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

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A. Neumaier

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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.
 
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ftr

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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

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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.
 
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dextercioby

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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

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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

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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

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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.
 
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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

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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

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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.
 
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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.
 
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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

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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.
 
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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

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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

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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.
 

Demystifier

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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.
 
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samalkhaiat

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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.
 

ftr

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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

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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.
 

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