Von Neumann entropy ($S(\rho )$)

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 unit-trace two-dimensional reduced state of a bipartite pure state, determinant $D$ gives eigenvalues $(1\pm \sqrt{1-4D})/2$. Their binary entropy quantifies entanglement. Natural logarithms give a maximum $\ln 2$; base-two logarithms give one bit.

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 nonzero eigenvalues, including multiplicities, of either reduced density operator of a bipartite pure state. A Schmidt decomposition $|\psi ⟩={\sum }_k\sqrt{p_k}|a_k{b}_k⟩$ gives ${\rho }_A={\sum }_k{p}_k|a_k⟩⟨a_k|$ and the corresponding expression for ${\rho }_B$ by partial trace. Thus unequal marginal spectra rule out a pure joint state. The amplitudes $\sqrt{p_k}$ are often called Schmidt coefficients; the spectrum here uses the probabilities $p_k$.

An orthonormal subsystem basis appearing in a Schmidt decomposition. Its vectors diagonalize the corresponding reduced density matrix. Degenerate Schmidt coefficients permit a freedom to rotate the paired bases together; zero-weight basis vectors complete the subsystem space without contributing to the state.

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 ${\rho }_2$.

The quantum conditional entropy is $H(A|B{)}_{\rho }=S({\rho }_{AB})-S({\rho }_B)$. Unlike its classical counterpart, it may be negative.

For a bipartite pure state with finite Von Neumann entropy, the joint Von Neumann entropy is zero, so its quantum conditional entropy is $-S({\rho }_A)$. The Schmidt decomposition makes the nonzero eigenvalues of ${\rho }_A$ the squared Schmidt coefficients. Their Shannon entropy vanishes exactly when there is one coefficient, equivalently the joint pure state is a product state. Thus negative quantum conditional entropy detects exactly the entangled bipartite pure states; it is only a sufficient criterion for entanglement of general mixed density operators.

For bipartite density operators,This is a consequence of Strong subadditivity of Von Neumann entropy.

If $\epsilon =\Vert {\rho }_{AB}-{\sigma }_{AB}{\Vert }_1/2$ and $d_A=\dim {\mathcal{H}}_A$, thenThe proof couples the two states through their positive and negative differences, then combines concavity of quantum conditional entropy with the entropy bound for a binary mixture.

For every state on ${\mathcal{H}}_A\otimes {\mathcal{H}}_B$,The upper bound follows from Subadditivity of Von Neumann entropy and the lower bound from the Araki–Lieb inequality.

The quantum relative entropy is $D(\rho \Vert \sigma )=\mathrm{Tr}[\rho (\log \rho -\log \sigma )]$ when the support of $\rho$ lies in that of $\sigma$.

For common classical weights, the matrix logarithm on each positive-weight block is $\log (p_i{\rho }_i)=(\log p_i)I+\log {\rho }_i$ on its support. The two logarithms of $p_i$ cancel in the quantum relative entropy, giving the displayed identity. Zero-weight blocks are omitted, including if their unweighted relative entropy is infinite. If a positive-weight block violates support inclusion, both sides are infinite. Applying the data-processing inequality for quantum relative entropy to the partial trace over the flag gives joint convexity of quantum relative entropy.

For density operators, Klein's inequality gives $D(\rho \Vert \sigma )\ge 0$, with equality exactly when $\rho =\sigma$. If the support of a positive operator $\rho$ is not contained in that of $\sigma$, the quantum relative entropy is $+\infty$.

Diagonalize two density operators with eigenvalues $r_i,s_j$, and put $q_{ij}=|⟨i|j⟩{|}^2$. This overlap matrix has row and column sums one. The scalar inequality $u\ln (u/v)\ge u-v$ yields $D(\rho \Vert \sigma )\ge {\sum }_{ij}{q}_{ij}(r_i-s_j)=0$ in natural units; division by $\ln 2$ gives bits. Equality forces $r_i=s_j$ whenever $q_{ij}>0$, so $\sigma |i⟩=r_i|i⟩$ and $\rho =\sigma$. Zero eigenvalues use limits, and failure of support inclusion gives infinite quantum relative entropy.

In bits, the quantum Pinsker inequality isThe trace-distance-preserving binary measurement converts the trace distance into classical total variation distance without loss. Apply the data-processing inequality for quantum relative entropy to that measurement, then classical Pinsker's inequality. The factor $\ln 2$ is omitted when relative entropy uses natural logarithms.

For an ensemble of density operators with average $\overline{\omega }={\sum }_j{p}_j{\omega }_j$,Expanding the trace definitions cancels the $\log {\omega }_j$ terms, and the remaining sum combines to a trace against $\overline{\omega }$. Positive-weight ensemble states have their support contained in that of the average, so the first sum on the right is finite in finite dimension. If the reference-state support condition fails, both sides are $+\infty$. The identity is associated with Donald's work on relative entropy.

The ensemble average $\overline{\omega }$ uniquely minimizes ${\sum }_j{p}_{j}D({\omega }_j\Vert \rho )$ over density operators $\rho$. Donald's identity separates this objective into a constant plus $D(\overline{\omega }\Vert \rho )$, and nonnegativity of quantum relative entropy proves the claim. The minimum is the Holevo quantity $S(\overline{\omega })-{\sum }_j{p}_{j}S({\omega }_j)$, equivalently the quantum mutual information of the associated classical-quantum state.

For every quantum channel $T$,For a partial trace, this says that discarding a subsystem cannot make two density operators more distinguishable in quantum relative entropy.

The quantum relative entropy is additive on tensor products:This follows by expanding $\log ({\sigma }_A\otimes {\sigma }_B)=\log {\sigma }_A\otimes I+I\otimes \log {\sigma }_B$ and its counterpart for $\rho$.

For a bipartite density operator ${\rho }_{AB}$ and a product reference state,The difference between the two sides is the nonnegative quantum mutual information of ${\rho }_{AB}$.

Repeated application of superadditivity of quantum relative entropy gives

If $\Vert {\rho }_n^{\prime }-{\rho }^{\otimes n}{\Vert }_1\to 0$, thenIn finite dimension this follows from multipartite superadditivity of quantum relative entropy, additivity of quantum relative entropy, contraction of trace distance under partial trace, and uniform continuity on the compact state space.

The quantum mutual information of a bipartite state isIts nonnegativity is equivalent to Subadditivity of Von Neumann entropy.

For a separable quantum state ${\rho }_{AB}={\sum }_k{p}_k{\sigma }_A^k\otimes {\tau }_B^k$, concavity of quantum conditional entropy gives $S(A|B{)}_{\rho }\ge {\sum }_k{p}_{k}S({\sigma }_A^k)\ge 0$. Consequently $I(A:B)=S(A)-S(A|B)\le S(A)$; swapping the parties gives the displayed bound. In finite dimension it is at most ${\log }_2\min (d_A,d_B)$. Thus more than one bit of mutual information witnesses entanglement of two qubits. Smaller positive mutual information need not be entanglement, because separable states can be classically correlated.

For a tripartite density operator, the quantum conditional mutual information is $I(A:C|B)=S(AB)+S(BC)-S(B)-S(ABC)$. It is nonnegative by Strong subadditivity of Von Neumann entropy. It measures the coherent information lost when a channel environment is discarded after an isometric dilation.

A local quantum channel on either subsystem cannot increase quantum mutual information. Indeed, apply the data-processing inequality for quantum relative entropy to $D({\rho }_{AB}\Vert {\rho }_A\otimes {\rho }_B)$; the channel maps the product of marginals to the product of the new marginals.

Let a uniformly random bit $A$ be copied into a three-state classical register $B$ with probability $\epsilon$, and otherwise replaced in $B$ by a flagged erasure independent of $A$. Then $I(A:B)=\epsilon$ bits. Projecting $B$ onto its two nonerased states and conditioning on success gives a perfectly correlated bit pair with mutual information one. The postselection branch is completely positive and trace decreasing, but renormalization after conditioning is not a deterministic quantum channel. This does not contradict data processing for quantum mutual information, which applies to trace-preserving operations or an unconditioned instrument including its outcome flag.

For a channel on $B$, use a Stinespring dilation $B\to B^{\prime }E$. Isometry invariance gives $I(A:B)=I(A:B^{\prime }E)$, so the loss on discarding $E$ is $I(A:B)-I(A:B^{\prime })=I(A:E\mid B^{\prime })$. This quantum conditional mutual information is nonnegative by Strong subadditivity of Von Neumann entropy. The identity holds for arbitrary mixed inputs and identifies the correlations lost to the discarded environment.

For any tripartite density operator, expanding the three quantum mutual information terms in Von Neumann entropy gives $I(X:BD)=I(X:B)+I(XB:D)-I(B:D)$. This algebraic identity allows separate bounds on two information contributions, while nonnegativity of quantum relative entropy allows the last term to be dropped in an upper bound.

The quantum conditional entropy has the relative-entropy representationwhere the minimum ranges over density operators on $B$.