# Prove that $f_\psi(x,p)$ is the Wigner Function of a pure state iff $H\star f_\psi= E f_\psi$

Given a pure state $$|\psi\rangle$$ with position wavefunction $$x\mapsto\psi(x)$$, define its Wigner function as $$f_\psi(x,p) = \frac{1}{2\pi} \int dy e^{-iyp} \psi(x+y/2)\psi^*(x-y/2) \equiv \frac{1}{2\pi}\int dy e^{-iyp}\tilde f_\psi(x,y),$$ where the last expression highlights that this is the Fourier transform of $$\tilde f_\psi(x,y)\equiv \psi(x+y/2)\psi^*(x-y/2)$$.

As far as I understand, if $$|\psi\rangle$$ is eigenstate of $$H$$ with energy $$E$$, i.e. $$H|\psi\rangle=E|\psi\rangle$$ and $$H(\mathfrak x,\mathfrak p)\psi=E\psi$$ (here $$\mathfrak{x,p}$$ denote the operators corresponding to $$x,p$$), then $$H\star f_\psi=Ef_\psi$$. On the other hand, if $$(x,p)\mapsto f(x,p)$$ is a real function such that $$H\star f=E f$$, then $$f$$ is the Wigner function of some state $$\psi$$ such that $$H(\mathfrak x,\mathfrak p)\psi=E\psi$$.

This is shown in (Curtright, Fairlie, Zachos, 2014) (Link to pdf), lemma 0.3, for the case of $$H(x,p)=\frac12 p^2+V(x)$$. I am, however, struggling to understand the proof that if $$H\star f=Ef$$ then $$f$$ is the Wigner of an eigenstate of $$H$$.

Writing $$f$$ via its Fourier transform as $$f(x,p)=\frac{1}{2\pi}\int dy e^{-iyp} \tilde f(x,y)$$, and using the identities $$(f\star g)(x,p) = f\left(x + \frac{i}{2}\partial_p, p - \frac{i}{2}\partial_x\right)g(x,p) = f(x,p) g\left(x - \frac{i}{2}\partial_p^L, p + \frac{i}{2}\partial_x^L\right),$$ we have $$(H\star f)(x,p) = \frac{1}{2\pi}\int dy \, H\Big(x +\frac{y}{2}, p-\frac{i}{2}\partial_x\Big) e^{-iyp}\tilde f(x,y),$$ $$(f\star H)(x,p) = \frac{1}{2\pi}\int dy \, H\Big(x -\frac{y}{2}, p+\frac{i}{2}\partial_x\Big) e^{-iyp}\tilde f(x,y).$$ I don't quite see how the conclusion is reached from this. In the treatment in the link above the authors seem to replace $$p\to i\partial_y$$, which however I'm not sure how to make sense of. Sure, if $$\partial_y$$ only acts on the exponential then $$i\partial_y e^{-iyp}=pe^{-iyp}$$, but why isn't the $$y$$ dependence of $$\tilde f$$ taken into consideration here?

Even granted the above replacement, we would reach the condition $$\int dy\, \left[H\left(x\pm\frac{y}{2},i\partial_y\mp\frac{i}{2}\partial_x\right) -E \right]e^{-iyp}\tilde f(x,p) = 0.$$ Why does this imply that $$\tilde f(x,y) = \psi^*(x-y/2)\psi(x+y/2)$$ for some $$\psi$$ eigenstate of $$H$$?

Congratulations for absorbing $$\hbar$$ in the variables, so it does not clutter the formal expressions. The short answer is that you are using the "wrong" Bopp shift! You wrote down the two of the four that are not useful here. Remember you wish to apportion the gradients to variables of the simplest functions you can spot, here of the ps, in this case the quadratic of the kinetic term of the Hamiltonian in the first argument, and the exponential in the second.
That is, you want to use, here, the equivalent option $$f(x,p) \star g(x,p) = f\left( x ,~ p-{i \over 2}{\stackrel{\rightarrow}{\partial}}_x \right )~ g\left (x,p +{i\over 2} {\stackrel{\leftarrow}{\partial}}_x\right ) ~.$$ so that $$H(x,p)\star f(x,p)= \\ = {1\over 2\pi} \left( \left( p- { i\over2} {\stackrel{\rightarrow}{\partial}}_x \right) ^2 \Big / 2m +V(x) \right) \int\! dy~ e^{-iy(p+{i\over2} {\stackrel{\leftarrow}{\partial}}_x)} \tilde{f}(xy)\\ = {1\over 2\pi} \int\! dy~ e^{-iyp} \left( \left( p-{ i\over2} {\stackrel{\rightarrow}{\partial}}_x\right) ^2 \Big / 2m +V(x+{y /2})\right) \tilde{f}(x,y) \\ = {1\over 2\pi} \int\! dy ~e^{-iyp} \left( \left( i{\stackrel{\rightarrow}{\partial}}_y +{ i\over2} {\stackrel{\rightarrow}{\partial}}_x\right) ^2 \Big / 2m + V (x+{y /2})\right) \tilde{f}(x,y) ,$$ where you recall, of course, the celebrated action of the Lagrange shift operator. To get to the last line, an integration by parts has been performed.
Finally, $$\int\! dy ~e^{-iyp} \left(-{1\over 2m} \left ( {\stackrel{\rightarrow}{\partial}}_y \pm {\stackrel{\rightarrow}{\partial}}_x/2\right ) ^2 + V(x\pm{ y/2})-E\right) \tilde{f} (x,y)=0.$$ Hence, solutions for $$\tilde f(x,y)$$ must be a product of functions of $$x-y/2$$ satisfying the "right-moving" TISE, $$({\mathfrak H} -E)\psi=0$$, times functions of $$x+y/2$$ in the kernel of the "left-moving" TISE operator; these operators automatically slip through the wrong direction solutions!
• I can't tell if you're being sarcastic in the first sentence =). I usually get rid of $\hbar$ and other similar "measurement units holding" symbols because I found them superfluous (you can always put them back in the final expression if you wish so by trivial dimensional analysis). Anyway, there is still something I'm missing. In the fourth row of the second equation, should it be $\partial_y^L$ rather than $\partial_y^R$? Otherwise I don't see how you replaced $p$ with it. Also, is the derivation specific for this form of the Hamiltonian, or does it work the same keeping a generic $H(x,p)$? – glS Sep 26 '20 at 17:49
• Not sarcastic at all! Unless you are taking classical limits, $\hbar$ is a drag! Yes and no! I replace p with $i\partial_y^L$, but then I integrate by parts, and the relative sign changes, as it acts on the R! Indeed, the statement holds for generic Hamiltonians, but more delicate maneuvers are warranted for velocity-dependent potentials; in fact, a less direct method hinging on the Weyl map. – Cosmas Zachos Sep 26 '20 at 18:20