PROFESSOR: It's not unsuspected,
because in some sense, energy
eigenvalues, what you need,
the first order energy's then
split the state.
The hope is that the second
order energy split the states,
so that's why you have
to go to second order.
So going to second order
is something we had to do.
There was no hope to do this
before we go to second order.
So let's go to second order,
and order lambda squared.
And we put n0l state in.
The left hand side,
happily, is 0.
So we have n0l from the right
hand side, en1 minus delta h
psi 1.
But again, psi 1 has two
pieces, a piece in the space
we had that we just calculated
plus a piece in the space of v
tilde n and 1 minus
delta h psi 1 vn
plus the energy, which
would be en2 in here.
And we hit with n0l,
so we pick an al0.
OK, that doesn't
look that terrible.
I don't know if you agree,
but it really doesn't.
Especially because a
few things are gone.
This term is zero.
Why?
Because state in the
degenerate subspace orthogonal
or to v hat.
Here I want to remind
you of what we did.
We did a long
computation to explain
that even though delta h is
only diagonal in the degenerate
subspace, when you
have a state here
in the degenerate subspace,
you can let delta h think of it
as acting as an eigenvalue.
But that's a great
thing, because if this
acts as an eigenvalue, this is
the first order of correction.
But all the first order of
corrections are the same,
so this here, delta
h and this, is
going to give the same
as en1 on this state,
and therefore this whole
state, happily, is all 0.
So it's again, this delta h--
we did it with a
resolution of the identity.
Hope you remember that argument.
If you don't, look
at it back later.
But with a resolution
of the identity,
we argued that delta h,
when acting in a state of vn
on the right, you
can assume that this
is an eigenstate of it.
So this whole term is zero.
So now we are in pretty
good shape, in fact.
The equation is not that bad.
The equation has become
minus n0l, delta h, psi 1,
on v hat plus en2 al0 equals 0.
And this psi 1, we've
already calculated it there.
So this is great.
You see, you should
realize that at this moment
we've solve the
problem, because we're
going to get from here something
complicated acting on a 0.
Because psi 1 has this
a1's, but the a1's
were given in terms of a0.
So something on a
0, something on a 0,
it's going to be an
eigenvalue equation for a0,
an eigenvector equation for a0.
So let me just finish it.
So I have to do a
little bit of algebra
with this left hand side.
I can put here this ap1
times the p0 states there.
So on the left it will be minus
the sum over p, n0l, delta h,
p0 times ap1 plus en2 al0, 0.
OK, now I just have to
copy that thing there.
And I better copy
it, because we really
need to see the final
result. It's not that messy.
So here it is.
I'll copy this thing.
I also can write
this as delta h nlp.
You recognize that thing.
So this will be the sum
over p of delta h nlp.
The sine, I will
take care of it,
times that thing over
there, 1 over ep0 minus en0.
That's another minus sign
that cancels this sign here,
and I have here the sum over
k equals 1 to n, delta h, pmk,
ak0 plus en2 al0 equals 0.
OK, so what do we do now?
Just rewrite it one more time
and it will all be clear.
So I'll write the k outside,
k big parentheses, p.
For the first term I'm
going to pull the k sum out
and we'll sum over p first.
So what is it?
Delta h nlp delta h pmk.
Those were the two things here.
And then we have
the denominator.
I'll change a sign
of the first term.
I will change its sign so that
the second term looks more
like an eigenvalue.
So I changed sign
here by changing
the order of things
in the denominator,
and then I put minus en2.
I don't have a sum
over k here, but we
can produce one by writing
delta kl ak0 equals 0.
Look, we got our answer.
If I call-- invent
a matrix mlk2, which
is precisely delta h nlp
delta h pmk over en minus ep
is sum over p.
This is a-- you see, you
sum over p, n is irrelevant,
you have a matrix here in lmk.
That's why we call it mlk.
And therefore this
whole equation
is the sum from k equals 1
to n of mlk2 minus en2 ak0--
ml2 en2 delta lk ak0 equals 0.
Or, if you wish, it's
just a matrix equation
of the form m2 minus en times
the identity on a vector
equals 0.
And this is an
eigenvalue equation.
So finally, here is the answer.
Let's say you compute this
matrix of order lambda squared
with this perturbation
and that is the matrix
that if you diagonalize, you
find the second order energy
corrections and you
find the eigenvectors.
And if you found
the eigenvectors,
you found the good basis.
So the problem has been
solved at this stage.
You now know that if
you fail to diagonalize,
to break the perturbation
of the first order,
you will have to diagonalize
a second order matrix.
And once you diagonalize it,
you now have the good basis,
and it will happen that if
all the levels get split
at second order,
we can calculate
fairly easily the rest of the
pieces of the [? states. ?]
So we'll leave it.
There you'll complete some
details, more elaborations
of that in the exercises, and
we'll go to hydrogen atom next.
