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