## 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

Jochen wrote:
Joy Christian wrote: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).
Joy Christian
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Location: Oxford, United Kingdom

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

minkwe wrote: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.

minkwe wrote: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 $\langle CHSH \rangle \leq 4$; 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.

minkwe wrote:4. You have claimed that the CHSH expression
$\langle A_iB_i\rangle + \langle C_iD_i\rangle + \langle C_iB_i\rangle - \langle A_iD_i\rangle$,
where $\{i\}$ represents a single set of particle pairs, is statistically equivalent to
$\langle A_iB_i\rangle + \langle C_jD_j\rangle + \langle C_kB_k\rangle - \langle A_lD_l\rangle$
where $\{i\}, \{j\}, \{k\}, \{l\}$ represent disjoint independent sets of particle pairs. Because according to you,
$\langle C_iD_i\rangle = \langle C_jD_j\rangle$ and $\langle C_iB_i\rangle = \langle C_kB_k\rangle$ and $\langle A_iD_i\rangle = \langle A_lD_l\rangle$

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 $P(A,B,C,D)$, 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 $\langle C_iD_i\rangle = \langle C_jD_j\rangle$
means that for every single individual pair of particles in the $\{i\}$ series which produced the outcome pair $(C_i, D_i)$, there is an equivalent pair of particles in the $\{j\}$ series, such that as the number of particles approaches infinity, it should be possible to find a function $f_{ij}$, which rearranges the sequence of $\{j\}$ outcomes to match the sequence of $\{i\}$, 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: $\langle X \rangle = \frac{n^+ - n^-}{n}$ for large number of measurements n, so in order to yield the same EV, there need to be approximately the same numbers $n^+$ of +1 outcomes and $n^-$ of -1 outcomes. However, equality of these numbers only holds for $n=\infty$, and for any two finite experiment runs i and j, both $n_i^+$ and $n_j^+$, as well as $n_i^-$ and $n_j^-$, 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
$f_{ik}$, $f_{il}$, $f_{jk}$ and $f_{jl}$. Such that after applying the functions, $A_i \equiv A'_j, C_k \equiv C'_l$ and $B_i \equiv B'_k$, 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 $n=\infty$ 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 $\{i\}, \{j\}, \{k\}, \{l\}$, precisely because after rearranging, you will get
$S' = \langle C_kB_k \rangle + \langle C_lD_l\rangle + \langle A_iB_i\rangle - \langle A_jD_j\rangle$
$S' = \langle C_kB_k \rangle + \langle C'_lD'_l\rangle + \langle A_iB_i\rangle - \langle A'_jD'_j\rangle$
Which can be factorized just like in point (1) to
$S' = \langle C_k(B_k + D'_l)\rangle + \langle A_i(B_i - D'_j)\rangle$
since $A_i \equiv A'_j, C_k \equiv C'_l$ and $B_i \equiv B'_k$

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
$S' = \langle C_k(B_k + D'_l)\rangle + \langle A_i(B_i - D'_j)\rangle \leq 2$ to be true, it must be the case that $B_k \equiv B'_i$ and $D'_l \equiv D'_j$. But both $D'_l$ and $D'_j$ 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 $D'_l$ agree with $D'_j$ will undo the previous rearrangements; the same for $B_k$ and $B_i$. Therefore for measurements on 4 independent disjoint sets to obey the the inequality$S' \leq 2$, 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 $S \leq 2$ 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

Posts: 79
Joined: Sat Jun 27, 2015 2:24 am

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

Joy Christian wrote:
Jochen wrote:
Joy Christian wrote: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.
Jochen

Posts: 79
Joined: Sat Jun 27, 2015 2:24 am

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

Jochen wrote: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?
Joy Christian
Research Physicist

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Location: Oxford, United Kingdom

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

Joy Christian wrote: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 $\mathrm{sign}(s^k\cdot a)=0$.
Jochen

Posts: 79
Joined: Sat Jun 27, 2015 2:24 am

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

Jochen wrote: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.

Jochen wrote:
minkwe wrote:1. Do you agree that the reason the trivial mathematical inequality
$Q = C_iB_i + C_iD_i + A_iB_i - A_iD_i = C_i(B_i+D_i) + A_i(B_i-D_i) \leq 2$ for 4 measurements measurements $\{A_i, B_i, C_i, D_i\}$ 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 $\{A_i, B_i, C_i, D_i\}$ of the set can be factorized therefore the maximum for each member of the set is 2 $Q = C_iB_i + C_iD_i + A_iB_i - A_iD_i = C_i(B_i+D_i) + A_i(B_i-D_i) \leq 2$, 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.

minkwe wrote:2. Do you agree that the 4 terms in the CHSH,
$\langle C_iB_i\rangle , \langle C_iD_i\rangle , \langle A_iB_i\rangle , \langle A_iD_i\rangle \leq 2$ 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.
minkwe

Posts: 1151
Joined: Sat Feb 08, 2014 10:22 am

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

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.
Last edited by Joy Christian on Wed Jul 08, 2015 1:48 am, edited 2 times in total.
Joy Christian
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Location: Oxford, United Kingdom

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

minkwe wrote: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.

minkwe wrote: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.

minkwe wrote: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.

minkwe wrote:
Jochen wrote: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 $\langle CHSH \rangle \leq 2$ continues to hold, contrary to what you seem to be claiming.

minkwe wrote:
minkwe wrote:2. Do you agree that the 4 terms in the CHSH,
$\langle C_iB_i\rangle , \langle C_iD_i\rangle , \langle A_iB_i\rangle , \langle A_iD_i\rangle \leq 2$ 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

Posts: 79
Joined: Sat Jun 27, 2015 2:24 am

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

Joy Christian wrote: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.
Jochen

Posts: 79
Joined: Sat Jun 27, 2015 2:24 am

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

Jochen wrote: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.

Jochen wrote: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.

Jochen wrote: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.

Jochen wrote:
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.

Jochen wrote: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!!!

Jochen wrote: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.
Joy Christian
Research Physicist

Posts: 2370
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

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

Joy Christian wrote:
Jochen wrote: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.

OK, then give me one set of (e,s) such that there does not exist a direction a such that A(a,e,s)=0. Of course, it's mathematically trivial that you can't find one, but since you assert that this is not the case, the onus is on you.

Joy Christian wrote:
Jochen wrote: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.

Then why are the components of the w vector explicitly dependent on a and b?
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)]

Joy Christian wrote:
Jochen wrote: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.

That wasn't my question.

Joy Christian wrote:
Jochen wrote: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.

Yes there is: you first compute which measurement directions a and b yield zero outcomes, and then neglect them. The problem is that without knowledge of the measurement directions, this computation could not be performed.

Jochen wrote: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.

The logic of your simulation, as it is, is as follows: first, a set of hidden variables (e,s) is produced. Then, the observed measurement outcomes are computed. Afterwards, those values of (e,s) for which A(a,e,s) is zero are excized. The remaining set of hidden variables (e,s) you then claim to be the ones that 'actually' occur in S^3. The problem is that you can then only decide which hidden variables 'actually' occur once you know the detector settings of a and b; these are, however, unknown at the source, i.e. at the point the decision for the hidden variable values is made. You can only retroactively decide which set of hidden variable values to admit; moreover, had the measurement directions of Alice and Bob been different, then the set of hidden variable values also would have been different. Hence, your hidden variables explicitly depend on the detector settings, and producing a violation of some Bell inequality in this way is of course perfectly trivial.

Joy Christian wrote: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.

OK, then provide a computation that can do without it! Of course, this isn't possible, and all that will come in reply is just more waffling; there is simply no way to compute your w vectors without knowing the detector settings and still produce a violation of the CHSH inequality.

Joy Christian wrote: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.

Then whatever you have programmed has nothing to do with S^3, since there are zero outcomes, period. You artificially exclude them afterwards, but this exclusion is only possible once the measurement directions of Alice and Bob are known.
Jochen

Posts: 79
Joined: Sat Jun 27, 2015 2:24 am

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

Jochen, we are done here. I have already answered all your questions many times before. I am off to Vatican City to convert the Pope to Islam. That would be easier.
Joy Christian
Research Physicist

Posts: 2370
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

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

Jochen wrote:
Joy Christian wrote:
Jochen wrote: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.

OK, then give me one set of (e,s) such that there does not exist a direction a such that A(a,e,s)=0. Of course, it's mathematically trivial that you can't find one, but since you assert that this is not the case, the onus is on you.

To quickly prove this point, recall that the function A(a,e,s) is:
$A(a,e,s)=\mathrm{sign}(a\cdot e)\,\mathrm{if}\,\,\,|a\cdot e|>-1+\frac{2}{\sqrt{1+\frac{3s}{\pi}}},$
$A(a,e,s)=0\,\,\,\,\,\,\,\,\mathrm{otherwise}$.

Thus, for any a such that $|a\cdot e|\leq-1+\frac{2}{\sqrt{1+\frac{3s}{\pi}}}$, there will be a zero outcome. The function $f(s)=-1+\frac{2}{\sqrt{1+\frac{3s}{\pi}}}$ is monotonically decreasing from 1 to 0 as s varies from 0 to pi. Hence, for any $s<\pi$, f(s) will be a finite positive number, and there exist an infinite number of a's such that $|a\cdot e|\leq f(s)$, for which thus A(a,e,s) = 0. For the case $s=\pi$, $f(s) = 0$, any a orthogonal to e will do.

Thus, for any pair (e,s), there exists an a such that A(a,e,s) = 0.
Jochen

Posts: 79
Joined: Sat Jun 27, 2015 2:24 am

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

Joy Christian wrote:Jochen, we are done here. I have already answered all your questions many times before. I am off to Vatican City to convert the Pope to Islam. That would be easier.

You haven't answered anything, and the answers you have provided are quite simply demonstrably false (see the above comment, for example).

And I mean, if any of your claims held water, it would be trivial to settle this once and for all: provide a way to compute the w vectors without knowledge of the detector settings. Yet you seem quite reluctant to do so. One can't help but wonder.
Jochen

Posts: 79
Joined: Sat Jun 27, 2015 2:24 am

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

Jochen wrote:
minkwe wrote: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 disagree with ???

I was merely trying to forestall confusion: while it's true that ...

There was no confusion, you do not disagree to anything in my point (1). You have not pointed to anything in my point (1) that is even remotely ambiguous, or disagreeable. My point (1) does not mention anything about experiments or hidden variables. It is a simple arithmetic fact that applies to any 4xN spreadsheet of numbers $\pm 1$, with columns labelled A,B,C,D, which you agree with. In any case, despite the initial "No" outburst, your answer to point (1) is actually Yes, so let us move on.

minkwe wrote:2. Do you agree that the 4 terms in the CHSH,
$\langle C_iB_i\rangle , \langle C_iD_i\rangle , \langle A_iB_i\rangle , \langle A_iD_i\rangle$ are therefore not independent, since they have been constructed from the same series of outcomes, simply recombined in pairs? 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?

Are the 4 terms in the CHSH expression independent or not? I'm not asking you about measurements, or experiments. I'm asking you based the simple algebra of the derivation of CHSH, and the way terms are related to each other. Is <AB> independent of <AD>, <CB>? Is <CD> independent of <CB>, <AD> etc?

Jochen wrote: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.

So then do you agree that <AB> is not independent of <AD>, since they share the same A column of data? And do you agree that there is a cyclical sharing of columns of data as explained in point (2): "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."

Jochen wrote: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 ...,

Again, this is completely irrelevant to point (2). I'm simply asking you if you agree that the 4 averages $\langle C_iB_i\rangle , \langle C_iD_i\rangle , \langle A_iB_i\rangle , \langle A_iD_i\rangle$ in the CHSH expression are independent of each other or not. Please review the derivation of the CHSH (https://en.wikipedia.org/wiki/Bell's_th ... inequality) and answer if you agree that the 4 terms in the final inequality are not independent.

Jochen wrote:... the rearranging you propose in general would not preserve the correlations, as already explained.

This is completely false, please read my point again carefully. There is never a separation of pairs in any of the points. All rearrangements/randomizations I've ever talked about involves moving pairs around. For example, I take a copy of the A and D columns together, so I now have a 2xN spreadsheet with columns labelled A,D. And then I randomize the rows of the 2xN spreadsheet, always keeping the pairs together, then the <AD> average does not change. Similarly, we can do the same for CB, CD, and AB pairs. The averages would be exactly the same. All I've done is replace $\frac{1}{N}\sum_{i}^{N} A_iB_i$, with a randomized variant $\frac{1}{N}\sum_{j}^{N} A_jB_j$ where $i \neq j$ but there exists a map $f_{ji}(j) = i$ which maps the new row indices $j$ to the original indices of the 4xN spreadsheet. And similar maps exist for the other pairs. In other words, although the A column of the randomized AD spreadsheet does not match the A column of the independently randomized AB spreadsheet, it is possible in principle to rearrange them using the corresponding mapping functions so that they match. Do you agree or disagree?

I believe point (2) is now very clearly explained. I'll await your response to it before responding to the other responses you provided earlier to the other points. Although you may find the need to revise them before I do so.
minkwe

Posts: 1151
Joined: Sat Feb 08, 2014 10:22 am

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

minkwe wrote:Are the 4 terms in the CHSH expression independent or not?

Of course they are not, if they are generated by a LHV model. What is your point?
Heinera

Posts: 767
Joined: Thu Feb 06, 2014 1:50 am

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

Heinera wrote:
minkwe wrote:Are the 4 terms in the CHSH expression independent or not?

Of course they are not, if they are generated by a LHV model. What is your point?

Heine, if you want to respond to my argument, you will have to address all 8 points viewtopic.php?f=6&t=168&start=230#p4731.

Do not think I missed the completely false trojan horse you injected above "... if they are generated by a LHV model..." . Everybody agrees that a coin is local realistic. If I toss a coin 1 million times and write on column X, the outcome above the table and on column Y the outcome below the table for the same coin, where H = +1 and T = -1, then the terms <X> and <Y> are not independent but if I toss two separate coins 1 million time and each time and write on column X the outcome above the table for coin 1, and on column Y the outcome below the table for coin 2, then the terms <X> and <Y> are independent even though coins are LHV. Your trojan horse collapses.

If you address all 8 points of my argument linked above, making sure you understand each one before responding, you will get the point. Otherwise, I will consider your posts as noise and simply ignore them.
minkwe

Posts: 1151
Joined: Sat Feb 08, 2014 10:22 am

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

minkwe wrote:There was no confusion, you do not disagree to anything in my point (1). You have not pointed to anything in my point (1) that is even remotely ambiguous, or disagreeable. My point (1) does not mention anything about experiments or hidden variables. It is a simple arithmetic fact that applies to any 4xN spreadsheet of numbers $\pm 1$, with columns labelled A,B,C,D, which you agree with. In any case, despite the initial "No" outburst, your answer to point (1) is actually Yes, so let us move on.

Do what you feel you must.

minkwe wrote:
Jochen wrote: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.

So then do you agree that <AB> is not independent of <AD>, since they share the same A column of data? And do you agree that there is a cyclical sharing of columns of data as explained in point (2): "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."

Yes, there are indeed dependencies in the expectation values---those are basically the correlations. But that doesn't mean that there must be some physical identity or continuity between the particle pairs in order to produce this dependence: it is given simply by the measurement outcomes being drawn from the same probability distribution. Remember the example of the dice.

minkwe wrote:
Jochen wrote:... the rearranging you propose in general would not preserve the correlations, as already explained.

This is completely false, please read my point again carefully. There is never a separation of pairs in any of the points. All rearrangements/randomizations I've ever talked about involves moving pairs around. For example, I take a copy of the A and D columns together, so I now have a 2xN spreadsheet with columns labelled A,D. And then I randomize the rows of the 2xN spreadsheet, always keeping the pairs together, then the <AD> average does not change. Similarly, we can do the same for CB, CD, and AB pairs.

But not independently of the way you rearranged the other pairs. Let's say there is perfect correlations between A and B, and perfect anticorrelation between A and D. Shuffling around the AB pair on its own, while keeping the D column fixed, can be used to create any kind of correlation at all between A and D. Hence, the only valid operations you can perform on the table of values are re-arrangings of rows.
Jochen

Posts: 79
Joined: Sat Jun 27, 2015 2:24 am

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

Jochen wrote:Do what you feel you must.

Are you suggesting you actually disagree with point (1) afterall? Then please point out what it is there that you disagree with -- point to a specific sentence or equation, rather than something you have dreamt up that is completely unrelated to my point (1).

Jochen wrote:
minkwe wrote:So then do you agree that <AB> is not independent of <AD>, since they share the same A column of data? And do you agree that there is a cyclical sharing of columns of data as explained in point (2): "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."

Yes, there are indeed dependencies in the expectation values

Thank you!

Jochen wrote:But that doesn't mean that there must be some physical identity or continuity between the particle pairs in order to produce this dependence.

Who said anything about point (2) implying some kind of physical identity or continuity? Please read it carefully again. Point (2) is stating an arithmetic fact. Which should be completely uncontroversial. The 4 terms in the CHSH are not independent of each other, and you agree to this. This means you generate new 4 new 2xN spreadsheets of data (AB, AD, CB, CD) by recombining columns from the 4 original columns of data (A,B,C,D). Independently randomize each new 2xN spreadsheet, keeping rows of pairs together will not change the averages <AB>, <AD>, <CB>, <CD>, even though similarly named columns of data in each 2xN spreadsheet would not match (due to independent randomization). Furthermore, it is in principle possible to reconstruct the original 4xN spreadsheet of data, starting from the 4 separate independent randomized spreadsheets (AB, AD, CB, CD).

Jochen wrote:
Jochen wrote:... the rearranging you propose in general would not preserve the correlations, as already explained.

minkwe wrote:This is completely false, please read my point again carefully. There is never a separation of pairs in any of the points. All rearrangements/randomizations I've ever talked about involves moving pairs around. For example, I take a copy of the A and D columns together, so I now have a 2xN spreadsheet with columns labelled A,D. And then I randomize the rows of the 2xN spreadsheet, always keeping the pairs together, then the <AD> average does not change. Similarly, we can do the same for CB, CD, and AB pairs.

But not independently of the way you rearranged the other pairs. Let's say there is perfect correlations between A and B, and perfect anticorrelation between A and D. Shuffling around the AB pair on its own, while keeping the D column fixed, can be used to create any kind of correlation at all between A and D.

Again you have completely failed to understand the point, I've explained many times in many different ways. Please read the explanations again more carefully:

a) I have original_4xN.csv which is a spreadsheet with 4 columns labelled A,B,C,D. This is the original 4xN spreadsheet.
b) I create 4 new blank spreadsheet files named AB_2xN.csv, AD_2xN.csv, CB_2xN.csv, CD_2xN.csv, with columns labelled (A,B), (A,D), (C,B), (C,D) respectively
c) I copy the column A from original_4xN.csv to the columns labelled A in AB_2xN.csv, AD_2xN.csv, then copy the column B from original_4xN.csv to the columns labelled B in AB_2xN.csv, CB_2xN.csv, and the same for C, and the same for D. At the end, I have 4 new files, each of which contains copies of two columns from original_4xN.csv.
d) Next I randomize the rows in AB_2xN.csv, then I independently randomize the rows in AD_2xN.csv, and repeat the same for CB_2xN.csv, CD_2xN.csv.
e) I now have 4 independently randomized 2xN spreadsheets. Yet the averages <AB>, <AD>, <CB>, <CD> calculated from AB_2xN.csv, AD_2xN.csv, CB_2xN.csv, CD_2xN.csv respectively will be exactly the same as the corresponding averages calculated using the columns of the original_4xN.csv, and the exact same dependencies present between the terms of the original file will be present between those terms from the randomized files. And even though the similarly named columns in the 2xN files no longer match each other or the original 4xN file, they (the 2xN files) could in principle be rearranged (un-randomized, keeping pairs together) so that the columns again match those in original_4xN.csv, and each other. This is point (2), again completely uncontroversial and not hard to understand. Your claim that independently randomizing the AB spreadsheet somehow changes the AD correlation is due to lack of understanding. Did you miss the clearly stated fact that the AB, AD spreadsheets contained copies [emphasis present in original].
minkwe

Posts: 1151
Joined: Sat Feb 08, 2014 10:22 am

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

OK, so let's get this over with. Point 1: I agree that for the hidden variable value assignments, every row is in fact bounded by 2. Point 2: I agree that if you separate out the data into your 2xN list, any reshuffling of rows does not change the expectation value (though of course, you can't then recombine the shuffled lists without undoing the shuffling, at least in such a way that the new 4xN list is just a row-shuffled form of the original). Can we now move on?
Jochen

Posts: 79
Joined: Sat Jun 27, 2015 2:24 am

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