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

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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 $\left( M_{\mu\nu},P_{\mu}\right)| \Omega \rangle = 0$.

• bhobba and Demystifier

vanhees71

Gold Member
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", J. Math. Mech. 14 (1965) 629. 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.

• physics loverq and Peter Morgan

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QFT, which I'd call the only physically sensible relativistic QT we know of
String theorists would disagree. vanhees71

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Since when is string theory physics? SCNR.

Demystifier

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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

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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?

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• vanhees71

Peter Morgan

Gold Member
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.

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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.

• vanhees71 and Demystifier

Demystifier

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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.

• dextercioby

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).

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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?

• Peter Morgan

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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

Peter Morgan

Gold Member
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

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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

Gold Member
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.

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Demystifier

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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".

• Mentz114

Peter Morgan

Gold Member
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.

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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.

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

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