I How "spooky action...." may work?

[wle]

Because random variables satisfy these properties by definition: $(AB)(x) = A(x)B(x)$, because this is how the $AB$ is defined.

You've lost me. What are $A$, $B$, and $x$ supposed to be in the context of Kolmogorov probability theory and what do they have to do with the assumptions $v(A + B) = v(A) + v(B)$ and $v(AB) = v(A) v(B)$ for commuting quantum observables $A$ and $B$ in the Kochen-Specker theorem?

The Kochen-Specker theorem proves it.

I don't think you've justified that. You haven't proved that the assumptions behind the Kochen-Specker theorem are equivalent to or follow from the axioms of Kolmogorov probability theory, and proofs of the Kochen-Specker theorem don't claim any such thing.

Hence, all theorems that are derived from Kolmogorov's axioms and all concepts that depend on this need to be adjusted to the new situation.

Likewise, theorems that don't use all of Kolmogorov's axioms are not necessarily restricted to Kolmogorov probability theory. Kolmogorov probability theory requires that the joint event $A \wedge B$ exists for all events $A$ and $B$, like you say. Bell's theorem, for example, does not require that all possible joint events exist.

This may be a valid causal explanation. However, the point is whether there can be causal explanations that don't violate the speed of light.

Why should that make a difference? If you accept that quantum correlations can be simulated by two computers communicating with each other faster than light, then you can certainly ask if two computers could simulate quantum correlations without communicating faster than light.

This is why I say your insistence on Kolmogorov probability theory isn't relevant. It is not difficult to program a computer to output random results in accord with the Born rule, and two computers allowed to communicate FTL could be programmed to simulate arbitrary quantum correlations. If you insist that we can only reason about causality, Reichenbach's principle, etc. within a certain mathematical framework, and that framework doesn't accommodate something I can simulate on a computer, then I'd say it's not a good framework to begin with.

It's the same if you look at the historical origins behind Bell's theorem. Essentially, Bell was aware that nonlocal hidden variable models like the de Broglie-Bohm interpretation could reproduce the predictions of quantum physics, and he was interested in the question of whether a local hidden variable model could achieve the same thing. So, similarly, if your framework for discussing causality doesn't accommodate the de Broglie-Bohm interpretation then it is not relevant to understanding Bell's theorem, at least not in the way Bell thought about it.

 

[rubi]

You've lost me. What are $A$, $B$, and $x$ supposed to be in the context of Kolmogorov probability theory and what do they have to do with the assumptions $v(A + B) = v(A) + v(B)$ and $v(AB) = v(A) v(B)$ for commuting quantum observables $A$ and $B$ in the Kochen-Specker theorem?

Is it really so hard to understand? The assumptions of the Kochen-Specker theorem require that the valuations of quantum observables follow the same rules as the valuations of classical random variables. Since the valuation of a classical random variable is given by $v(A) = A(x)$ and a the product of classical random variables is defined by $(AB)(x) := A(x)B(x)$, the requirements of Kochen-Specker follow (same for addition). Kochen-Specker says that one cannot represent quantum observables on a classical probability space without having to redefine multiplication of addition of random variables.

I don't think you've justified that. You haven't proved that the assumptions behind the Kochen-Specker theorem are equivalent to or follow from the axioms of Kolmogorov probability theory, and proofs of the Kochen-Specker theorem don't claim any such thing.

I (or rather Kochen and Specker themselves) have proven that not all quantum observables can be represented as classical random variables on a classical probability space. This is also not my personal claim, but it is standard knowledge that can be looked up in pretty much every book on quantum mechanics.

Likewise, theorems that don't use all of Kolmogorov's axioms are not necessarily restricted to Kolmogorov probability theory. Kolmogorov probability theory requires that the joint event $A \wedge B$ exists for all events $A$ and $B$, like you say. Bell's theorem, for example, does not require that all possible joint events exist.

Of course, Bell's theorem requires that, because it wants to make statements about all possible events. Otherwise it can only make statements like: "Among the events that are commuting with $A$ and $B$, none can be a common cause" or "No theory of local hidden variables for events commuting with $A$ and $B$ can reproduce all predictions of QM". Of course, for a classical probability theory, this is the same as Bell's theorem, since all classical random variables commute. But it would be a weak result for QM, since it doesn't allow to conclude the non-existence of a common cause or non-locality in QM, since in QM, there are also events not commuting with $A$ and $B$.

Why should that make a difference? If you accept that quantum correlations can be simulated by two computers communicating with each other faster than light, then you can certainly ask if two computers could simulate quantum correlations without communicating faster than light.

Two computers of course cannot do that, since they are classical objects. You need quantum objects to generate quantum statistics. The analogy with computers makes no sense here.

This is why I say your insistence on Kolmogorov probability theory isn't relevant.

Of course, it is highly relevant. It is really super trivial: Concepts that only work in Kolmogorov probability theory cannot be applied outside of Kolmogorov probability theory. Apparently, you deny this simple fact.

It is not difficult to program a computer to output random results in accord with the Born rule, and two computers allowed to communicate FTL could be programmed to simulate arbitrary quantum correlations. If you insist that we can only reason about causality, Reichenbach's principle, etc. within a certain mathematical framework, and that framework doesn't accommodate something I can simulate on a computer, then I'd say it's not a good framework to begin with.

You can of course simulate quantum physics on a computer, but you cannot have computers behave like quantum objects. There is no logical problem here. My whole point is that quantum theory is not a classical probability theory. This is really standard and well-known and it makes no sense to doubt it. Hence, concepts that require classical probability theory, just don't work anymore in the context of quantum mechanics. This is a fact of life. Of course, you can prefer Bohmian mechanics, but then you can only use the old concepts to make statements about Bohmian mechanics and not about quantum theory.

It's the same if you look at the historical origins behind Bell's theorem. Essentially, Bell was aware that nonlocal hidden variable models like the de Broglie-Bohm interpretation could reproduce the predictions of quantum physics, and he was interested in the question of whether a local hidden variable model could achieve the same thing.

Yes and of course he proved that no local hidden variable model can make the same predictions as QM. One cannot prove the theorem without the assumption of hidden variables. Hence, the theorem says nothing about theories without hidden variables, such as QM.

So, similarly, if your framework for discussing causality doesn't accommodate the de Broglie-Bohm interpretation then it is not relevant to understanding Bell's theorem, at least not in the way Bell thought about it.

First of all, it is not my framework, but the generally accepted framework of physics. What I am saying is generally agreed upon by all working physicists. Of course, dBB theory can be formulated within this framework. Then you just can't prove Bell's theorem anymore. However, dBB theory can also be formulated as a classical probability theory and hence, Bell's theorem applies. This is only possible, since dBB theory doesn't have all the quantum observables as random variables, since the KS theorem prohibits it. Not even Bohmians deny this.

 

[wle]

Is it really so hard to understand? The assumptions of the Kochen-Specker theorem require that the valuations of quantum observables follow the same rules as the valuations of classical random variables. Since the valuation of a classical random variable is given by $v(A) = A(x)$ and a the product of classical random variables is defined by $(AB)(x) := A(x)B(x)$, the requirements of Kochen-Specker follow (same for addition). Kochen-Specker says that one cannot represent quantum observables on a classical probability space without having to redefine multiplication of addition of random variables.

You haven't answered my question. In terms of the axioms of Kolmogorov probability theory, what are your $A$, $B$, and, especially, $x$ supposed to be?

Of course, Bell's theorem requires that, because it wants to make statements about all possible events.

No it doesn't. The mathematical assumption that Bell inequalities are derived from is that (bipartite) correlations can be expressed in the form $$P(ab \mid xy) = \int \mathrm{d} \lambda \rho(\lambda) P_{\mathrm{A}}(a \mid x; \lambda) P_{\mathrm{B}}(b \mid y; \lambda) \,,$$ where $a$, $b$, $x$, and $y$ are labels associated to measurement outcomes and choices, respectively. This does not require joint events to be defined except where quantum physics already says they exist.

Two computers of course cannot do that, since they are classical objects. You need quantum objects to generate quantum statistics. The analogy with computers makes no sense here.

Wrong. Computers can calculate the Born rule probability $P(a \mid \rho) = \mathrm{Tr}[M_{a} \rho]$ of obtaining a result $a$ from measuring the POVM $\{M_{a}\}_{a}$ on an initial state $\rho$. A computer can equally easily generate a random result with this probability given the state and measurement as inputs. The only limitations are technological: finite precision of floating point computations, quality of random number generators, and, for high-dimensional Hilbert spaces, processing speed and available memory.

With FTL communication, simulating a given set of bipartite quantum correlations $P(ab \mid xy) = \mathrm{Tr} \bigl[ (M_{a \mid x} \otimes N_{b \mid y}) \rho_{\mathrm{AB}} \bigr]$ on two distant computers would not be much more difficult. One way to do it is to express the probabilities as $P(ab \mid xy) = P_{1}(a \mid b, xy) P_{2}(b \mid y)$ where $P_{2}(b \mid y) = \sum_{a} P(ab \mid xy)$. Then program Bob's computer to accept $y$ as input, generate $b$ with probability $P_{2}(b \mid y)$, transmit $b$ and $y$ to Alice's computer, and finally output $b$. Likewise, Alice's computer would be programmed to accept $x$ as input, read $y$ and $b$ from Bob's computer, and output result $a$ with probability $P_{1}(a \mid b, xy)$.

Similar to what I say about Bell's theorem above, there is no requirement here that e.g. the probabilities $P_{2}(b \mid y)$ should admit a hidden variable model in the sense of Kochen-Specker or that joint events like $(b_{y}, b_{y'})$ for different inputs $y$ should be defined.

This is also not my personal claim, but it is standard knowledge that can be looked up in pretty much every book on quantum mechanics.

This is really standard and well-known and it makes no sense to doubt it.

First of all, it is not my framework, but the generally accepted framework of physics. What I am saying is generally agreed upon by all working physicists.

I am a working physicist and I'd say your interpretation of the Bell and Kochen-Specker theorems looks highly nonstandard and poorly supported to me. Please stop making claims along the lines "all physicists agree with this" or "all textbooks say this". They don't.

Concerning Kolmogorov, I think most researchers in quantum physics probably don't know offhand, or really much care, what the exact definition of Kolmogorov probability theory is, let alone "generally accept" it as a condition for discussing things like correlation or causality or Reichenbach's principle. So I doubt that a theorem stating that quantum physics is not a Kolmogorov probability theory would even have much impact in the physics community (certainly nothing like the impact of Bell's theorem).

 

[rubi]

You haven't answered my question. In terms of the axioms of Kolmogorov probability theory, what are your $A$, $B$, and, especially, $x$ supposed to be?

$A$ and $B$ are random variables on the probability space. $x$ is an element of the probability space. If you knew anything about probability theory at all, this should have been a triviality to you.

The valuation of classical random variables $v(A)=A(x)$ satisfies the assumptions of the KS theorem (for example: $v(AB) = (AB)(x) = A(x)B(x) = v(A)v(B)$), so if KS prove that no valuation function compatible with these assumptions can be defined for quantum observables, it means that the quantum observables can't be represented by classical random variables. This is the whole point of the KS theorem and the very motivation of Kochen and Specker for proving it in the first place.

No it doesn't. The mathematical assumption that Bell inequalities are derived from is that (bipartite) correlations can be expressed in the form $$P(ab \mid xy) = \int \mathrm{d} \lambda \rho(\lambda) P_{\mathrm{A}}(a \mid x; \lambda) P_{\mathrm{B}}(b \mid y; \lambda) \,,$$ where $a$, $b$, $x$, and $y$ are labels associated to measurement outcomes and choices, respectively. This does not require joint events to be defined except where quantum physics already says they exist.

Wrong. If the event algebra is not a sigma algebra, then your $\lambda$ will not encompass all possible events, but only such events that commute with $A$ and $B$. Hence, statements proved from this formula will only hold for events that commute with $A$ and $B$.

Wrong. Computers can calculate the Born rule probability $P(a \mid \rho) = \mathrm{Tr}[M_{a} \rho]$ of obtaining a result $a$ from measuring the POVM $\{M_{a}\}_{a}$ on an initial state $\rho$. A computer can equally easily generate a random result with this probability given the state and measurement as inputs. The only limitations are technological: finite precision of floating point computations, quality of random number generators, and, for high-dimensional Hilbert spaces, processing speed and available memory.

With FTL communication, simulating a given set of bipartite quantum correlations $P(ab \mid xy) = \mathrm{Tr} \bigl[ (M_{a \mid x} \otimes N_{b \mid y}) \rho_{\mathrm{AB}} \bigr]$ on two distant computers would not be much more difficult. One way to do it is to express the probabilities as $P(ab \mid xy) = P_{1}(a \mid b, xy) P_{2}(b \mid y)$ where $P_{2}(b \mid y) = \sum_{a} P(ab \mid xy)$. Then program Bob's computer to accept $y$ as input, generate $b$ with probability $P_{2}(b \mid y)$, transmit $b$ and $y$ to Alice's computer, and finally output $b$. Program Alice's computer to accept $x$ as input, read $y$ and $b$ from Bob's computer, and output result $a$ with probability $P_{1}(a \mid b, xy)$.

Similar to what I say about Bell's theorem above, there is no requirement here that e.g. the probabilities $P_{2}(b \mid y)$ should admit a hidden variable model in the sense of Kochen-Specker or that joint events like $(b_{y}, b_{y'})$ for different inputs $y$ should be defined.

This is not what I said. I said that two computers cannot simulate this without FTL communication. Of course you can simulate a quantum system on a computer if you simulate the whole system on one machine or simulate the systems individually on two machines with FTL communication. What two computers cannot do is to generate quantum correlations without FTL communication, using only local data. This is the challenge for Bell deniers. Of course, this is not possible (assuming there are no loopholes). The point is that this says nothing about what quantum objects can do. Computers sufficiently classical and hence aren't a good analogy to quantum objects.

I am a working physicist and I'd say your interpretation of the Bell and Kochen-Specker theorems looks highly nonstandard and poorly supported to me. Please stop making claims along the lines "all physicists agree with this" or "all textbooks say this". They don't.

I highly doubt that you are a working physicist. "My" interpretation is fully standard and evident to everyone who understands basic probability theory. KS says that QT cannot be embedded into a classical probability theory without changing the definition of multiplication and addition of random variables. This is definitely well established science. I have done my best to explain this to you, but you will not understand it if you don't invest at least a little bit of time into the study of probability theory and the KS theorem.

Concerning Kolmogorov, I think most researchers in quantum physics probably don't know offhand, or really much care, what the exact definition of Kolmogorov probability theory is, let alone "generally accept" it as a condition for discussing things like correlation or causality or Reichenbach's principle. So I doubt that a theorem stating that quantum physics is not a Kolmogorov probability theory would even have much impact in the physics community (certainly nothing like the impact of Bell's theorem).

I don't know where you learned physics from, but probability theory is an elementary part of physics education. Certainly, all physicists know probability theory. It is also fully standard that Bell's theorem requires the assumption of hidden variables. The fact that quantum theory is not a classical probability theory is well understood and there is a whole industry of research devoted to this fact. It is also not a new result, but known since half a century already. The impact is that almost 100 years after the discovery of QT, we are still discussing about interpretations of QT. If QT were just another classical probability theory, we wouldn't have any interpretational problems.

 

[wle]

$A$ and $B$ are random variables on the probability space. $x$ is an element of the probability space. The valuation of classical random variables $v(A)=A(x)$ satisfies the assumptions of the KS theorem (for example: $v(AB) = (AB)(x) = A(x)B(x) = v(A)v(B)$), so if KS prove that no valuation function compatible with these assumptions can be defined for quantum observables, it means that the quantum observables can't be represented by classical random variables. This is the whole point of the KS theorem and the very motivation of Kochen and Specker for proving it in the first place.

That doesn't support you. You are assuming that deterministic values $v(A)$ and $v(B)$ for different operators must be modeled as the same event $x$. This is effectively how I'd read the Kochen-Specker theorem if I try to translate it into the language of Kolmogorov probability theory. For example, if you consider the measurement bases $\mathcal{M}_{1} = \{ \lvert 0 \rangle, \lvert 1 \rangle, \lvert 2 \rangle \}$ or $\mathcal{M}_{2} = \bigl\{ \lvert 0 \rangle, \tfrac{\lvert 1 \rangle + \lvert 2 \rangle}{\sqrt{2}}, \tfrac{\lvert 1 \rangle - \lvert 2 \rangle}{\sqrt{2}}\bigr\}$, which share an eigenvector, then in terms of Kolmogorov probability theory the Kochen-Specker theorem assumes that getting the result $\lvert 0 \rangle$ when measuring in the basis $\mathcal{M}_{1}$ and getting the result $\lvert 0 \rangle$ when measuring in the basis $\mathcal{M}_{2}$ should both be modeled as the same event in the probability space.

I would not consider that the most general possible way to embed quantum physics in Kolmogorov probability theory.

(In fact, I'm having to guess to make sense of your post because what you're claiming doesn't even look well formed. For example in $v(A) = A(x)$ you have $A$ appearing as both a random variable and as a Hermitian operator.)

Wrong. If the event algebra is not a sigma algebra, then your $\lambda$ will not encompass all possible events, but only such events that commute with $A$ and $B$. Hence, statements proved from this formula will only hold for events that commute with $A$ and $B$.

I think you're making this up as you go along.

Since you insist on equating Kolmogorov probability theory with the Kochen-Specker theorem I'll add this: there's a simple special case that shows that the setting considered in Bell's theorem is not a special case of the Kochen-Specker theorem. Specifically, if you take a product state $\rho_{\mathrm{AB}} = \rho_{\mathrm{A}} \otimes \rho_{\mathrm{B}}$, the quantum correlation reduces to $$P(ab \mid xy) = \mathrm{Tr} \bigl[(M_{a \mid x} \otimes N_{b \mid y}) (\rho_{\mathrm{A}} \otimes \rho_{\mathrm{B}}) \bigr] = \mathrm{Tr} [ M_{a \mid x} \, \rho_{\mathrm{A}} ] \mathrm{Tr} [ N_{b \mid y} \, \rho_{\mathrm{B}} ] = P_{\mathrm{A}}(a \mid x) P_{\mathrm{B}}(b \mid y) \,.$$ This is (trivially) a Bell-local model in the sense I defined, regardless of what $P_{\mathrm{A}}(a \mid x) = \mathrm{Tr} [ M_{a \mid x} \, \rho_{\mathrm{A}} ]$ and $P_{\mathrm{B}}(b \mid y) = \mathrm{Tr} [ N_{b \mid y} \, \rho_{\mathrm{B}} ]$ are. In particular $P_{\mathrm{A}}(a \mid x)$ and $P_{\mathrm{B}}(b \mid y)$ may not admit contextual models satisfying the Kochen-Specker assumptions and the model would still count as local.

In general a given set of quantum correlations could admit only a local model (in the sense of Bell's theorem) or only a contextual model (in the sense of the Kochen-Specker theorem) or both or neither, so neither class of model is a subset of the other.

This is not what I said. I said that two computers cannot simulate this without FTL communication.

I never said you said that two computers cannot simulate this without FTL communication*. I described the FTL model to point out two things, neither of which you have addressed:

• If you accept that the FTL model is a valid explanation for quantum correlations then it is a perfectly natural question to ask whether a similar model could explain quantum correlations without FTL communication. Of course we already know the answer is "no" thanks to Bell's theorem.
• This question has nothing to do with the Kochen-Specker theorem or requiring that there is a joint probability space for everything or the like. The FTL simulation model I described need not necessarily admit a KS contextual model, for instance, so there is no reason to ask that there is a KS contextual model when you take away the FTL.

This is not a made up scenario. It is a fairly common way to think about Bell's theorem in the physics community and, in fact, there's a version of Bell's theorem (the "nonlocal game" approach) that is explicitly formulated this way.*If you're going to play "I said/you said" then you said "Two computers of course cannot do that" which leaves it ambiguous exactly what "that" is that you are denying, followed by "since they are classical objects. You need quantum objects to generate quantum statistics.", which is still wrong, since two computers using FTL communication is not a quantum object.

I highly doubt that you are a working physicist.

I work as a postdoctoral researcher in quantum information theory.

I have done my best to explain this to you, but you will not understand it if you don't invest at least a little bit of time into the study of probability theory and the KS theorem.

Oh please. Earlier in this thread you insisted, loudly and repeatedly, that there can't be a hidden-variable model for spin-1/2 until I pointed out that even the KS theorem doesn't apply to that. That immediately tells me you never invested your own "little bit of time" studying the theorem. If you actually look at a proof of the KS theorem (like the one on its Wikipedia page), it is actually quite easy to see why the kind of counterexample they construct cannot work for qubits.

 

[rubi]

(In fact, I'm having to guess to make sense of your post because what you're claiming doesn't even look well formed. For example in $v(A) = A(x)$ you have $A$ appearing as both a random variable and as a Hermitian operator.)

The fact that you are unable to make sense of the post shows that you haven't understood it. $A$ in that equation is a random variable on both sides of the equation. I will try one last time to explain the situation:

First we have the Kochen-Specker theorem:
In $d>2$, there is no valuation function (satisfying certain assumptions) on a certain subset of quantum observables.

The next theorem is a triviality:
On all random variables on a classical probability space, there exists a valuation function (satisfying said assumptions).
(If you are unable to prove this highly trivial theorem on your own, then see for example http://arxiv.org/abs/quant-ph/9803055v4)

Now, if we could represent the quantum observables as random variables on a classical probability space, then the second theorem would imply the existence of a valuation function. This is in contradiction with the KS theorem.

By the way, not even Bohmians have a problem admitting that not all quantum observables are represented as random variables in their theory. You are completely alone with the belief that this can be accomplished.

I think you're making this up as you go along.

No, this is also common knowledge. See for example the nice book that bhobba always quotes: https://www.amazon.com/dp/0387493859/?tag=pfamazon01-20
In order to get a probability space, you must restrict the event algebra of quantum mechanics (an orthomodular lattice) to a sublattice of commuting events. Otherwise the integral you have written down doesn't even make sense, because you don't even have an integration measure (the probabiliy functional on the full event algebra is not a measure). If you believe that you can define this integral using the probability functional on the quantum event algebra, then either explain how to do it or point me to a reference. If you cannot do this, then you should be very careful making such non-sensical statements.

Oh please. Earlier in this thread you insisted, loudly and repeatedly, that there can't be a hidden-variable model for spin-1/2 until I pointed out that even the KS theorem doesn't apply to that. That immediately tells me you never invested your own "little bit of time" studying the theorem. If you actually look at a proof of the KS theorem (like the one on its Wikipedia page), it is actually quite easy to see why the kind of counterexample they construct cannot work for qubits.

I agree, there is a counterexample in 2 dimensions. However, that doesn't invalidate the theorem, which holds for any dimension > 2. I have done my homework and studied all these things for many years. Even my own research is concerned with causality in quantum gravity. The fact that you don't even know what a random variable is clearly shows that you have no expertise in this subject. Random variables are the most basic concept of probability theory.

No. The most common reasons I've seen for discussing interpretations of quantum physics are the measurement problem (e.g. many-worlds interpretation, Bohmian mechanics, stochastic collapse models), "make it more intuitive" (e.g. Transactional interpretation) and attempts to redefine what should be expected from a scientific theory to include quantum physics (e.g. Qbism).

The measurement problem is the result of having a theory that is in conflict with classical probability theory. All interpretational problems on QM can be traced back to this fact. If QM were a classical probability theory, then the interpretational problems would be solved automatically.

 

[wle]

By the way, not even Bohmians have a problem admitting that not all quantum observables are represented as random variables in their theory. You are completely alone with the belief that this can be accomplished.

I have never expressed a belief that quantum physics can be fully embedded in Kolmogorov probability theory. I only disputed that this has been proved by the KS theorem. In fact in an earlier post I said this (emphasis added):

But let's say you're correct, and it's impossible to fully embed quantum physics in the language of Kolmogorov probability theory. In practice you may as well be correct anyway since quantum physics is not normally expressed in that language (whether there is a way to do it or not).

So this whole discussion has been a distraction as far as I'm concerned.

More importantly, I disputed that working in the framework of Kolmogorov probability or having a KS contextual model is necessary to understand Bell's theorem in the first place. You didn't address that at all.

No, this is also common knowledge. See for example the nice book that bhobba always quotes: https://www.amazon.com/dp/0387493859/?tag=pfamazon01-20
In order to get a probability space, you must restrict the event algebra of quantum mechanics (an orthomodular lattice) to a sublattice of commuting events. Otherwise the integral you have written down doesn't even make sense, because you don't even have an integration measure (the probabiliy functional on the full event algebra is not a measure). If you believe that you can define this integral using the probability functional on the quantum event algebra, then either explain how to do it or point me to a reference. If you cannot do this, then you should be very careful making such non-sensical statements.

If you're worried about problems with integration measures then, for the purpose of what I was saying, restricting the definition of local model to $$P(ab \mid xy) = \sum_{k} p_{k} P_{\mathrm{A}}(a \mid x; k) P_{\mathrm{B}}(b \mid y; k)$$ for a finite set of possible values of $k$, and restricting attention to a finite number of possible inputs $x$ and $y$ and outputs $a$ and $b$ (which is what is usually considered in the context of Bell's theorem anyway) works just fine for the point I was making. This way the integral changes to a sum and the definition only involves probabilities satisfying $p_{k}, P_{\mathrm{A}}(a \mid x; k), P_{\mathrm{B}}(b \mid y; k) \geq 0$ and $\sum_{k} p_{k} = \sum_{a} P_{\mathrm{A}}(a \mid x; k) = \sum_{b} P_{\mathrm{B}}(b \mid y; k) = 1$.

 

[rubi]

I have never expressed a belief that quantum physics can be fully embedded in Kolmogorov probability theory. I only disputed that this has been proved by the KS theorem.

Why don't you point out an error in my argument then? You doubt that it is correct, then you must point out an error. Simply doubting its correctness and feeling superior is not very scientific.

First we have the Kochen-Specker theorem:
In $d>2$, there is no valuation function (satisfying certain assumptions) on a certain subset of quantum observables.

The next theorem is a triviality:
On all random variables on a classical probability space, there exists a valuation function (satisfying said assumptions).
(If you are unable to prove this highly trivial theorem on your own, then see for example http://arxiv.org/abs/quant-ph/9803055v4)

Now, if we could represent the quantum observables as random variables on a classical probability space, then the second theorem would imply the existence of a valuation function. This is in contradiction with the KS theorem.

--

More importantly, I disputed that working in the framework of Kolmogorov probability or having a KS contextual model is necessary to understand Bell's theorem in the first place. You didn't address that at all.

Of course I addressed it. Apparently you missed it: Bell's theorem requires a probability space in which the $\lambda$'s live. In quantum theory, no such probability space exists. There only exist probability spaces for some subsets of commuting observables.

If you're worried about problems with integration measures then. for the purpose of what I was saying, restricting the definition of local model to $$P(ab \mid xy) = \sum_{k} p_{k} P_{\mathrm{A}}(a \mid x; k) P_{\mathrm{B}}(b \mid y; k)$$ for a finite set of possible values of $k$, $\sum_{k} p_{k} = 1$, and restricting attention to a finite number of possible inputs $x$ and $y$ and outputs $a$ and $b$ (which is what is usually considered in the context of Bell's theorem anyway) works just fine for the point I was making.

No, this is also a measure (the counting measure). Moreover, restricting is exactly what you cannot do if you want to prove something for all objects (i.e. if you want to prove that none of the $\lambda$'s or $k$'s in this case can serve as a common cause). I'll explain again what I mean:

An integral over a probability space $(\Omega,\Sigma,P)$ requires $\Sigma$ to be a sigma algebra and $P:\Sigma\rightarrow [0,1]$ to be a measure. However, in quantum theory, $P:O\rightarrow [0,1]$ is defined on an orthomodular lattice (see also quantum logic), rather than a sigma algebra. Only certain sublattices of this orthomodular lattice $O$ are sigma algebras, namely those that are formed by certain sets of commuting projectors. In the general setting of an orthomodular lattice, there is not even a definition of an integral. It just doesn't make sense to integrate over the full event algebra of quantum mechanics. It only makes sense to integrate over sublattices of commuting events that form a sigma algebra. For example the set of projectors of a single self-adjoint operator forms such a sublattice, hence we can compute expectation values of an observable in quantum theory. However, we don't have a probability space that encompasses all observables at our disposal. However, the strength of Bell's theorem comes from the fact that it makes a statement about all observables and not just some. No hidden variable can explain the correlations, not just some specific subset of hidden variables. Hence all theories with only commuting observables are excluded by the theorem. However, theories with non-commuting observables are not excluded.

It is simply a false assumption that you can represent all possible events of quantum theory on one probability space. There is no set of $\lambda$'s that encompasses all such events.

 

[wle]

Of course I addressed it.

No you didn't.

Take $P(ab \mid xy) = P_{\mathrm{A}}(a \mid x) P_{\mathrm{B}}(b \mid y)$ with $P_{\mathrm{A}}(a \mid x) = \lvert \langle u_{a, x} \vert \psi \rangle \rvert^{2}$ where $\lvert u_{a,x} \rangle, a \in \{1, \dotsc, 4\}, x \in \{1, \dotsc, 9\}$ is the (normalised) state in the $a$th row and $x$th column of the colourful table here, $\lvert \psi \rangle$ is any normalised four-dimensional state vector, and take $P_{\mathrm{B}}(1 \mid 1) = 1$ (i.e., the trivial case where Bob's inputs and outputs are restricted to $b, y \in \{1\}$).

• Pretty much anyone working in Bell nonlocality would count this as a (trivial) local model in the sense of Bell's theorem. (I am not going to argue with this. If you disagree, you are welcome to take this up with someone who works in the field, for instance one of the authors of this review article, which pretty much starts with the definition I used.)
• By construction, the correlations do not admit a contextual model in the sense of the Kochen-Specker theorem (since I took for Alice correlations resulting from measurements that are used in one proof of the Kochen-Specker theorem).

So not all Bell-local models must admit a KS-contextual model.

No, this is also a measure (the counting measure).

I'm tired of this. If you're going to tell me that I can't add a finite number of products of real numbers then I'm out.

It is simply a false assumption that you can represent all possible events of quantum theory on one probability space.

Straw man.

This does not require joint events to be defined except where quantum physics already says they exist.

Similar to what I say about Bell's theorem above, there is no requirement here that e.g. the probabilities $P_{2}(b \mid y)$ should admit a hidden variable model in the sense of Kochen-Specker or that joint events like $(b_{y}, b_{y'})$ for different inputs $y$ should be defined.

In particular $P_{\mathrm{A}}(a \mid x)$ and $P_{\mathrm{B}}(b \mid y)$ may not admit contextual models satisfying the Kochen-Specker assumptions and the model would still count as local.

I am not going to continue arguing with someone who cannot address what I actually say in my posts.

 

[rubi]

No you didn't.

Of course I did. You just ignored it. I specifically asked you to find an error in my proof. You just didn't even respond to it.

Take $P(ab \mid xy) = P_{\mathrm{A}}(a \mid x) P_{\mathrm{B}}(b \mid y)$ with $P_{\mathrm{A}}(a \mid x) = \lvert \langle u_{a, x} \vert \psi \rangle \rvert^{2}$ where $\lvert u_{a,x}, a \in \{1, \dotsc, 4\}, x \in \{1, \dotsc, 9\}$ is the (normalised) state in the $a$th row and $x$th column of the colourful table here, $\lvert \psi \rangle$ is any four-dimensional state vector, and take $P_{\mathrm{B}}(1 \mid 1) = 1$ (i.e., the trivial case where Bob's inputs and outputs are restricted to $b, y \in \{1\}$).

• Pretty much anyone working in Bell nonlocality would count this as a (trivial) local model in the sense of Bell's theorem. (I am not going to argue with this. If you disagree, you are welcome to take this up with someone who works in the field, for instance one of the authors of this review article, pretty much starts with the definition I used.)
• By construction, the correlations do not admit a contextual model in the sense of the Kochen-Specker theorem (since I took for Alice correlations resulting from measurements that are used in one proof of the Kochen-Specker theorem).

So not all Bell-local models must admit a KS-contextual model.

So what? You haven't introduced a hidden variable $\lambda$ into the model. So far it is only an expression for calculating the probability. You can't prove Bell's theorem without introducing $\lambda$'s. And introducing $\lambda$'s is precisely the problem, since you will be restricted to the use of $\lambda$'s commuting with the observables.

I'm tired of this. If you're going to tell me that I can't add a finite number of products of real numbers then I'm out.

Have you even read what I wrote? You can of course add finite numbers of products, but those finitely many terms will not be enough to hit every event that can occur in quantum theory. And not even an integral will hit every event. The reason for this is that the algebra of events is bigger than a sigma algebra.

Straw man.

Apparently you haven't understood the argument at all. Why then do you think you should even have an opinion as long as you haven't invested the time to understand the argument?

I am not going to continue arguing with someone who cannot address what I actually say in my posts.

Then explain to me why my previous post did not address it? The argument is completely trivial: A theorem that makes use of probability theory can only hold for theories that are formulated using probability theory.

 

[bhobba]

See for example the nice book that bhobba always quotes: https://www.amazon.com/dp/0387493859/?tag=pfamazon01-20

Not sure I actually quote it because its HARD - although I do mention it. It's for those that want a rigorous mathematical treatment from quantum logic where everything such as observable etc is defined rigorously.

However reading this thread one thing that struck me was this harping on about KS. IMHO the better thing to look at is Gleason from which KS is a simple corollary:
http://www.kiko.fysik.su.se/en/thesis/helena-master.pdf

The probability assumption of Gleason is simply defining a measure on the space which of course is all the Kolmogorov axioms are.

Thanks
Bill

 

[rubi]

However reading this thread one thing that struck me was this harping on about KS. IMHO the better thing to look at is Gleason from which KS is a simple corollary:
http://www.kiko.fysik.su.se/en/thesis/helena-master.pdf

Yes, given Gleason's theorem, one can prove KS quite easily. However, the proof of Gleason is much harder than modern proofs of KS.

The probability assumption of Gleason is simply defining a measure on the space which of course is all the Kolmogorov axioms are.

That's not completely right. In addition to the algebraic relations, the axioms of classical probability theory require the domain of the measure to be a sigma algebra. The difference between quantum theory and classical probability theory is exactly this circumstance. That's also nicely explained in Varadarajan's book. Otherwise, some central concepts like integrals or conditional probabilities can't be defined.

By looking at any proof of Bell's theorem, one can easily see that such concepts from probability theory are used extensively. Hence, Bell's theorem cannot be proved without assuming classical probability theory. It's pretty much a triviality, not even Ilja doubted this (which is why he objected to the idea that QT isn't a classical probability theory. Of course he had to fail, since this is established science.). This assumption is usually called "realism", although that is a pretty stupid name in my opinion. More preferable names would be classicality, hidden variables, non-contextuality or simplicity of the state space.

 

[bhobba]

That's also nicely explained in Varadarajan's book. Otherwise, some central concepts like integrals or conditional probabilities can't be defined.

Hard that book may be, but penetrating of the quantum formalism it most certainly is.

Thanks
Bill