minkwe wrote:You did not understand what I said. Again, you have a 4xN spreadsheet with columns labelled A,B,C,D. You press A, write the outcome on row 1, col 1, then press button B, and write the outcome in col 2 row 1, and next C, and next D. When you reach the end, you jump to row 2 col 1 and repeat ad nauseum. Your 4xN spreadsheet is a joint PD of outcomes P(A,B,C,D).
I agree that it does; however, it won't be the actual probability distribution of the values of A, B, C, and D. It will be a product distribution of the single-observable marginals, which by definition fails to capture any correlations between the observables. So, in arguing in this way, you create a PD that doesn't have correlations, in order to argue that there are no correlations. It's like the example of the two correlated coins: you throw both, note the value of the first, throw them again, note the value of the second, and so on; the resulting distribution will perfectly reproduce the probabilities of a single coin coming up heads, but it will be completely oblivious to the fact that whenever the first one comes up heads, the second one does, too. But those correlations are exactly what we are interested in; thus artificially creating a distribution that lacks them is simply misguided.
minkwe wrote:It is in fact a shifting of goal posts. If I give you 3 spin-half particles and instruct that you only measure two of them each time, we are back to the Bell's inequalities.
No, we aren't. Bell derived his original inequality for the case of a pair of particles (or more generally, a bipartite system), and showed that an entangled quantum system violates it; nobody ever argued that it should apply to any other case (in particular, it won't be violated for a separable 2-qubit system either). Bell's theorem follows by observing that in this case, this particular inequality is violated, and hence, not all quantum mechanical observations can be reproduced by local hidden variables. Bell only needs to show this for a single case in order to invalidate the claim that all quantum predictions can be recovered in such a theory.
The experimental setting is instrumental in the derivation of each Bell inequality; thus, showing that there are settings in which that particular inequality is not violated is in no sense a counterargument to Bell's theorem, which merely asserts that there is at least one setting in which it is violated.
minkwe wrote:Please think about what you are saying here. If I give you 3particles each time but forbid you from measuring more than two of the 3, would you say Bell's inequalities apply or not. Don't you see the contradiction?
There is no contradiction. If each of the pairs I measure is in the singlet state, then I will observe a violation; if it isn't, then I won't. This is exactly what Bell argued. The strong quantum correlation only comes about for the right state, which is a maximally entangled two-qubit state (well, less entangled states work as well, but won't yield the maximum violation). In particular, due to the monogamy of entanglement, any pair of a three-qubit system can only be maximally entangled if the state is separable with respect to the third qubit.
minkwe wrote:Again, you are ascribing to me arguments I've never made. It is you who is arguing that pre-determination is necessary in order to have a joint PD.
The logic is as follows: whenever there is a joint PD, then there is a local realistic model that I can write down in order to produce the same predictions. You say that there is always a joint PD. Thus, I showed a model where that is not the case (the three coins). Violation of a Bell inequality indicates there is no joint PD, and that thus, there is no local realistic model, since all local realistic models have a joint PD. You can view a local realistic model as a big urn, in which we have value assignments for all observables; at each measurement instance, one of these value assignments is drawn. No disturbance assures that the assignment does not change based upon a measurement; realism means that the model determines a value for each observable such that knowing the hidden variable, I could say with certainty what outcome any measurement produces.
Yes, there are statistical processes for which I can also write down a joint PD---this means that for these processes, we have an equivalent local realistic model. But there are also statistical processes for which this fails, for which there is no such model. The coins are one, simple, example; quantum mechanics is another.
minkwe wrote:Again, please read my argument carefully. This is simple logic. It does not matter whether a random process can "always" produce a joint PD. If you argue that pre-determination is necessary to obtain a joint PD, then you mean that pre-determination must always be present for a joint PD to be possible.
The other way around: any predetermined process has a joint PD, and any process with a joint PD can be written as a predetermined one. Quantum mechanics, and things like the coin example (where the marginal problem is insoluble) don't have a joint PD, and hence, can't be written as a predetermined process.
minkwe wrote:You said my use of the term was "at variance" without saying how, so I'm entitled to understand what you mean in order to respond to it. I still do not see anything in your response that justifies your claim that anything I said was "at variance". I'm using degrees of freedom in the standard way it is used in statistics and mathematics:
Wikipedia wrote:In statistics, the number of degrees of freedom is the number of values in the final calculation of a statistic that are free to vary. The number of independent ways by which a dynamic system can move, without violating any constraint imposed on it, is called number of degrees of freedom.
OK, so what does this have to do with the freedom of the experimenter?
minkwe wrote:Jochen wrote:This is exactly what the assumption of hidden variables amounts to: if there are such hidden variables, then the statistical properties of the unmeasured ones are the same as the measured ones, since the particle or whatever system could not have known beforehand which ones would be measured, and hence, could not have arranged for the unmeasured ones to differ in any relevant way.
Not at all. Hidden variables do not imply that at all. That is what I've been explaining to you all along, the measured ones are not from the same particles as the unmeasured ones in the inequality.
But they are drawn from the same ensemble; hence, measured or not, their statistical properties are the same. Again, think of an urn from which value-assignments are drawn (immediately before a measurement is made): there is some definite value that any measurement, were you to perform it, would yield; this does not depend on which measurement you actually perfom. Hence, whether I actually measure A and B or C and D does not make any difference---it couldn't, otherwise we would violate the assumption of the existence of values revealed by the measurement that are independent of which measurements are made.
minkwe wrote:The relationship between the measured and unmeasured ones in the same set of particles is not the same relationship between measured ones in one set, and measured ones in another set. The symmetry inherent in the structure of a coin that links the H to the T such that when you measure one you definitely does not get the other, does not exist between the H of one coin and the T of another coin.
No, but if you toss another from the same ensemble (or re-toss the original coin), then H and T will have the same probability of appearing as in the first coin. And if you toss two correlated coins, and then toss a second pair of correlated coins, then HH and TT will have the same probability of appearing in both runs. That is, if you divide an ensemble of a large number of correlated coins in two, then the correlation you get from the first subensemble will, to witin statistical errors, be equal to that in the second. And if you had to different ways of tossing a coin, such that a coin tossed way one yields either H or T, and tossed the second way yields either X or Y, and you use the first half of the ensemble of coin pairs to find out the correlation between to coins when tossed the H-T way, then you know that the second half of the ensemble will show the same correlation (to within statistical discrepancies), so that if you use the second half of the ensemble to estimate the X-Y correlation, you'd know both the H-T and X-Y correlation across the whole ensemble. This is what happens in Bell tests.
minkwe wrote:If I toss a coin, I get H, the counter-factual outcome is constrained by symmetry of the coin to be T. It must be in this case. But If I toss one coin and get H, there is absolutely nothing which constrains a second toss even of the same coin to be T.
And I'm not saying that. I'm merely saying that the second coin toss has the same probability of coming up H or T as the first one did.
minkwe wrote:Jochen wrote:I think I understand quite well what's going on there, and hence you should be able to anticipate my response: get rid of the nondetections in the first, or of the failures to produce particle pairs in the second, and then we can talk. A local realistic strategy should work just as well in the case of ideal detectors and sources.
You think you know but I think you don't. Please explain and justify why you think I should change the models before we can talk.
Because you violate an inequality that is known not to hold in the scenario you analyze. What's the problem?
minkwe wrote:1) You continue to claim that local HV theories must always produce joint PD, in which case you show me the disturbance or non-locality in my simulations
2) You admit that I'm right, that local HV theories do not always have to produce a joint PD, in which case Bell's inequalities do not apply to cases in which they do not, such as epr-simple/epr-clocked.
3) I point out to you that the violation of a Bell inquality in a scenario it does not apply to does not show anything regarding the question of a LHV-completion of quantum mechanics.