The most known Bell test is due to Clauser, Horne, Shimony, Holt (CHSH) [3], for two observers *A* and *B*. Each of them can choose the measurement between 1 and 2 (*A*) or 3 and 4 (*B*) the choice-dependent outcome *A*_{i} and *B*_{j} has values ± 1. Locality means that outcomes depend only on local choice, i.e. *A*_{13} = *A*_{14} = *A*_{1}, etc. This assumption is critical because sufficiently slow choice or readout makes it invalid – the observers can ultimately get informed about the other party’s choice. One usually refers here to relativistic bound of speed of light, see Fig. 1 but the examples below do not at all rely on relativity. They are just simple quantum models with assumed states and their evolution.

Figure 1 Spacetime picture of a simple Bell-type test of local realism. Free choices, denoted by small letters generate spatio-temporal lightcones, bounded by speed of light. The measurements, denoted by capital letters must be completed outside of the other party’s choice (*A* and *B*) or both (*C*).

Then realism requires existence of the joint positive probability *p*(*A*_{1},*A*_{2},*B*_{3},*B*_{4}), yielding the inequality
$$\begin{array}{}{\displaystyle -2\le \u3008{A}_{1}{B}_{3}\u3009+\u3008{A}_{1}{B}_{4}\u3009+\u3008{A}_{2}{B}_{3}\u3009-\u3008{A}_{2}{B}_{4}\u3009\le 2}\end{array}$$(1)

It follows from the fact that *A*_{1}(*B*_{3}+*B*_{4})+*A*_{2}(*B*_{3}−*B*_{4}) is always ± 2. Quantum rule, $\begin{array}{}{\displaystyle \u3008AB\cdots Z\u3009=\mathrm{T}\mathrm{r}\hat{A}\hat{B}\cdots \hat{Z}\hat{\rho}}\end{array}$ for the simultaneously measurable observables (Hermitian) $\begin{array}{}{\displaystyle \hat{A},\phantom{\rule{thinmathspace}{0ex}}\hat{B}}\end{array}$ etc., and the (positive, Hermitian, and normalized) state matrix $\begin{array}{}{\displaystyle \hat{\rho}}\end{array}$, allows to predict all here relevant correlations. Taking $\begin{array}{}{\displaystyle \hat{\rho}}\end{array}$ = |*ψ*〉〈*ψ*| with Bell states $\begin{array}{}{\displaystyle \sqrt{2}}\end{array}$|*ψ*〉 = |+−〉−|−+〉 in the basis |±_{A},±_{B}〉 and the observables $\begin{array}{}{\displaystyle {\hat{X}}_{i}={e}^{i{\varphi}_{i}}|+\u3009\u3008-|+{e}^{-i{\varphi}_{i}}|-\u3009\u3008+|}\end{array}$ (in the respective *X* = *A*,*B* subspace) we get *X*_{i} = ± 1 and 〈 *A*_{i}*B*_{j}〉 = −cos(*ϕ*_{i}−*ϕ*_{j}). For *ϕ*_{1,2,3,4} = (0,*π*/2,5*π*/4,3*π*/4), the inequality (1) is violated as the middle correlator reads 2$\begin{array}{}{\displaystyle \sqrt{2}}\end{array}$⋍ 2.84 > 2. Unfortunately, experimental implementation of this model is quite hard. Either the actual outcome has the third value *0* (usually meaning lost particle) or the state decoheres too quickly. In both cases the original inequality is in practice no longer violated. There are at least two counteractions. One is to condition the state by entanglement swapping [13]. The distinguish between main Bell (localized) uppercase states |±_{A,B}〉 and auxiliary communication (traveling) lowercase states |±_{a,b}〉 If observers first generate their local Bell states $\begin{array}{}{\displaystyle \sqrt{2}}\end{array}$|*ψ*_{X}〉 = |+_{X},−_{x}〉−|−_{X},+_{x}〉 for *X* = *A*,*B*, *x* = *a*,*b* (matching upper and lower case letter) and send the lower case *a*,*b* states to the observable $\begin{array}{}{\displaystyle \hat{C}}\end{array}$ = |*ϕ*〉〈*ϕ*| with $\begin{array}{}{\displaystyle \sqrt{2}}\end{array}$|*ϕ*〉 = |+_{a},–_{b}〉−|−_{a},+_{b}〉 then the outcome *C* = 1 heralds the Bell state between *A* and *B*. Then even very low efficiency of the heralding does not prevent the Bell test if *C* is local - independent of all choices, i.e. *C* = *C*_{13} = *C*_{14} = *C*_{23} = *C*_{24}. The procedure is preselection and not postselection and so violation of *C* = 1 conditioned (1) is still correct signature of violation of local realism.

Another solution is to replace CHSH inequality with Eberhard inequality [4] which takes better into account detection efficiency (0 outcomes). The setup is similar to the previous one only *X*_{i} = 0, 1 (the case −1 is reassigned also to 0). Positive *p* yields the inequality
$$\begin{array}{}{\displaystyle p({1}_{1},{1}_{3})-p({1}_{1},{0}_{4})-p({0}_{2},{1}_{3})-p({1}_{2},{1}_{4})\le 0}\end{array}$$(2)

It follows from the decomposition of probability *p*(1_{1},1_{3}) = *p*(1_{1},1_{2},1_{3},0_{4})+*p*(1_{1},0_{2},1_{3})+*p*(1_{1},1_{2},1_{3},1_{4}) while *p*(1_{1},0_{4})≥ *p*(1_{1},1_{2},1_{3},0_{4}), *p*(0_{2},1_{3})≥ *p*(1_{1},0_{2},1_{3}) and *p*(1_{2},1_{4})≥ *p*(1_{1},1_{2},1_{3},1_{4}). Applied to the ideal Bell state with −1 replaced by 0 we get *p*(1_{i},1_{j}) = (1 − cos(*ϕ*_{i}−*ϕ*_{j}))/4 and *p*(1_{i},0_{j}) = (1+\cos(*ϕ*_{i}−*ϕ*_{j}))/4 and the inequality is violated for the same set of angles with the left hand side reading 1/$\begin{array}{}{\displaystyle \sqrt{2}}\end{array}$−1/2⋍ 0.2 > 0.

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