Why does knowledge of $\langle 0| x^m p^n |0\rangle$ determine the ground state wave function? As part of an exercise in Banks' QFT book (3.1) it is claimed that for a quantum mechanical system a knowledge of $\langle 0| x^m p^n|0 \rangle$ for all $n,m$ allows us to calculate the expression for the ground state wave function $\psi_{0}(x)$. I fail to see why this is so. Since the above matrix elements are simply numbers, my guess is that they are the coefficients in some expansion of $\psi_{0}(x)$, but I can't see the precise relation.
 A: $\newcommand{\avg}[1]{\left<#1\right>}\newcommand{\braopket}[3]{\left<#1\middle|#2\middle|#3\right>}$I will not provide a full solution, but a hint, that might help to proceed (I might later extend this to a full answer). The statement is related to the fact, that a probility density can be reconstructed from its moments (that is averages of the form $\avg{x^n}$). That is given $\avg{x^m}$ for all $m$ you can reconstruct $P(x)$. This can be seen as follows. Consider the characteristic function 
$$\chi(\tau) = \int_{-\infty}^\infty dx\, e^{-i\tau x} P(x) = \avg{e^{-i\tau x}}$$
By Taylor expansion of the $e$-function we get:
$$\chi(\tau) = \sum_{n=0}^\infty \frac{\avg{(-i \tau x)^n}}{n!} = \sum_{n=0}^\infty \frac{(-i)^n}{n!} \tau^n \avg{x^n}.$$
That means, that $\chi(\tau)$ can be computed from the moments, on the other hand $\chi(\tau)$ is the Fourier transform of $P(x)$, so
$$ P(x) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{i\tau t} \chi(\tau).$$
Side note: According to the Hohenberg-Kohn theorems the ground state wave function is fully determined by the ground state particle density (in a 1d system this holds for all bound states as those can be chosen real and so $\psi(x) = \sqrt{n(x)}$ will give a solution). So it is actually enough to know $\braopket{0}{x^n}{0}$ for all $n$, from which you can easily derive $n(x)$ (the probability density of finding a particle).
