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?

  • 1
    $\begingroup$ What is $f_{mn}$? $\endgroup$
    – Alex
    Oct 27, 2023 at 13:53
  • $\begingroup$ Linked. Note this operator is not Hermitian. $\endgroup$ Oct 27, 2023 at 16:03

1 Answer 1


You have been misinformed!

Only diagonal members of the Wigner function matrix are provably real. You just illustrated off-diagonal elements are complex, in general, as your text must state.

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.

  • $\begingroup$ Thank you. That makes sense. Also I came across this equation while going through your book :) $\endgroup$
    – Abi
    Oct 29, 2023 at 11:47
  • $\begingroup$ It says that, in so many words.... cf (91). $\endgroup$ Oct 29, 2023 at 13:55

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