Density matrix

A finite-dimensional density operator $\rho$ represents a quantum state on a Hilbert space. It is Hermitian, positive semidefinite, and has trace one:The density matrix is its matrix in a chosen orthonormal basis. In dimension $N$ it has $N^2-1$ real parameters: a Hermitian matrix has $N^2$, and fixing its trace removes one; positivity restricts the allowed values.

A quantum state ensemble of normalized pure states $|{\psi }_i⟩$ with positive probabilities $p_i$ givesZero-weight terms can be omitted. Conversely, the spectral theorem gives such an ensemble for every density operator, using its orthonormal eigenvectors and nonnegative eigenvalues, with zero eigenvalues omitted. A pure state has $\rho =|\psi ⟩⟨\psi |$; a mixed state needs at least two nonzero eigenvalues. ^[ref]

For a projective measurement with orthogonal projections ${\Pi }_a$, the Born rule isThis is the average of the component pure state probabilities. Positivity makes each outcome probability nonnegative, and ${\sum }_a{\Pi }_a=I$ makes them sum to one. The same density operator therefore predicts every measurement on the system without requiring a unique preparation history. ^[ref]

For a closed system, Schrödinger picture evolution under a unitary operator iswhen ${\rho }_H={\rho }_S(0)$ denotes the time-independent Heisenberg picture state and $U(0)=I$. In the Schrödinger picture the density operator evolves; in the Heisenberg picture the observables evolve instead, with identical measurement predictions. This unitary operator preserves the eigenvalues and hence the purity of a density operator. ^[ref]

A historical source for the probabilistic formulation of quantum mechanics is von Neumann's 1927 paper Wahrscheinlichkeitstheoretischer Aufbau der Quantenmechanik. ^[ref]

For a chosen normalized level state $|k⟩$, its population is the probability $⟨k|\rho |k⟩=\mathrm{Tr}(|k⟩⟨k|\rho )$ of finding the system in that state. In an orthonormal level basis these diagonal entries are nonnegative and sum to one. Off-diagonal quantum coherence contains information absent from the population vector, so matching all level populations need not prepare a specified coherent superposition.

For a chosen orthonormal basis, the off-diagonal entries of a density operator encode quantum coherence between distinct basis states. A diagonal density operator is an incoherent mixture in that basis. This property is basis-dependent: diagonalizing the operator does not imply that it lacked coherence in the measurement or energy basis of interest. Relative phases affect these entries, while a global wavefunction phase cancels from the density operator. Dephasing damps the entries and can spoil stimulated Raman adiabatic passage by removing the ground-state interference that maintains the dark state.

This collective-unitary-invariant two-qubit state has singlet weight $p$ and equal weights on the three triplet Bell states. With $P_s$ the spin singlet state projector, its partial transpose has eigenvalues $(1-2p)/2$ once and $(1+2p)/6$ three times. It is entangled exactly when $p>1/2$. At $p=1/2$, it is the equally weighted mixture of six antiparallel product states along the three Pauli axes. In the interval $1/4\le p\le 1/2$, convex mixing of this boundary state with $I/4$ explicitly proves separability. The weight $p$ here is the singlet probability, not the alternative coefficient multiplying a pure singlet in a white-noise mixture.

Write $W_F=(I\otimes I-w{\sum }_j{\sigma }_j\otimes {\sigma }_j)/4$, with $w=(4F-1)/3\in (1/3,1]$. For positive matrices $X=A^{†}A=a_{0}I+a\cdot \sigma$ and $Y=B^{†}B=b_{0}I+b\cdot \sigma$, the success probability is $p=a_0{b}_0-wa\cdot b$. It is at least $a_0{b}_0-|a||b|\ge \sqrt{(a_0^2-|a{|}^2)(b_0^2-|b{|}^2)}=|\det A\det B|$. Squaring the middle comparison leaves $(a_0|b|-b_0|a|{)}^2\ge 0$.

Combine the single-copy Werner filtering normalization bound with concurrence scaling under invertible local filters. Every complete classical record of local operations and classical communication has a product filter $A\otimes B$; singular branches are separable. Thus every retained branch has concurrence at most that of the input. Coarse-graining successful records preserves the bound by convexity. Since concurrence of a two-qubit Werner state increases strictly with $F$, no successful single-copy protocol can output a Werner state with larger $F$. Several input copies permit collective distillation and are outside this statement.

For any normalized pure two-qubit vector, its squared overlap with a maximally entangled state is at most $(1+C)/2$: in a Schmidt decomposition the overlap is bounded by $(\sqrt{\lambda }+\sqrt{1-\lambda })/\sqrt{2}$. Averaging gives $C(W_F)\ge 2F-1$. Equality follows from the six-state ensemble $\sqrt{F}|{\Psi }_{-}⟩\pm \sqrt{1-F}|t_j⟩$, with equal weights, where $t_1={\Psi }_{+}$, $t_2={\Phi }_{-}$ and $t_3=i{\Phi }_{+}$. The coherences cancel and each coefficient determinant gives pure concurrence $2F-1$.

Average the four correlated local Pauli operations to remove coherences in the Bell states basis. A one-party Pauli operation can permute the largest of the four resulting weights into the singlet sector; that weight is at least $1/4$. Correlated common local unitaries then permute the three triplet projectors, while collective-unitary covariance of the two-qubit singlet keeps the singlet projector unchanged. Averaging all six permutations makes the triplet weights equal, producing a two-qubit Werner state. Shared randomness or local operations and classical communication coordinates these local operations. Common unitaries alone cannot increase an initially small singlet weight, because that weight is invariant under them.

The density operators on a finite-dimensional Hilbert space form a convex set. Its extreme points are exactly the rank-one projectors, hence exactly the pure states. To prove this, suppose $|\psi ⟩⟨\psi |={\sum }_i{a}_i{\rho }_i$ with $a_i>0$ and ${\rho }_i$ positive semidefinite. For $w\perp \psi$, $0={\sum }_i{a}_i⟨w|{\rho }_i|w⟩$ is a sum of nonnegative numbers, so every term vanishes. Diagonalizing ${\rho }_i$ shows that $⟨w|{\rho }_i|w⟩=0$ implies ${\rho }_{i}w=0$. Thus each ${\rho }_i$ is supported on the line spanned by $\psi$, and its unit trace forces ${\rho }_i=|\psi ⟩⟨\psi |$.

Conversely, a density operator of rank at least two has a spectral decomposition with at least two positive eigenvalues. If $\lambda$ is one of them, $0<\lambda <1$ and $\rho =\lambda |e⟩⟨e|+(1-\lambda )\sigma$, where $\sigma$ is the normalized remainder. These are distinct density operators, proving that $\rho$ is not an extreme point.

The unitary orbit consists exactly of density operators with the same spectrum. If distinct eigenvalues have multiplicities $n_j$, its stabilizer is ${\prod }_{j}U(n_j)$ and its real dimension is $N^2-{\sum }_j{n}_j^2$. A reachable compact group with Lie algebra $\mathfrak{g}$ is transitive on this orbit exactly when $\dim \mathfrak{g}-\dim (\mathfrak{g}\cap \{X:[X,\rho ]=0\})$ equals the orbit dimension. Global phases act trivially by conjugation.

A quantum state ensemble specifies a random preparation: normalized pure states $|{\psi }_i⟩$ are selected with probabilities $p_i\ge 0$, ${\sum }_i{p}_i=1$. Without an accessible preparation label, its measurement statistics are those of the density operator $\rho ={\sum }_i{p}_i|{\psi }_i⟩⟨{\psi }_i|$. Indeed every POVM effect $E$ has probability ${\sum }_i{p}_i⟨{\psi }_i|E|{\psi }_i⟩=\mathrm{Tr}(E\rho )$. Different ensembles can therefore have identical operational statistics: equal mixtures of computational-basis qubits or Hadamard-basis qubits both have density operator $I/2$. The ensemble decomposition contains information about the preparation, not additional information accessible from the system alone.

A convex roof extension extends a function $\mu$ on pure states to arbitrary density operators by minimizing its ensemble average over all pure state decompositions. It agrees with $\mu$ on pure states: every positive-weight state in a decomposition of a rank-one density operator has the same one-dimensional support. Concatenating nearly minimizing ensembles proves that $F$ is a convex function. Moreover, every convex function that is no larger than $\mu$ on pure states is no larger than $F$, by applying convexity to each decomposition. Infima need not be attained; an arbitrarily small additive error suffices when $F$ is finite.

If a convex roof extension is non-increasing on average for every pure input to a quantum instrument, then it is non-increasing on average for mixed inputs too. For an input ensemble $\{p_j,{\pi }_j\}$, let $q_{k|j}$ and ${\sigma }_{jk}$ be its conditional outcome probabilities and states. Linearity of each unnormalized outcome map gives $P_k{\rho }_k={\sum }_j{p}_j{q}_{k|j}{\sigma }_{jk}$. Convexity gives ${\sum }_k{P}_{k}F({\rho }_k)\le {\sum }_{j,k}{p}_j{q}_{k|j}F({\sigma }_{jk})\le {\sum }_j{p}_j\mu ({\pi }_j)$. Taking the infimum over input ensembles proves the result, even when conditional outputs of pure inputs are mixed, provided the averages are well defined.

For a two-qubit density operator, the real three-by-three matrix $T$ encodes the spin-product expectation values: unit measurement axes $\mathbf{a},\mathbf{b}$ have $E(\mathbf{a},\mathbf{b})={\mathbf{a}}^{T}T\mathbf{b}$. The tensor does not include local Bloch-vector data and alone need not specify the complete density operator. The CHSH optimum from the correlation tensor depends on the two largest eigenvalues of $T^{T}T$.

The maximally mixed density operator on an $N$-dimensional Hilbert space assigns equal weights to every vector of any orthonormal basis. It is invariant under every unitary operator. Its purity is $\mathrm{Tr}({\rho }_{\ast }^2)=1/N$, the minimum possible because the nonnegative eigenvalues sum to one and obey the Cauchy-Schwarz inequality. For $N>1$ it is a mixed state; for $N=1$ it is the sole pure state. A rank-$k$ projector has probability $k/N$ in this state.

Let $P$ be a rank-$k$ projector with $0<k<N$. No unitary operator can map every pure state whose $P$ probability equals $k/N$ to a state with the same fixed smaller probability. Average $\sqrt{k/N}|g⟩\pm \sqrt{1-k/N}|b⟩$ over good and bad basis vectors and both signs. The average density operator is the maximally mixed state, so every unitary preserves its average $P$ probability. This contradicts a strictly smaller probability for all states in the ensemble. A preparation tailored to one known starting state is a different resource and is used in exact amplitude amplification.

A mixed state is a density operator that is not a rank-one orthogonal projection. In finite dimension it is characterized by $\mathrm{Tr}({\rho }^2)<1$, so its purity of a density operator distinguishes it from a pure state.

Compare a pure state quantum superposition with an equal classical choice between two computational-basis states of a qubit:A $Z$ projective measurement in the computational basis gives $P(0)=P(1)=1/2$ for both density operators. An $X$ projective measurement in the Hadamard basis $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$ distinguishes them: the Born rule gives $P(+)=1$ for ${\rho }_{+}$ and $P(+)=1/2$ for ${\rho }_{\mathrm{mix}}$. These comparisons use separate copies of each preparation; measuring $Z$ first changes the state tested by a subsequent $X$ measurement.

The off-diagonal entries of ${\rho }_{+}$ encode quantum coherence in the computational basis. Such quantum coherence depends on the basis: ${\rho }_{+}$ is diagonal in the Hadamard basis but remains a pure state. Mixedness is independent of the basis, because a unitary change of basis preserves the eigenvalues and the purity of a density operator.

Even the quantum state ensemble is not unique:The off-diagonal entries of the two Hadamard basis projectors cancel. Thus the density operator retains precisely what measurements on the system can reveal, while an accessible preparation label carries additional information. A mixed state can also be the reduced density matrix of an entangled state, even when the whole system is in a pure state; classical ignorance about a selected component is not its only physical origin. ^[ref]

The density operator of a closed quantum system obeys

A purification of a density operator ${\rho }_A$ is a pure state $|\Psi {⟩}_{AE}$ on a larger system such that ${\mathrm{Tr}}_E|\Psi ⟩⟨\Psi |={\rho }_A$. A spectral decomposition ${\rho }_A={\sum }_j{p}_j|j⟩⟨j|$ gives the canonical construction $|\Psi ⟩={\sum }_j\sqrt{p_j}|j{⟩}_A|j{⟩}_E$.

If subsystem $S$ has pure state $|u⟩⟨u|$ in a normalized joint pure vector $|\Omega {⟩}_{SE}$, then $|\Omega {⟩}_{SE}=|u{⟩}_S\otimes |e{⟩}_E$. Indeed, for an orthonormal basis $\{v_j\}$ of $u^{\perp }$, each squared norm $\Vert (⟨v_j|\otimes I)|\Omega ⟩{\Vert }^2=⟨v_j|{\rho }_S|v_j⟩$ is zero. Therefore the joint vector has no component outside $\mathbb{C}u\otimes {\mathcal{H}}_E$. In particular, if Alice and Bob share a known pure entangled state, an adversary holding a purifying system has no correlations with it. This does not make every pure bipartite state useful for key distribution: a pure product state has no shared random correlations.

For $\rho ={\sum }_x{p}_x{\rho }_x$, choose component purifications $|u_x{⟩}_{AR}$ and an orthonormal flag register. The vector $|U⟩={\sum }_x\sqrt{p_x}|u_x⟩|x⟩$ is normalized and reduces to $\rho$ after tracing out $R$ and the flag. Orthogonal labels remove the cross terms. This construction turns mixture identities into pure-state overlap arguments and proves joint concavity of quantum fidelity.

Any two purifications of the same density operator on sufficiently large reference spaces differ by an isometry on the reference system, and by a unitary when the reference spaces have the same dimension.

The Hughston--Jozsa--Wootters theorem says that two pure-state ensembles represent the same density operator exactly when their subnormalized state vectors are related by an isometry, or by a unitary after padding the shorter ensemble with zero vectors.

For the nonzero spectral decomposition $\rho ={\sum }_{a=1}^r{\lambda }_a|e_a⟩⟨e_a|$, every pure-state quantum state ensemble of $\rho$ is specified by the displayed column linear isometry. To prove this, write $w_j=\sqrt{p_j}{\varphi }_j$. Positivity forces each $w_j$ into the support of $\rho$, and the coefficients $U_{ja}=⟨e_a|w_j⟩/\sqrt{{\lambda }_a}$ have orthonormal columns because ${\sum }_j|w_j⟩⟨w_j|=\rho$. Conversely this column identity reconstructs $\rho$; the squared vector norms are the probabilities and sum to one. Complete the columns to unitary bases after adding zero ensemble vectors to obtain the usual unitary-mixing formulation.

Uhlmann's theorem states that the fidelity of two density operators is the maximum absolute overlap between their purifications on a common reference space.

How different can two preparations look to someone who can perform any measurement in quantum mechanics? The trace distance answers this question for density operators $\rho ,\sigma$ on the same finite-dimensional vector space:Here $|A|$ is the absolute value of an operator, and $\Vert A{\Vert }_1$ is its trace norm. Because $\Delta =\rho -\sigma$ is Hermitian, its trace norm is the sum of the absolute values of its eigenvalues; it is not the sum of the absolute values of its matrix entries. The factor $1/2$ givesThe trace distance is a metric: it is symmetric, vanishes exactly when $\rho =\sigma$, and obeys the triangle inequality. Identical preparations have trace distance zero; preparations with orthogonal occupied eigenspaces have trace distance one and can be identified perfectly in a single measurement in quantum mechanics.

Operationally, the trace distance is the largest total variation distance between the outcome probability distributions obtainable by a POVM. Equivalently, the variational characterization of trace distance says it is the largest possible probability gap for one event in a binary measurement in quantum mechanics. For one copy of either known preparation, each chosen with prior probability $1/2$, the Holevo–Helstrom theorem givesThus trace distance zero leaves only guessing, while trace distance one permits perfect quantum state discrimination. The optimum ranges over measurements; different measurement bases can expose different amounts of distinguishability. ^[ref]

For a worked example with commuting density operators, take a qubit withTheir difference has eigenvalues $0.5,-0.5$, so $D=(|0.5|+|-0.5|)/2=0.5$. Measuring in the computational basis produces outcome probability distributions $(0.8,0.2)$ and $(0.3,0.7)$, whose total variation distance is also $0.5$. Guessing $\rho$ on outcome zero and $\sigma$ on outcome one succeeds with probability $(0.8+0.7)/2=0.75$. A projective measurement in the Hadamard basis instead produces $(0.5,0.5)$ for both preparations and gives no discrimination advantage.

The trace-distance contraction under quantum channels ensures that a common quantum channel, including the partial trace, cannot increase the trace distance. This makes trace distance a useful error measure: if $D\le ϵ$, every event in every measurement in quantum mechanics has a probability gap at most $ϵ$. Losing access to a subsystem can hide distinguishability, but cannot create more of it.

For qubit density operators, the trace distance has a particularly simple geometry. Write their Bloch vectors as $\mathbf{r},\mathbf{s}$ in the Bloch ball:where $I$ is the identity operator and $\mathbf{\sigma }=({\sigma }_x,{\sigma }_y,{\sigma }_z)$ denotes the Pauli matrices. This Bloch vector representation includes both pure states on the Bloch sphere and mixed states in the interior of the Bloch ball. ^[ref]

The Pauli matrix multiplication law gives $[(\mathbf{r}-\mathbf{s})\cdot \mathbf{\sigma }{]}^2=|\mathbf{r}-\mathbf{s}{|}^{2}I$. Hence the two eigenvalues of $\Delta$ are $\pm |\mathbf{r}-\mathbf{s}|/2$, and their absolute sum givesThe trace distance is half the straight-line Euclidean distance between the Bloch vectors, not the angular or surface distance along the Bloch sphere. Opposite Bloch vectors have trace distance one; the centre $I/2$ is at trace distance $1/2$ from every pure state.

For the pure states $|0⟩$ and $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$, the Bloch vectors are $\mathbf{r}=(0,0,1)$ and $\mathbf{s}=(1,0,0)$. Their straight-line separation is $\sqrt{2}$, soThese Bloch vectors are perpendicular, but the pure states are not orthogonal: $|⟨0|+⟩|=1/\sqrt{2}$. The Helstrom measurement for the zero and plus states measures along the difference direction $\mathbf{n}=(\mathbf{r}-\mathbf{s})/\sqrt{2}=(-1,0,1)/\sqrt{2}$, rather than along either input Bloch vector.

By the variational characterization of trace distance, maximize $\mathrm{Tr}[P\mathcal{E}(\rho -\sigma )]$ over $0\le P\le I$. The adjoint of a quantum channel is positive and unital, so $0\le {\mathcal{E}}^{\ast }(P)\le I$. Moving the channel onto $P$ therefore restricts, rather than enlarges, the effects available in the input variational problem. This proves the displayed bound for every pair of density operators.

For the pure qubit $\alpha |0⟩+\beta |1⟩$, subtracting the output of dephasing channel leaves a Hermitian matrix with zero diagonal and off-diagonal modulus $|\alpha \beta ||1-c|$. Its two eigenvalues are that number and its negative, giving the displayed trace distance. For real $c=1-2p$, this is $2p|\alpha \beta |$.

For two normalized pure states, the difference of their projectors has two eigenvalues $\pm \sqrt{1-|⟨u|v⟩{|}^2}$, giving that trace distance. Write a mixed density operator as $\rho ={\sum }_j{q}_j|u_j⟩⟨u_j|$. Convexity of the trace norm and weighted Cauchy-Schwarz inequality give $D(\rho ,|v⟩⟨v|)\le {\sum }_j{q}_j\sqrt{1-|⟨u_j|v⟩{|}^2}\le \sqrt{1-⟨v|\rho |v⟩}$. Thus a small probability of failing the pure-state projector test forces trace-distance proximity, including for correlated multipartite outputs.

For the Hermitian difference $X=\rho -\sigma$, define its positive part of a Hermitian operator $Q=X_{+}$ and negative part of a Hermitian operator $R=X_{-}$. Their supports are orthogonal and $|X|=Q+R$. Equal traces give $\mathrm{Tr}Q=\mathrm{Tr}R$, so the trace distance is both $\frac{1}{2}(\mathrm{Tr}Q+\mathrm{Tr}R)$ and $\mathrm{Tr}Q$. This explains why the positive spectral projector attains the variational characterization of trace distance.

For a density operator $\rho$ and pure state $P=|\psi ⟩⟨\psi |$, the diagonal absolute-sum bound for the trace norm in a basis containing $|\psi ⟩$ gives $D(\rho ,P)\ge 1-⟨\psi |\rho |\psi ⟩=1-F(\rho ,P{)}^2$. The quantum fidelity is unsquared. Block-diagonal states with respect to the target and its complement attain equality.

The variational characterization of trace distance turns a matrix distance into the best possible binary experiment. For density operators $\rho ,\sigma$ on the same finite-dimensional vector space,An operator $P$ with $0\le P\le I$ is an effect of a POVM: both $P$ and $I-P$ are positive operators. The Born rule assigns the first outcome probabilities $\mathrm{Tr}(P\rho )$ and $\mathrm{Tr}(P\sigma )$. Their signed gap is $\mathrm{Tr}(P\Delta )$, where $\Delta =\rho -\sigma$. Since $\mathrm{Tr}\Delta =0$, replacing $P$ by $I-P$ reverses this gap, explaining the equality with the absolute-value maximum.

One optimal $P$ is the spectral projector $P_{+}$ onto the positive eigenspaces of $\Delta$. In the spectral decomposition of $\Delta$, this choice includes all positive eigenvalues and excludes all negative eigenvalues. Their positive sum equals the trace distance by the spectral-parts formula for trace distance. Any zero eigenspace can be assigned to either outcome without changing the gap. This is the effect selected by the equal-prior Holevo–Helstrom theorem. ^[ref]

The same optimum has a classical description. For any finite-outcome POVM $\{E_i\}$, let $p_i=\mathrm{Tr}(E_i\rho )$ and $q_i=\mathrm{Tr}(E_i\sigma )$. ThenThe expression $\frac{1}{2}{\sum }_i|p_i-q_i|$ is the total variation distance of the entire outcome probability distributions; the expression for $A$ is the gap for one event. Grouping outcomes in $A$ gives the single effect $P_A={\sum }_{i\in A}{E}_i$. Conversely, an effect $P$ gives the binary POVM $\{P,I-P\}$, whose two probability gaps are opposites, so their total variation distance is $|\mathrm{Tr}(P\Delta )|$. Thus the optimum over all POVMs is already attained by one binary projective measurement. ^[ref]

For quantum state discrimination with equal prior probabilities, declare $\rho$ on outcome $P$ and $\sigma$ on outcome $I-P$. Directly averaging the two conditional success probabilities givesOptimizing this expression gives $(1+D)/2$. For the diagonal example in trace distance, $P=|0⟩⟨0|$ gives a gap $0.8-0.3=0.5$. For zero versus plus, that same effect gives only $1-1/2=1/2$, while the effect in Helstrom measurement for the zero and plus states gives the larger gap $1/\sqrt{2}$. The trace-distance-preserving binary measurement depends on the pair of preparations; a fixed measurement basis does not generally attain the trace distance for every pair.

For two density operators $\rho ,\sigma$, let $P_{+}$ project onto the positive spectral subspace of $\rho -\sigma$. The binary POVM $\{P_{+},I-P_{+}\}$ has output probability difference $(D,-D)$, where $D=D(\rho ,\sigma )$. Its measurement channel therefore preserves this pair's trace distance exactly. The optimizing quantum measurement depends on the pair; a single fixed measurement need not preserve every pair's distance.

The Von Neumann entropy of a density operator isIt vanishes for a pure state and measures mixedness and quantum entanglement in reduced states.

For density operators in dimension $d$, let $\delta =\frac{1}{2}\Vert \rho -\sigma {\Vert }_1$. The positive and negative parts of their difference have the same trace $\delta$. Normalize them as ${\tau }_{+},{\tau }_{-}$ and write $(\rho +\delta {\tau }_{-})/(1+\delta )=(\sigma +\delta {\tau }_{+})/(1+\delta )$. Applying the two entropy bounds for a quantum mixture to this common state proves the displayed bound. The function $g(\delta )=(1+\delta )h_2(\delta /(1+\delta ))$ is increasing and concave, with $g(0)=0$, which also controls average entropy errors in pure-source compression.

For density operators supported on mutually orthogonal subspaces, the eigenvalues of their mixture are $p_i{\lambda }_{ij}$, where ${\lambda }_{ij}$ are the eigenvalues of ${\rho }_i$. Substitution into the definition of Von Neumann entropy gives the displayed decomposition into Shannon entropy of the label and the mean entropy within each block. Zero eigenvalues use $0\log 0=0$.

A mixture of two orthogonal pure qubit states has Von Neumann entropy equal to the binary entropy of its weights. Its entropy difference from the first pure state is exactly $h_2(u)$ and its trace distance from that state is $u$. The elementary bound $-(1-u)\ln (1-u)\le u$ gives $h_2(u)\le u{\log }_2(e/u)$; entropy also never exceeds one bit. For small $u$, the leading term is $u{\log }_2(1/u)$.

If a density operator $\sigma$ on an $m\ge 2$ dimensional Hilbert space has overlap $t=⟨\psi |\sigma |\psi ⟩$ with a fixed pure state, then $S(\sigma )\le h(t)+(1-t){\log }_2(m-1)$. Dephase in a basis containing $|\psi ⟩$ and apply the entropy bound with one prescribed probability. The abstract state bound is attained by $\sigma =t|\psi ⟩⟨\psi |+(1-t)(I-|\psi ⟩⟨\psi |)/(m-1)$.

On a $d$-dimensional Hilbert space, the Von Neumann entropy is maximized uniquely by $I/d$. Indeed, nonnegativity of quantum relative entropy giveswith equality exactly when $\rho =I/d$.

If $\sigma ={\sum }_i{P}_i\rho P_i$ is a nonselective projective measurement, then $S(\sigma )\ge S(\rho )$. Equality holds exactly when $\rho$ is already block diagonal in the measurement decomposition, equivalently $[\rho ,P_i]=0$ for every $i$.

The pinching map is self-adjoint for the trace pairing, and $\log \mathcal{P}(\rho )$ is block diagonal. Thus $\mathrm{Tr}\rho \log \mathcal{P}(\rho )=\mathrm{Tr}\mathcal{P}(\rho )\log \mathcal{P}(\rho )$. The kernel of $\mathcal{P}(\rho )$ is contained in that of $\rho$, so the relative entropy is finite on its support. Expanding the quantum relative entropy gives the displayed identity. Its nonnegativity proves entropy increase, with equality exactly when the original state is already block diagonal.

For a bipartite pure state, the entanglement entropy is the Von Neumann entropy of either reduced density matrix. The two reduced states have the same nonzero eigenvalues, so their entropies agree.

For a bipartite pure state, entanglement entropy is the Von Neumann entropy of either reduced density matrix. When Alice makes a local measurement with pure conditional branches, Bob's reductions average to his original reduction. Concavity of Von Neumann entropy therefore gives ${\sum }_r{p}_{r}S({\rho }_{B,r})\le S({\rho }_B)$. When Bob measures, apply concavity to Alice's reductions. In each pure branch the two reduced entropies coincide by the Schmidt decomposition. Refining measurement outcomes to individual Kraus operators and iterating proves the inequality for arbitrary finite-round LOCC. Discarded local systems can be measured and their outcomes refined as well, so discarding does not evade the bound.

Every finite-dimensional bipartite pure state has a decomposition $|\psi ⟩={\sum }_{j=1}^r{s}_j|a_j⟩|b_j⟩$ with positive Schmidt coefficients $s_j$ and orthonormal families on each subsystem.

The operator Schmidt rank of a bipartite linear operator is its shortest expansion as a sum of product linear operators. Expanding in local orthonormal operator bases gives a coefficient matrix; its matrix rank is the operator Schmidt rank, by a singular value decomposition. It is unchanged by invertible local operator-basis changes. A rank-one projector onto a bipartite pure state of Schmidt rank $d$ has operator Schmidt rank $d^2$: its expansion contains all $d^2$ independent products of local matrix units.

If two parties share a pure resource of Schmidt rank $R$, a fully refined branch of their local operations has Kraus operator $K_{\mu \nu }={\sum }_{j=1}^R\sqrt{p_j}{A}_{\mu j}\otimes B_{\nu j}$. Hence its operator Schmidt rank is at most $R$. Local ancillas, locally adaptive readouts and refined discarded environments are included in the branch operators. Shared classical randomness only mixes such branches. Classical comparison of records does not increase their operator Schmidt rank.

For the pure state ${\lambda }_0|00⟩+{\lambda }_1|11⟩$ with real normalized Schmidt coefficients, the two-body Pauli correlators form $T=\mathrm{diag}(s,-s,1)$, $s=2{\lambda }_0{\lambda }_1$. Thus arbitrary local axes give $⟨(\mathbf{a}\cdot \sigma )\otimes (\mathbf{b}\cdot \sigma )⟩=s(a_x{b}_x-a_y{b}_y)+a_z{b}_z$. The negative $yy$ component follows from the phases in the Pauli operators.

The Schmidt coefficients of a normalized bipartite pure state are the nonnegative numbers $s_j$ in its Schmidt decomposition; they satisfy ${\sum }_j{s}_j^2=1$ and their squares are the nonzero eigenvalues of either reduced density matrix.

The Schmidt rank is the number of nonzero terms in a Schmidt decomposition, equivalently the rank of either reduced density matrix of a bipartite pure state.

The coefficient matrix of a bipartite pure state has rank equal to its Schmidt rank, by singular value decomposition. A local product operation $A\otimes B$ sends it to $AM{B}^{\mathsf{T}}$, whose rank cannot increase. A complete classical transcript of an LOCC protocol still selects one product Kraus operator, so every nonzero branch obeys the bound. Discarding the transcript gives a mixed output with Schmidt number at most the initial rank.

The Schmidt rank of a bipartite pure state cannot increase under local operations, even in a nonzero selected branch. A coefficient matrix $C$ changes under a local product operator to $AC{B}^T$, whose rank is at most that of $C$. For deterministic exact conversions the same conclusion follows from Nielsen's pure-state conversion theorem: at the input rank $r$, the majorization inequality forces all output coefficients after $r$ to vanish.

A normalized bipartite pure state of Schmidt rank $r$ has quantum fidelity at most $\sqrt{r/d}$ with a fixed maximally entangled state on $d\otimes d$. Its coefficient matrix has matrix rank $r$ and Hilbert-Schmidt norm one. Its overlap is the trace of that matrix divided by $\sqrt{d}$; trace duality and the Cauchy-Schwarz inequality bound this by $\sqrt{r/d}$. Uniform Schmidt coefficients on $r$ matching basis pairs attain equality.

The Schmidt number of a mixed bipartite state is the smallest $k$ for which it has a pure-state ensemble whose members all have Schmidt rank at most $k$. Schmidt number one is equivalent to separability.

For a bipartite density operator ${\rho }_{AB}$,Equality holds exactly when ${\rho }_{AB}={\rho }_A\otimes {\rho }_B$.

For a bipartite density operator ${\rho }_{AB}$,

For a tripartite density operator ${\rho }_{ABC}$,Equivalently, conditioning on an additional quantum system cannot increase quantum conditional entropy.

Apply data-processing inequality for quantum relative entropy to ${\rho }_{ABC}$ and ${\rho }_A\otimes {\rho }_{BC}$, tracing out $C$. The resulting comparison is $D({\rho }_{ABC}\Vert {\rho }_A\otimes {\rho }_{BC})\ge D({\rho }_{AB}\Vert {\rho }_A\otimes {\rho }_B)$. Expand the logarithms of the product marginals to obtain $S(AB)+S(BC)-S(B)-S(ABC)\ge 0$, exactly Strong subadditivity of Von Neumann entropy.

Every tripartite density operator satisfies $S(AC)+S(BC)\ge S(A)+S(B)$. Purify to $ABCD$ and apply Strong subadditivity of Von Neumann entropy to $ACD$, conditioned on $A$. The Schmidt decomposition gives $S(AD)=S(BC)$ and $S(ACD)=S(B)$, proving the inequality. The original tripartite state need not be pure.

For density operators ${\rho }_1,{\rho }_2$ and $0\le p\le 1$, Von Neumann entropy is concave:

For $0<b<1$ and distinct density operators, the Von Neumann entropy of $\rho =b{\rho }_1+(1-b){\rho }_2$ is strictly larger than the corresponding average. The difference is $bD({\rho }_1\Vert \rho )+(1-b)D({\rho }_2\Vert \rho )$, positive by nonnegativity of quantum relative entropy unless both states coincide. On the closed interval $0\le b\le 1$, equality holds exactly when $b=0$, $b=1$, or ${\rho }_1={\rho }_2$.

For a mixture of density operators, the lower bound follows from $S(\bar{\rho })-{\sum }_i{p}_{i}S({\rho }_i)={\sum }_i{p}_{i}D({\rho }_i\Vert \bar{\rho })\ge 0$, with equality precisely when all positive-weight states coincide. For the upper bound, resolve each state into pure eigenvectors and form the Gram matrix of the resulting weighted vectors. Its nonzero eigenvalues are those of $\bar{\rho }$, while its diagonal has probabilities $p_i{\lambda }_{ia}$. Averaging diagonal unitary conjugations deletes its off-diagonal entries; Concavity of Von Neumann entropy therefore gives the bound. Equality holds exactly when positive-weight component supports are mutually orthogonal.

The entropy of a binary mixture also satisfieswhere $H$ is binary entropy. This follows by applying the operator monotonicity of logarithm to $p{\rho }_1\le p{\rho }_1+(1-p){\rho }_2$ and its counterpart for ${\rho }_2$.