Eigenvalues of generators If I have a hamiltonian like $\omega\sigma_z$ and 2 Lindblad operators as $\gamma\sigma_-$ and $\gamma\sigma_+$ how can I find eigenvalues of generators?
I think I should put the general form of $\rho$ and these parameters in the Lindblad equation and find $\dot{\rho}$ and then write it like
$$\dot{\rho}=L[\rho]$$
 A: If I understood correctly, you are interested in the Liouvillian superoperator $L$, given by:
$$
L[\rho]=-i[\omega\sigma_z,\rho]+\gamma\left(\sigma_-\rho\sigma_+-\frac{1}{2}\{\sigma_+\sigma_-,\rho\}+\sigma_+\rho\sigma_--\frac{1}{2}\{\sigma_-\sigma_+,\rho\}\right).
$$
For finite-dimensional systems, finding its eigenvalues is quite straightforward. A brute-force, still quite immediate method consists in writing the $2\times 2$ density matrix of the qubit as 4-dimensional vector: if the density matrix reads
$$
\begin{pmatrix}
\rho_{00}&\rhọ_{01}\\
\rho_{10}&\rhọ_{11}\\
\end{pmatrix},
$$
construct the column vector $|\!\,|\rho\rangle =(\rho_{00},\rho_{01},\rho_{10},\rho_{11})^T$. Then, write the Liouvillian $L$ as a $4\times 4$ matrix acting on the vectorized density matrix $|\!\,|\rho\rangle$ (have a look at this reference for further details, or check this previous question). You can quickly verify that:
$$
L=\begin{pmatrix}
-\gamma&0&0&\gamma\\
0&-\gamma+2i\omega&0&0\\
0&0&-\gamma-2i\omega&0\\
\gamma&0&0&-\gamma\\
\end{pmatrix}.
$$
We finally find the eigenvalues:
$$
\lambda_1=0,\quad \lambda_2= -\gamma-2i \omega,\quad\lambda_3=-\gamma+2i\omega,\quad\lambda_4=-2\gamma.
$$
As we may expect, they are the eigenvalues describing the emission and absorption of a two-level atom in an infinite-temperature thermal bath. For the very simple case of a single qubit that we are considering, they can be also derived through the standard Bloch equations [1].
[1] Breuer & Petruccione, The theory of open quantum systems, Oxford University Press (2002).
