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I'm trying to understand how to obtain a set of Kraus operators from Lindblad master equations. For a $1$-qubit dephasing noise model, it is well-known that the set of Kraus operators is $\{ \sqrt{p}I, \sqrt{1-p}Z \}$ and Lindblad operator is $L = \sqrt{\gamma} Z$, where one can express $p$ in terms of $\gamma$. In fact, this reference provides a great explanation on it.

Now I want to extend this to multiple qubits. For $2$-qubits system subjected to a pure local dephasing noise model, Lindblad operators are $L_i = \sqrt{\gamma}Z_i$ where $Z_i$ is a Pauli-Z operator acting on $i$-th qubit (see e.g., this post). What is a set of Kraus operators for this model, and how one should derive it?

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In general, finding the Kraus operators is non-trivial, since "finding the Kraus operators" is essentially the same as "solving the Lindblad equation".

The solution of the Lindblad equation is a quantum map $\Lambda_t$ such that $\rho_t = \Lambda_t \rho_0$. (If the Lindblad equation is time-homogeneous, i.e., $\dot\rho_t = L \rho_t$ with constant $L$, then $\Lambda_t = \exp(Lt)$.) The Kraus operators can be found from the Choi representation of $\Lambda_t$ as follows.
Let $|+\rangle = \frac{1}{\sqrt d} \sum_n |n\rangle \otimes |n\rangle$ be the maximally entangled state on $\mathcal H \otimes \mathcal H$, where $\mathcal H$ is the Hilbert space of the system and $d$ its dimension. The Choi representation of $\Lambda$ ($t$ subscripts will be omitted from now on) is then $$ J(\Lambda) = \bigl( 1 \otimes \Lambda \bigr)\bigl(\, |+\rangle\langle+|\, \bigr) . $$ Its eigendecomposition can be brought into the form (see for example Norbert Schuch's answer here for details) $$ J(\Lambda) = \sum_i (1 \otimes A_i)\, |+\rangle\langle+|\, (1 \otimes A_i)^\dagger $$ and the $A_i$ here are the Kraus operators, $\rho_t = \sum A_i \rho_0 A_i^\dagger$.

In your concrete problem, if I understand it correctly, you have two subspaces $\mathcal H_1$ and $\mathcal H_2$, and quantum maps $\Lambda_1$ and $\Lambda_2$ acting independently on the subspaces. If $\{ A_i \}_i$ is the set of Kraus operators for $\Lambda_1$ and $\{ B_j \}_j$ the one for $\Lambda_2$, then $\{ A_i \otimes B_j \}_{i,j}$ are Kraus operators for $\Lambda_1 \otimes \Lambda_2$.

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