PROFESSOR: This
definition in which
the uncertainty of the
permission operator
Q in the state psi.
It's always important to
have a state associated
with measuring the uncertainty.
Because the uncertainty will be
different in different states.
So the state should
always be there.
Sometimes we write
it, sometimes we
get a little tired of writing
it and we don't write it.
But it's always implicit.
So here it is.
From the analogous discussion
of random variables,
we were led to this
definition, in which we
would have the expectation
value of the square
of the operator minus the
square of the expectation value.
This was always-- well, this
is always a positive quantity.
Because, as claim 1 goes, it can
be rewritten as the expectation
value of the square of the
difference between the operator
and its expectation value.
This may seem a little strange.
You're subtracting from
an operator a number,
but we know that numbers can be
thought as operators as well.
Operator of minus a
number acting on a state
is well defined.
The operator acts on the state,
the number multiplies a state.
So this is well defined.
And claim 1 is proven
by direct computation.
You certainly indeed prove.
You can expand what is
inside the expectation value,
so it's Q hat squared.
And then the double product
of this Q hat and this number.
Now, the number
and Q hat commute,
so it is really
the double product.
If you have A plus B times A
plus B, you have AB plus BA,
but if they commute
it's 2AB, so this
is minus 2 Q hat Q. Like that.
And then, the last term
is the number squared,
so it's plus Q squared.
And sometimes I don't
put the hats as well.
And all this is the
expectation value
of the sum of all these things.
The expectation value
of a sum of things
is the expectation
value of the first
plus the expectation
value of the second,
plus the expectation
value of the next.
So we can go ahead and do
this, and this is therefore
expectation value of Q
squared minus the expectation
value of this whole thing.
But now the expectation value
of a number times an operator,
the number can go out.
And this is a number,
and this is a number.
So it's minus 2 expectation
value of Q, number went out.
And then you're left with
expectation value of another Q.
And the expectation
value of a number
is just the number,
because then you're
left within the world of psi
star psi, which is equal to 1.
So here is plus Q hat squared.
And these two terms, the
second and the third,
are the same really.
They are both equal to
expectation value of Q squared.
They cancel a little bit,
and they give you this.
So indeed, this is equal
to expectation value
of Q squared minus expectation
value of Q squared.
So claim 1 is true.
And claim 1 shows in particular
that this number, delta Q
squared, in the expectation
value of a square of something,
is positive.
We'll see more
clearly in a second
when we have claim number 2.
And claim number 2
is easily proven.
That's another expression
for uncertainty.
For claim number 2, we will
start with the expectation
value of Q minus Q squared, like
this, which is the integral dx
psi star of x and t, Q minus
expectation value of Q, Q
minus expectation
value of Q, on psi.
The expectation value
of this thing squared
is psi star, the
operator, and this.
And now, think of this as an
operator acting on all of that.
This is a Hermitian operator.
Because Q hat is Hermitian, and
expectation value of Q is real.
So actually this real
number multiplying something
can be moved from
the wave function
to the starred wave
function without any cost.
So even though you might
not think of a real number
as a Hermitian operator, it is.
And therefore this whole
thing is Hermitian.
So it can be written as dx.
And now you have this
whole operator, Q minus Q
hat, acting on psi of x and t.
And conjugate.
Remember, the operator,
the Hermitian operator,
moves to act on psi, and
the whole thing [INAUDIBLE].
And then we have here
the other term left over.
But now, you see that you
have whatever that state is
and the state
complex conjugated.
So that is equal
to this integral.
This is the integral dx of the
norm squared of Q hat minus Q
hat psi of x and t squared,
which means that thing, that's
its complex conjugate.
So this completes
our verification
that these claims are
true, and allow us
to do the last step on
this analysis, which
is to show that if you
have an eigenstate of Q,
if a state psi is an eigenstate
of Q, there is no uncertainty.
This goes along with our
measurement postulate that
says an eigenstate
of Q, you measure Q
and you get the eigenvalue of
Q and there's no uncertainty.
In particular, we'll
do it here I think.
If psi is an eigenstate
of Q, so you'll
have Q psi equal lambda psi,
where lambda is the eigenvalue.
Now, this is a nice thing.
It's stating that the state
psi is an eigenstate of Q
and this is the eigenvalue,
but there is a little bit more
than can be said.
And it is.
It should not surprise
you that the eigenvalue
happens to be the expectation
value of Q on the state psi.
Why?
Because you can take this
equation and integrate dx times
psi star.
If you bring that in into
both sides of the equation
then you have Q psi equals
integral dx psi star psi,
and the lambda goes up.
Since my assumption whenever
you do expectation values,
your states are normalized,
this is just lambda.
And by definition, this is
the expectation value of Q.
So lambda happens to be equal
to the expectation value of Q,
so sometimes we can say
that this equation really
implies that Q hat psi is equal
to expectation value of Q psi
times psi.
It looks a little
strange in this form.
Very few people write
it in this form,
but it's important to recognize
that the eigenvalue is
nothing else but the
expectation value
of the operator of that state.
But if you recognize
that, you realize
that the state satisfies
precisely Q hat minus Q on psi
is equal to 0.
Therefore, if Q hat minus
Q on psi is equal to 0,
delta Q is equal to 0.
By claim 2.
Q hat minus Q expectation
value kills the state,
and therefore this is 0.
OK then.
The other way is also true.
If delta Q is equal to 0, by
claim 2, this integral is 0.
And since it's
the sum of squares
that are always positive, this
state must be 0 by claim 2.
And you get that Q minus
Q hat psi is equal to 0.
And this means that psi
is an eigenstate of Q.
So the other way
around it also works.
So the final
conclusion is delta Q
is equal to 0 is
completely equivalent of--
I'll put in the psi.
Psi is an eigenstate of Q.
So this is the main conclusion.
Also, we learned some
computational tricks.
Remember you have to
compute an expectation
value of a number,
uncertainty, you
have these various
formulas you can use.
You could use the
first definition.
Sometimes it may
be the simplest.
In particular, if the
expectation value of Q
is simple, it's the easiest way.
So for example, you can have
a Gaussian wave function,
and people ask you, what is
delta of x of the Gaussian wave
function?
Well, on this Gaussian
wave function,
you could say that
delta x squared
is the expectation value of x
squared minus the expectation
value of x squared.
What is the
expectation value of x?
Well, it would seem reasonable
that the expectation value of x
is 0.
It's a Gaussian
centered at the origin.
And it's true.
For a Gaussian it would be 0,
the expectation value of x.
So this term is 0.
You can also see 0
because of the integral.
You're integrating x
against psi squared.
Psi squared is even,
x is odd with respect
to x going to minus x.
So that integral
is going to be 0.
So in this case, the
uncertainty is just
the calculation of the
expectation value of x squared,
and that's easily done.
It's a Gaussian integral.
The other good thing
about this is that
even though we have not
proven the uncertainty
principle in all generality.
We've only [? multivated ?] it.
It's precise with
this definition.
So when you have
the delta x, delta p
is greater than or
equal to h bar over 2,
these things are computed
with those definitions.
And then it's precise.
It's a mathematically
rigorous result.
It's not just hand waving.
The hand waving is good.
But the precise result
is more powerful.
