What's wrong with this local realistic counter-example to Bell's theorem?

[Gordon Watson]

This thread is an offshoot of https://www.physicsforums.com/showthread.php?t=403210 -- Joy Christian's disproof of Bell.

This thread is a response to:

Maybe I'm simply not sophisticated enough, but there's a version of Bell's theorem which is so terribly elementary, that I don't see how you could "disprove" it. You could just as well try to disprove an elementary theorem in number theory or something.

The elementary version I'm talking about is the one in Sakurai

http://en.wikipedia.org/wiki/Sakurai%27s_Bell_inequality

The idea is simple: you pick 3 well-chosen axes in a couple of spin-1/2 analysers.
You consider that the population of pairs consists of 8 sub-populations, which are programmed to give (+++), (++-), (+-+) ... (---) for the 3 possible axes at Alice, and the opposite at Bob's. Each pair is randomly drawn from one of these 8 subpopulations, with a priori probabilities P1...P8. P1 + ... + P8 = 1 of course.

It is then shown that there cannot exist 8 positive numbers P1... P8 that will satisfy the statistical outcomes as predicted by quantum mechanics.

This proof is so simple that I don't consider it worth reading any paper that claims the opposite, honestly. You can just as well write a paper arguing that Pythagoras' theorem is wrong in Euclidean geometry, no ?

THE SAKURAI LINK (above) SHOULD BE STUDIED AND UNDERSTOOD. This thread also has its basis in the following [somewhat edited] exchange:

...<SNIP> ...

QM just gives mathematical functions which tell you the probability of some measurement result(s) given knowledge of some other measurement result(s).

Dear JesseM and vanesch, I agree with Jesse's point above.

Earlier in this thread, vanesch cited Sakurai and http://en.wikipedia.org/wiki/Sakurai%27s_Bell_inequality

I'd be pleased to see the QM probabilities for the eight (8) probabilities (P1-P8) in the cited text.

Can you provide them, please?

Thank you, JenniT

Those probabilities are for the various hidden-variable states, not for measurable outcomes (you can't measure more than one angle a, b, or c for a given particle). Since QM doesn't say anything about hidden-variable states which may or may not exist, only about measurable outcomes, QM does not assign probabilities to P1-P8. But what Bell shows is that we can imagine any possible combination of probabilities for P1-P8 in a hidden-variable theory (with the probabilities being in the range 0 ≤ Pn ≤ 1 and adding up to 1, of course), and the theory will always predict that the inequality P(a+, b+) ≤ P(a+, c+) + P(c+, b+) will be respected, but QM predicts this inequality is violated.

So the above led me to start this thread. For it seems to me that a local realistic counter-example can be put on an equal footing with QM in predicting the Bell inequality to be violated.

Most of this case is based on simple maths -- so errors can be easily spotted and agreed, such errors perhaps having important lessons about BT.

So in this OP we set out to:

A: Deliver P1-P8; see Sakurai http://en.wikipedia.org/wiki/Sakurai%27s_Bell_inequality which is our context.

B: Have them summing to unity.

C: Have them fully compatible with QM-style experiments; delivering accepted QM results.

D: Have them recognizing a topology [for want of a better word] associated with the spherical symmetry of the singlet state and measuring-device settings.

E: Have them challenging the basis of Bell's inequality.

F: Have them based on nothing more than high-school maths and logic; so no fancy maneuvers are involved -- and the discussion should be understood by most everyone.

NOTATION: The following short-hand notation is used.

Angles: ab denotes the angle between the orientations a and b, etc.

Functions: Cab denotes cos^2 s(ab), where s is the intrinsic spin of the relevant particle; here, as in Sakurai, s = 1/2; Sab denotes sin^2 s(ab); etc.

Reference orientation: Orientation c is taken as the reference orientation; so orientations a and b are defined with reference to c. This reference orientation arises from Bohr's responses to EPR, neatly captured in Jammer (The Philosophy of Quantum Mechanics: The Interpretations of Quantum Mechanics in Historical Perspective, 1974): "And just as the choice of a different frame of reference in relativity affects the result of a particular measurement, so also in quantum mechanics the choice of a different experimental setup has its effects on measurements, for it determines what is measurable."

Probabilities based on a local realistic hidden-variable analysis -- that theory not addressed here -- (P1-P8 identifiers after Sakurai, see above):

P1 = Cac.Cbc/2
P2 = Sac.Sbc/2
P3 = Cac.Sbc/2
P4 = Sac.Cbc/2
P5 = Sac.Cbc/2
P6 = Cac.Sbc/2
P7 = Sac.Sbc/2
P8 = Cac.Cbc/2

CONTINUATION: The reader should ascertain that the above Ps sum to one, yielding outcomes fully compatible with QM-style experiments; i.e., delivering accepted QM results.

PROVISO: Note that, in this example, the outcome-probabilities attaching to the ab settings are averages over the two ab possibilities. This follows from the topological fact re spatial relations here: ab may be constructed in two ways:

(1) ab = ac + bc.

(2) ab = ac - bc.

FOR DISCUSSION: It seems to me that this example of Bell's theorem is similar to a triangle-inequality where we can only ever measure two sides of any triangle (i.e., use a or b or c as the reference frame); see JesseM's comment above re QM and P1-P8. So any inference to a third side will be misleading. This is illustrated above where the ac and bc results are definitive because we took c as the datum; the ab result not so.

Which raises the question of the relevance of Bell's theorem to local realism? The above P1-P8 are based on a local realistic hidden-variable analysis; and yield relevant experimental outcomes; the HVs taken to be the orientation (in 3-space) of the total angular momentum of each particle, the particles pairwise correlated in the spherically symmetric singlet state.

And to this extent, the above analysis is on an equal footing with QM: Both predict that the Bell inequality will be violated.

E and OE,

JenniT

 

[DrChinese]

The criticism is the same for all such: it isn't realistic! ("So any inference to a third side will be misleading. ")

If it is, simply provide a dataset for us to look at. 0, 120, 240 degrees is always a good combo to supply. We will see if the QM predictions hold.

 

[JesseM]

NOTATION: The following short-hand notation is used.

Angles: ab denotes the angle between the orientations a and b, etc.

Functions: Cab denotes cos^2 s(ab), where s is the intrinsic spin of the relevant particle; here, as in Sakurai, s = 1/2; Sab denotes sin^2 s(ab); etc.

Reference orientation: Orientation c is taken as the reference orientation; so orientations a and b are defined with reference to c. This reference orientation arises from Bohr's responses to EPR, neatly captured in Jammer (The Philosophy of Quantum Mechanics: The Interpretations of Quantum Mechanics in Historical Perspective, 1974): "And just as the choice of a different frame of reference in relativity affects the result of a particular measurement, so also in quantum mechanics the choice of a different experimental setup has its effects on measurements, for it determines what is measurable."

Probabilities based on a local realistic hidden-variable analysis -- that theory not addressed here -- (P1-P8 identifiers after Sakurai, see above):

P1 = Cac.Cbc/2
P2 = Sac.Sbc/2
P3 = Cac.Sbc/2
P4 = Sac.Cbc/2
P5 = Sac.Cbc/2
P6 = Cac.Sbc/2
P7 = Sac.Sbc/2
P8 = Cac.Cbc/2

Not sure I follow your notation. Define the reference orientation c to be an angle of 0, and following DrChinese's suggestion, say b is an angle of 120 (relative to c) and a is an angle of 240. Then would P1 be equal to this?

Cos^2 [(1/2)(240)] * Cos^2 [(1/2)(120)] / 2

This would work out to 0.25*0.25/2 = 1/32. This would also be the value for P8 since it involves two Cos^2 terms as well (and with a=240, b=120 and c=0, Cos^2 of any combination, multiplied by s=1/2, is always 0.25). Meanwhile Sin^2 of any combination of angles would be 0.75, so P2 and P7 would be 0.75*0.75/2 = 9/32, while P3, P4, P5 and P6 would be 0.75*0.25/2 = 3/32. So the sum of all the probabilities would be 2*(1/32) + 2*(9/32) + 4*(3/32) = 2/32 + 18/32 + 12/32 = 32/32, which does mean that the probabilities all add up to 1 here.

But even if that matches what you meant, you haven't really answered my question about whether you agree that if Alice selects angle a and Bob selects angle b, then P(a+, b+) = P3 + P4, since P3 and P4 are the only probabilities whose corresponding hidden states (in the table from the Bell inequality[/url] page) have + in the "a" column of Alice's particle and + in the "b" column of Bob's particle. If you don't agree with that, then you are apparently not assuming P1-P8 denote probabilities of the same type of hidden states that Sakurai assumed, and you need to actually explain what each P1-P8 tells us about the hidden states and how those hidden states determine the outcome that Alice and Bob see for any given pair of polarizer angles they might choose.

On the other hand, if you do agree with P(a+, b+) = P3 + P4, and you also agree that if Alice chooses a and Bob chooses c we have P(a+, c+) = P2 + P4, and if Alice chooses c and Bob chooses b we have P(c+, b+) = P3 + P7, then given the numbers above we will have:

P(a+, b+) = P3 + P4 = 3/32 + 3/32 = 6/32
P(a+, c+) = P2 + P4 = 9/32 + 3/32 = 12/32
P(c+, b+) = P3 + P7 = 3/32 + 9/32 = 12/32

So, the inequality is not violated here; the above numbers indicate that the inequality

P(a+, b+) ≤ P(a+, c+) + P(c+, b+)

would be satisfied.

 

[Gordon Watson]

The criticism is the same for all such: it isn't realistic! ("So any inference to a third side will be misleading. ")

If it is, simply provide a dataset for us to look at. 0, 120, 240 degrees is always a good combo to supply. We will see if the QM predictions hold.

Thank you DrC, but I am not clear about your reference to "all such"; nor why you say that it is not realistic?

1. I gave the example because I am not aware that there is any other such example relating to Bell's theorem (BT). SO my example is definitely not intended to be in the same (poor) class as most critiques of BT! In most of those cases, I am sure that you and I will be as one, and on the same side. By which I mean that my example is different in that it provides specific values for P1-P8 that deliver the QM outcomes for any c-based reference frame -- c being any arbitrary real orientation consistent with a real test.

2. As for not being "realistic" -- I think it is, in that it delivers the correct QM outcomes -- with that interesting proviso. And it is based on real hidden-variables in that we do not determine the pristine orientation of the total spin of the subject particles. That is, our "measurements" perturb them -- hence they are "hidden elements of reality" --- "hidden" from us -- in my terms.

3. As for these outcomes; they flow at once from observation. I will expand on your comment in my response to JesseM -- next. But please note that my example is designed to question the validity of such data-set requirements -- in that they cannot be justified realistically. QM takes this view (I believe); and I take this view; my simple maths and example is intended to support this view. See next post from me.

Appreciating you response, I look forward to our discussions.

JenniT

 

[Gordon Watson]

<Underlining emphasis added by JenniT.>

Not sure I follow your notation. Define the reference orientation c to be an angle of 0, and following DrChinese's suggestion, say b is an angle of 120 (relative to c) and a is an angle of 240. Then would P1 be equal to this?

Cos^2 [(1/2)(240)] * Cos^2 [(1/2)(120)] / 2

This would work out to 0.25*0.25/2 = 1/32. This would also be the value for P8 since it involves two Cos^2 terms as well (and with a=240, b=120 and c=0, Cos^2 of any combination, multiplied by s=1/2, is always 0.25). Meanwhile Sin^2 of any combination of angles would be 0.75, so P2 and P7 would be 0.75*0.75/2 = 9/32, while P3, P4, P5 and P6 would be 0.75*0.25/2 = 3/32. So the sum of all the probabilities would be 2*(1/32) + 2*(9/32) + 4*(3/32) = 2/32 + 18/32 + 12/32 = 32/32, which does mean that the probabilities all add up to 1 here.

But even if that matches what you meant, you haven't really answered my question about whether you agree that if Alice selects angle a and Bob selects angle b, then P(a+, b+) = P3 + P4, since P3 and P4 are the only probabilities whose corresponding hidden states (in the table from the Bell inequality[/url] page) have a + in the "a" column of Alice's particle and a + in the "b" column of Bob's particle. If you don't agree with that, then you are apparently not assuming the probabilities denote the same type of hidden states that Sakurai assumed, and you need to actually explain what each P1-P8 tells us about the hidden states and how those hidden states determine the outcome that Alice and Bob see for any given pair of polarizer angles they might choose.

On the other hand, if you do agree with that, and you also agree that if Alice chooses a and Bob chooses c we have P(a+, c+) = P2 + P4, and if Alice chooses c and Bob chooses b we have P(c+, b+) = P3 + P7, then given the numbers above we will have:

P(a+, b+) = P3 + P4 = 3/32 + 3/32 = 6/32
P(a+, c+) = P2 + P4 = 9/32 + 3/32 = 12/32
P(c+, b+) = P3 + P7 = 3/32 + 9/32 = 12/32

So, the inequality is not violated here; the above numbers indicate that the inequality

P(a+, b+) ≤ P(a+, c+) + P(c+, b+)

would be satisfied.

Thanks Jesse, and you seem to understand the notation OK.

Note that you can simply sum the P1-P8 in their most general form as given by me [for each is a simple trigonometric combination]. So, no matter the angle/orientation combinations, the boundary condition -- SUM TO UNITY -- is satisfied in general. The point being that no error is committed at this first step.

Note also that, if Alice selects angle a and Bob selects angle b, there is no reference to c. My example is worked with c as the reference orientation (RO) to demonstrate this very point. So if you select a or b as the RO, then you will get -- for sure -- the values and answer that you seek. BUT the problem will move to another double-valued angle.

The point of my reference orientation follows from my view that BT is akin to a triangle-inequality and that we need to carefully track what it is that we measure. Whenever you refer to just two orientations, here a and b, there is no need for an RO since EITHER orientation automatically serves as such. BUT NOTE: You invoke a double-valued angle ab if you seek to refer to orientation c -- when c is the designated RO.

There is the ab = ac + bc. There is the ab = ac - bc. The calculation that you are doing is delivering the average value over both these ab values. So the calculation is realistic (DrC), delivering an important real value -- the average value over both ab values. It is NOT a wrong answer under the specified example; the example CAN deliver the response you seek if you select a or b as the RO.

The point here being: In QM this issue does not arise, because as you said, QM does not deliver P1-P8. The point of my local realistic counter-example to BT is that IT DOES deliver P1-P8; and it does deliver valid outcome distributions (probabilities). BUT the basis for those probabilities needs to be understood.

So, if I am not using the same hidden states as Sakurai, is it possible that Sakurai's states are impossible?

My example is intended to go beyond any early impossibility with respect to states. AND YET, (please note) in agreement with QM, at the same time provide a different (very elementary) perspective on the invalidity of BT.

In my approach, I am raising the question: Is BT a valid standard against which to judge local realism?

I suggest that it is not. Not just because it falls foul of QM -- but because it falls foul of this local realistic counter-example.

So, to be clear: Where you derive an unsatisfactory result -- so do I. But it is my hope that my accompanying analysis points the finger correctly at BT and not at LR! That's the question.

With thanks and appreciation, as always, I trust this clarifies our general agreement re BT-fundamentals -- but with new hope for LRs such as I,

noting that I've here used a double-valued LR,

JenniT

 

[JesseM]

Thanks Jesse, and you seem to understand the notation OK.

Note that you can simply sum the P1-P8 in their most general form as given by me [for each is a simple trigonometric combination]. So, no matter the angle/orientation combinations, the boundary condition -- SUM TO UNITY -- is satisfied in general. The point being that no error is committed at this first step.

Note also that, if Alice selects angle a and Bob selects angle b, there is no reference to c. My example is worked with c as the reference orientation (RO) to demonstrate this very point. So if you select a or b as the RO, then you will get -- for sure -- the values and answer that you seek. BUT the problem will move to another double-valued angle.

The point of my reference orientation follows from my view that BT is akin to a triangle-inequality and that we need to carefully track what it is that we measure. Whenever you refer to just two orientations, here a and b, there is no need for an RO since EITHER orientation automatically serves as such. BUT NOTE: You invoke a double-valued angle ab if you seek to refer to orientation c.

There is the ab = ac + bc. There is the ab = ac - bc. The calculation that you are doing is delivering the average value over both these ab values. So the calculation is realistic (DrC), delivering an important real value -- the average value over both ab values. It is NOT a wrong answer under the specified example; the example CAN deliver the response you seek if you select a or b as the RO.

None of this verbal explanation really makes sense to me, I don't know why the fact that Alice chose a and Bob chose b would somehow mean their angles a and b were no longer defined relative to the "reference orientation" c, or how you think this would affect any of the calculations. Instead of giving a lot of verbal explanation which isn't understandable to me and probably isn't understandable to DrChinese or anyone else besides you, can you just give some specific numerical example where you calculate P1-P8 (perhaps just repeating the numbers I gave above, assuming I didn't misunderstand your meaning there), and then show explicitly how you use these to calculate specific values for P(a+, b+), P(a+, c+) + P(c+, b+)?

The point here being: In QM this issue does not arise, because as you said, QM does not deliver P1-P8. The point of my local realistic counter-example to BT is that IT DOES deliver P1-P8; and it does deliver valid outcome distributions (probabilities).

HOW? You haven't said one word about how you want to calculate probabilities of different outcomes like (a+, b+).

So, if I am not using the same hidden states as Sakurai, is it possible that Sakurai's states are impossible?

Again there seems to be no explanation of the reasoning for anything you say. Why should it be "impossible" in a local hidden variables theory that the hidden variables associated with each particle would predetermine what response they are going to give to any specific angle like a, b, and c? This is why I asked you very specifically in [post=3150315]post 68 on the other thread[/post] whether you agreed with the idea that each hidden variable state would have a predetermined result for each of the three angles, but instead of giving me a direct answer you just made a vague statement that "I'm familiar with many of the Bell fundamentals" and then claimed your model would address everything, which it obviously hasn't since you are now questioning whether P1-P8 have the same meaning as they do in Sakurai, but offer no substitute meaning. P1 - P8 were defined to be the probabilities of the types of hidden states he listed, so if you're changing the meaning then you're obviously dealing with a completely different set of 8 probabilities then the ones he listed, so the burden is on you to explain what the symbols P1-P8 even mean. Speaking of which, why do you even think there should be 8 categories of hidden states as opposed to 9 or 9000, if you are not defining the categories to be combinations of predetermined +'s and -'s for each of the three angles? The only reason there were 8 was that there are only 8 distinct possible ways to assign + or - to each of the three angles a,b,c.

My example is intended to go beyond any early impossibility with respect to states. AND YET, (please note) in agreement with QM, at the same time provide a different (very elementary) perspective on the invalidity of BT.

You have provided zero evidence that you have a model which is "in agreement with QM" because you have given no method to calculate probabilities like P(a+, b+).

Think about it this way: if you believe you have a local realist counterexample to BT, then you should be able to simulate the way it works in the following manner. You will play the part of the "source", and on each trial you will fill two flash drives with data representing the simulated properties (hidden or measurable, whatever you want) of two particles emitted by the source (perhaps you will use some algorithm to assign the properties on each trial in a probabilistic way rather than doing it by hand). Then you will send one flash drive to Alice and another to Bob. They each plug their flash drive into their computer, then randomly choose whether to press the key "A", "B", or "C" representing their choice of simulated polarizer angle. The computer then takes the simulated polarizer angle, along with the data from the flash drive about the properties (hidden or measurable) of the particle reaching the polarizer, and uses these along with an algorithm representing the laws of physics (which you programmed into each computer earlier, so they can be whatever laws you want) to calculate as output either "+" or "-". If we assume you have no foreknowledge of which key Alice and Bob will press on each trial, do you think there's any way for you to "win" this game by matching the statistics seen in QM, so on any trial that Alice and Bob press the same key they are guaranteed to get opposite results, yet if we look at the trials where they picked different keys, we will find a violation of the inequality P(a+, b+) ≤ P(a+, c+) + P(c+, b+)?

 

[Gordon Watson]

None of this verbal explanation really makes sense to me, I don't know why the fact that Alice chose a and Bob chose b would somehow mean their angles a and b were no longer defined relative to the "reference orientation" c, or how you think this would affect any of the calculations. Instead of giving a lot of verbal explanation which isn't understandable to me and probably isn't understandable to DrChinese or anyone else besides you, can you just give some specific numerical example where you calculate P1-P8 (perhaps just repeating the numbers I gave above, assuming I didn't misunderstand your meaning there), and then show explicitly how you use these to calculate specific values for P(a+, b+), P(a+, c+) + P(c+, b+)?

Apologies; but had I hoped that verbal explanation would not be an essential part of the proceedings. I wanted to rely on the simple maths. That is why I tried to spell out the position fairly fully in the OP.

Note that that the C and S functions (from which P1-P8 are built) are respectively Cos^2 and Sin^2 functions; so all calculations can be generalized without limitation to specific examples. That is why I was surprised that you checked the SUM TO UNITY for one specific example when you could add the 4 sets of 2 probabilities; virtually in your head; once and for all.

I also gave the PROVISO re calculations that involve the ab angle. For, in the given example, ac and bc are defined explicitly; ab is not. That is: ab is is two-valued in the given example: ab = ac +bc or ab = ac - bc. (Edited for clarity.)

To see the result of interest to you, relabel you ab as ac and use my table of P1-P8. The correct results will then fall out for ab. But you will have moved the problem so that average values are now derived for one other angle. (Relabeling is offered as a short-cut way to study another example; i.e., with a different RO.)

The outcomes under a change in RO, under local realism (LR), are presented as LR endorsements of the Bohr/Jammer remark in the OP: That "a different experimental setup has its effects on measurements, for it determines what is measurable," noting that we measure across just one angle in physically testing such setups. [Expanded and edited, seeking clarity following comments by JesseM.]

To stick with the simplicity of my example, let us evaluate the conditions where it is designed to deliver, exactly, the QM result that you appear to doubt.

P(a+,c+) = P2 + P4 = Sac/2.
P(a+,c-) = P1 + P3 = Cac/2.
P(a-,c+) = P6 + P8 = Cac/2.
P(a-,c-) = P5 + P7 = Sac/2.

All being the CORRECT result!

All the P(b,c) combinations follow similarly.

BUT: IF you now want to evaluate P(a,b) combinations under the exemplified conditions, THEN only the average result falls out because, under the example, ab is two-valued (with c the RO).

P(a+,b+) = P3 + P4 = [Cac.Sbc + Sac.Cbc]/2; etc.

Another CORRECT result (but now an average) when (and because here) ab is deliberately two-valued: as in my example.

HOW? You haven't said one word about how you want to calculate probabilities of different outcomes like (a+, b+).

Sorry; I thought it would be clear. I thought it was clear that the QM results are equally a boundary-condition on my example. Therefore:

Since only 2 orientations, and one angle, are involved; just relabel the ab as ac and use the P1 and P2 that I provided.

Note that my example is specifically designed to focus on the difficulty that you are raising re ab (by invoking c as the RO). Note also that my argument is NOT with QM but with BT; and that the OP did assume that critics would put P1-P8 into play, in their most general format, and discover for themselves the basis of (and the need for) the PROVISO.

Again there seems to be no explanation of the reasoning for anything you say. Why should it be "impossible" in a local hidden variables theory that the hidden variables associated with each particle would predetermine what response they are going to give to any specific angle like a, b, and c? This is why I asked you very specifically in [post=3150315]post 68 on the other thread[/post] whether you agreed with the idea that each hidden variable state would have a predetermined result for each of the three angles, but instead of giving me a direct answer you just made a vague statement that "I'm familiar with many of the Bell fundamentals" and then claimed your model would address everything, which it obviously hasn't since you are now questioning whether P1-P8 have the same meaning as they do in Sakurai, but offer no substitute meaning. P1 - P8 were defined to be the probabilities of the types of hidden states he listed, so if you're changing the meaning then you're obviously dealing with a completely different set of 8 probabilities then the ones he listed, so the burden is on you to explain what the symbols P1-P8 even mean. Speaking of which, why do you even think there should be 8 categories of hidden states as opposed to 9 or 9000, if you are not defining the categories to be combinations of predetermined +'s and -'s for each of the three angles? The only reason there were 8 was that there are only 8 distinct possible ways to assign + or - to each of the three angles a,b,c.

Since my theory allows for an infinite number of HVs -- I defined them as orientations in 3-space -- it follows that, in my example, NO TWO particle-pairs will EVER have the same HV pair!

So I avoid the use of such terms as "predetermined" because -- as I thought was also clear: I allow that a "measurement interaction" delivers ONLY TWO possible outcomes from a non-repeating infinity of inputs. So what is predetermined? Random inputs; each PAIRWISE correlated; two outputs only, EACH SGM orientation randomly and independently selected at the last instant by Alice and Bob (which are, here, the names of the SGM operators, not the particles), just before a particle arrives for detection; though determined spin and gyroscopic transitions spring to mind!

You have provided zero evidence that you have a model which is "in agreement with QM" because you have given no method to calculate probabilities like P(a+, b+).

Sorry that you take that to be the case: I hope the above makes my position clearer. Especially your understanding that I am for QM -- and questioning BT.

I so far see no reason to withhold this guidance: If you find ANY conflict WHATSOEVER with QM -- in any calculation that you derive from my example -- you will have made a mistake.

I am not against QM; and I reject every facile attack on QM or BT known to me. I certainly invoke NO REQUIREMENT whatsoever for ANY loophole-based escape clauses.

Think about it this way: if you believe you have a local realist counterexample to BT, then you should be able to simulate the way it works in the following manner. You will play the part of the "source", and on each trial you will fill two flash drives with data representing the simulated properties (hidden or measurable, whatever you want) of two particles emitted by the source (perhaps you will use some algorithm to assign the properties on each trial in a probabilistic way rather than doing it by hand). Then you will send one flash drive to Alice and another to Bob. They each plug their flash drive into their computer, then randomly choose whether to press the key "A", "B", or "C" representing their choice of simulated polarizer angle. The computer then takes the simulated polarizer angle, along with the data from the flash drive about the properties (hidden or measurable) of the particle reaching the polarizer, and uses these along with an algorithm representing the laws of physics (which you programmed into each computer earlier, so they can be whatever laws you want) to calculate as output either "+" or "-". If we assume you have no foreknowledge of which key Alice and Bob will press on each trial, do you think there's any way for you to "win" this game by matching the statistics seen in QM, so on any trial that Alice and Bob press the same key they are guaranteed to get opposite results, yet if we look at the trials where they picked different keys, we will find a violation of the inequality P(a+, b+) ≤ P(a+, c+) + P(c+, b+)?

Well, apart from my difficulty in envisioning a program with infinite randomized inputs, pairwise correlated, that can mimic the singlet state ... And apart from my results BEING AT ONE WITH QM ... DO you not see that I AGREE THAT BT is violated ... That this is the point of my LR example?

BT: Violated, as you and I agree, by QM.

BT: Violated too, as I suggest, by LR.

PS: E & OE: If you see a conflict with QM in any calculation that you derive from my example, you will have made a mistake.

Sincerely hoping that this clarifies many of your concerns,

With best regards,

JenniT

 

[Gordon Watson]

ADDENDUM TO THE OP:

Boundary conditions on the given example:

1: P1-P8 sum to unity.

2. All the accepted results of QM may be derived from P1-P8, and are accepted. HOWEVER:

3. It is expected that the need for the PROVISO will be examined by the reader: checking the results for QM by means of general settings; or via settings of interest to the reader.

4. This is NOT an attack on QM.

5. This is NOT an attack on BT based on facile or loop-hole arguments.

6. The example is offered to suggest that BT may be acceptably rejected on the basis of Local Realism.

7. That is, acceptably rejected in line with QM's own rejection of BT.

8. The example is designed to show that Local Realism can deliver P1-P8 realistically and consistent with QM.

NB: It has been said that QM does not deliver P1-P8. I think that QM can; but in any case, my P1-P8 are (IMHO) not in conflict with QM.

JenniT

 

[JesseM]

Apologies; but had I hoped that verbal explanation would not be an essential part of the proceedings. I wanted to rely on the simple maths. That is why I tried to spell out the position fairly fully in the OP.

Note that that the C and S functions (from which P1-P8 are built) are respectively Cos^2 and Sin^2 functions; so all calculations can be generalized without limitation to specific examples. That is why I was surprised that you checked the SUM TO UNITY for one specific example when you could add the 4 sets of 2 probabilities; virtually in your head; once and for all.

I don't understand what you mean by "4 sets of 2 probabilities"--what would each "set" consist of?

edit: perhaps in this case you mean:
(P1 + P3) = cos^2(ac)/2 * (cos^2(bc) + sin^2(bc)) = cos^2(ac)/2
(P2 + P4) = sin^2(ac)/2 * (sin^2(bc) + cos^2(bc)) = sin^2(ac)/2
(P5 + P7) = sin^2(ac)/2 * (cos^2(bc) + sin^2(bc)) = sin^2(ac)/2
(P6 + P8) = cos^2(ac)/2 * (sin^2(bc) + cos^2(bc)) = cos^2(ac)/2

...and of course (cos^2(ac) + sin^2(ac))/2 + (sin^2(ac) + (cos^2(ac))/2 = 1/2 + 1/2 = 1. But it's pretty weird that you would immediately expect me to have seen this pattern (as suggested by your comment "That is why I was surprised that you checked the SUM TO UNITY for one specific example when you could add the 4 sets of 2 probabilities; virtually in your head; once and for all") in a long list of formulas for P1-P8 when you had given no indication of where these formulas were supposed to have come from, or made any mention of the fact that they could be grouped in such a way.

I also gave the PROVISO re calculations re the ab angle, if you invoke a third orientation.

You said something about the ab angle and the "reference orientation", but again you used words which were totally unclear to me and no specific numerical example.

To see the result of interest to you, relabel you ab as ac

Why would I want to "relabel" anything? And what "result of interest" would I be calculating? Again, I don't think these verbal explanations are going to be at all comprehensible without an example.

To stick with the simplicity of my example, let us evaluate the conditions where it is designed to deliver, exactly, the QM result that you appear to doubt.

P(a+,c+) = P2 + P4 = Sac/2.
P(a+,c-) = P1 + P3 = Cac/2.
P(a-,c+) = P6 + P8 = Cac/2.
P(a-,c-) = P5 + P7 = Sac/2.

All being the CORRECT result!

All the P(b,c) combinations follow similarly.

BUT: IF you now want to evaluate P(a,b) combinations under the exemplified conditions, THEN only the average result falls out because, under the example, ab is two-valued (with c the RO).

P(a+,b+) = P3 + P4 = Cac.Sbc + Sac.Cbc; etc.

Another CORRECT result (but now an average) when (and because here) ab is deliberately two-valued: as in my example.

I don't understand what "ab is two-valued" means, I assumed the notation ab just meant the difference between the angles a and b, taken in some standard order (either a-b or b-a, for the angles a=240 and b=120 the order is irrelevant anyway since Sin^2(120/2)=Sin^2(-120/2) and Cos^2(120/2)=Cos^2(-120/2)). So I don't understand why you would now claim that P3 = Cac.Sbc rather than Cac.Sbc/2 as in your original table. It appears you are claiming that the probability P3 of different hidden variable states can somehow vary depending on whether Bob chooses angle b or angle c, which is definitely not allowed since the idea is that each experimenter chooses their detection angles at random after the source has already emitted the particles, and the source doesn't have any precognitive abilities that would allow it to vary the probabilities of what hidden variables it would assign to each particle based on what angles the experimenters would later choose--this is known as the "no-conspiracy" assumption in Bell's proof, it sounds like your supposed counterexample is probably based on violating this.

Sorry; I thought it would be clear. I thought it was clear that the QM results are equally a boundary-condition on my example. Therefore:

Since only 2 orientations, and one angle, are involved; just relabel the ab as ac and use the P1 and P2 that I provided.

It's physically incoherent to suggest you can just "relabel" things and then calculate probabilities in the way you do above, the hidden states that the probabilities refer to are supposed to specify the predetermined results on all three angles, understood to be defined relative to some fixed coordinate system that doesn't change from one trial to another. So if you relabel depending on what angles Alice and Bob choose and calculate the probabilities, then if we translate this back into the fixed coordinate system, that means the probabilities that the source emits a particle pair with a given hidden state will actually vary depending on how Alice and Bob later set their polarizers! Again this would imply a sort of precognition on the part of the source, and a violation of the no-conspiracy assumption in Bell's proof.

Since my theory allows for an infinite number of HVs -- I defined them as orientations in 3-space -- it follows that, in my example, NO TWO particle-pairs will EVER have the same HV pair!

But again, the 8 states that are assigned probabilities P1-P8 are not meant to be the complete set of hidden variables, they are just broad categories that each possible hidden variable state should fall into. For example, we might imagine the complete set of variables as a function F(theta), which can be given any angle theta from 0 to 360, and it will give a +1 or -1 for each angle. Although there would be an infinite number of possible functions of this type, each possible function can be put into one of eight categories depending on what results it gives for the specific angles the experimenters have chosen to restrict themselves to, in my example a=240, b=120 and c=0. So for example if you have one function F1 which has F1(240)= +1, F1(120)= -1, and F1(0) = +1, and another function F2 which has F2(240)=+1, F2(120) = -1, and F2(0) = +1, then they would both go into the category (+ - +) in spite of the fact that they might differ for many other choices of angles, for example F1(175.31) = +1 and F2(175.31) = -1. So you might do billions of trials and have a different Fn for Alice's particle on each trial, but you could still label Alice's particle as falling into one of the eight categories of hidden states listed in the table on the Bell inequality[/url] page.

So I avoid the use of such terms as "predetermined" because -- as I thought was also clear: I allow that a "measurement interaction" delivers ONLY TWO possible outcomes from a non-repeating infinity of inputs.

"Predetermined" just means that immediately after the particles have been emitted, their hidden states are such that they predetermine what results they would give for each possible polarizer angle. It doesn't mean that on any given trial the source is predetermined to emit particles in a particular hidden-variable state, it just means that once the particles have already been emitted and their hidden-variable state is already well-defined, then from that point on it's inevitable what results the particles will give for any specific combination of polarizer angles chosen by Alice and Bob. If the results weren't predetermined immediately after emission in this way, there would seem to be no way to explain in local realist terms how it could be that whenever Alice and Bob choose identical polarizer angles, they are guaranteed with probability 1 to get opposite results. If there was any random element in terms of how the hidden variables carried by each particle interacted with the polarizer to produce a result, then if the particles have no way to communicate to coordinate their results, we would occasionally expect that even with the same detector angle the two experimenters would get non-opposite results. Do you disagree with any of this?

Think about it this way: if you believe you have a local realist counterexample to BT, then you should be able to simulate the way it works in the following manner. You will play the part of the "source", and on each trial you will fill two flash drives with data representing the simulated properties (hidden or measurable, whatever you want) of two particles emitted by the source (perhaps you will use some algorithm to assign the properties on each trial in a probabilistic way rather than doing it by hand). Then you will send one flash drive to Alice and another to Bob. They each plug their flash drive into their computer, then randomly choose whether to press the key "A", "B", or "C" representing their choice of simulated polarizer angle. The computer then takes the simulated polarizer angle, along with the data from the flash drive about the properties (hidden or measurable) of the particle reaching the polarizer, and uses these along with an algorithm representing the laws of physics (which you programmed into each computer earlier, so they can be whatever laws you want) to calculate as output either "+" or "-". If we assume you have no foreknowledge of which key Alice and Bob will press on each trial, do you think there's any way for you to "win" this game by matching the statistics seen in QM, so on any trial that Alice and Bob press the same key they are guaranteed to get opposite results, yet if we look at the trials where they picked different keys, we will find a violation of the inequality P(a+, b+) ≤ P(a+, c+) + P(c+, b+)?

Well, apart from my difficulty in envisioning a program with infinite randomized inputs, pairwise correlated, that can mimic the singlet state

But the program doesn't need to simulate all the hidden variables, only those that are directly relevant to determining the output given that Alice and Bob's only choices of angles are the ones represented by the three keys A, B, and C (they don't have an option to enter in some different angle). Again, if you accept that the hidden variable immediately after the particles are emitted must predetermine what result the particle would give for any possible polarizer angle they might later encounter, in that case the only thing that you'd need to load on the flash drive would be the three predetermined results for each of the three angles A, B, and C.

In any case you are free to load whatever data you want onto these flash drives, and imagine that their memory capacity is as large as you want (even infinite). Given that, do you think you can "win" at the game I describe above, assuming you have no precognition and cannot know in advance what button Alice and Bob are going to press on each trial as you decide what simulated hidden variables to load on the flash drives on that trial? Yes or no?

... And apart from my results BEING AT ONE WITH QM ... DO you not see that I AGREE THAT BT is violated ... That this is the point of my LR example?

Again, the problem is that the rule you use for figuring out the probabilities, specifically the confusing "relabeling", seems to imply that if we just want to know the probabilities for different combinations of predetermined answers for angles a,b,c defined relative to some fixed spatial coordinate system, the probabilities will actually depend on the future choice made by Alice and Bob about what angles to set their polarizers, which implies the source has a strange foreknowledge of the future choices of human experimenters when it assigns hidden variables to each particle pair, and violates Bell's no-conspiracy assumption.

 

[Gordon Watson]

I don't understand what you mean by "4 sets of 2 probabilities"--what would each "set" consist of? Again, it would really help if you would give specific numerical examples rather than just explanations in words.

SUM [P1-P8] = Cac.Cbc + Sac.Sbc + Cac.Sbc + Sac.Cbc

= (Cac + Sac)Cbc + (Cac + Sac)Sbc = Cbc + Sbc = 1

= (Cbc + Sbc)Cac + (Cbc + Sbc)Sac = Cac + Sac = 1

for any numerical example.

I am keen to provide the general case, confident that any numerical example of anyone's choosing will reveal the point that I'm seeking to make. You did a specific example for the sum to unity; I prefer to do the general case.

You said something about the ab angle and the "reference orientation", but again you used words which were totally unclear to me and no specific numerical example.

In the 3-orientation setup that constitutes the example, for any numerical examples consistent with the example, the given P1-P8 hold.

Why would I want to "relabel" anything? And what "result of interest" would I be calculating? Again, I don't think these verbal explanations are going to be at all comprehensible without an example.

I've edited that post to make it clearer that relabeling is just a short-cut way to study other tripartite examples; meaning ... rather than go back to the first principles used to derive the given P1-P8.

You seemed to be interested in the ab results, to which the PROVISO applies. The example was selected to have the problem focussed on ab outcomes and conclusions therefrom.

The point being that LR, as well as QM, points to a problem with BT; witness the ab-outcome averages. Witness the problem shift if you take an alternative RO.

I don't understand what "ab is two-valued" means, I assumed the notation ab just meant the difference between the angles a and b, taken in some standard order (either a-b or b-a, for the angles a=240 and b=120 the order is irrelevant anyway since Sin^2(120/2)=Sin^2(-120/2) and Cos^2(120/2)=Cos^2(-120/2)). So I don't understand why you would now claim that P3 = Cac.Sbc rather than Cac.Sbc/2 as in your original table. It appears you are claiming that the probability P3 of different hidden variable states can somehow vary depending on whether Bob chooses angle b or angle c, which is definitely not allowed since the idea is that each experimenter chooses their detection angles at random after the source has already emitted the particles, and the source doesn't have any precognitive abilities that would allow it to vary the probabilities of what hidden variables it would assign to each particle based on what angles the experimenters would later choose--this is known as the "no-conspiracy" assumption in Bell's proof, it sounds like your supposed counterexample is probably based on violating this.

I've fixed (via an edit) the typo re the 1/2; thank you. I am not claiming that P3 of different HV states can vary. I am providing the Ps for different outcomes under the defined tripartite setup; introducing the third orientation to be consistent with Bell and Sakurai; recognizing that we can only test for one angle (two orientations) in anyone physical test.

This notion that you can just "relabel" things is physically incoherent, the hidden states that the probabilities refer to are supposed to specify the predetermined results on all three angles, understood to be defined relative to some fixed coordinate system that doesn't change from one trial to another. So if you relabel depending on what angles Alice and Bob choose and calculate the probabilities, then if we translate this back into the fixed coordinate system, that means the probabilities that the source emits a particle pair with a given hidden state will actually vary depending on how Alice and Bob later set their polarizers! Again this would imply a sort of precognition on the part of the source, and a violation of the no-conspiracy assumption in Bell's proof.

Relabeling was offered as a short-cut; that's all. The Ps about the source emissions are not in my discussion; no two paired-emissions being the same in my example. It is the correlations that vary, depending on the SGM settings. I am not implying precognition by the source. I am implying that a triangle inequality needs careful study.

But again, the 8 states that are assigned probabilities P1-P8 are not meant to be the complete set of hidden variables, they are just broad categories that each possible hidden variable state should fall into. For example, we might imagine the complete set of variables as a function F(theta), which can be given any angle theta from 0 to 360, and it will give a +1 or -1 for each angle. Although there would be an infinite number of possible functions of this type, each possible function can be put into one of eight categories depending on what results it gives for the specific angles the experimenters have chosen to restrict themselves to, in my example a=240, b=120 and c=0. So for example if you have one function F1 which has F1(240)= +1, F1(120)= -1, and F1(0) = +1, and another function F2 which has F2(240)=+1, F2(120) = -1, and F2(0) = +1, then they would both go into the category (+ - +) in spite of the fact that they might differ for many other choices of angles, for example F1(175.31) = +1 and F2(175.31) = -1. So you might do billions of trials and have a different Fn for Alice's particle on each trial, but you could still label Alice's particle as falling into one of the eight categories of hidden states listed in the table on the Bell inequality[/url] page.

OK. I had taken the Table to be about outcome-Ps; so I had given those P-s, P1-P8, from my LR perspective. I was then seeking to draw conclusions from those P-s; that is, having given my LR basis from which to derive the other P-s (that flow from the 8), studying the implications of those other P-s.

[I have not thought to label any particle. I personally think of them as gyroscopes whose measured outcomes are related to deterministic gyroscopic dynamics; with continuous spin-trajectories terminating in a specific device orientation when perturbed by that device. Probably, if QM can't label them, neither can I -- I've not tried -- me thinking that these deeper-dynamics take us away from the lessons that are available from studying outcome-dynamics that we can -- I hope -- all agree upon.]

"Predetermined" just means that immediately after the particles have been emitted, their hidden states are such that they predetermine what results they would give for each possible polarizer angle. It doesn't mean that on any given trial the source is predetermined to emit particles in a particular hidden-variable state, it just means that once the particles have already been emitted and their hidden-variable state is already well-defined, then from that point on it's inevitable what results the particles will give for any specific combination of polarizer angles chosen by Alice and Bob. If the results weren't predetermined immediately after emission in this way, there would seem to be no way to explain in local realist terms how it could be that whenever Alice and Bob choose identical polarizer angles, they are guaranteed with probability 1 to get opposite results. If there was any random element in terms of how the hidden variables carried by each particle interacted with the polarizer to produce a result, then if the particles have no way to communicate to coordinate their results, we would occasionally expect that even with the same detector angle the two experimenters would get non-opposite results. Do you disagree with any of this?

No disagreement.

There is be no random element in my considerations. Conserved total-spin, related total-spin-orientations, identical device settings, it all follows:

Correlated tests on correlated objects yield correlated outcomes.

But the program doesn't need to simulate all the hidden variables, only those that are directly relevant to determining the output given that Alice and Bob's only choices of angles are the ones represented by the three keys A, B, and C (they don't have an option to enter in some different angle). Again, if you accept that the hidden variable immediately after the particles are emitted must predetermine what result the particle would give for any possible polarizer angle they might later encounter, in that case the only thing that you'd need to load on the flash drive would be the three predetermined results for each of the three angles A, B, and C.

I suspect that the difficulty in meeting your requirement arises from the difficulty shown in my LR example with outcomes. All the more so in spherical gyroscopic-style dynamics. Is the total-spin-axis transiting, under perturbation a-b-c, b-c-a, c-a-b? Are such deeper considerations required to evaluate my discussion re OUTCOMES?

In any case you are free to load whatever data you want onto these flash drives, and imagine that their memory capacity is as large as you want (even infinite). Given that, do you think you can "win" at the game I describe above, assuming you have no precognition and cannot know in advance what button Alice and Bob are going to press on each trial as you decide what simulated hidden variables to load on the flash drives on that trial? Yes or no?

No.

Again, the problem is that the rule you use for figuring out the probabilities, specifically the confusing "relabeling", seems to imply that if we just want to know the probabilities for different combinations of predetermined answers for angles a,b,c defined relative to some fixed spatial coordinate system, the probabilities will actually depend on the future choice made by Alice and Bob about what angles to set their polarizers, which implies the source has a strange foreknowledge of the future choices of human experimenters when it assigns hidden variables to each particle pair, and violates Bell's no-conspiracy assumption.

Relabeling was just offered as a short-cut.

The outcome-correlations and their probabilities will vary with the detector settings.

The emissions are randomly, but pair-wise, correlated; no two pairs emitted the same; independent of any device setting.

No conspiracy is permitted.

 

[JesseM]

You seemed to be interested in the ab results, to which the PROVISO applies. The example was selected to have the problem focussed on ab outcomes and conclusions therefrom.

But I have no idea the meaning of your "proviso"--why would we need to do any averaging, and what does it mean to say ab is "two-valued"? You didn't answer my questions about this. Do you agree that in the example where a=240, b=120, c=0, and where P1-P8 refer to categories of hidden variables which tell us the predetermined results for each angle, then P(a+, b+) would simply be P3 + P4 = 3/32 + 3/32 = 6/32? Whereas the correct quantum-mechanical prediction would be would be 1/2 * sin^2((240-120)/2) = 1/2 * 0.75 = 3/8 = 12/32.

I've fixed (via an edit) the typo re the 1/2; thank you. I am not claiming that P3 of different HV states can vary. I am providing the Ps for different outcomes under the defined tripartite setup; introducing the third orientation to be consistent with Bell and Sakurai; recognizing that we can only test for one angle (two orientations) in anyone physical test.

When you say you are "providing the Ps for different outcomes", by "outcome" do you just mean the predetermined results for all three angles determined by the hidden variables, which we can never actually determine from empirical measurements? That's what the Ps are supposed to represent in the table on the wikipedia page. Or are you trying to define the Ps to mean something different than the wikipedia page, something that could actually be calculated from empirical measurements where we can only test two angles for a given particle pair? If that's the case I don't understand what your Ps are supposed to represent and you haven't really spelled it out, but this would account for a lot of my earlier confusion about what you were trying to say (why I thought you were suggesting the source was precognitive, for example). In any case if you're defining the Ps this way I don't see the relevance to the derivation of the Bell inequality, since in that derivation you are supposed to treat each P as giving the probability that the source will emit particles with hidden variables that give Alice a given triplet of predetermined results such as a+,b+,c- (and predetermine the opposite results on all three angles for Bob). As long as you agree that once the particles are emitted, in a local realist theory their hidden variables must predetermine their results for all three angles--and it seems from your comments below that you do--then can we agree that, however you have defined the Ps in the past, from now on we shall define them to refer to the predetermined results so that they have the same meaning they are assigned in the derivation of the Bell inequality? If this means you have to change your formulas for P1-P8 that's fine, but the point is that there is no possible way combination of formulas you can choose such that P1-P8 will all be non-negative, will all add up to one, and will violate the Bell inequality.

"Predetermined" just means that immediately after the particles have been emitted, their hidden states are such that they predetermine what results they would give for each possible polarizer angle. It doesn't mean that on any given trial the source is predetermined to emit particles in a particular hidden-variable state, it just means that once the particles have already been emitted and their hidden-variable state is already well-defined, then from that point on it's inevitable what results the particles will give for any specific combination of polarizer angles chosen by Alice and Bob. If the results weren't predetermined immediately after emission in this way, there would seem to be no way to explain in local realist terms how it could be that whenever Alice and Bob choose identical polarizer angles, they are guaranteed with probability 1 to get opposite results. If there was any random element in terms of how the hidden variables carried by each particle interacted with the polarizer to produce a result, then if the particles have no way to communicate to coordinate their results, we would occasionally expect that even with the same detector angle the two experimenters would get non-opposite results. Do you disagree with any of this?

No disagreement.

OK, so you agree that in any local realist theory, in order to explain how we consistently get opposite results when the experimenters choose the same polarizer angle, we must postulate that as soon as a given particle pair is emitted, their hidden variables already predetermine what results they will give if they encounter a polarizer with angles a, b, or c? In that case, do you agree that if we had omniscient knowledge of the hidden variables associated with each particle on each trial, then on any given trial we could say that Alice's particle must have fallen into one of the eight categories (a+, b+, c+), (a+, b+, c-), etc.? If so, then in a very large number of trials an omniscient observer could define P1 to mean (number of trials with predetermined results a+, b+, c+)/(total number of trials), and likewise for P2-P8. Finally if we assume that the choices of polarizer angle made by Alice and Bob on each trial are uncorrelated with the hidden variables, then it should be true that if instead of looking at all trials we just look at the subset of trials where Alice chose a and Bob chose b, the fraction of trials in this subset where Alice's particle had predetermined results (a+, b+, c+) is still the same P1 as it was over all trials, and likewise for the probabilities of the other predetermined results. So in the subset where Alice chose a and Bob chose b, P(a+, b+) must be P3 + P4. Likewise if we look at the subset where Alice chose a and Bob chose c, P(a+, c+) must be P2 + P4. Finally if we look at the subset where Alice chose c and Bob chose b, P(c+, b+) must be P3 + P7.

If you disagree with any point here, please make clear where your disagreement lies.

In any case you are free to load whatever data you want onto these flash drives, and imagine that their memory capacity is as large as you want (even infinite). Given that, do you think you can "win" at the game I describe above, assuming you have no precognition and cannot know in advance what button Alice and Bob are going to press on each trial as you decide what simulated hidden variables to load on the flash drives on that trial? Yes or no?

No.

Well, since "winning" just consisted in reproducing the statistics seen in QM, does this mean you similarly agree there's no way the source can assign local hidden variables to the particles on each trial in a way that gives the same statistics seen in QM? In other words, have I convinced you that no local hidden variables theory will suffice to explain QM?

 

[vanesch]

Let's start from Sakurai's example, as given on the page:

http://en.wikipedia.org/wiki/Sakurai%27s_Bell_inequality

We have 2 spin-1/2 particles in a singlet state 1/sqrt(2) ( |+z>|-z> - |-z>|+z> )

The quantum-mechanical prediction for up/up detection when the angle between the two axis is alpha,
is given by:

P++QM (alpha) = 1/2 sin(alpha)^2.

We take axis a to be at 0 degrees, axis c to be at 22.5 degrees and axis b to be at 45 degrees.

This means, according to the above formula, that as a function of the choice, the probability to have ++ according to QM is:

choice: A and A ===> probability 1/2 sin(0)^2 = 0

choice A and C ==> probability 1/2 sin(22.5)^2 = 0.073223

choice A and B ==> probability 1/2 sin(45)^2 = 0.25

choice of B and C ==> probability 1/2 sin(22.5)^2 = 0.073223

Note that we have broken the Bell inequality:

P(A,B) < P(A,C) + P(C,B)

0.25 is NOT smaller than 0.0732 + 0.0732 = 0.1464...These are the QM predictions to obtain a ++ result in the case the above choices are made for the observed axes.

Now, you give me your 8 numbers as follows:

P1 = Cac.Cbc/2 =0.3643
P2 = Sac.Sbc/2 = 0.0107
P3 = Cac.Sbc/2 = 0.0625
P4 = Sac.Cbc/2 = 0.0625
P5 = Sac.Cbc/2 = 0.0625
P6 = Cac.Sbc/2 = 0.0625
P7 = Sac.Sbc/2 = 0.0107
P8 = Cac.Cbc/2 = 0.3643

I took for instance: P1 = cos(22.5)^2 * cos(22.5)^2 / 2 and so on.

They add indeed up to 1.

Now, the probability to have ++ on axes A and B is given by: P3 + P4 = 0.125 (different from quantum prediction: 0.25...)

The probability to have ++ on axes A and C is given by: P2 + P4 = 0.0732 (= quantum prediction 0.0732)

The probability to have ++ on axes B and C is given by: P3 + P7 = 0.07322 (= quantum prediction 0.0732...)

These are NOT entirely compatible with the predictions of quantum mechanics.

Indeed, quantum mechanics predicted not 0.125 but 0.25 for P++(ab).

And note that YOUR result for P(AB), namely 0.125 IS smaller than 0.0732 + 0.0732 and so your result DOES respect the inequality

(as it should, given that it is an inequality that should be respected for ALL sets of 8 numbers between 0 and 1, and adding up to 1).

EDIT: I edited my post because initially, I had angles ab = 22.5 and bc = 22.5 and ac = 45 degrees, but that is not a Bell-inequality violating system the way it was done in Sakurai.

 

[Gordon Watson]

Let's start from Sakurai's example, as given on the page:

http://en.wikipedia.org/wiki/Sakurai%27s_Bell_inequality

We have 2 spin-1/2 particles in a singlet state 1/sqrt(2) ( |+z>|-z> - |-z>|+z> )

The quantum-mechanical prediction for up/up detection when the angle between the two axis is alpha,
is given by:

P++QM (alpha) = 1/2 sin(alpha)^2.

We take axis a to be at 0 degrees, axis c to be at 22.5 degrees and axis b to be at 45 degrees.

This means, according to the above formula, that as a function of the choice, the probability to have ++ according to QM is:

choice: A and A ===> probability 1/2 sin(0)^2 = 0

choice A and C ==> probability 1/2 sin(22.5)^2 = 0.073223

choice A and B ==> probability 1/2 sin(45)^2 = 0.25

choice of B and C ==> probability 1/2 sin(22.5)^2 = 0.073223

Note that we have broken the Bell inequality:

P(A,B) < P(A,C) + P(C,B)

0.25 is NOT smaller than 0.0732 + 0.0732 = 0.1464...These are the QM predictions to obtain a ++ result in the case the above choices are made for the observed axes.

Now, you give me your 8 numbers as follows:

P1 = Cac.Cbc/2 =0.3643
P2 = Sac.Sbc/2 = 0.0107
P3 = Cac.Sbc/2 = 0.0625
P4 = Sac.Cbc/2 = 0.0625
P5 = Sac.Cbc/2 = 0.0625
P6 = Cac.Sbc/2 = 0.0625
P7 = Sac.Sbc/2 = 0.0107
P8 = Cac.Cbc/2 = 0.3643

I took for instance: P1 = cos(22.5)^2 * cos(22.5)^2 / 2 and so on.

They add indeed up to 1.

Now, the probability to have ++ on axes A and B is given by: P3 + P4 = 0.125 (different from quantum prediction: 0.25...)

The probability to have ++ on axes A and C is given by: P2 + P4 = 0.0732 (= quantum prediction 0.0732)

The probability to have ++ on axes B and C is given by: P3 + P7 = 0.07322 (= quantum prediction 0.0732...)

These are NOT entirely compatible with the predictions of quantum mechanics.

Indeed, quantum mechanics predicted not 0.125 but 0.25 for P++(ab).

And note that YOUR result for P(AB), namely 0.125 IS smaller than 0.0732 + 0.0732 and so your result DOES respect the inequality

(as it should, given that it is an inequality that should be respected for ALL sets of 8 numbers between 0 and 1, and adding up to 1).

EDIT: I edited my post because initially, I had angles ab = 22.5 and bc = 22.5 and ac = 45 degrees, but that is not a Bell-inequality violating system the way it was done in Sakurai.

Is this line correct?

The probability to have ++ on axes B and C is given by: P3 + P7 = 0.07322 (= quantum prediction 0.0732...)

I have: The probability to have ++ on axes B and C is given by P2 + P6. Alice axis given first. I'm guessing its a typo but want to clear before detailed reply.

 

[vanesch]

Is this line correct?

The probability to have ++ on axes B and C is given by: P3 + P7 = 0.07322 (= quantum prediction 0.0732...)

I have: The probability to have ++ on axes B and C is given by P2 + P6. Alice axis given first

I took over from the Sakurai page without really re-verifying it. I took it that you used the same conventions and letters as he did (or as the wiki page did).

Yes, you're right, you should read (C,B). Alice has the "c" and bob has the "b".


The defining "happenings" are the Alice happenings, along axes a, b and c respectively.

So (++-) means that if Alice was going to measure along axis a, she would get a + result, and if Alice was going to measure along c, she would get a - result. It ALSO means that if Bob was going to measure along a he would get a - result, and if he were to measure along the c axis, he'd get a + result.

So in order to get Alice + and Bob + result where Alice measures along a and bob along b,

we need to have (+, -, *) in our notation. + because Alice has + along a, - because Bob has + along b, and * because we don't care about the c state.

That's P3 (+,-,+) and P4(+,-,-). So P3 + P4 for P++(ab).

Now, P++(cb) is (*,-,+) which gives us P3 (+,-,+) and P7(-,-,+) P3 + P7 for P++(cb).

And you're right that I should have written P(cb). It doesn't change anything for the quantum predictions.

 

[Gordon Watson]

I took over from the Sakurai page without really re-verifying it. I took it that you used the same conventions and letters as he did (or as the wiki page did).

Yes, you're right, you should read (C,B). Alice has the "c" and bob has the "b".


The defining "happenings" are the Alice happenings, along axes a, b and c respectively.

So (++-) means that if Alice was going to measure along axis a, she would get a + result, and if Alice was going to measure along c, she would get a - result. It ALSO means that if Bob was going to measure along a he would get a - result, and if he were to measure along the c axis, he'd get a + result.

So in order to get Alice + and Bob + result where Alice measures along a and bob along b,

we need to have (+, -, *) in our notation. + because Alice has + along a, - because Bob has + along b, and * because we don't care about the c state.

That's P3 (+,-,+) and P4(+,-,-). So P3 + P4 for P++(ab).

Now, P++(cb) is (*,-,+) which gives us P3 (+,-,+) and P7(-,-,+) P3 + P7 for P++(cb).

And you're right that I should have written P(cb). It doesn't change anything for the quantum predictions.

I can't give detailed replies right now. But I want to avoid any more confusion.

I did intend to follow the Zakurai/wiki conventions. Did I not?

You say that I'm right; but I did not say what you are agreeing to.

I can work through the typos;

OOPs, power outage, more tomorrow.

 

[DrChinese]

But I want to avoid any more confusion...

I will repeat my main objection again: it's not realistic if you do not provide values for measurements which cannot be performed. That is the definition of "realistic".

The "confusion" issue is: to the extent anyone agrees with you, we are simply talking about the usual approach to Bell or a closely related equivalent variation. To the extent you assert you have provided a LR counter-example, we keep explaining that actually you have violated the requirement of L locality or R realism despite your words. You cannot just wave your hands and say you have accomplished this without pointing us to some new revelation. I see nothing novel in your approach at all, and it seems to follow your arguments presented in other threads.

Where's the beef? It would really be nice if you would show us something new to discuss rather than just say "I'm right unless you show me where am I wrong".

 

[Avodyne]

I would like to point out, as I did on the earlier thread, that the entire issue is one of semantics.

Bell's argument rules out theories where the result of a measurement of the spin of a spin-one-half particle along a particular axis $\hat a$ is given as a definite function $F(\hat a, \lambda)$, where $\lambda$ is a "hidden variable" (or set of hidden variables) that completely characterizes the spin state of the particle. The function $F$ can take on the values +1 and -1 only, corresponding to spin "up" or "down" along the axis $\hat a$. Bell shows that no theory of this type can agree with quantum mechanics (provided the experimenter has the "free will" to choose $\hat a$ however she wishes).

Theories of this type have been termed "locally realistic".

So, if you have theory that agrees with quantum mechanics, then it cannot be recast as a theory of this type. Period.

The only thing left to discuss is whether the terminology "locally realistic" should be restricted to theories of this type, or should be expanded to include other theories that are not of this type (and hence can, in principle, agree with quantum mechanics).

To me, this is an extremely uninteresting question. I really don't care what term should be chosen for the type of theory that Bell has excluded. It's a well-deined type, and it's excluded. That's all I need to know.

 

[Gordon Watson]

I will repeat my main objection again: it's not realistic if you do not provide values for measurements which cannot be performed. That is the definition of "realistic".

The "confusion" issue is: to the extent anyone agrees with you, we are simply talking about the usual approach to Bell or a closely related equivalent variation. To the extent you assert you have provided a LR counter-example, we keep explaining that actually you have violated the requirement of L locality or R realism despite your words. You cannot just wave your hands and say you have accomplished this without pointing us to some new revelation. I see nothing novel in your approach at all, and it seems to follow your arguments presented in other threads.

Where's the beef? It would really be nice if you would show us something new to discuss rather than just say "I'm right unless you show me where am I wrong".

Thank you DrC; and thanks also to JesseM and vanesch for engaging with my model.

I'll first address the confusion. Then let's see if the model has any merit re novelty.

So far the model delivers what is claimed in the OP. A point confirmed by vanesch's example where two QM-specific results were confirmed; the third "deviation from QM" is the correct QM average over the two-valued ab. This average being required, per the model, because there are two equivalent paths to the ab results.

So the model is strong in terms of QM results. It produces the correct result for any 2-angle setting, say ac and bc. It produces the correct average result for the related third angle ab. This average result arises because the model delivers a two-valued third angle ab. That is: ab = ac + bc or ac = ac - bc.

In short:

A. The model reproduces any related QM result correctly for any single-angle test (i.e., between two orientations). So for single angles it does no better than QM. Both deliver Pab or Pbc or Pca, etc; where Pac denotes the outcome distribution (for ++, +-, -+, --, with related probability P) for any single angle ac; etc.

B. The model delivers the correct two-angle result. E.g., Pac and Pbc where ac and bc are precisely defined; any other two-angle combination following from a change in the reference orientation RO, which is c in this example.

From lack of knowledge, and having not studied the situation, I see no reason that QM can't do this also. I have taken it, per JesseM, (and correctly interpreted him, I trust), that QM does not do this. But if QM cannot, then the model is offered as an advance on QM in this situation.

C. The model then delivers the average result for the related 2-valued angle ab. This, it seems, goes beyond QM. It is mandated in the model because of the two-way street (trajectory) that is, here, explicit.

So the model (delivering on its claims; evidently novel, and so in new territory) has me asking, DrC: Is it the case that Bell's theorem is based on equating an average result with the sum of two specific results?

That is: The model appears to go beyond QM (in this limited area). BUT not as far as Bell would wish for an LR theory. Has Bell gone too far?

That is, from the Sakurai/wiki example, Bell has:

(2) P(a+, b+) = P3 + P4.
Similarly, if Alice measures spin in a direction and Bob measures in c direction, the probability that both obtain +1/2 is
(3) P(a+, c+) = P2 + P4.
Finally, if Alice measures spin in c direction and Bob measures in b direction, the probability that both obtain the value +1/2 is
(4) P(c+, b+) = P3 + P7.​

NOW: From the model, we have ab as two-valued. So we cannot construct (it seems to me) the Bell impossibility equation rigorously. Reason: On what basis would we equate an average outcome over a two-valued angle to the sum of two outcomes for specific angles?

So, if this were a correctly presented LR case, Bell's impossibility equation could not be constructed as a case against LR?

Moreover: In that the model evidently delivers results beyond QM, this result has me asking: Why would local realists have to deliver Bell's requirement if a specification beyond two orientations and one specific angle (i.e., a specification beyond QM's limits) brings into play WITHOUT FAIL a two-valued angle?

Thinking out loud: The model is consistent with the specification in the OP; it delivers results wholly compatible with QM --- but also (evidently) beyond QM.

It does not meet Bell's requirement. Should it, or any local realist, have to?

PS: I personally think, NO!

Reason? The dynamics addressed in the model lie in a plane orthogonal to the line-of-flight axis. The model does not fully exploit the spherical symmetry of the entangled singlet state and consequent detections off the said plane. And so it raises (for me) the question: Is Bell's requirement akin to a triangle inequality wherein we can only measure two sides accurately (Pac, Pbc in the model) and cannot therefore correctly infer the third side Pab (except by averaging)?

Perhaps more to the point: Does my idea of a "gyroscopic-style trajectory of the total-spin axis in 3-space" (i.e., the trajectory brought into being by the particle/device interaction) have more than one way of reaching its destination (i.e., its alignment to one axis of the device).

What if I say that is does? What if I propose: It is the average value over those alternate paths that we are deriving?

That is:

In going beyond QM, but short of Bell, we reach the truth!

Would that be something new to discuss, DrC?

With sincerest thanks and appreciation to you and the terrible-two; be assured that I'm here to learn.

Also, so sorry; being a vegetarian, I have no beef.

Hoping that you'll see that it's an interesting model; for a vegetarian,

XXXOOO

JenniT

Addendum: Your question re being "realistic" has been on my mind since you first raised it. Having regard to the envisioned trajectories, and the model's explicit introduction of a bifurcation (meaning: two-ways to the same end) in such trajectories, I remain of the view that my model rightly goes BEYOND what we can test. Further, in my current view, it correctly predicts what we would find IF WE COULD do the tests that the model addresses.

It is, of course, not relevant (though true) that I derived it from my realist perspective.

I'd welcome your further comments on this point (about being realistic), especially in the context of the model that is the subject of this thread.

I hope that you are now satisfied, on your other main point: The model does deliver all the many QM results that it promised; without error.

XO

 

[JesseM]

So far the model delivers what is claimed in the OP. A point confirmed by vanesch's example where two QM-specific results were confirmed; the third "deviation from QM" is the correct QM average over the two-valued ab. This average being required, per the model, because there are two equivalent paths to the ab results.

First of all, I don't think you can legitimately call what you have a physical "model" if it's just some arbitrary looking-mathematical calculations that you are unwilling to explain the physical meaning of--I have repeatedly asked you why you are calculating an "average" or why you say ab is "two-valued" but you haven't answered. You also didn't answer my question about what you think P1-P8 are supposed to represent the probabilities of, if they don't represent the probabilities of particles being emitted with hidden values that given them different predetermined values for all three angles.

Also, it's not even clear that your arbitrary-looking rules for calculating the probabilities even give the right QM answers--since you didn't respond to my post #11, you never answered my question about whether P(a+, b+) would simply be P3 + P4 = 3/32 + 3/32 = 6/32, which is different from the quantum mechanical-calculation of 12/32.

So the model is strong in terms of QM results. It produces the correct result for any 2-angle setting, say ac and bc. It produces the correct average result for the related third angle ab. This average result arises because the model delivers a two-valued third angle ab. That is: ab = ac + bc or ac = ac - bc.

This is still meaningless to me as you have given no physical explanation of your terms. Is ab just an angle, or is it a result at some angle, or something else? Again a numerical example to fill in your abstractions would be greatly helpful. For example, if I pick the polarizer orientations a=70 degrees, b=30 degrees, c=0 degrees, then would ab just be the angular difference between a and b (40 degrees), or is it the probability of getting some combination of results like a+,b+, or something else?

A. The model reproduces any related QM result correctly for any single-angle test (i.e., between two orientations). So for single angles it does no better than QM. Both deliver Pab or Pbc or Pca, etc; where Pac denotes the outcome distribution (for ++, +-, -+, --, with related probability P) for any single angle ac; etc.

B. The model delivers the correct two-angle result. E.g., Pac and Pbc where ac and bc are precisely defined; any other two-angle combination following from a change in the reference orientation RO, which is c in this example.

From lack of knowledge, and having not studied the situation, I see no reason that QM can't do this also. I have taken it, per JesseM, (and correctly interpreted him, I trust), that QM does not do this. But if QM cannot, then the model is offered as an advance on QM in this situation.

My comment was about QM not delivering the probabilities P1-P8 when they are defined as on the wikipedia page, i.e. they give the probability that on a given trial the source will emit particles which have a given combination of three predetermined results (dependent on the hidden variables) for the three polarizer angles a,b,c. QM certainly delivers probabilities for all measurable outcomes like the probability of ++, +-, -+ and -- for any pair of polarizer angles like ab or ca.

C. The model then delivers the average result for the related 2-valued angle ab. This, it seems, goes beyond QM. It is mandated in the model because of the two-way street (trajectory) that is, here, explicit.

As usual you give no explanation of what you mean by "average result" or why you think it goes beyond QM, this thread would be a lot more productive if you would respond to my requests for clarification and numerical examples instead of just repeating the same impenetrable terminology which seems to make sense only to you.

Moreover: In that the model evidently delivers results beyond QM, this result has me asking: Why would local realists have to deliver Bell's requirement if a specification beyond two orientations and one specific angle (i.e., a specification beyond QM's limits) brings into play WITHOUT FAIL a two-valued angle?

Thinking out loud: The model is consistent with the specification in the OP; it delivers results wholly compatible with QM --- but also (evidently) beyond QM.

It does not meet Bell's requirement. Should it, or any local realist, have to?

PS: I personally think, NO!

If you think a local realist shouldn't have to meet Bell's requirement, then presumably you must disagree with one of the points I made here in post #11, which I asked you to tell me if you disagreed with:

OK, so you agree that in any local realist theory, in order to explain how we consistently get opposite results when the experimenters choose the same polarizer angle, we must postulate that as soon as a given particle pair is emitted, their hidden variables already predetermine what results they will give if they encounter a polarizer with angles a, b, or c? In that case, do you agree that if we had omniscient knowledge of the hidden variables associated with each particle on each trial, then on any given trial we could say that Alice's particle must have fallen into one of the eight categories (a+, b+, c+), (a+, b+, c-), etc.? If so, then in a very large number of trials an omniscient observer could define P1 to mean (number of trials with predetermined results a+, b+, c+)/(total number of trials), and likewise for P2-P8. Finally if we assume that the choices of polarizer angle made by Alice and Bob on each trial are uncorrelated with the hidden variables, then it should be true that if instead of looking at all trials we just look at the subset of trials where Alice chose a and Bob chose b, the fraction of trials in this subset where Alice's particle had predetermined results (a+, b+, c+) is still the same P1 as it was over all trials, and likewise for the probabilities of the other predetermined results. So in the subset where Alice chose a and Bob chose b, P(a+, b+) must be P3 + P4. Likewise if we look at the subset where Alice chose a and Bob chose c, P(a+, c+) must be P2 + P4. Finally if we look at the subset where Alice chose c and Bob chose b, P(c+, b+) must be P3 + P7.

If you disagree with any point here, please make clear where your disagreement lies.

Or to put it another way, do you disagree with any of the following points?

1. In any local realist theory, immediately after a given particle was emitted its hidden variables must give it some predetermined answer to whether it will give a + or - if it encounters a polarizer at any of the three angles a,b,c.

2. On each trial Alice's particle must fall into one of the eight categories of predetermined results listed on the table in the wikipedia page, so if we consider a very large number of trials there must be some objective truth about the frequency of each category which gives us values for P1-P8.

3. With a large number of trials, the fraction of trials where Alice's particle fell into a given category like + - + should be about the same over all trials as it is in the subset of trials where Alice and Bob picked a particular pair of detector angles like ac.

4. It should be true that P(a+, b+) = P3 + P4 and P(a+, c+) = P2 + P4 and P(c+, b+) = P3 + P7.

Please tell me the first point here that you disagree is a logical implication of local realism, or require clarification about the meaning of.

 

[JesseM]

I would like to point out, as I did on the earlier thread, that the entire issue is one of semantics.

Bell's argument rules out theories where the result of a measurement of the spin of a spin-one-half particle along a particular axis $\hat a$ is given as a definite function $F(\hat a, \lambda)$, where $\lambda$ is a "hidden variable" (or set of hidden variables) that completely characterizes the spin state of the particle. The function $F$ can take on the values +1 and -1 only, corresponding to spin "up" or "down" along the axis $\hat a$. Bell shows that no theory of this type can agree with quantum mechanics (provided the experimenter has the "free will" to choose $\hat a$ however she wishes).

Not all Bell inequalities assume that the hidden variables totally determine what the response to any given angle will be, see the CHSH inequality for example where the angles a,a' used on one side may be different from the angles b,b' on the other in which case there'd be no combination of settings where knowledge of the result on one side gives you total certainty about the result on the other. I like to think of the definition of local realism this way:

1. The complete set of physical facts about any region of spacetime can be broken down into a set of local facts about the value of variables at each point in that regions (like the value of the electric and magnetic field vectors at each point in classical electromagnetism)

2. The local facts about any given point P in spacetime are only causally influenced by facts about points in the past light cone of P, meaning if you already know the complete information about all points in some spacelike cross-section of the past light cone, additional knowledge about points at a spacelike separation from P cannot alter your prediction about what happens at P itself (your prediction may be a probabilistic one if the laws of physics are non-deterministic).

From this it follows that if two experimenters making measurements at a spacelike separation are guaranteed to get the same result when they choose the same angle, despite the fact that they pick which angle to use at random, then the results for each possible angle must have already predetermined by facts in their past light cone (like facts about the hidden variables associated with each particle) at some time T after the particles had been emitted but before they picked their angles. But this is just a derived consequence of the notion of local realism above, not a definition of local realism.

 

[vanesch]

Thank you DrC; and thanks also to JesseM and vanesch for engaging with my model.

I'll first address the confusion. Then let's see if the model has any merit re novelty.

So far the model delivers what is claimed in the OP. A point confirmed by vanesch's example where two QM-specific results were confirmed; the third "deviation from QM" is the correct QM average over the two-valued ab. This average being required, per the model, because there are two equivalent paths to the ab results.

So the model is strong in terms of QM results. It produces the correct result for any 2-angle setting, say ac and bc. It produces the correct average result for the related third angle ab. This average result arises because the model delivers a two-valued third angle ab. That is: ab = ac + bc or ac = ac - bc.

Here, things bug. Maybe this comes about because of a misunderstanding of the exact set-up, or about what exactly we are talking about, I don't know. There's no "double-valuedness" of any angle.

Consider the set-up as follows:

On Monday, Alice puts her analyser vertically and her detector clicks when she gets an "up" result. Bob puts his analyser at 45 degrees with the vertical (s times angle is then 22.5 degrees, right) towards the window of the room, and his detector clicks also when he gets an "up" result. An electronic circuit links Alice's and Bob's detector signals to a counter, which counts each time there is a simultaneous click on both detectors.
Carol starts the electron-pair source in the middle, and let's it generate 1 million electron-pairs during the afternoon.
Quantum mechanics predicts that at the end of the afternoon, the counter will read something like 73 000 counts.

On Tuesday, Alice leaves her installation in place, but Bob rotates his axis until it is horizontal (so s times angle is 45 degrees), with his "up" direction pointing towards the window.
When Carol starts the 1 million electron pair source again, quantum theory predicts that the counter will read 250 000 at the end of the afternoon.

On Wednesday, Alice, Bob and Carol go to a party.

On Thursday, Bob leaves his installation in the horizontal direction, but now Alice rotates her axis also in the direction of the window, over 45 degrees. Carol makes the source again run and produce 1 million electron pairs. Quantum mechanics predicts that the counter will read 73 000 counts.

The whole point is that the statistical mixture of the 1 million electron pairs is each time the same ; that the source didn't suffer any influence from the choice of settings.

So if on Friday, Alice and Bob randomly change their axes and we take data until we have 1 million events where Alice and Bob had aligned their axes as on Monday (so only considering those results when by coincidence Alice and Bob had their axes as on Monday), we expect to find statistically the same result as on Monday ; if we take data until we have 1 million events where Alice and Bob had aligned their axes as on Tuesday (so considering only those results where by coincidence Alice and Bob had their axes as on Tuesday), we expect to find the same result statistically as on Tuesday. And same for Thursday. Also, if by coincidence Alice and Bob put their axes parallel, we find that the counter reads 0.
The results will be the same if the source is generating statistically identical sets of events, independently of how the axes are set.


Now, if we are to explain the results of Alice and Bob in a LR way, we have to assume that each pair sent out by the source must fall in 1 of 8 categories.

In the first category are the pairs which would give us a click in Alice's counter when it is vertical, and no click in Bob's counter when it is vertical ; that it would give us a click in Alice's counter when it was at 45 degrees, and no click in Bob's counter when it was at 45 degrees, and again that it would give a click in Alice's counter when at 90 degrees, and no click in Bob's counter when it was at 45 degrees. We write it as (+ + +). So events in this category will always give a click in Alice's counter and never one in Bob's counter.

and so on for the 7 other categories.

Note that there are no other possibilities: the 8 categories cover entirely the possibilities of the electron pair behaviour. It is for instance not possible that a pair wouldn't give a click in any Alice counter nor in any Bob counter. If a pair doesn't give a click in a vertical Alice counter, then it MUST give a click in a vertical Bob counter. So if we know the behaviour of a pair at Alice, we know that the behaviour at Bob's is complementary.

So the 1 million events must be subdivided in these 8 categories, with:

P1 * 1000000 = N1 the number of pairs in the first class,
P2 * 1000000 = N2 the number of pairs in the second class

etc...

Well, the number of pairs that belong to those that were counted on Monday are those in class 2 AND those in class 4. Each of the pairs in one of these classes will make the counter count, so we have that:

N2 + N4 = 73 000 up to statistical errors.

The counts on Tuesday are N3 + N4 = 250 000 up to statistical errors

The counts on Thursday are N3 + N7 = 73 000 up to statistical errors.

Well, you can't find such (positive) numbers N2, N3, N4, and N7.

Simply because if you add the counts on Monday and those on Thursday,

N2 + N3 + N4 + N7 = 146 000

and the counts on Tuesday are only N3 + N4 and they are LARGER: 250 000.

There's no "double angledness" or whatever here. There are specific measurements, with specific outcomes, and you CAN'T explain them with a pre-determined mixture of events. That's the point.

 

[Gordon Watson]

Addendum (added as an edit):

I have withdrawn this post because I've just realized that we don't need to discuss the RO.

The RO is like an accounting + auditing device that I use in my LR analysis. It is central to how I develop the requisite probabilities, but discussion will be simplified if the RO remains behind the scenes.

I've kept the withdrawn post and will edit it anew tomorrow.

 

[Gordon Watson]

A local realistic counter-example to Bell's theorem?

IN AN EFFORT TO MINIMIZE new terms here, and at the same time move to terminology more into line with QM, the term Reference Orientation (RO) is no longer required in discussion here: Its implications will now be addressed by drawing attention to the consequences of working 'simultaneously' across 3 orientations in 2-space. That is, we are still left with the need to work with, and understand, bi-angles (for want of a better name).

As an example: ac is a bi-angle when we have the possibility of ac = ab + bc, and ac = ab – bc in our deliberations and equations when we want to focus on ab and bc (say); or ON ANY 2 well-defined adjacent angles.

The relevance of these by-products for Bell's theorem will be addressed about 2 posts on in this thread.

Now: To help understand the RO term, as used earlier in this thread: RO is the tool used to support critical LR (Local Realistic) analyses of BT (Bell's theorem); noting that QM does not analyze to the level of generality of the LR model presented here. That is, as I understand the QM position:

Where QM works with single angles, between 2 orientations, the model here works with the 4 angles that arise from working 'simultaneously' across 3 orientations in 2-space. The 2-space here being orthogonal to an entangled particle's line-of-flight axis.

A device was required to carefully track the 3 orientations and their related 4 angles. ROs are thus used like accounting + auditing devices to track novel LR analysis. The ROs are central to developing the requisite probabilities ("Bell-style probabilities"), recognizing the consequences of working 'simultaneously' with 3 orientations in 2-space --

[As an aside: Bi-angles may be thought (locally and realistically) to be possible guides to equivalent or related trajectories of extrinsic and intrinsic spin-orientation transitions in 3-space. That is, helping to mentally picture to ourselves, some 'explanatory-LR-dynamics' of what is going on in EPR-Bohm-Bell experiments.]

Second: To the 4 angles, termed, for convenience: 2 conventional angles and 1 bi-angle.

Consider any 3 orientations (a, b, c ), taken to be the principal axes of the 3 SGM devices in our study here. We want to define their relation by specifying the 2 angles between them as simply as possible. That is, we want to go beyond QM and its one angle between 2 orientations to a local realistic study using 2 angles between 3 orientations.

We can do this readily: Let the required angles be ab and bc. But then note: WE ARE FORCED (or so it seems) to work with 2 angles and a bi-angle. For we have: ab, bc, ac = ab + bc, ac = ab – bc. So we need to be aware of bi-angles, essentially by-products from our need for the two adjacent or over-lapping angles ab and bc.

EDIT: NOTE that the Probable distributions of the outcomes ( ++, +–, –+, –– ) ARE UNCHANGED IN OUR STUDY OVER ab AND bc (etc., the foci of our study) whether ac = ab + bc OR ac = ab – bc !

This, it seems to me, is too often NOT understood; Bell's theorem notwithstanding!

To be continued, mathematically, about 2 posts on ...

 

[Gordon Watson]

Edited re-post following decision to de-emphasize the RO (Reference Orientation).

First of all, I don't think you can legitimately call what you have a physical "model" if it's just some arbitrary looking-mathematical calculations that you are unwilling to explain the physical meaning of--I have repeatedly asked you why you are calculating an "average" or why you say ab is "two-valued" but you haven't answered.

I thought "model" ok, as short-hand for the counter-example. It does model/deliver any Bell-type probability that you require.

In that regard, I cautioned about jumping to conclusions if results differing from QM were anywhere derived from the model.

Where such valid deviations appear, they will be averages for the bi-angle ab, [if working on the OP setting; the bi-angle being dependent on the 2 key angles under study] quite consistent with the averages that QM would deliver from each angle considered separately. See OP re ab and its two values; ab (from the OP setting) now termed a bi-angle for convenience. The actual form of the P relating to this average (= half the average) will be shown more clearly in my next post; remembering now that bi-angles are by-products (even unnecessary fringe-benefits) of our focus on 2 other (related) angles.

BUT perhaps critical to BT.

The duality of ab (here) arises from my use of orientations. This may not be a consideration in QM, if you focus alone on single angles. But, when you focus on two clearly defined angles in EPRB style experiments, you need to be aware that two combinations of those angles MAY BE ASSOCIATED WITH IDENTICAL outcome distributions. This is a critical consideration: Do you see the point? My example relates to 3 orientations, 2 clear and required angles, in the one experiment.

You also didn't answer my question about what you think P1-P8 are supposed to represent the probabilities of, if they don't represent the probabilities of particles being emitted with hidden values that given them different predetermined values for all three angles.

In my example, P1-P8 represent the outcome probabilities; here, in relation to outcomes defined by reference to two specific angles. Distributions associated with bi-angles being, essentially, unnecessary by-products of no relevance to our principal findings BUT perhaps critical and bad for BT ... as will be shown in my next post.

Note re the above, and some below: You appear not to have understood the OP. For example, ab is specifically defined there; ab's 2-valued-ness (making it now a bi-angle, for short) is also addressed there.

May I suggest that any term that is not clear in the OP, or in its separate (later) addendum, be raised as an issue so that the foundations for our discussions here are put to rights, in the one place, for the benefit of all. I'm committed to correcting mistakes, typos, ... improving clarity.

I want the model to succeed or fail on its merits; not from lack of clarity.

In this regard, there may be items that you, from your grounding in QM, think should be there, in the OP. So let's discuss them too.

I see more clearly now the need to avoid new terms; and I see little reason for any more of them: Though bi-angles are, in my view, essential to our understanding of BT ... probing and questioning why anyone would accept the inequality he's so famous for when presented in its clearest form, as in the Zakurai-wiki.

Also, it's not even clear that your arbitrary-looking rules for calculating the probabilities even give the right QM answers--since you didn't respond to my post #11, you never answered my question about whether P(a+, b+) would simply be P3 + P4 = 3/32 + 3/32 = 6/32, which is different from the quantum mechanical-calculation of 12/32.

As mentioned in the OP, it would not be as you write here. The OP example, with ITS FOCUS on ac and bc, delivers one-half the average over the two-valued ab.

This average could also be derived from the QM average, calculated over each angle value; i.e., directly from QM's one-angle calculations, done twice, one for each value of the 2-valued ab.

This is still meaningless to me as you have given no physical explanation of your terms. Is ab just an angle, or is it a result at some angle, or something else? Again a numerical example to fill in your abstractions would be greatly helpful. For example, if I pick the polarizer orientations a=70 degrees, b=30 degrees, c=0 degrees, then would ab just be the angular difference between a and b (40 degrees), or is it the probability of getting some combination of results like a+,b+, or something else?

[Which is your RO, please?]

NO, no more. Now: Which 2 angles do you seek to focus on, please?

In a single bound, the MODEL can handle twice as many angles as QM. And all 3 pairs of angles in turn if you wish -- see next post -- with no departure whatsoever from accepted QM results -- EXCEPT for the model being wholly LOCAL and REALISTIC​

.

Please see the OP re ab. [Do you understand my need for, and use of, an RO?] No; now: Do you understand the origin of the bi-angle?

[[[OUT: Who is to decide and define what is UP and what is DOWN in a triple-orientation experiment? OR by reference only to angles? With this last, it seems to me impossible to be consistent with + and - allocations ... when dealt with 3 at a time?]]]

My comment was about QM not delivering the probabilities P1-P8 when they are defined as on the wikipedia page, i.e. they give the probability that on a given trial the source will emit particles which have a given combination of three predetermined results (dependent on the hidden variables) for the three polarizer angles a,b,c. QM certainly delivers probabilities for all measurable outcomes like the probability of ++, +-, -+ and -- for any pair of polarizer angles like ab or ca.

[PS: Pairs of angles??

Would this correction be OK? Pair of orientations, or specific (one-valued) angles.]

OK. So QM does not give the Ps in the Table, and the model does? So it seems my conclusion is correct: The model goes beyond QM; delivering accurate two-angle Ps; a third angle average-based P; as well as any P that QM can derive from the triple-orientation setting; as well as specifically calculating and identifying the Ps in the BI.

Please note this: The model provides the Probability Functions on which Bell's Theorem is based -- and raises, by the way, a serious question:

Studying the tabulations in the next post, where could any correction/adjustment be made to accommodate Bell's Inequality.

All the accepted QM results delivered with high-school maths. Related averages correctly derived for all bi-angles. All based on local realism.

As usual you give no explanation of what you mean by "average result" or why you think it goes beyond QM, this thread would be a lot more productive if you would respond to my requests for clarification and numerical examples instead of just repeating the same impenetrable terminology which seems to make sense only to you.

Where I say that ab is two valued, you could derive the QM result for each ab (given in OP), then see if the average of such relates to the value you originally derived. (What coefficent is required?) I did expect that critics would derive general results from the model. That way we can be more sure we don't miss an orientation combination that fails to deliver a valid QM result -- and proves an error in the model. [vanesch appears to have adopted this approach -- I still owe him a fuller response --

Dear vanesch, could be begin anew with the next more-comprehensive post?]

I am reluctant to lose the model's generality, suspecting that your confusion is associated with not understanding ab as a bi-angle. The model's OP specifically targeting ab, by focussing on ac and bc, BECAUSE ab was the foundation of the BI in Zakurai-wiki.

Please study my next post to see the mathematical cohesion that is sacrificed when one works with specifics instead of generalities.

If you think a local realist shouldn't have to meet Bell's requirement, then presumably you must disagree with one of the points I made here in post #11, which I asked you to tell me if you disagreed with:

Or to put it another way, do you disagree with any of the following points?

1. In any local realist theory, immediately after a given particle was emitted its hidden variables must give it some predetermined answer to whether it will give a + or - if it encounters a polarizer at any of the three angles a,b,c.

2. On each trial Alice's particle must fall into one of the eight categories of predetermined results listed on the table in the wikipedia page, so if we consider a very large number of trials there must be some objective truth about the frequency of each category which gives us values for P1-P8.

3. With a large number of trials, the fraction of trials where Alice's particle fell into a given category like + - + should be about the same over all trials as it is in the subset of trials where Alice and Bob picked a particular pair of detector angles like ac.

#1 and 2, please see my next post.

[With #3, the question is not clear to me. Also, I believe there is a subtlety: Which orientation, on whose detector, is defining + and -? The issue may not arise in QM, limited as it is two dealing with two orientations, as you say above; or focussing on angles. The model is designed to treat this matter rigorously -- am I mistaken in this basic concern?] please reconsider this question after studying my next post. The post is designed to focus on maths, with less word tangles.

4. It should be true that P(a+, b+) = P3 + P4 and P(a+, c+) = P2 + P4 and P(c+, b+) = P3 + P7.

Not at all. This is one point that the model addresses; and rejects. Please see the OP re:

PROVISO: Note that, in this example, the outcome-probabilities attaching to the ab settings are averages over the two ab possibilities. This follows from the topological fact re spatial relations here: ab may be constructed in two ways:

(1) ab = ac + bc.

(2) ab = ac - bc.​

SEE NEXT POST also.

Please tell me the first point here that you disagree is a logical implication of local realism, or require clarification about the meaning of.

I trust this question is somewhat answered as I progressed through your post?With sincere thanks and appreciation, as always,

JenniT

 

[JesseM]

Consider any 3 orientations (a, b, c ), taken to be the principal axes of the 3 SGM devices in our study here. We want to define their relation by specifying the 2 angles between them as simply as possible. That is, we want to go beyond QM and its one angle between 2 orientations to a local realistic study using 2 angles between 3 orientations.

We can do this readily: Let the required angles be ab and bc. But then note: WE ARE FORCED (or so it seems) to work with 2 angles and a bi-angle. For we have: ab, bc, ac = ab + bc, ac = ab – bc. So we need to be aware of bi-angles, essentially by-products from our need for the two adjacent or over-lapping angles ab and bc.

I really don't understand what you mean by "bi-angle" here, and nowhere in this post do you clearly explain the term. Look, if you want to talk about angles there's no need for some convoluted notion of defining them relative to one another and picking one as a "reference angle", just do what is always done when talking about angles in physics, and define them relative to some fixed coordinate system! You could have a long straight rod stretching from one experimenter to the other whose position never changes and which is taken to define the x-axis of your coordinate system, and then the angle of the polarizer could just be defined as the angle relative to the rod, and then if you started the polarizer out parallel to the rod you could just see how many degrees you have to rotate it counterclockwise before it reaches the desired orientation, and call that the "angle" of the desired orientation. In this case every orientation would have a well defined angle, like a=70, b=30 and c=10, and then a difference between two angles like ac could just be defined as one minus the other, so ac=a-c while ca=c-a and so forth. In this case it's clear that ac=ab+bc is true since (a-c)=(a-b)+(b-c), while ac=ab-bc is false since (a-c)=(a-b)-(b-c)=a-2b+c which doesn't work. Given my example angles above you can see that ac=70-10=60, ab=70-30=40, and bc=30-10=20, so clearly ac=ab+bc does work since 60=40+20, but ac=ab-bc doesn't since 60 is not equal to 40-20.

I really hope your entire argument doesn't reduce to an incoherent notation for labeling angles...if not, then please just phrase your argument in terms of the standard type of coordinate-based angular notation I describe above.

 

[vanesch]

I wrote a post but it got lost during a problem with the PF server, I write it here more or less again...

Where such valid deviations appear, they will be averages for the bi-angle ab, [if working on the OP setting; the bi-angle being dependent on the 2 key angles under study] quite consistent with the averages that QM would deliver from each angle considered separately. See OP re ab and its two values; ab (from the OP setting) now termed a bi-angle for convenience. The actual form of the P relating to this average (= half the average) will be shown more clearly in my next post; remembering now that bi-angles are by-products (even unnecessary fringe-benefits) of our focus on 2 other (related) angles.

This is wrong, as JesseM also pointed out several times. You seem to claim that the QM calculation for the outcome of certain angles is wrong (the outcome for the case where Alice measures along the vertical, and Bob measures along the horizontal towards the window to be more precise). The straightforward calculation yields that when Alice measures along the vertical, and Bob along the horizontal, 25% of the events give you an up-up result, and you claim that this should be only half of it, namely 12.5%, in agreement with your 8 values of P1...P8.

You seem to base this on the "fact" that you can obtain 90 degrees (so s times angle 45 degrees) somehow by adding 45 degrees (so s times angle 22.5) and 45 degrees (22.5) together, but also by subtracting them. This is already strange that you obtain 90 degrees by subtracting 45 and 45. But ok.

However, even more strange in your proposition is that the outcome of the "up up" measurement when Alice has the vertical direction, and Bob the horizontal direction towards the window, should in some way depend on whether we also consider to do experiments with axes along a 45-degree line towards the window or not.

I mean, in what way will the counter result on Tuesday be influenced by whether or not we did an experiment on Monday and will do one on Thursday ? Why should, IF we do these experiments on Monday and on Thursday, we obtain only 125 000 clicks on Tuesday, but when we perform exactly the same experiment on Tuesday WITHOUT doing the experiments on Monday and on Thursday, we would obtain 250 000 clicks for exactly the same setup ?

 

[Gordon Watson]

I wrote a post but it got lost during a problem with the PF server, I write it here more or less again...



This is wrong, as JesseM also pointed out several times. You seem to claim that the QM calculation for the outcome of certain angles is wrong (the outcome for the case where Alice measures along the vertical, and Bob measures along the horizontal towards the window to be more precise). The straightforward calculation yields that when Alice measures along the vertical, and Bob along the horizontal, 25% of the events give you an up-up result, and you claim that this should be only half of it, namely 12.5%, in agreement with your 8 values of P1...P8.

You seem to base this on the "fact" that you can obtain 90 degrees (so s times angle 45 degrees) somehow by adding 45 degrees (so s times angle 22.5) and 45 degrees (22.5) together, but also by subtracting them. This is already strange that you obtain 90 degrees by subtracting 45 and 45. But ok.

However, even more strange in your proposition is that the outcome of the "up up" measurement when Alice has the vertical direction, and Bob the horizontal direction towards the window, should in some way depend on whether we also consider to do experiments with axes along a 45-degree line towards the window or not.

I mean, in what way will the counter result on Tuesday be influenced by whether or not we did an experiment on Monday and will do one on Thursday ? Why should, IF we do these experiments on Monday and on Thursday, we obtain only 125 000 clicks on Tuesday, but when we perform exactly the same experiment on Tuesday WITHOUT doing the experiments on Monday and on Thursday, we would obtain 250 000 clicks for exactly the same setup ?

Please see my emphasis above:

Help please: I am not aware, EVER, of making any such claim.

 

[Gordon Watson]

I really don't understand what you mean by "bi-angle" here, and nowhere in this post do you clearly explain the term. Look, if you want to talk about angles there's no need for some convoluted notion of defining them relative to one another and picking one as a "reference angle", just do what is always done when talking about angles in physics, and define them relative to some fixed coordinate system! You could have a long straight rod stretching from one experimenter to the other whose position never changes and which is taken to define the x-axis of your coordinate system, and then the angle of the polarizer could just be defined as the angle relative to the rod, and then if you started the polarizer out parallel to the rod you could just see how many degrees you have to rotate it counterclockwise before it reaches the desired orientation, and call that the "angle" of the desired orientation. In this case every orientation would have a well defined angle, like a=70, b=30 and c=10, and then a difference between two angles like ac could just be defined as one minus the other, so ac=a-c while ca=c-a and so forth. In this case it's clear that ac=ab+bc is true since (a-c)=(a-b)+(b-c), while ac=ab-bc is false since (a-c)=(a-b)-(b-c)=a-2b+c which doesn't work. Given my example angles above you can see that ac=70-10=60, ab=70-30=40, and bc=30-10=20, so clearly ac=ab+bc does work since 60=40+20, but ac=ab-bc doesn't since 60 is not equal to 40-20.

I really hope your entire argument doesn't reduce to an incoherent notation for labeling angles...if not, then please just phrase your argument in terms of the standard type of coordinate-based angular notation I describe above.

It's probably my problem, via defective explanation. The results are derived from such angles as you specify. If what I offer in my formalism, as angles, are not angles, well I'm a fool whose prepared to put (and lose) money on it.

The following may be of interim help:

1. I thought this was OK; from earlier post:

Second: To the 4 angles, termed, for convenience: 2 conventional angles and 1 bi-angle.

Consider any 3 orientations (a, b, c ), taken to be the principal axes of the 3 SGM devices in our study here. We want to define their relation by specifying the 2 angles between them as simply as possible. That is, we want to go beyond QM and its one angle between 2 orientations to a local realistic study using 2 angles between 3 orientations.

We can do this readily: Let the required angles be ab and bc. But then note: WE ARE FORCED (or so it seems) to work with 2 angles and a bi-angle. For we have: ab, bc, ac = ab + bc, ac = ab – bc. So we need to be aware of bi-angles, essentially by-products from our need for the two adjacent or over-lapping angles ab and bc.

EDIT: NOTE that the Probable distributions of the outcomes ( ++, +–, –+, –– ) ARE UNCHANGED IN OUR STUDY OVER ab AND bc (etc., the foci of our study) whether ac = ab + bc OR ac = ab – bc !

This, it seems to me, is too often NOT understood; Bell's theorem notwithstanding!

2. In my model, the bi-angles (so-called) are by-products, of no consequence: The model reproducing every relevant QM result correctly, despite them.

3. BUT, and I've probably been ahead of myself and not made this clear: They appear in the model's outcomes in just those places that Bell might have wished they had not.

This is again ahead of what's been demonstrated; so I trust you'll hold fire until I put the model's results on line here. That will be "the next post" (that I implied above).

To repeat another way: I find bi-angles in Bell's results. Surprisingly. But they, for me, are relevant to the "inequality" that you (as I recall) emphasized, reducing effectively to:

X = X + Y; X and Y rational numbers, Y greater than 0.

4. If you disagree with the model's results, via its use of angles, I'll be very surprised.

I'll rush them up next -- in DRAFT form.

 

[JesseM]

I thought "model" ok, as short-hand for the counter-example. It does model/deliver any Bell-type probability that you require.

I don't know what you mean by "Bell type probability" since you have said that you aren't calculating the probabilities that the source will emit particles with different combinations of predetermined results for each angle; or did I misunderstand you there? You certainly haven't provided a "counterexample" to Bell in the form of a local realistic physical model, i.e. one where you can give us some local hidden variables associate with the particle and rules for how the variables together with the polarizer angle determine (in a probabilistic or deterministic way) the outcome of each measurement, with the rules obeying locality (so that all values of variables and other events can only be causally influenced by values/events in their past light cone). If you had an actual local realist physical model you would be able to use it to meet the challenge I offered earlier:

Think about it this way: if you believe you have a local realist counterexample to BT, then you should be able to simulate the way it works in the following manner. You will play the part of the "source", and on each trial you will fill two flash drives with data representing the simulated properties (hidden or measurable, whatever you want) of two particles emitted by the source (perhaps you will use some algorithm to assign the properties on each trial in a probabilistic way rather than doing it by hand). Then you will send one flash drive to Alice and another to Bob. They each plug their flash drive into their computer, then randomly choose whether to press the key "A", "B", or "C" representing their choice of simulated polarizer angle. The computer then takes the simulated polarizer angle, along with the data from the flash drive about the properties (hidden or measurable) of the particle reaching the polarizer, and uses these along with an algorithm representing the laws of physics (which you programmed into each computer earlier, so they can be whatever laws you want) to calculate as output either "+" or "-". If we assume you have no foreknowledge of which key Alice and Bob will press on each trial, do you think there's any way for you to "win" this game by matching the statistics seen in QM, so on any trial that Alice and Bob press the same key they are guaranteed to get opposite results, yet if we look at the trials where they picked different keys, we will find a violation of the inequality P(a+, b+) ≤ P(a+, c+) + P(c+, b+)?

But you said you wouldn't be able to win at this challenge. So please don't continue to assert you have a local realist model or a counterexample to Bell if you don't even understand the notion of "local realism" well enough to see what this would actually entail. As you know this forum is not meant to be a platform for people who think they have made some brilliant discovery which destroys some mainstream result, when you asked if I thought it would be appropriate to start a thread like this I offered the opinion that it would be OK if you were here in a spirit of learning and being willing to listen to explanations as to why your argument doesn't falsify Bell's theorem, if you aren't willing to do that and just want to confidently assert that you have done so, then I don't think the discussion should continue on this forum.

The duality of ab (here) arises from my use of orientations. This may not be a consideration in QM, if you focus alone on single angles. But, when you focus on two clearly defined angles in EPRB style experiments, you need to be aware that two combinations of those angles MAY BE ASSOCIATED WITH IDENTICAL outcome distributions. This is a critical consideration: Do you see the point? My example relates to 3 orientations, 2 clear and required angles, in the one experiment.

As I said in my previous post #25, I would like you to use the standard type of notation for angles, where individual angles are defined relative to some fixed coordinate angles and differences between two angles are defined in some fixed way, like ab=a-b. If you think the terminology of "bi-angles" still makes sense in this context, then please explain clearly what you mean, hopefully using a numerical example where we have definite angles for a,b,c and can thus calculate any angles like ab and ac. If you're just saying that the probabilities of certain observable results may be identical for some combinations of angles, then certainly I agree, for example if we pick a=240, b=120 and c=0 then the probability of ++ for any combination of these angles will be 3/8 according to equation (7) on the Bell inequality[/url] page. But if you're saying something more than that, please explain more clearly.

In my example, P1-P8 represent the outcome probabilities; here, in relation to outcomes defined by reference to two specific angles.

Can you be more specific? What precise outcome does P1 give the probability of? And likewise for P2, P3, P4, P5, P6, P7, and P8? And do you understand that this is not what P1-P8 are defined to mean on the Sakurai's Bell inequality page, that there they are supposed to be probabilities for different combinations of predetermined results determined by hidden variables, not probabilities of observable outcomes?

Also, if P1-P8 are simply probabilities of observable outcomes, then why is it that you still seem to calculate P(a+, c+) = P2 + P4 and P(c+, b+) = P3 + P7? These equations make perfect sense if P1-P8 are predetermined results as on the wiki page, but I have no idea why an observable outcome like P(a+,c+) would be the sum of the probabilities of two other observable outcomes.

Note re the above, and some below: You appear not to have understood the OP.

Of course, that's why I keep asking for clarification on your terms, and asking for numerical examples, and saying things like "this thread would be a lot more productive if you would respond to my requests for clarification and numerical examples instead of just repeating the same impenetrable terminology which seems to make sense only to you". You can't just refer me back to the OP because your explanations there were no use to me!

For example, ab is specifically defined there; ab's 2-valued-ness (making it now a bi-angle, for short) is also addressed there.

Not in any way that makes sense to me, just with some incomprehensible jargon that seems to be your own private language, not any standard mathematical terminology I'm familiar with. For example, you say "This follows from the topological fact re spatial relations here: ab may be constructed in two ways"...what is a "topological fact re spatial relations", and what does it mean to "construct" ab? Isn't ab just the angular difference between two polarizers at angles a and b, i.e. isn't it just a-b? That's what an angular difference is normally defined to mean by everyone I've ever seen talking about the "angle between" two things!

As mentioned in the OP, it would not be as you write here. The OP example, with ITS FOCUS on ac and bc, delivers one-half the average over the two-valued ab.

Why? What physical or geometric considerations lead you to think P(a+, c+) = P2 + P4 and P(c+, b+) = P3 + P7, but that P3 + P4 is only "one-half the average"? Again since you seem to be defining P1-P8 in a totally different way than on the wikipedia page, it would really be helpful if you would tell us what observable outcomes P1-P8 are each supposed to give the probability of.

This average could also be derived from the QM average, calculated over each angle value; i.e., directly from QM's one-angle calculations, done twice, one for each value of the 2-valued ab.

This is still meaningless to me as you have given no physical explanation of your terms. Is ab just an angle, or is it a result at some angle, or something else? Again a numerical example to fill in your abstractions would be greatly helpful. For example, if I pick the polarizer orientations a=70 degrees, b=30 degrees, c=0 degrees, then would ab just be the angular difference between a and b (40 degrees), or is it the probability of getting some combination of results like a+,b+, or something else?

[Which is your RO, please?]

NO, no more. Now: Which 2 angles do you seek to focus on, please?

Your notion of "focusing" on 2 angles or "reference angles" are similarly incomprehensible to me, I'm just talking about angles in the standard way that physicists always talk about angles, defining them relative to some fixed coordinate system, see post #25. As I requested there, I would like you to start using this sort of standard definition of angles as well, if your argument really revolves around saying there is something fundamentally flawed about defining angles relative to a fixed coordinate system and that we must use your incomprehensible alternative definitions, then your argument really is hopelessly crackpot and I am not interested in continuing.

In a single bound, the MODEL can handle twice as many angles as QM.

Really? What angles do you think QM can't handle? If you have trouble with the standard notation for defining angles, could you draw a picture of polarizers at different orientations and show graphically what angles in the picture you think QM can't describe?

EXCEPT for the model being wholly LOCAL and REALISTIC

Again it's pretty clear you don't understand what "local and realistic" even mean, if you admit you can't answer my simulation challenge above but still think you have a local and realistic "model". Again if you are here in the spirit of learning rather than just here to proclaim your glorious victory over Bell, please stop making such claims and admit there may be some fault in your understanding of the notion of local realism. For example, did you completely understand the definition of local realism I gave in my [post=3154224]post #20 to Avodyne[/post]? If not we might use that as a starting point.

Please see the OP re ab. [Do you understand my need for, and use of, an RO?] No; now: Do you understand the origin of the bi-angle?

No, I have no idea what either of these terms mean, that's why I keep asking for clarifications. Please explain using either a numerical or diagrammatic example, not just strange abstract verbal discussions and equations where you haven't explained the meaning of the terms (like your using equations with P1 or ab without explaining what these mean physically) or where the equations come from (like why you think it should be true that ab = ac + bc and ac = ac - bc).

[[[OUT: Who is to decide and define what is UP and what is DOWN in a triple-orientation experiment? OR by reference only to angles? With this last, it seems to me impossible to be consistent with + and - allocations ... when dealt with 3 at a time?]]]

If the experiment is with a Stern-Gerlach apparatus this is pretty straightforward since the apparatus creates a magnetic field and "up" would just mean in the direction of magnetic North while "down" would mean in the direction of magnetic South. With photons, I believe + or - tells you whether the photon passed through the polarizer and was detected by the detector behind it, or whether it was reflected by the polarizer and detected by the other detector (see the diagram of the "two-channel polarizer" on the CHSH inequality page)

My comment was about QM not delivering the probabilities P1-P8 when they are defined as on the wikipedia page, i.e. they give the probability that on a given trial the source will emit particles which have a given combination of three predetermined results (dependent on the hidden variables) for the three polarizer angles a,b,c. QM certainly delivers probabilities for all measurable outcomes like the probability of ++, +-, -+ and -- for any pair of polarizer angles like ab or ca.

[PS: Pairs of angles??

Would this correction be OK? Pair of orientations, or specific (one-valued) angles.]

If you use a standard notation for labeling orientations with angles, then each orientation has a unique angle. If your argument is that it is somehow impossible or forbidden to use this type of standard notation where there's a one-to-one relation between orientations and angles, then it seems to me you're just confused about basic geometry and at the very least I want to see a diagram of an orientation that you think cannot be assigned a unique angle using the standard procedure (make sure to draw both the orientation and the x-axis of the fixed coordinate system).

OK. So QM does not give the Ps in the Table, and the model does?

Um, how could you possible get that from my comment above? I said that QM only fails to give the Ps in the table if those Ps are understood to have the standard meaning of predetermined results for each possible angle[/b], but you have said this is not what the Ps in your model mean at all, with comments like (from post #10) "I am providing the Ps for different outcomes under the defined tripartite setup; introducing the third orientation to be consistent with Bell and Sakurai; recognizing that we can only test for one angle (two orientations) in anyone physical test." As I keep telling you, the Ps on the wiki page aren't supposed to represent measured outcomes at all, rather they represent predetermined facts about what the results will be for each of the 3 possible angles, these predetermined facts based on hidden variables which human experimenters have no way of knowing. What's more, in a more recent post I asked you:

Or to put it another way, do you disagree with any of the following points?

1. In any local realist theory, immediately after a given particle was emitted its hidden variables must give it some predetermined answer to whether it will give a + or - if it encounters a polarizer at any of the three angles a,b,c.

And in the post of yours I am responding to right now (post #24) you responded:

#1 and 2, please see my next post.

Your "next post" did not actually explain your disagreement...but look, if you disagree with point #1, then you disagree that there are predetermined answers for all three angles, so you cannot possibly be defining your Ps in the same way as the wikipedia page! On the wikipedia page, P1-P8 specifically refer to the probabilities of different combinations of hidden predetermined results for all three angles! How many times do I have to repeat this, and how many times do I have to ask you what you mean when you write P1-P8 before you will give me a straight answer? If they are supposed to represent probabilities of some observable results, then your comment "So QM does not give the Ps in the Table, and the model does" is clearly wrong, since as I already stated very clearly in the quote you were responding to there, "QM certainly delivers probabilities for all measurable outcomes."

So it seems my conclusion is correct: The model goes beyond QM; delivering accurate two-angle Ps; a third angle average-based P; as well as any P that QM can derive from the triple-orientation setting; as well as specifically calculating and identifying the Ps in the BI.

Huh? Again, if you disagree with my point #1 above, then you disagree that there are hidden "predetermined results" for each possible angle, so how can you possibly think you can "identify the Ps in the BI" when these are supposed to be the probabilities of different predetermined results for each possible angle?

Please note this: The model provides the Probability Functions on which Bell's Theorem is based

No, it certainly doesn't. You have denied that you think the hidden variables associated with each particle after emission give it predetermined results for each possible angle, and if you changed your mind and accepted this, it would be very trivial to show it must be true that P(a+, b+) = P3 + P4 and P(a+, c+) = P2 + P4 and P(c+, b+) = P3 + P7.

All the accepted QM results delivered with high-school maths. Related averages correctly derived for all bi-angles. All based on local realism.

Please stop behaving like a crackpot and confidently proclaiming you have overturned Bell while ignoring all possibility that all the criticisms (such as the criticism that you don't seem to even understand what 'local realism' means) have any validity. Again, if you're here to learn and try to understand why the argument might be flawed that's fine, if you're just here to announce your earth-shattering discovery please use another forum.

Where I say that ab is two valued, you could derive the QM result for each ab (given in OP)

What does "each ab" even mean, geometrically or physically? I know of only one ab, the angle between the polarizer (or Stern-Gerlach device) which is at angle a relative to some fixed axis, and the polarizer (or Stern-Gerlach device) which is at angle b relative to the same axis.

 

[JesseM]

(reply to post #24, continued)

If you think a local realist shouldn't have to meet Bell's requirement, then presumably you must disagree with one of the points I made here in post #11, which I asked you to tell me if you disagreed with:

Or to put it another way, do you disagree with any of the following points?

1. In any local realist theory, immediately after a given particle was emitted its hidden variables must give it some predetermined answer to whether it will give a + or - if it encounters a polarizer at any of the three angles a,b,c.

2. On each trial Alice's particle must fall into one of the eight categories of predetermined results listed on the table in the wikipedia page, so if we consider a very large number of trials there must be some objective truth about the frequency of each category which gives us values for P1-P8.

3. With a large number of trials, the fraction of trials where Alice's particle fell into a given category like + - + should be about the same over all trials as it is in the subset of trials where Alice and Bob picked a particular pair of detector angles like ac.

#1 and 2, please see my next post.

If you're saying you disagree with #1, are you changing your mind from our earlier exchange in post #9 and 10? There I said:

"Predetermined" just means that immediately after the particles have been emitted, their hidden states are such that they predetermine what results they would give for each possible polarizer angle. It doesn't mean that on any given trial the source is predetermined to emit particles in a particular hidden-variable state, it just means that once the particles have already been emitted and their hidden-variable state is already well-defined, then from that point on it's inevitable what results the particles will give for any specific combination of polarizer angles chosen by Alice and Bob. If the results weren't predetermined immediately after emission in this way, there would seem to be no way to explain in local realist terms how it could be that whenever Alice and Bob choose identical polarizer angles, they are guaranteed with probability 1 to get opposite results. If there was any random element in terms of how the hidden variables carried by each particle interacted with the polarizer to produce a result, then if the particles have no way to communicate to coordinate their results, we would occasionally expect that even with the same detector angle the two experimenters would get non-opposite results. Do you disagree with any of this?

To which you responded:

No disagreement.

Have you changed your mind, or did you misunderstand my question, or do you not actually disagree with point #1?

[With #3, the question is not clear to me.

OK, suppose the "large number of trials" in #3 is 900 trillion. And if we are omniscient beings who know all the hidden variables associated with Alice's particle on each trial, and therefore know what predetermined results each particle sent towards Alice had before it actually reached the polarizer, then we can know exactly how many trials in total there were where the particle sent towards Alice had some set of predetermined results, like (a+,b+,c-). Suppose this number is 90 trillion, which would mean P2 is 90 trillion out of 900 trillion, or 1/10. What 3 is saying is that if the hidden variables sent out by the source on each trial have no correlation with the choice of angles Alice and Bob made on that trial, then this same P2 should also give us the fraction that had predetermined results (a+,b+,c-) in the subset of trials where Alice picked angle a and Bob picked angle c. If Alice and Bob each select their angle randomly, then we'd expect a 1/3*1/3=1/9 chance of any given combination of polarizer angles on each trial, so out of 900 trillion total trials, we'd expect that there were about 100 trillion trials where Alice chose angle a and Bob chose angle c. So 3 is saying that within this subset P2 still gives the fraction of trials where Alice's predetermined results were (a+,b+,c-), so the total number of trials where Alice chose a and Bob chose c and Alice's predetermined results were (a+,b+,c-) should be (1/10)*100 trillion = 10 trillion.

To sum up, what 3 is saying is just that the probability the source will send out particles with a given combination of hidden results on a given trial is uncorrelated with what polarizer angles Alice and Bob happened to choose on that trial. If you agreed with #1 and #2 (which apparently you don't), then I don't see how you'd be able to disagree with #3 unless you thought the source had some sort of precognitive knowledge of what settings Alice and Bob were going to choose on that trial, after the two particles had already been emitted.

 

[vanesch]

Please see my emphasis above:

Help please: I am not aware, EVER, of making any such claim.

I don't know how to interpret your post #18 differently.

So my question to you then AGAIN:
Do you agree with the number of clicks as predicted by QM in my post # 21 on Monday, Tuesday and so on ?

Now, this thread was started by your claim to produce 8 numbers P1...P8 that can REPRODUCE these predictions. You've given 8 numbers, and they DO NOT predict the numbers predicted by QM (they do so for 2 of the 3 proposed experiments, and give HALF of the actual result QM result for the third one).

So is it then your numbers which, after all, don't do what you claimed they did (namely explain the QM results) ; or is it the QM predictions (or our way of calculating them) which are, according to you, wrong ?

In other words, on Tuesday, do you think we will see (according to our way of calculating QM results) 250 000 hits, or (according to your numbers), only 125 000 hits ?

Because you can't have it both ways, right ? You can't have your numbers be different from the QM predictions as worked out here and still claim that you have provided us with a set of numbers which agree with QM.

 

[Gordon Watson]

I don't know how to interpret your post #18 differently.

So my question to you then AGAIN:
Do you agree with the number of clicks as predicted by QM in my post # 21 on Monday, Tuesday and so on ?

Now, this thread was started by your claim to produce 8 numbers P1...P8 that can REPRODUCE these predictions. You've given 8 numbers, and they DO NOT predict the numbers predicted by QM (they do so for 2 of the 3 proposed experiments, and give HALF of the actual result QM result for the third one).

So is it then your numbers which, after all, don't do what you claimed they did (namely explain the QM results) ; or is it the QM predictions (or our way of calculating them) which is, according to you, wrong ?

Because you can't have it both ways, right ? You can't have your numbers be different from the QM predictions as worked out here and still claim that you have provided us with a set of numbers which agree with QM.

Jesse and vanesch; thank you; detailed mathematical response on the way.

JenniT

 

[Gordon Watson]

DRAFTRESULTS DERIVED from, and offered as, a local realistic counter-example to Bell's theorem.

JenniT

With apologies for the formatting; this DRAFT being hurried on-line to address serious concerns above.​

1. Pab++ denotes the Probability (P) of outcome ++ across the angle ab, where ab is a clearly-defined (unambiguous) angle; etc. C and S functions are defined in the OP.

2. For such angles, all the probability functions [below; lines (A)-(C)] agree with QM.

3. Equations (X)-(Z) are NOT such functions; they are seen here as arising from (unrecognized) bi-angles when the Bell-Zakurai Table is compared with those below. Thus bi-angles have their origin, here, in this analysis of BT. They represent an attempt to identify and classify the source of Bell's Inequality.

4. Further, (X)-(Z) appear to be required to derive a Bell Inequality (BI).

5. The challenge for Bell's supporters, it seems to me, is to derive a BI without recourse to (X)-(Z); or explain why such "grotesque" functions are justifiably equated to functions from (A)-(C). I suggest that a rational BI cannot now be constructed, in that each P can now be matched to a value for any 3-particle EPRB test; and any "equating" now challenged!

6. Alternatively, show how (A)-(C) might be modified to reproduce (again) all the correct QM results, and derive a rational inequality.

7. Since these results derive from local realism, I conclude: Bell's Theorem is not an impediment to local realism and a local-realistic view of the world.

8. These results, extending beyond QM in analytic detail; being wholly based on local realistic considerations; are the reason that I am reluctant to consider specific examples -- thereby losing the beauty, and the check on each result, that generality here provides.

E & OE. ROa: Angles ab, ac. Bi-angle: bc = ab + ac, or bc = ab – bc; so 2Pbc = the average over the bi-angle.

(A1) +++ ––– P1 = Cab.Cac/2. Pab++ = P3 + P4 = Sab/2. (X1) Pbc++ = P2 + P6 = (Cab.Sac + Sab.Cac)/2.
(A2) ++– ––+ P2 = Cab.Sac/2. Pab+– = P1 + P2 = Cab/2. (X2) Pbc+– = P1 + P5 = (Cab.Cac + Sab.Sac)/2.
(A3) +–+ –+– P3 = Sab.Cac/2. Pab–+ = P7 + P8 = Cab/2. (X3) Pbc–+ = P4 + P8 = (Sab.Sac + Cab.Cac)/2.
(A4) +–– –++ P4 = Sab.Sac/2. Pab–– = P5 + P6 = Sab/2. (X4) Pbc–– = P3 + P7 = (Sab.Cac + Cab.Sac)/2.
(A5) –++ +–– P5 = Sab.Sac/2. Pac++ = P2 + P4 = Sac/2.
(A6) –+– +–+ P6 = Sab.Cac/2. Pac+– = P1 + P3 = Cac/2.
(A7) ––+ ++– P7 = Cab.Sac/2. Pac–+ = P6 + P8 = Cac/2.
(A8) ––– +++ P8 = Cab.Cac/2. Pac–– = P5 + P7 = Sac/2.ROb: Angles ab, bc. Bi-angle: ac = ab + bc, or ac = ab – bc; so 2Pac = the average over the bi-angle.

(B1) +++ ––– P1 = Cba.Cbc/2. Pab++ = P3 + P4 = Sab/2. (Y1) Pac++ = P2 + P4 = (Cab.Sbc + Sab.Cbc)/2.
(B2) ++– ––+ P2 = Cba.Sbc/2. Pab+– = P1 + P2 = Cab/2. (Y2) Pac+– = P1 + P3 = (Cab.Cbc + Sab.Sbc)/2.
(B3) +–+ –+– P3 = Sba.Sbc/2. Pab–+ = P7 + P8 = Cab/2. (Y3) Pac–+ = P6 + P8 = (Sab.Sbc + Cab.Cbc)/2.
(B4) +–– –++ P4 = Sba.Cbc/2. Pab–– = P5 + P6 = Sab/2. (Y4) Pac–– = P5 + P7 = (Sab.Cbc + Cab.Sbc)/2.
(B5) –++ +–– P5 = Sba.Cbc/2. Pbc++ = P2 + P6 = Sbc/2.
(B6) –+– +–+ P6 = Sba.Sbc/2. Pbc+– = P1 + P5 = Cbc/2.
(B7) ––+ ++– P7 = Cba.Sbc/2. Pbc–+ = P4 + P8 = Cbc/2.
(B8) ––– +++ P8 = Cba.Cbc/2. Pbc–– = P3 + P7 = Sbc/2.ROc: Angles ac, bc. Bi-angle: ab = ca + cb, or ab = ca – cb; 2Pab = the average over the bi-angle.

(C1) +++ ––– P1 = Cca.Ccb/2. Pac++ = P2 + P4 = Sac/2. (Z1) Pab++ = P3 + P4 = (Cac.Sbc + Sac.Cbc)/2.
(C2) ++– ––+ P2 = Sca.Scb/2. Pac+– = P6 + P8 = Cac/2. (Z2) Pab+– = P1 + P2 = (Cac.Cbc + Sac.Sbc)/2.
(C3) +–+ –+– P3 = Cca.Scb/2. Pac–+ = P1 + P3 = Cac/2. (Z3) Pab–+ = P7 + P8 = (Sac.Sbc + Cac.Cbc)/2.
(C4) +–– –++ P4 = Sca.Ccb/2. Pac–– = P5 + P7 = Sac/2. (Z4) Pab–– = P5 + P6 = (Sac.Cbc + Cac.Sbc)/2.
(C5) –++ +–– P5 = Sca.Ccb/2. Pbc++ = P2 + P6 = Sbc/2.
(C6) –+– +–+ P6 = Cca.Scb/2. Pbc+– = P4 + P8 = Cbc/2.
(C7) ––+ ++– P7 = Sca.Scb/2. Pbc–+ = P1 + P5 = Cbc/2.
(C8) ––– +++ P8 = Cca.Ccb/2. Pbc–– = P3 + P7 = Sbc/2.PS: I will be away from the web for a week or more, starting tomorrow.

Happy studying,

JenniT

 

[Gordon Watson]

Is there a simple way to format tables on PF, please ... ??

while I'm still in the editing phase .. ??

 

[JesseM]

The following may be of interim help:

1. I thought this was OK; from earlier post:

Second: To the 4 angles, termed, for convenience: 2 conventional angles and 1 bi-angle.

Consider any 3 orientations (a, b, c ), taken to be the principal axes of the 3 SGM devices in our study here. We want to define their relation by specifying the 2 angles between them as simply as possible. That is, we want to go beyond QM and its one angle between 2 orientations to a local realistic study using 2 angles between 3 orientations.

We can do this readily: Let the required angles be ab and bc. But then note: WE ARE FORCED (or so it seems) to work with 2 angles and a bi-angle. For we have: ab, bc, ac = ab + bc, ac = ab – bc. So we need to be aware of bi-angles, essentially by-products from our need for the two adjacent or over-lapping angles ab and bc.

I still have no idea why you think "we are forced" to do this, you don't explain your reasoning at all. Look, below I've drawn a crude diagram showing how you can define the angle of any given SG orientation relative to some fixed x-axis, and likewise define the angle between any two SG devices geometrically (and also using the simple mathematical definition that if one has angle x and the other has angle y, the angle between them xy is just x-y). Do you see any problem with this system? On a diagram like this, do you think it would be possible to draw in an SG orientation that we couldn't label with a unique angle, or possible to draw two SG devices with orientations such that there wouldn't be a unique angle between them? If you do think there would be any problems with non-uniqueness, any chance you could supply your own analogous diagram showing a horizontal x-axis along with one SG device which you think couldn't be assigned a unique angle, or two SG devices for which you don't think there'd be a unique angle between them?

(and once again, a page giving the basics of what a SG device is and how it works can be found here...incidentally, realistically it's not the central axis of the SG device that's moved to different angles as in my diagram, but rather the device is rotated about that central axis as depicted here, but my diagram is just to illustrate the general concept of labeling orientations with angles so it was easier to draw that way)