A What should hidden variables explain?

[Ilja]

The two classes of theories are well-defined, and they contain nontrivial examples. Classical GR in its spacetime interpretation is an example of a fundamentally relativistic theory. The dBB interpretation is an example of a theory where relativistic symmetry is not fundamental, but derived, accidental, and holds only in quantum equilibrium. Ether theories like http://arxiv.org/abs/0908.0591 are an example where relativistic symmetry appears only in a large distance limit.

You want one of the first class, but nonetheless quantum? Sorry, I see no reason to believe that there exist consistent quantum theories which are fundamentally relativistic. RQFT is not. It derives, with a lot of "grit your teeth", relativistic symmetry for observable effects, that's all, and for GR already nothing helps.

 

[naima]

Isn't QFT relativistic?

 

[Ilja]

It is relativistic in the weak sense - it tells us that the observable effects are indistinguishable. It does not tell us if there is really no difference.

 

[PeterDonis]

By observation of the violation of Bell's inequality (Aspect or better) I'm able to prove that we have a theory where Bell's theorem cannot be proven. In a fundamentally relativistic theory, it can be proven.

This seems to me to be nothing but your personal definition of a "fundamentally relativistic" theory. I think most physicists would say that quantum field theory is "fundamentally relativistic", yet, as you yourself point out, one can show that QFT violates the Bell inequalities (and these violations have, as you say, been experimentally confirmed, so we know QFT is accurate in this respect). So I don't understand what you mean by "fundamentally relativistic" except as your own personal arbitrary label.

It is relativistic in the weak sense - it tells us that the observable effects are indistinguishable. It does not tell us if there is really no difference.

I don't know what you're talking about here. QFT has Lorentz symmetry as an exact symmetry. If we built a theory with Lorentz symmetry as only an approximate symmetry, it would not be QFT, it would be some different theory. And the QFT we currently have, with Lorentz symmetry as an exact symmetry, predicts violation of the Bell inequalities. So again it looks to me like you are just assigning arbitrary labels.

 

[stevendaryl]

This seems to me to be nothing but your personal definition of a "fundamentally relativistic" theory. I think most physicists would say that quantum field theory is "fundamentally relativistic", yet, as you yourself point out, one can show that QFT violates the Bell inequalities (and these violations have, as you say, been experimentally confirmed, so we know QFT is accurate in this respect). So I don't understand what you mean by "fundamentally relativistic" except as your own personal arbitrary label.

I think that this is a little subtle. In quantum field theory, there are three different types of objects:

1. Field operators
2. The state that they operate on
   
3. Measurement results

The field operators obey relativistic equations of motion, so if they were the only objects, then QFT would certainly be a completely relativistic theory. The state, though, is not something that "lives" in spacetime; it lives in Hilbert space, so it's not clear what it would mean to say that it is or is not relativistic. As for measurement results, there isn't a relativistic equation governing the evolution of measurement results, they are only probabilistically related to the state.

So it's not clear to me that the whole shebang is a relativistic theory, or even what it would mean to be a relativistic theory.

To me, a completely relativistic theory would be one with the following sort of character:

1. There is a notion of "state" defined on any spacelike slice of spacetime.
2. If you partition the slice into small localized neighborhoods, the state for the slice can be factored into localized states for each neigborhood.
3. The future state of a neighborhood depends only on the states of neighborhoods in the backwards lightcone.
4. The laws relating future states to past states does not depend on how spacetime is sliced up into spacelike slices evolving over time.

Something like that--it would probably take a lot of work to make it mathematically precise. But roughly speaking, this amounts to fields and particles evolving according to relativistic equations of motion. QFT doesn't have this kind of character, because the quantum state doesn't factor into localized states (because of entanglement), and also because we don't really have any kind of relativistic evolution laws for observables.

 

[PeterDonis]

it's not clear to me that the whole shebang is a relativistic theory, or even what it would mean to be a relativistic theory.

The obvious definition of a "relativistic theory" is that all predictions of measurement results are Lorentz invariant. QFT satisfies this definition.

To me, a completely relativistic theory would be one with the following sort of character:

1. There is a notion of "state" defined on any spacelike slice of spacetime.
2. If you partition the slice into small localized neighborhoods, the state for the slice can be factored into localized states for each neigborhood.
3. The future state of a neighborhood depends only on the states of neighborhoods in the backwards lightcone.
4. The laws relating future states to past states does not depend on how spacetime is sliced up into spacelike slices evolving over time.

The only one of these that seems relevant to me in order to call something a "relativistic theory" is #4; but even that is problematic because I don't agree with #1, and #4 depends on #1. Who cares about definitions on spacelike slices? Measurement results are local--they consist of some scalar invariant being assigned to some event (point in spacetime). AFAICT you don't even need the concept of "spacelike slices" to formulate a relativistic theory (e.g., QFT). You do need the concept of "spacelike separated events" if you want to check that operators at spacelike separated events commute, but even that has nothing to do with defining states on spacelike slices.

#3 above seems to me to relate to "causality", not "relativistic"; the only "relativistic" part is that you use the backwards light cone. But I don't see what would prevent one in principle from constructing a theory that didn't obey #3 but still made predictions that were Lorentz invariant.

 

[stevendaryl]

The obvious definition of a "relativistic theory" is that all predictions of measurement results are Lorentz invariant.

I wouldn't call that obvious. Why give a special role to "measurement results"? Isn't a measurement just a special case of a quantum interaction between two subsystems (the special case in which one subsystem is a measuring device)?

 

[stevendaryl]

#3 above seems to me to relate to "causality", not "relativistic"; the only "relativistic" part is that you use the backwards light cone. But I don't see what would prevent one in principle from constructing a theory that didn't obey #3 but still made predictions that were Lorentz invariant.

Yes, you're right. There might be a formulation of QM (time-symmetric QM, or something) that would allow one to compute probabilities for histories that wasn't formulated in terms of interactions propagating at light speed (or slower). There might be some way to formulate atemporal, distant correlations that are Lorentz-invariant. I wouldn't say that we have that with QFT yet, though. The rigorous part is the evolution equations for field operators, but that's not the whole story.

 

[atyy]

This seems to me to be nothing but your personal definition of a "fundamentally relativistic" theory. I think most physicists would say that quantum field theory is "fundamentally relativistic", yet, as you yourself point out, one can show that QFT violates the Bell inequalities (and these violations have, as you say, been experimentally confirmed, so we know QFT is accurate in this respect). So I don't understand what you mean by "fundamentally relativistic" except as your own personal arbitrary label.
I don't know what you're talking about here. QFT has Lorentz symmetry as an exact symmetry. If we built a theory with Lorentz symmetry as only an approximate symmetry, it would not be QFT, it would be some different theory. And the QFT we currently have, with Lorentz symmetry as an exact symmetry, predicts violation of the Bell inequalities. So again it looks to me like you are just assigning arbitrary labels.

The standard model is not known to have exact Lorentz symmetry. The standard model is usually considered an effective field theory.

 

[PeterDonis]

Why give a special role to "measurement results"?

Perhaps I should have said "direct observables". I'm not try to get into issues involved with measurement in QM. I'm just trying to distinguish the theoretical predictions that we actually compare with experimental data, from "internal" aspects of the theory that don't get directly compared with experiment. I'm saying that only the former have to be Lorentz invariant in a "relativistic" theory.

 

[PeterDonis]

The standard model is not known to have exact Lorentz symmetry.

I will agree that we can't rule out the presence of operators in the standard model that violate exact Lorentz symmetry; the best we can do is to constrain their magnitude based on experimental data.

However, AFAIK it is still true that a QFT with exact Lorentz symmetry (i.e., not including any operators that could violate that symmetry) predicts violation of the Bell inequalities. That's really the primary point I was trying to make in response to Ilja.

 

[stevendaryl]

Perhaps I should have said "direct observables". I'm not try to get into issues involved with measurement in QM.

But I think that when people doubt whether quantum field theory is relativistic, they are thinking specifically about measurements.

 

[PeterDonis]

I think that when people doubt whether quantum field theory is relativistic, they are thinking specifically about measurements.

I can't say what other people are thinking in this connection. I agree with you that there is no principled distinction in QM between "measurements" and other interactions. But there is a distinction between observables and theoretical quantities that aren't observables.

 

[Ilja]

I don't know what you're talking about here. QFT has Lorentz symmetry as an exact symmetry.

I disagree. It has Lorentz symmetry as a symmetry of observable effects. As a symmetry for observable effects, it is exact. As far as the theory is well-defined (see Haag's theorem for reasons to doubt it is.)

If we built a theory with Lorentz symmetry as only an approximate symmetry, it would not be QFT, it would be some different theory.

I doubt. QFT can be understood as well as defined as a limit of regularized theories, with each regularized theory having Lorentz symmetry only approximately. This is what is done in the conceptually simplest case - lattice regularizations. Given that the limit itself is problematic, one can understand QFT as well as a theory which describes a lattice regularization with a critical length so small that violations of Lorentz symmetry becomes unobservable. Given that we anyway do not believe that QFT holds below Planck length, what would be the difference between QFT as a (well-defined) lattice theory with $h=10^{-100} l_{Pl}$ and the (hypothetically existing despite Haag's theorem) theory with exact Lorentz symmetry?

And, anyway, the difference between fundamental and weak relativistic symmetry is not at all about exact or approximate Lorentz symmetry. It is allowing for a hidden preferred frame which is the key difference. The Lorentz ether has exact Lorentz symmetry.

 

[PeterDonis]

Given that we anyway do not believe that QFT holds below Planck length, what would be the difference between QFT as a (well-defined) lattice theory with $h=10^{-100} l_{Pl}$ and the (hypothetically existing despite Haag's theorem) theory with exact Lorentz symmetry?

Observably, nothing, by hypothesis, unless and until we were able to make measurements at distance scales of $10^{-100} l_{Pl}$. If measurements at that scale are taken to be impossible in principle, then there is no measurable difference at all between the two theories; they both predict the same observables. But they are still obviously different theories conceptually and mathematically. The differences just can never be experimentally tested. See below.

It is allowing for a hidden preferred frame which is the key difference.

This "hidden preferred frame" has nothing to do with observable effects; it's purely an internal aspect of the theory. As my other posts have made clear, I don't think a "relativistic" theory has to have exact Lorentz symmetry of purely internal aspects of the theory. It only has to have exact Lorentz symmetry of observable effects. More precisely, it has to have that as predicted by the theory. It might be impossible, as above, to distinguish exact Lorentz symmetry from some approximate version due to finite limits on measurement accuracy.

 

[Ilja]

This "hidden preferred frame" has nothing to do with observable effects;

It has, because its existence allows violations of Bell's inequality. Which a fundamentally Einstein-causal theory would not allow. And violations of BI are observable.

 

[stevendaryl]

It has, because its existence allows violations of Bell's inequality. Which a fundamentally Einstein-causal theory would not allow. And violations of BI are observable.

I think you've blurred the distinction between two different things:

1. Einstein causality
2. Lorentz invariance

No Einstein causal theory can violate Bell's inequalities. But it's much less clear that no Lorentz invariant theory can. Lorentz invariance does not directly imply that there can be no FTL effects. What you can prove is that FTL signal propagation, together with relativity, leads to a contradiction, because you could then set up a closed loop in which you receive a reply to a message before you send the message. The type of nonlocal correlation implied by violations of Bell's inequality does not allow FTL signals to be sent, so the proof that it leads to a contradiction with relativity fails.

 

[PeterDonis]

It has, because its existence allows violations of Bell's inequality.

I understand that this is what your preferred theory says. But experiment does not say this. Experiment cannot detect your "hidden preferred frame", so there is no way of showing experimentally that that is what allows violations of BI. And since there are theories that do not have a "hidden preferred frame" but still predict violations of BI, you cannot justify this claim on theoretical grounds either. All it is is your personal preference.

 

[Ilja]

I understand that this is what your preferred theory says. But experiment does not say this.

No, my preferred theory has nothing to do with this claim.

If you don't go back to mysticism, denying the existence of objective reality even if you see it (the EPR criterion of reality) as well as that observed correlations require causal explanations (Reichenbach's common cause principle, which distinguishes science from astrology), then you have simple theorems.

From Einstein causality one can derive Bell's inequality.

From a theory with a hidden preferred frame, which would allow hidden causal influences, you cannot derive Bell's inequality.

Bell's inequality is testable and tested.

 

[Ilja]

I think you've blurred the distinction between two different things:

1. Einstein causality
2. Lorentz invariance

No Einstein causal theory can violate Bell's inequalities. But it's much less clear that no Lorentz invariant theory can.

Mystical theories (theories which reject as the EPR principle of reality, as Reichenbach's common cause) can violate everything. But for a realistic, causal theory with fundamental Lorentz invariance (that means, where not only observables but everything should have Lorentz invariance) you can derive Einstein causality from the requirement that causality has to preserve Lorentz invariance.

What you can prove is that FTL signal propagation, together with relativity, leads to a contradiction, because you could then set up a closed loop in which you receive a reply to a message before you send the message.

And why would this be a problem if there is no causality? For mysticism causal loops are not a problem at all. There are only some correlations, that's all. Everything is somehow mystically correlated, this is sufficient, once the idea of a necessity of a causal explanation is rejected.

 

[stevendaryl]

Mystical theories (theories which reject as the EPR principle of reality, as Reichenbach's common cause) can violate everything. But for a realistic, causal theory with fundamental Lorentz invariance (that means, where not only observables but everything should have Lorentz invariance) you can derive Einstein causality from the requirement that causality has to preserve Lorentz invariance.

And why would this be a problem if there is no causality? For mysticism causal loops are not a problem at all. There are only some correlations, that's all. Everything is somehow mystically correlated, this is sufficient, once the idea of a necessity of a causal explanation is rejected.

Well, if you have two theories that are empirically indistinguishable, then I don't see how you can call one "mystical" and the other not. The correlations implied by quantum mechanics are not arbitrary, they are very specific. It may be emotionally unsatisfying to have a theory that violates Einstein causality, but to go from there to "all bets are off, we might as well embrace magic and voodoo" is an over-reaction.

 

[Ilja]

Well, if you have two theories that are empirically indistinguishable, then I don't see how you can call one "mystical" and the other not. The correlations implied by quantum mechanics are not arbitrary, they are very specific. It may be emotionally unsatisfying to have a theory that violates Einstein causality, but to go from there to "all bets are off, we might as well embrace magic and voodoo" is an over-reaction.

Ok, we have three theories here:

1.) Fundamental realistic relativity, which gives Einstein causality and, then, Bell's inequality. This theory is empirically falsified by the known experiments.
2.) Realistic and causal Lorentz ether, or dBB interpretation of QM. It allows hidden causal influences into the future as defined by the preferred time coordinate. It does not allow to prove Bell's inequality, thus, is not empirically falsified by a violation of Bell's inequality.
3.) The immunization of (1) against this empirical falsification, by rejection of realism (EPR criterion) and causality (Reichenbach's common cause).

(1) and (2) are empirically distinguishable, by the violation of Bell's inequality, and have been empirically distinguished. (3) is mystical.

Feel free to explain me what is different between magic and voodoo, as long as they make predictions (astrology certainly does). The only remaining difference is that the numbers predicted by quantum theory fit better than those predicted by astrology, or at least we scientists think so. If this is fine with you, ok. But there was another difference between science and astrology in the past: Namely that science has constructed models of reality, models which have explained the numbers by realistic, causal influences.

 

[PeterDonis]

From Einstein causality one can derive Bell's inequality.

From a theory with a hidden preferred frame, which would allow hidden causal influences, you cannot derive Bell's inequality.

So what? There are also theories that do not have a hidden preferred frame, from which you cannot derive Bell's inequality (they predict violations of it). So why should I care that "Einstein causality" allows you to derive Bell's inequality? (That assumes that we even have a rigorous definition of "Einstein causality" plus a theory that exhibits it. As stevendaryl has already pointed out, "Einstein causality" is not the same as Lorentz invariance. See below.)

Fundamental realistic relativity

What theory are you talking about?

(1) and (2) are empirically distinguishable, by the violation of Bell's inequality, and have been empirically distinguished

I have closed this thread for moderation. Please PM me specific references for what you mean by "fundamental realistic relativity" if you want the thread reopened.