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PROFESSOR: So let's
attempt to solve something.

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And that first step is going
to be nice and simple and very

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illuminating, as well.

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So we're going to do a
three step procedure.

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Three step procedure.

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Three step procedure.

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And basically,
we're going to use

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this equation, the
ordered lambda equation,

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to find the energy corrections.

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One, ordered lambda equation
for energy corrections.

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Two, ordered lambda equation
to find the component of n1k

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in v hat.

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And three, we're going to
look at the ordered lambda 2

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equation to find the
component of that same state

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in the degenerate subspace.

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So let's do the first step.

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And for that, we're going
to overlap that equation,

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this ordered lambda equation.

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So for step one, we're going
to hit the equation with n0l

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from the left.

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OK.

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n0l from the left.

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Here it comes.

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Boom.

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It's here.

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What happens now?

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Well, h0 on that
state, n0l, that

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is part of the
degenerate subspace.

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So it has energy n0 with respect
to the original unperturbed

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Hamiltonian.

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So you get n0 minus n0, it's 0.

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So the left hand side vanishes.

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Kills left hand side.

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So we just look at
the right hand side.

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And we got n0l Enk1 minus
delta H n0k equals 0.

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That was the right hand side.

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So far, so good.

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Nice and simple.

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From here, what do we get?

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Let's write it just by moving
one term to the left and one

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term to the right.

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n0k Enk1 delta lk.

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And this should be true for
all k and l in 1 up to n.

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Because those
equations are for all k

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and we could have
hit with all l.

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And now this is an
amazingly nice equation.

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It tells you a nice story.

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What is this story?

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The story is that look at this.

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This equation says that if you
want this to hold it better

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be that delta H is diagonal on
the basis that you're using.

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Because for k and l
running in n by n.

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So this says that delta H
is diagonal in the chosen

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basis of the subspace vn.

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Delta H is not
diagonal elsewhere.

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On the big space, it
may not be diagonal.

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But on the little degenerate
space, it better be diagonal.

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So what it says is that, if you
want to start your perturbation

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theory, you cannot use an
arbitrary basis of states.

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Maybe you chose it wrong.

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You have to start in
the degenerate subspace

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with a basis that makes
the perturbation diagonal.

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And that's what our
little example had.

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It says, even to
begin with, you should

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have started with the basis
that makes the perturbation

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diagonal.

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In that case, you will find
the formulas giving you

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the right answers.

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And then, once you have
the matrix being diagonal,

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you can take l equal to k.

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And then you'll find
on the right hand

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sides Enk1 is equal to n.

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l is equal to k,
so n0k delta H n0k.

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So the energy corrections
are the diagonal elements

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of this matrix.

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What you would expect.

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So we will write this as
delta H. We could put kk,

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but to remind you,
its degenerate.

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You put nk nk.

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So we've done the
first step and we've

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realized that this
perturbation is a little funny.

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You really have to get
started with the right basis.

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And what is the right basis?

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The basis that makes
delta H diagonal.

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We'll see that everything
is much simpler

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if the eigenvalues of this
matrix are all different.

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So the degeneracy is broken.

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Then things are easier.

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If the degeneracy is not
broken, then the degeneracy

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may be broken to
order lambda squared,

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which is more interesting
and it will be something

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we'll study next time.

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So that's all.