Single excitation jump law

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.