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.