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For the Schmidt-basis Pauli correlation tensor, choose Alice's axes ${\mathbf{e}}_z,{\mathbf{e}}_x$ and Bob's $({\mathbf{e}}_z\pm s{\mathbf{e}}_x)/\sqrt{1+s^2}$. The CHSH inequality expression is $2\sqrt{1+s^2}$. It exceeds two exactly when both Schmidt coefficients are nonzero, reaching $2\sqrt{2}$ for maximal quantum entanglement. A product state gives $s=0$ and saturates rather than violates the inequality.

For two-qubit pure states with smaller squared Schmidt coefficients $s<t$, optimal stochastic conversion of a two-qubit pure state succeeds independently with probability $s/t$. The asymptotic pure-state entanglement conversion rate of collective LOCC is instead $h_2(s)/h_2(t)$, where $h_2$ is binary entropy. The difference is strict: differentiating gives $\frac{d}{dx}[h_2(x)/x]={\log }_2(1-x)/x^2<0$. Hence $h_2(s)/s>h_2(t)/t$. Independent filtering loses entanglement in its product-state failure branches, whereas collective concentration and dilution preserve entanglement asymptotically.

Let $|{\Phi }_d⟩=d^{-1/2}{\sum }_{j=0}^{d-1}|j,j⟩$ and $1\le m\le d$. It can be converted deterministically by LOCC to $|{\Phi }_m⟩$. For each $m$-element subset $S$ of the local basis labels, let ${\Pi }_S$ be its projector and let Alice's Kraus operator be $K_S={\Pi }_S/\sqrt{\left(\genfrac{}{}{0ex}{}{d-1}{m-1}\right)}$. Every label occurs in $\left(\genfrac{}{}{0ex}{}{d-1}{m-1}\right)$ subsets, so ${\sum }_S{K}_S^{†}{K}_S=I$. Each outcome has probability $m/[d(\genfrac{}{}{0ex}{}{d-1}{m-1})]=1/\left(\genfrac{}{}{0ex}{}{d}{m}\right)$ and leaves the maximally entangled state $m^{-1/2}{\sum }_{j\in S}|j,j⟩$. Alice reports $S$, and both parties relabel these $m$ basis states to $0,\dots ,m-1$ by local unitary operations.

For $d=6$, $m=4$, only three outcomes are needed: partition the six labels into disjoint pairs $T_0,T_1,T_2$ and use $K_r=(I-{\Pi }_{T_r})/\sqrt{2}$. Each label is retained twice, proving completeness, and each outcome has probability $1/3$. Every branch leaves four equal Schmidt coefficients, hence two independent Bell pairs after binary relabelling.

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.

Every entangled pure state of two qubits violates a CHSH inequality for a suitable choice of local measurement axes. If its two Schmidt coefficients are $c_0,c_1>0$, the optimal construction already gives a value $2\sqrt{1+(2c_0{c}_1{)}^2}>2$.

Under a product unitary operator $U_A\otimes U_B$, the reduced density matrix of $A$ becomes $U_A{\rho }_A{U}_A^{†}$. The unitary on $B$ cancels under the partial trace. Unitary conjugation preserves eigenvalues, so the squared Schmidt coefficients of a bipartite pure state are unchanged. Conversely, two normalized bipartite pure states with the same squared Schmidt coefficients, including multiplicities and zero padding, have Schmidt decompositions carried into each other by local unitaries mapping their Schmidt bases. Thus reduced-state spectra completely classify bipartite pure states up to local unitary operations.

An operation that applies a unitary operator to one party's subsystem and the identity to the other is a local unitary operation. Several separated parties may each apply a known unitary, resulting in a product of their local unitaries. Such operations preserve the squared Schmidt coefficients across the relevant bipartition.

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.

An exact deterministic LOCC conversion of a bipartite pure state to another is possible exactly when the input vector of squared Schmidt coefficients is majorized by the output vector. Sort the two probability vectors decreasingly and pad with zeros to a common length before applying majorization. A maximally entangled state may therefore be converted to a product state, while the reverse conversion is forbidden. The criterion does not assert conversion by postselection or with a catalyst.

The Schmidt decomposition and local unitary operations put any entangled two-qubit pure state intoBoth Schmidt coefficients are nonzero. Alice performs a two-outcome generalized measurement with Kraus operatorsTheir effects sum to the identity, so this is a legitimate local measurement in quantum mechanics. On success,The normalized output is the Bell state ${\Phi }_{+}$. Bob's unitary matrix $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ maps it to the spin singlet state. Alice communicates her success flag; no shared quantum operation is needed. At $\lambda =1/2$ the filter is the identity and success is certain. This is the singlet target case of optimal stochastic conversion of a two-qubit pure state.

For dichotomic local outcomes $A_x,B_y\in \{-1,1\}$, the CHSH inequality isIt assumes a local hidden-variable theory with freely chosen settings independent of the hidden-variable distribution. For the spin singlet state, direct evaluation of its Pauli matrices correlations givesChoose $a_0=e_z$, $a_1=e_x$, $b_0=(e_z+e_x)/\sqrt{2}$, and $b_1=(e_z-e_x)/\sqrt{2}$. The four correlations are $-1/\sqrt{2},-1/\sqrt{2},-1/\sqrt{2},+1/\sqrt{2}$, respectively, giving

The success flag must be recorded and communicated before the subsequent independent choice of Bell-test settings. This conditioning is legitimate because setting-independent heralding preserves Bell locality: a putative local hidden-variable theory for the entire sequential experiment would, after successful filtering, still have local response functions with a setting-independent conditional hidden-variable distribution. More explicitly, if the hidden variable is $\xi$ and $h$ is the already determined success flag, conditioning changes its distribution to $\rho (\xi \mid h=1)$ but leavesClassical messages used in preparation can be included in $\xi$; none are exchanged during the spacelike Bell measurements. Such a conditioned model still obeys the CHSH inequality, contradicting the displayed singlet correlations. Therefore no local hidden-variable theory reproduces all sequential measurement correlations of any entangled two-qubit pure state. This argument does not discard outcomes on the basis of the later measurement settings.

The resource here is one copy of a two-qubit Werner state, with the physical range $1/2<F\le 1$. The impossibility must include successful conditional branches of local operations and classical communication; average monotonicity alone would not exclude probabilistic enhancement.

For a normalized two-qubit pure state $|\chi ⟩={\sum }_{i,j}{M}_{ij}|ij⟩$, define its concurrence as $C(\chi )=2|\det M|$. Its mixed-state concurrence of two qubits is the convex roof extensionCombining near-optimal decompositions proves convexity. A separable quantum state has zero concurrence, because it has a decomposition into product vectors with zero coefficient determinant.

First establish the concurrence of a two-qubit Werner state. A pure state with squared Schmidt coefficients $\lambda ,1-\lambda$ has $C=2\sqrt{\lambda (1-\lambda )}$. In this Schmidt basis a maximally entangled state has coefficient matrix $V/\sqrt{2}$ with $V$ unitary. Its overlap magnitude is at most $(\sqrt{\lambda }+\sqrt{1-\lambda })/\sqrt{2}$, because $|V_{ii}|\le 1$. Thus the singlet overlap satisfiesAveraging any decomposition of $W_F$ gives $C(W_F)\ge 2F-1$.

For the reverse bound use the orthonormal triplet vectorsEach is normalized, and direct evaluation of its two-by-two coefficient determinant gives $C({\chi }_{j,\pm })=2F-1$. Their equally weighted six-state mixture has cancelled singlet–triplet coherences and triplet weight $(1-F)/3$, so it is exactly $W_F$. Hence

Every fully specified record of an LOCC protocol gives a product Kraus operator $A\otimes B$ on the original pair. Local ancillas and discarded local outcomes can be included by refining the record. Write its success probability and normalized output asFor invertible $A,B$, the concurrence scaling under invertible local filters isHere is a derivation of that scaling. Each pure coefficient matrix transforms to $AM{B}^{\mathsf{T}}$, so its unnormalized determinant is multiplied by $\det A\det B$. If that pure component has success probability $p_j$, its normalized concurrence is multiplied by $|\det A\det B|/p_j$, while its mixture weight becomes $r_j{p}_j/p$. Every ensemble average therefore scales by $|\det A\det B|/p$. Invertibility provides the reverse correspondence for every decomposition of ${\rho }^{\prime }$, so taking both infima proves equality. If either local filter has rank one, the output is a separable quantum state and the desired bound is immediate.

It remains to show that the filter normalization cannot make this ratio exceed one. Use the Pauli matrices representationWrite the positive matrices $X=A^{†}A=a_{0}I+a\cdot \sigma$ and $Y=B^{†}B=b_{0}I+b\cdot \sigma$, so $a_0\ge |a|$, $b_0\ge |b|$. Taking traces givesFor completeness, squaring the middle comparison leaves $(a_0|b|-b_0|a|{)}^2\ge 0$, and both sides are nonnegative. This proves the single-copy Werner filtering normalization bound, and hence

Forgetting some record information or grouping several successful records mixes these branch outputs; convexity preserves the same bound. Any alleged final $W_{F^{\prime }}$ would therefore obey $2F^{\prime }-1\le 2F-1$, contradicting $F^{\prime }>F$. Thus even heralded single-copy LOCC cannot produce the specified improved Werner state. This proves that single-copy Werner filtering cannot increase concurrence. It does not forbid collective entanglement distillation from several noisy pairs; allowing additional shared entangled input copies would change the resource assumption.

By the Schmidt decomposition theorem, suitable local unitary operations put the normalized pure state into the formIt is entangled precisely when both Schmidt coefficients are nonzero. Put $s=2{\lambda }_0{\lambda }_1$, so $0<s\le 1$ for the case of interest. The Pauli operatorssatisfy $⟨Z\otimes Z⟩=1$ and $⟨X\otimes X⟩=s$ in this pure state: $Z\otimes Z$ fixes both basis terms, and $X\otimes X$ exchanges them.

Choose Alice's two observables as $A_0=Z$, $A_1=X$, and Bob's asThese are valid dichotomic projective measurements. Indeed, $XZ+ZX=0$ and $X^2=Z^2=I$, so $B_0^2=B_1^2=I$; each observable has outcomes $\pm 1$. For the CHSH inequality convention $|⟨A_0\otimes (B_0+B_1)+A_1\otimes (B_0-B_1)⟩|\le 2$, the quantum value isFor the original pure state, conjugate these four observables by the corresponding local unitary operators. Their outcomes and Pauli correlators are unchanged, and each party still measures only their own qubit. Thus every entangled pure two-qubit state violates CHSH for suitable local measurements, proving Gisin's theorem directly. The CHSH axes for an entangled pure two-qubit state approach a nonviolating configuration continuously as one Schmidt coefficient tends to zero; the strict violation therefore need not be large for weak entanglement.

Diagonalize the reduced density matrix on Alice's Hilbert space:For every positive eigenvalue, define a vector on Bob's Hilbert space byThe definition of the partial trace givesThus the $|b_j⟩$ are orthonormal. Extend the $|a_j⟩$ to an orthonormal basis of Alice's Hilbert space. If $|a⟩$ lies in the zero-eigenvalue subspace of ${\rho }_A$, thenThere is therefore no component of $|{\Psi }_{AB}⟩$ in those additional basis directions. Expansion in Alice's orthonormal basis yields the Schmidt decompositionIts positive Schmidt coefficients are $s_j=\sqrt{{\lambda }_j}$ and its Schmidt rank is $r$. Taking the two partial traces and using the orthonormality of both families givesHence the two reduced density matrices have exactly the same nonzero eigenvalues, including multiplicities. Their numbers of zero eigenvalues may differ when the local Hilbert spaces have different dimensions. In particular their Von Neumann entropies agree.

The coefficient matrix in the computational orthonormal bases, with Alice's index labelling rows, isSince the entries are real, the reduced density matrices areBoth have trace one and determinant $1/9$. Their characteristic polynomial is ${\lambda }^2-\lambda +1/9$, so the nonzero eigenvalues areBoth are positive. Thus the Schmidt rank, or the Schmidt number of this pure state, isIf “Schmidt numbers” denotes the individual positive Schmidt coefficients instead, those areTheir squares are ${\lambda }_{\pm }$; giving both the Schmidt rank and the Schmidt coefficients removes this terminology ambiguity. The pure state is entangled because its Schmidt rank exceeds one.

Yes: a deterministic LOCC protocol exists for every allowed value of the parameter. Here is an explicit implementation, so existence does not depend on merely quoting a pure-state conversion theorem.

First convert the input Bell state to $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$ by local unitary operations:Write $c=\cos \varphi$ and $s=\sin \varphi$. Alice performs the two-outcome measurement in quantum mechanics with Kraus operatorsThey define a physical measurement in quantum mechanics becauseActing on $|{\Phi }^{+}⟩$, the unnormalized branch vectors areBoth branches have probability $1/2$. Alice communicates the outcome to Bob. If it is one, both apply the Pauli X gate; this turns the normalized second branch into $c|00⟩+s|11⟩$. If it is zero, neither applies this correction. Finally Bob applies a Pauli X gate in either branch, obtainingEvery outcome produces the same target pure state, so this is deterministic local operations and classical communication, not a successful branch selected by postselection. This two-outcome LOCC dilution of a Bell pair remains valid at $\varphi =0$, where the output is a product state, and at $\varphi =\pi /4$, where it is another Bell state.

The target Schmidt coefficients are $c,s$, while those of the input are both $1/\sqrt{2}$. For an interior value $0<\varphi <\pi /4$, local unitary operations alone could not change these Schmidt coefficients; the local measurement in quantum mechanics is the step that reduces the entanglement.

Apply the Schmidt decomposition to the bipartite pure state. Local unitary operators put it in the form $|{\psi }_{\theta }⟩=\cos \theta |00⟩+\sin \theta |11⟩$, with $0<\theta \le \pi /4$ after interchanging basis vectors. Both Schmidt coefficients are nonzero because the state is entangled. Put $s=\sin 2\theta >0$.

For this state, direct application of the Pauli operators givesFor example ${\sigma }_x\otimes {\sigma }_x$ interchanges $|00⟩$ and $|11⟩$, giving $2\cos \theta \sin \theta$; the crossed products take these two kets into their orthogonal complement.

Choose the CHSH axes for an entangled pure two-qubit state:Their associated Bloch vectors have unit length, so each is a valid two-outcome projective measurement with projectors $(I\pm A_i)/2$ or $(I\pm B_j)/2$. The correlation combination isA local hidden-variable assignment with outcomes $\pm 1$ gives absolute value at most two: exactly one of $b_0+b_1$ and $b_0-b_1$ vanishes, while the other is $\pm 2$. Averaging these assignments preserves that bound, which is the CHSH inequality.

Therefore $\boxed{\mathcal{S}=2\sqrt{1+{\sin }^{2}2\theta }>2}$. For the original state $(U\otimes V)|{\psi }_{\theta }⟩$, conjugate Alice's observables by $U$ and Bob's by $V$. These conjugations rotate their measurement axes and preserve all the displayed correlations. Thus every entangled pure two-qubit state has the requested violating axes. The maximally entangled case gives $2\sqrt{2}$, while the unentangled limit merely saturates two.

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.