Probability density for position and momentum and the wave function

Question

I know that if a particle has a wave function $\Psi(x)$ at a time $t$ then the probability density for the position of the particle is given by $|\Psi(x)|^2$, and if $\phi(p)$ is the Fourier transform of $\Psi(x)$ (or a multiple of it) then $|\phi(p)|^2$ is the probability density for the momentum at time $t$.

My question is : Do the two functions $|\Psi(x)|^2$ and $|\phi(p)|^2$ give us enough Information to guess the function $\Psi$, meaning they would give us all the information about the particle at time $t$?

Intuitively, I would think that, as $x$ and $p$ are all that we can measure, it would be weird for $\Psi$ to depend on anything else than this probability densities, but I never managed to find a book or a pdf with an answer.

Answer 1

Sorry, this is only a partial answer.

Let's set up some notation, which is always useful. Let $\Psi(x)$ be the actual wavefunction of the particle and $\Phi(p)$ its Fourier transform. By assumption, we know $|\Psi(x)|$ and $|\Phi(p)$. Let us define $$\psi(x) = |\Psi(x)| ~~~~\mathrm{and}~~~~~ \phi(p) = |\Phi(p)|.$$ We know that $$\Psi(x) = e^{i\alpha(x)}\psi(x),$$ for some real function $\alpha(x)$, because the absolute value of a product is the product of the absolute values $|f(x)g(x)| = |f(x)|\cdot|g(x)|$.

Your question is then: can we determine $\alpha(x)$ given only $\psi(x)$ and $\phi(x)$?

In general the Fourier Transform of $\phi(x)$ will be different from the Fourier transform of $\Phi(x)$. This is because the Fourier transform of the product of two functions is the convolution of the Fourier transform of the functions: $$\mathcal{F}[f\cdot g] = \mathcal{F}[f]\star\mathcal{F}[g].$$

Now it's only a matter of coming up with a counterexample. Find two functions $\alpha(x)$ and $\beta(x)$ such that $e^{i\alpha(x)}\psi(x)$ and $e^{i\beta(x)}\psi(x)$ have the same absolute value for at least one value of $p$.

Answer 2

It depends what you mean by “guess” but in principle yes although you’d need infinitely many values of $\vert\Psi(x)\vert^2$ to proceed with 100% certainty.

The mathematical difficulty here is that, given $\vert\Psi(x)\vert^2$, $\Psi(x)$ is not unique: in particular $\Psi(x)$ and $e^{i\varphi}\Psi(x)$ given the same probability density. The physics however states that two functions that differ only by an overall phase are physically equivalent in the sense that they lead to the same predictions for all value of $\langle x^n\rangle$.

Now one still has to be quite careful: this is limited to expectations values of $\hat x$ or $\hat p$, and their powers. In particular, using supersymmetric quantum mechanics, it is possible to construct systems where a given function $\Psi(x)$ is an eigenfunction of two different Hamiltonians: thus, given a probability density $\vert\Psi(x)\vert^2$, one can reconstruct (a family of equivalent) $\Psi(x)$, but cannot infer anything more about the physics of variables other than $x$ or $p$. It is not possible to infer (for instance) the potential $V(x)$ or the energy levels from $\vert\Psi(x)\vert^2$.

Finally, if you had chosen the probability density of an operator with a discrete spectrum (such as spin for instance), then it is possible using quantum state tomography to reconstruct the state, provided you do enough measurements. There are systematic procedures to invert the experimental data and reconstruct the states.