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PROFESSOR: What we
want to understand

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now is really about
momentum space.

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So we can ask the
following question--

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what happens to the
normalization condition

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that we have for the
wave function when

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we think in momentum variables?

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So yes, I will do this first.

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So let's think of integral
dx psi of x star psi of x.

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Well, this is what we called
the integral, the total integral

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for x squared, the thing
that should be equal to 1

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if you have a probability
interpretation,

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for the wave function.

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And what we would
like to understand

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is what does it
say about phi of k?

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So for that, I have to
substitute what [? size ?]

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r in terms of k and try to
rethink about this integral how

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to evaluate it.

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So for example, here, I
can make a little note

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that I'm going to use a
variable of integration

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that I call k for
the first factor.

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And for this factor, I'm
going to use k prime.

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You should use different
variables of integration.

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Remember, psi of x is
an integral over k.

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But we should use different
ones not to get confused.

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So here we go--

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dx 1 over square root of 2 pi--

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integral-- we said
this one is over k.

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So it would be phi star--

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let me put the dk first--

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dk phi star of k e to
the minus ikx and dk.

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That's the first psi.

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This is psi star.

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And now we put psi.

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So this is the dk prime--

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phi of x dk prime phi of k
prime e to the ik prime x.

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It's the same x in
the three places.

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OK, at this moment,
you always have

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to think, what do I do next?

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There are all these
many integrals.

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Well, the integrals
over k, there's

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no chance you're going to be
able to do them apparently--

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not to begin with, because
they are abstract integrals.

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So k integrals have no chance.

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Maybe the integral that
we have here, the dx,

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does have a chance.

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So in fact, let me write
this as integral dk phi

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star of k integral dk
prime phi of k prime.

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And then I have 1 over
2 pi integral dx e

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to the ik prime minus kx.

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I think I didn't
miss any factor.

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And now comes to help this
integral representation

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of the delta function.

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And it's a little opposite
between the role of k and x.

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Here the integration
variable is over x.

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There is was over k.

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But the spirit of the equality
or the representation is valid.

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You have the 1 over 2 pi, a
full integral over a variable,

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and some quantity here.

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And this is delta
of k prime minus k.

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And finally, I can
do the last integral.

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I can do the integral,
say, over k prime.

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And that will just give me--
because a delta function,

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that's what it does.

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It evaluates the
integrand at the value.

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So you integrate over k prime--

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evaluates phi at k.

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So this is equal to integral
dk phi star of k phi of k.

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And that's pretty neat.

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Look what we found.

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We've found what is called
Parseval's theorem, which

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is that integral dx
of psi of x squared

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is actually equal to
integral dk phi of k squared.

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So it's called
Parseval theorem--

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Parseval's theorem.

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Sometimes in the
literature, it's also

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called Plancherel's theorem.

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I think it depends on the
generality of the identity--

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so Plancherel's theorem.

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But this is very nice
for us, because it

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begins to tell us there's,
yes indeed, some more

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physics to phi of k.

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Why?

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The fact that this integral
is equal to 1 was a key thing.

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Well, the fact that it
didn't change in time thanks

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to showing the equation
was very important.

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It's equal to 1.

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We ended up with a
probabilistic interpretation

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for the wave function.

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We could argue that this could
be a probability, because it

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made sense.

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And now we have a very
similar relation for phi of k.

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Not only phi of k represents
as much physics as psi of k,

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as psi of x, and it
not only represents

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the weight with which you
superimpose plane waves,

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but now it also satisfies
a normalization condition

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that says that the
integral is also

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equal to this integral,
which is equal to 1.

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It's starting to lead to
the idea that this phi of k

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could be thought
maybe as a probability

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distribution in this new
space, in momentum space.

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Now I want to make momentum
space a little bit more clear.

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And this involves a little
bit of moving around

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with constants,
but it's important.

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We've been using k all the time.

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And momentum is h bar k.

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But now let's put things
in terms of momentum.

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Let's do everything
with momentum itself.

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So let's put it here.

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So let's go to momentum space,
so to momentum language--

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to momentum language.

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And this is not difficult. We
have p is equal to h bar k.

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So dp over h bar is equal
to dk in our integrals.

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And we can think
of functions of k,

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but these are just other
functions of momentum.

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So I can make these replacements
in my Fourier relations.

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So these are the two equations
that we wrote there--

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are the ones we're aiming
to write in a more momentum

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language rather than
k, even though it's

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going to cost a few h
bars here and there.

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So the first equation
becomes psi of x equal 1

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over square root
of 2 pi integral.

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Well, phi of k is now
phi of p with a tilde.

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e to the ikx is e to
the ipx over h bar.

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And dk is equal
to dp over h bar.

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Similarly, for the second
equation, instead of phi of k,

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you will put phi tilde
of p 1 over square root

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of 2 pi integral psi of x e
to the minus ipx over h bar.

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And that integral doesn't
change much [? dx. ?]

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So I did my change of variables.

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And things are not completely
symmetric if you look at them.

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Here, they were
beautifully symmetric.

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In here, you have
1 over h bar here

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and no h bar floating around.

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So we're going to do
one more little change

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for more symmetry.

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We're going to redefine.

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Let phi tilde of p be replaced
by phi of p times square root

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of h bar.

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You see, I'm doing
this a little fast.

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But the idea is that this
is a function I invented.

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I can just call it
a little different,

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change its normalization
to make it look good.

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You can put whatever you want.

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And one thing I did,
I decided that I

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don't want to carry all
these tildes all the time.

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So I'm going to replace phi
tilde of p by this phi of p.

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And that shouldn't be
confused with the phi of k.

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It's not necessarily
the same thing,

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but it's simpler notation.

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So if I do that here, look--

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you will have a phi
of p, no tilde, and 1

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over square root of h bar.

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Because there will be
e square root of h bar

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in the numerator
and h bar there.

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So this first equation
will become psi of x 1

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over square root of 2 pi
h bar integral phi of p e

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to the ipx over h bar dp.

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And the second
equation, here, you

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must replace it by a phi and
a square root of h bar, which

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will go down to
the same position

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here so that the
inverse equation is

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phi of p, now 1 over
square root of 2 pi h bar

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integral psi of x e to the
minus ipx over h bar dx.

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So this is Fourier's theorem
in momentum notation,

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in which you're really
something over momenta.

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And you put all these h
bars in the right place.

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And we've put them
symmetrically.

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You could do otherwise.

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It's a choice.

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But look at the
evolution of things.

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We've started with a
standard theorem with k

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and x, then derived
a representation

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for the delta function,
derived Parseval's theorem,

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and finally, rewrote this
in true momentum language.

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Now you can ask what happens
to Parseval's theorem.

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Well, you have to keep track
of the normalizations what

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will happen.

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Look, let me say it.

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This left-hand side, when
we do all these changes,

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doesn't change at all.

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The second one, dk,
gets a dp over h bar.

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And this is becomes
phi tilde of p.

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But phi tilde of p then
becomes a square root,

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then it's phi of p.

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So you get two square roots
in the numerator and the dk

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that had a h bar
in the denominator.

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So they all disappear, happily.

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It's a good thing.

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So Parseval now
reads, integral dx psi

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squared of x equal
integral dk phi of k--

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phi of p, I'm sorry.

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I just doing p now.

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And it's a neat formula
that we can use.