# Wigner-Weyl ordering in exponential

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.

• These guys, K FUJII & T SUZUKI, Modern Physics Letters A19 No. 11, pp. 827-840 (2004) do the combinatorics, but there is a generic "mechanical/conceptual" method involving the invertibility of the Weyl map... I could sketch it for phase-space variables, but you'd have to translate it into oscillators... if you were interested. – Cosmas Zachos Jun 13 '19 at 16:32
• @CosmasZachos would certainly be interesting – Wouter Jun 13 '19 at 16:39
• thanks, the MPL you referred to confuses me a bit with the use of their increments – Wouter Jun 13 '19 at 16:57
• I'm not sure how to help with that... Try simple examples and maniacal use of the degenerate BCH expansion identity used there and in our book all the time. If you have a text you can trust, just Wigner transform your expression and the Weyl-transform the answer. Schematically, it should be an integral of the exponential of a linear expression, analogous, mutatis mutandis, i's, etc... to my answer's. – Cosmas Zachos Jun 13 '19 at 21:33

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.

• You added before that if the exponent is symmetrically ordered, all terms of its expansion are as well, but this is not the case, right? – Wouter Jun 13 '19 at 20:28
• So you're suggesting to just use the definition of the transform? – Wouter Jun 13 '19 at 20:40
• 1. Deleted comments are hasty nonsense: I ignored the implicit * product. Don't worry about it. In my lark example you see the point: A mere exponential reorders to a double integral of exponentials. 2. Yes, you Wigner transform, and then Weyl-transform that: this is Kubo's trick. 3. Expectation values imply density matrices. Here we just trace. As per the Exercise in our book, the Wigner transform of the parity operator is just a phase-space δ-function: the Weyl transform thereof is what some call the "quantizer", obscurely and inappropriately attaching Stratonovich's name to it. – Cosmas Zachos Jun 13 '19 at 21:26
• I'm not sure what you are after--your question is not clear, or uses conventions I am not familiar with. An operator has a c-number Weyl symbol, possibly your "Wigner field", here a δ-function. The Weyl transform of a symbol is, automatically (definition!), the Weyl-ordered form of the operator. I didn't get your last sentence. Perhaps Schleich's book does optical phase space. – Cosmas Zachos Jun 13 '19 at 21:59
• Yeah, it is gothic/fraktur x : $\mathfrak{x}\equiv \hat x$..... – Cosmas Zachos Jun 14 '19 at 14:20