Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 343 / 1 / b / Solution

The exact spectral condition is that $0$ be a simple eigenvalue of $\mathcal{L}$ and that every other eigenvalue satisfy $\mathrm{Re}\lambda <0$. Then $e^{t\mathcal{L}}\rho$ converges for every initial density operator to the unique stationary state ${\rho }_{\ast }$ satisfying $\mathcal{L}({\rho }_{\ast })=0$. This is the unique stationary state of a Lindbladian condition; the smallest nonzero value of $-\mathrm{Re}\lambda$ is the Lindbladian gap.

A standard operator criterion is irreducibility: the only subspaces invariant under $H$, all $L_{\alpha }$, and all $L_{\alpha }^{†}$ are the zero and full spaces, equivalently their common commutant consists only of scalar multiples of the identity. With no nonzero imaginary-axis eigenvalues, this makes the semigroup relaxing. The spectral statement is the safest general answer because uniqueness of a fixed point alone does not exclude persistent oscillatory modes.