Is Bell's Theorem Inconsistent with Quantum Mechanics?

[emuc]

I've published a paper on local hidden variables with surprising consequences for Bells Theorem. It is available on https://doi.org/10.1515/phys-2017-0106 The journal Open Physics is listed in T/R.

To the background of Bell's argument, the following comments: Since Bell published his theorem for the proof of distant action more than 50 years ago, no one has yet provided a proof of how this remote effect actually takes place.
What's wrong with Bell's argument?

Bell had argued that if nature allowed only local effects, the results of polarization measurements would only depend on polarizer position and a possible hidden parameter. He then concluded that the expectation values of different measurements must be in a certain context, namely, that they followed Bell's inequality. If, as often measured, nature violates Bell's inequality, then, according to Bell, it can not be based on local effects.

Bell has argued imprecisely. His theorem is valid only if the dependency of the polarization measurement results on polarizer position and hidden parameter is the only one possible. If other models are possible, which correctly predict the measured expectation values with entangled photon pairs, his theorem loses its generality. Bell has thus failed to prove the universality. A counter-example suffices to refute his theorem. I presented such a counterexample. The measurement results also depend on the polarization of the incoming photons.

 

[Mentz114]

I've published a paper on local hidden variables with surprising consequences for Bells Theorem. It is available on https://doi.org/10.1515/phys-2017-0106

[]
.

Interesting. I have not digested your paper but disagree strongly that entanglement does not imply non-locality.
If we take entanglement to mean that both objects in an entanled pair always have the same(opposite) value of the entangled property, this can only be maintained if locality is relaxed or dropped.

Verified by experiment, I've heard.

 

[DrChinese]

As you are new, you likely are not aware that your paper does not meet PhysicsForums guidelines. It expresses a personal point of view, and has not been published in a generally accepted peer-reviewed publication.

 

[Dale]

After discussion amongst the mentors we decided to allow discussion of this paper. Despite having a quite low impact, the journal is peer reviewed and does appear on the master journal list.

 

[DrChinese]

I've published a paper on local hidden variables with surprising consequences for Bells Theorem. It is available on https://doi.org/10.1515/phys-2017-0106 The journal Open Physics is listed in T/R... A counter-example suffices to refute his theorem. I presented such a counterexample. The measurement results also depend on the polarization of the incoming photons.

OK, it is true that a counter-example would invalidate Bell. So in your paper, you present 6 assumptions you feel leads to a successful model of entanglement. There are a couple of basic questions I have to start with.

1. Your M1 assumption is "A propensity state called p-state determines which polarizer output a photon will take. A photon in p-state α would pass a polarizer set to α with certainty." I believe that implies that so-called perfect correlations of entangled photons requires them to be in the same p-state. Is that correct? Since entangled photon pairs match (or mismatch depending on Type - I think the mismatch is your preferred format) at all common angle settings, presumably all possible angles have simultaneously well-defined p-states. Is that your assertion too? (After all, there is no statistical deviation of the (mis)matching - it's 100%. Even though the matched values themselves appear random presumably due to parameter λ.)

2. Your M3 assumption says: "It is assumed that a photon can be simultaneously in different p-states depending on the value of a parameter λ and a chosen direction relative to the polarization of the photon. That means some of the photons with polarization α are also in p-state β and thus pass a polarizer set to β with certainty. As p-state and polarization are different physical entities ambiguous polarization states are excluded." So now I am confused. Polarization of an entangled photon is a different physical entity than the p-state? Does the polarization determine the p-state? How are they related, if they are?

3. And you also make a bold statement with this: "Entangled photons are generated by a common source on the side 1 with the polarization φ1=0∘ and on the side 2 with the polarization φ2=90∘ or on the side 1 with the polarization φ1= 90∘ and on the side 2 with the polarization φ2 = 0∘. We now look at side 1 with the polarizer setting α/α + π/2: Incoming photons have the polarization φ1 = 0∘ or φ1 = 90∘." This goes against almost every theoretical treatment of entangled photon pairs, as they are usually presented as being in a superposition of polarization states. And your pairs do not appear to be a superposition - they are in one of two well-defined polarizations. Is that correct?

Based on your answers to the above, I will present to you some obvious counter-arguments to your model. I want to understand your model first.

 

[DrChinese]

And by the way with my #1 above: Your M5 assumption addresses this too, but I am not clear if that is addressing p-states or polarization. So I am trying to determine if entangled photons have many well-defined p-states that are appropriately (mis)matched.

 

[emuc]

OK, it is true that a counter-example would invalidate Bell. So in your paper, you present 6 assumptions you feel leads to a successful model of entanglement. There are a couple of basic questions I have to start with.

1. Your M1 assumption is "A propensity state called p-state determines which polarizer output a photon will take. A photon in p-state α would pass a polarizer set to α with certainty." I believe that implies that so-called perfect correlations of entangled photons requires them to be in the same p-state. Is that correct?

MY COMMENT: No, entangled photons share the same parameter lambda. Lambda controls (M2) in which p-state a photon is. See M4 for details.

Since entangled photon pairs match (or mismatch depending on Type - I think the mismatch is your preferred format) at all common angle settings, presumably all possible angles have simultaneously well-defined p-states. Is that your assertion too? (After all, there is no statistical deviation of the (mis)matching - it's 100%. Even though the matched values themselves appear random presumably due to parameter λ.)2. Your M3 assumption says: "It is assumed that a photon can be simultaneously in different p-states depending on the value of a parameter λ and a chosen direction relative to the polarization of the photon. That means some of the photons with polarization α are also in p-state β and thus pass a polarizer set to β with certainty. As p-state and polarization are different physical entities ambiguous polarization states are excluded." So now I am confused. Polarization of an entangled photon is a different physical entity than the p-state? Does the polarization determine the p-state? How are they related, if they are?

MY COMMENT: M1: the p-state determines which polarizer output a photon will take. P-state and polarization direction are the same if all photons from an ensemble are in the same p-state. This is true for delta = 0.

3. And you also make a bold statement with this: "Entangled photons are generated by a common source on the side 1 with the polarization φ1=0∘ and on the side 2 with the polarization φ2=90∘ or on the side 1 with the polarization φ1= 90∘ and on the side 2 with the polarization φ2 = 0∘. We now look at side 1 with the polarizer setting α/α + π/2: Incoming photons have the polarization φ1 = 0∘ or φ1 = 90∘." This goes against almost every theoretical treatment of entangled photon pairs, as they are usually presented as being in a superposition of polarization states. And your pairs do not appear to be a superposition - they are in one of two well-defined polarizations. Is that correct?

MY COMMENT: One method to create entangled photon pairs is parametric down conversion. The output consists of signal photons and idler photons with fixed polarization but perpendicular to each other. Say |H> and |V> These photons appear in two cones which intersect. We have thus a mixture of photon pairs in state |H>|V>. At the intersection of the aforementioned cones the polarization of the photons is unknown. The state of the pairs is entangled 1/sqrt(2)(|H>|V> - |V>|H>). This is taken into account by M3 which distinguishes the entangled state from a mixture. My model approach was treating the entangled photons like single photons and accounting for entanglement by M3.

 

[DrChinese]

OK, let's use your example of Type II entanglement for convenience. They will mismatch at same angle, and be orthogonal (perpendicular). So your assertion is that their polarization is unknown and entangled. Each pair. Now you say about M3: "Entangled photons are generated by a common source on the side 1 with the polarization φ1=0∘ and on the side 2 with the polarization φ2=90∘ or on the side 1 with the polarization φ1= 90∘ and on the side 2 with the polarization φ2 = 0∘. We now look at side 1 with the polarizer setting α/α + π/2: Incoming photons have the polarization φ1 = 0∘ or φ1 = 90∘."

I asked if they were in one of 2 well defined polarizations per above, and you did not answer that directly.

But I believe you are also asserting per your M5 that they (entangled pairs) have a special property that is NOT present in single photons. And I am trying to tie that property - as you describe it - to perfect correlations, which only appear in entangled photon pairs. Each pair.

In other words: is there a (perpendicular) p-state for 10 degrees? Another for 20 degrees, 30, ...? How else do we get "perfect" mismatches on both? Can you explain this in more detail?

 

[Jilang]

You state that
“It is assumed that a photon can be simultaneously in different p-states depending on the value of a parameter λ and a chosen direction relative to the polarization of the photon. That means some of the photons with polarization α are also in p-state β and thus pass a polarizer set to β with certainty. As p-state and polarization are different physical entities ambiguous polarization states are excluded.”
This definition is at odds with realism. The p state cannot be decided until the chosen direction is. I think it is generally accepted that locality can be preserved if realism is not. How does your model differ from that?

 

[emuc]

I asked if they were in one of 2 well defined polarizations per above, and you did not answer that directly.
MY COMMENT: They don’t have a specific polarization until a selection is done.

But I believe you are also asserting per your M5 that they (entangled pairs) have a special property that is NOT present in single photons. And I am trying to tie that property - as you describe it - to perfect correlations, which only appear in entangled photon pairs. Each pair.

In other words: is there a (perpendicular) p-state for 10 degrees? Another for 20 degrees, 30, ...? How else do we get "perfect" mismatches on both? Can you explain this in more detail?

MY COMMENT:
Having photon 1 in polarization state alpha
Then we obtain from equation 2 photon 1 is in p-state alpha
for -1<lambda<cos(2delta) with delta = Position of P1 – alpha

and from eq. 12 we obtain photon 2 is in p-state 90°+alpha for -1<lambda<cos(2delta) with
delta = Position of P2 –alpha-90°. So, you can calculate the p-state for each polarizer position and each value of lambda#

 

[emuc]

You state that
“It is assumed that a photon can be simultaneously in different p-states depending on the value of a parameter λ and a chosen direction relative to the polarization of the photon. That means some of the photons with polarization α are also in p-state β and thus pass a polarizer set to β with certainty. As p-state and polarization are different physical entities ambiguous polarization states are excluded.”
This definition is at odds with realism. The p state cannot be decided until the chosen direction is.
MY COMMENT: The p-state describes a propensity and determines which polarizer output will be taken by the photon. With a particular polarization of a (single) photon the p-state can be calculated (see M4)
I think it is generally accepted that locality can be preserved if realism is not. How does your model differ from that?
MY COMMENT: I do not exactly know what realism is. But I have an opinion what real is. Real is a description of facts which is able to predict experimental results.

 

[Nugatory]

@emuc, you really have to learn to use this forum's quote functionality.

 

[DrChinese]

1. MY COMMENT: They don’t have a specific polarization until a selection is done.

2. MY COMMENT: ...So, you can calculate the p-state for each polarizer position and each value of lambda

OK, thanks for this.

1. As I read this, and your commentary indicates that once photon 1 is detected after passing through P1, photon 2 takes on the polarization d − 90∘. By what form of action does that occur, if not by non-local action? You say specifically: "Upon hitting the detector D1-1 photon 1 has ceased to exist and the singlet state no longer describes the state of the photon pair. Photon 2 must thus be in another state."

You state there is "no physical action upon photon 2 after it had left the source". You have simply assumed that which you seek to prove! It is precisely this point which everyone struggles to make sense of. Clearly, your concept depends on the nature of the observation made on photon 1, which is exactly the EPR point:

"This makes the reality of P and Q depend upon the process of measurement carried out on the first system, which does not disturb the system in any way. No reasonable definition of reality could be expected to permit this."

All you have done in your work is echo this sentiment from EPR, 1935.2. As everyone (else) already knows:

IF, on the other hand, the other, photon 2 had p-states which can be pre-calculated - as you have just asserted here...
THEN there is NO data set - despite your assertions otherwise - that will actually provide results that match QM.

Clearly you cannot do that without knowing - in advance - what the setting of d1 and d2 are. To see such is the case, simply take the DrChinese challenge. Hopefully, you won't hesitate to spend a few minutes here in the process of overturning Bell's Theorem, one of the most important discoveries in science (until your paper).

Rules: Hand pick values of lambda for perhaps 10-12 runs. Use your formula to calc the photon 1 and photon 2 p-states for the 3 angles I selected below for your selected lambda, as either being H or V. You must maintain perfect correlations, so the H or V values at the same angle for any run must be opposite.

Columns:
Run #
Lambda restrict this to 1 or 2 values)
Photon1 at P1=0 degrees (H or V)
Photon1 at P1=120 degrees (H or V)
Photon1 at P1=240 degrees (H or V)
Photon2 at P2=0 degrees (H or V)
Photon2 at P2=120 degrees (H or V)
Photon2 at P2=240 degrees (H or V)

For each run you present to me, I will select the P1 and P2 settings (but not the same angles). I predict one of the following as results:

A. The match rate will be less than 70%. The quantum mechanical prediction is 75% (Type II PDC).
B. You will not present any runs for me to test.

B is the usual response for Bell-deniers. Quite simply: if there exist predetermined values for entangled photon polarization outcomes without any non-local action occurring... WHAT ARE THEY? This is what the DrChinese challenge is. No hand-waving allowed, just show me the values.

Your move.

 

[bhobba]

As I read this, and your commentary indicates that once photon 1 is detected after passing through P1, photon 2 takes on the polarization d − 90∘. By what form of action does that occur, if not by non-local action? You say specifically: "Upon hitting the detector detector D1-1 photon 1 has ceased to exist and the singlet state no longer describes the state of the photon pair. Photon 2 must thus be in another state."

Sure - you have broken entanglement so naturally since we know one but since it is correlated we know the other. But - so? That is not the issue with EPR - that the results are correlated is a given. No augment. The issue is according to QM it has different statistical properties than standard probability theory. But why is that surprising - we now know QM is simply a different generalized probability model than ordinary probability theory - its quite reasonable it will have properties different to standard probability:
https://arxiv.org/abs/1402.6562

In fact as the above says - that there is different statistics to the correlations is no surprise - many different generalized probability models have that - the issue is:
'Returning to the question of what distinguishes quantum mechanics as the fundamental theory of nature, it is therefore not sufficient to explain the existence of non-classical correlations, one also has to give reasonable arguments why these correlations are not stronger.'

Surprising isn't it - its not that they are different - its why they are different in the way they are - just what feature of QM determines that and why is nature like that.

All you have done in your work is echo this sentiment from EPR, 1935.

Yes.

Thanks
Bill

 

[facenian]

After discussion amongst the mentors we decided to allow discussion of this paper. Despite having a quite low impact, the journal is peer reviewed and does appear on the master journal list.

Does this mean that "predatory journals" are excuded from the master journal list?

 

[Stephen Tashi]

The issue is according to QM it has different statistical properties than standard probability theory.

I'll disagree on a technicality. "Standard probability theory" is not a specific probability model for a phenomena. The correct statement is that QM statisical properties disagree with certain conventional probability models. That doesn't demonstrate that we need a new type of probability theory to model entanglement experiments. We may merely need a new probability model (e.g. https://arxiv.org/abs/1406.4886 "CHSH Inequality: Quantum probabilities as classical conditional probabilities")

 

[Dale]

Does this mean that "predatory journals" are excuded from the master journal list?

Some predatory journals have slipped through on to the master journal list, so we do take a case by case look. If we believe that a journal is from a predatory publisher then we will reject it. The publisher of Open Physics appears to be a legitimate open access publisher, even specifically identified as such by Jeff Beall.

 

[Dale]

To see such is the case, simply take the DrChinese challenge.

@emuc I do encourage you to take the challenge.

@DrChinese it might be good to have a brief explanation of why the DrChinese challenge is relevant to these types of discussions.

 

[emuc]

You state there is "no physical action upon photon 2 after it had left the source". You have simply assumed that which you seek to prove! It is precisely this point which everyone struggles to make sense of. Clearly, your concept depends on the nature of the observation made on photon 1, which is exactly the EPR point:MY COMMENT: I've shown here there is a reasonable explanation for the experimental results namely that the state of those photon 2 with peer photon 1 selected by P1 exists from the origin of the photon pair. It was assumed no spooky action taking place. This is confirmed by the model presented in section 3.
Rules: Hand pick values of lambda for perhaps 10-12 runs. Use your formula to calc the photon 1 and photon 2 p-states for the 3 angles I selected below for your selected lambda, as either being H or V. You must maintain perfect correlations, so the H or V values at the same angle for any run must be opposite.

Columns:
Run #
Lambda restrict this to 1 or 2 values)
Photon1 at P1=0 degrees (H or V)
Photon1 at P1=120 degrees (H or V)
Photon1 at P1=240 degrees (H or V)
Photon2 at P2=0 degrees (H or V)
Photon2 at P2=120 degrees (H or V)
Photon2 at P2=240 degrees (H or V)

MY COMMENT:
Here are my calculations:
Header:
1:setting of P1; 2:p-state photon 1; 3:polarization of photon 2; 4:setting of P2= p-state photon 2; 5:delta; 6:cos(2delta);7:equation#; 8:value range of lambda
P2 + 90°
0°;0°;90°;0°;-90°;-1;12;-
0°;0°;90°;90°;-90°;-1;13;-1<lambda<1

0°;0°;90°;120°;30°;0,5;12;-1<lambda<0,5
0°;0°;90°;210°;30°;0,5;13;0,5<lambda<1

0°;0°;90°;240°;150°;0,5;12;-1<lambda<0,5
0°;0°;90°;330°;150°;0,5;13;0,5<lambda<1

120°;120°;210°;0°;-210°;0,5;12;-1<lambda<0,5
120°;120°;210°;90°;-210°;0,5;13;0,5<lambda<1

120°;120°;210°;120°;-90°;-1;12;-
120°;120°;210°;210°;-90°;-1;13;-1<lambda<1

120°;120°;210°;240°;30°;0,5;12;-1<lambda<0,5
120°;120°;210°;330°;30°;0,5;13;0,5<lambda<1

240°;240°;330°;0°;-330°;0,5;12;-1<lambda<0,5
240°;240°;330°;90°;-330°;0,5;13;0,5<lambda<1

240°;240°;330°;120°;-210°;0,5;12;-1<lambda<0,5
240°;240°;330°;210°;-330°;0,5;13;0,5<lambda<1

240°;240°;330°;240°;-90°;-1;12;-
240°;240°;330°;330°;-90°;-1;13;0,5<lambda<1

I have presented a value range for lambda instead of discrete values.
Probabilities can easily be obtained.
For cos(2delta)=0,5 photon 2 hits P2 with probability 1,5/2= 75%.

Is it that what you wanted?

 

[StevieTNZ]

Dear emuc

See https://www.physicsforums.com/threads/is-bells-theorem-correct.937619/#post-5927002

 

[DrChinese]

... Rules: Hand pick values of lambda for perhaps 10-12 runs. Use your formula to calc the photon 1 and photon 2 p-states for the 3 angles I selected below for your selected lambda, as either being H or V. You must maintain perfect correlations, so the H or V values at the same angle for any run must be opposite.

Columns:
Run #
Lambda restrict this to 1 or 2 values)
Photon1 at P1=0 degrees (H or V)
Photon1 at P1=120 degrees (H or V)
Photon1 at P1=240 degrees (H or V)
Photon2 at P2=0 degrees (H or V)
Photon2 at P2=120 degrees (H or V)
Photon2 at P2=240 degrees (H or V)

MY COMMENT:
Here are my calculations:
Header:
1:setting of P1; 2:p-state photon 1; 3:polarization of photon 2; 4:setting of P2= p-state photon 2; 5:delta; 6:cos(2delta);7:equation#; 8:value range of lambda
P2 + 90°
0°;0°;90°;0°;-90°;-1;12;-
0°;0°;90°;90°;-90°;-1;13;-1<lambda<1

0°;0°;90°;120°;30°;0,5;12;-1<lambda<0,5
0°;0°;90°;210°;30°;0,5;13;0,5<lambda<1

0°;0°;90°;240°;150°;0,5;12;-1<lambda<0,5
0°;0°;90°;330°;150°;0,5;13;0,5<lambda<1

120°;120°;210°;0°;-210°;0,5;12;-1<lambda<0,5
120°;120°;210°;90°;-210°;0,5;13;0,5<lambda<1

120°;120°;210°;120°;-90°;-1;12;-
120°;120°;210°;210°;-90°;-1;13;-1<lambda<1

120°;120°;210°;240°;30°;0,5;12;-1<lambda<0,5
120°;120°;210°;330°;30°;0,5;13;0,5<lambda<1

240°;240°;330°;0°;-330°;0,5;12;-1<lambda<0,5
240°;240°;330°;90°;-330°;0,5;13;0,5<lambda<1

240°;240°;330°;120°;-210°;0,5;12;-1<lambda<0,5
240°;240°;330°;210°;-330°;0,5;13;0,5<lambda<1

240°;240°;330°;240°;-90°;-1;12;-
240°;240°;330°;330°;-90°;-1;13;0,5<lambda<1

I have presented a value range for lambda instead of discrete values.
Probabilities can easily be obtained.
For cos(2delta)=0,5 photon 2 hits P2 with probability 1,5/2= 75%.

Is it that what you wanted?

No, reducing it to H or V is the challenge. And for the 3 d1/d2 angles I specified, 0/120/240 degrees. It's your formula, and I expect you calculate the specific determinate value you claim exists. You said there are p-states that determine the outcomes with certainty. Let's see those babies, no more hand-waving.

Run 1
For lambda=L1 (you select this)
P1=H@0 degrees, P1=V@120, P1=H@240
P2=V@0 degrees, P2=H@120, P2=V@240

Run 2
For lambda=L2 (you select this)
P1=V@0 degrees, P1=V@120, P1=H@240
P2=H@0 degrees, P2=H@120, P2=V@240

Etc. for perhaps 10 or so runs, enough to see the trend. In other words: You claim to have a formula overturning a result that has been generally accepted by the community. I want to see it in action. Surely you've run through this exercise already, if you have spent the time to write a paper on it. Where's your new contribution if your result cannot be applied?

 

[DrChinese]

@DrChinese it might be good to have a brief explanation of why the DrChinese challenge is relevant to these types of discussions.[/USER]

Good suggestion.

The purpose of the DrChinese challenge is to convert general assertions that "viable local realistic scenarios exist" into specific cases that can be analyzed separately from the author's words. So far, papers purporting to overturn Bell fall flat when the rubber meets the road, so to speak. This is no surprise, it is simply a consequence of Bell's Theorem. But this is its application.

With the DrChinese challenge, we get to see the precise point in the argument where things fall apart. Even if we disagree, we at least know where.

 

[DrChinese]

You state there is "no physical action upon photon 2 after it had left the source". You have simply assumed that which you seek to prove! It is precisely this point which everyone struggles to make sense of. Clearly, your concept depends on the nature of the observation made on photon 1, which is exactly the EPR point:MY COMMENT: I've shown here there is a reasonable explanation for the experimental results namely that the state of those photon 2 with peer photon 1 selected by P1 exists from the origin of the photon pair. It was assumed no spooky action taking place. This is confirmed by the model presented in section 3.

Look up "tautology". Again. You use an assumption to prove the same assumption.

It is no secret that once you know the choice of detection angle for Alice, you can design an algorithm for a QM compliant result for Bob. And that depends that result being an input variable for Bob's result. That is precisely what Bell explicitly said could not be allowed. It's in the first paragraph of Bell:

"It is the requirement of locality, or more precisely that the result of a measurement on one system be unaffected by operations on a distant system with which it has interacted in the past, that creates the essential difficulty."

Note that your formula is not separable. You use the result of a measurement on photon 1 to calculate the result of photon 2.

 

[Ian J Miller]

As I see its, this issue is complicated by the hidden assumption that spin, say, is determined at measurement. The whole problem with locality goes away if the spin is determined at formation. The problem is, between formation and measurement, there is no way of knowing what it is. Accordingly, I do not see there is any way of knowing about locality.

Deviations from Bell's inequality by polariser experiments are a completely different issue. Take the Aspect experiment. Bell's inequality requires plus/minus (or pass/fail) observations under three different conditions. That means three pairs of pass-fail are required, and they must be independent measurements. Now, the photons are entangled because the law of conservation of angular momentum applies to them. However, in Aspect, A is the original setup, B is the rotation of a polariser by 22.5 degrees, and C is the rotation by 45 degrees. That means that B+C- is simply A+B- setup rotated by 22.5 degrees. You cannot get independent variables by rotating the apparatus, any more than taking it down the other end of the bench or doing the experiment tomorrow. The logic violates Nöther's theorem. Actually, there is worse than that because to get only entangled pairs registered, the second polariser ONLY records photons arriving 19 ns (by memory) after the first. That means any + result measures the number of photon pairs in the determination and can be normalised to 1. When you do that, energy or probability is not conserved. This does not mean anything about locality - but i believe the logic underpinning the analysis is flawed. The experiment is a wonderful example of wave particle duality, even for individual photons, but nothing more.

 

[DrChinese]

As I see its, this issue is complicated by the hidden assumption that spin, say, is determined at measurement.

The whole problem with locality goes away if the spin is determined at formation. ...

Are you discussing emuc's paper with the first sentence?

And I don't follow what you are saying with the second sentence, as this is essentially refuted by Bell. Aren't you a follower of dBB? I.e. you reject locality anyway?

 

[Fra]

Bell has argued imprecisely. His theorem is valid only if the dependency of the polarization measurement results on polarizer position and hidden parameter is the only one possible. If other models are possible, which correctly predict the measured expectation values with entangled photon pairs, his theorem loses its generality. Bell has thus failed to prove the universality.

I think you are discussing the premises of the theorem right? not the theorem itself? So the issue is what is a reasonable mathematical representation of "local realism"? Let's characterize the basic logic first before bothering with details.

(A,B outcomes; a,b settings on the device, $\lambda$ is the hidden variable)

As I see it, the premise that the proof depends is essentially the local deterministic realism condition that there are functions f and go such that
$A=f(a,\lambda)$
$B=g(b,\lambda)$

Then we get, given $a,b,\lambda$
$\left<AB\right>=\int_\lambda f(a,\lambda) g(b,\lambda) P(AB|a,b,\lambda) dP(\lambda|a,b)$
and
$P(AB|a,b,\lambda)=1$

So $\left<AB\right>=\int_\lambda f(a,\lambda) g(b,\lambda) dP(\lambda|a,b)$

It is usually also assume that $\lambda$ is totally independent from a and b, something that can certainly be questioned. Not because i believe in superdeterminism, its just that I can't see how it does not follow from any principle of rational inference. [ On the contrary it find it a suspicious and a bit incoherent position to believe in determinism in the sense in question and at the same time think that $\lambda$ would be totally statistically independent from a and b? ]

(But if $\left<AB\right>=\int_\lambda f(a,\lambda) g(b,\lambda) dP(\lambda)$ the theorem follows mathematically)

So without obstructing the logic here with angles and photons, where in this inferential structure is your change? Anything that is changing the premises must change f,g or $P(\lambda|ab)$? Explicit examples are nice, but the can also clog there view of the inferences.

/Fredrik

 

[bhobba]

I'll disagree on a technicality. "Standard probability theory" is not a specific probability model for a phenomena. The correct statement is that QM statisical properties disagree with certain conventional probability models. That doesn't demonstrate that we need a new type of probability theory to model entanglement experiments. We may merely need a new probability model (e.g. https://arxiv.org/abs/1406.4886 "CHSH Inequality: Quantum probabilities as classical conditional probabilities")

QM in general is a different generalized probability model to ordinary probability theory - in fact the next simplest after ordinary probability theory. It allows continuous transformations between pure states - normal probability theory does not allow that. It does it because probabilities are not positive numbers - but complex numbers which of course the Kolmogerov axioms do not have.

That is the reason for Bell. If however you want to interpret it in a more classical manner - you can - but you need FTL ie non-locality. If there is another way of doing it - via conditional probabilities I don't know - I will let others more into it decide if your referenced paper stands up to it. But in essence it changes nothing - Bell rules out local realistic theories. Of course your linked paper states it plainly - it can't be used as an out for local realistic theories.

Thanks
Bill

 

[bhobba]

You state there is "no physical action upon photon 2 after it had left the source"

In QM the precise statement is it is silent on if such action exists. In QM we do not know if after it has left the source it even has a physical existence for such to be meaningful.

Thanks
Bill

 

[bhobba]

If we take entanglement to mean that both objects in an entanled pair always have the same(opposite) value of the entangled property, this can only be maintained if locality is relaxed or dropped..

Entangled objects do not have individual properties from the very definition of entanglement. In fact if you observe just one of the entangled objects its in an improper mixed state.

Thanks
Bill

 

[Stavros Kiri]

Entangled objects do not have individual properties from the very definition of entanglement. In fact if you observe just one of the entangled objects its in an improper mixed state.

Thanbkds
Bill

Although I personally would prefer separate short posts, even though consecutive and within short time, some mentors have advised that we include all in one post, even using editing if we have to. That keeps the forum more neat, perhaps, by reducing the total net number of posts.

 

[bhobba]

Although I personally would prefer separate short posts, even though consecutive and within short time, some mentors have advised that we include all in one post, even using editing if we have to. That keeps the forum more neat, perhaps, by reducing the total net number of posts.

I personally consider it on a case by case basis. I will often post something - then see I expressed it badly and change it. Other times I do a separate post. I don't think there is a hard and fast rule.

As a mentor myself I personally don't put that rule particularly high on my list of priorities - but the general consensus of mentors is - yes - one post is best.

Thanks
Bill

 

[Stavros Kiri]

I personally consider it on a case by case basis. I will often post something - then see I expressed it badly and change it. Other times I do a separate post. I don't think there is a hard and fast rule.

As a mentor myself I personally don't put that rule particularly high on my list of priorities - but the general consensus of mentors is - yes - one post is best.

Thanks
Bill

Makes sense too. But same story for long posts. Which one is better? I would say too "it depends ...".

 

[bhobba]

Makes sense too. But same story for long posts. Which one is better? I would say too "it depends ...".

Exactly. Like I said as a mentor it's not something I am particularly on the lookout for - just take a common-sense approach and you will be fine. If its an issue I definitely would not give an infringement, warning or anything like that - no crucifixion first offence around here for relatively minor things (major things like being an obvious Troll deserves, and is given, much stronger action) - simply a nicely worded PM suggesting a slight change in in positing style.

Thanks
Bill

 

[Stephen Tashi]

QM in general is a different generalized probability model to ordinary probability theory - in fact the next simplest after ordinary probability theory. It allows continuous transformations between pure states - normal probability theory does not allow that.

I'll disagree again because standard probability theory is not a specific probability model for a phenomena. I agree that the manipulations of QM can be formulated as a theory of probability together with a model for applying that theory to physics.

It does it because probabilities are not positive numbers - but complex numbers which of course the Kolmogerov axioms do not have.
That is the reason for Bell.

What is the reason and what do mean by "Bell"? Do you mean the mathematical theorem of Bell's inequality or the physical interpretation entanglement experiments in the light of that inequality?

If however you want to interpret it in a more classical manner - you can - but you need FTL ie non-locality. If there is another way of doing it - via conditional probabilities I don't know - I will let others more into it decide if your referenced paper stands up to it. But in essence it changes nothing - Bell rules out local realistic theories. Of course your linked paper states it plainly - it can't be used as an out for local realistic theories.

What the paper tries to show is that if you construct an accurate probability model for entanglement experiments, you get satisfactory numbers. The accurate model includes a representation of the possibilities that some measurements are not made on a given pair of entangled things. I agree that the interpretation of such a model is that there can be non-local effects. What I find interesting is that if a person tries to model precisely what happens in entanglement experiments with standard probability theory, he gets a model where the statistics collected in the experiment need not satisfy Bell's inequality. It shows that pretending measurements not made have outcomes produces a different mathematical model that using a non-measurement as a type of outcome.

 

[bhobba]

As I see its, this issue is complicated by the hidden assumption that spin, say, is determined at measurement.

Many do hold that view and even some textbooks say it. The correct answer is its silent on the issue. Once you understand that Bell is much easier - it simply puts bounds on the kind of interpretations you can have ie local realistic ones are out.

Thanks
Bill