- #36
ShayanJ
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What bias are you talking about?naima said:How can we accept such biased methodology to derive anything?
What bias are you talking about?naima said:How can we accept such biased methodology to derive anything?
I thought the definition of 'hidden variables' in this context was just that they contain information that is not contained in the quantum representation (ket or sum of operators). In Bell's words it implies the existence of 'a more complete specification of the state', where the additional information is represented by the symbol ##\lambda##.Shyan said:Its the definition of hidden variables to assign values to unmeasured quantities.
With the following clarifications every thing Shyan has said makes sense to me.naima said:It is often said that the Bell's theorem precludes local hidden variables. From a "modern" point of view one should never deduce conclusions from the existence of outputs in non commutative measurements.
It seems that the derivations of this theorem use such results.
Is there a proof which uses only the ##\lambda## in the case of possible measurements?
The "locality" (Einstein causality would be better) argument of this article isnaima said:We find the words "local" and "locality" in the DrChinese paper but is there a locality argument in the way to fill the table.
We have a game with two possible results: 1 and 1/3.
Do we need an experiment to show that Nature with its possible 0 results is not like that?
My attempt to explain it is http://ilja-schmelzer.de/realism/game.php. And I think that one should recognize what one gives up if one rejects it: The EPR criterion of reality: If we can, without distorting the system in any way, predict with certainty the result of the experiment, then there exists a corresponding element of reality.But there was a price to pay for such this experimental setup: we must add a SECOND assumption. That assumption is: A measurement setting for one particle does not somehow affect the outcome at the other particle if those particles are space-like separated. This is needed because if there was something that affected Alice due to a measurement on Bob, the results would be skewed and the results could no longer be relied upon as being an independent measurement of a second attribute. This second assumption is called "Bell Locality" and results in a modification to our conclusion above. In this modified version, we conclude: the predictions of any LOCAL Hidden Variables theory are incompatible with the predictions of Quantum Mechanics. Q.E.D.
I didn't mean they should be unmeasurable. naima was criticizing that Bell's theorem rules out only hidden variable theories that assign values to unmeasured observables, not unmeasurable observables. So he was actually criticizing that Bell's theorem rules out only hidden variables theories that assume counterfactual definiteness, but actually hidden variables are there exactly to retain counterfactual definiteness! That's why they were proposed!andrewkirk said:I thought the definition of 'hidden variables' in this context was just that they contain information that is not contained in the quantum representation (ket or sum of operators). In Bell's words it implies the existence of 'a more complete specification of the state', where the additional information is represented by the symbol ##\lambda##.
As I understand his paper, Bell does not say that ##\lambda## must be unmeasurable. Quite likely, if there were such non-local hidden variables, as in Bohm's theory, there would be no current way to measure them. But I am not aware of any theorem that says that such additional information is in principle unmeasurable. Is there such a theorem? Can we perhaps obtain such a theorem as a corollary of the No-Communication Theorem, as being able to measure non-local quantities could conceivably open the door to FTL communication.
The position that Bell's theorem rules out "only" hidden variables that assume counterfactual definiteness missed, of course, the whole point of Bell's theorem, which uses, in its first part, the EPR argument to prove that there has to be counterfactual definiteness for this particular experiment if we assume Einstein causality.Shyan said:naima was criticizing that Bell's theorem rules out only hidden variable theories that assign values to unmeasured observables, not unmeasurable observables. So he was actually criticizing that Bell's theorem rules out only hidden variables theories that assume counterfactual definiteness, but actually hidden variables are there exactly to retain counterfactual definiteness! That's why they were proposed!
You will have to explain.Ilja said:And the result of Bell's theorem is that we now have a possibility to distinguish, by observation, theories where relativistic symmetry is fundamental from theories where relativistic symmetry is only an approximate, non-fundamental symmetry. For the first class, we can prove Bell's inequalities, for the second class we cannot.
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. So, the theory cannot be fundamentally relativistic.naima said:You will have to explain.
Take the electron Dirac theory.
According to you, the Bell theorem can give you a tool: By observation, you will be able to prove the inequality if the theory is relativistic. (maybe it is the reverse, the violation of the inequality?)
Are you thinking of the Aspect experiment?
How can you say that there are two classes of theories without an example in each class?Ilja said:And the result of Bell's theorem is that we now have a possibility to distinguish, by observation, theories where relativistic symmetry is fundamental from theories where relativistic symmetry is only an approximate, non-fundamental symmetry. For the first class, we can prove Bell's inequalities, for the second class we cannot.
Ilja said: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.
Ilja said: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 said: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.
stevendaryl said: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.
stevendaryl said:To me, a completely relativistic theory would be one with the following sort of character:
- There is a notion of "state" defined on any spacelike slice of spacetime.
- If you partition the slice into small localized neighborhoods, the state for the slice can be factored into localized states for each neigborhood.
- The future state of a neighborhood depends only on the states of neighborhoods in the backwards lightcone.
- The laws relating future states to past states does not depend on how spacetime is sliced up into spacelike slices evolving over time.
PeterDonis said:The obvious definition of a "relativistic theory" is that all predictions of measurement results are Lorentz invariant.
PeterDonis said:#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.
PeterDonis said: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.
stevendaryl said:Why give a special role to "measurement results"?
atyy said:The standard model is not known to have exact Lorentz symmetry.
PeterDonis said:Perhaps I should have said "direct observables". I'm not try to get into issues involved with measurement in QM.
stevendaryl said:I think that when people doubt whether quantum field theory is relativistic, they are thinking specifically about measurements.
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.)PeterDonis said:I don't know what you're talking about here. QFT has Lorentz symmetry as an exact symmetry.
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?PeterDonis said:If we built a theory with Lorentz symmetry as only an approximate symmetry, it would not be QFT, it would be some different theory.
Ilja said: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?
Ilja said:It is allowing for a hidden preferred frame which is the key difference.
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.PeterDonis said:This "hidden preferred frame" has nothing to do with observable effects;
Ilja said: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.
Ilja said:It has, because its existence allows violations of Bell's inequality.
No, my preferred theory has nothing to do with this claim.PeterDonis said:I understand that this is what your preferred theory says. But experiment does not say this.
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.stevendaryl said:I think you've blurred the distinction between two different things:
No Einstein causal theory can violate Bell's inequalities. But it's much less clear that no Lorentz invariant theory can.
- Einstein causality
- 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.stevendaryl said: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.
Hidden variables are theoretical concepts used in physics to explain the behavior of particles at a subatomic level. They are not directly observable, but are thought to influence the outcomes of quantum experiments.
Hidden variables are proposed to explain the apparent randomness and unpredictability of quantum systems. They are meant to provide a more deterministic explanation for the behavior of particles, in contrast to the probabilistic nature of quantum mechanics.
Observable variables are directly measurable and can be observed in experiments, while hidden variables are not directly observable but are believed to influence the behavior of observable variables. Hidden variables are also used to explain phenomena that cannot be explained by observable variables alone.
The concept of hidden variables is often used in the debate between determinism and indeterminism in quantum mechanics. Hidden variables propose a deterministic explanation for the behavior of particles, while indeterminism suggests that the behavior of particles is inherently random and cannot be fully predicted.
The use of hidden variables in quantum mechanics is a highly debated topic in the scientific community. While some scientists believe that hidden variables provide a more complete explanation for quantum phenomena, others argue that they are unnecessary and that the probabilistic nature of quantum mechanics is sufficient to explain observations.