Understanding bell's theorem: why hidden variables imply a linear relationship?

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  • #426
wle
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Translation: even though you you may violate for a single triple, when you average over a large number, the violation will disappear.
If you mean that the violation will disappear when averaged over all the possible triples, then yes.

Translation: the inequality derived for a single set must apply to three different sets [...]
Absolutely not what I said. If this is what you're getting from what I've posted here, then you haven't understood the point I am making. At all.

Translation: the inequality derived for a single set must apply to data collected in three different measurements because if Alice and Bob collect data for three pairs which violate the inequality, they must also collect exactly the equivalent combinations of the other correlators which do not violate the inequality so that the averages from the three datasets must not violate the inequality.
I don't understand what this is even supposed to mean. If Alice and Bob measure the (ab, ac, bc) terms in that order, then they cannot also measure the (ac, bc, ab) terms in that order on the same three photon pairs. They've already done the experiment and they can't change history. As far as I can tell you are describing something completely meaningless and I don't recognise it as having anything to do with anything I said.
 
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  • #427
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billschnieder said:
Translation: the inequality derived for a single set must apply to three different sets because when violations occur in individual cases, there must be corresponding non-violations which "cancel-out" those violations so that the averages obey the inequality.
Absolutely not what I said. If this is what you're getting from what I've posted here, then you haven't understood the point I am making. At all.
If it mustn't then it mustn't and violation of the inequality is meaningless. But if as you argue violation of the inequality is meaningful, then the inequality must apply to whatever system you are getting the correlations from to show violation.

A system can not violate a law that does not apply to the system. If you claim a system has violated a law, you MUST also be claiming that the law SHOULD apply to the system. You don't appear to understand your own argument.

If Alice and Bob measure the (ab, ac, bc) terms in that order, then they cannot also measure the (ac, bc, ab) terms in that order on the same three photon pairs. They've already done the experiment and they can't change history.
That's the whole point! You are arguing that the inequality applies to the averages from their measurements because they MUST have measured all three for the same three photon pairs, or other photon pairs so identical that all the correlators compensate each other, so that the averages obey the inequality.

Didn't you start this line of argument by saying even though individual correlators may violate the inequalities, the averages will obey it? Now you are claiming not to understand why you argued that the averages will obey the inequality.
 
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  • #428
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Rs(-30,30) = Rs(-30,0) + Rs(0,30) - 2* Rs((-30,0)&(0, 30)), the equality
Rs((-30,0)&(0, 30)) >= 0 ..... (*)
therefore Rs(-30,30) <= Rs(-30,0) + Rs(0,30), your inequality
I agree with that, although I could have derived the inequality without using that equation.

therefore Rs((-30,0)&(0, 30)) = 0.5 * [Rs(-30,0) + Rs(0,30) - Rs(-30,30)] >= 0
Yes, I agree with that.

If Rp=Rs(-30,30)=0.75, Rq=Rs(-30,0)=0.5, and Rr=Rs(0, -30)=0.5 .... (*)
then Rs((-30,0)&(0, 30)) = 0.5 * [0.5 + 0.5 - 0.75] = -0.125 < 0
I think you mean that Rs(-30,0) and Rs(0,30) equal .25, not .5. So it's Rs((-30,0)&(0, 30)) = .5 (.25 + .25 -.75) = -.125

You have two contradictory assumptions (*). If Rs((-30,0)&(0, 30)) >= 0 as you assumed when you derived the inequality, then it must be the case that the three correlations Rp(-30,30), Rq(-30,0), and Rr(0,30) CAN NOT ALL BE EQUAL to the three correlations Rs(-30,30), Rs(-30,0) and Rs(0,30).
Yes, we have reached a contradiction, so at least one of the assumptions used in deriving the contradiction must be wrong. But let me give you a proof that three relative frequencies Rs(-30,0), Rs(0,30), and Rs(-30,30) must have the same value that they have for p, q, and r.

You have already agreed that Rp(-30,0)=Rq(-30,0)=Rr(-30,0), and similarly for (0,30) and (-30,30). Let us denote by Np(-30,0) the number of photon pairs in p for which M(-30,0), let us denote by Np_tot the total number of photon pairs in p, and let us make similar definitions for q, r, and s. Then we know that Np(-30,0)/Np_tot=Nq(-30,0)/Nq_tot=Nr(-30,0)/Nr_tot.

And then Rs(-30,0) = Ns(-30,0)/Ns = (Np(-30,0) + Nq(-30,0) + Nr(-30,0))/(Np_tot + Nq_tot + Nr_tot) = (Rp(-30,0)*Np_tot+Rq(-30,0)*Nq_tot+Rr(-30,0)*Rq_tot)/(Np_tot+Nq_tot+Nr_tot) = (Rp(-30,0)*Np_tot+Rp(-30,0)*Nq_tot+Rp(-30,0)*Rq_tot)/(Np_tot+Nq_tot+Nr_tot) = Rp(-30,0). And then we can use similar reasoning for (0,30) and (-30,30). What do you disagree with here?

In other words, you assumed that p, q, r were not disjoint (presence of Rs((-30,0)&(0, 30)) in the derivation), and then later assumed that they were disjoint --> violation.
How does the presence of Rs((-30,0) & (0,30)) indicate that I'm assuming that p, q, and r are not disjoint? I am definitely assuming that p, q, and r are disjoint.
 
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