&=\frac{1}{N^2} \sum_\alpha \hbox{Tr} \left[ \hat{E}_\alpha^\dag
   \hat{A}_\beta {\mathcal E}(\hat{E}_\alpha) \hat{A}_\delta \right],
\end{split}$$ where we are defining $\hat{E}_\alpha$ to be $|j\rangle\langle k|$, with $\alpha = (j,k)$. We know that we can write the orthonormal matrix basis $\{\hat{E}_\alpha\}$ in terms of the alternative orthonormal matrix basis $\{\hat{A}_\gamma/\sqrt{N}\}$ as $$\hat{E}_\alpha = \sum_\gamma U_{\alpha\gamma}(\hat{A}_\gamma/\sqrt{N}),$$ where $U$ is an $N^2 \times N^2$ unitary matrix. (In fact, one can check that $U_{\alpha\gamma} = (1/\sqrt{N})\delta_{2\gamma_1,j+k}
\omega^{-\gamma_2(j-k)}$.) It follows that we can replace the basis $\{\hat{E}_\alpha\}$ in Eq. (\[firstB\]) with $\{\hat{A}_\gamma/\sqrt{N}\}$. This gives us $$\langle \Psi_\beta|\hat{C}|\Psi_\delta\rangle = \frac{1}{N^3}\sum_\gamma \hbox{Tr}\left[\hat{A}_\gamma \hat{A}_\beta {\mathcal E}(\hat{A}_\gamma)\hat{A}_\delta \right],$$ which agrees with Eq. (\[BwithE\]). Thus ${\mathcal B}$ is the Choi operator written in the basis $|\Psi_\beta\rangle$.

The above analysis becomes simpler in the case of unitary evolution. In that case, we can get an expression for the transition probabilities directly from Eq. (\[PfromE\]): $$\label{unitaryP}
P_{\alpha\rho} = \frac{1}{N} \hbox{Tr}(\hat{A}_\alpha \hat{U} \hat{A}_\rho \hat{U}^\dag),$$ where $U$ is the unitary evolution operator. From Eq. (\[unitaryP\]) and Eq. (\[properties\]a), we can see that this $P_{\alpha\rho}$, regarded as a matrix with indices $\alpha$ and $\rho$, is an [*orthogonal*]{} matrix: $PP^T = I$. We also note that in this case ${\mathcal B}_{\beta\delta}$ has the simple form $$\label{unitaryB}
{\mathcal B}_{\beta\delta} = B_\beta {B}^*_\delta  