Edit : 2x2 simple system instead, simplification of the question.
I would like to study a system into its diagonal form, but this system is represented by a non-Hermitian Hamiltonian $\tilde{\mathcal{H}}$:
\begin{eqnarray} \tilde{\mathcal{H}} &=& \mathcal{H} +i \gamma \\ &=& \begin{bmatrix} \omega_0 & \Omega \\ \Omega & \omega_1 \end{bmatrix} + i\begin{bmatrix} \gamma_0 & 0 \\ 0 & \gamma_1 \end{bmatrix} \end{eqnarray} where $\mathcal{H}$ represents the unitary part of the Hamiltonian, $\gamma$ the dissipative part. Now, it seems that we have two options, leading to different results:
either we diagonalize the full Hamiltonian $\tilde{\mathcal{H}}$: \begin{eqnarray} \tilde{D} &=& \tilde{P}^{-1} \tilde{\mathcal{H}} \tilde{P} \\ &=& \begin{bmatrix} \omega_- + i\gamma_- & 0 \\ 0 & \omega_+ + i\gamma_+ \end{bmatrix} \end{eqnarray} such that the system is now in a fully diagonal form : $\Re(\tilde{D})$ will give the unitary evolution and $\Im(\tilde{D})$ the dissipative part.
OR, we only diagonalize $\mathcal{H}$ and we apply the rotation to the dissipative part : \begin{eqnarray} D &=& P^{-1} \mathcal{H}P = \begin{bmatrix} \omega_- & 0 \\ 0 & \omega_+ \end{bmatrix}\\ \gamma_P &=& P^{-1} \gamma P = \begin{bmatrix} \gamma_- & \gamma_\pm \\ \gamma_\mp & \gamma_+ \end{bmatrix} \end{eqnarray} Which leads to a dissipative matrix $\gamma_P$ which is not diagonal.
My questions is : which basis best represents the system? I am in the dark here, any intuitive help would be really appreciated.