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

Question

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 $H$?

Answer 1

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.

I'll leave it to you to use the Bopp relation for the star product in the right star-genvalue equation you are trying to evaluate.

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!

The reality condition, then, ensures the left- and right-mover functions are the very same function!