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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 $O(\textbf{r})$

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  • $\begingroup$ Seems to me you make an unwarranted assumption that $A$ and $B$ can commute under this $[\cdot]_w$ symbol. But is that the case? Isn't it just a notation? From the 1st definition you give, I don't see why it should be true. $\endgroup$ Oct 13, 2022 at 22:38
  • $\begingroup$ Symbols are c-numbers. The Weyl-ordered products you write are not symbols... $\endgroup$ Oct 13, 2022 at 22:42

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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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    $\begingroup$ Yes of course, my confusion came from the fact that I thought we were applying the Weyl quantisation to the product $A(\textbf{r})B(\textbf{r})$, when we were in fact taking the product of the operators which were acquired by the Weyl quantisation $\endgroup$ Oct 14, 2022 at 12:00
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    $\begingroup$ and i agree it is a great physics paper! $\endgroup$ Oct 14, 2022 at 12:05
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    $\begingroup$ An arguably memorable mnemonic of the theorem is: "The star product combines Weyl symbols of operators to the Weyl symbol of the product of the operators". $\endgroup$ Oct 14, 2022 at 13:38
  • $\begingroup$ Thanks for your help, much more comfortable with this formulation now $\endgroup$ Oct 15, 2022 at 15:22

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