Questioning Bell's Assumption on All Local Hidden Variable Theories

[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:
$$\begin{eqnarray} A^{1}_{\mathbf{a}} &=& +1 \,, \\ A^{1}_{\mathbf{b}} &=& +1 \,, \\ A^{2}_{\mathbf{b}} &=& -1 \,, \\ A^{2}_{\mathbf{c}} &=& -1 \,. \end{eqnarray}$$
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:
$$\begin{eqnarray} \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle + \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle - \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle &\geq& -1 \,, \\ \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle - \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle + \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle &\geq& -1 \,, \qquad (*) \\ \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle + \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle + \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle &\leq& +1 \,, \\ \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle - \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle - \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle &\leq& +1 \,. \qquad (*) \end{eqnarray}$$
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 $A^{1}_{\mathbf{b}}$:
$$\begin{eqnarray} \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle + \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle + \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle &\geq& -1 \,, \qquad (\#) \\ \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle - \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle - \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle &\geq& -1 \,, \\ \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle + \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle - \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle &\leq& +1 \,, \\ \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle - \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle + \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle &\leq& +1 \,. \end{eqnarray}$$
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:
$$\begin{eqnarray} \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle + \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle - \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle &\geq& -1 \,, \\ \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle - \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle + \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle &\geq& -1 \,, \qquad (*) \\ \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle + \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle + \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle &\leq& +1 \,, \\ \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle - \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle - \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle &\leq& +1 \,. \qquad (*) \end{eqnarray}$$
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