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PROFESSOR: The story
begins with a statement

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that we will verify to some
degree in the homework that

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is due on Wednesday
of vector operators.

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This operator Sz,
the one-line summary

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is that this
operator is a vector

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operator under
angular momentum--

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under the total
angular momentum.

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And as such, its matrix
elements will behave

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like the matrix elements of Jz.

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So how is that true?

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What does it mean to say that
S is a vector operator under J?

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It is to say that for
J, S is like a vector.

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And that is a concrete
statement that you

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should check whether it's true.

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The statement is that Ji Sj--

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here, this i and j
run from 1 to 3--

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is equal to ih bar
epsilon ijk Sk.

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That is the statement that
S is a vector operator.

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You may have seen this
in a previous course,

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in 805, where you might have
proven that X and P are vector

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operators for L.
And here, the proof

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is not all that
difficult. In fact,

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it's almost obvious
this is true, isn't it?

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Ji is Li plus Si.

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Li doesn't talk to Sj.

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But Si with Sj satisfy the
algebra of angular momentum

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that is precisely this one.

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So this is almost
no calculation.

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You will check-- you'll
remind yourself--

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that if you have a
vector, that L-- for L--

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X and P are vector operators.

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So for example, Li Pj is
ih bar epsilon ijk Pk.

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And this, you do by
calculating the commutator.

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But after you calculate
the commutator a few times,

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it's better to just remember,
oh, it's a vector operator.

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That's a good way of
thinking about this.

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OK.

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If you have vector
operators, they

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have very peculiar
properties sometimes.

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One that may sound a
little unmotivated,

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but it's very useful,
is the following.

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Suppose you form the double
commutator of J squared

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with the vector operator S.

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Here, you will find an identity.

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And to make it fun for
you, I will not tell you

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the number that appears here.

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It's some number.

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But with some
number here, this is

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identical to the following
SJ times J minus 1/2

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J squared S plus SJ squared.

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Well, it's important to
look at this equation,

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even to make sure
everything is in order.

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This is a vector equation,
so it's three equations.

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So J squared, despite all
this arrow, is a scalar,

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has no indices.

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It's J1 squared plus J2
squared plus J3 squared.

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S, on the other hand, has
an arrow, and it's a vector.

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So you could look
at this equation

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for the third component,
for the first component,

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for the second.

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So here is a vector,
the three things.

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Here is also a vector.

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It's S dot J and J.
So the vector index

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is carried by the J here.

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The vector index is
carried by S here--

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once to the left of J squared,
once to the right of J squared.

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Also, maybe you
should notice that SJ

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is the same thing as JS, not
because these operators commute

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but because SX and JX commute.

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Sy and Jy commute.

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And Sz and Jz commute.

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Different components
would not commute.

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But here, these ones do commute.

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So this is a formula you
will show by computation.

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I don't think there's a simple
way to derive this formula.

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But it's true and
false by computation.

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This formula implies a
result that is quite pretty.

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It's sometimes called
a projection lemma.

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So all we're doing is trying
to compute a matrix element,

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and we're forced to
consider a lot of structure.

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We're just trying to
show this simple matrix

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element, with an Sz here,
is proportional to mj.

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This is our goal.

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And we're going to do that.

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So suppose you take that
interesting equation

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and find its expectation
value on a state that

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is an eigenstate of j.

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So suppose you take a j mj and
put this whole equation inside

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this-- left-hand side,
then right-hand side--

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a state that is an
eigenstate of j.

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Now, that state may be an
eigenstate of other operators,

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as well.

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It doesn't matter.

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Now, look at your
left-hand side.

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It's a commutator.

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You have a j squared on
the left, a commutator

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on the right minus
commutator on the left, j

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squared on the right.

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In both cases, there will be
either a j squared near the bra

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or a j squared near the ket.

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Those two terms come
with opposite signs.

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Since those are eigenstates with
the same eigenvalues, that's

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what we're doing-- an
expectation value here,

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the left-hand side is 0.

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So the left-hand side
contributes nothing.

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So left-hand side is
0, on this thing, is 0.

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And let's look at
the right-hand side.

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It's equal to j mj S
dot J J jmj and minus--

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so that was the first term.

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Now, we have to compute this
thing on this Jm Jm state.

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Again, a J squared is either
to the left or to the right.

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Therefore, this
gives a number which

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is h squared j times j plus 1.

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This gives the same number,
as I showed you, on the bra.

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You have two terms.

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The factor of 1/2 cancels.

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And you're left just with the
expectation value of S, which

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is kind of what we wanted here.

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So this is minus h
squared j times j

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plus 1, which is the
expectation value of J

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squared times the expectation
value jmj of S on the jmj.

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OK.

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You have this term minus
that term equal to 0.

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So what have we learned?

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We have learned that this term
that we can call expectation

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value of S vector
on a j eigenstate

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is equal to the expectation
value of this quantity,

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S dot J J on the eigenstate
divided by this number, which

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turns out to be the
expectation value of J

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squared on that eigenstate.

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This formula looks like a
projection formula in which you

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say the expectation value
of S is the expectation

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value of the projection of
the vectored S onto the vector

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J. Remember, if you have,
for example, projection

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of a vector a into
a unit vector n,

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what is the projection of a
vector a into a unit vector n?

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Well, the projection
is a dot n times n.

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That's the component of
a along the vector n.

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But if n is not a unit vector,
the projection of a along b

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is a dot b times
b over b squared.

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Because, in fact, the
projection along a vector

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or along a unit vector
is the same thing.

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It's just a projection.

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And here, you have unit vectors.

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So this is the projection lemma.

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It's a very nice result--

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pretty striking, in fact.

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This result is also
mentioned in Griffiths.

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It doesn't give a derivation of
this result. It's just quoted.

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But it's a beautiful
and important result.

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It's conceptually interesting.

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It's valid for any
vector operator under J.

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And this will
answer our question.

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Because now, we can
use this formula

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to compute the matrix element.

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So what do we have for our case?

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We have that nljmj
Sz nljmj is what?

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Well, we have the expectation
value of Jz on this state.

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So it's going to
be h bar mj over h

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squared j times j plus 1.

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That's the denominator.

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And you still have here what
may look like a small challenge,

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or a big challenge--

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happily, it's a small challenge.

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S dot L mljmj.

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Here, this is called
the scalar operator.

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This is a variant
on the rotations.

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And scalar operators
are independent of mj.

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We got the mj dependence here.

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We want to claim
that this expectation

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value is proportional to mj.

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And we have the result here,
unless there is mj dependence

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here.

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But there is no mj dependence
here because, as I said,

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this has to do with the fact
that this is a scalar operator.

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So let's calculate this part to
finish this whole computation.

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How do you do that?

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Well, you have to remember
you have J equal to L plus S.

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So in here, we'll
do the following.

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I'm sorry.

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I had the confusion here.

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S dot J-- it's S dot J here.

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It's starting to look wrong.

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So I mean the dot
product of J and S.

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So here, I'll take l
equals to J minus S.

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And L squared is equal to
J squared plus S squared

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minus 2S dot J.
Here, it's important

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that S dot J and J
dot S are the same.

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And therefore, this S dot J is
1/2 of J squared plus S squared

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minus L squared.

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So with S dot J being
this, you see immediately

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what is this number.

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This is h bar mj h
squared j times j plus 1.

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And now, 1/2-- so I have a 1/2
here, I'll put it in front--

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J squared-- so this
is j times j plus 1,

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there's an h squared
that's in addition--

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h squared here--
j times j plus 1

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S squared, which
is 3/4, for it's

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been 1/2, minus L squared,
which is minus l times l plus 1.

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OK, almost there.

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Wow, this takes time.

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But we have a result.
So what is the result?

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The matrix element nljmj lz--

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oops, I'll put the whole
thing together-- lz plus 2Sz--

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back to the whole
matrix element.

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Remember, we had one piece--

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hmj and, now, this part.

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00:16:39,140 --> 00:16:43,270
So adding the hmj
to this new part,

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we have hmj 1 plus j times
j plus 1 minus l times l

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plus 1 plus 3/4 over
2j times j plus 1.

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00:17:01,520 --> 00:17:03,690
Phew.

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00:17:03,690 --> 00:17:04,460
OK.

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00:17:04,460 --> 00:17:04,970
I'm sorry.

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It's all here.

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I just copied that term,
hopefully without mistakes.

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So we have our matrix element.

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And that matrix element
in the top blackboard

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there gives us the splitting.

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It's probably a good time
to introduce notation.

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00:17:25,470 --> 00:17:28,830
And there's a
notation here where

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this is called g sub J of l.

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And it's called
the Landé g-factor.

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It's a g-factor in the sense
that affects the energy

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levels as if you were
modifying the magnetic moment

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of the particle.

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So this number tells
you how the level split.

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They split proportional to mj--

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all the various levels.

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And for the full
multiplate, the multiplate

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is an eigenstate of j
and an eigenstate of l.

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So throughout all the
states in the multiplate,

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this is a single number.

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And just, you have the hmj.

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End result is the
weak Zeeman splitting

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nljm is eh bar over 2mc times
B times gJ of l times mj.

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00:18:45,180 --> 00:18:50,430
And this number is
about 579 times 10

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to the minus 9 eV per Gauss.

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It's small.

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So wow.

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It took us some effort.

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But here we are.

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We have the weak-field Zeeman
splitting completely computed.