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PROFESSOR: Important
thing to do is to just try

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to understand one more thing.

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The creation and annihilation
operators-- what do they

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do to those states?

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You see, a creation operator
will I add one more a dagger,

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so somehow must change
phi n into phi n plus 1.

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A destruction operator with an a
will kill one of these factors,

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and therefore it will give
you a state with lower number

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of phi n minus 1.

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And we would like to know
the precise relations.

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So look at this.

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Let's do with an A on phi n.

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And we know it should be
roughly phi n minus 1.

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This is one destruction
operator, but we can do it.

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Look-- this is 1 over
square root of n.

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A times a dagger to the n phi 0.

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A with a dagger to the n phi 0,
we can replace by a commutator

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again.

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Commutator of a, a
dagger to the n phi 0.

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This is 1 over square
root of n factorial,

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and here we get a
factor of n times

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a dagger to the n minus 1 phi 0.

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You know, it's all a matter
of those commutators we

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on the left blackboard.

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But this state--
by definition, we

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have n square root
of n factorial.

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That's state, by definition,
is phi n minus 1 times

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square root of n
minus 1 factorial.

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See, by looking at this
definition and saying,

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suppose I have n minus 1, n
minus 1, this is phi n minus 1.

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So n minus 1 a
daggers on phi 0 is

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n minus 1 factorial square
root multiplied phi n minus 1.

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And now we can simplify this--

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square root of n
factorial and square root

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of n minus 1 factorial
gives you just a factor

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of square root of
n that with this n

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here, this square root
of n, phi n minus 1.

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Sop there we go--

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here is the first relation.

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A is really a lowering operator.

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It gives you an
eigenstate 1 less energy,

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but it gives it with a
factor of square root of n,

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that if you care
about normalizations,

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you better keep it.

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That factor is there because
the overall normalization

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of this equation was designed
to make the states normalized.

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Similarly, we can do
the other operation,

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which is what is a
dagger acting on phi n.

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This would be 1 over
square root of n factorial,

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but this time a dagger to
the n plus 1 on phi n, phi 0.

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Because you had already
a dagger to the n,

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and you put one more a dagger.

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But this thing is equal to what?

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This is equal to square root
of n plus 1 factorial times

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phi n plus 1.

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From the definition--

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I hope you're not getting dizzy.

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Lots of factors here.

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But now you see that the n
part of the factorial cancels,

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and you get that
a hat dagger phi

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n is equal to square root
of n plus 1 phi n plus 1.

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OK let's do an application.

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Suppose somebody asks
you to calculate example.

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The expectation
value of the operator

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x on phi n, the expectation
value of p on phi n.

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How much are they?

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OK.

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This, of course, in conventional
language, at first sight

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looks prohibitive.

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I would have to get those phi n
[? in ?] some Hermit polynomial

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hn, for which I don't know
the closed form expression.

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It's a very large
polynomial, jumps 2 by 2,

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there is exponentials, I
will have to do an integral.

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That's something that
we don't want to do.

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So how can we do it
without doing integrals?

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Well, this one's-- actually, you
can do without doing anything.

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You don't have to do integrals,
you don't have to calculate.

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The answers are kind
of obvious, if you

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think about it the right way.

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That's not the obvious part,
to think about the right way.

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But here it is.

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Look, what is this integral?

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This is the integral of x times
phi and of x, those are real,

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quantity squared.

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And the phi n's are either even
or odd, but the fight n squared

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are even.

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And x is odd.

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So this integral should be 0,
and we shouldn't even bother.

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That's it.

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Momentum.

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Expectation value
of the momentum.

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All these are stationary states.

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Cannot have momentum.

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If it had momentum, here
is the harmonic oscillator,

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here is the wave function.

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If it has momentum, half
an hour later it's here.

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It's impossible.

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This thing cannot have momentum.

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This must be 0 as well.

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OK.

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Now this one is
something you actually

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proved in the first test--

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the expectation value
of the momentum operator

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on a bound state with a
real wave function was 0.

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And you did it by integration--
but in fact you proved it

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in two ways, in momentum
space, in coordinate space,

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is [? back ?] the same thing.

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OK.

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So these ones were too easy.

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So let's try to
see if we can find

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something more difficult to do.

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Well, actually, before doing
that I will do them anyway

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with this notation.

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So what would I have here?

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I would have phi n x phi n.

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And I say, oh, I don't know
how to do things with x.

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That's a terrible thing.

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I would have to do integrals.

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But then you say, no.

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X-- I can write in terms
of a and a daggers.

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And a and a daggers you
know how to manipulate.

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So this is a formula
we wrote last time,

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and it's that x is equal to
square root of h over 2m omega,

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a plus a dagger.

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So x is proportional
to a plus a dagger.

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So here is a square root
of h, 2m omega, phi n,

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a plus a dagger, on phi n.

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Now, this is 0, and why is that?

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Because this term is
a acting on phi n.

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Well, we have it there--

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is square root of
n, phi n minus 1.

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And a dagger acting on
phi n is square root of n

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plus 1, phi n plus 1.

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But the overlap of phi
n minus 1 with phi n

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is 0, because all these
states with different energies

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are orthogonal.

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It's probably a
property I should

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have written somewhere here.

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Which is-- not only
they're well-normalized,

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but phi n phi m is delta nm.

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If the numbers are
different, it's zero.

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And you see this is
something intuitively clear.

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If you wish, I'll just
say here-- these are 0,

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and this is 0 because the
numbers are different.

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If you have, for
example, a phi 2

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and phi 3, or let's do
the phi 3 and a phi 2,

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then you have roughly a dagger,
a dagger, a dagger, phi 0,

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a dagger, a dagger, phi 0.

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And then is equal to phi 0,
three a's, and two a daggers.

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Correct?

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And now you say, OK,
this a is ready to kill

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what is on the right hand side.

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On the right side to it.

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But it can't because
there are a daggers.

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But that a is going to kill
at least one of the a daggers.

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So an a kills an a dagger.

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The second a will kill the
only a dagger that is left.

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And now you have an a
that is ready to go here,

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no obstacle whatsoever,
and kills the phi 0,

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so this is zero.

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So each time there are
some different number

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of eight daggers on the left
input and the right input,

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you get 0.

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If you have more a
daggers on the right,

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then move them to
the left, and now you

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will have more a's than a
daggers and the same problem

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will happen.

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The only way to get something
to work is they are the same.

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But this of course is
guaranteed by our older theorems

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that the--

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eigenstates, if
Hermitian operators

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with different eigenvalues
are orthogonal.

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So this is nice to check
things, but it's not something

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that you need to check.

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All right.

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So now let's say you
want to calculate

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the the uncertainty
of x in phi n.

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Well, the uncertainty
of x squared

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is the expectation
value of x squared

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and phi n minus the expectation
value of x on phi n.

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On this already we know is 0,
but now we have a computation

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worth our tools.

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Let's calculate the expectation
value of x squared in phi n.

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And if you had to do it
with Hermit polynomials,

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it's essentially
a whole days work.

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Maybe a little
less if you started

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using recursion relations and
invent all kinds of things

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to do it.

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It's a nightmare,
this calculation.

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But look how we do it here.

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We say, all right, this is
phi n x hat squared phi n.

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But x hat squared would be h
bar over 2m omega, phi n times

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a plus a dagger time
a plus a dagger phi n.

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Now I must decide what to
do, and one possibility

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is to try to be clever and
do all kinds of things.

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Now, you could do
several things here,

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and none is a lot
better than the other.

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And all of them
take little time.

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00:15:06,120 --> 00:15:10,650
You have to develop
a strategy here,

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00:15:10,650 --> 00:15:20,710
but this is sufficiently doable
that we can do it directly.

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00:15:20,710 --> 00:15:22,830
So what does it
mean doing directly?

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Just multiply those operators.

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So you have phi n times a
a plus a dagger a dagger

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00:15:33,810 --> 00:15:40,660
plus a a dagger plus a dagger a.

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All that on phi n.

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I just multiplied, and
now I try to think again.

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And I say oh, the first term
is to annihilation operators

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00:16:00,150 --> 00:16:01,500
acting on phi n.

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00:16:01,500 --> 00:16:05,440
The first is go give
you phi n minus 1.

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00:16:05,440 --> 00:16:11,420
Second is going to give me a phi
n minus 2 by the time it acts.

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00:16:11,420 --> 00:16:19,290
And a phi n minus 2 is
orthogonal to a phi n.

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00:16:19,290 --> 00:16:23,730
So this term cannot contribute.

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00:16:23,730 --> 00:16:29,740
You know, this term has
two more a's than this one.

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So as we just sort
of illustrated,

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00:16:32,790 --> 00:16:35,190
but it just doesn't match.

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00:16:35,190 --> 00:16:39,810
These two terms acting on phi n
would give you a phi n minus 2.

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00:16:39,810 --> 00:16:42,250
And that's orthogonal.

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00:16:42,250 --> 00:16:45,790
So this term cannot do anything.

219
00:16:45,790 --> 00:16:49,210
Nor can this,
because both raise.

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00:16:49,210 --> 00:16:52,570
So this will end up
as phi n plus two,

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00:16:52,570 --> 00:16:58,410
for example, using that
top property over there.

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00:16:58,410 --> 00:17:00,900
Over there-- the
box equation there.

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00:17:00,900 --> 00:17:04,290
If you have two a
daggers acting on phi n,

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00:17:04,290 --> 00:17:07,050
you will end up
with a phi n plus 2.

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00:17:07,050 --> 00:17:11,160
So this term also
doesn't contribute.

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00:17:11,160 --> 00:17:14,609
And that's progress-- the
calculation became half as

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difficult.

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00:17:18,280 --> 00:17:21,099
OK, that-- now we--

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maybe it's a little
more interesting.

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But again, you should
you should refuse

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to do a [? long ?] computation.

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00:17:29,030 --> 00:17:31,160
Whenever you're looking
at those things,

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you have the temptation
to calculate--

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00:17:33,380 --> 00:17:35,990
refuse that temptation.

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00:17:35,990 --> 00:17:41,100
Look at things and let it
become clear what's going on.

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00:17:41,100 --> 00:17:42,680
There are two terms here--

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a dagger and a dagger a.

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That's not even a
commutator, it's sort of

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00:17:47,720 --> 00:17:49,380
like an anti-commutator.

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00:17:49,380 --> 00:17:52,470
That's strange.

241
00:17:52,470 --> 00:17:57,460
But this, a dagger
a, is familiar.

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That's n.

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00:17:59,115 --> 00:18:00,650
The operator n.

244
00:18:00,650 --> 00:18:05,620
And we know the n eigenvalue, so
this is going to be very easy.

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This is n hat.

246
00:18:09,800 --> 00:18:12,440
The other one is not
n hat, because it's

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00:18:12,440 --> 00:18:14,720
in the wrong order.

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00:18:14,720 --> 00:18:18,005
N hat has a dagger a.

249
00:18:22,350 --> 00:18:31,290
But this operator can be
written as the commutator

250
00:18:31,290 --> 00:18:33,660
plus the thing in
reverse order--

251
00:18:33,660 --> 00:18:36,990
that equation we
had on top-- ab is

252
00:18:36,990 --> 00:18:39,660
equal to ab commutator plus ba.

253
00:18:39,660 --> 00:18:47,310
So this is equal to a a
dagger plus a dagger a.

254
00:18:50,538 --> 00:18:55,640
And this is 1.

255
00:18:55,640 --> 00:18:58,800
Plus another n hat.

256
00:18:58,800 --> 00:19:03,150
So look-- when you have
a and a dagger multiply,

257
00:19:03,150 --> 00:19:07,561
it's either n hat or
it's 1 plus n hat.

258
00:19:10,267 --> 00:19:17,470
And Therefore x squared
expectation value has become

259
00:19:17,470 --> 00:19:24,970
h bar over 2mw phi m, and
this whole parenthesis

260
00:19:24,970 --> 00:19:31,280
is 1 plus 2 n hat phi n.

261
00:19:34,200 --> 00:19:42,210
And this is h bar over 2mw,
phi n, and this is a number.

262
00:19:42,210 --> 00:19:45,960
Because phi in is
an n hat eigenstate.

263
00:19:45,960 --> 00:19:52,300
So it's 1 plus two little
n, phi n phi times 1

264
00:19:52,300 --> 00:19:55,544
plus two [? little ?] n.

265
00:19:55,544 --> 00:19:58,290
And here is our final answer--

266
00:19:58,290 --> 00:20:02,130
expectation value
of x squared is

267
00:20:02,130 --> 00:20:11,980
equal to h bar over and m
omega, n plus 1/2 phi n.

268
00:20:15,730 --> 00:20:18,010
This is a fairly
non-trivial computation.

269
00:20:21,310 --> 00:20:28,110
And that is, of course, because
the expectation value of x

270
00:20:28,110 --> 00:20:33,420
is equal to zero, is the
uncertainty or x squared.

271
00:20:33,420 --> 00:20:41,800
It grows, the state is bigger,
as the quantum number n grows.

272
00:20:41,800 --> 00:20:44,710
By a similar computation,
you can calculate

273
00:20:44,710 --> 00:20:46,320
that you will do
in the homework,

274
00:20:46,320 --> 00:20:50,650
the expectation value
of b squared and phi n,

275
00:20:50,650 --> 00:20:56,980
and then you will see how
much is delta x, delta p,

276
00:20:56,980 --> 00:21:01,120
on the [INAUDIBLE] on phi n.

277
00:21:01,120 --> 00:21:03,240
How much it is.