What should hidden variables explain?

[ShayanJ]

How can we accept such biased methodology to derive anything?

What bias are you talking about?

 

[naima]

I repeat once again.
In the paper we have 8 lines and 3 counters. each of them can be increased at each shot. their sum always increases at less by 1/3. and at the end we compute the mean value.
We say then let us see if Nature agrees.
We also need 3 counters and at each shot we increase one of them if we have ++ or -- for the random directions. At the end it is possible that they still contain 0 ans that the mean value is null.
Do you think that there is no methodology problem when you compare the mean values?

 

[andrewkirk]

Its the definition of hidden variables to assign values to unmeasured quantities.

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.

 

[Zafa Pi]

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?

With the following clarifications every thing Shyan has said makes sense to me.
First there are various Bell type theorems and second there are Bell inequalities. Let me give a concrete example:

Theorem: Let Ah, At, Bh, Bt each be +1 or -1. (these values may come about via some random process such as coin flipping)
If Ah•Bh = 1 then P(At•Bt = -1) ≤ P(At•Bh = -1) + P(Ah•Bt = -1).

Proof: P(At•Bt = -1) = P(At•Bt•Ah•Bh = -1) = P(At•Bh•Bt•Ah = -1) =
P({At•Bh = -1 and Bt•Ah = 1} or {At•Bh = 1 and Bt•Ah = -1}) =
P(At•Bh = -1 and Bt•Ah =1) + P(At•Bh = 1 and Bt•Ah = -1) ≤
P(At•Bh = -1) + P(Ah•Bt = -1) QED

The Theorem is a valid piece of mathematics. Within the Theorem there is an Inequality.

The numbers Ah, At, Bh, Bt could come about in the usual fashion:
#1 The physical set up for the Theorem:
Alice and Bob are 2 light minutes apart, and Eve is half way between them. Alice
has a fair coin (see probability appendix) and a device. Her device has 2 buttons
labeled h and t, a port to receive a signal from Eve. The device also has a screen
that will display “Eve’s signal received” when a signal from Eve is received. It will
also display either +1 or -1 if one of the buttons is pushed. Bob has the same
equipment and shows the same values, tho the internal workings of his device
may be different.
#2 The following experiment is performed:
Eve simultaneously sends a light signal to each of Alice and Bob. When Alice’s
device indicates Eve’s signal has been received she flips her coin. If it comes up
heads she pushes button h, otherwise button t, and then notes what the screen
displays. What Alice does takes less than 30 seconds. The same goes for Bob.
Here we assume locality - no faster than light influences.
#3 Notation:
If Alice flipped a head and pushed button h, we let Ah be the value her screen
would show. So Ah = 1 or -1 and is the result of some objective physical process.
Similarly we let At be the value if she had flipped a tail. We let Bh and Bt be the
analogous values for Bob.

Eve could send each of Alice and Bob a photon from an entangled pair so that
Ah•Bh = 1, and P(At•Bt = -1) = 3/4, while P(At•Bh = -1) + P(Ah•Bt = -1) = 1/2.
(the state of the pair is √½(|00> +|11>), Ah and Bh measure at 0 degrees, At at 30,º and Bt at -30º)

The Inequality is false. How could this be? No aspect of reality or a physical theory will invalidate a mathematical theorem. The hypothesis of the Theorem has all 4 numbers Ah, At, Bh, Bt existing at once and shows up in the proof. In each experiment, however only two of the numbers are observed, e,g. At and Bt. The existence of the other two is inferred by counterfactual definiteness = hidden variables = realism, and it is that inference which must be questioned (assuming locality).

 

[Ilja]

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?

The "locality" (Einstein causality would be better) argument of this article is

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.

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.

 

[ShayanJ]

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.

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!

 

[Ilja]

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!

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.

But hidden variable theories are certainly not proposed to retain some counterfactual definiteness. Instead, proponents of dBB as well as other hidden variable theories feel quite comfortable with these proposals, despite the fact that these proposals do not have counterfactual definiteness.

 

[naima]

I would agree that hidden variables theory (as i see them) do not imply
counterfactual definiteness.
My criticism to the Bell theorem is that it gives a scientific status to counterfactuality. We examine a proposed model which is not even falsifiable. Is there a statistical experiment proposed by their defenders to verify it?
If i write that 97.5 % of the ghosts in NY know that God is a green Dragon, who has to prove or to disprove it?

 

[Ilja]

Sorry, but it is reverse. Bell's inequality indeed examines something which, before this proof, was thought to be completely unfalsifiable: The thesis that relativistic symmetry is not only restricted to observables (something we can test, and do test, and what is not questioned by the established hidden variable theories) but more, namely that it is a fundamental insight, and that there cannot exist a preferred frame.

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.

Counterfactuality is only an intermediate step of Bell's proof. It is derived from the EPR argument and Einstein causality.

 

[haushofer]

Just my two cents: Hidden variable theories are most useful in an attempt to provide a clear ontology to QM, like the Bohmian interpretation.

 

[naima]

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.

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?
Take any other quantic theory with a non relativistic lagrangian.
We are in the second class of theories.
Will you use the same device to prove something? and to prove what? non violation?

 

[Ilja]

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?

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]

Can you give me an example of two quantic theories in the opposite cases?
How can the same observation device tell you which is relativistic?

 

[Ilja]

No, because I don't know any quantum theory which would allow to prove Bell's inequality.

What observation can tell is that a theory is not fundamentally relativistic. If cannot tell if it is.

 

[naima]

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.

How can you say that there are two classes of theories without an example in each class?
We are in a quantum physics forum not in thermodynamics. Please could you give me examples in quantum theory?

 

[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.