$\ket{\Psi(t)}$ naturally follows the equality . The idea is to draw a sequence of realizations of process $\Xi$, evaluate $\ket{\Phi_{\Xi}(t)}$ for each realization, an estimate of $\ket{\Psi}$ is then given by the average. For easier implementation, we define $$\label{eq:eta_def}
\eta(t) = \int_0^t\int_{\Omega}
  \lambda(s,\omega) \,\mathrm{d}\mu_{\omega} \,\mathrm{d}s, \qquad
{\lvert {\widetilde{\Phi}}_{\Xi}(t)\rangle}= \mathrm{e}^{-\eta(t)} \ket{\Phi_{\Xi}(t)}.$$ It can be easily seen that $\eta(t)$ can be obtained by solving $$\label{eq:eta}
\frac{\mathrm{d}\eta}{\mathrm{d}t} = \int_{\Omega}
  \lambda(t,\omega) \,\mathrm{d}\mu_{\omega}.$$ Thus, while evolving the state $\ket{\Phi_{\Xi}(t)}$, an additional scalar quantity $\eta(t)$ needs to be evolved simultaneously. In our application, the mark space $\Omega$ is a finite set, and therefore solving does not introduce much numerical cost. In , the time-dependent state ${\lvert {\widetilde{\Phi}}_{\Xi}(t)\rangle}$ can be considered as a trajectory in ${\mathcal{A}}$. To obtain this trajectory, we use the fact that between two adjacent marks, the trajectory satisfies $$\label{eq:tilde_Phi}
\frac{\mathrm{d}}{\mathrm{d}t} {\lvert {\widetilde{\Phi}}_{\Xi}(t)\rangle}= -{\mathrm{i}}H_{\mathrm{easy}}(t) {\lvert {\widetilde{\Phi}}_{\Xi}(t)\rangle},
  \qquad t \in (t_{m-1}, t_m), \quad m > 0.$$ When a mark is met, one just needs to apply the operator ${\widetilde{A}}(t_m,
\omega_m)$. An illustration of the trajectory ${\lvert {\widetilde{\Phi}}_{\Xi}(t)\rangle}$ is given in Figure \[fig:tPhi\].

![Illustration of the trajectory ${\lvert {\widetilde{\Phi}}_{\Xi}(t)\rangle}$. The bold lines denote the evolution of the equation , and the thin dashed lines denote the 