# Which position and momentum distributions arise from some wave function?

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.

• Possible duplicates: physics.stackexchange.com/q/81303/2451 and links therein. – Qmechanic Sep 17 '14 at 8:11
• 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. – Oscar Cunningham Sep 17 '14 at 8:43
• – John Rennie Sep 17 '14 at 9:20
• 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. – Kyle Kanos Dec 2 '19 at 12:44
• @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. – Oscar Cunningham Dec 2 '19 at 12:56

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 $g=|\mathscr{F}(\psi e^{i(\alpha+\beta t)})|^2=|\hat \psi|^2$

• 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. – Héctor Sep 20 '14 at 14:29
• @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 $(f,g)$. – Oscar Cunningham Sep 20 '14 at 14:37
• 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. – user59412 Sep 20 '14 at 14:48
• 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? – Oscar Cunningham Sep 20 '14 at 15:25
• 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. – user59412 Sep 20 '14 at 16:27