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

[DrChinese]

Its the same problem as what hidden variable is contained in two particles that tells them that they have collided with each other.
The particles contain no property like temperature, mass etc that gives them their location in space. So how do they know when they collide?

Are you talking about entanglement of observables? Particles do not need to "directly" interact (collide) to become entangled on a basis.

 

[Ilja]

Well, a measurement of course always depends on the state of the measured object and the measurement apparatus, and to measure something implies that the measured object must interact with the measurement apparatus independent of the religion you follow in your metaphysical worldview ;-)).

Fine. So you accept that Kochen-Specker is a theorem about theories where the result does not depend on the state of the measurement apparatus, but has to be predefined by the measured object, and is therefore not relevant for hidden variable theories at all?

 

[rubi]

In the minimal interpretation, QM does not pretend to describe reality. If an interpretation claims that no reality exists, I reject it as nonsensical. But the minimal interpretation does not make such claims, it simply does not give a description of reality.

Of course, QM describes reality. It's just that our naive picture of reality needs to be modified. Open minded people without philosophical prejudices about the world have no problem with that.

I never made such a claim. Reichenbach's principle claims the existence of causal explanations, like common causes. It also specifies what a common cause is.

There are rules of reasoning, which cannot be proven to be false by any observation, because to derive some nontrivial predictions - something which could be falsified by observation - has to use them. So, claiming that these rules are wrong would be simply the end of science as we know it. If we would take such a solution seriously, we would simply stop doing science, because it would be well-known that the methods we use are inconsistent. (Ok, also not a decisive argument - we do a lot of inconsistent things anyway.)

Whatever, there is a hierarchy, we have rules, hypotheses or so which make science possible, to reject them would make science meaningless. They are, of course, only human inventions too, but if they are wrong, doing science becomes meaningless. We would probably continue doing science, because humans like to continue to do things even if they have recognized that doing them is meaningless, which is what is named culture. But this culture named science would not be really science as it is today, an endeavor to understand reality, to find explanations, but like the atheist going to Church as part of his living in a formerly religious culture.

But this has not happened yet, at least for me doing science has yet some of its original meaning, and is an endeavor to understand reality, to find explanations which are consistent with the rules of logic, of consistent reasoning. And this requires that some ideas, like the rules of logic, of consistent reasoning, the existence of some external reality, and the existence of explanations, have to be true.

It is not only the point that giving them up would make science meaningless. It is also that there is no imaginable evidence which would motivate it. Because, whatever the conflict with observation, this would be always only an open scientific problem. And giving up science because there are open scientific problems? Sorry, this makes no sense. Science without open scientific problems would be boring.

None of this makes sense. Science doesn't depend on some any of this. We're making progress almost on a daily basis.

Of course, one could imagine a Nature so that some beings in this Nature would be unable in principle to invent a theory about it without logical contradictions.

There are no logical contradictions in QM. QM is fully consistent with classical logic. If you don't agree, provide a counterexample.

Simply wrong. There are causal explanations, they are even quite simple and straightforward, but violate Einstein causality. This is not really a big problem. Anyway, the other appearances of a similar symmetry (like for acoustic wave equations, where also Lorentz transformation with the speed of sound allow to transform solutions into other solutions of the wave equation) are known to be not fundamental.

Simply wrong. The absence of common cause explanations cannot be proven for QM.

It is you who claims QM is in conflict with logic, namely with the rules of probability theory, which are, following Jaynes, Probability theory - the logic of science, the rules of consistent plausible reasoning. Consistent reasoning is not at all about intuition.

No, I claim that QM is fully consistent with logic. What you call logic isn't actually logic, but rather a formalization of classical intuition. It is unreasonable to expect nature to work according to classical intuition.

The Kochen-Specker theorem is about hidden variable models and uses assumptions beyond only probability theory.

No, the assumptions actually formalize some concepts that must be obeyed by a classical probability theory (non-contextuality). No non-classical probability theory will violate them.

I think Kochen and Specker themselves pointed out that you can always construct a joint probability distribution for different measurements just by taking the probabilities given by the Born rule and multiplying them, like I pointed out in my earlier post. I think their stance was that this sort of thing didn't make a very satisfactory hidden variable model, but if the exercise is just to invent a joint probability distribution in order to express quantum physics in some axiomatic language that requires it then it looks to me like you could do it this way.

The probability distribution you get by taking the product measure will be inconsistent with certain functional relationships between random variables that must hold in a classical probability theory. No probability distributions of random variables can be consistent with certain QM statistics. That is the theorem.

I wouldn't consider QM a probability theory in the first place. It's a physics theory that uses elements from probability theory as well as other areas of mathematics in its formulation.

QM is a theory that assigns probabilities to certain events. It does this in a fashion, which is incompatible with classical probability theory. If you don't want to call it a generalized probability theory, fine. That doesn't change the mathematical content of my statement.

You're acting as if you think all of QM should be seen as a special case of Kolmogorov probability theory. Why would we want to do this, independently of whether it is even possible?

I don't, but Ilja needs to, if he wants to apply Reichenbach's principle to QM. Reichenbach's principle requires a classical probability theory to be applied.

Likely untrue. Apparently debunked in this paper, which was linked from a recent post here (could not find back the post, but did find back the paper):

http://arxiv.org/abs/1305.1280 "The Pilot-Wave Perspective on Spin" -Norsen

This paper doesn't construct a classical probability theory with spin observables modeled as random variables. (By the way, it is even possible for a single isolated spin 1/2 particle, but that is pretty much the only exception.)

Oh, I haven't seen this. This clarifies the issue. Kochen-Specker presumes non-contextuality, while the known hidden variable theories like dBB are contextual.

Non-contextuality is exactly the assumption that observables can be modeled by classical random variables on a probability space, hence it proves my point. Of course dBB must be contextual, since Kochen-Specker proved that it cannot be non-contextual if it wants to reproduce QM.

This is, translated into layman language, the point I have made many times: In a contextual theory, what is named "measurement" is something different, an interaction, and its result depends as of the state of the "measured system", as of the state of the "measurment device", so that it is not a property of the system which is "measured".

I know that hidden variables must be contextual. That is exactly my point. You cannot cook up a classical probability theory that reproduces all statistics that can be computed from quantum mechanics.

 

[Ilja]

Of course, QM describes reality. It's just that our naive picture of reality needs to be modified. Open minded people without philosophical prejudices about the world have no problem with that.

"Open minded" people without prejudices have also no problem to accept Buddhism as a description of reality. Sorry for being closed minded on this, but for me a realistic theory has to describe all what is supposed to exist in reality, and this should include all the things around us which nobody doubts really exist. Including some equations how they change their states.

None of this makes sense. Science doesn't depend on some any of this. We're making progress almost on a daily basis.

Of course we make progress - because we do not give up the search for realistic causal explanations. Everywhere except in fundamental physics.

Simply wrong. The absence of common cause explanations cannot be proven for QM.

It can be. We observe 100% correlations if A and B measure the same direction. The common cause explanation would be that some common cause $\lambda$ defines this measurement result. This common cause exists in the past, thus, with some probability distribution $\rho(\lambda)$ independent of a and b. And it defines the measurement results A and B. So that we have the functions $A (a,'lambda), B(b,\lambda)$ we need to prove Bell's inequality. Once Bell's inequality is violated, the common cause explanation is excluded.

No, I claim that QM is fully consistent with logic. What you call logic isn't actually logic, but rather a formalization of classical intuition.

No. The essential argument used by Jaynes is consistency, which is a sufficiently precise requirement, and not some diffuse intuition.

No, the assumptions actually formalize some concepts that must be obeyed by a classical probability theory (non-contextuality). No non-classical probability theory will violate them.

dBB violates them, it is contextual, despite the fact that it is a completely classical, consistent, realistic, causal, deterministic theory. So, with non-contextuality you add some philosophical prejudice to probability theory which is not part of it.

QM is a theory that assigns probabilities to certain events. It does this in a fashion, which is incompatible with classical probability theory. If you don't want to call it a generalized probability theory, fine. That doesn't change the mathematical content of my statement.

It is incompatible with non-contextuality. Not with classical probability theory. Don't mingle a particular confusion about what happens (naming interactions "measurements" and results of the interactions "measurement results" even if nothing indicates that the result depends only on one part of the interaction) with the fundamental laws of consistent plausible reasoning known as probability theory.

I don't, but Ilja needs to, if he wants to apply Reichenbach's principle to QM. Reichenbach's principle requires a classical probability theory to be applied.

There is no problem with this. I can always use consistent reasoning. And I know that the laws of consistent plausible reasoning are those of probability theory. Read Jaynes.

This paper doesn't construct a classical probability theory with spin observables modeled as random variables. (By the way, it is even possible for a single isolated spin 1/2 particle, but that is pretty much the only exception.)

It does not construct what you name "classical probability theory", and what other people name a non-contextual model, because nobody needs it and nobody thinks that it correctly describes reality.

Non-contextuality is exactly the assumption that observables can be modeled by classical random variables on a probability space, hence it proves my point.

Maybe I will start to name my ether theory "classical logic"? This would allow me to accuse everybody who rejects the ether of making logical errors, and prove that ether theory follows from logic alone? That would be similar to your attempt to give non-contextuality (a very strange assumption about results of interactions, which would be appropriate only for a very special subclass of interactions named measurements) the status of of an axiom of consistent plausible reasoning.

I know that hidden variables must be contextual. That is exactly my point. You cannot cook up a classical probability theory that reproduces all statistics that can be computed from quantum mechanics.

I can. Take dBB theory in quantum equilibrium. Read Bohm 1952 for the proof.

 

[rubi]

It can be. We observe 100% correlations if A and B measure the same direction. The common cause explanation would be that some common cause $\lambda$ defines this measurement result. This common cause exists in the past, thus, with some probability distribution $\rho(\lambda)$ independent of a and b. And it defines the measurement results A and B. So that we have the functions $A (a,'lambda), B(b,\lambda)$ we need to prove Bell's inequality. Once Bell's inequality is violated, the common cause explanation is excluded.

No, your probability distribution needn't exist. In a quantum world, the common cause $\lambda$ might not commute with the observables of $A$ and $B$, hence your joint probability distribution might not exist and so none of the remaining reasoning can be carried out without a common cause principle that works for non-commuting observables. It's that simple. Reichenbach's principle can't be applied in this situation. It's just not general enough.

I can. Take dBB theory in quantum equilibrium. Read Bohm 1952 for the proof.

dBB theory doesn't reproduce the statistics of spin independent of measurement contexts. QM predicts probability distributions for spin independent of a measurement context.

 

[wle]

No, the assumptions actually formalize some concepts that must be obeyed by a classical probability theory (non-contextuality). No non-classical probability theory will violate them.

How so? The Kochen-Specker theorem includes an assumption that (deterministic) values v associated with quantum observables (Hermitian operators) satisfy conditions like $v(A + B) = v(A) + v(B)$ and $v(AB) = v(A) v(B)$ for all commuting $A$ and $B$. Quantum observables are a concept specific to quantum physics that doesn't appear at all in probability theory.

No probability distributions of random variables can be consistent with certain QM statistics. That is the theorem.

I don't think you've justified that.

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 what? However you classify the type of probability theory quantum physics uses, we've been using it in physics since QM was first formulated back in the 1920s and 1930s and for the most part we don't think anything special of it. In particular, when we talk about quantum behaviour and correlations and we contrast this with various types of "classical" behaviour, this is not what we are talking about.

I don't, but Ilja needs to, if he wants to apply Reichenbach's principle to QM. Reichenbach's principle requires a classical probability theory to be applied.

I wouldn't agree with this. The main reason QM doesn't fit neatly into Kolmogorov probability theory is that we treat measurement choice as a free variable. I wouldn't consider this a good reason to shut down a discussion about whether certain types of causal explanation for quantum correlations are possible or not.

An example: if you allow sufficiently fast classical communication then it's possible to simulate arbitrary (even Bell-violating) quantum correlations. If you gave me a "magic" ethernet cable that could transmit data instantaneously then I could program two computers, communicating with each other using this cable and accepting measurement choices as inputs, to generate outputs in accord with correlations predicted by QM. I would in principle consider this sort of thing a valid candidate causal explanation for QM correlations.

 

[rubi]

How so? The Kochen-Specker theorem includes an assumption that (deterministic) values v associated with quantum observables (Hermitian operators) satisfy conditions like $v(A + B) = v(A) + v(B)$ and $v(AB) = v(A) v(B)$ for all commuting $A$ and $B$. Quantum observables are a concept specific to quantum physics that doesn't appear at all in probability theory.

Because random variables satisfy these properties by definition: $(AB)(x) = A(x)B(x)$, because this is how the $AB$ is defined. Same for addition. It has nothing to do with quantum observables, it's just the functional relationships between the valuations. Kochen-Specker assumes that the random variables that must represent the quantum observables in the classical probability theory must satisfy the usual functional relationshipts that classical random variables obey by definition. Kochen-Specker essentially says that you cannot embedd the quantum probabilities into classical probability theory without having to readjust the very definitions of multiplication and addition of random variables.

I don't think you've justified that.

The Kochen-Specker theorem proves it.

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 what? However you classify the type of probability theory quantum physics uses, we've been using it in physics since QM was first formulated back in the 1920s and 1930s and for the most part we don't think anything special of it. In particular, when we talk about quantum behaviour and correlations and we contrast this with various types of "classical" behaviour, this is not what we are talking about.

It is not problematic that QM uses a generalized way for computing probabilities. It just means that certain concepts that used to work in classical probability theory cannot be carried over to the quantum framework without modification (such as Reichenbach's principle). This doesn't change how we use quantum theory in any way. Most physicist even understand intuitively how to correctly use quantum mechanics without understanding mathematically what is different about it.

I wouldn't agree with this. The main reason QM doesn't fit neatly into Kolmogorov probability theory is that we treat measurement choice as a free variable

No, the reason for why QM doesn't fit into Kolmogorov probability theory is that the event algebra is a certain orthomodular lattice, rather than a sigma algebra. The probability "measure" assigns probabilities to elements of these algebras. While in a sigma algebra, there always exists a third element $A\wedge B$ (the "meet"), given the events $A$ and $B$ (namely $A\cap B$), this is no longer true for an orthomodular lattice. However, Kolmogorov's axioms of probability theory depend on the existence of $A\wedge B$. 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.

I wouldn't consider this a good reason to shut down a discussion about whether certain types of causal explanation for quantum correlations are possible or not.

An example: if you allow sufficiently fast classical communication then it's possible to simulate arbitrary (even Bell-violating) quantum correlations. If you gave me a "magic" ethernet cable that could transmit data instantaneously then I could program two computers, communicating with each other using this cable and accepting measurement choices as inputs, to generate outputs in accord with correlations predicted by QM. I would in principle consider this sort of thing a valid candidate causal explanation for QM correlations.

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. Ilja wants to exclude those by naively using concepts of classical probability theory, which are known to not even be well-defined in the context of quantum theory. Certainly, this is not valid reasoning.

 

[atyy]

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. Ilja wants to exclude those by naively using concepts of classical probability theory, which are known to not even be well-defined in the context of quantum theory. Certainly, this is not valid reasoning.

But that is what Bell's theorem says. As we have discussed, you can escape it by using a more general notion of cause than that used in Bell's theorem, which is fine. But in that case you should simply clarify your terminology.

 

[rubi]

But that is what Bell's theorem says.

Bell's theorem is a theorem about theories formulated in the language of classical probability theory. It's not a theorem about quantum theory. We just use it to conclude that quantum theory is different from classical probability by noting that QM can violate an inequality that is not violated by certain classical theories. Bell says that there is a class of theories (local hidden variable theories) that satisfy a certain inequality. QM is not in that class. Bell's theorem can't be used to conclude anything about theories that are not both local and hidden variable theories.

As we have discussed, you can escape it by using a more general notion of cause than that used in Bell's theorem, which is fine. But in that case you should simply clarify your terminology.

No, it's not a more general notion of cause. Cause means the same as always. What changes is how we can tell what consistutes a cause just by looking at statistics. However, if the statistics is not compatible with the classical probability theory anymore, then it is to be expected that one needs to adjust the method which is used to tell whether some statistics hints at a causal relationship or not. Specifically, using Reichenbach's principle makes no sense in the context of quantum statistics. It's just not applicable here. That just means that we have no clear criterion that tells us what consitutes a common cause. The notion of cause itself is not modified.

 

[bhobba]

Nobody knows what the right assumptions about reality are. It's your personal opinion that QM doesn't describe reality.

Indeed.



Thanks
Bill

 

[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