A Is there a local interpretation of Reeh-Schlieder theorem?

[samalkhaiat]

Do I miss something?

Yes and it is very trivial. I have already told you that shifting the energy by a constant does not change any thing. So, if you have the eignvalue equations \left( M_{\mu\nu}, P_{\mu}\right)|\Omega \rangle = (c_{\mu\nu},q_{\mu})| \Omega \rangle, you can always redefine the Poincare’ generators \left( M_{\mu\nu},P_{\mu}\right) \to \left( M_{\mu\nu}+ c_{\mu\nu},P_{\mu}+q_{\mu}\right) to obtain \left( M_{\mu\nu},P_{\mu}\right)| \Omega \rangle = 0.

 

[vanhees71]

What I think is curious about the quantized KG field, and might turn at least some heads that might not be turned by any of the above discussion, is that the real part of its 2-point Wightman function is the inverse Fourier transform of $\left(\sqrt{\vec k\cdot\vec k+m^2}\right)^{-1}$, where the operator $\sqrt{-\vec\partial\cdot\vec\partial+m^2}$ is said to be "anti-local" by mathematicans (see my Physics Letters A 338 (2005) 8–12, arXiv:quant-ph/0411156, and, much more definitively, I.E. Segal, R.W. Goodman, "Anti-locality of certain Lorentz-invariant operators", http://www.jstor.org/stable/24901461. This last might possibly be the reference for some physicists to be confronted with, because its conclusions are quite similar to Hegerfeldt's conclusions.)
At the end of the day, I think all the nonlocality can be attributed to boundary and initial conditions, which, not being dynamical, for some people makes it not nonlocality. After dark, however, this somewhat pushes us towards the introduction of some form of superdeterminism, so perhaps it's just that you take your choice of poison.

It's the other way around, and this argument can be found in standard textbooks like Peskin&Schröder: Because time evolution with using $\sqrt{\hat{\vec{p}}^2+m^2}$ as an Hamiltonian in a putative 1st-quantization formulation of relativistic QT (which I call relativistic QM) leads to non-locality and breaks causality even for free fields, one concludes that one has to include the negative-frequency modes into the came, and then the observation of a stable world, i.e., the boundedness of the Hamiltonian of particles from below, forces us to use the 2nd-quantization formulation, i.e., QFT, which I'd call the only physically sensible relativistic QT we know of. It also allows for microcausality and validity of the linked-cluster theorem for the S-matrix, which clearly shows that interactions in QFT are indeed described as local interactions. There are no "spooky actions at a distance" as Einstein thought in the modern formulation of QFT, aka the "Standard Model".

There's of course the usual caveat that there's no mathematically rigorous proof for the existence of interacting QFTs in (1+3)-space-time dimensions, and purists might say that we cannot be sure that QFT really works as expected and suggested by renormalized perturbative QFT.

 

[Demystifier]

QFT, which I'd call the only physically sensible relativistic QT we know of

String theorists would disagree.

 

[vanhees71]

Since when is string theory physics? SCNR.

 

[Demystifier]

It also allows for microcausality and validity of the linked-cluster theorem for the S-matrix, which clearly shows that interactions in QFT are indeed described as local interactions. There are no "spooky actions at a distance" as Einstein thought in the modern formulation of QFT, aka the "Standard Model".

The presence/absence of spooky action at a distance has nothing to do with the fact that QFT interactions are local.
- If one accepts a suitable orthodox/minimal interpretation of QT, then the spooky action at a distance is absent even in non-relativistic QM with nonlocal Coulomb-like potentials.
- If one accepts an ontic/hidden-variable interpretation of QT, then the spooky action at a distance is present even in local QFT.

 

[Demystifier]

Since when is string theory physics? SCNR.

Anything published in a physics journal is - physics.

 

[A. Neumaier]

nonlocal Coulomb-like potentials.

Any potential displays spooky acts at a distance!

If one accepts an ontic/hidden-variable interpretation of QT, then the spooky action at a distance is present even in local QFT.

Which ontic/hidden-variable interpretation of which local relativistic QFT do you have in mind?

 

[Peter Morgan]

It's the other way around, and this argument can be found in standard textbooks like Peskin&Schröder: Because time evolution with using $\sqrt{\hat{\vec{p}}^2+m^2}$ as an Hamiltonian in a putative 1st-quantization formulation of relativistic QT (which I call relativistic QM) leads to non-locality and breaks causality even for free fields, one concludes that one has to include the negative-frequency modes into the came, and then the observation of a stable world, i.e., the boundedness of the Hamiltonian of particles from below, forces us to use the 2nd-quantization formulation, i.e., QFT, which I'd call the only physically sensible relativistic QT we know of. It also allows for microcausality and validity of the linked-cluster theorem for the S-matrix, which clearly shows that interactions in QFT are indeed described as local interactions. There are no "spooky actions at a distance" as Einstein thought in the modern formulation of QFT, aka the "Standard Model".

There's of course the usual caveat that there's no mathematically rigorous proof for the existence of interacting QFTs in (1+3)-space-time dimensions, and purists might say that we cannot be sure that QFT really works as expected and suggested by renormalized perturbative QFT.

I agree with all of this, which is an entirely consistent way of discussing QFT, but it's a perspective that I consider to be laden with conventions. What might be called the "Einstein conventions" are also entirely consistent, and we can rigorously transform from one to the other (arguably this is what is done in my arXiv:1709.06711 for the free EM, Dirac, and complex KG quantum and random fields, which is being not discussed here on PF; I'll propose that the math of the exact transformations there implicitly defines what the Einstein conventions might be), but within the Einstein conventions there is a precise kind of Lorentz invariant nonlocality and other properties are transformed (including that the positivity of the quantum Hamiltonian operator becomes the positivity of the Hamiltonian function). The conventions you are pressing for, almost insisting upon, which might be crudely stated as the Correspondence Principle and all its consequences, have been supremely successful for the last 90 years, but I suggest that a significant part of the progress in our understanding of and in our ability to engineer using quantum physics over the last 30 years, say, has been through considering alternative conventions, in some of which the effective nonlocality of a state can be considered something of a resource.

 

[Demystifier]

A ny potential acts at spooky distance!

What is ny potential?

Which ontic/hidden-variable interpretation of which local relativistic QFT do you have in mind?

E.g. Bohmian interpretation of relativistic QFT. But that's not the topic of this thread, I already said a lot about those things in other threads.

 

[martinbn]

He means any, not a ny.

 

[Demystifier]

A ny potential acts at spooky distance!

It is important where the word "spooky" is put. The expression is "spooky action at a distance", not the "action at spooky distance". The expression is used by Einstein to describe features related to the problem of quantum measurement, wave-function collapse, (in)completeness of QM and such. He did not use this expression to describe Newtonian mechanics.

 

[A. Neumaier]

What is ny potential? E.g. Bohmian interpretation of relativistic QFT. But that's not the topic of this thread, I already said a lot about those things in other threads.

But you could at least link to key posts in these threads.

 

[A. Neumaier]

He did not use this expression to describe Newtonian mechanics.

But Newton himself was already dissatisfied with action at a distance, and general relativity has eliminated this spooky feature.

Quantum mechanics reintroduced immediate effects at a distance; they are inherent to Born's probability interpretation of $|\psi(x)|^2$. Einstein therefore objected to the probabilistic interpretation of QM.

But relativistic QFT has eliminated again the action at a distance. It uses Born's rule only for asymptotic scattering results in the rest frame of the scattering event, not for the interpretation of arbitrary observables. Indeed, observables defined by smeared relativistic fields do not have a meaningful operational Born interpretation.

 

[martinbn]

It is important where the word "spooky" is put. The expression is "spooky action at a distance", not the "action at spooky distance". The expression is used by Einstein to describe features related to the problem of quantum measurement, wave-function collapse, (in)completeness of QM and such. He did not use this expression to describe Newtonian mechanics.

Actually it is Newtonian gravity that has action at a distance (whether one finds it spooky or not). Quantum mechanics has no action at a distance (but I agree with Einstein that it is spooky).

 

[Demystifier]

But relativistic QFT has eliminated again the action at a distance. It uses Born's rule only for asymptotic scattering results in the rest frame of the scattering event, not for the interpretation of arbitrary observables. Indeed, observables defined by smeared relativistic fields do not have a meaningful operational Born interpretation.

Are you saying that relativistic QFT cannot explain the experiments that show violation of Bell inequalities?

 

[Demystifier]

Actually it is Newtonian gravity that has action at a distance (whether one finds it spooky or not).

Of course.

Quantum mechanics has no action at a distance (but I agree with Einstein that it is spooky).

Quantum mechanics has correlations at a distance. Would you agree that those correlations are spooky?

 

[Peter Morgan]

Quantum mechanics has correlations at a distance. Would you agree that those correlations are spooky?

There are correlations at a distance in classical equilibrium Gibbs states, even for local dynamics such as the KG equation, either because equilibrium states are the consequence of infinitely long-time relaxation processes or else just because of nonlocal boundary conditions, translation invariance, minimum free energy, or some equivalent constraint. The former suggests there is no spookiness, the latter could be said to be spooky, but which one uses depends on tractability during computation as much as on philosophical niceties.

 

[Demystifier]

There are correlations at a distance in classical equilibrium Gibbs states, even for local dynamics such as the KG equation, either because equilibrium states are the consequence of infinitely long-time relaxation processes or else just because of nonlocal boundary conditions, translation invariance, minimum free energy, or some equivalent constraint. The former suggests there is no spookiness, the latter could be said to be spooky, but which one uses depends on tractability during computation as much as on philosophical niceties.

Those correlations can be explained by local deterministic beables. The Bell-type correlations cannot. That's why the latter are much spookier.

 

[Peter Morgan]

Those correlations can be explained by local deterministic beables. The Bell-type correlations cannot. That's why the latter are much spookier.

Violations of Bell-type inequalities can be seen as more-or-less natural by noting that they are a consequence of incompatible measurements, noncommuting operators, or transformations between different basis elements (take your choice between those three). I take the nonlocality not to be at the mathematical center of the story. I suggest the following reference:

From this perspective, the violation of Bell-type inequalities is a signal that at least some of the measurements are incompatible, so that noncommuting operators must be used in models of those measurements, which in turn can be taken to be a signal that somewhere in the experimental procedure a transformation from one basis to another was implicitly or explicitly introduced. That's classically as straightforward as it is for quantum theory. Of course, if the change of basis was something like a Fourier transform, which is about as maximally nonlocal as a transformation can be, then yes there's nonlocality, but it's a relatively demystified (I often want to use the word) spookiness. In classical signal analysis, the Wigner function turns up as soon as one considers time-frequency distributions.
Experimental violations of Bell inequalities are most often associated with eigenspaces of noncommuting operators that act as generators of representations of the Euclidean rotation group (usually in conditions where nonrelativistic approximations are entirely adequate). Such, generated by the Poisson bracket, are classically as natural as they are for quantum theory.

 

[Demystifier]

Of course, if the change of basis was something like a Fourier transform, which is about as maximally nonlocal as a transformation can be, then yes there's nonlocality, but it's a relatively demystified (I often want to use the word) spookiness. In classical signal analysis, the Wigner function turns up as soon as one considers time-frequency distributions.

What Bell has shown is that quantum nonlocality is not like this classical non-spooky "nonlocality".

 

[Peter Morgan]

What Bell has shown is that quantum nonlocality is not like this classical non-spooky "nonlocality".

I'd more put it that Bell has shown that one or more of his assumptions about what happens when we perform a measurement and describe it classically are not satisfied in an experiment that violates a Bell inequality. Bell's assumptions about what a random field looks like are too strong. We can be on the same page if you have a look at this, perhaps,

arXiv:cond-mat/0403692, however my more recent papers on random fields have moved the discussion somewhat and put much more mathematical backbone into the relationship between quantum and random fields.

 

[ftr]

But relativistic QFT has eliminated again the action at a distance.

That does not seem to be the case by your own account(both of them, I will explain). Since you deny the physicality of "virtual particles" then we are left with just a mathematical relation hinting at nonlocality. If you say that these are real photons(as I gathered from your interpretation FAQ), then assume two particle world, then how did the particles know about each other AND their distance from each other (to send to each other the appropriate photon/s to represent their momentum that results ) unless their presence inherently detected by each other nonlocally. It seems that the particles interaction is via an effect that is due to the particles presence, hence, nonlocality.

This question is also to other posters.

 

[A. Neumaier]

Are you saying that relativistic QFT cannot explain the experiments that show violation of Bell inequalities?

No. I was saying what I wrote. To get from QFT the setting in which Bell experiments are performed requires already several approximations

 

[A. Neumaier]

Quantum mechanics has no action at a distance (but I agree with Einstein that it is spooky).

?

Interacting nonrelativistic 2-particle quantum systems are defined by an action principle and have obvious action at a distance.

 

[A. Neumaier]

Those correlations can be explained by local deterministic beables. The Bell-type correlations cannot. That's why the latter are much spookier.

But Einstein didn't know Bell's theorem, and hence couldn't have referred to this kind of spookiness.

 

[A. Neumaier]

That does not seem to be the case by your own account

Nonsense.

Bell nonlocality is only correlation at a distance and has nothing to do with action at a distance.

Photons transmitted in optical fibers (as used for long-distance Bell experiments) have nothing to do with virtual particles. They are effective quasiparticles moving not with the vacuum speed of light (as the QFT photons) but at lower speed.

 

[ftr]

Nonsense.

I really meant the coulomb potential as modeled in QFT as in ZEE's. but now I am also confused by your declaration "Any potential acts at spooky distance!".

 

[A. Neumaier]

now I am also confused by your declaration "Any potential acts at spooky distance!".

Two interacting (classical or quantum) particles correspond to a dynamics where the first particle at x immediately responds to the second far away particle at y by the force obtained as the gradient of the potential V(x-y). This is an action at a distance in Newton's sense (and can be cast in terms of the variation of a nonlocal action in Lagrange's sense). Thus it is spooky in Einstein's sense.

the coulomb potential as modeled in QFT as in ZEE

Well, this leads to fully local QED; the apparent nonlocality is an artifact of the gauge in which the theory is written.

Of course one can write any local theory in nonlocal terms, but this does not make it nonlocal, or change the fact that everything observable is local in the sense of extended causality.

 

[dextercioby]

Two interacting (classical or quantum) particles correspond to a dynamics where the first particle at x immediately responds to the second far away particle at y by the force obtained as the gradient of the potential V(x-y). This is an action at a distance in Newton's sense (and can be cast in terms of the variation of a nonlocal action in Lagrange's sense). Thus it is spooky in Einstein's sense.

Well, this leads to fully local QED; the apparent nonlocality is an artifact of the gauge in which the theory is written.

Of course one can write any local theory in nonlocal terms, but this does not make it nonlocal, or change the fact that everything observable is local in the sense of extended causality.

That is why one should always use retarded potential and fields. Which is something neglected at full quantum level and also at semiclassical one (the H atom in the 1928 Dirac theory).

 

[Peter Morgan]

That is why one should always use retarded potential and fields. Which is something neglected at full quantum level and also at semiclassical one (the H atom in the 1928 Dirac theory).

Do you really mean that one should never use the Feynman propagator?

 

[vanhees71]

Anything published in a physics journal is - physics.

Hm, that's a pretty uncritical view about publications!

 

[vanhees71]

That is why one should always use retarded potential and fields. Which is something neglected at full quantum level and also at semiclassical one (the H atom in the 1928 Dirac theory).

You have to carefully keep in mind what you like to calculate to decide which of the infinitely many Green's functions you have to use. For evaluating S-matrix elements in vacuum QFT you need the time-ordered propagator (which is identical with the Feynman propagator in vacuum and temperature-0 physics) and the LSZ reduction formalism. In linear-response theory you get the retarded propagator (aka Green-Kubo formulae).

 

[vanhees71]

It is important where the word "spooky" is put. The expression is "spooky action at a distance", not the "action at spooky distance". The expression is used by Einstein to describe features related to the problem of quantum measurement, wave-function collapse, (in)completeness of QM and such. He did not use this expression to describe Newtonian mechanics.

Well, Newtonian physics uses by construction action-at-a-distance interactions, and there's nothing inconsistent with that within non-relativistic physics. The only problem is that it is disproven by experiment. Rather Faraday's and Maxwell's ideas prevailed and has lead to the discovery of relativity in the late 19th to early 20th century.

 

[weirdoguy]

Since you deny the physicality of "virtual particles"

You would probably be in the middle of some serious textbook on QFT if you put all this energy in learning instead of low-key arguing about existence of virtual particles.

 

[ftr]

The only problem is that it is disproven by experiment.

What experiment is that. If anything it has been repeated that "virtual particles", Feynman diagrams, propagators .. etc. are all mathematical artifacts.

 

[ftr]

You would probably be in the middle of some serious textbook on QFT if you put all this energy in learning instead of low-key arguing about existence of virtual particles.

I wasn't arguing about "VP". Please read the post with the proper context. Anyway, had we only needed to read textbooks we would not need PF which has diverse functions like any other forum.

 

[PeterDonis]

What experiment is that.

All of the experiments that confirm General Relativity's predictions where they differ from those of Newtonian physics. Go back and read the post you responded to in its proper context, just as you are asking others to do in your next post.

If anything it has been repeated that "virtual particles", Feynman diagrams, propagators .. etc. are all mathematical artifacts.

Yes, in the sense that we don't directly observe any of these things, they are features of particular theoretical models that make accurate predictions, but those features of the models don't correspond to anything directly observable. But that does not mean that the models overall are not accurate, or that the things that are directly observed are not valid observations.

 

[PeterDonis]

I wasn't arguing about "VP". Please read the post with the proper context.

I'm not sure that helps any. Here is what @weirdoguy quoted from your previous post, preceded by the quote to which it was a response:

But relativistic QFT has eliminated again the action at a distance.

Since you deny the physicality of "virtual particles" then we are left with just a mathematical relation hinting at nonlocality.

Which has nothing whatever to do with what @A. Neumaier said. The "action at a distance" he referred to has nothing to do with virtual particles at all, much less whether they have "physicality".

 

[ftr]

Go back and read the post you responded to in its proper context

I understood it as forces between charges since he mentioned "Faraday's and Maxwell's ideas", maybe Vanhees can clarify.

AFAIK "VP" has everything to do with electrons interacting(QED). Maybe there is some miscommunication.

 

[PeterDonis]

I understood it as forces between charges

Classical action at a distance forces between charges, yes. That's what Faraday and Maxwell's ideas were about (to an extent--they actually thought of the classical electromagnetic field as existing everywhere and the force on one charge being due to the field from another charge, not a direct action at a distance). Nothing whatsoever to do with virtual particles.

AFAIK "VP" has everything to do with electrons interacting(QED).

In the QED model, yes. But the QED model, as has already been pointed out to you, is not an "action at a distance" model. There are no "forces between charges" in that sense in QED. QED is a quantum field theory. "Virtual particles" are an artifact of one particular way of doing calculations in quantum field theory.

 

[samalkhaiat]

Since when is string theory physics? SCNR.

Scattering amplitudes in the LHC are calculated using the Parke-Taylor generating function which is a special case of the Witten-RSV formula in Twistor String Theory. If that is not physics, what is?

 

[Demystifier]

Scattering amplitudes in the LHC are calculated using the Parke-Taylor generating function which is a special case of the Witten-RSV formula in Twistor String Theory. If that is not physics, what is?

Generalization of physics is not necessarily physics.

 

[samalkhaiat]

Generalization of physics is not necessarily physics.

String Theory is not a “generalization of physics”, because we were forced to it by the Veneziano amplitude.

 

[Demystifier]

String Theory is not a “generalization of physics”, because we were forced to it by the Veneziano amplitude.

String theory is not Veneziano amplitude. String theory is a generalization of Veneziano amplitude. There is no theorem which says that string theory is the only theory which can give Veneziano amplitude. Indeed, since Veneziano amplitude is an (approximative !) description of certain phenomena in nuclear physics, it can be obtained from QCD. And QCD, of course, is not string theory.

Besides (correct me if I'm wrong), I think that perturbative superstring theory does not give Veneziano amplitude. The bosonic perturbative string theory does, but no string theorist thinks that bosonic string theory describes the actual physics.

 

[samalkhaiat]

String theory is not Veneziano amplitude. String theory is a generalization of Veneziano amplitude. There is no theorem which says that string theory is the only theory which can give Veneziano amplitude. Indeed, since Veneziano amplitude is an (approximative !) description of certain phenomena in nuclear physics, it can be obtained from QCD. And QCD, of course, is not string theory.

Besides (correct me if I'm wrong), I think that perturbative superstring theory does not give Veneziano amplitude. The bosonic perturbative string theory does, but no string theorist thinks that bosonic string theory describes the actual physics.

Don’t make such remarks if you don’t know the technical details. You should at least read something about the history of the Dual Resonance Model and how it led to String Theory.

 

[Demystifier]

Don’t make such remarks if you don’t know the technical details. You should at least read something about the history of the Dual Resonance Model and how it led to String Theory.

Patronization is not an argument.

 

[A. Neumaier]

Generalization of physics is not necessarily physics.

But:

Anything published in a physics journal is - physics.

Or do you object to your own arguments?

 

[ftr]

There are no "forces between charges" in that sense in QED

I don't understand what you mean by that. Can you elaborate (I don't mind a reference or a technical explanation), thanks.

 

[Demystifier]

But:

Or do you object to your own arguments?

I am not criticizing the idea that string theory is physics, nor I am criticing the idea that string theory is not physics. Both ideas can be defended by good arguments. But I am criticing the particular arguments (for both ideas) that have been offered here. It is better to have a good argument for a wrong idea than to have a bad argument for a right idea.

 

[PeterDonis]

I don't understand what you mean by that.

I meant the same thing that @A. Neumaier meant when he responded to you in post #78.