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PROFESSOR: So B
strong field Zeeman.

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So what did we say we would do?

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We would have H0 plus e
over 2mc B Lz plus 2Lz

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B. This would be our
H0 check I call it,

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plus delta H fine structure 1.

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And we said what
we have to do now

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is strong field Zeeman is more
important than fine structure.

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So we first have to
get the strong field

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Zeeman figured out, with
the bare bones hydrogen

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atom, what it does.

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And then on those states, we
will do perturbation theory

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for fine splitting.

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So it is redoing fine splitting.

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And you say oh my
god, that's hard.

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We spent a whole
lecture doing that.

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Well, second time you do
things, they go a little faster.

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So it's not that bad.

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But here there is
something quite remarkable.

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This was supposed to be
your known Hamiltonian.

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And you say no, it's not known.

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I never solved this before.

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On the other hand, when
we had H0 plus delta

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H1, a fine structure,
we did struggle.

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And we found those states--

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approximate states.

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Here the situation happily
is surprisingly simple.

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And one reason for
that is the following.

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That again, perhaps
your initial impression

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this can be solved exactly.

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You don't even need perturbation
theory to add this term.

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What?

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Yes!

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This Hamiltonian
commutes with H0.

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Isn't that right?

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H0 is rotational invariant.

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So it commutes with any j.

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Uncertain H0 has
nothing for a spin.

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It's a one matrix there.

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So this commutes
with Hamiltonian.

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So it's possible that you can
diagonalize this completely.

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Simultaneous-- eigenstates
of the first part

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and the second part, so
simultaneous eigenstates

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from all of those!

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But the news is even better.

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Your uncoupled states--
uncoupled states n, l, m, l,

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m, s, those were eigenstates
except eigenstates of H0,

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the all good ole hydrogen atom.

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But actually they are
exact eigenstates of lZ,

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and exact eigenstates of sz.

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So they're exact eigenstate
of the Zeeman Hamiltonian.

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So they state are it.

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These are the exact
states of H0 hat!

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These are exact eigenstates--

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eigenstates of H0 check--

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I'm sorry.

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Its was not hat with
eigenenergies as follows.

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There are no mystery,
the eigenenergies,

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they're very simple.

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The eigenenergies are
e, n, l, ml, l, m, s.

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Exact are e, m, 0, the ones
that the hydrogen atom has

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that don't depend on any of
these other things plus eB

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over 2mc eh bar B
over 2mc ml plus 2ms.

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So this serves perfectly the
name of known Hamiltonian.

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That was not the
case for weak Zeeman.

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And weak Zeeman who had this
one and this one, and the other

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was the approximately
known Hamiltonian, to which

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we added the weak Zeeman.

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Here, it's this perfectly
known Hamiltonian, to which we

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have to now add fine structure.

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So maybe the last
thing that helps

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you visualize what's
going on is to understand

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what happens to the splittings.

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Because you're going to have
to fine structure splitting.

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And fine structure again,
you will have to ask,

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can I use non- degenerate
perturbation theory or can

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I not use it?

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So you need to know what
happened with the degeneracies

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after you add this term.

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Are all that the degeneracies
of the hydrogen atom

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broken by this term,
or do some survive?

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If they survive, is
fine splitting diagonal

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in those degenerate
subspaces or not?

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So the most important
thing is figuring out

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what are the degenerates spaces
after you've added this term.

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This is intuitively
what you have to do.

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If you approach this
problem OK, I now

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have to compute fine
structure on a new basis,

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and you have no idea
what the new basis is,

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you are proceeding a little
bit with your eyes covered.

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You should always try to make
things a little more concrete.

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So look at the n equal 2 states.

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You have l equals
0 and l equals 1.

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Here, you have six states.

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Remember, their spin.

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There's two states here.

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The two states of l
equals 0 have ms equal 1/2

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and ms equal to minus 1/2.

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So they're going to split.

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This number is going to be
either plus 1 or minus 1,

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and they're going to split.

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And here-- so this is plus 1 on
this factor here, this number,

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or minus 1 for that number.

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For this states of l
equals 1, ml for example,

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can be 1, and ms plus 1/2--

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so 1 plus 1 is 2.

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So there is a state of 2.

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Ml equals minus 2 minus
1, and this minus 1/2

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gives you minus 2.

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So that's another state.

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You can have more states.

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For example, if you take
ml equals 1 and ms equals

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2 minus 1/2, you get the 0.

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But you can also have ml equals
minus 1 and this plus 1/2--

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plus 1/2, which
also gives you 0.

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So here there's a degeneracy.

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There are six states.

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You will see that there
is one here and one there.

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So here is the nature
of the degeneracy.

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The six states have
split like that.

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There's a degeneracy
across l multiplets.

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And there's a degeneracy
within l multiplets.

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So two types of degeneracy.

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And that's what you will
have to consider when you

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think of the fine splitting.

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In fact, the problem
in the homework

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gives you some sort of trickery
to evaluate this expectation

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value with a little less work
than the traditional method,

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but still asks the
question whether you

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can use the generator or
non-degenerate perturbation

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theory.

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And that's an
interesting question.

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And this example, can help you
visualize a little better what

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kind of degeneracies you have.

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OK, we're concluding a chapter
in the history of [INAUDIBLE]

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six.

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We're done with
perturbation theory,

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and we're done with hydrogen
atom for the moment.