I How does Bell make this step in his proof?

[N88]

From drchinese website http://www.drchinese.com/David/Bell_Compact.pdf on page 406 there are 2 equations at the top. How do you get from the top one to the second one? There is a hint about using (1) but I think it cannot be done. You might be able to do it with other assumptions but I think they will be wrong too. Any suggestions?

 

[Nugatory]

I have taken the liberty of changing the thread title to reflect the question that's actually being asked.

 

[zonde]

There is a hint about using (1) but I think it cannot be done.

$A(\vec{a},\lambda)$ is either 1 or -1 so $[1-A(\vec{b},\lambda)A(\vec{c},\lambda)]$ is either 0 or 2. If for particular $\lambda$ it's 0 then expression in the first integral for the same $\lambda$ is 0 as well. If it's 2 then in first integral it's 2 or -2. So there is no way how module of the first integral can get bigger than the second integral.

 

[N88]

$A(\vec{a},\lambda)$ is either 1 or -1 so $[1-A(\vec{b},\lambda)A(\vec{c},\lambda)]$ is either 0 or 2. If for particular $\lambda$ it's 0 then expression in the first integral for the same $\lambda$ is 0 as well. If it's 2 then in first integral it's 2 or -2. So there is no way how module of the first integral can get bigger than the second integral.

Are you discussing how to get from the second equation to the third equation? My question is how to get from the first to the second.

 

[zonde]

My question is how to get from the first to the second.

You mean how to see that $A(\vec{a},\lambda)A(\vec{b},\lambda)-A(\vec{a},\lambda)A(\vec{c},\lambda)=-A(\vec{a},\lambda)A(\vec{b},\lambda)[A(\vec{b},\lambda)A(\vec{c},\lambda)-1]$ ?

 

[N88]

You mean how to see that $A(\vec{a},\lambda)A(\vec{b},\lambda)-A(\vec{a},\lambda)A(\vec{c},\lambda)=-A(\vec{a},\lambda)A(\vec{b},\lambda)[A(\vec{b},\lambda)A(\vec{c},\lambda)-1]$ ?

Thanks. Yes.

 

[zonde]

Thanks. Yes.

It's easy to see in opposite direction. Just notice that from (1) you get that $A(\vec{b},\lambda)^2=1$

 

[N88]

It's easy to see in opposite direction. Just notice that from (1) you get that $A(\vec{b},\lambda)^2=1$

But in the second equation, the first $A(\vec{b},\lambda)$ was tested by Bob when Alice's detector was set to $\vec{a}$. The second $A(\vec{b},\lambda)$ in the second equation can be tested by Bob when Alice's detector is at setting $\vec{c}$. How does Bob come to have the same particles to test to get $A(\vec{b},\lambda)^2=1$? Because doesn't that relation $A(\vec{b},\lambda)^2=1$ only hold for tests on the same λ?

Looks like there's another assumption somewhere?

 

[zonde]

But in the second equation, the first $A(\vec{b},\lambda)$ was tested by Bob when Alice's detector was set to $\vec{a}$. The second $A(\vec{b},\lambda)$ in the second equation can be tested by Bob when Alice's detector is at setting $\vec{c}$. How does Bob come to have the same particles to test to get $A(\vec{b},\lambda)^2=1$? Because doesn't that relation $A(\vec{b},\lambda)^2=1$ only hold for tests on the same λ?

Looks like there's another assumption somewhere?

$\lambda$ is independent of measurement settings. If follows from assumption that is stated at the top of page 404 (Alice's result does not depend on Bob's measurement setting and vice versa).

 

[DrChinese]

If (and this sequence works in either direction):
$-[A(\vec{a},\lambda)A(\vec{b},\lambda)-A(\vec{a},\lambda)A(\vec{c},\lambda)]=A(\vec{a},\lambda)A(\vec{b},\lambda)[A(\vec{b},\lambda)A(\vec{c},\lambda)-1]$

Then:
$-[A(\vec{a},\lambda)A(\vec{b},\lambda)-A(\vec{a},\lambda)A(\vec{c},\lambda)]=A(\vec{a},\lambda)A(\vec{b},\lambda)[-1 + A(\vec{b},\lambda)A(\vec{c},\lambda)]$

Then:
$A(\vec{a},\lambda)A(\vec{c},\lambda)=A(\vec{a},\lambda)A(\vec{b},\lambda)[A(\vec{b},\lambda)A(\vec{c},\lambda)]$

Then we are finally saying:
$1[A(\vec{a},\lambda)A(\vec{c},\lambda)]=A(\vec{b},\lambda)A(\vec{b},\lambda)[A(\vec{a},\lambda)A(\vec{c},\lambda)]=A(\vec{a},\lambda)A(\vec{b},\lambda)[A(\vec{b},\lambda)A(\vec{c},\lambda)]$

Which I think Bell took for granted we would follow, I guess with some reference to (1) - not sure. As zonde already mentioned, $A(\vec{b},\lambda) A(\vec{b},\lambda)$ is 1 classically. The above is a relationship that should be true if $\vec{a}, \vec{b}, \vec{c}$ are all simultaneously well defined. That assumption is essentially the assumption of realism - that $\vec{a}, \vec{b}, \vec{c}$ are well-defined at all times. If they are not, it is like dividing by zero and you get results that themselves are suspect. Ergo a Bell Inequality, which does not always hold.

 

[N88]

If (and this sequence works in either direction):
$-[A(\vec{a},\lambda)A(\vec{b},\lambda)-A(\vec{a},\lambda)A(\vec{c},\lambda)]=A(\vec{a},\lambda)A(\vec{b},\lambda)[A(\vec{b},\lambda)A(\vec{c},\lambda)-1]$
Then:

But here you have assumed on the RHS the equality that I want to derive from the LHS. My earlier post remains valid:

On RHS, the first $A(\vec{b},\lambda)$ was tested by Bob when Alice's detector was set to $\vec{a}$. The second $A(\vec{b},\lambda)$ in the second equation can be tested by Bob when Alice's detector is at setting $\vec{c}$. How does Bob come to have the same particles to test to get $A(\vec{b},\lambda)^2=1$? Because doesn't that relation $A(\vec{b},\lambda)^2=1$ only hold for tests on the same λ?

Looks like there's another assumption somewhere?[/QUOTE]

 

[N88]

$\lambda$ is independent of measurement settings. If follows from assumption that is stated at the top of page 404 (Alice's result does not depend on Bob's measurement setting and vice versa).

I agree. But to make sense of their test results, they need to be comparing paired results from tests on the same particle pair. And Bell and QM do not allow that you can test the same pair twice.

 

[zonde]

But to make sense of their test results, they need to be comparing paired results from tests on the same particle pair. And Bell and QM do not allow that you can test the same pair twice.

We can do that in Bell's model. In his model outcome of measurement is fully determined by measurement angle and $\lambda$ and of course we can copy value (or set of values) of lambda.
Of course we can't rewind back reality, change something and then look what is different. But our models are not copies of reality, they are just models. And we can do such things with models.

 

[rubi]

Of course we can't rewind back reality, change something and then look what is different. But our models are not copies of reality, they are just models. And we can do such things with models.

We can do it only in classical models. In quantum models, non-commuting observables don't simultaneously have well-defined outcomes, so that it's not possible to consistently assign values to unobserved quantities. This phenomenon is called contextuality. It's very apparent in the GHZ state, where any assignment of values to unbserved quantities is inconsistent.

 

[stevendaryl]

On RHS, the first $A(\vec{b},\lambda)$ was tested by Bob when Alice's detector was set to $\vec{a}$. The second $A(\vec{b},\lambda)$ in the second equation can be tested by Bob when Alice's detector is at setting $\vec{c}$. How does Bob come to have the same particles to test to get $A(\vec{b},\lambda)^2=1$? Because doesn't that relation $A(\vec{b},\lambda)^2=1$ only hold for tests on the same λ?

I'm not sure I understand your objection. Bell is assuming (for the purpose of showing that the assumption leads to a contradiction with the predictions of quantum mechanics) that when a twin-pair of particles is created, a value of \lambda is associated with each particle, and that Alice's measurement result is a deterministic function of \lambda. She deterministically gets the result A(\vec{a}, \lambda), where \lambda is the hidden variable associated with the twin pair, and \vec{a} is the axis she uses for the measurement. Similarly, Bob's measurement result is a different function of \lambda: He gets the result B(\vec{b}, \lambda), where \vec{b} is the axis he uses.

So A(\vec{a}, \lambda) and B(\vec{b}, \lambda) are assumed to be ordinary functions of two parameters. His proof then just amounts to showing that there are no such functions reproducing the predictions of quantum mechanics.

In reasoning about A and B, you can forget that they correspond to measurements, and just think of them as mathematical functions.

 

[zonde]

We can do it only in classical models. In quantum models, non-commuting observables don't simultaneously have well-defined outcomes, so that it's not possible to consistently assign values to unobserved quantities. This phenomenon is called contextuality.

Quantum contextuality basically means that certain measurement results of individual particles can't be predetermined by hidden variables. This is the same conclusion that Bell reaches with his theorem.
However it does not mean that we can't calculate statistical predictions for alternative scenarios.

 

[rubi]

However it does not mean that we can't calculate statistical predictions for alternative scenarios.

It means that we cannot calculate statistical predictions for scenarios in which non-commuting observables have simultaneously well-defined outcomes. Hence the answer to post #12 is that this is where Bell's assumptions contradict QM, which eventually allows for the inequality violation.

 

[zonde]

It means that we cannot calculate statistical predictions for scenarios in which non-commuting observables have simultaneously well-defined outcomes. Hence the answer to post #12 is that this is where Bell's assumptions contradict QM, which eventually allows for the inequality violation.

OP has doubts whether Bell reasoning does not make additional assumption. So what is your answer? Is there additional assumption that does not follow from assumptions he has stated?

 

[rubi]

OP has doubts whether Bell reasoning does not make additional assumption. So what is your answer? Is there additional assumption that does not follow from assumptions he has stated?

Yes, Bell's paper is not very explicit about the assumptions he makes. This is the reason for the big confusion about his assumptions. Bell never states explicitely that the existence of values for unobserved variables is an assumption. On the contrary, he believes that this follows from the EPR argument. That's why I was pointing at the GHZ state, which shows that the EPR argument itself is inconsistent.

 

[zonde]

Yes, Bell's paper is not very explicit about the assumptions he makes. This is the reason for the big confusion about his assumptions. Bell never states explicitely that the existence of values for unobserved variables is an assumption. On the contrary, he believes that this follows from the EPR argument. That's why I was pointing at the GHZ state, which shows that the EPR argument itself is inconsistent.

Right at the start Bell says: "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, that creates the essential difficulty."
Then later on he provides his reasoning how he arrives at "more complete specification of the state" described as lambda.
Then again he gives his assumption at the top of page 404 (that B's result does not depend on A's measurement setting and vice versa).

But as I understand you maintain position that Bell's reasoning is faulty and there can be independent measurements on two distant particles that with certainty will give (UP,DOWN) or (DOWN,UP) but not (UP,UP) nor (DOWN,DOWN) and yet it does not follow that we can meaningfully talk about UP or DOWN measurement for each particle separately. So it is reasonable option that combination of two independent parts taken together is meaningful while each independent part taken separately is not.

 

[DrChinese]

But here you have assumed on the RHS the equality that I want to derive from the LHS. My earlier post remains valid:

On RHS, the first $A(\vec{b},\lambda)$ was tested by Bob when Alice's detector was set to $\vec{a}$. The second $A(\vec{b},\lambda)$ in the second equation can be tested by Bob when Alice's detector is at setting $\vec{c}$. How does Bob come to have the same particles to test to get $A(\vec{b},\lambda)^2=1$? Because doesn't that relation $A(\vec{b},\lambda)^2=1$ only hold for tests on the same λ?

Looks like there's another assumption somewhere?

$\lambda$ is the same set of parameters or functions for settings a, b c. The "other" assumption is one I spelled out: realism. That a, b and c yield simultaneously well defined outcome functions. That assumption is the heart of the Bell proof, as can be seen.

You also say: "But to make sense of their test results, they need to be comparing paired results from tests on the same particle pair. And Bell and QM do not allow that you can test the same pair twice."

That statement is incorrect. Bell assumes you could have made a counterfactual measurement. That assumption leads to a contradiction with experiment (and with the predictions of QM). Yours is a red herring argument that is frequently made when discussing Bell. Bell assumes the classical position - in which case your objection wouldn't matter. The QM position is irrelevant to that view other than the fact that it makes a different prediction. But QM could be wrong and Bell would still set limits (the inequality), and that would still be contradicted by experiment. (Of course Bell didn't know that for sure back in the day, but we do.)

 

[rubi]

Right at the start Bell says: "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, that creates the essential difficulty."

Yes, this is where he misses the possibility of contextuality. It was later shown by Stapp and Eberhard that this is an essential assumption of every Bell-type theorem. Theorems like GHZ or Kochen-Specker show that it is violated by QM. From Bell's theorem, we can only conclude that contextuality or locality (or both) is violated by QM.

But as I understand you maintain position that Bell's reasoning is faulty

No, it's not faulty. He just doesn't spell out the assumptions very explicitely. Original papers are rarely a good source for learning something.

and there can be independent measurements on two distant particles that with certainty will give (UP,DOWN) or (DOWN,UP) but not (UP,UP) nor (DOWN,DOWN) and yet it does not follow that we can meaningfully talk about UP or DOWN measurement for each particle separately. So it is reasonable option that combination of two independent parts taken together is meaningful while each independent part taken separately is not.

Bell assumes that there is a value up/down for each particle and each value of the angle $\vec a$, even for those that aren't measured. That's the assumption that isn't made very explicit in the paper and which is violated by QM, independently of Bell's result.

 

[DrChinese]

Yes, Bell's paper is not very explicit about the assumptions he makes. This is the reason for the big confusion about his assumptions. Bell never states explicitely that the existence of values for unobserved variables is an assumption. On the contrary, he believes that this follows from the EPR argument. That's why I was pointing at the GHZ state, which shows that the EPR argument itself is inconsistent.

I agree with rubi on this. Although the 2 equations at the top page of page 406 is where the realism assumption is introduced, Bell hardly makes note of it. He casually says "let c be another unit vector" when Bell might have said:

"Let's assume there are an infinite number of other unit vectors, c, d, e, etc. that are also defined by $\lambda$. Those would correspond to elements of reality as defined in the EPR paper. Using vector c, we would have:"

Had he done that, it would have clarified things for many a subsequent reader. However, at the time he wrote there were only a relatively small group of persons he was making his argument to, and he expected they would see his reasoning as presented. Apparently that was correct.

 

[DrChinese]

Right at the start Bell says: "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, that creates the essential difficulty."
Then later on he provides his reasoning how he arrives at "more complete specification of the state" described as lambda.
Then again he gives his assumption at the top of page 404 (that B's result does not depend on A's measurement setting and vice versa).

But as I understand you maintain position that Bell's reasoning is faulty and there can be independent measurements on two distant particles that with certainty will give (UP,DOWN) or (DOWN,UP) but not (UP,UP) nor (DOWN,DOWN) and yet it does not follow that we can meaningfully talk about UP or DOWN measurement for each particle separately. So it is reasonable option that combination of two independent parts taken together is meaningful while each independent part taken separately is not.

As mentioned, I agree with rubi's take on Bell. Forget Bell's statement "It is the requirement of locality ... that creates the essential difficulty." The requirement of locality is not the essential difficulty, if anything it is the requirement of realism that is essential. That can be better seen in the GHZ proof, which does not depend on locality. The requirement of locality is of course needed for the Bell conclusion, as the EPR formulation required 2 separated particles and you must assume there is no interaction between them that allows $\lambda$ to include a remote measurement setting.

 

[zonde]

No, it's not faulty. He just doesn't spell out the assumptions very explicitely. Original papers are rarely a good source for learning something.

Bell assumes that there is a value up/down for each particle and each value of the angle $\vec a$, even for those that aren't measured. That's the assumption that isn't made very explicit in the paper and which is violated by QM, independently of Bell's result.

I will just quote Bell (emphasis mine): "Since we can predict in advance the result of measuring any chosen component of $\vec\sigma_2$, by previously measuring the same component of $\vec\sigma_1$, it follows that the result of any such measurement must actually be predetermined."

 

[DrChinese]

I will just quote Bell (emphasis mine): "Since we can predict in advance the result of measuring any chosen component of $\vec\sigma_2$, by previously measuring the same component of $\vec\sigma_1$, it follows that the result of any such measurement must actually be predetermined."

Exactly. The classical view is that a function of $\lambda$ at any and all vector settings are predetermines the outcome. That was the (wrong) EPR conclusion. They too assumed realism and locality. But they assumed a theory was possible that respected these and would replicate the predictions of QM. Obviously they couldn't know what we now know.

 

[zonde]

Exactly. The classical view is that a function of $\lambda$ at any and all vector settings are predetermines the outcome. That was the (wrong) EPR conclusion. They too assumed realism and locality. But they assumed a theory was possible that respected these and would replicate the predictions of QM. Obviously they couldn't know what we now know.

Well, Bell assumes locality and from that follows that measurements are independent no matter what measurement angle A or B chooses. And QM prediction says that perfect anti-correlations will be observed whenever A and B chooses the same measurement angle. There is no lambda yet.

 

[rubi]

Well, Bell assumes locality and from that follows that measurements are independent no matter what measurement angle A or B chooses. And QM prediction says that perfect anti-correlations will be observed whenever A and B chooses the same measurement angle. There is no lambda yet.

Forget the EPR argument. It might be intriguing, but it is falsified experimentally by GHZ. There are no elements of reality for unobserved spin directions, not even if you don't introduce $\lambda$. However, Bell clearly uses them in his proof.

 

[DrChinese]

And QM prediction says that perfect anti-correlations will be observed whenever A and B chooses the same measurement angle. There is no lambda yet.

Right. That is not the classical view (EPR) that Bell was writing about. QM does not posit predetermination of all outcomes. Just that a few will be perfectly correlated. The classical view goes much further. Specifically, that outcomes are predetermined at all possible measurement settings.

 

[zonde]

Forget the EPR argument. It might be intriguing, but it is falsified experimentally by GHZ. There are no elements of reality for unobserved spin directions, not even if you don't introduce $\lambda$. However, Bell clearly uses them in his proof.

Of course we all know that Bell inequalities are violated. As far as I know there is no loophole free experiment performed for GHZ but it's not very important.
Important question is from what assumptions one can construct Bell inequality. So that we know what assumptions have to be questioned when we learn about experimental violation of Bell inequalities. And this question can be discussed without much reference to experiments.

 

[zonde]

QM does not posit predetermination of all outcomes. Just that a few will be perfectly correlated.

Yes, of course. To get predetermination we need perfect correlation for matching angles and we have to assume independence of measurements at any angle.

 

[rubi]

Of course we all know that Bell inequalities are violated. As far as I know there is no loophole free experiment performed for GHZ but it's not very important.

As we have seen with the Bell tests, loopholes are likely going to be closed in the future. We currently have no reason to doubt the correctness of the quantum mechanical predictions.

Important question is from what assumptions one can construct Bell inequality. So that we know what assumptions have to be questioned when we learn about experimental violation of Bell inequalities. And this question can be discussed without much reference to experiments.

This question can be discussed without reference to experiments. We would spend 20 pages discussing what's wrong with the EPR argument. But we don't need to, since GHZ already proves that the EPR reasoning must be wrong, unless one believes in a loophole and the failure of quantum mechanical predictions.

 

[DrChinese]

Yes, of course. To get predetermination we need perfect correlation for matching angles and we have to assume independence of measurements at any angle.

To deduce predetermination, we need perfect correlation for matching angles, whether measured or not.

 

[N88]

Please: What is X?

We start here:

(1) $-[A(\vec{a},\lambda)A(\vec{b},\lambda)-A(\vec{a},\lambda)A(\vec{c},\lambda)]$; under an integral.

IF X [a mathematical expression, explained in words; preferably something suggested by Einstein or EPR, for (in his Introduction), that's where Bell says he starts from]

THEN (by Bell's equality):

(1) $=$ (2) $=A(\vec{a},\lambda)A(\vec{b},\lambda)[A(\vec{b},\lambda)A(\vec{c},\lambda)-1]$; under an integral. QED.

Thank you.

 

[atyy]

Forget the EPR argument. It might be intriguing, but it is falsified experimentally by GHZ. There are no elements of reality for unobserved spin directions, not even if you don't introduce $\lambda$. However, Bell clearly uses them in his proof.

Right. That is not the classical view (EPR) that Bell was writing about. QM does not posit predetermination of all outcomes. Just that a few will be perfectly correlated. The classical view goes much further. Specifically, that outcomes are predetermined at all possible measurement settings.

I'm not fond of the EPR elements of reality argument. However, Bell does show in one of his papers that the inequality remains the same even if the outcome is not predeternined at all possible measurement settings - where possible means "possible to the experimentalist".

 

[stevendaryl]

Please: What is X?

We start here:

(1) $-[A(\vec{a},\lambda)A(\vec{b},\lambda)-A(\vec{a},\lambda)A(\vec{c},\lambda)]$; under an integral.

IF X (a mathematical expression, explained in words; preferably something suggested by Einstein or EPR)

THEN (by Bell's equality):

(1) $=$ (2) $=A(\vec{a},\lambda)A(\vec{b},\lambda)[A(\vec{b},\lambda)A(\vec{c},\lambda)-1]$; under an integral. QED.

Thank you.

X is just algebra.

Once again:

Since A(\vec{b}, \lambda) = \pm 1, then A(\vec{b}, \lambda) A(\vec{b}, \lambda) = +1. So we can write:

- (A(\vec{a}, \lambda) A(\vec{b}, \lambda) - A(\vec{a},\lambda) A(\vec{c},\lambda)) = - (A(\vec{a}, \lambda) A(\vec{b}, \lambda) - A(\vec{a},\lambda) A(\vec{b}, \lambda) A(\vec{b}, \lambda) A(\vec{c},\lambda))

Now, we can factor out A(\vec{a},\lambda) A(\vec{b}, \lambda):

- (A(\vec{a}, \lambda) A(\vec{b}, \lambda) - A(\vec{a},\lambda) A(\vec{c},\lambda)) = - A(\vec{a}, \lambda) A(\vec{b}, \lambda) (1 - A(\vec{b}, \lambda) A(\vec{c},\lambda))

 

[dlgoff]

Just that a few will be perfectly correlated.

I'm sure this is obvious for QM experts, but could you maybe say a little about the "few"?

 

[N88]

X is just algebra.

Once again: Since

(D) A(\vec{b}, \lambda) = \pm 1, then A(\vec{b}, \lambda) A(\vec{b}, \lambda) = +1. So we can write:

(E) - (A(\vec{a}, \lambda) A(\vec{b}, \lambda) - A(\vec{a},\lambda) A(\vec{c},\lambda)) = - (A(\vec{a}, \lambda) A(\vec{b}, \lambda) - A(\vec{a},\lambda) A(\vec{b}, \lambda) A(\vec{b}, \lambda) A(\vec{c},\lambda))

Now, we can factor out A(\vec{a},\lambda) A(\vec{b}, \lambda):

(F) - (A(\vec{a}, \lambda) A(\vec{b}, \lambda) - A(\vec{a},\lambda) A(\vec{c},\lambda)) = - A(\vec{a}, \lambda) A(\vec{b}, \lambda) (1 - A(\vec{b}, \lambda) A(\vec{c},\lambda))

So A(\vec{a}, \lambda) and B(\vec{b}, \lambda) are assumed to be ordinary functions of two parameters. His proof then just amounts to showing that there are no such functions reproducing the predictions of quantum mechanics.

In reasoning about A and B, you can forget that they correspond to measurements, and just think of them as mathematical functions.

Here's the long answer: I see no connection with EPR. Though Bell claimed his formulation was connected to EPR, their idea was NOT this simplistic.

Here's the short answer: There are no such functions of two parameters in math or physics. But let's see if we can get around that dilemma.

I start with Thank You. It's my belief that you have reproduced Bell's argument (having been-there done-that myself); and it will be interesting to see if we can improve on it.

Accepting a solution by means of functions, then there must be three parameters, else they're not functions.

Say we assume A(\vec{a}, \lambda) and B(\vec{b}, \lambda) to be ordinary functions full-stop. Since a function cannot map the same inputs to two different values, you need to accept something like this: A(\vec{b}, \lambda+)= +1; A(\vec{b}, \lambda-)= -1.

So how do you now manage the physics of the problem? How do you ensure that, when you bring random A(\vec{b}, \lambda+)= +1 and random A(\vec{b}, \lambda-)= -1 together, their product is always one?

This earlier reply has a clue to the physical problem:

But in the second equation, the first $A(\vec{b},\lambda)$ was tested by Bob when Alice's detector was set to $\vec{a}$. The second $A(\vec{b},\lambda)$ in the second equation can be tested by Bob when Alice's detector is at setting $\vec{c}$. How does Bob come to have the same particles to test to get $A(\vec{b},\lambda)^2=1$? Because doesn't that relation $A(\vec{b},\lambda)^2=1$ only hold for tests on the same λ?

Looks like there's another assumption somewhere?

So here's the new question, under the function-based analysis with A(\vec{b}, \lambda+)= +1; A(\vec{b}, \lambda-)= -1:

How many \lambdas in (E) need be identical, to allow the reduction to (F)?

 

[stevendaryl]

So how do you now manage the physics of the problem? How do you ensure that, when you bring random A(\vec{b}, \lambda+)= +1 and random A(\vec{b}, \lambda-)= -1 together, their product is always one?

The claim made is that for a FIXED value of \lambda, it must be the case that

-(A(\vec{a}, \lambda) A(\vec{b}, \lambda) - A(\vec{a}, \lambda) A(\vec{c}, \lambda)) = A(\vec{a}, \lambda) A(\vec{b}, \lambda) (1 - A(\vec{a}, \lambda) A(\vec{c}, \lambda))

That has been proved. You can ask: Why was Bell interested in proving this purely mathematical fact? Well, it was a step in proving the impossibility of a certain kind of hidden-variables theory that reproduced the predictions of quantum mechanics. But that particular step is pure mathematics.

 

[zonde]

Accepting a solution by means of functions, then there must be three parameters, else they're not functions.

Say we assume A(\vec{a}, \lambda) and B(\vec{b}, \lambda) to be ordinary functions full-stop. Since a function cannot map the same inputs to two different values, you need to accept something like this: A(\vec{b}, \lambda+)= +1; A(\vec{b}, \lambda-)= -1.

So how do you now manage the physics of the problem? How do you ensure that, when you bring random A(\vec{b}, \lambda+)= +1 and random A(\vec{b}, \lambda-)= -1 together, their product is always one?

You don't multiply $A(\vec{b}, \lambda+)$ with $A(\vec{b}, \lambda-)$ that's the point. You multiply $A(\vec{b}, \lambda_1)$ with $A(\vec{b}, \lambda_1)$
and $A(\vec{b}, \lambda_2)$ with $A(\vec{b}, \lambda_2)$ and so on. You never multiply $A(\vec{b}, \lambda_1)$ with $A(\vec{b}, \lambda_3)$.

So if you object that this is possible you have to backtrack Bell's reasoning to the point where he introduces expression $A(\vec{a}, \lambda)=\pm1$

 

[zonde]

Here's the short answer: There are no such functions of two parameters in math or physics.

Yes, that's true. But we couldn't be sure about that without Bell and experiments.

 

[DrChinese]

I'm sure this is obvious for QM experts, but could you maybe say a little about the "few"?

Where theta = 0 degrees, 90 degrees, 180 degrees, 270 degrees.

 

[atyy]

It's my belief that you have reproduced Bell's argument (having been-there done-that myself); and it will be interesting to see if we can improve on it.

Many people do think Bell was not quite clear initially. For modern presentations, I like:

https://arxiv.org/abs/1503.06413
Causarum Investigatio and the Two Bell's Theorems of John Bell
Howard M. Wiseman, Eric G. Cavalcanti

https://arxiv.org/abs/1208.4119
The lesson of causal discovery algorithms for quantum correlations: Causal explanations of Bell-inequality violations require fine-tuning
Christopher J. Wood, Robert W. Spekkens

Bell was more concerned about classical reality, but a very important corollary of his theorem is that if there is no faster than light signalling, then a violation of Bell inequalities implies operational randomness. This development could, in principle, be used to certify that random numbers are truly random for any practical purpose:

https://arxiv.org/abs/0911.3427
Random Numbers Certified by Bell's Theorem
S. Pironio, A. Acin, S. Massar, A. Boyer de la Giroday, D. N. Matsukevich, P. Maunz, S. Olmschenk, D. Hayes, L. Luo, T. A. Manning, C. Monroe

 

[zonde]

To deduce predetermination, we need perfect correlation for matching angles, whether measured or not.

We speak about model of reality not reality, right?
Now what it takes to claim that in particular model measurements of A and B are independent?
My answer is that it should be possible to split from the main model two independent submodels that take as an input something that represents two separate but entangled particles and separate measurement angle for each submodel.

 

[Nicky665]

These guys, Hess/DeRaedt/Michelsen, say the major assumption in Bell's inequalities is counterfactual definitess

https://arxiv.org/pdf/1605.04889.pdf

"The major premise for the derivation of Bell’s inequality is
counterfactual definiteness, which in connection with Bell’s
use of setting variables restricts the domain of the variables
in the argument of Bell’s functions A to a subset NB of general
physical independent variables. NB does not include the
variables necessary to describe a general dynamics describing
many body interactions in the measurement equipment.
Using only the independent variables defined by NB, it is impossible
to find a violation of Bell’s inequality, which therefore
represents a demarcation between possible and impossible
experience [8], not between classical and quantum physics."

 

[Heinera]

These guys, Hess/DeRaedt/Michelsen, say the major assumption in Bell's inequalities is counterfactual definitess

Counterfactual definiteness (also known as "realism") is one of the two major assumptions in Bell's theorem, yes (the other being locality). Nothing particularly novel in that paper.