**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.



Moyal $\star$-product of $\star$-exponentials
The question to resolve $h$ from 
$e^f e^g=e^h$
amounts to the [Baker–Campbell–Hausdorff formula](https://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula) for the Moyal $\star$ product. One can use the same [formula](https://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula#An_explicit_Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula) with the $\star$ commutator:
$$[f,g]=f\star g-g\star f.$$
The first few terms are
$$h=f+g+\tfrac{1}{2}[f,g]+\tfrac{1}{12}([[g,f],g]+[[f,g],g])+\cdots.$$


