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Consider a particle in one dimension with wave function $\psi$. The probability density function describing how likely it is to find it in a given position is given by $f(x)=\left|\psi(x)\right|^2$. Similarly the distribution for momentum is given by $g(p)=\left|\hat{\psi}(p)\right|^2$ where $\hat\psi$ is the fourier transform of $\psi$.

Which pairs of functions $(f,g)$ arise in this way?

The question could be formalised mathematically by asking for which $f,g\in L^1_{\mathbb R_{\geq 0}}$ there exists a $\psi\in L^2_{\mathbb C}$ such that $f(x)=\psi(x)^2$ and $g(p)=\hat\psi(p)^2$ for almost all $x$ and $p$.


I'm not interested in the problem of reconstructing $\psi$ from $f$ and $g$, which is impossible because the same $f$ and $g$ can arise from different $\psi$s. Instead I want a way to check if a given pair $(f,g)$ arises from any $\psi$ whatsoever.


The uncertainty principle gives a partial answer to this question. It says that $f$ and $g$ can't arise from a wavefunction if the product of their second moments is too large. So one way of rephrasing my question would be to say that I'm looking for a strengthening of the uncertainty principle that is strong enough to rule out every $f$ and $g$ that don't arise from a wavefunction.

The Entropic Uncertainty Principle is an example of a strengthening of the usual Uncertainty Principle that rules out more $(f,g)$ pairs. It says that for all $\alpha>1/2$ and $\beta$ chosen such that $1/\alpha+1/\beta=2$ we have

$$H_\alpha(f)+H_\beta(g)\geq \log\left(\pi\hbar\right) + \frac 1 2 \left(\frac{\log\alpha}{\alpha-1}+\frac{\log\beta}{\beta-1}\right)$$

where $H$ is the Rényi entropy. For $\alpha=\beta=1$ we interpret the expression by taking limits to get $H_1(f)+H_1(g)\geq\log\left(\pi\hbar\right)+1$ where $H_1$ is the usual entropy.

This is stronger than the usual uncertainty principle since we always have that $$\sigma_f\geq\frac 1{\sqrt{2\pi}}\exp\left(H_1(f) - \frac 12\right)$$ and hence $$\sigma_f\sigma_g\geq\frac 1{2\pi}\exp\left(H_1(f)+H_1(g)-1\right)\geq \frac {\exp\left(\log\left(\pi\hbar\right)\right)}{2\pi}=\frac \hbar 2\text{.}$$

But it still doesn't rule out all inadmissible $(f,g)$ pairs. For example if we take $f$ and $g$ to both be the uniform distribution on an interval of length $L\geq\sqrt{e\pi\hbar}$ then the above relation is satisfied for all $\alpha$. And yet such $f$ and $g$ cannot arise from a wavefunction because Benedicks's theorem states that they can't both have finite support.

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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/81303/2451 and links therein. $\endgroup$ – Qmechanic Sep 17 '14 at 8:11
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    $\begingroup$ I found several references for that problem when trying to find an answer for mine. It's the "uniqueness" part of the problem for which I want the "existence" part. $\endgroup$ – Oscar Cunningham Sep 17 '14 at 8:43
  • $\begingroup$ Related: Conjugate Variables, Noether's Theorem and QM $\endgroup$ – John Rennie Sep 17 '14 at 9:20
  • $\begingroup$ What kind of answer are you expecting? It seems rather list-based (e.g., given X type system, you have f1,g1; given Y, you get f2,g2; etc). And it also seems something that you more or less have to calculate directly to know. $\endgroup$ – Kyle Kanos Dec 2 '19 at 12:44
  • $\begingroup$ @KyleKanos For example, the uncertainty principle gives a partial answer to the question. It says that $f$ and $g$ can't arise from a wavefunction if the product of their second moments is too large. So one way of rephrasing my question would be to say that I'm looking for a strengthening of the uncertainty principle that is strong enough to rule out every $f$ and $g$ that don't arise from a wavefunction. $\endgroup$ – Oscar Cunningham Dec 2 '19 at 12:56
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Fourier transform is an linear operator: $$\mathscr{F}(\psi(x))=\hat \psi (k) =\frac{1}{\sqrt{2\pi}}\int \psi (x) e^{-ikx} dx$$ $$\mathscr{F}(\alpha \psi (x)) =\frac{1}{\sqrt{2\pi}}\int \alpha \psi (x) e^{-ikx} dx=\alpha\hat \psi (k)$$

Uncertainty principle is not any constraint since transform pair $\psi $ and $\hat \psi$ automatically obey Heisenberg's inequality, that is, the mathematical uncertainty principle (google it!). When you change variable $k=p/\hbar$ Heisenberg's inequality turns into Heisenberg's uncertainty principle.

Now, since the moduli of complex number doesn't change if it's multiplied by one: $|\psi e^{i\alpha}|=|\psi|$. Again, since time is a dummy constant in Fourier transform, we even have: $|\psi e^{i(\alpha+\beta t)}|=|\psi|$. But multiplying by $e^{i\alpha x}$ is not allowed due to modulation property of Fourier transform.

So answer to your question might be a bit boring: if $\psi$ and derivative $\psi^{'}$ are square-integrable.

Now you simply have $f=|\psi e^{i(\alpha+\beta t)}|^2=|\psi|^2$ and $g=|\mathscr{F}(\psi e^{i(\alpha+\beta t)})|^2=|\hat \psi|^2$

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    $\begingroup$ I don't think this answer OP question. It's clear that given $\psi$ you can take fourier tranform to get a compatible pair of square integrable function. OP ask if given a pair of functions $(f,g)$ there are complex functions $\psi$ and $\hat\psi$ where their square are $f$ and $g$ and are related each other by fourier transform. $\endgroup$ – Héctor Sep 20 '14 at 14:29
  • $\begingroup$ @Héctor Yes, you're right. I want some way of knowing whether $(f,g)$ is possible, without having to check every $(\psi,\hat\psi)$ to see if it gives rise to $(f,g)$. $\endgroup$ – Oscar Cunningham Sep 20 '14 at 14:37
  • $\begingroup$ Since Fourier transform is an isomorphism there is a linear bijection between $\psi$ and $\hat \psi$. Now, since $|\psi e^{i(\alpha+\beta t)}|^2=|\psi|^2$ that isomorphism is up to constants $\alpha$ and $\beta$. EDIT: Oscar: can you give any counter-example? To my knowledge it's enough if $\psi$ and derivative $\psi^{'}$ are square-integrable. There are no constraints. $\endgroup$ – user59412 Sep 20 '14 at 14:48
  • $\begingroup$ An example of the question I'm trying to ask is: Can $f$ be the Cauchy distribution whilst $g$ is the Normal distribution? i.e. Is there some $\psi$ that leads to those two distributions in particular? $\endgroup$ – Oscar Cunningham Sep 20 '14 at 15:25
  • $\begingroup$ I see. All I know is that if $|\psi|^2$ is Normal distribution then $|\hat\psi|^2$ is also, so it cannot be Cauchy or any other. But generally $\psi$ gets very concoluted in Fourier transform, e.g. transformation of unit box-function is very waving. And a $\psi$ that has no symmetry at all is a mess. $\endgroup$ – user59412 Sep 20 '14 at 16:27

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