Axiomatization of quantum mechanics and physics in general ?

[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.

 

[billschnieder]

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

Okay so we can linearly combine them. But it is nontrivial for non-commuting observables. The issue is not that we can not combine them but that we don't expect a trivial substitution to work. 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.

You can factorize that linear combination and end up with an expression of the form
$\langle A_a(B_b - B_{b'}) + A_a'(B_b + B_{b'})\rangle$ which is an expression of the form
$\hat{A}\otimes\hat{B} + \hat{C}\otimes\hat{D}$
If it has a solution, should satisfy

$\hat{A}|\phi\rangle\otimes\hat{B}|\chi\rangle + \hat{C}|\phi\rangle\otimes\hat{D}|\chi\rangle = \alpha(|\phi\rangle\otimes|\chi\rangle)$
But the LHS cannot be factored since the $[\hat{A},\hat{C}] ≠0$ and $[\hat{B},\hat{D}] ≠0$ so there is no solution. It is a meaningless expression for a single system.

 

[atyy]

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".

No, that is simply not true in the quantum formalism. There is no ψ1 and ψ2. The quantum calculation simply assumes ψ. Within the quantum formalism <ψ|x+p|ψ> has the same value as <ψ|x|ψ>+<ψ|p|ψ>.

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.

I believe we agree, except we are using different terms. Here by "same system" you mean that the distribution over the hidden variable is the same for different measurement settings. However, by "different systems", I mean that the distribution over the hidden variable is the same for different measurement settings, but that the different measurement settings are performed on different trials. And yes, it is a known loophole that if the measurement settings depend on the hidden variable, or if the hidden variable depends on the measurement setting, then one can have a local variable explanation of a Bell inequality violation.

 

[billschnieder]

No, that is simply not true in the quantum formalism.

I think there is a problem somewhere. If I have two entangled spin-1/2 particles send one to Alice and the other to Bob and I measure their spins, I always get opposite results according to QM, yes?
Now 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?

I believe we agree, except we are using different terms. Here by "same system" you mean that the distribution over the hidden variable is the same for different measurement settings.

No, I mean the exact same set of particles, not some different set with similar distribution of hidden variables.

However, by "different systems", I mean that the distribution over the hidden variable is the same for different measurement settings, but that the different measurement settings are performed on different trials.

This is what I mean by different systems. The same properties but not the same individual. The same type of particle pair but not the same particle pair. The same "type of" wavefunction but not the same wavefunction. That is what I mean by ψ1 and ψ2 vs ψ.

And yes, it is a known loophole that if the measurement settings depend on the hidden variable, or if the hidden variable depends on the measurement setting, then one can have a local variable explanation of a Bell inequality violation.

I'm not talking about loopholes at all. I'm not even talking about distributions of hidden variables and what may depend on settings or not depend on settings.

 

[bhobba]

But it is nontrivial for non-commuting observables.

Sorry - but its utterly trivial.

Its a simple consequence of Born's rule - commuting, non commuting - no difference.

Thanks
Bill

 

[bhobba]

But does QM say that

Of course it does - its a basic property of Bell States:
http://en.wikipedia.org/wiki/Bell_state

Pick any Bell state - say the first one.

The result of the observation will be |1A>|1B> or |0A>|0B>. If A gets state 1 B must be state 1 and conversely.

Its a simple result of what entanglement means.

Thanks
Bill

 

[billschnieder]

Of course it does - its a basic property of Bell States:
http://en.wikipedia.org/wiki/Bell_state

Pick any Bell state - say the first one.

The result of the observation will be |1A>|1B> or |0A>|0B>. If A gets state 1 B must be state 1 and conversely.

Its a simple result of what entanglement means.

Thanks
Bill

Maybe you misunderstood, each member of a pair is entangled with the other the second pair is not entangled with the first pair.

 

[billschnieder]

Sorry - but its utterly trivial.

Its a simple consequence of Born's rule - commuting, non commuting - no difference.

Thanks
Bill

So you think that expression has a solution?

 

[bhobba]

So you think that expression has a solution?

I have zero idea what you are getting at.

I was addressing the linearity of expectations.

Thanks
Bill

 

[bhobba]

Maybe you misunderstood, each member of a pair is entangled with the other the second pair is not entangled with the first pair.

So?

The second pair when observed will do the same thing.

I think there is a problem somewhere. If I have two entangled spin-1/2 particles send one to Alice and the other to Bob and I measure their spins, I always get opposite results according to QM, yes? Now 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?

Yes it does say that. The same experiments give the same results - that applies to any area of science - its so trivial its rarely if ever stated.

Thanks
Bill

 

[kith]

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?

Expectation values always commute because they are numbers, so your math is messed up. Do you think what you say fits within standard QM? If yes, you should be able to use standard QM notation. If no, you should clarify where your symbols deviate from the standard notation.

In QM, the state is associated with a preparation procedure. If we apply the same preparation procedure to two systems, they are in the same state. Is your question motivated by the idea what happens if the preparation procedure is unsharp, so that it prepares slightly different states in different trials?

Would it be appropriate to call your objection the induction-is-impossible loophole?

 

[bhobba]

Because even though <ψ1|A|ψ1> always commutes with <ψ2|B|ψ2> <ψ|A|ψ> does not commute with <ψ|B|ψ>.

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

Well spotted - I didn't even notice it.

I looked at it, looked at it again, scratched my head, scratched it some more - still no closer to understanding his issue.

Thanks
Bill

 

[kith]

Should there be a natural distinction between classical and quantum mechanics for continuous variables?

There are many ways to state fundamental differences between classical mechanics and QM, so I would say that these are certainly fundamentally different theories. Sure, we can supplement QM with Bohmian hidden variables to make at least some of these differences go away, but then we don't compare classical mechanics and QM but classical mechanics and Bohmian mechanics.

I don't really like the statement that QM is an effective theory because it suggests that there's something fundamentally wrong with QM while there's no experimental hint of this. Even in areas where QM gets in trouble conceptually -like dealing with gravity- the alleged more fundamental Bohmian mechanics doesn't seem to lead to different solutions than QM but again provides "only" an unobservable supplement. (To be fair I have to say that I may simply not know enough about possibly observable consequences from Bohmian mechanics which are incompatible with QM. Also I put "only" in " because I do value Bohmian mechanics as a different way of looking at QM. I just don't think it is superior.)

I tend to think more along Copenhagen-inspired lines that every physical theory is somehow effective. So if QM is a very good effective theory, Bohmian mechanics is a slightly worse effective theory because it still has the outstanding experimental support of QM but also introduces additional unobservable elements.

 

[billschnieder]

So?

The second pair when observed will do the same thing.

A bell state has two particles not 4. Now you
Are saying a measurement of one particle of a Bell state affects a particle not part of *the* Bell state. I'm surprised you don't see a problem.

Yes it does say that. The same experiments give the same results - that applies to any area of science - its so trivial its rarely if ever stated.

But that is the problem what you are suggesting disagrees with experiment.

 

[bhobba]

A bell state has two particles not 4. Now you Are saying a measurement of one particle of a Bell state affects a particle not part of *the* Bell state

Where you get that from has me beat - I specifically stated otherwise.

You measure one entangled pair and get one result.

Measure another entangled pair and get another result.

QM describes the outcome of both independent of each.

Like I said this is basic science not just QM - its the idea of independent experiments giving reproducible results.

I think there is a problem somewhere. If I have two entangled spin-1/2 particles send one to Alice and the other to Bob and I measure their spins, I always get opposite results according to QM, yes? Now 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?

Lets be clear. What you wrote and I have put in bold is saying if I repeat the same experiment you get the same result - of course you do. Here same result is you will get the same correlations.

Thanks
Bill

 

[billschnieder]

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?

But does QM say that

Of course it does - its a basic property of Bell States:
http://en.wikipedia.org/wiki/Bell_state

That is why I suggested that you misunderstood me, because you were (are) implying that the result of one pair is correlated with the result from a different pair. This is not what is observed experimentally. The reproducibility which I'm not questioning is the fact that the correlation between the members of the first pair is exactly the same correlation between the members of the second pair. Each pair is a Bell state after all. But you surely don't mean that you can mix and match one member of each pair and still have a bell state. Then you end up with a situation in which once the first pair is measured, the outcomes of the rest will be well defined and then the randomness from pair to pair usually obtained in experiments will not happen. That is why I say you are contradicting experiment.