"Bell's inequality and the coincidence-time loophole"

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Unregistered Submission:

( January 13th, 2015 11:34pm UTC )

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As concerns the Bertlmann's socks discussion you refer to, it doesn't address the issue at all.

Say we have "experimental units", in order to derive the inequalities, you still need the fair-sampling assumption. That is you still need to assume that the "Lambda" which caused one experimental unit to respond "yes/no" is the same lambda which caused the "sister" experimental unit to respond "yes/no". Without this assumption, you can't derive the inequality under this kind of experiment. Like you say in the above paper, the ensemble used for each term must be identical for the inequality to apply, and remember we are talking about the ensembles as identified by the space of hidden parameters responsible for the observed outcomes.

With these kinds of experiment, like the Christensen experiment, it seems you are arguing that it doesn't matter if those two particles at the start of the pockel cells are true sister particles. But it does matter, otherwise we would not use a source at all, we would simply send a time signal and match whatever pairs of cosmic radiation we can within a time-window. So contrary to your claims, the experimental unit is not time-windows, it is the particles.

Say we have "experimental units", in order to derive the inequalities, you still need the fair-sampling assumption. That is you still need to assume that the "Lambda" which caused one experimental unit to respond "yes/no" is the same lambda which caused the "sister" experimental unit to respond "yes/no". Without this assumption, you can't derive the inequality under this kind of experiment. Like you say in the above paper, the ensemble used for each term must be identical for the inequality to apply, and remember we are talking about the ensembles as identified by the space of hidden parameters responsible for the observed outcomes.

With these kinds of experiment, like the Christensen experiment, it seems you are arguing that it doesn't matter if those two particles at the start of the pockel cells are true sister particles. But it does matter, otherwise we would not use a source at all, we would simply send a time signal and match whatever pairs of cosmic radiation we can within a time-window. So contrary to your claims, the experimental unit is not time-windows, it is the particles.

Richard Gill:

( January 14th, 2015 3:01pm UTC )

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The whole point of a loophole-free experiment is that you do *not* need the fair-sampling assumption. Who says that "lambda" is contained within the particles? It is contained in the whole physical system of measurement devices plus whatever comes to them from the source.

Unregistered Submission:

( January 14th, 2015 3:17pm UTC )

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Please read my point again. You do not eliminate lambda by simply picking time-slots. You simply attribute the lambda to time slots instead. However, that lambda, includes all the lambdas of the particles which contribute to the outcome you considered for the time-slot, in your analysis. Your suggested method of picking particles systematically discards lambdas.

That is why the experiment is done with entangled particle pairs and not random cosmic radiation.

That is why the experiment is done with entangled particle pairs and not random cosmic radiation.

Unregistered Submission:

( January 16th, 2015 11:57am UTC )

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"Your suggested method of picking particles systematically discards lambdas."

The phrase "discards lambdas" does not make sense. Lambdas are unknown and inherently unknowable (if they exist at all). One therefore cannot really do anything with them, and that includes discarding.

The phrase "discards lambdas" does not make sense. Lambdas are unknown and inherently unknowable (if they exist at all). One therefore cannot really do anything with them, and that includes discarding.

Peer 3:

( January 16th, 2015 5:04pm UTC )

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Unregistered Submitter:

In the above paper, the authors state (and I agree with them): "The problem here is that the ensemble on which the correlations are evaluated changes with the settings,while the original Bell inequality requires that they stay the same."

Note: they are talking about the ensembles of lambdas over which each term in the inequality is integrated during it's derivation. Now, obviously those lambdas are unknown and inherently unknowable, but Bell's inequalities only apply if the ensembles stay the same.

If you discard particles using any criteria of your choosing, you have no way of knowing that your applied method is not systematically removing certain properties. The point here is that the detection time of the particle is an experimental observable. If the hidden/unknown lambda introduces a delay in the detection time (which you know and measure), and you only consider particles which have no or little delay in measurement time , then you've effectively biased the ensemble of lambdas against those which introduce a delay. If the combination of lambda with setting "a" at Alice's end introduces a delay but the combination of the same lambda with setting "b" at Bob's end does not, then by considering only those which have no/little delay, you have introduced differences between the ensemble considered on Bob's and Alice's side. Any violation obtained this way is therefore null and void since the inequalities simply do not apply any more.

In the above paper, the authors state (and I agree with them): "The problem here is that the ensemble on which the correlations are evaluated changes with the settings,while the original Bell inequality requires that they stay the same."

Note: they are talking about the ensembles of lambdas over which each term in the inequality is integrated during it's derivation. Now, obviously those lambdas are unknown and inherently unknowable, but Bell's inequalities only apply if the ensembles stay the same.

If you discard particles using any criteria of your choosing, you have no way of knowing that your applied method is not systematically removing certain properties. The point here is that the detection time of the particle is an experimental observable. If the hidden/unknown lambda introduces a delay in the detection time (which you know and measure), and you only consider particles which have no or little delay in measurement time , then you've effectively biased the ensemble of lambdas against those which introduce a delay. If the combination of lambda with setting "a" at Alice's end introduces a delay but the combination of the same lambda with setting "b" at Bob's end does not, then by considering only those which have no/little delay, you have introduced differences between the ensemble considered on Bob's and Alice's side. Any violation obtained this way is therefore null and void since the inequalities simply do not apply any more.

Unregistered Submission:

( January 17th, 2015 11:33am UTC )

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"Any violation obtained this way is therefore null and void since the inequalities simply do not apply any more."

That some inequalities do not apply does not mean that no inequality applies. The original Bell's inequality is excellent for the theoretical purpose of proving incompatibility of a certain class of theories with quantum mechanics. Practicalities of experimental test of theories are different from those of theoretical comparisons, so different inequalities are the most useful, more or less like the inequality of theorem 1 in http://arxiv.org/pdf/1207.5103v5.pdf . The details of the experiment and analysis determine the correct details of the inequality (introduce a cofficent to the right side or something like that). One can derive inequalities for different experiment proposals or different analysis methods for the same experiment and then choose the experiment proposal or analysis method that gives the most promising inequality.

That some inequalities do not apply does not mean that no inequality applies. The original Bell's inequality is excellent for the theoretical purpose of proving incompatibility of a certain class of theories with quantum mechanics. Practicalities of experimental test of theories are different from those of theoretical comparisons, so different inequalities are the most useful, more or less like the inequality of theorem 1 in http://arxiv.org/pdf/1207.5103v5.pdf . The details of the experiment and analysis determine the correct details of the inequality (introduce a cofficent to the right side or something like that). One can derive inequalities for different experiment proposals or different analysis methods for the same experiment and then choose the experiment proposal or analysis method that gives the most promising inequality.

Peer 3:

( January 18th, 2015 6:17pm UTC )

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Yes, it is possible to derive different inequalities for different experimental arrangements. However, any such inequality can't be derived by adding expectation values or probabilities from different ensembles or probability measures. This includes the inequality you mention. If you know of any other valid inequality which involves probabilities from different measures, point it out. The authors of the above paper appear to know about this, since they write:

"The integrals on the right-hand side cannot easily be added when Λac′ =/= Λad′ , since we

are taking expectations over different ensembles Λac′ and Λad′ , with respect to different

probability measures".

"The integrals on the right-hand side cannot easily be added when Λac′ =/= Λad′ , since we

are taking expectations over different ensembles Λac′ and Λad′ , with respect to different

probability measures".

Unregistered Submission:

( January 18th, 2015 6:59pm UTC )

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Peer 3: ( January 18th, 2015 6:17pm UTC ): "Yes, it is possible to derive different inequalities for different experimental arrangements. However, any such inequality can't be derived by adding expectation values or probabilities from different ensembles or probability measures. This includes the inequality you mention. If you know of any other valid inequality which involves probabilities from different measures, point it out. The authors of the above paper appear to know about this, since they write:

"The integrals on the right-hand side cannot easily be added when Λac′ =/= Λad′ , since we

are taking expectations over different ensembles Λac′ and Λad′ , with respect to different

probability measures"."

They are right when they write "easily". That can be done but it is not easy.

"The integrals on the right-hand side cannot easily be added when Λac′ =/= Λad′ , since we

are taking expectations over different ensembles Λac′ and Λad′ , with respect to different

probability measures"."

They are right when they write "easily". That can be done but it is not easy.

Peer 3:

( January 19th, 2015 9:10pm UTC )

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Unregistered Submission: "They are right when they write "easily". That can be done but it is not easy."

It is a woeful mathematical error to add expectations taken over different probability measures. Of course it can be done, in the sense that you can put one number next to another and place a plus sign in between. However the result is completely meaningless. In other words, it cannot be done, if the sum is to remain meaningful. This is the sense in which I understand "easy". Technically, there is nothing difficult about adding one number to another.

In any case, the Bell inequalities are not valid if the expectation values are taken over different ensembles. This is the starting point which both the authors and I agree on.

It is a woeful mathematical error to add expectations taken over different probability measures. Of course it can be done, in the sense that you can put one number next to another and place a plus sign in between. However the result is completely meaningless. In other words, it cannot be done, if the sum is to remain meaningful. This is the sense in which I understand "easy". Technically, there is nothing difficult about adding one number to another.

In any case, the Bell inequalities are not valid if the expectation values are taken over different ensembles. This is the starting point which both the authors and I agree on.

Unregistered Submission:

( January 21st, 2015 2:22pm UTC )

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Peer 3: ( January 19th, 2015 9:10pm UTC ): "In any case, the Bell inequalities are not valid if the expectation values are taken over different ensembles."

That was not the claim. For this situation one can derive other inequalities that may differ from Bell's or Gill's (see theorem 1 in http://arxiv.org/pdf/1207.5103v5.pdf for the latter) but still may serve the same purpose.

That was not the claim. For this situation one can derive other inequalities that may differ from Bell's or Gill's (see theorem 1 in http://arxiv.org/pdf/1207.5103v5.pdf for the latter) but still may serve the same purpose.

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Richard Gill:

( January 14th, 2015 3:11pm UTC )

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"The only difference here is that instead of using the times directly, to compare with each other, you now have an external clock which you are comparing both sides to. That accomplishes nothing. You will still not be able to match events which are inherently delayed in an angle dependent manner. So how does this avoid the coincidence loophole? It doesn't."

It accomplishes the following: under local realism, you will not be able to violate CHSH (or other appropriate Bell inequality). So if you do violate CHSH, you are onto something.

Sure, it makes it harder for QM to violate CHSH. Quantum correlations, if they exist at all, will be attenuated, washed out (to some extent), by noise. But there is a big distance between 2 and 2 sqrt 2, so it is still logically possible for QM to "win" i.e. get a result significantly bigger than 2.

In other words, following the external clock rules out those damned pesky loopholes. So if we see a violation we know what it means.

It makes life harder for QM, but QM still has a chance.

Why do you suppose Bell introduced this idea in Bertlmann's socks and why do you suppose the experimentalists are adopting it nowadays?

It accomplishes the following: under local realism, you will not be able to violate CHSH (or other appropriate Bell inequality). So if you do violate CHSH, you are onto something.

Sure, it makes it harder for QM to violate CHSH. Quantum correlations, if they exist at all, will be attenuated, washed out (to some extent), by noise. But there is a big distance between 2 and 2 sqrt 2, so it is still logically possible for QM to "win" i.e. get a result significantly bigger than 2.

In other words, following the external clock rules out those damned pesky loopholes. So if we see a violation we know what it means.

It makes life harder for QM, but QM still has a chance.

Why do you suppose Bell introduced this idea in Bertlmann's socks and why do you suppose the experimentalists are adopting it nowadays?

Unregistered Submission:

( January 14th, 2015 3:31pm UTC )

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Nope, you have it backwards, it is not about QM losing or wining. The issue is not one of reducing the QM prediction from 2 root 2 down to 2. It is one of raising the the local realistic one. In the above paper, you admit that if the ensemble of hidden parameters is different the inequalities are not valid. It doesn't matter if you attribute the hidden parameters to "time-slots" or particles, it changes nothing. All it does is to obscure the fact that they are different. Systematically discarding lambdas is a very easy way for allowing that to happen. If nature is operating in way which introduces angle-dependent time delays between particle pairs then it is very difficult to "repair" that. Simply picking pairs which are close together in time, whether you use an external clock, or time tags does not guarantee that you have the same ensemble of lambdas. Similarly, if nature is operating in a manner in which some properties are less likely to be detected at certain angles, it will be impossible to avoid the detection loophole. You could have a 100% efficient detectors and yet still not be able to detect 100% of the emitted particles. This is why I say it is not accurate to suggest, as Larsson does, that the problem is due to "deficiencies in the experimental equipment".

Donald A. Graft:

( January 14th, 2015 3:43pm UTC )

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"Why do you suppose Bell introduced this idea in Bertlmann's socks and why do you suppose the experimentalists are adopting it nowadays?"

Because they were not aware that it could result in artifactual violation via data discarding, as Graft has conclusively demonstrated. However, when all the events are included, local reality is confirmed, just as we expect.

Because they were not aware that it could result in artifactual violation via data discarding, as Graft has conclusively demonstrated. However, when all the events are included, local reality is confirmed, just as we expect.

Unregistered Submission:

( January 14th, 2015 4:09pm UTC )

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Peer 1: ( January 14th, 2015 3:43pm UTC )

""Why do you suppose Bell introduced this idea in Bertlmann's socks and why do you suppose the experimentalists are adopting it nowadays?"

Because they were not aware that it would result in false violation via data discarding, as Graft has conclusively demonstrated."

Could you give a pointer to that demonstration?

""Why do you suppose Bell introduced this idea in Bertlmann's socks and why do you suppose the experimentalists are adopting it nowadays?"

Because they were not aware that it would result in false violation via data discarding, as Graft has conclusively demonstrated."

Could you give a pointer to that demonstration?

Donald A. Graft:

( January 14th, 2015 4:28pm UTC )

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Please refer to Graft's paper analyzing Christensen et al discussed on another thread. It's linked in the sidebar.

Richard Gill:

( January 14th, 2015 4:35pm UTC )

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It is not about efficiency of detectors. The experimental unit is the time-slot. The hidden variables "lambda" are everything (*) that determines what happens within that time-slot, as far as relevant for the two binary outputs (or ternary if you prefer, or something else if you prefer). Bias arises, raising the LR bound 2 upwards, if you discard experimental units in a way which may depend on interactions between hidden variables and detector settings. (One new, random detector setting, in each wing of the experiment, per time-slot).

Discarding "events" within a time-slot and choosing just to pair the first event in each, which might not belong to the same emission, might (if QM is true) bias the observed correlation *downwards*. If local realism is true, it might conceivably bias the correlation in either direction, but it *cannot* bias it *upwards* any further than the local realist limit of 2.

(*) "everything" = everything apart from the outcomes of the two coin tosses which are used to select, from outside, which measurement settings are applied in each wing of the experiment *during the time slot*.

Discarding "events" within a time-slot and choosing just to pair the first event in each, which might not belong to the same emission, might (if QM is true) bias the observed correlation *downwards*. If local realism is true, it might conceivably bias the correlation in either direction, but it *cannot* bias it *upwards* any further than the local realist limit of 2.

(*) "everything" = everything apart from the outcomes of the two coin tosses which are used to select, from outside, which measurement settings are applied in each wing of the experiment *during the time slot*.

Peer 2:

( January 14th, 2015 6:20pm UTC )

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Unregistered wrote:

"Similarly, if nature is operating in a manner in which some properties are less likely to be detected at certain angles, it will be impossible to avoid the detection loophole. You could have a 100% efficient detectors and yet still not be able to detect 100% of the emitted particles. This is why I say it is not accurate to suggest, as Larsson does, that the problem is due to "deficiencies in the experimental equipment".

This makes no sense, since the very defintion of 100% efficient detectors is that they are able to detect 100% of the emitted particles. Could you elaborate?

"Similarly, if nature is operating in a manner in which some properties are less likely to be detected at certain angles, it will be impossible to avoid the detection loophole. You could have a 100% efficient detectors and yet still not be able to detect 100% of the emitted particles. This is why I say it is not accurate to suggest, as Larsson does, that the problem is due to "deficiencies in the experimental equipment".

This makes no sense, since the very defintion of 100% efficient detectors is that they are able to detect 100% of the emitted particles. Could you elaborate?

Donald A. Graft:

( January 14th, 2015 6:23pm UTC )

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One possibility: If the photons are lost along the way they do not reach the detectors. For example, they may be absorbed in a beamsplitter and none of their energy (or an insufficient amount) reaches the TES.

Unregistered Submission:

( January 14th, 2015 6:48pm UTC )

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Peer 1: ( January 14th, 2015 4:28pm UTC ) : "Please refer to Graft's paper analyzing Christensen et al discussed on another thread."

You mean Donald A. Graft: "Analysis of the Christensen et al.

Clauser-Horne (CH)-Inequality-Based

Test of Local Realism" ( http://arxiv.org/pdf/1409.5158v5.pdf ) ?

Its conclusions does not mention any theoretical results. It's just an analysis of an experiment. The paper notes that the result seems to violate quantum mechanics but does not claim that it were statistically significant. Maybe the discussion in that other thread will reveal more about this.

You mean Donald A. Graft: "Analysis of the Christensen et al.

Clauser-Horne (CH)-Inequality-Based

Test of Local Realism" ( http://arxiv.org/pdf/1409.5158v5.pdf ) ?

Its conclusions does not mention any theoretical results. It's just an analysis of an experiment. The paper notes that the result seems to violate quantum mechanics but does not claim that it were statistically significant. Maybe the discussion in that other thread will reveal more about this.

Unregistered Submission:

( January 14th, 2015 8:44pm UTC )

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Richard Gill: The basis for grouping "experimental units" is that they have something in common that causes the observations. In the case of particle pairs, we talk of shared or correlated hidden properties due to conservation of angular momentum. These are the lambdas over which Bell integrated. That is why you ague in your paper (and I agree) that the ensembles of lambda over which the expectation values are calculated must be identical for each term in the inequality. However, if you define lambda as "everything which causes the outcome in the time-slot", then of course that includes the particle lambdas within the cell. And if you select just the first particle pair, then of course it is possible to systematically create a differences in the ensemble. It doesn't matter what you call lambda, what matters is that the ensembles are identical. Without this, it is not possible to add the probabilities to obtain inequalities. Adding probabilities from different ensembles is a meaningless exercise.

Peer 2: As Peer 1 has already answered, 100% efficiency means the the detector registers a count for every particle which reaches it. But some particles may not reach the detectors due not to defects, but due to the fact that some combination of analyzer/beamsplitter or other settings and hidden parameters may naturally prevent the particle from reaching the detector (eg cross product of two co-linear vectors, or sign(0)).

Peer 2: As Peer 1 has already answered, 100% efficiency means the the detector registers a count for every particle which reaches it. But some particles may not reach the detectors due not to defects, but due to the fact that some combination of analyzer/beamsplitter or other settings and hidden parameters may naturally prevent the particle from reaching the detector (eg cross product of two co-linear vectors, or sign(0)).

Donald A. Graft:

( January 14th, 2015 9:05pm UTC )

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Section 4.5 of the Graft paper describes an artifactual violation due to post-selection (data discarding). All the theory we need is embodied in the data analysis, specifically here, Graft's new method to analyze a clocked EPRB experiment. When all the data is included, no violation is obtained, decisively confirming local realism, restoring a consistent axiom set for physics, and agreeing with our intuitions. There is extensive discussion on the paper's thread. Did I perhaps misunderstand what you asked for?

Peer 3:

( January 14th, 2015 9:34pm UTC )

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Richard Gill: "Discarding "events" within a time-slot and choosing just to pair the first event in each, which might not belong to the same emission, might (if QM is true) bias the observed correlation *downwards*. If local realism is true, it might conceivably bias the correlation in either direction, but it *cannot* bias it *upwards* any further than the local realist limit of 2."

Donald Graft's analysis shows the opposite. Discarding events biases the correlation upward to violate the inequality. Not discarding any events reduces the correlation downward to no violation. Besides, in every case, discarding results always enables the inequalities to be violated not the reverse you are suggesting. I think you are mistaken here. What is raised by discarding events, is the "genuine" local realistic bound so that even though a violation is observed for the base inequality, a violation of the "genuine" inequality is not possible.

Donald Graft's analysis shows the opposite. Discarding events biases the correlation upward to violate the inequality. Not discarding any events reduces the correlation downward to no violation. Besides, in every case, discarding results always enables the inequalities to be violated not the reverse you are suggesting. I think you are mistaken here. What is raised by discarding events, is the "genuine" local realistic bound so that even though a violation is observed for the base inequality, a violation of the "genuine" inequality is not possible.

Donald A. Graft:

( January 14th, 2015 9:37pm UTC )

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"Not discarding any events reduces the correlation downward to no violation."

Including all the data on an equal basis does not reduce or bias the correlation; it just gives you the correct result, the one we would expect from marginal sampling of a correlated joint distribution. I know what you are trying to say, and appreciate it, but if we are not precise, it can be misrepresented. Only the data discarding leads to bias.

The mysterians are back to the "formal proof" but haven't demonstrated it or offered any alternative explanation for the effect in the actual experiment discovered by Graft.

Including all the data on an equal basis does not reduce or bias the correlation; it just gives you the correct result, the one we would expect from marginal sampling of a correlated joint distribution. I know what you are trying to say, and appreciate it, but if we are not precise, it can be misrepresented. Only the data discarding leads to bias.

The mysterians are back to the "formal proof" but haven't demonstrated it or offered any alternative explanation for the effect in the actual experiment discovered by Graft.

Unregistered Submission:

( January 16th, 2015 12:14pm UTC )

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"Including all the data on an equal basis does not reduce or bias the correlation"

Peer 1 seems to say here that one gets the same result, whether events are discarded or not.

Later he contradicts himself: "Only the data discarding leads to bias".

However, his note on being precise is right and good.

Peer 1 seems to say here that one gets the same result, whether events are discarded or not.

Later he contradicts himself: "Only the data discarding leads to bias".

However, his note on being precise is right and good.

Donald A. Graft:

( January 16th, 2015 5:10pm UTC )

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"Peer 1 seems to say here that one gets the same result, whether events are discarded or not."

I see no contradiction and I don't say you get the same result. I am saying that including all the data gets the right answer (no violation), while discarding data produces a false violation by introducing bias. I hope it is clear.

I see no contradiction and I don't say you get the same result. I am saying that including all the data gets the right answer (no violation), while discarding data produces a false violation by introducing bias. I hope it is clear.

Unregistered Submission:

( January 17th, 2015 11:06am UTC )

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Including all data means including those detections that cannot be reliably paired or identifed as unpaired. Including mispaired events biases correlations towards zero. If events can be selectively discarded so that the remaining events can be more reliably paired the correlation is less biased. The number of events is smaller in the latter case but can be compensated by a longer experiment.

Peer 3:

( January 17th, 2015 5:22pm UTC )

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A local realistic model could perfectly arrange for some particles to not be detected at certain angles. In this case, unpaired particles is part of the model. By assuming all results should be paired, you are systematically biasing the results and the violations you obtain are meaningless. They simply point to a breakdown of your assumption that all particles must be detected in pairs, nothing mystical. Similarly, a local realistic model could perfectly arrange for time-delays to exist for certain particles at certain angles. By assuming particle pairs should be detected close together in time, you will systematically bias against particles which aren't close together in time.

Simply performing a longer experiment will not make undetectable particles detectable. Neither will it make time-delays disappear, it is the wrong-headed approach to the problem.

Simply performing a longer experiment will not make undetectable particles detectable. Neither will it make time-delays disappear, it is the wrong-headed approach to the problem.

Unregistered Submission:

( January 17th, 2015 6:40pm UTC )

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"In this case, unpaired particles is part of the model."

Unpaired particles must be included in the model anyway, for they are seen in experiments. But they are not the problem. The result is biased by mispaired detections.

Unpaired particles must be included in the model anyway, for they are seen in experiments. But they are not the problem. The result is biased by mispaired detections.

Peer 3:

( January 17th, 2015 11:38pm UTC )

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Definitely, mispaired detections is also a problem. Including only the first pair does nothing to ensure that the pairing is appropriate, and the length of the experiment does not fix that.

Unregistered Submission:

( January 18th, 2015 12:38pm UTC )

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"Including only the first pair does nothing to ensure that the pairing is appropriate"

While nothing can ensure that pairing is correct, pairing of first events has a greater probability to be correct than pairing events randomly.

While nothing can ensure that pairing is correct, pairing of first events has a greater probability to be correct than pairing events randomly.

Peer 3:

( January 18th, 2015 3:57pm UTC )

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I haven't seen any demonstration of that. Picking the first pair does not improve any probability of pairing more than any other method of picking pairs. The authors claim that picking the first pair **remedies** the coincidence time loophole. This is simply untrue. Now you are claiming it improves the probability of avoiding the coincidence time loophole. On what basis?

On the contrary, picking the first pair only, introduces systematic bias. In fact, for some models, picking the first pair may cause you to have 100% mismatched pairs, irrespective of the length of the experiment. At least picking randomly, does not bias a specific time-sequence so you will at least have some fraction of correctly matched pairs for longer experiments. Therefore picking randomly is much much better than picking the first pair.

On the contrary, picking the first pair only, introduces systematic bias. In fact, for some models, picking the first pair may cause you to have 100% mismatched pairs, irrespective of the length of the experiment. At least picking randomly, does not bias a specific time-sequence so you will at least have some fraction of correctly matched pairs for longer experiments. Therefore picking randomly is much much better than picking the first pair.

Unregistered Submission:

( January 19th, 2015 10:14am UTC )

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"for some models, picking the first pair may cause you to have 100% mismatched pairs"

For testing such models, a different experiment is needed. For example, bad synchronization of the clocks at the two measument sites can cause 100% mismatched pairs. Experimenters must avoid such errors as much as possible, and the analysis of the experiment must evaluate the effect of the remaining errors.

As mismatched pairs give uncorrelated results, they tend to bias the observed correlation towards zero. Therefore comparison of the predicted correlation of each model to the observed correlation gives a limit (different for each model) to mismatching. A model that gives a negative limit cannot be correct.

For testing such models, a different experiment is needed. For example, bad synchronization of the clocks at the two measument sites can cause 100% mismatched pairs. Experimenters must avoid such errors as much as possible, and the analysis of the experiment must evaluate the effect of the remaining errors.

As mismatched pairs give uncorrelated results, they tend to bias the observed correlation towards zero. Therefore comparison of the predicted correlation of each model to the observed correlation gives a limit (different for each model) to mismatching. A model that gives a negative limit cannot be correct.

Unregistered Submission:

( January 21st, 2015 2:07pm UTC )

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Peer 1: ( January 14th, 2015 3:43pm UTC ): ""Why do you suppose Bell introduced this idea in Bertlmann's socks and why do you suppose the experimentalists are adopting it nowadays?"

Because they were not aware that it could result in artifactual violation via data discarding, as Graft has conclusively demonstrated. However, when all the events are included, local reality is confirmed, just as we expect."

Graft's demonstrations are discussed in https://pubpeer.com/publications/E0F8384FC19A6034E86D516D03BB38 . Although the discussion hasn't produced any clear donclusions, it is already clear that Graft's demostrations are not conclusive. What is (so far) not clear is whether they are demonstrations.

Because they were not aware that it could result in artifactual violation via data discarding, as Graft has conclusively demonstrated. However, when all the events are included, local reality is confirmed, just as we expect."

Graft's demonstrations are discussed in https://pubpeer.com/publications/E0F8384FC19A6034E86D516D03BB38 . Although the discussion hasn't produced any clear donclusions, it is already clear that Graft's demostrations are not conclusive. What is (so far) not clear is whether they are demonstrations.

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Richard Gill:

( January 15th, 2015 8:27am UTC )

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Thanks Peer 1 and Peer 3 for further elucidating your positions. You both seem concerned with estimating the correlation between events corresponding to pairs of "particles" which are emitted as a pair. I am interested in testing local realism, taking the experimental unit to be a time-slot, which is the easiest way of all to avoid the detection loophole and the coincidence loophole. As Bell remarks we should avoid use of words like "particle" and "quantum", and focus instead on macroscopic experimental arrangements and macroscopic (real world) events.

You do not seem to bring up anything new here, and I see no refutation of my position (and I still see misunderstanding of it) so I will have to leave it at that.

You do not seem to bring up anything new here, and I see no refutation of my position (and I still see misunderstanding of it) so I will have to leave it at that.

Peer 3:

( January 15th, 2015 12:12pm UTC )

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My concern is that you can't claim to avoid the coincidence loophole if you do not make sure the ensembles of lambda are the same for each probability in the inequality. It does not matter whether experimental units are particles or timeslots, or macroscopic arrangements. You must make sure the ensembles are the same, in terms of the lambdas. There is no other basis for the inequality to be applicable if such a key assumption is not respected.

This is the premise of your paper above: "The problem here is that the ensemble on which the correlations are evaluated changes with the settings,while the original Bell inequality requires that they stay the same."

This is the premise of your paper above: "The problem here is that the ensemble on which the correlations are evaluated changes with the settings,while the original Bell inequality requires that they stay the same."

Richard Gill:

( January 16th, 2015 6:47am UTC )

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The ensembles of lambdas are the same for each probability in the inequality, if you take the experimental unit to be the time-slot and do not discard any time-slots.

lambda is "whatever determines what happens in the time-slot". If a heap of particles arrive in one time-slot, it is what determines what happens to all those particles.

lambda is "whatever determines what happens in the time-slot". If a heap of particles arrive in one time-slot, it is what determines what happens to all those particles.

Peer 3:

( January 16th, 2015 1:38pm UTC )

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Richard Gill: "The ensembles of lambdas are the same for each probability in the inequality, if you take the experimental unit to be the time-slot and do not discard any time-slots."

This is simply not true. Say particles arriving in Alice and Bob's corresponding timeslots (A100, B100) each have particle properties in the following order of time arrival:

A100: λ2,λ5,λ1,λ3,λ4

B100: λ1,λ2,λ4,λ5,λ3

If you consider just the first event in the timeslot, You now have the timeslot lambda λ[A100] = λ2, and λ[B100] = λ1. You end up with different ensembles for calculating the terms of the inequality. If instead, as you have suggested elsewhere, you pick the particle in the middle, you still end up with different ensembles. The only way to make sure you have the same ensembles is to use everything in the timeslot like Donald Graft did, or perform an experiment in which there is only one pair emission per timeslot, and the time slot is long-enough to guarantee that both particles (and only them) are detected within their time slots. Then picking the first particle in the timeslot is equivalent to picking all particles in the timeslot, because there is only one.

This is simply not true. Say particles arriving in Alice and Bob's corresponding timeslots (A100, B100) each have particle properties in the following order of time arrival:

A100: λ2,λ5,λ1,λ3,λ4

B100: λ1,λ2,λ4,λ5,λ3

If you consider just the first event in the timeslot, You now have the timeslot lambda λ[A100] = λ2, and λ[B100] = λ1. You end up with different ensembles for calculating the terms of the inequality. If instead, as you have suggested elsewhere, you pick the particle in the middle, you still end up with different ensembles. The only way to make sure you have the same ensembles is to use everything in the timeslot like Donald Graft did, or perform an experiment in which there is only one pair emission per timeslot, and the time slot is long-enough to guarantee that both particles (and only them) are detected within their time slots. Then picking the first particle in the timeslot is equivalent to picking all particles in the timeslot, because there is only one.

Richard Gill:

( January 17th, 2015 9:22am UTC )

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All hidden variables carried by all particles to the measurement stations together, together with all hidden variables already in the measurement stations, together, are *one* hidden variable lambda. You need to think at a higher level, Peer 3. The experimental unit is time-slot. The hidden variable lambda is *everything* (except the measurement settings actually chosen by Alice and Bob) which determine the delivered binary output.

Unregistered Submission:

( January 24th, 2015 12:45pm UTC )

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The correct statistical terminology is "sampling unit", instead of "experimental unit".

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Peer 3:

( January 17th, 2015 2:49pm UTC )

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Richard Gill:

λ[A100] = *everything* (except settings) which causes the outcome *considered* for timeslot 100 at Alice's arm of the experiment.

λ[B100] = *everything* (except settings) which causes the outcome *considered* for timeslot 100 at Bob's arm of the experiment.

Each term of the inequality is obtained by integrating over all the lambdas for all the timeslots. λ[A100] is just the 100th one on Alice's side. The integration has a single lambda term attributed to both sides for each experimental unit, that is λn =(λ[An], λ[Bn]). The integration is over the ensembles of all such lambdas Λ, which produced all the binary outcomes *considered for that term* in the inequality. These ensembles must be the same for *every term*, for the inequality to be applicable as you argue in the above paper.

Your suggested method of picking pairs, introduces angle dependent differences between the lambdas as follows.

Say for setting pair "a" and "b" we have the following time sequence of "particle lambdas".

a:A: pλ2,pλ5,pλ1,pλ3,pλ4

b:B: pλ1,pλ2,pλ4,pλ5,pλ3

And for setting pair "c" and "d" we have the following time sequence of "particle lambdas".

c:A: pλ4,pλ3,pλ2,pλ1,pλ5

d:B: pλ3,pλ2,pλ1,pλ5,pλ4

By pickin just the first particle pair, the ensemble Λab, is now going to be different from the ensemble Λcd. The inequalities require that they are the same.

Both you and Jan-Ake Larsson are wrong to suggest that picking the first pair avoids the coincidence time loophole. It absolutely does not. On the contrary, considering all the pairs is much better.

λ[A100] = *everything* (except settings) which causes the outcome *considered* for timeslot 100 at Alice's arm of the experiment.

λ[B100] = *everything* (except settings) which causes the outcome *considered* for timeslot 100 at Bob's arm of the experiment.

Each term of the inequality is obtained by integrating over all the lambdas for all the timeslots. λ[A100] is just the 100th one on Alice's side. The integration has a single lambda term attributed to both sides for each experimental unit, that is λn =(λ[An], λ[Bn]). The integration is over the ensembles of all such lambdas Λ, which produced all the binary outcomes *considered for that term* in the inequality. These ensembles must be the same for *every term*, for the inequality to be applicable as you argue in the above paper.

Your suggested method of picking pairs, introduces angle dependent differences between the lambdas as follows.

Say for setting pair "a" and "b" we have the following time sequence of "particle lambdas".

a:A: pλ2,pλ5,pλ1,pλ3,pλ4

b:B: pλ1,pλ2,pλ4,pλ5,pλ3

And for setting pair "c" and "d" we have the following time sequence of "particle lambdas".

c:A: pλ4,pλ3,pλ2,pλ1,pλ5

d:B: pλ3,pλ2,pλ1,pλ5,pλ4

By pickin just the first particle pair, the ensemble Λab, is now going to be different from the ensemble Λcd. The inequalities require that they are the same.

Both you and Jan-Ake Larsson are wrong to suggest that picking the first pair avoids the coincidence time loophole. It absolutely does not. On the contrary, considering all the pairs is much better.

Peer 2:

( January 17th, 2015 4:11pm UTC )

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Peer 3 wrote:

"Your suggested method of picking pairs, introduces angle dependent differences between the lambdas as follows.

Say for setting pair "a" and "b" we have the following time sequence of "particle lambdas".

a:A: pλ2,pλ5,pλ1,pλ3,pλ4

b:B: pλ1,pλ2,pλ4,pλ5,pλ3

And for setting pair "c" and "d" we have the following time sequence of "particle lambdas".

c:A: pλ4,pλ3,pλ2,pλ1,pλ5

d:B: pλ3,pλ2,pλ1,pλ5,pλ4"

---

But for a local realistic model, the sequence of hidden variable "particle lambdas" cannot depend on angle settings, can it? The point with a local hidden variable model is that the hidden variable is independent of detector settings, since the particle can be emitted before the detector decision must be made in the experiment.

What you are invoking here is the locality loophole, and Larsson et. al. explicitly said they did not close that in their experiment.

"Your suggested method of picking pairs, introduces angle dependent differences between the lambdas as follows.

Say for setting pair "a" and "b" we have the following time sequence of "particle lambdas".

a:A: pλ2,pλ5,pλ1,pλ3,pλ4

b:B: pλ1,pλ2,pλ4,pλ5,pλ3

And for setting pair "c" and "d" we have the following time sequence of "particle lambdas".

c:A: pλ4,pλ3,pλ2,pλ1,pλ5

d:B: pλ3,pλ2,pλ1,pλ5,pλ4"

---

But for a local realistic model, the sequence of hidden variable "particle lambdas" cannot depend on angle settings, can it? The point with a local hidden variable model is that the hidden variable is independent of detector settings, since the particle can be emitted before the detector decision must be made in the experiment.

What you are invoking here is the locality loophole, and Larsson et. al. explicitly said they did not close that in their experiment.

Peer 3:

( January 17th, 2015 5:08pm UTC )

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Peer2: "But for a local realistic model, the sequence of hidden variable "particle lambdas" cannot depend on angle settings, can it?"

Yes they can! The time delays on each side is entirely based on the combination of particle lambdas and angle setting in a completely local realistic manner. The time delay for detecting a particle on each arm is separate local realistic function of two variables T_A(pLambda, a), T_B(pLambda, b), therefore time difference between both particles depends on both settings in a completely local realistic manner. This is what the coincidence time loophole is all about, there are many examples of simulations showing precisely that, Gill and Larsson present an example in the article above, in which they argue that:

"The problem here is that the ensemble on which the correlations are evaluated changes

with the settings, while the original Bell inequality requires that they stay the same."

If you do not understand how the sequence of hidden variables can depend on the angle settings for coincidence experiments, please read the article (http://arxiv.org/pdf/quant-ph/0312035v2.pdf), this discussion is about the fact that the method suggested by the authors for avoiding the "problem", does not avoid the problem at all. It has nothing to do with the locality loophole.

Yes they can! The time delays on each side is entirely based on the combination of particle lambdas and angle setting in a completely local realistic manner. The time delay for detecting a particle on each arm is separate local realistic function of two variables T_A(pLambda, a), T_B(pLambda, b), therefore time difference between both particles depends on both settings in a completely local realistic manner. This is what the coincidence time loophole is all about, there are many examples of simulations showing precisely that, Gill and Larsson present an example in the article above, in which they argue that:

"The problem here is that the ensemble on which the correlations are evaluated changes

with the settings, while the original Bell inequality requires that they stay the same."

If you do not understand how the sequence of hidden variables can depend on the angle settings for coincidence experiments, please read the article (http://arxiv.org/pdf/quant-ph/0312035v2.pdf), this discussion is about the fact that the method suggested by the authors for avoiding the "problem", does not avoid the problem at all. It has nothing to do with the locality loophole.

Peer 2:

( January 17th, 2015 5:45pm UTC )

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Yes, I agree (i.e., I was mistaken). A local realist model can in fact change the sequence of the "particle lambdas" upon encountering the detector.

But: The point is that this must be done based on the local detector setting *only*, and can not be influenced by the detector setting in the other wing. So we can regard the set of "particle lambdas" in a time slot as just a vector, and this vector is the "real" lambda, and one and the same "real" lambda must always give the same result for a particular setting in the local wing, independent of the setting in the other wing. And that is all it takes for Bell's theorem to hold.

But: The point is that this must be done based on the local detector setting *only*, and can not be influenced by the detector setting in the other wing. So we can regard the set of "particle lambdas" in a time slot as just a vector, and this vector is the "real" lambda, and one and the same "real" lambda must always give the same result for a particular setting in the local wing, independent of the setting in the other wing. And that is all it takes for Bell's theorem to hold.

Peer 3:

( January 17th, 2015 6:56pm UTC )

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Peer 2: "But: The point is that this must be done based on the local detector setting *only*, and can not be influenced by the detector setting in the other wing."

No question about that. The local realistic model is not the one that introduces the dependency on both settings. It is the data analysis (post-processing) that introduces the dependency on both settings, by picking just the first pair of particles, or by assuming that particle pairs must be detected close together in time. This is why using all the particles in each time slot is much better than using just the first pair as the authors suggest.

Peer 2: "So we can regard the set of "particle lambdas" in a time slot as just a vector, and this vector is the "real" lambda."

Yes you should (as I explained in my previous post), but this is not what the authors recommend. They recommend, picking just the first pair. I thought this was clearly explained in my previous posts.

Peer 2: "and one and the same "real" lambda must always give the same result for a particular setting in the local wing, independent of the setting in the other wing. And that is all it takes for Bell's theorem[sic] to hold."

Bell's inequalities only hold if the ensemble of lambdas used to calculate each term of the inequality is the same. That each "real" lambda only produces a specific result for a specific outcome is uncontroversial. The "problem" which the authors clearly identify, is that the ensemble is different for each term, so the inequality does not apply. The suggested method of picking just the first pair in each time-slot does nothing to solve the problem, it compounds it.

No question about that. The local realistic model is not the one that introduces the dependency on both settings. It is the data analysis (post-processing) that introduces the dependency on both settings, by picking just the first pair of particles, or by assuming that particle pairs must be detected close together in time. This is why using all the particles in each time slot is much better than using just the first pair as the authors suggest.

Peer 2: "So we can regard the set of "particle lambdas" in a time slot as just a vector, and this vector is the "real" lambda."

Yes you should (as I explained in my previous post), but this is not what the authors recommend. They recommend, picking just the first pair. I thought this was clearly explained in my previous posts.

Peer 2: "and one and the same "real" lambda must always give the same result for a particular setting in the local wing, independent of the setting in the other wing. And that is all it takes for Bell's theorem[sic] to hold."

Bell's inequalities only hold if the ensemble of lambdas used to calculate each term of the inequality is the same. That each "real" lambda only produces a specific result for a specific outcome is uncontroversial. The "problem" which the authors clearly identify, is that the ensemble is different for each term, so the inequality does not apply. The suggested method of picking just the first pair in each time-slot does nothing to solve the problem, it compounds it.

Peer 2:

( January 17th, 2015 7:18pm UTC )

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Peer 3 wrote:

"No question about that. The local realistic model is not the one that introduces the dependency on both settings. It is the data analysis (post-processing) that introduces the dependency on both settings, by picking just the first pair of particles, or by assuming that particle pairs must be detected close together in time."

But, and this is the crux of the matter, how can picking just the first particle in a predetermined time slot, which is a completely local operation, introduce a dependency on both settings? And in this experiment, there is no other selection than that.

"No question about that. The local realistic model is not the one that introduces the dependency on both settings. It is the data analysis (post-processing) that introduces the dependency on both settings, by picking just the first pair of particles, or by assuming that particle pairs must be detected close together in time."

But, and this is the crux of the matter, how can picking just the first particle in a predetermined time slot, which is a completely local operation, introduce a dependency on both settings? And in this experiment, there is no other selection than that.

Peer 3:

( January 17th, 2015 10:04pm UTC )

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Peer2: "But, and this is the crux of the matter, how can picking just the first particle in a predetermined time slot, which is a completely local operation, introduce a dependency on both settings? And in this experiment, there is no other selection than that."

Please read the comment I posted which your comment is a response to. The one posted at (January 17th, 2015 2:49pm UTC) which starts as follows:

"Richard Gill:

λ[A100] = *everything* (except settings) which causes the outcome *considered* for timeslot 100 at Alice's arm of the experiment. ..."

Your questions are answered in that post.

Please read the comment I posted which your comment is a response to. The one posted at (January 17th, 2015 2:49pm UTC) which starts as follows:

"Richard Gill:

λ[A100] = *everything* (except settings) which causes the outcome *considered* for timeslot 100 at Alice's arm of the experiment. ..."

Your questions are answered in that post.

Peer 3:

( January 17th, 2015 10:27pm UTC )

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Here is the explanation again, in case you still do not understand:

Say for setting pair "a", "b", "c", "d", we have the following time sequence of "particle lambdas".

a:A: pλ2,pλ5,pλ1,pλ3,pλ4

b:B: pλ1,pλ2,pλ4,pλ5,pλ3

c:A: pλ4,pλ3,pλ2,pλ1,pλ5

d:B: pλ3,pλ2,pλ1,pλ5,pλ4

Which means every time pλ4 sees setting "a", it is delayed more than pλ3 for the same setting, etc. Where Alice (A) measures for settings "a", and "c" randomly, and Bob (B) measures settings "b" and "d".

The ensembles used to calculate the 4 paired expectation values are

Λab, Λad, Λcb, Λcd, where Λab is the ensemble of lambdas **considered** for calculating the Eab = E(a,b|Λab) expectation value, etc

By selecting only the first pair of particles each time, we now have

Λab = (pλ1,pλ2)

Λad = (pλ2,pλ3)

Λcb = (pλ1,pλ4)

Λcd = (pλ3,pλ4)

I hope you now can see how selecting just the first pair of particles causes the ensembles to be different, and how the ensembles depend on both settings on both sides of the experiment.

Say for setting pair "a", "b", "c", "d", we have the following time sequence of "particle lambdas".

a:A: pλ2,pλ5,pλ1,pλ3,pλ4

b:B: pλ1,pλ2,pλ4,pλ5,pλ3

c:A: pλ4,pλ3,pλ2,pλ1,pλ5

d:B: pλ3,pλ2,pλ1,pλ5,pλ4

Which means every time pλ4 sees setting "a", it is delayed more than pλ3 for the same setting, etc. Where Alice (A) measures for settings "a", and "c" randomly, and Bob (B) measures settings "b" and "d".

The ensembles used to calculate the 4 paired expectation values are

Λab, Λad, Λcb, Λcd, where Λab is the ensemble of lambdas **considered** for calculating the Eab = E(a,b|Λab) expectation value, etc

By selecting only the first pair of particles each time, we now have

Λab = (pλ1,pλ2)

Λad = (pλ2,pλ3)

Λcb = (pλ1,pλ4)

Λcd = (pλ3,pλ4)

I hope you now can see how selecting just the first pair of particles causes the ensembles to be different, and how the ensembles depend on both settings on both sides of the experiment.

Peer 2:

( January 17th, 2015 11:37pm UTC )

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No, your example shows that

Λab = (pλ2,pλ1)

Λad = (pλ2,pλ3)

Λcb = (pλ4,pλ1)

Λcd = (pλ4,pλ3)

Λab = (pλ2,pλ1)

Λad = (pλ2,pλ3)

Λcb = (pλ4,pλ1)

Λcd = (pλ4,pλ3)

Peer 3:

( January 17th, 2015 11:44pm UTC )

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In my notation, Λab = (pλ1,pλ 2) = (pλ2,pλ1) is a set not a vector. So I'll suggest you attempt to understand rather than say "No", based on a misunderstanding. Do you now see how selecting just the first pair of particles causes the ensembles to be different, and how the ensembles depend on both settings?

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Peer 2:

( January 18th, 2015 12:27am UTC )

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Ok, I guess by "set" in this context you mean that ordering is insignificant. So this means that in your opinion, my four pairs in the previous post is equivalent to the four pairs in your post just prior to that?

Richard Gill:

( January 18th, 2015 6:42am UTC )

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"Picking the first particle" is not a selection. It is merely part of the definition of A(lambda, a). The "hidden variable" lambda is *everything* that goes into determining the measurement outcome in the time-slot on Alice's side. Particles, measurement device. A is the binary outcome which the experimenter, who built the measurement apparatus, has engineered to be "the" outcome. By the measurement device I mean also the local processing of a more complicated output, as long as it might just as well have been put "inside" the measurement black-box.

If the experimenter, for instance, wants to take the event nearest to the mid-point of the time slot as determining the binary outcome for the time-slot, then that is built into the definition of the function A.

In a CHSH type experiment there are two different settings in each wing of the experiment. A mathematical-physical theory is called local realist if it allows the mathematical existence of A1, A2, B1, B2 where: A1 is what Alice's outcome would have be if she chooses her setting number 1 (it doesn't depend on which setting is chosen by Bob), A2 is what Alice's outcome would be if she chooses her setting number 2 (it doesn't depend on which setting is chosen by Bob), B1 is what Bob's outcome would be if he chooses his setting number 1 (it doesn't depend on which setting is chosen by Alice), B2 is what Bob's outcome would be if he chooses his setting number 2 (it doesn't depend on which setting is chosen by Alice). In a good experiment, the actual settings are chosen by independent fair coin tosses (i.e., independent of the quadruple A1, A2, B1, B2). That this is possible, is called the no-conspiracy assumption.

As far as a CHSH type experiment is concerned, therefore, a local hidden variable theory is effectively a joint probability distribution of four binary random variables A1, A2, B1, B2. There are only 2^4 = 16 possible joint outcomes. Each of those 16 corresponds to a *deterministic* local realist model: the outcomes of both measurements on both particles are fixed in advance, and the outcomes on Alice's side don't depend on which measurement is actually chosen on Bob's side, and vice versa.

One may as well reduce lambda to a randomly chosen point in a finite set of 16 elements (not necessarily uniformly distributed!). The set of all possible local realist models corresponds to the set of all possible probability distributions over the set of 16 possible values of this set of four binary variables.

If the experimenter, for instance, wants to take the event nearest to the mid-point of the time slot as determining the binary outcome for the time-slot, then that is built into the definition of the function A.

In a CHSH type experiment there are two different settings in each wing of the experiment. A mathematical-physical theory is called local realist if it allows the mathematical existence of A1, A2, B1, B2 where: A1 is what Alice's outcome would have be if she chooses her setting number 1 (it doesn't depend on which setting is chosen by Bob), A2 is what Alice's outcome would be if she chooses her setting number 2 (it doesn't depend on which setting is chosen by Bob), B1 is what Bob's outcome would be if he chooses his setting number 1 (it doesn't depend on which setting is chosen by Alice), B2 is what Bob's outcome would be if he chooses his setting number 2 (it doesn't depend on which setting is chosen by Alice). In a good experiment, the actual settings are chosen by independent fair coin tosses (i.e., independent of the quadruple A1, A2, B1, B2). That this is possible, is called the no-conspiracy assumption.

As far as a CHSH type experiment is concerned, therefore, a local hidden variable theory is effectively a joint probability distribution of four binary random variables A1, A2, B1, B2. There are only 2^4 = 16 possible joint outcomes. Each of those 16 corresponds to a *deterministic* local realist model: the outcomes of both measurements on both particles are fixed in advance, and the outcomes on Alice's side don't depend on which measurement is actually chosen on Bob's side, and vice versa.

One may as well reduce lambda to a randomly chosen point in a finite set of 16 elements (not necessarily uniformly distributed!). The set of all possible local realist models corresponds to the set of all possible probability distributions over the set of 16 possible values of this set of four binary variables.

Peer 3:

( January 18th, 2015 1:56pm UTC )

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Obviously, none of this addresses the points raised. Please address the specific points. I've presented one possible model which is fully local realistic , and for which your suggested selection methods generate different ensembles.

It is a fact that "the problem here is that the ensemble on which the correlations are evaluated changes with the settings, while the original Bell inequality requires that they stay the same." You wrote this, and I agree with it 100%. Have you changed your mind?

We don't need to get distracted by "sets if all possible local realistic models" when only one surfices. I've one which invalidates your selection methods. One is enough. I would appreciate your response as to why your selection methods do not introduce different ensembles for the model presented.

It is a fact that "the problem here is that the ensemble on which the correlations are evaluated changes with the settings, while the original Bell inequality requires that they stay the same." You wrote this, and I agree with it 100%. Have you changed your mind?

We don't need to get distracted by "sets if all possible local realistic models" when only one surfices. I've one which invalidates your selection methods. One is enough. I would appreciate your response as to why your selection methods do not introduce different ensembles for the model presented.

Peer 3:

( January 18th, 2015 3:50pm UTC )

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Richard Gill: "In a CHSH type experiment there are two different settings in each wing of the experiment.

...

As far as a CHSH type experiment is concerned, therefore, a local hidden variable theory is effectively a joint probability distribution of four binary random variables A1, A2, B1, B2. There are only 2^4 = 16 possible joint outcomes."

This is a naive view of how the experiments are performed and probably contributes to the misunderstanding. Here are some facts about a local realistic model of a typical CHSH experiment:

1) No experimental unit is ever used more than once for calculating a correlation.

2) There are 2^8 possible joint outcomes for the CHSH inequality, not 2^4 as you suggested.

Specifically, for four angles "a,b,c,d" we have the 8 outcomes (Aab, Acb, Aad, Acd, Bab, Bad, Bcb, Bcd), each of those can take a value of +1 or -1 independently. Aab, simply means the outcome registered at Alice's side when she had setting "a" active, and Bob had setting "b" active, the relevant ensemble of lambdas here is Λab , representing the set of all lambdas considered when Alice had "a" active and Bob had "b" active . Again, and this is the crucial fact. Those 8 terms are not measured on the same experimental units. They are obtained in pairs. [Aab, Bab], [Acb, Bcb], [Aad, Bad], [Acd, Bcd]. They start with a single run in which they randomly switch settings on both arms of the experiments, they then select the subset for which setting pairs "a" and "b" were active, those outcomes are [Aab, Bab] with corresponding ensemble of lambdas Λab, they do the same for a total of 4 pairs of lists of outcomes, and corresponding lambdas Λab, Λcb, Λad, Λcd. Then they use those ensembles to calculate expectation values Eab, Ead, Ecb, Ecd.

These are the 4 ensembles which MUST be the same for the inequalities to apply, which you talk about in your paper. This means, you must sample those ensembles in a way that does not introduce angle dependent biases. I've presented an example of a local realistic model above for which all your suggested methods of picking pairs introduces angle dependent biases.

I'll appreciate a direct response to that.

I'm also interested in what Larsson has to say about this since he prescribed the same methods in a related paper and appear to have collaborated with the authors on a recently famous Bell-test experiment in which the authors followed exactly your recommendations about picking pairs. I believe the discussion here calls into question the analysis methods they used, and ultimately the conclusion they made.

...

As far as a CHSH type experiment is concerned, therefore, a local hidden variable theory is effectively a joint probability distribution of four binary random variables A1, A2, B1, B2. There are only 2^4 = 16 possible joint outcomes."

This is a naive view of how the experiments are performed and probably contributes to the misunderstanding. Here are some facts about a local realistic model of a typical CHSH experiment:

1) No experimental unit is ever used more than once for calculating a correlation.

2) There are 2^8 possible joint outcomes for the CHSH inequality, not 2^4 as you suggested.

Specifically, for four angles "a,b,c,d" we have the 8 outcomes (Aab, Acb, Aad, Acd, Bab, Bad, Bcb, Bcd), each of those can take a value of +1 or -1 independently. Aab, simply means the outcome registered at Alice's side when she had setting "a" active, and Bob had setting "b" active, the relevant ensemble of lambdas here is Λab , representing the set of all lambdas considered when Alice had "a" active and Bob had "b" active . Again, and this is the crucial fact. Those 8 terms are not measured on the same experimental units. They are obtained in pairs. [Aab, Bab], [Acb, Bcb], [Aad, Bad], [Acd, Bcd]. They start with a single run in which they randomly switch settings on both arms of the experiments, they then select the subset for which setting pairs "a" and "b" were active, those outcomes are [Aab, Bab] with corresponding ensemble of lambdas Λab, they do the same for a total of 4 pairs of lists of outcomes, and corresponding lambdas Λab, Λcb, Λad, Λcd. Then they use those ensembles to calculate expectation values Eab, Ead, Ecb, Ecd.

These are the 4 ensembles which MUST be the same for the inequalities to apply, which you talk about in your paper. This means, you must sample those ensembles in a way that does not introduce angle dependent biases. I've presented an example of a local realistic model above for which all your suggested methods of picking pairs introduces angle dependent biases.

I'll appreciate a direct response to that.

I'm also interested in what Larsson has to say about this since he prescribed the same methods in a related paper and appear to have collaborated with the authors on a recently famous Bell-test experiment in which the authors followed exactly your recommendations about picking pairs. I believe the discussion here calls into question the analysis methods they used, and ultimately the conclusion they made.

Richard Gill:

( January 19th, 2015 11:13am UTC )

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Peer 3, I think that your view is naive and mine is sophisticated!

CHSH can be derived without assuming any sampling from ensembles of values of hidden variables lambda. The only randomness needed to derive the inequality is that in the repeated random choices of settings. See my http://arxiv.org/abs/1207.5103

My definition of local realism follows that of Tsirelson in his essay on Citizendium: http://en.citizendium.org/wiki/Entanglement_%28physics%29

CHSH can be derived without assuming any sampling from ensembles of values of hidden variables lambda. The only randomness needed to derive the inequality is that in the repeated random choices of settings. See my http://arxiv.org/abs/1207.5103

My definition of local realism follows that of Tsirelson in his essay on Citizendium: http://en.citizendium.org/wiki/Entanglement_%28physics%29

Peer 3:

( January 19th, 2015 12:03pm UTC )

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Could you please address the specific issues raised, or if unable, ask your coauthor to do so. Otherwise it will be an admission that your recommendations are flawed. Experimentalists have relied on flawed recommendations!

Richard Gill:

( January 19th, 2015 12:43pm UTC )

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Peer 3, I am sorry if you have not understood how my answers do take care of all the issues you have raised. Please study the first sections of my paper http://arxiv.org/abs/1207.5103 carefully, take note of Tsirelson's definition of local realism, and study the text (about three paragraphs, at most two pages) around Figure 7 of Bell's paper on Bertlmann's socks. A discussion at a blackboard could be helpful in locating the mental block. I don't think it is going to happen here.

I already pointed out the existence of this PubPeer thread to my co-author. There is also a button on the right-hand side-bar "Invite others to the conversation". You could yourself also send him a personal email.

For me, as for Bell, the experimental unit is a time-slot. lambda stands for all the stuff inside the black-box of Bell's Figure 7. A(a, lambda) is a completely arbitrary binary outcome on Alice's side, depending on the setting "a" which is put into the black box by Alice, and on the "stuff" inside; the function A might include some local computer post-processing. So this can include a local, computer-driven selection of one of several "physical" binary outcomes registered during the time-slot. Similarly B(b, lambda). Since there are only two settings on each side, all we have to deal with are A_1 = A(a_1, lambda), A_2 = A(a_2, lambda), B_1 = B(b_1, lambda), B_2 = B(b_2, lambda). The flips of two coins determine whether the system outputs A_1 or A_2 on Alice's side, and B_1 or B_2 on the other side. There is no need to say anything more about what lambda is or where it resides (particles, system, measurement devices, local post-processing, source ...).

Since we can derive CHSH by only assuming local realism and that the measurement *settings* are repeatedly chosen by fair coin tosses, there is no need to make any assumptions about lambda being drawn repeatedly from the same ensemble. The key assumption is local realism, which entails the simultaneous (mathematical) existence of both outcomes of both possible measurements on both sides of the experiment, together with the no-conspiracy (or "freedom") assumption, which entails that we can freely choose which measurements to perform.

The detection loophole occurs when A(a, lambda) is ternary, not binary, with a third outcome standing for "no detection". Discarding experimental units for which there is no outcome on one side or the other leads us to look at conditional correlations given a detection on both sides. The proof of CHSH doesn't go through. You could say that the freedom assumption has been broken: one is not free to measure with either setting. One of the settings can prevent a measurement altogether.

One can however adapt CHSH in various ways but the result is that local realism has got more scope: at a sufficiently low detection rate it can easily "imitate" quantum correlations.

The coincide loophole occurs when there are two outcomes on each side: a binary measurement outcome together with a time-stamp (a detection time). Discarding experimental units for which the difference between the two time-stamps exceeds some threshold (called the coincidence window) gives local realism even more scope to imitate quantum correlations. Again, what has happened is that the "freedom" assumption has been broken: changing the setting might change whether or not we get a "coincidence".

My short answer is: forget your fixation on *particles*. Forget old-fashioned derivations of Bell's inequality - we have much better ones today. We have a stronger Bell theorem: less assumptions *and* stronger conclusions. The experimental unit is a time-slot. Read Bertlmann's socks carefully. Understand a modern proof of CHSH, without detection or coincidence loophole, first. Understand the modern proof of CHSH when the only assumed randomness is the randomness of the detector settings.

I already pointed out the existence of this PubPeer thread to my co-author. There is also a button on the right-hand side-bar "Invite others to the conversation". You could yourself also send him a personal email.

For me, as for Bell, the experimental unit is a time-slot. lambda stands for all the stuff inside the black-box of Bell's Figure 7. A(a, lambda) is a completely arbitrary binary outcome on Alice's side, depending on the setting "a" which is put into the black box by Alice, and on the "stuff" inside; the function A might include some local computer post-processing. So this can include a local, computer-driven selection of one of several "physical" binary outcomes registered during the time-slot. Similarly B(b, lambda). Since there are only two settings on each side, all we have to deal with are A_1 = A(a_1, lambda), A_2 = A(a_2, lambda), B_1 = B(b_1, lambda), B_2 = B(b_2, lambda). The flips of two coins determine whether the system outputs A_1 or A_2 on Alice's side, and B_1 or B_2 on the other side. There is no need to say anything more about what lambda is or where it resides (particles, system, measurement devices, local post-processing, source ...).

Since we can derive CHSH by only assuming local realism and that the measurement *settings* are repeatedly chosen by fair coin tosses, there is no need to make any assumptions about lambda being drawn repeatedly from the same ensemble. The key assumption is local realism, which entails the simultaneous (mathematical) existence of both outcomes of both possible measurements on both sides of the experiment, together with the no-conspiracy (or "freedom") assumption, which entails that we can freely choose which measurements to perform.

The detection loophole occurs when A(a, lambda) is ternary, not binary, with a third outcome standing for "no detection". Discarding experimental units for which there is no outcome on one side or the other leads us to look at conditional correlations given a detection on both sides. The proof of CHSH doesn't go through. You could say that the freedom assumption has been broken: one is not free to measure with either setting. One of the settings can prevent a measurement altogether.

One can however adapt CHSH in various ways but the result is that local realism has got more scope: at a sufficiently low detection rate it can easily "imitate" quantum correlations.

The coincide loophole occurs when there are two outcomes on each side: a binary measurement outcome together with a time-stamp (a detection time). Discarding experimental units for which the difference between the two time-stamps exceeds some threshold (called the coincidence window) gives local realism even more scope to imitate quantum correlations. Again, what has happened is that the "freedom" assumption has been broken: changing the setting might change whether or not we get a "coincidence".

My short answer is: forget your fixation on *particles*. Forget old-fashioned derivations of Bell's inequality - we have much better ones today. We have a stronger Bell theorem: less assumptions *and* stronger conclusions. The experimental unit is a time-slot. Read Bertlmann's socks carefully. Understand a modern proof of CHSH, without detection or coincidence loophole, first. Understand the modern proof of CHSH when the only assumed randomness is the randomness of the detector settings.

Unregistered Submission:

( January 19th, 2015 1:18pm UTC )

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"Understand a modern proof"

So you accept that your proof in this discussed paper is wrong? It appears that way because you decline to answer questions about the paper and now point to a "modern proof".

So you accept that your proof in this discussed paper is wrong? It appears that way because you decline to answer questions about the paper and now point to a "modern proof".

Peer 3:

( January 19th, 2015 1:31pm UTC )

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I'm hoping that Dr Larsson is reading this thread. I would appreciate his response to the issues I have raised, because the other author is either unwilling or unable to.

Richard Gill: "For me, as for Bell, the experimental unit is a time-slot."

See my response from ( January 17th, 2015 2:49pm UTC ) explaining why this is not a valid counter-argument to the point I'm making. I notice you haven't responded to it.

Richard Gill: "Since we can derive CHSH by only assuming local realism and that the measurement *settings* are repeatedly chosen by fair coin tosses, there is no need to make any assumptions about lambda being drawn repeatedly from the same ensemble."

In the above article, you state **"The problem here is that the ensemble on which the correlations are evaluated changes with the settings, while the original Bell inequality requires that they stay the same"** Have you changed your mind? If not, then how can you also believe that the sampling that produces the ensembles is irrelevant?

I have presented a specific example of a model which has the following features:

1) "time-slots" are the experimental unit, in the same way as you mean it (or should mean it)

2) "time-slot-lambda", corresponds to everything that contributes to the outcome considered for the time-slot including the particles, except for the settings.

3) specific particle-lambdas have locally determined time-delays for being detected

4) selecting the first particle pair in the time-slots, leads to different ensembles of "time-slot-lambda" contrary to your claim that it solves the coincidence time loophole.

Richard Gill: "For me, as for Bell, the experimental unit is a time-slot."

See my response from ( January 17th, 2015 2:49pm UTC ) explaining why this is not a valid counter-argument to the point I'm making. I notice you haven't responded to it.

Richard Gill: "Since we can derive CHSH by only assuming local realism and that the measurement *settings* are repeatedly chosen by fair coin tosses, there is no need to make any assumptions about lambda being drawn repeatedly from the same ensemble."

In the above article, you state **"The problem here is that the ensemble on which the correlations are evaluated changes with the settings, while the original Bell inequality requires that they stay the same"** Have you changed your mind? If not, then how can you also believe that the sampling that produces the ensembles is irrelevant?

I have presented a specific example of a model which has the following features:

1) "time-slots" are the experimental unit, in the same way as you mean it (or should mean it)

2) "time-slot-lambda", corresponds to everything that contributes to the outcome considered for the time-slot including the particles, except for the settings.

3) specific particle-lambdas have locally determined time-delays for being detected

4) selecting the first particle pair in the time-slots, leads to different ensembles of "time-slot-lambda" contrary to your claim that it solves the coincidence time loophole.

Unregistered Submission:

( January 19th, 2015 1:46pm UTC )

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Peer 3: ( January 19th, 2015 1:31pm UTC ): "I'm hoping that Dr Larsson is reading this thread. I would appreciate his response to the issues I have raised, because the other author is not being rational."

Reading the discussion so far reveals that Dr Gill's responses are rational and issues raised are properly answered.

Reading the discussion so far reveals that Dr Gill's responses are rational and issues raised are properly answered.

Peer 3:

( January 19th, 2015 2:30pm UTC )

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Richard Gill: "Since there are only two settings on each side, all we have to deal with are A_1 = A(a_1, lambda), A_2 = A(a_2, lambda), B_1 = B(b_1, lambda), B_2 = B(b_2, lambda). The flips of two coins determine whether the system outputs A_1 or A_2 on Alice's side, and B_1 or B_2 on the other side."

Again, I've explained previously why this is a naive view of what is being done in the experiment. There are 2^8 possible outcome combinations not 2^4. In fact, your A_1 can be considered a function of lambda as follows:

A(a_1, lambda) = A1(lambda)

B(b_1, lambda) = B1(lambda)

A(a_2, lambda) = A2(lambda)

B(b_2, lambda) = B2(lambda)

But lambda is completely unknown. If I make 8 measurements of random variables, I get 8 possible outcome combinations not 4, unless I can control lambda and make sure they are the same. In CHSH tests, 8 measurements are made (ie, 4 pairs), not 4 singles, therefore there are 2^8 possible outcome combinations. If at any point in your derivation you used the relationship A1B1 + A1B2 + A2B1 - A2B2 = A1(B1 + B2) + A2(B1-B2) <= 2, then you have assumed that the value of lambda was the same for every paired term in the expression. In that case, and only in that case, do you have 2^4 outcome combinations. Effectively, it amounts to assuming the ensembles generated by sampling are the same. Every derivation of Bell-like inequalities relies on relationships of that sort. This is why I agree with you when you argue in your paper that the ensembles must be the same for the inequality to apply. It has everything to do with how lambda is sampled, irrespective of whatever your experimental unit is.

However, the issue I have raised is that your recommended sampling methods, can introduce systematic variations between the ensembles, so that the inequalities no longer apply, so any violations obtained as a result are null and void.

Again, I've explained previously why this is a naive view of what is being done in the experiment. There are 2^8 possible outcome combinations not 2^4. In fact, your A_1 can be considered a function of lambda as follows:

A(a_1, lambda) = A1(lambda)

B(b_1, lambda) = B1(lambda)

A(a_2, lambda) = A2(lambda)

B(b_2, lambda) = B2(lambda)

But lambda is completely unknown. If I make 8 measurements of random variables, I get 8 possible outcome combinations not 4, unless I can control lambda and make sure they are the same. In CHSH tests, 8 measurements are made (ie, 4 pairs), not 4 singles, therefore there are 2^8 possible outcome combinations. If at any point in your derivation you used the relationship A1B1 + A1B2 + A2B1 - A2B2 = A1(B1 + B2) + A2(B1-B2) <= 2, then you have assumed that the value of lambda was the same for every paired term in the expression. In that case, and only in that case, do you have 2^4 outcome combinations. Effectively, it amounts to assuming the ensembles generated by sampling are the same. Every derivation of Bell-like inequalities relies on relationships of that sort. This is why I agree with you when you argue in your paper that the ensembles must be the same for the inequality to apply. It has everything to do with how lambda is sampled, irrespective of whatever your experimental unit is.

However, the issue I have raised is that your recommended sampling methods, can introduce systematic variations between the ensembles, so that the inequalities no longer apply, so any violations obtained as a result are null and void.

Peer 3:

( January 19th, 2015 2:37pm UTC )

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Unregistered submitter: "Reading the discussion so far reveals that Dr Gill's responses are rational and issues raised are properly answered."

Thank you for your opinion, maybe I missed the post where Dr Gill explained how exactly his recommended method of picking particle pairs avoids the coincidence-time loophole for the specific local realistic model I presented on ( January 17th, 2015 2:49pm UTC ). I'll appreciate if you could justify your opinion by pointing me to the post since it appears you have found a reasonable response to the issues.

Thank you for your opinion, maybe I missed the post where Dr Gill explained how exactly his recommended method of picking particle pairs avoids the coincidence-time loophole for the specific local realistic model I presented on ( January 17th, 2015 2:49pm UTC ). I'll appreciate if you could justify your opinion by pointing me to the post since it appears you have found a reasonable response to the issues.

Richard Gill:

( January 19th, 2015 2:39pm UTC )

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Peer 3, I think you still haven't read the first couple of sections of my paper http://arxiv.org/abs/1207.5103 where there is a proof of a strong (finite sample) version of CHSH (Theorem 1) without any lambdas. You haven't realised that the idea of a Bell-CHSH experiment is to test local realism, not to estimate correlations between outcomes of measurement of pairs of particles. You haven't taken any notice of Bell's remark that we should abolish the use of words like "particle" or "quantum".

Of course we get different results if we "select" the first measurement outcome in a time-slot, or the last, or the one closest to the mid-point. It doesn't matter. If local realism holds, and measurement settings are chosen repeatedly (per time-slot) anew and completely at random, we cannot violate CHSH (up to finite-sample statistical variation, of course).

The experimentalist who believes in the predictions of QM and who moreover wants to disprove LR by violating CHSH can choose whatever local processing of time-slot data he or she likes. In order that the result is a valid test of LR, he or she just has to make sure that one binary outcome is defined, locally on each side of the experiment, per time-slot.

If however some experimental units (time-slots) are rejected, e.g. because of the detector efficiency loophole or the coincidence loophole, then a violation of CHSH doesn't mean so much - one should use a modified CHSH taking account of the amount of bias which a given amount of "inefficiency" (post-selection of experimental units) can generate.

BTW The derivation given by Jan-Ake Larsson and myself in our 2004 paper used the traditional hidden variable lambda and a traditional line of argument as a pedagogical device: we want to make life simple for the reader who is familiar with the traditional proof going back to Bell (1964), but who is not familiar with the notation and language of modern probability theory. Already I had been working out an alternative proof exploiting more fully the randomness of settings and depending on a *minimal* definition of local realism, in order to take care of the memory loophole and various claims that Bell had not taken account of *time*: http://arxiv.org/abs/quant-ph/0110137 "Accardi contra Bell (cum mundi): The Impossible Coupling", http://arxiv.org/abs/quant-ph/0301059 "Time, Finite Statistics, and Bell's Fifth Position".

Of course we get different results if we "select" the first measurement outcome in a time-slot, or the last, or the one closest to the mid-point. It doesn't matter. If local realism holds, and measurement settings are chosen repeatedly (per time-slot) anew and completely at random, we cannot violate CHSH (up to finite-sample statistical variation, of course).

The experimentalist who believes in the predictions of QM and who moreover wants to disprove LR by violating CHSH can choose whatever local processing of time-slot data he or she likes. In order that the result is a valid test of LR, he or she just has to make sure that one binary outcome is defined, locally on each side of the experiment, per time-slot.

If however some experimental units (time-slots) are rejected, e.g. because of the detector efficiency loophole or the coincidence loophole, then a violation of CHSH doesn't mean so much - one should use a modified CHSH taking account of the amount of bias which a given amount of "inefficiency" (post-selection of experimental units) can generate.

BTW The derivation given by Jan-Ake Larsson and myself in our 2004 paper used the traditional hidden variable lambda and a traditional line of argument as a pedagogical device: we want to make life simple for the reader who is familiar with the traditional proof going back to Bell (1964), but who is not familiar with the notation and language of modern probability theory. Already I had been working out an alternative proof exploiting more fully the randomness of settings and depending on a *minimal* definition of local realism, in order to take care of the memory loophole and various claims that Bell had not taken account of *time*: http://arxiv.org/abs/quant-ph/0110137 "Accardi contra Bell (cum mundi): The Impossible Coupling", http://arxiv.org/abs/quant-ph/0301059 "Time, Finite Statistics, and Bell's Fifth Position".

Unregistered Submission:

( January 19th, 2015 3:02pm UTC )

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"However, the issue I have raised is that your recommended sampling methods, can introduce systematic variations between the ensembles, so that the inequalities no longer apply, so any violations obtained as a result are null and void."

And Graft has shown that this effect actually occurs in the Christensen et al experiment.

And Graft has shown that this effect actually occurs in the Christensen et al experiment.

Unregistered Submission:

( January 19th, 2015 3:06pm UTC )

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Lots of dodging and weaving there, but no answers to the questions posed.

"The experimentalist who believes in the predictions of QM and who moreover wants to disprove LR by violating CHSH can choose whatever local processing of time-slot data he or she likes."

Data may be discarded as needed to prove the desired hypothesis! This is not science.

"The experimentalist who believes in the predictions of QM and who moreover wants to disprove LR by violating CHSH can choose whatever local processing of time-slot data he or she likes."

Data may be discarded as needed to prove the desired hypothesis! This is not science.

Peer 3:

( January 19th, 2015 3:41pm UTC )

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Richard Gill: "Peer 3, I think you still haven't read the first couple of sections of my paper http://arxiv.org/abs/1207.5103 where there is a proof of a strong (finite sample) version of CHSH (Theorem 1) without any lambdas. You haven't realised that the idea of a Bell-CHSH experiment is to test local realism, not to estimate correlations between outcomes of measurement of pairs of particles. You haven't taken any notice of Bell's remark that we should abolish the use of words like "particle" or "quantum"."

I'm beginning to think you do not really understand the issues being raised here because you keep bringing up irrelevant generalities instead of addressing very specificificites. Now you don't want to talk about lambdas or particles at all. The issues specifically deal with how lambda is sampled, and your suggested methods of picking particle pairs. Given the challenge, I would understand reluctance to to discuss those issues, but that doesn't make it go away. It doesn't prevent experimentalists from relying on the recommendations you have made in the literature about the coincidence time loophole. The results of 2 high-profile experiments are now called to question because of those ideas which appear to be wrong and sadly, you are unwilling to face the challenge directly.

Richard Gill: "Of course we get different results if we "select" the first measurement outcome in a time-slot, or the last, or the one closest to the mid-point. It doesn't matter. If local realism holds, and measurement settings are chosen repeatedly (per time-slot) anew and completely at random, we cannot violate CHSH (up to finite-sample statistical variation, of course).

The experimentalist who believes in the predictions of QM and who moreover wants to disprove LR by violating CHSH can choose whatever local processing of time-slot data he or she likes."

Your very own paper argues that this statement is false. You state **"The problem here is that the ensemble on which the correlations are evaluated changes with the settings, while the original Bell inequality requires that they stay the same"** In other words, if the ensembles of **lambdas** are not the same, the CHSH can be easily violated by a local realistic model. This is in your paper, and examples have been provided which do just that. The issue is not just one of obtaining "different" results. It is one of obtaining a violation if the model includes angle-dependent time-delays. You suggest that this "difficulty" can be fixed by considering just the first pair of particles in a time-slot. This suggestion is seriously mistaken. After I pointed this out with detailed explanations, you then suggested a new method, of picking just the pair in the middle of the time slot. Your new method is just as bad as the first one.

I'm beginning to think you do not really understand the issues being raised here because you keep bringing up irrelevant generalities instead of addressing very specificificites. Now you don't want to talk about lambdas or particles at all. The issues specifically deal with how lambda is sampled, and your suggested methods of picking particle pairs. Given the challenge, I would understand reluctance to to discuss those issues, but that doesn't make it go away. It doesn't prevent experimentalists from relying on the recommendations you have made in the literature about the coincidence time loophole. The results of 2 high-profile experiments are now called to question because of those ideas which appear to be wrong and sadly, you are unwilling to face the challenge directly.

Richard Gill: "Of course we get different results if we "select" the first measurement outcome in a time-slot, or the last, or the one closest to the mid-point. It doesn't matter. If local realism holds, and measurement settings are chosen repeatedly (per time-slot) anew and completely at random, we cannot violate CHSH (up to finite-sample statistical variation, of course).

The experimentalist who believes in the predictions of QM and who moreover wants to disprove LR by violating CHSH can choose whatever local processing of time-slot data he or she likes."

Your very own paper argues that this statement is false. You state **"The problem here is that the ensemble on which the correlations are evaluated changes with the settings, while the original Bell inequality requires that they stay the same"** In other words, if the ensembles of **lambdas** are not the same, the CHSH can be easily violated by a local realistic model. This is in your paper, and examples have been provided which do just that. The issue is not just one of obtaining "different" results. It is one of obtaining a violation if the model includes angle-dependent time-delays. You suggest that this "difficulty" can be fixed by considering just the first pair of particles in a time-slot. This suggestion is seriously mistaken. After I pointed this out with detailed explanations, you then suggested a new method, of picking just the pair in the middle of the time slot. Your new method is just as bad as the first one.

Richard Gill:

( January 19th, 2015 5:05pm UTC )

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Peer 3, what you call "irrelevant generalities", I call the fundamental issues; what you call "very specificities", I call irrelevant details. I think I understand the issues which you bring up and I have tried to explain that they are non-issues. I failed to get the message across so you had better talk to someone else.

Indeed, I don't want to talk about particles at all. I follow Bell's advice from 1981.

Following that same advice I insist on treating the experimental unit as being a time-slot.

And I would prefer that we cleaned up the discussion by not introducing any lambda's either. It's a relic of 50 years ago (and when Bell was writing in 1964, as a 60's physicist he used the mathematical probability theory of 50 years before that!). All we need to bring into the discussion is the set of counterfactual measurement outcomes A1, A2, B1, and B2 which would have been obtained, had either of the two settings been in force in each measurement wing. Any definition of local realism *implies* that these four exist (exist in a mathematical sense - i.e. can be constructed within a mathematical physical model for the phenomenon we are studying). A *minimal* definition of local realism is to assert that these exist. No-conspiracy means that we can choose by fair coin tosses whether to see A1 or A2, and whether to see B1 or B2. This is enough to derive CHSH.

Once you have understood a clean modern proof of Bell's theorem under minimal assumptions, and of course in a "no-loophole" situation, you can go on to think about various deviations from the ideal experimental protocol. What is the effect of the detection loophole? What is the effect of the coincidence loophole?

I am not proposing to discard data in order to prove the theory I want to prove. Following Bell (1981), I propose we perform a test of local realism which does not have any loopholes. In other words, a test which could, in principle, allow us (or rather: nature) to disprove local realism. This will necessitate that we coarse-grain in some way or other the information collected in each wing of the experiment in each time slot. As long as this coarse-graining is local, it is legal.

In contrast to doing a test which could end up by violating some kind of Bell inequality but still leave us with hundreds of alternative *local realist* reasons for how that came about.

So actually this coarse-graining (which you call "discarding data") is actually a tool we can use to make the experiment more informative, more powerful, more relevant. To get more information out of it. "Less is more" (sometimes).

Indeed, I don't want to talk about particles at all. I follow Bell's advice from 1981.

Following that same advice I insist on treating the experimental unit as being a time-slot.

And I would prefer that we cleaned up the discussion by not introducing any lambda's either. It's a relic of 50 years ago (and when Bell was writing in 1964, as a 60's physicist he used the mathematical probability theory of 50 years before that!). All we need to bring into the discussion is the set of counterfactual measurement outcomes A1, A2, B1, and B2 which would have been obtained, had either of the two settings been in force in each measurement wing. Any definition of local realism *implies* that these four exist (exist in a mathematical sense - i.e. can be constructed within a mathematical physical model for the phenomenon we are studying). A *minimal* definition of local realism is to assert that these exist. No-conspiracy means that we can choose by fair coin tosses whether to see A1 or A2, and whether to see B1 or B2. This is enough to derive CHSH.

Once you have understood a clean modern proof of Bell's theorem under minimal assumptions, and of course in a "no-loophole" situation, you can go on to think about various deviations from the ideal experimental protocol. What is the effect of the detection loophole? What is the effect of the coincidence loophole?

I am not proposing to discard data in order to prove the theory I want to prove. Following Bell (1981), I propose we perform a test of local realism which does not have any loopholes. In other words, a test which could, in principle, allow us (or rather: nature) to disprove local realism. This will necessitate that we coarse-grain in some way or other the information collected in each wing of the experiment in each time slot. As long as this coarse-graining is local, it is legal.

In contrast to doing a test which could end up by violating some kind of Bell inequality but still leave us with hundreds of alternative *local realist* reasons for how that came about.

So actually this coarse-graining (which you call "discarding data") is actually a tool we can use to make the experiment more informative, more powerful, more relevant. To get more information out of it. "Less is more" (sometimes).

Richard Gill:

( January 19th, 2015 6:16pm UTC )

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Peer 3, your "specific model" of January 17th, 2015 2:49pm UTC does not prove anything. I'm not disputing that a different selection principle will give different correlations, and that they would be in general be different from the correlations which the physicist might be interested in, i.e. the correlations between particles which were originally emitted together as a pair. All that I'm saying is that if local realism is true then no local selection principle (no local coarse-graining) can generate correlations (computed with experimental unit = time-slot, outcome binary) which violate CHSH. If you have a counter-example to this statement then please go ahead and program it. Programme a loophole-free violation of CHSH under strict local realism.

Peer 3:

( January 19th, 2015 6:47pm UTC )

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Dr Gill: "Indeed, I don't want to talk about particles at all"

Are you still unable to understand that it does not matter if you talk about "particles" or "experimental units"?! The point remains: **"The problem here is that the ensemble on which the correlations are evaluated changes with the settings, while the original Bell inequality requires that they stay the same"**, this is your quote. It could be ensembles of "particles" or ensembles of "experimental units" or ensembles of whatever you like, it doesn't change anything. Don't you understand that yet?

Dr Gill: "And I would prefer that we cleaned up the discussion by not introducing any lambda's either. It's a relic of 50 years ago (and when Bell was writing in 1964, as a 60's physicist he used the mathematical probability theory of 50 years before that!)."

First, you said we should forget about particle lambdas and focus on time-slot lambdas, then I explained how that will not change the issue because time-slot lambdas includes particle lambdas, then now you are saying we shouldn't talk about lambdas at all. The paper we are discussing talks specifically about lambdas, how can we discuss the paper if you want to talk about something other that the arguments made in the paper? Here is a quote from the paper:

"The integrals on the right-hand side cannot easily be added when ΛAC′ =/= ΛAD′ , since we

are taking expectations over different ensembles ΛAC′ and ΛAD′ , with respect to different

probability measures.

The problem here is that the ensemble on which the correlations are evaluated changes

with the settings, while the original Bell inequality requires that they stay the same. In effect,

the Bell inequality only holds on the common part of the four different ensembles ΛAC′ , ΛAD′ ,

ΛBC′ , and ΛBD′ "

Here you are discussing lambdas and ensembles of lambdas, combine that with your claim that picking the first pair of **"particles"** in a time-slot avoids this problem. Now you don't want to explain why you believe those contradictory claims simultaneously.

Dr Gill: "As long as this coarse-graining is local, it is legal."

I've just explained to you multiple times why your suggested "coarse-graining" method does not achieve the goal you claim it does. Yet you keep repeating that it does without addressing the challenge specifically.

Are you still unable to understand that it does not matter if you talk about "particles" or "experimental units"?! The point remains: **"The problem here is that the ensemble on which the correlations are evaluated changes with the settings, while the original Bell inequality requires that they stay the same"**, this is your quote. It could be ensembles of "particles" or ensembles of "experimental units" or ensembles of whatever you like, it doesn't change anything. Don't you understand that yet?

Dr Gill: "And I would prefer that we cleaned up the discussion by not introducing any lambda's either. It's a relic of 50 years ago (and when Bell was writing in 1964, as a 60's physicist he used the mathematical probability theory of 50 years before that!)."

First, you said we should forget about particle lambdas and focus on time-slot lambdas, then I explained how that will not change the issue because time-slot lambdas includes particle lambdas, then now you are saying we shouldn't talk about lambdas at all. The paper we are discussing talks specifically about lambdas, how can we discuss the paper if you want to talk about something other that the arguments made in the paper? Here is a quote from the paper:

"The integrals on the right-hand side cannot easily be added when ΛAC′ =/= ΛAD′ , since we

are taking expectations over different ensembles ΛAC′ and ΛAD′ , with respect to different

probability measures.

The problem here is that the ensemble on which the correlations are evaluated changes

with the settings, while the original Bell inequality requires that they stay the same. In effect,

the Bell inequality only holds on the common part of the four different ensembles ΛAC′ , ΛAD′ ,

ΛBC′ , and ΛBD′ "

Here you are discussing lambdas and ensembles of lambdas, combine that with your claim that picking the first pair of **"particles"** in a time-slot avoids this problem. Now you don't want to explain why you believe those contradictory claims simultaneously.

Dr Gill: "As long as this coarse-graining is local, it is legal."

I've just explained to you multiple times why your suggested "coarse-graining" method does not achieve the goal you claim it does. Yet you keep repeating that it does without addressing the challenge specifically.

Peer 3:

( January 19th, 2015 6:49pm UTC )

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Dr Gill: " I'm not disputing that a different selection principle will give different correlations, and that they would be in general be different from the correlations"

Dr Gill: "All that I'm saying is that if local realism is true then no local selection principle (no local coarse-graining) can generate correlations (computed with experimental unit = time-slot, outcome binary) which violate CHSH."

Are you disputing the fact that the ensembles used to calculate the correlations will be different from term to term? Are you disputing that? I'm not asking you if the resulting correlations will be different. I'm asking you if the ensemble used to calculate the correlations will be different from term to term. Get the difference?

Dr Gill: "All that I'm saying is that if local realism is true then no local selection principle (no local coarse-graining) can generate correlations (computed with experimental unit = time-slot, outcome binary) which violate CHSH."

Are you disputing the fact that the ensembles used to calculate the correlations will be different from term to term? Are you disputing that? I'm not asking you if the resulting correlations will be different. I'm asking you if the ensemble used to calculate the correlations will be different from term to term. Get the difference?

Unregistered Submission:

( January 19th, 2015 8:20pm UTC )

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"If you have a counter-example to this statement then please go ahead and program it."

There's no need for that. The Christensen et al experiment is the counter-example.

There's no need for that. The Christensen et al experiment is the counter-example.

Unregistered Submission:

( January 20th, 2015 4:15pm UTC )

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"There's no need for that. The Christensen et al experiment is the counter-example."

(where "counter-example" means one to Gill's "All that I'm saying is that if local realism is true then no local selection principle (no local coarse-graining) can generate correlations (computed with experimental unit = time-slot, outcome binary) which violate CHSH.")

An experiment is never an example of model. A counter-example to the claim actually involves two models, one without and one with a local selection principle, and to show that the latter produces correlations which violate CHSH although the former does not.

(where "counter-example" means one to Gill's "All that I'm saying is that if local realism is true then no local selection principle (no local coarse-graining) can generate correlations (computed with experimental unit = time-slot, outcome binary) which violate CHSH.")

An experiment is never an example of model. A counter-example to the claim actually involves two models, one without and one with a local selection principle, and to show that the latter produces correlations which violate CHSH although the former does not.

Donald A. Graft:

( January 20th, 2015 5:00pm UTC )

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Please explain, then, why the effect occurs in the experiment, i.e., that including all the photon detections on an equal basis produces no violation, while post-selecting produces a violation. The reason for the effect has been described in an easily understandable way, and it is seen to be the result of nonlocal post-selection. While this is enough to deflate Larsson-Gill, I agree it would be frosting on the cake to show it with a computer simulation as well. I am writing up such a model.

Unregistered Submission:

( January 24th, 2015 12:58pm UTC )

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The detection loophole occurs when A(a, lambda) is ternary, not binary, with a third outcome standing for "no detection".

What you miss is that the detection loophole can be dealt with in exactly the same way: define A(a,lambda) equal to, say, +1, when there is no detection in the corresponding time window.

What you miss is that the detection loophole can be dealt with in exactly the same way: define A(a,lambda) equal to, say, +1, when there is no detection in the corresponding time window.

Enter your reply below (Please read the **How To**)

Peer 2:

( January 19th, 2015 4:22pm UTC )

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"The problem here is that the ensemble on which the correlations are evaluated changes with the settings, while the original Bell inequality requires that they stay the same"

While not outright wrong, this sentence can be made more precise, since Bell's theorem is more robust than that. E.g., "...changes with the setting *at the other wing of the experiment*" would be better in my opinion. Both the detection loophole and the coincidence loophole require that results from both wings are compared before the decision is made as to which events should be excluded.

But in general, the modern proofs of Bell's theorem, basically avoiding mentioning lambdas, are much better.

While not outright wrong, this sentence can be made more precise, since Bell's theorem is more robust than that. E.g., "...changes with the setting *at the other wing of the experiment*" would be better in my opinion. Both the detection loophole and the coincidence loophole require that results from both wings are compared before the decision is made as to which events should be excluded.

But in general, the modern proofs of Bell's theorem, basically avoiding mentioning lambdas, are much better.

Peer 3:

( January 19th, 2015 7:21pm UTC )

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Peer 2: "While not outright wrong, this sentence can be made more precise"

The statement is accurate, I fault the authors for many things, but this is not one of them. They were spot on in identifying that "... the ensemble on which the correlations are evaluated changes with the settings, while the original Bell inequality requires that they stay the same".

Peer 2: "But in general, the modern proofs of Bell's theorem, basically avoiding mentioning lambdas, are much better."

Just because so-called "modern-proofs" avoid mentioning lambdas, does not mean the issue is not present. They sneak it in in may different ways such as factorization of random variables sampled at different times on different pairs, under the guise of "counter-factual terms".

Case in point, Dr Gill has been claiming that there are 2^4 possible joint outcomes in a CHSH experiment, a claim which is clearly false. He thinks about counterfactual results and imagines that the the random variables are sampled only 4 times, ie A1, A2, B1, B2, and recombined in pairs to obtain A1B1, A2B1, A1B2, A2B2. He thinks because one could imagine what would have been obtained had the setting on the other side been b2 instead of b1, and since those settings at the other end should not change what Alice observes if her setting is fixed at a1, then naively, he thinks the value of A1 in the pair A1B1 should be the same value for A1 in the pair A1B2.

But this is simply a naive view of the experiment, where actual data is measured. A1B1 is sampled from one pair, A1B2 is sampled from a different pair. The A1 in A1B1 is free to be different from the A1 in A1B2. This gives us 2^8 possible outcome combinations. However, if you can somehow arrange for the pairs to be identical to each other in every respect at the time of measurement, then you can conclude that in fact, the A1 in A1B1 should be the same as the A1 in A1B2. It is only in this scenario that you have 2^4 possible outcome combinations. Since we are measuring correlations, we are taking averages over many pairs, that is why we now have ensembles. This means to have 2^4 possible outcome combinations, the ensembles must be the same.

If only Dr Gill will understand and admit this simple fact. By the way, notice I did not use "lambdas" or "particles" in the above description.

The statement is accurate, I fault the authors for many things, but this is not one of them. They were spot on in identifying that "... the ensemble on which the correlations are evaluated changes with the settings, while the original Bell inequality requires that they stay the same".

Peer 2: "But in general, the modern proofs of Bell's theorem, basically avoiding mentioning lambdas, are much better."

Just because so-called "modern-proofs" avoid mentioning lambdas, does not mean the issue is not present. They sneak it in in may different ways such as factorization of random variables sampled at different times on different pairs, under the guise of "counter-factual terms".

Case in point, Dr Gill has been claiming that there are 2^4 possible joint outcomes in a CHSH experiment, a claim which is clearly false. He thinks about counterfactual results and imagines that the the random variables are sampled only 4 times, ie A1, A2, B1, B2, and recombined in pairs to obtain A1B1, A2B1, A1B2, A2B2. He thinks because one could imagine what would have been obtained had the setting on the other side been b2 instead of b1, and since those settings at the other end should not change what Alice observes if her setting is fixed at a1, then naively, he thinks the value of A1 in the pair A1B1 should be the same value for A1 in the pair A1B2.

But this is simply a naive view of the experiment, where actual data is measured. A1B1 is sampled from one pair, A1B2 is sampled from a different pair. The A1 in A1B1 is free to be different from the A1 in A1B2. This gives us 2^8 possible outcome combinations. However, if you can somehow arrange for the pairs to be identical to each other in every respect at the time of measurement, then you can conclude that in fact, the A1 in A1B1 should be the same as the A1 in A1B2. It is only in this scenario that you have 2^4 possible outcome combinations. Since we are measuring correlations, we are taking averages over many pairs, that is why we now have ensembles. This means to have 2^4 possible outcome combinations, the ensembles must be the same.

If only Dr Gill will understand and admit this simple fact. By the way, notice I did not use "lambdas" or "particles" in the above description.

Peer 2:

( January 19th, 2015 8:36pm UTC )

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Ok, it seems to me that you have now left your objection about the coincidence loophole, and have instead switched to the 2洠vs 2渠objection, based on the premise that a particle can only be measured once. This objection was originally put forward by Karl Hess, i believe (someone correct me if I am wrong).

This is an objection that I don't want to waste time discussing.

This is an objection that I don't want to waste time discussing.

Peer 3:

( January 19th, 2015 8:52pm UTC )

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Peer 2: "Ok, it seems to me that you have now left your objection about the coincidence loophole,"

I have left nothing. Just because I point out an error you made in your attempt to defend Dr Gill's position, does not mean I've left the other objection. It remains unanswered.

Peer 2: "based on the premise that a particle can only be measured once."

It is not a premise, it is a fact. There are 2^8 possible outcome combinations not 2^4, do you deny this?

It didn't escape my notice that you failed to answer the earlier question of whether you now understood how differences could be created between the ensembles by following the methods recommended by Drs Gill and Larsson.

We now have 2 unanswered questions for you as well:

1. Do you understand how picking just the first pair of particles in a time-slot creates differences between the ensembles used for each term?

2. Do you agree that there are 2^8 possible outcome combinations for a CHSH *experiment*, unless the ensembles used for the 4 pairs are the same?

I have left nothing. Just because I point out an error you made in your attempt to defend Dr Gill's position, does not mean I've left the other objection. It remains unanswered.

Peer 2: "based on the premise that a particle can only be measured once."

It is not a premise, it is a fact. There are 2^8 possible outcome combinations not 2^4, do you deny this?

It didn't escape my notice that you failed to answer the earlier question of whether you now understood how differences could be created between the ensembles by following the methods recommended by Drs Gill and Larsson.

We now have 2 unanswered questions for you as well:

1. Do you understand how picking just the first pair of particles in a time-slot creates differences between the ensembles used for each term?

2. Do you agree that there are 2^8 possible outcome combinations for a CHSH *experiment*, unless the ensembles used for the 4 pairs are the same?

Peer 2:

( January 19th, 2015 10:00pm UTC )

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"1. Do you understand how picking just the first pair of particles in a time-slot creates differences between the ensembles used for each term?"

I already explained to you that to open the coincidence loophole, the post-selection must depend on a comparison of results from both wings. That is not the case here.

"2. Do you agree that there are 2^8 possible outcome combinations for a CHSH *experiment*, unless the ensembles used for the 4 pairs are the same?"

The relevance of this has been so thoroughly debunked in the literature that I have nothing further to add. Google is your friend.

I already explained to you that to open the coincidence loophole, the post-selection must depend on a comparison of results from both wings. That is not the case here.

"2. Do you agree that there are 2^8 possible outcome combinations for a CHSH *experiment*, unless the ensembles used for the 4 pairs are the same?"

The relevance of this has been so thoroughly debunked in the literature that I have nothing further to add. Google is your friend.

Peer 3:

( January 20th, 2015 12:29am UTC )

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Peer 2: "I already explained to you that to open the coincidence loophole, the post-selection must depend on a comparison of results from both wings. That is not the case here."

Wrong. It is in fact the case as I have I just explained to you in the post on from ( January 17th, 2015 10:27pm UTC ), and as Donald Graft has shown in his analysis. The post-selection causes the results to be dependent on the settings from both wings. You never answered whether you understood the explanation or not. I guess you still did not understand it then. If you did understand it, could you please explain in what way exactly that explanation fails?

Peer 2: "The relevance of this has been so thoroughly debunked in the literature that I have nothing further to add. Google is your friend."

So why don't you put yourself on the record and state which one it is. Are there 2^4 or 2^8 distinct outcome combinations in a CHSH experiment??? And while you are at it, please provide a citation of the article page-number that that you claim "debunks" this fact.

Wrong. It is in fact the case as I have I just explained to you in the post on from ( January 17th, 2015 10:27pm UTC ), and as Donald Graft has shown in his analysis. The post-selection causes the results to be dependent on the settings from both wings. You never answered whether you understood the explanation or not. I guess you still did not understand it then. If you did understand it, could you please explain in what way exactly that explanation fails?

Peer 2: "The relevance of this has been so thoroughly debunked in the literature that I have nothing further to add. Google is your friend."

So why don't you put yourself on the record and state which one it is. Are there 2^4 or 2^8 distinct outcome combinations in a CHSH experiment??? And while you are at it, please provide a citation of the article page-number that that you claim "debunks" this fact.

Richard Gill:

( January 20th, 2015 7:57am UTC )

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This 2^4 vs 2^8 objection is put forward by a lot of people (Karl Hess, Andrei Khrennikov, Guillaume Adenier, ...). I think (actually: I know) these people don't understand statistics. Caroline Thompson http://freespace.virgin.net/ch.thompson1/Papers/The%20Record/TheRecord.htm who was both a statistician and a physicist *and* a local realist understood this perfectly well. She got into heated debates with local realists who simply did not understand (statistical) logic.

In a double blind randomised clinical trial we compare two groups of patients, one who got one treatment, the other who got the other treatment. We may imagine for each patient what might have happened to them had they got either treatment, but we can only give them one. There is even the "intention to treat principle" which means that we do *not* leave out patients who stop the treatment half way. In other words, all the basic principles which allow a loophole-free test of local realism are the core principles of evidence-based medicine (blinding, randomisation, intention-to-treat).

Time for evidence based physics! Rutherford famously said "if you need statistics, you did the wrong experiment". Physicists' training in probability and statistics is woefully inadequate, to this day. And Rutherford's arrogance is unfortunately still part of the standard mind-set.

Peer 3: have you studied Theorem 1 of my paper http://arxiv.org/abs/1207.5103 yet? Randomisation allows us to sensibly compare four correlations each computed on a different (disjoint) subset of the N experimental units.

In a double blind randomised clinical trial we compare two groups of patients, one who got one treatment, the other who got the other treatment. We may imagine for each patient what might have happened to them had they got either treatment, but we can only give them one. There is even the "intention to treat principle" which means that we do *not* leave out patients who stop the treatment half way. In other words, all the basic principles which allow a loophole-free test of local realism are the core principles of evidence-based medicine (blinding, randomisation, intention-to-treat).

Time for evidence based physics! Rutherford famously said "if you need statistics, you did the wrong experiment". Physicists' training in probability and statistics is woefully inadequate, to this day. And Rutherford's arrogance is unfortunately still part of the standard mind-set.

Peer 3: have you studied Theorem 1 of my paper http://arxiv.org/abs/1207.5103 yet? Randomisation allows us to sensibly compare four correlations each computed on a different (disjoint) subset of the N experimental units.

Richard Gill:

( January 20th, 2015 8:08am UTC )

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Peer 3: you say "He [Richard Gill] thinks about counterfactual results and imagines that the the random variables are sampled only 4 times, ie A1, A2, B1, B2, and recombined in pairs to obtain A1B1, A2B1, A1B2, A2B2. He thinks because one could imagine what would have been obtained had the setting on the other side been b2 instead of b1, and since those settings at the other end should not change what Alice observes if her setting is fixed at a1, then naively, he thinks the value of A1 in the pair A1B1 should be the same value for A1 in the pair A1B2."

Assuming local realism, it is the case that for each time-slot, four random(?) variables A1, A2, B1, B2 are sampled just once. Just one of the two A's and just one of the two B's gets observed; the choices as to which gets observed are made by independent fair coin tosses. I am not so naive as to think that A1 in A1B1 in one time-slot is the same as A1 in A1B2 in another time-slot.

Now please study Theorem 1 in http://arxiv.org/abs/1207.5103 and the subsequent application to a Bell-CHSH type experiment in the next section. You will see that there is not even an assumption that A1, A2, B1, B2 are random variables, and no assumption that the time-slots generate independent and identically distributed copies thereof.

In earlier papers I have even better results which also allow for memory effects and which were designed to take account of issues raised by Luigi Accardi, Karl Hess and Walter Philipp and many others: http://arxiv.org/abs/quant-ph/0110137 "Accardi contra Bell (cum mundi): The Impossible Coupling" and http://arxiv.org/abs/quant-ph/0301059 "Time, Finite Statistics, and Bell's Fifth Position" but these use slightly more sophisticated tools from probability theory (martingale theory, to be precise). So I recommend you start with the simplest version, Theorem 1 in http://arxiv.org/abs/1207.5103, "Statistics, Causality and Bell's Theorem", invited paper for special issue on causality in "Statistical Science" Volume 29, Number 4 (2014), 512-528, http://projecteuclid.org/euclid.ss/1421330545

Assuming local realism, it is the case that for each time-slot, four random(?) variables A1, A2, B1, B2 are sampled just once. Just one of the two A's and just one of the two B's gets observed; the choices as to which gets observed are made by independent fair coin tosses. I am not so naive as to think that A1 in A1B1 in one time-slot is the same as A1 in A1B2 in another time-slot.

Now please study Theorem 1 in http://arxiv.org/abs/1207.5103 and the subsequent application to a Bell-CHSH type experiment in the next section. You will see that there is not even an assumption that A1, A2, B1, B2 are random variables, and no assumption that the time-slots generate independent and identically distributed copies thereof.

In earlier papers I have even better results which also allow for memory effects and which were designed to take account of issues raised by Luigi Accardi, Karl Hess and Walter Philipp and many others: http://arxiv.org/abs/quant-ph/0110137 "Accardi contra Bell (cum mundi): The Impossible Coupling" and http://arxiv.org/abs/quant-ph/0301059 "Time, Finite Statistics, and Bell's Fifth Position" but these use slightly more sophisticated tools from probability theory (martingale theory, to be precise). So I recommend you start with the simplest version, Theorem 1 in http://arxiv.org/abs/1207.5103, "Statistics, Causality and Bell's Theorem", invited paper for special issue on causality in "Statistical Science" Volume 29, Number 4 (2014), 512-528, http://projecteuclid.org/euclid.ss/1421330545

Peer 3:

( January 20th, 2015 5:35pm UTC )

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Dr Gill: "have you studied Theorem 1 of my paper http://arxiv.org/abs/1207.5103 yet? Randomisation allows us to sensibly compare four correlations each computed on a different (disjoint) subset of the N experimental units."

This is why the ensembles have to be the same. You cannot meaningfully add expectations over different probability measures. So we go back to your quote which I've give many times:

"The problem here is that the ensemble on which the correlations are evaluated changes

with the settings, while the original Bell inequality requires that they stay the same. "

The issue is here is that you have failed to explain how picking just the first pair of particles in a time-slot, causes the ensembles to be the same. I have given a local realistic example for which it does not.

Dr Gill: "This 2^4 vs 2^8 objection is ut forward by a lot of people (Karl Hess, Andrei Khrennikov, Guillaume Adenier, ...). I think (actually: I know) these people don't understand statistics. "

What relevance is it that other people have put forth this argument before? Why would you find the need to insult others who may have disagreed with you in the past, rather than answering the direct challenge relevant to your paper?

You said there are 2^4 possible outcome combinations in the CHSH experiment. The point is that you are flatly wrong, there are 2^8 and it is easy to verify. At first I thought you were simply mistaken, but now it appears you are trying to mislead.

If you have 4 types of coins labelled A1, B1, A2, B2, each with sides labelled +1 and -1. All the A1 coins are identical to every other A1 coin and same for B1, etc. You take pairs of each type of coin and give to 8 individuals grouped into 4 pairs. To the first pair you give coins of type (A1, B1) and to the 3 other pairs, you give coin-pairs (A1, B2), (A2, B1), (A2, B2). Each individual tosses their coin just once. How many possible outcome combinations do you have? 2^8, not 2^4!

Your error is to think naively that the outcome from the A1 coin tossed by the person in the (A1, B1) pair, is the same outcome obtained from the A1 coin tossed by the person in the (A1, B2) pair. Just because the coins are identical does not mean they are the same coin. This is your error. It is okay to admit a mistake.

By the way as concerns the main issue I have been raising, you still haven't explained how picking the first pair (or the pair in the middle) of the time-slot allows you to avoid the coincidence time loophole? It absolutely does not.

I believe I have explained the issues quite clearly, and I have nothing else to say until such a time as you chose to admit the following two points:

1) there are 2^8 possible outcome combinations in the CHSH experiment.

2) picking the first pair (or the pair in the middle) of the time-slot does not avoid the coincidence time loophole.

or explain why those two points are wrong.

This is why the ensembles have to be the same. You cannot meaningfully add expectations over different probability measures. So we go back to your quote which I've give many times:

"The problem here is that the ensemble on which the correlations are evaluated changes

with the settings, while the original Bell inequality requires that they stay the same. "

The issue is here is that you have failed to explain how picking just the first pair of particles in a time-slot, causes the ensembles to be the same. I have given a local realistic example for which it does not.

Dr Gill: "This 2^4 vs 2^8 objection is ut forward by a lot of people (Karl Hess, Andrei Khrennikov, Guillaume Adenier, ...). I think (actually: I know) these people don't understand statistics. "

What relevance is it that other people have put forth this argument before? Why would you find the need to insult others who may have disagreed with you in the past, rather than answering the direct challenge relevant to your paper?

You said there are 2^4 possible outcome combinations in the CHSH experiment. The point is that you are flatly wrong, there are 2^8 and it is easy to verify. At first I thought you were simply mistaken, but now it appears you are trying to mislead.

If you have 4 types of coins labelled A1, B1, A2, B2, each with sides labelled +1 and -1. All the A1 coins are identical to every other A1 coin and same for B1, etc. You take pairs of each type of coin and give to 8 individuals grouped into 4 pairs. To the first pair you give coins of type (A1, B1) and to the 3 other pairs, you give coin-pairs (A1, B2), (A2, B1), (A2, B2). Each individual tosses their coin just once. How many possible outcome combinations do you have? 2^8, not 2^4!

Your error is to think naively that the outcome from the A1 coin tossed by the person in the (A1, B1) pair, is the same outcome obtained from the A1 coin tossed by the person in the (A1, B2) pair. Just because the coins are identical does not mean they are the same coin. This is your error. It is okay to admit a mistake.

By the way as concerns the main issue I have been raising, you still haven't explained how picking the first pair (or the pair in the middle) of the time-slot allows you to avoid the coincidence time loophole? It absolutely does not.

I believe I have explained the issues quite clearly, and I have nothing else to say until such a time as you chose to admit the following two points:

1) there are 2^8 possible outcome combinations in the CHSH experiment.

2) picking the first pair (or the pair in the middle) of the time-slot does not avoid the coincidence time loophole.

or explain why those two points are wrong.

Unregistered Submission:

( January 20th, 2015 9:49pm UTC )

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Peer3 is 100 percent correct. However, you would never see Larsson and Gill admit that what Peer3 has presented is correct as they would have to retract too many of their papers. The paper this thread is about should be retracted.

Donald A. Graft:

( January 21st, 2015 4:38am UTC )

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Retraction should be only for serious error and fraud, not theoretical analysis. Larsson-Gill remains an important advance and stimulus to thinking, a part of the evolution of physics, a manifestation of the divine creative spark. Nobody knows the final answer, even if we suppose there is one.

Unregistered Submission:

( January 21st, 2015 2:43pm UTC )

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Richard Gill: ( January 20th, 2015 7:57am UTC ): "Peer 3: have you studied Theorem 1 of my paper http://arxiv.org/abs/1207.5103 yet? Randomisation allows us to sensibly compare four correlations each computed on a different (disjoint) subset of the N experimental units."

The inequality of that theorem can be used as a test to the hypothesis that all correlations are computed on samples from the same population (instead of at least one coming from a different population).

The inequality of that theorem can be used as a test to the hypothesis that all correlations are computed on samples from the same population (instead of at least one coming from a different population).

Unregistered Submission:

( January 24th, 2015 1:03pm UTC )

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It's high time to change this motto to "if you don't need statistics, then your understanding of the experiment is wrong".

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Richard Gill:

( January 21st, 2015 7:10am UTC )

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Peer 3: you said:

" I have nothing else to say until such a time as you chose to admit the following two points: (1) there are 2^8 possible outcome combinations in the CHSH experiment; (2) picking the first pair (or the pair in the middle) of the time-slot does not avoid the coincidence time loophole. Or explain why those two points are wrong."

I have already explained at length why both points are wrong.

Here is the explanation for point (2), yet again: Group all hidden variables of all particles arriving in one time-slot into one big hidden variable. Define A(a, lambda) to be the function which selects the outcome of measuring the first particle registered in Alice's wing of the experiment, etc. We are now back with a standard CHSH situation and can follow a conventional proof that local realism implies CHSH, for this particular local coarse-graining to a binary outcome of everything that gets registered during the time-slot.

Regarding point (1), I have explained why we only need to look at 2^4 outcomes per experimental unit: because of randomisation of the detector settings. Tell me the error in the proof of Theorem 1 of my recent paper. I've explained why it is applicable to our situation. Your claim that I make a certain naive error is totally unfounded. I do not assume what you (naively) say that I assume.

Regarding the possibility of some kind of mistake in Larsson and Gill, it is worth mentioning the following. In his recent book "Einstein was right" Karl Hess argues that Hess and Philipp had already discovered/invented the coincidence loophole and that Larsson and Gill shamelessly stole it from them. He also says that Saverio Pascazio (1986; PRA) had already described it. Pascazio's is a little known paper, I had not heard of it before I read Hess' book, and Hess and Philipp did not know about Pascazio and the coincidence loophole, nor about Pearle and the detection loophole, when they wrote their paper. Hess claims in his book that what he had always been on about, with Philipp, was this kind of loophole issue.

Hess does not say that our work is wrong: on the contrary. He said we stole it from him. He certainly would say there was something wrong with it, if he could, since he was furious with the criticism which Gill, Weihs, Zukowski and Zeilinger levelled at the Hess-Philipp model. Instead, together with Hans de Raedt and Kristel Michielsen he used the coincidence loophole to build simulation models of local realist violation of CHSH. They have studied our paper really carefully, and they cite it. They do not say it is wrong.

" I have nothing else to say until such a time as you chose to admit the following two points: (1) there are 2^8 possible outcome combinations in the CHSH experiment; (2) picking the first pair (or the pair in the middle) of the time-slot does not avoid the coincidence time loophole. Or explain why those two points are wrong."

I have already explained at length why both points are wrong.

Here is the explanation for point (2), yet again: Group all hidden variables of all particles arriving in one time-slot into one big hidden variable. Define A(a, lambda) to be the function which selects the outcome of measuring the first particle registered in Alice's wing of the experiment, etc. We are now back with a standard CHSH situation and can follow a conventional proof that local realism implies CHSH, for this particular local coarse-graining to a binary outcome of everything that gets registered during the time-slot.

Regarding point (1), I have explained why we only need to look at 2^4 outcomes per experimental unit: because of randomisation of the detector settings. Tell me the error in the proof of Theorem 1 of my recent paper. I've explained why it is applicable to our situation. Your claim that I make a certain naive error is totally unfounded. I do not assume what you (naively) say that I assume.

Regarding the possibility of some kind of mistake in Larsson and Gill, it is worth mentioning the following. In his recent book "Einstein was right" Karl Hess argues that Hess and Philipp had already discovered/invented the coincidence loophole and that Larsson and Gill shamelessly stole it from them. He also says that Saverio Pascazio (1986; PRA) had already described it. Pascazio's is a little known paper, I had not heard of it before I read Hess' book, and Hess and Philipp did not know about Pascazio and the coincidence loophole, nor about Pearle and the detection loophole, when they wrote their paper. Hess claims in his book that what he had always been on about, with Philipp, was this kind of loophole issue.

Hess does not say that our work is wrong: on the contrary. He said we stole it from him. He certainly would say there was something wrong with it, if he could, since he was furious with the criticism which Gill, Weihs, Zukowski and Zeilinger levelled at the Hess-Philipp model. Instead, together with Hans de Raedt and Kristel Michielsen he used the coincidence loophole to build simulation models of local realist violation of CHSH. They have studied our paper really carefully, and they cite it. They do not say it is wrong.

Jan-Åke Larsson:

( January 21st, 2015 10:07am UTC )

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Peer 3: "Your error is to think naively that the outcome from the A1 coin tossed by the person in the (A1, B1) pair, is the same outcome obtained from the A1 coin tossed by the person in the (A1, B2) pair. Just because the coins are identical does not mean they are the same coin. This is your error."

We do not assume that. What we assume is that it is meaningful to talk about the B2 outcome as existing even if it is not realized in experiment, even if we are in the (A1,B1) pair, given by the hidden variable. This gives a bound for the expectation value combination in CHSH, not the individual outcomes. A bound for the expectation value.

It is correct that the individual outcomes are not bounded, if you do four separate experiments, one for each pair of settings. I have explained this in Larsson, J.-Å., 2014. Loopholes in Bell inequality tests of local realism. J. Phys. A 47, 424003. doi:10.1088/1751-8113/47/42/424003, http://arxiv.org/abs/1407.0363 . Please read subsection 1.5 from start to end, but especially the text surrounding equation (15). There I state clearly that the bound does in fact not apply to individual experiments, just as you say, but to the expectation values.

I go on to say that performing several experiments will make the sample mean tend to the expectation value. In short, this is because the sample mean is an unbiased estimate of the expectation value. Again, the CHSH inequality applies to the expectation value, not the sample mean.

An experimenter would perform several individual experiments to make up the sample mean. With enough data, the sample mean will be close to the expectation value, and the deviation can be quantified. This means that an experimental violation, where the sample mean exceeds the bound, will actually tell us something about the possibility that the (theoretical, unknown) expectation value violates the CHSH inequality.

We do not assume that. What we assume is that it is meaningful to talk about the B2 outcome as existing even if it is not realized in experiment, even if we are in the (A1,B1) pair, given by the hidden variable. This gives a bound for the expectation value combination in CHSH, not the individual outcomes. A bound for the expectation value.

It is correct that the individual outcomes are not bounded, if you do four separate experiments, one for each pair of settings. I have explained this in Larsson, J.-Å., 2014. Loopholes in Bell inequality tests of local realism. J. Phys. A 47, 424003. doi:10.1088/1751-8113/47/42/424003, http://arxiv.org/abs/1407.0363 . Please read subsection 1.5 from start to end, but especially the text surrounding equation (15). There I state clearly that the bound does in fact not apply to individual experiments, just as you say, but to the expectation values.

I go on to say that performing several experiments will make the sample mean tend to the expectation value. In short, this is because the sample mean is an unbiased estimate of the expectation value. Again, the CHSH inequality applies to the expectation value, not the sample mean.

An experimenter would perform several individual experiments to make up the sample mean. With enough data, the sample mean will be close to the expectation value, and the deviation can be quantified. This means that an experimental violation, where the sample mean exceeds the bound, will actually tell us something about the possibility that the (theoretical, unknown) expectation value violates the CHSH inequality.

Peer 3:

( January 21st, 2015 2:11pm UTC )

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Thank you Dr Larsson for contributing to this matter.

Dr Larsson: "We do not assume that. "

You don't in the paper, but Dr Gill clearly does assume that because he says there are 2^4 possible outcome combinations in a CHSH **experiment**. As I have explained, that is not true, there are 2^8 outcome combinations.

Dr Larsson: "What we assume is that it is meaningful to talk about the B2 outcome as existing even if it is not realized in experiment, even if we are in the (A1,B1) pair"

This is the crucial point. It is meaningful to talk about counterfactual results. For example: There are 2^4 possible counterfactual results for a single pair of particles at 4 angles, there are 2^8 possible counterfactual results for 4 pairs of particles at 4 angles. It is impossible to measure a single particle at 4 angles, but it is possible to measure 4 pairs of particles at 4 angles. However, no counterfactual results are used to calculate the expectation value in an **experiment** actual results are used always and in that case, there are 2^8 possibilities as well, not 2^4 as Dr Gill argued above. This is not a question of what we can imagine. It is a question of imagining precisely what is measured and not erroneously equating two very different things.

Dr Larsson: "This gives a bound for the expectation value combination in CHSH, not the individual outcomes. A bound for the expectation value."

You always start the derivation of the CHSH for expectation values, by considering the individual outcomes and factoring terms within the integral, and integrating over an ensemble of lambdas. The algebra cannot proceed if the expectation values are over different ensembles. You agree with this as you state in your paper. That is why it is important for the ensembles to be the same. If they are not, the CHSH does not apply as you state correctly in your paper:

"The integrals on the right-hand side cannot easily be added when 롣⠽/= 롤⠬ since we

are taking expectations over different ensembles 롣⠡nd 롤⠬ with respect to different

probability measures.

The problem here is that the ensemble on which the correlations are evaluated changes with the settings,while the original Bell inequality requires that they stay the same."

Dr Larsson: "It is correct that the individual outcomes are not bounded, if you do four separate experiments, one for each pair of settings."

So then it appears you do agree with me that there are 2^8 outcome possibilities in the CHSH **experiment** not 2^4 as Dr Gill argues above.

Could you then please explain how the expectation values obtained by picking the first pair of particles from a time-slot avoids the coincidence-time loophole for the local-realistic scenario I presented above on ( January 17th, 2015 2:49pm UTC). I explained there, how by picking just the first pair in a time-slot, the ensembles change with the settings so that the inequalities do not apply.

Note: In your other recent paper which you cited above (http://arxiv.org/abs/1407.0363), you make the same and other questionable claims. I'll appreciate if you could address them:

1) Top paragraph on Page 30. You say that the coincidence-loophole can be avoided by picking the pair of samples closest to the "event-ready" indicator (aka, beginning of time-slot).

2) First sentense on Page 31, after the figure. You say "Then, the subensemble that gives coincidences for all three of A1B1, A1B2, and, A2B1 will also give a coincidence forA2B2. This will also enable inequality". Although it is true if recombining 4 hypothetical counterfactual measurements from a single pair of particles. This is obviously false in a CHSH **experiment** (see my example from January 17th, 2015 2:49pm UTC). I see a trend where you rely on counterfactual measurements of a single pair of particles to derive inequalities, generalize to ensembles, and the extrapolate to particles picked in time-slots. This is not valid.

3) Further down on page 31, You say: "Both the efficiency loophole and the coincidence loophole can be viewed as caused by deficiencies in the experimental equipment. Low overall experimental efficiency or coincidence probability can in principle be counteracted by improving the experimental setup."

This is mistaken, because if the way nature works is that some hidden parameters just can't be detected at certain angles, then you could have perfect equipment, and still have the detection loophole. Similarly, even if the experimental equipment is perfect, how would you know which particles belong with each other if the arrival times are variable, and "hidden" (see my example from January 17th, 2015 2:49pm UTC)?

Dr Larsson: "We do not assume that. "

You don't in the paper, but Dr Gill clearly does assume that because he says there are 2^4 possible outcome combinations in a CHSH **experiment**. As I have explained, that is not true, there are 2^8 outcome combinations.

Dr Larsson: "What we assume is that it is meaningful to talk about the B2 outcome as existing even if it is not realized in experiment, even if we are in the (A1,B1) pair"

This is the crucial point. It is meaningful to talk about counterfactual results. For example: There are 2^4 possible counterfactual results for a single pair of particles at 4 angles, there are 2^8 possible counterfactual results for 4 pairs of particles at 4 angles. It is impossible to measure a single particle at 4 angles, but it is possible to measure 4 pairs of particles at 4 angles. However, no counterfactual results are used to calculate the expectation value in an **experiment** actual results are used always and in that case, there are 2^8 possibilities as well, not 2^4 as Dr Gill argued above. This is not a question of what we can imagine. It is a question of imagining precisely what is measured and not erroneously equating two very different things.

Dr Larsson: "This gives a bound for the expectation value combination in CHSH, not the individual outcomes. A bound for the expectation value."

You always start the derivation of the CHSH for expectation values, by considering the individual outcomes and factoring terms within the integral, and integrating over an ensemble of lambdas. The algebra cannot proceed if the expectation values are over different ensembles. You agree with this as you state in your paper. That is why it is important for the ensembles to be the same. If they are not, the CHSH does not apply as you state correctly in your paper:

"The integrals on the right-hand side cannot easily be added when 롣⠽/= 롤⠬ since we

are taking expectations over different ensembles 롣⠡nd 롤⠬ with respect to different

probability measures.

The problem here is that the ensemble on which the correlations are evaluated changes with the settings,while the original Bell inequality requires that they stay the same."

Dr Larsson: "It is correct that the individual outcomes are not bounded, if you do four separate experiments, one for each pair of settings."

So then it appears you do agree with me that there are 2^8 outcome possibilities in the CHSH **experiment** not 2^4 as Dr Gill argues above.

Could you then please explain how the expectation values obtained by picking the first pair of particles from a time-slot avoids the coincidence-time loophole for the local-realistic scenario I presented above on ( January 17th, 2015 2:49pm UTC). I explained there, how by picking just the first pair in a time-slot, the ensembles change with the settings so that the inequalities do not apply.

Note: In your other recent paper which you cited above (http://arxiv.org/abs/1407.0363), you make the same and other questionable claims. I'll appreciate if you could address them:

1) Top paragraph on Page 30. You say that the coincidence-loophole can be avoided by picking the pair of samples closest to the "event-ready" indicator (aka, beginning of time-slot).

2) First sentense on Page 31, after the figure. You say "Then, the subensemble that gives coincidences for all three of A1B1, A1B2, and, A2B1 will also give a coincidence forA2B2. This will also enable inequality". Although it is true if recombining 4 hypothetical counterfactual measurements from a single pair of particles. This is obviously false in a CHSH **experiment** (see my example from January 17th, 2015 2:49pm UTC). I see a trend where you rely on counterfactual measurements of a single pair of particles to derive inequalities, generalize to ensembles, and the extrapolate to particles picked in time-slots. This is not valid.

3) Further down on page 31, You say: "Both the efficiency loophole and the coincidence loophole can be viewed as caused by deficiencies in the experimental equipment. Low overall experimental efficiency or coincidence probability can in principle be counteracted by improving the experimental setup."

This is mistaken, because if the way nature works is that some hidden parameters just can't be detected at certain angles, then you could have perfect equipment, and still have the detection loophole. Similarly, even if the experimental equipment is perfect, how would you know which particles belong with each other if the arrival times are variable, and "hidden" (see my example from January 17th, 2015 2:49pm UTC)?

Jan-Åke Larsson:

( January 21st, 2015 3:32pm UTC )

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You seem to be denying that the sample mean is an unbiased estimate of the expectation value. Is that what you are claiming?

Peer 3:

( January 21st, 2015 3:45pm UTC )

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Dr Larsson: "You seem to be denying that the sample mean is an unbiased estimate of the expectation value. Is that what you are claiming?".

I'm denying no such thing. I'm asking you to explain how by picking just the first pair of particles in a time-slot avoids the problem of introducing differences in the ensembles, when you already agree as explained in your paper that the ensembles must be the same for the inequalities to apply.

The "sample mean" point you and Dr Gill keep raising is completely irrelevant if the ensembles are different. You can only talk of sample means being unbiased estimates if you do not have 4 different systematically biased samples as is the case here. So why don't you start by explaining how your suggested method of picking pairs produces unbiased samples? I've given you one example which proves that it does not do what you claim it does.

I'm denying no such thing. I'm asking you to explain how by picking just the first pair of particles in a time-slot avoids the problem of introducing differences in the ensembles, when you already agree as explained in your paper that the ensembles must be the same for the inequalities to apply.

The "sample mean" point you and Dr Gill keep raising is completely irrelevant if the ensembles are different. You can only talk of sample means being unbiased estimates if you do not have 4 different systematically biased samples as is the case here. So why don't you start by explaining how your suggested method of picking pairs produces unbiased samples? I've given you one example which proves that it does not do what you claim it does.

Peer 3:

( January 21st, 2015 4:15pm UTC )

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Dr Gill: "I have already explained at length why both points are wrong."

You simply have not, as the record of the discussions here show. Besides point (1) is a mathematical fact. There are 2^8 possible outcome combinations in a CHSH experiment as I have shown. Given two possible outcomes (+1 or -1), sampled 8 times (4-pairs). That gives you 2^8 possible combinations -- basic mathematics.

Dr Gill: "Here is the explanation for point (2), yet again: Group all hidden variables of all particles arriving in one time-slot into one big hidden variable. Define A(a, lambda) to be the function which selects the outcome of measuring the first particle registered in Alice's wing of the experiment"

Apparently you missed my posts from ( January 17th, 2015 2:49pm UTC ) and ( January 17th, 2015 10:27pm UTC ) which explain why using "one big hidden variable" does absolutely nothing to avoid the issue I am raising here. You still have not addressed any of those issues.

You simply have not, as the record of the discussions here show. Besides point (1) is a mathematical fact. There are 2^8 possible outcome combinations in a CHSH experiment as I have shown. Given two possible outcomes (+1 or -1), sampled 8 times (4-pairs). That gives you 2^8 possible combinations -- basic mathematics.

Dr Gill: "Here is the explanation for point (2), yet again: Group all hidden variables of all particles arriving in one time-slot into one big hidden variable. Define A(a, lambda) to be the function which selects the outcome of measuring the first particle registered in Alice's wing of the experiment"

Apparently you missed my posts from ( January 17th, 2015 2:49pm UTC ) and ( January 17th, 2015 10:27pm UTC ) which explain why using "one big hidden variable" does absolutely nothing to avoid the issue I am raising here. You still have not addressed any of those issues.

Richard Gill:

( January 21st, 2015 4:18pm UTC )

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I wonder, Peer 3, if you can explain what you mean by "ensemble". I have already explained how by defining the experimental unit to be "time-slot" and by allowing A(a, lambda) and B(b, lambda) to be arbitrary functions of the settings (a and b) applied to the time-slot and all hidden variables in all relevant physical systems (lambda), no (setting and hidden variable influenced) post-selection is done.

Observed averages of A(a, lambda)B(b, lambda), over a number of time-slots, are then reasonable estimates of mean values of the same thing, if we assume that what happens in different time-slots are independent and identically distributed random variables (i.e. realisations of lambda in different time-slots are independent and identically distributed drawings from what you might call an "ensemble".

However such strong probabilistic assumptions, though conventionally made in the text-books, are not necessary at all for a valid test of local realism to result. See Theorem 1 in http://arxiv.org/abs/1207.5103. All we need is local realism and complete randomness of the setting choices, per time slot, independently of the hidden variables in the physical systems of particles, detectors, etc etc et.

Observed averages of A(a, lambda)B(b, lambda), over a number of time-slots, are then reasonable estimates of mean values of the same thing, if we assume that what happens in different time-slots are independent and identically distributed random variables (i.e. realisations of lambda in different time-slots are independent and identically distributed drawings from what you might call an "ensemble".

However such strong probabilistic assumptions, though conventionally made in the text-books, are not necessary at all for a valid test of local realism to result. See Theorem 1 in http://arxiv.org/abs/1207.5103. All we need is local realism and complete randomness of the setting choices, per time slot, independently of the hidden variables in the physical systems of particles, detectors, etc etc et.

Richard Gill:

( January 21st, 2015 4:58pm UTC )

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I did not miss your posts, Peer 3. I did not get any refutation of my answers to the issues you raise (which I have argued are non-issues), and you did not even follow up most of the points I raised at all, so we will have to leave it there. Please write up your critique, post a paper on arXiv, and submit it to a journal.

Jan-Åke Larsson:

( January 21st, 2015 5:01pm UTC )

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Above I use the strong assumptions to make it simpler to present the argument, to people who are used to the text-book presentation. It seems that Peer 3 does understand the terminology I used, even if he seems to object to using it. Weaker assumptions typically needs heavier mathematical machinery (here Richard refers to Hoeffding's inequality and supermartingales).

Peer 3 is trying to drive home a different point, not related to this. But he does not seem to accept that the Bell inequality applies to the theoretical unknown expectation values, and that experiments allow an experimenter to estimate each individual term in the inequality. This is why I asked the question above.

It seems that Peer 3 is confusing the fact that each experiment (each individual pair, each individual lambda) only can give values for one term (which I agree with), with his claim that using only the first detection in each time slot can give a false violation (which I do not believe).

I am trying to make him understand that these two are two different issues. The first can be handled by statistical methods (unbiased estimates or heavier machinery), while the second can be handled by joining all the different lambdas into one vector-lambda with the individual lambdas as "coordinates".

Unfortunately, Peer 3 does not seem to accept either argument. I do not see why, because the arguments that Richard, I, and also other people are making are good. I would agree with Richard that Peer 3 could write up a more complete argument in a paper, the format here may be too cut up in pieces.

Peer 3 is trying to drive home a different point, not related to this. But he does not seem to accept that the Bell inequality applies to the theoretical unknown expectation values, and that experiments allow an experimenter to estimate each individual term in the inequality. This is why I asked the question above.

It seems that Peer 3 is confusing the fact that each experiment (each individual pair, each individual lambda) only can give values for one term (which I agree with), with his claim that using only the first detection in each time slot can give a false violation (which I do not believe).

I am trying to make him understand that these two are two different issues. The first can be handled by statistical methods (unbiased estimates or heavier machinery), while the second can be handled by joining all the different lambdas into one vector-lambda with the individual lambdas as "coordinates".

Unfortunately, Peer 3 does not seem to accept either argument. I do not see why, because the arguments that Richard, I, and also other people are making are good. I would agree with Richard that Peer 3 could write up a more complete argument in a paper, the format here may be too cut up in pieces.

Donald A. Graft:

( January 21st, 2015 5:13pm UTC )

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"with his claim that using only the first detection in each time slot can give a false violation (which I do not believe)"

How then do you account for the effect actually observed in the analysis of the Christensen et al experiment described by Graft? And if a simulation model is published also showing it will you then revise your belief, or do you still maintain that it is impossible to show such a model? If you believe it is impossible, can you show us your proof?

How then do you account for the effect actually observed in the analysis of the Christensen et al experiment described by Graft? And if a simulation model is published also showing it will you then revise your belief, or do you still maintain that it is impossible to show such a model? If you believe it is impossible, can you show us your proof?

Richard Gill:

( January 21st, 2015 5:39pm UTC )

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Peer 1: it is easy to make simulation models of local realistic loophole free CHSH-type experiments which violate CHSH 50% of the time. ie the mean value of S is 2 and half the time the experiment gives a larger result and half the time it gives a smaller result. Statistical uncertainty means that with a finite amount of data from a real experiment one legitimate analysis can show a violation of CHSH and another legitimate analysis not. After all, there are different legitimate ways to define a local, binary outcome for each time-slot and different ways to statistically evaluate the result.

So Graft's empirical finding regarding Christensen et al's data is interesting but does not prove much one way or another.

I don't know what you mean by a "simulation model also showing it" but *if* Graft comes up with a simulation model which contradicts predictions of existing theorems then that would be very interesting indeed. You ask for proof, but the mathematical proofs are already known and published and pretty well established. Maybe Graft is onto something which so far everyone has overlooked. I personally doubt it. I don't see any evidence that existing theorems in this field have been disproved. As I mentioned, de Raedt and Hess actually cite Larsson and Gill and build on our results. (Hess even claims that we found our results in his papers with Philipp). They would be the first to tell the world that any of our theorems are false, if that was the case.

I've repeatedly asked Peer 3 to tell me what is wrong with Theorem 1 in http://arxiv.org/abs/1207.5103 but I get no response at all.

So Graft's empirical finding regarding Christensen et al's data is interesting but does not prove much one way or another.

I don't know what you mean by a "simulation model also showing it" but *if* Graft comes up with a simulation model which contradicts predictions of existing theorems then that would be very interesting indeed. You ask for proof, but the mathematical proofs are already known and published and pretty well established. Maybe Graft is onto something which so far everyone has overlooked. I personally doubt it. I don't see any evidence that existing theorems in this field have been disproved. As I mentioned, de Raedt and Hess actually cite Larsson and Gill and build on our results. (Hess even claims that we found our results in his papers with Philipp). They would be the first to tell the world that any of our theorems are false, if that was the case.

I've repeatedly asked Peer 3 to tell me what is wrong with Theorem 1 in http://arxiv.org/abs/1207.5103 but I get no response at all.

Peer 3:

( January 21st, 2015 10:50pm UTC )

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Dr Larsson: "It seems that Peer 3 does understand the terminology I used"

I understand the terminology very well. You accuse me of not understanding terminology, yet you do not say what terminology, or what misunderstanding, and how it changes the validity of my argument here. Please point out a specific misunderstanding from any of my posts on this topic.

Dr Larsson: "the second can be handled by joining all the different lambdas into one vector-lambda with the individual lambdas as "coordinates""

Why don't you answer the question I've asked directly? I have explained how by joining all the different lambdas into one vector-lambda with the individual lambdas as coordinates, you still end up with different "vector-lambda" ensembles. Please review my posts from ( January 17th, 2015 2:49pm UTC ) and ( January 17th, 2015 10:27pm UTC ) and please explain how joining lambdas could possibly avoid the issue in that counter-example :

*** How exactly does picking the first pair of particles avoid the coincidence loophole as you have claimed, in that counter-example??? ***

Dr Larsson: "But he does not seem to accept that the Bell inequality applies to the theoretical unknown expectation values"

You are badly mistaken, my argument does not involve whether inequalities apply to theoretical unknown expectation values. The point, which both of you continue to avoid is the following: in your papers, you claim the coincidence-time loophole can be avoided in CHSH **experiments**. You suggest a method of picking pairs which does not do what you claim it does. I have presented a simple counter-example which demonstrates this. Do you want to go on the record to say that your suggested method is full-proof, despite my explanations?

Dr Larsson: "It seems that Peer 3 is confusing the fact that each experiment (each individual pair, each individual lambda) only can give values for one term (which I agree with), with his claim that using only the first detection in each time slot can give a false violation (which I do not believe)."

I'm not confusing anything. At least you admit the fact that there are 2^8 outcome combinations in a CHSH experiment. Dr Gill can't bring himself to admit this fact yet. This issue would not have entered the discussion, had Dr Gill not brought it in by wrongly claiming there were 2^4. Yes, it is different issue from the pairing question but you are wrong that I'm confusing the two.

Now you say you do not "believe" picking the first pair of particles in a time-slot can allow violation of the inequality, a somewhat softer position compared to the earlier claim that it avoids it, and the recommendation that experimenters can avoid the loophole by doing such a procedure. I've given a counter-example, which shows this to be false. You continue to not believe it, without any explanation why my counter-example should fail (if at all).

Would you (and Dr Gill) go on the record to state that it is impossible for a local realistic model using time-delays to violate Bell's-inequalities if externally determined time-slots are used, and only the first pair of particles in the time-slot considered?

You say as much in your papers, but re-affirming your positions here will at least demonstrate you understand the arguments but continue to believe they do not impact your previous statements at all. If you would do that, I see no need to continue this discussion as my points have been explained clearly enough. If however, your arguments are simply a matter of posturing, you may be reluctant to go on the record again repeating the same arguments I claim are wrong.

I understand the terminology very well. You accuse me of not understanding terminology, yet you do not say what terminology, or what misunderstanding, and how it changes the validity of my argument here. Please point out a specific misunderstanding from any of my posts on this topic.

Dr Larsson: "the second can be handled by joining all the different lambdas into one vector-lambda with the individual lambdas as "coordinates""

Why don't you answer the question I've asked directly? I have explained how by joining all the different lambdas into one vector-lambda with the individual lambdas as coordinates, you still end up with different "vector-lambda" ensembles. Please review my posts from ( January 17th, 2015 2:49pm UTC ) and ( January 17th, 2015 10:27pm UTC ) and please explain how joining lambdas could possibly avoid the issue in that counter-example :

*** How exactly does picking the first pair of particles avoid the coincidence loophole as you have claimed, in that counter-example??? ***

Dr Larsson: "But he does not seem to accept that the Bell inequality applies to the theoretical unknown expectation values"

You are badly mistaken, my argument does not involve whether inequalities apply to theoretical unknown expectation values. The point, which both of you continue to avoid is the following: in your papers, you claim the coincidence-time loophole can be avoided in CHSH **experiments**. You suggest a method of picking pairs which does not do what you claim it does. I have presented a simple counter-example which demonstrates this. Do you want to go on the record to say that your suggested method is full-proof, despite my explanations?

Dr Larsson: "It seems that Peer 3 is confusing the fact that each experiment (each individual pair, each individual lambda) only can give values for one term (which I agree with), with his claim that using only the first detection in each time slot can give a false violation (which I do not believe)."

I'm not confusing anything. At least you admit the fact that there are 2^8 outcome combinations in a CHSH experiment. Dr Gill can't bring himself to admit this fact yet. This issue would not have entered the discussion, had Dr Gill not brought it in by wrongly claiming there were 2^4. Yes, it is different issue from the pairing question but you are wrong that I'm confusing the two.

Now you say you do not "believe" picking the first pair of particles in a time-slot can allow violation of the inequality, a somewhat softer position compared to the earlier claim that it avoids it, and the recommendation that experimenters can avoid the loophole by doing such a procedure. I've given a counter-example, which shows this to be false. You continue to not believe it, without any explanation why my counter-example should fail (if at all).

Would you (and Dr Gill) go on the record to state that it is impossible for a local realistic model using time-delays to violate Bell's-inequalities if externally determined time-slots are used, and only the first pair of particles in the time-slot considered?

You say as much in your papers, but re-affirming your positions here will at least demonstrate you understand the arguments but continue to believe they do not impact your previous statements at all. If you would do that, I see no need to continue this discussion as my points have been explained clearly enough. If however, your arguments are simply a matter of posturing, you may be reluctant to go on the record again repeating the same arguments I claim are wrong.

Peer 3:

( January 21st, 2015 11:11pm UTC )

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Dr Gill: "I've repeatedly asked Peer 3 to tell me what is wrong with Theorem 1 in http://arxiv.org/abs/1207.5103 but I get no response at all."

You get no response because we are discussing your errorneous claim that your recommended sampling method (of picking just the first pair in a time-slot) creates unbiased samples in a CHSH **experiment**, because a counter-example has been presented showing that it fails to do that.

Your theorem 1 is irrelevant to the issue being discussed because it simply assumes the sampling is unbiased. Not only that, it assumes falsely that there are 2^4 outcome combinations (what you call Nx4 spreadsheet), which immediately renders it irrelevant for the CHSH **experiment**. If you want to derive a better theorem, start with the correct number which is 2^8 add whatever assumptions you want to add to bring the number down to 2^4. You may notice that your equations (1) and (2) immediately fail once the fact that there are 2^8 outcome combinations is recognized. This may be the reason why you are reluctant to admit this fact here.

Your theorem 1 is not only irrelevant, it is completely useless for CHSH **experiments**. So when you say on page 3 that "My version is not an inequality about theoretical expectation values, but is a probabilistic inequality about experimentally observed averages" it is simply untrue.

Dr Gill: "Observed averages of A(a, lambda)B(b, lambda), over a number of time-slots, are then reasonable estimates of mean values of the same thing, if we assume that what happens in different time-slots are independent and identically distributed random variables "

The key sentence in the above is **if we assume that what happens in different time-slots are independent and identically distributed random variables**. My counter-example shows that you cannot assume that at all. In fact, your suggested method of picking pairs, systematically creates angle-dependent non-identically distributed random variables. This is the point you keep avoiding for some reason.

Dr Gill: "All we need is local realism and complete randomness of the setting choices, per time slot, independently of the hidden variables in the physical systems of particles, detectors, etc etc et."

**Would you go on the record to state that it is impossible for a local realistic model using time-delays to violate Bell's-inequalities if externally determined time-slots and completely random settings are used, and only the first pair of particles in the time-slot considered?**

You get no response because we are discussing your errorneous claim that your recommended sampling method (of picking just the first pair in a time-slot) creates unbiased samples in a CHSH **experiment**, because a counter-example has been presented showing that it fails to do that.

Your theorem 1 is irrelevant to the issue being discussed because it simply assumes the sampling is unbiased. Not only that, it assumes falsely that there are 2^4 outcome combinations (what you call Nx4 spreadsheet), which immediately renders it irrelevant for the CHSH **experiment**. If you want to derive a better theorem, start with the correct number which is 2^8 add whatever assumptions you want to add to bring the number down to 2^4. You may notice that your equations (1) and (2) immediately fail once the fact that there are 2^8 outcome combinations is recognized. This may be the reason why you are reluctant to admit this fact here.

Your theorem 1 is not only irrelevant, it is completely useless for CHSH **experiments**. So when you say on page 3 that "My version is not an inequality about theoretical expectation values, but is a probabilistic inequality about experimentally observed averages" it is simply untrue.

Dr Gill: "Observed averages of A(a, lambda)B(b, lambda), over a number of time-slots, are then reasonable estimates of mean values of the same thing, if we assume that what happens in different time-slots are independent and identically distributed random variables "

The key sentence in the above is **if we assume that what happens in different time-slots are independent and identically distributed random variables**. My counter-example shows that you cannot assume that at all. In fact, your suggested method of picking pairs, systematically creates angle-dependent non-identically distributed random variables. This is the point you keep avoiding for some reason.

Dr Gill: "All we need is local realism and complete randomness of the setting choices, per time slot, independently of the hidden variables in the physical systems of particles, detectors, etc etc et."

**Would you go on the record to state that it is impossible for a local realistic model using time-delays to violate Bell's-inequalities if externally determined time-slots and completely random settings are used, and only the first pair of particles in the time-slot considered?**

Donald A. Graft:

( January 22nd, 2015 1:50am UTC )

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The simulation must produce statistics like those of the Christensen et al experiment, including the fact that the two counting methods give contradictory results, i.e., including all the events does not violate the inequality, while picking just the first one per timeslot does violate it. It will also be shown that this (strangely) does not violate the theorems, for reasons I develop elsewhere. Infinitely long experiments exist only in the imagination; we must be sure that our experiments are long enough to converge properly, and we must have a way to know when they have converged. Convergence may be from above or below.

Richard Gill:

( January 22nd, 2015 1:58am UTC )

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Peer 3: indeed, it is impossible for a local realistic model using time-delays to violate Bell's-inequalities (*) if externally determined time-slots and completely random settings are used, and only the first pair of particles in the time-slot considered. This already follows from results of mine from 2001: http://arxiv.org/abs/quant-ph/0110137, "Accardi contra Bell (cum mundi): The Impossible Coupling". The purpose of that paper was to determine a minimal sample size for a public bet with my anti-Bellist friend Luigi Accardi, that his computer implementation of his "chameleon model" would fail to significantly violate CHSH when tested rigorously. Since then, my result has been refined in various ways, and is even nowadays being used by experimentalists in order to take care of the "memory loophole".

I would be very amazed and very excited if you could find a counter-example to the results in that paper. It sets out a protocol for a local realistic computer-simulation of a Bell-CHSH type experiment under a stringent experimental protocol. You (the local realist) are allowed to programme whatever you like as description of what goes on in and around each measurement station within each time slot. You may have any number of particles arriving and getting measured. You are allowed to make use of memory from past time-slots, also from the other wing of the experiment. You are merely obliged to commit yourself to some binary outcome in each side of the experiment, in each time-slot.

(*) We have to be precise here: there is of course statistical variation in the observed value of "S" of the order of 1 divided by square root of the number of experimental units. For instance, it is *possible* to observe S = 4. But if N is large the probability of this is vanishingly small. The bet with Accardi specified N = 25 000 and he would win if S was above the half-way mark between 2 and 2 sqrt 2. I calculated that my chance of losing the bet was less than 1 in a million.

I would be very amazed and very excited if you could find a counter-example to the results in that paper. It sets out a protocol for a local realistic computer-simulation of a Bell-CHSH type experiment under a stringent experimental protocol. You (the local realist) are allowed to programme whatever you like as description of what goes on in and around each measurement station within each time slot. You may have any number of particles arriving and getting measured. You are allowed to make use of memory from past time-slots, also from the other wing of the experiment. You are merely obliged to commit yourself to some binary outcome in each side of the experiment, in each time-slot.

(*) We have to be precise here: there is of course statistical variation in the observed value of "S" of the order of 1 divided by square root of the number of experimental units. For instance, it is *possible* to observe S = 4. But if N is large the probability of this is vanishingly small. The bet with Accardi specified N = 25 000 and he would win if S was above the half-way mark between 2 and 2 sqrt 2. I calculated that my chance of losing the bet was less than 1 in a million.

Jan-Åke Larsson:

( January 22nd, 2015 7:35am UTC )

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Peer 3: "Your theorem 1 is not only irrelevant, it is completely useless for CHSH **experiments**". You have given no argument to support this claim. I gave a counterargument in my comment at January 21st, 2015 10:07am UTC, pointing to Larsson, J.-A, 2014. Loopholes in Bell inequality tests of local realism. J. Phys. A 47, 424003. doi:10.1088/1751-8113/47/42/424003, http://arxiv.org/abs/1407.0363 . Again, please read subsection 1.5 from start to end, but especially the text surrounding equation (15).

There I state clearly that the bound does in fact not apply to individual experiments, just as you say, but to the expectation values. Also, I agree that the CHSH inequality does not apply to the sample mean. But with enough data, the sample mean will be close to the expectation value, and the deviation can be quantified. This means that an experimental violation, where the sample mean exceeds the bound, will actually tell us something about the possibility that the (theoretical, unknown) expectation value violates the CHSH inequality.

This means that Richard's theorem 1 is relevant and completely useful for CHSH **experiments**

There I state clearly that the bound does in fact not apply to individual experiments, just as you say, but to the expectation values. Also, I agree that the CHSH inequality does not apply to the sample mean. But with enough data, the sample mean will be close to the expectation value, and the deviation can be quantified. This means that an experimental violation, where the sample mean exceeds the bound, will actually tell us something about the possibility that the (theoretical, unknown) expectation value violates the CHSH inequality.

This means that Richard's theorem 1 is relevant and completely useful for CHSH **experiments**

Richard Gill:

( January 22nd, 2015 1:52pm UTC )

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Donald Graft: "Infinitely long experiments exist only in the imagination; we must be sure that our experiments are long enough to converge properly, and we must have a way to know when they have converged. Convergence may be from above or below." Exactly. Hence my finite N probability bound versions of CHSH on which an Unregistered Submitter has started a new thread: https://pubpeer.com/publications/D985B475C637F666CC1D3E3A314522

Peer 3:

( January 22nd, 2015 4:00pm UTC )

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Dr Larsson: " You have given no argument to support this claim. "

Note:

1) Theorem 1 assumes there are 2^4 possible outcome combinations in a CHSH experiment, which is false. It should be 2^8 as you acknowledge in your paper (see equation 14 of http://arxiv.org/abs/1407.0363). A theorem which is based on a clearly false assumption is useless. I've given Dr Gill a suggestion on how to fix it. Start with 2^8 and provide additional reasonable assumptions which should reduce it to 2^4. Until then Theorem 1 is a dead-issue.

2) Note that neither of you have yet responded directly to how picking the first-pair of particles creates unbiased samples for the counter-example from ( January 17th, 2015 2:49pm UTC ) and ( January 17th, 2015 10:27pm UTC ) which show how your pair-picking methods fail. Instead you both want to discuss other issues and avoid the key issue being raised.

Dr Larsson: " I gave a counterargument in my comment at January 21st, 2015 10:07am UTC, pointing to Larsson, J.-A, 2014. Loopholes in Bell inequality tests of local realism. J. Phys. A 47, 424003. doi:10.1088/1751-8113/47/42/424003, http://arxiv.org/abs/1407.0363 . Again, please read subsection 1.5 from start to end, but especially the text surrounding equation (15). "

I'm dissapointed that you are unable to understand the argument being made, as evidenced by your suggestion above that section 1.5 addresses the issue. It does not. I have read your paper, carefully, but it doesn't seem you have read any of the points I've raised carefully. You have latched onto something you believe is my argument and you keep repeating it without actually reading the point I'm making.

For example: In section 1.5 of your 1407.0363, you state that:

"Individual experimental outcomes are simply not bounded by the inequality. This is

natural because of the random nature of the experiment, but could be thought of as an

unexpected circumstance, a loophole".

You start with a truth, that individual experiments are not bounded by the inequality, but then you add a "lie" to it. The reason individual experiments are not bounded by the CHSH is not because of any experimental problem, or randomness. It is simply due to the fact that there are 2^8 outcome combinations for individual experiments, not 2^4. You sample 8 times, you get 2^8. You sample 4 times you get 2^4. There is no loophole or experimental inadequacy involved here.

ab + ad + cb - cd <= 2 if you sample 4 times (a,b,c,d = +/-1)

ab + cd + ef - gh <= 4 if you sample 8 times (a,b,c,d,e,f,g,h = +/-1)

It is simple arithmetic, you should not expect the second scenario to be bounded by the inequality from the first scenario.

Next you say "The problem here is the finite sample size used in experiment[sic], above a single sample for each setting combination." Again, this is wrong. Sample size does not play any role to the bounds of "individual experiments", and repeating the experiment a large number of times will not change the fact that you have 2^8 outcome combinations. The problem is simply the fact that the CHSH is derived assuming 2^4 outcome combinations.

Further, you say "This loophole has also been observed under a different guise, because of the following observation: The different terms in [Equation 14] use different values of the hidden variable [lambda(i)], while the CHSH proof uses the same [lambda] for all four measurement setting combinations. Therefore the CHSH inequality does not apply to that sum of measurement values -- which is completely true, and indeed the reason for the nonzero probability in equation"

I fully agree with this. In fact, this is the same point you have made in the earlier paper that **The problem here is that the ensemble on which the correlations are evaluated changes with the settings,while the original Bell inequality requires that they stay the same**

Then you say this: "However, although the CHSH inequality does not apply to the above combination of individual samples, it does apply to the distribution mean. And the sample mean will tend to the distribution mean as the sample size grows; the adherence to the CHSH inequality will become better and better."

This is the core of the issue. The above statement is false (unless you add the caveat: **Provided the samples are unbiased and drawn from the same population, or stationary process**) See the counter example I have provided. Given your method of picking-pairs, no matter now large the sample, the ensembles never converge to be the same. This is the issue you still have made no attempt to answer. The reason you probably fail to see this is because your statement contains a hidden assumption: you have assumed that the problem is one of "finite statistics with unbiased sampling". If that were the case, you would be correct that a larger sample would resolve it. However, the issue I'm raising here is the problem with your pair-picking method is one of "systematically biased sampling". That cannot be fixed by taking larger samples. Do you disagree with this?

If you still disagree, could you please explain how taking larger samples resolves the ensemble differences highlighted in ( January 17th, 2015 2:49pm UTC ) and ( January 17th, 2015 10:27pm UTC )?

Note:

1) Theorem 1 assumes there are 2^4 possible outcome combinations in a CHSH experiment, which is false. It should be 2^8 as you acknowledge in your paper (see equation 14 of http://arxiv.org/abs/1407.0363). A theorem which is based on a clearly false assumption is useless. I've given Dr Gill a suggestion on how to fix it. Start with 2^8 and provide additional reasonable assumptions which should reduce it to 2^4. Until then Theorem 1 is a dead-issue.

2) Note that neither of you have yet responded directly to how picking the first-pair of particles creates unbiased samples for the counter-example from ( January 17th, 2015 2:49pm UTC ) and ( January 17th, 2015 10:27pm UTC ) which show how your pair-picking methods fail. Instead you both want to discuss other issues and avoid the key issue being raised.

Dr Larsson: " I gave a counterargument in my comment at January 21st, 2015 10:07am UTC, pointing to Larsson, J.-A, 2014. Loopholes in Bell inequality tests of local realism. J. Phys. A 47, 424003. doi:10.1088/1751-8113/47/42/424003, http://arxiv.org/abs/1407.0363 . Again, please read subsection 1.5 from start to end, but especially the text surrounding equation (15). "

I'm dissapointed that you are unable to understand the argument being made, as evidenced by your suggestion above that section 1.5 addresses the issue. It does not. I have read your paper, carefully, but it doesn't seem you have read any of the points I've raised carefully. You have latched onto something you believe is my argument and you keep repeating it without actually reading the point I'm making.

For example: In section 1.5 of your 1407.0363, you state that:

"Individual experimental outcomes are simply not bounded by the inequality. This is

natural because of the random nature of the experiment, but could be thought of as an

unexpected circumstance, a loophole".

You start with a truth, that individual experiments are not bounded by the inequality, but then you add a "lie" to it. The reason individual experiments are not bounded by the CHSH is not because of any experimental problem, or randomness. It is simply due to the fact that there are 2^8 outcome combinations for individual experiments, not 2^4. You sample 8 times, you get 2^8. You sample 4 times you get 2^4. There is no loophole or experimental inadequacy involved here.

ab + ad + cb - cd <= 2 if you sample 4 times (a,b,c,d = +/-1)

ab + cd + ef - gh <= 4 if you sample 8 times (a,b,c,d,e,f,g,h = +/-1)

It is simple arithmetic, you should not expect the second scenario to be bounded by the inequality from the first scenario.

Next you say "The problem here is the finite sample size used in experiment[sic], above a single sample for each setting combination." Again, this is wrong. Sample size does not play any role to the bounds of "individual experiments", and repeating the experiment a large number of times will not change the fact that you have 2^8 outcome combinations. The problem is simply the fact that the CHSH is derived assuming 2^4 outcome combinations.

Further, you say "This loophole has also been observed under a different guise, because of the following observation: The different terms in [Equation 14] use different values of the hidden variable [lambda(i)], while the CHSH proof uses the same [lambda] for all four measurement setting combinations. Therefore the CHSH inequality does not apply to that sum of measurement values -- which is completely true, and indeed the reason for the nonzero probability in equation"

I fully agree with this. In fact, this is the same point you have made in the earlier paper that **The problem here is that the ensemble on which the correlations are evaluated changes with the settings,while the original Bell inequality requires that they stay the same**

Then you say this: "However, although the CHSH inequality does not apply to the above combination of individual samples, it does apply to the distribution mean. And the sample mean will tend to the distribution mean as the sample size grows; the adherence to the CHSH inequality will become better and better."

This is the core of the issue. The above statement is false (unless you add the caveat: **Provided the samples are unbiased and drawn from the same population, or stationary process**) See the counter example I have provided. Given your method of picking-pairs, no matter now large the sample, the ensembles never converge to be the same. This is the issue you still have made no attempt to answer. The reason you probably fail to see this is because your statement contains a hidden assumption: you have assumed that the problem is one of "finite statistics with unbiased sampling". If that were the case, you would be correct that a larger sample would resolve it. However, the issue I'm raising here is the problem with your pair-picking method is one of "systematically biased sampling". That cannot be fixed by taking larger samples. Do you disagree with this?

If you still disagree, could you please explain how taking larger samples resolves the ensemble differences highlighted in ( January 17th, 2015 2:49pm UTC ) and ( January 17th, 2015 10:27pm UTC )?

Richard Gill:

( January 22nd, 2015 4:27pm UTC )

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Peer 3, the assumption which reduces 2^8 to 2^4 is called "local realism", together with the assumption that settings are sampled completely at random and independently of the physical hidden variables in the systems and measurement devices (which is allowed by the no-conspiracy assumption).

You don't see how my Theorem 1 (or something similar) resolves your issue. Plenty of other people do seem to understand the points Larsson and I make. But you are not alone: see http://arxiv.org/abs/1412.6987 by Andrei Khrennikov, or http://arxiv.org/abs/quant-ph/0006014 by Giullaume Adenier, or http://arxiv.org/abs/1108.3583 by Karl Hess, Hans De Raedt, and Kristel Michielsen. These are all "2^8-ists", various different flavours. This has been discussed so many times and refuted so many times, and no doubt that will go on for ever. New people keep coming into the field. Great! They sometimes bring in important new ideas. I don't know where you are based but I invite you to visit me in person so we can discuss these things face to face, at a blackboard. It is clear that we are not communicating to one another via the PubPeer interface. Please send me a personal email and we can see if we can arrange something.

I can only suggest that you write up your ideas, post a paper on arXiv, and submit to a journal. Or construct a simulation model and post it on internet. Preferably both.

Your so-called "counter-example" is a story which explains why different ways to select events in each time-slot will lead to different correlations. It does not explain why this could lead to improper violation of CHSH, because it can't. It can't, because of Theorem 1 (or similar results). It's up to you to give *proof* of your claim, either by an explicit mathematical model or by a simulation programme, preferably by both. I think that you will find it hard to do so. But if you can show that Bell's theorem is false, that will be very exciting, it will be revolutionary.

You don't see how my Theorem 1 (or something similar) resolves your issue. Plenty of other people do seem to understand the points Larsson and I make. But you are not alone: see http://arxiv.org/abs/1412.6987 by Andrei Khrennikov, or http://arxiv.org/abs/quant-ph/0006014 by Giullaume Adenier, or http://arxiv.org/abs/1108.3583 by Karl Hess, Hans De Raedt, and Kristel Michielsen. These are all "2^8-ists", various different flavours. This has been discussed so many times and refuted so many times, and no doubt that will go on for ever. New people keep coming into the field. Great! They sometimes bring in important new ideas. I don't know where you are based but I invite you to visit me in person so we can discuss these things face to face, at a blackboard. It is clear that we are not communicating to one another via the PubPeer interface. Please send me a personal email and we can see if we can arrange something.

I can only suggest that you write up your ideas, post a paper on arXiv, and submit to a journal. Or construct a simulation model and post it on internet. Preferably both.

Your so-called "counter-example" is a story which explains why different ways to select events in each time-slot will lead to different correlations. It does not explain why this could lead to improper violation of CHSH, because it can't. It can't, because of Theorem 1 (or similar results). It's up to you to give *proof* of your claim, either by an explicit mathematical model or by a simulation programme, preferably by both. I think that you will find it hard to do so. But if you can show that Bell's theorem is false, that will be very exciting, it will be revolutionary.

Peer 3:

( January 22nd, 2015 5:32pm UTC )

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Dr Gill: "Plenty of other people do seem to understand the points Larsson and I make."

Perhaps you should consider that those people do not think your points make any sense at all. I would agree with them. You will never understand what we are saying if you simply dismiss us by saying "physicists do not understand statistics", instead of addressing specific arguments.

Dr Gill: "the assumption which reduces 2^8 to 2^4 is called "local realism", together with the assumption that settings are sampled completely at random and independently of the physical hidden variables in the systems and measurement devices (which is allowed by the no-conspiracy assumption".

The above is simply false:

ab + ad + cb - cd is less than or equal to 2 if you sample 4 times randomly and independently

ab + cd + ef - gh is less than or equal to 4 if you sample 8 times randomly and independently

This is true irrespective of local realism or any hidden variables. It is a mathematical fact of combinations.

If you have 4 types of local realistic coins labelled A1, B1, A2, B2, each with sides labelled +1 and -1. All the A1 coins are identical to every other A1 coin and same for B1, etc. You take pairs of each type of coin and give to 8 individuals grouped into 4 pairs. To the first pair you give coins of type (A1, B1) and to the 3 other pairs, you give coin-pairs (A1, B2), (A2, B1), (A2, B2). Each individual tosses their coin just once. You get 2^8, not 2^4!

There is only one possible assumption you could make to reduce 2^8 to 2^4:

**That for some reason the outcome from the A1 coin tossed by the person in the (A1, B1) pair, is always the same outcome obtained from the A1 coin tossed by the person in the (A1, B2) pair and the same for A2, B1 and B2. **

This is possible only only if

1) The coins of the same type are so heavily biased, that they always produce the same outcome when tossed, OR

2) There is some mystical superluminal or backward in time communication between all coins of the same type so that they can arrange to have the same outcome.

Now if this is what you call "local-realism" and "no-conspiracy", then indeed it reduces the combinations to 2^4. But then you see why I do not take your arguments seriously.

Note, that the expectation values of the outcomes observed for coin A1 tossed by the person in the (A1, B2) pair, will tend toward the same value as the expectation value of the A1 coin tossed by the person in the (A1, B1) pair, as the number of tosses increases. However, you will still always have 2^8 possible outcome combinations, not 2^4! Because the coins are of the same type NOT the same coin.

Perhaps you should consider that those people do not think your points make any sense at all. I would agree with them. You will never understand what we are saying if you simply dismiss us by saying "physicists do not understand statistics", instead of addressing specific arguments.

Dr Gill: "the assumption which reduces 2^8 to 2^4 is called "local realism", together with the assumption that settings are sampled completely at random and independently of the physical hidden variables in the systems and measurement devices (which is allowed by the no-conspiracy assumption".

The above is simply false:

ab + ad + cb - cd is less than or equal to 2 if you sample 4 times randomly and independently

ab + cd + ef - gh is less than or equal to 4 if you sample 8 times randomly and independently

This is true irrespective of local realism or any hidden variables. It is a mathematical fact of combinations.

If you have 4 types of local realistic coins labelled A1, B1, A2, B2, each with sides labelled +1 and -1. All the A1 coins are identical to every other A1 coin and same for B1, etc. You take pairs of each type of coin and give to 8 individuals grouped into 4 pairs. To the first pair you give coins of type (A1, B1) and to the 3 other pairs, you give coin-pairs (A1, B2), (A2, B1), (A2, B2). Each individual tosses their coin just once. You get 2^8, not 2^4!

There is only one possible assumption you could make to reduce 2^8 to 2^4:

**That for some reason the outcome from the A1 coin tossed by the person in the (A1, B1) pair, is always the same outcome obtained from the A1 coin tossed by the person in the (A1, B2) pair and the same for A2, B1 and B2. **

This is possible only only if

1) The coins of the same type are so heavily biased, that they always produce the same outcome when tossed, OR

2) There is some mystical superluminal or backward in time communication between all coins of the same type so that they can arrange to have the same outcome.

Now if this is what you call "local-realism" and "no-conspiracy", then indeed it reduces the combinations to 2^4. But then you see why I do not take your arguments seriously.

Note, that the expectation values of the outcomes observed for coin A1 tossed by the person in the (A1, B2) pair, will tend toward the same value as the expectation value of the A1 coin tossed by the person in the (A1, B1) pair, as the number of tosses increases. However, you will still always have 2^8 possible outcome combinations, not 2^4! Because the coins are of the same type NOT the same coin.

Unregistered Submission:

( January 22nd, 2015 5:53pm UTC )

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Tossing of coins is so different from preparing a pair of particles in singlet state that it seems to cause confusion istead of any clarification.

Donald A. Graft:

( January 22nd, 2015 6:04pm UTC )

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The theorems all assume that the experiment is infinitely long or "long enough". But real experiments are short, i.e., not long enough for full convergence to occur and thereby make the theorems applicable. Any experiment that is short in this way leaves an open space for artifactual violation. I am preparing a paper applying this idea to the Christensen et al experiment; it will show that the published experimental data does not converge, that the observed difference between the two counting methods is thereby easily explained, and that the reported violation is seen to be artifactual. Nature is seen to be local, just as we expect.

I intentionally here do not comment on the possibility of setting-dependent delays (in conjunction with picking only one event per timeslot) also being a possible mechanism for false violations, nor do I require that mechanism to account for the results of Christensen et al. Peer 3 will surely continue that debate, and it will be interesting to see the outcome. Does the Larsson-Gill selection method succeed in its goal of eliminating unfair sampling due to delay modulation?

Jan-Åke Larsson:

( January 22nd, 2015 7:41pm UTC )

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Peer 3: if I understand your latest response to me correctly, you are not actually saying that theorem 1 is false, which is what I first thought. Section 1.5 of my paper is a response to that.

This latest response from you instead seems to claim that theorem 1 does not apply in the situation when the result is given by the first lambda (that gives a detection) in a time slot with several lambdas that can be re-ordered depending on the setting.

I want to make clear that this is your claim, so that I do not argue against a different statement. OK?

This latest response from you instead seems to claim that theorem 1 does not apply in the situation when the result is given by the first lambda (that gives a detection) in a time slot with several lambdas that can be re-ordered depending on the setting.

I want to make clear that this is your claim, so that I do not argue against a different statement. OK?

Unregistered Submission:

( January 24th, 2015 1:07pm UTC )

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Correction: the sample mean is slightly biased, asymptotically unbiased, and strongly consistent.

Jan-Åke Larsson:

( January 26th, 2015 6:41am UTC )

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If you are doing an IID sequence of measurements (simplifying assumptions to reduce the amount of mathematical machinery), the expectation value of the sample mean is the expectation value of the random variable being sampled.

But my last point rather concerns what the actual claim of Peer 3 is.

But my last point rather concerns what the actual claim of Peer 3 is.

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Unregistered Submission:

( January 22nd, 2015 9:38pm UTC )

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Richard Gill Wrote: *Regarding the possibility of some kind of mistake in Larsson and Gill, it is worth mentioning the following. In his recent book "Einstein was right" Karl Hess argues that Hess and Philipp had already discovered/invented the coincidence loophole and that Larsson and Gill shamelessly stole it from them.*

After reading their paper http://iopscience.iop.org/0295-5075/57/6/775, and then reading your reponse, http://arxiv.org/pdf/quant-ph/0204169.pdf, and also your your follow-up paper (the subject of this thread), it appears rather obvious that they were right and the original inequalities did neglect time dependencies. Your second paper though seems to contract the your first, for it admits that time dependences are a serious problem for Bell's inequalities and you attempt to fix it by deriving new inequalities. One may question why you would want to fix a problem you already argued was inexistent and a "red-herring".

So I'm afraid any rational mind would conclude that you thrashed their paper, turned around and wrote the paper with Larsson pretty much agreeing with what they said.

After reading their paper http://iopscience.iop.org/0295-5075/57/6/775, and then reading your reponse, http://arxiv.org/pdf/quant-ph/0204169.pdf, and also your your follow-up paper (the subject of this thread), it appears rather obvious that they were right and the original inequalities did neglect time dependencies. Your second paper though seems to contract the your first, for it admits that time dependences are a serious problem for Bell's inequalities and you attempt to fix it by deriving new inequalities. One may question why you would want to fix a problem you already argued was inexistent and a "red-herring".

So I'm afraid any rational mind would conclude that you thrashed their paper, turned around and wrote the paper with Larsson pretty much agreeing with what they said.

Jan-Åke Larsson:

( January 23rd, 2015 8:02am UTC )

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Their paper http://iopscience.iop.org/0295-5075/57/6/775 does not use time dependencies of the type we consider in our paper. In that paper, they propose that the hidden variable depends on time. This would make the hidden variable a function of time, something which is covered by Bell's analysis, since Bell makes no assumptions on what the hidden variable is. Real numbers, points in n-dimensional space, functions of time, all are valid hidden variables.

Richard's paper http://arxiv.org/pdf/quant-ph/0204169.pdf is a response to four other papers in which Hess and Philipp claim to present a counterexample. I also went through their counterexample (with Philipp, ~2002), which took some considerable time due to the complicated model. That model does violate CHSH, but not because of inclusion of time dependence. The violation arises because the model is nonlocal, there is a dependence on the remote setting. This is not immediately clear because Hess and Phillip drop an index in the middle of the calculation.

Richard asked me later that year: "Hess and Phillip are claiming that time dependence can cause a false violation of CHSH. Their approach doesn't work, but is there any chance that this could happen?" After a day or so I had a working local hidden-variable model that gave a false violation (so the answer is "yes"), but not of the type Hess and Phillip proposed. In my example, there are time delays that are such that not all pairs gave coincidences. Knowing that there would be limits to a violation of this kind, we then derived a modified inequality.

So what happened was that Richard (et al) argued against Hess and Phillip's model (that doesn't work), and then started thinking about the actual problem. Which we solved. And published the results. This is science: you discuss with your peers, connect your own knowledge with new ideas, and produce new results.

Richard's paper http://arxiv.org/pdf/quant-ph/0204169.pdf is a response to four other papers in which Hess and Philipp claim to present a counterexample. I also went through their counterexample (with Philipp, ~2002), which took some considerable time due to the complicated model. That model does violate CHSH, but not because of inclusion of time dependence. The violation arises because the model is nonlocal, there is a dependence on the remote setting. This is not immediately clear because Hess and Phillip drop an index in the middle of the calculation.

Richard asked me later that year: "Hess and Phillip are claiming that time dependence can cause a false violation of CHSH. Their approach doesn't work, but is there any chance that this could happen?" After a day or so I had a working local hidden-variable model that gave a false violation (so the answer is "yes"), but not of the type Hess and Phillip proposed. In my example, there are time delays that are such that not all pairs gave coincidences. Knowing that there would be limits to a violation of this kind, we then derived a modified inequality.

So what happened was that Richard (et al) argued against Hess and Phillip's model (that doesn't work), and then started thinking about the actual problem. Which we solved. And published the results. This is science: you discuss with your peers, connect your own knowledge with new ideas, and produce new results.

Richard Gill:

( January 23rd, 2015 8:33am UTC )

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Well now we are talking about ancient history. The original Hess and Philipp work (two papers in PNAS, 2001, doi: 10.1073/pnas.251525098, doi: 10.1073/pnas.251524998) contained a lot of talk, and a very elaborate mathematical construction of a local hidden variables model which, if it was correct, would reproduce the singlet correlations in a loophole-free experiment (no post-selection, no conspiracy loohole, ...). The mathematical model was supposed to back up the verbal stuff, kind of "proof of concept".

There was no talk of post-selection of any kind. Their mathematical theorem (that their LHV model reproduced the singlet correlations) contradicted existing rigorous mathematical statements of Bell's theorem, and the notion that memory effects would spoil Bell's theorem (time dependence), contradicted a theorem of mine from 2001 http://arxiv.org/abs/quant-ph/0110137 about the impossibility of computer simulated local realist violation of CHSH in a loophole-free setting (but allowing time dependence, ie the memory loophole).

There was an error in the mathematical construction at the heart of doi: 10.1073/pnas.251525098. They had forgotten an index and not normalised a probability measure. My theorem is true, the memory loophole is not a loophole, "time" does not destroy Bell's theorem.

Their model was so complicated that they considered it an existence theorem. They were certainly not going to implement it as a computer simulation, and didn't recommend anyone to try.

So the mathematical part of their work was just yet another construction of a probability distribution of a variable lambda, and functions A(a, lambda) and B(b, lambda) taking values +/-1, such that ... . The proof was very difficult. Please study doi: 10.1073/pnas.251525098 and let us know if you can find the error! (There has to be one, right?) It is very well hidden.

Please read their 2001 PNAS papers and ask yourself if they are writing about the detection loophole or the coincidence loophole. Even with hindsight. Of course at that time the coincidence loophole had not been named. Hess writes himself that he only learnt about the detection loophole years later when he coincidentally met Pearle.

I tried to explain this many times to Karl (physicist) and Walter (pure mathematician) but they did not want to understand. Karl called me a pretzel and a third-rater and blocked my emails. They had short tempers and a lot of self-confidence, they were certain that their work was revolutionary. I was just some statistician (not a physicist, not even really a mathematician). Of course their work was controversial and they went after, and got, a lot of publicity, hence also a lot of criticism. The experts could not make sense of what they had written but knew it couldn't be correct. Perhaps because they are both originally from Austria, Zeilinger got involved.

I read that Hess and Philipp had disproved Bell's theorem, and that the Bohr-Einstein debate was now settled in favour of Einstein, in the science supplement of my Dutch daily newspaper. So it was a big event, world-wide! I wrote two papers with Weihs, Zukowski and Zeilinger. One of them in PNAS in 2002: doi: 10.1073/pnas.182536499. And started discussing a genuine time issue connected to post-selection (coincidence loophole) with Jan-Ake. As far as we knew nobody had thought about it before, but later it turned out that Saverio Pascazio had brought it up in a pretty much universally ignored PLA paper in 1986. It was great fun.

The stakes were high. Emotions ran high. Tempers were lost. You will not read anything in Hess' book about the embarrassing mistake in Philipp's math. Since Hess (like Einstein) is always right, and since Larsson-Gill is easy to read and evidently correct, it had to be what the Hess-Philipp papers had really been about all the time, but nobody had realised that ... until we had successfully decoded the papers and stolen the idea for ourselves.

There was no talk of post-selection of any kind. Their mathematical theorem (that their LHV model reproduced the singlet correlations) contradicted existing rigorous mathematical statements of Bell's theorem, and the notion that memory effects would spoil Bell's theorem (time dependence), contradicted a theorem of mine from 2001 http://arxiv.org/abs/quant-ph/0110137 about the impossibility of computer simulated local realist violation of CHSH in a loophole-free setting (but allowing time dependence, ie the memory loophole).

There was an error in the mathematical construction at the heart of doi: 10.1073/pnas.251525098. They had forgotten an index and not normalised a probability measure. My theorem is true, the memory loophole is not a loophole, "time" does not destroy Bell's theorem.

Their model was so complicated that they considered it an existence theorem. They were certainly not going to implement it as a computer simulation, and didn't recommend anyone to try.

So the mathematical part of their work was just yet another construction of a probability distribution of a variable lambda, and functions A(a, lambda) and B(b, lambda) taking values +/-1, such that ... . The proof was very difficult. Please study doi: 10.1073/pnas.251525098 and let us know if you can find the error! (There has to be one, right?) It is very well hidden.

Please read their 2001 PNAS papers and ask yourself if they are writing about the detection loophole or the coincidence loophole. Even with hindsight. Of course at that time the coincidence loophole had not been named. Hess writes himself that he only learnt about the detection loophole years later when he coincidentally met Pearle.

I tried to explain this many times to Karl (physicist) and Walter (pure mathematician) but they did not want to understand. Karl called me a pretzel and a third-rater and blocked my emails. They had short tempers and a lot of self-confidence, they were certain that their work was revolutionary. I was just some statistician (not a physicist, not even really a mathematician). Of course their work was controversial and they went after, and got, a lot of publicity, hence also a lot of criticism. The experts could not make sense of what they had written but knew it couldn't be correct. Perhaps because they are both originally from Austria, Zeilinger got involved.

I read that Hess and Philipp had disproved Bell's theorem, and that the Bohr-Einstein debate was now settled in favour of Einstein, in the science supplement of my Dutch daily newspaper. So it was a big event, world-wide! I wrote two papers with Weihs, Zukowski and Zeilinger. One of them in PNAS in 2002: doi: 10.1073/pnas.182536499. And started discussing a genuine time issue connected to post-selection (coincidence loophole) with Jan-Ake. As far as we knew nobody had thought about it before, but later it turned out that Saverio Pascazio had brought it up in a pretty much universally ignored PLA paper in 1986. It was great fun.

The stakes were high. Emotions ran high. Tempers were lost. You will not read anything in Hess' book about the embarrassing mistake in Philipp's math. Since Hess (like Einstein) is always right, and since Larsson-Gill is easy to read and evidently correct, it had to be what the Hess-Philipp papers had really been about all the time, but nobody had realised that ... until we had successfully decoded the papers and stolen the idea for ourselves.

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Richard Gill:

( January 23rd, 2015 11:05am UTC )

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Peer 3: you write (but I correct one term in the second line):

ab + ad + cb - cd is less than or equal to 2 if you sample 4 times randomly and independently

ab + cd + ef - gh is less than or equal to 4 if you sample 8 times randomly and independently

I think you mean that

(1) ab + ad + cb - cd is less than or equal to 2 if a, b, c, and d are four arbitrary numbers in the set {-1, +1}.

(2) ab + cd + de - gh is less than or equal to 4 if a, b, c, d, e, f, g, h are eight arbitrary numbers in the set {-1, +1}.

Forget about sampling randomly and independently. Let's save the probability theory (or statistics) for a moment.

Now you also agreed with me that if local realism is true, then one can define a, b, c, d to be the outcomes which would have been observed of Alice and Bob had chosen settings one or two: a = A1, b = B1, c = A2, d = B2. And these do satisfy (1).

Define S = setting chosen by Alice, T = setting chosen by Bob (taking values 1 or 2), and suppose they are completely random and independent of a, b, c, d. Use I(...) to denote an indicator random variable. Then the expectation value of

I((S, T) = (1, 1)) a b

+ I((S, T) = (1, 2)) a d

+ ((S, T) = (2, 1)) c b

- I((S, T) = (2, 2)) c d

is 1/4 times (ab + ad + cb - cd) which is therefore less than or equal to 1/2

Add N independent copies (the independence is in the coin tosses S, T; the numbers a, b, c, d can be different every time, they are completely arbitrary) and divide by N/4. The expectation value of the result is less than or equal to 2. Now we are just about home, since each of the four indicator variables will have been equal to 1 on about N/4 of the occasions. We have (almost) recovered the expectation value of the CHSH quantity S. For large N the difference is negligeable.

Please do think about independent fair coin tosses instead of singlet correlations for a while. Sure, it's confusing if you are not used to probability theory. You have to do a bit of work and a bit of abstract (pure mathematical) thinking. But there is a big pay-off.

ab + ad + cb - cd is less than or equal to 2 if you sample 4 times randomly and independently

ab + cd + ef - gh is less than or equal to 4 if you sample 8 times randomly and independently

I think you mean that

(1) ab + ad + cb - cd is less than or equal to 2 if a, b, c, and d are four arbitrary numbers in the set {-1, +1}.

(2) ab + cd + de - gh is less than or equal to 4 if a, b, c, d, e, f, g, h are eight arbitrary numbers in the set {-1, +1}.

Forget about sampling randomly and independently. Let's save the probability theory (or statistics) for a moment.

Now you also agreed with me that if local realism is true, then one can define a, b, c, d to be the outcomes which would have been observed of Alice and Bob had chosen settings one or two: a = A1, b = B1, c = A2, d = B2. And these do satisfy (1).

Define S = setting chosen by Alice, T = setting chosen by Bob (taking values 1 or 2), and suppose they are completely random and independent of a, b, c, d. Use I(...) to denote an indicator random variable. Then the expectation value of

I((S, T) = (1, 1)) a b

+ I((S, T) = (1, 2)) a d

+ ((S, T) = (2, 1)) c b

- I((S, T) = (2, 2)) c d

is 1/4 times (ab + ad + cb - cd) which is therefore less than or equal to 1/2

Add N independent copies (the independence is in the coin tosses S, T; the numbers a, b, c, d can be different every time, they are completely arbitrary) and divide by N/4. The expectation value of the result is less than or equal to 2. Now we are just about home, since each of the four indicator variables will have been equal to 1 on about N/4 of the occasions. We have (almost) recovered the expectation value of the CHSH quantity S. For large N the difference is negligeable.

Please do think about independent fair coin tosses instead of singlet correlations for a while. Sure, it's confusing if you are not used to probability theory. You have to do a bit of work and a bit of abstract (pure mathematical) thinking. But there is a big pay-off.

Peer 3:

( January 23rd, 2015 1:44pm UTC )

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Dr Gill: That you would proceed with just (ab + ad + cb - cd) even though the CHSH experiment has (ab + cd + ef - gh) tells me alot. I find it is a waste of time trying to engage you. Please attempt to at least understand the points being made because it is painful to repeat it over and over and you keep presenting irrelevant arguments which do not address the key issue.

If on the other hand you are ready to admit there are 2^8 possible outcome combinations in a CHSH experiment, that would be a good point to start then we can work through this piece by piece. That is the starting point. There is no reason we can't use this as a white board. First we state clearly what we agree on and then we go to the next step, bit by bit.

***Are you ready to agree there are here are 2^8 possible outcome combinations in a CHSH experiment?***

If you are not, ready to answer with a simple yes/no, then we are done here I'm afraid.

If on the other hand you are ready to admit there are 2^8 possible outcome combinations in a CHSH experiment, that would be a good point to start then we can work through this piece by piece. That is the starting point. There is no reason we can't use this as a white board. First we state clearly what we agree on and then we go to the next step, bit by bit.

***Are you ready to agree there are here are 2^8 possible outcome combinations in a CHSH experiment?***

If you are not, ready to answer with a simple yes/no, then we are done here I'm afraid.

Richard Gill:

( January 23rd, 2015 3:23pm UTC )

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You already had my answer: a simple no.

In one run (I mean one time-slot, or experimental unit) we observe four binary variables (two random settings and two outcomes). Under local realism these are certain functions of six binary variables (two random settings and four counterfactual outcomes under both possible settings in each wing of the experiment).

In one run (I mean one time-slot, or experimental unit) we observe four binary variables (two random settings and two outcomes). Under local realism these are certain functions of six binary variables (two random settings and four counterfactual outcomes under both possible settings in each wing of the experiment).

Peer 3:

( January 23rd, 2015 5:02pm UTC )

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***Are you ready to agree there are here are 2^8 possible outcome combinations in a CHSH experiment?***

Dr Gill: "You already had my answer: a simple no."

I guess this concludes my discussion with you. See equations (14) and (15) of Dr Larsson's paper http://arxiv.org/pdf/1407.0363v1 If you still do not understand that there are 2^8 possible outcome combinations, I'll leave it up to Dr Larsson to explain those equations to you.

Dr Gill: "You already had my answer: a simple no."

I guess this concludes my discussion with you. See equations (14) and (15) of Dr Larsson's paper http://arxiv.org/pdf/1407.0363v1 If you still do not understand that there are 2^8 possible outcome combinations, I'll leave it up to Dr Larsson to explain those equations to you.

Richard Gill:

( January 24th, 2015 6:27am UTC )

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Thank you Peer 3. I hope indeed that this concludes the discussion on the 2^8 versus 2^4 issue, which in my opinion is a non-issue. A red herring. Possibly it revolves merely around misunderstanding of notation.

The link you give to Larsson's excellent and comprehensive survey paper doesn't seem to work. But http://arxiv.org/pdf/1407.0363.pdf automatically gives the latest version. It is worth studying very carefully.

In one run (I mean one time-slot, or experimental unit) we observe four binary variables: two random settings and two outcomes. Under local realism these are certain functions of six binary variables: two random settings and four counterfactual outcomes under both possible settings in each wing of the experiment. If we would only assume realism (but not locality) then the four observed variables would be a function of 10: two settings and 8 counterfactual outcomes for each measurement under each pair of settings. That's your 8 for you.

In N runs, 4N binary variables are actually observed. Under local realism they are certain functions of 6N binary variables.

You count 8 by giving a measurement outcome a different name, depending on the setting in the other wing of the experiment. You are allowed to do that. So what?

Take a look at http://arxiv.org/pdf/quant-ph/0204169v1.pdf

Comment on "Exclusion of time in the theorem of Bell" by K. Hess and W. Philipp

R.D. Gill, G. Weihs, A. Zeilinger, M. Zukowski

Europhys. Lett. (2003) 61, 282-28

On page 3 we introduce 8 variables called X_ij and Y_ij where i and j take the values 1 and 2. We explain how locality implies that they can be reduced to a set of 4.

The link you give to Larsson's excellent and comprehensive survey paper doesn't seem to work. But http://arxiv.org/pdf/1407.0363.pdf automatically gives the latest version. It is worth studying very carefully.

In one run (I mean one time-slot, or experimental unit) we observe four binary variables: two random settings and two outcomes. Under local realism these are certain functions of six binary variables: two random settings and four counterfactual outcomes under both possible settings in each wing of the experiment. If we would only assume realism (but not locality) then the four observed variables would be a function of 10: two settings and 8 counterfactual outcomes for each measurement under each pair of settings. That's your 8 for you.

In N runs, 4N binary variables are actually observed. Under local realism they are certain functions of 6N binary variables.

You count 8 by giving a measurement outcome a different name, depending on the setting in the other wing of the experiment. You are allowed to do that. So what?

Take a look at http://arxiv.org/pdf/quant-ph/0204169v1.pdf

Comment on "Exclusion of time in the theorem of Bell" by K. Hess and W. Philipp

R.D. Gill, G. Weihs, A. Zeilinger, M. Zukowski

Europhys. Lett. (2003) 61, 282-28

On page 3 we introduce 8 variables called X_ij and Y_ij where i and j take the values 1 and 2. We explain how locality implies that they can be reduced to a set of 4.

Peer 3:

( January 24th, 2015 2:47pm UTC )

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Dr Gill, since you denied an easily verified mathematical fact, I do not see any point in engaging in a discussion.

There are 4 settings, each measured twice on different particle/experimental units every iteration. Since each outcome can be {+1,-1}, and you have 8 different particles, you 2^8 possible combinations of measured outcomes. This is shown on eqs. 14,15 of Larsson's paper. Nobody needs to be confused by "counterfactual" unmeasured results. Correlations in *experiments* do not include any counterfactual unmeasured outcomes. You are confused by the fact that you have **2^4 setting combinations**. Correlations are calculated from **outcome combinations** and there are 2^8 possibilities, because two different particles are measured at each specific setting.

This is a basic mathematical fact. So I won't attempt to respond to further misunderstandings and errors in your responses since you continue to deny this.

To anyone following, note that failure to respond does not mean Dr Gill's response has resolved the issue. He continues to dodge it.

There are 4 settings, each measured twice on different particle/experimental units every iteration. Since each outcome can be {+1,-1}, and you have 8 different particles, you 2^8 possible combinations of measured outcomes. This is shown on eqs. 14,15 of Larsson's paper. Nobody needs to be confused by "counterfactual" unmeasured results. Correlations in *experiments* do not include any counterfactual unmeasured outcomes. You are confused by the fact that you have **2^4 setting combinations**. Correlations are calculated from **outcome combinations** and there are 2^8 possibilities, because two different particles are measured at each specific setting.

This is a basic mathematical fact. So I won't attempt to respond to further misunderstandings and errors in your responses since you continue to deny this.

To anyone following, note that failure to respond does not mean Dr Gill's response has resolved the issue. He continues to dodge it.

Unregistered Submission:

( January 24th, 2015 2:52pm UTC )

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Peer 3: ( January 24th, 2015 2:47pm UTC ): "

Dr Gill, since you denied an easily verified mathematical fact, I do not see any point in engaging in a discussion."

There is still one thing you could do: enumerate those 2ˆ8 different possible outcomes so that we could have some idea what (if anything) you are talking about.

Dr Gill, since you denied an easily verified mathematical fact, I do not see any point in engaging in a discussion."

There is still one thing you could do: enumerate those 2ˆ8 different possible outcomes so that we could have some idea what (if anything) you are talking about.

Richard Gill:

( January 24th, 2015 3:07pm UTC )

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Peer 3: in order to have measured with all four setting combinations just once, you need 8 particles. We want to estimate four correlations. Those four correlations will each be experimentally determined using a different set of pairs of particles.

Do we all agree on that?

You think this is an issue, but you refuse to work through the math which shows it is not an issue, and you're certainly not going to waste your time by thinking about what local realism implies, and what repeatedly choosing new settings by external fair coin tosses adds to the picture.

So it seems to me that that is your problem that you do not understand the literature on Bell's theorem, not mine. Though to be sure, your "mental block" is quite common, I earlier gave a number of references to scientists who have put forward more or less the same objection.

You might like to take note of Donald Graft's statement on another thread: "What is wrong/naive in the paper [an earlier paper by himself] is the suggestion that all inequality violations (even for CH, etc.) are due only to lack of joint measurability of incompatible arrangements. I no longer believe this and I have shown in a later paper that the CH inequality is derived (by Clauser and Horne) without any requirement for joint measurability of incompatible arrangements, and that correctly applied the CH inequality can validly test aspects of locality." https://pubpeer.com/publications/E0F8384FC19A6034E86D516D03BB38

You can find Donald's paper "On Bell-Like Inequalities for Testing Local Realism" on his ResearchGate webpage https://www.researchgate.net/profile/Donald_Graft2 and on arXiv http://arxiv.org/abs/1404.4329

Do we all agree on that?

You think this is an issue, but you refuse to work through the math which shows it is not an issue, and you're certainly not going to waste your time by thinking about what local realism implies, and what repeatedly choosing new settings by external fair coin tosses adds to the picture.

So it seems to me that that is your problem that you do not understand the literature on Bell's theorem, not mine. Though to be sure, your "mental block" is quite common, I earlier gave a number of references to scientists who have put forward more or less the same objection.

You might like to take note of Donald Graft's statement on another thread: "What is wrong/naive in the paper [an earlier paper by himself] is the suggestion that all inequality violations (even for CH, etc.) are due only to lack of joint measurability of incompatible arrangements. I no longer believe this and I have shown in a later paper that the CH inequality is derived (by Clauser and Horne) without any requirement for joint measurability of incompatible arrangements, and that correctly applied the CH inequality can validly test aspects of locality." https://pubpeer.com/publications/E0F8384FC19A6034E86D516D03BB38

You can find Donald's paper "On Bell-Like Inequalities for Testing Local Realism" on his ResearchGate webpage https://www.researchgate.net/profile/Donald_Graft2 and on arXiv http://arxiv.org/abs/1404.4329

Peer 3:

( January 24th, 2015 9:25pm UTC )

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Unregistered submission: "There is still one thing you could do: enumerate those 2渠different possible outcomes so that we could have some idea what (if anything) you are talking about."

Okay then, first look at equation 14 in Larsson's paper, http://arxiv.org/pdf/1407.0363v1 do you see the 8 terms in that expression?

Each of those terms can take values 2 values {+1, -1} independently. That gives you 2^8 combinations. does that convince you, or do you want me to list all 256 possiblities?

Okay then, first look at equation 14 in Larsson's paper, http://arxiv.org/pdf/1407.0363v1 do you see the 8 terms in that expression?

Each of those terms can take values 2 values {+1, -1} independently. That gives you 2^8 combinations. does that convince you, or do you want me to list all 256 possiblities?

Peer 3:

( January 24th, 2015 11:22pm UTC )

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Gill: "You think this is an issue, but you refuse to work through the math which shows it is not an issue."

I have worked through all the math. It is definitely an issue. You would come to the same conclusion if you worked through the math carefully. This article may be able to help you or anyone who believes it is not an issue, in that process:

http://vixra.org/abs/1305.0129

In any case, the concluding message here is that you have failed to show how exactly picking just the first pair of particles in a time-slot avoids the coincidence time loophole and I have you on record saying it is impossible for the coincidence time loophole to persist if your pair-picking method is used. Be prepared to walk back those comments very soon.

I have worked through all the math. It is definitely an issue. You would come to the same conclusion if you worked through the math carefully. This article may be able to help you or anyone who believes it is not an issue, in that process:

http://vixra.org/abs/1305.0129

In any case, the concluding message here is that you have failed to show how exactly picking just the first pair of particles in a time-slot avoids the coincidence time loophole and I have you on record saying it is impossible for the coincidence time loophole to persist if your pair-picking method is used. Be prepared to walk back those comments very soon.

Richard Gill:

( January 25th, 2015 6:45am UTC )

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Good luck Peer 3! I look forward to your paper / computer simulation. We will see. We believe that we have shown exactly how picking just the first pair of particles in a time-slot avoids the coincidence time loophole, and this is all on record, and not anonymous either. (I would actually suggest picking the pair of particles nearest to the middle of the time-slot).

Furthermore, please let's stick to discussing the coincidence loophole on this thread.

I think that the 2^4 versus 2^8 issue (which is much older and much more basic) should be kept separate. Join the discussion about doi:10.1214/14-STS490 at https://pubpeer.com/publications/D985B475C637F666CC1D3E3A314522 and take a look at http://rpubs.com/gill1109/spreadsheet

PS I know Elemer Rosinger's preprints on viXra. And indeed I had nice discussions with him at Växjö last summer. I think he is a bit muddled, and no doubt he thinks the same about me. I repeat:**I look forward to your paper / computer simulation**.

Furthermore, please let's stick to discussing the coincidence loophole on this thread.

I think that the 2^4 versus 2^8 issue (which is much older and much more basic) should be kept separate. Join the discussion about doi:10.1214/14-STS490 at https://pubpeer.com/publications/D985B475C637F666CC1D3E3A314522 and take a look at http://rpubs.com/gill1109/spreadsheet

PS I know Elemer Rosinger's preprints on viXra. And indeed I had nice discussions with him at Växjö last summer. I think he is a bit muddled, and no doubt he thinks the same about me. I repeat:

Unregistered Submission:

( January 25th, 2015 1:06pm UTC )

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Peer 3: ( January 24th, 2015 9:25pm UTC ): "Unregistered submission: "There is still one thing you could do: enumerate those 256 different possible outcomes so that we could have some idea what (if anything) you are talking about."

Okay then, first look at equation 14 in Larsson's paper, http://arxiv.org/pdf/1407.0363v1 do you see the 8 terms in that expression?

Each of those terms can take values 2 values {+1, -1} independently. That gives you 2^8 combinations. does that convince you, or do you want me to list all 256 possiblities?"

The mentioned equation 14 is in another paper so what is its relevance here?

That equation has four terms with two factors each, referring to 8 independent tosses of coins in four separate experiments with a pair of coins. It is not clear, how this 8 would stay the same if instead of four we have some large numger of pairs.

Okay then, first look at equation 14 in Larsson's paper, http://arxiv.org/pdf/1407.0363v1 do you see the 8 terms in that expression?

Each of those terms can take values 2 values {+1, -1} independently. That gives you 2^8 combinations. does that convince you, or do you want me to list all 256 possiblities?"

The mentioned equation 14 is in another paper so what is its relevance here?

That equation has four terms with two factors each, referring to 8 independent tosses of coins in four separate experiments with a pair of coins. It is not clear, how this 8 would stay the same if instead of four we have some large numger of pairs.

Peer 3:

( January 25th, 2015 7:39pm UTC )

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Unregistered submission: "The mentioned equation 14 is in another paper so what is its relevance here?"

The relevance is that it illustrates how you get 2^8. But I asked already, if that equation did not help you to understand how you get 256 possibilities, I could list all of them for your benefit. Is that what you want? Or do you too want to deny the mathematical fact? Please make up your mind.

There are 4 angles (a, a', b, b'). Outcomes are measured at 4 pairs of angles (a, b), (a, b'), (a', b), (a',b') each on separate pairs of particles for a total of 8 independent outcomes, since each of those outcomes can be one of 2 values {+1, -1}, there 2^8 possible combinations of outcomes in such an experiment.

Unregistered submission: "That equation has four terms with two factors each, referring to 8 independent tosses of coins in four separate experiments with a pair of coins. "

Wrong. Please look at the equation again carefully. There are 8 terms. Do not confuse the fact that you have 4 coins, with the fact that there are 8 terms. This seems to be the common mistake. In a CHSH experiment you have 4 angles, but 8 terms. Same thing in that equation, there are 4 coins, but 8 terms. Each of the 8 terms can **independently** attain values {+1, -1} therefore 2^8. This is uncontroversial. It is a mathematical fact, you simply admit it and move on, okay.

I'll encourage you to study this paper: http://vixra.org/abs/1305.0129

The relevance is that it illustrates how you get 2^8. But I asked already, if that equation did not help you to understand how you get 256 possibilities, I could list all of them for your benefit. Is that what you want? Or do you too want to deny the mathematical fact? Please make up your mind.

There are 4 angles (a, a', b, b'). Outcomes are measured at 4 pairs of angles (a, b), (a, b'), (a', b), (a',b') each on separate pairs of particles for a total of 8 independent outcomes, since each of those outcomes can be one of 2 values {+1, -1}, there 2^8 possible combinations of outcomes in such an experiment.

Unregistered submission: "That equation has four terms with two factors each, referring to 8 independent tosses of coins in four separate experiments with a pair of coins. "

Wrong. Please look at the equation again carefully. There are 8 terms. Do not confuse the fact that you have 4 coins, with the fact that there are 8 terms. This seems to be the common mistake. In a CHSH experiment you have 4 angles, but 8 terms. Same thing in that equation, there are 4 coins, but 8 terms. Each of the 8 terms can **independently** attain values {+1, -1} therefore 2^8. This is uncontroversial. It is a mathematical fact, you simply admit it and move on, okay.

I'll encourage you to study this paper: http://vixra.org/abs/1305.0129

Jan-Åke Larsson:

( January 26th, 2015 7:26am UTC )

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Peer 3, your current argument isn't working. Just because there are eight independent values (if your random setting selection for some reason would give you the four different combinations) doesn't mean there is no limitation to their value. The sample means will eventually approach the true expectation values of the four products, so that an experimental violation would be evidence for a true violation.

But earlier I saw hints of a different argument: that the selection process ("the first pair in the time slot") would make the experiments use different lambdas. That these lambdas would never coincide, so that the CHSH inequality itself would be invalid. But this second argument does not work either.

Imagine a timeslot with several pairs. Assume for the moment that there are five pairs in a timeslot. There are now five lambdas (lambda1,lambda2,lambda3,lambda4,lambda5). Imagine now that if Alice chooses setting a1 then lambda1 will give the first outcome for her, and if she chooses setting a2 then lambda2 gives the first outcome for her. For Bob, the setting b1 makes lambda3 produce the first outcome, and the setting b2 makes lambda4 produce the first outcome. lambda5 only influences the later outcomes in the timeslot. I *think* your argument is that this would make a CHSH violation possible, because the outcomes are de facto given by different hidden variables so are completely independent.

But there are actually two separate arguments why this is impossible. One is that the four terms of CHSH (eight factors) actually measure correlation between the components. And if the outcomes are independent, there will be no correlation. So a model built up as above cannot give high correlation, or indeed violation.

OK, so the lambdas maybe are not independent, one could still believe that it is possible to introduce correlations. I would agree with Richard: do try this, and you will see how hard it is. I know because I have done it, repeatedly (I only aim to saturate the bound, not violate it).

However, there is an argument that tells you violation is impossible. What you need to do is to consider the vector lambda=(lambda1,lambda2,lambda3,lambda4,lambda5). I know Peer 3 doesn't "care about vectors", but note that I am free to use what mathematical tools I want to prove something. Now, above I wrote that Alices outcome A depended on lambda1 if a1 was chosen, and lambda2 if a2 was chosen. Then I can construct A'(a,lambda) to take the value A(a1,lambda1) if a=a1 and the value A(a2,lambda2) if a=a2. In the same way I can construct B'(b,lambda) to take the value B(b1,lambda3) if b=b1 and the value B(b2,lambda4) if b=b2. I now have a new hidden-variable model that is derived from the old one, with a vector lambda and A' and B' as outcomes.

The CHSH inequality applies to the new model, because the proof holds for the new model. It is of course important that the choice of lambda1 vs lambda2 at Alice does not depend on Bob's setting and vice versa. As long as this is true, the new construction lambda, A', B' is a *local* hidden-variable model. The experiment outcomes I use in the new inequality is the same data as I would have used in the old, if I select the first output in each timeslot.

This is why there is no problem with using the first output in each timeslot.

You can handle varying number of pairs in the timeslots too, and more complicated situations, but the mathematical machinery gets more complicated. I don't want to reproduce that here, given the archaic interface. Perhaps I'll include it in a paper somewhere.

But earlier I saw hints of a different argument: that the selection process ("the first pair in the time slot") would make the experiments use different lambdas. That these lambdas would never coincide, so that the CHSH inequality itself would be invalid. But this second argument does not work either.

Imagine a timeslot with several pairs. Assume for the moment that there are five pairs in a timeslot. There are now five lambdas (lambda1,lambda2,lambda3,lambda4,lambda5). Imagine now that if Alice chooses setting a1 then lambda1 will give the first outcome for her, and if she chooses setting a2 then lambda2 gives the first outcome for her. For Bob, the setting b1 makes lambda3 produce the first outcome, and the setting b2 makes lambda4 produce the first outcome. lambda5 only influences the later outcomes in the timeslot. I *think* your argument is that this would make a CHSH violation possible, because the outcomes are de facto given by different hidden variables so are completely independent.

But there are actually two separate arguments why this is impossible. One is that the four terms of CHSH (eight factors) actually measure correlation between the components. And if the outcomes are independent, there will be no correlation. So a model built up as above cannot give high correlation, or indeed violation.

OK, so the lambdas maybe are not independent, one could still believe that it is possible to introduce correlations. I would agree with Richard: do try this, and you will see how hard it is. I know because I have done it, repeatedly (I only aim to saturate the bound, not violate it).

However, there is an argument that tells you violation is impossible. What you need to do is to consider the vector lambda=(lambda1,lambda2,lambda3,lambda4,lambda5). I know Peer 3 doesn't "care about vectors", but note that I am free to use what mathematical tools I want to prove something. Now, above I wrote that Alices outcome A depended on lambda1 if a1 was chosen, and lambda2 if a2 was chosen. Then I can construct A'(a,lambda) to take the value A(a1,lambda1) if a=a1 and the value A(a2,lambda2) if a=a2. In the same way I can construct B'(b,lambda) to take the value B(b1,lambda3) if b=b1 and the value B(b2,lambda4) if b=b2. I now have a new hidden-variable model that is derived from the old one, with a vector lambda and A' and B' as outcomes.

The CHSH inequality applies to the new model, because the proof holds for the new model. It is of course important that the choice of lambda1 vs lambda2 at Alice does not depend on Bob's setting and vice versa. As long as this is true, the new construction lambda, A', B' is a *local* hidden-variable model. The experiment outcomes I use in the new inequality is the same data as I would have used in the old, if I select the first output in each timeslot.

This is why there is no problem with using the first output in each timeslot.

You can handle varying number of pairs in the timeslots too, and more complicated situations, but the mathematical machinery gets more complicated. I don't want to reproduce that here, given the archaic interface. Perhaps I'll include it in a paper somewhere.

Jan-Åke Larsson:

( January 26th, 2015 7:29am UTC )

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Editing destroys the lambdas, I don't believe that the Pubpeer people haven't fixed this yet. Use mathjax!

Unregistered Submission:

( January 26th, 2015 9:25am UTC )

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Peer 3: ( January 25th, 2015 7:39pm UTC ): "The relevance is that it illustrates how you get 2^8. But I asked already, if that equation did not help you to understand how you get 256 possibilities, I could list all of them for your benefit."

I can understand how one can get 2^8 possibilities or any other number in an irrelevant context, but how is that number relevant to an experiment with a pair of particles?

An experiment may use 8 detectors (although there are other ways) but that cannot be what you mean, as the outcomes are not independent.

I can understand how one can get 2^8 possibilities or any other number in an irrelevant context, but how is that number relevant to an experiment with a pair of particles?

An experiment may use 8 detectors (although there are other ways) but that cannot be what you mean, as the outcomes are not independent.

Jan-Åke Larsson:

( January 26th, 2015 10:42am UTC )

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Peer 3 is considering a sequence of four experiments at the four different setting combinations that occur in CHSH. There are two measurements in each experiment, each with two outcomes: (2^2)^4. Peer 3 thinks this is important and that it invalidates the CHSH inequality.

I think not. For very good reasons, as I give three different proofs valid in two different situations.

I think not. For very good reasons, as I give three different proofs valid in two different situations.

Peer 3:

( January 26th, 2015 1:38pm UTC )

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Jan-ūe Larsson: Thank you for explaining to Unregistered Submitter how 2^8 arises in a CHSH experiment. Except you erred in concluding that I "think" it invalidates the CHSH inequality. All I'm arguing from the beginning of this thread is that for 2^8 outcomes you have to make specific assumptions about the distributions of the 4 different ensembles. They must be the same. This is what we've been discussing here all along. Providing proofs which demonstrate that the CHSH inequality holds when the distributions are the same does not invalidate the argument I've been presenting, ie that the CHSH should not be expected to hold if the distributions are different.

Isn't this what you wrote in your paper, that differences between the ensembles allow you to violate the CHSH?

Isn't this what you wrote in your paper, that differences between the ensembles allow you to violate the CHSH?

Peer 3:

( January 26th, 2015 1:58pm UTC )

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Jan-Åke Larsson: "Peer 3, your current argument isn't working. Just because there are eight independent values (if your random setting selection for some reason would give you the four different combinations) doesn't mean there is no limitation to their value. The sample means will eventually approach the true expectation values of the four products, so that an experimental violation would be evidence for a true violation."

Please Jan, consider carefully my example from ( January 17th, 2015 2:49pm UTC ): let me repeat a snippet here. Assume a system with just 5 unique lambdas each of which has a different time delay for detection at a the given angles as follows, where the order of particle lambdas is sorted according to time delays:

a:A: pλ2,pλ5,pλ1,pλ3,pλ4

c':B: pλ1,pλ2,pλ4,pλ5,pλ3

b:A: pλ4,pλ3,pλ2,pλ1,pλ5

d':B: pλ3,pλ2,pλ1,pλ5,pλ4

Note that the time delays are completely local. A=Alice, B=Bob. Also note that this sequence will never change no matter how many particle pairs you measure, t is built-in.

Based on your method of picking pairs, the ensembles used to calculate the 4 paired expectation values Λac', Λad', Λbc', Λbd', where Λac is the ensemble of lambdas **considered** for calculating the Eac = E(a,c|Λac) expectation value, etc

By selecting only the first pair of particles each time, we now have

Λac' = {pλ1,pλ2}

Λad' = {pλ2,pλ3)}

Λbc' = {pλ1,pλ4}

Λbd' = {pλ3,pλ4}

Again note that N can be any number as large as you want, and still those ensembles will never contain any other lambdas than the ones shown. Now please, you are a reasonable guy, do you believe the CHSH applies to an experiment like this, especially after you already mentioned in your paper that:

**"The original CHSH inequality is no longer valid, and the reason can be seen in the start of the proof where one wants to add

...

The integrals on the right-hand side cannot easily be added when Λac′=/= Λad′ , since we

are taking expectations over different ensembles Λac′ and Λad′ , with respect to different

probability measures.

The problem here is that the ensemble on which the correlations are evaluated changes with the settings,while the original Bell inequality requires that they stay the same. In effect,

the Bell inequality only holds on the common part of the four different ensembles Λac′ , Λad′ ,

Λbc′ , and Λbd"**

As I show above:

1) The ensembles on which the correlations are evaluated changes with the settings, therefore the coincidence time loophole is still present!

2) There is no common part of the 4 ensembles on which any inequality might apply (or rather, according to your paper, you end up with an inequality with upper bound of 4 which is never violated).

Hopefully you understand now why your suggested pair-picking method does not do what you claim it does.

Please Jan, consider carefully my example from ( January 17th, 2015 2:49pm UTC ): let me repeat a snippet here. Assume a system with just 5 unique lambdas each of which has a different time delay for detection at a the given angles as follows, where the order of particle lambdas is sorted according to time delays:

a:A: pλ2,pλ5,pλ1,pλ3,pλ4

c':B: pλ1,pλ2,pλ4,pλ5,pλ3

b:A: pλ4,pλ3,pλ2,pλ1,pλ5

d':B: pλ3,pλ2,pλ1,pλ5,pλ4

Note that the time delays are completely local. A=Alice, B=Bob. Also note that this sequence will never change no matter how many particle pairs you measure, t is built-in.

Based on your method of picking pairs, the ensembles used to calculate the 4 paired expectation values Λac', Λad', Λbc', Λbd', where Λac is the ensemble of lambdas **considered** for calculating the Eac = E(a,c|Λac) expectation value, etc

By selecting only the first pair of particles each time, we now have

Λac' = {pλ1,pλ2}

Λad' = {pλ2,pλ3)}

Λbc' = {pλ1,pλ4}

Λbd' = {pλ3,pλ4}

Again note that N can be any number as large as you want, and still those ensembles will never contain any other lambdas than the ones shown. Now please, you are a reasonable guy, do you believe the CHSH applies to an experiment like this, especially after you already mentioned in your paper that:

**"The original CHSH inequality is no longer valid, and the reason can be seen in the start of the proof where one wants to add

...

The integrals on the right-hand side cannot easily be added when Λac′=/= Λad′ , since we

are taking expectations over different ensembles Λac′ and Λad′ , with respect to different

probability measures.

The problem here is that the ensemble on which the correlations are evaluated changes with the settings,while the original Bell inequality requires that they stay the same. In effect,

the Bell inequality only holds on the common part of the four different ensembles Λac′ , Λad′ ,

Λbc′ , and Λbd"**

As I show above:

1) The ensembles on which the correlations are evaluated changes with the settings, therefore the coincidence time loophole is still present!

2) There is no common part of the 4 ensembles on which any inequality might apply (or rather, according to your paper, you end up with an inequality with upper bound of 4 which is never violated).

Hopefully you understand now why your suggested pair-picking method does not do what you claim it does.

Peer 3:

( January 26th, 2015 2:04pm UTC )

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Jan-Åke Larsson: "The sample means will eventually approach the true expectation values of the four products, so that an experimental violation would be evidence for a true violation."

One other thing. Note that the sample means of the 4 ensembles in my example above will never ever approach the the value of the original (population) ensemble of pairs. No matter how many pairs you measure. But the experimental results will be perfectly reproducible. Now Richard Gill has already claimed that it is impossible to violate the CHSH using such techniques in combination with your pair-picking method. Do you want to go on the record as well to agree with him, even after my explanation above?

One other thing. Note that the sample means of the 4 ensembles in my example above will never ever approach the the value of the original (population) ensemble of pairs. No matter how many pairs you measure. But the experimental results will be perfectly reproducible. Now Richard Gill has already claimed that it is impossible to violate the CHSH using such techniques in combination with your pair-picking method. Do you want to go on the record as well to agree with him, even after my explanation above?

Jan-Åke Larsson:

( January 26th, 2015 2:16pm UTC )

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It is not so simple.

If you do four experiments, one at each setting combination, there is a nonzero chance that you get the total 4, which is above the bound. This is what I write in Larsson, 2014. Loopholes in Bell inequality tests of local realism. J. Phys. A 47, 424003. doi:10.1088/1751-8113/47/42/424003, http://arxiv.org/abs/1407.0363, section 1.5. But if you do eight or twelve or four thousand, and the system can be described with a local hidden variable model, the average will with higher and higer probability be below 2. This follows from mathematical statistics (IID or supermartingales, depending on your assumptions), see the same paper.

If you have postselection, say low detector efficiency, then the effect you describe can occur, see section 3.1-3.2 in that paper. The result will be an increased bound, like eqn (38).

But if you have time slots, where you select the first outcome as *the* outcome for the time slot, there is no modification to the bound. The bound stays the same because of the proof I gave above (January 26th, 2015 7:26am UTC), and in print in Larsson et al, 2014. Bell-inequality violation with entangled photons, free of the coincidence-time loophole. Phys. Rev. A 90, 032107. doi:10.1103/PhysRevA.90.032107, http://arxiv.org/abs/1309.0712, page 3, second column, last paragraph.

The selection of the first outcome in a time slot is not postselection, it is *coarse-graining*. This means that you have data for CHSH which is not +-1, but rather (+1,-1,-1,+1,...) The coarse-graining is to use the label +1 for all the events that have a +1 first. Think of a six-sided die used as a coin: label odd outcomes "head" and even outcomes "tail" (the lowest bit of the number). This coin is fair if the die is fair. In CHSH, the coarse-grained outcomes can be modelled with a local hidden variable model if the non-coarse-grained "actual" outcomes can. There is no possibility of creating a false violation.

If you do four experiments, one at each setting combination, there is a nonzero chance that you get the total 4, which is above the bound. This is what I write in Larsson, 2014. Loopholes in Bell inequality tests of local realism. J. Phys. A 47, 424003. doi:10.1088/1751-8113/47/42/424003, http://arxiv.org/abs/1407.0363, section 1.5. But if you do eight or twelve or four thousand, and the system can be described with a local hidden variable model, the average will with higher and higer probability be below 2. This follows from mathematical statistics (IID or supermartingales, depending on your assumptions), see the same paper.

If you have postselection, say low detector efficiency, then the effect you describe can occur, see section 3.1-3.2 in that paper. The result will be an increased bound, like eqn (38).

But if you have time slots, where you select the first outcome as *the* outcome for the time slot, there is no modification to the bound. The bound stays the same because of the proof I gave above (January 26th, 2015 7:26am UTC), and in print in Larsson et al, 2014. Bell-inequality violation with entangled photons, free of the coincidence-time loophole. Phys. Rev. A 90, 032107. doi:10.1103/PhysRevA.90.032107, http://arxiv.org/abs/1309.0712, page 3, second column, last paragraph.

The selection of the first outcome in a time slot is not postselection, it is *coarse-graining*. This means that you have data for CHSH which is not +-1, but rather (+1,-1,-1,+1,...) The coarse-graining is to use the label +1 for all the events that have a +1 first. Think of a six-sided die used as a coin: label odd outcomes "head" and even outcomes "tail" (the lowest bit of the number). This coin is fair if the die is fair. In CHSH, the coarse-grained outcomes can be modelled with a local hidden variable model if the non-coarse-grained "actual" outcomes can. There is no possibility of creating a false violation.

Donald A. Graft:

( January 26th, 2015 2:34pm UTC )

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"In CH, the coarse-grained outcomes can be modelled with a local hidden variable model if the non-coarse-grained "actual" outcomes can. There is no possibility of creating a false violation."

How then do you explain the different results in Christensen et al between using all the detection events (no violation) and picking only the first (violation)? It's an experimental result; you can't simply deny it. I believe this is the third time I have asked you. Did I miss your response?

How then do you explain the different results in Christensen et al between using all the detection events (no violation) and picking only the first (violation)? It's an experimental result; you can't simply deny it. I believe this is the third time I have asked you. Did I miss your response?

Richard Gill:

( January 26th, 2015 3:09pm UTC )

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This was my response a while back:

The two first events are close together in time. They likely belong to the same pair. When you pick any two events in the same time window maybe one belongs to one emission the other to another. This could explain why the correlations computed in the two different ways are rather different - why your way gives less strong correlations than Christensen's way.

The two first events are close together in time. They likely belong to the same pair. When you pick any two events in the same time window maybe one belongs to one emission the other to another. This could explain why the correlations computed in the two different ways are rather different - why your way gives less strong correlations than Christensen's way.

Jan-Åke Larsson:

( January 26th, 2015 3:13pm UTC )

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I didn't say the statistics would stay the same.

I said "There is no possibility of creating a false violation." If a violation is there, it is not false.

I said "There is no possibility of creating a false violation." If a violation is there, it is not false.

Jan-Åke Larsson:

( January 26th, 2015 3:39pm UTC )

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Peer 3, the proof I gave in my post (January 26th, 2015 7:26am UTC) applies exactly in the situation when

a:A: pλ2,pλ5,pλ1,pλ3,pλ4

c':B: pλ1,pλ2,pλ4,pλ5,pλ3

b:A: pλ4,pλ3,pλ2,pλ1,pλ5

d':B: pλ3,pλ2,pλ1,pλ5,pλ4

There is nothing to add. What I wrote there is a mathematical *proof* (without handwaving) that there cannot be a false violation from this hidden-variable model, even with the different time delays. I define the vector λtimeslot:=(pλ1,pλ2,pλ3,pλ4,pλ5), which I am allowed to do, then

Λac' = Λad' = Λbc' = Λbd' = {λtimeslot}

No change in the ensemble

a:A: pλ2,pλ5,pλ1,pλ3,pλ4

c':B: pλ1,pλ2,pλ4,pλ5,pλ3

b:A: pλ4,pλ3,pλ2,pλ1,pλ5

d':B: pλ3,pλ2,pλ1,pλ5,pλ4

There is nothing to add. What I wrote there is a mathematical *proof* (without handwaving) that there cannot be a false violation from this hidden-variable model, even with the different time delays. I define the vector λtimeslot:=(pλ1,pλ2,pλ3,pλ4,pλ5), which I am allowed to do, then

Λac' = Λad' = Λbc' = Λbd' = {λtimeslot}

No change in the ensemble

Peer 3:

( January 26th, 2015 5:11pm UTC )

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Jan, please note that there can be no violation only if you **consider** EVERY particle in the time slot. If you consider only the first particles in the timeslot, λtimeslot =/= (pλ1,pλ2,pλ3,pλ4,pλ5). Please read my example carefully.

Λac' = Λad' = Λbc' = Λbd' = {λtimeslot} is only valid if you consider ALL particles in each time-slot. If you consider just the first pair of particles, you get

Λac' = {pλ1,pλ2}

Λad' = {pλ2,pλ3)}

Λbc' = {pλ1,pλ4}

Λbd' = {pλ3,pλ4}

and therefore Λac' =/= Λad' =/= Λbc' =/= Λbd'

Do you deny this?

Λac' = Λad' = Λbc' = Λbd' = {λtimeslot} is only valid if you consider ALL particles in each time-slot. If you consider just the first pair of particles, you get

Λac' = {pλ1,pλ2}

Λad' = {pλ2,pλ3)}

Λbc' = {pλ1,pλ4}

Λbd' = {pλ3,pλ4}

and therefore Λac' =/= Λad' =/= Λbc' =/= Λbd'

Do you deny this?

Unregistered Submission:

( January 26th, 2015 5:25pm UTC )

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Peer 3: ( January 26th, 2015 1:38pm UTC ): "All I'm arguing from the beginning of this thread is that for 2^8 outcomes you have to make specific assumptions about the distributions of the 4 different ensembles. They must be the same."

That exactly is what local causality assumes: the way the source generates particle pairs is not affected by the way experimenter will later choose to measure. Without local causality this assumption needs not be made and the inequality may be violated.

That exactly is what local causality assumes: the way the source generates particle pairs is not affected by the way experimenter will later choose to measure. Without local causality this assumption needs not be made and the inequality may be violated.

Jan-Åke Larsson:

( January 26th, 2015 8:28pm UTC )

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Peer 3: "Λac' = Λad' = Λbc' = Λbd' = {λtimeslot} is only valid if you consider ALL particles in each time-slot."

No, in the local hidden variable model with A', B', and λtimeslot, there is *one* output at each site for each timeslot, irrespective of the value of λtimeslot. Therefore, Λac' = Λad' = Λbc' = Λbd' = {λtimeslot}

No, in the local hidden variable model with A', B', and λtimeslot, there is *one* output at each site for each timeslot, irrespective of the value of λtimeslot. Therefore, Λac' = Λad' = Λbc' = Λbd' = {λtimeslot}

Peer 3:

( January 26th, 2015 10:21pm UTC )

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Peer 2: "No, in the local hidden variable model with A', B', and λtimeslot, there is *one* output at each site for each timeslot"

You mean there is *one* output **considered** for each time slot. Isn't that what you explained when you said: "The selection of the first outcome in a time slot is not postselection, it is *coarse-graining*. This means that you have data for CHSH which is not +-1, but rather (+1,-1,-1,+1,...) The coarse-graining is to use the label +1 for all the events that have a +1 first. "

Why would you even recommend to select the "first" outcome if you only have *one*. What you are doing is considering the first of many, and ignoring the rest as if they don't exist. This is in fact what Christensen et al did, as discovered by Donald Graft.

You can't have your cake and eat it. You can't say λtimeslot represents just one output, and at the same time, say λtimeslot = (pλ1,pλ2,pλ3,pλ4,pλ5).

You mean there is *one* output **considered** for each time slot. Isn't that what you explained when you said: "The selection of the first outcome in a time slot is not postselection, it is *coarse-graining*. This means that you have data for CHSH which is not +-1, but rather (+1,-1,-1,+1,...) The coarse-graining is to use the label +1 for all the events that have a +1 first. "

Why would you even recommend to select the "first" outcome if you only have *one*. What you are doing is considering the first of many, and ignoring the rest as if they don't exist. This is in fact what Christensen et al did, as discovered by Donald Graft.

You can't have your cake and eat it. You can't say λtimeslot represents just one output, and at the same time, say λtimeslot = (pλ1,pλ2,pλ3,pλ4,pλ5).

Jan-Åke Larsson:

( January 26th, 2015 11:04pm UTC )

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The model is constructed so that the *one* output data in each site *is* the first of the many particles being detected.

I constructed it exactly from your example model. This is the coarse-grained representation of the many pairs.

In the construction, λtimeslot = (pλ1,pλ2,pλ3,pλ4,pλ5) is used for just one output at each site.

By the construction, the CHSH inequality applies to the coarse-grained model, so the CHSH inequality does apply to the data you get when you select the first in each timeslot.

The construction and proof is explicit in my posting from this morning. I don't know how to help you further.

I constructed it exactly from your example model. This is the coarse-grained representation of the many pairs.

In the construction, λtimeslot = (pλ1,pλ2,pλ3,pλ4,pλ5) is used for just one output at each site.

By the construction, the CHSH inequality applies to the coarse-grained model, so the CHSH inequality does apply to the data you get when you select the first in each timeslot.

The construction and proof is explicit in my posting from this morning. I don't know how to help you further.

Enter your reply below (Please read the **How To**)

Peer 2:

( January 26th, 2015 5:42pm UTC )

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Peer 3: Can you clarify some notation here? These aggregated lambdas of yours seem to be unordered sets. So, when you write {pλ1,pλ2}, does this mean that both wings can see both lambdas? In that case, your model is nonlocal, since the lambdas here depend on settings. If not the case, you must explain what the notation actually means.

Peer 3:

( January 26th, 2015 7:02pm UTC )

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Peer 2: "Can you clarify some notation here? These aggregated lambdas of yours seem to be unordered sets. So, when you write {pλ1,pλ2}, does this mean that both wings can see both lambdas? In that case, your model is nonlocal. If not the case, you must explain what the notation actually means."

Sure, I can clarify. It doesn't mean both wings necessarily see both lambdas, it simply means it is irrelevant whether they do or not. Here is a detailed description of the notation: A source emits a bunch of "entangled" particles in pairs. To each pair it imparts one of 5 hidden properties pλ1,pλ2, pλ3, pλ4, pλ5. Since this is a pulsed experiment, a bunch of these particles arrive at the stations and are recorded in time-slots, during each time-slot one of 4 settings is active (a, b) on Alice's side, and (c', d') on Bob's side. Each setting introduces time-delays to particles according to which properties they possess. The time delay introduced is a local function of the hidden particle lambda, and the local setting in force, such that the order of detection of particles within time-slots is as follows:

a: pλ2,pλ5,pλ1,pλ3,pλ4

c': pλ1,pλ2,pλ4,pλ5,pλ3

b: pλ4,pλ3,pλ2,pλ1,pλ5

d': pλ3,pλ2,pλ1,pλ5,pλ4

The outcomes in each timeslot are not all necessarily the same, they can be (+1,-1,-1,+1,-1), corresponding to the different particles based on what order they were detected. In each of those time-slots, there are multiple particle detections but we do not know which particle in Alice's timeslot should match which particle on Bob's timeslot because we do not have access to the time signature outlined above (it is hidden).

However, if a timeslot with detection sequence (pλ3,pλ2,pλ1,pλ5,pλ4) (ie setting d') has outcomes (+1,-1,-1,+1,-1), if you pick just the first outcome, it is equivalent to picking just pλ3 and in this case the big-lambda or time-slot lambda is = pλ3, and if at the other arm you were measuring with setting a (pλ2,pλ5,pλ1,pλ3,pλ4), and the outcome was (-1,-1,-1,+1,-1), by picking the first outcome -1, you would have picked time-slot lambda = pλ2. As I hope you see, the time-slot lambda is not represented by what particles arrived within the time-slot. It is represented by what data you **considered** in your calculations. In those calculations, to calculate the correlation at the angle pair (a, c'), you would multiply the +1 picked from the first time-slot, with the -1 picked from the second and place the result in one table. Next to that value, you can imagine a pair of particle lambdas which correspond to those values, in this case (pλ3,pλ2). At the end you would take the average of all those paired products and call that C(a,d'). Now if you compile a set of all such particle lambdas and their relative frequency on the list that gave you C(a,d'), this represents the ensemble Λad'. In the example above, Λad' = {pλ2,pλ3}, since on every row of your correlation table for the angle pair (a,d'), you would simply have (pλ3,pλ2).

Applying this to the 4 setting pairs, you end up with

Λac' = {pλ1,pλ2}

Λad' = {pλ2,pλ3)}

Λbc' = {pλ1,pλ4}

Λbd' = {pλ3,pλ4}

and therefore Λac' =/= Λad' =/= Λbc' =/= Λbd'

Note a few things:

1) Traditionally, the coincidence problem described in the paper above was to try match which particle at Alice corresponds to which particle at Bob, by recording and comparing the exact detection times of each particle, and picking as pairs, particles which were close together in time. For this example, that fails because the correct pairs are not necessarily close together and by assuming that they should be, you have systematically biased the ensembles **considered** for calculating the expectation values. The particle lambdas which are close to each other in time varies with both settings and therefore the inequality does not apply as the paper correctly argues.

2) Note that the new method of using time-slots and only considering the first particles in each time-slot still does not match the correct pairs to each other. Rather, as illustrated above, you still end up **considering** outcomes from different ensembles and the issue is not solved at all. Just because you know exactly which time-slot on Alice's side corresponds to which on Bob's side does not mean the coincidence problem has disappeared.

3) Picking instead the particle pair in the middle of the time-slot does absolutely nothing to resolve the issue. You could reduce the effect of the problem by picking randomly *at each arm*, because at least, you will increase the chance of picking correct pairs a small number of times. But by picking a particle at a fixed location, you might never pick the correct pairs, as the example above shows.

4) There is however a way in which time-slots can avoid the loophole and it is only when you have a single particle pair emission per time-slot, and the duration of the time-slot is longer than all possible delays which may be introduced during detection. In that situation, it is moot to even suggest picking a the first particle because there is just one on each arm. Only in that situation will the pairs be correctly matched.

Sure, I can clarify. It doesn't mean both wings necessarily see both lambdas, it simply means it is irrelevant whether they do or not. Here is a detailed description of the notation: A source emits a bunch of "entangled" particles in pairs. To each pair it imparts one of 5 hidden properties pλ1,pλ2, pλ3, pλ4, pλ5. Since this is a pulsed experiment, a bunch of these particles arrive at the stations and are recorded in time-slots, during each time-slot one of 4 settings is active (a, b) on Alice's side, and (c', d') on Bob's side. Each setting introduces time-delays to particles according to which properties they possess. The time delay introduced is a local function of the hidden particle lambda, and the local setting in force, such that the order of detection of particles within time-slots is as follows:

a: pλ2,pλ5,pλ1,pλ3,pλ4

c': pλ1,pλ2,pλ4,pλ5,pλ3

b: pλ4,pλ3,pλ2,pλ1,pλ5

d': pλ3,pλ2,pλ1,pλ5,pλ4

The outcomes in each timeslot are not all necessarily the same, they can be (+1,-1,-1,+1,-1), corresponding to the different particles based on what order they were detected. In each of those time-slots, there are multiple particle detections but we do not know which particle in Alice's timeslot should match which particle on Bob's timeslot because we do not have access to the time signature outlined above (it is hidden).

However, if a timeslot with detection sequence (pλ3,pλ2,pλ1,pλ5,pλ4) (ie setting d') has outcomes (+1,-1,-1,+1,-1), if you pick just the first outcome, it is equivalent to picking just pλ3 and in this case the big-lambda or time-slot lambda is = pλ3, and if at the other arm you were measuring with setting a (pλ2,pλ5,pλ1,pλ3,pλ4), and the outcome was (-1,-1,-1,+1,-1), by picking the first outcome -1, you would have picked time-slot lambda = pλ2. As I hope you see, the time-slot lambda is not represented by what particles arrived within the time-slot. It is represented by what data you **considered** in your calculations. In those calculations, to calculate the correlation at the angle pair (a, c'), you would multiply the +1 picked from the first time-slot, with the -1 picked from the second and place the result in one table. Next to that value, you can imagine a pair of particle lambdas which correspond to those values, in this case (pλ3,pλ2). At the end you would take the average of all those paired products and call that C(a,d'). Now if you compile a set of all such particle lambdas and their relative frequency on the list that gave you C(a,d'), this represents the ensemble Λad'. In the example above, Λad' = {pλ2,pλ3}, since on every row of your correlation table for the angle pair (a,d'), you would simply have (pλ3,pλ2).

Applying this to the 4 setting pairs, you end up with

Λac' = {pλ1,pλ2}

Λad' = {pλ2,pλ3)}

Λbc' = {pλ1,pλ4}

Λbd' = {pλ3,pλ4}

and therefore Λac' =/= Λad' =/= Λbc' =/= Λbd'

Note a few things:

1) Traditionally, the coincidence problem described in the paper above was to try match which particle at Alice corresponds to which particle at Bob, by recording and comparing the exact detection times of each particle, and picking as pairs, particles which were close together in time. For this example, that fails because the correct pairs are not necessarily close together and by assuming that they should be, you have systematically biased the ensembles **considered** for calculating the expectation values. The particle lambdas which are close to each other in time varies with both settings and therefore the inequality does not apply as the paper correctly argues.

2) Note that the new method of using time-slots and only considering the first particles in each time-slot still does not match the correct pairs to each other. Rather, as illustrated above, you still end up **considering** outcomes from different ensembles and the issue is not solved at all. Just because you know exactly which time-slot on Alice's side corresponds to which on Bob's side does not mean the coincidence problem has disappeared.

3) Picking instead the particle pair in the middle of the time-slot does absolutely nothing to resolve the issue. You could reduce the effect of the problem by picking randomly *at each arm*, because at least, you will increase the chance of picking correct pairs a small number of times. But by picking a particle at a fixed location, you might never pick the correct pairs, as the example above shows.

4) There is however a way in which time-slots can avoid the loophole and it is only when you have a single particle pair emission per time-slot, and the duration of the time-slot is longer than all possible delays which may be introduced during detection. In that situation, it is moot to even suggest picking a the first particle because there is just one on each arm. Only in that situation will the pairs be correctly matched.

Peer 2:

( January 26th, 2015 8:05pm UTC )

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Peer 2: "Sure, I can clarify. It doesn't mean both wings necessarily see both lambdas, it simply means it is irrelevant whether they do or not. "

But of course this is not irrelevant; if both wings can see both lambdas, the model is nonlocal.

If we now rather assume that each wing can only see *one* of the two lambdas, namely the first lambda picked in a time slot *in that wing*, then we have to order the sets of two lambdas, and your example then gives the following:

Λac' = (pλ2,pλ1)

Λad' = (pλ2,pλ3)

Λbc' = (pλ4,pλ1)

Λbd' = (pλ4,pλ3)

Note that I have only ordered your sets according to what lambdas Alice and Bob actually get; the contents of the sets remains the same. The first element is what Alice sees, the second element is what Bob sees.

But with this ordering, there is no longer any problem regarding the inequalities. Alice sees the same lambda irrespective of Bob's setting, and vice versa (check this yourself) This is all it takes for the theorem to go through, and there can be no coincidence loophole here.

But of course this is not irrelevant; if both wings can see both lambdas, the model is nonlocal.

If we now rather assume that each wing can only see *one* of the two lambdas, namely the first lambda picked in a time slot *in that wing*, then we have to order the sets of two lambdas, and your example then gives the following:

Λac' = (pλ2,pλ1)

Λad' = (pλ2,pλ3)

Λbc' = (pλ4,pλ1)

Λbd' = (pλ4,pλ3)

Note that I have only ordered your sets according to what lambdas Alice and Bob actually get; the contents of the sets remains the same. The first element is what Alice sees, the second element is what Bob sees.

But with this ordering, there is no longer any problem regarding the inequalities. Alice sees the same lambda irrespective of Bob's setting, and vice versa (check this yourself) This is all it takes for the theorem to go through, and there can be no coincidence loophole here.

Jan-Åke Larsson:

( January 26th, 2015 8:34pm UTC )

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I agree with Peer 2, but for clarity: if Alice's choice of lambda did depend on Bob's setting, the model would be nonlocal.

Peer 3:

( January 26th, 2015 8:43pm UTC )

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Peer 2: "But of course this is not irrelevant; if both wings can see both lambdas, the model is nonlocal."

No no no. You are surely misunderstanding something. We are talking about data analysis after the fact of the experiment. The time slot sees all the lambdas reaching it, which is exactly what the other sister timeslot sees. Each particle in a pair have the same lambda, one goes left, the other goes right. By picking just one of the outcomes in a timeslot and ignoring the rest, you are filtering out the lambdas and **considering** just the ones corresponding to the outcome you picked. See my previous post for a very detailed description. For each pair of outcomes you consider together to calculate < AC' > there is one pair of corresponding lambdas. It is irrelevant which lambda comes from which particle in the pair. What is important is that when you consider the set of all lambdas, it should be the same as the set of lambdas used for calculating < AD' >, < BC' > and < BD' >.

You can't claim the model is non-local because you decide to put two remote results next to each other and multiply them. Otherwise all models will be non-local.

No no no. You are surely misunderstanding something. We are talking about data analysis after the fact of the experiment. The time slot sees all the lambdas reaching it, which is exactly what the other sister timeslot sees. Each particle in a pair have the same lambda, one goes left, the other goes right. By picking just one of the outcomes in a timeslot and ignoring the rest, you are filtering out the lambdas and **considering** just the ones corresponding to the outcome you picked. See my previous post for a very detailed description. For each pair of outcomes you consider together to calculate < AC' > there is one pair of corresponding lambdas. It is irrelevant which lambda comes from which particle in the pair. What is important is that when you consider the set of all lambdas, it should be the same as the set of lambdas used for calculating < AD' >, < BC' > and < BD' >.

You can't claim the model is non-local because you decide to put two remote results next to each other and multiply them. Otherwise all models will be non-local.

Peer 2:

( January 26th, 2015 8:53pm UTC )

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Sorry, I don't understand anything of that. When you say "We are talking about data analysis after the fact of the experiment," this particular analysis could well have been done during the experiment (with no knowledage of what happens at the other wing); it could even be integrated as a real time part of the detector hardware.

Jan-Åke Larsson:

( January 26th, 2015 9:00pm UTC )

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If both wings can see both lambdas the model is still local.

It is only if one site receives knowledge of the remote setting through some mechanism that the model is nonlocal. One example is if the local choice of lambda depends on the remote setting.

But it doesn't in Peer 3's model. Which is what Peer 2 is pointing to. (You are talking past each other)

It is only if one site receives knowledge of the remote setting through some mechanism that the model is nonlocal. One example is if the local choice of lambda depends on the remote setting.

But it doesn't in Peer 3's model. Which is what Peer 2 is pointing to. (You are talking past each other)

Peer 3:

( January 26th, 2015 9:06pm UTC )

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Peer2 : "Note that I have only ordered your sets according to what lambdas Alice and Bob actually get; the contents of the sets remains the same. The first element is what Alice sees, the second element is what Bob sees.

But with this ordering, there is no longer any problem regarding the inequalities. Alice sees the same lambda irrespective of Bob's setting, and vice versa (check this yourself) This is all it takes for the theorem to go through, and there can be no coincidence loophole here."

You simply do not understand it. The ensembles that have to be the same are the joint ensembles Λac' , Λad' , Λbc' , Λbd'. Your ordering does not change a single thing do you honestly believe that your ordering makes Λac' = Λad' = Λbc' = Λbd'.

Please review the above paper and note what ensembles Λac', Λad', Λbc', Λbd' are.

But with this ordering, there is no longer any problem regarding the inequalities. Alice sees the same lambda irrespective of Bob's setting, and vice versa (check this yourself) This is all it takes for the theorem to go through, and there can be no coincidence loophole here."

You simply do not understand it. The ensembles that have to be the same are the joint ensembles Λac' , Λad' , Λbc' , Λbd'. Your ordering does not change a single thing do you honestly believe that your ordering makes Λac' = Λad' = Λbc' = Λbd'.

Please review the above paper and note what ensembles Λac', Λad', Λbc', Λbd' are.

Peer 2:

( January 26th, 2015 9:25pm UTC )

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Peer 3: "The ensembles that have to be the same are the joint ensembles Λac' , Λad' , Λbc' , Λbd'."

No, and this is where you are mistaken. Lambda can be completely mutilated during the detection process and be transformed into an entirely different lambda, as long as this depends on the local setting only, and not on the setting in the other wing. Because then this process can just be regarded as part of the local realist model, and it cannot make the model violate the inequality.

No, and this is where you are mistaken. Lambda can be completely mutilated during the detection process and be transformed into an entirely different lambda, as long as this depends on the local setting only, and not on the setting in the other wing. Because then this process can just be regarded as part of the local realist model, and it cannot make the model violate the inequality.

Peer 3:

( January 26th, 2015 10:07pm UTC )

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Please review the above paper: Specifically please answer that you understand and agree with the authors when they say on page 4:

**"The original CHSH inequality is no longer valid, and the reason can be seen in the start of the proof where one wants to add

...

The integrals on the right-hand side cannot easily be added when Λac′=/= Λad′ , since we

are taking expectations over different ensembles Λac′ and Λad′ , with respect to different

probability measures.

The problem here is that the ensemble on which the correlations are evaluated changes with the settings,while the original Bell inequality requires that they stay the same. In effect,

the Bell inequality only holds on the common part of the four different ensembles Λac′ , Λad′ ,

Λbc′ , and Λbd"**

Then if you understand and agree with the authors there, then please point out what you believe the common part of Λac′ , Λad′ ,Λbc′ , and Λbd' is, in the following which you wrote above:

Λac' = (pλ2,pλ1)

Λad' = (pλ2,pλ3)

Λbc' = (pλ4,pλ1)

Λbd' = (pλ4,pλ3)

**"The original CHSH inequality is no longer valid, and the reason can be seen in the start of the proof where one wants to add

...

The integrals on the right-hand side cannot easily be added when Λac′=/= Λad′ , since we

are taking expectations over different ensembles Λac′ and Λad′ , with respect to different

probability measures.

The problem here is that the ensemble on which the correlations are evaluated changes with the settings,while the original Bell inequality requires that they stay the same. In effect,

the Bell inequality only holds on the common part of the four different ensembles Λac′ , Λad′ ,

Λbc′ , and Λbd"**

Then if you understand and agree with the authors there, then please point out what you believe the common part of Λac′ , Λad′ ,Λbc′ , and Λbd' is, in the following which you wrote above:

Λac' = (pλ2,pλ1)

Λad' = (pλ2,pλ3)

Λbc' = (pλ4,pλ1)

Λbd' = (pλ4,pλ3)

Peer 2:

( January 26th, 2015 11:18pm UTC )

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I guess that in retrospect the authors would rather have written "The problem here is that the ensemble on which the correlations are evaluated changes with both settings in a non-local manner, albeit indirectly". The paper was written in 2004, and there has been some developments in how Bell-type theorems are presented since then. And both authors have been instrumental in that process.

But anyway, I've found it to be good advice never to cling to a few sentences in a paper. Try rather to think about the simple fundamentals.

But anyway, I've found it to be good advice never to cling to a few sentences in a paper. Try rather to think about the simple fundamentals.

Richard Gill:

( January 27th, 2015 4:57am UTC )

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Good advice, Peer 2.

I guess it's about time the authors wrote a new (short, pedagogical) paper. The discussion here is very valuable. Thanks, to everyone!

I guess it's about time the authors wrote a new (short, pedagogical) paper. The discussion here is very valuable. Thanks, to everyone!

Peer 3:

( January 27th, 2015 2:22pm UTC )

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Peer2: "I guess that in retrospect the authors would rather have written "The problem here is that the ensemble on which the correlations are evaluated changes with both settings in a non-local manner, albeit indirectly".

Is this an admission that if the above paper is accurate then the inequalities do not apply to the example I've presented using the pair-picking method suggested by the authors?

You did not answer my previous question of what the common subset is for the 4 pairs of angles:

Λac' = (pλ2,pλ1)

Λad' = (pλ2,pλ3)

Λbc' = (pλ4,pλ1)

Λbd' = (pλ4,pλ3)

I assume you now understand that it is a null set. So please let us follow your advice and not simply "cling to a few sentences in a paper", let us see what the paper says about what the correct inequality should be if there is no common ensemble.

Page 4:

- equation (8), the expectation is therefore conditioned on a null set. An expectation over a null set is meaningless.

- equation (10), delta is zero since the numerator is a null set.

- Theorem 2 is no longer valid, because it says: "The prerequisites (i–iii) of

Theorem 1 are assumed to hold except at a null set, as is" and we now know the set is null for the example above.

- Equation (12), ΛI is null, therefore equation (12) which has 4 expectations conditioned on a null set is invalid, and so is equation (15).

- Since we know that delta is zero from equation (10), it means equation 17 says 0 is greater than or equal to 4 - 3/gamma, which means gamma is less than or equal to 3/4.

- if gamma = 3/4, then equation (20) has 4 on the RHS. Any value of gamma lower than 3/4 produces a negative number on the RHS which is impossible since then you have the sum of two positive numbers being negative. Therefore gamma is = 3/4, and the upper bound for the inequality is 4.

So then this is not simply a matter of improper wording of a small snippet of text. It is a big problem. The paper actually proves that the original inequalities do not apply to this example if the pair-picking method suggested is used. Not only that, they also prove that the correct inequality which should apply here has an upper bound of 4. These are the simple fundamentals.

Combined with my simple example presented in this thread, it proves that the pair-picking method suggested by the authors fails to avoid the coincidence-time loophole as the authors claim.

Is this an admission that if the above paper is accurate then the inequalities do not apply to the example I've presented using the pair-picking method suggested by the authors?

You did not answer my previous question of what the common subset is for the 4 pairs of angles:

Λac' = (pλ2,pλ1)

Λad' = (pλ2,pλ3)

Λbc' = (pλ4,pλ1)

Λbd' = (pλ4,pλ3)

I assume you now understand that it is a null set. So please let us follow your advice and not simply "cling to a few sentences in a paper", let us see what the paper says about what the correct inequality should be if there is no common ensemble.

Page 4:

- equation (8), the expectation is therefore conditioned on a null set. An expectation over a null set is meaningless.

- equation (10), delta is zero since the numerator is a null set.

- Theorem 2 is no longer valid, because it says: "The prerequisites (i–iii) of

Theorem 1 are assumed to hold except at a null set, as is" and we now know the set is null for the example above.

- Equation (12), ΛI is null, therefore equation (12) which has 4 expectations conditioned on a null set is invalid, and so is equation (15).

- Since we know that delta is zero from equation (10), it means equation 17 says 0 is greater than or equal to 4 - 3/gamma, which means gamma is less than or equal to 3/4.

- if gamma = 3/4, then equation (20) has 4 on the RHS. Any value of gamma lower than 3/4 produces a negative number on the RHS which is impossible since then you have the sum of two positive numbers being negative. Therefore gamma is = 3/4, and the upper bound for the inequality is 4.

So then this is not simply a matter of improper wording of a small snippet of text. It is a big problem. The paper actually proves that the original inequalities do not apply to this example if the pair-picking method suggested is used. Not only that, they also prove that the correct inequality which should apply here has an upper bound of 4. These are the simple fundamentals.

Combined with my simple example presented in this thread, it proves that the pair-picking method suggested by the authors fails to avoid the coincidence-time loophole as the authors claim.

Jan-Åke Larsson:

( January 27th, 2015 4:19pm UTC )

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It is not a null set. All four are equal, of positive measure, and are in fact identical to the total set of all possible λtimeslot (possibly, except for a null set).

Just because you do not list some of the coordinates does not mean that the entire vector λtimeslot cannot be used in all four entries.

Just because you do not list some of the coordinates does not mean that the entire vector λtimeslot cannot be used in all four entries.

Peer 3:

( January 27th, 2015 10:43pm UTC )

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Jan-Åke Larsson: "It is not a null set. All four are equal, of positive measure, and are in fact identical to the total set of all possible λtimeslot (possibly, except for a null set)."

It surely is a null set as previously explained. You are only allowed to include particle lambdas which contribute to the outcome you considered. The entire vector can only be used if you are also using the entire vector of outcomes in the timeslot. You can't pick the first outcome only, and assume all the lambdas from all the other particles contributed to the outcome.

It surely is a null set as previously explained. You are only allowed to include particle lambdas which contribute to the outcome you considered. The entire vector can only be used if you are also using the entire vector of outcomes in the timeslot. You can't pick the first outcome only, and assume all the lambdas from all the other particles contributed to the outcome.

Richard Gill:

( January 28th, 2015 4:36am UTC )

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You can. Mathematically you can, and what counts is the mathematical derivation that you can't violate CH (up to statistical variation ).

Jan-Åke Larsson:

( January 28th, 2015 6:55am UTC )

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In your example, you don't have to, but you *can* include the whole vector. This is what matters. Since I *can* include the whole vector, I can carry through the whole CHSH proof.

I could in fact build a more complicated example where the first output at Alice depends on all five lambdas ("if lambda1 is in this range, the output is this, otherwise let lambda2 decide the output, ..."). Then I *must* include the whole vector. And of course, the CHSH proof still holds in this situation.

I could in fact build a more complicated example where the first output at Alice depends on all five lambdas ("if lambda1 is in this range, the output is this, otherwise let lambda2 decide the output, ..."). Then I *must* include the whole vector. And of course, the CHSH proof still holds in this situation.

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Donald A. Graft:

( August 4th, 2015 3:34pm UTC )

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I show in the below-cited paper that the Larsson-Gill analysis based on the "fair coincidences assumption" does nothing to exclude a simple local model.

http://arxiv.org/abs/1507.06231

The quantum mysterians are squirming!

http://arxiv.org/abs/1507.06231

The quantum mysterians are squirming!

Unregistered Submission:

( August 5th, 2015 11:14am UTC )

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You should add that such exclusion would be too far from the topic of the analysis (which is to show that, in certain situations, such models are not excluded).

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- - Analysis of the Christensen et al Clauser-Horne (CH)-Inequality-Based Test of Local Realism
- - Analysis of the Christensen et al Clauser-Horne (CH)-Inequality-Based Test of Local Realism
- - Analysis of the Christensen et al Clauser-Horne (CH)-Inequality-Based Test of Local Realism
- - Statistics, Causality and Bell's Theorem
- - Macroscopic Observability of Spinorial Sign Changes under 2π Rotations
- - Macroscopic Observability of Spinorial Sign Changes under 2π Rotations
- - Statistics, Causality and Bell's Theorem
- - Statistics, Causality and Bell's Theorem
- - Analysis of the Christensen et al Clauser-Horne (CH)-Inequality-Based Test of Local Realism

Richard Gill Wrote:

"Unregistered submitter: why it avoids the coincidence loophole: because there is no selective rejection of experimental units (paired time slots). The processing of data in each wing of the experiment, in each time slot, is local. If however you decide on the basis of comparing results in the two wings of the experiment to reject altogether some of the experimental units, you may very well introduce bias and that bias may very well depend on the measurement settings applied to each time slot.

The coincidence loohole, and the detection loophole, both arise through selectively discarding some of the experimental units, in a way which might depend on hidden variables and on the measurement setting applied to the experimental unit.

See "Bertlmann's socks", especially the discussion around Figure 7. Picking and reporting the first event in the time-slot is part of what happens (or may be thought to happen) entirely *within* the time limits and the space limits of the big box drawn in the figure. It's just a little computer chip built into the measurement device. The measurement device generates a binary outcome at a certain place and within a certain time limit."

This is why as asked you to be specific about the problem of coincidence. Say we have a local realistic model. As part of this model, the detection time at each station is a function of the hidden local particle parameter, and the local setting chosen at the station. This is all perfectly fine local realistically. You agree that such a system can easily violate the inequality as you discuss in the above paper.

The reason it can violate the inequality is because the time difference between sister particles is not a fixed number, but varies in a non-factorable way, even though all the delays were caused by only local effects. The problem is then that by selecting only particles which are closer together than an arbitrary time offset, you will be ignoring those which are further apart in time than the chosen offset. I think you will agree to this. The goal of coincidence matching is to faithfully group the detected pairs as close as possible to the emitted pairs, not simply a matter of grouping *any* pairs which happen to be close in time. But from the above, we know this is practically impossible.

Now Larsson claims in his survey paper that the problem of coincidence and detection loopholes is one of "deficiencies in the experimental equipment". This is false. If the way nature works is that some hidden parameters just can't be detected at certain angles, then you could have perfect equipment, and still have the detection loophole. Similarly, even if the experimental equipment is perfect, how would you know which particles belong with each other if the arrival times are variable, and "hidden"?

Larsson also claims that the coincidence loophole can be avoided if "coincidence is determined by closeness to the event-ready indication time, not by small time difference to the remote detection." In other words, he still assumes that the two particles will be detected close to the start of the "event-ready" signal, ie, close to each other. The only difference here is that instead of using the times directly, to compare with each other, you now have an external clock which you are comparing both sides to. That accomplishes nothing. You will still not be able to match events which are inherently delayed in an angle dependent manner. So how does this avoid the coincidence loophole? It doesn't.

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