we just got one term
from taking the derivative.
They both give you cosine.
But now there are two ways to
act with the two derivatives.
And this gives you a constant.
Now what's the phi dependence?
It's nothing, because
it's got m equals 0.
And so the only
dependence is on theta.
And we get a cos squared
whoops, there's a 3--
3 cos squared theta minus 1.
AUDIENCE: Do you mean Y2, 0?
PROFESSOR: Oh, shoot.
Thank you.
Yes, I mean Y2, 0.
Thank you.
And then if we continue
lowering to Y2 minus 2,
this is equal to,
again, a constant.
And the it's going to be the
same dependence on theta,
but a different dependence of
phi. e to the minus 2i phi,
sine squared theta.
OK?
Yeah.
AUDIENCE: Aren't these
not normalizable?
PROFESSOR: Why?
AUDIENCE: Oh, never mind.
PROFESSOR: Good.
So let me turn that
into a question.
The question is, are
these normalizable?
Yeah, so how would
we normalize them?
What's the check for if
they're normalizable or not?
AUDIENCE: [INAUDIBLE]
PROFESSOR: Yeah,
we integrate them
norm squared over a sphere.
Not over a volume,
just over a sphere.
Because they're only wave
functions on the sphere.
We haven't dealt with
the radial function.
We'll deal with that later.
That will come in
the next lecture.
Other questions?
Yeah?
AUDIENCE: Can you explain
one more time why m equals
0 doesn't have [INAUDIBLE]?
PROFESSOR: Yeah, why
m equals 0 doesn't
have any phi dependence?
If m equals 0 had
phi dependence,
then we know that
the eigenvalue of Llz
is what we get when we take a
derivative with respect to phi.
But if the Lz
eigenvalue is 0, that
means that when we act
with dd phi we get 0.
That means it can't
depend on phi.
Cool?
Other questions?
What I'm going to do next
is I'm going show you,
walk you through
some of these angular
momentum eigenstates,
graphically on the computer.
Before we do that, any questions
about the calculation so far?
OK.
So this mathematical package
I'll post on the web page.
And at the moment I think
it's doing the real part.
So what we're looking
at now is the real part.
Actually, let's look
at the absolute value.
Good.
So here we are, looking
at the absolute value of--
that's not what I wanted to do.
So what we're looking
at in this notation
is some horrible
parametric plot.
You don't really need to see
the mathemat-- oh, shoot.
Sorry.
You don't need to see
the code, particularly.
So I'm not going to worry about
it, but it will be posted.
So here we're looking at
the absolute value, the norm
squared of the spherical
harmonic Y. And the lower
eigenvalue here, lower
coordinate is the l.
And the upper is m.
So when l is 0, what we get--
here's what this
plot is indicating.
The distance away from the
origin in a particular angular
direction is the absolute
value of the wave function.
So the further away
from the origin
the colored point you see is,
the larger the absolute value.
And the color here is just to
indicate depth and position.
It's not terribly meaningful.
So here we see that
we get a sphere.
So the probability
density or the norm
squared of the wave function
of the spherical harmonic
is constant.
So that makes sense.
It's spherically symmetric.
It has no angular momentum.
As we start increasing
the angular momentum,
let's take the angular
momentum l is 1, m is 0 state,
now something
interesting happens.
The total angular momentum is 1.
And we see that there
are two spheres.
Let me sort of rotate this.
So there are two spheres,
and there's the z-axis
passing through them.
And so the probability
is much larger
around this lobe on the top
or the lobe on the bottom.
And it's 0 on the plane.
Now, that's kind
of non-intuitive
if you think, well,
Lz is large, so why
is it along the vertical?
Why is that true?
So here, this is the
Lz equals 0 state.
That means it carries no angular
momentum along the z-axis.
That means it's not rotating
far out in the xy plane.
So your probability of
finding it in the xy plane
is very small.
Because if it was
rotating in the xy plane
it would carry a large
angular momentum in Lz.
But Lz is 0.
So it can't be extended
out in the xy plane.
Cool?
On the other hand, it can
carry angular momentum
in the x or the y direction.
But if it carries
angular momentum
in the x direction
for example, that
means the system is
rotating around the z--
sorry.
If it carries angular
momentum in the x direction,
it's rotating in the zy plane.
So there's some
probability to find it out
of the plane in y and z.
But it can't be in the xy plane.
Hence it's got to be
in the lobes up above.
That cool?
So it's very useful to develop
an intuition for this stuff
if you're going to do
chemistry or crystallography
or any condensed
manner of physics.
It's just very useful.
So I encourage you to play
with these little applets.
And I'll post this
mathematics package.
But let's looked at what happens
now if we crank up the angular
momentum.
So as we crank up the
l angular momentum,
now we're getting this
sort of lobe-y thing, which
looks like some sort of
'50s sci-fi apparatus.
So what's going on there?
This is the 2, 0 state.
And the 2, 0 state has a 3
cos squared theta minus 1.
But cos squared
theta, that means
it's got two periods as it
goes from vertical to negative.
And if you take that
and you square it,
you get exactly this.
OK?
So they're using the cos squared
as a function of the angle
of declination from vertical.
And it's m equals 0, so you're
saying no dependence on phi.
Of course, that's
a little cheap.
Because the angular
dependence on phi
is just an overall phase.
So we're not going to see
it in the absolute value.
Everyone agree on that?
We're not going to
see the absolute--
OK.
So let's check that.
Let's take l2 and m2.
So now when l is 2 and m is
2, we just get this donut.
So what that's saying is, we've
got some angular momentum.
l is 2.
But all the angular momentum,
almost all of it, anyway,
is in the z direction.
And is that what
we're seeing here?
Well, yeah.
It seems like it's most
likely to find the particle,
the probability is greatest, out
in this donut around the plane.
Now, if it were Lz is equal
to l, it would be flat.
It would be a strictly
0 thickness pancake.
But we have some uncertainty
in what Lx and Ly are,
which is why we got
this fattened donut.
Everyone cool with that?
And if we crank up l and
we make-- yeah, right?
So you can see that you've
got some complicated shapes.
But as we crank up
l and crank up m,
we just get a thinner
and thinner donut.
And the fact that the donut's
getting thinner and thinner
is that l over L squared
that we did earlier.
Cool?
It's still a donut,
but it's getting
relatively thinner
and thinner, by virtue
of getting wider and wider.
And a last thing to show you
is let's take a look now at--
in fact, let's go to
the l equals 0 state.
Let's take a look
at the real part.
So now we're looking
at the real part.
And nothing much
changed for the Y0, 0.
But for Y2, 0, well,
still not much changed.
For Y2, 0 let's now--
this is sort of disheartening.
Nothing really has changed.
Why?
Because it's real, exactly.
So for Y2, 0, as long as m
is equal to 0, this is real.
There's no phase.
The phase information
contains information
about the Lz eigenvalue.
So we can correct this by
changing the angular momentum.
Let's-- oh shoot,
how do I do that?
I can't turn it
off at the moment.
OK, whatever.
So here we have a
large M. And now
we've got this very
funny lobe-y structure.
So this is the Y2, 2, which a
minute ago looked rotationally
symmetric.
And now it's not
rotationally symmetric.
It's this lobe-y structure,
where the lobes are-- remember,
previously we had a donut.
Now we have these lobes when
we look at the real part.
How does that make sense?
Well, we've got an
e to the i 2 phi.
And if we look at the real
part, that's cosine of 2 pi.
And so we're getting a cosine
function modulating the donut.
And if we look at the real
part, let's do the same thing.
Let's look at the
imaginary part.
And the imaginary part
of Y2, 0, we get nothing.
That's good, because
Y2, 0 was real.
That would have been bad.
But if we look at the
imaginary part of Y2, 2,
we get the corresponding
lobes, the other lobes,
so that cos squared
plus sine squared is 1.
Play with these.
Develop some intuition.
They're going to be
very useful for us
when we talk about hydrogen
and the structure of solids.
And I will see you next week.
[APPLAUSE]
