Quantum Mechanics with HV is Local!

[Heinera]

And what I am focused on at present is not just any old model, but quantum mechanics itself, and whether QM, properly developed and understood, might in fact itself come to be seen as a local, realistic, and complete hidden variables theory.
Jay

The standard formulation of QM is already local in the sense of Einstein's special relativity, because there is no possibility of super-luminal communication in QM (the no-communication theorem). Even HV-formulations of QM (like Bohmian mechanics) is local in this sense.

Locality in the sense of Bell is different from locality in the sense of special relativity. Many have problems with grasping this difference.

[Heinera]

For given values of a and , the RHS of (8) and (9) looks completely deterministic to me. Where is the extra source of randomness that can produce both +1 and -1 for the same values of a and ?

You were right that extra randomness. We took that out because it can give AB = +1 instead of AB = -1 when a = b.

OK, so now there is no extra source of randomness in and , and the only randomness comes from the (random) binary valued ? Then my original argument about 2 vs 4 combinations still stands.

[Joy Christian]

For given values of a and , the RHS of (8) and (9) looks completely deterministic to me. Where is the extra source of randomness that can produce both +1 and -1 for the same values of a and ?

You were right that extra randomness. We took that out because it can give AB = +1 instead of AB = -1 when a = b.

OK, so now there is no extra source of randomness in and , and the only randomness comes from the (random) binary valued ? Then my original argument about 2 vs 4 combinations still stands.

Your "argument" has no leg to "stand." Did you read my earlier reply to you? Or are you determined to ignore my reply because it is too detrimental to your belief system?

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[Joy Christian]

The standard formulation of QM is already local in the sense of Einstein's special relativity, because there is no possibility of super-luminal communication in QM (the no-communication theorem).

This is correct. Orthodox quantum mechanics harbors no-signaling non-locality only, and therefore it remains compatible with special relativity.

Even HV-formulations of QM (like Bohmian mechanics) is local in this sense.

This is only partly correct. Bohmian mechanics violates "parameter independence" and therefore it is not compatible with special-relativistic locality.

Locality in the sense of Bell is different from locality in the sense of special relativity.

This is essentially correct. Locality in the sense of Bell can be split up into two distinct parts: (1) parameter independence, and (2) outcome independence. It is very easy to understand these to senses of locality in terms of Bell's measurement functions, which are defined as A(a, h) and B(b, h), where a and b are freely chosen experimental parameters and the "hidden variable" h is a randomness shared between Alice and Bob. Then "parameter independence" is the fact that A(a, h) is independent of b, and likewise B(b, h) is independent of a. A violation of parameter independence thus violates special-relativistic locality. On the other hand, "outcome independence" is the fact that A(a, h) is independent of the outcome B --- and likewise B(b, h) is independent of the outcome A. Consequently, violation of "outcome independence" would be a violation of non-signaling locality that remains compatible with special relativity.

Nota bene: Orthodox quantum mechanics violates "outcome independence" but preserves "parameter independence" (thereby remaining compatible with special relativity).

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[Heinera]

Did you read my earlier reply to you? Or are you determined to ignore my reply because it is too detrimental to your belief system?

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Your earlier reply (in another thread for this paper; don't understand why you created a new one) was of course that there was a separate randomness in the functions and because s_1 could be both +a and -a. Quote:

If lambda = + 1 and the limit s_1 --> +a then A(a, lambda) = -1
If lambda = + 1 and the limit s_1 --> -a then A(a, lambda) = +1
If lambda = - 1 and the limit s_1 --> +a then A(a, lambda) = -1
If lambda = - 1 and the limit s_1 --> -a then A(a, lambda) = +1

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This is clearly no longer the case in your new version of the paper, so is now a deterministic function.

[FrediFizzx]

Did you read my earlier reply to you? Or are you determined to ignore my reply because it is too detrimental to your belief system?

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Joy said that not me.
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[Heinera]

Did you read my earlier reply to you? Or are you determined to ignore my reply because it is too detrimental to your belief system?

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Joy said that not me.
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Sorry, my fault. Sometimes these threads can get nested in a complicated way.

[FrediFizzx]

Did you read my earlier reply to you? Or are you determined to ignore my reply because it is too detrimental to your belief system?

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Joy said that not me.
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Sorry, my fault. Sometimes these threads can get nested in a complicated way.

No problem. Your comments are actually helpful. We are working on sorting things out now.
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[Joy Christian]

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Why do you guys keep ignoring my replies? They are right there. There is no problem to be solved. All you have to do is read my replies above.

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[gill1109]

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As I noted above, all inequalities are red herring. They are a worthless distraction from the actual physics and from what can be observed in the actual experiments.

The only physical question of interest in the Einstein-Bell debate is this: Given a pair of measurement functions A(a, h) and B(b, h) defined by Bell, where a and b are freely chosen experimental parameters and h is a shared randomness between Alice and Bob, can the average of their product equal to -a.b? In other words, can the following equality hold?

E(a, b) = 1/n Sum_k A(a, h_k) B(b, h_k) = -a.b

The answer is: Yes, it can. And the paper being discussed in this thread demonstrates how. A purely geometrical version of the model can be found here: https://arxiv.org/abs/1103.1879.

First of all, some further requirements need to be mentioned. (1) The functions A and B have to take values in the set of just two points {-1, +1}. (2), one needs to take a limit as n tends to infinity. (3) one requires that the equality in question holds for all a and all b simultaneously.

It seems that Christian is defining probability in the von Mises way as a limiting relative frequency.

As was mentioned earlier, if the h_k themselves also take values in the set {-1, +1}, and do that with equal probabilities (i.e., with limiting relative frequencies) 1/2, 1/2, then E(a, b) must be equal to -1, 0 or +1. This is because For half of the values of k, takes on one value, and for the remaining half it takes on another value (or the same value again).

[Joy Christian]

It seems that Christian is defining probability in the von Mises way as a limiting relative frequency.

Let alone defining, I haven't even mentioned "probability." Probability and statistics are perfect tools for obfuscating a simple computation of an average of the product of numbers, +/-1:

E(a, b) = 1/n Sum_k A(a, h_k) B(b, h_k) = -a.b.

Bell himself was guilty of using probability and statistics to obfuscate issues that hide an obvious physical problem in the proof of his so-called theorem: https://arxiv.org/abs/1704.02876.

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[gill1109]

It seems that Christian is defining probability in the von Mises way as a limiting relative frequency.

Let alone defining, I haven't even mentioned "probability." Probability and statistics are perfect tools for obfuscating a simple computation of an average of the product of numbers, +/-1:

E(a, b) = 1/n Sum_k A(a, h_k) B(b, h_k) = -a.b.

Bell himself was guilty of using probability and statistics to obfuscate issues that hide an obvious physical problem in the proof of his so-called theorem: https://arxiv.org/abs/1704.02876.

Indeed, you didn't even mention probability. And as far as I know, *nobody* (except you) ever wrote . So what are you talking about?

[FrediFizzx]

It seems that Christian is defining probability in the von Mises way as a limiting relative frequency.

Let alone defining, I haven't even mentioned "probability." Probability and statistics are perfect tools for obfuscating a simple computation of an average of the product of numbers, +/-1:

E(a, b) = 1/n Sum_k A(a, h_k) B(b, h_k) = -a.b.

Bell himself was guilty of using probability and statistics to obfuscate issues that hide an obvious physical problem in the proof of his so-called theorem: https://arxiv.org/abs/1704.02876.

Indeed, you didn't even mention probability. And as far as I know, *nobody* (except you) ever wrote . So what are you talking about?

Joy is talking about an average of a product of two functions that each produce +/- 1.



It is pretty simple.
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[Heinera]

No problem. Your comments are actually helpful. We are working on sorting things out now.
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Good. First step would be to introduce a hidden variable that can take on 4 different values, since that is obviously needed to produce the four different combinations of outcomes that QM predicts for a fixed pair of detector settings.

(of course, Bell's theorem says that will still not help. Even an infinite number of values won't help. But let's take that another day).

[Joy Christian]

No problem. Your comments are actually helpful. We are working on sorting things out now.
.

Good. First step would be to introduce a hidden variable that can take on 4 different values, since that is obviously needed to produce the four different combinations of outcomes that QM predicts for a fixed pair of detector settings.

(of course, Bell's theorem says that will still not help. Even an infinite number of values won't help. But let's take that another day).

In the model, the product AB can take both values, +1 and -1. If it didn't, then the correlation E(a, b) would not work out to be -a.b. But it does, as derived and proved in eqs. (12) to (19).

So the issues you have been raising are irrelevant.

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[Heinera]

In the model, the product AB can take both values, +1 and -1.
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Of course it can. But QM predicts more than that. So that is not sufficient.

[Joy Christian]

In the model, the product AB can take both values, +1 and -1.
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Of course it can. But QM predicts more than that. So that is not sufficient.

Quantum mechanics does not predict more than that. Quantum mechanics does not predict individual outcomes at all. Nor are individual outcomes observed in experiments. Only statistical averages of outcomes and the correlations -a.b between the outcomes are observed in the experiments.

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[Heinera]

In the model, the product AB can take both values, +1 and -1.
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Of course it can. But QM predicts more than that. So that is not sufficient.

Quantum mechanics does not predict more than that. Quantum mechanics does not predict individual outcomes at all. Nor are individual outcomes observed in experiments.

Huh? Of course individual outcomes are observed in experiments..

Only statistical averages of outcomes and the correlations -a.b between the outcomes are observed in the experiments.
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No, both QM predictions and experimental evidence says that all four combinations (+1,+1), (-1,-1). (+1,-1), and (-1,-1) will be observed for any fixed pair of settings (with exception of equal or exactly opposite settings). These are observables, and not correlations.

[Joy Christian]

Huh? Of course individual outcomes are observed in experiments..

Provide a reference of an experimental paper proving that individual outcomes like A = +1 or B = -1 are observed in an EPRB type experiment. You won't be able to.

No, both QM predictions and experimental evidence says that all four combinations (+1,+1), (-1,-1). (+1,-1), and (-1,-1) will be observed for any fixed pair of settings (with exception of equal or exactly opposite settings). These are observables, and not correlations.

That is incorrect. Only coincidences between A and B are observed in the experiments, amounting to measuring their product AB = either +1 or -1.

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[jreed]

Huh? Of course individual outcomes are observed in experiments..

Provide a reference of an experimental paper proving that individual outcomes like A = +1 or B = -1 are observed in an EPRB type experiment. You won't be able to.

No, both QM predictions and experimental evidence says that all four combinations (+1,+1), (-1,-1). (+1,-1), and (-1,-1) will be observed for any fixed pair of settings (with exception of equal or exactly opposite settings). These are observables, and not correlations.

That is incorrect. Only coincidences between A and B are observed in the experiments, amounting to measuring their product AB = either +1 or -1.

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Here's what Gregor Weihs' Phys. Rev paper on his experiment says:

7. COINCIDENCE EVALUATION
After a measurement run was completed, either for a certain time or for a maximal
number of data points on each side, the data were written to the computers’ hard drives
on each side. Anyone could then later examine the data and draw their own conclusions.
We decided to take the files and extract the photon time-tags for which there was a
coincident detection on the other side.

Notice that there are two computers, one for Alice and one for Bob. They each recorded individual measurements for Alice and Bob. Coincidences are worked out later by comparing the data on these two computers.