Kraus Operators from Lindblad equation One should be able to formulate the time evolution given by Lindblad equation in terms of Kraus Operators. But how does one do that in practise i.e given $H$ and Lindblad operators $L_\mu$, how does one go about constructing the corresponding Kraus Operators $M_\mu$? 
 A: This derivation closely follows this reference, in the "Proof of Lindblad's theorem" section. More specifically this is the reverse of what is done there.
Start by writing your Lindblad equation
$$
\frac{\mathrm d \rho}{\mathrm d t} = -i[H,\rho] + \sum_\mu L_\mu \rho L_\mu^\dagger - \frac{1}{2} \{L_\mu^\dagger L_\mu,\rho\}
$$
Discretize it with a time step $\Delta t$, with $\frac{\mathrm d \rho}{\mathrm d t}\approx \frac{\rho(t+\Delta t) - \rho(t)}{\Delta t}$, which in the limit of $\Delta t \rightarrow 0$ reduces back to your Lindblad equation. We then get:
$$
\rho(t+\Delta t) = (I +G\Delta t ) \rho (I + G^\dagger\Delta t) + \Delta t \sum_\mu L_\mu \rho L_\mu^\dagger,
$$
where $G = -\frac{1}{2}\sum_\mu L_\mu^\dagger L_\mu$.
Define now $M_u(\Delta t) = I + G\Delta t$ and $M_\mu(\Delta t) = \sqrt{\Delta t} L_\mu$ and we get the propagation of the density matrix with Kraus operators:
$$
\rho(t+\Delta t) = M_u(\Delta t) \rho M_u^\dagger(\Delta t) + \sum_\mu M_\mu(\Delta t) \rho M_\mu^\dagger(\Delta t)
$$
