follows, we will save a bit of space by defining the four-point structure function $$\Xi_{\alpha\beta\gamma\delta} = \frac{1}{N} \hbox{Tr}(\hat{A}_\alpha \hat{A}_\beta \hat{A}_\gamma \hat{A}_\delta).$$ Then Eq. (\[Pdef\]) becomes $$\label{PXi}
P_{\alpha\gamma} = \sum_{\beta\delta} \Xi_{\alpha\beta\gamma\delta} {\mathcal B}_{\beta\delta}.$$ We note for future reference that in addition to being invariant under cyclic permutations of its indices, $\Xi_{\alpha\beta\gamma\delta}$ also has the following symmetry: $$\label{Xisymmetry}
\Xi_{\gamma\beta\alpha\delta} = 
\Xi_{\alpha\beta\gamma\delta}^*$$ since switching $\alpha$ and $\gamma$ effectively reverses the order of the $A$ operators inside the trace.

We can extend the definition of $P_{\alpha\gamma}$ to the case of linear transformations ${\mathcal E}$ that are not necessarily completely positive. Let Eq. (\[probprocess\]) serve as the definition of $P_{\alpha\gamma}$ for such a transformation. Then from $\hat{\rho}' = {\mathcal E}(\hat{\rho})$ and Eq. (\[probprocess\]), we have $${\mathcal E}\Big(\sum_\beta W_\beta \hat{A}_\beta\Big) = \sum_\sigma W'_\sigma \hat{A}_\sigma
= \sum_{\sigma\tau} P_{\sigma\tau} W_\tau \hat{A}_\sigma.$$ Now inserting for the initial Wigner function the illegal state $W_\beta = \delta_{\beta\gamma}$, we get $$\label{EfromP}
{\mathcal E}(\hat{A}_\gamma) = \sum_\sigma P_{\sigma\gamma} \hat{A}_\sigma \; ,$$ from which it follows that $$\label{PfromE}
P_{\alpha\gamma} = \frac{1}{N} \hbox{Tr} [\hat{A}_\alpha {\mathcal E}(\hat{A}_\gamma)].$$ (It is perfectly fine to use the illegal state $W_\beta = \delta_{\beta\gamma}$ in this derivation. The map ${\mathcal E}$ is defined for all operators acting on the $N$-dimensional Hilbert space, 