I try to understand how I can convert a Lindbladian time evolution operator to the corresponding Kraus operator sum. Let's assume we have a time independent Hamiltonian $H$ and a set of time independent collapse operators $L_i$. Then, the time evolution is given by \begin{equation} \tilde{\rho}(t) = \mathrm{e}^{(\mathcal{G}+\mathcal{H})t}\tilde{\rho}\tag{1} \end{equation} where \begin{equation} \mathcal{H} = -\mathrm{i}(H \otimes \mathbb{1} + \mathbb{1} \otimes H)\tag{2} \end{equation} \begin{equation} \mathcal{G} = \sum_i L_i^* \otimes L_i - \frac{1}{2} \mathbb{1} \otimes (L_i^\dagger L_i ) - \frac{1}{2} ({L_i^*}^\dagger {L_i^*} ) \otimes \mathbb{1}\tag{3} \end{equation} and $\tilde{\rho}$ is the density martix in vector form. I would like to obtain a representation like \begin{equation} \rho = \sum_k A_k \rho(0) A_k^\dagger\tag{4} \end{equation}

For the case of a pure coherent evoltution, this is obious \begin{equation} A_0 = \mathrm{e}^{-\mathrm{i}H t} \qquad \text{and} \qquad A_i = 0 \quad \forall\quad i>0\tag{5} \end{equation}

How is this now in the generalized case? Does the relaxation in the system introduce only $A_i$ with $i>0$? How do I obtain the $A_i$?




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