Axiomatization of quantum mechanics and physics in general ?

[billschnieder]

Expectation values always commute because they are numbers, so your math is messed up.

Well spotted - I didn't even notice it.

So then what was Bell complaining about the nontriviality of additivity of expectation values?

The essential assumption can be criticized as follows. At first sight, the required additivity of expectation values seems very reasonable, and is rather the non-additivity of allowed values (eigenvalues) which requires explanation. Of course the explanation is well known: A measurement of a sum of noncommuting observables cannot be made by combining trivially the results of separate observations on the two[individual] terms -- it requires a quite distict experiment. ... But this explanation of the non-additivity of allowed values also establishes the nontriviality of the additivity of expectation values

That is why I said earlier:
Starting from
$E_\psi(x,z) = \langle\psi|(\sigma_L\cdot x)(\sigma_R\cdot z)|\psi\rangle = -x\cdot{z}$

$S^{\psi} = |\langle\psi|(\sigma_L\cdot{a})(\sigma_R\cdot{b}) - (\sigma_L\cdot{a})(\sigma_R\cdot{b'}) + (\sigma_L\cdot{a'})(\sigma_R\cdot{b}) + (\sigma_L\cdot{a'})(\sigma_R\cdot{b'})|\psi\rangle|$
Do all those spin observables commute?

 

[wle]

I'm not talking about the derivation. I'm talking about the demonstration of QM violation of the inequality. The part where expectations are linearly combined.

For example, how do you show that QM violates Bell's inequality. It is this calculation I'm talking about.

I think you're misunderstanding what the CHSH correlator is and what the local bound on it actually means.

Formally, the CHSH correlator is by definition the linear combination
S = \sum_{abxy} (-1)^{a + b + xy} P(ab \mid xy) \,,
where the variables x, y \in \{0, 1\} are Alice's and Bob's choice of measurement settings, a, b \in \{0, 1\} are the measurement results, and P(ab \mid xy) is the prior probability that Alice and Bob get the result (a, b) given that they chose the measurement pair (x, y) in a CHSH-type experimental setting.

For the purpose of what Bell actually proved originally -- that no locally causal theory can make the same predictions as quantum physics -- it's not an error that the P(ab \mid xy)s are all defined on the same particle pair because the CHSH correlator doesn't need to be a physically meaningful or measurable quantity. Quantum mechanics is a theory that is defined mathematically and Bell defined the class of locally causal theories mathematically. As such, they can be compared mathematically without ever needing to do an actual experiment. In this context, the CHSH correlator is just an intermediate variable in the proof of Bell's theorem and doesn't need to "mean" anything beyond that.

 

[billschnieder]

I think you're misunderstanding what the CHSH correlator is and what the local bound on it actually means. ...
For the purpose of what Bell actually proved originally -- that no locally causal theory can make the same predictions as quantum physics -- it's not an error that the P(ab∣xy)s are all defined on the same particle pair because the CHSH correlator doesn't need to be a physically meaningful or measurable quantity.

I don't think I am. I think you are misunderstanding what the issue is. The CHSH *is* indeed derived for a single system. The problem I see is that the demonstration of QM violation of the CHSH uses 4 expectation values with two possible interpretations of the QM substitution, either one not good for Bell. The two options are:
1) The first option is that the QM expression, like the CHSH correlator applies to a single system. This is problematic because then it would be making the same mistake as von Neumann which Bell criticized.

Yet the Von Neumann proof, if you actually come to grips with
it, falls apart in your hands. There’s nothing to it. It’s not
just flawed. . .it’s silly. When you translate his assumptions into
physical significance, they’re nonsense. You may quote me on
this. The proof of Von Neumann is not just false, it’s foolish

2) The second option is that QM expression, unlike the CHSH correlator applies to 4 different (but similar) systems. This is problematic because then, he won't be able to demonstrate violation, since the 4-system upper bound is 4.

 

[wle]

I don't think I am. I think you are misunderstanding what the issue is. The CHSH *is* indeed derived for a single system. The problem I see is that the demonstration of QM violation of the CHSH uses 4 expectation values with two possible interpretations of the QM substitution, either one not good for Bell. The two options are:
1) The first option is that the QM expression, like the CHSH correlator applies to a single system. This is problematic because then it would be making the same mistake as von Neumann which Bell criticized.

I'm not familiar with von Neumann's argument so I don't know offhand what the issue with it was. von Neumann may well have added terms from quantum mechanics together in a way that didn't make sense in the context of his argument. But there's nothing problematic with the CHSH correlator. Like I said in my previous post, it's just defined as the linear combination of probabilities

S = \sum_{abxy} (-1)^{a + b + xy} P(ab \mid xy) \,, \qquad a, b, x, y \in \{0, 1\} \,.
Alternatively, it can be written more compactly as the scalar product

S = \bar{I} \cdot \bar{P}
with the vector I defined by the components I_{abxy} = (-1)^{a + b + xy}. There's no problem evaluating this quantity (which is just a variable defined for convenience) for a given set of measurement operators on a given quantum state in quantum physics. The proof of Bell's theorem just uses that if \bar{I} \cdot \bar{P}_{1} \neq \bar{I} \cdot \bar{P}_{2} then necessarily \bar{P}_{1} \neq \bar{P}_{2}.

 

[bhobba]

I'm not familiar with von Neumann's argument so I don't know offhand what the issue with it was.

Its simple. If you assume expectations are additive then Born's Rule follows. Its not hard.

First its easy to check <bi|O|bj> = Trace (O |bj><bi|).

O = ∑ <bi|O|bj> |bi><bj| = ∑ Trace (O |bj><bi|) |bi><bj|

Now we use the linearity assumption ie expectations are additive and if f is that expectation

f(O) = ∑ Trace (O |bj><bi|) f(|bi><bj|) = Trace (O ∑ f(|bi><bj|)|bj><bi|)

Define P as ∑ f(|bi><bj|)|bj><bi| and we have f(O) = Trace (OP).

P, by definition, is called the state of the quantum system. The following are easily seen. Since f(I) = 1, Trace (P) = 1. Thus P has unit trace. f(|u><u|) is a positive number >= 0 since |u><u| is an effect. Thus Trace (|u><u| P) = <u|P|u> >= 0 so P is positive.

The trouble is while experiment shows expectations of quantum observables are additive, and its very intuitive anyway, it does not necessarily apply to hidden variables. That's the key point Bell showed. He was not the only one - but due to Von Neumann's reputation (of course he is correctly regarded as a mathematician/mathematical physicist of the highest calibre - many have him in the top 10 greatest of all time - as do I) they were ignored:
http://mpseevinck.ruhosting.nl/seevinck/Aberdeen_Grete_Hermann2.pdf

Stronger proofs came along later - Gleason's theorem probably being the deepest. The real key is non-contextuality - its also tied up with locality and Bells Theorem in a subtle way:
http://people.maths.ox.ac.uk/tillmann/CATlect2013SA4.pdf

Thanks
Bill

 

[wle]

I had a quick skim through Bell's article on the subject [Rev. Mod. Phys. 38, 447 (1996)]. If I've understood the issue I'd explain it as follows:

Suppose E(A \mid \Psi; \lambda) is the expectation value associated with an observable A given a state vector \lvert \Psi \rangle and some additional variables \lambda according to some hidden variable theory, presumably satisfying some condition like \int \mathrm{d} \lambda \rho(\lambda) E(A \mid \Psi; \lambda) = \langle \Psi \rvert A \lvert \Psi \rangle in order to recover quantum physics. The issue seems to be that von Neumann assumed a linearity condition along the lines of

E(a A + b B \mid \Psi; \lambda) = a E(A \mid \Psi; \lambda) + b E(A \mid \Psi; \lambda) \,,
while consistency with quantum physics would only require the weaker condition that linearity holds after averaging, i.e.,

\int \mathrm{d} \lambda \rho(\lambda) E(a A + b B \mid \Psi; \lambda) = a \int \mathrm{d} \lambda \rho(\lambda) E(A \mid \Psi; \lambda) + b \int \mathrm{d} \lambda \rho(\lambda) E(A \mid \Psi; \lambda) \,.

If that's the case, I'd agree that isn't necessarily justified, and it's equally clear to me that there is no such problem with Bell's theorem.

 

[wle]

Erratum:

E(a A + b B \mid \Psi; \lambda) = a E(A \mid \Psi; \lambda) + b E(A \mid \Psi; \lambda) \,,

\int \mathrm{d} \lambda \rho(\lambda) E(a A + b B \mid \Psi; \lambda) = a \int \mathrm{d} \lambda \rho(\lambda) E(A \mid \Psi; \lambda) + b \int \mathrm{d} \lambda \rho(\lambda) E(A \mid \Psi; \lambda) \,.

Those last terms should, of couse, be b E(B \mid \Psi; \lambda) rather than b E(A \mid \Psi; \lambda). (Why is there a 3 minute time limit on editing posts?!)

 

[bhobba]

If that's the case, I'd agree that isn't necessarily justified, and it's equally clear to me that there is no such problem with Bell's theorem.

.
Home run hit.

Spot on.

And indeed Bells Theorem has no such issue.

Thanks
Bill

 

[wle]

[Rev. Mod. Phys. 38, 447 (1996)].

... and that should be 1966.

 

[billschnieder]

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?

Bhobba, I know we've discussed this previously so let me clarify what I mean by "exact same system" as opposed to "different but similar": If we are talking about two particles in a Bell state, then "exact same system" means there are only two particles in the discussion, and we are simply adding up what the exact same particle pair would do at different settings for the exact same two particles (one pair) . "different but similar" means there are 8 particle pairs in the discussion and we are adding results from one particle pair at one pair of settings, to results of a different but similar particle pair at another setting pair etc.

So which interpretation is it? Or do you not think there is a difference between both.

 

[atyy]

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?

They do not apply to what you are calling "the exact same system". They apply to different pairs of particles, each drawn from the same ensemble.

 

[billschnieder]

They do not apply to what you are calling "the exact same system". They apply to different pairs of particles, each drawn from the same ensemble.

Not sure I follow what you mean by "the same ensemble". Do you simply mean that they are similar? Or do you mean that the 4 expectation values apply to the "exact same ensemble". The issue does not disappear because we start talking about "ensemble" as opposed to individual particle pairs.

Perhaps it is still not clear what the difference is between "the same" and "different but similar". If i = 1,..,N represents identity for N members of a set, where the members could be particle pairs, ensembles, or whatever. "the same" means the same i, "different but similar" means different i, but still a member of the set. In computer-science-speak, it would be the difference between "equality" and "identity" or the difference between "value" and "reference".

So my question in this context will be:
Are those expectation values applicable to the same system, or to different but similar systems (whatever your definition of system is, particle pairs or ensembles).

You could then also ask the same question, of von Neumann's essential assumption:
von Neumann: Any real linear combination of any two Hermitian operators represents an observable, and the same linear combination of expectation values is the expectation of the combination.
Was he talking about the same system or different but similar system.

 

[atyy]

Not sure I follow what you mean by "the same ensemble". Do you simply mean that they are similar? Or do you mean that the 4 expectation values apply to the "exact same ensemble". The issue does not disappear because we start talking about "ensemble" as opposed to individual particle pairs.

Perhaps it is still not clear what the difference is between "the same" and "different but similar". If i = 1,..,N represents identity for N members of a set, where the members could be particle pairs, ensembles, or whatever. "the same" means the same i, "different but similar" means different i, but still a member of the set. In computer-science-speak, it would be the difference between "equality" and "identity" or the difference between "value" and "reference".

So my question in this context will be:
Are those expectation values applicable to the same system, or to different but similar systems (whatever your definition of system is, particle pairs or ensembles).

The ensemble is an infinite number of pairs of particles, prepared in such a way that if I make a measurement on a large enough subset of pairs independently drawn from the infinite number of pairs, and then do this again on a different but large enough subset, the histogram of results will be essentially identical.

You could then also ask the same question, of von Neumann's essential assumption:
von Neumann: Any real linear combination of any two Hermitian operators represents an observable, and the same linear combination of expectation values is the expectation of the combination.
Was he talking about the same system or different but similar system.

The von Neuman proof and Bell's criticism of the implication von Neumann drew is irrelevant. Here we are talking about quantum mechanics. von Neumann and Bell were talking about hidden variables for quantum mechanics.

 

[billschnieder]

The ensemble is an infinite number of pairs of particles, prepared in such a way that if I make a measurement on a large enough subset of pairs independently drawn from the infinite number of pairs, and then do this again on a different but large enough subset, the histogram of results will be essentially identical.

But you are not answering my simple question. Do the expectation values each apply to the same large enough subset of pairs, or do they each apply to a different large-enough subsets of pairs from your "ensemble"? Or do you think there is no difference. That is my question.

 

[billschnieder]

The von Neuman proof and Bell's criticism of the implication von Neumann drew is irrelevant. Here we are talking about quantum mechanics. von Neumann and Bell were talking about hidden variables for quantum mechanics.

When we write
S = ⟨A(a)B(b)⟩+⟨A(a′)B(b′)⟩+⟨A(a′)B(b)⟩−⟨A(a)B(b′)⟩
And then substitute in expectation values from QM, we are talking about the linear combination of expectation values being the expectation of the linear combination and we are using it to imply that local realistic hidden variables do not agree with QM, are we not?

 

[atyy]

But you are not answering my simple question. Do the expectation values each apply to the same large enough subset of pairs, or do they each apply to a different large-enough subset of pairs from your "ensemble"? Or do you think there is no difference. That is my question.

First you have to let me know what you mean by these terms. Let's take the simpler case of one particle in a harmonic potential. The energy is $H = x^{2} + p^{2}$. Let's concentrate on pure states for simplicity. Let's consider an ensemble represented by $\psi$. On one realization of the ensemble I measure $x^{2}$ and find the expectation $X^{2} = \langle\psi|x^{2}|\psi\rangle$. On a second separate realization of the same ensemble represented by $\psi$, I measure $p^{2}$ and find the expectation $P^{2} = \langle\psi|p^{2}|\psi\rangle$. On a third separate realization of the same ensemble represented by $\psi$, I measure $H$ and find the expectation $E = \langle\psi|H|\psi\rangle$

1. Will I find that $E = X^{2} + P^{2}$?

2. In you terminology, did I measure $x^{2}$ and $p^{2}$ on the same or different "large-enough subset of pairs"?

 

[billschnieder]

First you have to let me know what you mean by these terms. Let's take the simpler case of one particle in a harmonic potential. The energy is $H = x^{2} + p^{2}$. Let's concentrate on pure states for simplicity. Let's consider an ensemble represented by $\psi$. On one realization of the ensemble I measure $x^{2}$ and find the expectation $X^{2} = \langle\psi|x^{2}|\psi\rangle$. On second separate realization of the same ensemble represented by $\psi$, I measure $p^{2}$ and find the expectation $P^{2} = \langle\psi|p^{2}|\psi\rangle$. On a third separate realization of the same ensemble represented by $\psi$, I measure $H$ and find the expectation $E = \langle\psi|H|\psi\rangle$

1. Will I find that $E = X^{2} + P^{2}$?
2. In you terminology, did I measure $x^{2}$ and $p^{2}$ on the same or different "large-enough subset of pairs"?

See the three underlined statements above? I am asking you whether when you write $S = ⟨A(a)B(b)⟩+⟨A(a′)B(b′)⟩+⟨A(a′)B(b)⟩−⟨A(a)B(b′)⟩$? the terms represent "the same realization" of the ensemble, or ⟨A(a)B(b)⟩ represents "one realization", while ⟨A(a′)B(b′)⟩ represents "a second separate realization.", etc in your terminology. Or do you believe it does not matter (ie, it can be both). The issue can not be directly translated to your example with position and momentum but to answer your (2) you measured the terms on different subsets.

 

[atyy]

See the three underlined statements above? I am asking you whether when you write $S = ⟨A(a)B(b)⟩+⟨A(a′)B(b′)⟩+⟨A(a′)B(b)⟩−⟨A(a)B(b′)⟩$? the terms represent "the same realization" of the ensemble, or ⟨A(a)B(b)⟩ represents "one realization", while ⟨A(a′)B(b′)⟩ represents "a second separate realization.", etc in your terminology. Or do you believe it does not matter (ie, it can be both). The issue can not be directly translated to your example with position and momentum but to answer your (2) you measured the terms on different subsets.

Still sticking with my example, how about the answer to (1) if the answer to (2) is that I measured the terms on different subsets?

 

[billschnieder]

Still sticking with my example, how about the answer to (1) if the answer to (2) is that I measured the terms on different subsets?

Okay, you start by saying:

Let's take the simpler case of one particle in a harmonic potential. The energy is $H = x^{2} + p^{2}$.

Implying it is a relationship which applies to one particle. Then you suddenly switch to ensembles while using the same symbols

Let's consider an ensemble represented by $\psi$. On one realization of the ensemble I measure $x^{2}$ and find the expectation $X^{2} = \langle\psi|x^{2}|\psi\rangle$. On a second separate realization of the same ensemble represented by $\psi$, I measure $p^{2}$ and find the expectation $P^{2} = \langle\psi|p^{2}|\psi\rangle$. On a third separate realization of the same ensemble represented by $\psi$, I measure $H$ and find the expectation $E = \langle\psi|H|\psi\rangle$

1. Will I find that $E = X^{2} + P^{2}$?

Your question is not clear because you are using the same symbols but they mean different things. H no longer applies to one particle but to an ensemble. Maybe if you answered the question and said why the answer was relevant to my question, I would appreciate the point you are making.

 

[atyy]

Okay, you start by saying:

Implying it is a relationship which applies to one particle. Then you suddenly switch to ensembles while using the same symbols

Your question is not clear because you are using the same symbols but they mean different things. H no longer applies to one particle but to an ensemble. Maybe if you answered the question and said why the answer was relevant to my question, I would appreciate the point you are making.

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?

 

[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.