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A finite-dimensional quantum dynamical semigroup is Davies irreducible when no proper nonzero subspace can support a state for all future times. Equivalently, if a projection $P$ satisfies ${\Phi }_t(P\rho P)=P{\Phi }_t(P\rho P)P$ for every state $\rho$ and every $t\ge 0$, then $P=0$ or $P=I$. This is a property of invariant supports, rather than merely uniqueness of a stationary density matrix.

For a two-level system with only spontaneous decay, $L=\sqrt{\gamma }|0⟩⟨1|$ and zero Quantum Hamiltonian, the ground-state line is invariant. The stationary state is unique but pure, so the dynamics is reducible. Adding a positive excitation rate $L^{\prime }=\sqrt{\kappa }|1⟩⟨0|$ removes that invariant line; the stationary populations are $\gamma /(\gamma +\kappa )$ and $\kappa /(\gamma +\kappa )$ and the dynamics is irreducible. In finite dimensions irreducibility ensures a unique faithful stationary state, which has full rank. Related notions of quantum irreducibility must be distinguished.^[ref]

Thomas Barthel and Yikang Zhang established algebraic tests for Davies irreducibility of Markovian open-system quantum evolution. Their criterion connects the Lindblad operators and effective drift operator with the algebra they generate.^[ref]

For a finite-dimensional irreducible system, the stationary density matrix is unique and full rank. The work distinguishes different notions of reducibility, preventing a convenient algebraic test from being applied to the wrong dynamical definition.

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 $H,L_{\alpha },L_{\alpha }^{†}$ 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.^[ref]

For a directly checkable counterexample without a faithful stationary state, take an orthonormal basis $|0⟩,|1⟩,|e⟩$, $H=0$, andEvery density matrix supported on $\mathrm{span}\{|0⟩,|1⟩\}$ 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.

Yikang Zhang, with Thomas Barthel, characterized Davies irreducibility in quantum dynamical semigroups. In their finite-dimensional criterion, the multiplicative algebra generated by the Lindblad operators and $K=iH+\frac{1}{2}{\sum }_a{L}_a^{†}{L}_a$ must be the full operator algebra, with $\hslash =1$.^[ref]

This makes dynamical accessibility an algebraic question. Changing a Hamiltonian term or a dissipator can change irreducibility even when much of the model is left in place.