friend said:
So I take it that p here is the momentum at x (not at x' ). I mean, of course it is, right?
No, there is no "x" in the expression \langle p|U(t)|x'\rangle. Roughly speaking, this is the probability amplitude that a particle initially located at x' will be found later to have a momentum p.
If we have a function of x, say f(x), then its Fourier transform is a momentum at that x, right?
No, that's not correct at all. The point of a Fourier transform is to represent a function f(x) as a linear combination of functions of the form e^{ikx}. You already know, I assume, that sines and cosines can be written as combinations of complex exponentials:
sin(\frac{2 \pi x}{L}) = \frac{1}{2i} e^{i\frac{2 \pi x}{L}} + \frac{-1}{2i} e^{-i\frac{2\pi x}{L}}
cos(\frac{2 \pi x}{L}) = \frac{1}{2} e^{i\frac{2 \pi x}{L}} + \frac{1}{2} e^{-i\frac{2\pi x}{L}}
Well, any function that is periodic with period L (that is, f(x + L) = f(x)) can be written as a combination of exponentials:
f(x) = A_0 + A_{-1} e^{-i \frac{2\pi x}{L}} + A_1 e^{+i \frac{2\pi x}{L}} + A_{-2} e^{-2 i \frac{2\pi x}{L}} + A_2 e^{+2i \frac{2\pi x}{L}} + ... = \sum_n A_n e^{n i \frac{2 \pi x}{L}}
The Fourier transform of f(x) is just the coefficients of the expansion of f: A_n. With a Fourier integral, we replace the sum by an integral, and instead of
f(x) = \sum_n A_n e^{i n \frac{2 \pi x}{L}}
we write:
f(x) = \int dk A(k) e^{i k x}
(The latter can sort of be thought of as the limit of the first, as L \rightarrow \infty). The Fourier transform of f(x) is A(k). That is just a number, (complex, in general). It is not a momentum. How Fourier transforms relate to momenta is this:
If a particle is initially in a state described by the wave function f(x), and you measure its momentum, then the probability that the result will be p \pm \delta p is proportional to |A(p/\hbar)|^2 \delta p. The Fourier transform of a wave function doesn't give a momentum, it gives a probability distribution on momenta.