using the expectations $$\mu^{jk}_t(X) = \mathbb{E}_{jk}[ X(t) ].$$ Note that for all $t \geq 0$ we have $$\mu_t^{00}(I)= 1 = \mu^{11}_t(I) , \ \ \ \mu^{01}_t(I)=0=\mu^{10}_t(I).$$

\[thm:master\] The master equation in Heisenberg form for the system when the field is in the single photon state $\vert 1_\xi \rangle$ is given by the system of equations $$\begin{aligned}
\dot{\mu}^{11}_t (X) &=& \mu^{11}_t(\mathcal{L}(X)) + \mu^{01}_t( S^\ast
[X,L] ) \xi^\ast(t) + \mu^{10}_t( [L^\ast, X] S ) \xi(t)  \notag \\
&& + \mu^{00}_t( S^\ast X S - S) \vert \xi(t) \vert^2,
\label{eq:rho-dyn-a-11} \\
\dot{\mu}^{10}_t (X) &=& \mu^{10}_t(\mathcal{L}(X)) + \mu^{00}_t( S^\ast [X,
L] ) \xi^\ast(t) ,  \label{eq:rho-dyn-a-10} \\
\dot{\mu}^{01}_t (X) &=& \mu^{01}_t(\mathcal{L}(X)) + \mu^{00}_t( [L^\ast,
X] S ) \xi(t) ,  \label{eq:rho-dyn-a-01} \\
\dot{\mu}^{00}_t (X) &=& \mu^{00}_t(\mathcal{L}(X)) .
\label{eq:rho-dyn-a-00}\end{aligned}$$ The initial conditions are $$\mu^{11}_0(X)= \mu^{00}_0(X)= \langle \eta, X \eta \rangle, \ \
\mu^{10}_0(X)= \mu^{01}_0(X)=0.$$

Equations (\[eq:rho-dyn-a-11\])- (\[eq:rho-dyn-a-00\]) are obtained by applying Lemma \[lemma:expectation-basic\] to the Heisenberg equation (\[eq:X-dyn\]). [$\Box$]{}

It is apparent from Theorem \[thm:master\] that the single photon expectation $\mu^{11}_t(X) = \mathbb{E}_{11}[ X(t)]$ cannot be determined by a single differential equation, and that instead a system of coupled equations is required, equation (\[eq:rho-dyn-a-11\])-(\[eq:rho-dyn-a-00\]). Note that the unitary matrix $S$ appearing in the Schrödinger equation (\[eq:unitary\]) does 