WEBVTT
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PROFESSOR: Expectation
values of operators.
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So this is, in a sense,
one of our first steps
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that we're going to take towards
the interpretation of quantum
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mechanics.
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We've had already that the
wave function tells you
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about probabilities.
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But that's not quite enough to
have the full interpretation
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of what we're doing.
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So let's think of operators
and expectation values
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that we can motivate.
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So for example, if you
have a random variable q,
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that can take values--
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so this could be a coin that
can take values heads and tails.
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It could be a pair of dice
that takes many values--
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can take values in the set q1
up to qn with probabilities
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p1 up to pn, then in
statistics, or 8044,
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you would say that this
variable, this random variable
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has an expectation value.
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And the expectation value--
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denoted by this angular symbols
over here, left and right--
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it's given by the sum
over i, i equal 1 to n,
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of the possible values
the random variable can
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take times the probabilities.
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It's a definition
that makes sense.
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And it's thought to be,
this expectation value is,
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the expected value,
or average value,
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that you would obtain if you
did the experiment of tossing
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the random variable many times.
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For each value of
the random variable,
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you multiply by the probability.
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And that's the number
you expect to get.
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So in a quantum system,
we follow this analogy
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very closely.
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So what do we have
in a quantum system?
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In the quantum system you
have that psi star of x and t.
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The x is the probability
that the particle
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is in x, x plus dx.
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So that's the probability
that the particle
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is going to be found
between x and x plus dx.
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The position of this particle
is like a random variable.
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You never know where you
are going to find it.
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But he has different
probabilities to find it.
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So we could now define in
complete analogy to here,
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the expectation value of
the position operator,
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or the expectation value of
the position, expectation
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value of x hat, or the
position, and say, well, I'm
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going to do exactly
what I have here.
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I will sum the products of the
position times the probability
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for the position.
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So I have to do
it as an integral.
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And in this integral,
I have to multiply
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the position times the
probability for the position.
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So the probability that the disc
takes a value of x, basically
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all that is in the interval
dx about x is this quantity.
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And that's the position
that you get when
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you estimate this probability.
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So you must sum the values
of the random variable
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times its probability.
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And that is taken in quantum
mechanics to be a definition.
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We can define the expectation
value of x by this quantity.
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And what does it
mean experimentally?
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It means that in quantum
mechanics, if you
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have a system represented
by a wave function,
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you should build many
copies of the system, 100
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copies of the system.
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In all of these copies,
you measure the position.
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And you make a
table of the values
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that you measure the position.
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And you measure them at the
same time in the 100 copies.
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There's an experimentalist
on each one,
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and it measures
the position of x.
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You construct the
table, take the average,
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and that's what this quantity
should be telling you.
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So this quantity, as you
can see, may depend on time.
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But it does give you
the interpretation
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of expected value
coinciding with a system,
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now the quantum mechanical
system, for which the position
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is not anymore a quantity
that is well defined
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and it's always the same.
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It's a random variable,
and each measurement
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can give you a different
value of the position.
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Quantum mechanically, this
is the expected value.
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And the interpretation
is, again if you
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measure many times, that is
the value, the average value,
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you will observe.
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But now we can do the
same thing to understand
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expectation values.
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We can do it with the momentum.
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And this is a little
more non-trivial.
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So we have also, just like we
said here, that psi star psi
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dx is the probability that
the particle is in there,
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you also have that
phi of p squared.
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dp is the probability to find
the particle with momentum
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in the range p p plus dp.
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So how do we define the
expected value of the momentum?
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The expected value
of the momentum
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would be given by, again, the
sum of the random variable,
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which is the momentum,
times the probability
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that you get that value.
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So this is it.
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It's very analogous
to this expression.
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But it's now with momentum.
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Well, this is a
pretty nice thing.
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But we can learn more about it
by pushing the analogy more.
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And you could say,
look, this is perfect.
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But it's all done
in momentum space.
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What would happen
if you would try
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to do this in position space?
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That is, you know how 5p
is related to psi of x.
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So write everything
in terms of x.
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I would like to see this
formula in terms of x.
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It would be a very
good thing to have.
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So let's try to do that.
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So we have to do a little bit
of work here with integrals.
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So it's not so bad.
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p phi star of p phi of p dp.
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And for this one,
you have to write it
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as an integral
over some position.
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So let me call it
over position x prime.
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This you will write this an
integral over some position x.
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And then we're going to try
to rewrite the whole thing
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in terms of coordinate space.
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So what do we have here?
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We have integral pdp.
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And the first phi star would
be the integral over dx prime.
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We said there is the
square root 2 pi h
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bar that we can't forget.
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5p, it would have an e to
the ip x prime over h bar,
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and a psi star of x prime.
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So I did conjugate
this phi star of p.
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I may have it here.
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Yes, it's here.
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I conjugated it and did
the integral over x prime.
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And now we have another one,
integral vx over 2 pi h bar e
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to the minus ipx over h
bar, and you have psi of x.
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Now there's a lot
of integrals there,
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and let's try to
get them simplified.
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So we're going to try to
do first the p integral.
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So let's try to clean up
everything in such a way
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that we have only p done first.
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So we'll have a 1 over 2 pi h
bar from the two square roots.
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And I'll have the two
integrals dx prime psi
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of x prime star the x psi of x.
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So again, as we
said, these integrals
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we just wrote them out.
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They cannot be done.
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So our only hope is to simplify
first the pdp integral.
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So here we would have
integral of dp times p times
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e to the ipx prime over h bar.
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And e to the minus
ipx over h bar.
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Now we need a little
bit of-- probably
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if you were doing this, it
would not be obvious what to do,
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unless you have some
intuition of what the momentum
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operator used to be.
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The momentum operator
used to be dvx, basically.
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Now this integral would
be a delta function
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if the p was not here.
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But here is the p.
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So what I should try to
do is get rid of that p
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in order to understand
what we have.
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So here we'll do integral dp.
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And look, output here
minus h bar over idvx.
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And leave everything
here to the right,
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e to the ipx prime over h bar
p to the minus ipx over h bar.
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I claim this is the same.
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Because this
operator, h over iddx,
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well, it doesn't act on x prime.
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But it acts here.
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And when it does, it will
produce just the factor of p
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that you have.
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Because the minus i and
the minus i will cancel.
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The h bar will cancel.
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And the ddh will
just bring down a p.
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So this is the way to have
this work out quite nicely.
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Now this thing is
inside the integral.
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But it could as well be
outside the integral.
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It has nothing to do with dp.
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So I'll rewrite this again.
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I'll write it as dx prime psi
of x prime star integral dx
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psi of x.
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And I'll put this here,
minus h bar over iddx,
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in front of the integral.
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The 1 over 2 pi h bar here.
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Integral dp e to the ipx
prime minus x over h bar.
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So I simply did a
couple of things.
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I moved that 1 over
2 pi h to the right.
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And then I said this derivative
could be outside the integral.
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Because it's an integral over p.
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It doesn't interfere with x
derivative, so I took it out.
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Now the final two steps,
we're almost there.
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The first step is to say,
with this it's a ddx.
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And yes, this is a
function of x and x prime.
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But I don't want to
take that derivative.
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Because I'm going to
complicate things.
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In fact, this is already
looking like a delta function.
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There's a dp dp.
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And the h bars that
actually would cancel.
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So this is a perfectly
nice delta function.
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You can change variables.
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Do p equal u times h bar and
see that actually the h bar
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doesn't matter.
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And this is just delta
of x prime minus x.
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And in here, you could
act on the delta function.
00:17:40.340 --> 00:17:44.180
But you could say, no, let
me do integration by parts
00:17:44.180 --> 00:17:45.390
and act on this one.
00:17:52.170 --> 00:17:54.670
When you do
integration by parts,
00:17:54.670 --> 00:17:58.560
you have to worry about
the term at the boundary.
00:17:58.560 --> 00:18:03.070
But if your wave functions
vanish sufficiently fast
00:18:03.070 --> 00:18:05.530
at infinity, there's no problem.
00:18:05.530 --> 00:18:08.350
So let's assume
we're in that case.
00:18:08.350 --> 00:18:11.980
We will integrate by parts and
then do the delta function.
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So what do we have here?
00:18:16.260 --> 00:18:26.203
I have integral dx prime psi
star of x prime integral dx.
00:18:26.203 --> 00:18:29.940
And now I have, because
of the sign of integration
00:18:29.940 --> 00:18:37.420
by parts, h over iddx of psi.
00:18:37.420 --> 00:18:42.470
And then we have the delta
function of x minus x prime.
00:18:49.400 --> 00:18:52.880
It's probably better
still to write
00:18:52.880 --> 00:19:02.540
the integral like this, dx
h bar over iddx of psi times
00:19:02.540 --> 00:19:11.890
integral dx prime psi star of x
prime delta of x minus x prime.
00:19:16.096 --> 00:19:20.720
And we're almost
done, so that's good.
00:19:25.590 --> 00:19:26.700
We're almost done.
00:19:26.700 --> 00:19:30.480
We can do the
integral over x prime.
00:19:30.480 --> 00:19:35.270
And it will elevate
that wave function at x.
00:19:38.590 --> 00:19:43.850
So at the end of the
day, what have we found?
00:19:43.850 --> 00:19:47.950
We found that p, the
expectation value of p,
00:19:47.950 --> 00:19:58.210
equal integral of p phi of
p squared dp is equal to--
00:19:58.210 --> 00:19:59.670
we do this integral.
00:19:59.670 --> 00:20:05.630
So we have integral dx,
I'll write it two times,
00:20:05.630 --> 00:20:15.275
h over i d psi dx of x and
t and psi star of x and t.
00:20:18.170 --> 00:20:21.550
I'm not sure I
carried that times.
00:20:21.550 --> 00:20:23.500
I didn't put the time anywhere.
00:20:23.500 --> 00:20:28.860
So maybe I shouldn't
put it here yet.
00:20:28.860 --> 00:20:30.780
This is what we did.
00:20:30.780 --> 00:20:35.160
And might as well write
it in the standard order,
00:20:35.160 --> 00:20:39.060
where the complex conjugate
function appears first.
00:20:45.300 --> 00:20:46.375
This is what we found.
00:20:48.880 --> 00:20:53.280
So this is actually very neat.
00:20:53.280 --> 00:20:57.810
Let me put the time back
everywhere you could put time.
00:20:57.810 --> 00:21:00.210
Because this is a
time dependent thing.
00:21:00.210 --> 00:21:09.480
So p, expectation value
is p phi of p and t
00:21:09.480 --> 00:21:21.960
squared dp is equal to
integral dx psi star of x.
00:21:21.960 --> 00:21:26.430
And now we have, if
you wish, p hat psi
00:21:26.430 --> 00:21:31.230
of x, where p hat
is what we used
00:21:31.230 --> 00:21:33.450
to call the momentum operator.
00:21:43.360 --> 00:21:46.000
So look what has happened.
00:21:46.000 --> 00:21:51.430
We started with this
expression for the expectation
00:21:51.430 --> 00:21:55.630
value of the momentum
justified by the probabilistic
00:21:55.630 --> 00:21:58.710
interpretation of phi.
00:21:58.710 --> 00:22:02.670
And we were led to
this expression, which
00:22:02.670 --> 00:22:05.620
is very similar to this one.
00:22:05.620 --> 00:22:10.440
You see, you have the psi
star, the psi, and the x there.
00:22:10.440 --> 00:22:15.050
But here, the momentum
appeared at this position,
00:22:15.050 --> 00:22:21.020
acting on the wave function
psi, not on psi star.
00:22:21.020 --> 00:22:24.360
And that's the way,
in quantum mechanics,
00:22:24.360 --> 00:22:30.270
people define expectation
values of operators in general.
00:22:30.270 --> 00:22:45.040
So in general,
for an operator q,
00:22:45.040 --> 00:22:50.680
we'll define the
expectation value of q
00:22:50.680 --> 00:23:05.440
to be integral dx psi
star of x and t q acting
00:23:05.440 --> 00:23:08.250
on the psi of x and t.
00:23:14.050 --> 00:23:22.280
So you will always do this
of putting the operator
00:23:22.280 --> 00:23:25.550
to act on the second part
of the wave function,
00:23:25.550 --> 00:23:28.190
on the second appearance
of the wave function.
00:23:28.190 --> 00:23:31.020
Not on the psi star,
but on the psi.
00:23:36.550 --> 00:23:41.950
We can do other examples of
this and our final theorem.
00:23:45.160 --> 00:23:47.380
This is, of course,
time dependent.
00:23:51.870 --> 00:23:58.190
So let me do one example and
our final time dependence
00:23:58.190 --> 00:24:00.565
analysis of this quantity.
00:24:22.410 --> 00:24:27.450
So for example, if you would
think of the kinetic operator
00:24:27.450 --> 00:24:30.110
example.
00:24:30.110 --> 00:24:42.030
Kinetic operator t is p squared
over 2m is a kinetic operator.
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How would you compute
its expectation value?
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Expectation value of the
kinetic operator is what?
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Well, I could do the
position space calculation,
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in which I think of the kinetic
operator as an operator that
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acts in position space where
the momentum is h bar over iddx.
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So then I would have integral
dx psi star of x and t.
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And then I would have minus
h squared over 2m d second dx
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squared of psi of x and t.
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So here I did exactly
what I was supposed
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to do given this formula.
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But you could do another
thing if you wished.
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You could say, look, I can
work in momentum space.
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This is a momentums operator p.
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Just like I defined the
expectation value of p,
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I could have the expectation
value of p squared.
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So the other
possibility is that you
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have t is equal to the
integral dp of p squared
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over 2m times phi of p squared.
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This is the operator.
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And this is the probability.
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Or you could write it
more elegantly perhaps.
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dp phi star of p t
squared over 2m phi of p.
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These are just
integrals of numbers.
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All these are numbers already.
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So in momentum
space, it's easier
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to find the expectation value
of the kinetic operator.
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In coordinate space,
you have to do this.
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You might even say, look,
this thing looks positive.
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Because it's p squared
of the number squared.
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In the center here,
it looks negative.
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But that's an illusion.
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The second derivative can
be partially integrated.
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One of the two
derivatives can be
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integrated to act on this one.
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So if you do partial
integration by parts,
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you would have integral
dx h squared over 2m.
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And then you would have d
psi dx squared by integration
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by parts.
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And that's clearly
positive as well.
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So it's similar to this.