text: # Wigner function of $|n\rangle\langle m|$
  
  For $$n=m$$, the Wigner function is given by, $$W_n(\alpha) = \frac{2}{\pi} (-1)^n \exp(-2 |\alpha|^2) L_n(4 |\alpha|^2),$$
  
  And for $$n \neq m$$, it is, $$f_{mn}=\sqrt{\frac{m!}{n!}} e^{i(m-n) \arctan\left( p/x\right) } \frac{\left( -1\right)^{m}}{\pi\hbar}\left ( \frac{x^{2}+p^{2}}{\hbar/2} \right ) ^ {\left( n-m\right) /2} L_{m}^{n-m}\left( \frac{x^{2}+p^{2}}{\hbar/2}\right) e^{-\left( x^{2}+p^{2}\right) /\hbar}$$ and in this case, there will be always a $$e^{i(m-n) \arctan\left( p/x\right)}$$ term. But the Wigner function is a real valued function. How to reconcile this?
  
  • What is $f_{mn}$?
  – Alex
  Oct 27, 2023 at 13:53
  • Linked. Note this operator is not Hermitian. Oct 27, 2023 at 16:03
  
  In general, Wigner transforms of Hermitian operators are real, $$g(x,p)= \hbar \int \!\! dy ~ e^{-iyp} \langle x+\hbar y/2| \hat G |x-\hbar y /2 \rangle , \implies \\ g(x,p)^*= \hbar \int \!\! dy ~ e^{iyp} \langle x+\hbar y/2| \hat G |x-\hbar y /2 \rangle ^* \\ =\hbar \int \!\! dy ~ e^{iyp} \langle x-\hbar y/2| \hat G ^\dagger |x+\hbar y /2 \rangle ,$$ so $$g=g^*$$ for $$\hat G= \hat G^\dagger$$, like WF diagonal elements, otherwise not. $$|m\rangle \langle n|$$ is not hermitian.