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For bipartite pure source and target states with positive target entanglement entropy, the optimal rate of LOCC conversion with vanishing output error is the ratio of their entanglement entropies. Here vanishing error means that the trace distance between the output and $|\varphi ⟩⟨\varphi {|}^{\otimes m}$ tends to zero; a recorded failure probability must also tend to zero. Entanglement concentration first provides $nS({\rho }_A^{\psi })-o(n)$ Bell pairs, and entanglement dilution needs $mS({\rho }_A^{\varphi })+o(m)$ such pairs. This achieves the ratio. Conversely, refine the output into pure branches. Average monotonicity of pure-state entanglement entropy bounds their average entropy by the input entropy; continuity bound for quantum conditional entropy, with trivial conditioning, makes the output entropy loss per target tend to zero. This forbids a larger rate. The convention permits approximations for finite blocks: exact deterministic conversion of every finite block can obey stricter majorization constraints. The reversible concentration construction is given in the original concentration paper.

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.

An auxiliary entangled pure state can enable an otherwise forbidden deterministic bipartite LOCC transformation while being returned exactly and uncorrelated with the output. Nielsen's pure-state conversion theorem reduces this to the displayed majorization condition on sorted squared Schmidt coefficients. It does not follow from a scalar comparison of entanglement entropies: tensoring changes which products occupy the largest cumulative-weight sums. An exact returned catalyst can be reused; merely restoring its marginal while allowing correlations is a different resource task.^[ref]

Entanglement dilution uses local operations and classical communication to prepare partially entangled states from maximally entangled states, such as Bell pairs. For many identical bipartite pure targets, the asymptotic cost in Bell pairs per target, allowing a vanishing approximation error, is their entanglement entropy in bits. The quantum typical subspace of the target's Schmidt coefficients has almost all its weight and dimension at most $2^{m(S+\delta )}$. A maximally entangled state of this dimension can be converted to the normalized truncated target by Nielsen's pure-state conversion theorem. The resulting state approaches the full target as the retained weight tends to one. Average monotonicity of pure-state entanglement entropy and entropy continuity give the converse lower bound. This is the reverse asymptotic task to entanglement concentration.

Taking the partial trace over either photon of the pure pair state givesThe two Schmidt coefficients are both $1/\sqrt{2}$, so the entanglement entropy is one bit for every $\phi$. A local phase gate on Alice maps $|{\Psi }^0⟩$ to $|{\Psi }^{\phi }⟩$, proving local-unitary equivalence. In strict four-vector Bell-basis terminology only $\phi =0,\pi$ give the standard $|{\Psi }^{+}⟩,|{\Psi }^{-}⟩$; every phase nevertheless gives a maximally entangled Bell-type state.

The eigenpolarizations of the requested observable areA polarizing beam splitter with its transmitted polarization axis at angle $\alpha /2$ to horizontal separates these two orthogonal states. Assign outcome $+1$ to its first detector and $-1$ to the other. Equivalently, use a fixed horizontal/vertical splitter preceded by a half-wave plate whose fast axis is at $\alpha /4$; the diagram shows this equivalent implementation. Bob uses $\beta /2$ for his splitter axis, or $\beta /4$ for the preceding half-wave plate.

The pair correlations are $⟨Z\otimes Z⟩=-1$, $⟨X\otimes X⟩=\cos \phi$, and $⟨Z\otimes X⟩=⟨X\otimes Z⟩=0$. Expanding the two observables therefore givesThe half-angle between the Bloch observable parameter and the physical linear-polarization analyser angle is essential.

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.

Fix paired orthonormal bases in the two $d$-dimensional Hilbert spaces. A maximally entangled state isIt is normalized, and the partial traces of its rank-one density operator are ${\rho }_A={\rho }_B=I/d$. Thus all its Schmidt coefficients are $1/\sqrt{d}$ and its entanglement entropy is ${\log }_{2}d$. No pure bipartite state of these dimensions can have larger entanglement entropy: nonnegativity of quantum relative entropy applied to a reduced state and $I/d$ gives $0\le D({\rho }_A\Vert I/d)={\log }_{2}d-S({\rho }_A)$. Hence it is maximally entangled.

Use base-two logarithms, so entanglement entropy is measured in bits. Write $h_2(p)=-p{\log }_{2}p-(1-p){\log }_2(1-p)$ for the binary entropy, with $0{\log }_{2}0=0$. Taking a partial trace over any two parties removes the off-diagonal terms: the two traced strings are orthogonal. Consequently all three one-party reduced density matrices of the first pure state areEach division into two groups is one party versus the other two. The Schmidt coefficients are $\sqrt{p}$ and $\sqrt{1-p}$, so the entanglement entropy across each of $A:BC$, $B:AC$ and $C:AB$ is $h_2(p)$. For the second pure state, set $p=1/2$: every one-party reduced density matrix is $I/2$, and all three entanglement entropies are one bit.

For every pair $XY$, the first pure state instead gives the reduced density matrixThis is a separable quantum state, explicitly a convex combination of two product states. Its relative entropy of entanglement is therefore zero: in $E_R(\rho )={\inf }_{\sigma \in \mathrm{Sep}}D(\rho \Vert \sigma )$, choose $\sigma =\rho$ to attain zero, while nonnegativity of quantum relative entropy excludes a negative value. The same argument with equal weights applies to the second pure state. ThusThe pairwise density matrices contain classical correlations, despite the entanglement of the global pure state; discarding a party loses the coherence between its two branches.

The entanglement entropy of $n$ independent source copies across any cut is $n{h}_2(p)$, whereas $f(n)$ copies of the target GHZ state have entanglement entropy $f(n)$. Average monotonicity of pure-state entanglement entropy under LOCC gives $P(n)f(n)\le n{h}_2(p)$ for exact successful forward branches. In a reversible asymptotic conversion the reverse direction similarly prevents a first-order deficit, since $n^{\prime }/n\to 1$. For asymptotically faithful outputs, a continuity bound for quantum conditional entropy with trivial conditioning changes these inequalities only by $o(n)$. Hence the reversible leading-order rate isThe zero pairwise relative entropies of entanglement are consistent with this conversion: no independent pairwise resource has to be supplied or removed.

There is a qualification to the literal finite-block formulation in the source. For $0<p<1$, the Schmidt rank of the exact state of $n^{\prime }$ source copies is $2^{n^{\prime }}$, whereas $f$ target GHZ states have Schmidt rank $2^f$. Every refined LOCC branch acts by a product of local linear operators and cannot increase this Schmidt rank. For $p\ne 1/2$, the displayed rate has $f<n^{\prime }$ for sufficiently large blocks when $n^{\prime }/n\to 1$, so an exact successful reverse conversion is impossible, irrespective of its claimed probability. The exact dilution obstruction from Schmidt rank is avoided in the usual asymptotic formulation by allowing outputs approaching the desired pure state. The leading-order rate above and the exact forward protocol below remain valid; neither establishes the source's stronger exact round-trip claim. At $p=1/2$ the two states are identical. At $p=0$ or $1$, the source is a known product state, the rate is zero, and local preparation supplies any desired number of source copies.

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.

Use the Schmidt decomposition of the bipartite pure state:The two reduced density matrices, obtained by partial trace, areTheir nonzero eigenvalues are identical, even if the two ambient dimensions differ. Zero eigenvalues contribute no Von Neumann entropy, soThis common Von Neumann entropy is the entanglement entropy of the bipartite pure state.

For $|J,J⟩$, tracing out subsystem 2 gives the reduced density matrixwhich is pure, so its entanglement entropy is zero.

For $|J,J-1⟩$, the two subsystem-2 factors in part (iii) are orthogonal. The partial trace therefore removes the cross terms and givesIts nonzero eigenvalues are $j_1/J$ and $j_2/J$, so the Von Neumann entropy isWhen $j_1=j_2$, both eigenvalues are $1/2$ and $S=\log 2$, the maximal entropy for a rank-two reduced state.

Cutting an open-boundary matrix product state across one virtual bond of dimension $\chi$ gives Schmidt rank at most $\chi$, so its entanglement entropy obeysA periodic interval cuts two bonds and obeys $S(A)\le 2\log \chi$. Since the boundary of a one-dimensional interval has a constant number of points, this is an area law.

For a projected entangled pair state, a bipartition crossing $n_{\partial }$ virtual bonds has Schmidt rank at most ${\chi }^{n_{\partial }}$. The tensor-network area-law bound is thereforeBecause $n_{\partial }$ is proportional to the lattice boundary area, every fixed-bond-dimension PEPS satisfies an area law.

Nielsen's pure-state conversion theorem concerns exact deterministic transformations of bipartite pure states by local operations and classical communication. Let $\lambda (\psi )$ be the decreasing vector of squared Schmidt coefficients, equivalently the eigenvalues of either party's reduced density matrix, padded with zeros when necessary. ThenThe majorization relation means ${\sum }_{j=1}^k{\lambda }_j(\psi )\le {\sum }_{j=1}^k{\lambda }_j(\chi )$ for every $k$, with equality for the total sum. The direction allows a maximally entangled state to become a product state, not the reverse. Local-unitary invariance of reduced-state spectra alone would require equality of all the weights; measurements and communicated outcomes permit this larger class of conversions. A comparison of entanglement entropies alone is not a sufficient test in higher dimension.^[ref]

For the supplied states, the weight vectors and cumulative sums areAt $k=2$, $0.8>0.75$, violating the required inequality. The desired conversion is impossible with certainty using LOCC alone. The reverse conversion also fails, already at $k=1$, so these states are incomparable under this deterministic ordering.

Taking a tensor product with the catalyst multiplies every original weight by each catalyst weight. Regrouping the two local systems on Alice's side and the two on Bob's side leaves a Schmidt decomposition across the Alice–Bob partition. After sorting, the resulting weight vectors areThe actual nonnegative Schmidt amplitudes are $\sqrt{x_j}$ and $\sqrt{y_j}$; the displayed numbers are their squares. The cumulative sums areEvery component obeys $X_k\le Y_k$, so $x\prec y$. HenceThe final catalyst is exactly its initial state and is uncorrelated with the desired output. It can therefore be returned and reused. This is entanglement catalysis; it is stronger than merely retaining the same catalyst marginal while leaving correlations with the output.^[ref]

For the maximally entangled two-level auxiliary state, the sorted input weights are $(0.2,0.2,0.2,0.2,0.05,0.05,0.05,0.05)$, while the output weights are $(0.25,0.25,0.125,0.125,0.125,0.125,0,0)$. Their first four sums are $0.8$ and $0.75$. This maximally entangled auxiliary state cannot catalyze the conversion while being returned unchanged.

Nevertheless, $(0.5,0.5)\prec (0.6,0.4)$, so it can first be converted deterministically into the nonmaximal catalyst. The desired conversion can then proceed, after which Alice and Bob can return $|\varphi ⟩$, rather than the borrowed maximally entangled state. Turning that returned pair alone back into a maximally entangled pair would increase its entanglement entropy and is forbidden by LOCC. No joint restoration procedure can both keep the exact desired $|{\psi }_2⟩$ and return the original maximally entangled auxiliary state deterministically: its composition with the preceding steps would contradict the failed four-weight inequality above.

A bipartite pure state is entangled when it is not a product state. Its Schmidt decomposition is $|\psi ⟩={\sum }_i\sqrt{{\lambda }_i}|i_A{i}_B⟩$, where ${\lambda }_i\ge 0$ and ${\sum }_i{\lambda }_i=1$. More than one nonzero coefficient means entanglement. The two reduced density matrices have the same nonzero spectrum, and their Von Neumann entropy $-{\sum }_i{\lambda }_i{\log }_2{\lambda }_i$ is the pure-state entanglement entropy. A maximally entangled pair of qubits has one bit of this entropy.

For mixed states, classical correlations are not sufficient: a separable quantum state is a convex mixture ${\sum }_j{p}_j{\rho }_A^{(j)}\otimes {\rho }_B^{(j)}$, and entanglement means that no such decomposition exists. A reduced entropy alone no longer measures mixed-state entanglement. An entanglement witness or a suitable separability criterion can detect it.

Local unitary transformations preserve the Schmidt spectrum. More general local operations and classical communication can discard entanglement but cannot freely create it from a separable input. Nielsen's pure-state conversion theorem gives the exact deterministic criterion: $|\psi ⟩\to |\varphi ⟩$ is possible precisely when ${\lambda }^{\psi }\prec {\lambda }^{\varphi }$. In decreasing order, every partial sum for the input is no larger than the corresponding output sum, with equal totals. The direction matters: a uniform spectrum can be converted to a concentrated one; a product state cannot deterministically become a Bell pair.

A probabilistic local filter can concentrate a partially entangled pair. For $|\psi ⟩=\sqrt{\lambda }|00⟩+\sqrt{1-\lambda }|11⟩$ with $1/2\le \lambda <1$, Alice's successful operator $F=\mathrm{diag}(\sqrt{(1-\lambda )/\lambda },1)$ produces equal amplitudes. The outcome is a Bell pair with probability $2(1-\lambda )$; a complementary operator supplies a complete measurement. Failure is heralded, so there is no deterministic creation of a more entangled pair. This illustrates entanglement concentration.

Many-copy entanglement distillation and entanglement dilution connect noisy or nonuniform resources with standard Bell pairs. Entanglement enables quantum teleportation and other protocols, but classical communication is still required; the no-communication theorem excludes message transmission merely by choosing a distant local measurement.

In the printed phased Bell basis, the actions on its four kets areEach matrix is purely imaginary and antisymmetric. Let $T$ have these four basis kets as columns in the computational basis. The infinitesimal generators of the standard rotation $U_A=\exp [-i\theta \mathbf{n}\cdot {\mathbf{\sigma }}_A/2]$ areUsing the displayed actions gives the phase-convention-specific identitiesThey are real antisymmetric matrices. A finite $SU(2)$ rotation on $A$ therefore acts on the magic-basis coefficient vector as $b\mapsto Ob$, with $O\in SO(4)$. ConsequentlyIt preserves concurrence and hence preserves the reduced eigenvalues and entanglement entropy. A general $U(2)$ operation adds a common phase to $b$; that changes $q$ by a phase of twice the angle but leaves $|q|$ unchanged. The real-rotation description of local transformations is a characteristic property of the magic basis.^[ref]

As an independent direct check, a local unitary sends ${\rho }_A$ to $U_A{\rho }_A{U}_A^{†}$, which is a unitary similarity transformation and cannot change its spectrum. This local-unitary invariance of entanglement entropy also explains why merely rotating one subsystem cannot create entanglement.

Fix the mode ordering with the $b^{†}$ operator before the $c^{†}$ operator, as in the given state. In the sector with one fermion in each subsystem, use the ordered basisThe state vector is $v=(1/2,1/\sqrt{2},0,1/2{)}^T$, whose squared norm is one. Thus the global density matrix is the rank-one projectorIt has trace one and satisfies ${\rho }_{BC}^2={\rho }_{BC}$, so the global state is pure.

Take the partial trace over $B$. The conditional vectors in $C$ are $v_1=(1/2,1/\sqrt{2}{)}^T$ for $B_1$ and $v_2=(0,1/2{)}^T$ for $B_2$. Therefore the reduced density matrix isIts determinant is $1/16$, giving eigenvaluesBoth are positive; in particular, $\mathrm{Tr}{\rho }_C^2=7/8<1$, so the reduced state is mixed despite global purity. The entanglement entropy is obtained by diagonalizing this reduced matrix and evaluatingFor a globally pure bipartite state this equals the entropy of either reduced subsystem. No numerical evaluation is needed.

The reduced density matrix has two eigenvalues equal to $1/2$. Its Von Neumann entropy, which is the entanglement entropy of the bipartite pure state, isThis is one bit when base-two logarithms are used. It is the maximum entanglement entropy for a pair of two-dimensional systems, while the joint Von Neumann entropy of the singlet state remains zero.

A bipartite density operator is a product state when it factors as ${\rho }_A\otimes {\rho }_B$. A pure product state factors as $|{\psi }_A⟩\otimes |{\psi }_B⟩$; its reduced density matrices have rank one and zero entanglement entropy. Mixed product states can have nonzero marginal Von Neumann entropy.

If a bipartition cuts $n_{\partial }$ tensor-network bonds of dimension at most $\chi$, its Schmidt rank is at most ${\chi }^{n_{\partial }}$ and its entanglement entropy is at most $n_{\partial }\log \chi$.