WEBVTT
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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
00:12:34.990 --> 00:12:36.100
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--
00:12:46.840 --> 00:12:49.030
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
00:13:07.210 --> 00:13:16.780
the the uncertainty
of x in phi n.
00:13:16.780 --> 00:13:21.280
Well, the uncertainty
of x squared
00:13:21.280 --> 00:13:23.980
is the expectation
value of x squared
00:13:23.980 --> 00:13:29.840
and phi n minus the expectation
value of x on phi n.
00:13:29.840 --> 00:13:34.720
On this already we know is 0,
but now we have a computation
00:13:34.720 --> 00:13:37.300
worth our tools.
00:13:37.300 --> 00:13:45.160
Let's calculate the expectation
value of x squared in phi n.
00:13:45.160 --> 00:13:50.470
And if you had to do it
with Hermit polynomials,
00:13:50.470 --> 00:13:55.890
it's essentially
a whole days work.
00:13:55.890 --> 00:13:57.870
Maybe a little
less if you started
00:13:57.870 --> 00:14:02.160
using recursion relations and
invent all kinds of things
00:14:02.160 --> 00:14:03.330
to do it.
00:14:03.330 --> 00:14:05.220
It's a nightmare,
this calculation.
00:14:05.220 --> 00:14:07.440
But look how we do it here.
00:14:07.440 --> 00:14:19.110
We say, all right, this is
phi n x hat squared phi n.
00:14:19.110 --> 00:14:28.140
But x hat squared would be h
bar over 2m omega, phi n times
00:14:28.140 --> 00:14:34.620
a plus a dagger time
a plus a dagger phi n.
00:14:41.080 --> 00:14:50.890
Now I must decide what to
do, and one possibility
00:14:50.890 --> 00:14:56.540
is to try to be clever and
do all kinds of things.
00:14:56.540 --> 00:14:58.420
Now, you could do
several things here,
00:14:58.420 --> 00:15:02.540
and none is a lot
better than the other.
00:15:02.540 --> 00:15:06.120
And all of them
take little time.
00:15:06.120 --> 00:15:10.650
You have to develop
a strategy here,
00:15:10.650 --> 00:15:20.710
but this is sufficiently doable
that we can do it directly.
00:15:20.710 --> 00:15:22.830
So what does it
mean doing directly?
00:15:22.830 --> 00:15:25.200
Just multiply those operators.
00:15:25.200 --> 00:15:33.810
So you have phi n times a
a plus a dagger a dagger
00:15:33.810 --> 00:15:40.660
plus a a dagger plus a dagger a.
00:15:40.660 --> 00:15:44.676
All that on phi n.
00:15:44.676 --> 00:15:50.415
I just multiplied, and
now I try to think again.
00:15:53.460 --> 00:16:00.150
And I say oh, the first term
is to annihilation operators
00:16:00.150 --> 00:16:01.500
acting on phi n.
00:16:01.500 --> 00:16:05.440
The first is go give
you phi n minus 1.
00:16:05.440 --> 00:16:11.420
Second is going to give me a phi
n minus 2 by the time it acts.
00:16:11.420 --> 00:16:19.290
And a phi n minus 2 is
orthogonal to a phi n.
00:16:19.290 --> 00:16:23.730
So this term cannot contribute.
00:16:23.730 --> 00:16:29.740
You know, this term has
two more a's than this one.
00:16:29.740 --> 00:16:32.790
So as we just sort
of illustrated,
00:16:32.790 --> 00:16:35.190
but it just doesn't match.
00:16:35.190 --> 00:16:39.810
These two terms acting on phi n
would give you a phi n minus 2.
00:16:39.810 --> 00:16:42.250
And that's orthogonal.
00:16:42.250 --> 00:16:45.790
So this term cannot do anything.
00:16:45.790 --> 00:16:49.210
Nor can this,
because both raise.
00:16:49.210 --> 00:16:52.570
So this will end up
as phi n plus two,
00:16:52.570 --> 00:16:58.410
for example, using that
top property over there.
00:16:58.410 --> 00:17:00.900
Over there-- the
box equation there.
00:17:00.900 --> 00:17:04.290
If you have two a
daggers acting on phi n,
00:17:04.290 --> 00:17:07.050
you will end up
with a phi n plus 2.
00:17:07.050 --> 00:17:11.160
So this term also
doesn't contribute.
00:17:11.160 --> 00:17:14.609
And that's progress-- the
calculation became half as
00:17:14.609 --> 00:17:18.280
difficult.
00:17:18.280 --> 00:17:21.099
OK, that-- now we--
00:17:21.099 --> 00:17:24.740
maybe it's a little
more interesting.
00:17:24.740 --> 00:17:27.200
But again, you should
you should refuse
00:17:27.200 --> 00:17:29.030
to do a [? long ?] computation.
00:17:29.030 --> 00:17:31.160
Whenever you're looking
at those things,
00:17:31.160 --> 00:17:33.380
you have the temptation
to calculate--
00:17:33.380 --> 00:17:35.990
refuse that temptation.
00:17:35.990 --> 00:17:41.100
Look at things and let it
become clear what's going on.
00:17:41.100 --> 00:17:42.680
There are two terms here--
00:17:42.680 --> 00:17:45.320
a dagger and a dagger a.
00:17:45.320 --> 00:17:47.720
That's not even a
commutator, it's sort of
00:17:47.720 --> 00:17:49.380
like an anti-commutator.
00:17:49.380 --> 00:17:52.470
That's strange.
00:17:52.470 --> 00:17:57.460
But this, a dagger
a, is familiar.
00:17:57.460 --> 00:17:59.115
That's n.
00:17:59.115 --> 00:18:00.650
The operator n.
00:18:00.650 --> 00:18:05.620
And we know the n eigenvalue, so
this is going to be very easy.
00:18:05.620 --> 00:18:06.954
This is n hat.
00:18:09.800 --> 00:18:12.440
The other one is not
n hat, because it's
00:18:12.440 --> 00:18:14.720
in the wrong order.
00:18:14.720 --> 00:18:18.005
N hat has a dagger a.
00:18:22.350 --> 00:18:31.290
But this operator can be
written as the commutator
00:18:31.290 --> 00:18:33.660
plus the thing in
reverse order--
00:18:33.660 --> 00:18:36.990
that equation we
had on top-- ab is
00:18:36.990 --> 00:18:39.660
equal to ab commutator plus ba.
00:18:39.660 --> 00:18:47.310
So this is equal to a a
dagger plus a dagger a.
00:18:50.538 --> 00:18:55.640
And this is 1.
00:18:55.640 --> 00:18:58.800
Plus another n hat.
00:18:58.800 --> 00:19:03.150
So look-- when you have
a and a dagger multiply,
00:19:03.150 --> 00:19:07.561
it's either n hat or
it's 1 plus n hat.
00:19:10.267 --> 00:19:17.470
And Therefore x squared
expectation value has become
00:19:17.470 --> 00:19:24.970
h bar over 2mw phi m, and
this whole parenthesis
00:19:24.970 --> 00:19:31.280
is 1 plus 2 n hat phi n.
00:19:34.200 --> 00:19:42.210
And this is h bar over 2mw,
phi n, and this is a number.
00:19:42.210 --> 00:19:45.960
Because phi in is
an n hat eigenstate.
00:19:45.960 --> 00:19:52.300
So it's 1 plus two little
n, phi n phi times 1
00:19:52.300 --> 00:19:55.544
plus two [? little ?] n.
00:19:55.544 --> 00:19:58.290
And here is our final answer--
00:19:58.290 --> 00:20:02.130
expectation value
of x squared is
00:20:02.130 --> 00:20:11.980
equal to h bar over and m
omega, n plus 1/2 phi n.
00:20:15.730 --> 00:20:18.010
This is a fairly
non-trivial computation.
00:20:21.310 --> 00:20:28.110
And that is, of course, because
the expectation value of x
00:20:28.110 --> 00:20:33.420
is equal to zero, is the
uncertainty or x squared.
00:20:33.420 --> 00:20:41.800
It grows, the state is bigger,
as the quantum number n grows.
00:20:41.800 --> 00:20:44.710
By a similar computation,
you can calculate
00:20:44.710 --> 00:20:46.320
that you will do
in the homework,
00:20:46.320 --> 00:20:50.650
the expectation value
of b squared and phi n,
00:20:50.650 --> 00:20:56.980
and then you will see how
much is delta x, delta p,
00:20:56.980 --> 00:21:01.120
on the [INAUDIBLE] on phi n.
00:21:01.120 --> 00:21:03.240
How much it is.