PROFESSOR: Important
thing to do is to just try
to understand one more thing.
The creation and annihilation
operators-- what do they
do to those states?
You see, a creation operator
will I add one more a dagger,
so somehow must change
phi n into phi n plus 1.
A destruction operator with an a
will kill one of these factors,
and therefore it will give
you a state with lower number
of phi n minus 1.
And we would like to know
the precise relations.
So look at this.
Let's do with an A on phi n.
And we know it should be
roughly phi n minus 1.
This is one destruction
operator, but we can do it.
Look-- this is 1 over
square root of n.
A times a dagger to the n phi 0.
A with a dagger to the n phi 0,
we can replace by a commutator
again.
Commutator of a, a
dagger to the n phi 0.
This is 1 over square
root of n factorial,
and here we get a
factor of n times
a dagger to the n minus 1 phi 0.
You know, it's all a matter
of those commutators we
on the left blackboard.
But this state--
by definition, we
have n square root
of n factorial.
That's state, by definition,
is phi n minus 1 times
square root of n
minus 1 factorial.
See, by looking at this
definition and saying,
suppose I have n minus 1, n
minus 1, this is phi n minus 1.
So n minus 1 a
daggers on phi 0 is
n minus 1 factorial square
root multiplied phi n minus 1.
And now we can simplify this--
square root of n
factorial and square root
of n minus 1 factorial
gives you just a factor
of square root of
n that with this n
here, this square root
of n, phi n minus 1.
Sop there we go--
here is the first relation.
A is really a lowering operator.
It gives you an
eigenstate 1 less energy,
but it gives it with a
factor of square root of n,
that if you care
about normalizations,
you better keep it.
That factor is there because
the overall normalization
of this equation was designed
to make the states normalized.
Similarly, we can do
the other operation,
which is what is a
dagger acting on phi n.
This would be 1 over
square root of n factorial,
but this time a dagger to
the n plus 1 on phi n, phi 0.
Because you had already
a dagger to the n,
and you put one more a dagger.
But this thing is equal to what?
This is equal to square root
of n plus 1 factorial times
phi n plus 1.
From the definition--
I hope you're not getting dizzy.
Lots of factors here.
But now you see that the n
part of the factorial cancels,
and you get that
a hat dagger phi
n is equal to square root
of n plus 1 phi n plus 1.
OK let's do an application.
Suppose somebody asks
you to calculate example.
The expectation
value of the operator
x on phi n, the expectation
value of p on phi n.
How much are they?
OK.
This, of course, in conventional
language, at first sight
looks prohibitive.
I would have to get those phi n
[? in ?] some Hermit polynomial
hn, for which I don't know
the closed form expression.
It's a very large
polynomial, jumps 2 by 2,
there is exponentials, I
will have to do an integral.
That's something that
we don't want to do.
So how can we do it
without doing integrals?
Well, this one's-- actually, you
can do without doing anything.
You don't have to do integrals,
you don't have to calculate.
The answers are kind
of obvious, if you
think about it the right way.
That's not the obvious part,
to think about the right way.
But here it is.
Look, what is this integral?
This is the integral of x times
phi and of x, those are real,
quantity squared.
And the phi n's are either even
or odd, but the fight n squared
are even.
And x is odd.
So this integral should be 0,
and we shouldn't even bother.
That's it.
Momentum.
Expectation value
of the momentum.
All these are stationary states.
Cannot have momentum.
If it had momentum, here
is the harmonic oscillator,
here is the wave function.
If it has momentum, half
an hour later it's here.
It's impossible.
This thing cannot have momentum.
This must be 0 as well.
OK.
Now this one is
something you actually
proved in the first test--
the expectation value
of the momentum operator
on a bound state with a
real wave function was 0.
And you did it by integration--
but in fact you proved it
in two ways, in momentum
space, in coordinate space,
is [? back ?] the same thing.
OK.
So these ones were too easy.
So let's try to
see if we can find
something more difficult to do.
Well, actually, before doing
that I will do them anyway
with this notation.
So what would I have here?
I would have phi n x phi n.
And I say, oh, I don't know
how to do things with x.
That's a terrible thing.
I would have to do integrals.
But then you say, no.
X-- I can write in terms
of a and a daggers.
And a and a daggers you
know how to manipulate.
So this is a formula
we wrote last time,
and it's that x is equal to
square root of h over 2m omega,
a plus a dagger.
So x is proportional
to a plus a dagger.
So here is a square root
of h, 2m omega, phi n,
a plus a dagger, on phi n.
Now, this is 0, and why is that?
Because this term is
a acting on phi n.
Well, we have it there--
is square root of
n, phi n minus 1.
And a dagger acting on
phi n is square root of n
plus 1, phi n plus 1.
But the overlap of phi
n minus 1 with phi n
is 0, because all these
states with different energies
are orthogonal.
It's probably a
property I should
have written somewhere here.
Which is-- not only
they're well-normalized,
but phi n phi m is delta nm.
If the numbers are
different, it's zero.
And you see this is
something intuitively clear.
If you wish, I'll just
say here-- these are 0,
and this is 0 because the
numbers are different.
If you have, for
example, a phi 2
and phi 3, or let's do
the phi 3 and a phi 2,
then you have roughly a dagger,
a dagger, a dagger, phi 0,
a dagger, a dagger, phi 0.
And then is equal to phi 0,
three a's, and two a daggers.
Correct?
And now you say, OK,
this a is ready to kill
what is on the right hand side.
On the right side to it.
But it can't because
there are a daggers.
But that a is going to kill
at least one of the a daggers.
So an a kills an a dagger.
The second a will kill the
only a dagger that is left.
And now you have an a
that is ready to go here,
no obstacle whatsoever,
and kills the phi 0,
so this is zero.
So each time there are
some different number
of eight daggers on the left
input and the right input,
you get 0.
If you have more a
daggers on the right,
then move them to
the left, and now you
will have more a's than a
daggers and the same problem
will happen.
The only way to get something
to work is they are the same.
But this of course is
guaranteed by our older theorems
that the--
eigenstates, if
Hermitian operators
with different eigenvalues
are orthogonal.
So this is nice to check
things, but it's not something
that you need to check.
All right.
So now let's say you
want to calculate
the the uncertainty
of x in phi n.
Well, the uncertainty
of x squared
is the expectation
value of x squared
and phi n minus the expectation
value of x on phi n.
On this already we know is 0,
but now we have a computation
worth our tools.
Let's calculate the expectation
value of x squared in phi n.
And if you had to do it
with Hermit polynomials,
it's essentially
a whole days work.
Maybe a little
less if you started
using recursion relations and
invent all kinds of things
to do it.
It's a nightmare,
this calculation.
But look how we do it here.
We say, all right, this is
phi n x hat squared phi n.
But x hat squared would be h
bar over 2m omega, phi n times
a plus a dagger time
a plus a dagger phi n.
Now I must decide what to
do, and one possibility
is to try to be clever and
do all kinds of things.
Now, you could do
several things here,
and none is a lot
better than the other.
And all of them
take little time.
You have to develop
a strategy here,
but this is sufficiently doable
that we can do it directly.
So what does it
mean doing directly?
Just multiply those operators.
So you have phi n times a
a plus a dagger a dagger
plus a a dagger plus a dagger a.
All that on phi n.
I just multiplied, and
now I try to think again.
And I say oh, the first term
is to annihilation operators
acting on phi n.
The first is go give
you phi n minus 1.
Second is going to give me a phi
n minus 2 by the time it acts.
And a phi n minus 2 is
orthogonal to a phi n.
So this term cannot contribute.
You know, this term has
two more a's than this one.
So as we just sort
of illustrated,
but it just doesn't match.
These two terms acting on phi n
would give you a phi n minus 2.
And that's orthogonal.
So this term cannot do anything.
Nor can this,
because both raise.
So this will end up
as phi n plus two,
for example, using that
top property over there.
Over there-- the
box equation there.
If you have two a
daggers acting on phi n,
you will end up
with a phi n plus 2.
So this term also
doesn't contribute.
And that's progress-- the
calculation became half as
difficult.
OK, that-- now we--
maybe it's a little
more interesting.
But again, you should
you should refuse
to do a [? long ?] computation.
Whenever you're looking
at those things,
you have the temptation
to calculate--
refuse that temptation.
Look at things and let it
become clear what's going on.
There are two terms here--
a dagger and a dagger a.
That's not even a
commutator, it's sort of
like an anti-commutator.
That's strange.
But this, a dagger
a, is familiar.
That's n.
The operator n.
And we know the n eigenvalue, so
this is going to be very easy.
This is n hat.
The other one is not
n hat, because it's
in the wrong order.
N hat has a dagger a.
But this operator can be
written as the commutator
plus the thing in
reverse order--
that equation we
had on top-- ab is
equal to ab commutator plus ba.
So this is equal to a a
dagger plus a dagger a.
And this is 1.
Plus another n hat.
So look-- when you have
a and a dagger multiply,
it's either n hat or
it's 1 plus n hat.
And Therefore x squared
expectation value has become
h bar over 2mw phi m, and
this whole parenthesis
is 1 plus 2 n hat phi n.
And this is h bar over 2mw,
phi n, and this is a number.
Because phi in is
an n hat eigenstate.
So it's 1 plus two little
n, phi n phi times 1
plus two [? little ?] n.
And here is our final answer--
expectation value
of x squared is
equal to h bar over and m
omega, n plus 1/2 phi n.
This is a fairly
non-trivial computation.
And that is, of course, because
the expectation value of x
is equal to zero, is the
uncertainty or x squared.
It grows, the state is bigger,
as the quantum number n grows.
By a similar computation,
you can calculate
that you will do
in the homework,
the expectation value
of b squared and phi n,
and then you will see how
much is delta x, delta p,
on the [INAUDIBLE] on phi n.
How much it is.
