Let us say we have a Lindblad equation for the density matrix
$$
\dot{\rho}=-\frac{i}{\hbar}[H, \rho]+\sum_{i=1}^{N^{2}-1} \gamma_{i}\left(L_{i} \rho L_{i}^{\dagger}-\frac{1}{2}\left\{L_{i}^{\dagger} L_{i}, \rho\right\}\right) \equiv \mathcal{L} \rho.
$$
Let us say that the Liouvillian $\mathcal{L}$ has a set of eigenvalues $\lambda_i$, i.e. $\mathcal{L}\rho_i = \lambda_i \rho_i$. It is well-known that the $\mathcal{L}$ is non-Hermitian, i.e. $\lambda_i \in \mathbb{C}$.
Now, one applies time-independent unitary transformation $U$ as follows
$$
U\dot{\rho}U^{\dagger}= U \mathcal{L} \rho U^{\dagger}.
$$
Let us denote $\tilde{\rho} \equiv U \rho U^{\dagger}$ and $\tilde{\mathcal{L}} \equiv U \mathcal{L} U^{\dagger}$. Thus, one can write
$$
\dot{\tilde{\rho} } = \tilde{\mathcal{L}}\tilde{\rho}.
$$
Let us say that the transformed Liouvillian $\tilde{\mathcal{L}}$ has a set of eigenvalues $\mu_i$, i.e. $\tilde{\mathcal{L}}\tilde{\rho}_i = \mu_i \tilde{\rho}_i$.
Is it true that the spectrum will not change, i.e. $\lambda_i = \mu_i$?
I know that the unitary transformation does not change the spectrum of the Hermitian operator, but I am not sure about a non-Hermitian one.
Best Answer
Let $\rho_i$ be a set of right eigenvectors of $\mathcal L$, ie $\mathcal L\rho_i = \lambda_i \rho_i$. Then, taking $\tilde \rho_i = U\rho_iU^\dagger$ we have : \begin{align} \tilde{\mathcal L}\tilde \rho_i &= (U\mathcal LU^\dagger )(U\rho_iU^\dagger) \\ &= U(\mathcal L\rho_i)U^\dagger\\ &= \lambda_i U\rho_i U^\dagger\\ &= \lambda_i \tilde\rho_i \end{align} therefore we can take $\mu_i = \lambda_i$.