Bell Imposed -0 and +0 Bounds on the CHSH Correlator

[Joy Christian]

***
I just realized that the argument put forward by Bell in his famous theorem is in fact far worse than anyone has hitherto realized. I just realized that the bounds Bell imposed on the CHSH correlator are in fact not -2 and +2, but -0 and +0. Let me explain:

The key step in Bell's argument (and I have already criticized this step differently in the Appendix D of this paper) is a transition from the sum of four averages to a single average of four possible events in spacetime. In the standard EPRB experiments one is interested in the following four averages over a large number of trials:

E(a, b) = << A(a)*B(b) >> ,

E(a, b' ) = << A(a)*B(b' ) >> ,

E(a', b) = << A(a' )*B(b) >> ,

and

E(a', b' ) = << A(a' )*B(b' ) >> ,

where A and B are equal to +1 or -1, and a, a', b, and b' are the four possible measurement settings, freely chosen by Alice and Bob at the two ends of the setup.

Here the averages of all individual outcomes are always found to vanish: << A(a) >> = << B(b) >> = << A(a' ) >> = << B(b' ) >> = 0, regardless of the setting choices.

The corresponding CHSH-correlator is then bounded by 4 (or 2\/2 if you do not neglect the crucial geometrical and topological properties of the physical space):

- 4 < E(a, b) + E(a, b' ) + E(a', b) - E(a', b' ) < + 4 .

The above bounds of +/-4 are very easy to derive --- see, for example, my derivation in the Appendix D of this paper: http://arxiv.org/pdf/1501.03393v6.pdf.

Bell and his followers, however, claim that the local-realistic bounds on the above CHSH string of averages are actually +/-2 by using the mathematical identity

<< A(a)*B(b) >> + << A(a)*B(b' ) >> + << A(a' )*B(b) >> - << A(a' )*B(b' ) >> = << A(a)*[ B(b) + B(b' ) ] >> + << A(a' )*[ B(b) - B(b' ) ] >> ........................ (1)

Now let us try to understand what the RHS of this identity is actually telling us. It is asking us to consider the averages of the events such as A(a)*[ B(b) + B(b' ) ].

But since b and b' are two mutually exclusive measurement directions that Bob could have chosen corresponding to two physically incompatible experiments, no event such as A(a)*[ B(b) + B(b' ) ] can possibly exist in any possible physical world. The probability for the existence of the events such as A(a)*[ B(b) + B(b' ) ] in spacetime is therefore exactly zero! Consequently, the average on the RHS of the identity (1) is also zero. But if the RHS of the identity (1) is identically zero, then so is its LHS.

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!



***

[FrediFizzx]

***
...
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!



***

That is a good one. That is new! Wow, let then try to wiggle out of that one.

[Joy Christian]

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

Yeah, it would be interesting to see how they try to wiggle out of it. To ensure everyone that I am not joking, let me recall that the average E of events is defined as



***

[Joy Christian]

***

Oops... I made a mistake in the definition of E. There should be no factor before the sum. It is already included in the probabilities. The correct definition is:



***

[FrediFizzx]

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]

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:

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.

***

[FrediFizzx]

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:

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.

***

Of course Heine (HR) is totally ignoring your result that CHSH is useless in evaluating a LHV model. All the experiments ever do is validate that the predictions of QM are correct. Absolutely nothing to do with local-realism.
.

[FrediFizzx]

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.

Triple I had to reproduce that here because it is probably the most simple example yet of why Bell's theory is nonsense and junk physics.

[Joy Christian]

***
My latest comments on Retraction Watch:

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.

***

[Joy Christian]

***
Hi Everyone,

The dispute over Bell's theorem seems to have been finally settled, or at least a precise point of disagreement has been pinpointed. Below I am reproducing my recent exchange with Richard Gill on Retraction Watch, following my post there I have reproduced just above. Let me summarize the essence of this exchange, bringing out the exact point of remaining disagreement. It should now be a simple Yes / No answer for anyone interested to decide for themselves who is right and who is wrong:

In the notations used in my posts above, 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.

Since the random variable X(lambda) can only take the values +2, 0, or -2, the average or expectation value of X is given by

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

The remaining point of disagreement is then the following:

Richard Gill claims that the three probabilities, p(X = +2), p(X = 0) and p(X = -2), are non-negative and add up to +1.

I, on the other hand, claim that the three 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.

So, identically zero, or not identically zero, that is the question: Yes / No.

If what I claim is true, then the conclusion of my initial post in this thread holds, and Bell's theorem is finally put to rest.

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”.

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.

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.

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.

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.

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]

***
My latest post at Retraction Watch:

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.

***

[thray]

Hi Joy,

I think this is reaffirmation of the analyticity of your framework.

Michel Fodje (Mikko) is correct that there must be probability of unity for a detection to be made. It is your substitution of "and" for "+" that conditions the result to an algebraic sum.

As an integral, detection is made over a continuous range of paired variables, correlated by the boundary conditions. If we avoid boundaries (Richard's preference), we cannot avoid probabilities. As I see that Richard agrees that the " ... sum of (an) integral equals the integral of sums ..." your result follows. He fudges by averaging (sum of integrals equals integral of sums) -- one can't do that without boundary conditions, so probability is the only option.

[FrediFizzx]

Hi Tom,

Mikko is not Michel.

[Joy Christian]

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.

***

[FrediFizzx]

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.

***

LOL! Don't hold your breath waiting for either of them to concede.

[thray]

Oh, heck. I'm sorry. Can I edit?

[Joy Christian]

***
My latest post on Retraction Watch:

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).

What this simple example shows is that an extra assumption has been smuggled-in by Bell and his followers from the back door, to derive the bounds of -2 and +2 on CHSH. An assumption that enters the derivation of the bounds only at a later stage, in the middle of their manipulation. It therefore goes entirely undetected at first sight. But the assumption is so absurd and so unphysical that it is no surprise at all that equally absurd bounds of -2 and +2 pop out in the end. Very sneaky indeed.

***

[minkwe]

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.

***

I've been away too long. But Bell's RHS definitely exposes the magicians tricks.

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]

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?

[minkwe]

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?

If you have the balls to do it, be my guest.