Reconstruction of “wavefunction” phases from $|\psi(x)|$ and $|\tilde \psi(p)|$

Question

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 $x$)?

Answer 1

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.

Answer 2

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.

Answer 3

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

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

and

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

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.

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$^1$ 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).

Answer 4

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.