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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.

The Schmidt decomposition theorem says that a normalized vector in the tensor product of two finite-dimensional Hilbert spaces can be writtenwhere both families are orthonormal. The numbers $\sqrt{{\lambda }_j}$ are its Schmidt coefficients. The Schmidt rank $r$ and coefficients, including multiplicities, are uniquely determined by the state; the local vectors may be changed within degenerate eigenspaces and by compensating phases. Zero coefficients may be appended to make $n_1$ terms, since $n_2\ge n_1$.

For a proof, form the reduced density matrix ${\rho }_1={\mathrm{Tr}}_2|\psi ⟩⟨\psi |$. It is a Hermitian matrix that is positive semidefinite, with trace one. The spectral theorem supplies an orthonormal basis $u_1,\dots ,u_{n_1}$ of eigenvectors with nonnegative eigenvalues ${\lambda }_j$. Define the unnormalized relative vectors in the second Hilbert space byCompleteness of the first orthonormal basis gives $|\psi ⟩={\sum }_j|u_j⟩|w_j⟩$. In any orthonormal basis $e_b$ of the second factor, the definition of the partial trace givesThus a zero eigenvalue gives $w_j=0$, and for each positive eigenvalue the vector $v_j=w_j/\sqrt{{\lambda }_j}$ has unit norm. Distinct such vectors are orthogonal by the same identity. Substituting them gives the required Schmidt decomposition, with at most $n_1$ nonzero terms.

Taking the other partial trace now givesTherefore the two reduced density matrices have identical nonzero eigenvalues, namely the squared Schmidt coefficients. This also proves uniqueness of the coefficient multiset and shows that the pure state is a product state exactly when $r=1$.

Write $a=|\alpha {|}^2$, $b=|\beta {|}^2=1-a$, so $1/2<a<1$ and $0<b<1/2$. Regrouping the four copies by party giveswhere $z(x)$ counts zeroes. Alice must measure only this count, preserving coherence inside each degenerate eigenspace, rather than measuring her four bits separately. Her projective measurement has projectors ${\Pi }_k={\sum }_{z(x)=k}|x⟩⟨x|$, and the Born rule givesAlice sends $k$ to Bob. Up to the common phase of ${\alpha }^k{\beta }^{4-k}$, their conditional pure state isThis is Schmidt projection entanglement concentration: its equal Schmidt coefficients give a maximally entangled state of Schmidt rank $d_k$.

For $k=0$ or $4$, $d_k=1$, and the branch is a product state. No Bell pair can be extracted using local operations and classical communication. The respective probabilities are $b^4$ and $a^4$.

For $k=1$ or $3$, $d_k=4$. List the four surviving strings in lexicographic order as $x_0,x_1,x_2,x_3$. For $k=1$ this list is $0111,1011,1101,1110$; for $k=3$ it is $0001,0010,0100,1000$. Each party applies a local unitary operation extending the basis relabellingSuch a unitary exists because these are four distinct orthonormal basis vectors mapped to four distinct basis vectors; extend the mapping to a permutation of all sixteen. Discarding the two fixed local ancillary bits leavesThus these branches give exactly two Bell pairs, with probabilities $4a{b}^3$ and $4a^{3}b$, and no further measurement in quantum mechanics is needed.

For $k=2$, the six strings in lexicographic order are $x_0=0011$, $x_1=0101$, $x_2=0110$, $x_3=1001$, $x_4=1010$, $x_5=1100$. A suitable local POVM gives two Bell pairs with certainty, rather than losing some of these branches to a simple subspace projection. Partition these labels into the three pairs $T_0=\{0,1\}$, $T_1=\{2,3\}$, $T_2=\{4,5\}$. Alice uses the Kraus operatorsOn the six-dimensional support, each label is retained by two operators, so ${\sum }_{r=0}^2{K}_r^{†}{K}_r={\Pi }_2$. Add $K_{\perp }=I-{\Pi }_2$ to define a complete measurement in quantum mechanics on Alice's four qubits; that extra outcome has probability zero on this branch. Each nonzero outcome has conditional probability $(1/2)(4/6)=1/3$, and gives the normalized stateAlice sends $r$ to Bob. Both parties relabel the four surviving strings in increasing order to $|00⟩,|01⟩,|10⟩,|11⟩$ on two output qubits, with two fixed ancillary bits, just as above. Every $r$ therefore gives two Bell pairs. The joint probability of $(k,r)=(2,r)$ is $p_2/3=2a^2{b}^2$ for each $r$.

For an explicit local implementation of this POVM, Alice can append an ancilla in a fixed state and apply an isometry on the six-dimensional support,It preserves inner products, extends to a unitary on system and ancilla, and measuring the ancilla realizes precisely the operators $K_r$. No nonlocal operation has been used. This is a deterministic reduction of maximally entangled Schmidt rank from six to four.

The complete output distribution for this strategy is consequentlyThe mean yield is $2(1-a^4-b^4)$ pairs per four-copy block. These are the maximum possible pair counts after the stipulated count measurement: any nonzero branch has Schmidt rank at most six, whereas three independent Bell pairs require Schmidt rank eight. In any refined LOCC branch the coefficient matrix is multiplied by local matrices on the left and right, which cannot increase its rank.

The Schmidt decomposition of a normalized bipartite pure state isThe two sets of local vectors are orthonormal. The Schmidt coefficients are the nonnegative numbers $\sqrt{{\lambda }_j}$, and the Schmidt number for this pure state, also called its Schmidt rank, is $r$, the number of nonzero coefficients. It equals the rank of either reduced density operator and of the coefficient matrix in any product bases. The general mixed-state Schmidt number specializes to this same value on a pure state.

Write a bipartite pure state in fixed product bases as $|\psi ⟩={\sum }_{ij}{M}_{ij}|i⟩|j⟩$. The Schmidt decomposition theorem, equivalently the singular value decomposition of $M$, says that its Schmidt rank equals $\mathrm{rank}M$. A branch of local operations is a product operator $A\otimes B$, so its coefficient matrix isThe elementary rank inequality $\mathrm{rank}(AM{B}^{\mathsf{T}})\le \mathrm{rank}M$ proves Schmidt-rank contraction under product operators. Normalizing a nonzero branch does not change matrix rank.

For an adaptive LOCC protocol, a complete classical transcript selects one local Kraus operator at each stage. Multiplying the local operators along that transcript still gives one product $A_t\otimes B_t$. ConsequentlyClassical communication changes which product operator is chosen, not this rank bound. If the overall output is a pure state, every nonzero branch must be proportional to that same vector, so its Schmidt rank also cannot increase. If the outcome is discarded and the output is mixed, pure-state Schmidt rank is not defined; the induced decomposition instead shows that its Schmidt number is at most the original rank. This includes probabilistic filtering as well as deterministic pure-state conversion.

Let $r$ be the input Schmidt rank, so ${\lambda }_j=0$ for $j>r$ and ${\sum }_{j=1}^r{\lambda }_j=1$. If $r=d$, the output rank is already at most $r$. Otherwise, Nielsen's pure-state conversion theorem and majorization at $k=r$ giveThus all output coefficients beyond $r$ vanish, provingThis is monotonicity of Schmidt rank under LOCC. It also holds separately in any nonzero postselected branch: represent the input amplitudes by a matrix $C$; a local branch maps it to $AC{B}^T$, whose rank cannot exceed the rank of $C$. The deterministic result requested here follows already from the majorization criterion.

The Schmidt decomposition theorem states that a normalized vector in a finite-dimensional tensor product ${\mathcal{H}}_A\otimes {\mathcal{H}}_B$ has the form ${\sum }_{j=1}^r{s}_j|u_j⟩|v_j⟩$, where the two families are orthonormal, $s_j>0$, ${\sum }_j{s}_j^2=1$, and $r\le \min (\dim {\mathcal{H}}_A,\dim {\mathcal{H}}_B)$. Extending each family to an orthonormal basis for two qubits givesThe nonzero Schmidt coefficients can be made positive by absorbing phases into the basis vectors. Both are positive exactly when the pure state is entangled. A product state has Schmidt rank one, so one coefficient is zero; the source's assertion of two positive coefficients for every pure state needs this exception.

Choose each local orthonormal basis independently as the computational basis. In that basis ${\sigma }_z=|0⟩⟨0|-|1⟩⟨1|$. This is a local change of coordinates, implemented by separate unitary matrices, rather than a physical restriction on the original quantum state. The associated Pauli operators supply the other two local axes. Up to an irrelevant common phase, each unitary change of qubit basis corresponds to a rotation of its Bloch sphere.

Put $s=2{\lambda }_0{\lambda }_1$. The Schmidt-basis Pauli correlation tensor is diagonal. The Pauli operators ${\sigma }_x\otimes {\sigma }_x$ exchange $|00⟩$ and $|11⟩$, while ${\sigma }_y\otimes {\sigma }_y$ do so with minus signs, and ${\sigma }_z\otimes {\sigma }_z$ leaves both fixed. HenceMixed components vanish: those containing one $z$ and one transverse Pauli operator map the occupied basis vectors outside their span, while the $xy$ and $yx$ matrix elements are purely imaginary and cancel for real Schmidt coefficients. By bilinearity, for arbitrary real vectors,For a CHSH inequality test take unit quantum measurement axesThese CHSH axes for an entangled pure two-qubit state giveThe local bound is two: for each hidden state, $A(B+B^{\prime })+A^{\prime }(B-B^{\prime })$ is $\pm 2$ when all four outcomes are $\pm 1$, and averaging cannot increase its absolute value. Thus every entangled pure two-qubit state violates a CHSH inequality, the content of Gisin's theorem. A product state has $s=0$ and does not violate it, so the unqualified final claim in the question is false for that case. The maximum $2\sqrt{2}$ occurs for equal Schmidt coefficients. This excludes local hidden-variable theories satisfying measurement independence, but does not permit faster-than-light signalling: the local reduced density matrix and its quantum measurement probabilities are unchanged by the remote choice of axis.

Write $|\zeta ⟩={\sum }_{i,j}{M}_{ij}|i{⟩}_A|j{⟩}_B$. Normalization says $\Vert M{\Vert }_{\mathrm{HS}}=1$, and the Schmidt rank equals the matrix rank of $M$. Its nonzero singular values are therefore $s_1,\dots ,s_r$ with ${\sum }_j{s}_j^2=1$.

The overlap with the maximally entangled state is $⟨{\varphi }^{+}|\zeta ⟩=\mathrm{Tr}(M)/\sqrt{d}$. The trace norm estimate of Q3(i), with test linear operator $I$, followed by the Cauchy-Schwarz inequality, givesFor two pure states, the unsquared quantum fidelity is the absolute value of their overlap. Hence the maximally entangled overlap bound from Schmidt rank isThe bound is attained by $|\zeta ⟩=r^{-1/2}{\sum }_{j=0}^{r-1}|jj⟩$. Thus uniform Schmidt coefficients on $r$ aligned basis pairs give the greatest possible overlap for that Schmidt rank.

For any normalized pure state on finite-dimensional Hilbert spaces ${\mathcal{H}}_A\otimes {\mathcal{H}}_B$, the Schmidt decomposition givesThe $u_j$ and $v_j$ form orthonormal sets in their respective spaces. The positive numbers $\sqrt{p_j}$ are the Schmidt coefficients, and $r$ is the Schmidt rank. Taking a partial trace gives the reduced density matricesThus the nonzero eigenvalues of the two reduced density matrices agree.

One construction is a singular value decomposition of the coefficient matrix $M$ in $|\psi ⟩={\sum }_{a,b}{M}_{ab}|a⟩|b⟩$. If $M=U\Sigma V^{†}$, the columns of $U$ give $u_j$, the complex conjugates of the columns of $V$ give $v_j$, and the singular values give $\sqrt{p_j}$. This also explains why the Schmidt coefficients are real and nonnegative.

The Schmidt decomposition states that every finite-dimensional bipartite pure state haswhere $s_j>0$, ${\sum }_j{s}_j^2=1$, and the two displayed families are orthonormal. The integer $r$ is the Schmidt rank.

To prove it, choose product bases and writeApply the singular value decomposition $C=US{V}^{†}$. Absorbing the columns of $U$ and the complex conjugates of the columns of $V$ into new orthonormal bases gives the stated sum, with the nonzero singular values as the Schmidt coefficients. Equivalently, $s_j^2$ are the common nonzero eigenvalues of the two reduced density matrices, so $r=\mathrm{rank}{\rho }_A=\mathrm{rank}{\rho }_B$.

The linear extension of the stated k-reduction map isIt suffices to consider a pure state $|v⟩$ of Schmidt rank $r\le k$, because positivity is preserved by sums. WriteThenFor every $|x⟩={\sum }_{ij}{x}_{ij}|ij⟩$, the Cauchy-Schwarz inequality givesThus the operator is a positive semidefinite operator. Applying this to every vector in a Schmidt-number-$k$ ensemble proves

The pure state $|\Psi {⟩}_{AB}$ is entangled when it cannot be written as a product state $|a{⟩}_A|b{⟩}_B$, equivalently when its Schmidt rank exceeds one. Its subsystem states are the reduced density matriceswhere the partial trace is characterized by $\mathrm{Tr}(M_A{\rho }_A)=⟨\Psi |M_A\otimes I_B|\Psi ⟩$ for every observable $M_A$.

Let Bob's measurement have Kraus operators $M_b$ satisfying ${\sum }_b{M}_b^{†}{M}_b=I_B$. If Alice does not learn Bob's random outcome, her state after the measurement isThis no-communication theorem means that Bob can change Alice's conditional state after she learns his outcome, but cannot change any local outcome distribution available to her alone. Standard nonrelativistic quantum mechanics is therefore operationally compatible with the prohibition of superluminal signalling in special relativity, despite its nonlocal conditional-state updates.

An exact nondisturbing state readout would destroy this protection if it reported the globally collapsed state on an absolute-time slice. For example, Alice and Bob may share Bell states. At a prearranged time Bob encodes a bit by measuring his qubit in either the computational or Hadamard basis. The usual projection postulate assigns Alice respectively one of $|0⟩,|1⟩$ or one of $|+⟩,|-⟩$. A device returning the exact pure state lets Alice identify which basis Bob chose without waiting for his outcome, producing a superluminal signal. Equivalently, Bob may choose whether to measure, and the device distinguishes the resulting proper pure state from the original improper maximally mixed local state.

A causal alternative is a quantum state readout device located at a spacetime point $x$ that reports the local quantum state under objective collapse: take reduced states on spacelike hypersurfaces through $x$ and let those hypersurfaces approach the past light cone of $x$. The result includes localized collapses in $J^{-}(x)$ and excludes spacelike-separated collapses. This assumes a fixed Minkowski spacetime, localized collapse events, ordinary local unitary dynamics between them, and outputs that may control only operations in their causal future. The postulate is logically consistent because it adds a classical record of this local state without changing the state or any standard measurement probability. It also cannot support an indirect signalling algorithm. Inductively through the algorithm's events, every readout at $x$ depends only on operations, collapses, and earlier readouts in $J^{-}(x)$; any operation selected from that output lies in the readout's future light cone. Composing readouts, unitary evolutions, and measurements therefore never carries a controllable dependence outside a future light cone, so relativistic causality is preserved.

Take $|L⟩=|0⟩$ and $|R⟩=|1⟩$. Initially the two masses are in the product stateThe branch-dependent Newtonian gravitational potential energy is $-G{m}^2/d_{ab}$. Under the stated approximation, only the $|RL⟩$ branch acquires an appreciable relative phase, so after time $t$,The determinant of its two-by-two coefficient matrix is $(1-e^{i\varphi })/4$, which is nonzero unless $\varphi$ is a multiple of $2\pi$. Thus the state generally has Schmidt rank two: the branch-dependent gravitational phase creates gravitationally induced entanglement.

An entanglement witness has a bound obeyed by every separable quantum state and violated by at least one entangled state. For a product state with Bloch vectors $\mathbf{r}$ and $\mathbf{s}$,by the Cauchy-Schwarz inequality. Convexity gives the same bound for every separable mixed state. Consequently $W>1$ certifies entanglement; in conventional operator form, one of $I\mp (X\otimes Z+Y\otimes Y)$ has negative expectation whenever the absolute-value criterion is violated.

For the state above, direct use of the Pauli matrices givesWith the supplied values,which is close to $\pi$. Hence $W\simeq 2.00$, and to the nearest integerThis is the operating principle of the Bose--Marletto--Vedral experiment.

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.