A question about Moyal product In Geometric quantization theory the Moyal product is one of main tools. We know Moyal product for the smooth functions $f$ and $g$ on $ℝ^{2n}$   takes the form
$f\star g = fg + \sum_{n=1}^{\infty} \hbar^{n} C_{n}(f,g)$
where each $C_n$  is a certain bidifferential operator of order $n$  with the following properties and also where  $\hbar$  is the reduced Planck constant.
1.$\quad f\star g = fg + \mathcal O(\hbar)$


*

*$\quad f\star g-g\star f = \mathrm i\hbar\{f,g\} + \mathcal O(\hbar^2) \equiv \mathrm i\hbar \{\{f,g\}\}$

*$\quad f\star 1=1\star f=f$
4.$ \quad \overline{f\star g} = \overline{g}\star \overline{f}$
My question is how can we find $C_n$. I am looking for a method , not explicit formula based on Poisson bivector. Also when can we use of this product?
 A: Let me give you the naïve picture at least, which I hope partially answers your question; from what I understand, in the actual research literature on deformation quantisation, i.e., in noncommutative geometry à la Kontsevich, one takes a somewhat more flexible but technically more delicate and sophisticated approach to the same basic strategy. In any event, the basic idea is to translate the properties of $\star$ into a system of equations in the $$C_k : C^\infty(\mathbb{R}^{2n},\mathbb{C}) \otimes_{\mathbb{C}} C^\infty(\mathbb{R}^{2n},\mathbb{C}) \to C^\infty(\mathbb{R}^{2n},\mathbb{C}),$$
which one could try to solve recursively.
For simplicity, write
$$
 C_0(f,g) := fg, \quad C_1(f,g) = i\{f,g\},
$$
so that
$$
 f \star g = \sum_n \hbar^n C_n(f,g).
$$
Let us look at the properties of $\star$ in turn:


*

*Associativity: By direct computation,
$$
 (f \star g) \star h = \sum_m \hbar^m \sum_{k=0}^m C_k(C_{m-k}(f,g),h), \quad f \star (g \star h) =  \sum_m \hbar^m \sum_{k=0}^m C_k(f,C_{m-k}(g,h))
$$
so that by your assumptions on $C_0$ and $C_1$, $(f \star g) \star h = f \star (g \star h)$ if and only if for all $m \geq 2$,
$$
  \sum_{k=0}^m C_k(C_{m-k}(f,g),h) = \sum_{k=0}^m C_k(f,C_{m-k}(g,h)).
$$

*Unitality: By direct computation,
$$
 f \star 1 = \sum_m \hbar^m C_m(f,1), \quad 1 \star f = \sum_m \hbar C_m(1,f),
$$
so that by your assumptions on $C_0$ and $C_1$, $f \star 1 = f = 1 \star f$ if and only if for all $m \geq 2$,
$$
 C_m(f,1) = 0 = C_m(1,f).
$$
However, since you're requiring $C_m$ to be a bidifferential operator of order $m$, this is automatic.

*Compatibility with complex conjugation: Again, by direct computation,
$$
 \overline{f \star g} = \sum_m \hbar^m \overline{C_m(f,g)}, \quad \overline{g} \star \overline{f} = \sum_m \hbar^m C_m(\overline{g},\overline{f}),
$$
so that by your assumptions on $C_0$ and $C_1$, $\overline{f \star g} = \overline{g} \star \overline {f}$ if and only if for all $m \geq 2$,
$$
 \overline{C_m(f,g)} = C_m(\overline{g},\overline{f}).
$$


So, constructing your deformation quantisation, at least in principle, proceeds inductively: 


*

*$C_0(f,g) = fg$ and $C_1(f,g) = i\{f,g\}$ are given.

*For $m \geq 2$, given that you've already found $C_0,\dotsc,C_{m-1}$, try to find an order $m$ bidifferential operator $C_m$ satisfying $\overline{C_m(f,g)} = C_m(\overline{g},\overline{f})$, solving the equation of bidifferential operators:
$$
 \forall f,g, \quad \sum_{k=0}^m C_k(C_{m-k}(f,g),h) = \sum_{k=0}^m C_k(f,C_{m-k}(g,h)),
$$
or more abstractly
$$
 \sum_{k=0}^m C_k \circ (C_{m-k} \otimes \operatorname{Id}) = \sum_{k=0}^m C_k \circ (\operatorname{Id} \otimes C_{m-k}).
$$

