The background

I want to solve the $\star$-genvalue equation

$$ H(x,p) \star \psi(x,p) = E~\psi(x,p),$$

where $\star$ denotes the Moyal star product given by

$$ \star \equiv \exp \left\lbrace \frac{i \hbar}{2} \left( \overset{\leftarrow}{\partial}_{x} \overset{\rightarrow}{\partial}_{p} - \overset{\leftarrow}{\partial}_{p} \overset{\rightarrow}{\partial}_{x} \right) \right\rbrace. $$

I consider the simplest possible case, the Hamiltonian of a free particle in one dimension which can be given by

$$ H(x,p) = \frac{p^2 }{2m}. $$

Since this is a finite polynomial on phase space, the Moyal product truncates naturally at the second order. As a consequence, the $\star$-genvalue equation reduces to a second order ordinary differential equation (ODE) of the form

$$ \left(- \frac{i\hbar}{2m} p~\partial_x - \frac{\hbar^2}{8m}~\partial_x^2\right) \psi(x,p)= \left( E - \frac{p^2}{2m} \right) \psi(x,p). $$

If we analyse this equation in the asymptotic limit $\hbar \to 0$, we find for general solutions $\psi(x,p)$ that the energy is $E = p^2 /(2m)$ as expected.

We might however solve the ODE quite easily at any order of $\hbar$. One obtains a linear combination of the two solutions

$$ \psi_\pm(x,p) = \exp \left\lbrace 2 i x \left( \sqrt{2m E} \pm p\right)/\hbar\right\rbrace = \exp \left\lbrace 2 i x \left( |p| \pm p\right)/\hbar\right\rbrace = \exp \left\lbrace 4 ipx\theta(\pm p)/\hbar\right\rbrace, $$

with the Heaviside step function $\theta(p)$. At this moment of the calculation I became amazingly confused about the meaning of what I just did. Apparently, $\star$-genfunctions can have some oscillating behaviour which will vanish when averaged over the scale of $\Delta x \Delta p \approx \hbar$. This reminds me of free wave propagation.

I tried to gain some insight from the corresponding Schrödinger equation $\hat{H} |{\psi}\rangle = E |{\psi}\rangle$. However, the Schrödinger equation would have to be promoted to an operator equation in order to make sense of its Wigner transformation.

My question

Is there a general relationship between the $\star$-genvalue above and the stationary Schrödinger equation in a way that you could obtain the solution for one equation from the other?


2 Answers 2


Basically there is, for the real part of the Wigner function, Lemma 3 (pp 27-29) of my book, CTQMPS. This is the basic exercise any student of that structure should do, whether instructed to, or not. Let's be cavalier with normalizations, since plane waves are not normalizable, anyway.

Write the constant eigenvalue E to be the convenient form $\frac{\hbar^2k^2}{2m}$, for constant k.

The eigenfunctions of the free Schroedinger equation, $\frac{\hbar^2(k^2+\partial_x^2)}{2m}\psi(x)=0$, are your usual plane waves, $a e^{ikx}+ b e^{-ikx}$.

The corresponding Wigner function then is, trivially, in cavalier normalizations, $$ f(x,p) \propto aa^* \delta (p-\hbar k) + bb^* \delta (p+\hbar k) + \delta(p)(ab^* e^{i2kx} + a^*b e^{-i2kx})~, $$ real, confined to just three parallel, flat, razor-thin lines in phase space. The zero momentum line is not dross: it is the heart of the construction--beats!

You immediately see that it solves the associated $\star$-genvalue equation $$ \left(- \frac{i\hbar}{2m} p~\partial_x - \frac{\hbar^2}{8m}~\partial_x^2 - \frac{\hbar^2k^2 -p^2}{2m} \right) f(x,p)=0~, $$ by virtue of the respective delta functions.

In fact, it also solves the c.c. equation, namely the $\star$-genvalue equation $f\star H= E f$, as the equivalence lemma dictates. In the p-outprojected piece, you discern the Schroedinger equation, for rescaled eigenvalues, if you'd really wish to go there.... but why? The point of these maps is to shed the Schroedinger formalism and Hilbert space, and use concepts of classical mechanics.

Your rush to enforce the classical $|p|=|k|$ was thus unwarranted. You are right, though, that the zero mode at p = 0 has no classical x dependence because of the complete delocalization enforced by the uncertainty principle, which is built into this formalism (cf. the relevant chapter of that text.) In fact, it is easy to see the p =0 line vanishes altogether in the classical limit. For a fixed E and vanishing $\hbar$, the exponentials $\exp (\pm i2x\sqrt{2mE}/\hbar)$ oscillate wildly and destructively to 0 as usual for the classical limit, and the middle line "trajectory" disappears, and properly did not present itself to you in your limit. Taking a=b=1 you may instantly see the negative values dictated by the $\cos(2xk)$ form, the hallmark of QM interference: beats.

To summarize: For any potential, not just the vanishing one, the crucial link between the solutions ψ of the Schroedinger equation and the $\star$-genvalue equation is the Wigner function, $$ f(x,p)=\frac{1}{2\pi}\int dy ~\psi^* \left (x-\frac{\hbar}{2}y\right ) e^{-iyp} \psi\left ( x+\frac{\hbar}{2}y\right ). $$ It then follows as per Lemma 3, that the Schroedinger equation solutions plugged into this yield real solutions of the $\star$-genvalue equation.

Conversely, real solutions of the $\star$-genvalue equation are perforce of the above bilinear form, where each ψ solves the Schroedinger equation.
(We largely worked out this explicit lemma for pedagogical purposes, as, given the invertible pair of Wigner-Weyl maps, it would be inconceivable for it to fail! But the public clamored for direct, line-by-line proofs.) In our case, this is even more direct: Fourier-transforming just p , $\tilde{f}(x,y)=\int dp e^{iy p} f(x,p)\propto aa^* e^{iy\hbar k} + bb^* e^{-iy\hbar k} + ab^* e^{i2kx}+a^*b e^{-i2kx}$, the above stargenvalue eqn collapses to $$ \int dy ~ \left ( (\partial_y\pm \frac{\hbar}{2}\partial_x)^2+ \hbar^2 k^2 \right) \tilde{f}(x,y) =0, $$ thus factorizing into left-right Schroedingers in light-cone variables, $\psi^*(x-\hbar y/2)\psi(x+\hbar y/2)$, alright.

  • $\begingroup$ Very nice and complete answer. Thank you very much! $\endgroup$ Commented Feb 21, 2018 at 7:49

You can look more about it here https://www.sciencedirect.com/science/article/pii/S0003491604000466 There is a theorem where $\psi\star\psi^\dagger=f_w$. Where $f_w$ is the Wigner function. With Liouville-von Neumann equation as the time evolution equation of the Wigner Function.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.