# L8.4 Expectation values of operators.

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