Is the Quantum Randi Challenge (QRC) valid?

Foundations of physics and/or philosophy of physics, and in particular, posts on unresolved or controversial issues

Re: Is the Quantum Randi Challenge (QRC) valid?

Postby gill1109 » Sat Feb 15, 2014 2:05 pm

"The analogy is very appropriate. Any statistician should recognize it immediately. You cannot use disjoint samples to disprove a claim about JOINT properties. If you disagree about this crucial point say so. You have repeatedly avoided this question."

You can use disjoint samples to disprove a claim about joint properties. The important thing is that they are random samples. The "proof" will not be a certain proof, but a statistical proof.

My paper carefully explains in detail how this all works out in the case of interest. Theorem 1 in Section 2 is an example of exactly this principe. Its proof is in the appendix. Every statistician understands that there are situations where it can be done.

Do you disagree with the theorem? The theorem and its proof has been carefully read by many very competent probabilists and statisticians. It is not "rocket science". In fact it is pretty elementary (finitary, counting) probability - no calculus, no limits.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby FrediFizzx » Sat Feb 15, 2014 2:13 pm

We are going around in circles here and the discussion of Gill's version of CHSH is off topic for this thread. This thread is about the QRC.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby minkwe » Sat Feb 15, 2014 2:14 pm

FrediFizzx wrote:Hmm... Well, unlike Gill's version of CHSH that NOTHING can violate, not even QM, the QRC can show violation if there is a connection between the a and b vectors. Do you have Mathematica? I have programmed the QRC with your parameters and the notebook file is here.

I am just trying to find what the exact mathematical flaw might be but I hear what you are saying.

Actually, not even the standard CHSH can be violated by anything if you are doing a correct mathematical calculation. Gill's version is not very different. In fact I believe Gill's paper doesn't add anything new to the conversation. Not even QM can violate the standard CHSH. The reason is because the terms in the CHSH represent a relationship between joint properties which can not be simultaneously measured (factual & counterfactual). I come to the same conclusion as many others that the CHSH can never be tested experimentally until such a time as we can recover measured particles and remeasure them. This is why all challenges like Gill's have some variant of a requirement which is similar to re-measuring already measured particles in order to obtain counterfactual results. Gill knows this very well that is why he is avoiding/diverting my questions on this central point.

All EPR-type experiments (and QM) are calculating expressions of the type <A1B1> + <A2B2'> + <A3'B3> - <A4'B4'>, and comparing the results with the CHSH which is an expression of the type <A1B1> + <A1B1'> + <A1'B1> - <A1'B1'>. The correct upper bound for the former expression is 4 not 2. Gills proof is deriving an expression of the latter type (upper-bound 2) not the former type (upper-bound 4).

The QRC and similar challenges ask you to produce A1, A1', B1, B1' which will violate the CHSH. This is an invalid challenge because QM and experiments do not produce that, they produce A1, B1, A2, B2', A3', B3, A4', B4'!

Besides, the claimed QM violation of the standard CHSH is not even a valid calculation as far as QM is concerned since the full expression implies that we can simply combine separate observations of non-commuting observables to obtain the joint measurement result. That this is a grave error has escaped many people including Gill but guess who first used this argument to disprove a so-called "proof" of no hidden variables? Bell himself!

Bell wrote:Of course the explanation is well known: A measurement of a sum of non-commuting observables cannot be made by combining trivially the results of separate observations on the two terms -- it requires a quite distinct experiment.
...
But this explanation of the non-additivity of allowed values also establishes the non-triviality of the additivity of expectation values.


By calculating the QM expectation values of each pair separately and then adding them trivially, everyone who claims violation of the CHSH is making this mistake. But note something, the reason why you can't do this, is the same reason why you can't make statements about joint properties by using disjoint measurements, just expressed differently.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby FrediFizzx » Sat Feb 15, 2014 2:24 pm

All EPR-type experiments (and QM) are calculating expressions of the type <A1B1> + <A2B2'> + <A3'B3> - <A4'B4'>, and comparing the results with the CHSH which is an expression of the type <A1B1> + <A1B1'> + <A1'B1> - <A1'B1'>. The correct upper bound for the former expression is 4 not 2. Gills proof is deriving an expression of the latter type (upper-bound 2) not the former type (upper-bound 4).


That is not correct. The correct form of CHSH is E(a, b) + E(a', b) + E(a, b') - E(a', b') which corresponds to <A1B1> + <A2B2> + <A3B3> - <A4B4> (no primes). Even though the a detector angles are the same in E(a, b) and E(a, b') the A's and B's are not.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby minkwe » Sat Feb 15, 2014 2:48 pm

Fred,
There is only a notational misunderstanding. What I have written is exactly correct. In an experiment, to calculate E(a,b), for a large number of measurements, we collect all the data pairs for the set of particle pairs where Alice measured at angle a and Bob measured at angle b. We multiply the results together and take the average. This is what I mean by <A1B1>. where the 1 represents the set of particles over which we are averaging (call it spreadsheet 1). A1 represents the column of spreadsheet 1 which contains the outcomes at angle a for the first set of particles, B1 represents the column of spreadsheet 1 which contains the outcomes at angle b for the same set. To calculate E(a', b), we collect all the data pairs for the set of particle pairs where Alice measured at angle a' and Bob measured at angle b. Note that in actual experiments, this is a completely different set of particle pairs which we now call set 2 (spreadsheet 2). The columns of this spreadsheet are labelled A2, and B2' (corresponding to settings a and b'. We now calculate E(a, b') = <A2B2'>. Do you understand now what I mean? Similarly for E(a', b) and E(a', b'), calculated as <A3'B3>, <A4'B4'> respectively.

This is why I say EPR-experiments only ever measure expressions of the form <A1B1> + <A2B2'> + <A3'B3> - <A4'B4'>. The upper bound of expressions of this type is 4 not 2 contrary to the CHSH.

The notation is that A4' represents the column in spreasheet 4 which corresponds to a setting of a'. I agree with you that comun A1 is different from column A2 even though they both use the setting a, this is because the set of particles measured is different (set 1 vs set 2). But columns A1 and A3' are different both in the settings used (a vs a') and the set of particles measured (set 1 vs set 3). Your notation of "<A1B1> + <A2B2> + <A3B3> - <A4B4> (no primes)" is confusing because it does not capture the fact that there are 4 settings being used (a, a', b, b'), two at Alice and two at Bob. The cells in our spreasheet of course always contain +/-1.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby FrediFizzx » Sat Feb 15, 2014 3:04 pm

Ok sorry, looked at your quote again and that is right if you are just using the primes to indicate where a' and b' are used. What you have wrong is the following quote.

The QRC and similar challenges ask you to produce A1, A1', B1, B1' which will violate the CHSH. This is an invalid challenge because QM and experiments do not produce that, they produce A1, B1, A2, B2', A3', B3, A4', B4'!


The QRC does not do that. Only Gill does that. QRC does the correct calculation of CHSH I believe. Take a look at the CHSH calculation in this PDF file.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby gill1109 » Sat Feb 15, 2014 4:21 pm

FrediFizzx wrote:We are going around in circles here and the discussion of Gill's version of CHSH is off topic for this thread. This thread is about the QRC.

The discussion applies equally well to the QRC, which is in fact a stronger test than CHSH. It is CHSH with an extra demand: one of the four correlations must equal 1.

Both Fred and Minkwe don't understand that different samples can be used to say something about a quantity which is not measured in either - to say something about the relation between the quantities measured in the two different samples.

Let me give a practical example. I want to say something about the mean of the difference between people's income in 2012 and their income in 2013. I assume that we are talking about a well defined, large, population of people who have an income in both years, and that we can find out what it is. I take a sample of 100 people and take their average income in 2012. I take another sample of 100 people and take their average income in 2013. The difference between these two averages is an estimate of the mean differences, over all the people in the population. Make the sample sizes "100" big enough and it can be as accurate as you like.

Take a spreadsheet with 800 rows. Split the rows completely at random into four sub-samples. The averages of any particular column per subsample will (with large probability!) all be close to one another and close to the average over the whole table.

Apparently Minkwe doesn't believe me. Ask any probabilist. Or do a little sampling experiment yourself, to test this experimentally.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby FrediFizzx » Sat Feb 15, 2014 4:48 pm

Take a look at the CHSH calculation in this PDF file.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby gill1109 » Sat Feb 15, 2014 5:03 pm

FrediFizzx wrote:Take a look at the CHSH calculation in this PDF file.


I took a look. Yes it is possible (and indeed easy) to violate Bell's original inequality (demanding perfect anti-correlation) about 50% of the time by a local realist mechanism. One example is not enough. QRC says you have to create software which, at N = 800, does it 99% of the time (or something like that, I forget the details), just like QM can.

Once you have figured out how to do it, post your programs on internet so everyone can convince themselves you are right, and you win undying fame.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby FrediFizzx » Sat Feb 15, 2014 5:23 pm

That is not what you were talking about. Please stay focused. You will see a real calculation of CHSH. That is how it is done. Not the way you express it.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby gill1109 » Sat Feb 15, 2014 5:33 pm

FrediFizzx wrote:That is not what you were talking about. Please stay focused. You will see a real calculation of CHSH. That is how it is done. Not the way you express it.


Sorry I have kind of lost the thread... focus on what?

I still have no idea what the question means "Is the Quantum Randi Challenge (QRC) valid?". The abstract of Vongehr's "Annals of Physics" article reads: "There are increasingly suggestions for computer simulations of quantum statistics which try to violate Bell type inequalities via classical, common cause correlations. The Clauser–Horne– Shimony–Holt (CHSH) inequality is very robust. However, we argue that with the Einstein–Podolsky–Rosen setup, the CHSH is inferior to the Bell inequality, although and because the latter must assume anti-correlation of entangled photon singlet states. We simulate how often quantum behavior violates both inequalities, depending on the number of photons. Violating Bell 99% of the time is argued to be an ideal benchmark. We present hidden variables that violate the Bell and CHSH inequalities with 50% probability, and ones which violate Bell 85% of the time when missing 13% anti- correlation. We discuss how to present the quantum correlations to a wide audience and conclude that, when defending against claims of hidden classicality, one should demand numerical simulations and insist on anti-correlation and the full amount of Bell violation."

Is there anything there you don't understand / don't agree with? Or do you have another question in mind?
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby FrediFizzx » Sat Feb 15, 2014 5:48 pm

We were talking about the difference between your CHSH expression and how it is done in the QRC. The QRC calculation is correct; yours is wrong. Dead wrong.

One more time; Is the QRC really a valid test for Bell models? Or is there a hidden subtle flaw in it? The anti-correlation part looks OK. The CHSH part looks OK. But don't know yet about the rest of it.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby minkwe » Sat Feb 15, 2014 8:28 pm

FrediFizzx wrote:Ok sorry, looked at your quote again and that is right if you are just using the primes to indicate where a' and b' are used. What you have wrong is the following quote.

The QRC and similar challenges ask you to produce A1, A1', B1, B1' which will violate the CHSH. This is an invalid challenge because QM and experiments do not produce that, they produce A1, B1, A2, B2', A3', B3, A4', B4'!


The QRC does not do that. Only Gill does that. QRC does the correct calculation of CHSH I believe. Take a look at the CHSH calculation in this PDF file.

Don't you see that on the section that calculates the CHSH each term is calculated over the same summation? They are summing over the exact same set of outcomes (j=1..n) for each term. If I understand correctly.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby FrediFizzx » Sat Feb 15, 2014 11:05 pm

No. There are "If" conditions there so it is only adding up the 1's for if the condition is met. If the condition is not met, a 0 is put in the summation. Take for example E_0 = 2N_E0/N_0 -1. Which we could say represents the E(a, b) part of CHSH. For that sum it is only adding up if a = b, etc. Of course that only happens when a and b are at 0 degrees. Then E_1 is taking a different set of when the angles are at a certain value. And so forth.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby gill1109 » Sun Feb 16, 2014 5:59 am

FrediFizzx wrote:We were talking about the difference between your CHSH expression and how it is done in the QRC.


Which expression from my paper is what you call "my CHSH expression"?
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby minkwe » Sun Feb 16, 2014 8:22 am

FrediFizzx wrote:No. There are "If" conditions there so it is only adding up the 1's for if the condition is met. If the condition is not met, a 0 is put in the summation. Take for example E_0 = 2N_E0/N_0 -1. Which we could say represents the E(a, b) part of CHSH. For that sum it is only adding up if a = b, etc. Of course that only happens when a and b are at 0 degrees. Then E_1 is taking a different set of when the angles are at a certain value. And so forth.

I see, so the sets used to calculate each term is disjoint from each other. Which means they are calculating an expression of the form <A1B1> + <A2B2'> + <A3'B3> - <A4'B4'>, for which the upper bound is 4. In that case it is a fair representation of what is possible in experiments but it is not calculating the true CHSH, which requires a single spreadsheet (that is, an expression of the type <A1B1> + <A1B1'> + <A1'B1> - <A1'B1'>.). That is why it can violate it.

The main point I'm making is that all the paradoxes originate from confusing The results from <A1B1> + <A2B2'> + <A3'B3> - <A4'B4'> with a condition based on <A1B1> + <A1B1'> + <A1'B1> - <A1'B1'>. The QRC does it by comparing the former expression based on 4 disjoint sets, with 2, which is the upper bound for the latter expression derived from a single set.

It is similar to claiming that the expression "women are shorter than men" is not true, by using height data from one set of randomly selected people without regard for gender, and gender data from a disjoint set of randomly selected people without regard for height. Its bad statistics.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby gill1109 » Sun Feb 16, 2014 12:37 pm

minkwe wrote:It is similar to claiming that the expression "women are shorter than men" is not true, by using height data from one set of randomly selected people without regard for gender, and gender data from a disjoint set of randomly selected people without regard for height. Its bad statistics.


This is amusing. Let me turn it around. Suppose we are interested in the average difference in height between a husband and wife. We could take a sample of married couples, and average the differences.

But ... we could also take a random sample of married couples, and average the heights of the women; we could take another, independent, random sample of married couples, and average the heights of the men. The difference between the two averages is a decent estimator of the average difference in height within married couples.

Statistics is a powerful tool.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby FrediFizzx » Sun Feb 16, 2014 1:02 pm

minkwe wrote: I see, so the sets used to calculate each term is disjoint from each other. Which means they are calculating an expression of the form <A1B1> + <A2B2'> + <A3'B3> - <A4'B4'>, for which the upper bound is 4. In that case it is a fair representation of what is possible in experiments but it is not calculating the true CHSH, which requires a single spreadsheet (that is, an expression of the type <A1B1> + <A1B1'> + <A1'B1> - <A1'B1'>.). That is why it can violate it.

The main point I'm making is that all the paradoxes originate from confusing The results from <A1B1> + <A2B2'> + <A3'B3> - <A4'B4'> with a condition based on <A1B1> + <A1B1'> + <A1'B1> - <A1'B1'>. The QRC does it by comparing the former expression based on 4 disjoint sets, with 2, which is the upper bound for the latter expression derived from a single set.


Something is not quite right with your analysis for the "true CHSH" as it has been mathematically proven that the upper bound on
E(a, b) + E(a', b) + E(a, b') - E(a', b') is 2sqrt(2). But I think I did notice that sometimes when QRC's CHSH is violated, it was like 3.9 so I think you are right about it is doing <A1B1> + <A2B2'> + <A3'B3> - <A4'B4'>. If that is the case then that is a mistake in the QRC. But perhaps not a serious mistake. I will check this out further. Thanks.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby minkwe » Sun Feb 16, 2014 1:10 pm

FrediFizzx wrote:Something is not quite right with your analysis for the "true CHSH" as it has been mathematically proven that the upper bound on
E(a, b) + E(a', b) + E(a, b') - E(a', b') is 2sqrt(2).


I do not agree with that, do you have a paper where that prove is presented, I'll like to break it apart and show you the flaw. Are you perhaps referring to the calculation in which each term is calculated using QM independently, or are you referring to the derivation of the CHSH. Note that you can not have two mathematical proofs disagreeing with each other so if you say there is a mathematical proof showing an upper bound of 2 root 2, and also a mathematical proof showing an upper bound of 2, something must be wrong somewhere.

But I think I did notice that sometimes when QRC's CHSH is violated, it was like 3.9 so I think you are right about it is doing <A1B1> + <A2B2'> + <A3'B3> - <A4'B4'>. If that is the case then that is a mistake in the QRC. But perhaps not a serious mistake. I will check this out further. Thanks.

Remember I've been saying the upper bound is 4, so 3.9 is pretty close to the bound. Do you agree that the upper bound of <A1B1> + <A2B2'> + <A3'B3> - <A4'B4'> is 4?
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby gill1109 » Sun Feb 16, 2014 1:16 pm

Vongehr's set-up is taken, without no alterations at all, straight from Bell (1965) - the guy who invented (in that landmark paper) the Bell inequalities, right!?

Since then almost everyone has agreed that this is a decent Gedankenexperiment for discriminating between QM and local realism. (Not everyone: Christian disagrees, and he was just the so many'th in a long list of rebels). It relies on the "Gedankentatsache" of perfect anti-correlation when the two measurement settings are the same. This theoretical fact is unfortunately impossible to realize *perfectly* in the real world. Hence CHSH invented a *weakening* of Bell's inequality (weakening: it is easier to beat) which made more sense for real world experiments. You do not demand something completely unfair.

Vongehr argues that for pedagogical purposes thought experiments are enough, and that computer simulations of thought experiments ought to be perfect, so demanding perfect anti-correlation at equal settings is not unfair. He argues that it does bring in a number of advantages.

FrediFizzx wrote:"Is it valid?" just means is it really a true Bell model tester? Nothing really perplexing about that. So far, it seems pretty limited since you can only have 3 different angles for a, a', b, and b'. Actually that is not quite right; you can have two angles for a and a' and two angles for b and b'. Is that sufficient for a true test?
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