TITLE: Obtaining the star product from the Weyl quantisation of the product of two symbols
QUESTION [1 upvotes]: It can be shown (Groenewold 1946)  that the Weyl quantisation of the product of two Weyl symbols is given by
$$ [A(\textbf{r})B(\textbf{r})]_{w}=\frac{1}{(2\pi)^{2}}\int_{\mathbb{R}^{4}}e^{i\boldsymbol{\Gamma}\cdot(\hat{\textbf{R}}-\textbf{r})} \left(A(\textbf{r})e^{-\frac{i\hbar}{2}(\overleftarrow{\partial}_{q}\overrightarrow{\partial}_{p}-\overleftarrow{\partial}_{p}\overrightarrow{\partial}_{q})}B(\textbf{r})\right)d\alpha d\beta dq dp $$ where $\boldsymbol{\Gamma}=(\alpha,\beta)$, $\textbf{r}=(q,p)$ and $\hat{\textbf{R}}=(\hat{Q},\hat{P})$ and $A(\textbf{r})$ and $B(\textbf{r})$ are the Weyl symbols of operators $\hat{A}$ and $\hat{B}$ respectively. The exponential can be split as follows,
$$[A(\textbf{r})B(\textbf{r})]_{w}=\hat{O}_{1}-i\hat{O}_{2}$$
where
\begin{align*}
    &\hat{O}_{1}=\frac{1}{(2\pi)^{2}}\int_{\mathbb{R}^{4}}e^{i\boldsymbol{\Gamma}\cdot(\hat{\textbf{R}}-\textbf{r})} A(q,p)\mathrm{Cos}\left(\frac{\hbar}{2}(\overleftarrow{\partial}_{q}\overrightarrow{\partial}_{p}-\overleftarrow{\partial}_{p}\overrightarrow{\partial}_{q})\right)B(q,p)d\alpha d\beta dq dp \\
    &\hat{O}_{2}=\frac{1}{(2\pi)^{2}}\int_{\mathbb{R}^{4}}e^{i\boldsymbol{\Gamma}\cdot(\hat{\textbf{R}}-\textbf{r})} A(q,p)\mathrm{Sin}\left(\frac{\hbar}{2}(\overleftarrow{\partial}_{q}\overrightarrow{\partial}_{p}-\overleftarrow{\partial}_{p}\overrightarrow{\partial}_{q})\right)B(q,p)d\alpha d\beta dq dp \\
\end{align*}
Unless I've completely misunderstood Weyl ordering,
In accordance with Weyl ordering,
$$ [A(\textbf{r})B(\textbf{r})]_{w}=\frac{1}{2}(\hat{A}\hat{B}+\hat{B}\hat{A})$$however, in his paper, Groenewold uses $[A(\textbf{r})B(\textbf{r})]_{w}=\hat{A}\hat{B}$ in order to say that $\hat{O}_{2}=\frac{i}{2}(\hat{A}\hat{B}-\hat{B}\hat{A})$, therefore recovering the form of the star product. I don't see how this would make sense, since it would imply that $[B(\textbf{r})A(\textbf{r})]_{w}=\hat{B}\hat{A}\neq[A(\textbf{r})B(\textbf{r})]_{w}$, however the scalar functions will commute with eachother before you make the Weyl quantisation.
Edit:
In case it helps anyone in the future, my confusion came from misinterpreting the expression. We actually have
$$ [A(\textbf{r})]_{w}[B(\textbf{r})]_{w}=\hat{A}\hat{B}\\
=\frac{1}{(2\pi)^{2}}\int_{\mathbb{R}^{4}}e^{i\boldsymbol{\Gamma}\cdot(\hat{\textbf{R}}-\textbf{r})} \left(A(\textbf{r})e^{-\frac{i\hbar}{2}(\overleftarrow{\partial}_{q}\overrightarrow{\partial}_{p}-\overleftarrow{\partial}_{p}\overrightarrow{\partial}_{q})}B(\textbf{r})\right)d\alpha d\beta dq dp $$
where the notation $[O(\textbf{r})]_{w}=\hat{O}$ means the Weyl quantisation of the classical function $O(\textbf{r})$

REPLY [1 votes]: You just might have completely misunderstood Weyl ordering and Groenewold's fundamental theorem of phase-space quantization.
His operator formula (4.27) is, instead,
$$ \hat A(\hat{\textbf{R}})\hat B(\hat{\textbf{R}})=\frac{1}{(2\pi)^{2}}\int_{\mathbb{R}^{4}}\!\!d\alpha d\beta dq dp ~~e^{i\boldsymbol{\Gamma}\cdot(\hat{\textbf{R}}-\textbf{r})} \left(A(\textbf{r})e^{\frac{i\hbar}{2}(\overleftarrow{\partial}_{q}~\vec\partial_{p}-\overleftarrow{\partial}_{p}\ \vec\partial_{q})}B(\textbf{r})\right),$$
where the left-hand-side is automatically Weyl reordered : that is, the messy product of $\hat q$ s and $\hat p$ s is equal to a reordering thereof to Weyl's perfect symmetrization of these operator variables if one used the fundamental commutation relation thereof to rearrange them.
But  one doesn't need to, since the right-hand-side is automatically Weyl ordered by construction—the operator exponential. I would put carets inside the funny expression $[AB]_w$ you used, but that would confuse things... In point of fact, Weyl ordering is a canard in this question. Nothing would go differently if one did not pay attention to Weyl ordering here,$^\natural$ or even if one did not know about, or had not noticed, Weyl ordering!
Taking A-B symmetric and antisymmetric parts of the left-hand-side operators for the anticommutator and commutator halved, respectively, in his subsequent (4.28) and (4.29), he identifies them with your $\hat O_1$ and $i\hat O_2$, respectively. It is one of the great physics papers of the 1940s.
$^\natural$It might be worthwhile to remind yourself of the terms you are using, by a simple example. Take $\hat A= 3\hat x$, and $\hat B= \hat p^2$. Then the Weyl symbol of $3\hat  x \hat p^2$ is $3x\star p^2=3xp^2+3i\hbar p$, since
$$
3\hat  x \hat p^2=\hat  x \hat p^2+ \hat p \hat  x  \hat p + \hat p^2 \hat  x +3i\hbar \hat p,
$$
the r.h.s. being Weyl reordered, equal to itself. This confuses some students used to normal ordering in QFT, where terms are discarded. No such thing here: the unordered and reordered terms are expressly equal!
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