The real puzzle

[minkwe]

Many Bell proponents continue to believe that it is impossible for QM correlations to be reproduced by local realistic theories. In other words, they believe Bell's theorem is true.

In a recent paper, Richard Gill (a Bell theorem believer) asks a relevant question:

"In the case of equal settings, how can it be that the outcomes are equal and opposite, if they were not predetermined at the source?"

I would add to that a related question:

"How can it be that two particles are "entangled" at the stations if there were not "entangled" at the source?"
OR
"Why do we need a single source (as opposed to two separate sources) to begin with if we do not believe the source imparts shared hidden properties to the particle pairs?"

Aside: Please discuss all Bet's and experiments in their relevant threads.

[Heinera]

"In the case of equal settings, how can it be that the outcomes are equal and opposite, if they were not predetermined at the source?"

In orthodox QM, one would say that with equal detector settings, the wave function for two entangled particles can only collapse to (1, -1) or (-1, 1) and not (1, 1) or (-1, -1).

"How can it be that two particles are "entangled" at the stations if there were not "entangled" at the source?"

But they are entangled at the source as well.

"Why do we need a single source (as opposed to two separate sources) to begin with if we do not believe the source imparts shared hidden properties to the particle pairs?"

We need a single source to create entangled particle pairs. Entanglement is a property of the pair (or rather, its wave function), not of each particle separately.

[gill1109]

I am not a Bell theorem believer.

Michel asks:


(1) "How can it be that two particles are "entangled" at the stations if they were not "entangled" at the source?"

and (he said "or", but I think it is another question)

(2) "Why do we need a single source (as opposed to two separate sources) to begin with if we do not believe the source imparts shared hidden properties to the particle pairs?"

My answers are, (1), no they can't be, unless of course in the mean time they each came into interaction with two other entangled particles, and their entanglement got swapped; and (2) I agree, we don't need a single source if ...

[minkwe]

In orthodox QM, one would say that with equal detector settings, the wave function for two entangled particles can only collapse to (1, -1) or (-1, 1) and not (1, 1) or (-1, -1).

I know that you "would say that", I'm ( or rather Richard's) asking how that can be without that property having been predetermined from the source?

"How can it be that two particles are "entangled" at the stations if there were not "entangled" at the source?"

But they are entangled at the source as well.

"Why do we need a single source (as opposed to two separate sources) to begin with if we do not believe the source imparts shared hidden properties to the particle pairs?"

We need a single source to create entangled particle pairs. Entanglement is a property of the pair (or rather, its wave function), not of each particle separately.

Again confirming the premise of the question that it must have been predetermined from the source. We need the source in order to impart (predetermine) the specific property on both particles.

I've read many papers arguing that entanglement is evidence that the world is irreducibly stochastic/probabilistic. I don't take them seriously for these reasons, it is a self defeating argument.

[minkwe]

I am not a Bell theorem believer.

You don't believe Bell's theorem: "No physical theory of local hidden variables can ever reproduce all of the predictions of quantum mechanics."?
I assumed from your papers and our discussions that you weren't a Bell denier. Unless you changed your mind in the interim?

Michel asks:

(1) "How can it be that two particles are "entangled" at the stations if they were not "entangled" at the source?"

Actually it was your question in your paper, I just borrowed it.

and (he said "or", but I think it is another question)
(2) "Why do we need a single source (as opposed to two separate sources) to begin with if we do not believe the source imparts shared hidden properties to the particle pairs?"

My answers are, (1), no they can't be, unless of course in the mean time they each came into interaction with two other entangled particles, and their entanglement got swapped; and (2) I agree, we don't need a single source if ...

So again, you are confirming the fact that hidden properties are being transfered to the pair as they interact with the source or other particles. Therefore it can not be that the world is irreducibly stochastic/probabilistic, what we see is predetermined by past interactions of the particles.

Based on your answers, I would say according to Einstein, you are a local realist too. You are in good company, it is the only logically consistent position.

[gill1109]

I am not a Bell theorem believer.

You don't believe Bell's theorem: "No physical theory of local hidden variables can ever reproduce all of the predictions of quantum mechanics."?
I assumed from your papers and our discussions that you weren't a Bell denier. Unless you changed your mind in the interim?

Michel asks:

(1) "How can it be that two particles are "entangled" at the stations if they were not "entangled" at the source?"

Actually it was your question in your paper, I just borrowed it.

and (he said "or", but I think it is another question)
(2) "Why do we need a single source (as opposed to two separate sources) to begin with if we do not believe the source imparts shared hidden properties to the particle pairs?"

My answers are, (1), no they can't be, unless of course in the mean time they each came into interaction with two other entangled particles, and their entanglement got swapped; and (2) I agree, we don't need a single source if ...

So again, you are confirming the fact that hidden properties are being transfered to the pair as they interact with the source or other particles. Therefore it can not be that the world is irreducibly stochastic/probabilistic, what we see is predetermined by past interactions of the particles.

Based on your answers, I would say according to Einstein, you are a local realist too. You are in good company, it is the only logically consistent position.

Regarding the last point, it depends whether the information being exchanged and transferred is quantum or classical. I find the word "hidden" not very useful.

Regarding Bell's theorem: there is some mathematics in Bell's work, and there are metaphysical positions which one might take, on consideration *together* (a) of the maths (logic, arithmetic, elementary probability theory and statistics) and (b) of the experimental results as known so far. In "Bertlmann's socks" Bell identified four options, and in correspondence with Santos he admitted that there exists a fifth, which I later entitled "Bell's fifth position". Since there still has not been a successful loophole free experiment, we still cannot rationally exclude any of the five possible conclusions! It's a matter of taste! Bell agrees! There is no theorem giving the answer!

So as you state the theorem, I think it is wrong. If five options are logically open, it cannot be a theorem that one particular one of them is true! Who started calling the damn thing a theorem, anyway? Not John Bell, for sure.

I am presently inclined to reject "realism" in order to keep "locality". It leads to an interpretation of QM (Belavkin's "eventum mechanics") which resolves the measurement problem. Physicists don't like it at present, because it has an arrow of time. In fact the arriw of time is nothing else than the collapse of entanglement. However this notion, that quantum decoherence *is* time's arrow is coming back into fashion, right now. There is now a relativistically invariant version of Belavkin's theory, thanks to Daniel Beddingham's recent work on the CSL (continuous spontaneous localuzation) model, which through the mathematical technique of "dilation" (jokingly called: an appeal to the church of the larger Hilbert space) can be seen as an eventum mechanics model. The metaphysical interpretation is different, the maths is identical. Just a question of whether you prefer to drop "locality" or "realism" (in a technical sense .. the word really causes more confusion than help, nowadays).

[Heinera]

"Why do we need a single source (as opposed to two separate sources) to begin with if we do not believe the source imparts shared hidden properties to the particle pairs?"

We need a single source to create entangled particle pairs. Entanglement is a property of the pair (or rather, its wave function), not of each particle separately.

Again confirming the premise of the question that it must have been predetermined from the source. We need the source in order to impart (predetermine) the specific property on both particles.

I've read many papers arguing that entanglement is evidence that the world is irreducibly stochastic/probabilistic. I don't take them seriously for these reasons, it is a self defeating argument.

Well, something is certainly predetermined from the source: It's called the initial state, aka the wave function of the two particles. If you want to call that a hidden variable, it's fine with me.

But a local hidden variable theory requires more than a hidden variable; it also requires locality.

With QM, when we apply the Born rule to the wave function of an entangled pair, we get a joint probability function P[x; a, b]for the outcomes x {(1, -1), (-1, 1), (-1, -1), (1, 1)} that depends on both detector settings a and b. The function looks like this:

P[(1, -1); a, b] = (1 + cos(a.b))/4

P[(-1, 1); a, b] = (1 + cos(a.b))/4

P[(-1, -1); a, b] = (1 - cos(a.b))/4

P[(1, 1); a, b] = (1 - cos(a.b))/4

What Bell's theorem says is that it is impossible to split the function P into three functions Source, DetectorA, and DetectorB so that the only allowed communication is one way from source to the detectors, and Source does not depend on a or b, DetectorA does not depend on b, and DetectorB does not depend on a.

This is a purely mathematical theorem, like the theorem that says it is impossible to trisect an arbitrary angle using only a compass and straightedge. Bell's theorem says that a particular function cannot be decomposed in a particular way. And that's why we say QM is non-local. Interpret that any way you like.

[Joy Christian]

With QM, when we apply the Born rule to the wave function of an entangled pair, we get a joint probability function P[x; a, b]for the outcomes x {(1, -1), (-1, 1), (-1, -1), (1, 1)} that depends on both detector settings a and b. The function looks like this:

P[(1, -1); a, b] = (1 + cos(a.b))/4

P[(-1, 1); a, b] = (1 + cos(a.b))/4

P[(-1, -1); a, b] = (1 - cos(a.b))/4

P[(1, 1); a, b] = (1 - cos(a.b))/4

What Bell's theorem says is that it is impossible to split the function P into three functions Source, DetectorA, and DetectorB so that the only allowed communication is one way from source to the detectors, and Source does not depend on a or b, DetectorA does not depend on b, and DetectorB does not depend on a.

This is a purely mathematical theorem, like the theorem that says it is impossible to trisect an arbitrary angle using only a compass and straightedge. Bell's theorem says that a particular function cannot be decomposed in a particular way. And that's why we say QM is non-local. Interpret that any way you like.

Then Bell's theorem is simply wrong, because precisely such a local split of Source, DetectorA, and DetectorB has been demonstrated in equations (A.9.33) to (A.9.40) of this paper: http://libertesphilosophica.info/blog/w ... hapter.pdf.

[Heinera]

Then Bell's theorem is simply wrong, because precisely such a local split of Source, DetectorA, and DetectorB has been demonstrated in equations (A.9.33) to (A.9.40) of this paper: http://libertesphilosophica.info/blog/w ... hapter.pdf.

Just like Pierre Wantzel's proof of the impossibility of trisecting an arbitrary angle has been "disproved" by just about any enterprising high school student since 1837. The theorem is still rock solid.

[FrediFizzx]

Then Bell's theorem is simply wrong, because precisely such a local split of Source, DetectorA, and DetectorB has been demonstrated in equations (A.9.33) to (A.9.40) of this paper: http://libertesphilosophica.info/blog/w ... hapter.pdf.

Just like Pierre Wantzel's proof of the impossibility of trisecting an arbitrary angle has been "disproved" by just about any enterprising high school student since 1837. The theorem is still rock solid.

What you have written is complete garbage.

I have always suspected that you are a [comment edited out] and not really as mathematically qualified as you claim to be. If you were, then you can easily see that what I have presented in the paper is exactly what all Bell-believers like you claim to be impossible. Note well that others, such as Michel and Fred for example, do not have any difficulty seeing this. Perhaps it is time for you to come clean and admit that you are dogmatically committed to something that is manifestly and demonstrably false.

[minkwe]

Regarding the last point, it depends whether the information being exchanged and transferred is quantum or classical.

Sounds circular to me, aka distinction without a difference.

In "Bertlmann's socks" Bell identified four options, and in correspondence with Santos he admitted that there exists a fifth, which I later entitled "Bell's fifth position". Since there still has not been a successful loophole free experiment, we still cannot rationally exclude any of the five possible conclusions! It's a matter of taste! Bell agrees!

The loopholes business is IMHO completely misguided and irrelevant. (loopholes reveal a hidden assumption, More on this later). But we can ask a silly question like "how can a square circle be triangular?" and use that to take all kinds of metaphysical positions about how geometry could be false, or circles do not exist etc.. Or, we could simply recognize the obvious absurdity in the original question, and accept that there is no conundrum to begin with

To arrive at Bell's theorem, Bell used:
- Inequalities derived for a specific system
- QM predictions for a completely different system

Then concludes that because the two do not match, the first system does not exist. Then we all line up like sheep to find a way out of the "non-conundrum" without asking the simple questions.

(a) of the maths (logic, arithmetic, elementary probability theory and statistics) and (b) of the experimental results as known so far.

But we've established already that the maths is for one system and the experimental results are for a completely different system (hidden assumption alert here. More on this later.)

So as you state the theorem, I think it is wrong. If five options are logically open, it cannot be a theorem that one particular one of them is true! Who started calling the damn thing a theorem, anyway? Not John Bell, for sure.

Let us write down the assumptions which lead to Bell's theorem:

1. Local Realist Variable theories MUST obey Bell-inequalities
2. The terms from QM mean the same as the terms in Bell's inequalities
3. The terms from Experiments mean the same as those in Bell's inequalities
Conclusion: Since QM and Experiments agree with each other and violate Bell-inequalities, QM can not be a Local Realist Variable theory.

As we've seen assumptions (1), (2) and (3) are all false. So why do you still think there is a conundrum?

I am presently inclined to reject "realism" in order to keep "locality".

That is the real puzzle. Why the haste to reject "realism" or "locality" when you have enough reason to simply reject any of (1,2,3) above?

Let me state the question in another way. Why are you so convinced that the QM expectation values represent population mean values which exist prior to measurement, even though QM says nothing about what exists prior to measurement? This is the hidden assumption I'm talking about.

And what about loopholes. If a loophole-ridden experiment agrees with QM, then QM must have the loophole too. For if QM is expected to agree with the experiment both when it has loopholes and when it does not have loopholes, then the loophole is irrelevant. So a loophole is relevant, then QM has it too. Therefore the question of whether it is possible to reproduce QM correlations in a Local Realist Variable theory can not be sensitive to the loophole argument.

[minkwe]

But a local hidden variable theory requires more than a hidden variable; it also requires locality.

Local hidden variables do not force you to measure all your relative frequencies on a single set of particle pairs.

P[(1, -1); a, b] = (1 + cos(a.b))/4

P[(-1, 1); a, b] = (1 + cos(a.b))/4

P[(-1, -1); a, b] = (1 - cos(a.b))/4

P[(1, 1); a, b] = (1 - cos(a.b))/4

It is impossible for the same particle to turn up as +1 and at the same time as -1 at a detector. The QM probabilities above when considered together are therefore necessarily for separate disjoint particles. Unlike the terms in the inequalities. The elephant in the room is the assumption implicit in your argument, which is the same one I pointed out to Richard: It is the assumption that the above probabilities simultaneously apply to a single population. The "conundrum" disappears as soon as you recognize that they are not. By definition, they are incompatible. And this is not even a quantum issue. Non-commuting observables also exist in classical physics.

What Bell's theorem says is that it is impossible to split the function P into three functions Source, DetectorA, and DetectorB so that the only allowed communication is one way from source to the detectors, and Source does not depend on a or b, DetectorA does not depend on b, and DetectorB does not depend on a.

But we already know that this is false.

Bell's theorem says that a particular function cannot be decomposed in a particular way. And that's why we say QM is non-local. Interpret that any way you like.

Of course by making the false assumption that local functions must be decomposed in a particular way. Is Malus' Law local in your opinion?

[Heinera]

But a local hidden variable theory requires more than a hidden variable; it also requires locality.

Local hidden variables do not force you to measure all your relative frequencies on a single set of particle pairs.

Correct, obviously not.

P[(1, -1); a, b] = (1 + cos(a.b))/4

P[(-1, 1); a, b] = (1 + cos(a.b))/4

P[(-1, -1); a, b] = (1 - cos(a.b))/4

P[(1, 1); a, b] = (1 - cos(a.b))/4

It is impossible for the same particle to turn up as +1 and at the same time as -1 at a detector.

Obviously correct.

The QM probabilities above when considered together are therefore necessarily for separate disjoint particles.

Obviously correct.

Unlike the terms in the inequalities.

What inequalities? There are no inequalities in my post.

The elephant in the room is the assumption implicit in your argument, which is the same one I pointed out to Richard: It is the assumption that the above probabilities simultaneously apply to a single population.

No, that's not an assumption. A pair of particles is measured only once (one particle at Alice and one at Bob), and each measurement will have a random outcome according to the probabilities I wrote down.

What Bell's theorem says is that it is impossible to split the function P into three functions Source, DetectorA, and DetectorB so that the only allowed communication is one way from source to the detectors, and Source does not depend on a or b, DetectorA does not depend on b, and DetectorB does not depend on a.

But we already know that this is false.

Do we, now? I have yet to see you produce such a decomposition. Your last attempt had three possible outcomes (-1, 0, 1) for each measurement. The theorem restricts the outcomes to two possibilities.

Is Malus' Law local in your opinion?

Obviously yes. Malus' law has nothing to do with entanglement. I think you should get your hands on a textbook on quantum optics.

[minkwe]

What inequalities? There are no inequalities in my post.

Playing games again Heinera, you know what inequalities.

The elephant in the room is the assumption implicit in your argument, which is the same one I pointed out to Richard: It is the assumption that the above probabilities simultaneously apply to a single population.

No, that's not an assumption. A pair of particles is measured only once (one particle at Alice and one at Bob), and each measurement will have a random outcome according to the probabilities I wrote down.

A single particle gives you +1 or -1, how can you obtain a probability from one measurement. You see, I too can play word games.

Do we, now? I have yet to see you produce such a decomposition. Your last attempt had three possible outcomes (-1, 0, 1) for each measurement. The theorem restricts the outcomes to two possibilities.

And in your opinion QM says (-1, 0, 1) is not allowed? On what basis do you reject my (-1, 0, 1) attempt. (hidden assumption alert)

Is Malus' Law local in your opinion?

Obviously yes. Malus' law has nothing to do with entanglement. I think you should get your hands on a textbook on quantum optics.

Who said anything about entanglement. I think you should get your hands on a textbook on reading comprehension.

[Heinera]

What inequalities? There are no inequalities in my post.

Playing games again Heinera, you know what inequalities.

If you are thinking about the inequalities in Richard's proof, I suggest you forget about them, since they were not to your liking. There must be at least a thousand different proofs of Bell's theorem published on the web. Google and find one that you can relate to.

No, that's not an assumption. A pair of particles is measured only once (one particle at Alice and one at Bob), and each measurement will have a random outcome according to the probabilities I wrote down.

A single particle gives you +1 or -1, how can you obtain a probability from one measurement. You see, I too can play word games.

Good to see you are joking here. One obviously doesn't obtain probabilities from one measurement, just like a single flip of a coin can't tell you if it it's fair or not. In fact, we don't obtain the QM probabilities from the outcomes; they are given by the theory. If a long run of an experiment would show different empirical probabilities, we would be in trouble, since it would be a big blow to the theory. Fortunately, that hasn't happened in 90 years.

Do we, now? I have yet to see you produce such a decomposition. Your last attempt had three possible outcomes (-1, 0, 1) for each measurement. The theorem restricts the outcomes to two possibilities.

And in your opinion QM says (-1, 0, 1) is not allowed? On what basis do you reject my (-1, 0, 1) attempt. (hidden assumption alert)

It is not allowed by the definitions of the theorem. Furthermore, an experiment has been performed on trapped ions where the outcomes where only (-1, 1), i.e., 100% detection.

Is Malus' Law local in your opinion?

Obviously yes. Malus' law has nothing to do with entanglement. I think you should get your hands on a textbook on quantum optics.

Who said anything about entanglement.

Then what was the relevance of the question?

[minkwe]

If you are thinking about the inequalities in Richard's proof, I suggest you forget about them, since they were not to your liking. There must be at least a thousand different proofs of Bell's theorem published on the web. Google and find one that you can relate to.

Each one just as flawed as the next.

And in your opinion QM says (-1, 0, 1) is not allowed? On what basis do you reject my (-1, 0, 1) attempt. (hidden assumption alert)

It is not allowed by the definitions of the theorem. Furthermore, an experiment has been performed on trapped ions where the outcomes where only (-1, 1), i.e., 100% detection.

First of all, the claims of 100% detection are false, read the paper. Secondly, QM does not say every particle must be observed. In fact QM does not say anything about a single particle.

Then what was the relevance of the question?

A polarized beam of particles is incident on a polarizer. The intensity measured is cos^2(theta). Is every particle detected? Can you represent that function as separate functions of source, particle and detector? Now add a second polarizer after it, are all particles detected, can you represent the intensity as separate functions of source particle and detectors?

[gill1109]

Bell did not prove a theorem. He was a physicist, not a mathematician.

The attack by many so-called Bell-deniers on the so-called Bell theorem (e.g. the "theorem" quoted by Michel Fodje) is a straw-man attack.

Bell analyzed a thought experiment and drew up a list of five possible alternative metaphysical conclusions one might draw about quantum mechanics and other physical theories. At the time, all five were logically open, i.e., it could be taken as a matter of taste which one you prefer. Truly metaphysics, not physics.

The experiment desired and described by Bell has still not been done. The five metaphysical options are still open.

Some experimentalists have said that the experiment desired by Bell (or effectively equivalent to it - it will be a 2x2x3 experiment, not a 2x2x2 Bell type experiment) seems achievable within the next five years.

The experiment, if successful, will not show that a mathematical inequality is violated. It will result in experimentally observed correlations which could only occur extremely rarely in any event-based local realistic computer imitation of the experiment.

If successful, the experiment would tend to rule out two of Bell's five options, still leaving three metaphysical options completely open ... as they always will be.

To date, computer simulations by de Raedt, Fodje, Christian and others have always been of the wrong experiment. Not of the experiment desired, specified, described, by Bell.

(The trapped ion experiment with 100% detection unfortunately has the two ions rather close to one another relative to the time it takes to get the two measurements done and the separation of the two sources of measurement settings).

[Xray]

I am not a Bell theorem believer.

…
[A] Since there still has not been a successful loophole free experiment, we still cannot rationally exclude any of the five possible conclusions! It's a matter of taste! Bell agrees! There is no theorem giving the answer!

[B] So as you state the theorem, I think it is wrong. If five options are logically open, it cannot be a theorem that one particular one of them is true! Who started calling the damn thing a theorem, anyway? Not John Bell, for sure.

…

Answer to [A]. There IS a THEORY that gives the answer; see viewtopic.php?f=6&t=49 Seems likely that several theorems might be lurking there.

Answer to [B]. According to Clauser in 2002: "Bell's [so-called] theorem" was first called such in CHSH (1969). [.] = my insertion.

[gill1109]

Answer to (B). According to Clauser in 2002: "Bell's theorem" was first called such in CHSH (1969).

Excellent! Bell (1980) disowns Bell's theorem. There is not one and one only possible inference. There are at least four possible inferences (standpoints, positions), all four at the time still open:

Bell, 1980, Bertlmann's Socks and the Nature of Reality
http://cds.cern.ch/record/142461/files/198009299.pdf

And later correspondence of Bell with Santos even added a fifth option:

Gill, 2003: Time, Finite Statistics, and Bell's Fifth Position
http://arxiv.org/abs/quant-ph/0301059

One can even say that Bell (1980) not only disowns the theorem, but also all Bell-CHSH experiments done to date, and all computer simulations of all Bell-CHSH experiments done to date. Read his paper, carefully!

We are making progress! We already agreed that mathematical inequalities cannot be violated, now we see that Bell's theorem is not Bell's theorem. And finally we see that all so-called Bell experiments to date, and all simulations of these past experiments, are irrelevant, too.

Of course Caroline Thompson and many others have been saying this for years:
http://freespace.virgin.net/ch.thompson1/Papers/The%20Record/TheRecord.htm
Following a line of enquiry established by Pearle (1970):
Philip M. Pearle (1970), "Hidden-Variable Example Based upon Data Rejection", Phys. Rev. D 2 (8): 1418–25

[gill1109]

Bell (1980) disowns Bell's theorem.

I should have said, Bell (1980) disowns the so-called "Bell's theorem".