## Evidence that QM does not violate Bell's inequalities

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### Re: Evidence that QM does not violate Bell's inequalities

Perhaps Guest should review this thread to find out what the discussion here is all about. Lambda is completely irrelevant to the topic of discussion here. Read specifically post #1 and then my last response to Heinera from Sat Aug 29th.

### Re: Evidence that QM does not violate Bell's inequalities

"minkwe" should like this: Bell (1981): "It is notable in this argument that nothing is said about the locality, or even localizability, of the variables lambda. They could well include, for example, QM state vectors"

### Re: Evidence that QM does not violate Bell's inequalities

Heinera wrote:I promised myself I wouldn't bother anymore, but weak willpower makes me try one more time to teach someone some elementary probability theory:

Promise yourself to learn some basic statistics before you speak next time. And before you respond to this thread any further, I strongly suggest you read post #1, very carefully

Heinera wrote:$\langle A_iB_i\rangle=\langle A_jB_j\rangle$
$\langle A_iD_i\rangle =\langle A_kD_k\rangle$
$\langle C_iB_i\rangle =\langle C_lB_l\rangle$
$\langle C_iD_i\rangle= \langle C_mD_m\rangle$

As clearly explained in post #1, the above is impossible. So again, go back to the first page and read the argument in post #1, and save yourself from further embarrassment. Make sure you do not miss the part where it is shown that the above is impossible. If for some reason you do not like me personally (evidently), such that you have a bias against valid arguments made by me, perhaps you may like to read a purely mathematics paper which pre-dates Bell, and shows exactly the same thing rigorously http://www.panix.com/~jays/vorob.pdf. Do not be deceived that your naive "probability reasoning" is valid. Such arguments are made all the time by novices who do not understand a shred of probability theory. You can read some more examples in Chapter 15 of Jaynes Book, Probability Theory: The Logic of Science. (http://www-biba.inrialpes.fr/Jaynes/cc15b.pdf)

In short, you have no clue what you are talking about. Just because you are friends with, and hang around a few old-school statisticians does not mean you understand probability theory.

Heinera wrote:The extra degrees of freedom is of no use to the modeler (or nature), since he/she can't make any use of them due to the random selection of settings.

Please also do yourself a favor and read-up on Bertrands paradox. Random sampling of "hidden variables" is an oxymoron. To randomly sample any variable you need to know ALL the parameters that influence its values and exactly how the values change. If you do not understand this, read this article, specifically related to Bell's inequality, in which a counterexample demonstrating this pitfall is presented

http://arxiv.org/abs/0907.0767

### Re: Evidence that QM does not violate Bell's inequalities

FrediFizzx wrote:I don't think that can be right for CHSH. The CHSH string is about two iterations (trials) minimum.

Well that was wrong. In a real experiment, there would have to be at least 4 iterations minimum. Keep that in mind when evaluating CHSH.

### Re: Evidence that QM does not violate Bell's inequalities

Heinera wrote:I promised myself I wouldn't bother anymore, but weak willpower makes me try one more time to teach someone some elementary probability theory:

Because the expected mean of a randomly selected sample from a larger population is the same as the mean of the population itself, we have the following equalities when the sets of j,k,l,m are randomly selected subsets of the set of i:

$\langle A_iB_i\rangle=\langle A_jB_j\rangle$
$\langle A_iD_i\rangle =\langle A_kD_k\rangle$
$\langle C_iB_i\rangle =\langle C_lB_l\rangle$
$\langle C_iD_i\rangle= \langle C_mD_m\rangle$

We now do the trivial substitution from

$\langle A_iB_i\rangle - \langle A_iD_i\rangle + \langle C_iB_i\rangle + \langle C_iD_i\rangle \leq 2$

to get

$\langle A_jB_j\rangle - \langle A_kD_k\rangle + \langle C_lB_l\rangle + \langle C_mD_m\rangle \leq 2$, which is the CHSH inequality.

I don't think that can be right for CHSH. The CHSH string is about two iterations (trials) minimum.
So $\langle A_iB_i\rangle$ and $\langle A_iD_i\rangle$ can't be in the same iteration for the first expression. But then in the second expression the A's are not the same in the first and second expectation terms, etc.

### Re: Evidence that QM does not violate Bell's inequalities

minkwe wrote:
Mikko wrote:The CHSH inequality is usually written
|E(a, b) − E(a, b′) + E(a′, b) + E(a′, b′)| ≤ 2
although some unimportant variation is possible.

Please go through derivation, then you will discover that your notation is incomplete.
The CHSH is
$\langle A_iB_i\rangle - \langle A_iD_i\rangle + \langle C_iB_i\rangle + \langle C_iD_i\rangle \leq 2$

Anyway, the right hand side has 2 as the upper bound, independently of any dependencies or anything.

The upper bound of the RHS is for the specific completely specified arrangement of terms shown on the LHS. That arrangement embodies the fact that the terms are dependent. Therefore it is nonsense to say independence/dependence does not matter.

Therefore, whenever the left side expression has a value that is greater than 2 the CHSH inequality is violated.

That is impossible. The specific arrangement of terms on the LHS (with complete notation) can never violate the RHS. Any suggestion otherwise is contrary to mathematics. How ever, by sloppy incomplete notation, it is easy to deceive yourself by calculating $\langle A_jB_j\rangle - \langle A_kD_k\rangle + \langle C_lB_l\rangle + \langle C_mD_m\rangle$ and mistaking it for the LHS of the CHSH. This amounts to changing the LHS, and expecting the RHS to stay the same. This new expression does not embody the same kind of dependences present in the LHS of the CHSH, therefore the corresponding RHS has a different upper bound. Tsirelsons inequality is a variant of this.

I promised myself I wouldn't bother anymore, but weak willpower makes me try one more time to teach someone some elementary probability theory:

Because the expected mean of a randomly selected sample from a larger population is the same as the mean of the population itself, we have the following equalities when the sets of j,k,l,m are randomly selected subsets of the set of i:

$\langle A_iB_i\rangle=\langle A_jB_j\rangle$
$\langle A_iD_i\rangle =\langle A_kD_k\rangle$
$\langle C_iB_i\rangle =\langle C_lB_l\rangle$
$\langle C_iD_i\rangle= \langle C_mD_m\rangle$

We now do the trivial substitution from

$\langle A_iB_i\rangle - \langle A_iD_i\rangle + \langle C_iB_i\rangle + \langle C_iD_i\rangle \leq 2$

to get

$\langle A_jB_j\rangle - \langle A_kD_k\rangle + \langle C_lB_l\rangle + \langle C_mD_m\rangle \leq 2$, which is the CHSH inequality.

Anyone who doesn't understand this is well advised to consult an introductory text on probability theory. The extra degrees of freedom is of no use to the modeler (or nature), since he/she can't make any use of them due to the random selection of settings.

### Re: Evidence that QM does not violate Bell's inequalities

Mikko wrote:The CHSH inequality is usually written
|E(a, b) − E(a, b′) + E(a′, b) + E(a′, b′)| ≤ 2
although some unimportant variation is possible.

Please go through derivation, then you will discover that your notation is incomplete.
The CHSH is
$\langle A_iB_i\rangle - \langle A_iD_i\rangle + \langle C_iB_i\rangle + \langle C_iD_i\rangle \leq 2$

Anyway, the right hand side has 2 as the upper bound, independently of any dependencies or anything.

The upper bound of the RHS is for the specific completely specified arrangement of terms shown on the LHS. That arrangement embodies the fact that the terms are dependent. Therefore it is nonsense to say independence/dependence does not matter.

Therefore, whenever the left side expression has a value that is greater than 2 the CHSH inequality is violated.

That is impossible. The specific arrangement of terms on the LHS (with complete notation) can never violate the RHS. Any suggestion otherwise is contrary to mathematics. How ever, by sloppy incomplete notation, it is easy to deceive yourself by calculating $\langle A_jB_j\rangle - \langle A_kD_k\rangle + \langle C_lB_l\rangle + \langle C_mD_m\rangle$ and mistaking it for the LHS of the CHSH. This amounts to changing the LHS, and expecting the RHS to stay the same. This new expression does not embody the same kind of dependences present in the LHS of the CHSH, therefore the corresponding RHS has a different upper bound. Tsirelsons inequality is a variant of this.

### Re: Evidence that QM does not violate Bell's inequalities

minkwe wrote:Eyes that are blind do not see. The terms from your model are independent, thus the upper bound is 4. You haven't violated anything.

The CHSH inequality is usually written
|E(a, b) − E(a, b′) + E(a′, b) + E(a′, b′)| ≤ 2
although some unimportant variation is possible.
Anyway, the right hand side has 2 as the upper bound, independently of any dependencies or anything. Therefore, whenever the left side expression has a value that is greater than 2 the CHSH inequality is violated.
If change the right side, it no longer is the CHSH inequality. E.g.,
|E(a, b) − E(a, b′) + E(a′, b) + E(a′, b′)| ≤ 2√2
is Tsirelson's inequality. For example, for certain experiments, quantum mechanics predicts a violation of CHSH inequality but not of Tsirelsons inequality.

### Re: Evidence that QM does not violate Bell's inequalities

Heinera wrote:
minkwe wrote:
Heinera wrote:We are talking about models here. With your deep insight, it should be no problem for you to make a local version of my trivial model.

Of course, you could rather claim that these experiments can be explained by a local theory, but the theory just can't be sufficiently formalized so that it can be turned into a simulation. This was Joy's initial position. Downside: you risk being laughed off the stage by the physics community.

Okay, enough of the rubbish. You haven't been able to produce a non-local model that violates Bell's inequality. You haven't been able to produce QM expectations for the terms in Bell's inequality. So you are now taking this thread off-topic. This thread is about the failure of QM and experiments to violate Bell's inequality. Start a new thread if you want to discuss other things.

Of course I have. The model is out there. You are just denying what everyone else can see with their own eyes.

Eyes that are blind do not see. The terms from your model are independent, thus the upper bound is 4. You haven't violated anything.

### Re: Evidence that QM does not violate Bell's inequalities

minkwe wrote:
Heinera wrote:We are talking about models here. With your deep insight, it should be no problem for you to make a local version of my trivial model.

Of course, you could rather claim that these experiments can be explained by a local theory, but the theory just can't be sufficiently formalized so that it can be turned into a simulation. This was Joy's initial position. Downside: you risk being laughed off the stage by the physics community.

Okay, enough of the rubbish. You haven't been able to produce a non-local model that violates Bell's inequality. You haven't been able to produce QM expectations for the terms in Bell's inequality. So you are now taking this thread off-topic. This thread is about the failure of QM and experiments to violate Bell's inequality. Start a new thread if you want to discuss other things.

Of course I have. The model is out there. You are just denying what everyone else can see with their own eyes.

### Re: Evidence that QM does not violate Bell's inequalities

Heinera wrote:We are talking about models here. With your deep insight, it should be no problem for you to make a local version of my trivial model.

Of course, you could rather claim that these experiments can be explained by a local theory, but the theory just can't be sufficiently formalized so that it can be turned into a simulation. This was Joy's initial position. Downside: you risk being laughed off the stage by the physics community.

Okay, enough of the rubbish. You haven't been able to produce a non-local model that violates Bell's inequality. You haven't been able to produce QM expectations for the terms in Bell's inequality. So you are now taking this thread off-topic. This thread is about the failure of QM and experiments to violate Bell's inequality. Start a new thread if you want to discuss other things.

### Re: Evidence that QM does not violate Bell's inequalities

The above quote from my 2013 reply in no way contradicts with my earlier reply to you:
Joy Christian wrote:
Heinera wrote:Of course, you could rather claim that these experiments can be explained by a local theory, but the theory just can't be sufficiently formalized so that it can be turned into a simulation. This was Joy's initial position.

This has never been my "position." My position since 2007 has been explained here, here, and here.

My position has been that a simulation is an implementation of a given model, not the model itself.

### Re: Evidence that QM does not violate Bell's inequalities

Well, this is the reply you gave to John Reed in the usenet group sci.physics.foundations (2013) when he tried to event-by-event simulate your theory:

Joy Christian wrote:Hi John,

I do not wish to discourage you, but this will be a rather negative
comment. Regardless of the details of your simulation, I can guarantee
you that you will not be able to get either the sin^2 curve for the probabilities or the cosine curve for the correlations in your simulation.
The reason is pure and simple: Bell's theorem. It is a mathematical
theorem that proves, ones and for all, that no procedure like what you
are considering will ever produce the cosine correlation. There are also
some corollaries of Bell's theorem, specifically aimed at simulation
attempts of the kind you are considering, which prove that you will not
be able to generate correlations or probabilities stronger than the
straight line you are getting. Of course, if you somehow let some
information pass on from one side to the other non-locally, or exploit
some kind of a loophole like the detector loophole, then you will be
able to get stronger correlations. One such example which exploits the
time-window loophole can be found here: http://rugth30.phys.rug.nl/eprbdemo/simulation.php

Best,

Joy

### Re: Evidence that QM does not violate Bell's inequalities

Heinera wrote:Of course, you could rather claim that these experiments can be explained by a local theory, but the theory just can't be sufficiently formalized so that it can be turned into a simulation. This was Joy's initial position.

This has never been my "position." My position since 2007 has been explained here, here, and here.

My position has been that a simulation is an implementation of a given model, not the model itself.

### Re: Evidence that QM does not violate Bell's inequalities

minkwe wrote:
Heinera wrote:With my model, it is impossible to compute a 4xN matrix as defined in Richard Gill's paper.

Are you referring to this paper https://pubpeer.com/publications/D985B4 ... 3E3A314522? I encourage anyone who believes a shred of anything in it to read that discussion. From appearances, even Gill himself has stopped promoting the paper. It is seriously flawed.

Rubbish.
If you were to come up with a local loophole-free model, such a 4xN matrix would be trivial to construct: For each i, I would simply compute your model four times, changing only the settings, and keeping everything else the same.

You've not learned any physics all these years. If you had read and understood post #1, you will also see that for the random experiment described it is impossible to generate a 4xN spreadsheet as well, even for a local realistic model. So why have you conveniently avoided the argument?

We are talking about models here. With your deep insight, it should be no problem for you to make a local version of my trivial model.

Of course, you could rather claim that these experiments can be explained by a local theory, but the theory just can't be sufficiently formalized so that it can be turned into a simulation. This was Joy's initial position. Downside: you risk being laughed off the stage by the physics community.

### Re: Evidence that QM does not violate Bell's inequalities

Heinera wrote:With my model, it is impossible to compute a 4xN matrix as defined in Richard Gill's paper.

Are you referring to this paper https://pubpeer.com/publications/D985B4 ... 3E3A314522? I encourage anyone who believes a shred of anything in it to read that discussion. From appearances, even Gill himself has stopped promoting the paper. It is seriously flawed.

I you were to come up with a local loophole-free model, such a 4xN matrix would be trivial to construct: For each i, I would simply compute your model four times, changing only the settings, and keeping everything else the same.

You've not learned any physics all these years. If you had read and understood post #1, you will also see that for the random experiment described it is impossible to generate a 4xN spreadsheet as well, even for a local realistic model. So why have you conveniently avoided the argument?

### Re: Evidence that QM does not violate Bell's inequalities

minkwe wrote:He does not deny that Bell's inequality is the following:
$\langle A_iB_i\rangle - \langle A_iD_i\rangle + \langle C_iB_i\rangle + \langle C_iD_i\rangle \leq 2$,
which makes use of the four terms $\langle A_iB_i\rangle ,\; \langle A_iD_i\rangle ,\; \langle C_iB_i\rangle ,\; \langle C_iD_i\rangle$ all defined for the same set of outcomes $A_i, B_i, C_i, D_i$

Oh, I do. Bell's inequality is the following:

$\langle A_jB_j\rangle ,\; \langle A_kD_k\rangle ,\; \langle C_lB_l\rangle ,\; \langle C_mD_m\rangle$.

minkwe wrote:He presents a simulation which is equivalent to the terms each calculated from 8 separate columns of outcomes: $A_j, B_j, A_k, D_k, C_l, B_l, C_m, D_m$ recombined into 4 independent paired spreadsheets, yielding the 4 terms $\langle A_jB_j\rangle ,\; \langle A_kD_k\rangle ,\; \langle C_lB_l\rangle ,\; \langle C_mD_m\rangle$. He then calculates the expression

$\langle A_jB_j\rangle - \langle A_kD_k\rangle + \langle C_lB_l\rangle + \langle C_mD_m\rangle$ and finds that the value is greater than 2. But this is to be expected because the terms in this expression are statistically independent.

As you can see in every proof of the CHSH or Bell's inequality (https://en.wikipedia.org/wiki/Bell%27s_ ... inequality for example), There are only 4 columns of data $A_i, B_i, C_i, D_i$, recombined into pairs such that the cyclic dependency of the paired terms is present. What is often missing is that the subscripts are left out which allows them to later confusingly or intentionally mislead by assuming there are only 4 columns of data in $A_j, B_j, A_k, D_k, C_l, B_l, C_m, D_m$, since without subscripts, all the As, Bs, Cs and Ds look alike and you can simply say $A, B, C, D$.

With my model, it is impossible to compute a 4xN matrix as defined in Richard Gill's paper.

I you were to come up with a local loophole-free model, such a 4xN matrix would be trivial to construct: For each i, I would simply compute your model four times, changing only the settings, and keeping everything else the same. Now I would have a matrix that perfectly replicates your model for those four settings. So I could then just draw a random pair of settings for each row, and read off the results. The correlations I get would be the exact same values your model would produce. And because the starting point was a 4xN matrix, the expectations for the CHSH expression would be 2 or lower.

And by the way, it is not very important to me if you understand this or not.

### Re: Evidence that QM does not violate Bell's inequalities

Heinera wrote:There is no problem. My model does not respect the bound of 2, because of its non-locality. All local models wil have a bound of 2.

This is complete and utter garbage and if I did not suspect that Heine may really be completely clueless about this, I would have concluded that he was lying through his teeth.

The dependence and independence we are talking about, have no relationship to locality/non-locality. I have already shown since post #1 that it is impossible to satisfy the conditions required to compare the experimental and QM expectations with Bell's inequalities -- even if the model/experiment is fully local realistic. Heine simply ignores the argument.

He does not deny that Bell's inequality is the following:
$\langle A_iB_i\rangle - \langle A_iD_i\rangle + \langle C_iB_i\rangle + \langle C_iD_i\rangle \leq 2$,
which makes use of the four terms $\langle A_iB_i\rangle ,\; \langle A_iD_i\rangle ,\; \langle C_iB_i\rangle ,\; \langle C_iD_i\rangle$ all defined for the same set of outcomes $A_i, B_i, C_i, D_i$

He presents a simulation which is equivalent to the terms each calculated from 8 separate columns of outcomes: $A_j, B_j, A_k, D_k, C_l, B_l, C_m, D_m$ recombined into 4 independent paired spreadsheets, yielding the 4 terms $\langle A_jB_j\rangle ,\; \langle A_kD_k\rangle ,\; \langle C_lB_l\rangle ,\; \langle C_mD_m\rangle$. He then calculates the expression

$\langle A_jB_j\rangle - \langle A_kD_k\rangle + \langle C_lB_l\rangle + \langle C_mD_m\rangle$ and finds that the value is greater than 2. But this is to be expected because the terms in this expression are statistically independent. Note that there are 8 columns of data in 4 independent pairs in the data from his simulation, whereas there are 4 columns of data with cyclic dependent pairing in Bell's inequality. Note also that the origin of the upper bound is entirely due to this fact. It is the dependencies or lack thereof that determine the upper bound of the inequality, so a proper understanding of it's presence or absence is crucial to any understanding of Bell's theorem.

As you can see in every proof of the CHSH or Bell's inequality (https://en.wikipedia.org/wiki/Bell%27s_ ... inequality for example), There are only 4 columns of data $A_i, B_i, C_i, D_i$, recombined into pairs such that the cyclic dependency of the paired terms is present. What is often missing is that the subscripts are left out which allows them to later confusingly or intentionally mislead by assuming there are only 4 columns of data in $A_j, B_j, A_k, D_k, C_l, B_l, C_m, D_m$, since without subscripts, all the As, Bs, Cs and Ds look alike and you can simply say $A, B, C, D$.

When presented with this obvious fact, Heine then proceeds first to make the incredible claim that he is indeed calculating
$\langle A_iB_i\rangle - \langle A_iD_i\rangle + \langle C_iB_i\rangle + \langle C_iD_i\rangle$
not
$\langle A_jB_j\rangle - \langle A_kD_k\rangle + \langle C_lB_l\rangle + \langle C_mD_m\rangle$

Heinera wrote:As shown in post #1, you are calculating the independent terms $E11_i,\; E12_j,\; E21_k,\; E22_l$ which are definitely not the terms in Bell's inequality.

My model does indeed produce $E11_i,\; E12_i,\; E21_i,\; E22_i$. Read the code again.[/quote]
Obviously anyone with a shred of statistics training should be able to see that if Heine is right that he is calculating $\langle A_iB_i\rangle - \langle A_iD_i\rangle + \langle C_iB_i\rangle + \langle C_iD_i\rangle2$, then it means $A_j = A_k = A_i,\; B_j = B_l = B_i,\; C_l =C_m = C_i,\; D_k = D_m = D_i$. Only then can his claim be true. An easy statistics test of this claim is to calculate the cross correlation between $(A_j,\; A_k)$ or $(B_j, B_l)$ or $(C_l, C_m)$, or $(D_k, D_m)$. If those columns of data are equivalent, then those correlations will be significantly different from zero. This is what Fred has shown in viewtopic.php?f=6&t=181&start=110#p5302 and found that the cross correlations if you substitute A_i for A_k becomes zero and the same for the rest of them. Therefore Heine is calculating $E11_i,\; E12_j,\; E21_k,\; E22_l$ not $E11_i,\; E12_i,\; E21_i,\; E22_i$ like he claims, and therefore the upper bound is 4 not 2.

So what has this got to do with non-locality or locality? Nothing whatsoever, as shown in the first post of this thread, if you start with 4 random statistically independent sets of outcome pairs from any source whatsoever (local or non-local), it is impossible to carry out the required row-permutations in order to demonstrate that all the equivalences $A_j \equiv A_k,\; B_j \equiv B_l,\; C_l \equiv C_m,\; D_k \equiv D_m$ are simultaneously true. Some of them might be the same but all of them can not be simultaneously the same as verly clearly explained in post #1. It does not matter whether we are dealing with a local or a non-local model. Therefore it is a mathematical error to assume that all the expectation values are simultaneously equal, ie
$\langle A_jB_j\rangle = \langle A_iB_i\rangle,\; \langle A_kD_k\rangle = \langle A_iD_i\rangle ,\; \langle C_lB_l\rangle = \langle C_iB_i\rangle ,\; \langle C_mD_m\rangle = \langle C_iD_i\rangle$
Some of them may be equal, but they can not all be simultaneously equal. This was clearly explained in post #1. Bell's believers simply ignore the argument.

You may have heard noises about sample of a population having the same value as population. You will note that when this bogus argument is presented, they always talk about a single value, not 4 values simultaneously. But they ignore one very crucial detail. The expectations in Bell's inequality are not equivalent to independent random samples. They are heavily dependent on each other. You can not replace them with independent random samples. Bell's followers repeatedly proclaim how Bell's theorem forces us to rethink the freedom of the experimenter, but it is the complete opposite. The problem is not that the experimenters are not free enough to measure what they want. The problem is that they are too free compared to the freedom assumed in Bell's inequality. Therefore the push to have even more perfectly random and perfectly free experimental situations is completely misguided as far as the inequalities are concerned.

Even worse, the problem described above is fatal. The only solution is to put Bell's theorem in the garbage bin where it belongs. See http://link.springer.com/article/10.100 ... 010-9461-z for a slightly different take of these ideas.

### Re: Evidence that QM does not violate Bell's inequalities

Heinera wrote:All local models wil have a bound of 2.

Not all local models. Here is a local model that has bound of $2\sqrt{2}$ : http://arxiv.org/pdf/1501.03393.pdf [see Eq. (26)].

### Re: Evidence that QM does not violate Bell's inequalities

FrediFizzx wrote:
Heinera wrote:The whole difference between a local and a non-local model is that for a local model the a in E11 must be the same as the a in E12, while in a non-local model this need not be the case, since Bob's different setting in E12 can give rise to a different outcome for a.

I think you found your problem. Or maybe you meant something different? To properly adhere to Bell-CHSH with a bound of 2, all models must have the same a in E11 and E12. Even the quantum experiments adhere to that.

There is no problem. My model does not respect the bound of 2, because of its non-locality. All local models wil have a bound of 2.

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