A new simulation of the EPR-Bohm correlations

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

Re: A new simulation of the EPR-Bohm correlations

Postby FrediFizzx » Thu Jul 02, 2015 5:22 pm

helgus wrote:Dear Jochen and other, see Richard Gill new paper http://arxiv.org/abs/1507.00106
He wants to contribute to the ongoing discussion.

Welcome helgus but sorry; Gill was banned from this site for spamming it with too much nonsense. Any further posts about his banning will be deleted.
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Re: A new simulation of the EPR-Bohm correlations

Postby minkwe » Thu Jul 02, 2015 5:39 pm

Ben,
You have to be careful when you read the literature on weak measurements. They rarely say what they mean or mean what they say.

See for example this experiment which we discussed a while back viewtopic.php?f=6&t=80&start=10#p3475

http://www.nature.com/ncomms/2014/14072 ... s5492.html
http://arxiv.org/abs/1312.3775


Pay particular attention to "pre-processing" and "post-processing" when you read it.

See also:
http://arxiv.org/abs/1408.0363
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Re: A new simulation of the EPR-Bohm correlations

Postby Heinera » Fri Jul 03, 2015 12:37 am

minkwe wrote:
minkwe wrote:How can inequalities which apply to ALL local HV theories have any exceptions which are local HV theories ???

You can't answer this question can you?

I can. The CHSH inequality applies to all LHV theories where all particles can be detected and correctly paired. Your models do not belong to this class. You should compute the CH-inequality instead.

I claim that Bell's inequalities do not apply to my simulation, nor to the predictions of QM, nor to the results of EPRB experiments. What is your response to that exactly?

See above.
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Re: A new simulation of the EPR-Bohm correlations

Postby Schmelzer » Fri Jul 03, 2015 2:43 am

Joy Christian wrote:
Schmelzer wrote:I see only a laughable collection of meaningless formulas of type which, it seems, even Joy does not try to defend anymore. What I don't understand are, for example, formulas containing terms like .

I am not your teacher. I am under no obligation of teaching you elementary mathematics.

LOL, you are not in the role of the teacher here, but in the role of the pupil who writes down completely meaningless nonsense. And, much worse, refuses to accept this and correct even his trivial errors.

is meaningless, because one could interpret it as . For every reasonable person it would be a simple thing to say, oh, ok, I was too lazy writing this formula, this should have been .

But you know why you don't use this simple way out of it. You know that there will be some followup questions about what limit is used in the formulas (A.9.18), (A.9.19).
Last edited by Schmelzer on Fri Jul 03, 2015 3:06 am, edited 1 time in total.
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Re: A new simulation of the EPR-Bohm correlations

Postby Joy Christian » Fri Jul 03, 2015 2:57 am

Joy Christian wrote:I am not your teacher. I am under no obligation of teaching you elementary mathematics. In any case, I have already answered your question once. Take it or leave it:

Equation (A.9.15) describes a limit of a geometric product of two different quaternions, which belong to S^3. And since S^3 remains closed under multiplication, the product is also a unit quaternion, as shown in equations (A.9.16) and (A.9.17). Also worth noting is that quaternionic S^3 is necessarily "flat", with constant but non-vanishing torsion. The limits in these equations simply describe the limits in which the scalar part of the product quaternion reduces to +/-1 while the bivector part reduces to zero, as shown in equations (A.9.17) to (A.9.19). The purpose of this demonstration is to show that the geometry and topology of the 3-sphere is highly non-trivial.

I will not be responding to any further nonsensical questions from Schmelzer until he learns some elementary mathematics from his local school teacher.
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Re: A new simulation of the EPR-Bohm correlations

Postby Jochen » Fri Jul 03, 2015 5:54 am

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.
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Re: A new simulation of the EPR-Bohm correlations

Postby Jochen » Fri Jul 03, 2015 6:37 am

Joy Christian wrote:Again, you are taking the toy model far too seriously. I brought it up to counter a specific claim, namely that AB is always equal to -1 in my one-page paper. The toy model explains why that claim is wrong and naïve within the context of my original model. Beyond that the toy model should be taken with a large pinch of salt. In the 3-sphere model it is quite evident that experimenters are completely free to choose their measurement directions at will. Their "free will" is not compromised.

OK, so how would the result that Alice had obtained if she had, in some given instance, chosen to measure a' instead of a, differ from the one she got when in fact measuring a? That is, what is the counterfactual prediction for different measurement directions?

How does one compute p(A=+1|a), the probability that Alice receives the outcome +1 if measuring in the direction a? How about p(AB=+1|a,b)?

Joy Christian wrote:Thus it is simply wrong to say that there is any restrictions on what a and b the experimenters can choose.

I didn't say that; my question is merely about how the measurement outcomes depend on the local directions they choose. In the toy model, the two components within the Möbius band don't play any role at all, right? So I'm just wondering about the situation in your full-fledged model.

Joy Christian wrote:In the Clifford-algebraic representation of S^3 the probability distribution of A and B factorizes but the correlator does not, because of the geometric product of A and B. And this is the stumbling block for many people who are not familiar with geometric algebra, and have not studied how I manage to have my cake and eat it too.

It certainly is a stumbling block for me. If I have the probability distribution, I can directly compute the correlation from there, no? In other words, what goes wrong in the elementary calculation if ?

Jochen wrote:And another question: what must be the case in order for Bell inequalities to not be violated in your setup? The correct correlation for the singlet state is -a.b, but for a separable state, you just get , which doesn't violate any inequalities. How is that represented within your scheme?

Good question. A separable state in my model is represented by spin bivectors and that are always orthogonal to each other, so that .

Joy Christian wrote:There are lots of inequalities floating around, and I am a lone researcher with rather limited resources. So, no, I haven't looked at many of these inequalities.

I'd be especially interested in their probability-based form. Perhaps if I understood how probabilities of measurement outcomes are calculated in your model, I could have a look myself, hence my asking.
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Re: A new simulation of the EPR-Bohm correlations

Postby Jochen » Fri Jul 03, 2015 6:46 am

Ben6993 wrote:But at what stage in the sequence of weak measurements on the same electron does the spooky chance decision arise that the particle is (say) left-handed? {I don't believe in such spookiness myself.} Presumably on the first weak measurement.

The idea of weak measurements is that they don't collapse the state to an eigenstate of any particular spin value; they merely slightly bias the distribution. So if at first you're in an equal superposition over all possible spin states, e.g. , after a weak measurement, you'll end up in something like , where but generally, . That is, on a following strong measurement, you'll have different chances of finding spin up or spin down, but the electron hasn't 'collapsed' to a state in which it has a definite spin value. Thus, there also is no first weak measurement that collapses the state; rather, if you increase the coupling strength, the distribution afterwards will get more and more sharply peaked on a spin eigenstate, up to some point where you're effectively doing a strong measurement.
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Re: A new simulation of the EPR-Bohm correlations

Postby Joy Christian » Fri Jul 03, 2015 8:41 am

Jochen wrote:OK, so how would the result that Alice had obtained if she had, in some given instance, chosen to measure a' instead of a, differ from the one she got when in fact measuring a? That is, what is the counterfactual prediction for different measurement directions?

I have given an explicit formula analogous to Bell's formula for Alice's results for all directions etc., both in the simulation and in the theoretical paper on which the simulation is based. Please see the formulae (24) and (25) in the paper, where of Bell's formulae (1) of 1964 corresponds to the pair in my paper [in the simulation instead]. I am actually surprised that you are asking this question after all the evidence I have presented already. If you are asking about how Alice's results are obtained in the Clifford-algebraic representation of S^3 (as in my one-page paper, for example), then I have already told you that individual measurement results are not directly computed (nor computable) in that representation. What can be proven has been proven in Eq. (B10) of this paper.

Jochen wrote:How does one compute p(A=+1|a), the probability that Alice receives the outcome +1 if measuring in the direction a? How about p(AB=+1|a,b)?

All 13 possible probabilities for various measurement outcomes for the EPR-B scenario have been calculated in this paper. In the Clifford-algebraic representation the probabilities are not convenient to calculate. There the main objective has been to calculate the strong correlations and individual averages such as and .

Jochen wrote:
Joy Christian wrote:In the Clifford-algebraic representation of S^3 the probability distribution of A and B factorizes but the correlator does not, because of the geometric product of A and B. And this is the stumbling block for many people who are not familiar with geometric algebra, and have not studied how I manage to have my cake and eat it too.

It certainly is a stumbling block for me. If I have the probability distribution, I can directly compute the correlation from there, no? In other words, what goes wrong in the elementary calculation if ?

As you can see from the calculations of the 13 probabilities and the correlation (42) in the above paper, in your equation you have tacitly assumed something and that's what goes wrong. Gill makes a similar mistake in his often repeated claims, which I have brought out in my "Reply to Gill." See eqs. (12) to (14) of my reply.
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Re: A new simulation of the EPR-Bohm correlations

Postby minkwe » Fri Jul 03, 2015 9:22 am

Jochen wrote:I agree that it does; however, it won't be the actual probability distribution of the values of A, B, C, and D.

Good that you agree. Who cares about unmeasured distributions? The only outcomes you can calculate with are the ones that are measured. So your so-called "actual probability distribution" is a fantasy. The point is that there is a joint PD P(A,B,C,D) of outcomes based on how the experiment was performed, irrespective of whatever mechanism is producing the outcomes, and the corollary is that even if the process producing the outcomes may have a fantastic unmeasurable "actual distribution", you could still obtain measurement outcomes which do not have a joint PD P(A,B,C,D) simply by choosing to measure the outcomes in a certain way, such as is done in EPRB experiments.

Jochen wrote:
minkwe wrote: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.

I take it then you believe that a system producing 3 correlated spin-half particles heading towards Alice, Bob and Cindy, with Cindy simply ignoring her particles, and Alice and Bob measuring theirs like a typical EPRB cannot violate a Bell inequality no matter the setting combination they use, and no matter the nature of the correlation between them? Yes or no?

Jochen wrote: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).

I'm not asking you if the system will generate the full correlation. I'm asking you if you agree that an entangled 3-particle system with each particle measured at angles ("a", "b", "c") can ever violate Bell's inequality or not. And if you believe the fact that it doesn't violate Bell's inequality can be used to conclude that there is no-disturbance present, and no non-locality present. If you believe that then, I'm asking you if you believe the claimed non-locality or disturbance only shows up for specific states, but decides to hide if you have more particles that can actually be measured.

Take for example the CHSH in which 2 spin-half particles in the singlet state are measured at 4 settings , in the pairs . If instead of producing just the 2 particles each time, our source produces 4 entangled particles heading towards 4 stations, Alice, Bob, Cindy, and Dave, such that the particles pair going to Alice and Bob are always in a singlet state, those going to Cindy and Dave are also always in a singlet state, and the particles going to Alice and Cindy are identical to each other, just like the particles going to Bob and Dave. Thus all the paired measurements are by themselves measurements on a pair of spin-half particles in a singlet state.

Now Alice, Bob, Cindy and Dave do their measurements each time, and write down the results (A,B,C,D) on our 4xN spreadsheet. Do you agree that they will never violate the CHSH inequality? If you do, then please tell me what expectation values they would observe according to QM for <AB>, <CB>, <AD>, <CD>. Does the CHSH inequality apply to this experiment? Does QM predict a violation? Would those values be different if Cindy and Dave simply ignored their particle pairs, so that Alice and Bob did all the measurements at different times and different settings on different ensembles?

Jochen wrote:You say that there is always a joint PD.

Nope, you did not understand. I say the presence or absence of a joint PD is based on how you do the measurement, not due to disturbance or non-locality, I've given you examples (tablets) in which even with locality and no-disturbance, you did not not have a joint PD of outcomes based on how you did the measurement. I have given examples (non-local state machine pressed in sequence ) in which a joint PD of outcomes was present even with disturbance or non-locality to prove this point. I have also given another example (see above) using 4 spin-1/2 particle particles in which a joint PD is present and the inequalities must be satisfied. I hope you will address that example because I think it shows clearly the problem with your argument.
Jochen wrote:
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.


So let me ask you again the questions I asked you are the beginning. A simple yes/no answer will resolve all the word games
1) Is locality required or necessary in order to have a joint PD? Yes or no.
2) Is pre-determination required or necessary in order to have a joint PD? Yes or no.
3) Is no-disturbance required or necessary in order to have a joint PD? Yes or no.

If you want to change your answers to no, no, no. Then we can put this issue to bed and proceed to other things.

Jochen wrote:
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?

I hope you will eventually show me what you claim is "at variance" in my use of "degrees of freedom"? If I toss a die such that it settles with the 5 exposed sides facing up(U), north(N), south(S), east(E) and west(W). The experimenter is free to pick any direction to read the outcome. Once a direction is picked, say N, the remaining 4 directions are counterfactual. A counterfactual experimenter is no longer free to pick any direction. They can only pick from the remaining U, S, E, W. But if we toss two separate dies, and ask two actual experimenters to pick a direction to read from, they are both free to pick any of the 5 possibilities. The fact that a throw of two separate dies has more degrees of freedom, allows the actual experimenters more freedom to pick directions than the combination of actual and counterfactual experimenters from a single die. Thus, you can not claim that the counterfactual results obtained will be the same as the actual ones simply because the actual ones are picked with more freedom from the experimenters than the counterfactual ones, which are severely restricted once the actual measurements have been made. Bell's inequalities is a relationship between 3 expectation values, 2 of which are restricted in this way. The expectations from experiments and QM are not restricted in this way. It is a subtle point and easily missed if not considered carefully. I think the example I gave above about 4 pairs of particles and the CHSH should make this crystal clear.

Jochen wrote:Because you violate an inequality that is known not to hold in the scenario you analyze. What's the problem?

The problem is that you continue to argue that the inequality should hold even in the scenario I have shown it does not apply to. You continue to argue that a joint PD must be possible if the model is a local HV one. There is no disturbance in my simulations, and there is no non-locality. The simulations are completely local realistic, and yet they reproduce the QM predictions very well. So it must be false that a joint PD can always be constructed if the model is local realistic and does not have any disturbance. If you disagree, show me the disturbance or non-locality that my simulations are using to produce their outcomes. They are designed to be run on separate unconnected computers.

All you have to do is admit that based on how the measurements are done, it is not always possible to reconstruct a joint PD, even if the model is local and realistic, just like is the case in the EPRB experiment. If you admit that, then we can put this issue to bed.
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Re: A new simulation of the EPR-Bohm correlations

Postby FrediFizzx » Fri Jul 03, 2015 10:03 am

Heinera wrote:...The CHSH inequality applies to all LHV theories where all particles can be detected and correctly paired. Your models do not belong to this class. You should compute the CH-inequality instead.

That is rather pointless since nothing can violate Bell-CHSH nor Bell-CH. Not even QM.
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Re: A new simulation of the EPR-Bohm correlations

Postby minkwe » Fri Jul 03, 2015 10:23 am

FrediFizzx wrote:
Heinera wrote:...The CHSH inequality applies to all LHV theories where all particles can be detected and correctly paired. Your models do not belong to this class. You should compute the CH-inequality instead.

That is rather pointless since nothing can violate Bell-CHSH nor Bell-CH. Not even QM.


Fred,
I'm looking forward to an answer from every Bell believer about this question. I hope Heine would also answer this question which I asked Jochen:

Consider the CHSH in which 2 spin-half particles in the singlet state are measured at 4 paired settings . If instead of producing just the 2 particles each time, our source produces 4 entangled particles to be measured at 4 stations A, B, C, D by 4 people Alice, Bob, Cindy, and Dave respectively. The particle pair going to Alice and Bob are always in a singlet state, and so is the pair to Cindy and Dave. The particles going to Alice and Cindy are identical to each other, and so are the particles going to Bob and Dave. Therefore all the paired measurements are by themselves measurements on a pair of spin-half particles in a singlet state.

Each iteration, Alice, Bob, Cindy and Dave do their measurements, and write down the results (A,B,C,D) on one row of a 4xN spreadsheet.

1) Does the CHSH inequality apply to this experiment?
2) Would their measured outcomes ever violate the CHSH inequality?
3) What expectation values does QM predict for their measurement outcomes for <AB>, <CB>, <AD>, <CD>?
4) How would these QM predictions be different if the Cindy and Dave stations were not present and Alice and Bob had to switch between settings (a, c) and (b, d) respectively to measure the separate terms on different ensembles?
5) If they are different, why are they different?
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Re: A new simulation of the EPR-Bohm correlations

Postby Guest » Fri Jul 03, 2015 11:01 am

Minkwe asked: "Consider the CHSH in which 2 spin-half particles in the singlet state are measured at 4 paired settings . If instead of producing just the 2 particles each time, our source produces 4 entangled particles to be measured at 4 stations A, B, C, D by 4 people Alice, Bob, Cindy, and Dave respectively. The particle pair going to Alice and Bob are always in a singlet state, and so is the pair to Cindy and Dave. The particles going to Alice and Cindy are identical to each other, and so are the particles going to Bob and Dave. Therefore all the paired measurements are by themselves measurements on a pair of spin-half particles in a singlet state."

This is a fantastic question. The singlet state is a maximally entangled state. If particles A and B are in the singlet state, and particles C and D are in the singlet state, then particles A and B are completely independent of particles C and D.
The statistics of measurements of all four particles will exhibit the singlet correlations between A and B, and the singlet correlations between C and D, and zero correlation between those two pairs.

The word "identical particles" doesn't exist in QM. One can talk about identical states. But not identical particles. There is a very famous "no-cloning" theorem.
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Re: A new simulation of the EPR-Bohm correlations

Postby Heinera » Fri Jul 03, 2015 1:28 pm

minkwe wrote: [...]The particles going to Alice and Cindy are identical to each other, and so are the particles going to Bob and Dave. [...]

What do you mean by the particles being "identical to each other"? You've heard about the no-cloning theorem in QM, or? (Ok, you've probably not)
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Re: A new simulation of the EPR-Bohm correlations

Postby minkwe » Fri Jul 03, 2015 5:19 pm

Guest wrote:This is a fantastic question. The singlet state is a maximally entangled state. If particles A and B are in the singlet state, and particles C and D are in the singlet state, then particles A and B are completely independent of particles C and D.



There is no copying or cloning involved. I'm asking what the answers to those questions would be if the source produced the particles just like described. If you think such a source is impossible, then QM cannot predict any correlations at all. But what would QM predict if such a source produced the particles just as described in the gedankenexperiment such that A and B can be described by a singlet state and so can C and D, and the outcomes of A and C are correlated in the sense that if a measurement is done on A at setting "a", a simultaneous measurement on C at the same setting will always produce exactly the same outcome as A. In other words for all values of a. Same between B and D. So it is not correct to say A and B are completely independent of C and D. I've specified exactly how they are correlated.

Guest wrote:The statistics of measurements of all four particles will exhibit the singlet correlations between A and B, and the singlet correlations between C and D, and zero correlation between those two pairs.

First of all, there is no "statistics" of measurements of "four particles". There is statistics of measurement of an ensemble of groups of four particles. Each quartet is measured just once. Secondly, if C has the exact same outcomes as A for all settings, then any theory must give the same values for the expectation value <AB> as for <CB>. The expectation value is not from a single pair. It is from a series of pairs. It is each specific pair that is in a singlet state, not the ensemble. So it is not true that the correlation <CB> and <AD> will be zero.
Last edited by minkwe on Fri Jul 03, 2015 6:05 pm, edited 3 times in total.
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Re: A new simulation of the EPR-Bohm correlations

Postby minkwe » Fri Jul 03, 2015 5:25 pm

Heinera wrote:What do you mean by the particles being "identical to each other"?

See above.

You've heard about the no-cloning theorem in QM, or? (Ok, you've probably not)

You've heard about quantum cloning, Gedankenexperiment, or? (Ok you've probably not). But cloning or no-cloning you should be able to give clear answers to all the questions. Or are you in the business of looking for peripheral rabbit trails to avoid the central issue?
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Re: A new simulation of the EPR-Bohm correlations

Postby Guest » Fri Jul 03, 2015 7:45 pm

minkwe wrote:
Guest wrote:This is a fantastic question. The singlet state is a maximally entangled state. If particles A and B are in the singlet state, and particles C and D are in the singlet state, then particles A and B are completely independent of particles C and D.



There is no copying or cloning involved. I'm asking what the answers to those questions would be if the source produced the particles just like described. If you think such a source is impossible, then QM cannot predict any correlations at all. But what would QM predict if such a source produced the particles just as described in the gedankenexperiment such that A and B can be described by a singlet state and so can C and D, and the outcomes of A and C are correlated in the sense that if a measurement is done on A at setting "a", a simultaneous measurement on C at the same setting will always produce exactly the same outcome as A. In other words for all values of a. Same between B and D. So it is not correct to say A and B are completely independent of C and D. I've specified exactly how they are correlated.

Guest wrote:The statistics of measurements of all four particles will exhibit the singlet correlations between A and B, and the singlet correlations between C and D, and zero correlation between those two pairs.

First of all, there is no "statistics" of measurements of "four particles". There is statistics of measurement of an ensemble of groups of four particles. Each quartet is measured just once. Secondly, if C has the exact same outcomes as A for all settings, then any theory must give the same values for the expectation value <AB> as for <CB>. The expectation value is not from a single pair. It is from a series of pairs. It is each specific pair that is in a singlet state, not the ensemble. So it is not true that the correlation <CB> and <AD> will be zero.

According to quantum mechanics there is no such source. This is due to the "monogamy of entanglement". If A and B are maximally entangled then they are jointly in a pure state. If C and D are maximally entangled then they are jointly in a pure state. Therefore, A, B, C and D together are in a pure state, and regarding (A, B) and (C, D), it must be the product state. Two independent singlet state pairs.

C will not have the same outcomes as A for all settings. D will not have the same outcomes as B for all settings. In fact, whatever settings you apply to measure particles A and C, the correlation between them will be zero. Similarly B and D.
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Re: A new simulation of the EPR-Bohm correlations

Postby Heinera » Sat Jul 04, 2015 12:22 am

minkwe wrote:You've heard about quantum cloning, Gedankenexperiment, or? (Ok you've probably not). But cloning or no-cloning you should be able to give clear answers to all the questions. Or are you in the business of looking for peripheral rabbit trails to avoid the central issue?

It is a Gedankenexperiment that relies on assumptions that in QM are not even in principle correct. So your five questions are pointless.
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Re: A new simulation of the EPR-Bohm correlations

Postby Ben6993 » Sat Jul 04, 2015 2:46 am

Hi Jochen, thanks for your reply of Fri Jul 03, 2015 5:46 am (here: http://www.sciphysicsforums.com/spfbb1/viewtopic.php?f=6&t=168&start=160#p4660)
[Sorry for not using quotes from your reply but my editor loses all the maths symbols during a copy and paste.]

Your reply does not seem to disagree with what I was trying to write as I did know that a weak measurement does not collapse the state. I will try again to make my point better. My aim was to try and locate at what point the spooky business of quantum random choice occurs {which I do not believe in}. Now I may be wrong, and my apologies to you if I am, but your reply seems to identify the spooky moment of random choice with the actual collapse of the state at a strong measurement.

If a sequence of many weak measurements is made, all measurements identical as far as possible, followed by a strong measurement in the same direction which gives an 'up' outcome [ abs(α) =1], then is the expectation that abs(α) would have tended to exceed abs(β) in the preceding weak measurements? In which case the spooky random choice could have occurred during the weak measurements?
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Re: A new simulation of the EPR-Bohm correlations

Postby Ben6993 » Sat Jul 04, 2015 3:00 am

Hi Michel,
Michel wrote on Fri Jul 03, 2015 12:39 am:
You have to be careful when you read the literature on weak measurements. They rarely say what they mean or mean what they say.

Thanks for the warnings about a sequence of weak measurements not really being made on the same particle. I am sceptical too, but am running with the idea to see where it would lead if it were true.
See for example this experiment which we discussed a while back viewtopic.php?f=6&t=80&start=10#p3475
http://www.nature.com/ncomms/2014/14072 ... s5492.html
http://arxiv.org/abs/1312.3775

I looked at those when you first posted the links and did not get much further than the drawings of cats. I seem to have developed an aversion to quantum cats.
Pay particular attention to "pre-processing" and "post-processing" when you read it.

I have been reading http://iopscience.iop.org/1367-2630/16/5/053015/article and that seems quite reasonable to my naive, but sceptical, eye. The post-selection routine, however,does raise doubts as to the chances of the sequence of measurements actually being on the same particle. Even if quantum computing turns out to be a red herring, research in it will no doubt have positive spin offs. I have no problem with an electron having a field-only effect which may possibly be measurable but does not collapse its field.
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