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PROFESSOR: Let us consider the
anharmonic oscillator, which

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means that you're taking the
unperturbed Hamiltonian to be

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the harmonic oscillator.

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00:00:18,910 --> 00:00:22,600
And now, you want to
add an extra term that

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will make this anharmonic.

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Anharmonic reflects the fact
that the perturbations are

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00:00:31,300 --> 00:00:35,200
oscillations of the system
are not exactly harmonic.

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00:00:35,200 --> 00:00:39,190
And in the harmonic oscillator,
the energy difference

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between levels is
always the same.

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That's a beautiful property
of the harmonic oscillator.

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That stops happening in
an anharmonic oscillator.

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The energy differences can vary.

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So the things, if you have
a transition from one level,

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first level to the ground state,
or second level to the ground

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state, one is not the
harmonic of the other

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because they're not exactly
twice as big as each other.

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So let's try to add an x to
the 4th perturbation, which

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is intuitively very clear.

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You have a potential.

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And you're adding
now an extra piece

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that behaves like x to the 4th.

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And it's going to make this
potential blow up faster.

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Now, in order to
get the units right,

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we need the length scale.

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There's a length scale d
in the harmonic oscillator.

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We're going to use it.

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One way to derive
that length scale is

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to recall from units that p,
a momentum, has units of h

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divided by length.

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So b squared has units of
h squared over d squared.

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There's an m.

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So that's an energy.

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And this is an energy too.

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So we can set it equal to
m omega squared d squared.

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From where we get, for
example, as 1 over d to the 4th

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is equal to m squared omega
squared over h squared.

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And d squared is equal
to h over m omega.

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So that's a length
scale, something

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with units of length in
the harmonic oscillator.

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So if we want to add
the perturbation that

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is x to the 4th, we'll
add a lambda delta

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H that is going to
be a lambda, which

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is unit freedom, something
that has units of energy.

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So you can put h bar
mega has units of energy.

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And then you can
put the operator x

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to the 4th divided
by d to the 4th.

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It's a perturbation
with units of energy.

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This is good.

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So there's a couple of
ways of thinking of it.

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You may remember that in
the harmonic oscillator

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x, the operator x, was given
by the square root of h

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over 2m omega, a plus a dagger.

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So this is d times a plus a
dagger over square root of 2.

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So this perturbation lambda
delta H is equal to lambda.

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H bar omega would be here.

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H bar omega.

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Now x/d is this a plus a dagger
divided by square root of 2.

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So you have a 4a plus
a dagger to the 4th.

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So that's lambda delta H.
That's your perturbation.

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So what do we want to do
with this perturbation?

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It's a nice perturbation.

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We want to compute
the corrections

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to the energy,
corrections to the states,

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and see how they go.

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We can do that easily.

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Let's compute first that
corrections to the ground state

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energy.

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These are calculations
that are not difficult,

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but they take some care.

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As any calculation,
it's based on just

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getting a lot of numbers right.

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So what is the ground
state energy correction?

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So we had a series of
formulas that I just erased.

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But this index k0
for our state, we

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can think of k as being the
eigenvalue of the number

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operator.

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So we'll get n.

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And remember in the
harmonic oscillator,

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we label states by 0, 1, 2, 3.

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And that's the eigenvalue
of the number operator

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in the harmonic oscillator.

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So the ground state energy
would correspond to 0,

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has a first order
correction that

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would be given by the
unperturbed ground state

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and delta H 0.

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So what is that?

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Delta H is h omega over 4.

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And we have 0 a plus
a dagger to the 4th 0.

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Easy enough.

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That's what we have to do.

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Now, the evaluation
of that matrix element

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is a simple exercise,
the kind of things

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you've been doing before.

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You have to expand and just
use the aa dagger computation

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relation that this aa
dagger is equal to 1.

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You said repeatedly
a kills the vacuum.

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a dagger kills the other
vacuum on the left.

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Please do it.

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If you feel you're
out of practice,

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the answer is the number 3 here.

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So e01 is equal to
3/4 h bar omega.

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So how does that
tell you anything?

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Well, better use
the full formula.

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So e0 of lambda is supposed to
be equal to the ground state

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energy unperturbed,
which was h omega over 2

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plus lambda times the first
order correction, which

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is 3/4 h bar omega plus
order lambda squared.

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So e0 of lambda
is h omega over 2,

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1 plus 3/2 lambda plus
order lambda squared.

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That's what we got.

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And that makes nice sense.

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Your ground state energy you
knew used to be h omega over 2.

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If you introduce
this perturbation,

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you don't expect it
to jump abruptly.

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It's going to grow
up with the lambda

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to the 4th term with the
x to the 4th term into 3/2

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lambda here, a small correction.

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So far, so good.

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What takes a bit more work
is doing a higher order

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correction.

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So let's do the
next one as well.

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So for this second order
correction, we have a formula.

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And let's see what it tells us.

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It tells us that the second
order correction to the energy

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is minus the sum over all states
that are not the ground states.

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So it's k different from 0.

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That's good enough.

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Delta H 0k squared
over Ek0 minus e00.

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So potentially, it's
an infinite sum.

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And that's the kind of thing
that is sometimes difficult.

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In many examples, you
have infinite sums.

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Sometimes you can
do an approximation,

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say a few terms is
all you need to do.

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But in this case,
happily, it's not

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going to be an infinite sum.

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So what do we have?

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Delta H0k is what we
need to calculate.

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The energy differences, we know.

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These are the unperturbed
harmonic oscillator energy

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differences.

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What we need is this object,
which is h bar omega over 4.

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Remember, what is delta H there
in that box formula there.

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And we have 0 a plus a
dagger to the 4th K, where

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K is a state with number K.

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If you wish, you could
say K is a dagger

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to the K over square root of K
factorial acting on the ground

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state.

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But sometimes you
don't need that.

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You actually don't need
to put the value of K.

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So our challenge here is to
see which values of K exist.

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For that, you can
think of this part

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of the term, the a plus a
dagger to the 4th acting

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on the vacuum.

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This is like, in
a sense, this term

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is nothing else but in saying
what happens to the wave

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function when you
multiply it by x to the 4,

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to the ground state
wave function.

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That's physically what
you're doing here.

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You're multiplying an x to
the 4th times the ground

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state wave function.

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So it should give you
an even wave function.

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So you would expect that x
to the 4th multiplied by 5 0

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should be proportional to 5
0, but maybe 5 2 and 5 4th.

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Should not have the odd ones
because this function is not

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odd.

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It's even.

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So as you express x to the 4th
5 0 in terms of the other ones,

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it should be that.

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So indeed, again, I ask you
to do a little computation.

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If you do 0 a plus a dagger
to the 4th, this is equal to--

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you can show.

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It gives you 4 factorial
square root the state 4

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plus 6 square root of
2 times the state 2

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plus 3 times times
the ground state.

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OK.

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So we have that look
let's continue therefore.

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Given that [INAUDIBLE] has
given us three states--

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one proportional to the
vacuum, one proportional

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to the state with occupation
number 2, and one with a state

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of occupation number four--

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the values of k
that are relevant,

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since k is different
from 0, are only 2 and 4.

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0 doesn't matter.

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And only 2 and 4
can couple here.

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So what do we get?

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00:14:12,240 --> 00:14:20,650
Delta H0 4 would be, from
this formula, h omega

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00:14:20,650 --> 00:14:25,570
over 4 times this inner
product when k is equal to 4.

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00:14:25,570 --> 00:14:31,940
So you get here square
root of 4 factorial,

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00:14:31,940 --> 00:14:38,130
which is h omega over 2
times square root of 6.

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00:14:38,130 --> 00:14:46,290
Delta H02 is equal
to h omega over 4.

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And the overlap of
this state with 2,

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which is 6 square root of 2.

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So this is h omega 3
square root of 2/2.

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OK.

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00:15:01,590 --> 00:15:04,140
So that's your first step.

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00:15:04,140 --> 00:15:07,180
When you have to
compute a perturbation,

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00:15:07,180 --> 00:15:10,650
you have to compute all the
relevant matrix elements.

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00:15:10,650 --> 00:15:13,290
And you have to think
what you can get.

199
00:15:13,290 --> 00:15:16,680
And only k equal 2
and 4 happen, so this

200
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is the sum of two terms.

201
00:15:19,780 --> 00:15:23,490
So the second order
correction to the energy

202
00:15:23,490 --> 00:15:31,110
is minus delta H0 2
squared divided by E2 0

203
00:15:31,110 --> 00:15:44,310
minus E0 0 minus delta H0 4
squared over E4 0 minus E0 0.

204
00:15:50,470 --> 00:15:52,270
So what is this?

205
00:15:52,270 --> 00:16:02,210
It's minus delta H0 2 squared
is h bar omega squared.

206
00:16:02,210 --> 00:16:05,470
9 times 2/4.

207
00:16:05,470 --> 00:16:08,380
9 times 2/4.

208
00:16:08,380 --> 00:16:10,540
And you divide by
this difference.

209
00:16:10,540 --> 00:16:16,230
And the difference between
E2 and E0 is 2 h bar omega.

210
00:16:16,230 --> 00:16:19,010
There's an h bar omega
difference every state.

211
00:16:19,010 --> 00:16:23,040
So you go from E0 to
E1 to E2, two of them.

212
00:16:23,040 --> 00:16:36,490
Minus delta H0 4 squared, which
is H omega squared times 6/4

213
00:16:36,490 --> 00:16:40,880
over 4 h bar omega.

214
00:16:40,880 --> 00:16:46,465
Well, this gives you
minus 21/8 h bar omega.

215
00:16:50,560 --> 00:16:56,740
And therefore, the
energy, this is the E0 2,

216
00:16:56,740 --> 00:17:05,560
so you can go back here
and write E0 lambda is

217
00:17:05,560 --> 00:17:08,619
equal to H omega over 2.

218
00:17:08,619 --> 00:17:12,220
1 plus 3/2 lambda.

219
00:17:12,220 --> 00:17:15,109
And then the next term
that we now calculated

220
00:17:15,109 --> 00:17:20,859
is 21/4 lambda squared.

221
00:17:20,859 --> 00:17:21,819
Very good.

222
00:17:21,819 --> 00:17:22,990
We work hard.

223
00:17:22,990 --> 00:17:26,859
We have lambda squared
correction to this energy.

224
00:17:26,859 --> 00:17:29,930
Now, this is not really trivial.

225
00:17:29,930 --> 00:17:33,310
There's no analytic way
to solve this problem.

226
00:17:33,310 --> 00:17:37,190
Nobody knows an analytic
way to solve this problem.

227
00:17:37,190 --> 00:17:43,540
So this is a nice
result. It's so famous

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00:17:43,540 --> 00:17:48,080
the problem that people have
written hundreds of papers

229
00:17:48,080 --> 00:17:49,870
on this.

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00:17:49,870 --> 00:17:53,730
The first people that did
a very detailed analysis

231
00:17:53,730 --> 00:17:56,320
was Bender and Wu.

232
00:17:56,320 --> 00:18:04,170
And they computed probably
100 terms in the series or so.

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00:18:04,170 --> 00:18:12,220
The next term turns out
to be 333/8 lambda cube.

234
00:18:12,220 --> 00:18:23,060
Followed by 30,855/64 lambda
to the 4th plus ordered lambda

235
00:18:23,060 --> 00:18:23,840
to 5th.

236
00:18:29,470 --> 00:18:36,060
Funnily, these people also
proved that this series

237
00:18:36,060 --> 00:18:39,180
has no radius of convergence.

238
00:18:39,180 --> 00:18:41,400
It never converges.

239
00:18:41,400 --> 00:18:44,060
It just doesn't converge at all.

240
00:18:44,060 --> 00:18:47,310
You would say, oh,
what a waste of time.

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00:18:47,310 --> 00:18:52,950
No, it's an asymptotic
expansion, this series.

242
00:18:52,950 --> 00:18:56,760
So it can be used.

243
00:18:56,760 --> 00:19:00,140
So it's a very funny thing.

244
00:19:00,140 --> 00:19:02,810
Series sometimes don't
converge so well.

245
00:19:02,810 --> 00:19:10,340
And this one is a peculiarly
non-convergent series.

246
00:19:10,340 --> 00:19:14,660
So what does it mean only that
it's an asymptotic expansion?

247
00:19:14,660 --> 00:19:19,010
Basically, the expansion
for the factorial

248
00:19:19,010 --> 00:19:21,890
that you use in statistical
mechanics all the time,

249
00:19:21,890 --> 00:19:26,330
Sterling's expansion, is
an asymptotic expansion.

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Quantum electro dynamics
with Feynman rules

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00:19:29,950 --> 00:19:34,070
is an asymptotic expansion.

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Very few things actually
converge in physics.

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That's the nature of things.

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00:19:41,720 --> 00:19:44,330
What it means is that
when you take lambda,

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00:19:44,330 --> 00:19:49,730
for example, to be 1 in 1,000,
a very small number, then

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00:19:49,730 --> 00:19:50,870
this is 1.

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00:19:50,870 --> 00:19:52,290
This will be very small.

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00:19:52,290 --> 00:19:54,650
This will still be small.

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00:19:54,650 --> 00:19:55,700
This will be small.

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00:19:55,700 --> 00:20:00,150
And eventually, the terms go
smaller, smaller, smaller,

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00:20:00,150 --> 00:20:01,970
smaller, smaller, smaller.

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00:20:01,970 --> 00:20:06,030
And eventually,
start going up again.

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00:20:06,030 --> 00:20:08,420
And if you have an
asymptotic expansion,

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you're supposed to
stop adding terms

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until they become the smallest.

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00:20:12,770 --> 00:20:16,730
And that will be a good
approximation to your function.

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00:20:16,730 --> 00:20:19,350
So this can be used.

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00:20:19,350 --> 00:20:20,330
You can use it.

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00:20:20,330 --> 00:20:22,550
You can do numerical
work with the states

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00:20:22,550 --> 00:20:25,220
and compare with
asymptotic expansion.

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00:20:25,220 --> 00:20:28,530
And it's a very nice thing.

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Another thing is that you can
compute the first correction

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00:20:32,570 --> 00:20:34,580
to the state.

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00:20:34,580 --> 00:20:36,470
I will leave that
as an exercise.

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00:20:36,470 --> 00:20:40,520
And in the notes, you
will find the expression

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00:20:40,520 --> 00:20:42,920
of the first correction
to this state.

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00:20:42,920 --> 00:20:46,190
It's not much work
because, after all, it

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00:20:46,190 --> 00:20:49,920
involves just this kind
of matrix elements.

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00:20:49,920 --> 00:20:53,060
So it's basically using
the matrix elements

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00:20:53,060 --> 00:20:55,630
to write something else.