Why does the star product satisfy the "Bopp Shift relations": $f(x,p)\star g(x,p)=f(x+\frac{i}{2}\partial_p,p-\frac{i}{2}\partial_x) g(x,p)$?

Question

In (Curtright, Fairlie, Zachos 2014), the authors mention (Eq. (14) in this online version) the following relation, known as "Bopp shifts": $$f(x,p)\star g(x,p)=f\left(x+\frac{i}{2}\partial_p,p-\frac{i}{2}\partial_x\right) g(x,p),\tag1$$ where the $\star$-product is defined as $$\star\equiv\exp\left[ \frac{i}{2}(\partial_x^L\partial_p^R - \partial_p^L \partial_x^R)\right],\tag2$$ and I'm denoting with $\partial_i^L,\partial_i^R$ the partial derivative $\partial_i$ applied to the left or right, respectively.

I'm trying to get a better understanding of where this comes from. As far as I understand, $f$ and $g$ are regular functions here (usually Wigner functions I suppose), so $f\star g$ should produce another "regular" function. If this is so, what do the derivatives in the argument of $f$ mean exactly? If I were to simply apply (2) to $f\star g$, I would naively get the following expression: $$f\star g = \sum_{s=0}^\infty \frac{(i/2)^s}{s!} \sum_{k=0}^s (-1)^k (\partial_x^{s-k}\partial_p^k f) (\partial_x^k \partial_p^{s-k}g). \tag3 $$ How is this compatible with (1)?

In fairness, if I were to very handwavily apply (2) to (1) without being too careful, I could think of $\partial_p^R$ and $\partial_x^R$ in the exponential as $c$ numbers, and the $\star$ operator as only acting on $f$, so that $\frac{i}{2}\partial_x^L\partial_p^R$ would be the operator enacting the translation $x\mapsto x+\frac{i}{2}\partial_p$, and similarly for the other term in the exponential. At a purely formal level this seems to make sense, but more concretely I'm not sure what the expression (1) is even supposed to represent, and how it is consistent with (3).

Answer 1

TL;DR: The underlying basic identity behind the Bopp shift is a Taylor expansion, which amounts to a translation/shift, $$e^{\hat{A}\partial_x}f(x)~=~f(x+\hat{A})\tag{A}$$ Here we assume the operator $\hat{A}$ does not depend on $x$.

Sketched proof:

$$\begin{align} (f\star g)(x,p)~=~&\left. e^{\frac{i\hbar}{2}(\partial_p\partial_{x^{\prime}}-\partial_x\partial_{p^{\prime}})}f(x^{\prime},p^{\prime})g(x,p)\right|_{x^{\prime}=x,p^{\prime}=p}\cr ~\stackrel{(A)}{=}~&\left. f\left(x^{\prime}+\frac{i\hbar}{2}\partial_p,p^{\prime}-\frac{i\hbar}{2}\partial_x\right)g(x,p)\right|_{x^{\prime}=x,p^{\prime}=p} \cr ~=~& f\left(x+\frac{i\hbar}{2}\partial_p,p-\frac{i\hbar}{2}\partial_x\right)g(x,p). \end{align}\tag{B}$$ $\Box$