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BARTON ZWIEBACH: We want to
understand now our observables.

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So we said these
are observables,

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so can we observe them?

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Can we have a state in which we
say, what is the value of Lx,

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the value of Ly,
and the value of Lz.

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Well, a little
caution is necessary

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because we have states and
we have position and momentum

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operator and they didn't
commute and we ended up

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that we could not
tell simultaneously

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the position and the
momentum of a state.

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So for this angular momentum
operators, they don't commute,

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so a similar situation
may be happening.

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So I want to explain,
for example, or ask,

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can we have simultaneous
eigenstates of Lx, Ly, and Lz?

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And the answer is no.

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And let's see why that happens.

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So let's assume we can have
simultaneous eigenstates

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and let's assume, for example,
that Lx on that eigenstate phi

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nought is some number
lambda x phi nought,

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and Ly and phi nought is
equal to lambda y phi nought.

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Well, the difficulty with
this is essentially--

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well, we could even say
that Lz on phi nought

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is equal to lambda z phi nought.

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So what is the complication?

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The complication are
those commutators.

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If you do Lx, Ly and
phi nought, you're

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supposed to get i h-bar
Lz and phi nought.

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And therefore, you're
supposed to get i h-bar lambda

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z times phi nought, because it's
supposed to be an eigenstate.

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But how about the
left hand side?

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The left hand side is LxLy
and phi nought minus LyLx

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and phi nought.

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When Ly acts, it produces a
lambda y, but then phi nought,

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and then when Lx acts,
it produces a lambda x,

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so this produces lambda x lambda
y phi nought minus lambda y

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lambda x phi nought,
which is the same thing,

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so the left hand side is 0.

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0 is equal to lambda z
phi nought, so you get a--

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lambda z must be 0.

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If you have a non-trivial
state, lambda z should be 0.

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By the other commutators--

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this can be attained or
applied to phi nought--

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would be 0 again, because
each term produces a number

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and the order doesn't matter.

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But then it would show
that lambda x is 0,

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and this will show
that lambda y is 0.

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So at the end of the day,
if these three things hold,

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then all of them are 0.

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Lambda x equals lambda y
equals lambda z equals 0.

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So you can have something
that is killed by all

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of the operators, but you cannot
have a non-trivial state with

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non-trivial eigenvalues
of these things.

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So we cannot have--

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we cannot tell what is Lx on
this state and Ly on this state

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simultaneously.

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Any of those two is too much.

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So if we can't tell
that, what can we tell?

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So what is the most we
can tell about this state

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Is our question now.

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We can tell maybe what
is its value of Lx,

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but then Ly and Lz
are undetermined.

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Or we can tell
what is Lz and then

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Lx and Ly are undetermined,
incalculable, impossible

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in principle to calculate them.

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So let's see what we
can do, and the answer

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comes from a rather
surprising thing, the fact

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that if you think about what
could commute with Lx, Ly, Lz,

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it should be a rotationally
invariant thing, because Lx,

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Ly, and Lz do rotations.

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So the only thing
that could possibly

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commute with this
thing is something

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that is rotationally invariant.

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The thing that could work out
is some thing that is invariant

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and there are rotations.

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Now we said, for example,
the magnitude of the vector R

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is invariant under rotation.

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You rotate the vector,
the demanded is invariant.

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So we can try the
operator L squared,

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which is proportionate
to the magnitude squared,

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so we define it to be
LxLx plus LyLy plus LzLz.

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And we tried, we tried to see
if maybe Lx commutes with L

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squared.

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Remember, we had a role for L
squared in this differential

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operator that had the
Laplacian, the angular part

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of the Laplacian was
our role for L squared,

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so L squared is
starting to come back.

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So let's see here--

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this is Lx-- now, I'll
write the whole thing--

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LxLx plus LyLy plus LzLz.

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Now, Lx and Lx
commute, so I don't

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have to bother with
this thing, that's 0.

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But the other ones
don't commute.

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So let's do the
distributive law.

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So this would be an Lx,
Ly Ly plus Ly Lx, Ly--

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this is from the first--

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plus Lx, Lz Lz plus Lz Lx, Lz.

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You know, if you don't put these
operators in the right order,

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you don't get the right answer.

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So I think I did.

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Yes.

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It's correct.

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Now you use the commutators
and hope for the best.

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So Lx, Ly is i h-bar LzLy.

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Lx, Ly is plus i h-bar LyLz.

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So far, no signs of
canceling, these two things

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are very different
from each other.

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They don't even appear
with a minus sign,

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so this is not a commutator,
but anyway, what is this?

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Lx with Lz.

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Well, you should always
think cyclically.

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Lz with Lx is i h-bar, so this
would be minus i h-bar LyLz,

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and this is again Lz with Lx
would have been i h-bar Ly,

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so this is minus i h-bar
LzLy, and it better cancel--

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yes.

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This term cancels with the first
and this term cancels with this

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and you get 0.

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That's an incredible
relief, because now you

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have a second operator that
is measurable simultaneously.

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You can get eigenstates that are
eigenstates of one of the L's--

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for example, Lx and L
squared, because they commute,

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and you won't have the
problems you have there.

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In fact, it's a general theorem
of linear algebra that--

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we'll see a little bit of
that in this course and you'll

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see it more completely in 805--

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that if you have two Hermitian
operators that commute,

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you can find a simultaneous
eigenstates of both operators.

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I mean, eigenstates that
are eigenstates of 1,

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and eigenstates of the second.

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Simultaneous eigenstates
are possible.

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So we can find simultaneous
eigenstates of these operators,

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and in fact, you could find
simultaneous eigenstates of Lx

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and L squared, but given
the simplicity of all this,

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it also means that Ly
commutes with L squared,

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and that Lz also
commutes with L squared.

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So you have a choice--

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you can choose Lx,
Ly, or Lz and L

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squared and try to form
simultaneous eigenstates

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from all these operators.

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Two of them.

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Let's study those operators
as differential operators

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a little bit.

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So x, y, and z are your
spherical coordinates

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and they are r sin theta
cos phi, r sin theta sin

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phi, and r cos theta.

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We're trying to calculate
the differential

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operators associated
with angular momentum

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using spherical coordinates.

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So r is x squared plus y
squared plus z squared.

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Theta is cosine minus 1 of z
over r and 5 is tan minus 1 y

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over z.

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And there's something very
nice about one angular momentum

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operator in spherical
coordinates,

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there is only one angular
momentum that is very simple--

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its rotations about z.

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Rotations about z
don't change the angle

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theta of spherical coordinates,
just change the angle phi.

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r doesn't change.

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The other rotation, the rotation
about x messes up phi and theta

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and all the others
are complicated,

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so maybe we can have some luck
and understand what is d/d-phi,

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the d/d-phi operator.

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Well, the d/d-phi operator
is d/dy dy/d-phi plus d/dx

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dx/d-phi--

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the rules of chain rule
for partial derivatives--

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plus d/dz dz/d-phi.

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But z doesn't depend on phi.

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On the other hand,
dy/d-phi is what?

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dy/d-phi, this becomes
a cos phi-- it's x.

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X d/dy.

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And dx/d-phi is minus y.

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And you say, wow,
x d/dy is like x

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py minus y px, that's a z
component of angular momentum!

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So indeed, Lz, which is h-bar
over i x d/dy minus y d/dx,

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h-bar over i is
because of the p's--

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x py minus y px.

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And this thing is d/d-phi,
so Lz, we discover,

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is just h-bar over i d/d-phi.

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A very nice equation
that tells you

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that the angular momentum
in the z direction

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is associated with its operator.

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So I have left us exercises to
calculate the other operators

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that are more messy, and
to calculate Lx Ly as well

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in terms of d/d-theta's
and d/d-phi's.

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And as you remember, angular
momentum has units of h-bar

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and angles have no units,
so the units are good

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and we should find that.

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So that calculation is
left as an exercise,

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but now you probably could
believe that L squared,

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which is LxLx plus LyLy plus
LzLz is really minus h squared

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1 over sin theta d/d-theta.

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No, it's-- not 1
over sin theta--

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00:16:34,860 --> 00:16:36,330
uh, yep.

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00:16:36,330 --> 00:16:38,470
1 sin theta d/d-theta--

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00:16:41,320 --> 00:16:52,760
sin theta d/d-theta plus 1 over
sin squared theta d second d

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phi squared.

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So the claim that they had
relating the angular momentum

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operator to the
Laplacian is true.

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But, you know, you
now see the beginning

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of how you calculate
these things,

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but it will be a simple and
nice exercise for you to do it.