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PROFESSOR: The fact is
that angular momentum is

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an observable, and as such
it deserves attention.

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There is an active way of
thinking of observables,

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and we have not developed
it that much in this course.

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But for example, with
a momentum operator

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you've learned that the
momentum operator can give you

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the differential operator.

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It's a derivative, and
derivatives tell you how

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to move, how a function varies.

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So with the momentum
operator, for example,

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you have the momentum
operator p hat,

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which is h bar over i d dx.

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And you could ask
the question of, OK,

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so the momentum operator
moves or takes a derivative,

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does the momentum
operator move a function?

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Does it generate a translation?

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And the answer is, yes.

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That's another way of thinking
of the momentum operator

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as a generator of translations.

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But how does it do it?

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This is a Hermitian operator,
and it takes a derivative.

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It doesn't translate
the function.

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But there is a universal
trick that if you exponentiate

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i times a Hermitian
operator, you

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get a new kind of operator
that actually, in this case,

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moves things.

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So we could think
of exponentiating

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e to the i p hat, and
for purposes of units

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I have to put a constant
with units of length,

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and an h bar here.

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And now you have the
exponential of an operator.

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That's good.

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That's a very
interesting operator,

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and we can ask what
does it do when

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you act on a wave function?

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It's an operator.

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And look, simplify by putting
what p is going to do.

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P is h over i d dx.

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So this is like a
d dx exponentiated

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acting on psi of x.

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And as an exponential,
it can be expanded

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in a Taylor series with
this funny object there,

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but it would be the
sum from n equals 0

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to infinity 1 over
n factorial a d dx.

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I will write it this
as normal derivatives,

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because we just have a function
of x, a d dx to the n psi of x.

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And you see that,
of course, this

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is psi of x plus a d
psi dx plus 1 over 2

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a squared, d second
psi dx squared.

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But this is nothing else but
the Taylor series for this.

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And there it is, the miracle.

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The e to the i momentum
generated translation.

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It really moves
the wave function.

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So that in a sense
is a deeper way

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of characterizing
the momentum operator

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as a generator of translations.

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With the angular
momentum operators,

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we will have that they
generate rotations.

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So I need a little bit
more mathematics here,

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because I have to deal with
three dimensions, a vector,

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and produce an exponential
that rotates the vector,

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so that it gives you the wave
function at a rotated point.

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But this will be the same story.

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Angular momentum will
generate rotations

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the same way as momentum
generates translations.

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And there is yet another story
that when you will appreciate

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the abstract properties
of angular momentum

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that some of them
will appear today,

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you will realize that in
addition of angular momentum

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that represent rotations
of objects doing things,

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there is another way of
having angular momentum.

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And that's spin
angular momentum.

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That mysterious property
of particles that have--

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even though they
have 0 size, they

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behave as if they were little
balls rotating and spinning.

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That spin angular momentum
has no ordinary wave functions

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associated to it, and
it's fractional sometimes.

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And the study of angular
momentum inspired

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by orbital angular
momentum associated

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with normal rotations,
will lead us

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to understand where spin
angular momentum comes about.

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So it's a gigantic
interesting subject,

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and we're beginning
with it today.

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So it's really quantum
mechanics in three dimensions,

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central potentials,
and angular momentum.

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And let's begin by
mentioning that if we

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are in three dimensions--
and many things

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with it so far in
this course, we always

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took the time to write
them in three dimensions.

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So we wrote this, for
example, as a generalization

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of the derivative form
of the momentum operator.

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Meaning there is a Px,
which is h bar over i d dx,

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Py h bar over i d dy, and
Pz equal h bar over i d dz.

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And we had commutators within Px
and x, Py and y, and Pz and z.

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There were always the same
commutators of the form

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x Px equal i h bar.

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Similar things here.

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With this we wrote the three
dimensional Schrodinger

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equation, which was minus h
squared over 2m, and instead

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of p squared three
dimensional, he

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would have a derivative if you
were doing in one dimension.

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For three dimensions
you have the Laplacian.

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And this time you
have a wave function

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that depends on the
vector x plus v of r--

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v of x.

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Should I write r?

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Let me write r vector.

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V of r psi of r
equal e psi of r.

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This is our time independent
Schrodinger equation.

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This corresponds to
the energy eigenstate,

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but in three dimensions.

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So this is the equation
we wish to understand,

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and our ability to
understand that equation

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in a simple and nice way
rests on a simplification.

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That is not always
true, but it's

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true under so many
circumstances that it's worth

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studying by itself.

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And it's the case when you
have a central potential,

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and by that we mean that
the potential is not

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quite the vector function
of r, but is just a function

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of the magnitude of r.

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That's a little bit
funny way of writing it,

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because I'm using
the same letter v,

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but I hope there's no confusion.

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I mean that the potential just
depends on the value of r.

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So what this means physically
is that over concentric spheres,

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the potential is constant.

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All over the surface of
spheres of constant radius,

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the potential is
constant, because it only

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depends on the radius.

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And this potential is there
for a spherically symmetric.

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You can rotate the
world, and the potential

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still looks the same, because
rotations don't change

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the magnitude of vectors.

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If you have a vector of
some length, you rotate it,

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it's the same length,
and therefore you

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remain on this sphere.

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So the central potential
are spherically symmetric.

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By that we mean they're
invariant under rotations.

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So this is the reason
why angular momentum will

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play an important role, because
precisely the angular momentum

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operators, in the fashion
we discussed a minute ago,

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generate rotation.

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So they will have
a nice interplay,

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to be developed in the following
lectures, with the Hamiltonian.

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So at this moment we
have a central potential,

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and let's assume
that's the case.

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And we need to
understand a little more

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of this differential equation.

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So let me write the formula for
the Laplacian of a function.

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It has a radial contribution.

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You know it's second
order derivatives.

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And it has a radial part,
and an angular part.

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The units are 1
over length squared.

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So you need, if you have an
angular part, all over here

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is going to be
angular, you still

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need the 1 over on r squared
here for the units to work out.

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So here it is, it's
slightly complicated.

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d d theta sine
theta d d theta of--

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well, put the psi, plus 1
over sine squared theta,

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d second d phi squared
all acting on psi.

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It's a complicated
operator, and here

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is some radial
derivatives, and here there

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are some angular derivatives.

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So you see, today's lecture
will have many steps,

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and you have to keep track
of where we're going.

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And what we're going
to do is, build up

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a structure that
allows us pretty much

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to forget about all this thing.

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That's our goal.

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And angular momentum will
play a role in doing this.

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So there are in fact two
things I want to justify,

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two facts to be justified.

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So I will erase this.

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The first fact is the relation
between this differential

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operator and angular momentum.

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So two facts to justify.

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The first is that minus h
squared 1 over sine theta

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d d theta sine theta d d theta
plus 1 over sine squared theta

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d second d phi squared.

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This whole thing can be viewed
as the differential operator

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version of angular momentum.

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Remember, d dx was a
differential operator version

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of momentum.

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So maybe this has to do with
angular momentum, and indeed

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this whole thing, remember,
units of angular momentum

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is h bar.

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Angular momentum is
length times momentum.

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And from the
certainty principle,

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you know that x times
p has units of h bar.

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So angular momentum
has units of h bar.

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So there's h bar squared here.

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So this must be angular
momentum squared.

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In fact, if you
think about that,

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angular momentum is x times p.

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So x times a derivative.

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So it's a first order
differential operator,

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but this is a second order one.

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So this could not be
just angular momentum.

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Anyway, angular
momentum is a vector.

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So this will turn
out to be, and we

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will want to justify L squared.

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The quantum version of the
angular momentum operator

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squared.

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And the other thing I want
to justify if I write--

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call this equation one.

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00:16:11,540 --> 00:16:16,460
So this is fact
one, and fact two,

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is that equation one
is relevant, when--

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00:16:28,280 --> 00:16:32,572
let me wait a second
to complete this.

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00:16:32,572 --> 00:16:37,280
This equation is an
equation for a particle

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moving in a potential,
a spherically

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symmetric potential.

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00:16:42,800 --> 00:16:46,760
It turns out that is
relevant under more

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general circumstances.

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00:16:49,160 --> 00:16:57,074
If you have two particles
whose potential energy--

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if you have two
particles you have

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a potential energy between them,
maybe it's a electromagnetic--

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if the potential
energy just depends

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on the distance
that separates them,

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this two body problem can
be reduced to a one body

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problem of this form.

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This is a fairly non-trivial
fact, and an absolutely

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00:17:23,839 --> 00:17:25,040
interesting one.

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00:17:25,040 --> 00:17:28,850
Because if you want to really
solve the hydrogen atom,

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you have an electron
and a proton.

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Now it turns out that the proton
is almost 2,000 times heavier

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than the electron.

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And therefore, you
could almost think

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that the proton
creates a potential

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in which the electron moves.

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00:17:44,400 --> 00:17:53,010
But similar analysis is valid
for neuron orbiting a nucleus.

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And in that case, the neuron is
still lighter than the proton,

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but not that much lighter.

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Or maybe for a quark and an
anti-quark orbiting each other.

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00:18:04,620 --> 00:18:08,410
Or an electron and a
positron orbiting each other,

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and this would be
valid and useful.

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So we need to somehow
explain that as well.

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00:18:16,590 --> 00:18:20,490
If you really want to
understand what's going on,

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00:18:20,490 --> 00:18:26,160
is that equation
one is relevant when

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we have a two body problem
with a potential function

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v of x1 x2.

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00:18:44,800 --> 00:18:47,530
The potential energy
given that configuration,

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x1 and x2 of the first
and second particle,

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is a function of
the separation only.

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00:18:58,660 --> 00:19:02,880
The absolute value or the length
of the vector, it's 1 minus x2.

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00:19:06,580 --> 00:19:09,610
This far we'll
get through today.

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This will be next lecture still.