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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.

Sidharth Ghoshal
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    have you tried simple examples, such a Gaussians or finite dimensional vectors? – user619894 Oct 22 '21 at 05:54
  • I'm trying the toy problem of $\psi(x) = \frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}} e^{ix}$ but I am not able to prove that up to multiplicative constant it is the ONLY solution which satisfies $|\psi(x)|^2 = \frac{1}{2\pi}e^{-x^2}$ and $|\psi(p)|^2 = \frac{1}{2\pi}e^{-(p+1)^2}$ – Sidharth Ghoshal Oct 23 '21 at 04:05
  • You can go through the motions of the problem and you end up with a system, where you do some intuitive pattern matching and find the solution but its hardly a rigorous proof of uniqueness. – Sidharth Ghoshal Oct 23 '21 at 04:06
  • Also that strategy ONLY works for this specific face, the general problem I have posed remains out of reach for me at the moment, though I'm consider assuming additional structure on $\psi$ (ex: sum of gaussians) to see where that leads – Sidharth Ghoshal Oct 23 '21 at 04:07
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    If you start with finite dimensional vectors, you can count the number of equations and unknowns to see of even in principle this is possible. Even if it is possible, There needn't be a clean algorithm, maybe just some kind of Newton iteration, which brings to mind another approach: suppose you are "near" the correct functions, can the first order approximation be computed? I am also reminded of this: https://en.wikipedia.org/wiki/Phase_retrieval – user619894 Oct 24 '21 at 12:49
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    Have you tried by replacing $\mathbb{R}$ by $\mathbb{Z}/N\mathbb{Z}$? – Plop Nov 02 '21 at 19:15

1 Answers1

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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.

Cosmas Zachos
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