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:

• 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?

There is an infinity of phase-space functions corresponding to differently ordered operators, as their ps and qs may be ordered in different ways with the intercalated * 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 hbar-dependence.