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PROFESSOR: We have a problem.

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The general perturbation
theory, again, end states.

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But this time, the
degeneracy is not broken.

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So degeneracy not
lifted at first order.

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Surprisingly, this subject
is, as you can see,

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you have to do things with care.

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It's not in any of the
textbooks that I know.

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In fact, I don't
know of any place

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where it is discussed in
detail or discussed anywhere,

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even though this result is
pretty useful and you need it,

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and I think people
improvise when they do this.

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Even this discussion
that with it here is not

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in any of your textbooks.

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In [INAUDIBLE],, the
advanced quantum mechanics,

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there's some discussion
and mentions this factor

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and how subtle it is, but
doesn't quite do everything

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till the bitter end.

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This case is harder.

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This is simple compared to
what we're going to do now.

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But we're going to sketch it
so that you see the picture.

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We're going to do
everything, but we're

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going to try to keep
the picture clear

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because it gets sophisticated.

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So what's the situation
that is happening.

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Physically, you have lambda
here, and you have the state.

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And to first order,
they don't split.

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So there are four states.

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And what we will assume is
that the first order, they

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don't split.

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So they sort of come together.

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And then to second
order, they will split.

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We will assume that the
second order, they will split.

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If they don't split the second
order, things get complicated.

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I'm actually curious about that.

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But I don't know of
any physics example

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that they don't split to
first nor to second order.

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You will have in the
homework things that

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don't split to first order.

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That really happens often.

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Then to second order,
if you have four states

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and then split, means
that the second order

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they'll go like this.

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And they'll start
looking like that

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maybe, something like that.

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So the slope is the
same to begin with,

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but then they split.

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OK.

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So degeneracy not
lifted at first order.

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So what is the problem here
that the degeneracy is not

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lifted to first order?

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The problem is physically that
this concept of a good basis,

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that basis that corresponds
to the continuous changing

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eigenstates.

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You did the perturbation
theory to first order,

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and you still don't know
what is the good basis.

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Because the good
basis is the one

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that the states
change continuously.

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So suppose that we had
an example where we've

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had in two dimensions the
conventional eigenstates,

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1 0, and 0 1, but the good
basis was the 1 1 and 1 minus 1.

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And you would say,
oh, they jumped.

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But you have to choose
the right basis.

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You do first order to
generate perturbation theory.

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And if the states split,
you have the good basis.

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But if they don't
split, you have

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no idea what is the good basis.

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So in that sense, you're
trying to go high,

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because we did first order.

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Now, you have to go to second
order of perturbation theory

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to see if you can find
what the good basis is.

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So we will have to diagonalize
something to second order

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to find the good basis.

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And once you have
the good basis,

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you sort of have to
start from the beginning.

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Now, I have a good basis.

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Let's do the perturbation
theory nicely.

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So if the degeneracy
not broken, we

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have that n0k delta H
and 0l is en1 delta kl.

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It is diagonal still.

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You do have a basis that
diagonalizes the perturbation.

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You still diagonalize it.

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But once you diagonalize it, all
the eigenvalues are the same.

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There's no account for ml here.

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All of them are the same.

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So what is the catch?

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The catch is that you now
need to find good eigenstates.

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So what is the good basis?

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We'll write it in
the following way.

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I will form linear combinations
k equal 1 to n of the n0k.

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And I will put
coefficients a, k, 0 here.

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And I will say, look, if you
give me some coefficients a, k,

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0, I know how to superimpose
now the N states.

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And let's hope that that's
one good basis vector.

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You know, I've written one
good basis vector if a, k

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it's a column vector of numbers.

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It gives me what is the
superposition of those

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degenerate states
that makes one vector.

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So I'll call this
a vector PSI 0.

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So I think of this a, k
not as unknown numbers,

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unknown vector that
specifies a good state.

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And moreover, if
I do things right,

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I should calculate these things.

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And I should find that my
equations end up giving me

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N of those good vectors.

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You see, if I specify a column
vector of constants here,

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I get one vector here,
a linear superposition.

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But I should get n
linear superpositions.

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Because at the end
of the day, there

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has to be n good
vectors in that basis.

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So in some sense, I hope
for an a0 with an index Ik,

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with I going from 1 to n,
in which this by varying I

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provides me k N column
vectors, each one of which

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gives me a state.

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So this gives me one state.

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If I had N of
those coefficients,

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I would get n state.

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So that's my hope.

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So suppose we get one
of those good states,

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how should it look in
perturbation theory?

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Well, the state PSI lambda is
going to be the state PSI 0.

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That's, in fact, the
zeroth order state.

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We don't know
those coefficients.

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That's our ignorance about
the zeroth order states now.

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PSI 0 plus lambda PSI 1 plus.

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And the energy's en lambda
is en 0, the 0 energy,

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plus the first order
energy correction

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plus the second order
energy corrections.

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And from this, the
Schrodinger equation

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that as usual says that
h of lambda PSI of lambda

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is equal to En of
lambda PSI of lambda

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gives you these equations that,
by now, you're pretty familiar.

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And we'll need a couple of them.

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So I will write them again.

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To order lambda 0, you get H0
minus En0 on PSI 0 equal 0.

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That's a trivial
equation because PSI 0

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is made up of states, all
of which have energy En0.

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So that's always a
trivial equation.

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Then to order lambda,
you have H0 minus En0

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on the first state.

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We've done this
already a few times.

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So I'm sure you don't have
trouble believing this.

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And to order lambda squared,
the last one that I'll write,

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En0 PSI 2 is equal to En1 minus
delta H PSI 1 plus En2 PSI 0.

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OK.

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Kind of tire of writing
these equations.

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OK.

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So let's see what we should do.

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So we can start
calculating things.

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And the way I'm going to do
it, for the benefit of time,

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is that I will
indicate the steps

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and what you do and
skip the algebra.

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It will all be in the notes.

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And in fact, the whole
calculation will,

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if you want to go to
the very end of it

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and exploit it to
a maximum, there's

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lots of things one can do.

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And some of that will
be left for homework.

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So our idea here is to
give you the road map

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and for you to see how
we're doing things.

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So let me see here.

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We can do a few things
with these equations.

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So the first
equation is trivial,

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so we don't bother with it.

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And the second
equation, it's kind of

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interesting to see what it does.

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We can find from this
equation a confirmation

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of what first order
perturbation theory does.

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If we inject here one
of the states in VN,

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those N0l's, you get
the statement, again,

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that delta H should
diagonalize the perturbation.

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So I think I'll skip that.

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We'll discuss it in the notes.

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But let's do the case when
this equation gives something

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non-trivial.

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So let's calculate p acting
on the order lambda equation.

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So when we had N0l, say, on
the order lambda equation,

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that is kind of familiar.

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We've already done it a
few times, so I'll skip it.

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But let's do the p
acting on this equation.

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So p0 H0 minus
En0 times PSI 1 is

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equal to p0 En1
minus delta H PSI 0.

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H0 is a known Hamiltonian.

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And this we know the energy.

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So this gives you Ep0 minus
En0 on p0 PSI 1 is equal to.

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Here is a state in v-hat.

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And this is a state in
the degenerate subspace.

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We wrote it up there.

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That's our answers.

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We don't know even the
zeroth order states.

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That's our problem.

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But this term is
orthogonal to that.

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So across the number,
it gives us 0.

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On the other hand,
here what do we get?

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We get minus the sum.

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Think of this state
being the sum that we

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wrote in the previous thing.

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The sum from k equal 1
to n delta Hp nk a0 k.

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I've substituted this formula
for the state PSI 0 in there.

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And now, we have
computed these overlaps

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because this is in principle
known in terms of a0.

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And now we have these
constants, these overlaps

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of the states outside
the general space

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with the first
correction and this part.

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So I'll write it this way.

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PSI 1 of v-hat is
equal to p ap1 p0.

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And what is this ap1?

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It's those coefficients
we're trying to find here.

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00:16:18,030 --> 00:16:22,220
So this is the component
of PSI 1 along p0.

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00:16:22,220 --> 00:16:26,130
So that component
should go here.

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And therefore, it's minus
1 over Ep minus En0 all.

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00:16:36,580 --> 00:16:48,701
And here we have sum from k
equals 1 to n delta H and Hp nk

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00:16:48,701 --> 00:16:49,201
ak0.

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OK.

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So we're pretty close to
our goal, believe it or not.

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00:17:05,800 --> 00:17:08,089
So what have we achieved?

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00:17:08,089 --> 00:17:11,140
Let's see, again, where we are.

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00:17:11,140 --> 00:17:14,290
We have no idea what
are the good states,

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so we introduced a0.

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a0 parameterizes our
ignorance of the good states.

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So we know, however, that
the first order correction

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doesn't lift the degeneracy.

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And therefore, I don't
have to recalculate that

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from the second equation.

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I just find the piece
of PSI 1 that is also

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determined from that equation.

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And now, we've done completely
the first order equation.

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And what did we learn?

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We learned that the first
order correction to the state

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is known if I know
the good basis.

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Because all this first order
correction to the state

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in the outside subspace in the
rest is determined if I know

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the a0's.

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But I don't know the a0's.

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00:18:06,800 --> 00:18:12,740
So we are calculating
things, but we have not

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solved our problem.

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But we run out of equations
with the order lambda.

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We've finished.

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No more lambda equation, so we
have to look into this equation

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now.