WEBVTT

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PROFESSOR: All right.

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So let's get now the state n1.

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So I want to make
a general remark.

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You have an equation like this.

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And you want to solve it.

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It's a vector equation.

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Operator and a vector
equal number and a vector--

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a more operator and a vector.

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To make sure you
have solved it, when

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you have a vector equation
you must make sure

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that every component-- you
can write a vector equation

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in the form vector equals zero.

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And then you must make sure that
every component of the vector

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is zero.

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What we did here
is we found what

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happens when I look at
the component along n0.

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And I figure out that,
whoops, this equation,

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when I look at the
component along n0,

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tells me what the energy is.

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So the rest of the
information of this equation

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arises when I look at
it along the components

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on the other states, not n0 but
the k states that we introduced

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from the beginning, the k0's
that run from one to infinity.

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So what we're
going to do is take

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that original--
this second equation

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and form k0 h0 minus Em0 m1 1
is equal to k0 Em1 minus delta H

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n0.

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So I now took the same equation
and I put it in a problem

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with k0.

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And I say, look, k
will be different

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from n, because
when we put k equal

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to n that already we've done.

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And we've learned all about it.

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And, in fact, n1 didn't appear.

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The state that we wanted
didn't appear at all.

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So now we do this
with arbitrary k.

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And we need to figure
out what this gives.

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So you have to look
at these things

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and try to remember a little
of the definitions with both.

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So h0, we know what
it gives from k0.

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It gives you a number,
the energy of that state.

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So this is another number.

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So that's great.

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This simplifies this
to ek0 minus En0

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times the overlap of k0 with n1.

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That's the left hand side.

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How about the right hand side?

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All right.

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Let's see what this is.

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First term.

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The En1 is a number,
so I must ask

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myself is what happens
when k0 meets n0?

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Well, those are our
original orthonormal states.

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And we said that k
is different from n.

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So this term is 0 with an En1.

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This is a number.

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And these two states
are orthogonal.

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So this term gives you a 0,
not because this number is 0,

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but because the overlap is 0.

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And I get here
minus k0 delta H n0.

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And it's good
notation to call this,

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to save writing, delta Hkn
It's a good name for it.

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It's the matrix k--

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the kn-th element of the matrix
delta H. And this is a number

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so I can solve k0 n1
is equal to minus delta

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Hkn divided by Ek0 minus En0.

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And this is true for
every k different for n.

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And here we find, for the first
time, our energy denominators.

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These energy denominators
are the things

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that are going to make life
interesting and difficult.

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And it answers the
question already

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that if you had
degenerate states,

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there would be some
k state that have

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the same energy as this one.

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And this blows up.

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And this is unsolvable
for this component.

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So you start
getting difficulties

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if you have degeneracies.

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As long as every k state--

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all the other states
of the spectrum

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have different energy from En,
nevermind if the other states

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are degenerate.

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They're not degenerate
with the state you care.

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You care just about one
state now, the n-th state.

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And if that's nondegenerate, all
these denominators are non-zero

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and you're OK.

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So here is the solution
for this thing.

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Now I can write the expressions
for the state and the energy.

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So let me do it.

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So I have this n1 like that.

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Now you can say the following.

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Let me do this very
deliberately first.

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n1 is equal to the sum
over all k of k0 k0 n1.

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This is the resolution
of the identity formula.

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That's the unit operator.

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You can always do that.

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And now you know that the
state n1 is orthogonal to n0.

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So this becomes the sum
over k different from n,

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because for k equal to n, these
are orthogonal of k0 k0 n1.

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And that's what we
calculated here.

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So what did we get?

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Therefore, the state n1, I can
substitute what we had there.

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It's the sum from k different
from n of k0 delta Hnk over Ek0

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minus En0.

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That's n1.

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I should have a minus sign.

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The minus sign is
there at the state n1.

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So the state n1 is a
complicated correction.

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It gets a little component
from every other state

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of the spectrum.

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And the coefficient depends
on the matrix element

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of your state with the state
you're contributing with.

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So you have the state n and
all the other states here.

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The amount of this state k
that enters into the correction

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is proportional to the matrix
element between n and k.

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If the matrix element is 0, that
state does not contribute here.

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And then there is the
energy denominator as well.

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So we're getting to the
end of this calculation.

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There's one more thing one
can do, which is to find--

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so I'm starting to
wrap up this, but still

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an important step
what we have to do.

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I'll get the second
order energy correction.

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What is our second
order energy correction?

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Our second order
energy correction

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can be found from the formula
on that blackboard, En2.

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We already found the first
order energy correction,

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which I happened to have
erased it right now.

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It was there.

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En2 is obtained by doing
n0 delta H times n1, which

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we already know.

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So I must do n0 delta H on that.

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So look what you get.

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You get minus the sum
over k different from n.

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Think of putting the n0 and the
delta H, they're all together.

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It's a [INAUDIBLE] so far.

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It's a delta H and n0.

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IT should go into n1.

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But the only state in n1 is k0.

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So here we have k0.

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And then we have delta
Hnk over Ek0 minus En0.

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OK.

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A little bit of work.

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So what is this?

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This is another matrix element.

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This is the matrix--

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OK.

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I'm sorry.

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Here do I have a mistake?

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Oh, yes, I have kn.

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I copied it wrong.

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It's kn.

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Yes.

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Yes.

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So here I have delta Hnk.

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but delta Hnk is this.

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If you complex conjugate--

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if you complex
conjugate delta Hkn,

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complex conjugate is k
delta H n complex conjugate,

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which changes the order.

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n delta H, which
is her mission k.

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And that's delta Hnk.

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So delta Hnk is equal
to delta Hkn star.

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And therefore the second
order energy correction

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has a nice formula.

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En2 is equal to minus the
sum over k different from n.

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Delta Hnk, which is the
star of that times this one,

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so you get delta Hnk absolute
value squared divided by Ekn.

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Ek0 minus En0.

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So we've done a lot of work.

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We've written the perturbation.

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Here is the answer.

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So far we have n of lambda
equal n0 plus lambda n1.

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n1 has been calculated.

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Energy is En0 plus lambda En1.

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That was calculated
what was just

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delta H in this state
plus lambda squared

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En2, which we have calculated.

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So this is as far as we will do
for nondegenerate perturbation

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theory.

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But we have found rather
interesting formulas.

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And we're going to spend
half of its lecture trying

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to understand them better.