A question about Moyal product

Question

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.$\quad f\star g = fg + \mathcal O(\hbar)$

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

2. $\quad f\star 1=1\star f=f$

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?

Answer 1

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:

1. 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 $m \geq 2$, $$ \sum_{k=0}^m C_k(C_{m-k}(f,g),h) = \sum_{k=0}^m C_k(f,C_{m-k}(g,h)). $$
2. 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.
3. 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:

1. $C_0(f,g) = fg$ and $C_1(f,g) = i\{f,g\}$ are given.
2. 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 $$ \sum_{k=0}^m C_k \circ (C_{m-k} \otimes \operatorname{Id}) = \sum_{k=0}^m C_k \circ (\operatorname{Id} \otimes C_{m-k}). $$