If the Weyl symbol $A_W$ of an operator $\hat{A}$ has a multiplicative inverse at every point of the phase-space, can I conclude that $\hat{A}$ is invertible?

  • 1
    $\begingroup$ Weyl symbols compose through the $\star$-product. What makes you suspect plain multiplication suffices for $\star$-invertibility, which is tantamount to operator invertibility? Are you looking for counterexamples? $\endgroup$ – Cosmas Zachos Feb 15 at 15:08
  • $\begingroup$ Thank you for your reply, @CosmasZachos. I know about the Moyal product composition, but I was just wondering if the answer to my question was positive. In case the idea does not hold, a counter-example would suffice. $\endgroup$ – ilp Feb 15 at 16:03
  • $\begingroup$ I frankly don't know the answer, but it would be hard to ascertain. I might look for a counterexample, AB=I but A⋆B=0, but it too might be tricky to cook up.., $\endgroup$ – Cosmas Zachos Feb 15 at 16:10
  • $\begingroup$ I will keep thinking about it. Perhaps something on the line you suggested could lead to a counter-example. Thanks again. $\endgroup$ – ilp Feb 15 at 16:13
  • $\begingroup$ You might mean that if $A_W(x,p)$ has no zeros, then it has no zero star-gen-values, $A_W\star f(x,p)=0$...? $\endgroup$ – Cosmas Zachos Feb 15 at 20:25

Your Answer

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

Browse other questions tagged or ask your own question.