Quantum mechanics is not weird, unless presented as such

[stevendaryl]

You've certainly misunderstood something here. The object of study in Bell's theorem is the joint probability $P(ab \mid xy)$ (according to some candidate theory) that Alice and Bob obtain results indexed by variables $a$ and $b$ given that they decide to do measurements indexed by variables $x$ and $y$. This is not restrictive. In particular, the joint probability distribution should be given by the Born rule according to quantum mechanics, i.e., have the form $$P(ab \mid xy) = \mathrm{Tr} \bigl[ (M_{a \mid x} \otimes N_{b \mid y}) \rho_{\mathrm{AB}} \bigr]$$ where in general the variables $x$ and $y$ are associated with POVMs $\mathcal{M}_{x} = \{M_{a \mid x}\}_{a}$ and $\mathcal{N}_{y} = \{N_{b \mid y}\}_{b}$. This is perfectly well defined even if the POVMs $\mathcal{M}_{x}$ for different $x$ and $\mathcal{N}_{y}$ for different $y$ are incompatible.

Well, the assumption that Bell makes that I think rubi is objecting to is factorizability:

$P(ab \mid xy) = \sum_\lambda P(\lambda) P(a\mid \lambda x) P(b\mid \lambda y)$

 

[rubi]

You've certainly misunderstood something here. The object of study in Bell's theorem is the joint probability $P(ab \mid xy)$ (according to some candidate theory) that Alice and Bob obtain results indexed by variables $a$ and $b$ given that they decide to do measurements indexed by variables $x$ and $y$. This is not restrictive. In particular, the joint probability distribution should be given by the Born rule according to quantum mechanics, i.e., have the form $$P(ab \mid xy) = \mathrm{Tr} \bigl[ (M_{a \mid x} \otimes N_{b \mid y}) \rho_{\mathrm{AB}} \bigr]$$ where in general the variables $x$ and $y$ are associated with POVMs $\mathcal{M}_{x} = \{M_{a \mid x}\}_{a}$ and $\mathcal{N}_{b} = \{N_{b \mid y}\}_{b}$. This is perfectly well defined even if the POVMs for different $x$ and $y$ are incompatible.

It is you who has misunderstood something. Alice's and Bob's observables commute and thus a joint distribution exists for them. However, Alice's observables $A_a$ don't commute among each other and neither do Bob's. It is completely uncontroversial that non-commuting observables can't be represented on a joint probability space. The probabilities won't add up to $1$ in general. (Also, using POVMs is completely unnecessary here.)

Edit: To put it differently: Bell assumes that $A_a$ and $B_b$ are random variables on a probability space $(\Lambda,\Sigma,\mu)$. Then you can take random vectors like $X=(A_1, A_2, B_1, B_2)$ and get joint probability distributions $P_X(A) =\mu(X^{-1}(A))$. The fact that the $A_a$ and $B_b$ are random variables on one space entails this already.

 

[wle]

However, Alice's observables $A_a$ don't commute among each other and neither do Bob's. It is completely uncontroversial that non-commuting observables can't be represented on a joint probability space.

Bell's theorem does not depend on an assumption here that is different from quantum mechanics. Like I said, Bell's theorem only assumes a priori that it is meaningful to talk about the conditional probabilities $P(ab \mid xy)$, according to some theory, of obtaining different results depending on different possible measurements. This in itself is not in conflict with quantum mechanics, like I said in my previous post. Bell does not assume, a priori, that there is a joint underlying probability distribution $P(a_{1}, a_{2}, \dotsc, b_{1}, b_{2}, \dotsc)$. In the end, it turns out that for any model satisfying the locality constraint that Bell arrives at (which stevendaryl posted) you can always construct a joint probability distribution for all the measurement outcomes, but this is a corollary of Bell's definition, not an additional assumption.

 

[rubi]

Bell's theorem does not depend on an assumption here that is different from quantum mechanics. Like I said, Bell's theorem only assumes a priori that it is meaningful to talk about the conditional probabilities $P(ab \mid xy)$, according to some theory, of obtaining different results depending on the choices of measurements. This is perfectly consistent with quantum mechanics, like I said in my previous post. Bell does not assume, a priori, that there is a joint underlying probability distribution $P(a_{1}, a_{2}, \dotsc, b_{1}, b_{2}, \dotsc)$. In the end, it turns out that for any model satisfying the locality constraint that Bell arrives at (which stevendaryl posted) you can always construct a joint probability distribution for all the measurement outcomes, but this is a corollary of Bell's definition, not an additional assumption.

Repeating it doesn't make it true.
Bell clearly assumes that the variables $A_a$, $B_b$ are random variables on one probability space (and thus joint probabilities exist). Only then can you write down Bell's factorization condition. Quantum mechanics clearly says that no joint probability distribution for all these variables exists. (QM is also not relevant for the proof of Bell's inequality.)
It feels like we're going in circles.

Do you deny that $X_x:\Lambda\rightarrow\{-1,1\}$ are random variables on a probability space $(\Lambda,\Sigma,\rho(\lambda)\mathrm d\lambda)$? I don't see how you can seriously deny that and if you do, then I don't know what else I can say. I don't agree.

 

[wle]

Repeating it doesn't make it true.
Bell clearly assumes that the variables $A_a$, $B_b$ are random variables on one probability space (and thus joint probabilities exist).

I've seen more than one version of the derivation of Bell's theorem even by Bell, and they don't simply assume the "random variables on one probability space" that you refer to. The closest I've seen to this is the functions $A(\vec{a}, \lambda)$ and $B(\vec{b}, \lambda)$ appearing in Bell's original 1964 paper and similar derivations, but even there: 1) these are deterministic mappings, not random variables, and 2) assuming locality, Bell inferred that these functions should exist, via the EPR argument, from the fact that quantum physics predicts perfectly correlated and anticorrelated results for certain measurement choices. He did not simply assume that they should exist a priori.

 

[rubi]

Repeating things you read on the internet doesn't make them true.

Your style of argumentation is really annoying. Can you please stop treating me like an idiot who just repeats things from the internet? I obtained my information from books and papers and I have worked hard to understand it. I'm not an amateur.

I've seen more than one version of the derivation of Bell's theorem even by Bell, and they don't simply assume the "random variables on one probability space" that you refer to. The closest I've seen to this is the functions $A(\vec{a}, \lambda)$ and $B(\vec{b}, \lambda)$ appearing in Bell's original 1964 paper and similar derivations, but even there: 1) these are deterministic mappings, not random variables

The maps $\lambda\mapsto A(a,\lambda)$ are clearly random variables. They map from one probability space to a measurable space. This makes them random variables by definition.

2) assuming locality, Bell inferred that these functions should exist, via the EPR argument, from the fact that quantum physics predicts perfectly correlated and anticorrelated results for certain measurement choices. He did not simply assume that they should exist a priori.

Locality is the assumption that $A_a$ does not depend on $b$ and vice versa. Locality does not entail that these variables must be random variables on the same probability space. This is an extra assumption.

I don't have any more time for this, since apparently, we don't even agree on the very basics of probability theory.

 

[wle]

Repeating it doesn't make it true.

Repeating things you read on the internet doesn't make them true.

Your style of argumentation is really annoying. Can you please stop treating me like an idiot who just repeats things from the internet? I obtained my information from books and papers and I have worked hard to understand it. I'm not an amateur.

Has it occurred to you to maybe do me the same courtesy?

Locality does not entail that these variables must be random variables on the same probability space. This is an extra assumption.

No, like I said, it is inferred from the EPR argument and the fact that quantum physics predicts perfect correlations. And this doesn't even matter since, if you find Bell's original argument based on EPR too handwavy, Bell described much more careful formulations of his theorem in the 1970s and 1980s which clearly don't depend on this "same probability space" assumption you keep bringing up.

I don't have any more time for this, since apparently, we don't even agree on the very basics of probability theory.

No, apparently we disagree on how Bell's theorem is derived.

 

[rubi]

Has it occurred to you to maybe do me the same courtesy?

Well, you kept making one wrong statement after another, while accusing me of having a misunderstanding. Naturally, I become annoyed.

No, like I said, it is inferred from the EPR argument and the fact that quantum physics predicts perfect correlations.

You can't infer from the EPR argument that the hidden variables must be non-contextual. This is a non-trivial assumption.

And this doesn't even matter since, if you find Bell's original argument based on EPR too handwavy, Bell described much more careful formulations of his theorem in the 1970s and 1980s which clearly don't depend on this "same probability space" assumption you keep bringing up.

Sooner or later, you will have to introduce random variables if you want to calculate the correlations that appear in the inequality. These random variables are always defined on the same probability space (I keep bringing it up, because it is crucial). Nevertheless, there are of course other approaches and they need to be treated differently. Khrennikov treats them in his book, but I don't want to start another topic as long as we haven't settled on the case of Bell's inequality yet.

 

[atyy]

You can't infer from the EPR argument that the hidden variables must be non-contextual. This is a non-trivial assumption.

Do you mean contextual, as what is normally meant when people discuss the Kochen-Specker theorem? https://en.wikipedia.org/wiki/Kochen–Specker_theorem

 

[rubi]

Do you mean contextual, as what is normally meant when people discuss the Kochen-Specker theorem? https://en.wikipedia.org/wiki/Kochen–Specker_theorem

I use it like Khrennikov, who uses it as follows: A theory is non-contextual if all observables can be modeled as random variables on one probability space, independent of the experimental setup. Otherwise, it is contextual. Kochen-Specker define non-contextuality for theories defined in the Hilbert space framework. However, if such theories were non-contextual according to KS, then they would also be non-contextual according to Khrennikov, so Khrennikov's definition is in a sense more general, as it allows for theories that are not necessarily modeled in the Hilbert space framework. For example, if a theory would exceed the Tsirelson bound, it would have to be contextual, but couldn't be modeled in a Hilbert space. (However, in general, theories that don't exceed the Tsirelson bound don't need to have a Hilbert space model either. At least I'm not aware of a proof.)

 

[atyy]

I use it like Khrennikov, who uses it as follows: A theory is non-contextual if all observables can be modeled as random variables on one probability space, independent of the experimental setup. Otherwise, it is contextual. Kochen-Specker define non-contextuality for theories defined in the Hilbert space framework. However, if such theories were non-contextual according to KS, then they would also be non-contextual according to Khrennikov, so Khrennikov's definition is in a sense more general, as it allows for theories that are not necessarily modeled in the Hilbert space framework. For example, if a theory would exceed the Tsirelson bound, it would have to be contextual, but couldn't be modeled in a Hilbert space. (However, in general, theories that don't exceed the Tsirelson bound don't need to have a Hilbert space model either. At least I'm not aware of a proof.)

OK, but it doesn't mean that contextuality can save locality. Bell's theorem shows that no local hidden variable theory, contextual or not, is consistent with quantum theory (the usual outs are retrocausation, superdeterminism, many-worlds - but contextuality is not one of them). Khrennikov's out is essentially to redefine "local hidden variable" so that it includes something weird like his suggestion of p-adic probabilities, which may be fine, but it's totally unclear how that would solve the measurement problem. It's a bit similar to consistent histories, whose claim to be local is not in contradiction to Bell's theorem, because it is not a realistic theory.

 

[rubi]

OK, but it doesn't mean that contextuality can save locality. Bell's theorem shows that no local hidden variable theory, contextual or not, is consistent with quantum theory (the usual outs are retrocausation, superdeterminism, many-worlds - but contextuality is not one of them).

I don't agree here. One can clearly point to the place where the non-contextuality assumption is made in the proof of Bell's inequality. Bell's theorem rules out a large class of hidden variable theories. Maybe we shouldn't call contextual theories hidden variable theories (I'm not sure about that), but Bell's locality definition can only be applied to non-contextual theories. Locality has no clear probabilistic definition in the case of contextual theories.

Khrennikov's out is essentially to redefine "local hidden variable" so that it includes something weird like his suggestion of p-adic probabilities, which may be fine, but it's totally unclear how that would solve the measurement problem. It's a bit similar to consistent histories, whose claim to be local is not in contradiction to Bell's theorem, because it is not a realistic theory.

I don't find his p-adic probability theory appealing either and I'm also not advocating (contextual) hidden variables. However, he is right with the idea that there is no apriori reason for why we should be able to model all observables on the same probability space, independent of the experimental setting. It is important to note that this doesn't change the class of theories that are ruled out by Bell's theorem, so we aren't talking about loopholes. I'm saying that the probabilistic definition of locality can't be applied in the contextual case, so we have no probabilistic definition of locality for contextual theories, such as QM.

 

[stevendaryl]

I don't agree here. One can clearly point to the place where the non-contextuality assumption is made in the proof of Bell's inequality.

But is this just true by definition? Bell assumes that probability distributions for two distant measurements must factor, once you've taken into account all the relevant information that is common to the two measurements. The definition of "non-contextual" amounts to the same thing, doesn't it? So "non-contextual" is just another word for Bell's factorizability condition. It's not that contextuality provides an explanation for violation of Bell's inequalities.

 

[rubi]

But is this just true by definition? Bell assumes that probability distributions for two distant measurements must factor, once you've taken into account all the relevant information that is common to the two measurements. The definition of "non-contextual" amounts to the same thing, doesn't it? So "non-contextual" is just another word for Bell's factorizability condition. It's not that contextuality provides an explanation for violation of Bell's inequalities.

If you state it in terms of probability, you just shift the introduction of non-contextuality a bit. You will have to introduce random variables in order to compute the correlations that appear in the inequality. You make the non-contextuality assumption the moment you say that these random variables live on the same probability space.

 

[atyy]

I don't agree here. One can clearly point to the place where the non-contextuality assumption is made in the proof of Bell's inequality. Bell's theorem rules out a large class of hidden variable theories. Maybe we shouldn't call contextual theories hidden variable theories (I'm not sure about that), but Bell's locality definition can only be applied to non-contextual theories. Locality has no clear probabilistic definition in the case of contextual theories.

I guess what is puzzling to me about your statement is that one thinks of Bohmian mechanics as contextual and a nonlocal hidden variable theory, so it is consistent with both the requirements of the Kochen-Specker theorem and the Bell theorem.

 

[stevendaryl]

If you state it in terms of probability, you just shift the introduction of non-contextuality a bit. You will have to introduce random variables in order to compute the correlations that appear in the inequality. You make the non-contextuality assumption the moment you say that these random variables live on the same probability space.

Well, Bell's reasoning, or at least his reasoning as interpreted by me, goes like this:

You assume that when Alice/Bob makes a measurement, his/her result depends only on the setting of his/her detector and facts about the particle being measured. So at the time of the measurement, there is some kind of probability function for Alice $P_A(\lambda, \vec{a}, \alpha)$ that gives the probability of getting a result $+1$ given that the particle has property $\lambda$ and her detector setting is $\vec{a}$ and $\alpha$ represents other facts about her detector above and beyond the setting. Similarly, there is a function $P_B(\lambda, \vec{b}, \beta)$ for Bob. The assumption of locality is captured by the fact that Alice's result can't depend on anything at Bob's location, and vice-versa.

At this point, where is there an assumption of non-contextuality? It seems to me that it is simply saying that Alice's result depends only on local information. Where does this business about whether random variables "live on the same probability space" come into play?

 

[stevendaryl]

I guess what is puzzling to me about your statement is that one thinks of Bohmian mechanics as contextual and a nonlocal hidden variable theory, so it is consistent with both the requirements of the Kochen-Specker theorem and the Bell theorem.

I'm having trouble reconciling the definition rubi is using for "contextuality" with the definition you are using. The way I understand "contextual" as applied to Bohmian mechanics is that a measurement of spin using something like a Stern-Gerlach device doesn't reveal a pre-existing property of the particle being measured. Instead, the result--spin-up or spin-down--is the result of collaboration between the particle and the measuring device. The two together determine the spin, not the particle itself. The problem with spin measurements being "emergent" in this sense is it's hard (impossible?) to explain how Alice's results could be perfectly anti-correlated with Bob's if the results are emergent unless there is some nonlocal interaction guaranteeing the perfect anti-correlation. Which is no problem for Bohm, since it's explicitly nonlocal, but is a problem for local hidden variables.

Rubi's definition of "contextual" is not about whether measurement results are revealing pre-existing properties of the particle being measured, but is simply a statement about probability distributions governing random variables. I don't see the connection.

 

[rubi]

I guess what is puzzling to me about your statement is that one thinks of Bohmian mechanics as contextual and a nonlocal hidden variable theory, so it is consistent with both the requirements of the Kochen-Specker theorem and the Bell theorem.

Well, contextual theories are not necessarily local (assuming we had a definition of locality for contextual theories). However, you have encountered a nice subtlety here. The original EPR state happens to have a non-contextual model and you can't derive Bell's inequality for it. This part of QM can be defined on one probability space. However, this is not true for the Bohm state, so even in BM, spin needs to stay contextual. I'm not sure how the KS definition applies here, since we are not in the Hilbert space framework, but maybe my knowledge of BM is just too narrow.

 

[rubi]

Well, Bell's reasoning, or at least his reasoning as interpreted by me, goes like this:

You assume that when Alice/Bob makes a measurement, his/her result depends only on the setting of his/her detector and facts about the particle being measured. So at the time of the measurement, there is some kind of probability function for Alice $P_A(\lambda, \vec{a}, \alpha)$ that gives the probability of getting a result $+1$ given that the particle has property $\lambda$ and her detector setting is $\vec{a}$ and $\alpha$ represents other facts about her detector above and beyond the setting. Similarly, there is a function $P_B(\lambda, \vec{b}, \beta)$ for Bob. The assumption of locality is captured by the fact that Alice's result can't depend on anything at Bob's location, and vice-versa.

At this point, where is there an assumption of non-contextuality? It seems to me that it is simply saying that Alice's result depends only on local information. Where does this business about whether random variables "live on the same probability space" come into play?

In order to derive Bell's inequality, you need to introduce the correlations $C(a,b)$ (because the inequality is formulated in terms of them). Correlations are always correlations between random variables. So you can't get around introducing random variables in order to arrive at Bell's inequality. And when you introduce them, you will have to decide, which probability spaces they live on. A probability theory without random variables can't be related to experiment, just like a physical theory without observables has no connection to experiments.

 

[stevendaryl]

In order to derive Bell's inequality, you need to introduce the correlations $C(a,b)$ (because the inequality is formulated in terms of them). Correlations are always correlations between random variables. So you can't get around introducing random variables in order to arrive at Bell's inequality. And when you introduce them, you will have to decide, which probability spaces they live on. A probability theory without random variables can't be related to experiment, just like a physical theory without observables has no connection to experiments.

But there is only one random variable, $\lambda$, that is determined at the moment of pair creation. So Bell naturally only uses a single probability distribution, $P(\lambda)$ the probability of producing hidden variable $\lambda$. So I don't understand this business about multiple probability spaces.

 

[rubi]

But there is only one random variable, $\lambda$, that is determined at the moment of pair creation. So Bell naturally only uses a single probability distribution, $P(\lambda)$ the probability of producing hidden variable $\lambda$. So I don't understand this business about multiple probability spaces.

Well, you are arguing for a fully deterministic world. Such a world cannot be local, since this is excluded by Bell's theorem. However, it may also be the case that there is an intrinsic element or randomness to the world and the amount of randomness may depend on the experimental setup (i.e. the angles in a Bell test). In that case, we cannot apriori say that the probabilistic contexts must be comptabile (see my post #480).

 

[stevendaryl]

But there is only one random variable, $\lambda$, that is determined at the moment of pair creation. So Bell naturally only uses a single probability distribution, $P(\lambda)$ the probability of producing hidden variable $\lambda$. So I don't understand this business about multiple probability spaces.

Since Khrennikov references Pitowski, let me just summarize Pitowski's local hidden variable model that seems (at first blush) to contradict Bell's theorem.

Pitowski defines a class of functions $S(\hat{a})$ from directions in space (parameterized by a unit vector $\hat{a}$) into $\{ +1, -1 \}$. Then he assumes that such a function is associated with each particle of a correlated twin pair. The idea is that any measurement along an axis $\hat{a}$ will deterministically give the result $S(\hat{a})$ for one particle, and $-S(\hat{a})$ for the other. The function $S$ is constructed to give the same probabilities as Quantum mechanics. That is, take (almost) any direction $\hat{a}$. Then take a random second direction $\hat{b}$ such that $\hat{a} \cdot \hat{b} = cos(\theta)$. (There is a whole circle of possible directions to choose from.) Then the measure of the set of $\hat{b}$ such that $S(\hat{a}) = S(\hat{b})$ is $cos^2(\frac{\theta}{2})$.

How is this consistent with Bell's inequality? Well, one way to try to prove that there is no such function $S$ is by considering three different axes, $\hat{a}, \hat{b}, \hat{c}$. For example, we can pick three directions such that the angle between any two of them is 120 degrees. Then we ask, according to this hidden-variables model, what is the probability that $S(\hat{a}) = S(\hat{b}) = S(\hat{c})$? It turns out that there is no consistent way to assign a probability to such a triple coincidence. So what is Pitowki's way out? The function $S$ that he constructs is non-measurable. That is, the set of all triples $\hat{a}, \hat{b}, \hat{c}$ such that the angles between any two is 120 degrees and such that $S$ gives the same value for all three is a nonmeasurable set. On the other hand, by construction, the set of all pairs $\hat{a}, \hat{b}$ such that the angle between them is $\theta$ and $S$ gives the same result on each is measurable.

So this sounds very similar, to my mind, to Khrennikov's business about not having a single probability space. You can define a measure on pairs of directions, but not on triples, so counterfactual reasoning about measurements not performed can't be carried out--you can't compute such counterfactual probabilities.

The criticism that Pitowski's model generated, and I don't know whether this applies to Khrennikov, or not, is this:
Forget about measure theory, and just count: Generate 100 twin pairs, and count up how many times it's the case that three axes $\hat{a},\hat{b}, \hat{c}$ all have the same result, according to the model. Bell's inequality implies the impossibility to assign relative frequencies to all possible measurement results in keeping with the predictions of quantum mechanics. The "out" of having nonmeasurable sets doesn't do anything for you, because even if certain measures are undefined, the corresponding relative frequencies have to exist--it's just a matter of counting.

 

[stevendaryl]

Well, you are arguing for a fully deterministic world.

No, I wasn't arguing for that. What I assumed, as I said in an earlier post, was:

1. There is a single random variable, $\lambda$, associated with the twin pair. This is chosen according to some probability distribution, $P(\lambda)$.
2. When a particle reaches Alice, she has already picked a measurement setting $\vec{a}$, and her device is already in some state $\alpha$. Then she will get result $+1$ according to some probability $P_A(\vec{a}, \alpha, \lambda)$ that depends on $\vec{a}$, $\alpha$ and $\lambda$.
3. Similarly, when the other particle reaches Bob, he will get result $+1$ according to some probability $P_B(\vec{b}, \beta, \lambda)$ that depends on $\vec{b}$, $\beta$ and $\lambda$, where $\vec{b}$ is his detector's setting, and $\beta$ is other facts about his detector.

There is no assumption of determinism here. But there is no way to reproduce the perfect anti-correlations predicted by QM unless Alice's and Bob's results are deterministic functions of $\lambda, \vec{a}$ and $\vec{b}$, or unless there are nonlocal interactions (so that $P_A$ may depend on facts about Bob, or $P_B$ may depend on facts about Alice).

 

[rubi]

So this sounds very similar, to my mind, to Khrennikov's business about not having a single probability space. You can define a measure on pairs of directions, but not on triples, so counterfactual reasoning about measurements not performed can't be carried out--you can't compute such counterfactual probabilities.

Unfortunately, I'm not familiar enough with this stuff to comment on this. Do you say that contextual theories escape non-contextuality necessarily by invoking non-measurable sets?

The criticism that Pitowski's model generated, and I don't know whether this applies to Khrennikov, or not, is this:
Forget about measure theory, and just count: Generate 100 twin pairs, and count up how many times it's the case that three axes $\hat{a},\hat{b}, \hat{c}$ all have the same result, according to the model. Bell's inequality implies the impossibility to assign relative frequencies to all possible measurement results in keeping with the predictions of quantum mechanics. The "out" of having nonmeasurable sets doesn't do anything for you, because even if certain measures are undefined, the corresponding relative frequencies have to exist--it's just a matter of counting.

Khrennikov also treats frequency approaches in his book and isolates non-contextuality assumptions there as well. However, I haven't studied this deeply enough to know, whether I agree with him or not.

No, I wasn't arguing for that. What I assumed, as I said in an earlier post, was:

1. There is a single random variable, $\lambda$, associated with the twin pair. This is chosen according to some probability distribution, $P(\lambda)$.
2. When a particle reaches Alice, she has already picked a measurement setting $\vec{a}$, and her device is already in some state $\alpha$. Then she will get result $+1$ according to some probability $P_A(\vec{a}, \alpha, \lambda)$ that depends on $\vec{a}$, $\alpha$ and $\lambda$.
3. Similarly, when the other particle reaches Bob, he will get result $+1$ according to some probability $P_B(\vec{b}, \beta, \lambda)$ that depends on $\vec{b}$, $\beta$ and $\lambda$, where $\vec{b}$ is his detector's setting, and $\beta$ is other facts about his detector.

There is no assumption of determinism here. But there is no way to reproduce the perfect anti-correlations predicted by QM unless Alice's and Bob's results are deterministic functions of $\lambda, \vec{a}$ and $\vec{b}$, or unless there are nonlocal interactions (so that $P_A$ may depend on facts about Bob, or $P_B$ may depend on facts about Alice).

Well, you are assuming that there are hidden variables and everything is determined by them and we just lack information. This is excluded by Bell and I don't deny this. However, there may be an intrinsic amount of randomness that varies depending on the measurement context. I don't see how this is covered by your requirements.

 

[stevendaryl]

Well, you are assuming that there are hidden variables and everything is determined by them and we just lack information. This is excluded by Bell and I don't deny this. However, there may be an intrinsic amount of randomness that varies depending on the measurement context. I don't see how this is covered by your requirements.

Well, I'm allowing for Alice's result to be a probabilistic function of the relevant parameters $P_A(\vec{a}, \alpha, \lambda)$. Why isn't that good enough to allow randomness that varies depending on the measurement context? When Alice performs her measurement, the "context" is just facts about her device: $\alpha$ and $\vec{a}$, and facts about the particle being measured, $\lambda$. What I'm assuming, though, is that Bob's choice of device setting is not part of Alice's context. But why should it be?

 

[rubi]

Well, I'm allowing for Alice's result to be a probabilistic function of the relevant parameters $P_A(\vec{a}, \alpha, \lambda)$. Why isn't that good enough to allow randomness that varies depending on the measurement context? When Alice performs her measurement, the "context" is just facts about her device: $\alpha$ and $\vec{a}$, and facts about the particle being measured, $\lambda$. What I'm assuming, though, is that Bob's choice of device setting is not part of Alice's context. But why should it be?

You only allow for "lack of information" type randomness. You describe a world, in which everything can have a definite value and we just don't know it. In a contextual world, there can be genuine randomness, while some facts can nevertheless be pre-determined.

 

[Jilang]

You are going round and round in circles here! The only way to reconcile the results is that if spin is pre-determined in one direction, but not the other two. Whatever direction you the choose to measure it in will lead to opposite results for an entangled pair (due to the first part) but the correlations will be be larger than you would expect (due to the second part).

 

[stevendaryl]

You are going round and round in circles here! The only way to reconcile the results is that if spin is pre-determined in one direction, but not the other two.

Alice can decide at the last moment which direction to measure spin relative to. So the solution, that it is predetermined in the direction that is actually measured, only makes sense if either Alice's choice is known ahead of time, or if somehow, Alice's choice is itself predetermined (the superdeterminism option--which is actually taken seriously by t'Hooft).

 

[Jilang]

Alice has only two choices; The northern hemisphere or the southern one. The predetermined axis will be in either one or the other.

 

[stevendaryl]

Alice has only two choices; The northern hemisphere or the southern one. The predetermined axis will be in either one or the other.

What? In EPR-type experiments, Alice chooses a direction in space to measure spin relative to. So there's a continuum of choices.

 

[stevendaryl]

What? In EPR-type experiments, Alice chooses a direction in space to measure spin relative to. So there's a continuum of choices.

Bell discussed a toy model for EPR correlations in which the "hidden variable" was a hemisphere, and Alice measured spin-up if she chose an axis in that hemisphere, and spin-down if she chose an axis not in that hemisphere. That model does not replicate the predictions of QM.

 

[wle]

You can't infer from the EPR argument that the hidden variables must be non-contextual. This is a non-trivial assumption.

Like I said, if you don't find Bell's original argument based on EPR convincing then he gave much clearer explanations in later decades that don't depend on EPR.

Sooner or later, you will have to introduce random variables if you want to calculate the correlations that appear in the inequality. These random variables are always defined on the same probability space (I keep bringing it up, because it is crucial).

I have no idea where you're getting this from. The situation considered by Bell is that you have some theory capable of (among other things) predicting the probabilities $P(ab \mid xy)$ of different possible outcomes given different possible measurement choices. The correlation terms that appear in Bell-1964 and CHSH are defined directly in terms of these: $$E_{xy} = P(00 \mid xy) - P(01 \mid xy) - P(10 \mid xy) + P(11 \mid xy) \,.$$ Bell's locality condition is $$P(ab \mid xy) = \int \mathrm{d} \lambda \rho(\lambda) P(a \mid x; \lambda) P(b \mid y; \lambda) \,.$$ This is sometimes just taken as the definition of Bell locality, though Bell gives a derivation and argument for this factorisation. Either way, inserting this into the definition of the correlation terms gets you the factorised expression $$E_{xy} = \int \mathrm{d} \lambda \rho(\lambda) \bigl( P(0 \mid x; \lambda) - P(1 \mid x; \lambda) \bigr) \bigl( P(0 \mid y; \lambda) - P(1 \mid y; \lambda) \bigr) \,.$$ This is just mathematics and it's an easy exercise from here to show that, for instance, the CHSH inequality $$E_{00} + E_{01} + E_{10} - E_{11} \leq 2$$ must hold for any conditional probability distribution compatible with Bell's locality condition given above.

This is a brief sketch of how you derive an inequality like CHSH as I understand the subject. In which part of this are you claiming Khrennikov's "single probability space" assumption appears?

 

[rubi]

Like I said, if you don't find Bell's original argument based on EPR convincing then he gave much clearer explanations in later decades that don't depend on EPR.

I'm aware of these derivations and they all depend on the assumption nevertheless.

This is a brief sketch of how you derive an inequality like CHSH as I understand the subject. In which part of this are you claiming Khrennikov's "single probability space" assumption appears?

It appears here:

$$E_{xy} = P(00 \mid xy) - P(01 \mid xy) - P(10 \mid xy) + P(11 \mid xy) \,.$$

You have written this in an unsuspicious looking way, but really it is the integral of $A_x B_y$ over the single probability space on which the variables are defined. It only looks so unsuspicious, because the variables are $\{-1,1\}$ valued and the expectation value reduces to a sum.

 

[wle]

It appears here:

$$E_{xy} = P(00 \mid xy) − P(01 \mid xy)− P(10 \mid xy)+P(11 \mid xy) \,.$$

You have written this in an unsuspicious looking way, but really it is the integral of $A_x B_y$ over the single probability space on which the variables are defined. It only looks so unsuspicious, because the variables are $\{-1,1\}$ valued and the expectation value reduces to a sum.

How so? The definition of $E_{xy}$ I gave you is exactly that: a definition. It doesn't involve any assumption at all.

The expression isn't even specific to Bell locality. Case in point: in quantum mechanics $P(ab \mid xy) = \mathrm{Tr} \bigl[ (P_{a \mid x} \otimes Q_{b \mid y}) \rho \bigr]$ for (for instance) projection operators $P_{a \mid x}$ and $Q_{b \mid y}$, so, substituting this in the definition of $E_{xy}$ I gave, for quantum mechanics you would get $$\begin{eqnarray*}
E_{xy} &=& \mathrm{Tr} \bigl[ (P_{0 \mid x} \otimes Q_{0 \mid y}) \rho \bigr] - \mathrm{Tr} \bigl[ (P_{0 \mid x} \otimes Q_{1 \mid y}) \rho \bigr]
- \mathrm{Tr} \bigl[ (P_{1 \mid x} \otimes Q_{0 \mid y}) \rho \bigr] + \mathrm{Tr} \bigl[ (P_{1 \mid x} \otimes Q_{1 \mid y}) \rho \bigr] \\
&=& \mathrm{Tr} \bigl[ \bigl( (P_{0 \mid x} - P_{1 \mid x}) \otimes (Q_{0 \mid x} - Q_{1 \mid x}) \bigr) \rho \bigr] \\
&=& \mathrm{Tr} \bigl[ (A_{x} \otimes B_{y}) \rho \bigr]
\end{eqnarray*}$$ for Hermitian operators $A_{x} = P_{0 \mid x} - P_{1 \mid x}$ and $B_{y} = Q_{0 \mid y} - Q_{1 \mid y}$. As I'm sure you know, in this case, with the correct state and measurements, it's possible to attain $E_{00} + E_{01} + E_{10} - E_{11} = 2 \sqrt{2}$. So, clearly, the definition I gave cannot even implicitly be assuming anything that contradicts quantum mechanics.

 

[rubi]

How so? The definition of $E_{xy}$ I gave you is exactly that: a definition. It doesn't involve any assumption at all.

You can't just define a quantity and claim that it represents a correlation. A correlation is a well-defined notion from probability theory. In order to apply the general concept of a correlation to a specific case, you must clearly define what probability spaces you're working with and how your random variables are defined. You are trying to escape this duty by being sloppy about the math. If you are really interested in the exact assumptions that go into the theorem, you should strive for maximal mathematical rigor and clearly expose all mathematical details, even those that you feel are unnecessary. By applying this to Bell's assumptions, you will end up with your expression. If you do it for a contextual theory, you will get a different expression. That's just the way it is.

The expression isn't even specific to Bell locality. Case in point: in quantum mechanics $P(ab \mid xy) = \mathrm{Tr} \bigl[ (P_{a \mid x} \otimes Q_{b \mid y}) \rho \bigr]$ for (for instance) projection operators $P_{a \mid x}$ and $Q_{b \mid y}$, so, substituting this in the definition of $E_{xy}$ I gave, for quantum mechanics you would get $$\begin{eqnarray*}
E_{xy} &=& \mathrm{Tr} \bigl[ (P_{0 \mid x} \otimes Q_{0 \mid y}) \rho \bigr] - \mathrm{Tr} \bigl[ (P_{0 \mid x} \otimes Q_{1 \mid y}) \rho \bigr]
- \mathrm{Tr} \bigl[ (P_{1 \mid x} \otimes Q_{0 \mid y}) \rho \bigr] + \mathrm{Tr} \bigl[ (P_{1 \mid x} \otimes Q_{1 \mid y}) \rho \bigr] \\
&=& \mathrm{Tr} \bigl[ \bigl( (P_{0 \mid x} - P_{1 \mid x}) \otimes (Q_{0 \mid x} - Q_{1 \mid x}) \bigr) \rho \bigr] \\
&=& \mathrm{Tr} \bigl[ (A_{x} \otimes B_{y}) \rho \bigr]
\end{eqnarray*}$$ for Hermitian operators $A_{x} = P_{0 \mid x} - P_{1 \mid x}$ and $B_{y} = Q_{0 \mid y} - Q_{1 \mid y}$.

That is the quantum mechanical expression and it doesn't take the same form as the one you wrote, precisely, because we are dealing with a contextual theory.

As I'm sure you know, in this case, with the correct state and measurements, it's possible to attain $E_{00} + E_{01} + E_{10} - E_{11} = 2 \sqrt{2}$. So, clearly, the definition I gave cannot even implicitly be assuming anything that contradicts quantum mechanics.

Actually you have proven yourself wrong here, because with your expression, you can achieve at most $2$, rather than $2\sqrt{2}$, so the quantum mechanical expression for $E_{xy}$ must necessarily be different form the one you gave.