is in fact $4\infty$ as $r=4$. Consequently, one can put 1 instead of $i^\infty$ or $(-i)^\infty$\].

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The Moyal $\star$-Product {#the-moyal-star-product}
-------------------------

According to the WWM correspondence, the symbol of an operator $\widehat{F}$ is given by $$F[{\bf b}^*,{\bf b} ] = W^{-1}(\widehat{F}) = {\rm tr}
\bigg\{\widehat{\Omega}[{\bf b}^*,{\bf b}] \widehat{F} \bigg\}.
\label{invmap}$$ The Moyal $\star$-product in this case can be defined similarly as it has been done in (\[treinta\]). Let $F_1[{\bf b}^*,{\bf b}]$ and $F_2[{\bf b}^*,{\bf b}]$ be functionals over the Dirac phase space defined by: ${\cal Z}_D = \{(
{\pi_{\psi}}_{\alpha}(\vec{x}),\psi_{\alpha}(\vec{x}))_{\vec{x}\in
\mathbb{R}^3}\} = \{(i \hbar {\bf b}^*(\vec{p},r), ({\bf
b}(\vec{p},r))_{r=1,\cdots,4}\}$ and let $\widehat{F}_1$ and $\widehat{F}_2$ be their corresponding operators. Then by a similar computation to that done in Sec. [\[SP\]]{} we finally get $$\big(F_1 \star F_2\big)[{\bf b}^*,{\bf b}] = F_1[{\bf b}^*,{\bf
b}] \exp\bigg({i\hbar\over 2} \buildrel{\leftrightarrow} \over
{\cal P}_D\bigg) F_2[{\bf b}^*,{\bf b}], \label{fmoyalp}$$ where $$\buildrel{\leftrightarrow}\over {\cal P}_D := -{i \over \hbar}
\sum_{r=1}^{4}\int {d^3p }\bigg( {{\buildrel{\leftarrow}\over
{\delta}}\over \delta {\bf b}(\vec{p},r)}
{{\buildrel{\rightarrow}\over {\delta}}\over \delta {\bf
b}^*(\vec{p},r)} + {{\buildrel{\leftarrow}\over {\delta}}\over
\delta {\bf b}^*(\vec{p},r)} {{\buildrel{\rightarrow}\over
{\delta}}\over \delta {\bf b}(\vec{p},r)}  \bigg),
\label{dpoisson}$$ is the Poisson operator. In terms of the standard Grassmann variables $b$, $b^*$, $d$ and $d^*$, it can be rewritten as $$\buildrel{\leftrightarrow}\over {\cal P}_D = - 