by Joy Christian » Wed Dec 21, 2016 4:58 am
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The following image is from the last page of my very first paper on Bell's theorem, written way back in 2007:
https://arxiv.org/abs/quant-ph/0703179.
Here the standard scores A(a), A(a'), B(b) and B(b') in eq. (22) are all bivectors, so they are non-commuting numbers. In fact, their commutators in eq. (23) are + or -2 in value, with the product of the two commutators being 2 x 2 = 4. The non-vanishing of the commutator [A(a), A(a')] means that in practice A(a) and A(a') cannot be measured simultaneously. They are only counterfactually possible standardized measurement outcomes along a and a'. And likewise for the commutator [B(b), B(b')].
More precisely, Alice's results A(a) and A(a’) commute with the space-like separated results B(b) and B(b’) of Bob, but A(a) does not commute with A(a’), nor does B(b) with B(b’). Which shows that measurements by the same detector along different vectors such as a and a' are only counterfactually possible. If the standard scores did commute and the commutators did vanish, then the function F in eq. (23) would reduce to |2|, giving the upper bounds of |2| on the CHSH correlator. But since they do not commute, the value of F is 2\/2, giving the quantum mechanical bounds on the CHSH correlator (quantum mechanics without quantum mechanics, if you like).
But now compare the above eqs. (22) and (23) with the eqs. (A.4) and (A.5) below from the Appendix of
my latest paper. The difference is that now A(a), A(a'), B(b) and B(b') are all scalar numbers, +/-1, and therefore they all commute with each other. In that case my argument below shows that one can easily derive the bounds of -2 and +2 on the CHSH correlator
without assuming locality, if we continue to take into account the fact that Alice cannot measure A(a) and A(a') simultaneously, and nor can Bob measure B(b) and B(b') simultaneously. But of course they can measure any of the space-like separated results A(a) and B(b), or B(b') simultaneously.
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The following image is from the last page of my very first paper on Bell's theorem, written way back in 2007: https://arxiv.org/abs/quant-ph/0703179.
[img]http://libertesphilosophica.info/blog/wp-content/uploads/2016/12/fibell.png[/img]
Here the standard scores A(a), A(a'), B(b) and B(b') in eq. (22) are all bivectors, so they are non-commuting numbers. In fact, their commutators in eq. (23) are + or -2 in value, with the product of the two commutators being 2 x 2 = 4. The non-vanishing of the commutator [A(a), A(a')] means that in practice A(a) and A(a') cannot be measured simultaneously. They are only counterfactually possible standardized measurement outcomes along a and a'. And likewise for the commutator [B(b), B(b')].
More precisely, Alice's results A(a) and A(a’) commute with the space-like separated results B(b) and B(b’) of Bob, but A(a) does not commute with A(a’), nor does B(b) with B(b’). Which shows that measurements by the same detector along different vectors such as a and a' are only counterfactually possible. If the standard scores did commute and the commutators did vanish, then the function F in eq. (23) would reduce to |2|, giving the upper bounds of |2| on the CHSH correlator. But since they do not commute, the value of F is 2\/2, giving the quantum mechanical bounds on the CHSH correlator (quantum mechanics without quantum mechanics, if you like).
But now compare the above eqs. (22) and (23) with the eqs. (A.4) and (A.5) below from the Appendix of [url=http://libertesphilosophica.info/blog/wp-content/uploads/2016/11/Fatal.pdf]my latest paper[/url]. The difference is that now A(a), A(a'), B(b) and B(b') are all scalar numbers, +/-1, and therefore they all commute with each other. In that case my argument below shows that one can easily derive the bounds of -2 and +2 on the CHSH correlator [color=#FF0000]without assuming locality[/color], if we continue to take into account the fact that Alice cannot measure A(a) and A(a') simultaneously, and nor can Bob measure B(b) and B(b') simultaneously. But of course they can measure any of the space-like separated results A(a) and B(b), or B(b') simultaneously.
[img]http://libertesphilosophica.info/blog/wp-content/uploads/2016/11/Appendix1.png[/img]
[img]http://libertesphilosophica.info/blog/wp-content/uploads/2016/11/Appendix2.png[/img]
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