Finite reduced dynamics, memory and quantum trajectories — Examples

The Hermitian matrix $Z\otimes Z$ has the orthonormal eigenbasis of product $Z$ eigenvectors and eigenvalues $\pm 1$ by direct multiplication, consistent with [F7]. For this exact matrix and environment preparation, [F1] defines the matrix exponential by its power series and derives $U_t={\sum }_{b=\pm 1}{V}_b(t)\otimes \mathrm{\mid }b⟩⟨b\mathrm{\mid }$, where $V_b(t)=\mathrm{diag}(e^{-ibgt},e^{ibgt})$. By [F4], each diagonal phase has modulus one, so these blocks and $U_t$ are unitary. The environment state has equal weight on $b=\pm 1$, so the reduced map is ${\mathrm{\Phi }}_t(X)=\frac{1}{2}(V_{+1}X{V}_{+1}^{\ast }+V_{-1}X{V}_{-1}^{\ast })$; multiplying the two diagonal matrices shows that the off-diagonal entries are multiplied by $\kappa (t)=\frac{1}{2}(e^{-2igt}+e^{2igt})=\cos (2gt)$, while the diagonal entries are unchanged.

By [F3] choose $\theta \in [0,2\pi )$ with $\cos \theta =0$ and $\sin \theta =1$, and put $s=\theta \mathrm{/}(2g)>0$. Then [F4] gives $e^{i\theta }=i$, while [F5] and [F6] with $e^{i\theta }{e}^{-i\theta }=e^0=1$ give $e^{-i\theta }=-i$; another application of [F5] yields $\kappa (s)=0$ and $\kappa (2s)=\frac{1}{2}(e^{-2i\theta }+e^{2i\theta })=-1$.

Let $X_P=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ and ${\rho }_{\pm }=(I\pm X_P)\mathrm{/}2$. Since $X_P=X_P^{\ast }$, $X_P^2=I$, and $\mathrm{Tr}{X}_P=0$, each ${\rho }_{\pm }$ is a positive trace-one matrix. Steps 1.1 and 1.2 give ${\mathrm{\Phi }}_s({\rho }_{+})={\mathrm{\Phi }}_s({\rho }_{-})=I\mathrm{/}2$, but ${\mathrm{\Phi }}_{2s}({\rho }_{+})={\rho }_{-}$ and ${\mathrm{\Phi }}_{2s}({\rho }_{-})={\rho }_{+}$. Thus ${\mathrm{\Phi }}_{2s}({\rho }_{+})\mathrm{\ne }{\mathrm{\Phi }}_s({\mathrm{\Phi }}_s({\rho }_{+}))$, so the semigroup law in [F2] fails. If a channel $V_{2s,s}$ satisfied ${\mathrm{\Phi }}_{2s}=V_{2s,s}\circ {\mathrm{\Phi }}_s$, it would map the equal intermediate states to equal outputs, contradicting their distinct final states; by [F1], every ${\mathrm{\Phi }}_t$ is a channel, so the family is not CP-divisible by [F2].

For $L=\sqrt{\gamma }\mathrm{\mid }0⟩⟨1\mathrm{\mid }$, the no-jump generator is $K=\mathrm{diag}(0,-i\omega -\gamma \mathrm{/}2)$. Hence $R(t)\mathrm{\mid }1⟩=e^{-(i\omega +\gamma \mathrm{/}2)t}\mathrm{\mid }1⟩$ and $R(t)\mathrm{\mid }0⟩=\mathrm{\mid }0⟩$. For the initial state $\mathrm{\mid }1⟩⟨1\mathrm{\mid }$, the no-jump branch weight is $\mathrm{Tr}(R(t)\mathrm{\mid }1⟩⟨1\mathrm{\mid }R(t{)}^{\ast })=e^{-\gamma t}$, equal to the excited population in [F1].

For $\gamma >0$ and a one-jump history at $s\in (0,T)$, $R(T-s)LR(s)\mathrm{\mid }1⟩=\sqrt{\gamma }{e}^{-(i\omega +\gamma \mathrm{/}2)s}\mathrm{\mid }0⟩$, since $L\mathrm{\mid }1⟩=\sqrt{\gamma }\mathrm{\mid }0⟩$ and $R(T-s)\mathrm{\mid }0⟩=\mathrm{\mid }0⟩$. Its branch-weight density from [F2] is $\gamma e^{-\gamma s}$. Because $L\mathrm{\mid }0⟩=0$ and $R(u)\mathrm{\mid }0⟩=\mathrm{\mid }0⟩$, every branch with at least two jumps has zero weight. The normalized ensemble [F3] therefore gives ${\int }_0^T\gamma e^{-\gamma s}ds=1-e^{-\gamma T}$: the only nonzero contributions are the no-jump weight $e^{-\gamma T}$ from step 1.1 and the one-jump integral. Letting $T\to \mathrm{\infty }$, [F4] identifies this as the real exponential and [F5] gives $e^{-\gamma T}\to 0$ because $\gamma >0$; hence the first-jump density has total mass one. Each first jump leaves the normalized state $\mathrm{\mid }0⟩⟨0\mathrm{\mid }$. If $\gamma =0$, then $L=0$, all positive-jump branches vanish, and the no-jump weight is one.