PROFESSOR: Suppose you define
now, one state called phi 1
as a dagger acting on phi 0.
You could not define
any interesting state
with a acting on phi 0
because a kills phi 0,
so you try phi 0 like this.
Now you could ask, OK,
what energy does it have?
Is it an energy eigenstate?
Well it is an energy eigenstate
if it's a number eigenstate.
And we can see if it's a
number eigenstate by acting
with the number operator.
So N phi 1 is equal
to N a dagger phi 0.
OK.
Here comes trick.
Maybe it's too much to even
call it a trick, number one.
This thing you look
at it and you say,
I want to sort of simplify
this, learn something about it.
If this is supposed to be
an eigenstate of N hat,
I have to make it
happen somehow.
Now n hat kills phi 0.
So if I would have a term a
dagger times N hat near phi 0,
it would be 0.
So I claim, and this
is a step that I
want you to be able
to do also quickly,
that I can replace this by
the commutator of these two
operators.
The product is replaced
by the commutators.
Why?
Aren't products simpler
than commutators?
No.
We have formulas
for commutators.
And products are, in
general, more complicated.
And why is this correct?
And you say, well, it is correct
because this has two terms.
The term I want
minus a dagger N hat.
But the term a dagger N hat is
0 because N hat kills phi 0.
So I can do that because
this is N dagger a hat, which
is what I had, minus a
hat dagger N on phi 0.
And this term is 0.
So you would have put
a 2 here or a 3 here,
or any number even.
But the right one to
put is the commutators.
So that's this.
And now this commutator
is already known.
That's why we computed it.
It's just a dagger.
So this is a dagger phi 0,
and that's what we call phi 1.
So N hat on phi 1 is phi 1.
N hat has eigenvalue 1 on phi 1.
So N is equal to 1.
That's the eigenvalue.
It is an eigenstate.
It is an energy eigenstate.
In fact how much
energy, E, is h bar
omega times N, which is 1, plus
1/2, which is 3/2 h bar omega?
And look what this is.
This is the reason this is
called a creation operator.
Because by acting on the ground
state, what people sometimes
call the vacuum, the
lowest energy state,
the vacuum is called the
lowest energy state, by acting
on the vacuum you get a state.
I mean, you've created
a state, therefore.
How is this concretely done?
Remember you had
phi 0 of x, what
it is, and a dagger over there
is x minus ip over m omega.
So this is x minus--
or minus h bar
over m omega d dx.
So you can act on it.
It may be a little messy.
But that's it.
It's a very closed
form expression.
Now, phi 0 was defined, the
ground state such that it's
a normalized state.
This means the integral of
phi 0 multiplied with phi 0
over x is 1.
That's how we had
the ground state.
You could ask, if I've
defined phi 1 this way,
is simply normalized?
So I'll try it.
And now you could say, oh, this
is going to be a nightmare.
Normalizing phi 0
is difficult. Now I
have to act with a
dagger, which means act
with x, take derivatives.
It's going to grow twice as big.
Then I'm going to have to
square it and integrate it.
It looks very bad.
The good thing is those with
these a's and a daggers,
you have to compute
anything, pretty much.
See how we do it.
I want to know how much
is phi 1 with phi 1.
Is it 1?
And it's normalized or not?
Then I say, look, phi 1 is a
dagger phi 0, a dagger phi 0.
So far so good.
But I just know
things about phi 0.
So let's clear up one phi 0.
At least I can move
the a dagger as an a.
So this is phi 0
a a dagger phi 0.
Can I finish the
computation in this line?
Yes, I think we can.
Phi 0.
a with a dagger,
same story as before.
a would kill phi 0.
So you can replace
that by a commutator.
Commutator of a
with a dagger phi 0.
But the commutator of
a with a dagger is 1,
so this is phi 0 phi
0 and it's equal to 1.
Yes, it is properly normalized.
So that's the nice thing
about these a's and a daggers.
Just start moving them around.
You have to get practice.
Where should you move it?
Where should you put it?
When you replay something by
a commutator, when you don't.
It's a matter of practice.
There's no other way.
You have to do a lot
of these commutators
to get a feeling
of how they work
and what you're supposed to do.
Let's do another state.
Let's try to do phi 2.
I'll put a prime because
I'm not sure this is going
to work out exactly right.
And this time, I'll put an a
dagger a dagger on the vacuum.
Two a daggers, two creation
operators on the vacuum.
And now I want to see if
this is an energy eigenstate.
Well, this is a dagger
squared on the vacuum.
So let's ask, is N hat--
is phi 2 prime an
eigenstate of N hat?
Well I would have N hat on
a dagger squared on phi 0.
Again, by now you know, I should
replace this by a commutator
because N hat kills the phi
0, so N hat with a dagger
squared phi 0.
And that commutator
has been done.
It's two times a
hat dagger squared,
two times a dagger squared
on phi 0, which is 2 phi 2.
That's what we call
the state phi 2 prime.
I'm sorry.
So again, it is an
energy eigenstate.
Is it normalized?
Well, let's try it.
Phi 2 prime phi 2 prime is
equal to a dagger a dagger.
Let me not put the hats.
I'm getting tired of them.
a dagger a dagger phi 0.
Now I move all of them.
This a dagger becomes
an a, the next a dagger
becomes an a here.
So this is phi 0 a a a
dagger a dagger phi 0.
Wow, this looks a
little more complicated.
Because we don't want to
calculate that thing, really.
We definitely don't want
to start writing x and p's.
But, you know, you decide.
Take it one at a time.
This a is here and wants
to act on this thing.
And then this other
a will, but let's
just concentrate on the
first a that wants to act.
a would kill phi 0, so we
can replace this whole thing
by a commutator.
So this is phi 0.
The first a is still there,
but the second, we'll
replace it by the
commutator, this commutator.
I've replaced this product, the
product of a times this thing,
by the commutator of
those two operators.
And then I say, oh
look, you've done that.
a with a dagger to the k
is k a dagger k minus 1.
So I'll write it here.
This will be a
factor of 2 phi 0 a.
And this is supposed to be
now a dagger to one power
less, so it's just
a dagger phi 0.
So this is supposed
to be 2a dagger.
So that's what I did.
And again, this a
wants to act on phi 0
and it's just
blocked by a dagger,
but you can replace
it by a commutator.
a with a dagger phi 0.
And this is therefore a 1,
so this whole result is a 2.
So this phi 2 prime, yes, it
is the next excited state.
Two creation operators
on the ground state.
Energy and eigenvalues too.
You had N equal zero
eigenvalue for the ground state
1 for phi 1, 2 for phi 2 prime.
But it's not
properly normalized.
Well, if the normalization gives
you 2, then you should define
phi 2 as 1 over the
square root of 2
a dagger a dagger on phi 0.
And that's proper.
So it's time to go general.
The n-th excited state, we
claim is given by an a dagger
a dagger, n of them, acting on
phi 0 with a coefficient 1 over
square root of--
we might think it's
n, but it's actually,
you can't tell at this far--
this one is n factorial.
That's what you need.
That is the state.
And what is the
number of this state?
What is the number
eigenvalue on phi n?
Well, it is 1 over square
root of n factorial.
The number acting on the a
daggers, the n of them, phi 0.
You can replace
by the commutator,
which then is 2 times already.
So it's N commutator with a
dagger to the little n phi 0
times 1 over square root of n.
And how much is this commutator?
Over there.
This is N times a
dagger to the n phi 0.
So between these
three factors, you're
still getting n phi to the n.
So the number for this
state is little n.
It is an energy eigenstate.
The N eigenvalue is little n.
And the energy is h bar omega.
The eigenvalue of N hat,
which is little n plus 1/2.
So it is the energy
eigenstate of number little n.
This is the definition.
And the last thing you may want
to check is the normalization.
Let me almost check it here.
No, I will check it.
Let's say I think this
is a full derivation.
Phi n with phi n would
be two factors of those,
so I would have 1 over n
factorial a dagger a dagger,
n of them on phi 0, a
dagger a dagger, n of them
again on phi 0.
So then that's equal to
1 over n factorial phi
0 a a, lots of a's, n of them,
n a daggers, phi 0, like that.
That's what it is.
We had to move all
the a daggers that
were acting on the left
input of the integral,
or the inner product,
all the way to the right.
And that's it.
So now comes this step.
And I think you can
see why it's working.
Think of moving the
first a all the way here.
Well, you can replace the
first a with a commutator.
But that a with lots of a
daggers, with n a daggers,
would give you a
factor of n, with n
a daggers will give you a factor
of n times one a dagger less.
So to move the first a,
there are n a daggers
and you get one factor
of n from this a.
But for the next a, there's
now n minus 1 a daggers,
so this time you get a factor
of n minus 1 when you move it.
From the next one, there's going
to be n minus 2 a daggers, so n
minus 2.
All of them all the way up to
one, cancels this n factorial,
and that's equal to 1.
