PROFESSOR: Today we'll
talk about observables
and Hermitian operators.
So we've said that an operator,
Q, is Hermitian in the language
that we've been working
so far, if you find
that the integral,
dx psi 1 Q psi 2,
is actually equal to
the integral dx of Q,
acting this time of
Psi 1 all star psi 2.
So as you've learned already,
this requires some properties
about the way
functions far away,
at infinity, some integration
by parts, some things to manage,
but this is the
general statement
for a large class of
functions, this should be true.
Now we want to, sometimes,
use a briefer notation
for all of this.
And I will sometimes
use it, sometimes not,
and you do whatever you feel.
If you like to use
this notation, us it.
So here's the definition.
If you put up Psi 1,
Psi 2 and a parentheses,
this denotes a number,
and in fact denotes
the integral of psi 1
star of x, psi 2 of x dx.
So whatever you put
in the first input
ends up complex motivated.
When you put in
the second input,
it's like that,
it's all integrated.
This has a couple of
obvious properties.
If you put a number times
psi 1 times psi 2 like this,
the number will appear,
together with psi 1,
and will complex conjugated.
So it can go out as
a star psi 1 psi 2.
And if you put the number
on the second input,
it comes out as is.
Because the second
input is not complex
conjugated in the definition.
With this definition,
a Hermitian operator,
Q is Hermitian, has
a nice look to it.
It becomes kind of
natural and simple.
It's the statement that if
you have psi 1, Q psi 2,
you can put the Q
in the first input.
Q psi 1 psi 2.
This second term in
the right hand side
is exactly this integral here.
And the first tern
in the left hand side
is the left hand side
of that condition.
So it's just maybe a
briefer way to write it.
So when you get tired of writing
integral dx of the first,
the second, you can use this.
Now with distance last
time, the expectation
values of operators.
So what's the expectation value
of Q in some state psi of x?
And that is denoted as
these braces here and of psi
is equal to the integral of psi.
The expectation value depends
on the state you live in
and it's psi Q psi.
Or if you wish, dx in
written notation psi
Q. I should put the
hats everywhere.
This is the expectation
value of Q. I'm sorry,
I missed here a star.
So so far, so good.
We've reviewed what a
Hermitian operator is,
what an expectation value is,
so let's begin with some claims.
Claim number one.
The expectation value
of Q, with Q Hermitian.
So everywhere here,
Q will be Hermitian.
The expectation
value of Q is real.
A real number, it belongs
to the real numbers.
So that's an important thing.
You want to figure out the
expectation value of Q,
you have a psi star,
you have a psi.
Well, it'd better be real
if we're going to think,
and that's the goal
of this discussion,
that Hermitian operators
are the things you
can measure in
quantum mechanics,
so this better be real.
So let's see what this is.
Well, Q psi, that's
the expectation value.
If I complex conjugate
it, I must complex
conjugate this whole thing.
Now if you want to complex
conjugate an integral,
you can complex
conjugate the integrand.
Here it is.
I took this right hand
side here, the integrand.
I copied it, and now I
complex conjugated it.
That's what you mean by complex
conjugating an integral.
But this is equal, integral dx.
Now I have a product
of two functions here.
Psi star and Q that
has acted on psi.
So that's how I think.
I never think of
conjugating Q. Q is
a set of operations
that have acted on psi
and I'm just going
to conjugate it.
And the nice thing
is that you never
have to think of what is Q
star, there's no meaning for it.
So what happens here?
Priority of two functions,
the complex conjugate
of the first--
now if you [INAUDIBLE]
normally something twice,
you get the function back.
And here you've got Q psi star.
But that, these are functions.
You can move around.
So this Q hat psi star Q psi.
And so far so good.
You know, I've done
everything I could have done.
They told to come to
complex conjugate this,
so I complex conjugated it
and I'm still not there.
But I haven't used that
this operator is Hermitian.
So because the
operator is Hermitian,
now you can move the Q from this
first input to the second one.
So it's equal to integral
dx psi star Q psi.
And oh, that was the
expectation value of Q on psi,
so the star of this number is
equal to the number itself,
and that proves the
claim, Q is real.
So this is our first claim.
The second claim that is
equally important, claim two.
The eigenvalues of the
operator Q are real.
So what are the
eigenvalues of Q?
Well you've learned, with
the momentum operator,
eigenvalues or
eigenfunctions of an operator
are those special functions
that the operator acts on them
and gives you a number
called the eigenvalue times
that function.
So Q, say, times,
psi 1, if psi 1
is a particularly
nice choice, then it
will be equal to some number.
Let me quote Q1 times psi1.
And there, I will say
that Q1 is the eigenvalue.
That's the definition.
And psi1 is the eigenvector,
or the eigenfunction.
And the claim is that that
number is going to real.
So why would that be the case?
Well, we can prove
it in many ways,
but we can prove it kind of
easily with claim number one.
And actually gain
a little insight,
cold calculate the
expectation value of Q
on that precise state, psi 1.
Let's see how much is it.
You see, psi 1 is
a particular state.
We've called it an
eigenstate of the operator.
Now you can ask, suppose
you live in psi 1?
That's who you are,
that's your state.
What is the expectation
value of this operator?
So we'll learn more about
this question later,
but we can just do it, it's the
integral of dx psi 1 Q psi 1.
And I keep forgetting
these stars,
but I remember them
after a little while.
So at this moment, we can
use the eigenvalue condition,
this condition
here, that this is
equal to dx psi 1 star Q1 psi 1.
And the Q1 can go out,
hence Q 1 integral dx of psi
1 star psi 1.
But now, we've proven,
in claim number one,
that the expectation
value of Q is always real,
whatever state you take.
So it must be real if you
take it on the state psi 1.
And if the expectation
value of psi 1 is real,
then this quantity, which
is equal to that expectation
value, must be real.
This quantity is the
product of two factors.
A real factor here--
that integral is not only
real, it's even positive--
times Q1.
So if this is real, then
because this part is real,
the other number must be real.
Therefore, Q1 is real.
Now it's an
interesting observation
that if your eigenstate,
eigenfunction
is a normalized eigenfunction,
look at the eigenfunction
equation.
It doesn't depend on what
precise psi 1 you have,
because if you put psi 1
or you put twice psi 1,
this equation still holds.
So if it hold for psi 1, if psi
1 is called an ideal function,
3 psi 1, 5 psi 1, minus psi
1 are all eigenfunctions.
Properly speaking
in mathematics,
one says that the
eigenfunction is
the subspace generated by
this thing, by multiplication.
Because everything is accepted.
But when we talk
about the particle
maybe being in the
state of psi 1,
we would want to normalize it,
to make psi 1 integral squared
equal to 1.
In that case, you would
obtain that the expectation
value of the operator
on that state
is precisely the eigenvalue.
When you keep measuring
this operator, this state,
you keep getting the eigenvalue.
So I'll think about the
common for a normalized psi
1 as a true state that you
use for expectation values.
In fact, whenever we
compute expectation values,
here is probably a
very important thing.
Whenever you compute
an expectation value,
you'd better normalize the
state, because otherwise,
think of the expectation value.
If you don't normalize the
state, you the calculation
and you get some
answer, but your friend
uses a wave function three
times yours and your friend
gets now nine times your answer.
So for this to be a
well-defined calculation,
the state must be normalized.
So here, we should really say
that the state is normalized.
Say one is the ideal
function normalized.
And this integral would be equal
to Q1 belonging to the reals.
And Q1 is real.
So for a normalized psi
1 or how it should be,
the expectation value
of Q on that eigenstate
is precisely equal
to the eigenvalue.
