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?


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.