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A Bell test compares experimental probabilities or expectation values with a Bell inequality. In a two-setting photon polarization test, each station chooses a polarizer orientation and records an outcome. A coincidence circuit identifies candidate pairs using an agreed timing rule. The choice of which trials enter the statistic is part of the experiment: correlations calculated only from detected pairs can differ from correlations of all emitted pairs. The Polarization CHSH count model separates ideal expectation values from this experimental selection problem.

Bell's theorem shows that no theory assigning outcomes through local hidden variables can reproduce every quantum correlation.

The force of the Bell theorem is that a disagreement about a local hidden-variable theory becomes a constraint on frequencies in a Bell test. Quantum entanglement alone does not guarantee a violation for every measurement in quantum mechanics: the quantum state, analyzer settings and treatment of missed events all matter. A violation of the CHSH inequality tests the conjunction of Bell local causality, measurement independence and the stated outcome and sampling assumptions; it remains compatible with quantum no-signalling.

For a specified Bell test statistic, fair sampling assumes that conditioning on successful detections preserves the correlations needed to represent the intended ensemble. Constant detector efficiency in an independently specified loss model can ensure this; constant observed count rates alone do not establish independence from the hidden state. The Polarization CHSH count model explicitly assumes independent losses.

In 1998, Jian-Wei Pan, Dik Bouwmeester, Harald Weinfurter and Anton Zeilinger created two photon pairs by two passages of a pulsed pump through a crystal using spontaneous parametric down-conversion. Interfering the inner photons at a nonpolarizing beam splitter implemented a partial Bell-basis measurement. Narrow optical filters and spatial filtering removed distinguishing information; changing the outer polarizer bases checked the conditional correlations. Fourfold detections selected the entanglement swapping events. The reported interference supported conditional quantum entanglement, but the paper reserved a statistically significant Bell inequality violation for improved visibility; it was not a demonstration closing the detection loophole.^[ref]

The ideal calculation makes the distinction between conditional and unconditional quantum states explicit. Start with spin singlet states $|{\psi }^{-}{⟩}_{12}|{\psi }^{-}{⟩}_{34}$, where $|{\psi }^{-}⟩=(|HV⟩-|VH⟩)/\sqrt{2}$ and $H,V$ denote orthogonal photon polarizations. Taking the inner product with $⟨{\psi }^{-}{|}_{23}$ gives $-|{\psi }^{-}{⟩}_{14}/2$: expanding the four terms leaves precisely the two antisymmetric outer terms. That Bell-basis measurement outcome has probability $1/4$ and leaves the outer pair in a spin singlet state. Ignoring the inner outcome instead leaves the outer density matrix $I_4/4$, a maximally mixed state. A conditional preparation flag is useful for later Bell tests only with the timing and setting conditions in setting-independent heralding preserves Bell locality.

Let $H,V$ be orthogonal linear polarizations, and prepare the Bell state $|{\Phi }^{+}⟩=(|HH⟩+|VV⟩)/\sqrt{2}$. The $+1$ analyzer vector is $|a⟩=\cos a|H⟩+\sin a|V⟩$; its orthogonal companion has outcome $-1$. Direct inner products with the Bell state give $P_{++}=P_{--}={\cos }^2(a-b)/2$ and $P_{+-}=P_{-+}={\sin }^2(a-b)/2$. Thus the expectation value of the outcome product is $E(a,b)=\cos 2(a-b)$. These are physical polarizer angles; their corresponding Bloch sphere angles are doubled.

For a stipulated isotropic admixture, use the density matrixThe maximally mixed state contributes zero correlation, so $P_{st}(a,b)=[1+stv\cos 2(a-b)]/4$ for $s,t\in \{-1,1\}$. Each local marginal is $1/2$, independently of the other setting, illustrating quantum no-signalling. In this model $v$ is the interferometric visibility of the coincidence fringe; other noise models need not have one common visibility for every basis.

Choose $a_0=0^{\circ }$, $a_1=45^{\circ }$, $b_0=22.5^{\circ }$ and $b_1=-22.5^{\circ }$. The four correlations in the displayed CHSH inequality are $v/\sqrt{2},v/\sqrt{2},v/\sqrt{2},-v/\sqrt{2}$. Consequently $S=2\sqrt{2}v$, exceeding the local bound exactly when $v>1/\sqrt{2}$. Perfect correlation at a single common setting would not yield this comparison between incompatible measurement in quantum measurements.

To connect probabilities with expected counts, stipulate $M$ emitted pairs per setting pair and independent detection efficiencies ${\eta }_A,{\eta }_B$, independent also of settings, outcomes and hidden variables. The expected joint counts are $n_{st}=M{\eta }_A{\eta }_B{P}_{st}$. Their normalized correlation isFor original teaching inputs $M=1000$, ${\eta }_A=0.8$, ${\eta }_B=0.75$ and $v=0.8$, the $(a_0,b_0)$ expected counts are $n_{++}=n_{--}\simeq 234.85$ and $n_{+-}=n_{-+}\simeq 65.15$. They total $600$ and give $E_{\det }\simeq 0.56569$, hence $S\simeq 2.26274$. Fractional expected counts are not measured data. Efficiency cancels here because the loss model was specified to be independent; this algebra supplies no justification for discarding failures in an arbitrary Bell test with a detection loophole.