Questioning Bell's Assumption on All Local Hidden Variable Theories

[stevendaryl]

Hmm, then I don't understand yours.

Okay, well I will expand a little.

The general idea behind local hidden variables is that in the case of maximal available knowledge, the probability of an event should be conditional only on local information (information available in the backward light cone). So how that applies to the EPR experiment is this:

Let A be the event in which Alice measures spin-up. Let \alpha be the angle giving the orientation of her detector (for simplicity, fixed in the x-y plane, with \alpha=0 being the x-axis). Let \lambda be the hidden variable associated with the production of a twin-pair. Let \omega be the hidden variable describing the details of Alice's detector (plus the immediate environment, including Alice herself). Then we assume that there is a probability function

P(A | \alpha \wedge \lambda \wedge \omega)

Similar, for Bob at the other detector, there is a probability function

P(B | \beta \wedge \lambda \wedge \nu)

where B is the event at which Bob measures spin-up, and \beta is the angle giving the orientation of his detector, and \nu is the hidden variable describing the details of Bob's detector.

Now, if everything is local, and \alpha, \beta, \lambda, \omega, \nu is the complete list of relevant facts, then Bell's assumption is that joint probabilities should factor as follows:

P(A \wedge B | \alpha \wedge \beta)<br /> = \sum P(\lambda) P(\omega) P(\nu) P(A | \alpha \wedge \lambda \wedge \omega)<br /> P(B | \beta \wedge \lambda \wedge \nu)

(Pardon my lazy notation in which I use P(\lambda) to mean "The probability that the hidden variable associated with the twin pair has value \lambda", etc.)

Now, the prediction of quantum mechanics for this case is
P(A \wedge B | \alpha \wedge \beta) = \frac{1}{2} sin^2(\frac{1}{2}(\beta - \alpha))

My claim is that you can prove that this is only possible if in fact

P(A | \alpha \wedge \lambda \wedge \omega) = P(A | \alpha \wedge \lambda)
P(B | \beta\wedge \lambda \wedge \nu) = P(A | \beta \wedge \lambda)

In other words, the hidden variables due to the details of the detector state are irrelevant.

With a little more work, we can also prove that

P(A | \alpha \wedge \lambda) = zero or one.
P(B | \beta \wedge \lambda) = zero or one.

And then with a little more work, we can prove that there is no solution; there are no probability distributions


P(A | \alpha \wedge \lambda \wedge \omega)
P(B | \beta\wedge \lambda \wedge \nu)
P(\lambda)
P(\omega)
P(\nu)

that give the correct quantum mechanical prediction for
P(A \wedge B | \alpha \wedge \beta)

(Bell's argument is in terms of correlations, not joint probabilities, but you can make a similar argument in terms of joint probabilities.)

 

[zonde]

Now, the prediction of quantum mechanics for this case is
P(A \wedge B | \alpha \wedge \beta) = \frac{1}{2} sin^2(\frac{1}{2}(\beta - \alpha))

My claim is that you can prove that this is only possible if in fact

P(A | \alpha \wedge \lambda \wedge \omega) = P(A | \alpha \wedge \lambda)
P(B | \beta\wedge \lambda \wedge \nu) = P(A | \beta \wedge \lambda)

Hmm, I can understand your point if I replace full QM prediction with reduced QM prediction only about perfect correlations (otherwise it looks like you are claiming you can disprove Bell's theorem).
And I say you are assuming that \omega and \nu are independent. And what I was saying is that it does not work if we assume that \omega and \nu are not independent.

 

[stevendaryl]

Hmm, I can understand your point if I replace full QM prediction with reduced QM prediction only about perfect correlations (otherwise it looks like you are claiming you can disprove Bell's theorem).
And I say you are assuming that \omega and \nu are independent. And what I was saying is that it does not work if we assume that \omega and \nu are not independent.

No, I'm talking about the full QM prediction, and I'm not in any way disproving Bell's theorem. I'm outlining a proof of Bell's theorem (except, as I said, for joint probabilities, instead of correlations). As I said, there is no probability distribution on the assumed hidden variables that match the predictions of QM for the EPR experiment. That's what Bell proved.

The possibility of \nu and \omega not being independent seems to be a loophole that I had not considered. I don't think that it would change anything, substantially, because you could then factor out the information that the two detectors had in common, and lump that into \lambda. But I'll have to think about it.

 

[stevendaryl]

No, I'm talking about the full QM prediction, and I'm not in any way disproving Bell's theorem. I'm outlining a proof of Bell's theorem (except, as I said, for joint probabilities, instead of correlations). As I said, there is no probability distribution on the assumed hidden variables that match the predictions of QM for the EPR experiment. That's what Bell proved.

The possibility of \nu and \omega not being independent seems to be a loophole that I had not considered. I don't think that it would change anything, substantially, because you could then factor out the information that the two detectors had in common, and lump that into \lambda. But I'll have to think about it.

Maybe you're confused by the structure of the argument. I'm not saying "There are probability distributions P(\lambda), P(\nu), P(\omega) etc., such that the predictions of QM are satisfied." I'm saying: "Let's assume that there are such probability distributions, and then see what follows from that." What follows from that is, in fact, a contradiction, because there are no such probability distributions.

 

[zonde]

No, I'm talking about the full QM prediction, and I'm not in any way disproving Bell's theorem. I'm outlining a proof of Bell's theorem (except, as I said, for joint probabilities, instead of correlations). As I said, there is no probability distribution on the assumed hidden variables that match the predictions of QM for the EPR experiment. That's what Bell proved.

Your original statement was: hidden variables + perfect correlations => counterfactual definiteness.
It does not involve full QM prediction. If you are talking about full QM prediction then you have lost the topic.

Maybe you're confused by the structure of the argument. I'm not saying "There are probability distributions P(\lambda), P(\nu), P(\omega) etc., such that the predictions of QM are satisfied." I'm saying: "Let's assume that there are such probability distributions, and then see what follows from that." What follows from that is, in fact, a contradiction, because there are no such probability distributions.

I am not confused. Given context that quoted part is not very confusing. But taken out of context it is clear that particular part has sloppy wording. And as I wanted to replay to this particular part I pointed this out.

The possibility of \nu and \omega not being independent seems to be a loophole that I had not considered. I don't think that it would change anything, substantially, because you could then factor out the information that the two detectors had in common, and lump that into \lambda. But I'll have to think about it.

Yes, it would not change anything. Or at least it would not change conclusions.
Not sure you can lump that into \lambda. I'll have to think about it too.

 

[stevendaryl]

Your original statement was: hidden variables + perfect correlations => counterfactual definiteness.

Yes. If you add in the other predictions of quantum mechanics, you get a stronger implication:

hidden variables + other predictions of EPR => contradiction

It does not involve full QM prediction.

Right. You do not need the full QM prediction in order to conclude that

hidden variables + perfect correlations (in the case of parallel or anti-parallel detector orientations) implies counterfactual definiteness.

In the case where you don't take the hidden variables of the detector into account, you have

P(A \wedge B | \alpha \wedge \beta) = \sum P(\lambda) P(A | \alpha \wedge \lambda) P(B | \beta \wedge \lambda)

In the specific case where \beta = \pi - \alpha, given the quantum prediction for joint probabilities, this becomes:
\dfrac{1}{2} = \sum P(\lambda) P(A | \alpha \wedge \lambda) P(B | (\beta = \pi - \alpha) \wedge \lambda)

But we also know:

P(A | \alpha) = \sum P(\lambda) P(A | \alpha \wedge \lambda) = \dfrac{1}{2}

since for any angle, the probability of a single detector measuring spin-up at that angle is 1/2. So putting these together, we get:
\sum P(\lambda) P(A | \alpha \wedge \lambda) P(B | (\beta = \pi - \alpha) \wedge \lambda) = \sum P(\lambda) P(A | \alpha \wedge \lambda)

We can rearrange this as follows:

\sum P(\lambda) P(A | \alpha \wedge \lambda) (1 - P(B | (\beta = \pi - \alpha) \wedge \lambda)) = 0

If you have a sum of terms that are all nonnegative, then the only way that they can add up to zero is if each term is equal to zero. Therefore, it must be:

P(\lambda) P(A | \alpha \wedge \lambda) (1 - P(B | (\beta = \pi - \alpha) \wedge \lambda)) = 0

We can insist that P(\lambda) &gt; 0, because there is no reason for the sum to range over values of \lambda that never occur. So we haveP(A | \alpha \wedge \lambda) (1 - P(B | (\beta = \pi - \alpha) \wedge \lambda)) = 0

What we can conclude from this is that

If P(A | \alpha \wedge \lambda) &gt; 0, then P(B | (\beta = \pi - \alpha) \wedge \lambda) = 1

A similar analysis leads to the conclusion:

If P(B | \beta \wedge \lambda) &gt; 0, then P(A| (\alpha= \pi - \beta) \wedge \lambda) = 1

Together, these facts imply that for any \alpha, either

P(A | \alpha \wedge \lambda) = 0 or P(A | \alpha \wedge \lambda) = 1.

For any \beta, eitherP(B | \beta\wedge \lambda) = 0 or P(B | \beta \wedge \lambda) = 1.

So in fact, A and B are deterministic functions of \alpha, \lambda and \beta, \lambda, respectively.

So local hidden variables + perfect correlations → determinism (and counterfactual definiteness).

 

[audioloop]

there are 3 conjectures at the core of Bell's theorem: free will, no signaling, outcome independence.

 

[zonde]

hidden variables + perfect correlations (in the case of parallel or anti-parallel detector orientations) implies counterfactual definiteness.

If I ask myself what is necessary to speak about counterfactual definiteness then the answer seems to be: perfect correlations and measurement independence (locality in particular context).
And I can explain my conclusion that way - to speak about counterfactual definiteness or ask "what if" type questions and test(!) them we have to be able to clone physical situation. Perfect correlations between two recordings of measurements demonstrates that we can clone physical situation if we agree to assume independence in this case (because we can't demonstrate independence, the only way is to demonstrate dependence - the we should drop that assumption).

So we don't have to assume counterfactual definiteness in Bell's theorem - if follows from QM prediction and locality.

On the other hand hidden variables actually is mathematical model of counterfactual definiteness so it presupposes counterfactual definiteness.

 

[stevendaryl]

On the other hand hidden variables actually is mathematical model of counterfactual definiteness so it presupposes counterfactual definiteness.

I don't think that's correct. Hidden variables don't presuppose counterfactual definiteness. Only deterministic hidden variables theories imply counterfactual definiteness. A nondeterministic hidden variables theory doesn't imply or presuppose counterfactual definiteness.

On the other hand, nondeterministic hidden variables are not compatible with perfect correlations. That's why I said:

Hidden variables + Perfect correlations → Counterfactual definiteness

 

[wle]

One last try to clarify that interesting illustration:So, Bell's inequality violation incl. the condition that A^{1}_{\mathbf{b}}(w) = - A^{2}_{\mathbf{b}}(w) is correctly reproduced.

They didn't test Bell's inequality. If you put the correct labelling on then the inequality they tested was
\langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle + \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle + \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle \geq - 1 \,.
This inequality does not hold for locality if \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{b}} \rangle = -1. It is trivial to construct a counterexample. Just take:
<br /> \begin{eqnarray}<br /> A^{1}_{\mathbf{a}} &amp;=&amp; +1 \,, \\<br /> A^{1}_{\mathbf{b}} &amp;=&amp; +1 \,, \\<br /> A^{2}_{\mathbf{b}} &amp;=&amp; -1 \,, \\<br /> A^{2}_{\mathbf{c}} &amp;=&amp; -1 \,.<br /> \end{eqnarray}<br />
Notice that A^{1}_{\mathbf{b}} = - A^{2}_{\mathbf{b}}. Calculate the correlator and you get
A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} + A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} + A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} = -3 \ngeq -1 \,,
so you can violate the inequality deterministically using a local model for which \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{b}} \rangle = -1. This is not news.

 

[audioloop]

I don't think that's correct. Hidden variables don't presuppose counterfactual definiteness.

i agree.
proof: Bohm model

and

http://www.tau.ac.il/~vaidman/lvhp/m105.pdf
"the core of the controversy is that quantum counterfactuals about the results of measurements of observables, and especially “elements of reality” are understood as attributing values to observables which are not observed. But this is completely foreign to quantum mechanics. Unperformed experiments have no results! “Element of reality” is just a shorthand for describing a situation in which we know with certainty the outcome of a measurement if it is to be performed, which in turn helps us to know how weakly coupled particles are influenced by the system. Having “elements of reality” does not mean having values for observables. The semantics are misleading since “elements of reality” are not “real” in the ontological sense"

 

[harrylin]

[..] What they claim to violate is Bell's original 1964 inequality. [..] Specifically, in their notation, and putting the locations back on (Lille = 1, Lyon = 2), the Bell inequality uses the assumption that A^{1}_{\mathbf{b}}(w) = - A^{2}_{\mathbf{b}}(w) [sign corrected].
[...] it just means that the correct way to state Bell's inequality should really be something like
\langle A^{1}_{\mathbf{a}}(w) A^{2}_{\mathbf{b}}(w) \rangle + \langle A^{1}_{\mathbf{a}}(w) A^{2}_{\mathbf{c}}(w) \rangle + \langle A^{1}_{\mathbf{b}}(w) A^{2}_{\mathbf{c}}(w) \rangle \geq -1 \quad \text{given that} \quad \langle A^{1}_{\mathbf{b}}(w) A^{2}_{\mathbf{b}}(w) \rangle = -1 \,. [sign corrected]
[..]

They didn't test Bell's inequality. If you put the correct labelling on then the inequality they tested was
\langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle + \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle + \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle \geq - 1 \,.
This inequality does not hold for locality if \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{b}} \rangle = -1. It is trivial to construct a counterexample. Just take:
<br /> \begin{eqnarray}<br /> A^{1}_{\mathbf{a}} &amp;=&amp; +1 \,, \\<br /> A^{1}_{\mathbf{b}} &amp;=&amp; +1 \,, \\<br /> A^{2}_{\mathbf{b}} &amp;=&amp; -1 \,, \\<br /> A^{2}_{\mathbf{c}} &amp;=&amp; -1 \,.<br /> \end{eqnarray}<br />
Notice that A^{1}_{\mathbf{b}} = - A^{2}_{\mathbf{b}}. Calculate the correlator and you get
A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} + A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} + A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} = -3 \ngeq -1 \,,
so you can violate the inequality deterministically using a local model for which \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{b}} \rangle = -1. This is not news.

I found your first analysis useful; it's unclear to me how you think that their illustration does not contain the essentials of Bell's original inequality as you indicated earlier, and which I again cited here with correction...

 

[wle]

I found your first analysis useful; it's unclear to me how you think that their illustration does not contain the essentials of Bell's original inequality as you indicated earlier, and which I again cited here with correction...

Nothing in my post needed correction. Bell derived some inequalities for the case where \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{b}} \rangle = -1. You can alternatively derive some similar but not identical inequalities for the case where \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{b}} \rangle = +1. The particular inequality that de Raedt et. al. considered is derived assuming \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{b}} \rangle = +1, and there is simply no reason to expect it should be satisfied if \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{b}} \rangle = -1.

Specifically, if you assume \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{b}} \rangle = -1, you can derive the following four inequalities:
<br /> \begin{eqnarray}<br /> \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle + \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle - \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle &amp;\geq&amp; -1 \,, \\<br /> \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle - \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle + \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle &amp;\geq&amp; -1 \,, \qquad (*) \\<br /> \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle + \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle + \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle &amp;\leq&amp; +1 \,, \\<br /> \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle - \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle - \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle &amp;\leq&amp; +1 \,. \qquad (*)<br /> \end{eqnarray}<br />
The second and fourth of these inequalities, which I've marked (*), are the ones Bell derived in 1964. Specifically, they're equivalent to Eq. (15) of Bell's 1964 paper [1]. The other two can easily be derived in an analogous manner (or, alternatively, just by flipping the sign of A^{2}_{\mathbf{c}}).

If you instead set \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{b}} \rangle = +1, you get four slightly different inequalities, which you can basically all derive by flipping the sign on A^{1}_{\mathbf{b}}:
<br /> \begin{eqnarray}<br /> \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle + \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle + \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle &amp;\geq&amp; -1 \,, \qquad (\#) \\<br /> \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle - \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle - \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle &amp;\geq&amp; -1 \,, \\<br /> \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle + \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle - \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle &amp;\leq&amp; +1 \,, \\<br /> \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle - \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle + \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle &amp;\leq&amp; +1 \,.<br /> \end{eqnarray}<br />
The first of these, (#), is the one that de Raedt et. al. tested. It can only be derived as a local bound if you set A^{1}_{\mathbf{b}} = A^{2}_{\mathbf{b}}, so violating it with A^{1}_{\mathbf{b}} = - A^{2}_{\mathbf{b}} isn't news.

(As an aside, all of the inequalities I've written here can be derived as special cases of the eight possible CHSH inequalities, just with one of the terms fixed to +1 or -1.)
[1] J. S. Bell, Physics 1 3 195--200 (1964).

 

[harrylin]

[..]
Specifically, if you assume \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{b}} \rangle = -1, you can derive the following four inequalities:
<br /> \begin{eqnarray}<br /> \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle + \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle - \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle &amp;\geq&amp; -1 \,, \\<br /> \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle - \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle + \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle &amp;\geq&amp; -1 \,, \qquad (*) \\<br /> \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle + \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle + \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle &amp;\leq&amp; +1 \,, \\<br /> \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle - \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle - \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle &amp;\leq&amp; +1 \,. \qquad (*)<br /> \end{eqnarray}<br />
The second and fourth of these inequalities, which I've marked (*), are the ones Bell derived in 1964. Specifically, they're equivalent to Eq. (15) of Bell's 1964 paper [1]. [..]

Oops I had overlooked the lack of a minus sign in that Boole inequality of De Raedt. Thanks for pointing that out!
I'll dig into that and after that I'll post a comment in the thread on that paper of De Raedt (will also mention that here).

 

[Ilja]

On the other hand, nondeterministic hidden variables are not compatible with perfect correlations. That's why I said:

Hidden variables + Perfect correlations → Counterfactual definiteness

This also presupposes something like locality.

Else, one of the two measurements, the one which happens first in some absolute time, may have an arbitrary result, but, then, send an FTL message to the other part which fixes the other measurement result.

 

[stevendaryl]

This also presupposes something like locality.

Else, one of the two measurements, the one which happens first in some absolute time, may have an arbitrary result, but, then, send an FTL message to the other part which fixes the other measurement result.

Right. My mistake. It should be something like:

Perfect correlations + Hidden variables/realism + Einstein locality → Determinism/counterfactual definiteness

 

[harrylin]

OK I now commented on the last contribution by Wie here:
https://www.physicsforums.com/showthread.php?p=4465579
In a nutshell, it doesn't seem to make an essential difference; it remains an illustration that local realism can break Bell's inequality, or so it looks to me.

And concerning Morrobay's comments, it's regretfully not clear to me. But note the missing absolute sign.

 

[stevendaryl]

OK I now commented on the last contribution by Wie here:
https://www.physicsforums.com/showthread.php?p=4465579
In a nutshell, it doesn't seem to make an essential difference; it remains an illustration that local realism can break Bell's inequality, or so it looks to me.

And concerning Morrobay's comments, it's regretfully not clear to me. But note the missing absolute sign.

To me, Bell's inequality is not important in itself. It is simply a step in proving that the joint probabilities predicted by quantum mechanics in the twin-pair EPR experiment are not explainable in terms of a particular type of model.

Given two distant experimenters, Alice and Bob, let

P(A \wedge B | \alpha \wedge \beta) be the joint probability that Alice, got result A when her detector had setting \alpha and Bob got result B when his detector had setting \beta. A locally realistic hidden variables explanation of this joint probability would have

1. A set \Lambda of possible values for the hidden variable.
2. A probability distribution P(\lambda) for the values of the hidden variable.
3. A conditional probability P(A | \lambda \wedge \alpha) for Alice to get result A, given that the hidden variable has value \lambda and Alice's detector setting is \alpha
4. A conditional probability P(B | \lambda \wedge \beta) for Bob to get result B, given that the hidden variable has value \lambda and Bob's detector setting is \alpha

The explanation succeeds provided that:

P(A \wedge B | \alpha \wedge \beta) = \sum_\lambda P(\lambda) P(A | \lambda \wedge \alpha) P(B | \lambda \wedge \beta)

Using the prediction of quantum mechanics for the twin-pair EPR experiment to compute P(A \wedge B | \alpha \wedge \beta), we can show that there is no local, realistic hidden variables explanation of the sort described by 1-4 above. To me, Bell's inequality is only of interest as a step in establishing this.

 

[morrobay]

OK I now commented on the last contribution by Wie here:
https://www.physicsforums.com/showthread.php?p=4465579
In a nutshell, it doesn't seem to make an essential difference; it remains an illustration that local realism can break Bell's inequality, or so it looks to me.

And concerning Morrobay's comments, it's regretfully not clear to me. But note the missing absolute sign.

In this form of the inequality: 1 + {A¹_bA²_c} + A¹_aA²_c} ≥ {A¹_aA²_b
spin values are multiplied to get overall value with:
A¹_a = -1
A¹_b = -1
A²_b = -1
A²_c = +1
Inequality is dis proven, 1 -1 -1 ( is not ≥ ) 1

In this form of the inequality : 1 + P(b+c-) + P(a+c-) ≥ P(a+b-)
overall value of inequality is with addition: 1 + [ P₁+P₅] + [ P₁+P₈] ≥ [P₁+P₂]
::A:::::::::::::::::::::B
a b c:::::::::::::::::a b c
+ + +:::::::::::::::::- - - P₁
+ + -:::::::::::::::::- - + P₂
+ - -:::::::::::::::::- + + P₃
- - -:::::::::::::::::+ + + P₄
- + +::::::::::::::::;+ - - P₅
- - +:::::::::::::::::+ + - P₆
- + -:::::::::::::::::+ - + P₇
+ - +:::::::::::::::::- + - P₈

P₁ converts to P₂, b+c- to a+b- during measurement
P₈ converts to P₂, a+c- to a+b- during measurement
Then inequality is dis proven

 

[morrobay]

In this form of the inequality: 1 + {A¹_bA²_c} + A¹_aA²_c} ≥ {A¹_aA²_b
spin values are multiplied to get overall value with:
A¹_a = -1
A¹_b = -1
A²_b = -1
A²_c = +1
Inequality is dis proven, 1 -1 -1 ( is not ≥ ) 1

In this form of the inequality : 1 + P(b+c-) + P(a+c-) ≥ P(a+b-)
overall value of inequality is with addition: 1 + [ P₁+P₅] + [ P₁+P₈] ≥ [P₁+P₂]
::A:::::::::::::::::::::B
a b c:::::::::::::::::a b c
+ + +:::::::::::::::::- - - P₁
+ + -:::::::::::::::::- - + P₂
+ - -:::::::::::::::::- + + P₃
- - -:::::::::::::::::+ + + P₄
- + +::::::::::::::::;+ - - P₅
- - +:::::::::::::::::+ + - P₆
- + -:::::::::::::::::+ - + P₇
+ - +:::::::::::::::::- + - P₈

P₁ converts to P₂, b+c- to a+b- during measurement
P₈ converts to P₂, a+c- to a+b- during measurement
Then inequality is dis proven

Note the above, if correct, applies to individual cases. For the total , the correlation function:

P(a.b) = 1/N Ʃ(A_iB_i)

A(ah₁) = ± 1
B(bh₂) = ± 1
++ = +, -- = +, +- = -, -+ = -
P(a,b) = # coincidences - # anti- coincidences