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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.

A unitary operator $U:H\to K$ is a surjective linear isometry between complex inner product spaces. It preserves every inner product, not just the norm of a selected vector:Its adjoint operator is its inverse function, soFor an operator on one space, both identities have the same identity operator on the right. In finite-dimensional orthonormal bases, its matrix is a unitary matrix; the real counterpart is an orthogonal transformation.

Surjectivity matters in infinite dimension. An isometric embedding can satisfy $U^{\ast }U=I$ without satisfying $U{U}^{\ast }=I$: a proper image leaves some vectors outside its range. A unitary operator instead gives a reversible change of orthonormal coordinates, preserving all lengths and orthogonality.