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PROFESSOR: So
angular momentum, we

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need to deal with
angular momentum,

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and the inspiration
for it is classical.

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We have L is r cross p.

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Classically.

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So let's try to just
use that information

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and write the various operators.

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And in fact, we're
lucky in this case.

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The operators that we
would write inspired

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by the classical definition
are good operators

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and will do the job.

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So what do we have?

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Lx, if you remember
the cross-product rule,

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that would be y Pz minus z Py.

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Now you can think of
this thing as a cyclic,

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like a circle when you have x
and Px, y and Py, and z and Pz.

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Things are cyclically symmetric.

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There's no real difference
between this core.

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And so you can go
cyclically here.

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So you say, let's go
cyclical on this index.

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Ly is equal to the next
cyclic to y in that direction

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is z Px minus x Pz.

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And Lz is equal to
x Py minus y Px.

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And these things, I'll think
of them as the operators.

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Let's put hats to everything.

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The first thing I can
wonder with a little bit

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of trepidation is maybe
I got the ordering wrong.

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Should I have written--

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here classically,
you put r cross p,

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and then the order of these
two terms doesn't matter.

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Does it matter
quantum mechanically?

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Happily, it doesn't matter
because y and Pz commute.

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z and Py commute,
so you could even

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have written them the other
way, and they are good.

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All of them are ambiguous.

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You could have even
written them the other way,

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and they would be fine.

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But now these are operators.

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And moreover, they are
Hermitian operators.

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Hermitian.

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Let's see.

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Lx dagger.

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Well, the dagger
of two operators

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you would do Pz
dagger y hat dagger--

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recall the dagger
changes the order--

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minus Py dagger z dagger.

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Now, p and x's are all
for Hermitian operators,

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so this is Pz y minus Py z.

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And we use, again, that y and
Pz commute, and z and Py commute

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to put it back in
the standard form.

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And that's, again, Lx.

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So it is an Hermitian operator.

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And so is Ly and Lz.

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That means these
operators are observables.

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That's all you need for the
operator to be an observable.

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And that's a very good thing.

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So these operators
are observables.

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Li's are observable.

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But they're funny properties.

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With these operators,
they're not

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all that simple in some ways.

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So next we have these operators.

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Whenever you have
quantum operators,

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the thing you do next is compute
their commutators Just like we

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did with x and p, we wanted to
know what that commutator is.

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We want to know what is the
commutator of this L operator.

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So we'll do Lx with Ly.

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Try to compute the commutators.

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So Lx is y Pz.

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Let me forget the hat,
so basically minus z Py.

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And Ly is z Px minus x Pz.

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Here is a y.

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The y commutes with everything
here, so the y doesn't get.

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The Pz gets stuck with the z
and doesn't care about this.

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So this term just
talks to that term.

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And here the z Py, the Py
doesn't care about anybody

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here, but this z, well,
doesn't care about that z,

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but it does care about this Pz.

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So the only
contribution, there could

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have been four terms out of
this commutator, but only two

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are relevant.

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So let's write them down.

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y Pz with z Px and minus,
it's a plus z Py x Pz.

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Well, you can start
peeling off things.

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You can think of this as a
single operator with this too,

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and it will fail to
commit with the first.

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So you have y Pz z Px.

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That's all this commutator gets.

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And the same thing here.

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This fails to commute just
with Pz, so the x can go out,

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x z Py Pz.

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And then here the y
actually can go out,

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doesn't care about this
z, goes out on the left.

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Not that it matters
much here, but that's

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how using the commutator
identities does.

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And this Py can go out and
let's go out on the right, z Pz.

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And basing this
on this identity,

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we just have A BC
commutator and then

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AB C commutators, how
things distribute.

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Now, this is minus i h
bar, and this is i h bar.

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So here we get i h
bar x Py minus y Px.

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See everything came out
in the right position.

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And you recognize
that operator as Lz.

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So this commutator
here has given you

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Lx with Ly equal i h bar Lz.

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It's a very interesting
and fascinating property

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that somehow you're
doing this commutator,

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it could have been
a mess, but it

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combined to give you another
angular momentum operator.

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Now, it looks like a
miracle, but physically,

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it's not that miraculous.

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It actually has to do with
the concept of symmetry.

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Symmetry transformations.

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If you have a symmetry
transformation

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and you do commutators within
those symmetry operators,

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you must get an operator that
corresponds to that symmetry,

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or you must get a symmetry
at the very least.

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So if we say that the potential
has very close symmetry,

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that suggests that
when you do operations

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with these operators
that generate rotation,

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you should get
some rotation here.

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And alternatively,
although, again,

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this is suggestive, it
can be made very precise,

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when you do rotations
in different order,

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you don't get the
same thing at the end.

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Everybody knows
if you have a page

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and you do one rotation
and then the other

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as opposed to the other
and then the first one,

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you don't get the same thing.

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Rotations do not commute.

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A single rotation does
commute in one direction,

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but rotations in different
directions don't commute.

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That is the reason
for this equation.

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And this equation, as we
said, everything is cyclic.

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so you don't have to
work again to argue

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that then Ly Lz, going cyclic,
must be equal to i h bar Lx.

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And that Lz Lx
must be i h bar Ly.

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And this is called the quantum
algebra of angular momentum.

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In fact, it is so important
that this algebra appears

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in all fields of
physics and mathematics,

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and all kinds of things show up.

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This algebra is related to
the algebra of generators

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of the group SU2, Special
Unitary Transformations in Two

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Dimensions.

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It is related to
the orthogonal group

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in three dimensions
where you rotate things

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in three-dimensional space.

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It is here, the algebra of
operators and in a sense,

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it's a deeper result
than the derivation.

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It is one of those cases when
you start with something very

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concrete and you suddenly
discover a structure

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that is rather universal.

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Because we started with very
concrete representation of L's

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in terms of y P's
and all these things.

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But then they form a
consistent unit by themselves.

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So sometimes there
will be operators

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that satisfy these
relations, and they

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don't come from x's and P's,
but still they satisfy that.

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And that's what happens
with spin angular momentum.

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The spin angular
momentum operators will

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be denoted with Sx,
for example, and Sy

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will have i h bar spin
in the z direction,

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and the others will follow.

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But nevertheless
nobody will ever

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be able to write spin
as something like that

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because it's not,
but spin exists.

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And it's because
this structure is

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more general than the situation
that allowed us to discover it.

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It's a lot more general
and a lot more profound.

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So in fact, mathematicians don't
even mention angular momentum.

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They say, let's study.

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The subject of Lie
algebra is the subject

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of classifying all possible
consistent commutation

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relations.

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And this is the first
non-trivial example they have,

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and they studied the
books on this algebra.