Lecture 8: Uncovering momentum space. Expectation values and their time dependence.

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L8.1 Fourier transforms and delta functions (13:58)

L8.2 Parseval identity (15:50)

L8.3 Three-dimensional Fourier transforms (06:04)

L8.4 Expectation values of operators (28:15)

L8.5 Time dependence of expectation values (7:37)

L8.1 Fourier transforms and delta functions (13:58)

MITOCW | watch?v=8abBLKEZLaI
BARTON
ZWIEBACH:
Today’s subject is momentum space. We’re going to kind of discover the relevance of
momentum space. We’ve been working with wave functions that tell you the probabilities for
finding a particle in a given position and that’s sometimes called coordinate space or position
space representations of quantum mechanics, and we just talked about wave functions that
tell you about probabilities to find a particle in a given position.
But as we’ve been seeing with momentum, there’s a very intimate relation between
momentum and position, and today we’re going to develop the ideas that lead you to think
about momentum space in a way that is quite complimentary to coordinate space. Then we
will be able to talk about expectation values of operators and we’re going to be moved some
steps into what is called interpretation of quantum mechanics. So operators have expectations
values in quantum mechanism-- they are defined in a particular way and that will be something
we’re going to be doing in the second part of the lecture.
In the final part of the lecture, we will consider the time dependence of those expectation
values, which is again, the idea of dynamics-- if you want to understand how your system
evolves in time, the expectation value-- the things that you measure-- may change in time and
that’s part of the physics of the problem, so this will tie into the Schrodinger equation.
So we’re going to begin with momentum space. So we’ll call this uncovering momentum
space. And for much of what I will be talking about in the first half of the lecture, time will not
be relevant-- so time will become relevant later. So I will be writing wave functions that don’t
show the time, but the time could be put there everywhere and it would make no difference
whatsoever.
So you remember these Fourier transform statements that we had that a wave function of x
we could put time, but they said let’s suppress time. It’s a superposition of plane waves. So
there is a superposition of plane waves and we used it last time to evolve wave packets and
things like that. And the other side of Fourier’s theorem is that phi of k can be written by a
pretty similar integral, in which you put psi of x, you change the sign in the exponential, and of
course, integrate now over x.
Now this is the wave function you’ve always learned about first, and then this wave function, as
you can see, is encoded by phi of k as well. If you know phi of k, you know psi of x. And so this
phi of k has the same information in principle as psi of x-- it tells you everything that you need
to know. We think of it as saying, well, phi of k has the same information as psi of x. And the
other thing we’ve said about phi of k is that it’s the weight with which you’re superposing plane
waves to reconstruct psi of x. The Fourier transform theorem is a representation of the wave
function in terms of a superposition of plane waves, and here, it’s the coefficient of the wave
that accompanies each exponential. So phi of k is the weight of the plane waves in the
superposition.
So one thing we want to do is to understand even deeper what phi of k can mean. And so in
order to do that, we need a technical tool. Based on these equations, one can derive a way of
representing this object that we call the delta function. Delta functions are pretty useful for
manipulating objects and Fourier transforms, so we need them. So let’s try to obtain what is
called the delta function statement.
And this is done by trying to apply these two equations simultaneously. Like, you start with psi
of x, it’s written in terms of phi of k, but then what would happen if you would substitute the
value of phi of k in here? What kind of equation you get? What you get is an equation for a
delta function.
So you begin with psi of x over square root of 2 pi, and I’ll write it dk e to the ikx, and now, I
want to write phi of k. So I put phi of k-- 1 over square root of 2 pi integral psi, and now I have
to be a little careful. In here in the second integral, x is a variable of integration, it’s a dummy
variable. It doesn’t have any physical meaning, per se-- it disappears after the integration.
Here, x represents a point where we’re evaluating the wave function, so I just cannot copy the
same formula here because it would be confusing, it should be written with a different x and xprime. Because that x certainly has nothing to do with the x we’re writing here.
So here it is, we’ve written now this integral. And let’s rewrite it still differently. We’ll write as
integral dx-prime. We’ll change the orders of integration with impunity. If you’re trying to be
very rigorous mathematically, this is something you worry about. In physics problems that we
deal with, it doesn’t make a difference. So here we have dx-prime, psi of x-prime, then a 1
over 2 pi integral dk e to the ik x minus x-prime.
So this is what the integral became. And you look at it and you say, well, here is psi of x, and
it’s equal to the integral over x-prime of psi of x-prime times some other function of x and xprime. This is a function of x minus x-prime, or x and x-prime if you wish. It doesn’t depend on
k, k is integrated. And this function, if you recognize it, it’s what we call a delta function. It’s a
function that, multiplied with an integral, evaluates the integrand at a particular point. So this is
a delta function, delta of x-prime minus x.
That is a way these integrals then would work. That is, when you integrate over x-prime–
when you have x-prime minus something, then the whole integral is the integrand evaluated at
the point where we say the delta function fires. So the consistency of these two equations
means that for all intents and purposes, this strange integral is a representation of a delta
function. So we will write it down. I can do one thing here, but-- it’s kind of here you see the psi
of x-prime minus x, but here you see x minus x-prime.
But that’s sign, in fact, doesn’t matter. It’s not there. You can get rid of it. Because if in this
integral, you can let K goes to minus k, and then the dk changes sign, the order of integration
changes sign, and they cancel each other. And the effect is that you change the sign in the
exponents, so if you let k goes to minus k, the integral just becomes 1 over 2 pi integral dk e to
the ik x-prime minus x. So we’ll say that this delta function is equal to this thing, exploiting this
sign ambiguity that you can always have. So I’ll write it again in the way most people write the
formula, which is at this moment, switching x and x-prime. So you will write this delta of x
minus x-prime is 1 over 2 pi integral from infinity to minus infinity dk e to the ik x minus xprime.
So this is a pretty useful formula and we need it all the time that we do Fourier transforms as
you will see very soon. It’s a strange integral, though. If you have x equal to x-prime, this is 0
and you get infinity. So it’s a function, it’s sort of 0-- when x minus x-prime is different from 0,
somehow all these waves superimpose to 0. But when x is equal to x-prime, it blows up. So it’s
does the right thing, it morally does the right thing, but it’s a singular kind of expression and we
therefore manipulate it with care and typically, we use it inside of integrals.
So it’s a very nice formula, we’re going to need it, and it brings here for the first time in our
course, I guess, the delta functions. And this is something if should-- if you have not ever play
without the functions, this may be something interesting to ask in recitation or you can try to
prove, for example, just like we show that delta of minus x is the same as delta of x, the delta
of a being a number times x is 1 over absolute value of a times delta of x. Those are two
simple properties of delta functions and you could practice by just trying to prove them-- for
example, use this integral representation to show them.

L8.2 Parseval identity (15:50)

MITOCW | watch?v=i-bP2OkQxUI
PROFESSOR: What we want to understand now is really about momentum space. So we can ask the
following question-- what happens to the normalization condition that we have for the wave
function when we think in momentum variables?
So yes, I will do this first. So let’s think of integral dx psi of x star psi of x. Well, this is what we
called the integral, the total integral for x squared, the thing that should be equal to 1 if you
have a probability interpretation, for the wave function. And what we would like to understand
is what does it say about phi of k?
So for that, I have to substitute what [? size ?] r in terms of k and try to rethink about this
integral how to evaluate it. So for example, here, I can make a little note that I’m going to use
a variable of integration that I call k for the first factor. And for this factor, I’m going to use k
prime.
You should use different variables of integration. Remember, psi of x is an integral over k. But
we should use different ones not to get confused.
So here we go-- dx 1 over square root of 2 pi-- integral-- we said this one is over k. So it would
be phi star-- let me put the dk first-- dk phi star of k e to the minus ikx and dk. That’s the first
psi. This is psi star.
And now we put psi. So this is the dk prime-- phi of x dk prime phi of k prime e to the ik prime
x. It’s the same x in the three places.
OK, at this moment, you always have to think, what do I do next? There are all these many
integrals. Well, the integrals over k, there’s no chance you’re going to be able to do them
apparently-- not to begin with, because they are abstract integrals. So k integrals have no
chance. Maybe the integral that we have here, the dx, does have a chance.
So in fact, let me write this as integral dk phi star of k integral dk prime phi of k prime. And then
I have 1 over 2 pi integral dx e to the ik prime minus kx. I think I didn’t miss any factor.
And now comes to help this integral representation of the delta function. And it’s a little
opposite between the role of k and x. Here the integration variable is over x. There is was over
k.
But the spirit of the equality or the representation is valid. You have the 1 over 2 pi, a full
But the spirit of the equality or the representation is valid. You have the 1 over 2 pi, a full
integral over a variable, and some quantity here. And this is delta of k prime minus k.
And finally, I can do the last integral. I can do the integral, say, over k prime. And that will just
give me-- because a delta function, that’s what it does. It evaluates the integrand at the value.
So you integrate over k prime-- evaluates phi at k. So this is equal to integral dk phi star of k
phi of k.
And that’s pretty neat. Look what we found. We’ve found what is called Parseval’s theorem,
which is that integral dx of psi of x squared is actually equal to integral dk phi of k squared. So
it’s called Parseval theorem-- Parseval’s theorem.
Sometimes in the literature, it’s also called Plancherel’s theorem. I think it depends on the
generality of the identity-- so Plancherel’s theorem.
But this is very nice for us, because it begins to tell us there’s, yes indeed, some more physics
to phi of k. Why? The fact that this integral is equal to 1 was a key thing.
Well, the fact that it didn’t change in time thanks to showing the equation was very important.
It’s equal to 1. We ended up with a probabilistic interpretation for the wave function. We could
argue that this could be a probability, because it made sense.
And now we have a very similar relation for phi of k. Not only phi of k represents as much
physics as psi of k, as psi of x, and it not only represents the weight with which you
superimpose plane waves, but now it also satisfies a normalization condition that says that the
integral is also equal to this integral, which is equal to 1. It’s starting to lead to the idea that this
phi of k could be thought maybe as a probability distribution in this new space, in momentum
space.
Now I want to make momentum space a little bit more clear. And this involves a little bit of
moving around with constants, but it’s important. We’ve been using k all the time. And
momentum is h bar k.
But now let’s put things in terms of momentum. Let’s do everything with momentum itself. So
let’s put it here. So let’s go to momentum space, so to momentum language-- to momentum
language.
And this is not difficult. We have p is equal to h bar k. So dp over h bar is equal to dk in our
integrals. And we can think of functions of k, but these are just other functions of momentum.
So I can make these replacements in my Fourier relations.
So these are the two equations that we wrote there-- are the ones we’re aiming to write in a
more momentum language rather than k, even though it’s going to cost a few h bars here and
there. So the first equation becomes psi of x equal 1 over square root of 2 pi integral.
Well, phi of k is now phi of p with a tilde. e to the ikx is e to the ipx over h bar. And dk is equal
to dp over h bar.
Similarly, for the second equation, instead of phi of k, you will put phi tilde of p 1 over square
root of 2 pi integral psi of x e to the minus ipx over h bar. And that integral doesn’t change
much [? dx. ?]
So I did my change of variables. And things are not completely symmetric if you look at them.
Here, they were beautifully symmetric. In here, you have 1 over h bar here and no h bar
floating around.
So we’re going to do one more little change for more symmetry. We’re going to redefine. Let
phi tilde of p be replaced by phi of p times square root of h bar.
You see, I’m doing this a little fast. But the idea is that this is a function I invented. I can just
call it a little different, change its normalization to make it look good. You can put whatever you
want.
And one thing I did, I decided that I don’t want to carry all these tildes all the time. So I’m going
to replace phi tilde of p by this phi of p. And that shouldn’t be confused with the phi of k. It’s not
necessarily the same thing, but it’s simpler notation.
So if I do that here, look-- you will have a phi of p, no tilde, and 1 over square root of h bar.
Because there will be e square root of h bar in the numerator and h bar there. So this first
equation will become psi of x 1 over square root of 2 pi h bar integral phi of p e to the ipx over
h bar dp.
And the second equation, here, you must replace it by a phi and a square root of h bar, which
will go down to the same position here so that the inverse equation is phi of p, now 1 over
square root of 2 pi h bar integral psi of x e to the minus ipx over h bar dx.
So this is Fourier’s theorem in momentum notation, in which you’re really something over
momenta. And you put all these h bars in the right place. And we’ve put them symmetrically.
You could do otherwise. It’s a choice.
But look at the evolution of things. We’ve started with a standard theorem with k and x, then
derived a representation for the delta function, derived Parseval’s theorem, and finally, rewrote
this in true momentum language.
Now you can ask what happens to Parseval’s theorem. Well, you have to keep track of the
normalizations what will happen. Look, let me say it.
This left-hand side, when we do all these changes, doesn’t change at all. The second one, dk,
gets a dp over h bar. And this is becomes phi tilde of p.
But phi tilde of p then becomes a square root, then it’s phi of p. So you get two square roots in
the numerator and the dk that had a h bar in the denominator. So they all disappear, happily.
It’s a good thing.
So Parseval now reads, integral dx psi squared of x equal integral dk phi of k-- phi of p, I’m
sorry. I just doing p now. And it’s a neat formula that we can use.

L8.3 Three-dimensional Fourier transforms (06:04)

MITOCW | watch?v=MJM1AzpB6Y4
PROFESSOR: We got here finally in terms of position and in terms of momentum. So this was not an accident
that it worked for position and wave number. It works with position and for momentum. And
remember, this phi of p now has interpretation of the weight that is associated with a plane
wave of momentum p, and you’re summing over P in here. So we’ll do the natural thing that
we did with x. We’ll interpret phi of p squared-- phi of p squared-- dp is the probability.
Find the particle with momentum in the range p, p plus dp. Just the same way as we would
say that psi squared of x, dx is the probability to find the particle between x and x plus dx. So
this is allowed now by the conservation of probability and this, therefore, makes sense. It’s a
postulate, though. It’s not something that can derive. I can just argue that it’s consistent to
think in that way, and that’s the way we finally promote this phi of p, which did encode psi of x.
Phi of b has the same information as phi of x. phi of p is the weight of the superposition but,
finally, it’s given a probabilistic interpretation. It represents a probability to find the particle with
some momentum.
So this is what is going to allow us to do expectation values in a minute. But I want to close off
this discussion by writing for you the three-dimensional versions of these equations. 3D
version of Fourier transform. So this is what we want to rewrite. So what would it be? It’s psi of
the vector x. Since you’re going to have three integrals-- because you’re going into it over
three components of momentum-- this factor appears three times. So actually, it’s 2 pi h bar to
the three halfs integral phi of vector p into the i vector p dot product vector x h bar d cube p,
and phi of p vector the inverse theorem-- same factor, we keep the nice symmetry between x
and p-- psi of x vector negative exponent same dot product but negative exponent d cube x.
So these are the three dimensional versions of your x versus p. And there is a threedimensional version of Parseval. So oh there’s a three dimensional version of the delta
function. Just like we had a delta function here-- a delta function in three dimensional space
would be delta cubed x minus x prime would be one over 2 pi cubed integral d cube k e to the i
k vector x minus x prime vector.
It’s all quite analogous. I think you should appreciate that you don’t have to memorize them or
anything like that. They won’t be in any formula sheet, but they are very analogous
expressions. Parseval also works in the same way. And you have-- just as you would imagine-

• that the integral all over three dimensional space of psi of x squared is equal to the integral
  over three dimensional momentum space of phi of p squared. So the three results-- the
  Fourier theorem the delta function and Parseval hold equally well.

L8.4 Expectation values of operators (28:15)

MITOCW | watch?v=XQKV-hpsurs
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.

L8.5 Time dependence of expectation values (7:37)

MITOCW | watch?v=AnzhigYawy8
PROFESSOR: It’s a statement about the time dependence of the expectation values. It’s a pretty fundamental
theorem. So here it goes.
You have d dt of the expectation value of Q. This is what we want to evaluate. We Now this
would be d dt of integral psi star of x and t, Q psi of x and t. And the d dt acts on the two of
them. So it gives you integral partial psi star dt Q psi of x and t plus psi star Q partial psi dt.
And this is the integral over the x.
You’ve seen that kind of stuff. And what is it? Well, integral dx, this is this Schrodinger
equation, d psi star dt is i over h bar, H psi star. From the Schrodinger equation. Then you
have the Q psi of x and t. On this term, you will have a very similar thing. Minus i over H bar
this time, psi star QH psi of x and t.
So we use the Schrodinger equation in the form, i d psi dt-- i H bar d psi dt-- equal H psi. I
used it twice. So then, it’s actually convenient to multiply here by i H bar d dt of Q. So I
multiplied by i H bar, and I will cancel the i and the H bar in this term, minus them this term. So
we’ll have d cube x psi star Q H hat psi minus H hat psi star cube psi.
OK. Things have simplified very nicely. And there’s just one more thing we can do. Look, this is
the product of Q and H. But by hermiticity, H in here can be brought to the other side to act on
this wave function. So this is actually equal to the integral d cube dx psi star QH hat psi minus-

• the H can go to the other side-- psi star H cube psi.
  But then, what do we see there? We recognize a commutator. This commutator is just like we
  did for x and p, and we started practicing how to compute them. They show up here. And this
  is maybe one of the reasons commutators are so important in quantum mechanics. So what
  do we have here? i H bar d dt of the expectation value of Q is equal to the integral dx of psi
  star, QH minus HQ psi. This is all of x and t.
  Well, this is nothing else but the commutator of Q and H. So our final result is that iH bar d dt
  of the expectation value of Q is equal to-- look. It’s the expectation value of the commutator.
  Remember, expectation value for an operator-- the operator is the thing here-- so this is
  nothing else than the expectation value of Q with H.
  This is actually a pretty important result. It has all the dynamics of the physics in the
  observables. Look, the wave functions used to change in time. Due to their change in time, the
  expectation values of the operators change in time. Because this integral can’t depend on
  time.
  But here what you have succeeded is to represent the change in time of the expectation
  value-- the change in time of the position that you expect you find your particle-- in terms of
  the expectation value of a commutator with a Hamiltonian.
  So if some quantity commutes with a Hamiltonian, its expectation value will not change in time.
  If you have a Hamiltonian, say with a free particle, well, the momentum commutes with this.
  Therefore the expected value of the momentum, you already know, since the momentum
  commutes with H. This is 0. The expected value of this is 0. And the expected value of the
  momentum will not change, will be conserved.
  So conservation laws in quantum mechanics have to do with things that commute with the
  Hamiltonian. And it’s an idea we’re going to develop on and on.

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