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Using the Moyal product between two delta functions in $(x,p)$-space one gets $$ \delta(x) \star \delta(p) = \frac{1}{\pi} e^{2ixp}. $$

However, $\delta(-x)=\delta(x)$ and last time I checked $e^{2ixp} \neq e^{-2ixp}$.

What's going on here?

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Recall the star operator also depends on x, $$ \delta (x) ~ \star ~ \delta(p) = \delta (x) ~ \exp{\left( \frac{i \hbar}{2} \left(\overleftarrow {\partial }_x \overrightarrow{\partial }_p-\overleftarrow{\partial }_p \overrightarrow {\partial }_x \right) \right)} ~ \delta(p) ={2\over h} \exp \left (2i{xp\over\hbar}\right ). $$ Consequently, flipping the sign of x complex conjugates it, even though it leaves the delta function invariant.

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  • $\begingroup$ Ah, of course, thanks! $\endgroup$ – PhysSE is Cancer Apr 24 at 21:08

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