= Unique stationary state of a Lindbladian
For a finite-dimensional <Lindbladian> $\mathcal L$, convergence to one stationary <density matrix> for every initial state holds exactly when zero is a simple <eigenvalue> and every other <eigenvalue> has strictly negative real part. The decaying spectral modes then leave only the stationary component.
A scalar <commutant of an operator algebra> generated by $H,L_\alpha,L_\alpha^\dagger$ is not sufficient on its own. A useful sufficient criterion requires an additional hypothesis: a faithful stationary state, meaning a stationary <density matrix> that is a <positive-definite matrix>. Under that hypothesis, a commutant consisting only of scalar multiples of the <identity operator> implies uniqueness.<Criteria for Davies Irreducibility of Markovian Quantum Dynamics by Yikang Zhang and Thomas Barthel>[Section 2, Theorem 2.2, explicit faithful-state hypothesis; Example 2.1 distinguishes a trivial commutant from uniqueness]{ref}
For a directly checkable counterexample without a faithful stationary state, take an <orthonormal basis> $|0\rangle,|1\rangle,|e\rangle$, $H=0$, and
$$
L_0=\sqrt\gamma|0\rangle\langle e|,\qquad L_1=\sqrt\gamma|1\rangle\langle e|,\qquad\gamma>0.
$$
Every <density matrix> supported on $\operatorname{span}\{|0\rangle,|1\rangle\}$ is stationary: both jump terms and the <anticommutator> terms vanish there. Nevertheless, the jumps, their adjoints and their products generate all nine <matrix units>, so their commutant is scalar. There are therefore infinitely many stationary states despite that scalar commutant. The missing faithful-state assumption changes the conclusion.
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