I learned that Moyal's evolution equation is the equation for the time-evolution of Wigner's quasi-probability distribution. However, I couldn't say I perfectly understand the meaning of this equation.

Here are questions to understand related concepts.

  • Does every solution to Moyal's evolution equation become Wigner's quasi-probability distribution?
  • If not, are there some equations for such purpose? i.e. Are there equations whose solutions are Wigner's quasi-probability distributions for a given Hamiltonian $H$?
  • If it's yes, could these reproduce every solution from Schr"odinger's equation? (This is I think, true, because we can make the Wigner's quasi-probability distribution directly, and since the equation are designed for such solutions, it trivially holds.)

I think these are simple questions, and maybe obvious. But, I'm not 100% sure about the answers.

EDIT: Maybe the original questions are somewhat useless. This may be what Zachos try to point out.

  • Some solutions to Moyal's evolution equation are not physically correct, and hence it will not be Wigner's quasi-probability distribution.

and so the answer to the first question is negative.

However, we could ask this :

  • Could Moyal's evolution equation predict results as Schroediger's equation do? What it means : for given system, investigating Scroedinger's equation give us a $\psi$ and hence we know all about the system in QM level. This question is for the case of Moyal's evolution equation.

  • And is this process independent to the Schroedinger's equation? What it means : Could we get the same physics without knowing $\psi$ and not solving Schroedinger's equation, but just from solving Moyal's evolution equation of given system?

What I guess is that this is all positive as Zachos mentioned in the comment of his answer.

  • $\begingroup$ @CosmasZachos Thanks for the link, however I can't catch the relationship of the paper to my question. Could you explain it? In fact, what I understood for Moyal's equation, there is already given Hamiltonian, and maybe it's somehow compatible to the Hamiltonian of Schroedinger's one $\endgroup$
    – ChoMedit
    Nov 21, 2022 at 15:11
  • $\begingroup$ Hence, the Schroedinger's one and maybe Moyal's one always generate physical solution.. $\endgroup$
    – ChoMedit
    Nov 21, 2022 at 15:12
  • 1
    $\begingroup$ Apologies, I linked the wrong paper. The booklet I had in mind was this. Moyal's and Schroedinger's equations say nothing about the physicality or not of their solutions! They describe the time evolution of a function which you must ensure is physical at the origin of time! If you put in nonphysical garbage IC you get time evolved unphysical garbage. To ensure physicality, read up on section 0.9 in the booklet, and equation (6), with extreme prejudice. $\endgroup$ Nov 21, 2022 at 15:18

1 Answer 1


This booklet addresses your questions in Chapter 0.9 and specifically in equation (6). The Moyal equation is just the von Neumann evolution equation for the density matrix in a different, phase-space, representation. The von Neumann equation amounts to the Schrödinger equation, i.e. they specify time development of functions, with absolutely no information on what such functions are. If the initial condition inputs are physical wavefunctions $\psi(x,t_0)$ or Wigner distributions (WF) $f(x,p,t_0)$ with all the requisite requirements for physicality (normalization, reality for the WF, eqn (6), etc), the evolution output is a physical function at time t. If the inputs are are garbage, the outputs are most likely garbage: the Hamiltonian bears no relation to the physicality of the solutions.

You are thus probably asking about the requisite conditions for a phase space function to be physical at a specific moment, e.g. the origin of time. For a real normalized phase-space function to be of the form $$ f(x,p)=\frac{1}{2\pi}\int\! dy~\psi^* \left (x-\frac{\hbar}{2} y \right )~e^{-iyp} ~ \psi \left (x+\frac{\hbar}{2} y \right ), $$ so bona-fide physical and normalized, you need to ensure its cross-spectral density, its Fourier transform, "left-right" factorizes, $$ \tilde{f}(x,y)=\int dp ~e^{ipy} f(x,p) ~ = ~ g^{*}_L (x-\hbar y/2) ~g_R (x+\hbar y/2)~. $$ That is,
$$ {\partial^2 ~~~\ln \tilde{f} \qquad \qquad \phantom{a} \over \partial(x-\hbar y/2)~\partial(x+\hbar y/2)} =0 ~, \tag{6} $$ so that, for real $f$, ~ $g_L=g_R$. You may thus work out the Schrödinger wavefunction from this cross-spectral density; the map is bijective.

This all holds for pure states; for mixed states, you have to parse and reorganize the above argument.

Further comments as per question edit

Could Moyal's evolution equation predict results as Schrödiger's equation does? What it means : for a given system, investigating Scroedinger's equation give us a 𝜓 and hence we know all about the system in QM. This question is for the case of Moyal's evolution equation.

Yes! Just as TDSE tells you everything about the future development of a complex normalized function as IC, just so, the Moyal equation tells you everything about a real normalized f satisfying (6) as IC, now for all times in the future.

And is this process independent of Schroedinger's equation? What this means: Could we get the same physics without knowing 𝜓 and not solving Schroedinger's equation, but just from solving Moyal's evolution equation of given system?

A qualified yes: as emphasized in Lemma 0.3 of Ch 0.10, much more important than Moyal's equation is the fundamental "stargenfunction" equation of the *-spectrum of the Hamiltonian!

The *-spectral resolution of the Hamiltonian essentially solves the problem, and its determination does not rely on wavefunctions 𝜓!

Once this star spectrum is known, (which is explicitly illustrated for the oscillator in Ch 0.13), the future development of a bona-fide WF at the origin of time is known completely for all points in the future. The IC WF is *-resolvable in these stargenfunctions, assuming it obeys all the above conditions, and propagates trivially as illustrated by Moyal's equation, just as 𝜓 does for the TDSE.

Indeed, this is the celebrated autonomy of the formulation: the formulation is self-standing, and equivalent, to the Hilbert space formulation (TDSE).

No knowledge of Schrödinger's equation is assumed. It would yield the same physics results on a planet where this formulation arose first, and then the TDSE of Hilbert space after it, the opposite of what happened on our plant!

(Truth be told, Hilbert space and the TDSE are easier to handle than Moyal's equation, so chemistry would be less advanced on that planet than in ours, unless these aliens were really smarter than us-- hardly a surprise...)

  • If you were mathematically inclined, all you have to appreciate is that the WF is the (invertible) Winger transform of the density matrix, and the Moyal equation of the von Neumann equation. You are then solving the von Neumann equation in an unfamiliar representation (phase space), Feynman's point of view.
  • $\begingroup$ Thanks for answer, however I still wonder how can we get $f$ systematically, independently from $\psi$. I mean, in many case, the only possible way to compute $\psi$ is from solving Schroedinger's equation to find it. Could we do the similar things for finding $f$? I think to determine something physical would be done after we compute some candidates. $\endgroup$
    – ChoMedit
    Nov 21, 2022 at 16:15
  • $\begingroup$ I thought the criteria suggested in booklet telling when $f$ has the form of Wigner's distribution seems weak, because we still don't know how the distribution is related to the Hamiltonian of the given system. $\endgroup$
    – ChoMedit
    Nov 21, 2022 at 16:16
  • $\begingroup$ The Hamiltonian has little to do with physics: it only specifies time evolution; but it might have a nice set of eigenfunctions or stargenfunctions. Read on to Ch 0.10 for the stargenvalue time independent solutions, then Chapter 0.13 for solving for the oscillator in a universe where the Schroedinger wavefunction was never known!! Perhaps you need to refocus your question. $\endgroup$ Nov 21, 2022 at 16:33
  • $\begingroup$ Those chapters could be the answer to my question! Thanks. Indeed, those chapters seem to explain what I want to know. $\endgroup$
    – ChoMedit
    Nov 21, 2022 at 16:44
  • $\begingroup$ No problem, thanks for verification and good answer. I've realized that you are one of the author of the booklet. It's honor to hear your answer. $\endgroup$
    – ChoMedit
    Nov 24, 2022 at 0:40

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