Definitions and assumptions
On a 2n-dimensional phase-space with coordinates $x$ and $p$, the Moyal product can be written explicitly as
$$g(x,p) \star h(x,p) = g(x,p) e^{\frac{i}{2}\left( \overleftarrow{\partial_x} \cdot \overrightarrow{\partial_p} - \overrightarrow{\partial_x} \cdot \overleftarrow{\partial_p}\right)} h(x,p) \, .$$
We define the $\star$-exponential $e_\star$ as $$e_\star^{f(x,p)}=\sum_{n=0}^{\infty}\frac{f^{\star n}}{n!}$$
or more precisely as the solution to the differential equation $$\frac{d}{dt} F(x,p;t)=H(x,p)\star F(x,p;t) \, \, \, \& \, \, \, F(x,p;0)=1,$$ i.e. $F(x,p;t)=e_\star^{-itH(x,p)}$.
Problem
Suppose we have two phase-space functions, $f(x,p)$ and $g(x,p)$. Is it always possible to find a unique $h(x,p)$ such that $$e_\star^{f(x,p)} \star e_\star^{g(x,p)} = e_\star^{h(x,p)} \, \, ?$$
I tried to fiddle with it and I searched far and wide trying to dig out an answer or something that might point me to the answer or (even better) a procedure for calculating $h(x,p)$, but without any luck. Any ideas? Have you seen something like this?
EDIT: Conjecture
Since $$f(x,p) \star g(x,p) = \hat{f}(x,p) g(x,p) = f(\hat{x},\hat{p}) g(x,p),$$ where $\hat{x}$ and $\hat{p}$ are non-commuting coordinates and $\hat{f}(x,p)$ is the operator defined by changing $(x,p) \to (\hat{x},\hat{p})$, it seems to me that $$e_\star^f \star e_\star^g = e^{\hat{f}} e^{\hat{g}} = e^{\hat{f} + \hat{g} + \frac{1}{2} [\hat{f} ,\hat{g}] + \dots}$$ where the RHS is just the BCH formula. Since $[\hat{f} , \hat{g}] = [f \overset{\star}{,} g]$, it looks like $$ e_\star^f \star e_\star^g = e_\star^{f + g + \frac{1}{2} [f \overset{\star}{,} g] + \dots} $$ but I'm not so sure about it, I might be missing something.
