# Density matrix from Wigner distribution

Density matrix or Wigner function can be defined from the other with Fourier (or inverse) transformation. equivalently the value of W(q,p) can be seen as the mean value of the displaced parity operator U(q,p). Introducing the trace norm, the scalar product of two operators A and B is Tr(A B). If the U(q,p) give an othonormal basis it becomes straightforward to write: $$\rho = \int \int Tr(\rho U(q,p)) U(q,p) dq dp$$ It is the equivalent of $$|v \rangle = \Sigma \langle i|v\rangle |i\rangle$$ So the density matrix is retrieved in another equivalent way. Question: How can we prove the orthonormality of these displaced parity operators (or prove the equivalent equality)?

Edit: I have the beginning of an answer. The $U(q,p) = U(\alpha)$ are $D(\alpha) (-1)^{a^\dagger a} D(-\alpha)$ So the "scalar product of two displaced parity operators is: $$TrD(\beta) (-1)^{a^\dagger a} D(-\beta) D(\alpha) (-1)^{a^\dagger a} D(-\alpha)$$ $$= Tr(-1)^{a^\dagger a} D(-\beta) D(\alpha) (-1)^{a^\dagger a} D(-\alpha)D(\beta)$$ So when we have twice the same operator we get $Tr (-1) ^{2a^\dagger a}$ So they have the same norm. IF they are different we get $Tr (-1)^{a^\dagger a} U(\alpha - \beta)$ Here i need help.

Working in Fock space with creation and annihilation operators is a bit self-defeating, unless you were suitably adroit. Sticking to standard phase space operators $\hat{x}, \hat{p}$ and using the standard CBH identity for QM gets you there directly.
In any case, your $U$ is Royer's $\Pi$ of (5'') which instantly gets you his (4), your top expression, even though he spends the rest of his short note fussing it and projecting its consequences.
Nevertheless, as well-known, it is all a lift from Kubo's historic paper, Theorem 3, your expression, based on his Theorems 2 & 1, where your $U$ is now his $\Delta$, which some Russian-influenced authors sometimes choose to call the "quantizer". The orthonormality of these $\Delta$s is in Kubo's (2.15, 2.16, 2.17), but, frankly, it cannot outrange the self-evident.