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An upper bound follows directly from the Schmidt-tail entanglement monotonewhere ${\rho }_A^{\chi }$ is the reduced density matrix of a bipartite pure state. On these two-qubit states, this is the smaller squared Schmidt coefficient, so $E_2(\psi )=s$ and $E_2(\varphi )=t$.

Here is the average-monotonicity argument needed for all LOCC strategies. The largest eigenvalue is convex: for any unit vector $v$, $⟨v|{\sum }_r{p}_r{\rho }_r|v⟩\le {\sum }_r{p}_r{\lambda }_{\max }({\rho }_r)$, and maximizing the left side gives the claim. Hence $1-{\lambda }_{\max }$ is concave. If Alice makes a local measurement, Bob's conditional reductions average to his original reduction. Concavity therefore gives ${\sum }_r{p}_r{E}_2({\chi }_r)\le E_2(\chi )$. If Bob measures, use Alice's reductions instead. The two reductions of a pure state have the same nonzero eigenvalues by the Schmidt decomposition. Refining all measurement outcomes into individual Kraus operators keeps conditional states pure; iterating the inequality covers adaptive rounds of classical communication. The same bound persists for limits of such protocols.

If success produces the target with probability $p$, its contribution to the final average is $pt$. Failure branches have nonnegative $E_2$. ThusThis is a bound on arbitrary local protocols, not merely on a particular filtering measurement.

Nielsen's pure-state conversion theorem gives the exact deterministic LOCC criterion. Let the Schmidt decompositions be $|\psi ⟩={\sum }_j\sqrt{{\lambda }_j}|a_j{b}_j⟩$ and $|\varphi ⟩={\sum }_j\sqrt{{\mu }_j}|a_j^{\prime }{b}_j^{\prime }⟩$. The vectors $\lambda ,\mu$ consist of squared Schmidt coefficients, equivalently the eigenvalues of either reduced density operator. Order each in decreasing order and pad with zeros to a common length $d$.

Then deterministic conversion is possible if and only ifThis is majorization, with the input vector majorized by the output vector. The direction matters: a maximally entangled state has a uniform vector, which is majorized by a product state's vector $(1,0,\dots )$, so entanglement can be discarded by LOCC. The criterion is for certain exact conversion, without catalysts or postselection on a successful branch.

Choose a Stinespring dilation $V:A⟶B_{1}E$ of the quantum channel. Applying it to the input purification gives the pure vectorIts $R{B}_1$ marginal is the output state in the definition of coherent information. Complementary subsystems of a pure state have the same nonzero eigenvalues, by the Schmidt decomposition. Thus $S(B_1)=S(RE)$ and $S(R{B}_1)=S(E)$. It follows thatHere $S(R\mid E)$ denotes quantum conditional entropy, evaluated on the complementary output $RE$. This is coherent information as an environment conditional entropy. Equivalently $I_c=-S(R\mid B_1)$; the environment expression has the positive sign, while the receiving-system expression has the negative sign.

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.

Take a purification of a density operator $|\Psi {⟩}_{ABCD}$ of ${\rho }_{ABC}$. Apply Strong subadditivity of Von Neumann entropy to the reduced state on $ACD$, with $A$ the conditioning system:Complementary subsystems of a pure state have the same Von Neumann entropy, by the Schmidt decomposition. Therefore $S(AD)=S(BC)$ and $S(ACD)=S(B)$. Substitution givesThis is weak monotonicity of quantum entropy. The purification of a density operator is only a proof device; no purity assumption is imposed on ${\rho }_{ABC}$.

The coefficient matrix in the computational basis isFor a bipartite pure state, the partial traces give ${\psi }_A=M{M}^{†}$ and ${\psi }_B=M^T\overline{M}$. Therefore the requested reduced density matrices areThe columns of $M$ are orthogonal. Their norms are $\sqrt{2/3}$ and $\sqrt{1/3}$, so normalizing them gives the orthonormal setWith $|v_0⟩=|0⟩$ and $|v_1⟩=|1⟩$, a Schmidt decomposition isExpanding this expression reproduces the minus sign in the $|01⟩$ coefficient. The two reduced density matrices both have eigenvalues $2/3$ and $1/3$, consistent with these Schmidt coefficients.

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.

Use GHZ preparation with Hadamard and controlled-Z gates for $n\ge 2$. Apply a Hadamard gate to the first qubit, then a controlled-NOT gate from that qubit to each of the other $n-1$ qubits. Every controlled-NOT can be written in the allowed gate set asIndeed conjugating the target $Z$ by $H$ makes the controlled phase into a controlled $X$. The rightmost gate acts first. Consequently the Clifford circuit uses $1+3(n-1)$ allowed gates and givesFor every chosen qubit $j$, tracing out the other qubits yields ${\rho }_j=\frac{1}{2}(|0⟩⟨0|+|1⟩⟨1|)=I/2$. Equivalently, across that qubit versus the rest, the displayed state is a Schmidt decomposition with two nonzero coefficients $1/\sqrt{2}$. It is therefore entangled across every such bipartition.

Entanglement does not prevent the polynomial single-output simulation from part (iii). The simulation follows the measured observable rather than claiming that the intermediate states remain product states. When $n=1$, there is no remaining subsystem and the requested entanglement is impossible; the construction and claim require $n\ge 2$.

For a purification $|{\Psi }_{\rho }{⟩}_{RA}$ of $\rho$, the entanglement fidelity isUsing a Kraus representation $\Lambda (X)={\sum }_k{A}_{k}X{A}_k^{†}$ givesIn a Schmidt decomposition of the purification,Therefore

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 $r=\mathrm{rank}{\rho }_A=\mathrm{rank}{\rho }_B$.

Take Schmidt decompositions of the two purifications across $A:R$. Their squared Schmidt coefficients and their $A$-side eigenspaces are fixed by the same reduced state ${\rho }_A$. The reference-side Schmidt vectors are two orthonormal families, so a unitary $U_R$ maps one family to the other, including arbitrary choices inside degenerate subspaces. Hence the unitary freedom of purification givesIf the reference supports have different dimensions, the corresponding statement uses an isometry.

The Schmidt decomposition is $|\psi ⟩={\sum }_j{s}_j|a_j⟩|b_j⟩$. Its reduced states are ${\rho }_A={\sum }_j{s}_j^2|a_j⟩⟨a_j|$ and ${\rho }_B={\sum }_j{s}_j^2|b_j⟩⟨b_j|$. They have the same nonzero eigenvalues, so

The Schmidt decomposition of an entangled pure two-qubit state has two nonzero terms. Absorb both complex phases into the local basis vectors and interchange the labels of the second qubit to obtainIn this state the only nonzero same-axis two-qubit Pauli correlators areAll mixed-axis correlators vanish. Expanding $(\mathbf{a}\cdot \mathbf{\sigma })\otimes (\mathbf{b}\cdot \mathbf{\sigma })$ therefore gives

For the four stated vectors this becomesIt follows immediately thatEvery separable quantum state obeys the corresponding CHSH inequality with upper bound two. Set $s=2c_0{c}_1>0$, choose ${\beta }^{\prime }=\pi -\beta$, and take $\tan \beta =s$ with $0<\beta <\pi /2$. The displayed expression is thenThus every entangled pure two-qubit state has local measurement correlations that no separable state can reproduce, which is Gisin's theorem. In an ideal Bose--Marletto--Vedral experiment, optimized local measurements can therefore witness any nonzero pure-state entanglement generated during the gravitational interaction. If gravity is the only interaction between the masses, such a violation shows that the mediator cannot be described by a purely classical local variable under the assumptions of the proposal; experimentally, control of decoherence and nongravitational forces is essential to that inference.

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.

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.

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.