normalization condition (\[newPcondition\]a) implies that the trace of ${\mathcal B}$ is unity. So the sole non-zero eigenvalue of ${\mathcal B}$ must be 1; that is, ${\mathcal B}$ must be a one-dimensional projection operator. Now, if ${\mathcal B}$ is a one-dimensional projection, then there is essentially only a single $\hat{B}$ operator in Eq. (\[transformation\]). (There could be several $\hat{B}_j$’s, but they would all be proportional to each other.) In that case the sum condition (\[Kraus\]), which, as we have seen, follows from the normalization condition (\[newPcondition\]a), implies that this $\hat{B}$ is unitary.

Continuous Hamiltonian evolution
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In the preceding section, we were interested in a single discrete transformation taking $\hat{\rho}$ to $\hat{\rho}'$. We now consider a continuous transformation governed by the von Neumann equation: $$\label{vonN}
\frac{d\hat{\rho}}{dt} = -\frac{i}{\hbar}[\hat{H},\hat{\rho}],$$ where $\hat{H}$ is the Hamiltonian, which we assume to be constant. Let $H_\alpha$ be the expansion coefficients of $\hat{H}$ in the phase point operators $\hat{A}_\alpha$: $$\hat{H} = \sum_\alpha H_\alpha \hat{A}_\alpha\, ,\hspace{2mm} \hbox{so that}\; H_\alpha = \frac{1}{N} \hbox{Tr}(\hat{H}\hat{A}_\alpha).$$ Then we can rewrite Eq. (\[vonN\]) in phase space language as $$\frac{dW_\alpha}{dt} = \frac{2}{\hbar} \sum_{\beta\gamma} \hbox{Im}(\Gamma_{\alpha\beta\gamma}) H_\beta W_\gamma  \; ,$$ where $\Gamma_{\alpha\beta\gamma}$ is defined in Eq. (\[Gamma\]) and we have used the fact that $\Gamma_{\alpha\gamma\beta} = 