PROFESSOR: Expectation
values of operators.
So this is, in a sense,
one of our first steps
that we're going to take towards
the interpretation of quantum
mechanics.
We've had already that the
wave function tells you
about probabilities.
But that's not quite enough to
have the full interpretation
of what we're doing.
So let's think of operators
and expectation values
that we can motivate.
So for example, if you
have a random variable q,
that can take values--
so this could be a coin that
can take values heads and tails.
It could be a pair of dice
that takes many values--
can take values in the set q1
up to qn with probabilities
p1 up to pn, then in
statistics, or 8044,
you would say that this
variable, this random variable
has an expectation value.
And the expectation value--
denoted by this angular symbols
over here, left and right--
it's given by the sum
over i, i equal 1 to n,
of the possible values
the random variable can
take times the probabilities.
It's a definition
that makes sense.
And it's thought to be,
this expectation value is,
the expected value,
or average value,
that you would obtain if you
did the experiment of tossing
the random variable many times.
For each value of
the random variable,
you multiply by the probability.
And that's the number
you expect to get.
So in a quantum system,
we follow this analogy
very closely.
So what do we have
in a quantum system?
In the quantum system you
have that psi star of x and t.
The x is the probability
that the particle
is in x, x plus dx.
So that's the probability
that the particle
is going to be found
between x and x plus dx.
The position of this particle
is like a random variable.
You never know where you
are going to find it.
But he has different
probabilities to find it.
So we could now define in
complete analogy to here,
the expectation value of
the position operator,
or the expectation value of
the position, expectation
value of x hat, or the
position, and say, well, I'm
going to do exactly
what I have here.
I will sum the products of the
position times the probability
for the position.
So I have to do
it as an integral.
And in this integral,
I have to multiply
the position times the
probability for the position.
So the probability that the disc
takes a value of x, basically
all that is in the interval
dx about x is this quantity.
And that's the position
that you get when
you estimate this probability.
So you must sum the values
of the random variable
times its probability.
And that is taken in quantum
mechanics to be a definition.
We can define the expectation
value of x by this quantity.
And what does it
mean experimentally?
It means that in quantum
mechanics, if you
have a system represented
by a wave function,
you should build many
copies of the system, 100
copies of the system.
In all of these copies,
you measure the position.
And you make a
table of the values
that you measure the position.
And you measure them at the
same time in the 100 copies.
There's an experimentalist
on each one,
and it measures
the position of x.
You construct the
table, take the average,
and that's what this quantity
should be telling you.
So this quantity, as you
can see, may depend on time.
But it does give you
the interpretation
of expected value
coinciding with a system,
now the quantum mechanical
system, for which the position
is not anymore a quantity
that is well defined
and it's always the same.
It's a random variable,
and each measurement
can give you a different
value of the position.
Quantum mechanically, this
is the expected value.
And the interpretation
is, again if you
measure many times, that is
the value, the average value,
you will observe.
But now we can do the
same thing to understand
expectation values.
We can do it with the momentum.
And this is a little
more non-trivial.
So we have also, just like we
said here, that psi star psi
dx is the probability that
the particle is in there,
you also have that
phi of p squared.
dp is the probability to find
the particle with momentum
in the range p p plus dp.
So how do we define the
expected value of the momentum?
The expected value
of the momentum
would be given by, again, the
sum of the random variable,
which is the momentum,
times the probability
that you get that value.
So this is it.
It's very analogous
to this expression.
But it's now with momentum.
Well, this is a
pretty nice thing.
But we can learn more about it
by pushing the analogy more.
And you could say,
look, this is perfect.
But it's all done
in momentum space.
What would happen
if you would try
to do this in position space?
That is, you know how 5p
is related to psi of x.
So write everything
in terms of x.
I would like to see this
formula in terms of x.
It would be a very
good thing to have.
So let's try to do that.
So we have to do a little bit
of work here with integrals.
So it's not so bad.
p phi star of p phi of p dp.
And for this one,
you have to write it
as an integral
over some position.
So let me call it
over position x prime.
This you will write this an
integral over some position x.
And then we're going to try
to rewrite the whole thing
in terms of coordinate space.
So what do we have here?
We have integral pdp.
And the first phi star would
be the integral over dx prime.
We said there is the
square root 2 pi h
bar that we can't forget.
5p, it would have an e to
the ip x prime over h bar,
and a psi star of x prime.
So I did conjugate
this phi star of p.
I may have it here.
Yes, it's here.
I conjugated it and did
the integral over x prime.
And now we have another one,
integral vx over 2 pi h bar e
to the minus ipx over h
bar, and you have psi of x.
Now there's a lot
of integrals there,
and let's try to
get them simplified.
So we're going to try to
do first the p integral.
So let's try to clean up
everything in such a way
that we have only p done first.
So we'll have a 1 over 2 pi h
bar from the two square roots.
And I'll have the two
integrals dx prime psi
of x prime star the x psi of x.
So again, as we
said, these integrals
we just wrote them out.
They cannot be done.
So our only hope is to simplify
first the pdp integral.
So here we would have
integral of dp times p times
e to the ipx prime over h bar.
And e to the minus
ipx over h bar.
Now we need a little
bit of-- probably
if you were doing this, it
would not be obvious what to do,
unless you have some
intuition of what the momentum
operator used to be.
The momentum operator
used to be dvx, basically.
Now this integral would
be a delta function
if the p was not here.
But here is the p.
So what I should try to
do is get rid of that p
in order to understand
what we have.
So here we'll do integral dp.
And look, output here
minus h bar over idvx.
And leave everything
here to the right,
e to the ipx prime over h bar
p to the minus ipx over h bar.
I claim this is the same.
Because this
operator, h over iddx,
well, it doesn't act on x prime.
But it acts here.
And when it does, it will
produce just the factor of p
that you have.
Because the minus i and
the minus i will cancel.
The h bar will cancel.
And the ddh will
just bring down a p.
So this is the way to have
this work out quite nicely.
Now this thing is
inside the integral.
But it could as well be
outside the integral.
It has nothing to do with dp.
So I'll rewrite this again.
I'll write it as dx prime psi
of x prime star integral dx
psi of x.
And I'll put this here,
minus h bar over iddx,
in front of the integral.
The 1 over 2 pi h bar here.
Integral dp e to the ipx
prime minus x over h bar.
So I simply did a
couple of things.
I moved that 1 over
2 pi h to the right.
And then I said this derivative
could be outside the integral.
Because it's an integral over p.
It doesn't interfere with x
derivative, so I took it out.
Now the final two steps,
we're almost there.
The first step is to say,
with this it's a ddx.
And yes, this is a
function of x and x prime.
But I don't want to
take that derivative.
Because I'm going to
complicate things.
In fact, this is already
looking like a delta function.
There's a dp dp.
And the h bars that
actually would cancel.
So this is a perfectly
nice delta function.
You can change variables.
Do p equal u times h bar and
see that actually the h bar
doesn't matter.
And this is just delta
of x prime minus x.
And in here, you could
act on the delta function.
But you could say, no, let
me do integration by parts
and act on this one.
When you do
integration by parts,
you have to worry about
the term at the boundary.
But if your wave functions
vanish sufficiently fast
at infinity, there's no problem.
So let's assume
we're in that case.
We will integrate by parts and
then do the delta function.
So what do we have here?
I have integral dx prime psi
star of x prime integral dx.
And now I have, because
of the sign of integration
by parts, h over iddx of psi.
And then we have the delta
function of x minus x prime.
It's probably better
still to write
the integral like this, dx
h bar over iddx of psi times
integral dx prime psi star of x
prime delta of x minus x prime.
And we're almost
done, so that's good.
We're almost done.
We can do the
integral over x prime.
And it will elevate
that wave function at x.
So at the end of the
day, what have we found?
We found that p, the
expectation value of p,
equal integral of p phi of
p squared dp is equal to--
we do this integral.
So we have integral dx,
I'll write it two times,
h over i d psi dx of x and
t and psi star of x and t.
I'm not sure I
carried that times.
I didn't put the time anywhere.
So maybe I shouldn't
put it here yet.
This is what we did.
And might as well write
it in the standard order,
where the complex conjugate
function appears first.
This is what we found.
So this is actually very neat.
Let me put the time back
everywhere you could put time.
Because this is a
time dependent thing.
So p, expectation value
is p phi of p and t
squared dp is equal to
integral dx psi star of x.
And now we have, if
you wish, p hat psi
of x, where p hat
is what we used
to call the momentum operator.
So look what has happened.
We started with this
expression for the expectation
value of the momentum
justified by the probabilistic
interpretation of phi.
And we were led to
this expression, which
is very similar to this one.
You see, you have the psi
star, the psi, and the x there.
But here, the momentum
appeared at this position,
acting on the wave function
psi, not on psi star.
And that's the way,
in quantum mechanics,
people define expectation
values of operators in general.
So in general,
for an operator q,
we'll define the
expectation value of q
to be integral dx psi
star of x and t q acting
on the psi of x and t.
So you will always do this
of putting the operator
to act on the second part
of the wave function,
on the second appearance
of the wave function.
Not on the psi star,
but on the psi.
We can do other examples of
this and our final theorem.
This is, of course,
time dependent.
So let me do one example and
our final time dependence
analysis of this quantity.
So for example, if you would
think of the kinetic operator
example.
Kinetic operator t is p squared
over 2m is a kinetic operator.
How would you compute
its expectation value?
Expectation value of the
kinetic operator is what?
Well, I could do the
position space calculation,
in which I think of the kinetic
operator as an operator that
acts in position space where
the momentum is h bar over iddx.
So then I would have integral
dx psi star of x and t.
And then I would have minus
h squared over 2m d second dx
squared of psi of x and t.
So here I did exactly
what I was supposed
to do given this formula.
But you could do another
thing if you wished.
You could say, look, I can
work in momentum space.
This is a momentums operator p.
Just like I defined the
expectation value of p,
I could have the expectation
value of p squared.
So the other
possibility is that you
have t is equal to the
integral dp of p squared
over 2m times phi of p squared.
This is the operator.
And this is the probability.
Or you could write it
more elegantly perhaps.
dp phi star of p t
squared over 2m phi of p.
These are just
integrals of numbers.
All these are numbers already.
So in momentum
space, it's easier
to find the expectation value
of the kinetic operator.
In coordinate space,
you have to do this.
You might even say, look,
this thing looks positive.
Because it's p squared
of the number squared.
In the center here,
it looks negative.
But that's an illusion.
The second derivative can
be partially integrated.
One of the two
derivatives can be
integrated to act on this one.
So if you do partial
integration by parts,
you would have integral
dx h squared over 2m.
And then you would have d
psi dx squared by integration
by parts.
And that's clearly
positive as well.
So it's similar to this.
