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

[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