= Solution
The exact spectral condition is that $0$ be a simple eigenvalue of $\mathcal L$ and that every other <eigenvalue> satisfy $\operatorname{Re}\lambda<0$. Then $e^{t\mathcal L}\rho$ converges for every initial <density operator> to the unique stationary state $\rho_*$ satisfying $\mathcal L(\rho_*)=0$. This is the <unique stationary state of a Lindbladian> condition; the smallest nonzero value of $-\operatorname{Re}\lambda$ is the <Lindbladian gap>.
A standard operator criterion is irreducibility: the only subspaces invariant under $H$, all $L_\alpha$, and all $L_\alpha^\dagger$ 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.
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