If the particle number is $\hat{a}^\dagger\hat{a}\leftrightarrow|\alpha_w|^2-1/2 $, it can be mapped on the Wigner fields by assuming symmetric ordering:$|\alpha_w|^2\leftrightarrow\hat{a}^\dagger\hat{a}+\hat{a}\hat{a}^\dagger$.
My question is: is there a clear way to work with functions of this? I would be interested in how the parity operator
$$\hat{\Pi}=\exp(i\pi\hat{a}^\dagger\hat{a})$$
can be mapped on the $\alpha_w$ fields.
Of course, it would be possible to expand the exponential and perturbatively reorder the leading terms, but perhaps there is a better approach.
Answer
I will perorate on the R Kubo 1964 trick that generically Weyl-orders absolutely any operator systematically, albeit formally. I will rely on Ch 18 of our booklet, including the Exercise at its end, using gothic characters for operators, and mindful of the fundamental algebraic isomorphism with your oscillators, $[\hat a, \hat a ^\dagger ]=1 \leftrightarrow [i\mathfrak{p}/\hbar, ~\mathfrak{x}]=1 $. The correspondence is fleshed out in this WP page and this one.
The key point is that the c-number "Weyl symbol" kernel g(x,p) of any operator $\mathfrak {G}$, in indifferent/arbitrary ordering, is provided by the Wigner map, $$ g(x,p) =\frac{\hbar}{2\pi} \int d\tau d\sigma ~ e^{i(\tau p + \sigma x)} \operatorname{Tr~}\left ( e^{-i(\tau {\mathfrak p} + \sigma {\mathfrak x})} {\mathfrak G} \right ) \\ = \hbar \int dy~ e^{-iyp} \left \langle x +\frac{\hbar}{2}y \right | {\mathfrak G}({\mathfrak x},{\mathfrak p}) \left | x-\frac{\hbar}{2}y \right \rangle . $$ The Weyl symbol is then pluggable into the Weyl map formula (the inverse of the above!) which defines symmetrized Weyl order, $$ {\mathfrak G}({\mathfrak x},{\mathfrak p}) =\frac{1}{(2\pi)^2}\int d\tau d\sigma dx dp ~g(x,p) \exp \Bigl ( i\tau ({\mathfrak p}-p)+i\sigma ({\mathfrak x}-x) \Bigr ) , $$ so you are done—provided you can take all traces and perform all integrals involved.
In practice, I doubt anyone uses it extensively, but it is an "in-principle Weyl-symmetrizer" undergirded by the force of theorem.
- As a lark, and cavalierly with over-all normalizations, let us evaluate the Weyl-ordering of $\bbox[yellow]{\exp (-\pi \mathfrak {xp}/\hbar)}= -i \exp \left ( \frac{-\pi}{2\hbar}(\mathfrak {xp} +\mathfrak {px}) \right )$, by utilizing its Weyl symbol, $$ g(x,p)\propto \int dy~ e^{-iyp} \left \langle x +\frac{\hbar}{2}y \right |\exp (-\pi \mathfrak {xp}/\hbar) \left | x-\frac{\hbar}{2}y \right \rangle \propto \bbox[yellow]{\delta(x) \delta(p)} . $$ (Recall that $\mathfrak{p}|z\rangle= i\hbar \partial_z |z\rangle$, trivial to prove; so the pseudo-dilatation operator merely flips the sign of the space argument of the ket, $\exp (-i\pi z\partial_z)|z\rangle=|-z\rangle$; and thus nets $\delta(x)$ in the dot product.)
Plug this into the Weyl map formula, to net the manifestly Weyl-ordered expression, $$ \bbox[yellow]{\int d\tau d\sigma \exp \Bigl ( i\tau {\mathfrak p}+i\sigma {\mathfrak x}\Bigr )} , $$ in fact, the integral of the generating function of all Weyl-ordered polynomials.
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