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A Bell inequality is a constraint on observable correlations satisfied by the specified class of local hidden-variable models. In finite-setting scenarios such constraints bound linear combinations of probabilities or expectation values. The CHSH inequality is one example, following from Bell factorization, measurement independence and bounded outcomes. A violation rules out the conjunction of the assumptions defining that model class; it does not alone identify a unique false premise.

For spacelike-separated measurement regions, a sufficient specification $\lambda$ of their relevant past should screen each outcome from additional information in the other wing. In a two-setting experiment a Bell-local model has $P(A,B\mid a,b,\lambda )=P(A\mid a,\lambda )P(B\mid b,\lambda )$. Together with measurement independence and bounded outcomes, this implies the CHSH inequality. Bell local causality is stronger than quantum no-signalling and is not an assumption that every outcome was predetermined.

For outcomes in $\{0,\dots ,d-1\}$, let $[x{]}_d$ be their nonnegative modular representative. A deterministic local hidden-variable model satisfiesThe unreduced chain telescopes to $-1$, so its reduced nonnegative sum is at least $d-1$ for each assignment. For binary outcomes and two settings, it is a form of the CHSH inequality: $I_{2,2}=2-(E_{11}+E_{21}+E_{22}-E_{12})/2$.

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.

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 $2\sqrt{1+(2c_0{c}_1{)}^2}>2$.

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.

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.

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.

For a spin singlet state, quantum spin measurements along unit vectors $\mathbf{a},\mathbf{b}$ predict $E(\mathbf{a},\mathbf{b})=-\mathbf{a}\cdot \mathbf{b}$. Take ${\mathbf{a}}_0=\mathbf{z}$, ${\mathbf{a}}_1=\mathbf{x}$, ${\mathbf{b}}_0=(\mathbf{z}+\mathbf{x})/\sqrt{2}$ and ${\mathbf{b}}_1=(\mathbf{z}-\mathbf{x})/\sqrt{2}$. These giveThe violation therefore excludes the whole conjunction of the sufficient assumptions, for any model reproducing these correlations. It does not identify which premise fails without further commitments about causal explanation, state completeness and measurement setting generation.

A defensible verdict is to retain ordinary independent setting preparation and reject Bell local causality, specifically its classical screening-off factorization. In standard quantum theory, taking the pure entangled state as the complete state preserves parameter independence: each local marginal is $1/2$, regardless of the distant setting. Yet outcome independence fails. For equal spin axes, $P(A=+,B=-)=1/2$, whereas $P(A=+)P(B=-)=1/4$. The distant outcome changes the conditional prediction even after the quantum state is specified. This failure is compatible with quantum no-signalling, since conditioning on that outcome requires learning a result which cannot itself be controlled.

Adding further past variables does not restore the entire Bell conjunction while retaining the violating correlations; the CHSH inequality already allowed arbitrary such $\lambda$. Alternatively, one may reject measurement independence, permit retrocausal setting dependence in the relevant past state, or adopt another non-Bell-local description. Those are logically possible responses, but each must explain its replacement probability and causal model rather than treating the inequality as a failed algebraic theorem. A deterministic theory is not exempt from this requirement.

The assumption of measurement independence is supported by the purpose and design of independently varied settings, but cannot be proved for every conceivable hidden variable merely by inspecting observed setting frequencies. Conversely, a rejection of Bell factorization does not alone establish a controllable superluminal signal or select a unique ontology. The reasoned verdict is against Bell's screening-off locality, conditional on the other stated premises; the unconditional conclusion concerns their conjunction.

Consider two spacelike-separated measurement regions $R_A$ and $R_B$, with locally chosen settings $a,b\in \{0,1\}$ and recorded outcomes $A,B\in \{-1,1\}$. A relevant past region supplies a sufficient specification $\lambda$ of the common preparation and other hidden physical information needed to screen the outcomes. This is not necessarily just a quantum state written at the source, nor a conditioning on arbitrary records which already contain the future choices. The spacetime shielding and sufficiency requirements are substantive causal assumptions.

Use Bell local causality: once the relevant past is sufficiently specified, learning facts in the remote measurement region changes no local outcome probability. In particular, requirewhere conditional probabilities are defined. The probability chain rule givesThis factorization combines parameter independence with outcome independence. Neither probability is required to be zero or one: the argument includes stochastic local hidden-variable theories.

Also assume measurement independence, $\rho (\lambda \mid a,b)=\rho (\lambda )$, so all four setting pairs average over one normalized nonnegative past-variable distribution. This is a statistical condition on the setting procedure and relevant hidden information, not a proof of a metaphysical doctrine of free will. Finally the recorded outcomes must represent the same ensemble of trials. One can include nondetections in a specified bounded-outcome rule; if instead one discards events depending on setting or outcome, a further sampling assumption is required. Setting-dependent postselection is not covered silently by the argument.

Let ${\alpha }_a(\lambda )={\sum }_{A}AP(A\mid a,\lambda )$ and ${\beta }_b(\lambda )={\sum }_{B}BP(B\mid b,\lambda )$. They lie in $[-1,1]$, and the factorization givesFor each $\lambda$,Averaging with the same $\rho$ proves the CHSH inequalityNo separate assumption of simultaneous predetermined values for incompatible quantum observables was needed. The common probability model and its screening relations did the work.

A Reichenbach common cause principle can motivate the same calculation: a complete classical common cause in the past screens the two outcomes, while each setting affects only its own wing. Together with setting-independent sampling this gives the displayed factorization. It is not enough to provide a different screening variable for each pair of settings: those constructions need not form one compatible common-cause model for the four correlations.

A stochastic Einstein locality formulation instead constrains objective chances by facts in the appropriate causal past. To use it as a sufficient premise here, specify both local dependence on settings and a joint screening condition for the two outcomes. Merely requiring local marginal chances not to depend on a remote setting gives parameter independence, which is weaker than factorization. Thus the precise stochastic locality version matters; one should not infer a Bell inequality from every condition loosely described as no superluminal influence.

This is the dephased Bell-state mixture with coherence parameter $2x$. The Pauli correlation tensor isso spin measurements along unit vectors have $E(\mathbf{a},\mathbf{b})={\mathbf{a}}^{T}T\mathbf{b}$. ChooseThen $E(a,b)-E(a,b^{\prime })=2/\sqrt{1+4x^2}$ and $E(a^{\prime },b)+E(a^{\prime },b^{\prime })=8x^2/\sqrt{1+4x^2}$. The CHSH inequality expression is thereforeThese axes also attain the CHSH optimum from the correlation tensor, since the two largest eigenvalues of $T^{T}T$ are $1$ and $4x^2$. At $x=1/2$ the value is $2\sqrt{2}$; at $x=0$ the explicit LHV model in part (c) applies. Thus exactly $0<x\le 1/2$ has correlations incompatible with a local hidden-variable description.

Arbitrarily small positive coherence suffices in this particular one-parameter family, with optimally chosen axes; the violation above two tends to zero quadratically as $x\to 0$. This does not imply that every entangled mixed state violates a CHSH inequality.

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.

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 the spin singlet state, measurements along the same axis are perfectly anticorrelated. By measuring either spin component of particle $B$, one can therefore predict with certainty the corresponding result for distant particle $A$. The Einstein–Podolsky–Rosen criterion of reality then says that every such spin component of $A$ is an element of physical reality, because the choice at $B$ can be made without disturbing $A$. Since the quantum state assigns no simultaneous sharp values to noncommuting spin components, the EPR argument concludes that the wavefunction is not a complete description, provided this locality premise is accepted.

Let $\lambda$ denote a proposed complete hidden state, and let $a,b$ be the chosen axes.

For two axes on each side, every $\lambda$ obeysMeasurement independence permits averaging the same distribution of $\lambda$ for all four setting pairs, producing the CHSH inequality

The singlet prediction isChoose coplanar unit vectorsThenand thereforeThus the predictions of quantum theory violate the conjunction of outcome determinism, parameter independence, and measurement independence. This is Bell theorem; the calculation alone does not select which premise a deeper theory must abandon.

The Einstein–Podolsky–Rosen criterion of reality says that if, without disturbing a system, one can predict a physical quantity with certainty, then an element of physical reality corresponds to that quantity.

For two settings on each wing and outcomes in $\{-1,1\}$, the CHSH inequality for every local hidden-variable theory is

Choose all four axes in one plane, at anglesTheir relevant unsigned separations areThe specified correlation function therefore gives $E(a,b)=E(a,b^{\prime })=E(a^{\prime },b)=1$ and $E(a^{\prime },b^{\prime })=-1$. Hencethe algebraic maximum and thus a maximal violation of the CHSH inequality.

Suppose local elements of reality predetermined outputs $x(a,\lambda ),x(a^{\prime },\lambda ),y(b,\lambda ),y(b^{\prime },\lambda )\in \{-1,1\}$. For every hidden state $\lambda$,equals $2$ or $-2$, because exactly one bracket vanishes and the other equals $\pm 2$. Averaging over $\lambda$ proves the CHSH inequality. Part (b) instead gives $4$, so no such locally predetermined response functions can reproduce the device. This is the standard Bell theorem obstruction.

A Popescu–Rohrlich box is a bipartite no-signalling box attaining the algebraic maximum of the CHSH inequality.