I Quantum mechanics is not weird, unless presented as such

[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.

 

[craigi]

Does quantum mechanics have to be weird?

It sells much better to the general public if it is presented that way, and there is a long history of proceeding that way.

But in fact it is an obstacle for everyone who wants to truly understand quantum mechanics, and to physics students who have to unlearn what they were told as laypersons.

I think that while the discussion about the Interpretations of Quantum Mechanics is still such a controversial subject then yes QM is weird even to the intiated. Sure the maths is there and is solid, but we can't yet properly bridge the gap between what the equations tell us and the correct way to fully apply it to reality. QM still stands out there on its own as the subject in physics which blows the mind of the undergrad and has never been truly reconcilled with our experience. Delving into the maths allows us to hold the technical knowledge to make predictions in laboratory conditions but we don't yet have a way to perceive the subject which stops it being weird. I think I'm in good company in believing that the study of the Interpretations of QM is highly significant in understanding the scope of the problem of Quantum Gravity.

The fact that on this forum, Interpretations of QM is still so often deferred to the realms of philosophy while we don't have solutions to the Preferred Basis Problem (and its ilk) and QG are testament to the fact that we all still find it weird.

From an educational perspective, I fully understand that for the purpose of motivation, QM is presented as mysterious from the outset with the Double Slit Experiment, but there is no route through the subject which can avoid the question of how the quantum world gives rise to our everyday experience, and we just don't have all the information to explain it. Personally, I can't buy the arguments that, for any given interpretation, all that remains to be done is "dotting the i's and crossing the t's", because each of those come from a presumption that the originating interpretation is correct, which only has subjective merit.

In my experience physicists are naturally depth first learners as opposed to breadth first learners and QM is taught depth first to avoid the complexity of the Interpretations of QM but the bright physicist naturally generates questions on how to interpret the subject and these questions should be addressed even without any definitive answer.

There are still questions to be answered and research to be performed, to explain how the microscopic world and macroscropic world co-exist. I firmly believe that until we have the answers to these questions then we should be mindful that we while we can make any particular interpretation work, with some unknowns, we cannot presume any interpretation to be correct and that is the source of the weirdness.

 

[wle]

You can't just define a quantity and claim that it represents a correlation.

I can define anything I want, since names don't have any intrinsic value in themselves. Maybe you and Khrennikov like to reserve the word "correlation" for something different than what's used in some simple Bell inequalities like CHSH. If that's the case then good for you, but that doesn't say anything about Bell's theorem.

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.

Your reply doesn't even make any sense. I took the definition that I wrote for $E_{xy}$ and that you quoted and I substituted in the Born rule to get the quantum mechanical expression for $E_{xy}$.

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.

Huh? Given only the definition $E_{xy} = P(00 \mid xy) - P(01 \mid xy) - P(10 \mid xy) + P(11 \mid xy)$, the algebraic bound on the CHSH expression is 4. Only Bell-local models are limited to 2.

 

[rubi]

I can define anything I want, since names don't have any intrinsic value in themselves. Maybe you and Khrennikov like to reserve the word "correlation" for something different than what's used in some simple Bell inequalities like CHSH. If that's the case then good for you, but that doesn't say anything about Bell's theorem.

Well, you can define whatever you want, but I'm telling you what it means: It is the correlation that you get by assuming that all random variables live on a single probability space. It's not me and Khrennikov, but rather the whole mathematics and physics community that defines correlations the way I told you and you will find it in every single book on the topic. I'm telling you that your definition is just a special case of the general definition. Again, if you want to understand subtleties, you have to be rigorous about everything. Your presentation of the proof would not be acceptable to a probability theorist. The first thing he'd ask you is: "What probability spaces are you working with?"

Your reply doesn't even make any sense. I took the definition that I wrote for $E_{xy}$ and that you quoted and I substituted in the Born rule to get the quantum mechanical expression for $E_{xy}$.

Woops, I misread. I somehow thought you had already included the locality condition and performed the integral. The combination of your $E_{xy}$ and the locality condition implies a specific probability space. You can only split it up that way, because the measures happen to be product measures. (You can also have a locality condition in the contextual case! See Khrennikovs paper.) However, you can always split up mathematical expressions into two parts. That doesn't mean that they refer to one thing.

Huh? Given only the definition $E_{xy} = P(00 \mid xy) - P(01 \mid xy) - P(10 \mid xy) + P(11 \mid xy)$, the algebraic bound on the CHSH expression is 4. Only Bell-local models are limited to 2.

Yes, again, I had the complete definition of $E_{xy}$ in mind. The point is that the quantum mechanical expression cannot coincide with Bell's.

 

[atyy]

I can define anything I want, since names don't have any intrinsic value in themselves. Maybe you and Khrennikov like to reserve the word "correlation" for something different than what's used in some simple Bell inequalities like CHSH. If that's the case then good for you, but that doesn't say anything about Bell's theorem.

Your reply doesn't even make any sense. I took the definition that I wrote for $E_{xy}$ and that you quoted and I substituted in the Born rule to get the quantum mechanical expression for $E_{xy}$.

Huh? Given only the definition $E_{xy} = P(00 \mid xy) - P(01 \mid xy) - P(10 \mid xy) + P(11 \mid xy)$, the algebraic bound on the CHSH expression is 4. Only Bell-local models are limited to 2.

Labelling the outcomes as $+1$ and $-1$, Wikipedia gives $E(a,b) = \int{\underline{A}(a,\lambda)\underline{B}(b,\lambda})\rho(\lambda)d\lambda$, which is why rubi says it is a correlation or an expectation.

https://en.wikipedia.org/wiki/CHSH_inequality

However, like you, I don't see how the contextuality assumption enters, since the quantity can be directly computed in quantum mechanics.

 

[rubi]

Let me make it clear by stating the assumptions completely:
In an experiment, we measure correlations $E_{ab}$ and we can ask, whether there is a probabilistic model that explains these explanation. So we are looking for the following:
1. A probability space $(\Gamma_{ab},\Sigma,\mu_{ab})$
2. Random variables $A_a : \Gamma_{ab}\rightarrow \{-1,1\}$ and $B_b : \Gamma_{ab}\rightarrow \{-1,1\}$
We want $E_{ab}= \int_{\Gamma_{ab}} A_a(\gamma) B_b(\gamma)\mathrm d\mu_{ab}(\gamma)$.

Bell assumes $\Gamma_{ab} = \Lambda$, $\mathrm d\mu_{ab}=\rho(\lambda)\mathrm d\lambda$, $A_a : \Lambda\rightarrow \{-1,1\}$ and $B_b : \Lambda\rightarrow \{-1,1\}$, so $E_{ab}=\int_\Lambda A_a(\lambda) B_b(\lambda)\mathrm \rho(\lambda)\mathrm d\lambda$ and this already includes his locality condition.

A contextual theory would be: $\Gamma_{ab} = \Lambda\times\Lambda_a\times\Lambda_b$, $A_a : \Lambda\times\Lambda_a\times\Lambda_b\rightarrow \{-1,1\}$ and $B_b : \Lambda\times\Lambda_a\times\Lambda_b\rightarrow \{-1,1\}$ and the locality condition would be $A_a(\lambda,\lambda_a,\lambda_b) = A_a(\lambda,\lambda_a)$ and $B_b(\lambda,\lambda_a,\lambda_b) = B_b(\lambda,\lambda_b)$. So $E_{ab}=\int_{\Lambda\times\Lambda_a\times\Lambda_b} A_a(\lambda,\lambda_a) B_b(\lambda,\lambda_b)\mathrm d\mu_{ab}(\lambda,\lambda_a,\lambda_b)$.

These are two different ways to define a probabilistic model that explains the correlations. In Bell's case, the random variables all live on one probability space, while in the contextual case, they live on many different spaces, depending on $a$ and $b$.

 

[wle]

Again, if you want to understand subtleties, you have to be rigorous about everything. Your presentation of the proof would not be acceptable to a probability theorist. The first thing he'd ask you is: "What probability spaces are you working with?"

I think I've been quite a bit more rigorous than you in this thread. Among other things I gave you an outline of Bell's theorem and invited you to point out exactly where your "single probability space" assumption is being introduced. All you've done since then is take issue with a simple definition at the beginning, before I even mentioned Bell locality, and lecture me about rigour.

However, like you, I don't see how the contextuality assumption enters, since the quantity can be directly computed in quantum mechanics.

Indeed, I don't think rubi is even reading my posts.

 

[rubi]

I think I've been quite a bit more rigorous than you in this thread. Among other things I gave you an outline of Bell's theorem and invited you to point out exactly where your "single probability space" assumption is being introduced. All you've done since then is take issue with a simple definition at the beginning, before I even mentioned Bell locality, and lecture me about rigour.

You don't need to give me an outline of the proof, since I already have seen it. You lack rigor for the reason that you don't specify your assumptions at a level of rigor that a mathematician would find acceptable and this is also the reason for why you are unable to see where the assumptions are made. What you call "a simple definition" is the crucial point of the argument, so I naturally put emphasis on it.

The point is that Bell uses a very specific probabilistic model to explain the correlations and he shows that this model is not consistent with QM. However, Bell's model is not the only model one could pick. You only see this if you are rigorous about all parts of the argument.

 

[atyy]

You don't need to give me an outline of the proof, since I already have seen it. You lack rigor for the reason that you don't specify your assumptions at a level of rigor that a mathematician would find acceptable and this is also the reason for why you are unable to see where the assumptions are made. What you call "a simple definition" is the crucial point of the argument, so I naturally put emphasis on it.

The point is that Bell uses a very specific probabilistic model to explain the correlations and he shows that this model is not consistent with QM. However, Bell's model is not the only model one could pick. You only see this if you are rigorous about all parts of the argument.

But wle's point is that the expectation value does exist within quantum mechanics itself, so defining the expectation value does not constitute an assumption of non-contextuality.

 

[rubi]

But wle's point is that the expectation value does exist within quantum mechanics itself, so defining the expectation value does not constitute an assumption of non-contextuality.

That's a misunderstanding, the QM expectation value is a way to obtain the numerical values of the correlations. A probabilistic model is a theory that explains the correlations. The QM expectation value doesn't yet include a hidden variable. Bell wants to explain the correlations by postulating that the random variables live on a single probability space. However, one can think of a different, more complicated probabilistic model that explains the correlations by allowing contextuality (i.e. several probability spaces). Bell doesn't consider the possibility to explain the correlations this way in his proof. The violation of his inequality only falsifies his specific probabilistic model, not the contextual one (see my post #530).

Edit: In other words: Bell assumes a specific probabilistic model to explain the correlations and he finds that it satisfies an inequality. This inequality is violated, so his model cannot explain the correlations. Since his proof assumed his specific probabilistic model, it cannot be used to argue against the other one. One would have to prove an individual inequality that is satisfied by the contextual model or find some other argument to falsify it.

Edit2: So here is a challenge for you and wle: Prove that the contextual model in post #530 is incompatible with the predictions of QM.

And by the way, it is not important for this argument, whether wle splits up his model for the correlations into two parts or not. His model is exactly the same model as the first one I gave in post #530, just phrased differently.

 

[wle]

You lack rigor for the reason that you don't specify your assumptions at a level of rigor that a mathematician would find acceptable

Bell's theorem is physics, not mathematics. Your post #530 likewise doesn't impress me as a physicist, for instance, since it completely ignores the context that Bell's theorem was derived in. In particular, the variable $\lambda$ appearing in Bell's theorem is intended to represent initial conditions that you know or could know, according to some theory, that could help you eventually make predictions about the outcomes in an experiment. This means it should have a well defined value independently of the measurements pretty much by definition. Even quantum physics provides an object for this purpose -- the initial quantum state -- that is well defined independently of the measurements that are eventually performed.

I shouldn't even have to argue this since the point seems to me to already have been made: Khrennikov's article was published several years ago and it has not generally impacted the way we think about Bell's theorem.

this is also the reason for why you are unable to see where the assumptions are made

Your post #530 undermines your point as far as I am concerned. First of all, if you thought the "hidden assumption" was that the variable $\lambda$ in Bell's theorem is well defined independently of the measurements, you could have just said so. I don't think the mathematical jargon in your post makes that clearer at all. Second, like I point out, introducing different "probability spaces" for the variable $\lambda$ associated with different measurements doesn't make a whole lot of sense if you consider Bell's theorem in context, so it is not something I would have flagged by being "more rigorous".

 

[rubi]

Bell's theorem is physics, not mathematics.

Physics also needs to use valid mathematics. Being mathematically precise about everything just improves clarity. You wanted me to expose, where Bell assumes a single probability space. I then described it rigorously, so not even you seem to question the assumption anymore, and now you are accusing me for using rigorous mathematics.

Your post #530 likewise doesn't impress me as a physicist, for instance, since it completely ignores the context that Bell's theorem was derived in. In particular, the variable $\lambda$ appearing in Bell's theorem is intended to represent initial conditions that you know or could know, according to some theory, that could help you eventually make predictions about the outcomes in an experiment. This means it should have a well defined value independently of the measurements pretty much by definition. Even quantum physics provides an object for this purpose -- the initial quantum state -- that is well defined independently of the measurements that are eventually performed.

Bell's theorem is not relevant for the contextual model. The point of the contextual model is to provide an alternative explanation for the correlations. It also has a variable $\lambda$ that is common to all measurements and that serves to describe pre-determined information. As I pointed out earlier, the purpose of my argument was to argue for the possibility of a common cause explanation of the correlations, which can in principle be achieved by the contextual model through the variable $\lambda$. Thus locality is not falsified by QM. I don't care, whether the contextual model is not a classical deterministic theory. After all, QM isn't one either. I only care about saving locality. Since the contextual model has not yet been falsified, locality is not falsified either, whether you like it or not. In order to argue against locality, you would have to argue against the contextual model, rather than questioning that Bell assumes a single probability space, which you did since the beginning of the argument.

I shouldn't even have to argue this since the point seems to me to already have been made: Khrennikov's article was published several years ago and it has not generally impacted the way we think about Bell's theorem.

If all physicists are like you, then this is not surprising. You were already rejecting his argument aggresively, before you had even attempted to understand it. Also, the paper is not even 10 years old. This is a very short amount of time for research level physics to reach a larger audience.

Your post #530 undermines your point as far as I am concerned. First of all, if you thought the "hidden assumption" was that the variable $\lambda$ in Bell's theorem is well defined independently of the measurements, you could have just said so.

This is not the "hidden assumption". The hidden variable $\lambda$ is well defined independently of the measurements even in the contextual model, because it is a variable that is shared by all the probability spaces. The "hidden assumption" is the fact that the $X_x$ live on one single probability space, and it is also not "hidden", since it is clearly visible even in informal arguments (unless one tries to hide it explicitely).

I don't think the mathematical jargon in your post makes that clearer at all.

The "mathematical jargon" exposes the underlying probability structure precisely, while informal presentations never emphasize it. The point of science is to be precise about every subtlety, rather than to sweep things under the carpet.

Second, like I point out, introducing different "probability spaces" for the variable $\lambda$ associated with different measurements doesn't make a whole lot of sense if you consider Bell's theorem in context, so it is not something I would have flagged by being "more rigorous".

As I said, the variable $\lambda$ is shared by all probability spaces. The point of the contextual theory is that $\lambda$ does not solely determine the correlations. There can be additional $\lambda_x$ for every context.

 

[wle]

As I said, the variable $\lambda$ is shared by all probability spaces. The point of the contextual theory is that $\lambda$ does not solely determine the correlations. There can be additional $\lambda_x$ for every context.

I think that misses the point of Bell's theorem, like I said.

 

[atyy]

As I said, the variable $\lambda$ is shared by all probability spaces. The point of the contextual theory is that $\lambda$ does not solely determine the correlations. There can be additional $\lambda_x$ for every context.

Doesn't this traditionally come under the outs called "superdeterminism" or "free will"? It seem the same as what is discussed eg. under the "locality loophole" on p51 of http://arxiv.org/abs/1303.2849.

 

[rubi]

I think that misses the point of Bell's theorem, like I said.

Well, it misses the point of Bell's theorem, because it is supposed to miss the point of Bell's theorem. Bell wants to exclude deterministic hidden variables and the violation of the inequality shows that he was successful. It's not about finding a loophole in Bell's argument, I happily reject deterministic hidden variables. The point of the contextual model is to offer an alternative explanation to deterministic hidden variables, while still maintaining locality. The question is: Is there an apriori reason to exclude contextual probabilistic models like the one I described in post #530? If not, then either we can show that these models are incompatible with QM as well (which I doubt), or we are unable to claim that QM violates locality.

Doesn't this traditionally come under the outs called "superdeterminism" or "free will"? It seem the same as what is discussed eg. under the "locality loophole" on p51 of http://arxiv.org/abs/1303.2849.

I don't think it is the same as superdeterminism, since it doesn't require any kind of fine-tuning and it also doesn't doubt the free will of the experimenters. The locality loophole in that article seems to be concerned with deterministic hidden variables as well. I think the point of all loopholes is to reject the conclusions of Bell's theorem, in order to save local realism. The contextual models don't attempt to save local realism. Instead, they offer an alternative to local realism. (By local realism, I mean Bell's probabilistic model.)

 

[atyy]

I don't think it is the same as superdeterminism, since it doesn't require any kind of fine-tuning and it also doesn't doubt the free will of the experimenters. The locality loophole in that article seems to be concerned with deterministic hidden variables as well. I think the point of all loopholes is to reject the conclusions of Bell's theorem, in order to save local realism. The contextual models don't attempt to save local realism. Instead, they offer an alternative to local realism. (By local realism, I mean Bell's probabilistic model.)

It seems to me that if there are additional $\lambda_{x}$, and $x$ is the measurement choice, then the measurement choice is not independent of the preparation, so it is a violation of free will.

Edit: Another example which I think makes clear that what you are talking about is freedom of choice is Scheidl's http://arxiv.org/abs/0811.3129: "In other words, the probability distribution of the hidden variables is therefore independent of the setting choices: ρ(λ|a,b) = ρ(λ) for all settings a and b. Without this independence, there is a loophole for local realistic theories which has not been addressed by any experiment to date."