Wigner map of the product of two operators Does anyone know how to prove that for the product of two operators $\hat{A}\hat{B}$ the Weyl-Wigner correspondence reads
$$
(AB)(x,p) = A\left (x-\frac{\hbar}{2i}\frac{\partial}{\partial p}, p+\frac{\hbar}{2i}\frac{\partial}{\partial x}\right )B(x,p),
$$
without using the star product? Here $B(x,p)$ represents the usual  Wigner  map image
$$
B(x,p) = \int_{-\infty}^\infty d\xi e^{-ip\xi/\hbar}\langle x + \xi/2|\hat{B}|x-\xi/2 \rangle.
$$
I think one has to consider the operators as a function of $\hat{x}$ and $\hat{p}$ and then do a Taylor expansion, but I'm not sure.
 A: Your first line is the Bopp-shift definition of the star product, and it is proven directly as in Groenewold's original paper inventing it, always mindful of Lagrange's translation operator, $e^{\epsilon \partial_z} f(z)= f(z+\epsilon)$, of course. However, it is crucial to indicate  the derivatives in A strictly only act on the arguments of B! That is the reason authors either define primed phase space coordinates for the trailing expression, and unprime them at the very end, or they use the far more sensible arrow notation (we use). To the extent you appreciate that, I will stick to your notation.
$$
  A\left(x+i\frac{\hbar}{2}\frac{\partial}{\partial p}, p-i\frac{\hbar}{2}\frac{\partial}{\partial x}\right )B(x,p)\\
= \int_{-\infty}^\infty d\zeta \langle x + \zeta/2+\frac{i\hbar}{2}\partial_p '|\hat{A}|x-\zeta/2 +\frac{i\hbar}{2}\partial_p '\rangle  e^{-ip\zeta/\hbar +\frac{\zeta}{2} \partial_x }
~~ \int_{-\infty}^\infty d\xi e^{-ip'\xi/\hbar}\langle x + \xi/2|\hat{B}|x-\xi/2 \rangle\\
=\int  d\zeta   d\xi ~~\langle x + \zeta/2+\xi /2  |\hat{A}|x-\zeta/2 +\xi/2  \rangle  e^{-i(p /\hbar)(\zeta+\xi) }
 \langle x + \xi/2-\zeta/2|\hat{B}|x-\xi/2 -\zeta/2\rangle.
$$
In the last line, we set $p'=p$, since there cannot be any ambiguity anymore. This is the only place with momentum dependence.
You then define light-cone coordinates $\rho\equiv \zeta+\xi$ and $\sigma\equiv x+(\xi -\zeta)/2$, whose Jacobian is unity, so that the above integral readily collapses to
$$
\int  d\rho   d\sigma ~~\langle x + \rho/2  |\hat{A}|\sigma  \rangle  e^{-i(p /\hbar)(\rho) }
\langle \sigma|\hat{B}|x-\rho/2  \rangle=  \int  d\rho  ~e^{-ip \rho/\hbar }\langle x + \rho/2  |\hat{A} \hat{B}|x-\rho/2  \rangle=(AB)(x,p).  
$$
This is Groenewold's fundamental isomorphism theorem (1946), and, no matter what language you use, you find yourself doing the same symplectic manipulation again and again, and yet again. (In our booklet, this is the only theorem!) 
