Axiomatization of quantum mechanics and physics in general ?

[Fredrik]

What about relativistic quantum mechanics logic ?

The difference between non-relativistic QM and special relativistic QM is just a choice between the Galilean group and the Poincaré group. You postulate that there's a homomorphism from one of these groups into the group of automorphisms of the lattice. I you choose the former group, the result is non-relativistic QM. If you choose the latter, the result is special relativistic QM.

Axiomatic QFT (which is relativistic QM) is a whole new ball game.

I view relativistic quantum field theories as theories defined within the framework of special relativistic QM (as defined above). I believe that there are also non-relativistic QFTs, but I have never studied one.

The axioms of (axiomatic) QFT are supposed to define what a quantum field theory is, so yes, it's definitely a very different game.

 

[bhobba]

I view relativistic quantum field theories as theories defined within the framework of special relativistic QM (as defined above). I believe that there are also non-relativistic QFTs, but I have never studied one.

I have off and on been studying QFT over the years from various books such as Zee and others.

Mathematically - yea - I got it. But physically it didn't gel.

Then I came across thew following book recently released:
https://www.amazon.com/dp/019969933X/?tag=pfamazon01-20

The Kindle price was pretty good so I took a punt.

Really good at explaining what it means.

Its not a well known fact, bu5t still true, that QM can be reformulated as a the3ory of creation and anhilation operators:
http://math.bu.edu/people/mak/Styer Am J Phys 2002.pdf

See interpretation F.

What that book does is explain that view step by step then shows how it applies to QFT so you immediately know what the formalism means.

Thanks
Bill

 

[kith]

But does Hardy fail to mention the assumption of non-contextuality? Or is it in there, and just in a more natural or "reasonable" way, as he intends?

What does it even mean for an inherently probabilistic theory to be (non-)contextual?

 

[atyy]

What does it even mean for an inherently probabilistic theory to be (non-)contextual?

My understanding about non-contextuality in the context of Gleason's theorem comes from the comments by Peres in http://books.google.com/books?id=IjCNzbJYONIC&dq=peres+quantum&source=gbs_navlinks_s.

If understand Peres correctly, he says Gleason's theorem assumes that if Pu and Pv are orthogonal projectors, then <Pu + Pv> = <Pu> + <Pv>. However, there isn't a unique way to write P = Pu + Pv = Px + Py, so the assumption is that <Pu> + <Pv> = <Px> + <Py>. The assumption is non-trivial since measurements of Pu and Pv usually require different experimental setups from those that measure Px and Py.

Hardy doesn't even assume Hilbert spaces or anything, but he does derive the Born rule. Most people believe his derivation is correct and complete, so it's most likely that he has not left out an assumption such as non-contextuality. Rather, he has other axioms which do the work, and I'm wondering which of his axioms do that, and whether one can understand them as equivalent to non-contextuality in Gleason's.

 

[microsansfil]

The difference between non-relativistic QM and special relativistic QM is just a choice between the Galilean group and the Poincaré group. You postulate that there's a homomorphism from one of these groups into the group of automorphisms of the lattice. I you choose the former group, the result is non-relativistic QM. If you choose the latter, the result is special relativistic QM.

"A principle of modern mathematics holds in this lesson: when you are dealing with an entity S with a measure of structure, try to determine its group of automorphisms, the group of transformations of its components that preserve the structural relations. You can expect to gain a deep understanding of the constitution of S in this way. "Hermann Wey

http://en.wikipedia.org/wiki/Representation_theory_of_the_Poincaré_group

Does this mathematical logic view could be useful in the search for unification of General relativity and Quantum mechanics ?

Patrick

 

[Fredrik]

"A principle of modern mathematics holds in this lesson: when you are dealing with an entity S with a measure of structure, try to determine its group of automorphisms, the group of transformations of its components that preserve the structural relations. You can expect to gain a deep understanding of the constitution of S in this way. "Hermann Wey

http://en.wikipedia.org/wiki/Representation_theory_of_the_Poincaré_group

Does this mathematical logic view could be useful in the search for unification of General relativity and Quantum mechanics ?

I know almost nothing about that. I think that loop quantum gravity is an attempt to develop "general relativistic quantum mechanics" in a way that's similar to what I was talking about, but I don't really know.

 

[kith]

If understand Peres correctly, he says Gleason's theorem assumes that if Pu and Pv are orthogonal projectors, then <Pu + Pv> = <Pu> + <Pv>. However, there isn't a unique way to write P = Pu + Pv = Px + Py, so the assumption is that <Pu> + <Pv> = <Px> + <Py>. The assumption is non-trivial since measurements of Pu and Pv usually require different experimental setups from those that measure Px and Py.

So the assumption is called non-contextuality because for a single observable Pu+Pv, I can use all kinds of different experimental setups -where different setups correspond to different bases of the eigenspace of Pu+Pv- to measure it's expectation value?

I don't have access to Peres at the moment and an immediate follow-up question is how does this relate to Bohmian mechanics and it's contextuality? By what mathematical elements are observables even represented in Bohmian mechanics?

Hardy doesn't even assume Hilbert spaces or anything, but he does derive the Born rule. Most people believe his derivation is correct and complete, so it's most likely that he has not left out an assumption such as non-contextuality. Rather, he has other axioms which do the work, and I'm wondering which of his axioms do that, and whether one can understand them as equivalent to non-contextuality in Gleason's.

I haven't seen his derivation but since the first four axioms also apply to classical probability theory, it certainly has to do with the fifth. I would also be interested in seeing how the Hilbert space formalism and Hardy's formulation are related exactly.

 

[atyy]

So the assumption is called non-contextuality because for a single observable Pu+Pv, I can use all kinds of different experimental setups -where different setups correspond to different bases of the eigenspace of Pu+Pv- to measure it's expectation value?

I don't have access to Peres at the moment and an immediate follow-up question is how does this relate to Bohmian mechanics and it's contextuality? By what mathematical elements are observables even represented in Bohmian mechanics?

I don't understand this issue very well. As I understand it, the non-contextuality in Gleason's theorem seems to have nothing to do with hidden variables, since it is just about measures on states in Hilbert space.

However, by some corollary of Gleason's, there is apparently a link between the non-contextuality there and in hidden variable theories. It is discussed by in this link given by bhobba to an article by Bell http://fy.chalmers.se/~delsing/QI/Bell-RMP-66.pdf, and also in this proof by Busch of a Gleason-like theorem, but pertaining to POVMs instead of POMs http://arxiv.org/abs/quant-ph/9909073. Maybe bhobba or Fredrik can explain in more detail here?

I haven't seen his derivation but since the first four axioms also apply to classical probability theory, it certainly has to do with the fifth. I would also be interested in seeing how the Hilbert space formalism and Hardy's formulation are related exactly.

Here's Hardy's first derivation http://arxiv.org/abs/quant-ph/0101012. The fifth axiom just says that there's a continuous reversible transformation between pure states. Isn't that met by classical mechanics in phase space?

 

[kith]

Just a quick note:

The fifth axiom just says that there's a continuous reversible transformation between pure states. Isn't that met by classical mechanics in phase space?

Hardy's states are probability vectors (p₁,...,p_n). For pure states, all but one of the p_i are zero so the pure states correspond to the elements of the n-dimensional standard basis and there's no continuous transformation between them.

However, I am not sure if Hardy's formulation of QM works for infinite dimensional systems like a particle in space.

 

[atyy]

Just a quick note:

Hardy's states are probability vectors (p₁,...,p_n). For pure states, all but one of the p_i are zero so the pure states correspond to the elements of the n-dimensional standard basis and there's no continuous transformation between them.

However, I am not sure if Hardy's formulation of QM works for infinite dimensional systems like a particle in space.

Perhaps that is why Hardy's derivation doesn't work for continuous variables. For discrete variables, it is obvious that classically, a particle is either in one box or the other, whereas in the quantum case, it can be in a superposition of being in both boxes. As I understand it, the Chiribella et al derivation is also only of finite dimensional quantum mechanics.

On the other hand the Mackey-Piron-Soler approach, according to bhobba, gets finite and infinite dimensional quantum mechanics. The other "reformulation" I know that can get infinite dimensional QM is the Leifer-Spekkens http://arxiv.org/abs/1107.5849. That paper only deals with the finite dimensional case, but in the comments here he says the extension to the infinite dimensional should be straightforward http://mattleifer.info/2011/08/01/the-choi-jamiolkowski-isomorphism-youre-doing-it-wrong/.

 

[bhobba]

What does it even mean for an inherently probabilistic theory to be (non-)contextual?

Non contextuality means the probability measure does not depend on what resolution of the identity a projection operator is part of. Its a very natural condition to impose mathematically because its simply expressing basis independence - you would not expect a measure to depend of your basis - after all that is a pretty basic property of vector spaces - the important geometric stuff like length or angle doesn't depend on basis. But physically it has profound implications because a resolution of the identity corresponds to an actual measurement apparatus.

Its the key ingredient in Gleason's proof - there are others such as the strong superposition principle - but that is the key one.

In the geometric approach to QM you start with a logistic then show its observables (one can define things like observables, states, even probability measures on observables, etc in a logistic - see Chapter 3 of Varadarajan) are isomorphic to a Hilbert space. One then uses Gleason to show that probability measure is the Born rule.

The issue though is when you model something using it. Are all the things that determine how the system behaves observables? There may be hidden variables and they may be contextual. In BM for example the pilot wave is explicitly contextual:
http://philsci-archive.pitt.edu/3026/1/bohm.pdf

Don't necessarily agree with that link saying BM is not using hidden variables - but non contextual ones. The pilot wave is hidden - you can't ever directly observe it.

But of course Dymystifyer is the expert on BM around here - not me.

Thanks
Bill

 

[bhobba]

As I understand it, the non-contextuality in Gleason's theorem seems to have nothing to do with hidden variables, since it is just about measures on states in Hilbert space.

You are correct - it doesn't SEEM to. But if there are contextual variables, hidden or otherwise, then Gleason's breaks down.

Gleason is basically a stronger version of Kochen-Speker - in fact Kochen-Speker is a simple corollary to Gleason.

Thanks
Bill

 

[bhobba]

However, I am not sure if Hardy's formulation of QM works for infinite dimensional systems like a particle in space.

Personally in discussing foundational issues I stick to finite vector spaces.

I view the infinite dimensional case via the Rigged Hilbert Space formalism where the dual is simply introduced for mathematical convenience. Here the dual I am referring to is the dual to the space of all sequences of finite length. In the weak convergence of that space any linear functional is the limit of a sequence of the space the functionals are defined on.

Thanks
Bill

 

[kith]

I don't understand this issue very well. As I understand it, the non-contextuality in Gleason's theorem seems to have nothing to do with hidden variables, since it is just about measures on states in Hilbert space.

Well, let's look at KS first. I am not sure if I am on the right track but my current understanding is this (following Peres).

KS says that if we have a Hilbert space of dimension $N ≥ 3$, we cannot find a function $v$ which consistently assigns a probability of $0$ and $1$ to all projectors acting on this Hilbert space. Namely, if we combine a projector with $N-1$ other commuting projectors with $\sum_i P_i=1$ we cannot have $\sum_i v(P_i)=1$ for all choices of the $N-1$ projectors.

Let's say we have an observable $Q$ with possible experimental outcomes $a$, $b$ and $c$ and associated projectors $P_a$, $P_b$ and $P_c$ (these commute and sum to the identity). In order to assign a true but hidden value to our quantity we could assign a probability $v$ of $0$ or $1$ to each of the projectors. Let's say that the true value of $Q$ is $b$, so $v(P_a)=v(P_c)=0$ and $v(P_b)=1$.

KS now implies that there exists a similar observable $R$ where the probabilities either don't sum to one or are inconsistent with previously assigned probabilities. The situation could be something like this: $Q$ and $R$ share the projector $P_a$ but instead of $P_b$ and $P_c$, $R$ is associated with $P_{\beta}$ and $P_{\gamma}$. Consistency with other observables $Q', Q'', ...$ forces us to either choose $v(P_{\beta}) = v(P_{\gamma}) = 0$ or $v(P_a)=1$. So either the sum of probabilities for $R$ is equal to $0$ or the sum for $Q$ is equal to $2$. Both options imply that $v$ isn't a probability in the first place, so this kind of hidden variable assignment is ruled out.

The problem goes away if the probability associated with $P_a$ depends on the context, i.e when we allows different probabilities $v(P_a; P_b,P_c)=0$ and $v(P_a; P_{\beta}, P_{\gamma})=1$ for different observables $Q$ and $R$.

 

[kith]

Perhaps that is why Hardy's derivation doesn't work for continuous variables.

Has it really been shown that it doesn't work or did he simply restrict his discussion to the finite / countably infinite case?

 

[kith]

Gleason is basically a stronger version of Kochen-Speker - in fact Kochen-Speker is a simple corollary to Gleason.

I haven't followed the logic of the proofs in detail but I tend to agree. KS says that the non-contextual probability distribution cannot look a certain way while Gleason says how exactly it has to look.

 

[atyy]

Well, let's look at KS first. I am not sure if I am on the right track but my current understanding is this (following Peres).

KS says that if we have a Hilbert space of dimension $N ≥ 3$, we cannot find a function $v$ which consistently assigns a probability of $0$ and $1$ to all projectors acting on this Hilbert space. Namely, if we combine a projector with $N-1$ other commuting projectors with $\sum_i P_i=1$ we cannot have $\sum_i v(P_i)=1$ for all choices of the $N-1$ projectors.

Let's say we have an observable $Q$ with possible experimental outcomes $a$, $b$ and $c$ and associated projectors $P_a$, $P_b$ and $P_c$ (these commute and sum to the identity). In order to assign a true but hidden value to our quantity we could assign a probability $v$ of $0$ or $1$ to each of the projectors. Let's say that the true value of $Q$ is $b$, so $v(P_a)=v(P_c)=0$ and $v(P_b)=1$.

KS now implies that there exists a similar observable $R$ where the probabilities either don't sum to one or are inconsistent with previously assigned probabilities. The situation could be something like this: $Q$ and $R$ share the projector $P_a$ but instead of $P_b$ and $P_c$, $R$ is associated with $P_{\beta}$ and $P_{\gamma}$. Consistency with other observables $Q', Q'', ...$ forces us to either choose $v(P_{\beta}) = v(P_{\gamma}) = 0$ or $v(P_a)=1$. So either the sum of probabilities for $R$ is equal to $0$ or the sum for $Q$ is equal to $2$. Both options imply that $v$ isn't a probability in the first place, so this kind of hidden variable assignment is ruled out.

The problem goes away if the probability associated with $P_a$ depends on the context, i.e when we allows different probabilities $v(P_a; P_b,P_c)=0$ and $v(P_a; P_{\beta}, P_{\gamma})=1$ for different observables $Q$ and $R$.

Thanks! I found this explanation by Spekkens which seems to match what you wrote http://arxiv.org/abs/quant-ph/0406166v3: "Traditionally, a noncontextual hidden variable model of quantum theory is one wherein the measurement outcome that occurs for a particular set of values of the hidden variables depends only on the Hermitian operator associated with the measurement and not on which Hermitian operators are measured simultaneously with it. For instance, suppose A,B and C are Hermitian operators such that A and B commute, A and C commute, but B and C do not commute. Then the assumption of noncontextuality is that the value predicted to occur in a measurement of A does not depend on whether B or C was measured simultaneously. The Bell-Kochen-Specker theorem shows that a hidden variable model of quantum theory that is noncontextual in this sense is impossible for Hilbert spaces of dimension three or greater."

Intuitively, this seems like a generalization of the idea that canonically conjugate observables like momentum and position do not have simultaneous existence, since it depends on non-commuting observables, even if only indirectly.

 

[atyy]

Has it really been shown that it doesn't work or did he simply restrict his discussion to the finite / countably infinite case?

Originally I thought that his method did not work for the infinite dimensional case, because axiom five which is the distinction between finite dimensional classical and quantum theories, clearly holds for classical continuous variables. However, according to his discussion of the issue in section 9 of http://arxiv.org/abs/quant-ph/0101012, his axioms still rule out classical continuous variables, because of axiom 3: "A system whose state is constrained to belong to an M dimensional subspace (i.e. have support on only M of a set of N possible distinguishable states) behaves like a system of dimension M."

So I guess it is unknown whether his axioms work or not for the continuous case.

 

[kith]

However, according to his discussion of the issue in section 9 of http://arxiv.org/abs/quant-ph/0101012, his axioms still rule out classical continuous variables, because of axiom 3: "A system whose state is constrained to belong to an M dimensional subspace (i.e. have support on only M of a set of N possible distinguishable states) behaves like a system of dimension M."

This sounds a bit unphysical to me. Such a finite-dimensional subspace of phase space is a set of measure zero (loosely speaking a collection of delta functions). It seems strange to say that classical statistical mechanics violates this axiom because such sets of measure zero are not considered physical there anyway. So I would say that this axiom is irrelevant in classical statistical mechanics because there are no physical subspaces with a smaller dimension than the whole phase space.

Another interesting thought in section 9 is that superpositions smooth out the discontinuities of a possibly quantized space which may reduce the discomfort associated with this notion.

 

[atyy]

This sounds a bit unphysical to me. Such a finite-dimensional subspace of phase space is a set of measure zero (loosely speaking a collection of delta functions). It seems strange to say that classical statistical mechanics violates this axiom because such sets of measure zero are not considered physical there anyway. So I would say that this axiom is irrelevant in classical statistical mechanics because there are no physical subspaces with a smaller dimension than the whole phase space.

Another interesting thought in section 9 is that superpositions smooth out the discontinuities of a possibly quantized space which may reduce the discomfort associated with this notion.

Should there be a natural distinction between classical and quantum mechanics for continuous variables? For position and momentum, there's Bohmian mechanics which is a classical way of viewing quantum mechanics. There's also Montina's arument http://arxiv.org/abs/0711.4770 that hidden variables for a finite dimensional quantum system must be continuous, if the dynamics are Markovian.

So I tend to think of quantum mechanics as a very good effective theory, because the true underlying variables are usually much more inconvenient.

 

[billschnieder]

Interesting discussion. Although you guys haven't mentioned Bell's theorem but think it is relevant to the issue here. Specifically to demonstrate violation of the CHSH, it is often written that

S = E(a, b) − E(a, b′) + E(a′, b) + E(a′ b′) ≤ 2

E(a,b) = -E(a, b′) = E(a′, b) = E(a′ b′) = 1/√2
∴ S = 2√2 > 2 → Violation.

However, there is an ambiguity:
Possibility 1: All 4 terms are observables on single system. This is actually the assumption used in the derivation. In this case, although the a measurement commutes with the b measurement, E(a,b) does not commute with the E(a',b) and requires quite a different experimental arrangement to measure and it won't be proper to just add the add the separate individual terms from separate systems (the von Neuman error).
Possibility 2: Each term is an observable of a different but similarly prepared system. This allows S to be the linear combination of individual results but because of different degrees of freedom, the derivation of the inequality becomes problematic.

It therefore seems non-contextuality is relevant to the issue of hidden variables, both von Neuman's approach and Bell's. To measure each term, E(a,b) for example you post select a set of particle pairs using coincidence at (a,b) settings. Then to measure E(a, b') you have two possibilities. You could post select within the first set, all those pairs for which there is also (a,b') coincidence, but this is non-trivial since the b measurement does not commute with b' measurement, but will not necessarily give you the same result as if you post select a completely different set of particle pairs with coincidence at (a,b').

Did Bell make the same mistake as von Neuman then? It looks like it.

 

[atyy]

Did Bell make the same mistake as von Neuman then? It looks like it.

No, that is not correct. The Bell derivation is just classical probability, see http://arxiv.org/abs/1208.4119 (particularly Fig. 19, 25-27).

 

[billschnieder]

No, that is not correct. The Bell derivation is just classical probability, see http://arxiv.org/abs/1208.4119 (particularly Fig. 19, 25-27).

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.

 

[bhobba]

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

Its a basic calculation following from the principles of QM eg:
http://en.wikipedia.org/wiki/Bell's...re_violated_by_quantum_mechanical_predictions

Of relation to this thread where QM is developed using the geometrical approach based on quantum logic the reason is the different logic of QM.

Thanks
Bill

 

[billschnieder]

Its a basic calculation following from the principles of QM eg:
http://en.wikipedia.org/wiki/Bell's...re_violated_by_quantum_mechanical_predictions

Of relation to this thread where QM is developed using the geometrical approach based on quantum logic the reason is the different logic of QM.

Thanks
Bill

Thanks for the link. I'm talking specifically about the expression after the "so that" in the section you quoted above. Filling in the part they left out, we get something like the following:

Starting from
$E_\psi(x,z) = \langle\psi|(\sigma_L\cdot x)(\sigma_R\cdot z)|\psi\rangle = -x\cdot{z}$

For expectation values for a single system, that expression from wikipedia becomes something like:
$S^{\psi} = |\langle\psi|(\sigma_L\cdot{a})(\sigma_R\cdot{b}) - (\sigma_L\cdot{a})(\sigma_R\cdot{b'}) + (\sigma_L\cdot{a'})(\sigma_R\cdot{b}) + (\sigma_L\cdot{a'})(\sigma_R\cdot{b'})|\psi\rangle|$
Which has no solution.
However, the reason many people think it works is because they think it is equivalent to
expectation values for 4 independent similarly prepared systems which is:
$S^{1234} = |\langle\psi_1|(\sigma_L\cdot{a})(\sigma_R\cdot{b})|\psi_1\rangle - \langle\psi_2|(\sigma_L\cdot{a})(\sigma_R\cdot{b'})|\psi_2\rangle + \langle\psi_3|(\sigma_L\cdot{a'})(\sigma_R\cdot{b})|\psi_3\rangle + \langle\psi_4|(\sigma_L\cdot{a'})(\sigma_R\cdot{b'})|\psi_4\rangle|\psi \rangle|$

 

[bhobba]

Thanks for the link. I'm talking specifically about the expression after the "so that" in the section you quoted above.

Cant follow your concern - looks like a trivial substitution to me.

Thanks
Bill

 

[billschnieder]

Cant follow your concern - looks like a trivial substitution to me.

Thanks
Bill

Which of the two expressions I gave above is the correct representation of this "trivial" substitution?
$S^{\psi} = |\langle\psi|(\sigma_L\cdot{a})(\sigma_R\cdot{b})|\psi\rangle - \langle\psi|(\sigma_L\cdot{a})(\sigma_R\cdot{b'})|\psi\rangle + \langle\psi|(\sigma_L\cdot{a'})(\sigma_R\cdot{b})|\psi\rangle + \langle\psi|(\sigma_L\cdot{a'})(\sigma_R\cdot{b'})|\psi\rangle|$

$S^{1234} = |\langle\psi_1|(\sigma_L\cdot{a})(\sigma_R\cdot{b})|\psi_1\rangle - \langle\psi_2|(\sigma_L\cdot{a})(\sigma_R\cdot{b'})|\psi_2\rangle + \langle\psi_3|(\sigma_L\cdot{a'})(\sigma_R\cdot{b})|\psi_3\rangle + \langle\psi_4|(\sigma_L\cdot{a'})(\sigma_R\cdot{b'})|\psi_4\rangle|$

Note that in $S^\psi$ the 4 spin correlation observables do not commute so their linear combination is not an observable. It is a questionable substitution.

 

[bhobba]

Which of the two expressions I gave above is the correct representation of this "trivial" substitution?

You mentioned after 'so that'.

Just before that we have some rather easy to show identities that are 1/root 2 or -1/root 2.

Substitute them into the equation after and you have 2 root 2 > 2.

The equations you wrote down don't bear any relation to it - at least as far as I can see.

You will have to provide a LOT more detail of exactly what you are getting at.

Thanks
Bill

 

[atyy]

Note that in $S^\psi$ the 4 spin correlation observables do not commute so their linear combination is not an observable. It is a questionable substitution.

You will have to provide a LOT more detail of exactly what you are getting at.

I don't see any problem either, and I don't really understand billschneider's concern. However, I have a guess that what billschenider is saying is that the quantum expression T = <ψ|A+B|ψ> can be interpreted in two ways. First we can treat O = A + B as a single observable, and say that T = <ψ|O|ψ> should be measured by a single apparatus that measures O. However, we can also treat A and B as separate observables and measure <ψ|A|ψ> and <ψ|B|ψ> separately, then add them up to get T. The quantum formalism says that both physically different procedures yield the same value of T.

As I understand it, the Bell inequality as a derivation based on classical probability assumes the second interpretation: each term is a measurement with a different physical setup, and we add the results up.

However, it is interesting to consider the first interpretation, from the quantum point of view. Do A and B have to commute in order for O to exist as a quantum observable that can be measured by a single physical setup? I would say no. An example is the energy of the simple harmonic oscillator E = p² + x².

 

[billschnieder]

Just before that we have some rather easy to show identities that are 1/root 2 or -1/root 2.

Substitute them into the equation after and you have 2 root 2 > 2.

Ok, let me start by referring to some arguments from Bell's paper which you mentioned earlier

http://fy.chalmers.se/~delsing/QI/Bell-RMP-66.pdf

Consider now the proof of von Neumann that dispersion free states, and so hidden variables are impossible. His essential assumption is: Any real linear combination of any two Hermitian operators represents an observable, and the same linear combination of expectation values is the expectation value of the combination. This is true for quantum mechanical states; it is required by von Neumann of the hypothetical dispersion free states also.
...
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

Now looking at the "so that" expression from Wikipedia, as you said yourself, it is a trivial addition of expectation values. I'm saying according to Bell's own argument against von Neumann, the addition of expectation values in QM is non-trivial, especially for non-commuting observables. So the claim that the the "so that" expression gives you 2√2, is suspect.

 

[billschnieder]

However, I have a guess that what billschenider is saying is that the quantum expression T = <ψ|A+B|ψ> can be interpreted in two ways. First we can treat O = A + B as a single observable, and say that T = <ψ|O|ψ> should be measured by a single apparatus that measures O. However, we can also treat A and B as separate observables and measure <ψ|A|ψ> and <ψ|B|ψ> separately, then add them up to get T. The quantum formalism says that both physically different procedures yield the same value of T.

Yes, it should give the same result for the same system, if A and B commute. But in this case we are dealing with non-commuting observables. If you measure A on one system ψ1 and B on a different system ψ2, you will not necessarily get the same result as what you should expect if you had measured A and B on the same system. Because even though <ψ1|A|ψ1> always commutes with <ψ2|B|ψ2> <ψ|A|ψ> does not commute with <ψ|B|ψ>. If I derive an inequality for a single system which contains both A and B, it will not be correct to trivially substitute in values from different systems, would it?

Therefore:
$\langle A(a) B(b) \rangle + \langle A(a') B(b') \rangle + \langle A(a') B(b) \rangle - \langle A(a) B(b') \rangle$
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.

So what is the problem (you say), why don't we just pick the first and be done with it?
Because Bell's derivation assumes a single system. Note that the experiments are performed in accordance with (1).

 

[atyy]

Yes, it should give the same result for the same system, if A and B commute. But in this case we are dealing with non-commuting observables. If you measure A on one system ψ1 and B on a different system ψ2, you will not necessarily get the same result as what you should expect if you had measured A and B on the same system. Because even though <ψ1|A|ψ1> always commutes with <ψ2|B|ψ2> <ψ|A|ψ> does not commute with <ψ|B|ψ>. If I derive an inequality for a single system which contains both A and B, it will not be correct to trivially substitute in values from different systems, would it?

There is no ψ1 and ψ2, just ψ. The issue of commuting/non-commuting is a not relevant. But even if it were, it is not true that the quantum formalism predicts different values for <ψ|A+B|ψ> and <ψ|A|ψ> + <ψ|B|ψ>.

Therefore:
$\langle A(a) B(b) \rangle + \langle A(a') B(b') \rangle + \langle A(a') B(b) \rangle - \langle A(a) B(b') \rangle$
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.

So what is the problem (you say), why don't we just pick the first and be done with it?
Because Bell's derivation assumes a single system. Note that the experiments are performed in accordance with (1).

Bell's derivation assumes what you call (1).

 

[atyy]

@billschneider, if you are not talking about the quantum prediction, and asking whether a theory with local variables in which the different measurement settings also correspond to different hidden variables can explain violations of a Bell inequality, then yes, that is a known loophole to even an ideal Bell test. In http://arxiv.org/abs/1208.4119 some varieties of local variable explanations invoking correlations between measurement settings and the hidden variable are termed "superdeterminism" and "retrocausation" (Fig. 26, 27).

 

[bhobba]

There is no ψ1 and ψ2, just ψ. The issue of commuting/non-commuting is a not relevant. But even if it were, it is not true that the quantum formalism predicts different values for <ψ|A+B|ψ> and <ψ|A|ψ> + <ψ|B|ψ>.

Exactly.

Here we are considering the quantum formalism in which the addition of expectation values is true.

Its not like Von-Neuman's proof where he assumed it for all variables - including hidden ones. The observables here are NOT hidden.

Thanks
Bill

 

[billschnieder]

There is no ψ1 and ψ2, just ψ. The issue of commuting/non-commuting is a not relevant.

commuting/non-commuting is irrelevant if you have ψ1 and ψ2 (isolated systems). But it is surely relevant if you have just ψ (one system). I can prepare one system, measure x precisely, then prepare a different one very similarly and measure p precisely. The two observables would commute, that won't be the case if you have just ψ. It is a distinction between "similar" and "the same".

Bell's derivation assumes what you call (1).

This is not true. You can easily verify that the expression:

ab - a'b + a'b + a'b' <= 2

is only valid "the same" system because you can factorize a(b-b') + a'(b+b') and show that whenever (b-b') = 0, (b+b) = -2 or 2 and vice-versa for a,b,a'b' = {+1, -1}. This is not true for separate systems because you cannot factorize the expression (a1b2 - a2'b2 + a3'b3 + a4'b4') and the RHS is necessarily 4.