Boole vs. Bell - the latest paper of De Raedt et al

[wle]

Now, it looks to me that your representation here above of Bell's original inequality is still not quite right: an absolute sign is lacking. According to my copy, Bell's eq.15 for locations 1 and 2 is (rearranged):

$$\lvert \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle - \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle \rvert - \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle \leq +1 \,.$$

This is the same thing as I wrote, simply because if you measure some quantity X in an experiment then either \lvert X \rvert = X or \lvert X \rvert = -X. So the inequality as you write it above is equivalent to two linear inequalities being satisfied:
$$\begin{eqnarray} \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle - \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle - \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle &\leq& 1 \,, \\ - \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{b}} \rangle + \langle A^{1}_{\mathbf{a}} A^{2}_{\mathbf{c}} \rangle - \langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{c}} \rangle &\leq& 1 \,. \end{eqnarray}$$
Up to an overall sign these are the same inequalities as the ones I marked (*) in the post you were quoting. For anticorrelated \mathbf{b} outcomes there are two (and only two) additional inequalities that should always be satisfied that are imposed by locality.

Here are the fictive measurement results once more, for locations 1-3 on even and odd days:

... Even ...|.. Odd
L ...1...2...3.|..1...2...3
A_a +1 +1 +1.| -1. -1. -1
A_b +1. -1 +1.| -1 +1. -1
A_c. -1. -1. -1.|+1 +1 +1

Computing from the results for location pair (1,2), I obtain as outcomes: +1, -1.
That location pair does not break Bell's inequality, the average is 0.

Are you sure you've done this correctly? For starters your odd table is the same as the even table except with all the signs flipped, so you should always get the same correlator value in each case simply because you're always taking products of pairs of terms.

As an aside, if you want to apply Bell's inequality the way you wrote it in the quote above, then you shouldn't be calculating the LHS separately for the even and odd days and then averaging them. You calculate the average of the separate terms individually and should only take the absolute value of the first two at the very end (though in this case it shouldn't affect the end result, because you should get the same thing on the even and odd days anyway).

For location pair (1,3), I obtain as outcomes: +3, +3. Average +3.
If I'm not mistaken, this pair very strongly breaks Bell's inequality!

In your table you also always have A^{1}_{\mathbf{b}} = +A^{3}_{\mathbf{b}} while Bell's inequality can only be derived assuming A^{1}_{\mathbf{b}} = - A^{3}_{\mathbf{b}}. So you're violating Bell's inequality in a context where there's no particular reason it should hold in the first place.

 

[harrylin]

"Spooky action at a distance" is not a mechanism, it's just a description of the experimental evidence.

Once more, IMHO it's an illusion, similar to the "experimental evidence" of the doctors in Lille and Lyon.

I wouldn't say that. I don't find it at all plausible to be an actual explanation of what's going on in EPR type experiments. For one thing, you could generate twin pairs hours or even days apart. It's just not plausible that one of the experimenters might accidentally get a particle from a different twin pair (at least not in any consistent way).

It's a trick, it's not a serious model.

Once more, I don't fancy their partilcate models much; such mechanicms look rather articficial to me compared to wave models.

 

[audioloop]

So there are papers saying that Bell is wrong, because it's easy for a local hidden variables theory to violate the inequalities. Then there are other papers saying that Bell is wrong because nothing can violate the inequalities, not even quantum mechanics.

any logical possibility i think.



.

 

[stevendaryl]

Once more, IMHO it's an illusion, similar to the "experimental evidence" of the doctors in Lille and Lyon.

What's an illusion? I see orthodox quantum mechanics as simply a "recipe" for computing results. It's not a mechanism for how those results come about. So I don't know what you are calling an illusion.

 

[harrylin]

What's an illusion? I see orthodox quantum mechanics as simply a "recipe" for computing results. It's not a mechanism for how those results come about. So I don't know what you are calling an illusion.

Yes indeed quantum mechanics is a recipe for computing results. Just as in your magic trick example, the inferred "spooky action at a distance" is an interpretation of the observation of such results, as a consequence of looking at the phenomena in a certain way. The inferred cutting in two of a girl at the stage is an interpretation of the observation of a magic trick; and that interpretation is the result of the illusionist setting the stage and the observer following the line of thinking that the illusionist suggests.

 

[harrylin]

[..]
Are you sure you've done this correctly? For starters your odd table is the same as the even table except with all the signs flipped, so you should always get the same correlator value in each case simply because you're always taking products of pairs of terms.

Well seen! I did it twice but repeated the same error... the very first number is wrong. See next.

As an aside, if you want to apply Bell's inequality the way you wrote it in the quote above, then you shouldn't be calculating the LHS separately for the even and odd days and then averaging them. You calculate the average of the separate terms individually and should only take the absolute value of the first two at the very end (though in this case it shouldn't affect the end result, because you should get the same thing on the even and odd days anyway).

Very right! So, we get then for the average of all locations not 4/3 but 3/3=1. And that means that whille it is broken between two locations, on the average of all locations, DeRaedt's illustration doesn't break that inequality.

In your table you also always have A^{1}_{\mathbf{b}} = +A^{3}_{\mathbf{b}} while Bell's inequality can only be derived assuming A^{1}_{\mathbf{b}} = - A^{3}_{\mathbf{b}}. So you're violating Bell's inequality in a context where there's no particular reason it should hold in the first place.

Ah yes, well seen -again!
Is this particular inequality really "harder" than the other one? That would be surprising for me... I'll have another go at it.

 

[wle]

Very right! So, we get then for the average of all locations not 4/3 but 3/3=1. And that means that whille it is broken between two locations, on the average of all locations, DeRaedt's illustration doesn't break that inequality.

How do you mean? No (relevant) Bell inequality can be violated from the table you gave, whether you condition on the even/odd days or not.

Is this particular inequality really "harder" than the other one?

Which inequality are you talking about? For perfectly correlated or anticorrelated A^{i}_{\mathbf{b}} and A^{j}_{\mathbf{b}}, there are four inequalities imposed by Bell locality that I listed for you in [POST=4459297]this post[/POST]. You just need to be careful that you are testing one of the "right" inequalities, since they differ for A^{i}_{\mathbf{b}} = + A^{j}_{\mathbf{b}} and A^{i}_{\mathbf{b}} = - A^{j}_{\mathbf{b}}.

One way not to make any mistake here, by the way, is simply to test all the possible CHSH inequalities, since the three-term Bell inequalities are just special cases of them anyway:
$$\begin{eqnarray} -2 \leq & +\, \langle A^{i}_{\mathbf{a}} A^{j}_{\mathbf{b}} \rangle + \langle A^{i}_{\mathbf{a}} A^{j}_{\mathbf{c}} \rangle + \langle A^{i}_{\mathbf{b}} A^{j}_{\mathbf{c}} \rangle - \langle A^{i}_{\mathbf{b}} A^{j}_{\mathbf{b}} \rangle & \leq 2 \,, \\ -2 \leq & +\, \langle A^{i}_{\mathbf{a}} A^{j}_{\mathbf{b}} \rangle + \langle A^{i}_{\mathbf{a}} A^{j}_{\mathbf{c}} \rangle - \langle A^{i}_{\mathbf{b}} A^{j}_{\mathbf{c}} \rangle + \langle A^{i}_{\mathbf{b}} A^{j}_{\mathbf{b}} \rangle & \leq 2 \,, \\ -2 \leq & +\, \langle A^{i}_{\mathbf{a}} A^{j}_{\mathbf{b}} \rangle - \langle A^{i}_{\mathbf{a}} A^{j}_{\mathbf{c}} \rangle + \langle A^{i}_{\mathbf{b}} A^{j}_{\mathbf{c}} \rangle + \langle A^{i}_{\mathbf{b}} A^{j}_{\mathbf{b}} \rangle & \leq 2 \,, \\ -2 \leq & -\, \langle A^{i}_{\mathbf{a}} A^{j}_{\mathbf{b}} \rangle + \langle A^{i}_{\mathbf{a}} A^{j}_{\mathbf{c}} \rangle + \langle A^{i}_{\mathbf{b}} A^{j}_{\mathbf{c}} \rangle + \langle A^{i}_{\mathbf{b}} A^{j}_{\mathbf{b}} \rangle & \leq 2 \,. \end{eqnarray}$$
These inequalities are known to be a tight characterisation of the set of correlations compatible with Bell's definition of locality for two parties with binary inputs and outputs. This means that if you have a two-party probability distribution that satisfies all of these inequalities, then a local explanation for it is known to be possible. Conversely, if any one of them is violated, then a local explanation is ruled out.

These inequalities explicitly include the \langle A^{i}_{\mathbf{b}} A^{j}_{\mathbf{b}} \rangle term and always hold regardless of its value (provided, of course, that -1 \leq \langle A^{i}_{\mathbf{b}} A^{j}_{\mathbf{b}} \rangle \leq 1, which is just part of the definition of the correlators). By explicitly setting \langle A^{i}_{\mathbf{b}} A^{j}_{\mathbf{b}} \rangle = +1 or \langle A^{i}_{\mathbf{b}} A^{j}_{\mathbf{b}} \rangle = -1 you can recover the three-term inequalities I listed for you [POST=4459297]here[/POST].

It is pointless to try to violate a relevant three-term inequality with De Raedt's example. This is simply because, as I've just explained, they're special cases of CHSH, and De Raedt et. al. explicitly and openly state that their example never violates a CHSH inequality. Unlike with the three-term inequalities, the CHSH inequalities are always applicable and so De Raedt et. al. never get an opportunity to mislead themselves by testing a "wrong" one.

 

[harrylin]

[..] Which inequality are you talking about?

The original one of Bell, eq.15 compared the one of Boole(?) that De Raedt provided. Once more, you were very right to point out that the two are not identical.

[..]
It is pointless to try to violate a relevant three-term inequality with De Raedt's example. This is simply because, as I've just explained, they're special cases of CHSH, and De Raedt et. al. explicitly and openly state that their example never violates a CHSH inequality. [..]

[Edited]
I doubted your suggestion that it makes an important difference. However, it had escaped my attention that Bell's inquality is a special case of the CHSH inequalites. Consequently according to the paper that illustration (as well as basic variants, as I now verified) cannot break that inequality. In other words, Bell's inequality is stronger than the one illustrated.

That's now perfectly clear to me - thanks again!

 

[morrobay]

While the above probability and statistics is way overhead and while the ± is trivial ( providing it is set up correctly) is this correct ? From (A¹_aA²_b) - (A¹_aA²_c) - (A¹_bA²_c) ≤ +1
With (A¹_bA²_b) = -1 , A¹_b = +1
Then the inequality ; 1 + (A¹_bA²_c) + (A¹_aA²_c ≥ (A¹_aA²_b) is dis proven with 1 + (b+c-) + (a+c-) ≥ (a-b-) values
From [-++ +--] + [+-+ -+-}≥ [ -+- +-+] substituting in above , products equal : 1 - 1 -1 is not ≥ +1

 

[harrylin]

While the above probability and statistics is way overhead and while the ± is trivial ( providing it is set up correctly) is this correct ? From (A¹_aA²_b) - (A¹_aA²_c) - (A¹_bA²_c) ≤ +1
With (A¹_bA²_b) = -1 , A¹_b = +1
Then the inequality ; 1 + (A¹_bA²_c) + (A¹_aA²_c ≥ (A¹_aA²_b) is dis proven with 1 + (b+c-) + (a+c-) ≥ (a-b-) values
From [-++ +--] + [+-+ -+-}≥ [ -+- +-+] substituting in above , products equal : 1 - 1 -1 is not ≥ +1

Sorry, your notation is too cryptic for me. What do you mean with b- if not -b, and what does [-++ +--] stand for?

 

[morrobay]

While the above probability and statistics is way overhead and while the ± is trivial ( providing it is set up correctly) is this correct ? From (A¹_aA²_b) - (A¹_aA²_c) - (A¹_bA²_c) ≤ +1
With (A¹_bA²_b) = -1 , A¹_b = +1
Then the inequality ; 1 + (A¹_bA²_c) + (A¹_aA²_c ≥ (A¹_aA²_b) is dis proven with 1 + (b+c-) + (a+c-) ≥ (a-b-) values
From [-++ +--] + [+-+ -+-}≥ [ -+- +-+] substituting in above , products equal : 1 - 1 -1 is not ≥ +1

::A::::::::::B::
-++... +-- = b+c-
+-+... -+- = a+c-
-+-... +-+ = a-b-
Can the inequality be dis proven from individual case above. If not then what three
streams P₁ P₂ P₃ would disprove it
And how would you apply correlation function to show dis proof.
q(∅₁∅₂) = N same(∅₁∅₂) - N different(∅₁∅₂) / N same + N different

Actually I am trying to follow post # 181 and some other posts by wle and post 151 by harrylin who are showing how inequalities are violated, and I am not sure what the exact steps are, thanks.

 

[harrylin]

::A::::::::::B::
-++... +-- = b+c-
+-+... -+- = a+c-
-+-... +-+ = a-b-
Can the inequality be dis proven from individual case above. If not then what three
streams P₁ P₂ P₃ would disprove it
And how would you apply correlation function to show dis proof.
q(∅₁∅₂) = N same(∅₁∅₂) - N different(∅₁∅₂) / N same + N different

Actually I am trying to follow post # 181 and some other posts by wle and post 151 by harrylin who are showing how inequalities are violated, and I am not sure what the exact steps are, thanks.

A quick first reply: post #151 was corrected in posts #186 - #188.
That simple illustration is useful but not so convincing because it only shows the principle of breaking that kind of inequalities. More is needed to break Bell's inequality.

 

[Ilja]

Yes, I quite agree that is an assumption of Bell. A correct one, of course! And this is not coming from the quantum mechanical side, it is coming from the realism side. As I have said many times before: if the above is NOT a concise requirement, then what DOES IT MEAN TO BE REALISTIC?

Note: virtually anything LESS than the above is essentially returning to the standard QM position.

Bell himself has often clarified that the assumption of realism is much weaker than that the results of the possible spin measurements are predefined. That these results have to be predefined is, instead, a conclusion of the first part of the game, namely of the EPR argument. And the EPR argument already needs not only realism, but also Einstein causality.

What it means to be realistic can be easily explained by the example of the Bohmians. Bohmians are realists. So, if you believe in de Broglie-Bohm theory, you believe in a realistic theory. As a consequence, there cannot be any observable contradiction between quantum theory and realism - dBB in quantum equilibrium and QT make the same predictions.

 

[Ilja]

Here, the illustration that the point made in the last post is not only about DrChinese, but also about Raedt et al themself.

Raedt et al write in http://arxiv.org/pdf/0901.2546v2.pdf:

However, it is well-known that Bell’s assumptions to prove his inequalities are equivalent to the statement that there exists a three-variable joint probability that returns the probabilities of Bell. No additional (metaphysical) assumptions about the nature of the model, other than the assignment of non negative real values to pairs and triples are required to arrive at this conclusion.

The point is that the assumption that there exists a three-variable joint probability that returns these probabilities is not an assumption made by Bell, but a conclusion. The conclusion of the first, EPR part of the argument.

To derive this "assumption" via the EPR argument from something else, we need these additional metaphysical assumptions, not only of classical realism, but also of Einstein causality.

I have not found the exact quote that already Bell has made this point himself too, but found a quite similar point in "Bertlmann's socks and the nature of reality":

It is important to note that to the limited degree to which determinism plays a role in the EPR argument, it is not assumed but inferred.

Together with what follows, this makes the point sufficiently clear.

 

[rkastner]

Bell himself has often clarified that the assumption of realism is much weaker than that the results of the possible spin measurements are predefined. That these results have to be predefined is, instead, a conclusion of the first part of the game, namely of the EPR argument. And the EPR argument already needs not only realism, but also Einstein causality.

What it means to be realistic can be easily explained by the example of the Bohmians. Bohmians are realists. So, if you believe in de Broglie-Bohm theory, you believe in a realistic theory. As a consequence, there cannot be any observable contradiction between quantum theory and realism - dBB in quantum equilibrium and QT make the same predictions.

This is interesting because the traditional Bohmian picture involves 'guiding waves' that are not spacetime entities. Yet these guiding waves somehow steer particles in spacetime. According to this definition of realism, there are real entities that don't live in spacetime that nonlocally steer particles that do. I have never seen an explanation of how this is supposed to be accomplished in realist terms. On the other hand, I do provide a realist account of how pre-spacetime entities can be the basis of spacetime events, in my possibilist extension of Cramer's TI, 'PTI'. It involves embracing the wavelike pre-spacetime reality of quantum objects as offer (and confirmation) waves but letting go of the 'hidden variables' of particle positions. In PTI, particles are just actualized transactions, which give rise to spacetime events. The transactions are where the 'particles' come from -- they are not hidden variables. I think this is more straightforward and it has a smooth transition to the relativistic realm, in contrast to the Bohmian picture which has difficulties with relativistic domain.

 

[bahamagreen]

I'm still back around page 8... but I don't suppose there is such a crystal that always emits six photons at a time, three photons in one direction and three in the opposite direction, each time randomly making the three going one way all have the same spin and opposite of the spin of the three going the other way... to make the triple possibilites actual triplicates... (or maybe pretending such a crystal exists)... would this make the challenge dataset then realistic?

 

[Ilja]

... in my possibilist extension of Cramer's TI, 'PTI'... I think this is more straightforward and it has a smooth transition to the relativistic realm, in contrast to the Bohmian picture which has difficulties with relativistic domain.

Cramer's transactional interpretation gives up classical causality and classical concepts of time. I see no sufficient justifications for this. To give up such really fundamental concepts, one needs really strong evidence. Everything else is, in my humble opinion, mysticism.

The Bohmian picture has only a minor problem with the relativistic domain: It needs a hidden preferred frame. Not a big deal. It was, last but not least, the original "Lorentz ether" interpretation of relativity.

Except for die-hard positivists who reject the existence of everything they cannot observe. In a world where we cannot even observe the NSA with sufficient accuracy this seems to be a quite unnatural restriction.

 

[Dale]

This has degenerated into the usual "my interpretation is better than yours" discussion. Thread closed.