PROFESSOR: So
angular momentum, we
need to deal with
angular momentum,
and the inspiration
for it is classical.
We have L is r cross p.
Classically.
So let's try to just
use that information
and write the various operators.
And in fact, we're
lucky in this case.
The operators that we
would write inspired
by the classical definition
are good operators
and will do the job.
So what do we have?
Lx, if you remember
the cross-product rule,
that would be y Pz minus z Py.
Now you can think of
this thing as a cyclic,
like a circle when you have x
and Px, y and Py, and z and Pz.
Things are cyclically symmetric.
There's no real difference
between this core.
And so you can go
cyclically here.
So you say, let's go
cyclical on this index.
Ly is equal to the next
cyclic to y in that direction
is z Px minus x Pz.
And Lz is equal to
x Py minus y Px.
And these things, I'll think
of them as the operators.
Let's put hats to everything.
The first thing I can
wonder with a little bit
of trepidation is maybe
I got the ordering wrong.
Should I have written--
here classically,
you put r cross p,
and then the order of these
two terms doesn't matter.
Does it matter
quantum mechanically?
Happily, it doesn't matter
because y and Pz commute.
z and Py commute,
so you could even
have written them the other
way, and they are good.
All of them are ambiguous.
You could have even
written them the other way,
and they would be fine.
But now these are operators.
And moreover, they are
Hermitian operators.
Hermitian.
Let's see.
Lx dagger.
Well, the dagger
of two operators
you would do Pz
dagger y hat dagger--
recall the dagger
changes the order--
minus Py dagger z dagger.
Now, p and x's are all
for Hermitian operators,
so this is Pz y minus Py z.
And we use, again, that y and
Pz commute, and z and Py commute
to put it back in
the standard form.
And that's, again, Lx.
So it is an Hermitian operator.
And so is Ly and Lz.
That means these
operators are observables.
That's all you need for the
operator to be an observable.
And that's a very good thing.
So these operators
are observables.
Li's are observable.
But they're funny properties.
With these operators,
they're not
all that simple in some ways.
So next we have these operators.
Whenever you have
quantum operators,
the thing you do next is compute
their commutators Just like we
did with x and p, we wanted to
know what that commutator is.
We want to know what is the
commutator of this L operator.
So we'll do Lx with Ly.
Try to compute the commutators.
So Lx is y Pz.
Let me forget the hat,
so basically minus z Py.
And Ly is z Px minus x Pz.
Here is a y.
The y commutes with everything
here, so the y doesn't get.
The Pz gets stuck with the z
and doesn't care about this.
So this term just
talks to that term.
And here the z Py, the Py
doesn't care about anybody
here, but this z, well,
doesn't care about that z,
but it does care about this Pz.
So the only
contribution, there could
have been four terms out of
this commutator, but only two
are relevant.
So let's write them down.
y Pz with z Px and minus,
it's a plus z Py x Pz.
Well, you can start
peeling off things.
You can think of this as a
single operator with this too,
and it will fail to
commit with the first.
So you have y Pz z Px.
That's all this commutator gets.
And the same thing here.
This fails to commute just
with Pz, so the x can go out,
x z Py Pz.
And then here the y
actually can go out,
doesn't care about this
z, goes out on the left.
Not that it matters
much here, but that's
how using the commutator
identities does.
And this Py can go out and
let's go out on the right, z Pz.
And basing this
on this identity,
we just have A BC
commutator and then
AB C commutators, how
things distribute.
Now, this is minus i h
bar, and this is i h bar.
So here we get i h
bar x Py minus y Px.
See everything came out
in the right position.
And you recognize
that operator as Lz.
So this commutator
here has given you
Lx with Ly equal i h bar Lz.
It's a very interesting
and fascinating property
that somehow you're
doing this commutator,
it could have been
a mess, but it
combined to give you another
angular momentum operator.
Now, it looks like a
miracle, but physically,
it's not that miraculous.
It actually has to do with
the concept of symmetry.
Symmetry transformations.
If you have a symmetry
transformation
and you do commutators within
those symmetry operators,
you must get an operator that
corresponds to that symmetry,
or you must get a symmetry
at the very least.
So if we say that the potential
has very close symmetry,
that suggests that
when you do operations
with these operators
that generate rotation,
you should get
some rotation here.
And alternatively,
although, again,
this is suggestive, it
can be made very precise,
when you do rotations
in different order,
you don't get the
same thing at the end.
Everybody knows
if you have a page
and you do one rotation
and then the other
as opposed to the other
and then the first one,
you don't get the same thing.
Rotations do not commute.
A single rotation does
commute in one direction,
but rotations in different
directions don't commute.
That is the reason
for this equation.
And this equation, as we
said, everything is cyclic.
so you don't have to
work again to argue
that then Ly Lz, going cyclic,
must be equal to i h bar Lx.
And that Lz Lx
must be i h bar Ly.
And this is called the quantum
algebra of angular momentum.
In fact, it is so important
that this algebra appears
in all fields of
physics and mathematics,
and all kinds of things show up.
This algebra is related to
the algebra of generators
of the group SU2, Special
Unitary Transformations in Two
Dimensions.
It is related to
the orthogonal group
in three dimensions
where you rotate things
in three-dimensional space.
It is here, the algebra of
operators and in a sense,
it's a deeper result
than the derivation.
It is one of those cases when
you start with something very
concrete and you suddenly
discover a structure
that is rather universal.
Because we started with very
concrete representation of L's
in terms of y P's
and all these things.
But then they form a
consistent unit by themselves.
So sometimes there
will be operators
that satisfy these
relations, and they
don't come from x's and P's,
but still they satisfy that.
And that's what happens
with spin angular momentum.
The spin angular
momentum operators will
be denoted with Sx,
for example, and Sy
will have i h bar spin
in the z direction,
and the others will follow.
But nevertheless
nobody will ever
be able to write spin
as something like that
because it's not,
but spin exists.
And it's because
this structure is
more general than the situation
that allowed us to discover it.
It's a lot more general
and a lot more profound.
So in fact, mathematicians don't
even mention angular momentum.
They say, let's study.
The subject of Lie
algebra is the subject
of classifying all possible
consistent commutation
relations.
And this is the first
non-trivial example they have,
and they studied the
books on this algebra.
