Bell experiment would somehow prove non-locality and information FTL?

[DrChinese]

3. Unbeknown to Alice, her dichotomic polariser-analyser (detector) is yoked to the box such that her freely-chosen detector-setting (principal axis at unit-vector a) becomes also the setting of the principal axis of the box.

I do not understand the point of this twist. This is simply the same effect as if her setting remained at 0 degrees all of the time, while Bob varies his setting. So what if Alice's setting is chained to the reference of the source, and she doesn't know it?

Can you explain how this changes the setup or the results? Otherwise, I think it confuses the issues for most readers.

 

[Hurkyl]

Do you mean ''dependent''?

But P(B=+|ab) = one-half for any a; which is independence (as it should be)?

That is: P(B=+|ab) = P(B=+|b) = one-half. Yes?

Thanks, wm

Yes, dependent. Hrm, did I misread the problem? Oh well, I'll think about it later.

 

[wm]

wm,

The math you have is exactly correct. The conclusion you draw from it is not.

Your (3) is a standard presentation of the Bell Inequality.

Your (4) is a standard presentation of the predictions of QM, as relates to (3).

This shows that QM is incompatible with the Bell Inequality. It does NOT show that Bell's Inequality is violated for classical situations.

Doc, thanks for this; but I'm not clear on your NOT. I tried to use no QM at all in deriving (4); that is, I tried to leave QM out of the whole affair by reducing the experiment to be only classical mechanics (CM) at the level (say) of high-school physics.

But the fact that basic CM and QM agree does not mean that I have used QM. (NB: I agree that some ''CM'' situations will be compatible with BI.) So:

1. Why do you imply (as I read it) that I have used QM?

2. And doesn't my example show a classical situation that provides the basis for refuting many Bellian inequalities in classical settings?

3. My personal belief is that BT is no part of QM. I think Peres at least agrees. Is it accepted and valid to say ''BT is no part of QM''?

Thanks, wm

 

[wm]

I do not understand the point of this twist. This is simply the same effect as if her setting remained at 0 degrees all of the time, while Bob varies his setting. So what if Alice's setting is chained to the reference of the source, and she doesn't know it?

Can you explain how this changes the setup or the results? Otherwise, I think it confuses the issues for most readers.

Doc, There is no change in RESULTS, the outcome being exactly (mathematically) equivalent; the simplest classical refutation (PS: in my old terms, pending a reply to my last post) of BI that I yet know of. BUT:

I thought it best not to have Alice (it could alternatively be Bob) reduced to a cipher/zombie. AND I thought of building a simple class-room demonstration (electric light bulb, swapping mono-polarisers, measure intensities, etc). So I then wanted a student at each end to be able to independently change each detector (with the mechanism hidden); to essentially provide a realistic simulation of Aspect's experiment with the exact same correlations.

It also makes the CM maths to require just a bit more thought.

SO: Considering this, and in that I have already been given a few suggestions for improvement, do you think I should go with the simplified version here? OR: Should we encourage readers to think the given model through? (Since they can readily be helped and it's not that difficult.)

Thanks, wm

 

[wm]

Perhaps I was not clear. Alice and Bob are intended to be synonymous with the entangled photons and their measurement. So I am providing a description of quantum objects. Yes, I totally agree that these are only "pristine" (to use your term) once. So no disagreement there.

But you say that Bob is not affected when the "sledgehammer" is applied to Alice. Oh, but that is NOT true at all! Bob absolutely acts as if he was given the same sledgehammer as Alice! That is the non-local collapse of the wave function, and it is definitely and demonstrably instantaneous. If this did not happen, we wouldn't have anything interesting to discuss in this thread.

Doc, I really appreciate your hanging in there on this! Which brings me to two issues:

1. I think that I have a better explanation of this whole sorry business. BUT it's essentially ''personal research''. YET it would surely benefit by public discussion here. SO do you think it possible that a special discussion could be opened up, something between the existing PF rules of the open discussion we're having now and the ''personal research'' section.

That is: It would be open public discussion but the Admins would have agreed that it was not totally crackpot stuff? It might carry a warning re ''speculative''? OR: Is there somewhere else on the web better suited for discussion of such stuff?

2. Now, returning to the topic: You say above (slight edit)

wm says that Bob is not affected when the "sledgehammer" is applied to Alice. Oh, but that is NOT true at all! Bob absolutely acts as if he was given the same sledgehammer as Alice! That is the non-local collapse of the wave function, and it is definitely and demonstrably instantaneous.

Question: You appear to be conjoining a quite-magical long-distance physical effect with that ''non-local collapse of the wave-function'' with [sic] something ''definitely and demonstrably instantaneous''. Could you elaborate, please?

Like: If ''demonstrably'' has its old meaning, and if ''seeing is believing'', it seems that a clear and definitive case can be made for your position? Yet my impression is that each of the three conjoined phrases is problematic?

(PS: I'll likely open a new thread on a related subject: LOCALITY or Local QM.)

Thanks again, wm

 

[DrChinese]

Doc, thanks for this; but I'm not clear on your NOT. I tried to use no QM at all in deriving (4); that is, I tried to leave QM out of the whole affair by reducing the experiment to be only classical mechanics (CM) at the level (say) of high-school physics.

But the fact that basic CM and QM agree does not mean that I have used QM. (NB: I agree that some ''CM'' situations will be compatible with BI.) So:

1. Why do you imply (as I read it) that I have used QM?

2. And doesn't my example show a classical situation that provides the basis for refuting many Bellian inequalities in classical settings?

3. My personal belief is that BT is no part of QM. I think Peres at least agrees. Is it accepted and valid to say ''BT is no part of QM''?

Thanks, wm

wm,

You introduced QM when you used the cos^2 formula (Malus, 1807). You may think that is 'associated' with classical optics, and of course it is. But classical optics does not have an analogue to entanglement, and generally a different formula applies.

At any rate, it does not matter. QM says it must be cos^2, and that formula is incompatible with local realism - as you aptly demonstrate. Any classical theory that posits cos^2 is *also* incompatible with local realism.

 

[DrChinese]

Question: You appear to be conjoining a quite-magical long-distance physical effect with that ''non-local collapse of the wave-function'' with [sic] something ''definitely and demonstrably instantaneous''. Could you elaborate, please?

Like: If ''demonstrably'' has its old meaning, and if ''seeing is believing'', it seems that a clear and definitive case can be made for your position? Yet my impression is that each of the three conjoined phrases is problematic?

(PS: I'll likely open a new thread on a related subject: LOCALITY or Local QM.)

Thanks again, wm

Referring to instantaneous collapse being demonstrable...

Please keep in mind that entanglement is not the only way to demonstrate instantaneous collapse. Anytime a measurement is performed showing a particle is "here", that means it is also not "there". For example, you can send a single photon through a beamsplitter and have the outputs go to 2 boxes, and then move them miles apart without observing the photon in any way (in theory at least). When you open one to see if a photon is present, you change the state of the box far away too.

After a while, you end up realizing that entanglement is really no different. Either way, an observation collapses the particle's state description into a eigenstate. In an entangled setup, there is simply more than one particle. Mind you, I don't pretend to understand what "collapse" actually is at a physical level any more than anyone else. As far as I know, this is a mystery and an open question.

 

[heusdens]

It's obviously not a refutation of Bell's theorem. The question is if it's a classical violation of Bell's theorem. It's not, because one of the hypotheses (parameter independence) is violated: P(B=+|ab) is dependent on a.

I've read somewhere (in a paper/website that explains the Bell Inequality) that (for the Bell inequality) it does not matter wether parameters (= properties?) are dependent or not.

 

[wm]

wm,

You introduced QM when you used the cos^2 formula (Malus, 1807). You may think that is 'associated' with classical optics, and of course it is. But classical optics does not have an analogue to entanglement, and generally a different formula applies.

At any rate, it does not matter. QM says it must be cos^2, and that formula is incompatible with local realism - as you aptly demonstrate. Any classical theory that posits cos^2 is *also* incompatible with local realism.

Doc, you seem to be saying that I'll introduce QM if I cite Newton's Law of Gravitation! Or: QM began with Malus (1807) and not Planck (1900). So could you elaborate on:

1. How it is that cos^2 is incompatible with local realism, please?

2. How my experiment demonstrates the failure of local realism?

3. Also: What is the different formula that you refer to?

4. Do you acept that my experiment REBUTS the general implication of negative probabilities with Bell's theorem?

Thanks, as always, wm

 

[wm]

Referring to instantaneous collapse being demonstrable...

Please keep in mind that entanglement is not the only way to demonstrate instantaneous collapse. Anytime a measurement is performed showing a particle is "here", that means it is also not "there". For example, you can send a single photon through a beamsplitter and have the outputs go to 2 boxes, and then move them miles apart without observing the photon in any way (in theory at least). When you open one to see if a photon is present, you change the state of the box far away too.

After a while, you end up realizing that entanglement is really no different. Either way, an observation collapses the particle's state description into a eigenstate. In an entangled setup, there is simply more than one particle. Mind you, I don't pretend to understand what "collapse" actually is at a physical level any more than anyone else. As far as I know, this is a mystery and an open question.

Dear Doc, Re the difficulties with understanding: You may be conjoining contradictory concepts. To wit -- what "collapse" actually is at a physical level --

I'm sure Heisenberg himself thought that ''collapse'' was mathematical and NOT physical. In a sense: Just plain old Bayesian updating.

So: Are you aware of objections to (or weaknesses in) the Heisenberg view? And: Do you know where can I read on those who believe that ''collapse'' is physical; contradicting Heisenberg?

wm

 

[DrChinese]

Doc, you seem to be saying that I'll introduce QM if I cite Newton's Law of Gravitation! Or: QM began with Malus (1807) and not Planck (1900). So could you elaborate on:

1. How it is that cos^2 is incompatible with local realism, please?

2. How my experiment demonstrates the failure of local realism?

3. Also: What is the different formula that you refer to?

4. Do you acept that my experiment REBUTS the general implication of negative probabilities with Bell's theorem?

Thanks, as always, wm

1. You have demonstrated that, a la Bell. That is what Bell's Theorem shows us, that QM makes predictions (cos^2) which are incompatible with local realism.

2. A true Bell test will support cos^2 and will lead to the rejection of local realism.

3. Some local realists posit alternative formulae, although some stick with cos^2 and say Bell's Theorem is wrong. Or the experiments are flawed. etc.

The usual "naive" alternative formulae are one of the following:

a. 1/4 + (cos^2 theta)/2 , which varies from .75 to .25 and is a result of applying Malus to 2 independent classical particles. This formula is not at odds with the Bell Inequality.
b. A straight line function going from 1 to 0 over 90 degrees. This formula is not at odds with the Bell Inequality. You can see a graph of this on my Negative Probability page.

4. cos^2 plus local realism leads to the prediction of negative probabilities. Bell paved the way for this conclusion. The reason there are no negative probabilities is that either locality or realism must be rejected.

Malus' Law was discovered before QM, of course, due to observation. QM makes the same prediction, as you would hope QM would match observation. The issue is that QM does not assert local realism, and Malus never knew of Bell (since they lived in different centuries).

 

[DrChinese]

Dear Doc, Re the difficulties with understanding: You may be conjoining contradictory concepts. To wit -- what "collapse" actually is at a physical level --

I'm sure Heisenberg himself thought that ''collapse'' was mathematical and NOT physical. In a sense: Just plain old Bayesian updating.

So: Are you aware of objections to (or weaknesses in) the Heisenberg view? And: Do you know where can I read on those who believe that ''collapse'' is physical; contradicting Heisenberg?

wm

The collapse issue has nothing to do with Bell's Theorem itself. It is a question of interpretation of QM. Some folks consider the collapse issue to be the central flaw to Orthodox QM. I don't claim that collapse is physical (or not); only that whatever it is, it occurs instantaneously (FTL). The MWI and BM camps often cite this as a problem that they can "fix".

 

[JesseM]

OK, I think I'm closer to understanding the scenario you're proposing now. But a few more questions:

1. Let's modify a typical Aspect/Bell test (using photons) in the following way, retaining no significant connection between Alice's detector (oriented a, orthogonal to the line-of-flight axis) and Bob's (oriented b, orthogonal to the line-of-flight axis). (a and b are unit vectors, freely chosen.)

If both Alice's detector and Bob's director are "orthogonal to the line-of-flight axis", how is it that they have a choice of which angle to measure? Do you just mean that their own polarization sheets are directly facing the source, but that they can still rotate them about the line-of-flight axis, like if I were holding a piece of paper so you were viewing it head-on, but at the same time I was free to rotate the paper so that the letters on it could be right-side-up, sideways, upside-down, etc.?

2. We place the Aspect-style singlet-source in a box. The LH and RH sides of the box (facing Alice and Bob respectively) each contain a dichotomic-polariser, the principal axis of which is aligned with the principal axis (say) of the box. (We thus have a classical source of correlated photon-pairs in identical but unknown states of linear polarisation.)

OK, from your previous answers I get the idea that the point of the dichotomic-polariser is to ensure that even though the source is yoked to Alice's detector, she will still get a random combination of +'s and -'s on her measurements, rather than 100% +'s. But are the dichotomic-polarisers on each side some sort of electronic devices that can be "yoked" to one another, so that each pair of photons always have the same linear polarization, or could the photon received by Alice be a different polarization from its twin received by Bob?

3. Unbeknown to Alice, her dichotomic polariser-analyser (detector) is yoked to the box such that her freely-chosen detector-setting (principal axis at unit-vector a) becomes also the setting of the principal axis of the box.

This, I think, is the step where you violate one of the assumptions behind Bell's theorem, and for some setups it will be impossible to make it work without FTL signals between Alice's detector and the box, so you won't be able to replicate all possible EPR-style experiments using only classical devices. But more on this later, first I want to make sure I'm understanding what you're proposing.

5. From classical analysis, and in a fairly obvious notation, the following equations hold:

(1) P(AB = S|ab) = cos^2(a, b),
(2) P(AB = D|ab) = sin^2(a, b);

Since I'm not so good on optics, can you go into a little more detail on the classical analysis involved? if an ordinary classical EM wave with a linear polarization in a certain direction is sent through a polarization filter at an angle A relative to it, what fraction of the wave (in terms of energy, perhaps, since for a wave with a single frequency the energy will be proportional to the number of photons when we go into quantum physics) will get through as a function of the angle? And in the quantum version of the same situation, what is the probability a photon known to have polarization in that direction (because it passed through a previous polarization filter) will make it through a polarization filter at angle A? Is it just cos^2(A) in both cases?

If so, I think I'd disagree with DrChinese that this experiment relies on any specifically quantum properties of photons. After all, instead of real photons the source could just send out classical signals describing the polarization of simulated photons, and depending on how Alice and Bob set their simulated polarization filters, the computer could calculate the probability each simulated photon makes it through using the equations above, and use a random number generator to display a + or - based on this probability.

7. However: For the differential-direction set {(a, b) = 67.5°, (a, c) = 45°, (b, c) = 22.5°} we have from (1) and (2):

Can you state explicitly what individual angles you're assuming for the three possible detector settings a, b, and c?

 

[DrChinese]

If so, I think I'd disagree with DrChinese that this experiment relies on any specifically quantum properties of photons.

Disagree with me...?

Well, the 2 issues are:

a) Are you performing a Bell-type test using entangled photons? If not, then I'm not interested anyway and there is no disagreement.

b) Assuming you are doing a Bell test: Are you looking to predict the outcome? If so, you will use the cos^2 function unless... you want to get the wrong answer. In which case again, I'm not interested anyway and there is no disagreement.

All said in fun.

 

[wm]

OK, I think I'm closer to understanding the scenario you're proposing now. But a few more questions: If both Alice's detector and Bob's director are "orthogonal to the line-of-flight axis", how is it that they have a choice of which angle to measure? Do you just mean that their own polarization sheets are directly facing the source, but that they can still rotate them about the line-of-flight axis, like if I were holding a piece of paper so you were viewing it head-on, but at the same time I was free to rotate the paper so that the letters on it could be right-side-up, sideways, upside-down, etc.?

Yes; the unit-vector defining the detector-orientation is orthogonal to the line-of-flight axis. So to define a unit-vector on the piece of paper with the letter on it, put an arrow on your piece of paper. Thus the detector/piece-of-paper orientation may range 0 to 2 PI. (The line-of-flight is akin to your specified line-of-sight.)

OK, from your previous answers I get the idea that the point of the dichotomic-polariser is to ensure that even though the source is yoked to Alice's detector, she will still get a random combination of +'s and -'s on her measurements, rather than 100% +'s.

Yes; that's it.

But are the dichotomic-polarisers on each side some sort of electronic devices that can be "yoked" to one another, so that each pair of photons always have the same linear polarization, or could the photon received by Alice be a different polarization from its twin received by Bob?

I think we should keep it simple and take the dichotomic polarisers to be made from calcite crystals. (The manufacture is sort of indicated by their representation in diagrams as rectangles with a single diagonal.) So we have crystals in holders that can be easily handled and mechanically yoked.

In my example the polarisations are identical in each pair. This keeps it simple.

This, I think, is the step where you violate one of the assumptions behind Bell's theorem, and for some setups it will be impossible to make it work without FTL signals between Alice's detector and the box, so you won't be able to replicate all possible EPR-style experiments using only classical devices. But more on this later, first I want to make sure I'm understanding what you're proposing.

OK, but: It would be good if you could briefly expand on what you have in mind. Then I can be thinking about it. But NB: Re FTL, I generally try to avoid the impossible or the really-difficult.

Since I'm not so good on optics, can you go into a little more detail on the classical analysis involved? if an ordinary classical EM wave with a linear polarization in a certain direction is sent through a polarization filter at an angle A relative to it, what fraction of the wave (in terms of energy, perhaps, since for a wave with a single frequency the energy will be proportional to the number of photons when we go into quantum physics) will get through as a function of the angle? And in the quantum version of the same situation, what is the probability a photon known to have polarization in that direction (because it passed through a previous polarization filter) will make it through a polarization filter at angle A? Is it just cos^2(A) in both cases?

Yes; it's Malus' Law; just check the net.

If so, I think I'd disagree with DrChinese that this experiment relies on any specifically quantum properties of photons. After all, instead of real photons the source could just send out classical signals describing the polarization of simulated photons, and depending on how Alice and Bob set their simulated polarization filters, the computer could calculate the probability each simulated photon makes it through using the equations above, and use a random number generator to display a + or - based on this probability.

1. I'd sure like to hear your point of view because I'm not yet clear on what the Doc is meaning.

2. I think it may be simpler than you say. Let me know off-thread if you're interested in a computer version.

Can you state explicitly what individual angles you're assuming for the three possible detector settings a, b, and c?

Let's take the simplest: a = 0, b = 67.5, c = 45 will do.

Regards, wm

 

[wm]

Dear DrChinese,

I asked: How is cos^2 incompatible with local realism?

You replied (slight edit = A & B added):

A. You have demonstrated that, a la Bell. B.That is what Bell's Theorem shows us, that QM makes predictions (cos^2) which are incompatible with local realism.

Re A: My result refutes Bell (via his inequalities) while meeting the boundary conditions of Bell's theorem.

(PS: That ''a la Bell'' looks like an attempt to bring me into the current Bell camp when I am far outside it --- Bell himself disliked his own theory!)

1. The refutation follows from Bell (and his supporters) predicting a certain inequality and me showing that it is, in general, FALSE.

2. My meeting the boundary conditions is supported by Peres (Quantum Theory, 1995, page 162). Quote: ''Bell's theorem ... applies to any physical system with dichotomic variables, whose values are arbitrarily called 1 and -1.'' That's also the basis of the CHSH Inequality, which is (in general) equally false (as I see it.)

3. My experiment is (surely) totally local and (even classically) realistic. How do you see it differently?

Re B: My result made NO use of QM. (How could it?) My result could have been derived as early as 1807!

Now, as previously discussed, there are many definitions of realism. But you have again left it undefined. Here's the crux: cos^2 is incompatible with local Bellian-realism. So why not drop Bellian-realism and join Einstein's camp on locality?

And there's the clue: I CANNOT refute Bell when the system is compatible with Bellian-realism. That is, with systems like dirty-socks, down-hill skiers, students ... systems (it seems) where (except for destructive measurements) pre-measurement beables are unperturbed by observation.

I conclude: My cos^2 is incompatible with Bellian-realism; so Bellian-realism is, in general, FALSE. In that this appears to fix some of your ongoing uncertainties: What do you think?

Fun, peace, joy, thanks, wm

 

[wm]

Dear Doc,

My question was: Do you acept that my experiment REBUTS the general implication of negative probabilities with Bell's theorem?

Your reply was:

cos^2 plus local realism leads to the prediction of negative probabilities. Bell paved the way for this conclusion. The reason there are no negative probabilities is that either locality or realism must be rejected.

Now I am trying to follow your reasoning and understand your meaning.

Here's one of my difficulties: My experiment is

1. Classically local.

2. Employs (old-fashioned) classical realism.

3. Uses Malus' (1807) cos^2 Law.

4. Yields ONLY positive probabilities.

So why is it not a counter-example to your claim?

Perhaps you mean: ''Negative probabilities'' arise in association with Bellian-realism? (That I could understand.)

wm

 

[vanesch]

There's a lot of confusion here.

To all who are concerned (especially wm):

the P(A,B) = 1/2 cos^2(th_a - th_b)
or, equivalently, because P(B) = 1/2,

P(A | B) = cos^2(th_a - th_b)

relationship is NOT Malus' law, despite its formal resemblance. This is a very common misunderstanding by "Bell theorem refuters".

Malus' law says the following:
the intensity of a beam with polarization angle th_beam, after a polarizer with angle th_pol, is given by: I_after/I_before = cos^2(th_beam - th_pol).

Mind you that what enters here is the angle of the polarization of the beam, at the polarizer, and the polarizer angle itself.

In photon speak, this comes down to saying that the probability for a photon which has polarization th_beam to go through a polarizer with angle th_pol, is given by P = cos^2(th_beam - th_pol).

But this has a priori nothing to do with two angles at two different places, applied at two different beams. In the case of entangled photons, we could simply think of them to have "equal polarization angles", and calculate the joint probability of BOTH photons, INDEPENDENTLY, to get through the polarizers. But this joint probability is not given by the expression:
1/2 cos^2(th_a - th_b), but rather by P(A) x P(B) = cos^2(th_a - th_pol1) x cos^2(th_b - th_pol2) (where we can set th_pol1 = th_pol2). THIS is Malus' law, applied to the case at hand. In Malus' law, we have to enter an angle of a polarizer, and an angle of a photon. Not two angles of polarizers.

And if you work this out for a uniform distribution of polarizing angles, you do NOT obtain the result P(A,B) = 1/2 cos^2(th_a - th_b).

So, despite its resemblance, 1/2 cos^2(th_a - th_b) is NOT Malus' law.

Another remark: with Newtonian gravity, it is not difficult to violate Bell's inequalities. The reason is that Newtonian gravity is not local.

 

[wm]

There's a lot of confusion here.

To all who are concerned (especially wm):

the P(A,B) = 1/2 cos^2(th_a - th_b)
or, equivalently, because P(B) = 1/2,

P(A | B) = cos^2(th_a - th_b)

relationship is NOT Malus' law, despite its formal resemblance. This is a very common misunderstanding by "Bell theorem refuters".

Malus' law says the following:
the intensity of a beam with polarization angle th_beam, after a polarizer with angle th_pol, is given by: I_after/I_before = cos^2(th_beam - th_pol).

Mind you that what enters here is the angle of the polarization of the beam, at the polarizer, and the polarizer angle itself.

In photon speak, this comes down to saying that the probability for a photon which has polarization th_beam to go through a polarizer with angle th_pol, is given by P = cos^2(th_beam - th_pol).

But this has a priori nothing to do with two angles at two different places, applied at two different beams. In the case of entangled photons, we could simply think of them to have "equal polarization angles", and calculate the joint probability of BOTH photons, INDEPENDENTLY, to get through the polarizers. But this joint probability is not given by the expression:
1/2 cos^2(th_a - th_b), but rather by P(A) x P(B) = cos^2(th_a - th_pol1) x cos^2(th_b - th_pol2) (where we can set th_pol1 = th_pol2). THIS is Malus' law, applied to the case at hand. In Malus' law, we have to enter an angle of a polarizer, and an angle of a photon. Not two angles of polarizers.

And if you work this out for a uniform distribution of polarizing angles, you do NOT obtain the result P(A,B) = 1/2 cos^2(th_a - th_b).

So, despite its resemblance, 1/2 cos^2(th_a - th_b) is NOT Malus' law.

Another remark: with Newtonian gravity, it is not difficult to violate Bell's inequalities. The reason is that Newtonian gravity is not local.

Dear Vanesch,

With respect, I think there is some confusion on your part.

Malus Law was used to derive the relationships that you criticise. So they are a consequence of Malus Law. As you rightly point out: They themselves are NOT Malus Law directly.

PS: I'm not aware that they were claimed to be Malus Law. Were they, please?

I must go now, but I will decipher your notation to see if anything else needs comments.

(The Newtonian gravity was meant as a fun shot at one of DrChinese's positions.)

Thank you, in haste, wm

 

[vanesch]

3. Unbeknown to Alice, her dichotomic polariser-analyser (detector) is yoked to the box such that her freely-chosen detector-setting (principal axis at unit-vector a) becomes also the setting of the principal axis of the box.

Right. It is the mechanics of the box which transmits the setting of Alice to Bob. But this means that Alice cannot change her axis in a space-like way independent from Bob's. In other words, Alice's setting information has been transmitted to Bob's side.

If you were to place Alice on Alpha-centauri, and Bob on the moon, the box would be about 4 lightyears long, and the elastic waves in the box wouldn't go faster than lightspeed. Now, a typical EPR setting makes Alice change her settings exactly FASTER than this, in order to keep the decisions at spacelike distances.

If you keep your "rigid box" thing, you have of course not satisfied the Bell independence condition. In the case the box is really rigid, this is a non-local effect, and in the other case, you've just had to wait for the elastic wave to arrive.

If you do not use this "box transmission" in some way in your model, then you have to show me how you arrive at:

5. From classical analysis, and in a fairly obvious notation, the following equations hold:

(1) P(AB = S|ab) = cos^2(a, b),
(2) P(AB = D|ab) = sin^2(a, b);

because that is NOT what you arrive at using Malus' law on both sides.

 

[wm]

Right. It is the mechanics of the box which transmits the setting of Alice to Bob.

Sort of. I would say: It is the mechanics of the linkage (the yoking) between Alice's detector and the box which results in the transmission of the setting of Alice to Bob.

But this means that Alice cannot change her axis in a space-like way independent from Bob's.

I think rather that Alice CAN change her detector-axis in a space-like way, independent of Bob. NB: There is no linkage between the detectors of Alice and Bob, and they are space-like separated. Alice sits beside her detector AND the box; Bob is far away.

In other words, Alice's setting information has been transmitted to Bob's side.

Yes; that is correct. But note: This is intentional; not here to be a general subversion of Bell's theorem BUT to provide a wholly classical system which breaches a wide range of Bellian inequalities. Years ago I was challenged to refute the generality of the CHSH Inequality; this design was the result.

If you were to place Alice on Alpha-centauri, and Bob on the moon, the box would be about 4 lightyears long, and the elastic waves in the box wouldn't go faster than lightspeed. Now, a typical EPR setting makes Alice change her settings exactly FASTER than this, in order to keep the decisions at spacelike distances.

The design is such that Alice sits in Paris with her detector AND the box on her desk; the related elastic waves are those we commonly experience in short mechanical linkages in our everyday laboratory experience.

Bob is on Alpha-Centauri. (ALSO, as noted above, it is not meant to be a definitive and general EPRB situation. Just a wholly classical refutation of many Bellian inequalities; contrary to the claim that Bellian Inequalities could not be classically refuted.)

If you keep your "rigid box" thing, you have of course not satisfied the Bell independence condition. In the case the box is really rigid, this is a non-local effect, and in the other case, you've just had to wait for the elastic wave to arrive.

As it stands, Alice and Bob are able to chose independently any setting that they choose. Also the + or - results they observe are independent of any setting that they choose.

If you do not use this "box transmission" in some way in your model, then you have to show me how you arrive at:

(1) P(AB = S|ab) = cos^2(a, b),
(2) P(AB = D|ab) = sin^2(a, b);

because that is NOT what you arrive at using Malus' law on both sides.

Yes, that is correct. If the Alice-to-box link is broken then a DIFFERENT set of equations arises from Malus' Law. But (of course) they again reproduce exactly the results that would be found experimentally with the newly established system..

(If both the link AND the box are removed then we have Aspect's Experiment. Then we would apply Malus' Law in a way akin to QM -- akin to DrChinese's recent post -- to derive the experimental correlation locally and realistically. But that must be discussed elsewhere; I understand.)

Hoping this helps, and appreciating your comments, (in haste again), wm

 

[vanesch]

Yes, that is correct. If the Alice-to-box link is broken then a DIFFERENT set of equations arises from Malus' Law. But (of course) they again reproduce exactly the results that would be found experimentally with the newly established system..

You should show me how you derive your correlation cos^2(th_a - th_b). Because, as I said, this does NOT follow from a standard application of Malus' law at both sides. As Dr. Chinese pointed out, when you apply Malus' law on both sides, you find an "attenuated" form of correlation, which is compatible with the Bell inequalities, and NOT cos^2.

 

[DrChinese]

Dear Vanesch,

With respect, I think there is some confusion on your part.

Malus Law was used to derive the relationships that you criticise. So they are a consequence of Malus Law. As you rightly point out: They themselves are NOT Malus Law directly.

PS: I'm not aware that they were claimed to be Malus Law. Were they, please?

I must go now, but I will decipher your notation to see if anything else needs comments.

(The Newtonian gravity was meant as a fun shot at one of DrChinese's positions.)

Thank you, in haste, wm

wm,

Vanesch is absolutely right, and I alluded to the same thing in an earlier post about the alternative formulae that some local realists put forth. Unfortunately, I am quite guilty of what Vanesch refers to because I reference Malus without supplying the entire story. Again, it comes back to the basics that I keep re-iterating.

1. IF you assume locality, realism (sometimes called hidden variables) and the QM formula (cos^2 theta) for predicting coincidences for entangled photon experiments, you end up with Bell's Inequality. You know this is correct, because you derived this relationship for yourself.

2. Tests of Bell's Inequality support the QM formula for predicting coincidences, within a very small margin of error. IF you accept these tests as valid, then you must reject either locality OR realism.

3. It is perfectly acceptable to conclude, as you do, that realism should be rejected. One of the incentives for doing this is to keep relativity in a position as a fundamental law of nature.

4. IF you create a classical test which respects both locality and realism, you cannot also have an algorithm that violates Bell's Inequality. Even if you could somehow do this, you still would not disprove Bell's Theorem, which states (in my words):

No physical theory of local Hidden Variables can ever reproduce all of the predictions of Quantum Mechanics.

5. I think if we stick to the above issues, and ignore negative probabilities, "perverse" applications of Malus, etc., we can be more constructive and it will be a lot easier to see the true elements of the debate.

-DrC

 

[JesseM]

OK, just trying to make sure I understand the math here:

7. However: For the differential-direction set {(a, b) = 67.5°, (a, c) = 45°, (b, c) = 22.5°} we have from (1) and (2):

(4) LHS (3) = 0.85 - 0.5 - 0.15 = 0.2.

Now, you said you were using the angles a = 0, b = 67.5, c = 45. So, calculating P(BC = S|bc), P(AC = D|ac) and P(AB = S|ab) explicitly, we have:

P(BC = S|bc): Since Alice's detector is b, 67.5, there's a 50% chance the signal is oriented at 67.5 and a 50% chance it's oriented at -22.5. So with Bob at 45, we have P(Bob sees + | signal 67.5, Bob 45) = cos^2(22.5) = 0.85, and P(Bob sees - | signal -22.5, Bob 45) = sin^2(67.5) = 0.85. Since Alice is guaranteed to see a + at 67.5 and a - at -22.5, this means the probability they get the same answer in this situation is 0.85.

P(AC = D|ac): 50% chance signal is at 0 and 50% chance it's at 90. With Bob at 45, we have P(Bob sees - | signal 0, Bob 45) = sin^2(45) = 0.5, and P(Bob sees + | signal 90, Bob 45) = cos^2(45) = 0.5. Since Alice sees + at 0 and - at 90, this means the probability they get different answers is 0.5.

P(AB = S|ab): 50% chance signal is at 0 and 50% chance it's at 90. With Bob at 67.5, we have P(Bob sees + | signal 0, Bob 67.5) = cos^2(67.5) = 0.15, and P(Bob sees - | signal 90, Bob 67.5) = sin^2(22.5) = 0.15. Since Alice sees + at 0 and - at 90, this means the probability they get the same answer is 0.15.

OK, I think I understand all the important features of your experiment now. As I said, although you did suggest using entangled photons, you're not relying on any specifically quantum phenomena to get the apparent violation of Bell's theorem, since you could just as easily have the source send out ordinary classical signals to Alice and Bob's computers telling them the "simulated source angle" (which would be based on Alice's choice of simulated detector angle, either equal to it or 90 degrees from it), and then the computer would calculate the probability that they see a + based on cos^2(the difference between the simulated source angle and simulated detector angle).

Similarly, you could also do this experiment using ordinary classical electromagnetic waves of a single frequency (as from a laser)--as long as the delay between the time Alice switches the angle of her polarizing filter and the time the yoked source reorients its own angle is negligible, then if Alice and Bob's computers just measure the fraction of the energy that's getting through the polarizing filter when it's set at a given angle, and that fraction becomes the probability they'll see a + on their screen for that trial, you'll get the same results, with no need to worry about entanglement at all.

Now that I think I understand your proposal, I'll explain why it isn't a genuine violation of Bell's theorem in a followup post.

 

[JesseM]

You should show me how you derive your correlation cos^2(th_a - th_b). Because, as I said, this does NOT follow from a standard application of Malus' law at both sides. As Dr. Chinese pointed out, when you apply Malus' law on both sides, you find an "attenuated" form of correlation, which is compatible with the Bell inequalities, and NOT cos^2.

See above, I think it works if you just assume that the probability Bob sees a photon get through is equal to cos^2(angle of source polarization - angle of Bob's polaroid), and likewise that the probability Alice sees a photon get through is equal to cos^2(angle of source polarization - angle of Alice's polaroid), with the condition that the source is "yoked" to Alice's polaroid so there's a 50% chance it's parallel to hers and a 50% chance it's at 90 degrees relative to hers. Of course, the problem is that this yoked condition actually violates one of the basic assumptions behind Bell's theorem, namely that any properties of what it emits, "hidden" or otherwise, should be statistically independent of Alice and Bob's choice of detector settings on each trial.

 

[JesseM]

But if the photons leave the box before Alice makes her choice of detector setting, how is that possible without the effects of Alice's choice traveling backwards in time?

Good point; I should add to the effect that: Alice's re-orientation time is short in relation to the detector dwell time. So the mismatch -- the number of ''prior-orientation'' photons in transit -- has little effect on the overall probability distribution.

The problem is that by making this assumption, you're sidestepping the whole reason that violations of Bell's inequalities rules out local realism! Obviously your setup wouldn't work at all if Alice was regularly switching her detector settings, and the time x in seconds between switches was smaller than the distance y in light-seconds from Alice to the source--in this case, under a local theory, your "yoking" scheme can't work because by the time the source learns of each new detector setting, Alice has already switched to a different (random) detector setting. Now, you'd have to ask someone more familiar with experimental tests of the Bell inequalities to learn if the condition above has in fact been met; but there's no denying that quantum theory still predicts the Bell inequalities would be violated in this situation, and it's precisely this prediction that poses the fundamental problem for anyone trying to come up with a local realistic theory to match QM's predictions.

Notice that when I was stating the assumptions behind Bell's theorem in post #133, I included the bolded part below:

do you agree or disagree that if we have two experimenters with a spacelike separation who have a choice of 3 possible measurements which we label A,B,C that can each return two possible answers which we label + and - (note that these could be properties of socks, downhill skiers, whatever you like), then if they always get opposite answers when they make the same measurement on any given trial, and we try to explain this in terms of some event in both their past light cone which predetermined the answer they'd get to each possible measurement with no violations of locality allowed (and also with the assumption that their choice of what to measure is independent of what the predetermined answers are on each trial, so their measurements are not having a backwards-in-time effect on the original predetermining event, as well as the assumption that the experimenters are not splitting into multiple copies as in the many-worlds interpretation), then the following inequalities must hold:

When I wrote this I was thinking of a well-known loophole in Bell's theorem, which I think would be explicitly ruled out with a statistical independence condition in any fully rigorous proof of the theorem. The loophole is that if the source is somehow able to "anticipate" the detector settings Alice and Bob choose ahead of time, then it can adjust the hidden variables based on this in such a way that Bell inequalities can be violated. This should be pretty obvious in the case where the source has foreknowledge of both Alice and Bob's detector settings--if they both choose the same detector, like B B, then the source just has to send out photons whose "hidden" states are identical for that setting, like {A+,B+,C-} and {A-,B+,C-}. On the other hand, if they choose different settings, like A C, the source can send out photons whose hidden states on those settings are different (like {A+, B-, C-} and {A+, B+, C-}) 3/4 of the time, and photons whose hidden states on those settings are identical (like {A-, B-, C+} and {A+, B-, C-}) 1/4 of the time, thus violating the Bell inequality which says that when Alice and Bob choose different detector settings, the probability they get the same answer must be greater than or equal to 1/3.

This idea has certainly been discussed before--for example, one of the main ideas of Huw Price's book https://www.amazon.com/dp/0195117980/?tag=pfamazon01-20 was that it might be possible to come up with a hidden-variables explanation for QM if we allow backwards causation, i.e. Alice and Bob's future choices having an effect on the source in the past. Of course, any theory allowing such backwards causation might not be deemed "local" even if one-way causes were restricted to stay within the cause's light cone, since an event A could have an effect on an event B in its past light cone, and B could have an effect on an event C in its future light cone, such that there was a spacelike separation between A and C. There's also a variation on this idea which I've seen discussed, which is that in a fully deterministic universe, Alice and Bob's choices would already be implicit in the state of the universe before or at the moment the source emits the photons, so that the source could have foreknowledge without technically violating locality. But since a wide variety of causes throughout Alice's past light cone may affect her decision even in a deterministic universe, it's hard to see how any reasonable theory would allow the source to "deduce" her future choice (especially since some of the events in Alice's past light cone may lie outside the source's past light cone), and this seems a bit like retrocausation by another name (in a deterministic universe it is also conceivable that the hidden variables of the particles emitted by the source on a given trial would somehow force Alice and Bob to make particular choices of settings, or that the initial conditions of the universe would be chosen in such a way as to insure this sort of correlation between the source and Alice and Bob's choice of detector settings on each trial, but this all seems to imply a weird cosmic conspiracy that wouldn't arise in a natural way from any simple fundamental physics equations).

Anyway, the point is that I have seen this idea--that there is a possible loophole in that the source might "know" in advance the detector settings and adjust its output accordingly--discussed in a number of places, so it seems to be well-understood that one of the key assumptions behind Bell's theorem must be the statistical independence of the source's output on a given trial and Alice and Bob's detector settings on the same trial. You haven't discovered anything fundamentally new with your example, and in any case, without some form of "retrocausation" the type of local hidden variables explanation in your example won't be able to cover all cases (again, it won't cover cases where Alice and Bob randomly switch detector settings at regular time-intervals, and the time x in seconds between switches is smaller than the distance y in light-seconds from them to the source).

 

[JesseM]

To add to my previous post, and to support my claim that a modern rigorous derivation of the Bell inequalities would include a condition about the source not "anticipating" the detector settings, check out this paper:

http://www.citebase.org/fulltext?format=application%2Fpdf&identifier=oai%3AarXiv.org%3Aquant-ph%2F0312176

In particular, look at section D on p.6, where the "No conspiracy" assumption is discussed.

 

[wm]

In response to my:

Yes, that is correct. If the Alice-to-box link is broken then a DIFFERENT set of equations arises from Malus' Law. But (of course) they again reproduce exactly the results that would be found experimentally with the newly established system,

vanesch wrote:

You should show me how you derive your correlation cos^2(th_a - th_b). Because, as I said, this does NOT follow from a standard application of Malus' law at both sides. As Dr. Chinese pointed out, when you apply Malus' law on both sides, you find an "attenuated" form of correlation, which is compatible with the Bell inequalities, and NOT cos^2.

Dear vanesch; I would say that my original agreement (my paragraph at top) remains OK. So: Do we differ?

For I'm not clear on what you want me to show you; or, rather, I am not clear on the experimental setting that you require the correlation for?

Perhaps, with the deepest respect, you are confusing two very different experimental situations. (OR perhaps I implied somewhere that your correlation cos^2(th_a - th_b) applies to the DIFFERENT case?)

1. The correlation that you cite above IS derived from Malus' Law, AND is the correct result for the related experiment. (Let us identify that experiment -- which is the experiment described in post #166 https://www.physicsforums.com/showpost.php?p=1216247&postcount=166 -- as wmX1.) That is: That correlation -- cos^2(th_a - th_b) = cos^2 (a, b) -- would certainly be confirmed by experiment on a wmX1 setting.


2. In the second case, when you destroy or remove the mechanical Alice-to-box linkage, we have a very DIFFERENT experiment. Let us call it wmX2. BUT, again using Malus' Law on both sides, we can derive the correct (experimentally confirmed) correlation.

3. wmX2 certainly provides an attentuated correlation when compared to the correlation for wmX1; a correlation now consistent with a Bellian Inequality: For Alice is now on the same footing as Bob, her one-to-one correlation with the box (the source) gone; the overall correlation reduced (as you say).

4. To be clear: IF THE BOX REMAINS IN PLACE (as in wmX1 and wmX2), application of Malus' Law to both sides of the box will yield the correct (experimentally confirmed) correlation.

Do we differ?

Regards, wm

 

[JesseM]

Disagree with me...?

Well, the 2 issues are:

a) Are you performing a Bell-type test using entangled photons? If not, then I'm not interested anyway and there is no disagreement.

Well, what do you mean by "a Bell-type test"? He's suggesting an experiment similar to Bell's in that each experimenter can choose between three possible detector settings which will each give yes/no answers, and looking at the correlation depending on whether the experimenters use the same setting or different settings. But the "yoking" of the source to one experimenter's setting on each trial violates one of the basic assumptions that must be used when proving Bell's theorem rigorously (and as I mentioned in an earlier post to wm, this sort of yoking would be impossible under a local theory if the time between the experimenter switching detector settings was sufficiently small and the distance between the experimenter and the source was sufficiently large).

b) Assuming you are doing a Bell test: Are you looking to predict the outcome? If so, you will use the cos^2 function unless... you want to get the wrong answer. In which case again, I'm not interested anyway and there is no disagreement.

But the cos^2 rule would still work if you weren't talking about the probability of individual photons being detected, but instead talking about the reduction in intensity of classical EM waves of a single frequency as a polarized beam from the source passes through an experimenter's polarization filter, depending on the angle between their filter and the angle of the filter at the source (a setup like the one shown on http://scholar.hw.ac.uk/site/physics/topic6.asp?outline= ). And of course, a computer could use this ratio between (intensity of light getting through source's filter) and (intensity of light getting through experimenter's filter) as a probability to display a + on the experimenter's screen. The probabilities of getting a + vs. a - would be exactly the same, given the experiment wm described, as if + and - corresponded to individual entangled photons making it through/not making it through the experimenter's filter. So, the fact that he finds a violation of a Bell inequality does not really have anything to do with quantum physics, it's just a consequence of this "yoking".

 

[DrChinese]

4. To be clear: IF THE BOX REMAINS IN PLACE (as in wmX1 and wmX2), application of Malus' Law to both sides of the box will yield the correct (experimentally confirmed) correlation.

Do we differ?

I still do not understand the point of your twist. How does it change anything? Unless you can explain this point, I am about ready to drop out of the discussion. (Your argument so far amounts to a repeat of Bell's Theorem, which we were already happy with.)

And about the application of Malus' Law to both sides: keep in mind that Malus' Law can be applied in 2 completely different ways:

1. The way it is applied by QM: Assuming that the wave function for Alice & Bob's particles collapse on first observation of either, and we now know the polarization of the other; so we can predict the rate of correlation as cos^2(theta). This will agree with experiment.

2. The way a local realist would naturally want to apply it: entanglement is not real and there is a specific polarization for the 2 photons that is simply the same value; therefore you apply Malus on both sides (independently, then integrate) and get the .25+cos^2(theta)/2 formula. This formula deviates wildly from experimental results.

Once you follow these points - two very different ways of looking at cos^1 - we can drill into any questions about the specifics of these.

 

[DrChinese]

Well, what do you mean by "a Bell-type test"? He's suggesting an experiment similar to Bell's in that each experimenter can choose between three possible detector settings which will each give yes/no answers, and looking at the correlation depending on whether the experimenters use the same setting or different settings. But the "yoking" of the source to one experimenter's setting on each trial violates one of the basic assumptions that must be used when proving Bell's theorem rigorously (and as I mentioned in an earlier post to wm, this sort of yoking would be impossible under a local theory if the time between the experimenter switching detector settings was sufficiently small and the distance between the experimenter and the source was sufficiently large).

...So, the fact that he finds a violation of a Bell inequality does not really have anything to do with quantum physics, it's just a consequence of this "yoking".

I don't get the point of the "yoking". Assuming we are using entangled photons, you get a violation of the Bell Inequality. This is true whether or not we rigorously rule out locality violation. The only difference is that when the yoking is present, the results are not rigorous. That's a step backward, so that is why I don't understand the twist. What do we get for it?

(It is almost like saying, Alice always selects A and Bob can select A, B or C. I think... )

 

[JesseM]

And about the application of Malus' Law to both sides: keep in mind that Malus' Law can be applied in 2 completely different ways:

1. The way it is applied by QM: Assuming that the wave function for Alice & Bob's particles collapse on first observation of either, and we now know the polarization of the other; so we can predict the rate of correlation as cos^2(theta). This will agree with experiment.

wm is not directly applying Malus' law to the correlation between Alice's result and Bob's result. wm's example just uses Malus' law to find the probability that each one individually sees a photon when they orient their polarizers at a certain angle, given that the angle of the polarizer at the source is "yoked" to Alice's polarizer (using ordinary classical signals, so there must be a built-in delay) in such a way that it has a 50% chance of being at the same angle and a 50% chance of being at a 90-degree angle relative to hers (the random element is included just to ensure that she sees a random mix of + and - results, rather than all +'s). In other words, the two angles being fed into the cos^2(angle_1 - angle_2) are not the angles of Alice and Bob's polarizers, they are the angles of Bob's polarizer and the source's polarizer. So any entanglement is irrelevant, if you were just using classical EM waves it would still be true that the light reaching Bob is reduced in intensity by a factor of cos^2(angle_1 - angle_2). And of course in QM, intensity is proportional to the number of photons making it through, or the probability an individual photon makes it through.

2. The way a local realist would naturally want to apply it: entanglement is not real and there is a specific polarization for the 2 photons that is simply the same value;

Yes, this is what is assumed in wm's example.

therefore you apply Malus on both sides (independently, then integrate) and get the .25+cos^2(theta)/2 formula.

That would be true if the source's polarization was statistically independent of Alice's choice of how to orient her polarizer, as would be assumed in a proof of Bell's theorem, but it's not true in wm's example because of the "yoking" he assumes. Of course if the yoking is done by classical means, there will be a delay which will cause wm's classical experiment to deviate somewhat from quantum predictions, and the classical setup will be wholly unable to replicate quantum predictions in the case where Alice is rapidly switching measurement angles in such a way that by the time a classical signal about each new angle has made it to the source, she has already switched to a different angle. In this case, all the Bell inequalities will be obeyed in wm's classical experiment, so this type of "yoking" will not allow a local hidden variables theory to replicate quantum predictions about entanglement in general.

 

[wm]

wm,

Vanesch is absolutely right, and I alluded to the same thing in an earlier post about the alternative formulae that some local realists put forth. Unfortunately, I am quite guilty of what Vanesch refers to because I reference Malus without supplying the entire story. Again, it comes back to the basics that I keep re-iterating.

1. IF you assume locality, realism (sometimes called hidden variables) and the QM formula (cos^2 theta) for predicting coincidences for entangled photon experiments, you end up with Bell's Inequality. You know this is correct, because you derived this relationship for yourself.

Doc: NOT quite TRUE; NOT quite CORRECT. The photons emerging from the box are dis-entangled. The calculations are clearly based on these NON-ENTANGLED photons; as I believe they must be if I want to give the CORRECT CLASSICAL derivation that I did give.

NOTE (and this may be the point of confusion here): I used an Aspect-style source of entangled photons, sandwiched between dichotomic linear polarisers, to create the NON-entangled photons that the calculations apply to. That's why I said earlier that Malus (1807) could have derived the result if he'd assumed the existence of such photon-pairs.

2. Tests of Bell's Inequality support the QM formula for predicting coincidences, within a very small margin of error. IF you accept these tests as valid, then you must reject either locality OR realism.

Interesting theory, Doc; IF .. THEN? (Might run better on the Pop-psychology Forum (perhaps)):

For here is the result of testing your theory:

1. I accept, without reservation, the validity of the results of Bell-Inequality-Tests (eg, Aspect's with photons).

2. I DO NOT reject locality or realism (properly defined)!

3. PS: Confident that Nature is local, I reject, without reservation, Bellian-realism.

3. It is perfectly acceptable to conclude, as you do, that realism should be rejected. One of the incentives for doing this is to keep relativity in a position as a fundamental law of nature.

Yes; and thank you. But PLEASE: With realism coming in so many flavours, there is no need to reject REALISM as such; only certain brands. Thus, having no taste for Bellian Realism, that's the flavour I reject.

PS: Per citation at foot, Bell had similar taste-buds: It seems John Bell had no taste at all (or a certain distaste) for his own cooking: ie, Bell's Theorem?

4. IF you create a classical test which respects both locality and realism, you cannot also have an algorithm that violates Bell's Inequality.

BUT I just did exactly that! What am I missing here, please?

Even if you could somehow do this, you still would not disprove Bell's Theorem, which states (in my words):

No physical theory of local Hidden Variables can ever reproduce all of the predictions of Quantum Mechanics.

Well; I just did ''this'' (the classical experiment). As for disproving Bell's theorem, Doc: You know where my website is!

5. I think if we stick to the above issues, and ignore negative probabilities, "perverse" applications of Malus, etc., we can be more constructive and it will be a lot easier to see the true elements of the debate.
-DrC

I am happy to accept your position. As I see it, you move us to the central issue: No physical theory of local Hidden Variables can ever reproduce all of the predictions of Quantum Mechanics.

However, since that subject might be a hi-jacking of this thread, should you open a new one?

Pending that for now, let me say:

John Bell was not happy with this representation of his theorem (as I understand it).

Neither am I happy with his theorem! Convinced that Nature is local, I just go further and reject it.

To quote a leading authority in the field: ''To those for whom nonlocality is anathema, Bell's Theorem finally spells the death of the hidden variables program. But not for Bell. None of the no-hidden-variables theorems persuaded him that hidden variables were impossible.(Mermin, Rev. Mod. Phys. 65 page 814, 1993) [emphasis added].

Cheers, wm

 

[wm]

HEY JESSE, have you MISQUOTED wm. (Which certainly doesn't help his case at all.) Look at this:

Originally Posted by wm
2. The way a local realist would naturally want to apply it: entanglement is not real and there is a specific polarization for the 2 photons that is simply the same value;

That just doesn't look like anything that wm would say. Can you find the quote; or FIX IT please? Thanks.

 

[JesseM]

I don't get the point of the "yoking". Assuming we are using entangled photons, you get a violation of the Bell Inequality.

But with the yoking, you're making the entanglement irrelevant so it can be viewed as a classical problem, and of course wm is trying to show that violations of the Bell inequality can occur in classical situations. Think of it this way--the experiment would work exactly the same way if you took Alice out of the picture entirely and just used single photons, and you had the source itself randomly pick three angles A, B, and C, and then used some other event like an atomic decay to decide whether to have its polarizer be at the 0 degrees relative to the angle it had chosen or at 90 degrees relative to it. If you mark every trial where it picks 0 degrees as + and every trial where it picks 90 degrees as -, then the polarizer's recorded series of +'s and -'s are 100% guaranteed to be exactly the same as the series Alice would have recorded if you were using pairs of entangled photons and the source was yoked to Alice's choice of measurement angle (assuming you ignore the possibility of delays in the yoking which might allow occasional photons to be emitted between the time Alice changes her angle and the time the source updates its own angle in response). So the violation of Bell's inequality that wm derives would also occur if you were looking at correlations between the +'s and -'s recorded by the source (which were based on something like the decay of a radioactive atom which was not entangled with the photon emitted by the source) and the +'s and -'s recorded by Bob. Also, as noted above, you could get the exact same violation if you were assuming classical EM waves rather than photons, with Bob's computer choosing the probability to display a + on each trial based on the reduction in intensity of the light passing through his polarizing filter. But these are not genuine violations of Bell's theorem, because Bell's theorem only says the inequalities must be obeyed under certain conditions, and one of the conditions is the statistical independence of the properties of signals/particles emitted by the source and experimenters' choice of what to measure (not true in the case where the source is yoked to Alice's measurements via an ordinary classical signal), while another condition is that there be a spacelike separation between each pair of measurements (not true in the case of the radioactive decay and Bob's measurement, since the decay actually has a causal effect on Bob's measurement in terms of the source's polarizer angle depending on it).