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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.

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.

Use logarithms to base two and set $0{\log }_{2}0=0$. For a density matrix with eigenvalues $({\lambda }_j)$, the Von Neumann entropy isThus it is the Shannon entropy of the spectrum. In dimension $d$, each term is nonnegative, and entropy vanishes exactly when one eigenvalue is one, namely for a pure state. The Jensen inequality applied to the strictly concave function $-x{\log }_{2}x$ gives $S(\rho )\le {\log }_{2}d$, with equality precisely at the maximally mixed state $I/d$. Unitary conjugation preserves eigenvalues, so $S(U\rho U^{†})=S(\rho )$. Product eigenvalues giveThe finite-dimensional entropy is continuous because the eigenvalues depend continuously on the matrix and $-x\log x$ is continuous also at zero.

Define the partial trace over ${\mathcal{K}}_2$ by choosing an orthonormal basis $(e_b)$ and settingEquivalently it is uniquely characterized by $\mathrm{Tr}({\rho }_{1}A)=\mathrm{Tr}(\rho (A\otimes I))$ for every operator $A$ on ${\mathcal{K}}_1$. This identity proves basis independence. Define ${\rho }_2$ similarly, with $\mathrm{Tr}({\rho }_{2}B)=\mathrm{Tr}(\rho (I\otimes B))$. Hermiticity is preserved by the displayed sum. For each $u\in {\mathcal{K}}_1$,Taking the trace in a product orthonormal basis gives $\mathrm{Tr}{\rho }_1=\mathrm{Tr}\rho =1$, and the same proof works for ${\rho }_2$. Thus both reductions are density matrices. This is the partial trace positivity from product vectors argument.

For a bipartite pure state $\rho =|\psi ⟩⟨\psi |$, expand $\psi$ in a product orthonormal basis with coefficient matrix $C$. A singular value decomposition gives the Schmidt decompositionwith orthonormal families on both sides. Taking the two partial traces makes all cross terms vanish:The spectra may have different numbers of zeros but have identical nonzero eigenvalues. Consequently the reduced entropies coincide:This common value is the entanglement entropy. In infinite-dimensional trace-class systems the same partial-trace and Schmidt arguments apply, and the equality remains valid with the value $+\infty$ allowed; the bound ${\log }_{2}d$ is specifically finite-dimensional.

For completeness, the basic mixing and correlation properties also admit short proofs. Define quantum relative entropy by $D(\tau \Vert \sigma )=\mathrm{Tr}\tau ({\log }_2\tau -{\log }_2\sigma )$ when supports are included, and by $+\infty$ otherwise. Here is its nonnegativity proof without merely citing an operator inequality. Diagonalize $\tau ,\sigma$ with eigenvalues $r_i,s_j$, and let $q_{ij}$ be the squared overlaps of their eigenbases. Both row and column sums of $q$ are one. The scalar inequality $r\ln (r/s)\ge r-s$, with zero cases taken by limits, givesEquality forces $r_i=s_j$ on every nonzero overlap, hence $\tau =\sigma$. For $\overline{\rho }={\sum }_i{p}_i{\rho }_i$, expanding the trace definition yieldswhich proves Concavity of Von Neumann entropy.

The complementary entropy bounds for a quantum mixture areTo prove the upper bound, write the spectral pure-state decompositions of the ${\rho }_i$ as one ensemble with weights $w_{ij}=p_i{\lambda }_{ij}$. Form the column matrix with columns $\sqrt{w_{ij}}|{\psi }_{ij}⟩$. Its two Gram operators have the same nonzero spectrum: one is $\overline{\rho }$, and the other, $G$, has diagonal $w$. Relative entropy against $\mathrm{diag}w$ gives $0\le D(G\Vert \mathrm{diag}w)=H(w)-S(G)$. Since $H(w)=H(p)+{\sum }_i{p}_{i}S({\rho }_i)$ and $S(G)=S(\overline{\rho })$, the bound follows. Zero-weight columns are omitted.

Finally the support of $\rho$ is contained in $\mathrm{supp}{\rho }_1\otimes \mathrm{supp}{\rho }_2$. Indeed, a vector in a marginal kernel gives a sum of nonnegative product-vector expectations equal to zero; positivity forces the corresponding product vectors into the kernel of $\rho$. The operator identity $\log ({\rho }_1\otimes {\rho }_2)=\log {\rho }_1\otimes I+I\otimes \log {\rho }_2$ on this support and the partial-trace identities giveThis proves Subadditivity of Von Neumann entropy, with equality exactly for a product state. Purify ${\rho }_{12}$ by adjoining a third system, using its spectral decomposition. Equality of pure-state marginal entropies gives $S({\rho }_{13})=S({\rho }_2)$ and $S({\rho }_3)=S({\rho }_{12})$. Subadditivity on systems $1,3$ then gives $S({\rho }_2)\le S({\rho }_1)+S({\rho }_{12})$; swapping $1,2$ gives the Araki–Lieb inequality $\boxed{|S({\rho }_1)-S({\rho }_2)|\le S({\rho }_{12})}$. These finite-dimensional proofs establish the usual basic entropy bounds, rather than attributing all mixed-state behavior to pure-state Schmidt decompositions.

For a density matrix $\rho$ on a $d$-dimensional Hilbert space, its Von Neumann entropy, in bits, iswhere the ${\lambda }_a$ are its eigenvalues and $0{\log }_{2}0=0$. Each summand is nonnegative, proving $S\ge 0$, with equality exactly when one eigenvalue is one and the rest zero, namely for a pure state.

We first prove the nonnegativity of quantum relative entropy needed below. Diagonalize $\rho$ with eigenvalues ${\lambda }_a$ and eigenvectors $u_a$, and $\sigma$ with eigenvalues ${\mu }_b$ and eigenvectors $v_b$. Put $B_{ab}=|⟨u_a,v_b⟩{|}^2$ and $q_a=⟨u_a|\sigma |u_a⟩={\sum }_b{B}_{ab}{\mu }_b$. Each row of $B$ sums to one. Concavity of the ordinary logarithm gives ${\sum }_b{B}_{ab}{\log }_2{\mu }_b\le {\log }_2{q}_a$. ConsequentlyThe final Gibbs inequality follows directly from ${\sum }_{{\lambda }_a>0}{\lambda }_a\log (q_a/{\lambda }_a)\le \log {\sum }_{{\lambda }_a>0}{q}_a\le 0$. Support failure gives $D=+\infty$; zero eigenvalues can otherwise be handled by restriction or regularization. Equality requires $q_a={\lambda }_a$ and equality in the logarithm inequality on every positive-weight row. Thus $u_a$ is a $\sigma$ eigenvector of eigenvalue ${\lambda }_a$ wherever ${\lambda }_a>0$. The remaining diagonal entries of $\sigma$ vanish and positivity makes the remaining rows and columns zero. Hence equality holds exactly when $\rho =\sigma$.

Taking $\sigma =I/d$ gives $D(\rho \Vert I/d)={\log }_{2}d-S(\rho )$, provingThe lower equality is attained exactly by pure states and the upper exactly by the maximally mixed state $I/d$.

For $\bar{\rho }={\sum }_i{p}_i{\rho }_i$, each positive-weight state has support contained in that of $\bar{\rho }$. Expanding the quantum relative entropy givesThis proves Concavity of Von Neumann entropy, with equality precisely when every positive-weight ${\rho }_i$ is the same state.

For the upper entropy bound for a quantum mixture, diagonalize each component, ${\rho }_i={\sum }_a{\lambda }_{ia}|v_{ia}⟩⟨v_{ia}|$, and set $q_{ia}=p_i{\lambda }_{ia}$. Form the rectangular matrix $B$ whose columns are $\sqrt{q_{ia}}{v}_{ia}$. Then $B{B}^{†}=\bar{\rho }$, while the Gram matrix $\Gamma =B^{†}B$ has diagonal $q_{ia}$. The two matrices have the same nonzero eigenvalues: $B$ and $B^{†}$ map corresponding positive-eigenvalue eigenspaces isomorphically. Thus $S(\Gamma )=S(\bar{\rho })$.

If there are $L$ columns, let $W=\mathrm{diag}(1,\omega ,\dots ,{\omega }^{L-1})$, with $\omega =e^{2\pi i/L}$. ThenAll the summands have the same Von Neumann entropy. The just-proved concavity therefore implies $S(\bar{\rho })=S(\Gamma )\le S(\mathrm{diag}\Gamma )=H(q)$. Finally,We have proved both requested bounds:For the upper equality, strictness in the concavity argument forces $\Gamma$ to be diagonal. Its entries between two positive-weight component eigenvectors then force those vectors to be orthogonal. Thus upper equality holds exactly when the supports of the positive-weight component states are mutually orthogonal. In that case their classical label is perfectly distinguishable, explaining the extra $H(p)$ bits.

Introduce an orthonormal classical-label basis $|i⟩$ and the classical-quantum stateIts reduced density operators are ${\Gamma }_X={\sum }_i{p}_i|i⟩⟨i|$ and ${\Gamma }_A=\overline{\rho }={\sum }_i{p}_i{\rho }_i$. If ${\lambda }_{ij}$ are the eigenvalues of ${\rho }_i$, the eigenvalues of ${\Gamma }_{XA}$ are $p_i{\lambda }_{ij}$. Thus the entropy of an orthogonal quantum mixture iswhere $H(p)=-{\sum }_i{p}_i{\log }_2{p}_i$ is the Shannon entropy. Apply the proved Subadditivity of Von Neumann entropy to $X$ and $A$ and cancel $H(p)$. This gives Concavity of Von Neumann entropy:Zero-probability terms may simply be omitted. The equality condition from the preceding solution implies that equality holds precisely when all the positive-probability states are the same: the block equations $p_i{\rho }_i=p_i\overline{\rho }$ follow from ${\Gamma }_{XA}={\Gamma }_X\otimes {\Gamma }_A$.

Introduce an orthonormal classical label register $X$ and the classical-quantum stateThe blocks have orthogonal supports because their label vectors are orthogonal, regardless of whether the original ${\rho }_i$ have orthogonal supports. Part (a) gives $S(XQ)=H(p)+{\sum }_i{p}_{i}S({\rho }_i)$. The marginals are ${\omega }_X={\sum }_i{p}_i|i⟩⟨i|$ and ${\omega }_Q={\sum }_i{p}_i{\rho }_i$, so $S(X)=H(p)$. Apply the Subadditivity of Von Neumann entropy and cancel $H(p)$:Thus $S({\sum }_i{p}_i{\rho }_i)\ge {\sum }_i{p}_{i}S({\rho }_i)$. This is the Concavity of Von Neumann entropy.

Use the usual asymptotic convention: the output approaches the desired $m(n)$-copy pure target in trace distance, and any recorded failure probability tends to zero. Define the binary entropyThe entanglement entropies per source and target are $E_{\psi }=h_2(s)$ and $E_{\varphi }=h_2(t)$. The asymptotic pure-state entanglement conversion rate isBoth achievability and the upper bound matter here.

For entanglement concentration, expand $n$ sources in their correlated computational basis. Every pair of strings of Hamming weight $k$ has the same amplitude $(1-s{)}^{(n-k)/2}{s}^{k/2}$. Alice measures only this weight, without resolving the individual string. Bob's string has the same weight automatically. Conditional on $k$, the normalized state is maximally entangled on $D_k=\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ correlated strings. Its outcome probability is $\left(\genfrac{}{}{0ex}{}{n}{k}\right)s^k(1-s{)}^{n-k}$, a binomial distribution. By the weak law of large numbers, $k/n\to s$ in probability; Stirling's formula then gives $n^{-1}{\log }_2{D}_k\to h_2(s)$. A uniform Schmidt vector on $D_k$ entries is majorized by a uniform vector on $2^{\lfloor {\log }_2{D}_k\rfloor }$ entries padded with zeros. Thus Nielsen's pure-state conversion theorem converts the conditional state into that many Bell pairs exactly. For any fixed $\delta >0$, one obtains at least $n(E_{\psi }-\delta )$ Bell pairs with probability tending to one.

For entanglement dilution, consider the Schmidt strings of $m$ targets with probabilities $q_x$. The quantum typical subspace retains strings satisfyingThe retained probability $Q_m$ tends to one by the weak law applied to the independent variables $-{\log }_2{q}_{x_j}$. The number of retained strings is at most $2^{m(E_{\varphi }+\delta )}$, because their probabilities sum to at most one. Normalize the target after truncation to those strings. Its squared overlap with the full target is $Q_m\to 1$. A uniform Schmidt vector of dimension $2^L$, with $L=\lceil m(E_{\varphi }+\delta )\rceil$, is majorized by the truncated target's Schmidt vector padded with zeros: the sum of its largest $k$ entries is at least $k/2^L$. Hence $L$ Bell pairs prepare the truncated target exactly by LOCC. Its trace distance from the true target is $\sqrt{1-Q_m}\to 0$. Combining concentration and dilution achieves any rate $R<E_{\psi }/E_{\varphi }$. Choosing margins tending sufficiently slowly to zero attains the ratio as a limiting rate.

For the converse, average monotonicity of pure-state entanglement entropy follows from Concavity of Von Neumann entropy: a measurement by one party leaves the other party's reduction unchanged on average, so its average conditional entropy cannot increase. Refine local measurement records, including local discarded systems, to obtain pure output branches $\{p_r,|{\chi }_r⟩\}$ on the $m$ output qubits of each party. Iterating the concavity argument givesLet $|{\Phi }_m⟩=|\varphi {⟩}^{\otimes m}$ and let the squared-overlap error of the average output ${\rho }_{\mathrm{out}}$ be $\epsilon =1-⟨{\Phi }_m|{\rho }_{\mathrm{out}}|{\Phi }_m⟩\to 0$. Define $d_r=\sqrt{1-|⟨{\Phi }_m|{\chi }_r⟩{|}^2}$, the pure-state trace distance. Concavity of the square root gives ${\sum }_r{p}_r{d}_r\le \sqrt{\epsilon }$. Partial trace cannot increase trace distance, so the reduced states are also within $d_r$. The continuity bound for quantum conditional entropy, with trivial conditioning and local dimension $2^m$, impliesConsequentlywhich excludes a limiting rate exceeding $E_{\psi }/E_{\varphi }$.

The vanishing-error convention is essential. If exact deterministic conversion were required for every finite block, majorization would also require the largest squared Schmidt coefficient to satisfy $(1-s{)}^n\le (1-t{)}^m$, henceThis can be strictly smaller than the entropy ratio: at $s=0.1,t=0.2$, the two bounds are approximately $0.4722$ and $0.6496$. Thus the boxed rate answers the standard asymptotic task, rather than imposing unmentioned exact zero-error conditions on every finite block.

Discarding the outcome of the complete projective measurement givesFor $m$ projectors, let $\omega =e^{2\pi i/m}$ and $U={\sum }_j{\omega }^j{P}_j$. Then the pinching identity isThe Concavity of Von Neumann entropy and its invariance under unitary operators implyEquality holds exactly when every conjugate in the average is the same, equivalentlyThus equality holds when the input already has no coherence between distinct measurement subspaces.

The quantity ${\chi }^{\ast }(\Lambda )$ is the one-use Holevo information of the channel: by the Holevo--Schumacher--Westmoreland theorem, its regularization gives the classical capacity, and without entangled inputs across uses it is the asymptotically achievable classical rate using collective output measurements.

If an ensemble contains a mixed ${\rho }_x={\sum }_j{q}_{j|x}|{\psi }_{xj}⟩⟨{\psi }_{xj}|$, refine its label to $(x,j)$ with probability $p_x{q}_{j|x}$. The average channel output is unchanged, while Concavity of Von Neumann entropy givesThus refinement cannot decrease the Holevo quantity, so the maximum may be restricted to pure input states.

For a density operator $\rho$ on a finite-dimensional Hilbert space, the Von Neumann entropy iswith $0\log 0=0$. Its concavity says that, for $0\le p\le 1$,