Evidence that QM does not violate Bell's inequalities

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

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

So Guest and Heinera are demonstrating selective amnesia or something similar. I will advice them to go read the first page of this thread and then answer the question I asked.

minkwe wrote:Here is the question: Please demonstrate that QM violates Bell's inequality
$|\langle A_iB_i\rangle - \langle A_iC_i\rangle| - \langle B_iC_i\rangle \leq 1$
by providing , the QM predictions for the terms
$\langle A_iB_i\rangle, \; \langle A_iC_i\rangle, \; \langle B_iC_i\rangle$
minkwe

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

minkwe wrote:So Guest and Heinera are demonstrating selective amnesia or something similar. I will advice them to go read the first page of this thread and then answer the question I asked.

minkwe wrote:Here is the question: Please demonstrate that QM violates Bell's inequality
$|\langle A_iB_i\rangle - \langle A_iC_i\rangle| - \langle B_iC_i\rangle \leq 1$
by providing , the QM predictions for the terms
$\langle A_iB_i\rangle, \; \langle A_iC_i\rangle, \; \langle B_iC_i\rangle$

I already told you in another thread: This expression only makes sense for LHV-models, and not for nonlocal models where the outcome depend on settings in both wings. Nor is this the inequality we are testing against in experiments. Since the expectation of a random subset of values from a population is the same as the population expectation, we can replace the inequality with

$|\langle A_jB_j\rangle - \langle A_kC_k\rangle| - \langle B_lC_l\rangle \leq 1$

This is the inequality we are testing against. Since the first inequality only makes sense for LHV-models, the second inequality only applies to LHV-models as well.
Heinera

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

Heinera wrote:I already told you in another thread: This expression only makes sense for LHV-models, and not for nonlocal models where the outcome depend on settings in both wings.

Then you mean that QM does not make a prediction for it. And therefore it is nonsense to say QM violates it. Good that you finally bring yourself to admit the fact that it is nonsense to say QM violates Bell's inequality.

Then let us move on to experiments.
Please demonstrate that Experiments violate Bell's inequality
$|\langle A_iB_i\rangle - \langle A_iC_i\rangle| - \langle B_iC_i\rangle \leq 1$
by providing , the empirical values for the terms
$\langle A_iB_i\rangle, \; \langle A_iC_i\rangle, \; \langle B_iC_i\rangle$

While doing that, include a reference to the article in which those measurements were made.

Heinera wrote:Nor is this the inequality we are testing against in experiments.

Let me get this straight, are you now admitting that experiments can't violate Bell's inequality either? If that is not your claim, please provide the empirical terms for Bell's inequality, not some invalid gobbledygook you hope to sneak in through faulty statistical reasoning.

heinera wrote:Since the expectation of a random subset of values from a population is the same as the population expectation, we can replace the inequality with

$|\langle A_jB_j\rangle - \langle A_kC_k\rangle| - \langle B_lC_l\rangle \leq 1$

This is the inequality we are testing against.

No! you cannot do that magicians bait-and-switch trick. Please provide a detailed step-by-step derivation. The terms in Bell's inequality do not represent statistically independent outcomes of a population. You can not replace them with values from random statistically independent outcomes with or without LHV. This is basic statistics and has nothing do with any physics. When will you learn this?? You did not even bother to read the first page of this thread where I demonstrate that your claims are completely false.

Here is the relevant part again. Please actually read it for once:

minkwe wrote:Bell's inequality is the following:
$|\langle A_iB_i\rangle - \langle A_iC_i\rangle| - \langle B_iC_i\rangle \leq 1$,
which makes use of the three terms $\langle A_iB_i\rangle,\; \langle A_iC_i\rangle,\; \langle B_iC_i\rangle$ all defined for the same set of outcomes $A_i, B_i, C_i$.

The terms from experiments are each calculated from 3 separate sets of outcomes: $A_i, B_i, A_j, C_j, B_k, C_k$ yielding the three terms $\langle A_iB_i\rangle,\; \langle A_jC_j\rangle,\; \langle B_kC_k\rangle$. As I have argued in the other thread, these terms do not have the same values as those in the original inequality, therefore we are not allowed to use the QM predictions for these terms in Bell's inequality as is customarily done to demonstrate violation by QM or Experiments. In other words, the common procedure of calculating
$|\langle A_iB_i\rangle - \langle A_jC_j\rangle| - \langle B_kC_k\rangle$,

from QM and experimental data to proclaim violation of Bell's inequality is a farce. One other thing I argued in the other threads is that, we could be allowed to use those terms if and only if we can demonstrate that they are statistically equivalent through rearrangements. This ensures that $A_i \equiv A_j, B_i \equiv B_k, C_j \equiv C_k$ must hold. Only under this condition are we allowed to use the QM prediction, and experimental results in Bell's inequality, and only under this condition do the original inequalities, derived on a single set, apply to data measured on disjoint sets. As I argued in the other thread, given 3 statistically independent sets of N outcome pairs, it is not possible to do such rearrangement in order to demonstrate the above equivalences. To summarize why, note that we could perform row permutations of the $A_j,C_j$ set of outcomes so that the $A_j$ column matches the $A_i$ column of the $A_i,B_i$ set. Then we could do row permutations of the $B_k,C_k$ set of outcomes so that the $B_k$ column matches the $B_i$ column of the $A_i,B_i$ set. One more rearrangement and we are home, since only the $C_k$ and $C_j$ pairs have not been matched. But, since only row permutations are allowed, any rearrangements to make $C_k$ match $C_j$ will undo the match between $B_k$ and $B_i$. Therefore, it is not possible for 3 statistically independent sets of outcome pairs to satisfy $A_i \equiv A_j, B_i \equiv B_k, C_j \equiv C_k$ simultaneously. Therefore the inequality does not apply, and the QM predictions for $|\langle A_iB_i\rangle - \langle A_jC_j\rangle| - \langle B_kC_k\rangle$ can not possibly be the same as the QM prediction for $|\langle A_iB_i\rangle - \langle A_iC_i\rangle| - \langle B_iC_i\rangle$. Therefore it is not surprising that QM or even LHV theories would have different predictions for those terms.
minkwe

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

minkwe wrote:
heinera wrote:Since the expectation of a random subset of values from a population is the same as the population expectation, we can replace the inequality with

$|\langle A_jB_j\rangle - \langle A_kC_k\rangle| - \langle B_lC_l\rangle \leq 1$

This is the inequality we are testing against.

No! you cannot do that magicians bait-and-switch trick. Please provide a detailed step-by-step derivation. The terms in Bell's inequality do not represent statistically independent outcomes of a population. You can not replace them with values from random statistically independent outcomes with or without LHV. This is basic statistics and has nothing do with any physics.

A random sampla drawn from a population with expectation $\mu$ also has expectation $\mu$. If I have a large set with mean $\mu$, and randomly divide it into 3 or 4 subsets, each subset vil have a mean very close to $\mu$. This is elementary probability theory. It is you who are invoking magic by believing that these means will miraculously be very different.
Heinera

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

Heinera wrote:
minkwe wrote:
heinera wrote:Since the expectation of a random subset of values from a population is the same as the population expectation, we can replace the inequality with

$|\langle A_jB_j\rangle - \langle A_kC_k\rangle| - \langle B_lC_l\rangle \leq 1$

This is the inequality we are testing against.

No! you cannot do that magicians bait-and-switch trick. Please provide a detailed step-by-step derivation. The terms in Bell's inequality do not represent statistically independent outcomes of a population. You can not replace them with values from random statistically independent outcomes with or without LHV. This is basic statistics and has nothing do with any physics.

A random sampla drawn from a population with expectation $\mu$ also has expectation $\mu$. If I have a large set with mean $\mu$, and randomly divide it into 3 or 4 subsets, each subset vil have a mean very close to $\mu$. This is elementary probability theory. It is you who are invoking magic by believing that these means will miraculously be very different.

There is no miracle. We we are talking about 3 non-independent simultaneous views of a single ensemble, not a single expectation. As long as you keep arguing from a single expectation $\mu$, and extending it to multiple random expectations, you will never understand your problem. Unfortunately, you are still are unable to read or understand the argument which I copied for your convenience in my previous post. Please read it carefully. Statistically independent sets of outcomes cannot at the same time be statistically dependent.

Is it too hard for you to see that the terms$\langle A_jB_j\rangle, \langle A_kC_k\rangle, \langle B_lC_l\rangle$ from your bogus inequality are statistically independent, the terms from Bell's inequality $\langle A_iB_i\rangle, \langle A_iC_i\rangle, \langle B_iC_i\rangle$ are not statistically independent. The argument which you continue to ignore demonstrates that you can not equate them like you are doing with bogus probability reasoning. I demonstrate conclusively that your assumption that they are equal is false.

Now where is the evidence that experiments violate Bell's inequality???

$\langle A_iB_i\rangle, \; \langle A_iC_i\rangle, \; \langle B_iC_i\rangle$
While doing that, include a reference to the article in which those measurements were made.
And make sure you are providing the empirical terms for Bell's inequality, not some invalid gobbledygook you hope to shift the goal post to, through bait and switch.
minkwe

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

minkwe wrote:
Heinera wrote:
minkwe wrote:Is it too hard for you to see that the terms$\langle A_jB_j\rangle, \langle A_kC_k\rangle, \langle B_lC_l\rangle$ from your bogus inequality are statistically independent,

They are no more or less statistically independent than the four subsets i described in my previous post. Only a fool would believe the four subsets would have very different means.
Heinera

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

huh? Have you even read the argument? From the first post, repeated in my previous post to you? Only an idiot will continue to make arguments which are obviously false, without making an effort to read the post which demonstrates this fact.
minkwe

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

Heinera wrote:
minkwe wrote:
Heinera wrote:

They are no more or less statistically independent than the four subsets i described in my previous post. Only a fool would believe the four subsets would have very different means.
minkwe wrote:huh? Have you even read the argument? From the first post, repeated in my previous post to you? Only an idiot will continue to make arguments which are obviously false, without making an effort to read the post which demonstrates this fact.

In $\langle A_jB_j\rangle, \langle A_kC_k\rangle, \langle B_lC_l\rangle$, the sets of j,k, and l are random disjoint subsets of the set of i in $\langle A_iB_i\rangle, \langle A_iC_i\rangle, \langle B_iC_i\rangle$.. Thus the terms will have approximately the same values in both inequalities for large N. This is truly elementary probability theory.
Heinera

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

minkwe wrote:Now where is the evidence that experiments violate Bell's inequality???

It is in the reports of those experiments. The reports tell what was done and what happened.

However, the topic of this discussion is not experiments. It is quantum mechanics instead. More specifically, whether quantum mechanics predicts violation of Bell's inequalities. Apparently none of those who might know is motivated to reveal it here. Elsewhere we may read that it does. Easy to check with a simulation.
Mikko

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

Mikko wrote:
minkwe wrote:Now where is the evidence that experiments violate Bell's inequality???

It is in the reports of those experiments. The reports tell what was done and what happened.

However, the topic of this discussion is not experiments. It is quantum mechanics instead. More specifically, whether quantum mechanics predicts violation of Bell's inequalities. Apparently none of those who might know is motivated to reveal it here. Elsewhere we may read that it does. Easy to check with a simulation.

The topic is about both QM and quantum experiments. Since the Bell inequalities are mathematically impossible to violate, the debate is really over. But go ahead and keep "shifting the goal posts" to show "violation". You are only fooling yourself.
FrediFizzx
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Re: Evidence that QM does not violate Bell's inequalities

Heinera wrote:I already told you in another thread: This expression only makes sense for LHV-models, and not for nonlocal models where the outcome depend on settings in both wings. Nor is this the inequality we are testing against in experiments. Since the expectation of a random subset of values from a population is the same as the population expectation, we can replace the inequality with

$|\langle A_jB_j\rangle - \langle A_kC_k\rangle| - \langle B_lC_l\rangle \leq 1$

This is the inequality we are testing against. Since the first inequality only makes sense for LHV-models, the second inequality only applies to LHV-models as well.

Nope. This is the inequality that you are testing against.

$|\langle A_jB_j\rangle - \langle A_kC_k\rangle| - \langle B_lC_l\rangle \leq 3$

Since the terms are independent.
FrediFizzx
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Re: Evidence that QM does not violate Bell's inequalities

FrediFizzx wrote:
Heinera wrote:I already told you in another thread: This expression only makes sense for LHV-models, and not for nonlocal models where the outcome depend on settings in both wings. Nor is this the inequality we are testing against in experiments. Since the expectation of a random subset of values from a population is the same as the population expectation, we can replace the inequality with

$|\langle A_jB_j\rangle - \langle A_kC_k\rangle| - \langle B_lC_l\rangle \leq 1$

This is the inequality we are testing against. Since the first inequality only makes sense for LHV-models, the second inequality only applies to LHV-models as well.

Nope. This is the inequality that you are testing against.

$|\langle A_jB_j\rangle - \langle A_kC_k\rangle| - \langle B_lC_l\rangle \leq 3$

Since the terms are independent.

For an LHV-model, the expectations are not independent. They are all expectations of disjoint subsets drawn randomly from a common population, and in this common population, the terms are not independent.
Heinera

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

Heinera wrote:They are all expectations of disjoint subsets drawn randomly from a common population ...

A complete and utter hogwash.
Joy Christian
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Re: Evidence that QM does not violate Bell's inequalities

Heinera wrote:For an LHV-model, the expectations are not independent.

According to Bell they are not independent. But that is the whole point. Bell was wrong. They are independent for an LHV model.
FrediFizzx
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Re: Evidence that QM does not violate Bell's inequalities

Joy Christian wrote:
Heinera wrote:They are all expectations of disjoint subsets drawn randomly from a common population ...

A complete and utter hogwash.

Because?
Heinera

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

Heinera wrote:
Joy Christian wrote:
Heinera wrote:They are all expectations of disjoint subsets drawn randomly from a common population ...

A complete and utter hogwash.

Because?

According to Bell they are not independent. But that is the whole point. Bell was wrong. They are independent for an LHV model.
FrediFizzx
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Re: Evidence that QM does not violate Bell's inequalities

Heinera wrote:In $\langle A_jB_j\rangle, \langle A_kC_k\rangle, \langle B_lC_l\rangle$, the sets of j,k, and l are random disjoint subsets of the set of i in $\langle A_iB_i\rangle, \langle A_iC_i\rangle, \langle B_iC_i\rangle$.. Thus the terms will have approximately the same values in both inequalities for large N. This is truly elementary probability theory.

Like I thought, you have no answer to the very clear arguments which I've provided since the very first post of this thread, debunking this nonsense. You think because you append "This is truly elementary probability theory" to a statement that has absolutely nothing to do with probability theory, it changes the fact that it is junk? You simply pretend the argument does not exist. Here it is again, for your enjoyment:

minkwe wrote:Bell's inequality is the following:
$|\langle A_iB_i\rangle - \langle A_iC_i\rangle| - \langle B_iC_i\rangle \leq 1$,
which makes use of the three terms $\langle A_iB_i\rangle,\; \langle A_iC_i\rangle,\; \langle B_iC_i\rangle$ all defined for the same set of outcomes $A_i, B_i, C_i$.

The terms from experiments are each calculated from 3 separate sets of outcomes: $A_i, B_i, A_j, C_j, B_k, C_k$ yielding the three terms $\langle A_iB_i\rangle,\; \langle A_jC_j\rangle,\; \langle B_kC_k\rangle$. As I have argued in the other thread, these terms do not have the same values as those in the original inequality, therefore we are not allowed to use the QM predictions for these terms in Bell's inequality as is customarily done to demonstrate violation by QM or Experiments. In other words, the common procedure of calculating
$|\langle A_iB_i\rangle - \langle A_jC_j\rangle| - \langle B_kC_k\rangle$,

from QM and experimental data to proclaim violation of Bell's inequality is a farce. One other thing I argued in the other threads is that, we could be allowed to use those terms if and only if we can demonstrate that they are statistically equivalent through rearrangements. This ensures that $A_i \equiv A_j, B_i \equiv B_k, C_j \equiv C_k$ must hold. Only under this condition are we allowed to use the QM prediction, and experimental results in Bell's inequality, and only under this condition do the original inequalities, derived on a single set, apply to data measured on disjoint sets. As I argued in the other thread, given 3 statistically independent sets of N outcome pairs, it is not possible to do such rearrangement in order to demonstrate the above equivalences. To summarize why, note that we could perform row permutations of the $A_j,C_j$ set of outcomes so that the $A_j$ column matches the $A_i$ column of the $A_i,B_i$ set. Then we could do row permutations of the $B_k,C_k$ set of outcomes so that the $B_k$ column matches the $B_i$ column of the $A_i,B_i$ set. One more rearrangement and we are home, since only the $C_k$ and $C_j$ pairs have not been matched. But, since only row permutations are allowed, any rearrangements to make $C_k$ match $C_j$ will undo the match between $B_k$ and $B_i$. Therefore, it is not possible for 3 statistically independent sets of outcome pairs to satisfy $A_i \equiv A_j, B_i \equiv B_k, C_j \equiv C_k$ simultaneously. Therefore the inequality does not apply, and the QM predictions for $|\langle A_iB_i\rangle - \langle A_jC_j\rangle| - \langle B_kC_k\rangle$ can not possibly be the same as the QM prediction for $|\langle A_iB_i\rangle - \langle A_iC_i\rangle| - \langle B_iC_i\rangle$. Therefore it is not surprising that QM or even LHV theories would have different predictions for those terms.

minkwe

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

Heinera wrote:
Joy Christian wrote:
Heinera wrote:They are all expectations of disjoint subsets drawn randomly from a common population ...

A complete and utter hogwash.

Because?

Because there is no subset drawn from a common population. To claim there is, is to fail to understand what "counterfactually possible" means in the present context.
Joy Christian
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Re: Evidence that QM does not violate Bell's inequalities

minkwe wrote:
Heinera wrote:In $\langle A_jB_j\rangle, \langle A_kC_k\rangle, \langle B_lC_l\rangle$, the sets of j,k, and l are random disjoint subsets of the set of i in $\langle A_iB_i\rangle, \langle A_iC_i\rangle, \langle B_iC_i\rangle$.. Thus the terms will have approximately the same values in both inequalities for large N. This is truly elementary probability theory.

Like I thought, you have no answer to the very clear arguments which I've provided since the very first post of this thread, debunking this nonsense. You think because you append "This is truly elementary probability theory" to a statement that has absolutely nothing to do with probability theory, it changes the fact that it is junk? You simply pretend the argument does not exist. Here it is again, for your enjoyment:

Your "clear arguments" was thoroughly debunked by Jochen. No need to repeat that (use the search function in this forum instead) If you think statistics and probability theory is nonsense, or that it does not apply in this case, then so be it. Just stay away from the casinos, or you'll soon find yourself to be a very poor man.
Heinera

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

Heinera wrote:
minkwe wrote:
Heinera wrote:In $\langle A_jB_j\rangle, \langle A_kC_k\rangle, \langle B_lC_l\rangle$, the sets of j,k, and l are random disjoint subsets of the set of i in $\langle A_iB_i\rangle, \langle A_iC_i\rangle, \langle B_iC_i\rangle$.. Thus the terms will have approximately the same values in both inequalities for large N. This is truly elementary probability theory.

Like I thought, you have no answer to the very clear arguments which I've provided since the very first post of this thread, debunking this nonsense. You think because you append "This is truly elementary probability theory" to a statement that has absolutely nothing to do with probability theory, it changes the fact that it is junk? You simply pretend the argument does not exist. Here it is again, for your enjoyment:

Your "clear arguments" was thoroughly debunked by Jochen. No need to repeat that (use the search function in this forum instead) If you think statistics and probability theory is nonsense, or that it does not apply in this case, then so be it. Just stay away from the casinos, or you'll soon find yourself to be a very poor man.

LOL! Here we go again... The fact remains that Bell was wrong. LHV models can have independent expectation terms. Another fact is that it is mathematically impossible to "violate" the inequalities. There is no getting around that one.
FrediFizzx
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