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PROFESSOR: Suppose you define
now, one state called phi 1

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as a dagger acting on phi 0.

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You could not define
any interesting state

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with a acting on phi 0
because a kills phi 0,

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so you try phi 0 like this.

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Now you could ask, OK,
what energy does it have?

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Is it an energy eigenstate?

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Well it is an energy eigenstate
if it's a number eigenstate.

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And we can see if it's a
number eigenstate by acting

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with the number operator.

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So N phi 1 is equal
to N a dagger phi 0.

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OK.

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Here comes trick.

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Maybe it's too much to even
call it a trick, number one.

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This thing you look
at it and you say,

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I want to sort of simplify
this, learn something about it.

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If this is supposed to be
an eigenstate of N hat,

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I have to make it
happen somehow.

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Now n hat kills phi 0.

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So if I would have a term a
dagger times N hat near phi 0,

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it would be 0.

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So I claim, and this
is a step that I

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want you to be able
to do also quickly,

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that I can replace this by
the commutator of these two

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operators.

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The product is replaced
by the commutators.

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Why?

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Aren't products simpler
than commutators?

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No.

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We have formulas
for commutators.

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And products are, in
general, more complicated.

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And why is this correct?

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And you say, well, it is correct
because this has two terms.

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The term I want
minus a dagger N hat.

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But the term a dagger N hat is
0 because N hat kills phi 0.

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So I can do that because
this is N dagger a hat, which

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is what I had, minus a
hat dagger N on phi 0.

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And this term is 0.

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So you would have put
a 2 here or a 3 here,

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or any number even.

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But the right one to
put is the commutators.

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So that's this.

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And now this commutator
is already known.

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That's why we computed it.

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It's just a dagger.

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So this is a dagger phi 0,
and that's what we call phi 1.

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So N hat on phi 1 is phi 1.

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N hat has eigenvalue 1 on phi 1.

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So N is equal to 1.

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That's the eigenvalue.

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It is an eigenstate.

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It is an energy eigenstate.

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In fact how much
energy, E, is h bar

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omega times N, which is 1, plus
1/2, which is 3/2 h bar omega?

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And look what this is.

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This is the reason this is
called a creation operator.

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Because by acting on the ground
state, what people sometimes

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call the vacuum, the
lowest energy state,

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the vacuum is called the
lowest energy state, by acting

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on the vacuum you get a state.

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I mean, you've created
a state, therefore.

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How is this concretely done?

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Remember you had
phi 0 of x, what

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it is, and a dagger over there
is x minus ip over m omega.

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So this is x minus--

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or minus h bar
over m omega d dx.

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So you can act on it.

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It may be a little messy.

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But that's it.

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It's a very closed
form expression.

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Now, phi 0 was defined, the
ground state such that it's

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a normalized state.

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This means the integral of
phi 0 multiplied with phi 0

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over x is 1.

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That's how we had
the ground state.

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You could ask, if I've
defined phi 1 this way,

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is simply normalized?

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So I'll try it.

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And now you could say, oh, this
is going to be a nightmare.

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Normalizing phi 0
is difficult. Now I

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have to act with a
dagger, which means act

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with x, take derivatives.

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It's going to grow twice as big.

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Then I'm going to have to
square it and integrate it.

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It looks very bad.

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The good thing is those with
these a's and a daggers,

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you have to compute
anything, pretty much.

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See how we do it.

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I want to know how much
is phi 1 with phi 1.

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Is it 1?

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And it's normalized or not?

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Then I say, look, phi 1 is a
dagger phi 0, a dagger phi 0.

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So far so good.

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But I just know
things about phi 0.

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So let's clear up one phi 0.

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At least I can move
the a dagger as an a.

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So this is phi 0
a a dagger phi 0.

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Can I finish the
computation in this line?

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Yes, I think we can.

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Phi 0.

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a with a dagger,
same story as before.

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a would kill phi 0.

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So you can replace
that by a commutator.

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Commutator of a
with a dagger phi 0.

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But the commutator of
a with a dagger is 1,

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so this is phi 0 phi
0 and it's equal to 1.

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Yes, it is properly normalized.

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So that's the nice thing
about these a's and a daggers.

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Just start moving them around.

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You have to get practice.

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Where should you move it?

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Where should you put it?

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When you replay something by
a commutator, when you don't.

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It's a matter of practice.

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There's no other way.

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You have to do a lot
of these commutators

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to get a feeling
of how they work

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and what you're supposed to do.

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Let's do another state.

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Let's try to do phi 2.

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I'll put a prime because
I'm not sure this is going

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to work out exactly right.

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And this time, I'll put an a
dagger a dagger on the vacuum.

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Two a daggers, two creation
operators on the vacuum.

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And now I want to see if
this is an energy eigenstate.

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Well, this is a dagger
squared on the vacuum.

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So let's ask, is N hat--

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is phi 2 prime an
eigenstate of N hat?

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Well I would have N hat on
a dagger squared on phi 0.

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Again, by now you know, I should
replace this by a commutator

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because N hat kills the phi
0, so N hat with a dagger

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squared phi 0.

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And that commutator
has been done.

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It's two times a
hat dagger squared,

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two times a dagger squared
on phi 0, which is 2 phi 2.

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That's what we call
the state phi 2 prime.

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I'm sorry.

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So again, it is an
energy eigenstate.

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Is it normalized?

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Well, let's try it.

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Phi 2 prime phi 2 prime is
equal to a dagger a dagger.

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Let me not put the hats.

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I'm getting tired of them.

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a dagger a dagger phi 0.

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Now I move all of them.

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This a dagger becomes
an a, the next a dagger

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becomes an a here.

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So this is phi 0 a a a
dagger a dagger phi 0.

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Wow, this looks a
little more complicated.

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Because we don't want to
calculate that thing, really.

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We definitely don't want
to start writing x and p's.

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But, you know, you decide.

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Take it one at a time.

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This a is here and wants
to act on this thing.

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And then this other
a will, but let's

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just concentrate on the
first a that wants to act.

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a would kill phi 0, so we
can replace this whole thing

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by a commutator.

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So this is phi 0.

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The first a is still there,
but the second, we'll

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replace it by the
commutator, this commutator.

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I've replaced this product, the
product of a times this thing,

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by the commutator of
those two operators.

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And then I say, oh
look, you've done that.

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a with a dagger to the k
is k a dagger k minus 1.

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So I'll write it here.

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This will be a
factor of 2 phi 0 a.

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And this is supposed to be
now a dagger to one power

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less, so it's just
a dagger phi 0.

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So this is supposed
to be 2a dagger.

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So that's what I did.

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And again, this a
wants to act on phi 0

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and it's just
blocked by a dagger,

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but you can replace
it by a commutator.

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a with a dagger phi 0.

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And this is therefore a 1,
so this whole result is a 2.

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So this phi 2 prime, yes, it
is the next excited state.

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Two creation operators
on the ground state.

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Energy and eigenvalues too.

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You had N equal zero
eigenvalue for the ground state

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1 for phi 1, 2 for phi 2 prime.

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But it's not
properly normalized.

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Well, if the normalization gives
you 2, then you should define

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phi 2 as 1 over the
square root of 2

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a dagger a dagger on phi 0.

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And that's proper.

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So it's time to go general.

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The n-th excited state, we
claim is given by an a dagger

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00:13:49,090 --> 00:13:57,850
a dagger, n of them, acting on
phi 0 with a coefficient 1 over

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square root of--

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00:14:00,030 --> 00:14:02,510
we might think it's
n, but it's actually,

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you can't tell at this far--
this one is n factorial.

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That's what you need.

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That is the state.

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00:14:15,650 --> 00:14:19,040
And what is the
number of this state?

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What is the number
eigenvalue on phi n?

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00:14:24,660 --> 00:14:29,810
Well, it is 1 over square
root of n factorial.

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00:14:29,810 --> 00:14:43,360
The number acting on the a
daggers, the n of them, phi 0.

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00:14:43,360 --> 00:14:45,550
You can replace
by the commutator,

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00:14:45,550 --> 00:14:48,280
which then is 2 times already.

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00:14:48,280 --> 00:14:57,595
So it's N commutator with a
dagger to the little n phi 0

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00:14:57,595 --> 00:15:00,820
times 1 over square root of n.

203
00:15:00,820 --> 00:15:03,760
And how much is this commutator?

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Over there.

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00:15:05,500 --> 00:15:12,550
This is N times a
dagger to the n phi 0.

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00:15:12,550 --> 00:15:15,370
So between these
three factors, you're

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00:15:15,370 --> 00:15:19,640
still getting n phi to the n.

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00:15:19,640 --> 00:15:24,860
So the number for this
state is little n.

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00:15:24,860 --> 00:15:27,590
It is an energy eigenstate.

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00:15:27,590 --> 00:15:32,190
The N eigenvalue is little n.

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00:15:32,190 --> 00:15:37,260
And the energy is h bar omega.

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00:15:37,260 --> 00:15:46,100
The eigenvalue of N hat,
which is little n plus 1/2.

213
00:15:46,100 --> 00:15:56,310
So it is the energy
eigenstate of number little n.

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00:15:56,310 --> 00:16:00,060
This is the definition.

215
00:16:00,060 --> 00:16:04,850
And the last thing you may want
to check is the normalization.

216
00:16:04,850 --> 00:16:09,420
Let me almost check it here.

217
00:16:09,420 --> 00:16:10,410
No, I will check it.

218
00:16:10,410 --> 00:16:13,320
Let's say I think this
is a full derivation.

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00:16:13,320 --> 00:16:17,430
Phi n with phi n would
be two factors of those,

220
00:16:17,430 --> 00:16:27,510
so I would have 1 over n
factorial a dagger a dagger,

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00:16:27,510 --> 00:16:34,190
n of them on phi 0, a
dagger a dagger, n of them

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00:16:34,190 --> 00:16:37,100
again on phi 0.

223
00:16:37,100 --> 00:16:41,490
So then that's equal to
1 over n factorial phi

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00:16:41,490 --> 00:16:57,120
0 a a, lots of a's, n of them,
n a daggers, phi 0, like that.

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00:16:57,120 --> 00:16:59,860
That's what it is.

226
00:16:59,860 --> 00:17:03,760
We had to move all
the a daggers that

227
00:17:03,760 --> 00:17:08,319
were acting on the left
input of the integral,

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00:17:08,319 --> 00:17:12,040
or the inner product,
all the way to the right.

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And that's it.

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00:17:15,300 --> 00:17:19,690
So now comes this step.

231
00:17:19,690 --> 00:17:25,410
And I think you can
see why it's working.

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00:17:25,410 --> 00:17:29,460
Think of moving the
first a all the way here.

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00:17:29,460 --> 00:17:33,210
Well, you can replace the
first a with a commutator.

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00:17:33,210 --> 00:17:38,490
But that a with lots of a
daggers, with n a daggers,

235
00:17:38,490 --> 00:17:42,040
would give you a
factor of n, with n

236
00:17:42,040 --> 00:17:46,340
a daggers will give you a factor
of n times one a dagger less.

237
00:17:46,340 --> 00:17:51,460
So to move the first a,
there are n a daggers

238
00:17:51,460 --> 00:17:56,080
and you get one factor
of n from this a.

239
00:17:56,080 --> 00:18:01,270
But for the next a, there's
now n minus 1 a daggers,

240
00:18:01,270 --> 00:18:06,250
so this time you get a factor
of n minus 1 when you move it.

241
00:18:06,250 --> 00:18:10,330
From the next one, there's going
to be n minus 2 a daggers, so n

242
00:18:10,330 --> 00:18:11,680
minus 2.

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00:18:11,680 --> 00:18:16,840
All of them all the way up to
one, cancels this n factorial,

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00:18:16,840 --> 00:18:19,300
and that's equal to 1.