# Obtaining the star product from the Weyl quantisation of the product of two symbols

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 $$O(\textbf{r})$$

• 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. Oct 13, 2022 at 22:38
• Symbols are c-numbers. The Weyl-ordered products you write are not symbols... Oct 13, 2022 at 22:42

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!

• 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 Oct 14, 2022 at 12:00
• and i agree it is a great physics paper! Oct 14, 2022 at 12:05
• 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". Oct 14, 2022 at 13:38
• Thanks for your help, much more comfortable with this formulation now Oct 15, 2022 at 15:22