Binney's interpretation of Violation of Bell Inequalities

[stevendaryl]

Of formalizing the assumption that a system(the universe, whatever) is in a certain state at a certain time, this is QM's first postulate, and I'd say it covers the assumption you mentioned above as the starting point of Bell's theorem.

Well, I disagree. I don't think that the QM state can be interpreted as the "state of the universe at a certain time", precisely because of scenarios such as EPR. Initially, you have the state in a superposition $|\psi\rangle = \frac{1}{\sqrt{2}}(|U D\rangle - |D U \rangle)$. Alice measures the spin of the first particle and finds it spin-up. So what is the state of the system now? If you say that it collapses to $|U D\rangle$, well that's a nonlocal change. If you say that it doesn't collapse, then you have Many Worlds.

 

[Derek Potter]

Well, I disagree. I don't think that the QM state can be interpreted as the "state of the universe at a certain time", precisely because of scenarios such as EPR. Initially, you have the state in a superposition $|\psi\rangle = \frac{1}{\sqrt{2}}(|U D\rangle - |D U \rangle)$. Alice measures the spin of the first particle and finds it spin-up. So what is the state of the system now? If you say that it collapses to $|U D\rangle$, well that's a nonlocal change. If you say that it doesn't collapse, then you have Many Worlds.

Many Worlds as popularized depicts it as parallel universes which certainly makes THE state of THE universe pretty meaningless. But in the improved version of Relative State Formulation there remains just one universe in a pure state. The pure state can, of course be decomposed into a superposition of orthogonal observer-worlds. EPR is explained locally in such a model.

 

[Derek Potter]

The classical idea of causality in SR is that the causes of an event are entirely in its past light cone - that idea is not preserved in QFT. What is preserved in QFT is a weaker sense of causality - that classical information cannot be sent faster than light.

Can QFT can be reformulated in MW terms so that it remains local and SR-causal?

 

[TrickyDicky]

Well, I disagree. I don't think that the QM state can be interpreted as the "state of the universe at a certain time",

I said state of the system, it's anyone's choice what that system is.

precisely because of scenarios such as EPR. Initially, you have the state in a superposition $|\psi\rangle = \frac{1}{\sqrt{2}}(|U D\rangle - |D U \rangle)$. Alice measures the spin of the first particle and finds it spin-up. So what is the state of the system now? If you say that it collapses to $|U D\rangle$, well that's a nonlocal change.

You are calling this collapse nonlocal, I prefer to call it simply non-unitary, because as I explained in #46 its consequences on entangled two-states systems can be described without any reference to locality, reality or causality. And yes this is in contradiction with the rest of the formalism as everyone knows.
Also note that the original EPR involved entanglement of continuous observables that have been shown(see Popper's experiment) not to exhibit correlations contrary to what happens in the two-state systems like spin or polarization that Bell's theorem exemplify.

 

[Pat71]

Also note that the original EPR involved entanglement of continuous observables that have been shown(see Popper's experiment) not to exhibit correlations contrary to what happens in the two-state systems like spin or polarization that Bell's theorem exemplify.

Doesn't Strekalov's "Ghost Interference" experiment of 1995 counter this claim? PRL74p3600.pdf

"...Another experiment which (unknowingly) implements Popper's test in a conclusive way, has actually been carried out. Its results are in contradiction with Popper's prediction..." (Tabish Qureshi, "Analysis of Popper's experiment and its realization," 2012): ptp.oxfordjournals.org/content/127/4/645.short

Or, similarly: www.arxiv.org/abs/1206.1432

 

[atyy]

Can QFT can be reformulated in MW terms so that it remains local and SR-causal?

I don't know. However, the Many-Worlds interpretation certainly escapes being nonlocal via the Bell theorem, because the Bell theorem assumes that only one outcome is realized. That is among the well-known loopholes including superdeterminism and retrocausation, which I usually don't mention just to be concise.

 

[TrickyDicky]

Doesn't Strekalov's "Ghost Interference" experiment of 1995 counter this claim? PRL74p3600.pdf

"...Another experiment which (unknowingly) implements Popper's test in a conclusive way, has actually been carried out. Its results are in contradiction with Popper's prediction..." (Tabish Qureshi, "Analysis of Popper's experiment and its realization," 2012): ptp.oxfordjournals.org/content/127/4/645.short

Or, similarly: www.arxiv.org/abs/1206.1432

No, it counters Popper's original claim that QM's indeterminacy(HUP) was falsified by a certain outcome of his thought experiment. Both Shih's 1999 and more recent 2015(see http://phys.org/news/2015-01-popper-againbut.html) papers agreed on this too.

 

[TrickyDicky]

It's not clear to me either, and I'm interested in all possible alternatives to explaining EPR. If anything, you're quite laconic, can you point to a paper which shares your views? Or if you really have a personal new idea, but maybe don't have enough skills to fully expand on it, at least lay down a formal pointer to your interpretation in which noncommutative algebra in some way eliminates the ftl action problem?

Edited scenario as the initial description allowed confusion

That correlations of two-state systems is unrelated to locality considerations can be easily shown with this example: Take a book and call the cover the +> state and the back cover the -> state when the book is oriented parallel to the x axis, and similarly when it is oriented parallel to the y and z axes, now you can truly claim that after an arbitrary 3-dimensional rotation of the book(involving rotations in different axes) the +> and -> states when measured in the diffrent orientations for x, and z axes are correlated in a way that violates Bell's inequalities. No so called "local realistic hidden variables theory" can explain the correlation of the cover and the back of the book. Spooky?

 

[Derek Potter]

the up> and down> states are correlated in a way that violates Bell's inequalities. No so called "local realistic hidden variables theory" can explain the correlation of the cover and the back of the book. Spooky?

Neither spooky nor true. 100% anti-correlation does not violate the Bell inequality.

 

[stevendaryl]

You are calling this collapse nonlocal, I prefer to call it simply non-unitary, because as I explained in #46 its consequences on entangled two-states systems can be described without any reference to locality, reality or causality.

If the state collapses for Bob as well as for Alice when Alice performs her measurements, then the collapse is nonlocal, affecting distant parts of space simultaneously. If the state only collapses for Alice, then it can't be a state of the universe, it must be relative to the observer.

 

[stevendaryl]

That correlations of two-state systems is unrelated to locality considerations can be easily shown with this example: Take a book and call the cover the up> state and the back cover the down> state, now you can truly claim that after an arbitrary 3-dimensional rotation of the book(involving rotations in different axes) the up> and down> states are correlated in a way that violates Bell's inequalities. No so called "local realistic hidden variables theory" can explain the correlation of the cover and the back of the book. Spooky?

What you're describing IS a local realistic hidden variables theory, and it doesn't violate Bell's inequalities.

 

[Derek Potter]

Hardly hidden !

 

[stevendaryl]

What you're describing IS a local realistic hidden variables theory, and it doesn't violate Bell's inequalities.

To expand on this, Bell's inequalities involve two adjustable parameters, $\alpha$ and $\beta$, which can take on values $a_1, a_2, b_1, b_2$ and two possible measurements $A$ and $B$, each of which produces a result $\pm 1$. You perform measurements at a variety of settings in order to come up with 4 numbers:

1. $E(a_1, b_1)$
2. $E(a_1, b_2)$
3. $E(a_2, b_1)$
4. $E(a_2, b_2)$

where $E(a,b)$ is the average value of the product $AB$ of measurement results, restricted to those measurements for which $\alpha = a$ and $\beta = b$. Locality comes into play when you have a situation where the settings $\alpha$ and $\beta$ that can be made independently at spacelike separations, and where the measurements $A$ and $B$ also take place at spacelike separations. In that situation, Bell showed that a local realistic theory predicts a certain inequality holding among the 4 numbers.

In your (Tricky Dicky's) toy model, it's clear that you are interpreting the measurement as seeing the front versus back of the book. But to apply Bell's inequality to the model, you have to two independent parameters $\alpha$ and $\beta$ that can be chosen at a spacelike separation. Presumably these are supposed to be the rotations of the book along two different axes? But that can't be done independently, at a spacelike separation.

So I don't see how your toy model relates to Bell's inequality at all.

 

[TrickyDicky]

Neither spooky nor true. 100% anti-correlation does not violate the Bell inequality.

See edited scenario, there is no 100% anti-correlation. The example just shows that rotations in 3D are non-commutative and the Bell inequalities follow a commutative algebra so they must be logically violated by non-commutative rotations.

 

[TrickyDicky]

In your (Tricky Dicky's) toy model, it's clear that you are interpreting the measurement as seeing the front versus back of the book. But to apply Bell's inequality to the model, you have to two independent parameters $\alpha$ and $\beta$ that can be chosen at a spacelike separation. Presumably these are supposed to be the rotations of the book along two different axes? But that can't be done independently, at a spacelike separation.

So I don't see how your toy model relates to Bell's inequality at all.

The math of the theorem just refers to outcomes and whether their correlation can be explained with the inequalities or not with Bell inequalities.
Would you see how the toy model relates to Bell if I allowed for the book to have arbitrary spacelike thickness(ignoring problems with born rigidity for the sake of the thought experiment as Bell's theorem doesn't have it as assumption)?

 

[stevendaryl]

There is no 100% anti-correlation, of course one has to allow to you can correlate the cover in two different rotated axes
See edited scenario, there is no 100% anti-correlation. The example just shows that rotations in 3D are non-commutative and the Bell inequalities follow a commutative algebra so they must be logically violated by non-commutative rotations.

As I said, Bell's inequality was developed for a scenario in which there are two adjustable parameters $\alpha$ and $\beta$ that can be made independently at a spacelike separation, and two measurements $A$ and $B$ that can also be made at a spacelike separation. I don't see how your toy model has anything to do with Bell's inequality. What are the adjustable parameters? If they are the choices of rotation angles along two different axes, then they certainly cannot be made independently at a spacelike separation.

I also don't know what you mean by saying that "Bell's inequalities follow a commutative algebra". I don't see that there is algebra involved, at all. Where in the derivation of his inequality is there an assumption made about commutativity?

 

[TrickyDicky]

I edited #84 and answered in #85 while you posted.

 

[stevendaryl]

The math of the theorem just refers to outcomes and whether their correlation can be explained with the inequalities or not with Bell inequalities.
Would you see how the toy model relates to Bell if I allowed for the book to have arbitrary spacelike thickness(ignoring problems with born rigidity for the sake of the thought experiment as Bell's theorem doesn't have it as assumption)?

I could see how it relates to Bell if you could point out two adjustable parameters that can be made independently at a spacelike separation.

His inequalities are about a situation where you have two observers, far from each other, traditionally called "Alice" and "Bob" (by me, anyway).

1. Alice chooses a detector setting, either $a_1$ or $a_2$
2. Alice performs a measurement, getting result $A = +1$ or $A = -1$
3. Meanwhile, far away, Bob chooses a detector setting, either $b_1$ or $b_2$
4. He gets a measurement result $B = \pm 1$

You perform the experiment many times, and get statistics for $E(a,b)$,which is the average of the product $AB$ of the two results, restricted to those trials where Alice chooses $a$ and Bob chooses $b$

So, if you want to relate your toy model to Bell, you need to say what are the settings that Alice and Bob choose. In order to say that your toy model violates Bell's inequality, you need to compute the values $E(a,b)$ for your model.

 

[TrickyDicky]

As I said, Bell's inequality was developed for a scenario in which there are two adjustable parameters $\alpha$ and $\beta$ that can be made independently at a spacelike separation, and two measurements $A$ and $B$ that can also be made at a spacelike separation. I don't see how your toy model has anything to do with Bell's inequality. What are the adjustable parameters? If they are the choices of rotation angles along two different axes, then they certainly cannot be made independently at a spacelike separation.

I also don't know what you mean by saying that "Bell's inequalities follow a commutative algebra". I don't see that there is algebra involved, at all. Where in the derivation of his inequality is there an assumption made about commutativity?

Axes x, y and z are spacelike so measurement referred to them are independent of spacelike separation.
The Bell's inequalities imply that + and - outcomes are related by commutative relations.

 

[TrickyDicky]

These are the inequalities:Number of <+ <+ results with apparatus A on x and apparatus B on y ≤ number of <+ <+ results with apparatus A on x and apparatus B on z plus number of <+< - result of apparatus A on y and apparatus B on z.

I call apparatus A and B Alice and Bob respectively simultaneously ascertaining if they are seeing the back or the cover of the book.

 

[DrChinese]

You make it sound like in that experiment it's arbitrary two photons that are entangled. But it's not. These two photons have to be similar enough to produce Hong-Ou Mandel interference. And in that particular experiment that you linked both photon pairs are produced from the same pulse of the same pump laser. And then of course there are four types of entanglemet that you can get. So there is quite a number of loopholes in your argument.

I am sure you are familiar with experiments such as the following but I point it out for those that are not:

http://arxiv.org/abs/1209.4191
Entanglement Between Photons that have Never Coexisted

http://arxiv.org/abs/0809.3991
High-fidelity entanglement swapping with fully independent sources

The pairs do not need to be produced from the same laser (regardless of the particular experiment I referenced earlier). They need only be phase locked together. It doesn't matter how many types of entanglement are produced, you know which one to expect. In fact you can can choose to cause it if you like.

My point is that a lot of folks picture a situation where the photons start from a common source so "naturally" they exhibit correlations - there is a common cause. But these pairs of photons have NO common point of overlap since they are never in a common light cone. And they do not exhibit the perfect correlations unless they are entangled, something which can be done anywhere and at anytime before or after detection.

 

[stevendaryl]

Axes x, y and z are spacelike so measurement referred to them are independent of spacelike separation.
The Bell's inequalities imply that + and - outcomes are related by commutative relations.

Huh? I was talking about settings being chosen at a spacelike separation.

Let's get concrete. Alice is at one location. 1/4 of a mile away, Bob is at another location. Alice chooses her setting $\alpha$. She makes a measurement (which presumably means she looks at the book and sees whether the front side or back side is visible to her).

Bob chooses a setting $\beta$. He makes an observation.

What are the settings $\alpha$ and $\beta$? If they are angles of rotation of a normal-sized book, then it's not possible for Alice and Bob to both make their settings if they are far apart. The book is only in one place.

So that's exactly what Bell's inequalities have to do with locality. You can't get a violation of Bell's inequality using a toy classical model, provided that the settings for the two observers are chosen at a spacelike separation. If they aren't chosen at a spacelike separation, then you can get a model that violates Bell's inequalities. For example, in your book example, if Bob chooses an angle to rotate the book about some axis, and Alice chooses an angle, and they both broadcast their angles to the person holding the book, who carries out their commands, then the resulting correlations may very well violate Bell's inequalities.

 

[TrickyDicky]

Huh? I was talking about settings being chosen at a spacelike separation.

Let's get concrete. Alice is at one location. 1/4 of a mile away, Bob is at another location. Alice chooses her setting $\alpha$. She makes a measurement (which presumably means she looks at the book and sees whether the front side or back side is visible to her).

Bob chooses a setting $\beta$. He makes an observation.

What are the settings $\alpha$ and $\beta$? If they are angles of rotation of a normal-sized book, then it's not possible for Alice and Bob to both make their settings if they are far apart. The book is only in one place.

So that's exactly what Bell's inequalities have to do with locality. You can't get a violation of Bell's inequality using a toy classical model, provided that the settings for the two observers are chosen at a spacelike separation. If they aren't chosen at a spacelike separation, then you can get a model that violates Bell's inequalities. For example, in your book example, if Bob chooses an angle to rotate the book about some axis, and Alice chooses an angle, and they both broadcast their angles to the person holding the book, who carries out their commands, then the resulting correlations may very well violate Bell's inequalities.

But can't you get the same effect as spacelike separation(since the spacelike separation reflects the demand of impossibility of info transmission between Bob and Alice) by taking sequential 90º rotations of the book and two machines(one for each side of the book and for each sequential rotated angle) that register the information about cover position from their side encrypted so that no information can be shared between the machines?

My claim would be that after collecting all the info registered by the two machines once it is put together and decoded, the results obtained would not respect the inequality in #90.

With the following correspondences: in the z axis the cover in position up is +>, and in position down is ->, for the y-axis the cover at the right position is +> and at the left position is ->, and for the x-axis the cover at the fron is +> and at the back is ->.

 

[Derek Potter]

But can't you get the same effect as spacelike separation(since the spacelike separation reflects the demand of impossibility of info transmission between Bob and Alice) by taking sequential 90º rotations of the book and two machines(one for each side of the book and for each sequential rotated angle) that registers the information about cover position from their side encrypted so that no information can be shared between the machines?

No. Alice's information about turning her side passes through the book and turns Bob's side too. To simulate blocking that information flow, you need to break the Born rigidity of the book by allowing the sides to rotate separately. Good luck patching your toy model to allow that.

 

[TrickyDicky]

Alice's information about turning her side passes through the book and turns Bob's side too.

True, but is that information about angle relevant? In the standard setting with particles the angles in which Bob and Alice measure spin of each particle can be planned beforehand, no?

 

[Derek Potter]

In the standard setting with particles the angles in which Bob and Alice measure spin of each particle can be planned beforehand, no?

Yes they can be planned ahead and the Bell Inequality will still be violated. But that is not in the least bit significant because the required conditions for Bell's Theorem are not met.

True, but is that information about angle relevant?

Yes, the angle information is crucial - though I should have said rotation, not angle. My bad. It is crucial because without it, Alice can turn the book from z+ to z-. But the back still faces z-, meaning that the book now has two back covers and no front one.

 

[Derek Potter]

I feel that maybe I am not understanding your proposed toy model. If so, may I ask you to spell it out in detail? Otherwise the thread will drag on for ever, picking over details, modifying the toy model and generally losing track of the point.

In particular it would help if you say exactly what it is you are trying to illustrate. Originally you were saying that EPR correlations can be explained by the non-commutivity of Alice and Bob's observations. The trade-off between entanglement and non-commutivity is dealt with in the paper by Michael Seevinck and Jos Uffink but for your statement to be relevant you would have to say why the non-commutivity of separated observations is a local explanation i.e. you need to explain how the non-commutivity of remote observations arises under assumptions of local causality. Otherwise you are just saying that it happens, which we know, and can be formulated in different ways, which is interesting but does not explain anything. The onus is on you to justify the claim that it does explain rather than simply describe.

Hopefully you can answer that quicker than it took to ask it

Anyway, if we can get that cleared up, I really would like to know exactly what the model is: Who rotates the book, i.e. from what place is it rotated? What rotations are allowed? What exactly are the observations, and in particular what happens to rotations that end up with the book edgewise-on? How is spacelike separation simulated without introducing unintended Born rigidity that would violate relativity if the separation was truly spacelike? Pre-arrangement might circumvent the rigidity but then the model is not local in the sense required by Bell's Theorem.

If there are no fatal flaws in the model, can you then demonstrate the violation of the Bell Inequality numerically - as I asked you before - please?

I'm not trying to put you to a lot of work, I just want to get it clear what you're talking about.

Thanks.

 

[TrickyDicky]

In particular it would help if you say exactly what it is you are trying to illustrate. Originally you were saying that EPR correlations can be explained by the non-commutivity of Alice and Bob's observations. The trade-off between entanglement and non-commutivity is dealt with in the paper by Michael Seevinck and Jos Uffink but for your statement to be relevant you would have to say why the non-commutivity of separated observations is a local explanation i.e. you need to explain how the non-commutivity of remote observations arises under assumptions of local causality. Otherwise you are just saying that it happens, which we know, and can be formulated in different ways, which is interesting but does not explain anything. The onus is on you to justify the claim that it does explain rather than simply describe.

Maybe I can only claim that it describes but the "book toy model" was intended to explain that if the property of non-commutativity can be made geometrical like it happens with euclidean rotations, it could help understand how one can dodge the locality-nonlocality dichotomy, since it is obvious that geometrical properties are somewhat local and non-local at the same time.

But I wanted the setting of the book experiment to be such that the cover and the back of the book are separated for all practical purposes, thus the machines, registering encrypted info, rotating sequentially in a predetermined set of angles in such a manner that machine A can register its + and - independently form machine B, it is obvious they can't do it simultaneously like in the real experiment with particles for the reasons you pointed out, but I think it can be done sequentially and register the statistics in the different angles independently. In this sense the toy model wouldn't be showing anything different than the experiment with photons or electrons, because if the experiment is done correctly the info gathered from machine A can be taken independently from the info gathered from machine B, the whole point being that it can be done in a book(or a coin for that matter) because the information is geometrical.

Anyway, if we can get that cleared up, I really would like to know exactly what the model is: Who rotates the book, i.e. from what place is it rotated? What rotations are allowed? What exactly are the observations, and in particular what happens to rotations that end up with the book edgewise-on? How is spacelike separation simulated without introducing unintended Born rigidity that would violate relativity if the separation was truly spacelike? Pre-arrangement might circumvent the rigidity but then the model is not local in the sense required by Bell's Theorem.

The machines rotate the book sequentially on any angles programmed on them, each machine follows one side of the book so there is no edge-on, and for each angle it registers either up vs down, left vs right, or front vs rear depending on the axis spatial orientation of the machine.
One could argue the model is neither local nor nonlocal, but the key point here is that it be able to allow independent gathering of the info for machine A and B (thus the pre-arrangement, sequentiality, and encryption of what the machines record to be put together, decoded and analyzed at a late time, thsi should make it equivalent to the spatial separation of the experiment with particles.

If there are no fatal flaws in the model, can you then demonstrate the violation of the Bell Inequality numerically - as I asked you before - please?

I should say that I'm not sure if there are fatal flaws in the setup because I thought it up on the fly as an example of something that seemed obvious to me and I have been "often wrong, never in doubt" (hopefully not always wrong, as Landau claimed of cosmologist's assertions) around here before.
A numerical demonstration would I think give similar results as the polarization experiment with photons, I'll look into it if I have the time.

 

[Derek Potter]

Maybe I can only claim that it describes but the "book toy model" was intended to explain that if the property of non-commutativity can be made geometrical like it happens with euclidean rotations, it could help help understand how one can dodge the locality-nonlocality dichotomy, since it is obvious that geometrical properties are somewhat local and non-local at the same time.

But I wanted the setting of the book experiment to be such that the cover and the back of the book are separated for all practical purposes, thus the machines, registering encrypted info, rotating sequentially in a predetermined set of angles in such a manner that machine A can register its + and - independently form machine B, it is obvious they can't do it simultaneously like in the real experiment with particles for the reasons you pointed out, but I think it can be done sequentially and register the statistics in the different angles independently. In this sense the toy model would be showing anything different than the experiment with photons or electrons, because if the experiment is done correctly the info gathered from machine A can be taken independently from the info gathered from machine B, the whole point being that it can be done in a book(or a coin for that matter) because the information is geometrical.

The machines rotate the book sequentially on any angles programmed on them, each machine follows one side of the book so there is no edge-on, and for each angle it registers either up vs down, left vs right, or front vs rear depending on the axis spatial orientation of the machine.
One could argue the model is neither local nor nonlocal, but the key point here is that it be able to allow independent gathering of the info for machine A and B (thus the pre-arrangement, sequentiality, and encryption of what the machines record to be put together, decoded and analyzed at a late time, thsi should make it equivalent to the spatial separation of the experiment with particles.

I should say that I'm not sure if there are fatal flaws in the setup because I thought it up on the fly as an example of something that seemed obvious to me and I have been "often wrong, never in doubt" (hopefully not always wrong, as Landau claimed of cosmologist's assertions) around here before.
A numerical demonstration would I think give similar results as the polarization experiment with photons, I'll look into it if I have the time.

Okay well I have no idea what any of that means. If someone else can make sense of it, I'll leave it to them.

 

[TrickyDicky]

Okay well I have no idea what any of that means.

Would you say that a geometrical feature like say equality of right angles implying isotropy of the space and that figures may be moved to any location and still be congruent is a local or a nonlocal property? Does the fact that figures with arbitrary spacelike separation can be measured to be congruent if they are put together again mean that measuring one figure is affecting the measure of the other figure making it congruent with itself?
Now this particular geometrical property is commutative so it would not violate the inequalities, but a non-commutative geometrical property like euclidean rotations is not commutative and does violate the inequalities, the spatial separation(local-vs nonlocal) is not the important feature for such properties because they are pervasive(ubiquitous) and independent of spatial separation of the objects on which it is observed.
Is this more understandable?

 

[stevendaryl]

Would you say that a geometrical feature like say equality of right angles implying isotropy of the space and that figures may be moved to any location and still be congruent is a local or a nonlocal property? Does the fact that figures with arbitrary spacelike separation can be measured to be congruent if they are put together again mean that measuring one figure is affecting the measure of the other figure making it congruent with itself?
Now this particular geometrical property is commutative so it would not violate the inequalities, but a non-commutative geometrical property like euclidean rotations is not commutative and does violate the inequalities, the spatial separation(local-vs nonlocal) is not the important feature for such properties because they are pervasive(ubiquitous) and independent of spatial separation of the objects on which it is observed.
Is this more understandable?

It isn't to me. As I said, Bell's inequalities are about a very specific situation: Alice has a device that is capable of producing a measurement result of $\pm 1$. The device has (at least) two possible settings, $a$ or $a'$. Bob similarly has a device with two possible settings, $b$ or $b'$. For many rounds, you perform the following procedure and collect statistics:

On round number $n$,

1. Alice chooses a setting $a_n$.
2. She performs a measurement, and gets a result $A_n$
3. Bob chooses a setting $b_n$.
4. Bob gets result $B_n$

Then we compute: $E(a,b) =$ the average, over all rounds $n$ such that $a_n = a$ and $b_n = b$, of $A_n \cdot B_n$

That's the context for which Bell derived his inequality. If you allow for Alice's choice $a_n$ to affect Bob's result, $B_n$, and vice-verse, then there is no reason to expect the inequality to hold. That's where locality comes in: without assuming locality, there is no reason to assume that the inequality holds. Locality in the context of Bell's theorem means a very specific thing: that Alice's choice cannot affect Bob's result, and vice-verse. So your musings about whether congruence of geometric figures is local or nonlocal don't seem to be related to Bell's notion of locality.

Bell defined a hidden variables theory to be an explanation of the correlations along the following lines:

1. Each round, there is a hidden variable $\lambda_n$ influencing the results. The variable may take on different values on different rounds (hence the subscript).
   
2. There are deterministic functions $F_A(a, b, \lambda)$ and $F_B(a, b, \lambda)$ such that $A_n = F_A(a_n, b_n, \lambda_n)$ and $B_n = F_B(a_n, b_n, \lambda_n)$
3. Each round, $\lambda_n$ is chosen randomly according to some probability distribution $P(\lambda)$

Those three define what Bell means by a "hidden variables theory". The special case of a "local" hidden variables theory makes the additional assumption that $F_A$ does not depend on $b$ and $F_B$ does not depend on $a$. That is,

$A_n = F_A(a_n, \lambda_n)$ and $B_n = F_B(b_n, \lambda_n)$

That's the critical assumption that allows him to derive his inequality. Right off the bat, I don't see how his derivation has anything to do with whether $a$ and $b$ are described by a commutative or noncommutative algebra.

Now, what someone has shown is that QM only predicts a violation of Bell's inequality in the case where the two measurements corresponding to Alice's settings $a$ and $a'$ are described by noncommuting operators, and the two measurements corresponding to Bob's settings $b$ and $b'$ are described by noncommuting operators. But that's a fact about quantum mechanics. Bell's derivation doesn't (as far as I can see) assume anything at all about whether things commute or not. The noncommutativity is about the two choices that Alice (or Bob) might make, not about Alice's measurements versus Bob's. Alice's measurements do commute with Bob's measurements.

 

[stevendaryl]

Bell defined a hidden variables theory to be an explanation of the correlations along the following lines:

1. Each round, there is a hidden variable $\lambda_n$ influencing the results. The variable may take on different values on different rounds (hence the subscript).
   
2. There are deterministic functions $F_A(a, b, \lambda)$ and $F_B(a, b, \lambda)$ such that $A_n = F_A(a_n, b_n, \lambda_n)$ and $B_n = F_B(a_n, b_n, \lambda_n)$
3. Each round, $\lambda_n$ is chosen randomly according to some probability distribution $P(\lambda)$

Those three define what Bell means by a "hidden variables theory". The special case of a "local" hidden variables theory makes the additional assumption that $F_A$ does not depend on $b$ and $F_B$ does not depend on $a$. That is,

$A_n = F_A(a_n, \lambda_n)$ and $B_n = F_B(b_n, \lambda_n)$

I should say that the above form for $F_A$ and $F_B$ are not the most general form compatible with local realism. More generally, the outcome $A$ might also depend on an additional hidden variable $\alpha_n$ that is local to Alice, and Bob's outcome $B$ might depend on a hidden variable $\beta_n$ that is local to Bob. But the fact that in an EPR-type experiment, there are perfect correlations at some settings implies that there can't be any dependency on additional local variables.

 

[TrickyDicky]

It isn't to me. As I said, Bell's inequalities are about a very specific situation: Alice has a device that is capable of producing a measurement result of $\pm 1$. The device has (at least) two possible settings, $a$ or $a'$. Bob similarly has a device with two possible settings, $b$ or $b'$. For many rounds, you perform the following procedure and collect statistics:

On round number $n$,

1. Alice chooses a setting $a_n$.
2. She performs a measurement, and gets a result $A_n$
3. Bob chooses a setting $b_n$.
4. Bob gets result $B_n$

Then we compute: $E(a,b) =$ the average, over all rounds $n$ such that $a_n = a$ and $b_n = b$, of $A_n \cdot B_n$

That's the context for which Bell derived his inequality. If you allow for Alice's choice $a_n$ to affect Bob's result, $B_n$, and vice-verse, then there is no reason to expect the inequality to hold. That's where locality comes in: without assuming locality, there is no reason to assume that the inequality holds. Locality in the context of Bell's theorem means a very specific thing: that Alice's choice cannot affect Bob's result, and vice-verse. So your musings about whether congruence of geometric figures is local or nonlocal don't seem to be related to Bell's notion of locality.

Bell defined a hidden variables theory to be an explanation of the correlations along the following lines:

1. Each round, there is a hidden variable $\lambda_n$ influencing the results. The variable may take on different values on different rounds (hence the subscript).
   
2. There are deterministic functions $F_A(a, b, \lambda)$ and $F_B(a, b, \lambda)$ such that $A_n = F_A(a_n, b_n, \lambda_n)$ and $B_n = F_B(a_n, b_n, \lambda_n)$
3. Each round, $\lambda_n$ is chosen randomly according to some probability distribution $P(\lambda)$

Those three define what Bell means by a "hidden variables theory". The special case of a "local" hidden variables theory makes the additional assumption that $F_A$ does not depend on $b$ and $F_B$ does not depend on $a$. That is,

$A_n = F_A(a_n, \lambda_n)$ and $B_n = F_B(b_n, \lambda_n)$

That's the critical assumption that allows him to derive his inequality. Right off the bat, I don't see how his derivation has anything to do with whether $a$ and $b$ are described by a commutative or noncommutative algebra.

Now, what someone has shown is that QM only predicts a violation of Bell's inequality in the case where the two measurements corresponding to Alice's settings $a$ and $a'$ are described by noncommuting operators, and the two measurements corresponding to Bob's settings $b$ and $b'$ are described by noncommuting operators. But that's a fact about quantum mechanics. Bell's derivation doesn't (as far as I can see) assume anything at all about whether things commute or not. The noncommutativity is about the two choices that Alice (or Bob) might make, not about Alice's measurements versus Bob's. Alice's measurements do commute with Bob's measurements.

Is $F_A× F_B$ the same as $F_B× F_A$ for the inequalities ?

 

[stevendaryl]

Is $F_A× F_B$ the same as $F_B× F_A$ for the inequalities ?

Yes, because they are real numbers, not operators.

 

[TrickyDicky]

Yes, because they are real numbers, not operators.

Got it, thanks!