Axiomatization of quantum mechanics and physics in general ?

[wle]

bhobba & wle,
I think you are still missing the point. Perhaps if I ask you both a simple question: From the wikipedia page you cited earlier:

$S = \langle A(a) B(b) \rangle + \langle A(a') B(b') \rangle + \langle A(a') B(b) \rangle - \langle A(a) B(b') \rangle = \tfrac{4}{\sqrt{2}} = 2 \sqrt{2} > 2$

My question is very simple: Do each of the expectation values in that expression apply to the exact same system, or does it apply to different but similar systems?

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. In the usual situation considered in quantum mechanics, that means one system of two entangled particles.

 

[Alien8]

My question is very simple: Do each of the expectation values in that expression apply to the exact same system, or does it apply to different but similar systems?

Ooops. Accidental post, writing in progress...

 

[billschnieder]

I am using standard quantum mechanical language. One particle means an ensemble, each member of which is one particle. What is your answer to (1) if your answer to (2) was that I measured on different subsets?

News to me that "one particle" means the same thing as "ensemble of particles" in QM. But if that is what you meant, then of course you will find $E=X^2+P^2$. What is your point exactly that you think is relevant to my own question?

 

[billschnieder]

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. In the usual situation considered in quantum mechanics, that means one system of two entangled particles.

OK, thanks. If you believe the expression applies to the same system, then do have a problem with the following:

$E_ψ(x,z)=⟨ψ|(σL⋅x)(σR⋅z)|ψ⟩=−x⋅z$
$S_ψ = E_ψ(a,b) - E_ψ(a,b') + E_ψ(a',b) + E_ψ(a',b')$
$S_ψ = ⟨ψ|(σL⋅a)(σR⋅b)|ψ⟩−⟨ψ|(σL⋅a)(σR⋅b')|ψ⟩+⟨ψ|(σL⋅a')(σR⋅b)|ψ⟩+⟨ψ|(σL⋅a')(σR⋅b')|ψ⟩$
$S_ψ = ⟨ψ|(σL⋅a)(σR⋅b)−(σL⋅a)(σR⋅b′)+(σL⋅a′)(σR⋅b)+(σL⋅a′)(σR⋅b′)|ψ⟩$
$S_ψ = ⟨ψ|(σL⋅a)[(σR⋅b)−(σR⋅b′)]+(σL⋅a′)[(σR⋅b)+(σR⋅b′)]|ψ⟩$

 

[atyy]

News to me that "one particle" means the same thing as "ensemble of particles" in QM. But if that is what you meant, then of course you will find $E=X^2+P^2$. What is your point exactly that you think is relevant to my own question?

Just trying to figure out how you are using language, and whether we can agree on basic quantum mechanics.

I think my example is relevant because it is an equation $0 = E - X^{2} - P^{2}$, where the three terms on the RHS are expectations of non-commuting observables, just as in CHSH there are 4 expectations of non-commuting observables which are added. In this case, the value predicted by quantum mechanics is 0. In the CHSH case, the value predicted is $2\sqrt{2}$.

 

[Alien8]

Couldn't edit my original post after 15min limit.

My question is very simple: Do each of the expectation values in that expression apply to the exact same system, or does it apply to different but similar systems?

So you are basically asking what E(X1,Y1), E(X1,Y2), E(X2,Y1) and E(X2,Y2) in CHSH inequality have in common to justify them being a part of the same equation (same system). According to Wikipedia the "system" you are talking about seems to be a triangle, or two triangles actually, and it's the triangle inequality which relates the four expectation values and sets the boundary below 4.
http://en.wikipedia.org/wiki/CHSH_inequality#Derivation_of_the_CHSH_inequality

Why and how exactly a geometrical inequality applies to probabilities is beyond me, but looking at Wikipedia I can tell you it follows from Cauchy–Schwarz inequality. Unfortunately that doesn't explain much because it says that the inequality as applied to probabilities has this formula: $|E(XY)|^2 \leq E(X^2)E(Y^2)$, rather than this: $|X+Y| \leq |X| + |Y|$.

 

[billschnieder]

I think my example is relevant because it is an equation $0 = E - X^{2} - P^{2}$, where the three terms on the RHS are non-commuting observables, just as in CHSH there are 4 non-commuting terms which are added. In this case, the value predicted by quantum mechanics is 0. In the CHSH case, the value predicted is $2\sqrt{2}$.

But you just said $X^{2}$ $P^{2}$ are measured on different realizations of the ensemble, so how can they be non-commuting? When measured on different realizations of the ensemble they all commute, since you are not talking about simultaneous measurements on the same realization of the ensemble, so your example is not relevant as you think it is.

In any case, what is your answer to my question then?

 

[atyy]

But you just said $X^{2}$ $P^{2}$ are measured on different realizations of the ensemble, so how can they be non-commuting? When measured on different realizations of the ensemble they all commute, since you are not talking about simultaneous measurements on the same realization of the ensemble, so your example is not relevant as you think it is.

In any case, what is your answer to my question then?

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.

 

[wle]

OK, thanks. If you believe the expression applies to the same system, then do have a problem with the following:

$E_ψ(x,z)=⟨ψ|(σL⋅x)(σR⋅z)|ψ⟩=−x⋅z$
$S_ψ = E_ψ(a,b) - E_ψ(a,b') + E_ψ(a',b) + E_ψ(a',b')$
$S_ψ = ⟨ψ|(σL⋅a)(σR⋅b)|ψ⟩−⟨ψ|(σL⋅a)(σR⋅b')|ψ⟩+⟨ψ|(σL⋅a')(σR⋅b)|ψ⟩+⟨ψ|(σL⋅a')(σR⋅b')|ψ⟩$
$S_ψ = ⟨ψ|(σL⋅a)(σR⋅b)−(σL⋅a)(σR⋅b′)+(σL⋅a′)(σR⋅b)+(σL⋅a′)(σR⋅b′)|ψ⟩$
$S_ψ = ⟨ψ|(σL⋅a)[(σR⋅b)−(σR⋅b′)]+(σL⋅a′)[(σR⋅b)+(σR⋅b′)]|ψ⟩$

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.

 

[billschnieder]

So you are basically asking what E(X1,Y1), E(X1,Y2), E(X2,Y1) and E(X2,Y2) in CHSH inequality have in common to justify them being a part of the same equation (same system). According to Wikipedia the "system" you are talking about seems to be a triangle, or two triangles actually, and it's the triangle inequality which relates the four expectation values and sets the boundary below 4.
http://en.wikipedia.org/wiki/CHSH_inequality#Derivation_of_the_CHSH_inequality

That is a good point. The final step in the proof of Bell's theorem involves performing a QM calculation to obtain a result which is then claimed to violate the inequality. If Bell's theorem is to be considered an important theorem, then that final step should also be considered very important. I was told earlier that the calculation is a trivial substitution but according to Bell's own critique of von Neumann, you shouldn't expect the substitution to be trivial for non-commuting observables.
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.

 

[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.

$|E(a,b) - E(a,b')| \leq 2 \pm [E(a',b) + E(a',b')]$ 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?