Questioning Bell's Assumption on All Local Hidden Variable Theories

[stevendaryl]

How can we explain this in QM terms??

Seriously, nice post stevendaryl. For a long time I’ve been trying to figure out exactly when the shared wavefunction of the two entangled photons collapse/decohere (or split the universe) to set the indefinite state (unknown polarization) to a definite state (polarized)??

Is it when one or the other passes through a filter?
Or, is it when one or the other ‘splash’ into the detector?
Or, is it some other “total-setup-view” criterion [which I think is DrC’s view]?

Well, for all practical purposes, it doesn't matter when the collapse happens. You could even imagine it happening 10 years after the measurement was made. The Many World's Interpretation basically amounts to taking the limit: the collapse never happens. Or alternatively, you can imagine that at the end of history, someone finally compiles a "history of the universe" that documents everything that ever happened, and it is that historian's observations that collapse the wavefunction. It doesn't make any difference, because of decoherence.

 

[stevendaryl]

I'm only a novice with this stuff but I'd guess that it is when the photon is detected, i.e. when it interacts with the detector and thus the energy transforms into an electrical signal.

That's the moment when it becomes practically impossible to observe interference effects between different states (because there can't be any measurable inference involving macroscopic objects). That doesn't actually mean that a collapse has happened, but it means that you if you assume that it has happened, you won't be proved wrong.

 

[DevilsAvocado]

Well, for all practical purposes, it doesn't matter when the collapse happens.

Oops, my fault... seems like we’re going down in Nugatory’s dreadful “rabbit hole”...

Let’s rephrase the question:

Exactly when is it not longer possible to perform an[other] action on one entangled photon that have an influence on its partner?


[the obvious answer is the detector/measurement, but I know DrC will maybe not agree on this one... ]

 

[DevilsAvocado]

I'm only a novice with this stuff but I'd guess that it is when the photon is detected, i.e. when it interacts with the detector and thus the energy transforms into an electrical signal.

That’s the obvious conclusion, but as you see – the daunting “rabbit hole” is approaching at the collapsed horizon...

 

[morrobay]

This is perhaps an overly picky point, but a hidden variables theory does NOT by itself say that the particle is really spin up or spin down all along. Instead, it allows for the spin outcome to be a possibly nondeterministic function of the state of the detector and the state of the particle being detected. But the existence of perfect correlations (in the case of distant detectors measuring along the same axis) implies that a hidden variables explanation can't possibly agree with experiment unless that outcome is a deterministic function of the hidden variable and the detector orientation. So the claim that the spin had a definite value all along is a conclusion from the local hidden variables assumption, it's not assumed.

In post #24 you said that EPR assumed there were definite values all along.
Thats my understanding of the EPR assumption.That well defined states exist independent of observation. And measuement on one particle does not affect its entangled twin.
And these ± spin values are from detector measurements along the same axis: One permutation of eight being
xyz-------------------------------xyz
-+-.........+-+
From here the inequality n[x+,y+] ≤ n[x+,z+] + n[y-,z-] which QM shows to be violated with
1/2(sin(∅/2)² ≤ 1/2(sin(∅/2)² + 1/2sin(∅/2)²

 

[morrobay]

In post #24 you said that EPR assumed there were definite values all along.
Thats my understanding of the EPR assumption.That well defined states exist independent of observation. And measuement on one particle does not affect its entangled twin.
And these ± spin values are from detector measurements along the same axis: One permutation of eight being
xyz-------------------------------xyz
-+-.........+-+
From here the inequality n[x+,y+] ≤ n[x+,z+] + n[y-,z-] which QM shows to be violated with
1/2(sin(∅/2)² ≤ 1/2(sin(∅/2)² + 1/2sin(∅/2)²

This is perhaps an overly picky point, but a hidden variables theory does NOT by itself say that the particle is really spin up or spin down all along. Instead, it allows for the spin outcome to be a possibly nondeterministic function of the state of the detector and the state of the particle being detected. But the existence of perfect correlations (in the case of distant detectors measuring along the same axis) implies that a hidden variables explanation can't possibly agree with experiment unless that outcome is a deterministic function of the hidden variable and the detector orientation. So the claim that the spin had a definite value all along is a conclusion from the local hidden variables assumption, it's not assumed.

So there seems to be a difference between the assumptions in an EPR hidden variable theory:
That a particle has well defined spin before measurement. And the above hidden variable theory
where the particle does not have spin all along. A Bell assumption ?. Because if it allows for the
spin outcome to be a nondetermiistic function of the state of the detector and the state of
the particle being detected then this is getting very close to non realism ?

 

[stevendaryl]

So there seems to be a difference between the assumptions in an EPR hidden variable theory:
That a particle has well defined spin before measurement. And the above hidden variable theory
where the particle does not have spin all along. A Bell assumption ?. Because if it allows for the
spin outcome to be a nondetermiistic function of the state of the detector and the state of
the particle being detected then this is getting very close to non realism ?

As I said, that the spin result existed before the measurement is a conclusion, rather than an assumption. You can start off with a function

$P(\alpha | \lambda)$ = the probability of Alice measuring spin-up at angle $\alpha$, given that the hidden variable has value $\lambda$. But the perfect anti-correlation (in the spin-1/2 case) between the two particles implies that
$P(\alpha | \lambda)$ must be either 0 or 1---that is, it has to be deterministic.

 

[Nugatory]

So there seems to be a difference between the assumptions in an EPR hidden variable theory:
That a particle has well defined spin before measurement. And the above hidden variable theory
where the particle does not have spin all along. A Bell assumption ?. Because if it allows for the
spin outcome to be a nondeterministic function of the state of the detector and the state of
the particle being detected then this is getting very close to non realism ?

Stevendaryl's point is that a theory cannot be local AND non-deterministic AND predict perfect anti-correlation when the detectors are exactly opposite. We know we get perfect anti-correlation, so we're allowed non-local non-deterministic theories, local deterministic theories, and non-local deterministic theories. Bell's theorem allows us to test for and reject the local deterministic theories, which are the ones that EPR considered.

 

[audioloop]

So there seems to be a difference between the assumptions in an EPR hidden variable theory:
That a particle has well defined spin before measurement. And the above hidden variable theory
where the particle does not have spin all along. A Bell assumption ?. Because if it allows for the
spin outcome to be a nondetermiistic function of the state of the detector and the state of
the particle being detected then this is getting very close to non realism ?

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"


.

 

[audioloop]

the bad of EPR oversimplifications and misuses...

Was Einstein Really a Realist ?
https://www3.nd.edu/~dhoward1/Was Einstein Really a Realist.pdf

 

[harrylin]

Bell admitted that the above calculation (or at least what follows after) demands what I understand to be counterfactual definiteness as follows in his paper on Bertlmann's socks:

"it may be that it is not permissible to regard the experimental settings a and b in the analyzers as independent variables, as we did" - and then he included examples of "apparently random [...] devices".

That topic together with a (too?) simple counter example of Bell's (originally Boole's) example was also discussed here:
https://www.physicsforums.com/showthread.php?t=499002 [..]

[..]
That's a long thread. Could you summarize what you think is a counterexample to what?

Yes, sorry for the delay! I had to refresh my memory first.

De Raedt attempted to give a counter example to Bell's derivation method. His simple counter example is given on p.25, 26 of http://arxiv.org/abs/0901.2546 :

In this second variation of the investigation, we let only two
doctors, one in Lille and one in Lyon perform the examina-
tions. The doctor in Lille examines randomly all patients of
types a and b and the one in Lyon all of type b and c each one
patient at a randomly chosen date. The doctors are convinced
that neither the date of examination nor the location (Lille or
Lyon) has any influence and therefore denote the patients only
by their place of birth. After a lengthy period of examination
they find
Γ(w) = Aa (w)Ab (w) + Aa (w)Ac (w) + Ab (w)Ac (w) = −3

They further notice that the single outcomes of Aa (w), Ab (w)
and Ac (w) are randomly equal to ±1. [..]
a single outcome manifests itself randomly in one city and [..]
the outcome in the other city is then always of opposite sign

Perhaps the weakest point of that example is that the freely chosen detector position of Bell tests with anti-correlation is not fully matched by it. And it is still unclear to me if that is impossible to implement in an example, or only difficult to do. Consequently, the question is for me still open if Bell's assumptions about local realism were valid or not.

 

[wle]

Yes, sorry for the delay! I had to refresh my memory first.

De Raedt attempted to give a counter example to Bell's derivation method. His simple counter example is given on p.25, 26 of http://arxiv.org/abs/0901.2546 :

In this second variation of the investigation, we let only two
doctors, one in Lille and one in Lyon perform the examina-
tions. The doctor in Lille examines randomly all patients of
types a and b and the one in Lyon all of type b and c each one
patient at a randomly chosen date. The doctors are convinced
that neither the date of examination nor the location (Lille or
Lyon) has any influence and therefore denote the patients only
by their place of birth. After a lengthy period of examination
they find
Γ(w) = Aa (w)Ab (w) + Aa (w)Ac (w) + Ab (w)Ac (w) = −3

They further notice that the single outcomes of Aa (w), Ab (w)
and Ac (w) are randomly equal to ±1. [..]
a single outcome manifests itself randomly in one city and [..]
the outcome in the other city is then always of opposite sign

That's not a counter-example.

What they claim to violate is Bell's original 1964 inequality. Bell's original inequality is something of an odd duckling in the zoology of Bell inequalities in that it relies on an extra (but entirely observable) assumption. Specifically, in their notation, and putting the locations back on (Lille = 1, Lyon = 2), the Bell inequality uses the assumption that $A^{1}_{\mathbf{b}}(w) = A^{2}_{\mathbf{b}}(w)$. This is observable, since it implies that $\langle A^{1}_{\mathbf{b}}(w) A^{2}_{\mathbf{b}}(w) \rangle = 1$, and it just means that the correct way to state Bell's inequality should really be something like
$$\langle A^{1}_{\mathbf{a}}(w) A^{2}_{\mathbf{b}}(w) \rangle + \langle A^{1}_{\mathbf{a}}(w) A^{2}_{\mathbf{c}}(w) \rangle + \langle A^{1}_{\mathbf{b}}(w) A^{2}_{\mathbf{c}}(w) \rangle \geq -1 \quad \text{given that} \quad \langle A^{1}_{\mathbf{b}}(w) A^{2}_{\mathbf{b}}(w) \rangle = 1 \,.$$
Their counter-example isn't a counter-example because it has $\langle A^{1}_{\mathbf{b}}(w) A^{2}_{\mathbf{b}}(w) \rangle = -1$. Incidentally, if you try to read the inequality above in the same way as other Bell inequalities (i.e. without imposing a condition like $A^{1}_{\mathbf{b}}(w) = A^{2}_{\mathbf{b}}(w)$), then it's easy to see that its local bound is actually -3 (the same as the algebraic bound) instead of -1.

So they've demonstrated a "violation" of the 1964 Bell inequality in a way that breaks a necessary and verifiable condition for it to hold as a test of locality in the first place. Their approach simply wouldn't work for any other Bell inequality, such as CHSH, that doesn't rely on a condition like this.

Later, in section VII.B, they give another "counter-example" that similarly only works because they define their model such that $E(\mathbf{b}, \mathbf{b}) = 4/\pi - 1 \neq 1$. Of course, as they themselves point out, their model is incapable of violating the CHSH inequality.

On a side note, a concluding remark toward the end of section VII.A

Because no $\lambda$ exists that would lead to a violation except a $\lambda$ that depends on the index pairs $(a, b)$, $(a, c)$ and $(b, c)$ the simplistic conclusion is that either elements of reality do not exist or they are non-local. The mistake here is that Bell and followers insist from the start that the same element of reality occurs for the three different experiments with three different setting pairs. This assumption implies the existence of the combinatorial-topological cyclicity that in turn implies the validity of a non-trivial inequality but has no physical basis. Why should the elements of reality not all be different? Why should they, for example not include the time of measurement? There is furthermore no reason why there should be no parameter of the equipment involved. Thus the equipment could involve time and setting dependent parameters such as $\lambda_{\mathbf{a}}(t)$, $\lambda_{\mathbf{b}}(t)$, $\lambda_{\mathbf{c}}(t)$ and the functions $A$ might depend on these parameters as well

reveals some basic confusions about what Bell's theorem actually implies and the assumptions underpinning it. Basically, if $S$ is some Bell correlator that you could measure, with a local bound $T_{\text{local}}$, then the authors seem to be reading Bell's theorem as implying a logical or algebraic constraint on $S$:
$$S \leq T_{\text{local}} \,.$$
For certain simple correlation inequalities this holds, as the authors say, if you assume that the hidden variable $\lambda$ is the same each time you do the test. But obviously we don't want to assume that, and what Bell's theorem actually proves is more like a bound on the expectation value of $S$, i.e. something more like
$$\langle S \rangle \leq T_{\text{local}} \,.$$
This means that, according to locality, it is entirely possible to do a Bell test and measure a value for $S$ that violates the local bound. It's just that the chance of this happening rapidly becomes very small if you do the test on (say) a very large number of entangled particles. So if you do a Bell test on a very large number of particles and get something significantly above the local bound, the idea is that you can rule out locality with very high confidence.

Finally, the authors suggest that the hidden variable $\lambda$ should be allowed to depend explicitly on time and possibly on the detector settings. Of course, letting $\lambda$ depend explicitly on time doesn't really affect Bell's theorem (it holds regardless of the probability distribution $\rho(\lambda)$ explicitly appearing in proofs of Bell inequalities, so allowing the probability distribution to change in time won't accomplish much). For a properly performed Bell test, letting $\lambda$ depend on the detector settings is normally argued away on the basis of the so-called "free will" or "no conspiracies" assumption (it could also occur if you allow retrocausality, which is not included in the definition of Bell-locality).

 

[audioloop]

reveals some basic confusions about what Bell's theorem actually implies

a interpretation of bell, aforegoing , bells interpretation of einstein..

 

[morrobay]

That's not a counter-example.



Finally, the authors suggest that the hidden variable $\lambda$ should be allowed to depend explicitly on time and possibly on the detector settings. Of course, letting $\lambda$ depend explicitly on time doesn't really affect Bell's theorem (it holds regardless of the probability distribution $\rho(\lambda)$ explicitly appearing in proofs of Bell inequalities, so allowing the probability distribution to change in time won't accomplish much). For a properly performed Bell test, letting $\lambda$ depend on the detector settings is normally argued away on the basis of the so-called "free will" or "no conspiracies" assumption (it could also occur if you allow retrocausality, which is not included in the definition of Bell-locality).

Then am I understanding that when detector settings are parallel: P(α|λ| is deterministic.
And when detector settings are not parallel then P(α|λ| is non-deterministic, depending on state of particle at time of measurement in relation to detector setting ? If so can this still be a local realistic assumption ?

 

[harrylin]

[..] What they claim to violate is Bell's original 1964 inequality. Bell's original inequality is something of an odd duckling in the zoology of Bell inequalities in that it relies on an extra (but entirely observable) assumption. [..] the correct way to state Bell's inequality should really be something like
$$\langle A^{1}_{\mathbf{a}}(w) A^{2}_{\mathbf{b}}(w) \rangle + \langle A^{1}_{\mathbf{a}}(w) A^{2}_{\mathbf{c}}(w) \rangle + \langle A^{1}_{\mathbf{b}}(w) A^{2}_{\mathbf{c}}(w) \rangle \geq -1 \quad \text{given that} \quad \langle A^{1}_{\mathbf{b}}(w) A^{2}_{\mathbf{b}}(w) \rangle = 1 \,.$$ [..]
So they've demonstrated a "violation" of the 1964 Bell inequality in a way that breaks a necessary and verifiable condition for it to hold as a test of locality in the first place. [..]

In words, [Edit:] I first thought that you mean, perhaps: an entangled particle spin measurement at location 1 must give the same result as the measurement of that same particle spin for the same angle at location 2. How can one measure the state of the same elementary particle in two locations without having interfered with it??
However, that's not what the suffix b means in this example, nor in that of Bell. The condition about entangled particles, which is projected on this example about patients, is that an entangled particle spin measurement at location 1 must give the opposite result as the measurement of the spin of its corresponding particle for the same angle at location 2. It is this verified condition that is exactly reproduced in their illustration about patients, although in a rather artificial way.

[..] On a side note, a concluding remark toward the end of section VII.A
reveals some basic confusions about what Bell's theorem actually implies and the assumptions underpinning it. Basically, if $S$ is some Bell correlator that you could measure, with a local bound $T_{\text{local}}$, then the authors seem to be reading Bell's theorem as implying a logical or algebraic constraint on $S$ [..] But obviously we don't want to assume that, and what Bell's theorem actually proves is more like a bound on the expectation value of $S$ [..]

They definitely read Bell's theorem as possible correlations between data, or in other words, mathematical expectations and correlations related to the data - as they explained in the introduction.

if you do a Bell test on a very large number of particles and get something significantly above the local bound, the idea is that you can rule out locality with very high confidence.

Yes, that's certainly the idea.

For a properly performed Bell test, letting $\lambda$ depend on the detector settings is normally argued away on the basis of the so-called "free will" or "no conspiracies" assumption (it could also occur if you allow retrocausality, which is not included in the definition of Bell-locality).

It appears that the authors considered non-conspiracy solutions.

 

[harrylin]

Then am I understanding that when detector settings are parallel: P(α|λ| is deterministic.
And when detector settings are not parallel then P(α|λ| is non-deterministic, depending on state of particle at time of measurement in relation to detector setting ? If so can this still be a local realistic assumption ?

It can still be deterministic and it has at least one local realistic solution in accord with some experiments, and which they presented elsewhere (they refer to that in the text). But it's not clear for me if the result can be entirely in accord with QM.
There is a thread on that attempt:
https://www.physicsforums.com/showthread.php?t=369286

 

[wle]

However, that's not what the suffix b means in this example, nor in that of Bell. The condition about entangled particles, which is projected on this example about patients, is that an entangled particle spin measurement at location 1 must give the opposite result as the measurement of the spin of its corresponding particle for the same angle at location 2.

That's the scenario that Bell originally considered: he derived local bounds given that $\langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{b}} \rangle = -1$. But you can also derive local bounds for the case where $\langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{b}} \rangle = +1$. You get different inequalities in each case, and the particular inequality,
$$\langle A_{\mathbf{a}} A_{\mathbf{b}} \rangle + \langle A_{\mathbf{a}} A_{\mathbf{c}} \rangle + \langle A_{\mathbf{b}} A_{\mathbf{c}} \rangle \geq - 1 \,,$$
considered by de Raedt et. al. is derived assuming $A^{1}_{\mathbf{b}} = A^{2}_{\mathbf{b}} = A_{\mathbf{b}}$. (Note that this is not the inequality Bell derived in 1964.)

So they don't have a counter-example. They took an inequality that is derived assuming $A^{1}_{\mathbf{b}} = A^{2}_{\mathbf{b}}$, and showed that it is violated by a local model in which $A^{1}_{\mathbf{b}} = - A^{2}_{\mathbf{b}}$. In fact, they explicitly point this out later as the reason their model violates the inequality:

The date index does not matter for the products since both signs are reversed on even and odd days leaving the products unchanged. Including the city labels the doctors realize that $A^{1}_{\mathbf{b}}(w, n) = − A^{2}_{\mathbf{b}}(w, n)$, totally against their expectations.

 

[zonde]

To me, counterfactual definiteness is stronger than the hidden variables assumption. In terms of my notation, it would be the additional assumption that

$P(o_1|f_1, f_2, \ldots) =$0 or 1[/itex]

That is, the outcome is definite if the underlying causal factors are known. Hidden variables don't seem to imply counterfactual definiteness by themselves, but together with the perfect correlations of EPR, they do.

You attribute hidden variables only to entangled particles. But they can be attributed to experimental equipment too i.e. say equipment becomes entangled at some stage. And in that case counterfactual definiteness is not attributable to properties of entangled particles even in case of perfect correlations.

 

[stevendaryl]

You attribute hidden variables only to entangled particles. But they can be attributed to experimental equipment too i.e. say equipment becomes entangled at some stage. And in that case counterfactual definiteness is not attributable to properties of entangled particles even in case of perfect correlations.

Well, the details of the spin-1/2 twin-pair experiment places severe limits on the effect of hidden variables in the equipment. If one detector measures spin-up along an axis, you know with absolute certainty that the other detector will NOT measure spin-up along that axis. So that places pretty strict limits on how much hidden variables in the detector can affect the measurement results.

 

[wle]

You attribute hidden variables only to entangled particles.

They're not really being attributed to anything. You're simply looking at some scenario where you see correlations between events occurring in two spacelike separated regions, and you imagine trying to explain those correlations as arising from some past interaction or common origin. The hidden variables just represent some -- any -- initial conditions in the overlap of the past light cones that you think might explain the correlations you see according to some candidate theory that respects locality.

Incidentally, determinism is not necessary as an assumption for deriving Bell inequalities. It isn't even necessary to argue that determinism follows from the fact quantum physics predicts perfect correlations in certain circumstances. It just isn't needed. (In case this wasn't already clear to anyone.)

 

[zonde]

Well, the details of the spin-1/2 twin-pair experiment places severe limits on the effect of hidden variables in the equipment. If one detector measures spin-up along an axis, you know with absolute certainty that the other detector will NOT measure spin-up along that axis. So that places pretty strict limits on how much hidden variables in the detector can affect the measurement results.

Think carefully. Let's say there are no hidden variables associated with particles. In that case is there something that can determine "absolutely" certain outcome of measurement?

 

[zonde]

They're not really being attributed to anything. You're simply looking at some scenario where you see correlations between events occurring in two spacelike separated regions, and you imagine trying to explain those correlations as arising from some past interaction or common origin.

You are missing what was the statement I was replaying to.
The statement was:
hidden variables + perfect correlations => counterfactual definiteness

 

[stevendaryl]

Think carefully. Let's say there are no hidden variables associated with particles. In that case is there something that can determine "absolutely" certain outcome of measurement?

I'm sorry, I don't understand that comment.

 

[zonde]

I'm sorry, I don't understand that comment.

Hmm, then I don't understand yours.

Well, the details of the spin-1/2 twin-pair experiment places severe limits on the effect of hidden variables in the equipment. If one detector measures spin-up along an axis, you know with absolute certainty that the other detector will NOT measure spin-up along that axis. So that places pretty strict limits on how much hidden variables in the detector can affect the measurement results.

With detector you mean set of equipment that determines spin-up/spin-down? Or just that piece of equipment that gives "click" at particular moment? In other words detector one is at Alice's end and other detector is at Bob's end, right? Or rather detector one gives "click" for spin-up particles and other detector gives "click" for spin-down particles?
And if you talk about details of experiment then it would be more useful to take something closer to real experiments.

 

[harrylin]

One last try to clarify that interesting illustration:

[..] 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 $A^{1}_{\mathbf{b}}(w) = A^{2}_{\mathbf{b}}(w)$. This is observable, since it implies that $\langle A^{1}_{\mathbf{b}}(w) A^{2}_{\mathbf{b}}(w) \rangle = 1$, and it just means that the correct way to state Bell's inequality should really be something like
$$\langle A^{1}_{\mathbf{a}}(w) A^{2}_{\mathbf{b}}(w) \rangle + \langle A^{1}_{\mathbf{a}}(w) A^{2}_{\mathbf{c}}(w) \rangle + \langle A^{1}_{\mathbf{b}}(w) A^{2}_{\mathbf{c}}(w) \rangle \geq -1 \quad \text{given that} \quad \langle A^{1}_{\mathbf{b}}(w) A^{2}_{\mathbf{b}}(w) \rangle = 1 \,.$$
Their counter-example isn't a counter-example because it has $\langle A^{1}_{\mathbf{b}}(w) A^{2}_{\mathbf{b}}(w) \rangle = -1$.
[..]

That's the scenario that Bell originally considered: he derived local bounds given that $\langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{b}} \rangle = -1$. But you can also derive local bounds for the case where $\langle A^{1}_{\mathbf{b}} A^{2}_{\mathbf{b}} \rangle = +1$. You get different inequalities in each case, and the particular inequality,
$$\langle A_{\mathbf{a}} A_{\mathbf{b}} \rangle + \langle A_{\mathbf{a}} A_{\mathbf{c}} \rangle + \langle A_{\mathbf{b}} A_{\mathbf{c}} \rangle \geq - 1 \,,$$
considered by de Raedt et. al. is derived assuming $A^{1}_{\mathbf{b}} = A^{2}_{\mathbf{b}} = A_{\mathbf{b}}$. (Note that this is not the inequality Bell derived in 1964.)

So they don't have a counter-example. They took an inequality that is derived assuming $A^{1}_{\mathbf{b}} = A^{2}_{\mathbf{b}}$, and showed that it is violated by a local model in which $A^{1}_{\mathbf{b}} = - A^{2}_{\mathbf{b}}$.
[..]

So, Bell's inequality violation incl. the condition that $A^{1}_{\mathbf{b}}(w) = - A^{2}_{\mathbf{b}}(w)$ is correctly reproduced. However they did not "derive" their simple illustration - that doesn't make sense! Instead, they modified Boole's example to illustrate that it is possible to obtain such weird results with local realism.

NB: -1 or 1 for the product doesn't matter much, as was discussed in the thread about that paper (there someone else argued just the contrary as you! - www.physicsforums.com/showthread.php?p=3361187). I did not try it, but I suppose that De Raedt's example can be modified to break the inequality for the in that example less weird result of equal outcomes for each pair, and still break the inequality. See also this example, with footnote 3: http://www.felderbooks.com/papers/bell.html

 

[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:

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

(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 $P(\lambda) > 0$, because there is no reason for the sum to range over values of $\lambda$ that never occur. So we have$P(A | \alpha \wedge \lambda) (1 - P(B | (\beta = \pi - \alpha) \wedge \lambda)) = 0$

What we can conclude from this is that

If $P(A | \alpha \wedge \lambda) > 0$, then $P(B | (\beta = \pi - \alpha) \wedge \lambda) = 1$

A similar analysis leads to the conclusion:

If $P(B | \beta \wedge \lambda) > 0$, 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 $\beta$, either$P(B | \beta\wedge \lambda) = 0$ 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:
$$\begin{eqnarray} A^{1}_{\mathbf{a}} &=& +1 \,, \\ A^{1}_{\mathbf{b}} &=& +1 \,, \\ A^{2}_{\mathbf{b}} &=& -1 \,, \\ A^{2}_{\mathbf{c}} &=& -1 \,. \end{eqnarray}$$
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.