# Convert a Lindbladian time evolution operator to the Kraus operator sum representation

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 $$A_i$$?

Thanks!

Peter