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Considering the Moyal star product

$$f(x,p) \star g(x,p) = f(x,p) \exp\left(\frac{i\hbar}{2} (\overleftarrow{\partial_x} \overrightarrow{\partial_p} - \overleftarrow{\partial_p} \overrightarrow{\partial_x})\right) g(x,p),$$

which plays a central role in the phase-space-representation of quantum mechanics. I am aware, that one can use a relation known as "Lagrange’s representation of the shift operator", $e^{a \partial_x} f(x) = f(x+a)$, to write:

$$ f(x,p) \star g(x,p) = f(x,p) \exp\left(\frac{i\hbar}{2} (\overleftarrow{\partial_x} \overrightarrow{\partial_p} - \overleftarrow{\partial_p} \overrightarrow{\partial_x})\right) g(x,p) = f(x+\frac{i\hbar}{2}\partial_p,p - \frac{i\hbar}{2}\partial_x) g(x,p). $$

I can prove Lagrange’s representation of the shift operator by developing the exponential as a power series and identifying a Taylor expansion of the function $f$ in the resulting expression. However, when doing so, as far as I can see, one implicitly assumes $f$ to be analytical.

What happens if $f$ is not analytical and in particular: What if $f$ is a Coulomb potential $1/x$?

Is it true that

$$ (1/x) \star g(x,p) = (1/(x+\frac{i\hbar}{2}\partial_p)) g(x,p)? $$

And if so, how does one interpret the expression $1/(x+\frac{i\hbar}{2}\partial_p)$? Is it just the inverse operator of $x + \frac{i\hbar}{2}\partial_p$ or is it something more complicated?

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Moyal apparently did not know about this product, discovered by Groenewold, instead, apparently during WWII. You repeat the standard arguments of popular texts on phase-space quantization, and apparently require further conditions/restrictions/assurances for arbitrary arguments $g(x,p)$. In this field, one eyeballs the analyticity landscape and stretches/extends the available algorithms to comport with the circumstances at hand.

Your discussion may be assuming that the high p behavior of g is bounded, so $$ (1/x) \star g(x,p) = \sum_n \frac{1}{x} \left (1-i((\hbar/2)\frac{\partial_p}{x} -(\hbar/2)^2\frac{\partial_p^2}{x^2}+...\right)g(x,p) $$

In that case, $1/(x+i(\hbar/2)\partial_p)$ is, indeed, just the inverse operator of $x + i\hbar\partial_p/2$. Your open answer is too general to answer.

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  1. In the context of deformation quantization & the Groenewold-Moyal star product, all quantities are technically formal power series in an indeterminate/variable/parameter $\hbar$. In particular, OP's expression $$\frac{1}{x+\frac{i\hbar}{2}\partial_p}~=~\frac{1}{x}\frac{1}{1+\frac{i\hbar}{2x}\partial_p}~=~\frac{1}{x}\sum_{n\in\mathbb{N}_0}\left(\frac{\hbar}{2ix}\partial_p\right)^n\tag{1}$$ represents a formal power series (with operator-valued coefficients). In plain English: the left-hand side of eq. (1) should be viewed as a shorthand notation for the right-hand side.

  2. Yes, OP's expression (1) is the inverse of the operator $x+\frac{i\hbar}{2}\partial_p$ in the ring of formal power series (with operator-valued coefficients).

  3. At the end of the calculation $\hbar$ is put equal to the reduced Planck constant. Convergence of the formal power series depends on further details.

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A more tolerant way to compute star products is to use the integral form of the star product $$ f(\alpha)\star g(\alpha) = 4 \int f(\alpha_1) g(\alpha_2) \exp[2\alpha_1(\alpha_2-\alpha) +2\alpha(\alpha_1-\alpha_2) +2\alpha_2(\alpha-\alpha_1)]\ \text{d}^2\alpha_1 \text{d}^2\alpha_2 , $$ where $$ \alpha = \frac{1}{\sqrt{2}} (x + i p) $$ is a complex variable. Provided that the integral is well-defined it can handle situations where one of the two functions is non-analytic.

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