A Does the Bell theorem assume reality?

[Demystifier]

A large portion of physicists thinks that Bell's theorem shows that reality does not exist. Another large portion of physicists thinks that reality is not an assumption of Bell's theorem, so that Bell's theorem just proves nonlocality, period. A third large portion of physicists thinks that both reality and locality are assumptions of Bell's inequalities, so that the Bell theorem proves that either reality or locality (or both) are wrong. So who is right?

I think the best answer to this question is given by Roderich Tumulka in http://de.arxiv.org/abs/1501.04168 . Among many other papers on this subject, the Tumulka's paper stands out by clearly distinguishing 4 different notions of "reality", which he calls (R1), (R2), (R3) and (R4), and analyzing each of them separately. He concludes that only (R4) is the assumption of Bell's theorem. But he also points out that (R4) is the mildest form of realism, that it is very hard to abandon it, and that he (Tumulka) takes it for granted, just as Bell did. By taking (R4) for granted, he concludes that Bell's theorem does not assume reality.

I absolutely agree that (R4) is the only reality assumption in the Bell theorem. I also agree that it is very hard to abandon it and hence that it is quite natural to take it for granted. Nevertheless, I do not think that it is absolutely impossible to abandon it and absolutely necessary to take it for granted. Hence, for the sake of completeness, in my "Solipsistic hidden variables" paper http://de.arxiv.org/abs/1112.2034 I explore the logically consistent (even if philosophically unappealing) possibility of an interpretation in which (R4) is explicitly abandoned, to better understand in what sense nonlocality could be removed (or at least reduced) by an explicit rejection of reality.

What I would like to see a discussion about, is whether others agree that (R4) is really the only assumption of reality that is relevant to the Bell theorem and whether it is reasonable to question the validity of that assumption.

 

[stevendaryl]

I absolutely agree that (R4) is the only reality assumption in the Bell theorem.

Well, the derivation of the Bell inequalities start with something like (R3).

 

[DrChinese]

The referenced paper offers some similar arguments to Norsen's line of reasoning on this subject. Essentially, that there is no realism assumption in Bell, and so you must conclude that nature is nonlocal if Bell Inequalities are violated.

On the other hand: I have said repeatedly that the realism assumption is present in Bell, it is simply not labeled as such. The realism requirement is the requirement that there is a counterfactual C in addition to A and B which can be measured. It is introduced after his (14). "Let c be a unit vector..." and works with Bell's (2).

The separability requirement is:
P(A,B|a,b,λ)=P(A|a,λ)P(B|b,λ)

Which also means:
P(A,C|a,c,λ)=P(A|a,λ)P(C|c,λ)
P(B,C|b,c,λ)=P(B|b,λ)P(C|c,λ)

But the last 2 are only good if you accept that there is (at least) an A, a B and a C - simultaneously. Which is Bell's realism. You just can't get to Bell's (15) without this assumption. It would have been better if this had been more explicit, but I guess he wrote what he thought made sense for his limited audience at the time.

 

[DarMM]

I agree with @stevendaryl , Bell is assuming (R3). This is the way Leifer presents the theorem in his review paper of $\psi$-ontology theorems for example. When proved in the ontological models framework $\lambda = \psi$ is explicitly included (Beltrametti–Bugajski model, ontological framework encoding of the Dirac-VonNeumann/"Textbook" interpretation).

I think abandoning (R4) is possible, Brukner and Zeilinger's Fundamental Randomness view, QBism and Rovelli's Relational QM do it for example, although they do it in different ways from each other and your own solipsistic hidden variables model.

Linking in with our discussion on the Fraichiger-Renner paper, the typical view of the Masanes version of the result is that if you abandon (R3) you have to abandon (R4) as well, i.e. if you have no hidden variables you have to be perspectival.

 

[DrChinese]

In the referenced paper, the (R3) requirement is:
There is some (“hidden”) variable λ that influences the outcome in a probabilistic way, as represented by the probability P(A, B|a, b, λ).

But it really includes these 3 to work out in Bell - this is usually ignored but to me it is the crux of the realism assumption:

P(A,B|a,b,λ)
P(A,C|a,c,λ)
P(B,C|b,c,λ)

We are assuming the existence of a counterfactual.

 

[atyy]

I think the best answer to this question is given by Roderich Tumulka in http://de.arxiv.org/abs/1501.04168 . Among many other papers on this subject, the Tumulka's paper stands out by clearly distinguishing 4 different notions of "reality", which he calls (R1), (R2), (R3) and (R4), and analyzing each of them separately. He concludes that only (R4) is the assumption of Bell's theorem. But he also points out that (R4) is the mildest form of realism, that it is very hard to abandon it, and that he (Tumulka) takes it for granted, just as Bell did. By taking (R4) for granted, he concludes that Bell's theorem does not assume reality.

I think Tumulka assumes R3. However, in R3 says that $\lambda$ is not necessarily a "hidden" variable - it may be an "unhidden" variable like the wave function.

 

[zonde]

What I would like to see a discussion about, is whether others agree that (R4) is really the only assumption of reality that is relevant to the Bell theorem and whether it is reasonable to question the validity of that assumption.

Yes, R4 is the only assumption of reality that is relevant to the Bell theorem.
R3 is not a statement about reality. Bell singles out QM predictions of perfect correlations. Adding locality to that QM prediction he arrives at predeterminism of measurement outcomes and from that he concludes that "more complete specification of the state" is possible. Well, you have to express this specification in some communicable form. What possible communicable form this specification could have that would not be convertible into some set of variables?

Speaking about validity of R4, it can be questioned within philosophical discussion, but it can not be questioned within philosophical framework of scientific approach. Of course we do not believe this assumption blindly. We make plenty of redundant information and then cross check that information for consistency. From another side there is no alternative to this belief that comes even close to usefulness of this belief.

 

[Demystifier]

I think Tumulka assumes R3. However, in R3 says that $\lambda$ is not necessarily a "hidden" variable - it may be an "unhidden" variable like the wave function.

Yes, but I think that nobody argues like this: "R3 is wrong, ergo QM is local." It would be nonsense, because saying that R3 is wrong would be saying that probability is not determined by the wave function.

 

[Demystifier]

Well, the derivation of the Bell inequalities start with something like (R3).

Yes, but see my reply to atyy above.

 

[Demystifier]

I agree with @stevendaryl , Bell is assuming (R3).

Yes, but see my reply to atyy above.

 

[stevendaryl]

Yes, but I think that nobody argues like this: "R3 is wrong, ergo QM is local." It would be nonsense, because saying that R3 is wrong would be saying that probability is not determined by the wave function.

So in other words, R3 is just TRUE, since there is a function that allows us to compute joint probabilities given an additional parameter (the wave function, for example).

The way that I've seen Bell's inequality derived follows along this outline

1. There is a probability distribution depending on the settings and the hidden variable $\lambda$: $P(A,B|a,b,\lambda)$.
2. (Factorizability): Assume that $P(A,B|a,b,\lambda) = P(A|a,b,\lambda) P(B|a,b,\lambda)$.
3. (Locality): Assume that $P(A|a,b,\lambda) = P(A|a,\lambda)$ and $P(B|a,b,\lambda) = P(B|b,\lambda)$
4. (Perfect correlations/anticorrelations): Perfect correlation/anti-correlation in the case where $a=b$ implies determinism.

Step 1 is just true (the example of $\lambda$ being the wave function shows that it's true). Step 3 is basically a definition of locality. Step 4 is provable. So the only controversial step is 2. Is that the assumption of realism?

 

[stevendaryl]

This might be obvious, but you can show that violations of Bell's inequality don't by themselves imply nonlocality.

Here's a toy Many-Worlds type model that is local and that has the same predictions as QM for the EPR experiment. Many-Worlds itself is not actually local, in my opinion, because it relies on a wave function that is a function on configuration space, while a local realistic model would have all functions defined on physical space.

Here's how the toy local explanation for EPR works:

1. Let's define a "history" to be a sequence of records of the form $(a,A,b,B)$ where $a$ and $b$ are Alice's and Bob's chosen detector orientations, and $A$ and $B$ are their measurement results (spin-up or spin-down).
2. Let's assume a "black box" device with the following property: You input a pair of orientations, $a$ and $b$, and the device outputs either "SAME" or "DIFFERENT". The probability of the box producing "SAME" is $sin^2(\frac{\theta}{2})$ where $\theta$ is the angle between $a$ and $b$. The probability of the box producing "DIFFERENT" is $cos^2(\frac{\theta}{2})$. Boxes are identical, in that if there are two such boxes given the same inputs, they will always produce the same output. (Note this sounds like it's nonlocal, but it can be implemented using a pseudo-random number generator, if all boxes use the same generator.)
3. Let's assume that every time Alice measures her particle's spin, some strange mechanism (God maybe) creates an exact duplicate of Alice, with all of her memories. One of the copies gets result "spin-up" and the other copy gets results "spin-down". These copies of Alice do not interact with each other in any way.
4. Similarly for Bob. When he makes a measurement, two Bobs are created that get opposite results.
5. Alices and Bobs with different histories cannot interact, either.
6. The state of Alice after she has made a measurement, but before she has found out what Bob's measurement is is described by a record $(h, a, A)$, where $h$ is a history, and $a$ is her last detector orientation, and $A$ is her last result. Similarly, Bob's state before they interact is described by a record $(h,b, B)$.
7. If the black box at the current time gives "SAME" for the pair $a,b$, then Alice in state $(h, a, A)$ can only interact with Bob in state $(h, b, B)$ if $A=B$. If the black box gives "DIFFERENT", then Alice can only interact with Bob if $A \neq B$.

So in this model, what happens on Alice's end is purely local: She picks an orientation, she splits in two, and each copy gets the opposite result. Similarly, what happens on Bob's end is purely local. But if one of the Alices tries to communicate with one of the Bobs, she is only able to communicate with one that has the correct statistics. So for each of the Alices and each of the Bobs, it will seem that there is only one Alice and one Bob, and their correlations are the same as predicted by QM.

 

[Demystifier]

So the only controversial step is 2. Is that the assumption of realism?

No, it's also a part of the assumption of locality. The assumption of reality (R4) is the zeroth assumption:
0. There are measurement outcomes A and B which exist separately.

But note that if one denies 0. that A and B exist separately, then one cannot talk about separate quantities $P(A|...)$ and $P(B|...)$, in which case the controversial step 2. is wrong. So even though your 2. is not an assumption of reality per se, it is a consequence of the assumptions of reality and locality.

Finally note that some interpretations do deny 0. For instance, the relational interpretation says that A and B can only exist together, not separately, which is why the quantites $P(A|...)$ and $P(B|...)$ do not make sense. I don't think that the relational interpretation makes sense (after all, someone can measure A without measuring B), but loosely speaking that's what the relational interpretation says.

 

[Demystifier]

So in this model, what happens on Alice's end is purely local: She picks an orientation, she splits in two

But after the splitting, the two copies don't live in the same universe. So there must exist some sort of a multiverse. The mere existence of a multiverse implies that we cannot have locality. Instead, we have to deal with some sort of multi-locality. Since there are two copies of Alice, Alice is not local but bi-local.

 

[DrChinese]

If the referenced paper - or any paper - seeks to demonstrate what the true assumptions of Bell are: it should be able to use that assumption to get to the Bell result. I don't believe that is possible with (R4). You need a counterfactual assumption of some kind.

 

[Demystifier]

The referenced paper offers some similar arguments to Norsen's line of reasoning on this subject. Essentially, that there is no realism assumption in Bell, and so you must conclude that nature is nonlocal if Bell Inequalities are violated.

That's not exactly what Tumulka says. He says that there is an assumption of realism, but that this assumption is only (R4), which is so mild that it can be taken for granted.

But the last 2 are only good if you accept that there is (at least) an A, a B and a C - simultaneously. Which is Bell's realism. You just can't get to Bell's (15) without this assumption. It would have been better if this had been more explicit, but I guess he wrote what he thought made sense for his limited audience at the time.

But this assumption of realism is the same as Tumulka's (R4), and I think that he is quite explicit that this kind of realism is assumed. To take an assumption for granted does not mean that this assumption is not used. It means that this assumption is not questioned.

 

[Demystifier]

If the referenced paper - or any paper - seeks to demonstrate what the true assumptions of Bell are: it should be able to use that assumption to get to the Bell result. I don't believe that is possible with (R4). You need a counterfactual assumption of some kind.

I believe that my analysis #13 of the @stevendaryl 's analysis #11 explains how (R4) gives the Bell result.

 

[DrChinese]

That's not exactly what Tumulka says. He says that there is an assumption of realism, but that this assumption is only (R4), which is so mild that it can be taken for granted. ...

But this assumption of realism is the same as Tumulka's (R4), and I think that he is quite explicit that this kind of realism is assumed. To take an assumption for granted does not mean that this assumption is not assumed. It means that this assumption is not questioned.

"(R4): Every experiment has an unambiguous outcome, and records and memories of that outcome agree with what the outcome was at the space-time location of the experiment."

I don't think this is realism at all. There is nothing counterfactual assumed, and is in fact the opposite of what is needed. You need an A, B and C - not all of which can be measured simultaneously - to get the Bell result.

 

[zonde]

You need an A, B and C - not all of which can be measured simultaneously - to get the Bell result.

Local model can produce predictions A, B and C without assuming anything counterfactual. You just perform three calculations with the same $\lambda$ and different measurement angles.

 

[Demystifier]

"(R4): Every experiment has an unambiguous outcome, and records and memories of that outcome agree with what the outcome was at the space-time location of the experiment."

I don't think this is realism at all. There is nothing counterfactual assumed, and is in fact the opposite of what is needed. You need an A, B and C - not all of which can be measured simultaneously - to get the Bell result.

So which of the Tumulka's (R1)-(R4) would you call realism actually needed to get the Bell result? Or do you think that we need a fifth notion of reality (R5)? If so, can you give a clear definition of (R5)?

 

[stevendaryl]

No, it's also a part of the assumption of locality. The assumption of reality (R4) is the zeroth assumption:
0. There are measurement outcomes A and B which exist separately.

I disagree. Factorizability is not related to locality. I was thinking that you were the one who mentioned the concept of ... sorry, I can't remember the name of the principle. It was something along the lines of assuming that if two measurement results are correlated, then there must be some common causal influence on both. If you take into account the common causal influence, then probabilities will factor. It has nothing to do with locality.

For example, if the probability that identical twins are both basketball players is correlated, that means for a random pair of twins A and B, the probability that they both play basketball is unequal to the product of the probability that each separately plays basketball. So $P(A,B) \neq P(A) P(B)$. But the assumption of common causal influence implies that there exists some number of causal factors influencing basketball playing, maybe genetics, maybe parents, maybe what school they go to, etc., such that if you control for all those factors, the probabilities will factor. Letting $\lambda$ be the vector of all such factors, we would have: $P(A,B | \lambda) = P(A|\lambda) P(B|\lambda)$

That is not at all a locality assumption.

 

[stevendaryl]

But after the splitting, the two copies don't live in the same universe.

Maybe philosophically, they don't. But the whole model can be implemented within a single universe. You can simulate the whole thing using computers in our universe.

 

[Demystifier]

You can simulate the whole thing using computers in our universe.

That's irrelevant. You can simulate Bohmian mechanics by a classical local computer in a single universe, but it doesn't make Bohmian mechanics local.

 

[stevendaryl]

That's irrelevant. You can simulate Bohmian mechanics by a classical local computer in a single universe, but it doesn't make Bohmian mechanics local.

I don't agree. You can't simulate Bohmian mechanics without violating locality---without the simulation of Alice's measurements relying on facts about Bob's measurements.

 

[stevendaryl]

I disagree. Factorizability is not related to locality. I was thinking that you were the one who mentioned the concept of ... sorry, I can't remember the name of the principle. It was something along the lines of assuming that if two measurement results are correlated, then there must be some common causal influence on both. If you take into account the common causal influence, then probabilities will factor. It has nothing to do with locality.

For example, if the probability that identical twins are both basketball players is correlated, that means for a random pair of twins A and B, the probability that they both play basketball is unequal to the product of the probability that each separately plays basketball. So $P(A,B) \neq P(A) P(B)$. But the assumption of common causal influence implies that there exists some number of causal factors influencing basketball playing, maybe genetics, maybe parents, maybe what school they go to, etc., such that if you control for all those factors, the probabilities will factor. Letting $\lambda$ be the vector of all such factors, we would have: $P(A,B | \lambda) = P(A|\lambda) P(B|\lambda)$

That is not at all a locality assumption.

I Googled it and the common cause assumption is "Reichenbach Common Cause Principle".

 

[Demystifier]

I disagree...
Letting $\lambda$ be the vector of all such factors, we would have: $P(A,B | \lambda) = P(A|\lambda) P(B|\lambda)$

That is not at all a locality assumption.

I think you are right. In view of what you just said, the assumption 2. is just a law of probability. Nothing controversial at all.

 

[Demystifier]

I don't agree. You can't simulate Bohmian mechanics without violating locality---without the simulation of Alice's measurements relying on facts about Bob's measurements.

Well, it depends on what one means by "simulate". I meant, I can solve Bohmian equations of motion on the computer.

 

[stevendaryl]

I think you are right. In view of what you just said, the assumption 2. is just a law of probability. Nothing controversial at all.

Well, if all 4 steps are uncontroversial, then the conclusion must be uncontroversial, right? But people do deny that QM is nonlocal.

I think that some people do deny Reichenbach's principle.

 

[Demystifier]

Well, if all 4 steps are uncontroversial, then the conclusion must be uncontroversial, right? But people do deny that QM is nonlocal.

I think that some people do deny Reichenbach's principle.

Yes, some people deny it.

 

[stevendaryl]

Well, it depends on what one means by "simulate".

Sure. But the sense in which you can have a local simulation of Bohmian mechanics is very different. It might be local in the real world, but in the simulated world, it's nonlocal because the part of the state having to do with Alice's measurement is influenced by the part of the state having to do with Bob's measurement.

In the simulation I'm talking about, the simulation of Alice can be done on a separate computer than the simulation of Bob. The one computer does not make use of any information from the other computer. Except when Alice sends a message to Bob, and in that case, the information for how to process the message can be encoded in the message itself.

So I would say that it's all pretty damn local. What it violates is the assumption that measurement results have a single outcome.

 

[Demystifier]

So I would say that it's all pretty damn local. What it violates is the assumption that measurement results have a single outcome.

With multiple outcomes, I think "multi-local" is a much better word to describe it than simply "local".

 

[DrChinese]

Local model can produce predictions A, B and C without assuming anything counterfactual. You just perform three calculations with the same $\lambda$ and different measurement angles.

No way. You cannot use formulae involving AB, BC and AC in an equation together unless you assume all 3 have simultaneous validity. You certainly can only measure one pair (AB, BC or C) at a time. You need something counterfactual as an assumption.

The entire idea of the referenced paper is to discuss possible assumptions which go into Bell. One thing I am surprised is not more often mentioned is rotational invariance.

 

[DrChinese]

So which of the Tumulka's (R1)-(R4) would you call realism actually needed to get the Bell result? Or do you think that we need a fifth notion of reality (R5)? If so, can you give a clear definition of (R5)?

I guess my posts were intended to say that Bell did not assume precisely any of (R1), (R2) or (R3). However, you could probably assume (R1) or (R2) and get to it. For (R3), you need to extend it to include simultaneous validity of the 3 expressions:

P(A,B|a,b,λ)
P(A,C|a,c,λ)
P(B,C|b,c,λ)

I realize that the author is trying to locate the "minimal" assumption(s) required to get the Bell result. Others have attempted this too. But I think it never hurts to identify what Bell actually did. Clearly, his inequality (15) depends on the relationships between 3 outcomes that do NOT vary according to which pair is being measured. Any way you wish to express that could be considered (R5). And that step is first introduced immediately after (14), and not before.

 

[DrChinese]

@Demystifier: And if I recall correctly: the Bohmian interpretation does NOT presume counterfactual definiteness. I seem to recall us discussing that previously.

 

[Demystifier]

@Demystifier: And if I recall correctly: the Bohmian interpretation does NOT presume counterfactual definiteness. I seem to recall us discussing that previously.

That is correct. But note that BM is non-local whenever the wave function is entangled, even in cases for which non-locality is not necessary true in the sense of a Bell-like theorem. An example would be the Bell state with spin measurements in z-direction only.

 

[wle]

• (Factorizability): Assume that $P(A,B|a,b,\lambda) = P(A|a,b,\lambda) P(B|a,b,\lambda)$.

The derivations I've seen use Bayes' theorem here, which implies $$P(A, B| a, b, \lambda) = P(A | a, b, \lambda) P(B | A, a, b, \lambda) \,.$$ The locality assumption used after this translates to $P(A | a, b, \lambda) = P(A | a, \lambda)$ and $P(B | A, a, b, \lambda) = P(B | b, \lambda)$.

 

[zonde]

No way. You cannot use formulae involving AB, BC and AC in an equation together unless you assume all 3 have simultaneous validity. You certainly can only measure one pair (AB, BC or C) at a time. You need something counterfactual as an assumption.

Let's say you are in laboratory and you have prepared a state. Your assistant after some time will perform one of the three measurements by his own choice (or using PRNG "choice"). According to protocol you have to produce predictions for all three measurements before any measurement is performed. Let's say you produce them. All three predictions should be unambiguous and valid otherwise your model is not a valid theory. All three predictions should unambiguously exist at the same time - factually. There is nothing counterfactual in such situation.

 

[stevendaryl]

Let's say you are in laboratory and you have prepared a state. Your assistant after some time will perform one of the three measurements by his own choice (or using PRNG "choice"). According to protocol you have to produce predictions for all three measurements before any measurement is performed. Let's say you produce them. All three predictions should be unambiguous and valid otherwise your model is not a valid theory. All three predictions should unambiguously exist at the same time - factually. There is nothing counterfactual in such situation.

In discussions of EPR and Bell's theorem and so forth, you have a number of "trials", where a trial consists of the creation of a particle/antiparticle pair. For each pair of particles, you can only measure spins along a pair of orientations. Alice measures the electron spin along axis $a$ and Bob measures the positron spin along axis $b$. Counterfactuality comes into play if you ask: "What result would Bob have gotten if he had measured his particle along axis $c$ instead?" It is not possible to measure spins along three different orientations.

You can certainly choose to measure along orientations $a, b$ for some trials, and along orientations $a, c$ for other trials. But if we're talking about the possibility of hidden variables affecting the outcomes, we don't know how the variables change from one trial to the next.

 

[zonde]

In discussions of EPR and Bell's theorem and so forth, you have a number of "trials", where a trial consists of the creation of a particle/antiparticle pair. For each pair of particles, you can only measure spins along a pair of orientations. Alice measures the electron spin along axis $a$ and Bob measures the positron spin along axis $b$. Counterfactuality comes into play if you ask: "What result would Bob have gotten if he had measured his particle along axis $c$ instead?" It is not possible to measure spins along three different orientations.

Yes, I understand that. But Bell inequality is calculated for certain hypothetical models. So instead of asking: "What result would Bob have gotten if he had measured his particle along axis $c$ instead?", we ask: "What prediction the model will produce if instead of measurement axis $b$ we will use measurement axis $c$ instead?"
We speak about the map not the territory so to say.

 

[stevendaryl]

The derivations I've seen use Bayes' theorem here, which implies $$P(A, B| a, b, \lambda) = P(A | a, b, \lambda) P(B | A, a, b, \lambda) \,.$$

I think we're talking about slightly different things. The factoring that you are talking about is always true, for any $\lambda$. Reichenbach's Principle claims that there always must be some choice of $\lambda$ such that the factoring looks like this:

$$P(A, B| a, b, \lambda) = P(A | a, b, \lambda) P(B | a, b, \lambda)$$

(no $A$ in the second term).

The distinction is illustrated by my example with twins playing basketball. I have some way to randomly pick a pair of identical twins out of the population. Let $P(A)$ be the probability that the first twin plays basketball. Let $P(B)$ be the probability that the second twin plays basketball. Let $P(A,B)$ be the probability that they both play basketball. Most likely, twins are alike in their basketball playing abilities, or at least more alike than any two random people. So $P(A,B) \neq P(A) P(B)$.

Without understanding anything at all about basketball playing ability, you can, using pure logic, write:

$P(A, B) = P(A) P(B | A)$

That's basically true by definition of conditional probability. So that kind of factoring isn't actually telling us anything about the root causes of basketball ability. On the other hand, let's suppose that we identify a bunch of factors that might come into play. Let $\lambda_1$ be genetics, let $\lambda_2$ be the schools they attend, let $\lambda_3$ be some characterization of their homelife (do they have siblings, do they have two parents, are the parents rich, etc.). Then some collection of such parameters would be a causal explanation of the correlation if:

$P(A, B | \overrightarrow{\lambda}) = P(A | \overrightarrow{\lambda}) P(B | \overrightarrow{\lambda})$

Reichenbach's principle, as formalized as factorability of probabilities, says that if you knew enough about the causes of basketball-playing ability, then it should no longer be necessary to know whether twin $A$ plays basketball to accurately predict whether twin $B$ plays basketball.

(Actually, I realize this example doesn't quite work, because the mere fact that one twin plays basketball might influence the other twin. We can account for this by saying that at some point, before the twins ever play basketball for the first time, we separate the twins and separately try them out on different sports, and decide independently which sport they like the best.)

It's only AFTER you have factored probability distributions by coming up with a complete set of causal factors is it the case that you can apply locality considerations. If you haven't already factored it, then imposing locality is a mistake. We can show this with the basketball players.

We take two twins and take them to distant locations and measure their basketball playing ability. Then using pure logic, we write:

$P(A,B) = P(A) P(B | A)$

If we then say: the basketball tests are far apart, so one twin playing basketball cannot influence the other twin's abilities. So we assume that

$P(B | A) = P(B)$

But that assumption is FALSE. Even though $A$ and $B$ are far apart, that doesn't mean that they are uncorrelated, and therefore it doesn't mean that $P(B | A) = P(B)$

 

[stevendaryl]

Yes, I understand that. But Bell inequality is calculated for certain hypothetical models. So instead of asking: "What result would Bob have gotten if he had measured his particle along axis $c$ instead?", we ask: "What prediction the model will produce if instead of measurement axis $b$ we will use measurement axis $c$ instead?"
We speak about the map not the territory so to say.

Right. It's the models that support or don't support counterfactual claims.

 

[zonde]

It's the models that support or don't support counterfactual claims.

Models always support counterfactual claims, because models don't care if you calculate from it prediction or retrodiction. Models are supposed to be independent from actual reality.

 

[stevendaryl]

Models always support counterfactual claims, because models don't care if you calculate from it prediction or retrodiction. Models are supposed to be independent from actual reality.

No, models don't always support counterfactual claims. A stochastic model, for instance, doesn't have a definite answer to the question: What measurement result would I have gotten, if I had measured this rather than that.

 

[stevendaryl]

No, models don't always support counterfactual claims. A stochastic model, for instance, doesn't have a definite answer to the question: What measurement result would I have gotten, if I had measured this rather than that.

To have definite answers to counterfactual questions, the model needs to be deterministic but not superdeterministic.

 

[zonde]

No, models don't always support counterfactual claims. A stochastic model, for instance, doesn't have a definite answer to the question: What measurement result would I have gotten, if I had measured this rather than that.

Any model would have some scope of things it can say. The point is that if the model have something to say about the future then then it's exactly the same thing it would say about the past or possible past. Just because you know measurement result does not mean that you can get exact result as retrodiction out of stochastic model. There is no difference between prediction and retrodiction even for stochastic model.

 

[stevendaryl]

Any model would have some scope of things it can say. The point is that if the model have something to say about the future then then it's exactly the same thing it would say about the past or possible past. Just because you know measurement result does not mean that you can get exact result as retrodiction out of stochastic model. There is no difference between prediction and retrodiction even for stochastic model.

I guess I agree, but I don't know how that relates to Dr. Chinese' point about counterfactuality.

 

[DrChinese]

EPR said in an entangled system: it would be possible to predict with certainty any possible observable on one (Alice), by measuring the other (Bob) of the pair. That implied that Bob's result was in fact predetermined. The question then is whether all possible observables of Bob were simultaneously predetermined. Only in a world in which there was a subjective realism - Bob's reality is shaped by the choice of measurement basis by Alice - would that NOT be true. They claimed, as an article of faith, such subjective realism - observer shape reality - would be an unreasonable position. So even in 1935, the question on the table was whether there was counterfactual realism.

Bell converted their general conjecture to a specific mathematical argument. He did it by assuming - quietly - that there were in fact predetermined values for non-commuting observables. He showed that these predetermined values could not be consistent with quantum mechanical predictions. And his "quiet" introduction of that conjecture occurs right after Bell's (14). It is NOT, as some believe, in Bell's (2). (2) codifies the argument that the results of Alice and Bob are independent (separable). You could call that the assumption of locality.

You cannot get the Bell result without some kind of counterfactual assumption. You can call it realism or objective reality or whatever, but there is an assumption. And it occurs after (14) - and not before. Whether or not it is (R1) or one of the others, I can't personally say. But I think it is a mistake to consider these independent of Bell. Because the question becomes whether you can assume something OTHER than what Bell assumed, and still get the Bell result.

 

[Demystifier]

You cannot get the Bell result without some kind of counterfactual assumption. You can call it realism or objective reality or whatever, but there is an assumption. And it occurs after (14) - and not before. Whether or not it is (R1) or one of the others, I can't personally say.

Lorentzo Maccone explains it in http://de.arxiv.org/abs/1212.5214 as follows:
"Let us define “counterfactual-definite” [14, 15] a theory whose experiments uncover properties that are preexisting. In other words, in a counterfactual-definite theory it is meaningful to assign a property to a system (e.g. the position of an electron) independently of whether the measurement of such property is carried out. [Sometime this counterfactual definiteness property is also called “realism”, but it is best to avoid such philosophically laden term to avoid misconceptions.]"

Maccone then shows that this counterfactual definiteness is one of the assumptions of the Bell theorem, showing that some kind of "reality" is an assumption of Bell theorem. But Maccone is not sophisticated enough to say whether this reality corresponds to (R1), (R2), (R3) or (R4). As far as as I am aware, the Tumulka's paper is the only paper that clearly distinguishes those 4 versions of reality.

Finally, let me add that (R4) is the same as the Axiom 1 in my "Bohmian mechanics for instrumentalists" linked in my signature below. Unfortunately I didn't cite Tumulka because at the time of writing I was not aware of that paper. I plan to change it in a revised version of the paper.

 

[DrChinese]

Lorentzo Maccone explains it in http://de.arxiv.org/abs/1212.5214 as follows:
"Let us define “counterfactual-definite” [14, 15] a theory whose experiments uncover properties that are preexisting. In other words, in a counterfactual-definite theory it is meaningful to assign a property to a system (e.g. the position of an electron) independently of whether the measurement of such property is carried out. [Sometime this counterfactual definiteness property is also called “realism”, but it is best to avoid such philosophically laden term to avoid misconceptions.]"

Maccone then shows that this counterfactual definiteness is one of the assumptions of the Bell theorem, showing that some kind of "reality" is an assumption of Bell theorem. But Maccone is not sophisticated enough to say whether this reality corresponds to (R1), (R2), (R3) or (R4). As far as as I am aware, the Tumulka's paper is the only paper that clearly distinguishes those 4 versions of reality.

Tumulka:
"(R4): Every experiment has an unambiguous outcome, and records and memories of that outcome agree with what the outcome was at the space-time location of the experiment."

That doesn't sound counterfactual in any sense. I read it as: The outcome of a performed experiment is what we see and record it as. I certainly don't believe Tumulka can recreate the Bell result from that, without one of the others.

 

[DarMM]

Yes, but I think that nobody argues like this: "R3 is wrong, ergo QM is local." It would be nonsense, because saying that R3 is wrong would be saying that probability is not determined by the wave function.

As far as I know, not quite, though it is a subtle thing.

If $\lambda = \psi$ then you have the Beltrametti–Bugajski model, which Leifer describes as "the orthodox interpretation of quantum theory into the language of ontological models". This is saying that $\psi$ is an ontic state of the theory and then Bell's theorem just amounts to a proof that the wavefunction if real (and there isn't multiple worlds) is a nonlocal object.

The way out of this is to reject R3 not by saying that probability is not determined by the wavefunction, but by saying the wavefunction isn't an element of reality/ontic, so the purported proof that it is nonlocal doesn't mean the world is physically nonlocal. This is what AntiRealist/Participatory realist interpretations do (e.g. Copenhagen).