When is the Liouvillian superoperator not diagonalisable? An $n \times n$ matrix, $L$, is diagonalisable if it has $n$ linearly independent eigenvectors.
I've recently been working with open quantum systems and come across the non-hermitian, Liouvillian superoperator. My question is, when does the Liouvillian have linearly independent eigenvectors and is there a physical meaning behind this.
I'm guessing it has something to do with the eigenvalues of $L$, as I know that eigenvectors corresponding to distinc eigenvalues are linearly independent.
 A: The "normal" situation is the one where all eigenvalues of the Liouvillian superoperator are distinct. Like you said, it is diagonalizable in this case.
The case where the Liouvillian is not diagonalizable is indeed interesting. In this context, it is usually considered to be a function of one or more system parameters. When the parameters are varied, some eigenvalues might suddenly coincide. (That happens along lower-dimensional sub-manifolds of the parameter space, which is why I called everything else the "normal" situation initially.) At points where eigenvalues coincide, it can happen that the eigenvectors coincide also, i.e., they are linearly dependent. Such points in the parameter space are called quantum exceptional points (EPs). EPs are an active area of research; this article might be a good introduction: https://arxiv.org/abs/1909.11619
The behavior of an open quantum system at an EP is comparable to a (classical) damped harmonic oscillator at critical damping. Note that if all eigenvalues are distinct, the Lindblad equation describes exponential decay of the system state towards a steady state, the decay rates are given by the real parts of the non-zero EVs (which are all negative). At EPs, this decay is not exponential, but of the form $e^{\lambda t} P(t)$ for some polynomial $P$.
