A classical challenge to Bell's Theorem?

[billschnieder]

6. So I ("imagining, as usual") answer your good question, "Do you [dear photon i] have properties λ_i which exist as part of your "identity" which ultimately interact with Bob's device to result in a measurement outcome +1?":

Yes, of course!​

Then you have no reservations with EPR's elements of reality for that is precisely what they mean. (cf The moon is there even when you are not looking).

7. Then, re this from you: "The EPR point of view is that if a theory can predict +1 with certainty, then the theory is complete with respect to the hidden element of reality λ_i which corresponds to +1."

7a. I suggest that this is NOT accurate at all (see PS): and may be misleading you! See how your question (to me, as photon) differs re +1? FOR I THINK this says something quite different; or maybe I'm missing something?

I guess you are missing something then, though it is not yet clear to me what.

7b. Can you not say: λ_i is the specific HV that delivers the output +1 during the Q_i particle/device interaction?

Yes of course. But what in your mind is the difference between HV and "element of reality".

8. Note this re your quote in #7 above, EPR say (with no mention of "theory" but with focus on existence): "If, without any way disturbing a system, we can predict with certainty, the value of a physical quantity, then there exists an epr corresponding to this physical quantity."

Remember that the goal of the EPR paper was to discuss the completeness of a certain theory. Their objective was not to discuss existence.

9. So, is it not the case that my prediction is disturbance-free wrt the subject system?

I don't get where you are going with this. Please enlighten.

10. So does it not follow, does it not remain the case, that the epr represented by b⁺ is brought into existence by the particle-device interaction; and not otherwise? Is it not the case that λ_i, the pre-interaction HV, is transformed (during the interaction) to become the previously-non-existent (the now-post-interaction existent) b⁺?

I think you may have been misled about what EPR elements of reality are. λ_i is an element of reality.

11. In a nutshell: Does it not remain the case that EPR's "corresponding" is just plain WRONG?

No, I do not buy that. I think we need a simple analogy so you could show me clearly what you think EPR elements of reality are and why you think it is wrong. For example:

We have a tablet with two well defined chemicals X and Y (aka elements of reality). In addition we have two glasses of different liquids A and B. In addition we have a theory which predicts with certainty the following *observables*:

a) if you place the tablet into liquid A, and drink it, it will taste sweet (X interacts with A to produce the sweetness).
b) if you place the tablet into liquid B, and drink it, it will taste bitter (Y interacts with B to produce the bitterness).

It is obvious that each observable (a) or (b) above *corresponds* to an element of reality. The two elements of reality (X,Y) in the particle are simultaneously well defined even before any experiment is performed. The prediction of the *observables* are certain. This is exactly what EPR were talking about.

YET! The *observables* (a) and (b) are not, and can NEVER be simultaneously actual, simply because you can only place your tablet into one of the two liquids. Once you place your tablet, you destroy the tablet. Therefore, the fact that a realist says elements of reality are well defined even when experiments are not performed, does not mean the results of all possible *observables* which can correspond to those observables are also simultaneously actual.

 

[Gordon Watson]

Bill, I'm in a meeting but just wanted to say THANKS for your effort. I'll get back to you. In meantime could you comment on the BLUE bit in my post. It might help me be more succinct. (I confirm that the moon is there when you're not looking! And I'm happy that our views might be closer than I thought -- even if it's my view that changes. ) Thanks. GW

 

[billschnieder]

*** How about: The OUTPUT +1 "corresponds to" the element of reality b⁺! For both +1 (the device output) and b⁺ (the particle's new property) represent/denote/signal/tell-us of a photon linearly-polarised-parallel-wrt-the-principal-axis-of-the-device (s = 1) or of a spin-half particle spin-up-wrt-the-principal-axis-of-the-device (s = 1/2).

Why in your opinion shouldn't it correspond to λ_i, isn't b⁺ simply a deterministic result of interaction of λ_i with the device? I guess I'm not getting why you insist on a strong distinction between λ_i and b⁺ as concerns EPR.

 

[billschnieder]

Bill, I'm in a meeting but just wanted to say THANKS for your effort. I'll get back to you. In meantime could you comment on the BLUE bit in my post. It might help me be more succinct. (I confirm that the moon is there when you're not looking! And I'm happy that our views might be closer than I thought -- even if it's my view that changes. ) Thanks. GW

Hi Gordon
I think our views are the same. The only difference I see is that you think your view differs from EPR and I think my view agrees with EPR. But from what you are saying, I'm more sure we agree that "our" views agree :!).

 

[Gordon Watson]

Why in your opinion shouldn't it correspond to λ_i, isn't b⁺ simply a deterministic result of interaction of λ_i with the device? I guess I'm not getting why you insist on a strong distinction between λ_i and b⁺ as concerns EPR.

? In Bell's terms, the entangled particles [with pre-test λ_i] are UNpolarised. The post-test particle with (now) b⁺ is POLARISED! BIG DIFFERENCE!

 

[billschnieder]

? In Bell's terms, the entangled particles [with pre-test λ_i] are UNpolarised. The post-test particle with (now) b⁺ is POLARISED! BIG DIFFERENCE!

Again I remind you to focus on one particle. What does polarization mean for this one particle? Isn't it simply a direction of it's spin vector? Don't all particles have spin vectors which have direction? (Polarized or not, entangled or not) Now you have a device which re-orients the spin-vector λi of a particle passing through it into a fixed direction b⁺. Isn't b⁺ simply a value for the new direction of λi after transformation? It is still "spin-vector" after all, so the element of reality continues to be spin-vector. It's new *value* happens to be b⁺. How do you suppose this new direction came to be? Did the device flip a die and randomly pick a direction, or does it have a mechanism which maps old values of λi to one of b⁺ or b^-? Isn't it true that the outcome +1 or -1 is determined ultimately by λi, and the nature of the interaction between λi and the device. Said "nature" which must be existing independently before any such interaction ever took place. This "nature" is what EPR call "elements of reality". Particles do have'em, devices do have'em.

So to my mind, I do not see the big difference at least as far as EPR's epr is concerned.

 

[Gordon Watson]

Focussing on one particle: What does polarisation mean for this one particle? It is simply a direction of its spin vector. Don't all particles have spin vectors which have direction: polarised or not, entangled or not? Now you have a device which re-orients the spin-vector λi of a particle passing through it into a fixed direction b⁺. Isn't b⁺ simply a value for the new direction of λi after transformation? It is still a "spin-vector" after all, so the element of reality continues to be a spin-vector. Its new *value* happens to be b⁺. How do you suppose this new direction came to be? The device has a mechanism which maps old values of λi to one of b⁺ or b^-! The outcome +1 or -1 is determined ultimately by λi and the nature of the interaction between λi and the device.

Said "nature" must be existing independently before any such interaction ever took place.

This "nature" is what EPR call "elements of reality". Particles do have them, devices do have them.

So, to my mind, I do not see a big difference in our views, at least as far as EPR's eprs are concerned.

NB: EDITED and emphasised by GW.

Bill, I've edited your post (as above) to remove some typos and put it closer to what I believe we both accept: our core agreement so far being that we each accept Einstein-locality without reservation.

However, the emphasised pieces above represent a point-of-view that I suspect has never occurred to me: so I have some homework to do.

It may take a day or so due my current situation. So, anticipating that I might have to yield to your position (but still doubting it), here's some homework for you:

As your own time permits, please comment on the formatting, the physical significance, the final results, including the short-cut in (6) which you should recognise, whatever, etc. We begin with Q $\in$ {W, X, Y, Z} as discussed earlier in this thread; primes indicate items in Bob's locale; their removal (where appropriate) from HVs is based on the initial correlation of each particle-pair via their $\lambda$ and $\lambda'$ relations; all analysis is classical; Einstein-locality is maintained throughout.

Device/particle interactions are denoted $\delta_{\textbf{b}}' \lambda$, etc. Note that we retain the prime in $\delta_{\textbf{b}}'$ because we may sometimes need to refer to $\delta_{\textbf{b}}$ and $\delta_{\textbf{a}}'$, etc., when discussing the results for identical device settings.

The chances are that there are errors in what follows, including formatting, etc. But the gist of our discussion so far is here, for sure:

$A({\textbf{a}}, \lambda)_Q = (\int d\lambda \;(\delta_{\textbf{a}} \lambda \rightarrow \left \{{\textbf{a}}^+ , {\textbf{a}}^- \right \})\;cos[2s\cdot({\textbf{a}}, \lambda)])_Q = \pm 1.\;\;\;\;(1)$

$B({\textbf{b}}, \lambda')_Q = (\int d\lambda'\;(\delta_{\textbf{b}}' \lambda' \rightarrow \left \{{\textbf{b}}^+ , {\textbf{b}}^- \right \})\;cos[2s\cdot ({\textbf{b}}, \lambda')])_Q \;\;\;\;(2)$

$= ((-1)^{2s} \cdot \int d\lambda\;(\delta_{\textbf{b}}' \lambda \rightarrow \left \{{\textbf{b}}^+ , {\textbf{b}}^- \right \})\;cos[2s\cdot ({\textbf{b}}, \lambda)])_Q = \pm 1. \;\;\;\;(3)$

$E(AB)_Q = (\int d\lambda\;\rho (\lambda )\;AB)_Q\;\;\;\;(4)$

$= ((-1)^{2s} \cdot\int d\lambda\;\rho (\lambda )\cdot\int d\lambda\; (\delta_{\textbf{a}} \lambda \rightarrow \left \{{\textbf{a}}^+ , {\textbf{a}}^- \right \})\;cos[2s\cdot({\textbf{a}}, \lambda)] \cdot \int d\lambda\;(\delta_{\textbf{b}}' \lambda \rightarrow \left \{{\textbf{b}}^+ , {\textbf{b}}^- \right \})\;cos[2s\cdot ({\textbf{b}}, \lambda)])_Q \;\;\;\;(5)$

$= [(-1)^{2s}]_Q \cdot[ 2 \cdot P(B^+|Q,\,A^+) - 1].\;\;\;\;(6)$

$E(AB)_W = (cos[2 ({\textbf{a}}, {\textbf{b}})])/2.\;\;\;\;(7)$

$E(AB)_X = - ({\textbf{a}}. {\textbf{b}})/2.\;\;\;\;(8)$

$E(AB)_Y = cos[2 ({\textbf{a}}, {\textbf{b}})].\;\;\;\;(9)$

$E(AB)_Z = - {\textbf{a}}. {\textbf{b}}.\;\;\;\;(10)$



(E. & O. E.)​

 

[Gordon Watson]

Again I remind you to focus on one particle. What does polarization mean for this one particle? Isn't it simply a direction of it's spin vector? Don't all particles have spin vectors which have direction? (Polarized or not, entangled or not) Now you have a device which re-orients the spin-vector λi of a particle passing through it into a fixed direction b⁺. Isn't b⁺ simply a value for the new direction of λi after transformation? It is still "spin-vector" after all, so the element of reality continues to be spin-vector. It's new *value* happens to be b⁺. How do you suppose this new direction came to be? Did the device flip a die and randomly pick a direction, or does it have a mechanism which maps old values of λi to one of b⁺ or b^-? Isn't it true that the outcome +1 or -1 is determined ultimately by λi, and the nature of the interaction between λi and the device. Said "nature" which must be existing independently before any such interaction ever took place. This "nature" is what EPR call "elements of reality". Particles do have'em, devices do have'em.

So to my mind, I do not see the big difference at least as far as EPR's epr is concerned.

The above is Bill's view of EPR-eprs, and I have yet to understand it in detail. However, losing sleep, I sense even more difficulties here (for me) than I did with my first reading of EPR and their eprs. At that first reading, years ago, EPR's definition of eprs jumped off the page (to me, so it may be my problem) as unsatisfactory.

EPR's definition (1935) of an EPR-epr: "If, without any way disturbing a system, we can predict with certainty (i.e., with probability equal to unity) the value of a physical quantity, then there exists an element of physical reality [an epr] corresponding to this physical quantity." (My emphasis, identifying two of my problem areas; my [.].)

I interpret "corresponding" in its conventional sense = "according, agreeing, conforming, fitting, matching, tallying."

In my view, EPR's definition of eprs relates to "inferred pre-existence": the inference that the predicted physical quantity existed PRIOR to the test interaction.

(The word "test" is used by me in place of "measurement" -- which I avoid where possible because of its tendency, imho, to mislead when compared to classical measurements. In QM a "measurement" generally perturbs the "measured" system.)

My alternative view relates to two concepts:

"Explicit local causality": "If, without any way disturbing a system p'(λ'), we can predict with certainty the result B of its test at δ_b, which may be a disturbance, then existents λ' and b locally mediate this result. That is B = B(λ', b)."

PLUS

"Explicit local determinism: "The 'prediction with certainty' implies that the new particle property, say b⁺, is determined locally by the interaction of λ' and b. The corresponding device-output B⁺ is equally locally determined."

Here p'(λ') is the particle p' that flies to Bob, its HV = λ'. δ_b is Bob's test-device, its principal axis oriented b. B is the usual test-outcome from Bell (1964), as is B = B(λ', b). B⁺ is the device-output corresponding to the particle property b⁺.

THUS in my view, the existents λ' and b (say, a field-orientation, by way of its connection with device δ), via their interaction, bring into existence the property b⁺ (say) which particle p', now designated p'(b⁺) post-test, NOW possesses ... B⁺ being the corresponding device-output.

SO my view arrives at the SAME function as Bell (1964). However, Bell's subsequent use of this function (1964, his unidentified eqn following his eqn (14)) may indicate that he had in mind EPE eprs -- which I did not; and do not.

SOS, please: critical, alternative, helpful comments invited!

 

[billschnieder]

EPR's definition (1935) of an EPR-epr: "If, without any way disturbing a system, we can predict with certainty (i.e., with probability equal to unity) the value of a physical quantity, then there exists an element of physical reality [an epr] corresponding to this physical quantity." (My emphasis, identifying two of my problem areas; my [.].)

I interpret "corresponding" in its conventional sense = "according, agreeing, conforming, fitting, matching, tallying."

In my view, EPR's definition of eprs relates to "inferred pre-existence": the inference that the predicted physical quantity existed PRIOR to the test interaction.

It is not the "physical quantity" that is inferred to pre-exist. It is the epr that is inferred to pre-exist. Purely from the plain-English parsing of the phrase, you can see that they distinguish "physical quantity" from "element of physical reality". They ascribe pre-existence only to the epr, not the physical quantity, and what is being predicted is the physical quantity not the epr. Take a step back, read the sentence again and see that this is the case, noting the use of the phrases "physical quantity" and "element of physical reality".

(The word "test" is used by me in place of "measurement" -- which I avoid where possible because of its tendency, imho, to mislead when compared to classical measurements. In QM a "measurement" generally perturbs the "measured" system.)

It doesn't matter as far as EPR is concerned, whether measurement perturbs the measured system or not because a complete theory should include in it all aspects of the perturbation if one happens.

My alternative view relates to two concepts:
...
THUS in my view, the existents λ' and b (say, a field-orientation, by way of its connection with device δ), via their interaction, bring into existence the property b⁺ (say) which particle p', now designated p'(b⁺) post-test, NOW possesses ... B⁺ being the corresponding device-output.

It sounds to me like you are just using a complicated way of saying the results are "contextual". There is no problem with that, contextual theories were well known and accepted prior to EPR (cf photoelectric effect) and EPR surely did not exclude them. Maybe you thought that EPR did not support contextual theories. But I think it is because you may have misunderstood the meaning of their "corresponding".

 

[Gordon Watson]

It is not the "physical quantity" that is inferred to pre-exist. It is the epr that is inferred to pre-exist.

EDIT: Agreed.

Purely from the plain-English parsing of the phrase, you can see that they distinguish "physical quantity" from "element of physical reality". They ascribe pre-existence only to the epr, not the physical quantity, and what is being predicted is the physical quantity not the epr. Take a step back, read the sentence again and see that this is the case, noting the use of the phrases "physical quantity" and "element of physical reality".

EDIT: OK; thanks.

It doesn't matter as far as EPR is concerned, whether measurement perturbs the measured system or not because a complete theory should include in it all aspects of the perturbation if one happens.

EDIT: I totally agree with this.

It sounds to me like you are just using a complicated way of saying the results are "contextual". There is no problem with that, contextual theories were well known and accepted prior to EPR (cf photoelectric effect) and EPR surely did not exclude them. Maybe you thought that EPR did not support contextual theories. But I think it is because you may have misunderstood the meaning of their "corresponding".

EDIT: I think this is the bit that you should expand for my benefit, please: contextual, corresponding: using the elements of reality (via their symbols) that I used in my posting above to make your point. Thanks.

...

Bill, thanks! I'm again breaking my vow: again posting in a meeting. BUT your effort here looks very good ... and I'm keen to get the monkey off my back. [END OF EDITS. NO More HERE. See next post from me.] GW

 

[billschnieder]

As your own time permits, please comment on the formatting, the physical significance, the final results, including the short-cut in (6) which you should recognise, whatever, etc. We begin with Q $\in$ {W, X, Y, Z} as discussed earlier in this thread; primes indicate items in Bob's locale; their removal (where appropriate) from HVs is based on the initial correlation of each particle-pair via their $\lambda$ and $\lambda'$ relations; all analysis is classical; Einstein-locality is maintained throughout.

Device/particle interactions are denoted $\delta_{\textbf{b}}' \lambda$, etc. Note that we retain the prime in $\delta_{\textbf{b}}'$ because we may sometimes need to refer to $\delta_{\textbf{b}}$ and $\delta_{\textbf{a}}'$, etc., when discussing the results for identical device settings.

...

1) I feel that you focus too much on the transformation $\lambda \rightarrow \left \{{\textbf{a}}^+ , {\textbf{a}}^- \right \}$, even though it does not contribute much to the end-result (I believe, though I may be wrong).
2) (3) is problematic as it is not obvious what justification you have to multiply with $(-1)^{2s}$. I prefer the other approach in which you derive the general result and then provide physical justification for how you calculate $P(B^+|Q,\,A^+)$ or $P(B^+|Q,\,A^-)$. If you emphasize the transformation at this point instead, I think it would make your approach easier to follow and understand (and defend).

 

[Gordon Watson]

1) I feel that you focus too much on the transformation $\lambda \rightarrow \left \{{\textbf{a}}^+ , {\textbf{a}}^- \right \}$, even though it does not contribute much to the end-result (I believe, though I may be wrong).

Thanks Bill: That focus was intended to support Bell's call (2004: 118, 1st para.) for greater physical precision: "to have the 'jump' in the equations and not just the talk."

It anticipates your important remark above: "It doesn't matter as far as EPR is concerned, whether measurement perturbs the measured system or not because a complete theory should include in it all aspects of the perturbation if one happens". With which I agree; and so tried to classically deliver.

Note that it is that classical transformation that crucially satisfies Einstein-locality AND THUS critically delivers the bottom line (your end-result): at the same time creating new eprs? These seemed to me important points to capture (and emphasise) early on. Which leads to ...

Remember: The recent equations presume familiarity with what went on earlier in this thread; except now I'm (mostly) into latex! But you are right in so far as any newcomer to them is concerned. And that can be fixed; the fix being helped by such critiques as yours.

But note also: The earlier equations were at the "beginner level" in that all they required was agreement that certain classical equations existed. They started with that presumption. But, given the "lack of submissions" here, and some rude remarks, I thought that question should be answered in the "advanced" case: hence the strange "kick-start" that you allude to.

2) (3) is problematic as it is not obvious what justification you have to multiply with $(-1)^{2s}$.

Fair enough, certainly confusing to anyone starting there: But that comes in (for advanced students like you ) to allow generalisation of the equations over Q. Recall that, in the beginning, we only covered the simpler wholly-classical V $\in$ {W, X}: whereas Q $\in$ {W, X, Y, Z} = the classical and entangled cases together: though still all handled wholly classically. That came in in response to questions; and more rude remarks.

I prefer the other approach in which you derive the general result and then provide physical justification for how you calculate $P(B^+|Q,\,A^+)$ or $P(B^+|Q,\,A^-)$. If you emphasize the transformation at this point instead, I think it would make your approach easier to follow and understand (and defend).

OK, thanks. I'll look at that. But remember: the general equations were not meant to, and do not, render the early wholly-classical cases and presentation obsolete. It is still all wholly classical.

...
1. Bill, please see EDITS that I made in my last post here (above). I'm sure your comments on "contextual" and "corresponding" via the context that I suggested would be a big help to me.

PS: I much appreciate the time and helpful effort you are putting in here, and trust my (possible) capitulation re EPR-eprs will be a suitable reward!

GW

 

[billschnieder]

EDIT: I think this is the bit that you should expand for my benefit, please: contextual, corresponding: using the elements of reality (via their symbols) that I used in my posting above to make your point. Thanks.

Contextual means "particle epr" + "device epr" + "new epr's created due to interaction" → outcome. Or in short "system eprs" → outcome. Where system includes particle, device and their interaction.

Non-contextual means "particle epr" → outcome. If you know everything about the particle, then you know for certain the outcome.

In both cases if you can predict the outcome with certainty, then there exists an epr in the system which corresponds to it ("system" in the non-contextual case being merely the particle). Note however that if you can not predict with certainty, it does not mean an "epr" does not exist? That would be a syllogistic fallacy. Suppose you decide to model only the particle in the contextual case and do not include the device. Since the outcome originates from both particle and device contextually, knowing everything about the particle will not allow you to predict with certainty the outcome. You will obtain only probabilistic results (cf QM). But that does not mean particle epr does not exist. It simply means your theory is not complete, which was the point of EPR.

 

[harrylin]

[..] In brief, the mechanics goes thus: The HV-carrying particles, their HVs pair-wise correlated by recognised mechanisms, separate and fly to Alice and Bob. Interaction with the respective devices leads to a local transformation of each HV, most clearly seen in W where photons (initially pair-wise linearly-polarised identically) are transformed into pairs with different linear-polarisations. (Representing a fact accepted early in the foundations of QM: a "measurement" perturbs the measured object.) ... ... ...

Since the classical analysis is straight-forward, and Einstein-local (but see below), I suggest you study it and then see how it applies to your interest in Herbert's Paradox and its mechanics.[..]

... Either I'm "missing" something, or you are! For, as I read it, the model that you propose here is exactly the simple model that Bell started with (although the polarization angles are in fact perpendicular to each other), and which has been shown not to work. That was the starting point of all the "Bell theorem" discussions, see for example the illustration starting on page 5 in Bell's "socks" paper http://cdsweb.cern.ch/record/142461?ln=en. It's quite the same for electrons, except that electrons are found to have "spin up" and "spin down". Similar examples are given for polarized light on internet. Did you try to simulate it? It's easy to do with Excel for example.

 

[Gordon Watson]

... Either I'm "missing" something, or you are! For, as I read it, the model that you propose here is exactly the simple model that Bell started with (although the polarization angles are in fact perpendicular to each other), and which has been shown not to work. That was the starting point of all the "Bell theorem" discussions, see for example the illustration starting on page 5 in Bell's "socks" paper http://cdsweb.cern.ch/record/142461?ln=en. It's quite the same for electrons, except that electrons are found to have "spin up" and "spin down". Similar examples are given for polarized light on internet. Did you try to simulate it? It's easy to do with Excel for example.

Hi Harald. Let's start with what you might have missed: The starting point you mention is quite primitive; I think Bell says it's the first he thought of; there is one model here, Q; Q is the first I thought of;* Q applies to four experiments; Q $\in$ {W, X, Y, Z); each experiment is well-defined; Q works; I prefer a working-model to simulations; the maths etc at https://www.physicsforums.com/showpost.php?p=3895681&postcount=252 might help;** the move from (5)-(6) follows workings given earlier in the thread; it's best DIY; I'll happily show you how if you like; (7)-(10) give the results for the four experiments; it's all based on classical analysis; answers to some of Bill's helpful points are here https://www.physicsforums.com/showpost.php?p=3896722&postcount=257; E & O E.

Hoping this helps, for starters. Let me know where it doesn't. I'm happy to help some more. And I will be adding further some simpler* more-helpful** maths*** today, DV. Thanks, GW

* The maths in https://www.physicsforums.com/showpost.php?p=3895681&postcount=252 is quite primitive. It's based on those I first used to obtain the results. (Crazy, I know! But OK! So historic interest only; strictly! But based on the situation as I saw it at the time.)

** The "might help" is a correct choice of words; a reflection on them; not you. So see next.

*** Probably best to wait for these; for I am sure they will help. They will be based on the much simpler analysis used earlier in this thread.

 

[Gordon Watson]

Contextual means "particle epr" + "device epr" + "new epr's created due to interaction" → outcome. Or in short "system eprs" → outcome. Where system includes particle, device and their interaction.

Non-contextual means "particle epr" → outcome. If you know everything about the particle, then you know for certain the outcome.

In both cases if you can predict the outcome with certainty, then there exists an epr in the system which corresponds to it ("system" in the non-contextual case being merely the particle). Note however that if you can not predict with certainty, it does not mean an "epr" does not exist? That would be a syllogistic fallacy. Suppose you decide to model only the particle in the contextual case and do not include the device. Since the outcome originates from both particle and device contextually, knowing everything about the particle will not allow you to predict with certainty the outcome. You will obtain only probabilistic results (cf QM). But that does not mean particle epr does not exist. It simply means your theory is not complete, which was the point of EPR.

OK, thanks! Do you have a simple example of: "particle epr" → outcome? And are you saying that model Q is EPR-complete? See next:-

Returning to a clear post of yours (#244): the goal is to see if we agree about the nature of EPR eprs.

A: Imagine yourself as the i-th photon, leaving the source flying toward a device. The EPR question is, "Do you have properties $\lambda_i$ which exist as part of your "identity" which ultimately interact with Bob's device to result in a measurement outcome $+1$?"

B: The EPR point of view is that if a theory can predict $+1$ with certainty, then the theory is complete with respect to the hidden element of reality $\lambda_i$ which corresponds to $+1$

Re A, the i-th photon answers: "YES, I have such properties." (Score 5.)

Re B: The theory in this thread can predict $+1$ with certainty. So (in your terms) the theory here is EPR-complete with respect to the hidden element of reality $\lambda_i$ which corresponds to $+1$. (Score 5 from first sentence.)

Score = 10/10. BUT is it not the case that: The hidden element of reality that corresponds to $+1$ is $\textbf{b}^+$, not $\lambda_i$?

AS FOLLOWS? If I phrase the position in strict EPR (1935) terms, guided by my understanding of your terminology: We predict with certainty the value ($+1$) of a physical quantity ($\phi$, the orientation of linear polarisation), so (says EPR) there exists an epr ($\textbf{b}^+$) corresponding to $\phi$.

THEN we're (seemingly) back to my problem: We've not yet conducted the experiment; we've only predicted; so $\textbf{b}^+$ is definitely not yet existing?

OR am I to read EPR this way? The "the prediction with certainty" is only established when the experiment is run! Then $\textbf{b}^+$ exists!??

 

[DrChinese]

... Either I'm "missing" something, or you are! For, as I read it, the model that you propose here is exactly the simple model that Bell started with (although the polarization angles are in fact perpendicular to each other), and which has been shown not to work.

I have been trying unsuccessfully to explain this point to Gordon. Further, PDC crystals can be made to generate entangled photons pairs that are NOT entangled on the polarization basis. They have known polarization (depending on whether it is Type I or Type II), which is Gordon's premise. These do NOT produce entangled state statistics (of course), they produce product (separable) state statistics. I already provided that well known formula, which does not violate a Bell Inequality.

So simply performing the experiment invalidates his premise. Which is another reason why the mathematical treatment is superfluous.

 

[Gordon Watson]

I have been trying unsuccessfully to explain this point to Gordon. Further, PDC crystals can be made to generate entangled photons pairs that are NOT entangled on the polarization basis. They have known polarization (depending on whether it is Type I or Type II), which is Gordon's premise. These do NOT produce entangled state statistics (of course), they produce product (separable) state statistics. I already provided that well known formula, which does not violate a Bell Inequality.

So simply performing the experiment invalidates his premise. Which is another reason why the mathematical treatment is superfluous.*

DrC, with apologies for your lack of success, but those PDCs sound a bit like W to me. I could be wrong. Could you have a look and let me know, please? Thanks, more soon, GW.

* Some of those 'superfluous' maths addressed some earlier claims by you, reflecting the situation at the time. New situation now, new addressing on the way.

 

[billschnieder]

Score = 10/10. BUT is it not the case that: The hidden element of reality that corresponds to $+1$ is $\textbf{b}^+$, not $\lambda_i$?

We may be confusing several things:
a) The real world. What actually exists out there.
b) The ontology of a theory, ie the things in the theory which represent the real world (what the theory assumes the real world is like as a basis for the theory).

Of course, $\textbf{b}^+$ is an element of reality in the ontology of the theory, and the outcome corresponds to it. But this does not mean $\lambda_i$ is not an element of reality. In fact it also is an element of reality in the ontology of the theory and the outcome also corresponds to it. It appears you are using A → B, to argue that "not A" → "not B". The fact that $\textbf{b}^+$ is an element of reality in the theory does not mean λi is not an element of reality in the theory. It appears here you have set up a theory which includes both b+ and λ_i as ontological entities and then you are trying to argue that b+ but not λ_i is an element of reality as defined by EPR. But note what EPR wrote:

A comprehensive definition of reality is, however, unnecessary for our purpose. We shall be satisfied with the following criterion, which we regard as reasonable. If, without in any way disturbing a system, we can predict with certainty (i.e, with probability equal to unity), the value of a physical quantity, then there exists an element of physical reality corresponding to this physical quantity. It seems to us that this criterion, while far from exhausting all possible ways of recognizing a physical reality, at least provides us with one such way, whenever the conditions set down in it occur. Regarded not as a necessary, but merely as a sufficient, condition of reality, this criterion is in agreement with classical as well as quantum-mechanical ideas of reality.

AS FOLLOWS? If I phrase the position in strict EPR (1935) terms, guided by my understanding of your terminology: We predict with certainty the value ($+1$) of a physical quantity ($\phi$, the orientation of linear polarisation), so (says EPR) there exists an epr ($\textbf{b}^+$) corresponding to $\phi$.

No. If you are following strict EPR, then all you can say is that there exists and element of reality corresponding to the outcome. Nothing in EPR helps you to say exactly what that element of physical reality is with respect to your theory. It will be up to you as part of your theory to posit what the ontology of the theory is and once you have done that, you can't say part of the ontology is real and others are not because that would be a contradiction.

THEN we're (seemingly) back to my problem: We've not yet conducted the experiment; we've only predicted; so $\textbf{b}^+$ is definitely not yet existing?

Doesn't matter. The photon doesn't exist yet either, does that mean it does not represent a real thing in your theoretical prediction? It is part of the ontology of your theory. If you are predicting with certainty, then $\textbf{b}^+$ is also part of the theory and thus a component of the prediction whether the experiment has been performed or not, and so is $\lambda_i$. In any case, according to the theory, when ever the experiment is actually performed $\textbf{b}^+$ will pre-exist the outcome $+1$.

OR am I to read EPR this way? The "the prediction with certainty" is only established when the experiment is run! Then $\textbf{b}^+$ exists!??

Now I too am confused about what the confusion is all about. Prediction is a theoretical exercise, is it not?

The way you should read EPR is as follows:

1) If a theory can predict an outcome with certainty, ie probability 1, then an element of physical reality exists in the real world corresponding to that outcome, without regard for how the theory actually represents this element of physical reality in its ontology.

2) a theory is only complete if it can predict with certainty, every outcome in it's domain of applicability. That is, every element of physical reality is represented in the ontology of the theory.

 

[billschnieder]

... Either I'm "missing" something, or you are! For, as I read it, the model that you propose here is exactly the simple model that Bell started with (although the polarization angles are in fact perpendicular to each other), and which has been shown not to work.

I think there is an interesting result in this thread but it is being obscured by the way in which it was presented mixing the physics with the math. This is how I see it. We have two functions A() and B() with outcomes ±1 (just like Bell's functions) and we want to calculate the expectation value of the product A()*B(), ie E(AB) (just like Bell's integral). Let us start from Bell's equation (2), his integral

$E(AB) = \int_{\Lambda} A(\textbf{a}, \lambda ) \cdot B(\textbf{b}, \lambda )\;\rho (\lambda )d\lambda = \int_{\Lambda} (A^{\lambda}_a \cdot B^{\lambda}_b) \;\rho (\lambda )d\lambda$

For the sake of illustration, assume λ is discrete. Therefore

$E(AB)_V = \sum_{\lambda} (A^{\lambda}_a\cdot B^{\lambda}_b)P(\lambda|V), \;\;\;\; \lambda \in [\lambda_1, \lambda_2, \lambda_3, \lambda_4]$
$= A^{\lambda_1}_a\cdot B^{\lambda_1}_b\cdot P(\lambda_1|V) + A^{\lambda_2}_a\cdot B^{\lambda_2}_b\cdot P(\lambda_2|V) + A^{\lambda_3}_a\cdot B^{\lambda_3}_b\cdot P(\lambda_3|V) + A^{\lambda_4}_a\cdot B^{\lambda_4}_b\cdot P(\lambda_4|V)$

Now suppose
$A^{\lambda_1}_a = A(a,\lambda_1) = +1 \;,\; B^{\lambda_1}_b = B(b,\lambda_1) = +1 \;,\; A^{\lambda_2}_a = A(a,\lambda_2) = +1 \;,\; B^{\lambda_2}_b = B(b,\lambda_2) = -1 \;,\;$
$A^{\lambda_3}_a = A(a,\lambda_3) = -1 \;,\; B^{\lambda_3}_b = B(b,\lambda_3) = +1 \;,\; A^{\lambda_4}_a = A(a,\lambda_4) = -1 \;,\; B^{\lambda_4}_b = B(b,\lambda_4) = -1$

It therefore follows that
$P(\lambda_1|V) = P(A^+_aB^+_b|V) \;,\; P(\lambda_2|V) = P(A^+_aB^-_b|V) \;,\; P(\lambda_3|V) = P(A^-_aB^+_b|V) \;,\; P(\lambda_4|V) = P(A^-_aB^-_b|V) \;,\;$

And it is immediately obvious that

$E(AB)_V = \sum_{\lambda} (A^{\lambda}_a\cdot B^{\lambda}_b)P(\lambda|V), = \sum_{ij} (A^i_a\cdot B^j_b)P(ij|V), \;\;\;\; ij \in [++, +-, -+, --]$
$= A^+_a\cdot B^+_b\cdot P(A^+_aB^+_b|V) + A^+_a\cdot B^-_b\cdot P(A^+_aB^-_b|V) + A^-_a\cdot B^+_b\cdot P(A^-_aB^+_b|V) + A^-_a\cdot B^-_b\cdot P(A^-_aB^-_b|V)$

Therefore,
$E(AB)_V = P(A_a^+B_b^+|V) - P(A_a^+B_b^-|V) - P(A_a^-B_b^+|V) + P(A_a^-B_b^-|V)$
$= P(A_a^+|V)P(B_b^+|V,\,A^+) - P(A_a^+|V)P(B_b^-|V,\,A_a^+) - P(A_a^-|V)P(B_b^+|V,\,A_a^-) + P(A_a^-|V)P(B_b^+|V,\,A_a^-)$
$= P(A_a^+|V)\left [P(B_b^+|V,\,A_a^+) - P(B_b^-|V,\,A_a^+) \right ]- P(A_a^-|V) \left [P(B_b^+|V,\,A_a^-) - P(B_b^-|V,\,A_a^-) \right ]$

Since:
$P(A_a^+|V) = P(A_a^-|V) = \frac{1}{2}$, for random variables
and
$P(B_b^+|V,\,{A_a^+}) + P(B_b^-|V,\,{A_a^+}) = 1$
$P(B_b^+|V,\,{A_a^-}) + P(B_b^-|V,\,{A_a^-}) = 1$
Therefore substituting above, we get
$E(AB)_V = 2 \cdot P(B_b^+|V,\,A_a^+) - 1$
$E(AB)_V = -2 \cdot P(B_b^+|V,\,A_a^-) + 1$
These are general results which follow directly from Bell's integral through simple probability algebra. The interesting question is then to look at a specific experiment V, and calculate on the basis of that experiment E(AB)_v and calculating either $P(B_b^+|V,\,A_a^+)$ or $P(B_b^+|V,\,A_a^-)$.

In other words, $P(B_b^+|V,\,A_a^+)$ is the answer to the question: "For the experiment V, what is the probability of recording a +1 at station B given that a +1 was also recorded at station A?" and $P(B_b^+|V,\,A_a^-)$ is the answer to the question: "For the experiment V, what is the probability of recording a +1 at station B given that a -1 was also recorded at station A?"

Gordon has provide a general way of answering these types of questions for all the types of experiments performed so far. In a sense it is a generalization of Malus law. This makes sense because Malus law was answering similar questions, just in a specific situation where we have two polarizers in series ie "what is the probability that the photon passes the second polarizer B, given that it passed through the first one A"

Using this more "general" law, as Gordon showed, we can reproduce the results of all the experiments and the QM predictions for the experiment, including the classical one which does not violate Bell's inequalities but also including the ones which do violate Bell's inequalities.

 

[DrChinese]

DrC, with apologies for your lack of success, but those PDCs sound a bit like W to me. I could be wrong. Could you have a look and let me know, please?

Yes, you defined:

W (the classical OP experiment) is Y [= Aspect (2004)] with the source replaced by a classical one (the particles pair-wise correlated via identical linear-polarisations).

X (a classical experiment with spin-half particles) is Z [= EPRB/Bell (1964)] with the source replaced by a classical one (the particles pair-wise correlated via antiparallel spins).

Y = Aspect (2004).

Z = EPRB/Bell (1964).

So a single Type I PDC produces identical outputs H>H> from a V> input (or if rotated 90 degrees, V>V> output from a H> input). These will of course provide product state statistics which will not violate a Bell Inequality. As a general rule, the formula is:

.25+.5(cos^2(a-b)) when the source is randomly oriented

However, in this case you would have an additional factor because the source is not randomly oriented. You would have to consider the orientation of the fixed crystal relative to both a and b. That doesn't change the essential result.

 

[Gordon Watson]

Yes, you defined:

W (the classical OP experiment) is Y [= Aspect (2004)] with the source replaced by a classical one (the particles pair-wise correlated via identical linear-polarisations).

X (a classical experiment with spin-half particles) is Z [= EPRB/Bell (1964)] with the source replaced by a classical one (the particles pair-wise correlated via antiparallel spins).

Y = Aspect (2004).

Z = EPRB/Bell (1964).

So a single Type I PDC produces identical outputs H>H> from a V> input (or if rotated 90 degrees, V>V> output from a H> input). These will of course provide product state statistics which will not violate a Bell Inequality. As a general rule, the formula is:

.25+.5(cos^2(a-b)) when the source is randomly oriented

However, in this case you would have an additional factor because the source is not randomly oriented. You would have to consider the orientation of the fixed crystal relative to both a and b. That doesn't change the essential result.

Thanks DrC. What is their result, please; and is the paper on-line? G

 

[Gordon Watson]

We may be confusing several things:
a) The real world. What actually exists out there.
b) The ontology of a theory, ie the things in the theory which represent the real world (what the theory assumes the real world is like as a basis for the theory).

Of course, $\textbf{b}^+$ is an element of reality in the ontology of the theory, and the outcome corresponds to it. But this does not mean $\lambda_i$ is not an element of reality. In fact it also is an element of reality in the ontology of the theory and the outcome also corresponds to it. It appears you are using A → B, to argue that "not A" → "not B". The fact that $\textbf{b}^+$ is an element of reality in the theory does not mean λi is not an element of reality in the theory. It appears here you have set up a theory which includes both b+ and λ_i as ontological entities and then you are trying to argue that b+ but not λ_i is an element of reality as defined by EPR. But note what EPR wrote:




No. If you are following strict EPR, then all you can say is that there exists and element of reality corresponding to the outcome. Nothing in EPR helps you to say exactly what that element of physical reality is with respect to your theory. It will be up to you as part of your theory to posit what the ontology of the theory is and once you have done that, you can't say part of the ontology is real and others are not because that would be a contradiction.


Doesn't matter. The photon doesn't exist yet either, does that mean it does not represent a real thing in your theoretical prediction? It is part of the ontology of your theory. If you are predicting with certainty, then $\textbf{b}^+$ is also part of the theory and thus a component of the prediction whether the experiment has been performed or not, and so is $\lambda_i$. In any case, according to the theory, when ever the experiment is actually performed $\textbf{b}^+$ will pre-exist the outcome $+1$.


Now I too am confused about what the confusion is all about. Prediction is a theoretical exercise, is it not?

The way you should read EPR is as follows:

1) If a theory can predict an outcome with certainty, ie probability 1, then an element of physical reality exists in the real world corresponding to that outcome, without regard for how the theory actually represents this element of physical reality in its ontology.

2) a theory is only complete if it can predict with certainty, every outcome in it's domain of applicability. That is, every element of physical reality is represented in the ontology of the theory.

Bill, with great respect, but some of this is so wrong I can't believe it! (And I generally find that you're not like that.) BUT:

Where the heck did this come from??

"It appears here you have set up a theory which includes both b+ and λ_i as ontological entities and then you are trying to argue that b+ but not λ_i is an element of reality as defined by EPR."

To give you the benefit of the doubt: MAYBE I've made some bad wording somewhere here?

So, where, please??

To be very clear, can you not see (and say) this?

"It appears here you (GW) have set up a theory which includes both b+ and λ_i as ontological entities and then you argue that both b+ and λ_i are elements of reality as defined by EPR."

I think the EPR issue is whether you and I see EPR's eprs the same way. Looks like we might soon! So your wrongheadedness (said in the nicest possible way ), might be the breakthrough on that score we need?

More soon; probably in a new post, but maybe also via edit; but don't wait to see them IFF this little post maybe changes some other words and confusions of yours! Thanks, GW

 

[Gordon Watson]

I think there is an interesting result in this thread but it is being obscured by the way in which it was presented mixing the physics with the math. This is how I see it. We have two functions A() and B() with outcomes ±1 (just like Bell's functions) and we want to calculate the expectation value of the product A()*B(), ie E(AB) (just like Bell's integral). Let us start from Bell's equation (2), his integral

$E(AB) = \int_{\Lambda} A(\textbf{a}, \lambda ) \cdot B(\textbf{b}, \lambda )\;\rho (\lambda )d\lambda = \int_{\Lambda} (A^{\lambda}_a \cdot B^{\lambda}_b) \;\rho (\lambda )d\lambda$

For the sake of illustration, assume λ is discrete. Therefore

$E(AB)_V = \sum_{\lambda} (A^{\lambda}_a\cdot B^{\lambda}_b)P(\lambda|V), \;\;\;\; \lambda \in [\lambda_1, \lambda_2, \lambda_3, \lambda_4]$
$= A^{\lambda_1}_a\cdot B^{\lambda_1}_b\cdot P(\lambda_1|V) + A^{\lambda_2}_a\cdot B^{\lambda_2}_b\cdot P(\lambda_2|V) + A^{\lambda_3}_a\cdot B^{\lambda_3}_b\cdot P(\lambda_3|V) + A^{\lambda_4}_a\cdot B^{\lambda_4}_b\cdot P(\lambda_4|V)$

Now suppose
$A^{\lambda_1}_a = A(a,\lambda_1) = +1 \;,\; B^{\lambda_1}_b = B(b,\lambda_1) = +1 \;,\; A^{\lambda_2}_a = A(a,\lambda_2) = +1 \;,\; B^{\lambda_2}_b = B(b,\lambda_2) = -1 \;,\;$
$A^{\lambda_3}_a = A(a,\lambda_3) = -1 \;,\; B^{\lambda_3}_b = B(b,\lambda_3) = +1 \;,\; A^{\lambda_4}_a = A(a,\lambda_4) = -1 \;,\; B^{\lambda_4}_b = B(b,\lambda_4) = -1$

It therefore follows that
$P(\lambda_1|V) = P(A^+_aB^+_b|V) \;,\; P(\lambda_2|V) = P(A^+_aB^-_b|V) \;,\; P(\lambda_3|V) = P(A^-_aB^+_b|V) \;,\; P(\lambda_4|V) = P(A^-_aB^-_b|V) \;,\;$

And it is immediately obvious that

$E(AB)_V = \sum_{\lambda} (A^{\lambda}_a\cdot B^{\lambda}_b)P(\lambda|V), = \sum_{ij} (A^i_a\cdot B^j_b)P(ij|V), \;\;\;\; ij \in [++, +-, -+, --]$
$= A^+_a\cdot B^+_b\cdot P(A^+_aB^+_b|V) + A^+_a\cdot B^-_b\cdot P(A^+_aB^-_b|V) + A^-_a\cdot B^+_b\cdot P(A^-_aB^+_b|V) + A^-_a\cdot B^-_b\cdot P(A^-_aB^-_b|V)$

Therefore,
$E(AB)_V = P(A_a^+B_b^+|V) - P(A_a^+B_b^-|V) - P(A_a^-B_b^+|V) + P(A_a^-B_b^-|V)$
$= P(A_a^+|V)P(B_b^+|V,\,A^+) - P(A_a^+|V)P(B_b^-|V,\,A_a^+) - P(A_a^-|V)P(B_b^+|V,\,A_a^-) + P(A_a^-|V)P(B_b^+|V,\,A_a^-)$
$= P(A_a^+|V)\left [P(B_b^+|V,\,A_a^+) - P(B_b^-|V,\,A_a^+) \right ]- P(A_a^-|V) \left [P(B_b^+|V,\,A_a^-) - P(B_b^-|V,\,A_a^-) \right ]$

Since:
$P(A_a^+|V) = P(A_a^-|V) = \frac{1}{2}$, for random variables
and
$P(B_b^+|V,\,{A_a^+}) + P(B_b^-|V,\,{A_a^+}) = 1$
$P(B_b^+|V,\,{A_a^-}) + P(B_b^-|V,\,{A_a^-}) = 1$
Therefore substituting above, we get
$E(AB)_V = 2 \cdot P(B_b^+|V,\,A_a^+) - 1$
$E(AB)_V = -2 \cdot P(B_b^+|V,\,A_a^-) + 1$
These are general results which follow directly from Bell's integral through simple probability algebra. The interesting question is then to look at a specific experiment V, and calculate on the basis of that experiment E(AB)_v and calculating either $P(B_b^+|V,\,A_a^+)$ or $P(B_b^+|V,\,A_a^-)$.

In other words, $P(B_b^+|V,\,A_a^+)$ is the answer to the question: "For the experiment V, what is the probability of recording a +1 at station B given that a +1 was also recorded at station A?" and $P(B_b^+|V,\,A_a^-)$ is the answer to the question: "For the experiment V, what is the probability of recording a +1 at station B given that a -1 was also recorded at station A?"

Gordon has provide a general way of answering these types of questions for all the types of experiments performed so far. In a sense it is a generalization of Malus law. This makes sense because Malus law was answering similar questions, just in a specific situation where we have two polarizers in series ie "what is the probability that the photon passes the second polarizer B, given that it passed through the first one A"

Using this more "general" law, as Gordon showed, we can reproduce the results of all the experiments and the QM predictions for the experiment, including the classical one which does not violate Bell's inequalities but also including the ones which do violate Bell's inequalities.

Bill, having just "snapped" (and maybe your head off ), I appreciate your workings here. I've not checked them in detail BUT your job (maybe all our jobs, especially other critics) will be a little easier as soon as I get the simplified maths here (probably today). It is along the line that you favoured in an earlier post. Thanks again, more soon, GW

 

[Gordon Watson]

We may be confusing several things:
a) The real world. What actually exists out there.
b) The ontology of a theory, ie the things in the theory which represent the real world (what the theory assumes the real world is like as a basis for the theory).

Looks good! Like: The ontology of a good theory maps the ontology of the real world to our epistemological advance and advantage. That's why it's a good theory.

Of course, $\textbf{b}^+$ is an element of reality in the ontology of the theory, and the outcome corresponds to it. But this does not mean $\lambda_i$ is not an element of reality. In fact it also is an element of reality in the ontology of the theory and the outcome also corresponds to it.

Is it not better to say, because simpler and more accurate:

"The particle outcome $\textbf{b}^+_i$, and corresponding device output $+1_i$, are both brought into being by the device/particle interaction $\delta_b\lambda_i$."​

??

It appears you are using A → B, to argue that "not A" → "not B". The fact that $\textbf{b}^+$ is an element of reality in the theory does not mean λi is not an element of reality in the theory. It appears here you have set up a theory which includes both b+ and λ_i as ontological entities and then you are trying to argue that b+ but not λ_i is an element of reality as defined by EPR.

Appearances can be deceiving: This serious inaccuracy is addressed in an earlier response from me. (Me apologising, just in case but doubting that it's my fault!)

But note what EPR wrote:

"A comprehensive definition of reality is, however, unnecessary for our purpose. We shall be satisfied with the following criterion, which we regard as reasonable. If, without in any way disturbing a system, we can predict with certainty (i.e, with probability equal to unity), the value of a physical quantity, then there exists an element of physical reality corresponding to this physical quantity. It seems to us that this criterion, while far from exhausting all possible ways of recognizing a physical reality, at least provides us with one such way, whenever the conditions set down in it occur. Regarded not as a necessary, but merely as a sufficient, condition of reality, this criterion is in agreement with classical as well as quantum-mechanical ideas of reality. GW emphasis added."

Thanks for this, for sure. But I still am of the view that their use of "corresponding" muddies the waters for some (like me, for sure); though I will take another look and see if it is maybe wholly my problem; me much preferring maths to words! Moreover, in our recent exchanges some welcome clarity appears to be emerging (for me). More anon.

No. If you are following strict EPR, then all you can say is that there exists and element of reality corresponding to the outcome. Nothing in EPR helps you to say exactly what that element of physical reality is with respect to your theory. It will be up to you as part of your theory to posit what the ontology of the theory is and once you have done that, you can't say part of the ontology is real and others are not because that would be a contradiction.

Fair enough. I'm in complete agreement. AND I DO NOT!

Doesn't matter. The photon doesn't exist yet either, does that mean it does not represent a real thing in your theoretical prediction? It is part of the ontology of your theory. If you are predicting with certainty, then $\textbf{b}^+$ is also part of the theory and thus a component of the prediction whether the experiment has been performed or not, and so is $\lambda_i$. In any case, according to the theory, when ever the experiment is actually performed $\textbf{b}^+$ will pre-exist the outcome $+1$.

Bill, I believe you're getting it, and I apologise that I was not clearer re my position earlier.

Now I too am confused about what the confusion is all about. Prediction is a theoretical exercise, is it not?

The way you should read EPR is as follows:

1) If a theory can predict an outcome with certainty, ie probability 1, then an element of physical reality exists in the real world corresponding to that outcome: without regard for how the theory actually represents this element of physical reality in its ontology.

2) A theory is only complete if it can predict with certainty, every outcome in its domain of applicability. That is, every relevant element of the relevant physical reality is represented in the ontology of the theory.

Bill, THIS I can accept (after some friendly edits with no change of meaning). And I think I do implement it?

But what about the critics? DrC? ttn? Mermin? Shimony? Will they, do they anywhere, accept your wording? For it looks both innocent and valid to me.

GW

 

[billschnieder]

Is it not better to say, because simpler and more accurate:

"The particle outcome $\textbf{b}^+_i$, and corresponding device output $+1_i$, are both brought into being by the device/particle interaction $\delta_b\lambda_i$."​

But outcome refers to what the experimenter sees ($+1_i$). $\textbf{b}^+_i$ is hidden to him. I see not need to distinguish "outcome" and "device output". It just creates confusion.

The part I still do not understand is that it seems as though you are going to great lengths to distinguish yourself from EPR but I'm lost as to what you think you gain from that. Am I wrong?

Thanks for this, for sure. But I still am of the view that their use of "corresponding" muddies the waters for some (like me, for sure);

.
No problem, though I'm not sure I fully understand what bothers you so much about their choice of the word "corresponding". What meaning do you think "corresponding" forces you to conclude from their statements? In other words, why do you think EPR point of view, as you understand it, is wrong

 

[Gordon Watson]

But outcome refers to what the experimenter sees ($+1_i$). $\textbf{b}^+_i$ is hidden to him. I see not need to distinguish "outcome" and "device output". It just creates confusion.

The part I still do not understand is that it seems as though you are going to great lengths to distinguish yourself from EPR but I'm lost as to what you think you gain from that. Am I wrong?
.
No problem, though I'm not sure I fully understand what bothers you so much about their choice of the word "corresponding". What meaning do you think "corresponding" forces you to conclude from their statements? In other words, why do you think EPR point of view, as you understand it, is wrong

NB: I am (personally) IN the ontology of the THEORY:- Device and particle are both, equally, eprs there: and so equally open to my view there.

Does that make sense; or just more confusing?

I been in that mode all along, from day one, and NOW suspect it is, in part, contributing to the "possible differences" that we have been discussing.

AND: No, not at all am I distinguishing myself from EPR. Sneak preview of re-formatting might help (see next post).

I am on my way to re-interperting "Corresponding" thanks to some of your better comments. Thanks for that.

More coming. GW

EDIT: In case of past confusions from me: Device/particle interactions are denoted $\delta_{\textbf{b}}' \lambda$, etc. We retain the prime in $\delta_{\textbf{b}}'$ to facilitate reference to $\delta_{\textbf{b}}$ and $\delta_{\textbf{a}}'$, etc., when discussing the results for identical device settings.

 

[Gordon Watson]

$Q_{\textit{0.1}}$

$[Q_{\textit{0.0}} \;with \; footnotes\;to\;facilitate \;critical\;comments.]$

With appreciative acknowledgment to

ThomasT for probing questions and billschnieder for prodigious help all-over!

$Q \in \{W, X, Y, Z\}.\;\;\;\;(1)^1$

$A({\textbf{a}}, \lambda)_Q \equiv \pm 1.\;\;\;\;(2)^2$

$B({\textbf{b}}, \lambda')_Q = ((-1)^{2s} \cdot B({\textbf{b}}, \lambda)_Q \equiv \pm 1. \;\;\;\;(3)^3$

$E(AB)_Q \equiv ((-1)^{2s} \cdot \int d\lambda\;\rho (\lambda )\;AB)_Q \;\;\;\;(4)^4$

$=((-1)^{2s})_Q \cdot \int d\lambda \;\rho(\lambda) \;[P(A^+B^+|Q) -P(A^+B^-|Q)-P(A^-B^+|Q)+P(A^-B^-|Q)]\;\;\;\;(5)^5$

$= [(-1)^{2s}]_Q \cdot[ 2 \cdot P(B^+|Q,\,A^+) - 1].\;\;\;\;(6)^6$

$E(AB)_W = (cos[2 ({\textbf{a}}, {\textbf{b}})])/2.\;\;\;\;(7)^7$

$E(AB)_X = - ({\textbf{a}}. {\textbf{b}})/2.\;\;\;\;(8)^8$

$E(AB)_Y = cos[2 ({\textbf{a}}, {\textbf{b}})].\;\;\;\;(9)^9$

$E(AB)_Z = - {\textbf{a}}. {\textbf{b}}.\;\;\;\;(10)^{10}$
​

Footnotes:

1. Re (1): The generality of Q, coupled with the general applicability of Malus' initiative (his Method), enables this wholly classical analysis to go through. Q embraces:

W (the classical OP experiment) is Y [= Aspect (2004)] with the source replaced by a classical one (the particles pair-wise correlated via identical linear-polarisations).

X (a classical experiment with spin-half particles) is Z [= EPRB/Bell (1964)] with the source replaced by a classical one (the particles pair-wise correlated via antiparallel spins).

Y = Aspect (2004).

Z = EPRB/Bell (1964).​

2. Re (2): $\equiv$ identifies relations drawn from Bell (1964). (2) & (3) correctly represent Einstein-locality: a principle maintained throughout this classical analysis.

3. Re (3): Primes indicate items in Bob's locale; their removal from HVs (when convenient) is based on the initial correlation of each particle-pair via their $\lambda$ and $\lambda'$ relations. Note that Bell (1964) does not distinguish between $\lambda$ and $\lambda'$: and we introduce s = intrinsic spin. $(-1)^{2s}$ thus arises from Q embracing spin-1/2 and spin-1 particles: in some ways a complication, it brings out the unity of the classical approach used here.

4. Re (4): Integrating over $\lambda$, with $\lambda'$ eliminated: hence the coefficient, per note at #3.

5. Re (5): $P$ denotes Probability. $A^+$ denotes $A = +1$, etc. The expansion is from classical probability theory: causal-independence and logical-dependence carefully distinguished. The probability-coefficients $+1, -1, -1, +1,$ respectively, represent the Einstein-local (causally-independent) values for the relevant $A\cdot B$ product.

The reduction (5)-(6) follows, (A1)-(A4), each step from classical probability theory; $\int d\lambda \;\rho(\lambda) = 1$. From (5):

$P(A^+B^+|Q) -P(A^+B^-|Q)-P(A^-B^+|Q)+P(A^-B^-|Q)\;\;\;\;(A1)$

$=P(A^+|Q)P(B^+|Q,A^+)-P(A^+|Q)P(B^-|Q,A^+)-P(A^-|Q)P(B^+|Q,A^-)+P(A^-|Q)P(B^-|Q,A^-)\;\;\;\;(A2)$

$=[P(B^+|Q,A^+)-P(B^-|Q,A^+)-P(B^+|Q,A^-)+P(B^-|Q,A^-)]/2\;\;\;\;(A3)$

$= 2 \cdot P(B^+|Q,\,A^+) - 1.\;\;\;\;(A4)$

$Since, \;in \;(A2), \;with \;random \;variables: P(A^+|Q)=P(A^-|Q)=P(B^+|Q)= P(B^-|Q) = 1/2.\;\;\;\;(A5)$​

6. Re (6): (6), or variants, allows the application of Malus' Method, as follows: Following Malus' example (ca 1810), we would study the results of experiments and write equations to capture the underlying generalities: here $P(B^+|Q,\,A^+)$. However, since no Q is experimentally available to us, we here derive (from theory), the expected observable probabilities: representing observations that could be made from real experiments, after Malus. Footnotes #7-10 below show the observations that lead from (6) to (7)-(10).

7. Re (7): Within Malus' capabilities, W would show (from observation):

$P(B^+|W,\,A^+) = [cos^2 (\textbf{a}, \textbf{b}) + 1/2]/2= ([cos^2 [s \cdot (\textbf{a}, \textbf{b})] + 1/2]/2)_W$ $(A6)$ in modern terms: whence (7), from (6).

Alternatively, he could derive the same result (without experiment) from his famous Law.

8. Re (8): Within Stern & Gerlach's capabilities, X would show (from observation):

$P(B^+|X,\,A^+) = ([cos^2 [s \cdot (\textbf{a}, \textbf{b})] + 1/2]/2)_X = [cos^2 [(\textbf{a}, \textbf{b})/2] + 1/2]/2$ $(A7)$: whence (8), from (6).

Alternatively, they could derive the same result (without experiment) by including their discovery, $s$, in Malus' Law.

9. Re (9): Conducted by Aspect (2004), Y would show (from observation):

$P(B^+|Y,\,A^+) = cos^2 [s\cdot(\textbf{a}, \textbf{b})]_Y = cos^2 (\textbf{a}, \textbf{b})$ $(A8)$: whence (9), from (6).

To see this, Aspect (2004: (3)) has (in our notation):

$P(A^+B^+|Y) = [cos^2 (\textbf{a}, \textbf{b})]/2 = P(A^+|Y)P(B^+|Y, A^+) = P(B^+|Y, A^+)/2$ $(A9)$, from $(A5)$; whence $P(B^+|Y, A^+) = cos^2 (\textbf{a}, \textbf{b}).\;(A8)$

10. Re (10): Analysed by Bell (1964), Z would show (from observation):

$P(B^+|Z,\,A^+) = cos^2 [s\cdot(\textbf{a}, \textbf{b})]_Z = cos^2 [(\textbf{a}, \textbf{b})/2]$ $(A10)$: whence (10), from (6).

To see this, unlike Aspect (2004), Bell (1964) does not derive subsidiary probabilities. Instead, Bell (1964: (3)) has (in our notation):

$E(AB)_Z = -(\textbf{a}. \textbf{b}) = -[ 2 \cdot P(B^+|Z,\,A^+) - 1]$ $(A11)$, from $(A5)$; whence $P(B^+|Z, A^+) = cos^2 [(\textbf{a}, \textbf{b})/2].\;(A10)$

References:

Aspect (2004): http://arxiv.org/abs/quant-ph/0402001

Bell (1964): http://www.scribd.com/doc/51171189/Bell-1964-Bell-s-Theorem


With questions, typos, improvements, critical comments, etc., most welcome,

GW

 

[Gordon Watson]

$E(AB)_Z = -(\textbf{a}. \textbf{b}) = -[ 2 \cdot P(B^+|Z,\,A^+) - 1]$ $(A11)$, from $(A5)$; whence $P(B^+|Z, A^+) = cos^2 [(\textbf{a}, \textbf{b})/2].\;(A10)$

CORRECTION: Should read:​

$E(AB)_Z = -(\textbf{a}. \textbf{b}) = -[ 2 \cdot P(B^+|Z,\,A^+) - 1]$ $(A11)$, from $(6)$, with $s$ = $1/2$; whence $P(B^+|Z, A^+) = cos^2 [(\textbf{a}, \textbf{b})/2].\;(A10)$

GW

PS: The story thus far: At Bill's prompting, clearly-observable (hence "reasonably"-ontological)-elements of the real-world have been separated from the clearly-ontogolical elements of the theory. The former eprs, widely observable, are things like Mermin's Red $(A^+, B^+)$ and Green $(A^-, B^-)$ lights, with their frequencies of occurrence and joint-occurrence.

We have attempted to put these "reasonable" eprs and their relations in mathematical form so that less words are henceforth required in our discussions. (The qualifier "reasonable" is intended to eliminate discussions of "what really exists" -- for we accept that much of physical reality is veiled from us.)

The next move (it seems to me) is to enunciate the clearly-ontogolical elements of the theory and their dynamical interactions. To show how the set of clearly-observable-eprs relate to the set of "proposed-eprs" -- the actual etrs (ETRs, elements of theoretical reality) advanced in the theory.

 

[Gordon Watson]

...
.
No problem, though I'm not sure I fully understand what bothers you so much about their choice of the word "corresponding". What meaning do you think "corresponding" forces you to conclude from their statements? In other words, why do you think EPR point of view, as you understand it, is wrong

As I recall, I did not yet say that EPR were "wrong". I'm confident of this because, until their meaning is clear to me, I'm in no position to pass judgment. Rather I said that I was "still of the view that their use of "corresponding" muddies the waters for some (like me, for sure)."

So let's try this:

1. In our (you and I) working with Y (after Aspect 2004), I predict with certainty the value $+1$ (equals: "There will be a Green light" ... as we agreed, say) of a physical quantity $q$.

2. SO (after EPR) there exists (otherwise I could not have predicted so accurately) an epr $e$ corresponding to $q$.

3. What are $e$ and $q$, please?

PERHAPS it is your view that question #3 should follow this:

2a. SO (after EPR??) there exists (when the prediction is confirmed) an epr $e$ corresponding to $q$.

3a. Then, what are $e$ and $q$ in this case, please?

 

[Gordon Watson]

A CALL to the CRITICS!​

Following Bill's suggestion, I've separated the maths from the physics.

The maths is here: https://www.physicsforums.com/showpost.php?p=3898144&postcount=273 ($Q_{\textit{0.1}}$).

I expect to post the accompanying one-to-one classical physics next week-end, DV. It will be $Q_{\textit{0.2}}$.

However, given that important one-to-one relation, it follows that most weaknesses in the case here are already posted at $Q_{\textit{0.1}}$.

So please do not wait for next-week-end to express your concerns: I would like to have most issues properly covered in $Q_{\textit{0.2}}$.


Any critic:

If your identifier is not listed below, chances are that I've missed your comments, or I'm confident that they do not apply. Please do not hesitate to remedy the position. For those who are looking for the "TRICK" etc., ... please note this FACT:

Since Planck's constant is (thus far) nowhere explicit, why should anything beyond classical physics be thus far required?​

billschnieder:

Our agreement re EL means that we're not discussing total non-sense. Many thanks for that!

However, I am coming to the view that we differ re EPR (which has a small consequence re BT; more anon), and I'm thinking that I might be able to sharpen your independent view re BT (see gill below); though I make no claim to have studied your position (yet) in detail; more anon.​

Delta Kilo:

Your maths enthusiasms are appreciated, and valued: but we cannot both be right!​

DrC:

I'm thinking that you might now be able to move towards the EL (Einstein-local) camp? For you'll have seen (in $Q_{\textit{0.1}}$) one of your favourite equations: but in its correct context. What more must I do?​

gill1109:

Thanks for raising CHSH, to which I've not yet replied: However, in that I derive CHSH from an Identity, are you not puzzled when an Identity is breached by valid experiments? Which is close to Bill's valid concern, I suspect?

More importantly: In the classical physics here, which matches the classical maths, you will have already seen physically-significant integrals like this (though I've now added the implicit driver, making it explicit):

$((2s\cdot h/4\pi) \int d\lambda \;(\delta _{a}\lambda \rightarrow \left \{ a^+, a^- \right \})\;cos[2s \cdot (a, \lambda )])_Q = (\pm1)\cdot (s\cdot h/2\pi)_Q.$**​

I call this physical-process (= "physical-function") a dynamic-iteration. However, the point of this dynamical-process is that it terminates when the trig argument is 0 or ∏. So please note the "push-me/pull-you dynamic" in moving to such an argument: one of $a^+$ xor $a^-$ is a certain terminus, the other impossible, for the $\lambda$ under test!

**Note that a wholly classical analysis leads to the view that spin should expressed in integer units of $h/4\pi$; for the size-2 "b-gger-factor (applied to $s$) would not be required!​

harrylin:

I believe that there is enough here already for you to tackle Herbert's Paradox!​

ThomasT:

Trust you are back and doing the math? With more questions?​

ttn:

Your comments would be welcome.​

 

[billschnieder]

So let's try this:

1. In our (you and I) working with Y (after Aspect 2004), I predict with certainty the value $+1$ (equals: "There will be a Green light" ... as we agreed, say) of a physical quantity $q$.

2. SO (after EPR) there exists (otherwise I could not have predicted so accurately) an epr $e$ corresponding to $q$.

3. What are $e$ and $q$, please?

According to EPR $e$ is an element of reality and $q$ is a physical quantity, which is an outcome( in this case $+1$). Is that what you are asking, otherwise I do not understand the question.

PERHAPS it is your view that question #3 should follow this:

2a. SO (after EPR??) there exists (when the prediction is confirmed) an epr $e$ corresponding to $q$.

No. Experimental confirmation does not come in, it is assumed. Saying you predict something with certainty, already contains the assumption that experiments will validate it.

Honestly, I still do not understand (1) what your issue is with EPR's "corresponding", and (2) why it is important. Despite your many attempts at explaining it, my feeble mind still does not grasp it for some reason. It may help me if you use a simple example as the "tablets and solutions" one I proposed earlier:

We have tablets with well defined chemicals λ_i. We have a machine which produces pairs of tablets with a random chemical, except each of the pair is identical to the other. One of each pair is sent to Alice at station "A" and the other to Bob at station "B". Alice and Bob each have a large collection of liquids at their disposal. Each liquid in their collection is such that placing any tablet in any liquid results in a either a sweet taste (designated +1) or a bitter taste (designated -1). You could assume a tablet reacts with a liquid to produce either compound T⁺ which tastes sweet or compound T^- which tastes bitter. We designate the liquid chosen by Alice "a", and the liquid chosen by Bob "b". Alice and Bob have the freedom to pick any of the liquids, but each specific tablet can only be tested once since the reaction destroys it.​

The similarity to the EPR scenario is evident so we should be able to proceed to clarify things with this example. Please, using this example, first explain YOUR view. What in the above is an element of physical reality and what is not according to your view. Then, using the same example, explain what you understand the EPR view to be and why you think there might be an issue with it. I'll greatly appreciate if you could help me understand your view with this example. Thanks! :shy:

 

[Gordon Watson]

According to EPR $e$ is an element of reality and $q$ is a physical quantity, which is an outcome( in this case $+1$). Is that what you are asking, otherwise I do not understand the question.


No. Experimental confirmation does not come in, it is assumed. Saying you predict something with certainty, already contains the assumption that experiments will validate it.

Honestly, I still do not understand (1) what your issue is with EPR's "corresponding", and (2) why it is important. Despite your many attempts at explaining it, my feeble mind still does not grasp it for some reason. It may help me if you use a simple example as the "tablets and solutions" one I proposed earlier:

We have tablets with well defined chemicals λ_i. We have a machine which produces pairs of tablets with a random chemical, except each of the pair is identical to the other. One of each pair is sent to Alice at station "A" and the other to Bob at station "B". Alice and Bob each have a large collection of liquids at their disposal. Each liquid in their collection is such that placing any tablet in any liquid results in a either a sweet taste (designated +1) or a bitter taste (designated -1). You could assume a tablet reacts with a liquid to produce either compound T⁺ which tastes sweet or compound T^- which tastes bitter. We designate the liquid chosen by Alice "a", and the liquid chosen by Bob "b". Alice and Bob have the freedom to pick any of the liquids, but each specific tablet can only be tested once since the reaction destroys it.​

The similarity to the EPR scenario is evident so we should be able to proceed to clarify things with this example. Please, using this example, first explain YOUR view. What in the above is an element of physical reality and what is not according to your view. Then, using the same example, explain what you understand the EPR view to be and why you think there might be an issue with it. I'll greatly appreciate if you could help me understand your view with this example. Thanks! :shy:

I'm certain that your feeble mind far exceeds mine ! BUT your shyness nowhere matches mine when it comes to discussing non-locality and related issues in the context of classical settings! For, imho, we needs must ever remember that Planck's constant, not zero, lurks among us!

Maybe it's just a silly/anxious avoidance phobia of mine: induced by so many classical examples that fail ... thereby erroneously strengthening the hand of Einstein's detractors! (I suspect the lack of attention here arises from unresolved issues attaching to many of those same examples.) ALL of which means that I'm "presently" choosing to side-step your neat classical example ... with sincere apologies (I should have said so) ... and stick with EPR a bit longer:-

$e$ and $q$ were given, essentially as blanks, for you to fill-in! Now that you've turned that back on me, I'll do $q$: trusting that we agree, at least up to that point:

$+1$ is a surrogate for $(+1)(s\cdot h/2\pi)$.

To me, that means: $q$ is the physical quantity $s\cdot h/2\pi$ = $s\cdot$(the current unit measure of spin angular momentum in QM). And $+1$ is its value.

Whence, anticipating your clarification: $e$ is the epr corresponding to the physical quantity $s\cdot h/2\pi$, which means ... ?

Your move!

PS: My next is sealed.

 

[billschnieder]

Maybe it's just a silly/anxious avoidance phobia of mine: induced by so many classical examples that fail ... thereby erroneously strengthening the hand of Einstein's detractors! (I suspect the lack of attention here arises from unresolved issues attaching to many of those same examples.) ALL of which means that I'm "presently" choosing to side-step your neat classical example ... with sincere apologies (I should have said so) ... and stick with EPR a bit longer:-

The simple example is aimed at clarifying *conceptual* issues which appear to be stumbling blocks. Ignoring such examples will surely rob us of their illuminating benefits. Besides, we are not leaving EPR behind, rather we are just putting it aside for a little while to clarify conceptual issues (one step back, three steps forward ). If you must side-step it, at least clearly state the issue we are trying to address here to help me focus my thoughts because I'm lost. I thought we were trying to understand what EPR meant by "epr" and why you have reservations about their view.

$e$ and $q$ were given, essentially as blanks, for you to fill-in!

With what? With the objective of showing what? For the purpose of what? In other words, I thought we were trying to understand EPR's epr conceptually? It appears here we are looking for the details of a specific model.

Now that you've turned that back on me, I'll do $q$: trusting that we agree, at least up to that point:

$+1$ is a surrogate for $(+1)(s\cdot h/2\pi)$.

To me, that means: $q$ is the physical quantity $s\cdot h/2\pi$ = $s\cdot$(the current unit measure of spin angular momentum in QM). And $+1$ is its value.

What do you measure in a real experiment? $s\cdot h/2\pi$ or $+1$? What did you predict with certainty? $s\cdot h/2\pi$ or $+1$? The physical quantity is the same as the answer to both questions. My quess will be $+1$, although it could be both.

Whence, anticipating your clarification: $e$ is the epr corresponding to the physical quantity $s\cdot h/2\pi$, which means ... ?

Again, I do not understand what you are looking for. $e$ also corresponds to $+1$.

I think we have a communication problem here and all of the above discussion may be unnecessary for resolving the conceptual difference that persists (maybe) between our views. You stated earlier you have reservations about EPR's use of "corresponding". Please if you can, say what it is about EPR's "corresponding" you disagree with. It would help this discussion significantly. What do you think they meant by it?

 

[Gordon Watson]

... Again, I do not understand what you are looking for. $e$ also corresponds to $+1$.

I think we have a communication problem here and all of the above discussion may be unnecessary for resolving the conceptual difference that persists (maybe) between our views...

An epr corresponds to $+1$? Have you expressed your view correctly? For here's what I expected you to say; here's what I sealed, as my view:

"The epr corresponding to $s\cdot h/2\pi$ is the unit spin angular momentum for the particle-type under test".

PS: So I'm now OK with EPR's definition of an epr in the above context! Does that now mean we agree? I'd also welcome your commenting on how this fits with your classical example. Thanks.