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quantum statistical mechanics describes equilibrium and fluctuations with density matrices and traces over a quantum Hilbert space. The canonical equilibrium state is $e^{-\beta H}/Z$, while particle exchange introduces $H-\mu N$ in the grand canonical ensemble. Noncommutativity and particle exchange distinguish it from a classical phase-space distribution.

A quantum superposition is a linear combination of quantum state vectors in a complex Hilbert space. In a specified orthonormal basis, a normalized pure state has the displayed expansion, where $a_j$ are probability amplitudes. A nonzero linear combination can always be normalized; an arbitrary sum of normalized vectors need not already have norm one. Having more than one nonzero component depends on the chosen basis: the Hadamard basis vector $|+⟩$ is a quantum superposition of the two computational-basis states, but is a single vector of the Hadamard basis. ^[ref].

The distinction from classical ignorance is quantitative. Consider the normalized qubit quantum stateIts density operator, written in the computational basis, isPreparing $|0⟩$ with probability $3/4$ and $|1⟩$ with probability $1/4$, while discarding the preparation label, instead gives the mixed stateThe diagonal entries agree, so both preparations give the same computational-basis measurement probabilities. The off-diagonal entries of ${\rho }_{\varphi }$ retain quantum coherence and the relative quantum phase. In the Hadamard basis, the Born rule givesThe density matrix representation retains this distinction between a pure state and a mixed state. ^[ref].

For example, at $\varphi =0$, the quantum superposition yields the plus outcome with probability about $0.9330$, whereas the mixed state yields it with probability $0.5$. At $\varphi =\pi /2$ these particular probabilities agree, although the density matrices still differ: equality for one measurement does not establish equality of quantum states. The probability amplitude figure shows the comparison throughout a full cycle of relative quantum phase.

This is why a quantum superposition is useful: quantum gates act linearly on its probability amplitudes, and unitary operators preserve the full pure state information, including relative quantum phases. A Hadamard gate can turn otherwise hidden quantum coherence into different outcome probabilities. The beautiful feature is that the same basis weights can support different predictions when amplitudes recombine. The expansion specifies a preparation and its measurement predictions; it does not assign every unmeasured outcome a simultaneous classical value. See projective measurement for the separate question of the conditional state after a measurement.

Quantum theory supplies a common language for quantum mechanics, measurement in quantum mechanics, and quantum information: prepare a quantum state, transform it, and ask which experimental outcomes occur. In a finite-dimensional idealization, a pure state is a normalized vector of complex numbers. Its entries are probability amplitudes; the Born rule turns their squared magnitudes into probabilities. Between measurements, an isolated system evolves by a unitary operator, preserving its norm and inner products. A quantum circuit makes this evolution explicit as a sequence of quantum gates.

The extra information carried by quantum phases is essential. Relative quantum phases affect how probability amplitudes add when different contributions reach the same output. Their interference can enhance one outcome and cancel another. A global phase, multiplying the entire quantum state by the same complex factor of magnitude one, changes neither these relative phases nor observable predictions. The Hadamard gate below provides a small experiment in which this distinction becomes a deterministic change of output.

Quantum coherence describes this capacity to interfere relative to a specified orthonormal basis. In a density matrix, it appears in off-diagonal entries; a diagonal mixed state retains the populations but lacks that basis coherence. Thus matching the probabilities of one measurement need not match the predictions of a later experiment. Baumgratz, Cramer, and Plenio made this basis-dependent viewpoint precise in their resource framework for quantum coherence. ^[ref].

This makes quantum information an operational question: which distinctions can a preparation, an available unitary operator, and measurement in quantum mechanics reveal? Amplitude amplification illustrates how carefully arranged interference converts a weakly represented answer into a likely observable result. Preserving the relevant quantum coherence throughout the computation is part of the physical task. ^[ref].

Fix a detected position and polarization of waves. Write the electric field operator as $\hat{E}(t)={\hat{E}}^{(+)}(t)+{\hat{E}}^{(-)}(t)$, where the positive-frequency part contains annihilation operators and ${\hat{E}}^{(-)}=({\hat{E}}^{(+)}{)}^{†}$. An ideal absorbing photodetector responds towith density matrix $\rho$. The diagonal gives the single-count rate up to the detector coupling. Multiple absorptions require higher, normal-ordered field products. Roy J. Glauber used these correlations to develop the quantum account of optical coherence.^[ref]

The ordering follows from a detection amplitude. If absorption takes an initial pure state $|i⟩$ to an unobserved final state $|f⟩$, its field factor is $⟨f|{\hat{E}}^{(+)}|i⟩$. Squaring its complex modulus and summing over a complete orthonormal basis of final states gives $⟨i|{\hat{E}}^{(-)}{\hat{E}}^{(+)}|i⟩$. Averaging over initial preparations replaces this expression by a trace. Thus a detector that absorbs cannot count the Fock vacuum, although the vacuum has quantum fluctuations.

For one selected normal mode, ${\hat{E}}^{(+)}(t)=\mathcal{E}u(t)a$. All Fock states with nonzero occupation, coherent states, and thermal light in this same mode havewhere both denominators are nonzero. Its magnitude is one even when the expected value of the field vanishes. Perfect first-order optical coherence therefore does not identify a coherent state. Pair detection supplies additional information.

A quantum Trajectory Theory unravelling represents a Lindblad equation by an ensemble of stochastic pure states. In units $\hslash =1$, define the non-Hermitian effective HamiltonianFor a normalized state $|\psi ⟩$, channel $\alpha$ has instantaneous jump rate $r_{\alpha }=⟨\psi |L_{\alpha }^{†}{L}_{\alpha }|\psi ⟩$. During a small interval $dt$, that jump occurs with probability $r_{\alpha }dt$ and sends the state to $L_{\alpha }|\psi ⟩/\sqrt{r_{\alpha }}$. A zero rate means that channel cannot jump. Otherwise, with total no-jump probability $1-{\sum }_{\alpha }{r}_{\alpha }dt$, evolve with $I-i{H}_{\mathrm{eff}}dt$ and renormalize.

To verify the ensemble evolution, put $\rho =|\psi ⟩⟨\psi |$. The conditional normalizations cancel their probability weights, givingExpanding recovers the Lindblad equation. Different monitoring choices can give different trajectories with the same ensemble density matrix. For spontaneous emission, $L=\sqrt{\Gamma }|g⟩⟨e|$ gives jump rate $\Gamma |⟨e|\psi ⟩{|}^2$ and resets the atom to its ground state.^[ref]

A qubit, or quantum bit, is a quantum system whose state space is a two-dimensional complex Hilbert space. Choose an orthonormal basis $|0⟩,|1⟩$, called the computational basis. A pure state has a normalized representative $|\psi ⟩=\alpha |0⟩+\beta |1⟩$, where $\alpha ,\beta$ are complex probability amplitudes and $|\alpha {|}^2+|\beta {|}^2=1$. General qubit states, including mixed states, require a density matrix. ^[ref].

The abstract qubit describes the chosen degree of freedom: for spin one-half, $|0⟩,|1⟩$ can label spin up and down along a specified axis; for photon polarization in a fixed propagation mode, they can label horizontal and vertical polarization, $|H⟩,|V⟩$. These implementations share the same two-dimensional linear algebra, but their preparation and measurement in quantum mechanics use different apparatus. A qubit therefore need not be an entire two-level particle. ^[ref].

The usefulness of a qubit lies in controlling quantum superpositions and their relative quantum phases, then converting them into outcome probabilities with quantum gates and an appropriate measurement. One ideal computational-basis measurement returns one label, zero or one. It does not reveal both probability amplitudes, their relative quantum phase, or all the bits of a larger register's expansion. Quantum state tomography instead estimates a qubit state using many repetitions of the preparation and suitable measurement settings. ^[ref].

An actual preparation and measurement example is Dederick and Beck's 2014 teaching-laboratory experiment. A 405-nm laser pumped two nonlinear crystals, producing signal and idler photons near 810 nm. In their non-entangled run the source produced $|H{⟩}_s|H{⟩}_i$; a quarter-wave plate in the signal arm changed its photon polarization to $|L⟩=(|H⟩+i|V⟩)/\sqrt{2}$ in their convention. The plate supplies a relative quantum phase between polarization components. An idler detection selected the signal trials; signal detectors measured horizontal/vertical, diagonal, and circular polarization bases on separate runs. Before the polarization analyzer, a half-wave plate at $22.5$ degrees selected the diagonal basis, and a quarter-wave plate at $45$ degrees selected the circular basis. Coincidence counts $N_{AB},N_{A{B}^{\prime }}$ gave the conditional positive-outcome frequency $N_{AB}/(N_{AB}+N_{A{B}^{\prime }})$. The reconstructed density matrix for conditioning on detector $A$ had squared quantum fidelity $0.96$ with $|L⟩$ and purity of a density operator $0.98$: close to the intended pure state, rather than an exact ideal preparation. ^[ref].

The model diagram explains why those settings matter. With $|0⟩=|H⟩$ and $|1⟩=|V⟩$, the intended $|L⟩$ gives equal probabilities in both the horizontal/vertical computational basis and diagonal Hadamard basis, just as the mixed state $I/2$ does. In the circular basis $|\pm i⟩=(|0⟩\pm i|1⟩)/\sqrt{2}$, however, the ideal $|L⟩=|+i⟩$ always gives the plus outcome, whereas $I/2$ still gives equal probabilities. These are deductions from the Born rule, not the paper's measured counts. Unequal detector efficiencies and finite samples must be accounted for when interpreting experimental frequencies; the ideal comparison assumes equally efficient detection.

Choose the Pauli matrices with ${\sigma }_z=\mathrm{diag}(1,-1)$, and let $\omega >0$ be an angular frequency. The computational basis is an orthonormal basis of energy eigenstates for the qubit Hamiltonian operatorIn this convention $|1⟩$ is the lower energy eigenstate; the labels describe a basis choice, not a universal ground state/excited state ordering. Start in the Hadamard basis pure state $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$. ThenThe first factor in the second expression is a global phase. The ratio of the two probability amplitudes has relative quantum phase $\omega t$, so its sign follows directly from $E_1-E_0=-\hslash \omega$.

Multiplying this density matrix by each Pauli matrix and taking its trace gives the Bloch vector:The Bloch vector travels counterclockwise around the equator of the Bloch sphere, from $+x$ toward $+y$ when viewed from $+z$. A computational-basis measurement always has $P_Z(0)=P_Z(1)=1/2$, while the Hadamard basis and ${\sigma }_y$ eigenbasis measurements giveAt $\omega t=\pi /2$, the quantum state is $|+i⟩=(|0⟩+i|1⟩)/\sqrt{2}$ up to a global phase, giving $P_Y(+)=1$. At $\omega t=\pi$, it is $|-⟩$ up to a global phase, giving $P_X(+)=0$. Both retain equal energy measurement probabilities.

At $t=2\pi /\omega$, the time-evolution operator is $U(t)=-I$, so $|\psi (t)⟩=-|+⟩$. The physical quantum state ray, its density operator, and every measurement's probabilities have returned. The representative vector returns exactly only at $t=4\pi /\omega$, when $U(t)=I$. The minus sign is a global phase for this isolated evolution; comparing an evolved branch with an unevolved branch through a controlled unitary gate can make it a relative quantum phase, as explained in global phase of an uncontrolled quantum gate is unobservable.

A qubit state specifies measurement probabilities, rather than a pre-existing answer to every possible measurement. For a pure state, normalized vectors differing by a global phase represent the same physical state:Equivalently, the pure state is a ray: the one-dimensional complex linear subspace spanned by any nonzero representative. Dividing a nonzero vector by its norm fixes its length, but leaves the global phase free. The density operator is unchanged because $(e^{i\gamma }|\psi ⟩)(e^{i\gamma }|\psi ⟩{)}^{†}=|\psi ⟩⟨\psi |$. In contrast, multiplying only one nonzero component by $e^{i\varphi }$ changes a relative quantum phase and can change measurement probabilities. ^[ref].

After choosing a global phase, every pure state of a qubit can be writtenAt $\theta =0$ or $\pi$, the parameter $\varphi$ does not distinguish physical states: only one component survives. Elsewhere it is the relative quantum phase. The corresponding unit Bloch vector is $\mathbf{r}=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$ on the existing Bloch sphere. Its angles describe a space of states; for a photon polarization qubit they are not the photon's position or direction of travel. ^[ref].

The Born rule in the computational basis gives $P(0)={\cos }^2(\theta /2)$ and $P(1)={\sin }^2(\theta /2)$. These probabilities discard $\varphi$. In the Hadamard basis $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$, adding or subtracting the probability amplitudes instead givesThis last equality follows by substituting the angular representative. In the basis $|\pm i⟩=(|0⟩\pm i|1⟩)/\sqrt{2}$, the same calculation givesThus both the real and imaginary parts of quantum coherence can be observed, using different measurement settings. For example, $(|0⟩+|1⟩)/\sqrt{2}$ and $(|0⟩-|1⟩)/\sqrt{2}$ both give $P(0)=P(1)=1/2$, but the first gives $P_X(+)=1$ and the second $P_X(+)=0$. A global phase could never cause this difference.

For a general qubit state, write its density matrix in the computational basis asThese are exactly the Hermitian, unit-trace, positive semidefinite constraints. Positivity means every eigenvalue is nonnegative, not that every matrix entry is nonnegative. The general definition is given in density matrix. ^[ref].

For this two-by-two matrix, the explicit constraint follows from its determinant, $\det \rho =p(1-p)-|c{|}^2$: with trace one, a nonnegative determinant and real eigenvalues rule out a negative eigenvalue. Its eigenvalues areThe pure states have $|c{|}^2=p(1-p)$ and eigenvalues $1,0$; the mixed states have strict inequality and two positive eigenvalues. For a normalized representative, $p=|\alpha {|}^2$ and $c=\alpha {\beta }^{\ast }$. A statistical preparation $\rho ={\sum }_j{w}_j|{\psi }_j⟩⟨{\psi }_j|$, with $w_j\ge 0$ and ${\sum }_j{w}_j=1$, averages probabilities, rather than adding probability amplitudes. Such a preparation is mixed if the positive-weight vectors do not all represent the same ray. A reduced density matrix of an entangled pair can also be mixed even when the pair has a pure state. ^[ref].

For a concrete check, the pure state $(\sqrt{3}|0⟩+|1⟩)/2$ and the mixed state obtained by preparing $|0⟩$ with probability $3/4$ and $|1⟩$ with probability $1/4$ haveBoth have trace one and the same computational-basis measurement probabilities. Their eigenvalues are respectively $(1,0)$ and $(3/4,1/4)$; the respective values of their purity of a density operator $\mathrm{Tr}({\rho }^2)$ are $1$ and $5/8$. Their Hadamard basis plus probabilities are respectively $1/2+\sqrt{3}/4\approx 0.9330$ and $1/2$. By contrast, setting $p=3/4,c=1/2$ would give determinant $-1/16$ and eigenvalue $(2-\sqrt{5})/4<0$: normalized diagonal entries alone do not make a valid qubit state.

The same constraints become especially beautiful in the Bloch ball representation. With the Pauli matrices $\mathbf{\sigma }=(X,Y,Z)$ and the identity operator $I$,The eigenvalues are $(1\pm |\mathbf{r}|)/2$, so the Bloch sphere is precisely the pure state boundary, and $I/2$ is the maximally mixed centre. For each axis $j$, the Born rule gives $P_j(\pm )=(1\pm r_j)/2$. This unifies a geometric picture, matrix positivity, and experimental frequencies in one description. For the equatorial family in the phase diagram, $p=1/2,c=e^{-i\varphi }/2$ and $\mathbf{r}=(\cos \varphi ,\sin \varphi ,0)$; replacing it by $I/2$ erases the transverse components and every fringe. See quantum state tomography for how independent ensembles measured along the three axes estimate these components. ^[ref].

Ramsey interferometry uses two coherent pulses with a free interval between them: the first creates a quantum superposition, the interval accumulates relative quantum phase, and the second converts that quantum phase into measurement probabilities. Its attraction is that an ordinary population readout can reveal the quantum coherence hidden from an energy measurement during the free interval.

Here is an ideal qubit model with an explicit pulse convention. Let $R_y(\theta )=\exp (-i\theta {\sigma }_y/2)$, prepare $|0⟩$, apply $R_y(\pi /2)$, evolve freely with $U_{\varphi }=\exp (-i\varphi {\sigma }_z/2)$, and apply the inverse pulse $R_y(-\pi /2)$ before a computational-basis measurement. The first pulse gives $|+⟩$. Direct matrix multiplication givesThus the Born rule gives $P(0)={\cos }^2(\varphi /2)$ and $P(1)={\sin }^2(\varphi /2)$. Before the second pulse, the same computational-basis measurement would give $1/2$ for both outcomes for every $\varphi$. At $\varphi =2\pi /3$, the second pulse changes these probabilities to $1/4$ and $3/4$. If the two pulses have the same sign instead, the output labels interchange: the pulse convention is part of the prediction.

In a pulse experiment the phase is measured relative to the oscillator supplying the coherent pulses. In a rotating frame, write the free Hamiltonian operator as $H_{\mathrm{rot}}=\hslash \delta {\sigma }_z/2$ with signed angular frequency difference $\delta$ in the chosen labeling convention. A free interval $T$ then accumulates $\varphi =\delta T$; changing the second pulse's phase changes the analysis direction. This comparison makes Ramsey interferometry useful for measuring frequency differences and diagnosing loss of quantum coherence.

For example, suppose the first pulse has prepared a density matrix with equal populations and transverse Bloch vector length $0\le V\le 1$:The same inverse analysis pulse yields $P(0)=(1+V\cos \varphi )/2$. The pure state case $V=1$ gives the full ideal fringe; the mixed state $I/2$ gives the flat result $1/2$. This is a controlled mathematical comparison, not a fit to an experiment. Experimental contrast alone does not identify quantum decoherence: imperfect preparation, pulse errors and readout errors can also reduce it.

The phase difference between nonzero amplitudes in a specified basis, for example between $a$ and $b$ in $a|1⟩+b|2⟩$. Unlike a global phase, it affects quantum interference when the two basis components are recombined. It is encoded in the off-diagonal density matrix elements and depends on the phase convention for the basis states.

Single-quantum coherence is a component of NMR coherence with coherence order $p=+1$ or $p=-1$. It connects basis states whose total spin projections differ by one unit. Under the convention $U(\varphi )=e^{-i\varphi F_z}$, an order-$p$ component acquires $e^{-ip\varphi }$.^[ref]

For one spin one-half nucleus, $I_{+}$ and $I_{-}$ are the positive- and negative-order components. Transverse magnetization is represented by their combinations,A Hermitian density matrix contains conjugate off-diagonal entries, so it can encode two real transverse components while containing both coherence-order signs.

An original state example usesIts eigenvalues are0.8 and0.2 for every $\varphi$, and its populations remain one half. Taking the traces against $I_x,I_y$ givesAt $\varphi =\pi /2$, the transverse expectation has moved from $+x$ to $+y$ without changing the populations. This is a constructed quantum-state comparison, not a measured thermal polarization.

Conventional transverse NMR detection can access the appropriate single-quantum contribution. The complex receiver signal chooses a phase convention for combining its two measured components. This is why a statement about a detected phase sign should specify its convention, while the operator transformation above remains explicit.^[ref]

Multispin single-quantum terms need not be merely isolated-spin magnetization: correlations can accompany the projection change. A coherence order identifies the transformation under a chosen common rotation, rather than uniquely specifying every spin correlation or the pulse sequence that generated it.

For a spin one-half system, an ideal projective measurement along a unit vector $\mathbf{n}$ measures ${\sigma }_{\mathbf{n}}=\mathbf{n}\cdot \mathbf{\sigma }$. Its eigenvalues are $\pm 1$, corresponding to physical spin angular momentum components $\pm \hslash /2$, and its orthogonal projections are ${\Pi }_{\pm }=(I\pm {\sigma }_{\mathbf{n}})/2$. For density matrix $\rho =(I+\mathbf{r}\cdot \mathbf{\sigma })/2$, the Born rule and the Pauli matrix multiplication law give the displayed probabilities. Thus a spin measurement along one direction tests only that component of the Bloch vector; measurements along three independent directions determine the entire density matrix.

For a differentiable family of density matrices ${\rho }_{\theta }$, the symmetric logarithmic derivative is a Hermitian operator $L_{\theta }$ satisfyingIf ${\rho }_{\theta }={\sum }_j{p}_j|j⟩⟨j|$, its determined matrix entries are $(L_{\theta }{)}_{jk}=2⟨j|{\partial }_{\theta }{\rho }_{\theta }|k⟩/(p_j+p_k)$ whenever $p_j+p_k>0$. Entries wholly in the kernel are undetermined when a solution exists and do not affect $\mathrm{Tr}({\rho }_{\theta }{L}_{\theta }^2)$. Rank-changing boundary points need extra care.

For ${\rho }_{\theta }=\mathrm{diag}(\theta ,1-\theta )$, $0<\theta <1$, one gets $L_{\theta }=\mathrm{diag}(1/\theta ,-1/(1-\theta ))$ and Quantum Fisher information $F_Q=1/[\theta (1-\theta )]$. This recovers ordinary Fisher information for a binary observation. In noncommuting families the symmetrized equation supplies the operator version needed to quantify local sensitivity.^[ref]

For linear operators $A$ on $V$ and $B$ on $W$, the tensor product operator is characterized byFor bounded linear operators on Hilbert spaces, it extends continuously to the Hilbert tensor product. If both factors are positive semidefinite operators, so is their tensor product operator. In finite dimensions an orthonormal eigenbasis in each factor gives a tensor-product basis with nonnegative eigenvalues $a_i{b}_j$.

The positivity assertion also holds without an eigenbasis. Take the positive square root of an operator of each factor and put $C=A^{1/2}\otimes B^{1/2}$. Then $A\otimes B=C^{†}C$, so its quadratic form is $\Vert Cz{\Vert }^2\ge 0$. This is the positivity used in the necessary direction of the positive partial transpose criterion, where each product term in a separable density matrix remains positive after transposing one factor.

Approach of a physical system to thermal equilibrium, usually through interactions with other degrees of freedom. Equality of a reduced density matrix with a quantum Gibbs state at one instant does not alone demonstrate a thermalization mechanism or equilibration over time.

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.

The zero-total-$Z$ eigenspace of two spin one-half systems stores one logical qubit. Both basis states satisfy $(ZI+IZ)|\psi ⟩=0$, so every collective dephasing rotation acts as the identity operator on their span. Also $ZZ|01⟩=-|01⟩$ and $ZZ|10⟩=-|10⟩$, so simultaneous Pauli Z gates preserve every encoded density matrix. For independent rotations, the logical amplitudes instead acquire phases $e^{-i({\varphi }_1-{\varphi }_2)/2}$ and $e^{i({\varphi }_1-{\varphi }_2)/2}$, which generally alter their relative phase. The code is not a two-dimensional decoherence-free subspace for arbitrary collective rotations about all axes; the invariant two-spin singlet alone is only one-dimensional.

For normalized pure states $|u⟩,|v⟩$, a two-member quantum state ensemble gives the maximally mixed state of a qubit exactly when its members are orthogonal and equally probable. To prove necessity, take expectation values of its density matrix in the two members: $1/2=p+(1-p)|⟨u|v⟩{|}^2\ge p$ and $1/2=1-p+p|⟨u|v⟩{|}^2\ge 1-p$. Hence $p=1/2$ and the overlap vanishes. Sufficiency is the resolution of the identity in an orthonormal basis.

After a computational-basis projective measurement, an observer who lacks the result assigns a diagonal density matrix. If the observer's preparation prior has mean $q=\mathbb{E}|\alpha {|}^2$, the assignment is the displayed matrix. No outcome record and no stated prior do not logically force $q=1/2$; that value follows from a symmetric or maximum-entropy ignorance convention. An observer who knows the preparation instead uses $q=|\alpha {|}^2$.

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.