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

[lugita15]

The relative frequencies (lets use R) Rs(-30,30), Rs(-30,0), Rs(0,30) all calculated within the same set "s" can not all be equal to the corresonding relative frequencies of Rp(-30,30), Rq(-30, 0), Rr(0,30) each calculated from three disjoint sets p, q and r respectively. Do you agree?

No, I don't agree with that at all. I believe that Rs(-30,0)=Rp(-30,0)=Rq(-30,0)=Rr(-30,0), and similarly for (0,30) and (-30,30).

And why do I believe that? Because if the relative frequency of something is the same for three different sets, then it is the same for their union, a statement I thought you agreed with.

 

[wle]

You've described one scenario in which they could obtain a violation but as you know that is not the only one.

If you're referring to the fact that I only considered one set of lists, then it was only intended as an example, and it's clear you're going to get similar results with any other set of lists anyway.

Besides, if Alice measures 123 for one pair, how do you make sure that they also must measure 213 and 321 in order to compensate for the violation?

They don't. If they happen to measure 123 then, in this particular scenario they get a very violent Bell violation despite the fact the model is local. That's their end result. If they were planning on writing a paper, then that's what they publish: they got 3. Same if they measured 231. But not if they measured 312, 132, 213, or 321. So if they pick the measurements completely randomly out of this set then according to this local model there's a 1 in 3 chance that they observe a Bell violation, and a 2 in 3 chance that they don't. This is in an experiment where the Bell correlator is estimated from measurement results on three photon pairs. The point is, it's not going to be a 1 in 3 chance for an experiment performed on, say, 10,000 photon pairs.

So I'm not sure I get your argument, it appears you are agreeing with me but it is written as a disagreement.

I'm saying that yes, a Bell inequality can be violated even according to a local theory, but it's not a problem if the chance of that happening becomes vanishingly small when the experiment is performed on a very large number of photon pairs.

If you're doing a Bell test on 10,000 entangled photon pairs and quantum physics says 1.5, does it really matter whether locality says < 1 or < 1.0001 except for some extremely low probability?

 

[billschnieder]

No, the constraint on the statistics is derived from a very black box definition of locality that Bell introduced. From that point of view, entangled photons are interesting only as an example of a system we can manipulate that can produce statistics that violate those constraints.

That is where you are mistaken, the constraints on the statistics exist precisely because those same constraints exist in the individual pairs. Let us verify this using Bell's original derivation http://www.drchinese.com/David/Bell_Compact.pdf and you can follow along:

Bell's expression (2) is an integral of the paired product of individual outcomes, then on Page 406, leading up to equation 15, he writes P(a,b) - P(a,c) on the LHS but note what is happening on the RHS, he is factoring out individual outcomes in precisely the way I did earlier for individual pairs. First he factors out A(a,λ) within the integral, and then he factors out A(b,λ) and as a result of this factorization, he creates the A(b,λ)A(c,λ) product within the integral which is then integrated to obtain the P(b,c) term. The factoring is a pivotal step! Obviously, the relationship between the statistics P(a,b), P(a,c) and P(b,c) is derived by considering the relationship between the individual outcomes being integrated.

 

[billschnieder]

This factorization step already implies that there are only three "lists" of outcomes [A(a,λi)], [A(b, λi)], and [A(c,λi)], which have been recombined and integrated over to obtain the terms P(a,b), P(a,c) and P(b,c). This is of course what Bell intended in the derivation because he simply assumes that another vector c exists along side a and b.

However, if you now want to use 3 correlations P(a1,b1), P(a2,c2) and P(b3,c3), obtained by recombining 6 lists [A(a1,λi)], [A(b1, λi)], [A(a2,λj)], [A(c2, λj)], [A(b3,λk)], and [A(c3,λk)], while maintaining the validity of the inequality derived for 3 lists, you MUST make one of the following two assumptions:

1. assume that [A(a1,λi)]=[A(a2, λj)] and [A(b1,λi)]=[A(b3, λk)] and [A(c2,λi)]=[A(c3,λk)]
2. assume that, whenever b3c3 = -1, there must be a corresponding "tuning" such that |a1b1 + a2c2| < 2 and when ever |a1b1 + a2c2| = 2, there must be a corresponding "tuning" such that b3c3 = 1.

These are the only two situations under which the inequality derived for a single set as Bell did, would apply to three different sets. It is not a choice, you do not have to like it, but by taking correlations obtained from three disjoint sets and comparing with the inequality, you ARE IN FACT making one or both of those assumptions whether you see it or not.

If you still disagree, I will ask you to derive the inequality starting with 3 different disjoint sets of photons or 6 lists of outcomes[A(a1,λi)]=[A(a2, λj)] and [A(b1,λi)]=[A(b3, λk)] and [A(c2,λi)]=[A(c3,λk)].

If you can derive the inequality from this starting point, without making one of those two assumptions then you will have proven that I am wrong on this point. If you prefer you can start with the statistics P(a1,b1), P(b3,c3) and P(a2,c2) and derive the inequalities. But we both know that you can't, because it is not possible. You will arrive at an inequality with 3 on the RHS, and you know this.

 

[billschnieder]

No, I don't agree with that at all. I believe that Rs(-30,0)=Rp(-30,0)=Rq(-30,0)=Rr(-30,0), and similarly for (0,30) and (-30,30).

At least now it is clear on which point we disagree. Or is it?

And why do I believe that? Because if the relative frequency of something is the same for three different sets, then it is the same for their union a statement I thought you agreed with.

I agree with this statement. But this is not the same statement as what we disagree with, nor does this statement prove or disprove what we disagree with. Can't you see that?

I said:

The relative frequencies Rs(-30,30), Rs(-30,0), Rs(0,30) all calculated within the same set "s" CAN NOT ALL BE EQUAL to the corresponding relative frequencies of Rp(-30,30), Rq(-30, 0), Rr(0,30) each calculated from three disjoint sets p, q and r respectively.

You are saying that the relative frequencies Rs(-30,30), Rs(-30,0), Rs(0,30) all calculated within the same set "s" ARE ALL EQUAL to the corresponding relative frequencies of Rp(-30,30), Rq(-30, 0), Rr(0,30) each calculated from three disjoint sets p, q and r respectively.


I think an example of such a set is in order.

 

[lugita15]

I agree with this statement. But this is not the same statement as what we disagree with, nor does this statement prove or disprove what we disagree with. Can't you see that?

No, I'm afraid I can't see that. The logic is really simple. We know that Rp(-30,0)=Rq(-30,0)=Rr(-30,0), so the relative frequency of M(-30,0) is the same for p, q, and r. Therefore it's the same for their union. So Rs(-30,0) is also equal to the same thing. (I can prove this if you like.) Where am I going wrong?

You are saying that the relative frequencies Rs(-30,30), Rs(-30,0), Rs(0,30) all calculated within the same set "s" ARE ALL EQUAL to the corresponding relative frequencies of Rp(-30,30), Rq(-30, 0), Rr(0,30) each calculated from three disjoint sets p, q and r respectively..

Yes, and the reason I'm saying that is because you agreed that Rp(-30,0), Rq(-30,0), and Rr(-30,0) are the same, and similarly for (0,30) and (-30,30).

 

[BenjaminTR]

Ok, how about this. Bill, if you think an LHV theory can match the predictions of QM, let's see a set of particles that does it.
QM predicts that 75% of pairs will differ when set at (-30,30), 25% will differ when measured at (-30,0) and 25% will differ when measured at (0,30). We want three disjoint sets that all obey these predictions separately, and thus whose union obeys them. Thus, consider 12 pairs, divided into groups a, b, and c, each with 4 pairs. For each pair, you just need to say whether it would give 1 or -1 at -30 degrees, 1 or -1 at 0 degrees and 1 or -1 at 30 degrees.
In other words, we just need 12 ordered triples from {-1,1}x{-1,1}x{-1,1} such that 9 of them differ between first and third place, 3 differ between first and second, and 3 differ between second and third. Then pick groups a, b and c so that 3 of their pairs differ between first and third place, 1 differ between first and second, and 1 differ between second and third.
Should be easy.

 

[billschnieder]

No, I'm afraid I can't see that.

Rs(-30,30) = Rs(-30,0) + Rs(0,30) - 2* Rs((-30,0)&(0, 30)), the equality
Rs((-30,0)&(0, 30)) >= 0 ... (*)
therefore Rs(-30,30) <= Rs(-30,0) + Rs(0,30), your inequality

therefore Rs((-30,0)&(0, 30)) = 0.5 * [Rs(-30,0) + Rs(0,30) - Rs(-30,30)] >= 0


If Rp=Rs(-30,30)=0.75, Rq=Rs(-30,0)=0.5, and Rr=Rs(0, -30)=0.5 ... (*)
then Rs((-30,0)&(0, 30)) = 0.5 * [0.5 + 0.5 - 0.75] = -0.125 < 0

You have two contradictory assumptions (*). If Rs((-30,0)&(0, 30)) >= 0 as you assumed when you derived the inequality, then it must be the case that the three correlations Rp(-30,30), Rq(-30,0), and Rr(0,30) CAN NOT ALL BE EQUAL to the three correlations Rs(-30,30), Rs(-30,0) and Rs(0,30). This is obvious because if the sets p, q, r are disjoint then R((-30,0)&(0, 30)) is meaningless, since M(-30,0) is measured in one set and M(0, 30) is measured in a disjoint set from the first. So your argument boils down to the tautological conclusion that the meaningless relative frequency is negative. In other words, you assumed that p, q, r were not disjoint (presence of Rs((-30,0)&(0, 30)) in the derivation), and then later assumed that they were disjoint --> violation.

 

[billschnieder]

Ok, how about this. Should be easy.

It should be easy for you to derive Bell's inequalities starting from 3 disjoint sets. How about that?

QM predicts that 75% of pairs will differ when set at (-30,30), 25% will differ when measured at (-30,0) and 25% will differ when measured at (0,30).

Wrong. QM does not predict that for the same set. QM predicts that for 3 disjoint sets, and those percentages do not violate the inequality derived from 3 disjoint sets. If you disagree, derive the inequality directly from 3 disjoint sets and show that such an inequality is still violated by QM.

The elephant in the room is this. Why do you always use the QM prediction for three disjoint sets when comparing with the inequality derived for a single set. Why don't you use the QM prediction for a single set while comparing with an inequality from a single set, or why don't you compare the QM prediction for 3 disjoint sets with the equivalent inequality derived for 3 disjoint sets? Why? Why? Why?

You may find the answer here:
Foundations of Physics Letters, Vol 15, No 5 (2002)
http://arxiv.org/pdf/quant-ph/0211031

 

[morrobay]

It should be easy for you to derive Bell's inequalities starting from 3 disjoint sets. How about that?



Wrong. QM does not predict that for the same set. QM predicts that for 3 disjoint sets, and those percentages do not violate the inequality derived from 3 disjoint sets. If you disagree, derive the inequality directly from 3 disjoint sets and show that such an inequality is still violated by QM.

The elephant in the room is this. Why do you always use the QM prediction for three disjoint sets when comparing with the inequality derived for a single set. Why don't you use the QM prediction for a single set while comparing with an inequality from a single set, or why don't you compare the QM prediction for 3 disjoint sets with the equivalent inequality derived for 3 disjoint sets? Why? Why? Why?

You may find the answer here:
Foundations of Physics Letters, Vol 15, No 5 (2002)
http://arxiv.org/pdf/quant-ph/0211031

The other elephant in the room is that n[y-z+] + n[x-y-] ≥ n[x-z+] is derived from values with parallel detector settings where P(α|λ| is considered deterministic. But then the inequality is compared against outcomes for non parallel detector settings: 45° , 90°, 135° where P(α|λ| can be non deterministic with respect to detector setting and state of particle at time of measurement.

 

[wle]

These are the only two situations under which the inequality derived for a single set as Bell did, would apply to three different sets. It is not a choice, you do not have to like it, but by taking correlations obtained from three disjoint sets and comparing with the inequality, you ARE IN FACT making one or both of those assumptions whether you see it or not.

You're substituting the point I made with a strawman. I am not making any such assumptions. That should have been clear to you since I explicitly agreed that the Bell correlator in a Bell test could violate the local bound you'd find in a textbook, even according to a local model.

But so what? Suppose it can be shown that the chance of a significant Bell violation becomes vanishingly unlikely in a Bell test involving measurements on thousands or millions of entangled pairs. Suppose I'm doing a CHSH test and expecting something close to the quantum bound of 2 \sqrt{2}. Does it really matter that locality says "the probability of getting more than 2 + \varepsilon is less than a billion for some very small \varepsilon", instead of just "you should get less than 2"? That is the question you should be asking yourself.

 

[billschnieder]

I am not making any such assumptions.

You are. You have no other legitimate reason to compare correlations from 3 sets with an inequality derived from a single set. If you reject the only possible assumptions that allow you to make that comparison, then you can't continue to make the comparison. Period!

That should have been clear to you since I explicitly agreed that the Bell correlator in a Bell test could violate the local bound you'd find in a textbook, even according to a local model. But so what?

So you are not allowed to use the inequality, unless you assume that the model is tuned to avoid situations where violations occur. Derive the inequality from 3 sets and show that it is still violated.

 

[BenjaminTR]

[...]
Wrong. QM does not predict that for the same set. QM predicts that for 3 disjoint sets, and those percentages do not violate the inequality derived from 3 disjoint sets. If you disagree, derive the inequality directly from 3 disjoint sets and show that such an inequality is still violated by QM.

[...]

It's true the predictions are for three disjoint sets. However, if the expected fraction of disagreeing pairs measured at (-30,0) is 1/4, the LHV theory must either say that 1/4 of all photon pairs would disagree if measured in that direction, or that the photons we measure are systematically biased, i.e. no fair sampling. Without one of these two assumptions, the expected value is not 1/4. Ruling out conspiracies, we have to pick the first option, that 1/4 of the pairs would disagree at (-30,0), whether we measure that or not.

 

[wle]

You are. You have no other legitimate reason to compare correlations from 3 sets with an inequality derived from a single set. If you reject the only possible assumptions that allow you to make that comparison, then you can't continue to make the comparison. Period!

If you think I am making such assumptions, then instead of making vague accusations I invite you to point out where, exactly, I use such an assumption in for instance my [POST=4432843]post #393[/POST], and how the actual argument I actually made is affected by it.

So you are not allowed to use the inequality, unless you assume that the model is tuned to avoid situations where violations occur. Derive the inequality from 3 sets and show that it is still violated.

You are demanding something completely unnecessary. It is not necessary for the inequality to hold for all 3 sets. If you insist that that alone makes experimental Bell tests meaningless then you are knocking down a strawman.

Incidentally, you still haven't given a direct response to a question I've posed you at least three times now:

Suppose it can be shown that the chance of a significant Bell violation becomes vanishingly unlikely in a Bell test involving measurements on thousands or millions of entangled pairs. Suppose I'm doing a CHSH test and expecting something close to the quantum bound of 2 \sqrt{2}. Does it really matter that locality says "the probability of getting more than 2 + \varepsilon is less than a billion for some very small \varepsilon", instead of just "you should get less than 2"? That is the question you should be asking yourself.

 

[BenjaminTR]

[...]If Rp=Rs(-30,30)=0.75, Rq=Rs(-30,0)=0.5, and Rr=Rs(0, -30)=0.5 ... (*)then Rs((-30,0)&(0, 30)) = 0.5 * [0.5 + 0.5 - 0.75] = -0.125 < 0You have two contradictory assumptions (*).

[...]

This is the point of the proof. If you assume LHV and the QM correlations, you get absurd results like negative probabilities. They cannot both be right, as you have just shown.

 

[billschnieder]

This is the point of the proof. If you assume LHV and the QM correlations, you get absurd results like negative probabilities. They cannot both be right, as you have just shown.

In other words, you have assumed what you claim to be proving, that's the whole point.

 

[billschnieder]

If you think I am making such assumptions, then instead of making vague accusations I invite you to point out where, exactly, I use such an assumption in for instance my [POST=4432843]post #393[/POST], and how the actual argument I actually made is affected by it.

Ok then:

Then it is entirely possible that the inequality S_{123} \leq 1 is violated, but not for instance all three of S_{123} \leq 1, S_{231} \leq 1, and S_{312} \leq 1 simply because the condition (*) above implies

S_{123} + S_{231} + S_{312} \leq 3 \,.
The average over the six possible tests, which is also the expectation value of the full Bell correlator, satisfies the inequality: \langle S_{ijk} \rangle \leq 1.

You admit that I'm right that Bell's inequality can be violated for a 3 separate pairs even for a locally causal theory. But then you argue that when you combine many such triples, the violations will be canceled out and you will not have a violation.

In other words you are saying everytime S_{123} violates the inequality, somehow, Alice and Bob must also measure the corresponding S_{123}, S_{213} AND S_{312} so that the averages \langle S_{ijk} \rangle \leq 1 still obey the inequality. How is that not tuning? You brought up this argument yourself, how is it a strawman?

 

[billschnieder]

You are demanding something completely unnecessary. It is not necessary for the inequality to hold for all 3 sets.

It is necessary because you claim that the data from three sets should obey the inequality and then when it doesn't you make spooky conclusions. But if the data from three sets should not be required to obey the inequality, your spooky conclusions are unfounded. So if you insist that the data from three sets must not violate the inequality, you must also be able to derive the inequality from three sets. That you do not see this obvious logic is surprising indeed.

Incidentally, you still haven't given a direct response to a question I've posed you at least three times now:

Suppose it can be shown that the chance of a significant Bell violation [using correlations obtained from three disjoint sets of photons]* becomes vanishingly unlikely in a Bell test involving measurements on thousands or millions of entangled pairs.

*[emphasis added]

I've already answered your question, perhaps not the way you liked. I've asked you to show it without assuming tuning, rather than just supposing that it can be shown. Yet you say it is unnecessary. Derive the inequality from 3 disjoint sets and show that the chance of it being violated becomes vanishingly unlikely when millions/billions of entangled pairs are measured. You can't do it, unless you make one of the two assumptions I mentioned. It is clear what you have to do to clear this up.

 

[billschnieder]

Is anyone going to answer this question:

Why oh why do you always use the QM prediction for three disjoint sets when comparing with the inequality derived for a single set. Why don't you use the QM prediction for a single set while comparing with an inequality from a single set, or why don't you compare the QM prediction for 3 disjoint sets with the equivalent inequality derived for 3 disjoint sets? Why? Why? Why?

 

[DrChinese]

Is anyone going to answer this question:

Why oh why do you always use the QM prediction for three disjoint sets when comparing with the inequality derived for a single set. Why don't you use the QM prediction for a single set while comparing with an inequality from a single set, or why don't you compare the QM prediction for 3 disjoint sets with the equivalent inequality derived for 3 disjoint sets? Why? Why? Why?

Everybody can answer why. You just dismiss the answer.

Local realism means that when examining any single stream of entangled particle pairs, you can pick any pair of angles (of 3 angles identified by Bell's Inequality) to determine attributes of the stream. According to EPR: it would be unreasonable to require that all 3 were simultaneously measurable, as you seem to want to.

I really believe this thread is going around in circles. I fail to see where you have added anything other than non-standard viewpoints anytime recently.

 

[BenjaminTR]

[...]

In other words you are saying everytime S_{123} violates the inequality, somehow, Alice and Bob must also measure the corresponding S_{123}, S_{213} AND S_{312} so that the averages \langle S_{ijk} \rangle \leq 1 still obey the inequality. [...]

They are appealing to the law of large numbers, which is a valid form of probabilistic reasoning, not to the gambler's fallacy, which is not valid. In the quoted passage, you describe the gambler's fallacy. No one is advocating that.

 

[wle]

In other words you are saying everytime S_{123} violates the inequality, somehow, Alice and Bob must also measure the corresponding S_{123}, S_{213} AND S_{312} so that the averages \langle S_{ijk} \rangle \leq 1 still obey the inequality.

I never said any such thing. I don't even know what you're talking about because what you're describing is completely meaningless.

The point is just this: you don't know in advance what measurements Alice and Bob are going to perform. he measurements are supposed to be chosen at random at each round (on each "pair", if you prefer).

For a Bell experiment involving three pairs, you can easily construct a local model in which Alice and Bob measure a violation if they happen to measure (say) the "ab" term on the first pair, the "ac" term on the second, and the "bc" term on the third. But the same model that predicts a violation for that particular sequence of measurements is also going to predict a non violation for different sequences of measurements. The reason for this is that the local bound holds on the Bell correlator if you average over all the possible measurements that Alice and Bob could perform.

So you cannot construct a local model in which it is guaranteed that Alice and Bob will see a Bell violation regardless of the measurements they will eventually end up carrying out. This already means you are going to have a hard time explaining why we consistently observe violations in actual Bell experiments.

Furthermore, it is certainly possible to show that the probability of a significant Bell inequality violation becomes vanishingly small according to local causality if the test is carried out on a large number of entangled pairs. This can be done even under very paranoid assumptions, i.e. if you allow Alice's and Bob's measurement choices and outcomes on each pair to influence future pairs. I don't know if this has been done explicitly in a paper dedicated solely to Bell tests, but the statistical machinery is in place and well known to researchers in the nonlocality community. You can find an example of the sort of analysis I am alluding to in section A.2 of this paper's appendix.

 

[billschnieder]

The funny thing is, the argument being made that QM violates the inequalities and therefore hidden variables are impossible, is essentially the same argument von Neumann made, which Bell himself demolished using the same argument I'm making here. For some reason Bell did not realize his new no-hidden variables argument was essentially the same as von Neumann's.

See section III of Bell's paper on the subject:
http://www.mugur-schachter.net/docsupload/autresPublications/autresPublications_doc2.pdf

Consider now the proof of von Neumann that dispersion free states, and so hidden variables, are impossible. His essential assumption is: Any real linear combination of any two Hermitian operators represents an observable, and the same linear combination of expectation values is the expectation value of the combination. This is true for quantum mechanical states; it is required by von Neumann of the hypothetical dispersion free states also. In the two-dimensional example of Sec. 11, the expectation value must then be a linear function of \alpha and \beta. But for a dispersion free state (which has no statistical character) the expectation value of an observable must equal one of its eigenvalues. The eigenvalues (2) are certainly not linear in \beta. Therefore, dispersion free states are impossible. If the state space has more dimensions, we can always consider a two-dimensional subspace; therefore, the demonstration is quite general.

The essential assumption can be criticized as follows. At first sight the required additivity of expectation values seems very reasonable, and it is rather the nonadditivity of allowed values (eigenvalues) which requires explanation. Of course the explanation is well known: A measurement of a sum of noncommuting observables cannot be made by combining trivially the results of separate observations on the two terms -- it requires a quite distinct experiment. For example the measurement of \sigma_x, for a magnetic particle might be made with a suitably oriented Stern Gerlach magnet. The measurement of \sigma_y, would require a different orientation, and of (\sigma_x + \sigma_y) a third and different orientation. But this esplanation of the nonadditivity of allowed values also establishes the nontriviality of the additivity of espectation values. The latter is a quite peculiar property of quantum mechanical states, not to be expected a priori. There is no reason to demand it individually of the hypothetical dispersion free states, whose function it is to reproduce the measurable peculiarities of quantum mechanics when averaged over.

Thus the formal proof of von Neumann does not justify his informal conclusion: "...". It was not the objective measurable predictions of quantum mechanics which ruled out hidden variables. It was the arbitrary assumption of a particular (and impossible) relation between the results of incompatible measurements either of which might be made on a given occasion but only one of which can in fact be made.

...

The danger in fact was not in the explicit but in the implicit assumptions. It was tacitly assumed that measurement of an observable must yield the same value independently of what other measurements may be made simultaneously. Thus as well as P(\phi_3) say, one might measure either P(\phi_2) or P(\psi_2), where and \phi_2 and \psi_2 are orthogonal to \phi_3 but not to one another. These different possibilities require different experimental arrangements; there is no a priori reason to believe that the results for P(\phi_3) should be the same. The result of an observation may reasonably depend not only on the state of the system (including hidden variables) but also on the complete disposition of the apparatus;
​

 

[billschnieder]

I never said any such thing.

You sure did, in the quote I quoted, and you are saying the same thing below.

For a Bell experiment involving three pairs, you can easily construct a local model in which Alice and Bob measure a violation if they happen to measure (say) the "ab" term on the first pair, the "ac" term on the second, and the "bc" term on the third. But the same model that predicts a violation for that particular sequence of measurements is also going to predict a non violation for different sequences of measurements. The reason for this is that the local bound holds on the Bell correlator if you average over all the possible measurements that Alice and Bob could perform.

Translation: even though you you may violate for a single triple, when you average over a large number, the violation will disappear.

Translation: the inequality derived for a single set must apply to three different sets because when violations occur in individual cases, there must be corresponding non-violations which "cancel-out" those violations so that the averages obey the inequality.

Translation: the inequality derived for a single set must apply to data collected in three different measurements because if Alice and Bob collect data for three pairs which violate the inequality, they must also collect exactly the equivalent combinations of the other correlators which do not violate the inequality so that the averages from the three datasets must not violate the inequality.

 

[billschnieder]

Everybody can answer why. You just dismiss the answer.

Local realism means that when examining any single stream of entangled particle pairs, you can pick any pair of angles (of 3 angles identified by Bell's Inequality) to determine attributes of the stream. According to EPR: it would be unreasonable to require that all 3 were simultaneously measurable, as you seem to want to.

You still haven't answered why you always compare correlations from three distinct sets with an inequality from a single set. There surely is an inequality for three distinct sets, but you never use that. Why? Because the RHS is 3 and QM will not violate it. It is possible to calculate the correct QM correlations for a single set using QM, as is done in the article I cited:

Foundations of Physics Letters, Vol 15, No 5 (2002)
http://arxiv.org/pdf/quant-ph/0211031

And use those to compare with the inequality from a single set, and you never do that. Why? Because the correct QM correlations from a single set do not violate the single set inequality.

 

[wle]

Translation: even though you you may violate for a single triple, when you average over a large number, the violation will disappear.

If you mean that the violation will disappear when averaged over all the possible triples, then yes.

Translation: the inequality derived for a single set must apply to three different sets [...]

Absolutely not what I said. If this is what you're getting from what I've posted here, then you haven't understood the point I am making. At all.

Translation: the inequality derived for a single set must apply to data collected in three different measurements because if Alice and Bob collect data for three pairs which violate the inequality, they must also collect exactly the equivalent combinations of the other correlators which do not violate the inequality so that the averages from the three datasets must not violate the inequality.

I don't understand what this is even supposed to mean. If Alice and Bob measure the (ab, ac, bc) terms in that order, then they cannot also measure the (ac, bc, ab) terms in that order on the same three photon pairs. They've already done the experiment and they can't change history. As far as I can tell you are describing something completely meaningless and I don't recognise it as having anything to do with anything I said.

 

[billschnieder]

Translation: the inequality derived for a single set must apply to three different sets because when violations occur in individual cases, there must be corresponding non-violations which "cancel-out" those violations so that the averages obey the inequality.

Absolutely not what I said. If this is what you're getting from what I've posted here, then you haven't understood the point I am making. At all.

If it mustn't then it mustn't and violation of the inequality is meaningless. But if as you argue violation of the inequality is meaningful, then the inequality must apply to whatever system you are getting the correlations from to show violation.

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

If Alice and Bob measure the (ab, ac, bc) terms in that order, then they cannot also measure the (ac, bc, ab) terms in that order on the same three photon pairs. They've already done the experiment and they can't change history.

That's the whole point! You are arguing that the inequality applies to the averages from their measurements because they MUST have measured all three for the same three photon pairs, or other photon pairs so identical that all the correlators compensate each other, so that the averages obey the inequality.

Didn't you start this line of argument by saying even though individual correlators may violate the inequalities, the averages will obey it? Now you are claiming not to understand why you argued that the averages will obey the inequality.

 

[lugita15]

Rs(-30,30) = Rs(-30,0) + Rs(0,30) - 2* Rs((-30,0)&(0, 30)), the equality
Rs((-30,0)&(0, 30)) >= 0 ... (*)
therefore Rs(-30,30) <= Rs(-30,0) + Rs(0,30), your inequality

I agree with that, although I could have derived the inequality without using that equation.

therefore Rs((-30,0)&(0, 30)) = 0.5 * [Rs(-30,0) + Rs(0,30) - Rs(-30,30)] >= 0

Yes, I agree with that.

If Rp=Rs(-30,30)=0.75, Rq=Rs(-30,0)=0.5, and Rr=Rs(0, -30)=0.5 ... (*)
then Rs((-30,0)&(0, 30)) = 0.5 * [0.5 + 0.5 - 0.75] = -0.125 < 0

I think you mean that Rs(-30,0) and Rs(0,30) equal .25, not .5. So it's Rs((-30,0)&(0, 30)) = .5 (.25 + .25 -.75) = -.125

You have two contradictory assumptions (*). If Rs((-30,0)&(0, 30)) >= 0 as you assumed when you derived the inequality, then it must be the case that the three correlations Rp(-30,30), Rq(-30,0), and Rr(0,30) CAN NOT ALL BE EQUAL to the three correlations Rs(-30,30), Rs(-30,0) and Rs(0,30).

Yes, we have reached a contradiction, so at least one of the assumptions used in deriving the contradiction must be wrong. But let me give you a proof that three relative frequencies Rs(-30,0), Rs(0,30), and Rs(-30,30) must have the same value that they have for p, q, and r.

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

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

In other words, you assumed that p, q, r were not disjoint (presence of Rs((-30,0)&(0, 30)) in the derivation), and then later assumed that they were disjoint --> violation.

How does the presence of Rs((-30,0) & (0,30)) indicate that I'm assuming that p, q, and r are not disjoint? I am definitely assuming that p, q, and r are disjoint.

 

[Dale]

Closed pending moderation