Suppose you have two functions $f,g: \mathbb{R} \rightarrow \mathbb{C}$ and you have their product and convolution
$$ h_1(t) = fg$$
$$ h_2(t) = \int_{-\infty}^{\infty} f(t-\tau)g(\tau) d\tau $$
Is it possible to express $f,g$ solely in terms of $h_1$, $h_2$? I have reason to believe no, but it should be possible up to a constant factor that is on the unit circle.
I would like to ask: how to go about doing so?
Why I think it is possible:
I believe if you know the entire momentum probability distribution $|\phi(p,t)|^2$ and position probability distribution $|\psi(x,t)|^2$ of an object you should be able to recover the wave function of the object up to a multiplicative factor of a root of unity.
If we define the physicists "fourier transform" as
$$ \mathcal{F}_x[f] = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty}f(x) e^{\frac{ipx}{\hbar}} dx$$
Then we can state the famous momentum position relationship as
$$ \psi(x,t) = F_x[\phi(p,t)]$$
$$ \phi(p,t) = F_p[\psi(x,t)]$$
It then follows that:
$$ \sqrt{2\pi\hbar} \mathcal{F}_x^{-1} [|\phi(p,t)|^2] = \psi(x,t) \star \psi^*(x,t)$$
$$ |\psi(x,t)|^2 = \psi(x,t)\psi^*(x,t) $$
Where $\star$ indicates convolution. I'm taking a bet that it should be possible to solve for $\psi, \psi^*$ here up to a multiplicative root of unity, although I am not clear how to do so.
Best Answer
This is a comment, squatting in an answer slot, for space purposes. You only have one, nightmarish, equation to solve for f, $$ h_2(t) = \int_{-\infty}^{\infty} e^{-\tau \partial_t}f(t)~~{h_1(\tau) \over f(\tau)} d\tau ,$$ but your bet is risky...
This is the celebrated Pauli problem.