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PROFESSOR: We got here
finally in terms of position

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and in terms of momentum.

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So this was not an accident that
it worked for position and wave

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number.

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It works with position
and for momentum.

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And remember, this phi of
p now has interpretation

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of the weight that is associated
with a plane wave of momentum

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p, and you're summing
over P in here.

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So we'll do the natural
thing that we did with x.

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We'll interpret
phi of p squared--

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phi of p squared--

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dp is the probability.

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Find the particle with momentum
in the range p, p plus dp.

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Just the same way as we would
say that psi squared of x, dx

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is the probability to find
the particle between x and x

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plus dx.

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So this is allowed now by the
conservation of probability

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and this, therefore,
makes sense.

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It's a postulate, though.

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It's not something
that can derive.

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I can just argue that
it's consistent to think

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in that way, and that's
the way we finally

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promote this phi of p,
which did encode psi of x.

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Phi of b has the
same information

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as phi of x. phi of p is the
weight of the superposition

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but, finally, it's given a
probabilistic interpretation.

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It represents a probability
to find the particle

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with some momentum.

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So this is what is going to
allow us to do expectation

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values in a minute.

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But I want to close
off this discussion

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by writing for you the
three-dimensional versions

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of these equations.

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3D version of Fourier transform.

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So this is what we
want to rewrite.

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So what would it be?

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It's psi of the vector x.

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Since you're going to
have three integrals--

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because you're going into
it over three components

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of momentum-- this factor
appears three times.

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So actually, it's 2 pi h bar
to the three halfs integral phi

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of vector p into the i vector
p dot product vector x h

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bar d cube p, and phi of p
vector the inverse theorem--

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same factor, we keep the nice
symmetry between x and p--

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psi of x vector negative
exponent same dot product

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but negative exponent d cube x.

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So these are the three
dimensional versions

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of your x versus p.

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And there is a three-dimensional
version of Parseval.

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So oh there's a three
dimensional version

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of the delta function.

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Just like we had a
delta function here--

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a delta function in
three dimensional space

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would be delta cubed
x minus x prime would

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be one over 2 pi cubed
integral d cube k e to the i k

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vector x minus x prime vector.

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It's all quite analogous.

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I think you should appreciate
that you don't have to memorize

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them or anything like that.

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They won't be in
any formula sheet,

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but they are very
analogous expressions.

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Parseval also works
in the same way.

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And you have-- just
as you would imagine--

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that the integral all over
three dimensional space of psi

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of x squared is equal to
the integral over three

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dimensional momentum
space of phi of p squared.

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So the three results--

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the Fourier theorem the
delta function and Parseval

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hold equally well.