Consider a "wavefunction" $\psi(x)$, which has a Fourier transform $\tilde \psi(p)$

Suppose that we know, for each $x$, $|\psi(x)|^2$, and that we know, for each $p$, $|\tilde \psi(p)|^2$.

Have we enough information to reconstruct the "wavefunction" $\psi(x)$, that is, obtain the phase of the "wavefunction" for all $x$ (up to a global phase, we are only interesting to the relative phases between the $x$)?

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    $\begingroup$ Readers might also like this related paper. $\endgroup$ Oct 19, 2013 at 11:51
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    $\begingroup$ This is related to the phase retrieval problem in diffraction. There you have the intensity of diffracted light---analogous to your $|\tilde{\psi}(p)|^2$, but instead of knowing the modulus of $\psi$ everywhere, you know the phase of the analogous thing---the density of scatterers. Specifically, you know the phase is zero, since the density is real. So basically, your porblem is like the phase retrieval problem except that you know the modulus of $\psi$ instead of the phase. Since the information is comparable, I would guess this is solvable. $\endgroup$ Oct 19, 2013 at 13:09
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    $\begingroup$ Related question on MO.SE: mathoverflow.net/q/188931/13917 $\endgroup$
    – Qmechanic
    Dec 5, 2014 at 9:53
  • $\begingroup$ By the way, this question was originally posed by Pauli $\endgroup$
    – lcv
    Dec 5, 2019 at 13:24
  • $\begingroup$ Here is the reference: W. Pauli, in Quantentheorie, edited by H. Geiger and K. Scheel, Handbiich der Physik Vol. 24 (Springer, Berlin, 1933), Pt. 1, p. 98 $\endgroup$
    – lcv
    Dec 5, 2019 at 13:31

4 Answers 4


No. Consider any state with a momentum wavefunction symmetric about zero. It's position-space and momentum-space norm-squared probability distributions are not changed by time-reversal, even though the wavefunction clearly is.

Here is an explicit example. Take the four Gaussian wavepacket of mean positions $x_0$ or $-x_0$, mean momenta $p_0$ or $-p_0$, and spatial spread $\sigma$:

$\psi_{(\pm_x , \pm_p)} (x) \propto e^{-(x \mp_x x_0))^2/4\sigma^2 - i x (\pm_p p_0) }$.

(Here, $\pm_x$ and $\pm_p$ are two binary variables taking the values of plus or minus. $\mp_x$ and $\mp_p$ are their opposites.) Now consider the two superpositions

$\phi_{\mathrm{away}} = \psi_{(+,+)} + \psi_{(-,-)} $,

$\phi_{\mathrm{toward}} = \psi_{(+,-)} + \psi_{(-,+)} $.

$\phi_{\mathrm{away}}$ is the superposition of two wavepackets separated by $2 x_0$ traveling away from each other with relative speed $2 p_0/m$. $\phi_{\mathrm{toward}}$ is the same with the packets traveling toward each other. One can check that $\vert \phi_{\mathrm{toward}} (x)\vert^2 = \vert \phi_{\mathrm{away}} (x) \vert^2$ and $\vert \tilde{\phi}_{\mathrm{toward}} (p) \vert^2 = \vert\tilde{\phi}_{\mathrm{away}} (p) \vert^2$.

I do not know if there are examples other than with time-reversal.


An intuitive dimensional reason why it couldn't work: a state vector in $\mathbb C^{N+1}$ is described by 2N real coordinates (one complex dimension is irrelevant), and so is its Fourier transform. If we only consider the normalized squared moduli of the components, we have 2N real numbers as well, so if these would actually be independent we should be able to reconstruct the original vector.

However, they are not independent: the most famous dependence between the squared modulus of a function and that of its Fourier transform is the Heisenberg uncertainty relation (of which analogues exist in a finite dimensional setting, if you want to remain in a setting where dimension counting is straightforward).

Another one is provided by the Paley-Wiener theorem, which implies that a compactly supported function has a Fourier transform that is not identically zero on any open set.

The smallest counterexamples occur in a 2-dimensional state space: the Fourier transform of $\begin{pmatrix}a\\ b\end{pmatrix}$ is $\begin{pmatrix}a + b\\ a - b\end{pmatrix}$ (up to a multiplicative constant), so e.g. $\begin{pmatrix}1\\ i\end{pmatrix}$ and $\begin{pmatrix}1\\ -i\end{pmatrix}$ describe different states but they have equal moduli and so do their Fourier transforms.

  • $\begingroup$ Yes, your counter-example is a discrete version of mine. (I'm happy to combine our answers, although I don't know how that's done.) A natural follow-up question is whether the pair (|f(x)|^2, |f(p)|^2) is sufficient to uniquely determine a state up to time-reversal. Based on your dimensional argument, I suspect it is (perhaps with additional discrete symmetries undetermined). $\endgroup$ Oct 21, 2013 at 12:25
  • $\begingroup$ very clear - fantastic simplicity! @JessRiedel I agree that unique determination up to time reversal seems likely. Shall think about that some more - sounds like you are too. $\endgroup$ Oct 21, 2013 at 13:25

Comments to the question (v1):

I) Reconstruction of phases from modulus$^1$ $|f(x)|$ of a signal $f(x)$ and modulus $|\tilde{f}(k)|$ of its Fourier transformed (FT) signal

$$\tag{1} \tilde{f}(k) ~:=~ \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} \!dx~ e^{-ikx} f(x)$$

is an interesting and likely a well-studied engineering problem, either for continuous or discrete Fourier transformation.

II) Example: A Gaussian signal

$$\tag{2} f(x)~=~Ae^{-\frac{a}{2}x^2+bx}, \qquad A,a,b\in \mathbb{C}, \qquad {\rm Re}(a)~>~0, $$

with Fourier transform

$$\tag{3} \tilde{f}(k)~=~\frac{A}{\sqrt{a}}\exp\left[-\frac{(k+ib)^2}{2a}\right] ~=~\frac{A}{\sqrt{a}}\exp\left[-\frac{\bar{a}}{2|a|^2}(k+ib)^2\right] ~=~\frac{A}{\sqrt{a}}\exp\left[-\frac{\bar{a}}{2|a|^2}(k^2+2ibk-b^2)\right];$$

with modulus

$$\tag{4} |f(x)| ~=~ |A| \exp\left[-\frac{{\rm Re}(a)}{2}x^2+{\rm Re}(b)x\right], $$


$$\tag{5} |\tilde{f}(k)| ~=~ \frac{|A|}{\sqrt{|a|}} \exp\left[-\frac{{\rm Re}(a)k^2-2{\rm Im}(\bar{a}b)k-{\rm Re}(\bar{a}b^2)}{2|a|^2}\right], $$

respectively. It is interesting that if one additionally knows that the signal is of Gaussian form (2), then it is possible from (4) and (5) to reconstruct the constant $a$ up to possibly a sign ambiguity of ${\rm Im}(a)$; the constant $A$ up to a phase; and the constant $b$ is unique for given choice of $a$.

III) The above Gaussian example induces hope that modulus of a signal and modulus of its FT are sufficiently complementary information such that reconstruction is possible up to possibly a finite number of self-consistent solutions, and modulo an overall global phase.

IV) We speculate that it may in practice be possible to reconstruct a self-consistent signal from its modulus and modulus of its FT via an iterative fixed-point algorithm$^2$: First Fourier transform the bare modulus of the signal; next multiply the phases of the result with the initially given FT modulus; then inverse FT; next multiply the phases of the result with the initially given modulus; then FT; and so forth, until a self-consistent fixed-point configuration is reached on each side of the FT.


$^1$ In signal analysis the modulus is also called amplitude, magnitude, or absolute value.

$^2$ Update: It turns out that this algorithm exists and is known as Gerchberg-Saxton algorithm (hat tip: WetSavannaAnimal aka Rod Vance).

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    $\begingroup$ You're spot on about the reconstruction - what you've "speculated" is real and known as the Gerchberg–Saxton algorithm. I don't know the theory of convergence of such things - we don't have a contraction mapping but an isometry for the FT - but I do know in practice this kind of thing works extremely well and I'm sure the convergence would have been studied. I'd be surprised if it weren't pretty universal - I was stunned how easy it is to get this working with even very noisy signals (isometry clearly helps here). $\endgroup$ Oct 19, 2013 at 23:25

Answer is no. Consider two waves: $\psi_{1} (x,t)= \psi (x,t)e^{i\alpha t} \space$ and $\space \psi_{2} (x,t)= \psi (x,t)e^{i\beta t}$. Surely those waves are different, but $|\psi_{1} (x,t)|=|\psi_{2} (x,t)|=| \psi (x,t)|$. So one cannot determine any wave in full detail from physical measurements.

  • $\begingroup$ Global phases are irrelevant, and every phase retrieval procedure deals with relative ones, as it is stated in this question. $\endgroup$
    – Andrestand
    Apr 29, 2015 at 8:54

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