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

[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;
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[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