Bopp operators and Wigner-Weyl representation I am learning about the Wigner-Weyl transformations to move a $c$-number Lindblad operator $A(x,p)$ back into operator form. As far as I know, to move back and forth normally requires a four variable integral transform. However, a colleague has pointed out that there is a simpler method to achieve the transform from operator to c-number, the so-called Bopp operators:
$$\hat{x}=x+\frac{i\hbar}{2}\frac{\partial}{\partial p},\qquad \hat{p}=p-\frac{i\hbar}{2}\frac{\partial}{\partial x}.$$
Using this substitution is a quick method to the Weyl symbol (for example, with $\hat{x}\hat{p}$ straight substitution leads to the correct WW form).
I have two questions:


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*If I would like to go from phase-space to operator form, is it as simple as substituting $x=\hat{x}-\frac{i\hbar}{2}\frac{\partial}{\partial p}$?

*The example given is written as $\hat{x}\hat{p}\hat{1}$, so the derivatives operate on the identity operator and give $0$. However, in the c-number function $A(x,p)$ there is no unity operator. When the derivative operates to the left, what does it operate on?
 A: Indeed, the Bopp shift is a clumsy Lagrange translation operator transcription of the celebrated $\star$ product, a 4-variable integral transform, cf. eqns (12-15) in Ref. 1 (14-17 in the linked online version).
There is an infinity of phase-space functions corresponding to differently ordered operators, as their ps and xs may be ordered in different ways with the intercalated $\star$s enforcing noncommutativity; or, equivalently, their Bopp shifts acting in differently ordered sequences. They thus all coincide to vanishing order in $\hbar$, but differ in their ℏ-dependence.
Identity operators in Hilbert space map to constants in phase space and vice versa.
When you have long  (longer than two!) strings  of operators, it is trivial to insert (associative!) $\star$-products between Weyl symbols, but unless you are veeeery careful with associative groupings, the Bopp trick will fail, and is not worth it.
The Wigner transform of the operator you wrote down, x̂ p̂ 1̂,  is, hence, directly, xp+iħ/2. The formal reason is
$$
\hat x \hat p \mapsto x\star p \star 1=\left (x+{i\hbar\over 2}\partial_p\right ) p= xp+  {i\hbar\over 2}.
$$
Note the p-derivative term is idle, so only the leading, conventional, term 1 survives as a contribution of the identity.
References:

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*Thomas L. Curtright, David B. Fairlie, & Cosmas K. Zachos,  A Concise Treatise on Quantum Mechanics in Phase Space, World Scientific, 2014. The PDF file is available here.

