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Let $\psi(x)$ be a quantum wavefunction on $\mathbb{R}$, that is, a complex function such that $\int_{-\infty}^{\infty} dx |\psi(x)|^2 = 1$. Let $\widetilde{\psi}(p)$ be its Fourier transform: $\widetilde{\psi}(p) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} dx e^{-ipx} \psi(x)$. Let $$ \langle x^d \rangle = \int_{-\infty}^{\infty} dx |\psi(x)|^2 x^d, $$ $$ \langle p^d \rangle = \int_{-\infty}^{\infty} dp |\widetilde{\psi}(p)|^2 p^d. $$ If we are only interested in $d=1,2$ then a complete list of constraints is known, namely $$\langle x^2 \rangle \geq \langle x \rangle^2$$ $$\langle p^2 \rangle \geq \langle p \rangle^2$$ $$(\langle x^2 \rangle - \langle x \rangle^2 ) ( \langle p^2 \rangle - \langle p \rangle^2 ) \geq \frac{1}{4},$$ the last of which is Heisenberg's uncertainty principle. It is not obvious that this set of constraints is complete, i.e. that any list of four numbers $\langle x \rangle, \langle x^2 \rangle, \langle p \rangle, \langle p^2 \rangle$ satisfying the above three inequalities is achievable by some normalized wavefunction $\psi(x)$. But this is proven in the quantum optics literature.

My question is: what if, in addition to $\langle x \rangle, \langle x^2 \rangle, \langle p \rangle, \langle p^2 \rangle$ we also include $\langle x^4 \rangle$ in our list of "moments". Can we find a complete list of constraints for these?

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    $\begingroup$ What have you tried? Have you tried simply cranking out a calculation by a method analogous to the one used for the lower moments? You write this as a question about quantum mechanics, but it seems to me more like a more general question about Fourier transforms. $\endgroup$
    – user21349
    Oct 10, 2016 at 22:40
  • $\begingroup$ Yes, I think it is really a harmonic analysis question. wlog you can take $\langle x \rangle = \langle p \rangle = 0$. This leaves three quantities: $\langle x^2 \rangle, \langle p^2 \rangle, \langle x^4 \rangle$. Two obvious inequalities are: the uncertainty principle $\langle x^2 \rangle \langle p^2 \rangle \geq 1/4$ and $\langle x^4 \rangle \geq \langle x^2 \rangle^2$. However, this list is definitely not complete. In particular, if the lower moments saturate the uncertainty principle then $\langle x^4 \rangle$ is uniquely determined. $\endgroup$
    – StephenJ
    Oct 11, 2016 at 0:57

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I have recently discovered some references which address this question:

Moments of the Wigner Distribution and a Generalized Uncertainty Principle
R. Simon and N. Mukunda
arXiv:quant-ph/9708037, 1997
(and references therein)

Uncertainty Principles Revisited
Kathi K. Selig
Electronic Transactions on Numerical Analysis Vol. 14, pg. 165-177, 2002

Moments of non-Gaussian Wigner distributions and a generalized uncertainty principle: I. The single-mode case
J. Solomon Ivan, N. Mukunda, and R. Simon
J. Phys. A:Math. Theor. 45 (2012) 195305

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