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 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.
