WEBVTT

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PROFESSOR: All
right, spin orbit.

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We have it still there,
delta H spin orbit, orbit.

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Well, we know what V is.

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So that derivative, dv dr can
be taken care of, the 1 over r.

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That gives you e squared over 2m
c squared 1 over r cubed S, L.

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And you remember that S, L,
from 805, J squared minus S

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squared minus L squared.

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The reason that's why you
tend to do addition of angular

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momentum, because S and
L, calculating the matrix

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elements of this
thing is very easy,

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in that base is because
every state there

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has a fixed value of S squared,
a fixed value of L squared.

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And j squared plus a
couple of possibilities.

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So we must work with
the couple basis.

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Basis.

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And therefore, we can attempt to
find E1 of a N, L, J, MJ, spin

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orbit, equal e squared over 2m
squared c squared and LJ, MJ,

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S, L, R cubed.

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The R cube has to stay
inside the expectation value,

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because the expectation
value includes

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integration over space.

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So this is a very important.

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N, L, J, MJ.

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And again the useful
question, the couple basis

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will have have degeneracies.

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All the states are
degenerate there.

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So this time, we fixed n,
because the degeneracies

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happen only when you fix n.

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So do we have the
right to do this,

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to use the formula
form perturbation,

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the non-degenerate perturbation
theory to do this calculation?

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And the answer is yes, because
the perturbation S.L over R

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cubed commutes with with
L squared, with J squared,

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and with Jz.

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you need all that because
you can have degeneracies

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by having different L values.

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And that would be taken
care by this operator that

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has different eigenvalues
when L is different.

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You can have the degeneracies
involving different J values.

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This would be taken
by this operator.

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And you can have
degeneracies when

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m has different value so that
involves the Jz operator.

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So you really need
a perturbation

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that commutes with all of them.

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And why does it commute
with all of them?

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You can see it in several ways.

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Let's do L squared.

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L squared is Casimir,
it commutes with any LI.

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It doesn't even think
about S because it doesn't

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know anything about S.
So it commutes with S.

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So L squared with any
LI and commutes with S.

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And L squared is
an invariant, it

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commute with r squared
because r squared

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is rotational invariant.

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So everything commute with that.

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In order to do the
other ones, you

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can also think in terms of
this matrix J squared over here

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and S squared and
do all of them.

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You should do it and convince
yourself that they all commute.

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So we can do this.

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If we can do it, it's
good because then we

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can evaluate these quantities.

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So let's do a little
of the the evaluation.

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So this E1nljmj is equal to.

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Let's evaluate the S dot L
part by using 1/2 j squared

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s squared minus L squared.

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So that gives you a factor
of h bar squared over 2,

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with this 2 over there.

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So you get e squared
h squared over 4m

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squared c squared, J times
J plus 1 minus l times l

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plus 1 minus 3/4, times
nljmj, 1 over r cubed, mljmj.

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OK.

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That should be
clear from the fact

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that you have a j
squared minus s squared.

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That's a 3/4 spin is
always 1/2, and l squared.

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This is known.

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It's equal, in fact, to
nlml 1 over cubed nlml.

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Which is equal, let
me discuss that again.

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It's a little-- a0l,
l plus 1, l plus 1/2.

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OK.

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So this is a known result. It's
one of those expectation values

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that you can get from
Feynman, Hellman,

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or for other
recursion relations.

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And this is always computed in
the original uncoupled basis.

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But we seem to need it
in the coupled basis.

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So again, are we in trouble?

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No.

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This is actually the same.

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And it is the same only
because this answer

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doesn't depend on ml.

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Because this states involve
various combinations of ml

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and ms. But doesn't
depend on ml,

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so the answer is
really the same.

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So these things are
really the same.

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Happily, that
simplifies our work.

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And now we have E1
nljm is equal to En 0--

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this is yet another notation,
this is the ground state energy

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here--

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mc squared n, j, j plus 1
minus l, l plus 1 minus 3/4,

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over l, l plus 1/2, l plus 1.

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OK this is spin orbit.