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Joy Christian wrote:***
...
Therefore, if we were to strictly follow the logic of Bell and his followers, then the correct bounds on the CHSH correlator are in fact not -2 and +2, but -0 and +0:

- 0 < E(a, b) + E(a, b' ) + E(a', b) - E(a', b' ) < + 0 .

Needless to say, the bounds of -0 and +0 are frequently "violated" by all kinds of physics, classical or quantum! We are inundated with non-localities and non-realities!



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FrediFizzx wrote:That is a good one. That is new! Wow, let them try to wiggle out of that one.

FrediFizzx wrote:Now Gill is claiming that A(a)B(b) and A(a)B(b') are not events. Here comes more obfuscation. Wiggle, wiggle, wiggle.

Joy Christian wrote:
The quantity A(a)*[B(b) + B(b’)] — which is a function of possible measurement events which could be observed by Alice and Bob — cannot possibly exist in any possible physical world, just as one cannot be in New York and Miami at exactly the same time. In other words, the probability of A(a)*[B(b) + B(b’)] ever occurring in any possible physical world is exactly zero. Now the proof of Bell’s theorem relies on the average or expectation value of the quantity A(a)*[B(b) + B(b’)]. But an average or expectation value of A(a)*[B(b) + B(b’)] very much depends on the probability of it ever occurring in at least some possible world; which, as noted, is exactly zero. Therefore the average or expectation value of the quantity A(a)*[B(b) + B(b’)] is also exactly zero. Thus Bell’s logic necessitates that the CHSH correlator must lie between -0 and +0. There is no way out of this conclusion.

Joy Christian wrote:

FrediFizzx wrote:Now Gill is claiming that A(a)B(b) and A(a)B(b') are not events. Here comes more obfuscation. Wiggle, wiggle, wiggle.

A part of my answer to him may be worth repeating here:

Joy Christian wrote:
The quantity A(a)*[B(b) + B(b’)] — which is a function of possible measurement events which could be observed by Alice and Bob — cannot possibly exist in any possible physical world, just as one cannot be in New York and Miami at exactly the same time. In other words, the probability of A(a)*[B(b) + B(b’)] ever occurring in any possible physical world is exactly zero. Now the proof of Bell’s theorem relies on the average or expectation value of the quantity A(a)*[B(b) + B(b’)]. But an average or expectation value of A(a)*[B(b) + B(b’)] very much depends on the probability of it ever occurring in at least some possible world; which, as noted, is exactly zero. Therefore the average or expectation value of the quantity A(a)*[B(b) + B(b’)] is also exactly zero. Thus Bell’s logic necessitates that the CHSH correlator must lie between -0 and +0. There is no way out of this conclusion.

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Joy Christion on RW wrote:The point I am making is very simple to understand. What is the probability of a single dice landing on both 3 and 6 at the same time? If you know the answer to this question, then you also know the answer to my question: What is the probability of A(a)*[B(b)+B(b’)] occurring in any possible physical world? The answer is the same for both questions.

Joy Christian wrote:
Different views are being put forward by different people. Jay is very good at sorting those out. Therefore I will stick to my point of view and try to summarize here, as clearly as possible, one of my several refutations of Bell’s theorem that is being currently discussed:

I think no one is (or should be) disputing the fact that A(a)*B(b) and A(a)*B(b’) are two only counterfactually possible events in space time. Each represents a simultaneous click of the space-like separated detectors of Alice and Bob. But they cannot both exist simultaneously, because they involve clicks of the detectors along two mutually exclusive directions b and b’ that Bob could have chosen to align his detector. The resulting experiments are then physically incompatible experiments. The question then is: Is it legitimate to even consider a hybrid quantity like A(a)*B(b)+A(a)*B(b’) = A(a)*[B(b)+B(b’)] as a physically meaningful quantity? Note that I am not stressing the fact that it is not simultaneously observable. Of course it is not simultaneously observable. But that is the least of the problems with writing a sum like A(a)*B(b)+A(a)*B(b’) = A(a)*[B(b)+B(b’)] of two only counterfactually possible events A(a)*B(b) and A(a)*B(b’) in space and time. Since such a quantity cannot possibly exist in any possible physical world, what on earth does such a quantity even mean? It is a Unicorn.

But I am willing to go along with Gill and Parrott for the sake of argument and think of it simply as a random variable in a probability space under consideration. The question then is: What is the probability associated with the quantity A(a)*[B(b)+B(b’)] ? Clearly, to anyone who is remotely respectful of the fact that we are concerned about actually performable physical experiments in a possible physical world, the probability associated with the quantity A(a)*[B(b)+B(b’)] is the same as that associated with a single dice landing on both 3 and 6 at the same time. I think any schoolchild would agree that that probability is identically zero in any possible world. If anyone claims that the probability associated with A(a)*[B(b)+B(b’)] is not zero, then please enlighten me of what value that probability has, and why it is not zero.

Now Bell’s theorem is proved by considering the average of the quantities such as A(a)*[B(b)+B(b’)]. So let me write down a simple expression for this average (as far as I can see, replacement of average with expectation value will not change the essence of my argument):

E( A(a)*[B(b)+B(b’)] ) = Sum_k { ( A(a)*[B(b)+B(b’)]^k ) x ( Probability of A(a)*[B(b)+B(b’)]^k ) } ,

where, although not explicated, the application of the law of lager numbers is understood.

But since the Probability of A(a)*[B(b)+B(b’)]^k is identically zero, it is evident that the average (or expectation value) E( A(a)*[B(b)+B(b’)] ) is also identically zero. It therefore follows from the much discussed mathematical identity that the lower and upper bounds imposed on CHSH by Bell’s logic and mathematics are not -2 and +2, but -0 and +0. QED.

I hope I do not have to spell out the funny implications of this conclusion.

Richard Gill wrote:

There are just three possible events and just three probabilities associated with the random variable X = A(a)*[B(b)+B(b’)].

According to local hidden variables, A(a)*[B(b)+B(b’)] is just some function of the hidden variable lambda. We don’t get to observe lambda and we don’t get to observe X. But if local hidden variables are true, then lambda and X(lambda) do exist. And X can take the values -2, 0 and 2. The set of all possible values of lambda can therefore be split into three disjoint subsets: the subset of all lambda where X(lambda) = -2, the subset where X(lambda) = 0, and the subset where X(lambda) = +2.

Those three subsets can be called “events”. The event where X = 2, etc. Which of the three events actually happens in any single trial is not observed by Alice and Bob. But when nature does pick a value of lambda, it will fall in just one of these three subsets, and it does that with probabilities which I’ll denote p(+2), p(0) and p(-2) respectively.

Those three probabilities are equal to the integrals of rho(lambda) d lambda over each of the three subsets. The three probabilities are nonnegative and add up to +1.

Finally, E(X) = 2 p(2) – 2 p(-2).

Christian seems to confuse “event” and “random variable” and still does not have the good formula for “expectation value”.

Joy Christian wrote:
I disagree with Gill on several counts. In a local hidden variable theory lambdas of course do exist by definition, but the function X(lambda) = A(a, lambda)*[B(b, lambda)+B(b’, lambda)] does not have any physical meaning. Nor can X take the values -2, 0 and +2 in a physically meaningful sense.

There are only two particles at disposal to Alice and Bob for each run of the experiment, not three. Such pairs of particles can be identify with the hidden variable lambda^k, or just with the index k as I have done in my Appendix D. Now X(lambda) defined by Gill is a function of three possible detection events in spacetime, or of three clicks recorded by the two detectors detecting the two particles at disposal to Alice and Bob, for each lambda (or k). These clicks of the two detectors are recorded as the numbers A(a), B(b) and B(b’), where B(b) and B(b’) are only counterfactually possible numbers. Therefore there is no physical sense in which X can take the values -2, 0 and +2. Another way of saying the same thing is that the probabilities of the function X(lambda) taking the values -2, 0, and +2 are identically zero: p(+2) = 0, p(0) = 0 and p(-2) = 0, unless of course Alice and Bob belong to a world in which the probability of finding a single dice landing on both 3 and 6 at the same time is non-vanishing. But since all three of these probabilities must be identically zero, the average or expectation value of X is also identically zero. Consequently the CHSH correlator must lie between -0 and +0.

I should stress that I have no problem with what Gill has written if X(lambda) is defined by involving only two detection events, such as x(lambda) = A(a, lambda)*B(b, lambda) or x(lambda) = A(a, lambda)*B(b’, lambda). These two definitions have perfectly reasonable physical meanings. But an expression like X(lambda) = A(a, lambda)*[B(b, lambda)+B(b’, lambda)] is physically self-contradictory even for any local hidden variable theory. It is not demanded by either Einstein’s or Bell’s conception of local realism.

Richard Gill wrote:
Nobody is saying that X has some physical meaning. But if we believe in local hidden variables, then lambda exists, the function X exists, and X(lambda) exists. Once nature has chosen lambda, X(lambda) is determined. Even if Alice and Bob have gone home and switched off their detectors.

Joy Christian wrote:
I am sure X(lambda) can be assumed to exist as a function. But the probabilities of it taking the values -2, 0, and +2 are identically zero: p(+2) = 0, p(0) = 0 and p(-2) = 0. Unless of course Alice and Bob belong to a world in which the probability of finding a tossed coin on heads and tails at the same time is non-vanishing.

Richard Gill wrote:
A(a, lambda), B(b, lambda) and B(b’, lambda) all exist, and are all equal to +/-1, whatever Alice and Bob actually measure. Clearly X(lambda) can only be -2, 0, or +2.

Joy Christian wrote:
I agree entirely: “A(a, lambda), B(b, lambda) and B(b’, lambda) all exist [at least counterfactually], and are all equal to +/-1 [by construction], whatever Alice and Bob actually measure. Clearly X(lambda) can only be -2, 0, or +2.”

But that is not going to rescue Bell’s theorem. Because the probabilities p(-2), p(0) and p(+2) of X(lambda) taking the possible values -2, 0, and +2 are all identically zero: p(+2) = 0, p(0) = 0 and p(-2) = 0. Unless of course the probability of my being in New York and Miami at exactly the same time can be non-vanishing in the local-realistic world we live in.

Joy Christian wrote:

Joy Christian wrote:
But that is not going to rescue Bell’s theorem. Because the probabilities p(-2), p(0) and p(+2) of X(lambda) taking the possible values -2, 0, and +2 are all identically zero: p(+2) = 0, p(0) = 0 and p(-2) = 0.

It seems to me that the precise point of disagreement over the validity of Bell’s theorem has now been pinpointed. I would therefore like to prove my claim above more explicitly:

In the notation introduced by Gill in response to my previous posts, let X = A(a)*[B(b)+B(b’)] be the random variable of interest and let p(X) be the probability of X taking the values +2, 0, or -2. The average or expectation value of X is then given by

E(X) = Sum_i [ X_i p(X_i) ] = +2 p(X = +2) – 2 p(X = -2) . ……………… (1)

Now Gill claims that the probabilities, p(X = +2), p(X = 0) and p(X = -2), are all non-negative and add up to +1. I, on the other hand, claim that the probabilities, p(X = +2), p(X = 0) and p(X = -2), are all identically zero:

p(X = +2) = 0, p(X = 0) = 0 and p(X = -2) = 0 . ………………………….. (2)

If my claim is true, then at least one of my objections to the validity of Bell’s theorem stands.

To prove my claim (2) above, let me now rewrite X more explicitly as

X(a, b, b’; k) := u(a, b; k) + v(a, b’; k) := A(a; k)*B(b; k) + A(a; k)*B(b’; k) , ….. (3)

where k is the hidden variable, or an initial state, or a run, or a given particle pair. Written this way, it is now very clear that the random variables u and v are only counterfactually realizable. In other words, they define two incompatible experiments for any given k.

The question now is: What is the probability of occurrence for the “event” X defined in (3)?

Now at least in my view, this probability is given by p(u and v) — i.e., the probability of X occurring is the same as the probability of u and v occurring simultaneously. But as already noted, u and v represent two mutually incompatible experiments that cannot be realized simultaneously in any possible physical world, classical or quantum. Therefore it is clear that

p(u and v) = 0 , …………………………………………………………………….. (4)

regardless of the values u and v may take. Note that I have interpreted “+” in (3) as “and” in (4). I will maintain this equivalence between “+” and “and” unless persuaded otherwise.

FrediFizzx wrote:Hi Tom,

Mikko is not Michel.

Joy Christian wrote:

FrediFizzx wrote:Hi Tom,

Mikko is not Michel.

I was about to point that out to Tom. Michel usually signs as minkwe, so it could be confusing. But what a terrible confusion that is!

In any case, Tom, it is all over. We are now just waiting for both Gill and Trump to concede defeat. It is just a matter of hours now.

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Joy Christian wrote:
Sometimes abstract ideas and complex notations obscure simple facts. So let me explain what I find problematic in Gill’s claim using a homely example. Consider an ordinary dice. Its six faces are marked with 1, 2, 3, 4, 5 and 6 dots. The “hidden variable space” in this homely example is the space of all possible faces on which the dice can land — or equivalently, the set of all possible outcomes, {1, 2, 3, 4, 5, 6}, of a throw of the dice. We can write down these six numbers representing the number of dots on the dice on a piece of paper, and that is then our abstract “hidden variable space” analogous to what Gill is talking about. Now we can ask: What is the probability of the dice landing on 3? Or: What is the probability of the dice landing on 5? Etc. But following Bell, Gill wants to ask a different question: What is the probability of the dice landing on (3+5)? To me, at least, it is quite obvious that the probability of the dice landing on (3+5) is identically zero — not non-negative adding up to +1 as Gill claims.

The hypothesis of local hidden variable theory is supposed to be about the world we live in, not about some impossible world that assigns non-negative probabilities to a dice landing on a non-existent face like (3+5).

Joy Christian wrote:

FrediFizzx wrote:Hi Tom,

Mikko is not Michel.

I was about to point that out to Tom. Michel usually signs as minkwe, so it could be confusing. But what a terrible confusion that is!

In any case, Tom, it is all over. We are now just waiting for both Gill and Trump to concede defeat. It is just a matter of hours now.

***

minkwe wrote:
New challenge for the Bell Test business people, perform the RHS experiment and demonstrate that you get the same results as the LHS. It should be easier since you have just two expectation values to measure.

<< A(a)*[ B(b) + B(b' ) ] >> , << A(a' )*[ B(b) - B(b' ) ] >>

Heinera wrote:

minkwe wrote:
New challenge for the Bell Test business people, perform the RHS experiment and demonstrate that you get the same results as the LHS. It should be easier since you have just two expectation values to measure.

<< A(a)*[ B(b) + B(b' ) ] >> , << A(a' )*[ B(b) - B(b' ) ] >>

Is it OK if I do it with fragments of small, exploding balls?