A new simulation of the EPR-Bohm correlations

[minkwe]

You never obtain correlations from a single system; you get a single outcome from a single system. Correlations are statistical quantities, and hence, obtained by re-measuring identically prepared systems.

Huh? Just more double-talk

No. I've told you the QM predictions for a two-particle system in the state , and the QM predictions for a four-particle system in the state . These are different systems; they produce different predictions.

You need to eat your own dog food!

Don't you see that the 4 paired correlations in the CHSH expression are not independent but the QM predictions you now claim are the correct ones for the CHSH are completely independent! Don't you see that what you are now providing could not possibly be the correct correlations for the CHSH scenario!?

They're what QM predicts, and what is experimentally observed. I don't know what's difficult about that.

Nope. The QM predictions are for independent systems, the measurements are performed on 4 independent disjoint sets of particle pairs. The QM predictions are for
Have you read and understood Adenier's paper yet?

Please read carefully, this argument has been thoroughly debunked in this post viewtopic.php?f=6&t=168&start=200#p4709 and this one viewtopic.php?f=6&t=168&start=160#p4662

What, exactly, do you believe is in those posts that refutes the argument?

Please you will have to pay attention this time because I won't explain this one more time.

1. Do you agree that the reason the trivial mathematical inequality

for 4 measurements measurements on a single particle pair can be extended to averages, is precisely because the prescribed factorization is possible for every row in the series of 4xN outcomes from N particle pairs? In other words, factorization inside the integral, or inside the summation like Bell used in the derivation, means precisely that every occurrence of the set is favorable, which means the sets by themselves are factorable. That is:


Which is not surprising because we start out with 4 columns of data (the joint PD of outcomes ABCD), then combine them in Pairs CB, CD, AB, AD, then calculate the averages <CB>, <CD>, <AB>, <AD>.

2. Do you agree that the 4 terms in the CHSH,

are therefore not independent, since they they have been constructed from the same series of outcomes, simply recombined in pairs? In other words, since every two terms shares one column of outcomes there is a cyclic dependency between them. Even if you would randomize each paired sequence after recombining, it should still be possible in principle to rearrange them so that the similarly labelled columns match exactly. That is the A column of the AB pair should match the A column of the AD pair, etc in a cyclical manner, back to the B column of CB matching the B column of AB. Do you agree or disagree?

3. Do you agree that for 4 independent particle pairs , the expression

is false, and the correct upper bound should be , and by the same logic of the argument in point (1) above, the correct upper bound for averages is

4. You have claimed that the CHSH expression

,
where represents a single set of particle pairs, is statistically equivalent to

where represent disjoint independent sets of particle pairs. Because according to you,
and and

5. Do you agree that the expression

means that for every single individual pair of particles in the series which produced the outcome pair , there is an equivalent pair of particles in the series, such that as the number of particles approaches infinity, it should be possible to find a function , which rearranges the sequence of outcomes to match the sequence of , so that we have the same numbers of +1's and -1's and the same pattern of occurrences of those numbers? Yes or no?

6. Do you agree that from point (4) above (which you believe), it follows that there must also exist other functions

, , and . Such that after applying the functions, and , where the prime , represents the fact that the sequence of outcomes has been rearranged using the appropriate function? That is, the equivalent sets of outcomes are for all practical purposes "identical". Yes or no?

7. Do you agree that if points (5) and (6) are true, then you can apply the same argument from point (1),

to measurements performed on 4 disjoint sets of particle pairs , precisely because after rearranging, you will get


Which can be factorized just like in point (1) to

since and

8. Do you see now that in order for

to be true, it must be the case that and . But both and have already been rearranged independently of each other, and since any rearrangement will shuffle both outcomes in the set of pairs any new rearrangement to make agree with will undo the previous rearrangements; the same for and . Therefore for measurements on 4 independent disjoint sets to obey the the inequality, you need several different independent sorting functions to be dependent. In other words, the assumption that 4 independent sets of particle pairs should obey the inequality is a contradiction and your claim in point (4) above fails. The upper bound of 2, depends on the fact that the terms are dependent. It is a contradiction to claim that the same upper bound applies to independent sets

[Jochen]

Come on now. This is getting quite tedious. Either you actually read the simulation in full (there are only about 20 crucial lines with English commentary) or you are never going to get this. I am not repeating the answer to your question which I have given at least 10 times by now. Please read the answers I have already given you.

I'm sorry to press the point, but I'm just trying to square your insistence that there are no zero outcomes with the fact that the function you use in your simulation to compute the measurement outcomes of Alice is

.

Plotted for the case of and , with Alice's measurement direction a parametrized by an angle , this yields:


Now, this seems to mean that for , Alice's measurement outcome is +1, while for , she gets the -1 outcome. Is this correct?

My question is then what happens if she chooses a measurement setting . There, A(a,e,s) = 0. There are three possibilities:

1. She does not observe an outcome, i.e. her measurement produces no result. This you claim to not be the case, since you say the detectors are 100% effective.
2. She can't measure along those directions.
3. Those cases don't occur, i.e. if Alice chooses her measurement in one of these directions, then the HV can't be and .

Now, your answers so far seem to be in favour of the third option. However, this induces an explicit dependence of the HV on the measurement setting; i.e. what Alice chooses to measure determines the HV-value.

So, what's happening here?

[Joy Christian]

So, what's happening here?

What is happening is that you are refusing to leave your flatland. Do your analysis within S^3, defined by the metric {g, t}. Then we may have a basis to talk further.

[Jochen]

You never obtain correlations from a single system; you get a single outcome from a single system. Correlations are statistical quantities, and hence, obtained by re-measuring identically prepared systems.

Huh? Just more double-talk

No. I've told you the QM predictions for a two-particle system in the state , and the QM predictions for a four-particle system in the state . These are different systems; they produce different predictions.

You need to eat your own dog food!

I'm not sure what's your issue here if you insist on being cryptic. Are you confused that I speak about statistical predictions for a single system, while insisting that they can't be checked on a single system? If so, this is again just basic QM: since it's a statistical theory, it only makes statistical predictions, i.e. expectation values or probabilities. So, for a system in some state , the most we can say about an observable O is its expectation value , but not what any given measurement will produce (in the general case). We can also not use the same system to make a large enough number of measurements to estimate the statistical properties, since after a measurement of O, the system will be in an eigenstate of O (this is the so-called collapse postulate), and hence, every further measurement will just again yield the same observed value for O. Hence, we need a large number of copies of the system in the state in order to estimate . Does this help?

Nope. The QM predictions are for independent systems, the measurements are performed on 4 independent disjoint sets of particle pairs. The QM predictions are for

The predictions of QM are for a system in some given state . In order to check these predictions, since they are of a statistical sort, a large number of systems in that state are needed; the observations made on this ensemble will approximate the QM prediction in the limit of infinitely many measurements.

Have you read and understood Adenier's paper yet?

No matter where your error comes from, it remains an error all the same (and the fact that it's apparently been submitted to J. Math. Phys. but wasn't accepted isn't exactly encouraging either). Make your own argument, but don't expect me to review everything that you could dig up and throw in my way.

1. Do you agree that the reason the trivial mathematical inequality

for 4 measurements measurements on a single particle pair can be extended to averages, is precisely because the prescribed factorization is possible for every row in the series of 4xN outcomes from N particle pairs?

No, the inequality can be extended to averages because no average of a set of values can't be greater than the maximum value of that set. So you have a set of values, all below two, and immediately you know that the average value will also be below two.

3. Do you agree that for 4 independent particle pairs , the expression

is false, and the correct upper bound should be , and by the same logic of the argument in point (1) above, the correct upper bound for averages is

For independent particle pairs, it is indeed the case that they are not bound by the value 2, but from there, it does not follow that hence, . The observed measurement outcomes are drawn from the same probability distribution, and (as you know) from that fact alone the CHSH bound of two follows.

Consider two independent dice, together with the observables T=value on top of the die after a throw, and B=value on the bottom after a throw. For each throw of a single die, then B + T = 7. This also holds in the average: . But for a single throw of both, the value on top of one plus the value at the bottom of the other is not 7, but bounded above by 12. So if you keep throwing both die, and record the value on top of one, and the value at the bottom of the other, does it then follow that ? (Hint: the answer is no.)

Since your argumentation breaks down here, there's no point in adressing the rest. If you understand why , then you'll also understand why .

[Jochen]

So, what's happening here?

What is happening is that you are refusing to leave your flatland. Do your analysis within S^3, defined by the metric {g, t}. Then we may have a basis to talk further.

So then tell me how to do the analysis of the following case in S^3: First, on Alpha Centauri, four years ago, the source sent out a particle w with and . Today, Alice has two choices for her measurement setting: and . Since you said before that every particle is detected, both must yield some measurement outcome. I can see what happens if she chooses (in both R^3 and S^3, I guess): she will observe the value -1. But what happens if she chooses ?

[Jochen]

Have you read and understood Adenier's paper yet?

Even Adenier himself seems to have reconsidered his stance, cf. the very first sentence of the abstract of a later paper:

In spite of many attempts, no local realistic model seems to be able to reproduce EPR-Bell type correlations, unless non ideal detection is allowed.

[Joy Christian]

So, what's happening here?

What is happening is that you are refusing to leave your flatland. Do your analysis within S^3, defined by the metric {g, t}. Then we may have a basis to talk further.

So then tell me how to do the analysis of the following case in S^3: First, on Alpha Centauri, four years ago, the source sent out a particle w with and . Today, Alice has two choices for her measurement setting: and . Since you said before that every particle is detected, both must yield some measurement outcome. I can see what happens if she chooses (in both R^3 and S^3, I guess): she will observe the value -1. But what happens if she chooses ?

There is no such initial state w = (e,s) in S^3.

[FrediFizzx]

Have you read and understood Adenier's paper yet?

Even Adenier himself seems to have reconsidered his stance, cf. the very first sentence of the abstract of a later paper:

In spite of many attempts, no local realistic model seems to be able to reproduce EPR-Bell type correlations, unless non ideal detection is allowed.

Of course, since Nature is tricking them. Ya gotta think S^3.

[Jochen]

There is no such initial state w = (e,s) in S^3.

So then, and please just give me a straight answer on this: what initial states are there in S^3? Can you just pick out a state?

[minkwe]

Jochen,
I will wait for a direct response to every point in my previous post before I respond any further to you.

Re Adenier, it is interesting that your only defense is that "it seems" Adenier changed his mind. Lies, Adenier did not change his mind as you can see by reading an even later paper than the one you referred to:
http://arxiv.org/pdf/0705.1477.pdf

Then your next excuse was that his paper was not published. Lies, Adenier's paper has been published, you didn't look hard enough.

Now please address all the points in my argument and state precisely where you claim the argument fails, if you can. I don't have any more time to waste trying to teach you statistics so please if you think you have a claim against my argument, address every point of it, rather than ignore them with hand-waving.

[Joy Christian]

There is no such initial state w = (e,s) in S^3.

So then, and please just give me a straight answer on this: what initial states are there in S^3? Can you just pick out a state?

Straight answer: The initial states are given by the w vectors in this simulation. They represent those (e, s) pairs that belong to S^3, as I have explained already.

I can give you, not just one or two, but all of the w vectors. They can be listed, for example in Excel or Word, by adding the following few lines in my R code:

Code: Select all

w = cbind(o, p, q)

write.csv(w, file = "AliceSpins.csv")  ## Alice's N spin vectors

write.csv(-w, file = "BobSpins.csv")  ## Bob's N spin vectors

In the theoretical model these very same w vectors are given by the vectors s^k appearing in Eq. (B10) of this paper (sorry for different notations in different papers).

[Jochen]

Jochen,
I will wait for a direct response to every point in my previous post before I respond any further to you.

A tempting offer. Anyway, since I've already pointed out a major misconception that invalidates your argument, I suppose it can't hurt to look at the rest of it. First, however, your accusations towards me:

Re Adenier, it is interesting that your only defense is that "it seems" Adenier changed his mind. Lies, Adenier did not change his mind as you can see by reading an even later paper than the one you referred to:
http://arxiv.org/pdf/0705.1477.pdf

My assertion (by quote) was that he seems to now believe that only models that don't possess ideal detection are viable as a local realistic QM completion. The model in that paper does not possess ideal detection, so what I said was completely true. Also, his earlier attempt was not based on nonideal detection; hence, it's fair to infer that he has changed his mind on the issue.

Then your next excuse was that his paper was not published. Lies, Adenier's paper has been published, you didn't look hard enough.

I didn't say it wasn't published, I said it was submitted to J. Math. Phys. (as per the arXiv comment), but appears to have never been published there. The only place I could find it published is in a conference proceedings, which is not surprising since it was presented as a talk there. So again, your accusation of lying has no base in actual fact, and I'd like to see you retract it.

Now please address all the points in my argument and state precisely where you claim the argument fails, if you can.

I've shown where the argument fails: from the fact that detections on separate particles are not bound by the value 2, but rather by 4, it does not follow that ; I've even given you an easy to understand example of just why that reasoning fails. But fair enough, in order to further reduce your wiggle room, I'll also address your other points.

4. You have claimed that the CHSH expression

,
where represents a single set of particle pairs, is statistically equivalent to

where represent disjoint independent sets of particle pairs. Because according to you,
and and

Yes I've claimed that, but you can scratch the 'according to you', since it's just according to basic statistics: given the same probability distribution , then indeed the correlation between two different experimental runs will be the same, because it's just derived from that distribution.

5. Do you agree that the expression

means that for every single individual pair of particles in the series which produced the outcome pair , there is an equivalent pair of particles in the series, such that as the number of particles approaches infinity, it should be possible to find a function , which rearranges the sequence of outcomes to match the sequence of , so that we have the same numbers of +1's and -1's and the same pattern of occurrences of those numbers? Yes or no?

This is just what the expectation value means: for large number of measurements n, so in order to yield the same EV, there need to be approximately the same numbers of +1 outcomes and of -1 outcomes. However, equality of these numbers only holds for , and for any two finite experiment runs i and j, both and , as well as and , will generally differ. Hence, in (almost) all real world experiments, such a function won't exist.

6. Do you agree that from point (4) above (which you believe), it follows that there must also exist other functions

, , and . Such that after applying the functions, and , where the prime , represents the fact that the sequence of outcomes has been rearranged using the appropriate function? That is, the equivalent sets of outcomes are for all practical purposes "identical". Yes or no?

Again, only for the case, which is never attained in experiments.

7. Do you agree that if points (5) and (6) are true, then you can apply the same argument from point (1),

to measurements performed on 4 disjoint sets of particle pairs , precisely because after rearranging, you will get


Which can be factorized just like in point (1) to

since and

There's two problems here. The first one is the erroneous nature of the argument in point 1, which I've already noted. The second is that even if the reordering functions did exist (which they won't in general), then you can't perform such reorderings and expect everything to stay the same: the correlations are generally not preserved under such a reordering. Consider two fair coins: if they are uncorrelated, you will get two columns of (approximately) as many Hs and Ts in each case. Now, you can reorder the columns: putting all the Hs in the same row, as well as all the Ts, for instance: but then, it will look as if the coins are perfectly correlated. Alternatively, you could put all the Hs of one in the same row with all the Ts of the other (since they're fair coins, there will be equally as many of each), and vice versa, and hence, obtain perfect anticorrelation.

8. Do you see now that in order for

to be true, it must be the case that and . But both and have already been rearranged independently of each other, and since any rearrangement will shuffle both outcomes in the set of pairs any new rearrangement to make agree with will undo the previous rearrangements; the same for and . Therefore for measurements on 4 independent disjoint sets to obey the the inequality, you need several different independent sorting functions to be dependent. In other words, the assumption that 4 independent sets of particle pairs should obey the inequality is a contradiction and your claim in point (4) above fails. The upper bound of 2, depends on the fact that the terms are dependent. It is a contradiction to claim that the same upper bound applies to independent sets

The only rearrangements you can validly perform on the measurement tables are row permutations; all other rearrangements will alter the correlations. This corresponds to re-labelling the experiment runs, and will not change anything about the expectation values. A Bell test experiment yields a meassurement table like the following:

Code: Select all

 Alice |  Bob 
 A | C | B | D
 +1|   |   | -1
 -1|   | -1|   
   | -1|   | +1
.
.
.
   | -1| -1|   
 +1|   | +1|   
 +1|   |   | +1


Whenever you rearrange a single column, all other columns have to be rearranged in the same way; else, you change the correlations. So you can't, for example, just rearrange the rows for C and D, since D also has correlations with A, and C with B; thus, you can only permute the rows. But this doesn't change anything. Is this clearer now?

[Jochen]

There is no such initial state w = (e,s) in S^3.

So then, and please just give me a straight answer on this: what initial states are there in S^3? Can you just pick out a state?

Straight answer: The initial states are given by the w vectors in this simulation. They represent those (e, s) pairs that belong to S^3, as I have explained already.

I can give you, not just one or two, but all of the w vectors. They can be listed, for example in Excel or Word, by adding the following few lines in my R code:

Code: Select all

w = cbind(o, p, q)

write.csv(w, file = "AliceSpins.csv")  ## Alice's N spin vectors

write.csv(-w, file = "BobSpins.csv")  ## Bob's N spin vectors

In the theoretical model these very same w vectors are given by the vectors s^k appearing in Eq. (B10) of this paper (sorry for different notations in different papers).

OK, that's progress. These w vectors of course explicitly depend on Alice's and Bob's measurement directions:

Code: Select all

o = x[t(a,e,s) & t(b,e,s)]
p = y[t(a,e,s) & t(b,e,s)]
q = z[t(a,e,s) & t(b,e,s)]

The function t is given by:

Code: Select all

t = function(u,v,s){abs(sign(g(u,v,s))) > 0} # Ensures sign(g) = +1 or -1,

and selects only those pairs (e,s) for which there is a measurement outcome different from zero. If hence these w vectors are the initial ensemble from which the source selects particle pairs, then this ensemble depends explicitly on the choice of measurement of Alice and Bob, and thus, since the source could have sent out the particles an arbitrary amount of time ago, must function in a retrocausal way. This of course also introduces a dependence of Bob's outcomes on Alice's measurement choices, since Bob would not receive a particle characterized by some (e,s) if Alice's choice of measurement direction a would yield to a zero outcome.

In particular, these w vectors would also not be computable in my proposed two-boxes instantiation, since in order to do so, joint knowledge of both Alice's and Bob's measurement directions would be necessary.

Of course, this then means that the model is in no sense in conflict with Bell's theorem: it is explicitly retrocausal, and hence, implicitly nonlocal, just as required for observing a violation of a Bell inequality.

[Joy Christian]

OK, that's progress. These w vectors of course explicitly depend on Alice's and Bob's measurement directions.

No they do not. You should not keep falling back to your preconceived flatland-based ideas that have already been addressed and dispelled by me.

Where do you see "explicit dependence" on measurement directions in the s^k vectors in Eq. (B10) of this paper? Don't ignore this question as you did before.

Where do you see any of the nonsense you claim in this explicit calculation?

Code: Select all

        A = +sign(g(a,e,s))  # Alice's measurement results A(a, e, s) = +/-1
         
        B = -sign(g(b,e,s))  # Bob's measurement results B(b, e, s) = -/+1
       
        Cuu = length((A*B)[A > 0 & B > 0])   # Coincidence count of (+,+) events
       
        Cdd = length((A*B)[A < 0 & B < 0])   # Coincidence count of (-,-) events
       
        Cud = length((A*B)[A > 0 & B < 0])   # Coincidence count of (+,-) events
       
        Cdu = length((A*B)[A < 0 & B > 0])   # Coincidence count of (-,+) events
       
        corrs[i,j] = (Cuu + Cdd - Cud - Cdu) / (Cuu + Cdd + Cud + Cdu)


The w vectors must respect the metric {g(u,v,s), t(u,v,s)} if they are to be within S^3, where u and v are any arbitrary vectors within S^3. As I pointed out to you before, the appearance of the specific vectors a and b is incidental. It is merely an artefact of the embedding space R^4, inevitably used to build the simulation.

Moreover, the simulation is simply an implementation of the analytical model presented in the above paper, as explicitly shown in this theoretical paper.

Thus you are confusing the flight simulator with the Jumbo Jet itself.

You are not allowed to use any concept external to S^3 to deduce what you wish to deduce. You have to see the entire model from within S^3. There is nothing outside of S^3 as far as my model is concerned.

It should be clear by now that your remaining commentary has nothing whatsoever to do with my actual model, but only with your deep-seated prejudices.

Nevertheless, if you wish to convince me that what you are claiming has an element of truth in it, then you must at least answer the question I asked you before:

Where do you see "explicit dependence" on measurement directions in the s^k vectors in Eq. (B10) of this paper?

[Jochen]

Where do you see "explicit dependence" on measurement directions in the s^k vectors in Eq. (B10) of this paper? Don't ignore this question as you did before.

Where are these vectors calculated in that paper? I can't say what they depend on without knowing how they are formed. The way they are presented, the dependence is in restricting the outcomes to +/-1: only those vectors which yield such an outcome, given the measurement directions, are chosen.

In your simulation, you explicitly compute them using both the measurement directions of Alice and Bob; so if that's different from your theoretical model, then your simulation does not implement the model and should not be used to bolster claims made about it.

I was working under the assumption that the simulation faithfully replicates your model. Thus, the explicit way to calculate the vectors, as you do, is by using the measurement directions of Alice and Bob.

Where do you see any of the nonsense you claim in this explicit calculation?

Code: Select all

        A = +sign(g(a,e,s))  # Alice's measurement results A(a, e, s) = +/-1
         
        B = -sign(g(b,e,s))  # Bob's measurement results B(b, e, s) = -/+1
       
        Cuu = length((A*B)[A > 0 & B > 0])   # Coincidence count of (+,+) events
       
        Cdd = length((A*B)[A < 0 & B < 0])   # Coincidence count of (-,-) events
       
        Cud = length((A*B)[A > 0 & B < 0])   # Coincidence count of (+,-) events
       
        Cdu = length((A*B)[A < 0 & B > 0])   # Coincidence count of (-,+) events
       
        corrs[i,j] = (Cuu + Cdd - Cud - Cdu) / (Cuu + Cdd + Cud + Cdu)


Well, the first two lines are erroneous, as neither A nor B is a +/-1 valued function, but produces the value 0; moreover, for each pair of (e,s) there is a measurement direction such that A(a,e,s) and B(a,e,s) are equal to zero. There are in fact infinitely many in any plane for s<pi, and the two vectors orthogonal to e in the case s=pi.

The w vectors must respect the metric {g(u,v,s), t(u,v,s)} if they are to be within S^3, where u and v are any arbitrary vectors within S^3. As I pointed out to you before, the appearance of the specific vectors a and b is incidental. It is merely an artefact of the embedding space R^4, inevitably used to build the simulation.

Good, then it must be possible to get rid of this dependence, and calculate the vectors w without knowledge of the measurement directions a and b. Can you show a calculation doing that?

Moreover, the simulation is simply an implementation of the analytical model presented in the above paper, as explicitly shown in this theoretical paper.

So if it is in fact a faithful implementation of your model, then your model does not constitute a refutation of Bell's theorem, but a confirmation of it.

Thus you are confusing the flight simulator with the Jumbo Jet itself.

If the simulation says the jet will crash and burn, then I'll indeed probably not board it.

You are not allowed to use any concept external to S^3 to deduce what you wish to deduce.

No. All the data which is needed in order to compute the outcomes of Alice's and Bob's measurements are important; you can't simply use an appeal to S^3 to sweep the dependence of Alice's measurement outcomes on Bob's measurement settings under the rug.

It should be clear by now that your remaining commentary has nothing whatsoever to do with my actual model, but only with your deep-seated prejudices.

I'm not arguing from prejudices; everything I've said is well justified by the actual computations your simulation performs.

Nevertheless, if you wish to convince me that what you are claiming has an element of truth in it, then you must at least answer the question I asked you before:

Where do you see "explicit dependence" on measurement directions in the s^k vectors in Eq. (B10) of this paper?

See above. By restricting the vectors such that only those that yield a measurement outcome of +/-1 are chosen, the set of vectors depends on the measurement directions of the experimenters. Otherwise, one could always choose an a such that .

[minkwe]

My assertion (by quote) was that he seems ... I didn't say it wasn't published, I said it was submitted to J. Math. Phys. (as per the arXiv comment), but appears to have never been published there. .... So again, your accusation of lying has no base in actual fact, and I'd like to see you retract it.

You were presented with an argument, and instead of answering it, you resorted to ad-hominem, first trying to discredit the argument by suggesting that it was rejected from a journal. A fact which you know very well is not dispositive. You did not mention the fact that the article had been significantly revised, presented at a conference and published in the proceedings of that conference and as you know very well (or should know), it is unscientific to publish the same paper more than once. Secondly, you tried to suggest that the argument must be false because, the author changed his mind. Not only is that not dispositive, were it true, but you don't point to any relevant evidence. You pointed to an article discussing a different subject matter, quite irrelevant. Therefore, while I would not call you a liar (I never did), the message you were intentionally conveying was misleading (aka lies), and I was justified in concluding that the intention was to deceive.

1. Do you agree that the reason the trivial mathematical inequality

for 4 measurements measurements on a single particle pair can be extended to averages, is precisely because the prescribed factorization is possible for every row in the series of 4xN outcomes from N particle pairs?

No, the inequality can be extended to averages because no average of a set of values can't be greater than the maximum value of that set. So you have a set of values, all below two, and immediately you know that the average value will also be below two.

Please read carefully. You answer No, and then proceed to provide a winding answer which is essentially Yes. What part of the argument do you disgagree with ???
Since each member of the set can be factorized therefore the maximum for each member of the set is 2 , therefore the average of the set can not exceed the maximum of any member of the set. This is uncontroversial, why do you pretend to disagree with this when in fact you agree ??? Therefore, based on your reply, I will consider that you do agree with point (1) of my argument. If you disagree, please point out the sentence or phrase of point (1) that you claim is wrong or that you disagree with.

2. Do you agree that the 4 terms in the CHSH,

are therefore not independent, since they have been constructed from the same series of outcomes, simply recombined in pairs? In other words, since every two terms share one column of outcomes there is a cyclic dependency between them. Even if you would randomize each paired sequence after recombining, it should still be possible in principle to rearrange them so that the similarly labelled columns match exactly. That is the A column of the AB pair should match the A column of the AD pair, etc in a cyclical manner, back to the B column of CB matching the B column of AB. Do you agree or disagree?

You have not addressed point (2) of my argument, copied above for your convenience. I will wait for your response to point (2) before I address the remaining points.

[Joy Christian]

An often repeated claim by the Bell-devotees in this thread is that there are "0 outcomes" in the simulation I have presented at the beginning of the thread:

This claim is completely vacuous. But not surprisingly, Bell-devotees are simply unable to see the vacuity of their claim. So now I have revised the simulation slightly and included explicit calculations showing that "0 outcomes" simply do not exist within S^3. Here is the essential part of the code, specifying A(a, L) = +/-1 = B(b, L):

Code: Select all

        A = +sign(g(a,e,s))  # Alice's measurement results A(a, e, s) = +/-1
         
        B = -sign(g(b,e,s))  # Bob's measurement results B(b, e, s) = -/+1
       
        Cuu = length((A*B)[A > 0 & B > 0])   # Coincidence count of (+,+) events
       
        Cdd = length((A*B)[A < 0 & B < 0])   # Coincidence count of (-,-) events
       
        Cud = length((A*B)[A > 0 & B < 0])   # Coincidence count of (+,-) events
       
        Cdu = length((A*B)[A < 0 & B > 0])   # Coincidence count of (-,+) events
       
        corrs[i,j] = (Cuu + Cdd - Cud - Cdu) / (Cuu + Cdd + Cud + Cdu)


The above calculation produces the strong EPR-B correlation, as can be seen in the simulation. But the persistent claim is that A(.) and B(.) also produce "0 outcomes."

So now, in the revised version, I have added the following new lines in the code, for the benefit of the flatlanders:

Code: Select all

(Cou = length((A*B)[A == 0 & B > 0]))  # Number of (0,+) events within S^3
                     
(Cod = length((A*B)[A == 0 & B < 0]))  # Number of (0,-) events within S^3
       
(Cuo = length((A*B)[A > 0 & B == 0]))  # Number of (+,0) events within S^3
       
(Cdo = length((A*B)[A < 0 & B == 0]))  # Number of (-,0) events within S^3

(Coo = length((A*B)[g(a,e,s) & A == 0 & B == 0])) # Number of (0,0) events

(CoB = length(A[g(a,e,s) & A == 0]))   # Number of A = 0 events within S^3

(CAo = length(B[g(b,e,s) & B == 0]))   # Number of B = 0 events within S^3


We can now see in the simulation that the above calculations explicitly prove Cou = Cod = Cuo = Cdo = Coo = 0.

Thus there are simply no "0 outcomes" within S^3, contrary to the claim by the Bell-devotees. The simulation thus decisively refutes the so-called "theorem" by Bell.

[Jochen]

You were presented with an argument, and instead of answering it, you resorted to ad-hominem, first trying to discredit the argument by suggesting that it was rejected from a journal. A fact which you know very well is not dispositive.

I was presented with a link to a paper, and noted that it failed to pass peer review at J. Math. Phys., i.e. that its claims had been scrutinized and found not to be sufficient to warrant publication there. I did not say that this invalidates the paper's claims, but merely that it doesn't inspire confidence.

Secondly, you tried to suggest that the argument must be false because, the author changed his mind. Not only is that not dispositive, were it true, but you don't point to any relevant evidence. You pointed to an article discussing a different subject matter, quite irrelevant.

No, I suggested that the author now considers the argument to be erroneous, because in a later publication, he claims that no model of the kind he suggests there can violate Bell inequalities.

Therefore, while I would not call you a liar (I never did), the message you were intentionally conveying was misleading (aka lies), and I was justified in concluding that the intention was to deceive.

Calling statements I made lies is calling me a liar, since somebody who lies is a liar. Calling statements I made lies falsely is a character attack.

No, the inequality can be extended to averages because no average of a set of values can't be greater than the maximum value of that set. So you have a set of values, all below two, and immediately you know that the average value will also be below two.

Please read carefully. You answer No, and then proceed to provide a winding answer which is essentially Yes. What part of the argument do you disgagree with ???

I was merely trying to forestall confusion: while it's true that if we had full access to the hidden variable outcome assignments, every line would be bounded by two, in a real experiment, we don't have access to every outcome, but only to the subset of simultaneously measured observables; hence, in a measurement table drawn up from a real experiment, such as the example I gave above, we only know two out of the four values, and thus, don't have access to the value of the complete row. But nevertheless, the inference that continues to hold, contrary to what you seem to be claiming.

2. Do you agree that the 4 terms in the CHSH,

are therefore not independent, since they have been constructed from the same series of outcomes, simply recombined in pairs? In other words, since every two terms share one column of outcomes there is a cyclic dependency between them. Even if you would randomize each paired sequence after recombining, it should still be possible in principle to rearrange them so that the similarly labelled columns match exactly. That is the A column of the AB pair should match the A column of the AD pair, etc in a cyclical manner, back to the B column of CB matching the B column of AB. Do you agree or disagree?

[/list]

You have not addressed point (2) of my argument, copied above for your convenience. I will wait for your response to point (2) before I address the remaining points.

Well, I'm not sure I understand your point here. The A column of the AB pair is the same as the A column of the AD pair, so they'll trivially agree. Or do you mean to divide the A column into two, one in which B was measured simultaneously, and one in which D was measured simultaneously? If so, then again, there will in general not be perfect agreement between the numbers of +1 and -1 outcomes in both columns, due to the finite length of the experiment; and even if there were, the rearranging you propose in general would not preserve the correlations, as already explained.

[Jochen]

An often repeated claim by the Bell-devotees in this thread is that there are "0 outcomes" in the simulation I have presented at the beginning of the thread:

This claim is completely vacuous.

It's not. I've installed R on my system and run your own code, and lo and behold, the lists A and B are full of 0 outcomes. I've not yet figured out the Rpub thing, but you could simply show the first 100 or so entries in each list in your code, and you'll see that there are, indeed, 0 outcomes. On my system, the first 100 entries of the list A were:

Code: Select all

[1] -1  1  0 -1 -1  0 -1 -1 -1 -1  1 -1  1  0 -1  1  1  1  0 -1 -1 -1 -1  1 -1 -1 -1
 [28]  0 -1  0  1  1  1 -1  0  0 -1  0 -1  1  0  0 -1  0  0 -1  1  1  0  1  0  1  1  0
 [55]  1  1  0  0  1 -1  1  1  1 -1  1  0  0 -1  0  0  1 -1  1 -1  1 -1 -1 -1  0 -1 -1
 [82]  1 -1  1  1  0 -1  1 -1 -1  1 -1 -1 -1 -1  1  0  1  1  1

You can also simply add the line length((A*B)[A*B==0]), which will correctly count the number of times either A or B (or both) is zero.

I've also explicitly shown that for each pair (e,s), there exist measurement directions a such that A(a,e,s) = 0. Do you deny this is the case? Also, do you deny that in order to compute the w vectors in your simulation, you explicitly need the measurement directions a and b? And do you deny that the function t is there to get rid of the 0 outcomes (it's explicitly claimed that t is there in order to ensure that sign(g) = +/-1)?

The problem is with these lines:

Code: Select all

(Coo = length((A*B)[g(a,e,s) & A == 0 & B == 0])) # Number of (0,0) events

(CoB = length(A[g(a,e,s) & A == 0]))   # Number of A = 0 events within S^3

(CAo = length(B[g(b,e,s) & B == 0]))   # Number of B = 0 events within S^3

You use knowledge of the measurement directions in order to eliminate the unwanted 0 elements; but of course, this knowledge is not present at the source, where the hidden variables are set.

[Joy Christian]

I've also explicitly shown that for each pair (e,s), there exist measurement directions a such that A(a,e,s) = 0. Do you deny this is the case?

Yes.

Also, do you deny that in order to compute the w vectors in your simulation, you explicitly need the measurement directions a and b?

Yes.

And do you deny that the function t is there to get rid of the 0 outcomes (it's explicitly claimed that t is there in order to ensure that sign(g) = +/-1)?

t is not used at all in the main computation of the correlation. Correlations are computed in the simulation in three different ways, for illustrative purposes.

The problem is with these lines:

Code: Select all

(Coo = length((A*B)[g(a,e,s) & A == 0 & B == 0])) # Number of (0,0) events

(CoB = length(A[g(a,e,s) & A == 0]))   # Number of A = 0 events within S^3

(CAo = length(B[g(b,e,s) & B == 0]))   # Number of B = 0 events within S^3

There is no problem with these lines.

You use knowledge of the measurement directions in order to eliminate the unwanted 0 elements; but of course, this knowledge is not present at the source, where the hidden variables are set.

This is double talk. At the detector the detector direction a is known at which the outcome A = +/-1 are observed. There are no "unwanted 0 elements" in S^3.

Besides, as I have explained many times before, the appearance of the vectors a or b and e in the metric g(a,e,s) is purely incidental. Reread my previous replies!!!

You can also simply add the line length((A*B)[A*B==0]), which will correctly count the number of times either A or B (or both) is zero.

This will give you a count that has nothing whatsoever to do with my 3-sphere model. The correct counts within S^3 are given in my simulation.

You are stuck in your flatland and refusing to listen to what I have explained to you over and over again. In S^3 there are no "0 outcomes", period.