WEBVTT
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PROFESSOR: Last time we
discussed the differential
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equation.
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I'll be posting notes very soon.
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Probably this
afternoon, at some time.
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And last time, we solved
the differential equation,
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we found the energy
eigenstates, and then turned
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into an algebraic
analysis in which we
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factorized the Hamiltonian.
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Which meant,
essentially, that you
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could write the Hamiltonian--
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up to an overall constant that
doesn't complicate matters--
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as the product of an a dagger a.
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And that was very useful
to show, for example,
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that any energy eigenstate
would have to have energy
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greater than h omega over 2.
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We call this a
dagger a the number
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operator n, which is
a Hermitian operator.
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Recall that the dagger
of a product of operators
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is the reverse order product
of the daggered operators.
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So the dagger of a
dagger a is itself.
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And then a was related to x
and p, and so was a dagger.
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Recall that x and
p are Hermitian.
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And there are overall constants
here that were wrote last time,
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but now they're not that urgent.
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And a and a dagger, the
commutator is equal to one.
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That was very useful.
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Finally, we also show that
while the energy of any state
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would have to be greater
than h omega over 2,
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if you had a state that
is killed by a hat,
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it would have the
lowest allowed energy--
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which is h omega over 2.
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And, therefore, that
is the ground state.
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And we looked at this
differential equation,
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and we found this
Gaussian wave function.
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And it's a first order
differential equation.
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And, therefore, it
has just one solution.
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And, therefore, there is
just one ground state,
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and it's a bound state.
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And, of course, you wouldn't
expect more than one ground
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state, because there's no
degeneracies in the bound state
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spectrum of a
one-dimensional potential.
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So we found one ground state was
phi 0, and it's killed by a--
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which means that
it's killed by n-hat,
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because a is to
the right in n-hat.
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So the a finds phi
0 and just kills it.
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Now, the other thing to
note is that the Hamiltonian
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is really, pretty much, the
same thing as the number
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operator multiplied by
something with units of energy.
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The number operator has no
units, because a and a dagger
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have no units.
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And that's very useful.
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So it's like a dimensionless
version of the energy.
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And, certainly, if you
have an eigenstate of h
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it must be an eigenstate of n.
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And the eigenvalue of n--
if we call it capital n.
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Therefore, you can
imagine this equation
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acting on an eigenstate--
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which happens to be an
eigenstate of n or of h.
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On the left-hand side you
would read the energy,
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and on the right-hand side
you would read the eigenvalue
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of the n operator.
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So that gives you a very
nice simple expression.
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You see that the energy is the
number plus a 1/2 multiplied
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by h-bar [INAUDIBLE].
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So that's pretty
much the content
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of what we reached last time.
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And now we have to
complete the solution.
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And the plan for today is
to complete the solution,
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familiarize ourselves
with these operators,
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learn how to work with a
harmonic oscillator with them.
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And then we'll leave
the harmonic oscillator
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for the time being--
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let you do some
exercises with it--
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but turn to scattering states.
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So the second part
of today's lecture
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we'll be talking about
scattering states.
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OK.
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So when we look at this
thing and you have a number
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operator-- which encodes
the Hamiltonian--
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it's a good idea to
try to understand
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how it interacts with the other
operators that you have here.
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And a good question,
whenever you have operators,
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is the commutator.
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So you can ask, what is
the commutator of n with a?
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And this commutator
is going to show up.
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But it's basically
that kind of thing.
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If you have a and a dagger, you
ask, what is the commutator?
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If you have n, you ask,
what is the commutator
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with the other thing?
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So n with a would be the
commutator of a dagger with a--
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a like that.
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And sometimes I will
not write the hats
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to write things more quickly.
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Now, in this commutator,
you can move the a out,
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and you have a dagger a a.
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And a dagger a is minus 1.
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Because aa dagger is 1.
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So this is minus a.
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So that's pretty nice.
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It's simple.
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How about n with a dagger?
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Well, this would be a
dagger a with a dagger.
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A dagger with a dagger commute.
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So this a dagger can go
out, and you get aa dagger.
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And that's 1, so
you get a dagger.
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So it's a nice kind of
computation relation.
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You would have
commutation reserve x
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with p given a constant.
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Now n with a gives
a number times a.
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N commutated with a dagger
gives a number times a dagger.
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And those numbers are
pretty significant,
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so I'll write this again.
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N with a is minus a.
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And n with a dagger
is plus a dagger.
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This is part of the reason--
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as we will see soon--
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that the name of a--
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which we call
destruction operator,
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because it destroys the vacuum--
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it's sometimes called
lowering operator,
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because it comes with
a negative sign here.
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And we'll see a better
reason for that name.
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A dagger is sometimes
called the creation operator
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or the raising operator, because
it increases some number,
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as you will see.
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And here it's reflected
by these plots.
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But we need a little
more than that.
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We need a little more
commutators than this.
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So for example, if I would
have the commutator of a
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with a dagger to the k--
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you can imagine this.
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You have to become very
used and very comfortable
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with these
commutation relations.
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And sometimes the only way to
do that is to just do examples.
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So I'm doing this with a k here.
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Maybe-- when you
review this lecture--
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you should do it with k
equals 2 or with k equals 3,
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and do it a few times.
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Until you're comfortable
with these things,
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and you know what identities
you've been using.
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If this was a little
quick, then go more slowly
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and make absolutely
sure you know
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how to do those commutators.
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In here, I'm going
to say what happens.
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You have an a and
you have to move it
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across a string of a-hats.
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Now, moving an a
cross an a-hat--
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because of the commutator--
gives you a factor of 1,
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but it destroys the
a and the a-hat.
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As you move the a
across the a-hats--
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because this is a
with all the a-hats,
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here, minus the a-hats
times the a there.
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So if you could just move
the a all the way across--
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then you cancel with this.
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What you get is
what happens when
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you're moving it to across.
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And you're moving across
a string of those.
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So each time you try to
move on a across an a-hat,
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you get this factor of
1 and you kill the a
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and you kill one a dagger.
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So this answer
will not have an a,
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and it will have
one less a dagger.
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So a dagger to the k, minus 1.
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And then I would argue-- and
you should do it more slowly--
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that you have to go
across k of those.
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And each time you
get a factor of 1,
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and you lose the a
and the a dagger.
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So at the end you get a k.
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You should realize
that this is not
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all that different from the
kind of commutators you had.
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Like, with x to the n.
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This was very similar--
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it might be a good time to
review how that was done--
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in which that pretty much gives
you an x to the n minus 1,
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times a factor of n,
because p is a derivative.
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You could almost think of a
as the derivative with respect
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to a dagger.
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And then this
commutator would be 1.
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So this is true.
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And there is also--
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if you want-- an a dagger
with a to the n or a to the k.
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This would give you--
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if you had just one of them
you would get a minus sign,
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because a dagger with a is that.
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But the same thing holds, you're
going to get one less a-hat.
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So a-hat to the k minus 1.
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A factor of k--
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because k times you're going to
move an a dagger across an a.
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And a minus because you're
getting a dagger commutator
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with a, as opposed to a
commutator with a dagger--
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which is 1.
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So these are two very
nice and useful equations
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that you should be
comfortable with.
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Now, this implies that you can
do more with an n operator.
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So n with a-hat to the k, this
time will be minus k a-hat
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to the k.
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It doesn't change
the number of a-hats,
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because you're now making
commutators with a dagger a.
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So each time you have
this commuted with one
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a-hat, the a dagger
and the a give you 1,
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but you have another a back.
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So the power is the same.
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The sign comes from this sign.
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AUDIENCE: Shouldn't
the n there have a hat?
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PROFESSOR: Yes, it
should have a hat.
00:13:45.065 --> 00:13:46.255
I'm sorry.
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Yes.
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And, similarly, n
a-hat dagger to the k.
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This is k a-hat dagger to the k.
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So what happened before,
that n-hat operator leaves
00:14:16.100 --> 00:14:18.600
the a same but puts a number--
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leaves the a dagger the
same and puts a number.
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Here, you see it
happening again.
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N with a collection--
with a string of a-hats--
00:14:30.840 --> 00:14:34.770
gives you the same
string, but the number.
00:14:34.770 --> 00:14:37.230
And with a collection
of a daggers
00:14:37.230 --> 00:14:40.410
gives you the same collection
of a daggers with a number.
00:14:40.410 --> 00:14:45.090
And the number happens to be
the number of a's or the number
00:14:45.090 --> 00:14:46.890
of a daggers.
00:14:46.890 --> 00:14:51.610
So that's the reason it's
called the number operator,
00:14:51.610 --> 00:14:56.180
because the eigenvalues are the
number of creation operators
00:14:56.180 --> 00:14:59.580
or the number of
destruction operators.
00:14:59.580 --> 00:15:04.120
I was a little glib by
calling it the eigenvalue.
00:15:04.120 --> 00:15:07.300
But it almost looks like
an eigenvalue equation,
00:15:07.300 --> 00:15:10.590
which have an operator,
another operator, and a number
00:15:10.590 --> 00:15:12.490
times the second operator.
00:15:12.490 --> 00:15:15.480
It is not exactly an
eigenvalue equation, though,
00:15:15.480 --> 00:15:17.790
because with eigenvalues
you would just have
00:15:17.790 --> 00:15:21.040
this acting on the second one.
00:15:21.040 --> 00:15:25.660
But the fact that
this case appear here
00:15:25.660 --> 00:15:31.000
are the reason these
are number operators.
00:15:31.000 --> 00:15:35.110
So it was a little
quick for many of you.
00:15:35.110 --> 00:15:36.880
Some of you may have
seen this before.
00:15:36.880 --> 00:15:41.020
It was a little slow,
but the important thing
00:15:41.020 --> 00:15:43.130
is after a couple
of days from now,
00:15:43.130 --> 00:15:49.750
or by Friday, you find all
this very straightforward.