example, consider the set of transition probabilities for a single qubit defined as follows: $$\label{example}
P_{\alpha\gamma} = \frac{1}{2} - \delta_{\alpha+\gamma, \zeta} \; ,$$ where $\zeta$ is the ordered pair $(1,1)$. That is, $P$ has the value $1/2$ unless the transition is to the opposite corner of the $2 \times 2$ phase space, in which case $P$ has the value $-1/2$. These values of $P$ are properly normalized. To check whether they represent an actual quantum transformation, we use Eq. (\[inverse\]) and Eq. (\[A2\]) to find ${\mathcal B}_{\beta\delta}$. The result is $${\mathcal B} = \frac{1}{4} {\left(\begin{array}{cccc}1 & 1 & 1 & -1 \\ 1 & 1 & -1 & 1 \\ 1 & -1 & 1 & 1 \\ -1 & 1 & 1 & 1\end{array}\right)},$$ where the vertical and horizontal indices are interpreted as $\beta$ and $\delta$, respectively, each index taking the values $(0,0), (0,1), (1,0), (1,1)$ in that order. This matrix has the eigenvalues $(1/2, 1/2, 1/2, -1/2)$ and is therefore not positive semidefinite. So the transition probabilities defined in Eq. (\[example\]) do not correspond to a possible transformation on a qubit. In fact, one can show from Eq. (\[EfromP\]) that they correspond to the transpose operation, which is 