I What is the physical significance of Bell's math?

[morrobay]

We can picture it this way:

The quantum version does not allow the prediction of Bob's result based on information about Region 5.

Alice and Bobs measurement results in zone 1 and 2 would be determined by the spin 1/2 particles spins at the source in zone 5 and detector settings at time of measurement:
A(a,λ) =±1
B(b,λ) =±1
For a given a and b (that can be chosen after particles leave source in zone 5) and λ. These unique outcomes are determined. (Bell) So if that is an example of quantum non locality it is not synonymous with FTL which would be a superluminal signal from spacelike separated zones 1 and 2 during measurements.
Suppose Alice measures + or - Then probability for double detection at Bob is (sin θ/2)² and for opposite results (cos θ/2 )²
Since θ = setting (a - b) then it seems that the measurement results at A and B are local .Ie. Ontic spin and interaction with detector settings a and b *
If Bell inequality is violated, of the three possibilities: non locality (FTL), non realism, contextual I would favor contextual*

 

[Boing3000]

It doesn't. However, I did define locality as no FTL

Actually you don't. Locality mean "having a location", which in every one's vocabulary, is identical (not synonymous) to having a space-time coordinate.
Incidentally, it happens that every single entity ever observed to have a location/coordinate, do obey the laws of relativity, that happens to include no FLT.
So it is tautological to associate no FLT with locality.

BTW if someone would ever want to include FLT somewhere in the equations, it would have to provide the exact means of computing space-time interval in that framework. Sadly, this is a waste of time because Bell's prove that there is no such interval.
Now I suppose you can always provide a theory where those interval would be smaller then the margin of error of experimental apparatus. But then the simple fact that Alice or Bob would make the measurement first would make a testable difference. In the non-local side of Bell's inequality it doesn't even matter.

so what does that make non locality mean?

It means no coordinate. The state/value is unique and span the whole universe, it is everywhere at once, and testable only at particles sharing this value (we called them entangled, not fiddledeedoomed)
It is not a feature of QM, but a feature of nature, that QM happens to describe successfully.

I am under the impression that the Bohmian pilot wave is also a non-local.

 

[Zafa Pi]

Entanglement is a feature of wave functions, but the notion of nonlocal information that I described is independent of whether quantum mechanics is the ultimate theory, or not.

Entangled particles are created in the lab, they part of nature, independent of whether quantum mechanics is the ultimate theory, or not.

 

[stevendaryl]

Entangled particles are created in the lab, they part of nature, independent of whether quantum mechanics is the ultimate theory, or not.

Maybe this depends on how you define "entanglement", but the definition I've used is that two particles are entangled if they are described by a two-particle wave function that cannot be "factored" into a product of one-particle wave functions. So it's a theoretical description of what goes on in EPR. If QM is false, then so is the description of the particles as "entangled".

 

[stevendaryl]

Alice and Bobs measurement results in zone 1 and 2 would be determined by the spin 1/2 particles spins at the source in zone 5 and detector settings at time of measurement:
A(a,λ) =±1
B(b,λ) =±1

That's true. That's the "superdeterminism". If it's possible to know Alice's and Bob's choices ahead of time, then you can violate Bell's inequality. But in the figure, Alice's choice may potentially depend on facts about Region 3 as well as Region 5. Bob's choice may depend on Region 4. So there is no reason to believe that the information about Alice's and Bob's future settings are determined in Region 5.

For a given a and b (that can be chosen after particles leave source in zone 5) and λ. These unique outcomes are determined. (Bell) So if that is an example of quantum non locality it is not synonymous with FTL which would be a superluminal signal from spacelike separated zones 1 and 2 during measurements.
Suppose Alice measures + or - Then probability for double detection at Bob is (sin θ/2)² and for opposite results (cos θ/2 )²
Since θ = setting (a - b) then it seems that the measurement results at A and B are local .Ie. Ontic spin and interaction with detector settings a and b *
If Bell inequality is violated, of the three possibilities: non locality (FTL), non realism, contextual I would favor contextual*

If I understand you correctly, I agree that for a fixed choice of Alice's and Bob's settings, one can come up with a hidden-variable explanation for their results. But the combination "Bob's setting plus Bob's result" is something that is completely nondetermined by anything in Region 5, even though Alice can find out something about it in Region 1 (that a particular combination did not/will not occur).

 

[Zafa Pi]

Maybe this depends on how you define "entanglement", but the definition I've used is that two particles are entangled if they are described by a two-particle wave function that cannot be "factored" into a product of one-particle wave functions. So it's a theoretical description of what goes on in EPR. If QM is false, then so is the description of the particles as "entangled".

You are giving the QM definition, and I agree with it. However, via various processes (e.g. down conversion) entangled photons are created in nature, and with the measuring devices (polarization analyzers) give the same weird results predicted by QM., thus independent of whether quantum mechanics is the ultimate theory, or not.

 

[stevendaryl]

syn·on·y·mous
səˈnänəməs/
adjective

1. (of a word or phrase) having the same or nearly the same meaning as another word or phrase in the same language.
   "aggression is often taken as synonymous with violence"
   • closely associated with or suggestive of something.
     "his deeds had made his name synonymous with victory"

By that definition of "synonymous", I don't think that any two of "entanglement", "FTL" or "local" are synonymous.
It doesn't.

However, I did define locality as no FTL

Fine. I don't think that's a helpful definition, though. I agree that QM does not provide a way for people to communicate FTL. I think everybody agrees with that. If that's the end of the matter for you, then there is no need to further discuss Bell and EPR.

 

[stevendaryl]

You are giving the QM definition, and I agree with it. However, via various processes (e.g. down conversion) entangled photons are created in nature, and with the measuring devices (polarization analyzers) give the same weird results predicted by QM., thus independent of whether quantum mechanics is the ultimate theory, or not.

Yes, I would just say that the predictions of QM in EPR-type experiments have been verified, so any successor theory must make the same predictions for those experiments. I would not say that a successor theory must have entangled particles.

 

[Zafa Pi]

If Bell inequality is violated, of the three possibilities: non locality (FTL), non realism, contextual I would favor contextual

Can you give me a short, simple example or explanation that distinguishes nonrealism (= nonCFD) from contextual?

 

[Zafa Pi]

By that definition of "synonymous", I don't think that any two of "entanglement", "FTL" or "local" are synonymous.
It doesn't.

Neither do I. This is disheartening, you've missed which two concepts I've claimed were synonymous. I presume you didn't read either post #91 or #100. I feel crushed.

You don't care for my definition of local. What do you prefer?

 

[morrobay]

Can you give me a short, simple example or explanation that distinguishes nonrealism (= nonCFD) from contextual?

No , nonCFD is equated with or subset of contextuality : Measurement outcomes are produced from particle/detector field interactions.

 

[Zafa Pi]

No , nonCFD is equated with or subset of contextuality : Measurement outcomes are produced from particle/detector field interactions.

Cool, I understand your whole sentence. So now all you have to do to make me a happy quantum camper is to give me an example of contextuality that isn't nonCFD.
The classic example of contextually is the Mermin-Peres quantum pseudo-telepathy, but that is also nonCFD.
If you can't provide me with such an example, I hereby threaten to say they are synonymous, just like I did with Bell non locality and measurements on entangled entities, thus earning further derision from PF luminaries.

 

[PeterDonis]

Everyone, please bear in mind that this is an I level thread and some math is expected. Throwing around ordinary language terms without giving them precise mathematical definitions is a recipe for endless discussion with no resolution, and eventually either a closed thread or a warning. This thread is getting close to the latter point now.

 

[morrobay]

Cool, I understand your whole sentence. So now all you have to do to make me a happy quantum camper is to give me an example of contextuality that isn't nonCFD.
The classic example of contextually is the Mermin-Peres quantum pseudo-telepathy, but that is also nonCFD.
If you can't provide me with such an example, I hereby threaten to say they are synonymous, just like I did with Bell non locality and measurements on entangled entities, thus earning further derision from PF luminaries.

For a quantitative example/explanation let's let one of the Masters or PhD's in physics take your question. My background is in Biology/Chemistry and I am not qualified to go any farther with this.

 

[Zafa Pi]

For a quantitative example/explanation let's let one of the Masters or PhD's in physics take your question. My background is in Biology/Chemistry and I am not qualified to go any farther with this.

I notice a while back there was a B level thread on exactly this topic. However, no answers were provided.

 

[stevendaryl]

Neither do I. This is disheartening, you've missed which two concepts I've claimed were synonymous. I presume you didn't read either post #91 or #100.

Post #91 is where you said

Your definition of non-locality in #74 is synonymous with the existence of entangled particles

I specifically said that I disagreed with that statement. You also said:

Do the correlations that are manifest in measuring entangled pairs require FTL phenomena

I specifically said that I don't think so.

So I read your post #91 and specifically responded to it. So don't feel crushed.

You don't care for my definition of local. What do you prefer?

I wrote a very detailed post, complete with a diagram that I spent a good part of an hour creating, several years ago, in an attempt to answer that question. Let me try one more time.

A "nonlocal correlation" is a correlation between distant (spacelike separated) events. A local theory (or model) explains or implements all nonlocal correlations in terms of a chain of two or more local correlations. So in a local theory, if A is correlated with B, then either A is in the causal past of B, or B is in the causal past of A, or there is a set of events C1, C2, ... such that all the Cs are in the causal past of both A and B and A is only correlated with B through its correlations with C1, C2, ...

The events C1, C2, are elements of the model---they aren't necessarily directly observable.

What's an example of a nonlocal model, then? Well, here's a toy example. Suppose there is a pair of coins, and I propose the following theory about those coins: The nth flip of one coin always produces the opposite result of the nth flip of the other coin.

That's a falsifiable theory about those coins. You can certainly test it by just flipping the coins a bunch of times. But it's a nonlocal theory, since it predicts a correlation between possibly distant coins and does not explain or implement the correlation in terms of local correlations.

In this case, I could come up with a different, local theory that made the same predictions as the first theory. But the original theory is nonlocal.

I believe that in the same way, quantum mechanics in the standard way that it is applied is nonlocal. Einstein et al believed that there might be a local theory that made the same predictions as quantum mechanics, but Bell proved them wrong.

 

[Zafa Pi]

So I read your post #91 and specifically responded to it. So don't feel crushed.

When I said, "Neither do I. This is disheartening, you've missed which two concepts I've claimed were synonymous. I presume you didn't read either post #91 or #100."
I was referring to your comment,

By that definition of "synonymous", I don't think that any two of "entanglement", "FTL" or "local" are synonymous.

And it appeared that you were misquoting me. I'm sorry I made you go through that again. I did understand your post #74. I feel uncrushed.

It seems that your definition of local (post #116) is the negation of your nonlocal, as it should be. Whereas my definition of nonlocal is the negation of local (post #54).
That our definitions disagree is ok with me. We've each said what we dislike about the other's definitions.

You did clear up Bell's definition for me that he made in OP's cited paper, to wit:
"It is the requirement of locality, or more precisely that the result of a measurement on one system be unaffected by operations on a distant system with which it has interacted in the past", which I found vague.

I'm sorry you hate CFD, I hope it doesn't feel crushed.

 

[stevendaryl]

I'm sorry you hate CFD, I hope it doesn't feel crushed.

It's just that I don't think it clarifies anything. A local, nondeterministic theory violates CFD, so violating CFD is not a big deal, it seems to me, and it doesn't do anything to understand the difference between a quantum theory and a classical theory.

 

[N88]

Everyone, please bear in mind that this is an I level thread and some math is expected. Throwing around ordinary language terms without giving them precise mathematical definitions is a recipe for endless discussion with no resolution, and eventually either a closed thread or a warning. This thread is getting close to the latter point now.

Thanks for this. The OP was intended to be a straight-forward (but highly relevant) question and I do not see that it has been satisfactorily answered. In my view, without Bell's assumption, his theorem fails. If he was mimicking EPR, then EPR fails. But I do not see that the failure of EPR would mean the failure of local realism.

Maybe a clearer example of my concern is this attachment from Isham's "Lectures on Quantum Theory" (1995). More recent examples exist.

In Isham's equation (9.33), he sums over the same set of particles on 4 occasions! It would never occur to me to do that: for, if I did , I would expect to get an outcome that applied to objects that I could meaningfully test 4 times. Classical objects would meet this criteria; quantum objects would not. So back to my question in the OP please.

 

[PeterDonis]

In Isham's equation (9.33), he sums over the same set of particles on 4 occasions! It would never occur to me to do that: for, if I did , I would expect to get an outcome that applied to objects that I could meaningfully test 4 times.

They don't have to be the same objects; they just have to be taken from an ensemble of objects all prepared in the same state.

 

[N88]

They don't have to be the same objects; they just have to be taken from an ensemble of objects all prepared in the same state.

It seems to me that this answer leaves me with my original difficulty with Bell's workings. I agree that, given 4N objects prepared in the same state, the average of each of the 4 terms in RHS (9.33) would be unchanged when each is tested over N runs. For example, under EPRB, the first such RHS term would reduce to -a.b; and so on.

But (9.33) is based on the a_n appearing again in the second term on RHS; the b_n appearing again in the third term on RHS; etc. So the problem (that I am wrestling with) goes back to eqn (9.32). There we see 4 pairs of terms created from 4 terms; I say that there should be 8 terms. For how is it possible to match the a in the first pair with the a result in the second? For in one run it may be +1, in the next -1.

This seems to be recognised by Isham in the 2nd paragraph: "The central realist assumption we are testing is that each particle has a definite value at all times in any direction of spin. We let a_n denote 2/h times the value of a⋅S possessed by particle 1 in the n'th element of the collection. Thus a_n = ±1 if a⋅S = ±h/2".

How can particle 1 in the n'th element be tested twice?

NB: As a local-realist, that "central realist assumption" has no place in my thinking: for surely it is false?

 

[PeterDonis]

How can particle 1 in the n'th element be tested twice?

It can't, but the "central realist" assumption Isham is describing amounts to the claim that particle 1's spin direction is a hidden variable which has a definite value whether it is measured or not, and can therefore be used to predict/model what the results would have been to measurements that were not actually made. Other sources call this "counterfactual definiteness" or similar names.

Whether that assumption makes sense to you is of course your own choice. But mathematically it is perfectly clear and Isham is simply calculating its implications.

 

[N88]

It can't, but the "central realist" assumption Isham is describing amounts to the claim that particle 1's spin direction is a hidden variable which has a definite value whether it is measured or not, and can therefore be used to predict/model what the results would have been to measurements that were not actually made. Other sources call this "counterfactual definiteness" or similar names.

Whether that assumption makes sense to you is of course your own choice. But mathematically it is perfectly clear and Isham is simply calculating its implications.

OK; thank you. Referring then to the OP: Do we agree then, that this is the same assumption that Bell made to move from his (14a) to his (14b)?

In other words: Is Isham (1995) simply calculating the same implications as in Bell (1964)?

If that were the case, I'm surprised that so many claim that Bell is definitive against "local realism" in general -- without spelling out that Bell's theorem is based upon (and therefore limited to) that naive (in my view) "central realist" assumption.

PS: Not wishing to change the subject but seeking to be very clear on this pesky subject: I understand that my views are compatible with QFT. So again, at that advanced level, Bell's theorem seems to be irrelevant?

 

[PeterDonis]

Do we agree then, that this is the same assumption that Bell made to move from his (14a) to his (14b)?

As far as I can tell, yes.

I'm surprised that so many claim that Bell is definitive against "local realism" in general -- without spelling out that Bell's theorem is based upon (and therefore limited to) that naive (in my view) "central realist" assumption.

How would you mathematically express "local realism" as opposed to "central realism"?

 

[PeterDonis]

As far as I can tell, yes.

It's worth noting, though, that Bell defines "local realism" in his paper by an earlier equation, the one that says the probability $P(a, b, \lambda)$ must factorize into $P(a, \lambda) P(b, \lambda)$ (I can't remember exactly which equation this is in his paper and don't have it handy to check). The equations you are talking about are derived from that original assumption. And that original assumption is already violated by the QM probabilities.

 

[N88]

… How would you mathematically express "local realism" as opposed to "central realism"?

Thanks for the good question: My preliminary opinion follows, based on Bell (1964), which we denote by E (for the crucial EPRB experiment).

Locality: A(a, λ_i) = ±1; B(b, -λ_i) = ±1. (1)

Therefore: B(b, -λ_i) = -A(b, λ_i). (2)

Realism: Bayes' Rule (which is never false) is relevant to (1) since A and B are independent per (1), but correlated per (2) and the correlated elements in (1).

Therefore: P(AB|E) = P(A|E)⋅P(B|EA) = P(B|E)⋅P(A|EB). (3)

As you say, and I agree: Bell's original probability assumption -- which is not (3) -- is already violated by the QM probabilities.

Note that (3) is a logical implication from (2): it should not be confused with (and has no relation to) causation, signalling, piloting, etc. Thus, for me, (3) is fundamental to a realist view of the world in the presence of correlations; especially when such correlations derive from the conservation of total angular momentum in the generation and emission of each and every particle-pair in EPRB. In my view, it is unrealistic to negate (3) given (1) and (2).

IMHO, putting it another way: To reason contrary to (3), as Bell does, is to not make proper use of the entanglement brought about by the conservation rules that apply in EPRB.

To be clear re logical implication: If Alice and Bob know that I am distributing numbered pairs of Red ribbons and numbered pairs of Blue ribbons, logical implication allows each to know what the other receives (by colour and number). Here, again, there is no causation involved -- just initial correlation at the source (which is me in this classical analogy). In the quantum case (EPRB), I know how the emitted-pairs are more cleverly correlated.

 

[PeterDonis]

Locality: A(a, λi) = ±1; B(b, -λi) = ±1. (1)

I don't understand what this means. The $\lambda_i$ are supposed to be hidden variables, which means we don't know their values, so how can this be tested?

Also, why should the values A and B only be 1 or -1? They are supposed to be probabilities, right?

Also, even leaving that aside, why does this express locality?

Therefore: B(b, -λi) = -A(b, λi). (2)

I don't see how this follows from (1).

Note that (3) is a logical implication from (2)

I don't see how this follows either. Bayes' rule is completely general.

 

[N88]

I don't understand what this means. The $\lambda_i$ are supposed to be hidden variables, which means we don't know their values, so how can this be tested?

We know the results A and B from experiment; we know a and b. We don't know the λ's beyond knowing that they are anti-correlated.

Also, why should the values A and B only be 1 or -1? They are supposed to be probabilities, right?

A and B are the results (outcomes), per Bell (1964), eqn (1). They are not probabilities.

Also, even leaving that aside, why does this express locality?

Outcome A is in Alice's location, as is the detector-setting a; as is each particle-property λ of each particle that she tests. Same for Bob. So the outcome events (the analysers locally signalling ±1) are local and spacelike separated..

I don't see how this follows from (1).

See similar at Bell (1964), eqn (13).

I don't see how this follows either. Bayes' rule is completely general.

Bayes' Rule is general, and applicable here. And here it remains in its non-reduced form because the outcomes are correlated.

 

[morrobay]

Can you give me a short, simple example or explanation that distinguishes nonrealism (= nonCFD) from contextual?

CFD is derived from following:
Outcomes are determined : A (a,b,λ) = ± 1, B( a,b,λ) = ± 1
Locality: A (a,λ) = ± 1 , B (b,λ) = ± 1
The perfect anti correlations along same axis : A (b,λ) = - B ( b,λ)
Then it follows: E ( a,b ) = - ∫ d λ p (λ) A ( a,λ) A (b λ)
That includes determinism and is a classical concept..

Contextuality is an orthodox QM viewpoint: The measuring apparatuses define the conditions and outcomes. Outcomes are not pre- encoded in measured object but only arise in interaction of object and detector fields. In accord with Kochen - Specker Theorem
CFD is only compatible with the orthodox view when detectors are aligned. However when settings a, b, and c are included as random variables on classic probability space - they are static properties and invalid contextually.
So I can only distinguish CFD from contextuality.
Reference, equations 1-7 https://arxiv.org/pdf/quant-ph/0606084.pdf

 

[stevendaryl]

This seems to be recognised by Isham in the 2nd paragraph: "The central realist assumption we are testing is that each particle has a definite value at all times in any direction of spin. We let a_n denote 2/h times the value of a⋅S possessed by particle 1 in the n'th element of the collection. Thus a_n = ±1 if a⋅S = ±h/2".

How can particle 1 in the n'th element be tested twice?

For the sake of clarity, let me reproduce the equations we're talking about:

g_n = a_n b_n + a_n b_n' + a_n' b_n - a_n' b_n' (9.32)

\frac{1}{N} |\sum_n g_n| = \frac{1}{N} |\sum_n a_n b_n + a_n b_n' + a_n' b_n - a_n' b_n'| (9.33)

What I think you're complaining about is that, since Alice (one of the experimenters) can only measure at most one of a_n or a_n' and Bob (the other experimenter) can only measure at most one of b_n or b_n', there is no way for them to measure g_n, since it involves all 4 values. Is that your complaint?

But at the end, we're averaging over all runs, so the fact that not all 4 measurements came from the same run does not make a difference (unless one of a number of loopholes is exploited---I'll get to those later).

First of all, Bell is assuming that the measurement results are deterministic functions of the hidden variables and the measurement choices. So in each run of the experiment, all 4 variables--a_n, a_n', b_n, b_n'--have definite values, even though Alice and Bob can only measure two of them.

So taking into account your complaint, what is actually measured, through many rounds of the experiment, is not

\langle g \rangle \equiv \frac{1}{N} \sum_n g_n

What is actually measured is four separate numbers:

1. \langle a b \rangle \equiv \frac{1}{N_1} \sum_{n,1} a_n b_n
2. \langle a b' \rangle \equiv \frac{1}{N_2} \sum_{n,2} a_n b_n'
3. \langle a' b \rangle \equiv \frac{1}{N_3} \sum_{n,3} a_n' b_n
4. \langle a' b' \rangle \equiv \frac{1}{N_4} \sum_{n,4} a_n' b_n'

Where

• \sum_{n,1} is the sum over all n such that on round number n, Alice measured a_n while Bob measured b_n,
• N_1 is the number of such rounds
• \sum_{n,2} is the sum over all n such that on round number n, Alice measured a_n while Bob measured b_n',
  
• N_2 is the number of such rounds
• etc.

So we don't actually measure \langle g \rangle. But the point is that under certain assumptions about random variables, we will have:
\langle g \rangle = \langle a b \rangle + \langle a b' \rangle + \langle a' b \rangle - \langle a' b' \rangle

What are those assumptions? Basically, that each round, the four numbers a_n, b_n, a_n', b_n' are produced with consistent probabilities, independent of the round number, and independent of which of the four Alice and Bob actually measure.

 

[stevendaryl]

PS: Not wishing to change the subject but seeking to be very clear on this pesky subject: I understand that my views are compatible with QFT. So again, at that advanced level, Bell's theorem seems to be irrelevant?

Some people have said this, and I think it's wrong. QFT is not a local realistic theory any more than nonrelativistic QM is. It violates Bell's inequality in the same way that nonrelativistic QM does. So whatever implications Bell's theorem has for nonrelativistic QM, it has the same implications for QFT, namely that there is no local realistic theory that makes the same predictions (well, unless you go for something weird like Many-Worlds or superdeterminism or back-in-time causality).

 

[PeterDonis]

We don't know the λ's beyond knowing that they are anti-correlated.

How do we know they are anti-correlated? Bell does not assume that; in fact he assumes the opposite (see last comment below).

A and B are the results (outcomes), per Bell (1964), eqn (1). They are not probabilities.

Ah, ok. But then eqn (2) of that paper expresses locality in terms of things we can actually observe; we don't observe the hidden variables $\lambda$.

See similar at Bell (1964), eqn (13).

Bell's eqn (13) is

$$
A(a, \lambda) = - B(a, \lambda)
$$

Note the key difference: his $\lambda$ has the same sign on both sides of his equation. Your $\lambda$ has opposite signs on the two sides of your equation. So your equation contradicts Bell's, yet they are both supposed to be derived from the same premises. Why is yours right and his wrong?

 

[N88]

For the sake of clarity, let me reproduce the equations we're talking about:

g_n = a_n b_n + a_n b_n' + a_n' b_n - a_n' b_n' (9.32)

\frac{1}{N} |\sum_n g_n| = \frac{1}{N} |\sum_n a_n b_n + a_n b_n' + a_n' b_n - a_n' b_n'| (9.33)

What I think you're complaining about is that, since Alice (one of the experimenters) can only measure at most one of a_n or a_n' and Bob (the other experimenter) can only measure at most one of b_n or b_n', there is no way for them to measure g_n, since it involves all 4 values. Is that your complaint?

That is not my "complaint" -- please see next.

But at the end, we're averaging over all runs, so the fact that not all 4 measurements came from the same run does not make a difference (unless one of a number of loopholes is exploited---I'll get to those later).

First of all, Bell is assuming that the measurement results are deterministic functions of the hidden variables and the measurement choices. So in each run of the experiment, all 4 variables--a_n, a_n', b_n, b_n'--have definite values, even though Alice and Bob can only measure two of them.

So taking into account your complaint, what is actually measured, through many rounds of the experiment, is not

\langle g \rangle \equiv \frac{1}{N} \sum_n g_n

What is actually measured is four separate numbers:

1. \langle a b \rangle \equiv \frac{1}{N_1} \sum_{n,1} a_n b_n
2. \langle a b' \rangle \equiv \frac{1}{N_2} \sum_{n,2} a_n b_n'
3. \langle a' b \rangle \equiv \frac{1}{N_3} \sum_{n,3} a_n' b_n
4. \langle a' b' \rangle \equiv \frac{1}{N_4} \sum_{n,4} a_n' b_n'

Where

• \sum_{n,1} is the sum over all n such that on round number n, Alice measured a_n while Bob measured b_n,
• N_1 is the number of such rounds
• \sum_{n,2} is the sum over all n such that on round number n, Alice measured a_n while Bob measured b_n',
  
• N_2 is the number of such rounds
• etc.

So we don't actually measure \langle g \rangle. But the point is that under certain assumptions about random variables, we will have:
\langle g \rangle = \langle a b \rangle + \langle a b' \rangle + \langle a' b \rangle - \langle a' b' \rangle

What are those assumptions? Basically, that each round, the four numbers a_n, b_n, a_n', b_n' are produced with consistent probabilities, independent of the round number, and independent of which of the four Alice and Bob actually measure.

In my terms, which I trust are accurate: Without any assumptions, what is actually derived from measurements is four separate expectations:

• (1) \langle a b \rangle \equiv \frac{1}{N_i} \sum^{N_i}_1 a_i b_i
• (2) \langle a b' \rangle \equiv \frac{1}{N_j} \sum^{N_j}_1 a_j b'_j
• (3) \langle a' b \rangle \equiv \frac{1}{N_k} \sum^{N_k}_1 a'_k b_k
• (4) \langle a' b' \rangle \equiv \frac{1}{N_l} \sum^{N_l}_1 a'_l b'_l

So, please: What assumptions are you making to modify these equations?

My problem (not so much a complaint) is that I can see no valid basis for modifying these equations; which are valid both classically and quantum mechanically.

From the OP, it is my belief that Bell's work [and the given mathematical assumption that he makes in moving from his (14a) to his (14b) in the context of EPRB (a QM setting)], limits his results to those delivered in classical settings.

 

[stevendaryl]

In my terms, which I trust are accurate: Without any assumptions, what is actually derived from measurements is four separate expectations:

• (1) \langle a b \rangle \equiv \frac{1}{N_i} \sum^{N_i}_1 a_i b_i
• (2) \langle a b' \rangle \equiv \frac{1}{N_j} \sum^{N_j}_1 a_j b'_j
• (3) \langle a' b \rangle \equiv \frac{1}{N_k} \sum^{N_k}_1 a'_k b_k
• (4) \langle a' b' \rangle \equiv \frac{1}{N_l} \sum^{N_l}_1 a'_l b'_l

So, please: What assumptions are you making to modify these equations?

Didn't I just go through that in my post? We use

\langle g \rangle = \langle a b\rangle + \langle a b'\rangle + \langle a' b\rangle - \langle a' b'\rangle

It's a valid move if each "run" of the experiment is independent, and the probabilities are constant.

My problem (not so much a complaint) is that I can see no valid basis for modifying these equations; which are valid both classically and quantum mechanically.

Nobody's modifying anything, we're just using the mathematics of probability to derive a fact about the correlations.

 

[N88]

Didn't I just go through that in my post? We use

\langle g \rangle = \langle a b\rangle + \langle a b'\rangle + \langle a' b\rangle - \langle a' b'\rangle

It's a valid move if each "run" of the experiment is independent, and the probabilities are constant.Nobody's modifying anything, we're just using the mathematics of probability to derive a fact about the correlations.

The reason that I spelt out the valid equations (1)-(4) is because I want to be clear as to what you are doing. I believe your response makes my questioning even clearer: You seem to be agreeing that we can use my equations (1)-(4) in

\langle g \rangle = \langle a b\rangle + \langle a b'\rangle + \langle a' b\rangle - \langle a' b'\rangle. (5)

But, as in the Isham pages that I posted (p.182; and as is customary), how do you now maintain that:

|\langle g \rangle| ≤ 2; (6)

since each expectation in my equations (1)-(4) lies in the range ±1?

That is: |\langle g \rangle| ≤ 4; (7)

with a maximum of 2√2 under EPRB. That's why it's not clear to me when you say: "Nobody's modifying anything, we're just using the mathematics of probability to derive a fact about the correlations." What "mathematics of probability" please?

 

[stevendaryl]

The reason that I spelt out the valid equations (1)-(4) is because I want to be clear as to what you are doing. I believe your response makes my questioning even clearer: You seem to be agreeing that we can use my equations (1)-(4) in

\langle g \rangle = \langle a b\rangle + \langle a b'\rangle + \langle a' b\rangle - \langle a' b'\rangle. (5)

But, as in the Isham pages that I posted (p.182; and as is customary), how do you now maintain that:

|\langle g \rangle| ≤ 2; (6)

You're asking about a mathematical proof that has been proved many times. What exactly did you not understand about that proof?

Really, you're questioning something that is mathematically provable. So it's really mathematics, rather than physics.

 

[PeterDonis]

each expectation in my equations (1)-(4) lies in the range ±1?

But they are not independent. $|g| \le 4$ assumes they are independent. When you take into account the correlations between them, given the assumptions, you have $|g| \le 2$.

 

[N88]

How do we know they are anti-correlated? Bell does not assume that; in fact he assumes the opposite (see last comment below).

Ah, ok. But then eqn (2) of that paper expresses locality in terms of things we can actually observe; we don't observe the hidden variables $\lambda$.

Bell's eqn (13) is

$$
A(a, \lambda) = - B(a, \lambda)
$$

Note the key difference: his $\lambda$ has the same sign on both sides of his equation. Your $\lambda$ has opposite signs on the two sides of your equation. So your equation contradicts Bell's, yet they are both supposed to be derived from the same premises. Why is yours right and his wrong?

Mine is right and Bell's is right because they are each correct and equivalent. Using your example
$$A(a, \lambda) = - B(a, \lambda) = B(a, -\lambda').$$ I associate the HVs with the conservation of total angular momentum and choose to be clear about which HV (in Bell's notation) that I am referring to. Thus:

$$ \lambda_i + \lambda'_i = 0 $$where, on the i-th run of the experiment, the former is the HV that Alice receives and the latter is the HV that Bob receives. It's my understanding that, in QM, the equivalent expression is:

$$ \sigma_1+ \sigma_2 = 0 $$where the subscripts denote Alice's particle and Bob's particle, respectively.

 

[stevendaryl]

You're asking about a mathematical proof that has been proved many times. What exactly did you not understand about that proof?

Really, you're questioning something that is mathematically provable. So it's really mathematics, rather than physics.

Once again, we have:

g_n = a_n b_n + a_n b_n' + a_n' b_n - a_n' b_n'

We rearrange it as follows:

g_n = a_n (b_n + b_n') + a_n' (b_n - b_n')

Since b_n and b_n' are each \pm 1, it follows that either

• Case A: b_n + b_n' = 0 (if they are opposite signs), or

• Case B: b_n - b_n' = 0 (if they are the same sign).

So in Case A,

g_n = a_n' (b_n - b_n') = 2 a_n' b_n = \pm 2

In Case B,

g_n = a_n (b_n + b_n') = 2 a_n b_n = \pm 2

So in either case, g_n = \pm 2.

If for every single n, g_n = \pm 2, then obviously the average value of g_n must be in the range -2 \leq \langle g \rangle \leq +2.

We agreed earlier that

\langle g \rangle = \langle a b \rangle + \langle a b' \rangle + \langle a' b \rangle - \langle a' b' \rangle

So putting those two facts together, we get Bell's inequality (or actually, the CHSH inequality)

-2 \leq \langle a b \rangle + \langle a b' \rangle + \langle a' b \rangle - \langle a' b' \rangle \leq +2

 

[N88]

But they are not independent. $|g| \le 4$ assumes they are independent. When you take into account the correlations between them, given the assumptions, you have $|g| \le 2$. Emphasis added.

But my equations (1)-(4) are independent. And the limit of 2√2 is achievable with such independent equations.

It is your assumptions that I am seeking to understand: the assumptions that you need to make to those equations (1)-(4) in order to establish the limit of 2 that you accept.

This all goes back to the OP where I sought to understand the physical significance of Bell's mathematical assumption that linked his (14a) to his (14b).

 

[PeterDonis]

on the i-th run of the experiment, the former is the HV that Alice receives and the latter is the HV that Bob receives

Then you are certainly not making the same assumptions as Bell. In Bell's model, the HVs are the same for both Alice and Bob; that's the whole point. The only difference is how they choose to orient their detectors.

 

[N88]

Then you are certainly not making the same assumptions as Bell. In Bell's model, the HVs are the same for both Alice and Bob; that's the whole point. The only difference is how they choose to orient their detectors.

Are you sure? In EPRB, the case that he studies in Bell (1964), the particles are anti-correlated.

In some tests with photons, the HV are the same; under the same rule re conservation of angular momentum.

 

[PeterDonis]

In EPRB, the case that he studies in Bell (1964), the particles are anti-correlated.

The observed spins, given that both Alice and Bob choose the same orientation for their detectors, are anti-correlated. That is not the same as the hidden variables being anti-correlated. The hidden variables contain all of the things that could affect either of the measurements; there is no separation into "hidden variables that affect Alice's measurement" and "hidden variables that affect Bob's measurement". If Alice's and Bob's measurements are perfectly anti-correlated, then there will be one set of HVs $\lambda$ that effect that anti-correlation. In other words, as Bell writes, $A(a, \lambda) = - B(a, \lambda)$: the same $\lambda$, the same measurement direction $a$, but opposite results A and B.

 

[stevendaryl]

Are you sure? In EPRB, the case that he studies in Bell (1964), the particles are anti-correlated.

I think that you have been misunderstanding the role of \lambda. It is supposed to represent the information that is shared between the two particles. The assumption is that Alice's result is a function A(\alpha, \lambda) that depends on the shared information, \lambda, and Alice's setting, \alpha. Bob's result is a function B(\beta, \lambda) taht depends on \lambda and Bob's setting, \beta. The anti-correlation is accomplished by the fact that if \alpha = \beta, then

B(\alpha, \lambda) = - A(\alpha, \lambda)

It's not that they have different values of \lambda---the whole point of \lambda is to explain the correlations/anti-correlations in terms of shared state information.

 

[N88]

Once again, we have:

g_n = a_n b_n + a_n b_n' + a_n' b_n - a_n' b_n'

We rearrange it as follows:

g_n = a_n (b_n + b_n') + a_n' (b_n - b_n')

Since b_n and b_n' are each \pm 1, it follows that either

• Case A: b_n + b_n' = 0 (if they are opposite signs), or

• Case B: b_n - b_n' = 0 (if they are the same sign).

So in Case A,

g_n = a_n' (b_n - b_n') = 2 a_n' b_n = \pm 2

In Case B,

g_n = a_n (b_n + b_n') = 2 a_n b_n = \pm 2

So in either case, g_n = \pm 2.

If for every single n, g_n = \pm 2, then obviously the average value of g_n must be in the range -2 \leq \langle g \rangle \leq +2.

We agreed earlier that

\langle g \rangle = \langle a b \rangle + \langle a b' \rangle + \langle a' b \rangle - \langle a' b' \rangle

So putting those two facts together, we get Bell's inequality (or actually, the CHSH inequality)

-2 \leq \langle a b \rangle + \langle a b' \rangle + \langle a' b \rangle - \langle a' b' \rangle \leq +2

Thank you for this detail. But please note: to achieve this latest analysis, you have NOT used the FOUR independent equations that I gave you.

That means they have been modified. It is the reasoning/justification behind these modifications that I am seeking to understand, as spelt out in the OP. Bell does it via a mathematical assumption. I am seeking to understand its physical significance.

Before we proceed, I'd better be clear about this point: Do you see that you cannot derive your result from my four valid equations? That is, they are defining equations that are valid, both classically and quantum mechanically.

 

[N88]

The observed spins, given that both Alice and Bob choose the same orientation for their detectors, are anti-correlated. That is not the same as the hidden variables being anti-correlated. The hidden variables contain all of the things that could affect either of the measurements; there is no separation into "hidden variables that affect Alice's measurement" and "hidden variables that affect Bob's measurement". If Alice's and Bob's measurements are perfectly anti-correlated, then there will be one set of HVs $\lambda$ that effect that anti-correlation. In other words, as Bell writes, $A(a, \lambda) = - B(a, \lambda)$: the same $\lambda$, the same measurement direction $a$, but opposite results A and B.

I am using the alternative technique that Bell specifically approved: see the second paragraph following Bell (1964), eqn (3). This technique is more realistic and physically significant to me. And it changes nothing in the analyses.

 

[stevendaryl]

Thank you for this detail. But please note: to achieve this latest analysis, you have NOT used the FOUR independent equations that I gave you.

I don't know what you're talking about. We agreed that:

• \langle g \rangle = \langle a b \rangle + \langle a b' \rangle + \langle a' b \rangle - \langle a' b' \rangle

I proved that

• -2 \leq \langle g \rangle \leq +2

It follows that

• -2 \leq \langle a b \rangle + \langle a b' \rangle + \langle a' b \rangle - \langle a' b' \rangle \leq +2

I really have no idea what your point is. I guess I should give up, because I'm getting very frustrated.

Before we proceed, I'd better be clear about this point: Do you see that you cannot derive your result from my four valid equations?

No, I do not agree with that. I really have no idea what you are talking about. I'm using the exact same mathematics that I thought we both had agreed with.

 

[PeterDonis]

I am using the alternative technique that Bell specifically approved: see the second paragraph following Bell (1964), eqn (3).

No, you're not, because Bell explicitly says that that technique uses the same equations: "this possibility is contained in the above, since $\lambda$ stands for any number of variables and the dependencies thereon of $A$ and $B$ are unrestricted". So using this "alternative technique" doesn't change any of the equations, yet you are changing them.

 

[N88]

...

No, I do not agree with that. I really have no idea what you are talking about. I'm using the exact same mathematics that I thought we both had agreed with.

I am sorry for your frustration. If you do not agree with my statement, then please do it. That is, please: start with four equations that we agree with and derive a conclusion that we disagree with. NB: These are not the same equations that you have been using.

The four equations, valid classically and under QM are (quoting my earlier post #133): Without any assumptions, what is actually derived from measurements is four separate expectations:

• (1) \langle a b \rangle \equiv \frac{1}{N_i} \sum^{N_i}_1 a_i b_i
• (2) \langle a b' \rangle \equiv \frac{1}{N_j} \sum^{N_j}_1 a_j b'_j
• (3) \langle a' b \rangle \equiv \frac{1}{N_k} \sum^{N_k}_1 a'_k b_k
• (4) \langle a' b' \rangle \equiv \frac{1}{N_l} \sum^{N_l}_1 a'_l b'_l

To be clear, please show how these 4 equations, in the CHSH format (which I assume is what you're using), cannot exceed 2. Alternatively, what is the physical significance of the changes that you require in (1)-(4) to justify that limit of 2?

Thanks.

 

[N88]

No, you're not, because Bell explicitly says that that technique uses the same equations: "this possibility is contained in the above, since $\lambda$ stands for any number of variables and the dependencies thereon of $A$ and $B$ are unrestricted". So using this "alternative technique" doesn't change any of the equations, yet you are changing them.

Please see the next sentence re initial values. The consequence is that I make no functional change to Bell's equations; I simply use those initial values to arrive at his identical equations, or their equivalents. Specifically, I use:
$$A(a, \lambda) = - B(a, \lambda).$$