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PROFESSOR: It's not unsuspected,
because in some sense, energy

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eigenvalues, what you need,
the first order energy's then

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split the state.

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The hope is that the second
order energy split the states,

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so that's why you have
to go to second order.

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So going to second order
is something we had to do.

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There was no hope to do this
before we go to second order.

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So let's go to second order,
and order lambda squared.

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And we put n0l state in.

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The left hand side,
happily, is 0.

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So we have n0l from the right
hand side, en1 minus delta h

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psi 1.

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But again, psi 1 has two
pieces, a piece in the space

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we had that we just calculated
plus a piece in the space of v

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tilde n and 1 minus
delta h psi 1 vn

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plus the energy, which
would be en2 in here.

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And we hit with n0l,
so we pick an al0.

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OK, that doesn't
look that terrible.

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I don't know if you agree,
but it really doesn't.

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Especially because a
few things are gone.

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This term is zero.

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Why?

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Because state in the
degenerate subspace orthogonal

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or to v hat.

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Here I want to remind
you of what we did.

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We did a long
computation to explain

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that even though delta h is
only diagonal in the degenerate

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subspace, when you
have a state here

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in the degenerate subspace,
you can let delta h think of it

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as acting as an eigenvalue.

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But that's a great
thing, because if this

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acts as an eigenvalue, this is
the first order of correction.

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But all the first order of
corrections are the same,

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so this here, delta
h and this, is

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going to give the same
as en1 on this state,

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and therefore this whole
state, happily, is all 0.

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So it's again, this delta h--

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we did it with a
resolution of the identity.

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Hope you remember that argument.

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If you don't, look
at it back later.

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But with a resolution
of the identity,

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we argued that delta h,
when acting in a state of vn

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on the right, you
can assume that this

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is an eigenstate of it.

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So this whole term is zero.

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So now we are in pretty
good shape, in fact.

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The equation is not that bad.

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The equation has become
minus n0l, delta h, psi 1,

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on v hat plus en2 al0 equals 0.

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And this psi 1, we've
already calculated it there.

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So this is great.

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You see, you should
realize that at this moment

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we've solve the
problem, because we're

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going to get from here something
complicated acting on a 0.

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Because psi 1 has this
a1's, but the a1's

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were given in terms of a0.

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So something on a
0, something on a 0,

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it's going to be an
eigenvalue equation for a0,

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an eigenvector equation for a0.

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So let me just finish it.

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So I have to do a
little bit of algebra

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with this left hand side.

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I can put here this ap1
times the p0 states there.

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So on the left it will be minus
the sum over p, n0l, delta h,

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p0 times ap1 plus en2 al0, 0.

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OK, now I just have to
copy that thing there.

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And I better copy
it, because we really

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need to see the final
result. It's not that messy.

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So here it is.

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I'll copy this thing.

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I also can write
this as delta h nlp.

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You recognize that thing.

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So this will be the sum
over p of delta h nlp.

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The sine, I will
take care of it,

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times that thing over
there, 1 over ep0 minus en0.

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That's another minus sign
that cancels this sign here,

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and I have here the sum over
k equals 1 to n, delta h, pmk,

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ak0 plus en2 al0 equals 0.

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OK, so what do we do now?

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Just rewrite it one more time
and it will all be clear.

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So I'll write the k outside,
k big parentheses, p.

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For the first term I'm
going to pull the k sum out

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and we'll sum over p first.

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So what is it?

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Delta h nlp delta h pmk.

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Those were the two things here.

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And then we have
the denominator.

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I'll change a sign
of the first term.

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I will change its sign so that
the second term looks more

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like an eigenvalue.

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So I changed sign
here by changing

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the order of things
in the denominator,

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and then I put minus en2.

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I don't have a sum
over k here, but we

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can produce one by writing
delta kl ak0 equals 0.

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Look, we got our answer.

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If I call-- invent
a matrix mlk2, which

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is precisely delta h nlp
delta h pmk over en minus ep

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is sum over p.

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This is a-- you see, you
sum over p, n is irrelevant,

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you have a matrix here in lmk.

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That's why we call it mlk.

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And therefore this
whole equation

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is the sum from k equals 1
to n of mlk2 minus en2 ak0--

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ml2 en2 delta lk ak0 equals 0.

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Or, if you wish, it's
just a matrix equation

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of the form m2 minus en times
the identity on a vector

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equals 0.

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And this is an
eigenvalue equation.

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So finally, here is the answer.

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Let's say you compute this
matrix of order lambda squared

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with this perturbation
and that is the matrix

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that if you diagonalize, you
find the second order energy

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corrections and you
find the eigenvectors.

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And if you found
the eigenvectors,

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you found the good basis.

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So the problem has been
solved at this stage.

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You now know that if
you fail to diagonalize,

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to break the perturbation
of the first order,

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you will have to diagonalize
a second order matrix.

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And once you diagonalize it,
you now have the good basis,

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and it will happen that if
all the levels get split

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at second order,
we can calculate

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fairly easily the rest of the
pieces of the [? states. ?]

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So we'll leave it.

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There you'll complete some
details, more elaborations

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of that in the exercises, and
we'll go to hydrogen atom next.