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PROFESSOR: So case A--
weak Zeeman effect.

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So what are our states here?

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We discovered that and we know
those are the coupled basis

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states.

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And the states of H0
tilde eigenstates--

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they are approximate
eigenstates--

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are the states n, l, j, mj.

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And the energies were
energies dependent on n and j.

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So the 1S 1/2, 2S 1/2,
2P 1/2, and 2P 3/2.

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Roughly, to remind
you of what happened,

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the original states
were shifted,

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and we used the j
quantum number in here.

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And those are our multiplets.

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Those are our multiplets.

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And we have a lot of
degeneracy as usual.

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So this is degeneracy
here and degeneracy here

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because a multiplet
P 3/2 is j equal 3/2.

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And that's four states.

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Here you have two
states, and here you

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have two states as well.

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So quite a bit of states
that are degenerate.

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So in principle, when we
do the Zeeman splitting,

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we may have to consider the
full matrix n, l, j, mj, delta H

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Zeeman, nl prime j m prime j.

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So what are our degeneracies?

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Our degeneracies are when
you have a given value of j.

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So a degenerate subspace
can have different l's--

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for example, here--
but the same j,

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and therefore different mj's.

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Or within a given j multiplet,
it might have different mj's.

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So this is the scope
of the degeneracy.

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And in principle, we may have to
diagonalize a matrix like that

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by looking at the
degenerate spaces.

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If you're doing
the level two, you

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would have to discuss
these four states here.

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You would have to discuss
this other four states.

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Happily, we don't
have to do that much

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because, as usual, delta H
Zeeman is proportional to l z

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plus 2Sz.

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And this commutes with l
squared with delta H Zeeman.

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l squared commutes
with any l operator.

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It certainly commutes
with any S operator.

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They don't even
talk to each other.

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And therefore, l
squared commutes

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with l Zeeman, which
means that when

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l is different from l prime,
this matrix element has

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to vanish.

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This is our remark from
perturbation theory long,

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long ago.

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You have another
operator for which

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the states have different
eigenvalues, commutes

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with your perturbation.

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The matrix element
of the perturbation

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must vanish between
those states.

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So we don't have
eigenstates like that.

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And when l is equal
to l prime already--

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so we focus on l
equals to l prime--

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we only need to worry
within multiplets.

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So you have n, l, j, mj, delta H
Zeeman now, and l, j, mj prime.

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It's an issue of mj prime now.

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But Zeeman thing
commutes with Jz.

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Jz commutes with delta H Zeeman.

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Jz is Lz plus Sz,
and z components

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in angular momentum--
two identical components

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always commute, of course.

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So Jz commutes with
delta H Zeeman.

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So this thing will vanish
unless m is equal to m prime.

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And that's great
because you're back

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to nondegenerate
perturbation theory.

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The whole matrix,
this Zeeman thing

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could have turned out to
be complicated matrices.

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No.

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It's perfectly
diagonal in this basis.

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There's nothing to
worry about here,

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except that it's still not easy
to compute, as we will see.

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So what do we need to compute?

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We'll have the first
order corrections

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due to Zeeman on the n,
l, j, mj basis is equal--

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well, the Zeeman Hamiltonian
had an e over 2mc.

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So let's put it there.

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e.

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Let's put the B close to the e.

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2mc.

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And now we have to do n l
j mj Lz plus 2Sz n l j mj.

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Perfectly diagonal.

86
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And that's nondegenerate
perturbation theory.

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It's going to give us
all the energies we want.

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All the splittings we want.

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So basically what's going to
happen, as you can see here,

90
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is that the things down mix.

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Everything is there.

92
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Honestly, these two
levels are going to split.

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These two levels
are going to split.

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These four levels
are going to split.

95
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Everything is going
to split here.

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The remarkable thing
of this formula,

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and it's going to keep us busy
for about 10, 15 more minutes,

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is that this thing, this matrix
element, is proportional to mj.

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So the states split proportional
to the m quantum number.

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The state with the m equal 3/2
will split three times as much

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as the state with m equal 1/2.

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And that's not obvious here.

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It's a remarkable
result. It's part

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of what's called the
Wigner-Eckart theorem,

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something that you study in
graduate quantum mechanics.

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But we're going to see a bit
of it, the beginning of it,

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in this computation.

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And it's a fairly
remarkable result.

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So the remarkable result here--

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remarkable-- is
that E [INAUDIBLE]

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are proportional to mj.

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And that defines
a linear splitting

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because it's linearly
proportional to the magnitude

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of the magnetic field and
divides the states nicely.

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So we want to understand
this matrix element.

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And there's a little
thing we can do.

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Notice that Lz plus 2Sz
is equal to Jz plus Sz.

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You can take one of the
Sz's and complete Jz,

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and you're left with that.

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So the matrix element n l j m
Lz plus 2Sz n l j mj is equal--

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if you have a Jz, that gives
you just something proportional

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to Hmj plus n l
j mj Sz n l j mj.

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So OK.

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A little bit of
the mystery maybe

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seems to you at least
consistent here.

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I said this matrix element turns
out to be proportional to mj.

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And certainly,
this piece, having

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to do with the J component
here, is proportional to mj.

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The mystery that remains
is why this matrix element

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would be proportional to mj.