Axiomatization of quantum mechanics and physics in general ?

[billschnieder]

No, I don't see a problem with it.

If you're concerned with part of the calculation doing something like \langle \psi \rvert A \lvert \psi \rangle + \langle \psi \rvert B \lvert \psi \rangle = \langle \psi \rvert (A + B) \lvert \psi \rangle, that's perfectly justified and follows from the fact that observables are represented by linear operators in quantum mechanics.

What if you found out that the expression $S_ψ=⟨ψ|(σL⋅a)[(σR⋅b)−(σR⋅b')]+(σL⋅a')[(σR⋅b)+(σR⋅b')]|ψ⟩$ can only have a solution if $(σL⋅a)$is collinear with $(σL⋅a')$, and $(σR⋅b)−(σR⋅b')$ is collinear with $(σR⋅b)+(σR⋅b')$, which is definitely not the case in the angle settings in play for EPRB experiment, will you still have no problem with it?

 

[wle]

What if you found out that the expression $S_ψ=⟨ψ|(σL⋅a)[(σR⋅b)−(σR⋅b')]+(σL⋅a')[(σR⋅b)+(σR⋅b')]|ψ⟩$ can only have a solution if $(σL⋅a)$is collinear with $(σL⋅a')$, and $(σR⋅b)−(σR⋅b')$ is collinear with $(σR⋅b)+(σR⋅b')$, which is definitely not the case in the angle settings in play for EPRB experiment, will you still have no problem with it?

What is that even supposed to mean? You've writted down a mathematical expression defined in terms of a state vector and certain Hermitian operators. Given the state vector and operators, you can compute its value quite easily whether or not all the operators commute. Where did you get the idea it doesn't have a solution?

 

[Alien8]

I'm simply trying to find out how the "trivial substitution" is interpreted, since it appears to me that the correct interpretation is at odds with the CHSH derivation, and the interpretation which is in agreement with the derivation does not make sense.

Yes, I only think you're being too abstract to pin-point the origin of the problem. I think the root of the answer to your question, and my question, is in understanding the justification to apply the triangle inequality. It's the only "reason", function, logic, system or mechanism that binds it all together, everything else are just variables.

By "substitution" do you mean the terms in the inequality are being substituted by actual measurement numbers?

Like I said, there is no problem with the substitution if we interpret is as corresponding to 4 isolated systems, but as you can see above, the inequality is different, S <= 4. The problem only arises if you interpret the substitution as pertaining to the same system for which S <=2. So when it is said that QM violates the S <= 2 inequality, it is suspect because if we carry that argument, we would have to treat the substitution as pertaining to the same system and we end up with an expression that has no solution because it is impossible to find an eigenvector for that specific combination of observables.

|E(a,b) - E(a,b&#039;)| \leq 2 \pm [E(a&#039;,b) + E(a&#039;,b&#039;)] is only true for E(\alpha,\beta) = cos2(\beta - \alpha). It also does not apply to arbitrary independent expectation values. All angles need to be independent to get four independent expectation values {-1,+1}, which should indeed be bound to 4, but then the equation would look like this: |E(a,b) - E(c,d)| \leq 4 \pm [E(e,f) + E(g,h)].

 

[bhobba]

By "substitution" do you mean the terms in the inequality are being substituted by actual measurement numbers?

I have taken leave of discussing this because his issue has me utterly beat.

It's related to what I posted earlier in the thread:
http://en.wikipedia.org/wiki/Bell's...re_violated_by_quantum_mechanical_predictions

Its a simple matter of working through the equalities and substituting.

Nothing else is involved - yet he doubts it.

In my experience when it reaches discussion of something so utterly trivial you won't get anywhere - its a waste of time.

Thanks
Bill

 

[billschnieder]

In CHSH the observables are measured on different subsets, the same as $X^{2}$, $P^{2}$ and $E$. So if you consider the example I gave to have no problem, there is no problem in CHSH also.

Exactly, there is no problem if the CHSH terms are measured on different subsets because in that case they do not commute either. That is why a while back I said

Therefore:
⟨A(a)B(b)⟩+⟨A(a′)B(b′)⟩+⟨A(a′)B(b)⟩−⟨A(a)B(b′)⟩\langle A(a) B(b) \rangle + \langle A(a') B(b') \rangle + \langle A(a') B(b) \rangle - \langle A(a) B(b') \rangle
Can be interpreted in two ways:
1. Each term represents a measurement on an separate isolated systems.
2. Each term represents observables on the same system.

This mean you are picking interpretation (1) where there is no problem with the QM calculation. But unfortunately the CHSH is derived under interpretation (2), where there would be a commutation problem if you would select it. For example wle thinks it should be (2).

 

[atyy]

That is why a while back I said

This mean you are picking interpretation (1) where there is no problem with the QM calculation. But unfortunately the CHSH is derived under interpretation (2), where there would be a commutation problem if you would select it. For example wle thinks it should be (2).

Your use of the term "non-commuting" is very non-standard. But let's talk terminology later.

CHSH is derived under interpretation (1). Each of the 4 terms in CHSH represents measurements made on different subsets. They are often said to be "on the same system" in the sense that the wave function or ensemble is the same in each case. But this is just a difference in terminology.

So to make the analogy, in $R = E -X^{2} - P^{2}$, to get $R=0$ each term on the RHS is measured on a different subset, and then added together. Quantum mechanically, we get the prediction by calculating each term separately, then adding them together. However, because of the mathematical structure of quantum mechanics, we can add all the operators together first ie. $R = \langle\psi|H - x^{2} - p^{2}|\psi\rangle$ even though the operators are non-commuting. If you like, this is just a valid mathematical trick to get the correct quantum mechanical prediction.

Similarly in CHSH, the terms are separate. For calculating the quantum mechanical prediction for a CHSH experiment, we can add all the operators together, then take the expectation of the sum of operators. If you don't like it, you can evaluate each term separately, and then perform the addition. The result is mathematically the same. Physically, of course, the experiment is to measure each term on a different subset.

 

[billschnieder]

What is that even supposed to mean? You've writted down a mathematical expression defined in terms of a state vector and certain Hermitian operators. Given the state vector and operators, you can compute its value quite easily whether or not all the operators commute. Where did you get the idea it doesn't have a solution?

Because it is impossible to find an eigenvector for that expression. Essentially you have an operator of the form
$\hat{A}\otimes\hat{B} + \hat{C}\otimes\hat{D}$ for which $[\hat{A},\hat{C}] ≠0$ and $[\hat{B},\hat{D}] ≠0$ since you said we are dealing with a single system. If an eigenvector exists, say $|\phi\rangle\otimes|\chi\rangle$ then
$[\hat{A}\otimes\hat{B} + \hat{C}\otimes\hat{D}]|\phi\rangle\otimes|\chi\rangle = \alpha(|\phi\rangle\otimes|\chi\rangle)$
$\hat{A}|\phi\rangle\otimes\hat{B}|\chi\rangle + \hat{C}|\phi\rangle\otimes\hat{D}|\chi\rangle = \alpha(|\phi\rangle\otimes|\chi\rangle)$
Which only has solutions of $\hat{A}|\phi\rangle$ and $\hat{C}|\phi\rangle$ or $\hat{B}|\chi\rangle$ and $\hat{D}|\chi\rangle$ are collinear. But they are not. Therefore that combination can not be an observable for a single system.

 

[billschnieder]

CHSH is derived under interpretation (1). Each of the 4 terms in CHSH represents measurements made on different subsets.

This is not correct. The 4 terms in the CHSH are derived from a single subset. I can back this up with published references and derivations if you are in doubt. I think even wle agrees with me:

The same system. In the proof of Bell's theorem, the CHSH correlator is a function of the joint probability distribution which is entirely defined for a single system.

However, if I convince you that the CHSH is derived for a single system, not separate realizations of a similar system, would you agree that there is an issue?

 

[atyy]

This is not correct. The 4 terms in the CHSH are derived from a single subset. I can back this up with published references and derivations if you are in doubt. I think even wle agrees with me:

They are not, by definition. The terms E(a,b) and E(a,b') are by definition different measurement settings and hence different measurements.

However, if I convince you that the CHSH is derived for a single system, not separate realizations of a similar system, would you agree that there is an issue?

Yes.

 

[wle]

Because it is impossible to find an eigenvector for that expression. Essentially you have an operator of the form
$\hat{A}\otimes\hat{B} + \hat{C}\otimes\hat{D}$ for which $[\hat{A},\hat{C}] ≠0$ and $[\hat{B},\hat{D}] ≠0$ since you said we are dealing with a single system. If an eigenvector exists, say $|\phi\rangle\otimes|\chi\rangle$ then
$[\hat{A}\otimes\hat{B} + \hat{C}\otimes\hat{D}]|\phi\rangle\otimes|\chi\rangle = \alpha(|\phi\rangle\otimes|\chi\rangle)$
$\hat{A}|\phi\rangle\otimes\hat{B}|\chi\rangle + \hat{C}|\phi\rangle\otimes\hat{D}|\chi\rangle = \alpha(|\phi\rangle\otimes|\chi\rangle)$
Which only has solutions of $\hat{A}|\phi\rangle$ and $\hat{C}|\phi\rangle$ or $\hat{B}|\chi\rangle$ and $\hat{D}|\chi\rangle$ are collinear. But they are not. Therefore that combination can not be an observable for a single system.

Huh? The expression $\langle \psi \rvert (\ldots) \lvert \psi \rangle$ that you wrote is a number, not an eigenvalue/vector problem. There's no need to find eigenvectors for it at all.

That said, although this isn't especially important here, if $A$, $B$, $C$, and $D$ are Hermitian operators (i.e., observables), then $A \otimes B + C \otimes D$ is also Hermitian and thus has real-valued eigenvalues and orthogonal eigenstates. It's just that the eigenstates are generally entangled. For instance, for the optimal CHSH measurements, up to a choice of basis you can write the operator appearing in the quantum CHSH expression as $$\sqrt{2} \bigl( \sigma_{z} \otimes\sigma_{z} + \sigma_{x} \otimes \sigma_{x} \bigr) \,.$$ This has two nonzero eigenvalues, $2 \sqrt{2}$ and $- 2 \sqrt{2}$, associated respectively with the eigenstates $$\begin{eqnarray}
\lvert \Phi^{+} \rangle &=& \frac{1}{\sqrt{2}} \bigl( \lvert 0 \rangle \lvert 0 \rangle + \lvert 1 \rangle \lvert 1 \rangle \bigr) \,, \\
\lvert \Psi^{-} \rangle &=& \frac{1}{\sqrt{2}} \bigl( \lvert 0 \rangle \lvert 1 \rangle - \lvert 1 \rangle \lvert 0 \rangle \bigr) \,.
\end{eqnarray}$$ If you think about it, that's actually exactly the sort of thing you should want from a good Bell correlator.

 

[billschnieder]

Huh? The expression $\langle \psi \rvert (\ldots) \lvert \psi \rangle$ that you wrote is a number, not an eigenvalue/vector problem. There's no need to find eigenvectors for it at all.

Of course I'm talking about the expression $(\ldots)$. Isn't that observable which can be represent by $\hat{A}\otimes\hat{B} + \hat{C}\otimes\hat{D}$ where $\hat{A}$, $\hat{B}$, $\hat{C}$, and $\hat{D}$ are Hermitian operators. So it is an eigenvalue/vector problem. Doesn't a meaningful number exist for $\langle \psi \rvert (\ldots) \lvert \psi \rangle$ only if it is possible to find an eigenvalue for $(\ldots)$, ie $[\hat{A}\otimes\hat{B} + \hat{C}\otimes\hat{D}]$?

$$\sqrt{2} \bigl( \sigma_{z} \otimes\sigma_{z} + \sigma_{x} \otimes \sigma_{x} \bigr) \,.$$

But what you have is an operator of the form $\hat{A}\otimes\hat{A} + \hat{B}\otimes\hat{B}$, quite a bit different from $\hat{A}\otimes\hat{B} + \hat{C}\otimes\hat{D}$.

 

[wle]

Of course I'm talking about the expression $(\ldots)$. Isn't that observable which can be represent by $\hat{A}\otimes\hat{B} + \hat{C}\otimes\hat{D}$ where $\hat{A}$, $\hat{B}$, $\hat{C}$, and $\hat{D}$ are Hermitian operators. So it is an eigenvalue/vector problem.

No, it's not. Just read the expression applying basic definitions from linear algebra. $\lvert \psi \rangle$ is a vector. $A \otimes B + C \otimes D$ is a linear operator. A linear operator acting on a vector by definition produces another vector, so $(A \otimes B + C \otimes D) \lvert \psi \rangle = \lvert \chi \rangle$ is a vector. Finally, $\langle \psi \rvert$ is a covector, and a covector acting on a vector produces a number (the inner product). So $\langle \psi \rvert (A \otimes B + C \otimes D) \lvert \psi \rangle = \langle \psi | \chi \rangle$ is a number.

For a given vector and operators, the expression $\langle \psi \rvert (A \otimes B + C \otimes D) \lvert \psi \rangle$ has a well defined value that you can evaluate without ever needing to diagonalise anything.

But what you have is an operator of the form $\hat{A}\otimes\hat{A} + \hat{B}\otimes\hat{B}$, quite a bit different from $\hat{A}\otimes\hat{B} + \hat{C}\otimes\hat{D}$.

You seem to be missing the point here in three ways:

1. While $2 \sqrt{2} (\sigma_{z} \otimes \sigma_{z} + \sigma_{x} \otimes \sigma_{x})$ isn't the most general expression of the form $A \otimes B + C \otimes D$, it is what you get if you substitute in the optimal measurements for CHSH. This is the example that $S = 2 \sqrt{2}$ is obtained with, and I've explicitly given you the eigenstates for it.
2. That example generalises anyway. If $A$, $B$, $C$, and $D$ are Hermitian, then $A \otimes B + C \otimes D$ is also Hermitian and is thus diagonalisable. (All Hermitian operators are diagonalisable, with real eigenvalues and orthogonal eigenstates.)
3. As I explained above, you don't even need to diagonalise anything to compute $\langle \psi \rvert (A \otimes B + C \otimes D) \lvert \psi \rangle$ anyway.

 

[atyy]

For a given vector and operators, the expression $\langle \psi \rvert (A \otimes B + C \otimes D) \lvert \psi \rangle$ has a well defined value that you can evaluate without ever needing to diagonalise anything.

If I understand bilschneider correctly, his concern isn't with algebra. He wants to know whether the expectations $\langle \psi \rvert A \otimes B \lvert \psi \rangle$ and $\langle \psi \rvert C \otimes D \lvert \psi \rangle$ are from measurements on the same realization or different realizations of the ensemble represented by $\lvert \psi \rangle$. His contention (correct, I believe) is that in the Bell tests they are made on different realizations of the same ensemble.

 

[billschnieder]

However, if I convince you that the CHSH is derived for a single system, not separate realizations of a similar system, would you agree that there is an issue?

Yes.

OK, according to the derivation (http://en.wikipedia.org/wiki/Bell's_theorem#Derivation_of_CHSH_inequality)
$A=A(a, \lambda), A'=A(a', \lambda), B=B(b, \lambda), B'=B(b', \lambda)$
$\begin{align} \rho(a,b) + \rho(a,b') + \rho(a',b) - \rho(a',b')&= \int_\Lambda AB\rho +\int_\Lambda AB'\rho +\int_\Lambda A'B\rho -\int_\Lambda A'B'\rho \\ &= \int_\Lambda (AB+AB'+A'B-A'B')\rho\\ &= \int_\Lambda (A(B+B') + A'(B-B')) \rho\\ &\leq 2 \end{align}$
The heart of the derivation is the 4th line above:

$AB+AB'+A'B-A'B'= A(B+B')+A'(B-B') \le 2.$
That factorization can not be done for different realizations of the same ensemble

Alternatively, For a single set $q$ of N particle pairs, with N sufficiently large
$S_q = \frac{1}{N} \sum_{i=1}^{N} A(a,\lambda_i)B(b,\lambda_i) - \frac{1}{N} \sum_{i=1}^{N} A(a,\lambda_i)B(b',\lambda_i) + \frac{1}{N} \sum_{i=1}^{N} A(a',\lambda_i)B(b,\lambda_i) + \frac{1}{N} \sum_{i=1}^{N} A(a',\lambda_i)B(b',\lambda_i)$
Which I can easily factorize like
$S_p = \frac{1}{N} \sum_{i=1}^{N} (A(a,\lambda_i)[B(b,\lambda_i) - B(b',\lambda_i)] + A(a',\lambda_i)[B(b,\lambda_i) + B(b',\lambda_i)])$
$A, B$ can only take values $\pm 1$, therefore whenever $B(b,\lambda_i) - B(b',\lambda_i)$ is 2, $B(b,\lambda_i) + B(b',\lambda_i)$ must be 0. The possible values within the sum are -2, 0, 2. Therefore $|S_p| \le 2$

For a 4 different sets $r,s,t,u$ of M,N,O,P particle pairs respectively, you instead have:
$S_{rstu} = \frac{1}{M} \sum_{i=1}^{M} A(a,\lambda_i)B(b,\lambda_i) - \frac{1}{N} \sum_{j=1}^{N} A(a,\lambda_j)B(b',\lambda_j) + \frac{1}{O} \sum_{k=1}^{O} A(a',\lambda_k)B(b,\lambda_k) + \frac{1}{P} \sum_{l=l}^{P} A(a',\lambda_l)B(b',\lambda_l)$

Which we can't factorize any further. In addition, Each of terms can independently attain the extrema of [-1, +1]. Therefore $|S_{rstu}| \le 4$.

 

[atyy]

OK, according to the derivation (http://en.wikipedia.org/wiki/Bell's_theorem#Derivation_of_CHSH_inequality)
$A=A(a, \lambda), A'=A(a', \lambda), B=B(b, \lambda), B'=B(b', \lambda)$
$\begin{align} \rho(a,b) + \rho(a,b') + \rho(a',b) - \rho(a',b')&= \int_\Lambda AB\rho +\int_\Lambda AB'\rho +\int_\Lambda A'B\rho -\int_\Lambda A'B'\rho \\ &= \int_\Lambda (AB+AB'+A'B-A'B')\rho\\ &= \int_\Lambda (A(B+B') + A'(B-B')) \rho\\ &\leq 2 \end{align}$
The heart of the derivation is the 4th line above:

$AB+AB'+A'B-A'B'= A(B+B')+A'(B-B') \le 2.$
That factorization can not be done for different realizations of the same ensemble

Alternatively, For a single set $q$ of N particle pairs, with N sufficiently large
$S_q = \frac{1}{N} \sum_{i=1}^{N} A(a,\lambda_i)B(b,\lambda_i) - \frac{1}{N} \sum_{i=1}^{N} A(a,\lambda_i)B(b',\lambda_i) + \frac{1}{N} \sum_{i=1}^{N} A(a',\lambda_i)B(b,\lambda_i) + \frac{1}{N} \sum_{i=1}^{N} A(a',\lambda_i)B(b',\lambda_i)$
Which I can easily factorize like
$S_p = \frac{1}{N} \sum_{i=1}^{N} (A(a,\lambda_i)[B(b,\lambda_i) - B(b',\lambda_i)] + A(a',\lambda_i)[B(b,\lambda_i) + B(b',\lambda_i)])$
$A, B$ can only take values $\pm 1$, therefore whenever $B(b,\lambda_i) - B(b',\lambda_i)$ is 2, $B(b,\lambda_i) + B(b',\lambda_i)$ must be 0. The possible values within the sum are -2, 0, 2. Therefore $|S_p| \le 2$

For a 4 different sets $r,s,t,u$ of M,N,O,P particle pairs respectively, you instead have:
$S_{rstu} = \frac{1}{M} \sum_{i=1}^{M} A(a,\lambda_i)B(b,\lambda_i) - \frac{1}{N} \sum_{j=1}^{N} A(a,\lambda_j)B(b',\lambda_j) + \frac{1}{O} \sum_{k=1}^{O} A(a',\lambda_k)B(b,\lambda_k) + \frac{1}{P} \sum_{l=l}^{P} A(a',\lambda_l)B(b',\lambda_l)$

Which we can't factorize any further. In addition, Each of terms can independently attain the extrema of [-1, +1]. Therefore $|S_{rstu}| \le 4$.

CHSH has two parts: LHS and RHS. The LHS is the thing that is assembled from measurements. The RHS is what is predicted assuming local realism (loosely speaking, since that doesn't seem to be the issue). So there is no RHS in the quantum case. The quantum case is a calculation of the LHS. So in quantum mechanics we don't even get to the RHS of the first line. The quantum mechanical prediction is a prediction for the LHS.

 

[atyy]

The heart of the derivation is the 4th line above:

$AB+AB'+A'B-A'B'= A(B+B')+A'(B-B') \le 2.$
That factorization can not be done for different realizations of the same ensemble

The LHS already assumes they are different realizations of the same ensemble. To proceed to the RHS, we assume local realism. After the first line of the RHS, it is algebra.

 

[billschnieder]

CHSH has two parts: LHS and RHS.

In the proof above, the RHS is simply an expansion of the LHS. The CHSH is not just |S| it is the inequality $|S| \le 2$. In the proof they are trying calculate the upper bound for $|S|$. In the QM calculation we are simply calculating the value for |S|. In experiments they are simply measuring |S|. Once you have S from all those places, you can then compare the value you get with the upper bound to see if there is agreement or not. But you can calculate or measure |S| by taking terms from different realizations or you can calculate it from the same realization. What I'm showing above is that the proof which culminates in $|S| \le 2$ uses the assumption that the terms are calculated from the same realization. Because you can not factorize different realizations like is done in the proof. And if you can't factorize, you end up with $|S| \le 4$ which is not violated by QM or experiments.

 

[atyy]

In the proof above, the RHS is simply an expansion of the LHS. The CHSH is not just |S| it is the inequality $|S| \le 2$. In the proof they are trying calculate the upper bound for $|S|$. In the QM calculation we are simply calculating the value for |S|. In experiments they are simply measuring |S|. Once you have S from all those places, you can then compare the value you get with the upper bound to see if there is agreement or not. But you can calculate or measure |S| by taking terms from different realizations or you can calculate it from the same realization. What I'm showing above is that the proof which culminates in $|S| \le 2$ uses the assumption that the terms are calculated from the same realization. Because you can not factorize different realizations like is done in the proof. And if you can't factorize, you end up with $|S| \le 4$ which is not violated by QM or experiments.

But you did not show it. The key assumption is the RHS of the first line. All subsequent steps are just algebra, including the factorization. The LHS and the first line already assume that each term is measured on a different realization. So if you are happy with both the LHS and the RHS of the first line, then the subsequent lines follow, including the factorization.

 

[billschnieder]

The LHS already assumes they are different realizations of the same ensemble. To proceed to the RHS, we assume local realism. After the first line of the RHS, it is algebra.

The algebra you are talking about is algebra for a single realization. In the expression $AB - AB' + A'B + A'B' \le 2$ for the same realization, the terms are not independent, you can factorize $A(B-B') + A'(B+B')$, Once the value of $(B+B')$ is determined, the other $B-B'$ is also determined. But you can not factorize different realizations in that way, because each term is free to vary independently of all the others. The number of degrees of freedom is 4 times larger. That is why you have the maximum of 2 for the same realization and 4 for different realizations. I do not see how local realism allows you to reduce the upper bound for different realizations to 2. It certainly will not reduce the number of degrees freedom. That is why I asked earlier:

if I take a different pair of entangled spin-1/2 particles similar to the first pair, and measure them both at exactly the same angles as the first two, according to QM, I still get the exact same relationship between them, opposite results again.But does QM say that The first particle from the first pair will have exactly opposite results from the second particle of the second pair?

Local realism does not allow me to say that the relationship between one particle and it's twin should be the same as the relationship between one particle in one pair and another particle in a different pair.

 

[atyy]

The algebra you are talking about is algebra for a single realization. In the expression $AB - AB' + A'B + A'B' \le 2$ for the same realization, the terms are not independent, you can factorize $A(B-B') + A'(B+B')$, Once the value of $(B+B')$ is determined, the other $B-B'$ is also determined.

It is not for a single realization, because it is within the integral over $\rho$. The thing that makes it factorizable is the assumption that $\rho$ is the same regardless of whether one chooses (a,b) or (a,b') or (a',b') or (a',b).

 

[billschnieder]

But you did not show it. The key assumption is the RHS of the first line. All subsequent steps are just algebra, including the factorization. The LHS and the first line already assume that each term is measured on a different realization. So if you are happy with both the LHS and the RHS of the first line, then the subsequent lines follow, including the factorization.

It is not for a single realization, because it is within the integral over $\rho$. The thing that makes it factorizable is the assumption that $\rho$ is the same regardless of whether one chooses(a,b) or (a,b') or (a',b') or (a',b).

But you start with 4 integrals and then combine them to one, and then you do algebra inside the integral, that means you are doing algebra for the same realization. The algebra inside the integral applies to every instance of the variable you are integrating over.

Besides, from that wikipedia page:

The CHSH inequality is seen to depend only on the following three key features of a local hidden variables theory: (1) realism: alongside of the outcomes of actually performed measurements, the outcomes of potentially performed measurements also exist at the same time;

If you are saying the CHSH is for 4 different realizations (which I disagree), which of those represent outcomes of potentially performed measurements?

 

[billschnieder]

In short, you seem to be saying for
For a 4 different realizations with sets of particle pairs $r,s,t,u$ of M,N,O,P particle pairs respectively:
$S_{rstu} = \frac{1}{M} \sum_{i=1}^{M} A(a,\lambda_i)B(b,\lambda_i) - \frac{1}{N} \sum_{j=1}^{N} A(a,\lambda_j)B(b',\lambda_j) + \frac{1}{O} \sum_{k=1}^{O} A(a',\lambda_k)B(b,\lambda_k) + \frac{1}{P} \sum_{l=l}^{P} A(a',\lambda_l)B(b',\lambda_l)$

Which we factorize because r=s=t=u and M=N=O=P and i=j=k=l so that the algebra proceeds. Is that what you mean by the "realism assumption"? Because that can only happen for the same realization.

 

[atyy]

But you start with 4 integrals and then combine them to one, and then you do algebra inside the integral, that means you are doing algebra for the same realization. The algebra inside the integral applies to every instance of the variable you are integrating over.

In short, you seem to be saying for
For a 4 different realizations with sets of particle pairs $r,s,t,u$ of M,N,O,P particle pairs respectively:
$S_{rstu} = \frac{1}{M} \sum_{i=1}^{M} A(a,\lambda_i)B(b,\lambda_i) - \frac{1}{N} \sum_{j=1}^{N} A(a,\lambda_j)B(b',\lambda_j) + \frac{1}{O} \sum_{k=1}^{O} A(a',\lambda_k)B(b,\lambda_k) + \frac{1}{P} \sum_{l=l}^{P} A(a',\lambda_l)B(b',\lambda_l)$

Which we factorize because r=s=t=u and M=N=O=P and i=j=k=l so that the algebra proceeds. Is that what you mean by the "realism assumption"? Because that can only happen for the same realization.

The factorization follows from the first line you wrote:

$\rho(a,b) + \rho(a,b') + \rho(a',b) - \rho(a',b')= \int_\Lambda AB\rho +\int_\Lambda AB'\rho +\int_\Lambda A'B\rho -\int_\Lambda A'B'\rho$

In principle, if you wanted to prevent factorization, you would instead write

$\rho(a,b) + \rho(a,b') + \rho(a',b) - \rho(a',b')= \int_\Lambda AB\rho_{a,b} +\int_\Lambda AB'\rho_{a,b'} +\int_\Lambda A'B\rho_{a',b} -\int_\Lambda A'B'\rho_{a',b'}$.

But since you wrote the same $\rho$ in all 4 terms, the factorization follows. So to complain about the factorization, it should be the first line, not any of the subsequent steps.

If you are saying the CHSH is for 4 different realizations (which I disagree), which of those represent outcomes of potentially performed measurements?

CHSH is for 4 different realizations. I do understand there's another presentation in terms of "outcomes of potentially performed measurements", but I don't understand that well. So let's discuss that later.

 

[Alien8]

$AB+AB'+A'B-A'B'= A(B+B')+A'(B-B') \le 2.$
That factorization can not be done for different realizations of the same ensemble

Can you name the principle of mathematics you are referring to and what is the definition of "realization" and "ensemble"?

For a 4 different sets $r,s,t,u$ of M,N,O,P particle pairs respectively, you instead have:
$S_{rstu} = \frac{1}{M} \sum_{i=1}^{M} A(a,\lambda_i)B(b,\lambda_i) - \frac{1}{N} \sum_{j=1}^{N} A(a,\lambda_j)B(b',\lambda_j) + \frac{1}{O} \sum_{k=1}^{O} A(a',\lambda_k)B(b,\lambda_k) + \frac{1}{P} \sum_{l=l}^{P} A(a',\lambda_l)B(b',\lambda_l)$

Which we can't factorize any further. In addition, Each of terms can independently attain the extrema of [-1, +1]. Therefore $|S_{rstu}| \le 4$.

They are not independent sets. It's not (a-b), (c-d), (e-f), (g-h), it's (a-c), (a-d), (b-c), and (b-d). Whether or not they can independently attain -1 or +1 depends on expectation value function E(x,y) = ?. For E(x,y) = cos2(y-x), E(a-c), E(a-d), E(b-c) and E(b-d) can not independently attain -1 or +1, so instead of 4 the boundary for E(x,y) = cos2(y-x) is 2.83.

 

[Alien8]

While $2 \sqrt{2} (\sigma_{z} \otimes \sigma_{z} + \sigma_{x} \otimes \sigma_{x})$ isn't the most general expression of the form $A \otimes B + C \otimes D$, it is what you get if you substitute in the optimal measurements for CHSH.

What is "optimal measurement" you are referring to?

 

[Alien8]

In the proof above, the RHS is simply an expansion of the LHS. The CHSH is not just |S| it is the inequality $|S| \le 2$. In the proof they are trying calculate the upper bound for $|S|$. In the QM calculation we are simply calculating the value for |S|. In experiments they are simply measuring |S|. Once you have S from all those places, you can then compare the value you get with the upper bound to see if there is agreement or not.

Exactly.

What I'm showing above is that the proof which culminates in $|S| \le 2$ uses the assumption that the terms are calculated from the same realization.

Yes. And the assumption you are talking about, what makes them belong to the same "realization", is the triangle inequality. Is there any particular reason you're hesitant to consider this? CHSH derivation on Bell's_theorem Wikipedia page starts with what only comes at the end of the actual derivation:

Factorization is of little consequence if your question is what are those four terms doing together in the first place. Look at the main article for CHSH inequality and note there is no any 'less or equal' symbol until the triangle inequality is applied. There would be no any inequality without the triangle inequality, so if you can explain the justification how and why it applies to CHSH experimental setup you will answer the question how and why are those four terms supposed to be a part of the same system. Can you explain?
http://en.wikipedia.org/wiki/CHSH_inequality#Derivation_of_the_CHSH_inequality

 

[wle]

What is "optimal measurement" you are referring to?

The (qubit) measurements that result in the maximal quantum violation of the CHSH inequality. In a suitable basis you can take these to be $$\begin{eqnarray}
A &=& \frac{1}{\sqrt{2}} \bigl( \sigma_{z} + \sigma_{x} \bigr) \,, \\
A' &=& \frac{1}{\sqrt{2}} \bigl( \sigma_{z} - \sigma_{x} \bigr)
\end{eqnarray}$$ and $$\begin{eqnarray}
B &=& \sigma_{z} \,, \\
B' &=& \sigma_{x} \,.
\end{eqnarray}$$ For the CHSH Bell operator, this works out to $$\begin{eqnarray}
\mathcal{S} &=& A \otimes B + A \otimes B' + A' \otimes B - A' \otimes B' \\
&=& (A + A') \otimes B + (A - A') \otimes B' \\
&=& \sqrt{2} \, \sigma_{z} \otimes \sigma_{z} + \sqrt{2} \, \sigma_{x} \otimes \sigma_{x} \,.
\end{eqnarray}$$

 

[kith]

$AB+AB'+A'B-A'B'= A(B+B')+A'(B-B')$
That factorization can not be done for different realizations of the same ensemble.

How would you show this mathematically? The question here clearly is how to assign the correct mathematical symbols and their dependencies in the beginning. Terms like "realization" or "ensemble" are only relevant for this initial assignment. Afterwards, we are talking about numbers, functions and their algebra, and your notions get insignificant for the correctness of a statement.

If there's no disagreement regarding the initial assignment of symbols and you agree with the first line but disagree with the quoted line, this can only mean that the shorthand notation doesn't capture the subtlety you are after. So please show were you think that things go wrong mathematically by using the full notation.

 

[Abc2020ro]

You cannot do that. Mathematics only deals with quantity. By axiomatizing the natural world, you are failing to take into account the qualitative part of it. And one huge example is consciousness. Not to mention related phenomena, such as free will.

 

[billschnieder]

Can you name the principle of mathematics you are referring to and what is the definition of "realization" and "ensemble"?

I'm using those words because after a lengthy discussion with atty(a few pages back), he used them to describe what I meant. I would not normally use those words to describe it. To me if you assign each individual photon pair a unique identifier say $i$, then when I say a realization of the experiment, I mean that you have one set say $p$ of N particle pairs $i = 1..N$. If I now have a different realization, I mean you now have a completely different set say $q$ of M particle pairs $i=N+1..N+M, etc. None of the$i's$in$p$exist in$q$, even though the system producing the particle pairs may be generating them such that the probability distribution of hidden variables in$p$and in$q$are the same. An inequality derived entirely within$p$, is not the same thing as an inequality derived from one part of$p$and a different part of$q$etc. Just as the$AB = -1## condition when angles are the same does not apply for particles from two separate pairs.

 

[billschnieder]

Sorry, should have previewed ...

I'm using those words because after a lengthy discussion with atty(a few pages back), he used them to describe what I meant. I would not normally use those words to describe it. To me if you assign each individual photon pair a unique identifier say $i$, then when I say a realization of the experiment, I mean that you have one set say $p$ of N particle pairs $i = 1..N$. If I now have a different realization, I mean you now have a completely different set say $q$ of M particle pairs $i=N+1..N+M$, etc. None of the $i's$ in $p$ exist in $q$, even though the system producing the particle pairs may be generating them such that the probability distribution of hidden variables in $p$ and in $q$ are the same.

An inequality derived entirely within $p$, is not the same thing as an inequality derived from one part of $p$ and a different part of $q$ etc. Just as the $AB = -1$ condition when angles are the same does not apply for particles from two separate pairs.

 

[billschnieder]

They are not independent sets. It's not (a-b), (c-d), (e-f), (g-h), it's (a-c), (a-d), (b-c), and (b-d). Whether or not they can independently attain -1 or +1 depends on expectation value function E(x,y) = ?. For E(x,y) = cos2(y-x), E(a-c), E(a-d), E(b-c) and E(b-d) can not independently attain -1 or +1, so instead of 4 the boundary for E(x,y) = cos2(y-x) is 2.83.

Well, let x = (a-c), y=(a-d), z=(b-c), and then (b-d) = x+y-z. So you are right it is not completely independent and that will affect the upper bound so you would have an expression like E(x) - E(y) + E(z) +E(x+y-z). However, you can still evaluate this expression in two ways. You could calculate E(x) from one set of particles, E(y) from a different set of particles, E(z) from yet a different set of particles and E(x+y-z) from yet another set. no two particle pairs in any set belonging to any other set. In this sense, the sets are independent, even though the results are not entirely independent owning to the E(x+y-z). However, you could also take a single set of particles, and evaluate all 4 expressions on the exact same set, every particle pair contributing to every term. This is the sense in which I'm referring to "dependence" independence. There is "more independence" so to speak for 4 separate sets compared to the same set.

Each should have a different inequality. Besides, isn't the realism assumption that the same set of particles have all those properties simultaneously? It won't be a realism assumption if we say different sets each have one property simultaneously.

You are also right that it may be easier to make the point starting from the beginning of the full CHSH derivation.

 

[DrChinese]

...Besides, isn't the realism assumption that the same set of particles have all those properties simultaneously?

Yes, this is the EPR realism assumption: the properties do not need to be simultaneously predictable as long as each one could be predicted with certainty individually. (Of course, I would also say that it would also be CONSISTENT with "a realism assumption if we say different sets each have one property simultaneously.")

 

[bhobba]

You cannot do that. Mathematics only deals with quantity. By axiomatizing the natural world, you are failing to take into account the qualitative part of it. And one huge example is consciousness. Not to mention related phenomena, such as free will.

Conciousness or free will has nothing to do with QM in nearly every interpretation - garbled half truths from some popularisations not withstanding.

QM is perfectly axiomatiseable - and with a breathtaking elegance in the Geometrical approach - although mathematically very non trivial - translation - its hard.

Thanks
Bill

 

[Abc2020ro]

Conciousness or free will has nothing to do with QM in nearly every interpretation - garbled half truths from some popularisations not withstanding.

QM is perfectly axiomatiseable - and with a breathtaking elegance in the Geometrical approach - although mathematically very non trivial - translation - its hard.

Thanks
Bill

It cannot be, since is not the final theory. LOL :w:w:w:w:w

 

[bhobba]

It cannot be, since is not the final theory. LOL :w:w:w:w:w

What has that got to with anything? Classical mechanics is not the final theory yet its perfectly axiomizeable.

But leaving that aside - how do you know the final theory will not be a quantum theory? The current most likely candidate is string theory and its a quantum theory.

Thanks
Bill

 

[atyy]

OK, according to the derivation (http://en.wikipedia.org/wiki/Bell's_theorem#Derivation_of_CHSH_inequality)
$A=A(a, \lambda), A'=A(a', \lambda), B=B(b, \lambda), B'=B(b', \lambda)$
$\begin{align} \rho(a,b) + \rho(a,b') + \rho(a',b) - \rho(a',b')&= \int_\Lambda AB\rho +\int_\Lambda AB'\rho +\int_\Lambda A'B\rho -\int_\Lambda A'B'\rho \\ &= \int_\Lambda (AB+AB'+A'B-A'B')\rho\\ &= \int_\Lambda (A(B+B') + A'(B-B')) \rho\\ &\leq 2 \end{align}$
The heart of the derivation is the 4th line above:

$AB+AB'+A'B-A'B'= A(B+B')+A'(B-B') \le 2.$
That factorization can not be done for different realizations of the same ensemble

Alternatively, For a single set $q$ of N particle pairs, with N sufficiently large
$S_q = \frac{1}{N} \sum_{i=1}^{N} A(a,\lambda_i)B(b,\lambda_i) - \frac{1}{N} \sum_{i=1}^{N} A(a,\lambda_i)B(b',\lambda_i) + \frac{1}{N} \sum_{i=1}^{N} A(a',\lambda_i)B(b,\lambda_i) + \frac{1}{N} \sum_{i=1}^{N} A(a',\lambda_i)B(b',\lambda_i)$
Which I can easily factorize like
$S_p = \frac{1}{N} \sum_{i=1}^{N} (A(a,\lambda_i)[B(b,\lambda_i) - B(b',\lambda_i)] + A(a',\lambda_i)[B(b,\lambda_i) + B(b',\lambda_i)])$
$A, B$ can only take values $\pm 1$, therefore whenever $B(b,\lambda_i) - B(b',\lambda_i)$ is 2, $B(b,\lambda_i) + B(b',\lambda_i)$ must be 0. The possible values within the sum are -2, 0, 2. Therefore $|S_p| \le 2$

For a 4 different sets $r,s,t,u$ of M,N,O,P particle pairs respectively, you instead have:
$S_{rstu} = \frac{1}{M} \sum_{i=1}^{M} A(a,\lambda_i)B(b,\lambda_i) - \frac{1}{N} \sum_{j=1}^{N} A(a,\lambda_j)B(b',\lambda_j) + \frac{1}{O} \sum_{k=1}^{O} A(a',\lambda_k)B(b,\lambda_k) + \frac{1}{P} \sum_{l=l}^{P} A(a',\lambda_l)B(b',\lambda_l)$

Which we can't factorize any further. In addition, Each of terms can independently attain the extrema of [-1, +1]. Therefore $|S_{rstu}| \le 4$.

OK, let's see if I can try to clarify this in a different way. Here we are not talking about quantum mechanics, just classical probability. So we can just talk about flipping a coin. Let's consider a coin with heads ($H=1$) or tails ($H=0$), and let the same coins also have each side coloured either red ($R=1$) or blue ($R=0$). Also, let heads always be red, and tails always be blue.

Let
$P(H=1)=0.5, P(H=0)=0.5$
$P(R=1)=0.5, P(R=0)=0.5$.

Then we define
$E(H)= \sum_{H}HP(H) = (1 X 0.5) + (0 X 0.5) = 0.5$
$E(R)= \sum_{R}RP(R)= (1 X 0.5) + (0 X 0.5) = 0.5$
$Y = E(H) - E(R) = 0$

How in experiments do we get $E(H)$ and $E(R)$? We assume we have a large number of coins $N_{T}$. To get an experimental estimate of $E(H)$ we randomly draw a large subset of $M$ coins, toss each one, measure whether it lands head or tails, and form the sum $\hat{E}(H) = \frac{1}{M} \sum_{i=1}^{M} H(i)$. To get an experimental estimate of $E(R)$ we randomly draw a different large subset of $N$ coins, toss each one, measure whether it lands red or blue and form the sum $\hat{E}(R) = \frac{1}{N} \sum_{j=1}^{N} R(j)$.

Because the number of trials for each measurement is finite, say $M = 100, N=99$. I could get the result $\hat{E}(H)=0.44, \hat{E}(R)=0.49, \hat{Y}=-0.05$, which is different from the predicted value of $Y=0$.

To get closer to the predicted value, what I need to do is increase the number of trials say $M = 100000, N=99999$. I could get the result $\hat{E}(H)=0.5005, \hat{E}(R)=0.5003, \hat{Y}=0.0002$, which is different from the predicted value of $Y=0$, but much closer.

It is true that in the classical case, we can imagine measuring heads and colour at the same time, but there is no need to. If we were to measure heads and colour at the same time, we would for this example in fact get $\hat{Y}=0$, the exact predicted Y value even for a finite number of trials. However there is no need to measure heads and colour on the same subset, since by increasing the number of trials, we can get closer and closer to the predicted Y value.

One could object that with the measurement on the same subset, we always get for this example exactly the predicted value, whereas by measuring on different subsets we don't get exactly the predicted value. By analogy, could one say that the Bell tests are consistent with local reality, but because we measured on different subsets, and because we have been very unlucky, what we consider a large number of trials simply isn't large enough? Yes. In fact, the general issue is the number of trials, not whether they are measured on the same or different subsets. The criterion one chooses to accept or reject a hypothesis is arbitrary. In some of these Bell tests, the deviation from any local deterministic theory is more than 20 standard deviations. But because the cut-off criterion is subjective, one could reject 20 standard deviations as sufficient.

 

[Alien8]

Well, let x = (a-c), y=(a-d), z=(b-c), and then (b-d) = x+y-z. So you are right it is not completely independent and that will affect the upper bound so you would have an expression like E(x) - E(y) + E(z) +E(x+y-z). However, you can still evaluate this expression in two ways. You could calculate E(x) from one set of particles, E(y) from a different set of particles, E(z) from yet a different set of particles and E(x+y-z) from yet another set. no two particle pairs in any set belonging to any other set. In this sense, the sets are independent, even though the results are not entirely independent owning to the E(x+y-z).

Yes. If we measure independently two distances X and Y and the only thing they have in common is maximum length of 1, then all we can say is X+Y <= 2. But if they are two sides of the same triangle XYZ, then we can also say X+Y <= Z.

Basically you are asking what E1(a,b), E2(a,b'), E3(a',b) and E4(a',b') have in common beside {-1,+1} limit. If each E limit was independent and the only common rule they must follow, then the boundary for E1−E2+E3+E4 would be 4, so there must be something else, some other common rule they obey or system they belong to.

They share the same E(x,y) function and there is a proportionality between |a-b| - |a'-b| = |a-b'| - |a'-b'|. But that's only relation between input parameters, it doesn't explain which common system those input variables are supposed to belong to, or in other words - why the choice of (a-b), (a-b'), (a'-b) and (a'-b') instead of (a-a'), (a-b'), (b-b') and (b-a') for example.

So what is it? The only other common rule or system applied in the derivation I see is the triangle inequality, and to be applied to CHSH setup those angles therefore have to somehow correspond to some triangles. I don't know how this relation between angles, triangles and probabilities works, but beside {-1,+1} limit that's the only common thing they share together, so that must be where the answer to all our questions is.

You are also right that it may be easier to make the point starting from the beginning of the full CHSH derivation.

http://en.wikipedia.org/wiki/CHSH_inequality#Derivation_of_the_CHSH_inequality

where A and B are the average values of the outcomes. Since the possible values of A and B are −1, 0 and +1, it follows that:

Then, if a, a′, b and b′ are alternative settings for the detectors,

At the beginning there is only a and b, then there is suddenly b' in the first line of step (6), and then in the second line a' materializes out of thin air as well. I think the question begins with the step (6), according to what logic, physics, or mathematical principle is justified.

 

[atyy]

Here is another proof of CHSH by Richard Gill http://arxiv.org/abs/1207.5103 (see section 2). In this case, it is made clear that the 4 terms correspond to *disjoint* subsets of size N/4 each, with a total size of N.

For finite N, there is some probability that a local deterministic theory will violate the Bell inequalities. Taking this into account, the Bell inequalities are not hard bounds for a local deterministic theory, but rather only something that a local deterministic theory is likely to satisfy with a probability given by Eq (3). As N approaches infinity, the traditional Bell inequality as a hard bound is recovered as given in Eq (4) .

 

[Alien8]

Here is another proof of CHSH by Richard Gill http://arxiv.org/abs/1207.5103 (see section 2). In this case, it is made clear that the 4 terms correspond to *disjoint* subsets of size N/4 each, with a total size of N.

1.) AB + AB' + A'B - A'B' <= 2

2.) E(a, b) − E(a, b′) + E(a′, b) + E(a′, b′) <= 2.

They are very different inequalities, one is dealing with binary outcomes -1 or +1, the other with expectation values of decimal range from -1.0 to +1.0. The paper says the first one is CHSH inequality, but Wikipedia says it's the second one, which is what makes sense. Binary outcomes inequality, the first one, is general and completely undefined relative to experimental settings, it can not be violated by anything as long as 1+1+1-1 = 2, so I see no reason to even mention it.

I believe this is how proper derivation goes, as the main CHSH Wikipedia article says:

where A and B are the average values of the outcomes. Since the possible values of A and B are −1, 0 and +1, it follows that:

Then, if a, a′, b and b′ are alternative settings for the detectors,

http://en.wikipedia.org/wiki/CHSH_inequality#Derivation_of_the_CHSH_inequalityStep (6) basically starts with: E1 - E2 = E1 - E2, which is completely pointless observation, but then it goes on to conclude something like: E1 - E2 = E3 - E4. Out of the blue. How do you start with E1 in step (4) and then end up with E1, E2, E3 and E4 all together in step (6)? Being alternative settings for the detectors doesn't imply or explain anything. So what physics, logic, or mathematics can justify placing those for expectation values together in such a relationship?

 

[Fredrik]

It cannot be, since is not the final theory. LOL :w:w:w:w:w

Perhaps this comment is influenced by the popular belief that axioms are "self-evidently true" or "obviously true" statements. The modern view of axioms is very different from this. Axioms are not obvious truths, or even objective truths. A list of axioms simply defines a branch of mathematics. That's all. The axioms are true in that branch, because the branch is by definition the part of mathematics where the axioms are true. Every axiom is false in some other branch of mathematics.

However, a theory of physics isn't defined by axioms in this sense. It's defined by a set of assumptions that I used to call "axioms" until a few years ago. A. Neumaier had a strong negative reaction to how I used that word in a discussion here. I decided that he was right about that. There's no reason to call them "axioms". So I call them "correspondence rules" now. I think almost everyone is OK with that term. The purpose of a set of correspondence rules is to tell us how to interpret some piece of mathematics as predictions about results of experiments.

We can certainly define a branch of mathematics using axioms, and a theory of physics using correspondence rules, without having any idea what the final theory might be, or if there even is one.

 

[atyy]

They are very different inequalities, one is dealing with binary outcomes -1 or +1, the other with expectation values of decimal range from -1.0 to +1.0. The paper says the first one is CHSH inequality, but Wikipedia says it's the second one, which is what makes sense. Binary outcomes inequality, the first one, is general and completely undefined relative to experimental settings, it can not be violated by anything as long as 1+1+1-1 = 2, so I see no reason to even mention it.

Yes. In the derivation I linked to, it is assumed that given the hidden variable and measurement setting, there is no variability, ie. the outcome is either +1 or -1 with certainty. This is the assumption of "local determinism". However, one can certainly imagine that given the hidden variable and measurement setting, there is variability, ie. the outcome is sometimes +1 and sometimes -1 with probability $p(A,B|a,b,\lambda) = p(A|a,\lambda)p(B|b,\lambda)$. However, it turns out that this second, and more general case of "local random or deterministic variables" can be rewritten as a "local deterministic" model by introducing additional hidden variables. For this reason, the two different proofs of CHSH are equivalent. You can find a description of the equivalence in http://arxiv.org/abs/1303.3081 (Proposition 2.1 in section 2.2.2).

 

[wle]

Here is another proof of CHSH by Richard Gill http://arxiv.org/abs/1207.5103 (see section 2). In this case, it is made clear that the 4 terms correspond to *disjoint* subsets of size N/4 each, with a total size of N.

For finite N, there is some probability that a local deterministic theory will violate the Bell inequalities. Taking this into account, the Bell inequalities are not hard bounds for a local deterministic theory, but rather only something that a local deterministic theory is likely to satisfy with a probability given by Eq (3). As N approaches infinity, the traditional Bell inequality as a hard bound is recovered as given in Eq (4) .

There are really two related but different theorems here. Bell's theorem (at least, originally) is a mathematical demonstration that the predictions of local theories and quantum physics are different. The objects being compared are the sets of joint probabilities predicted by quantum physics (the "quantum set" $\mathcal{Q}$) and by local causal theories (the "local set" or "local polytope" $\mathcal{L}$). These can be defined by $$\boldsymbol{P} \in \mathcal{Q} \Leftrightarrow P(ab \mid xy) = \mathrm{Tr}\bigl[M^{(x)}_{a} \otimes N^{(y)}_{b} \bigr] \,,$$ in which $\rho$ is a density operator and $M^{(x)}_{a}$ and $N^{(y)}_{b}$ are POVM elements, and $$\boldsymbol{P} \in \mathcal{L} \Leftrightarrow P(ab \mid xy) = \int \mathrm{d}\lambda \rho(\lambda) P(a \mid x; \lambda) P(b \mid y; \lambda) \,.$$ One simple way of showing that these are different sets is by comparing the maximal possible values of the CHSH correlator $$S = \boldsymbol{I} \cdot \boldsymbol{P} = \sum_{abxy} (-1)^{a + b + xy} P(ab \mid xy) \,.$$ The well known result is that $$\max_{\boldsymbol{P} \in \mathcal{L}} \boldsymbol{I} \cdot \boldsymbol{P} = 2 \,,$$ compared with $$\max_{\boldsymbol{P} \in \mathcal{Q}} \boldsymbol{I} \cdot \boldsymbol{P} = 2 \sqrt{2} \,,$$ which is only possible if there are probability distributions in the quantum set $\mathcal{Q}$ that aren't in the local set $\mathcal{L}$. In this case, the CHSH correlator is defined as a function of a joint probability distribution, so it's for just one realisation (e.g. one entangled particle pair).

In an actual Bell experiment, Bell's definition of locality is being tested against reality. If you want do this rigorously (though in practice, nobody seems to bother), this means recasting Bell's theorem in the form of a hypothesis test and doing some additional statistical analysis. Part of this is defining what the "experimental Bell correlator" that is going to be measured is, since the mathematical correlator defined for a single realisation isn't a directly measurable quantity. Gill describes one (but by no means the only possible) way of doing that which is close to what's done in most Bell experiments.

 

[Alien8]

Gill describes one (but by no means the only possible) way of doing that which is close to what's done in most Bell experiments.

Gill says for any four numbers A, A', B, B' each equal to either -1 or + 1, then: AB + AB' + A'B - A'B' = -2 or +2

There are no any relative angles and detector settings here, no any locality or non-locality assumptions, no any probabilities, no any theories involved in this inequality what so ever. It's a statement about numbers, like 1 + 1 = 2, it's not relevant to any physics or reality. It's about arbitrary combinations of the four variables having either value of -1 or +1, where any possible combination plugged in that equation will always yield either -2 or +2, that's all there is to it. QM can not violate that inequality any more than it can make 1 + 1 = 3. We can not compare QM and other kinds of predictions with that inequality, so what's the point of it?

 

[atyy]

Gill says for any four numbers A, A', B, B' each equal to either -1 or + 1, then: AB + AB' + A'B - A'B' = -2 or +2

There are no any relative angles and detector settings here, no any locality or non-locality assumptions, no any probabilities, no any theories involved in this inequality what so ever. It's a statement about numbers, like 1 + 1 = 2, it's not relevant to any physics or reality. It's about arbitrary combinations of the four variables having either value of -1 or +1, where any possible combination plugged in that equation will always yield either -2 or +2, that's all there is to it. QM can not violate that inequality any more than it can make 1 + 1 = 3. We can not compare QM and other kinds of predictions with that inequality, so what's the point of it?

Gill presents two equations he calls "CHSH". The one you are referring to is Eq 2. The one with measurement settings is Eq 4.

 

[billschnieder]

In this case, it is made clear that the 4 terms correspond to *disjoint* subsets of size N/4 each, with a total size of N.

For finite N, there is some probability that a local deterministic theory will violate the Bell inequalities.

I don't understand how you could calculate a probability that any local deterministic theory will violate Bell inequalities, without clearly defining the space of "local deterministic theories". For a given theory (and Gill gives one), yes I can imagine one easily checking the probability that it will violate the inequalities but how do you do that for "any local deterministic theory". It seems to me our interest is in the latter probability not the former one.

In the above paper, he says:

When N is large one would expect <AB>obs to be close to <AB>, and the same for the other three averages of observed products.
Hence, equation (2) should remain approximately true when we replace the averages of the four products over all N rows with the averages of the four products in each of four disjoint subsamples of expected size N/4 each.

N is the size of the spreadsheet with 4 columns. And he is saying that we have a certain distribution of numbers {+1,-1} in the spreadsheet, and if we divide that spreadsheet up into 4 disjoint parts, we will still have approximately the same averages?! Isn't he making a certain assumption about how the numbers are distributed in the spreadsheet to begin with?

We could start from the experimental situation, in which we have not one Nx4 spreadsheet but 4 different 2xN spreadsheets. Let us try to derive the inequality from this scenario, and make all the necessary assumptions we could want to make about local determinism and realism to end up with 2 in the RHS. Instead of 4 numbers A, B, A', B'. In this case, we now have 8 numbers A1, B1, A2, B2', A3', B3, A4', B4', so that we instead have

$A_1B_1 + A_2B'_2 + A'_3B_3 - A'_4B'_4 \le 4$

What assumptions do we have to apply to this in order to end up with 2 on the RHS? I can think of one. We could say $A_1 = A_2, A'_3 = A'_4, B'_2 = B'_4, B_1 = B_3$, which translating from the numbers to spreadsheets of numbers, it means the corresponding columns are identical, not just that the have the same ratios of {+1, -1} but that the pattern of changing back and forth is identical, or can be made identical by rearranging. This is a condition that will allow us to factorize the terms from 4 disjoint sets. For that to be the case, the source will have to know what set each pair will end up in, or the distributions will have to so uniform at all angle settings that a single set will not be able to reproduce the experimentally observed expectation value for one angle pair.

 

[wle]

I don't understand how you could calculate a probability that any local deterministic theory will violate Bell inequalities, without clearly defining the space of "local deterministic theories". For a given theory (and Gill gives one), yes I can imagine one easily checking the probability that it will violate the inequalities but how do you do that for "any local deterministic theory". It seems to me our interest is in the latter probability not the former one.

I haven't looked at the details of Gill's method, but I know a simple way of doing this that could be done in an experiment. (It's described in appendix A.2 of this paper; I don't know if it was proposed earlier.) The idea is based around defining an estimator $S_{k}$ for the Bell correlator on the $k$th realisation (i.e., the $k$th particle pair, if you want to assume the things being measured are actually particles). I'll describe how this works for the CHSH correlator, though the method can just as well be used for any Bell correlator that can be defined as a linear function of the probability distribution for a single realisation. The procedure, for the $k$th realisation, is:

1. Alice and Bob pick random measurements $x_{k}, y_{k} \in \{0, 1\}$ with probabilities $P(x_{k}) = P(y_{k}) = 1/2$, such that $P(x_{k} y_{k}) = P(x_{k}) P(y_{k}) = 1/4$.
2. They record the outcomes $a_{k}, b_{k} \in \{0, 1\}$ from the results of their measurements.
3. The estimator $S_{k}$, which Alice and Bob will be able to compute later when they compare their results, is defined in terms of these by $$S_{k} = 4 (-1)^{a_{k} + b_{k} + x_{k} y_{k}} \,.$$

Defined this way, $S_{k}$ is a random variable that can take only the values +4 and -4. If you write down its expectation value, that works out to $$\begin{eqnarray}
\langle S_{k} \rangle &=& \sum_{abxy} 4 (-1)^{a + b + xy} P (abxy) \\
&=& \sum_{abxy} 4 (-1)^{a + b + xy} P(ab \mid xy) P(xy) \\
&=& \sum_{abxy} (-1)^{a + b + xy} P(ab \mid xy) \,.
\end{eqnarray}$$The last line is exactly what's considered in most derivations of the CHSH expectation value, so the same results hold. In particular, $-2 \leq \langle S_{k} \rangle \leq 2$ according to any locally causal model and $- 2 \sqrt{2} \leq \langle S_{k} \rangle \leq 2 \sqrt{2}$ according to quantum physics.

If Alice and Bob repeat this $N$ times, the CHSH estimator for the whole experiment can just be defined as the average for each of the realisations in the obvious way: $$S = \frac{1}{N} \sum_{k = 1}^{N} S_{k} \,.$$ This is adding a list of random variables of values +4 or -4, but since their expectation values are all bounded by 2 for any local causal model, the probability with which a local causal model can predict a significant violation becomes very low for a large number $N$ of realisations. (If you need an upper bound on the probability with which that can happen, the paper I linked to above explains how to do that using the Azuma-Hoeffding inequality.)

 

[Alien8]

We could start from the experimental situation, in which we have not one Nx4 spreadsheet but 4 different 2xN spreadsheets. Let us try to derive the inequality from this scenario, and make all the necessary assumptions we could want to make about local determinism and realism to end up with 2 in the RHS. Instead of 4 numbers A, B, A', B'. In this case, we now have 8 numbers A1, B1, A2, B2', A3', B3, A4', B4', so that we instead have

$A_1B_1 + A_2B'_2 + A'_3B_3 - A'_4B'_4 \le 4$

What assumptions do we have to apply to this in order to end up with 2 on the RHS?

AB + AB' + A'B - A'B' = -2 or +2 has nothing to do with any locality, determinism or realism. There is no any assumptions related to that equation, it's entirely defined by its purely mathematical premise, which is that every possible combination of four variables A, B, C, D, were each can arbitrarily be either -1 or +1, when multiplied, added and subtracted in this particular order: AC + AD + BC - BD, will always yield either -2 or +2. That's all, numbers and mathematics, nothing else.

It can not be AB + CD + EF - GH, it has to be AC + AD + BC - BD because that's the particular combination which produces -2 or +2 result. It's not an assumption, it's mathematical truth, just a matter of choice. But that is not the inequality used in experiments, it has no any bearing to locality or determinism. We should be talking about proper CHSH inequality and relative angles: E(a,c) − E(a,d) + E(b,c) + E(b,d), then ask why it is not: E(a,b) − E(c,d) + E(e,f) + E(g,h).

 

[billschnieder]

The procedure, for the $k$th realisation, is:

1. Alice and Bob pick random measurements $x_{k}, y_{k} \in \{0, 1\}$ with probabilities $P(x_{k}) = P(y_{k}) = 1/2$, such that $P(x_{k} y_{k}) = P(x_{k}) P(y_{k}) = 1/4$.
2. They record the outcomes $a_{k}, b_{k} \in \{0, 1\}$ from the results of their measurements.
3. The estimator $S_{k}$, which Alice and Bob will be able to compute later when they compare their results, is defined in terms of these by $$S_{k} = 4 (-1)^{a_{k} + b_{k} + x_{k} y_{k}} \,.$$

Defined this way, $S_{k}$ is a random variable that can take only the values +4 and -4.

Yes, I would expect $-4 \le S_{k} \ge +4$. How you get from this to $-2 \le \langle S_{k} \rangle \ge +2$ is what the problem is.

If Alice and Bob repeat this $N$ times, the CHSH estimator for the whole experiment can just be defined as the average for each of the realisations in the obvious way: $$S = \frac{1}{N} \sum_{k = 1}^{N} S_{k} \,.$$ This is adding a list of random variables of values +4 or -4, but since their expectation values are all bounded by 2 for any local causal model

I don't follow. S is already a result of 4 different realizations, but then you appear to be averaging more than one S. The inequality is about what you can say for any $S$, not what you can say for averages $\langle S \rangle$, no? What is proved in the CHSH is $-2 \le S_{k} \ge +2$ not $-2 \le \langle S_{k} \rangle \ge +2$, the former is a sufficient but not necessary condition for the latter. I do not see how even proving the latter implies the former.

 

[billschnieder]

It can not be AB + CD + EF - GH, it has to be AC + AD + BC - BD because that's the particular combination which produces -2 or +2 result. It's not an assumption, it's mathematical truth, just a matter of choice. But that is not the inequality used in experiments, it has no bearing to locality or determinism. We should be talking about proper CHSH inequality and relative angles: E(a,c) − E(a,d) + E(b,c) + E(b,d), then ask why it is not: E(a,b) − E(c,d) + E(e,f) + E(g,h).

While I agree with you that it is a mathematical truth, you have to remember what is actually measured in experiments. There is no such thing as E(a,b) ... in an experiment. All you have are 8 lists of numbers in 4 pairs. for each we multiply each member of a pair in each list, add up all the products in each list and average it, then we call that E(a,b), it's actually $\langle AB\rangle$ for angles $a,b$, we do the same thing for the remaining 4 pairs. At the end we combine the 4 expressions we obtained, calculate $S$ and then compare that with an inequality. The issue is what is the correct inequality to use for this kind of result. Should we use an inequality we derived by assuming we had just 4 lists (A,B,C,D) which we recombined into 4 pairs (AB, AD, BC, BD), or should we use an inequality we derived by assuming we had 8 lists in 4 pairs (AB, CD, EF, GH). You are saying we can of course assume that since the 8 lists were obtained from just 4 angles, then we have just 4 lists. But we can't just assume that, there is more producing the outcomes than just angles. Having the same angles doesn't make the two systems have the same degrees of freedom. We could of course conduct an experiment in which we measure just 4 lists, no need to pair them at all. Just measure one single list at A, others at B,C,D. then recombine them to make the pairs. Why don't we do that? If we can combine 4 separate lists of pairs, why shouldn't we be able to combine 4 separate lists of singles? I suspect it is the same reason. If I throw a coin, knowing that it landed heads tells me clearly that it did not land tails. But If I through two identical coins, knowing that one coin landed heads, tells me absolutely nothing about what the other coin did or did not do.