A Why is Quantum Field Theory Local?

[LarryS]

TL;DR Summary

Why is Quantum Field Theory local?

Under the Schrodinger Picture, nonrelativistic Quantum Mechanics for a fixed number of particles is highly nonlocal, e.g. Quantum Entanglement.

But Quantum Field Theory is local. Why is that? Is it because QFT was created to accommodate SR, which, as a classical theory, is local?

As always, thanks in advance.

 

[PeterDonis]

Quantum Field Theory is local.

Not in the sense you are using the term "local", since you are saying that entanglement means "nonlocal", and QFT includes entanglement, since it includes non-relativistic QM as a special case and makes all of the same predictions for that case.

QFT is "local" in the sense that spacelike separated measurements, including those on entangled particles, must commute--their results must not depend on the order in which they are made (since the ordering of spacelike separated measurements is not invariant). But this is a different meaning of "local" from the one you are using (your meaning of "local" is basically "correlations satisfy the Bell inequalities", which is violated by measurements on entangled quantum particles).

 

[Demystifier]

Another sense in which QFT is local is that its Hamiltonian (or Lagrangian) is local.

 

[martinbn]

Another sense in which QFT is local is that its Hamiltonian (or Lagrangian) is local.

What's a local Hamiltonian?

 

[Demystifier]

What's a local Hamiltonian?

Roughly, a Hamiltonian that can be written in the form
$$H=\int d^3x\,{\cal H}(x)$$
where ${\cal H}(x)$ is Hamiltonian density. It's not perfectly precise, but I'll not nitpick unless you insist.

 

[martinbn]

Roughly, a Hamiltonian that can be written in the form
$$H=\int d^3x\,{\cal H}(x)$$
where ${\cal H}(x)$ is Hamiltonian density. It's not perfectly precise, but I'll not nitpick unless you insist.

Is it easy to write an example of a local and of a nonloval Hamiltonian?

 

[Demystifier]

Is it easy to write an example of a local and of a nonloval Hamiltonian?

Yes.

Local:
$$H=\int d^3x\, \frac{1}{2}\pi(x)\pi(x)$$
where $\pi(x)$ is canonical momentum density associated with the field $\phi(x)$. (For simplicity I gave an example which does not have a $\phi$-dependent term and does not describe a Lorentz covariant theory. Contrary to a widespread prejudice, locality and Lorentz covariance are independent requirements.)

Nonlocal:
$$H=\int d^3x\int d^3x' \, \frac{1}{2}\pi(x)G(x,x')\pi(x')$$
where $G(x,x')$ is not proportional to $\delta^3(x-x')$. For example,
$$G(x,x')=\frac{const}{|x-x'|}$$

 

[martinbn]

Yes.

Local:
$$H=\int d^3x\, \frac{1}{2}\pi(x)\pi(x)$$
where $\pi(x)$ is canonical momentum density associated with the field $\phi(x)$. (For simplicity I gave an example which does not have a $\phi$-dependent term and does not describe a Lorentz covariant theory. Contrary to a widespread prejudice, locality and Lorentz covariance are independent requirements.)

Nonlocal:
$$H=\int d^3x\int d^3x' \, \frac{1}{2}\pi(x)G(x,x')\pi(x')$$
where $G(x,x')$ is not proportional to $\delta^3(x-x')$. For example,
$$G(x,x')=\frac{const}{|x-x'|}$$

Well, they are both in the form $$H=\int d^3x\,\cal{H}(x)$$

I was looking for an example, where you can explicitly write down the operator.

 

[vanhees71]

Summary:: Why is Quantum Field Theory local?

Under the Schrodinger Picture, nonrelativistic Quantum Mechanics for a fixed number of particles is highly nonlocal, e.g. Quantum Entanglement.

But Quantum Field Theory is local. Why is that? Is it because QFT was created to accommodate SR, which, as a classical theory, is local?

As always, thanks in advance.

As already stated by other answers in this thread the confusion reflected in this question is due to the different meaning different communities of scientists and philosophers put into the words "local" and "non-local".

I'm a high-energy /particlenuclear physicist, and for me relativistic QFT is "local", because it's constructed to be local. The meaning in this case is that all local observables, represented by field operators $\hat{O}(t,\vec{x})$ (in the Heisenberg picture, which is the most natural picture of time evolution to discuss this issue) commute with the energy-density operator $\hat{\mathcal{H}}(t,\vec{x})$ at space-like separated arguments, i.e.,
$$[\hat{\op{O}}(x),\hat{\op{\mathcal{H}}(y)]=0 \quad \text{if} \quad (x-y)^2<0,$$
where I'm using the west-coast convention with $\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)$, such that four vectors are space-like separated if their Minkowski product with themselves is negative. This implies physically that no measurement done at a place can have an effect on any event (particularly also any other measurement) which is space-like separated from this measurement. This for sure realizes causality in the relativistic sense by construction, i.e., there's no way to communicate faster than light in any way. That's why it's also called microcausality condition.

It is usually realized by constructing field-operators that transform in the same local way as do the corresponding classical fields under Poincare transformations, fulfilling canonical equal-time commutation (bosons) or anti-commutation relations and then writing down a Lagrange density which is Poincare invariant and depends only on the fields and their first derivative at one spacetime point.

In addition it also ensures the unitarity of the S-matrix and the linked-cluster theorem (see Weinberg, QT of fields, vol. I for the details).

Of course also an in this sense local relativistic QFT necessarily implies the existence of entangled states, but this has nothing to do with "non-locality", but it rather describes "inseparability" (as Einstein put it), i.e., it implies correlations between far distant parts of a single (!) quantum system when measured at far-distant places.

[Edit: Erased interpretational statements which don't belong to this scientific part of the QM forum]

The confusion comes into the discussion, because some physicists and many philosophers name the long-range correlations due to entanglement (i.e., "inseparability") "non-locality", but that has of course a different meaning than the "locality of interactions/transformation properties of field operators under Poincare transformations" in relativistic local QFTs.

 

[Demystifier]

Well, they are both in the form $$H=\int d^3x\,\cal{H}(x)$$

That's why I said that it was not perfectly precise. I should have said something like "... where ${\cal H}(x)$ only depends on fields and its derivatives at $x$". But in the examples it should be clear what I meant.

I was looking for an example, where you can explicitly write down the operator.

What do you mean by that, can you give an example of an "explicitly written down operator"?

 

[DrChinese]

1. This in turn implies that the correlations cannot be due to the mutual influence of the measurements at far distant places when the measurement events are at space-like separated places in spacetime.

2. The confusion comes into the discussion, because some physicists and many philosophers name the long-range correlations due to entanglement (i.e., "inseparability") as "non-locality", but that has of course another meaning than the "locality of interactions" in relativistic local QFTs.

1. Again, this sentence represents an opinion is not shared by most physicists doing work on entanglement. Of course, this is not a popularity contest. The OP has a right to know, and I suspect he has followed discussions on this point previously from his question.

2. Virtually the entirety of the physics community calls the correlations evidence of "quantum non-locality". No one understands the mechanism enough to say with certainty anything is happening that is FTL; certainly I don't.

I won't quote the community or otherwise comment further because we have already been down this ridiculous rabbit hole too many times already. Bell tests demonstrate quantum non-locality, in fact that's the generally accepted definition... period.

Cheers,

-DrC

 

[PeterDonis]

Virtually the entirety of the physics community calls the correlations evidence of "quantum non-locality".

The statement that "QFT is local" is also a very common statement, and, as has been posted previously in this thread, in order to reconcile this statement with the statement that Bell inequality violations mean "nonlocality", one has to recognize that the term "local" is being used in two different senses.

 

[martinbn]

What do you mean by that, can you give an example of an "explicitly written down operator"?

Your examples have integrals of $\pi(x)$, for which you haven't given an explicit formula. As for examples take the Laplace operator, it is written down explicitly as $\sum\frac{\partial^2}{\partial x_i^2}$.

 

[vanhees71]

1. Again, this sentence represents an opinion is not shared by most physicists doing work on entanglement. Of course, this is not a popularity contest. The OP has a right to know, and I suspect he has followed discussions on this point previously from his question.

2. Virtually the entirety of the physics community calls the correlations evidence of "quantum non-locality". No one understands the mechanism enough to say with certainty anything is happening that is FTL; certainly I don't.

I won't quote the community or otherwise comment further because we have already been down this ridiculous rabbit hole too many times already. Bell tests demonstrate quantum non-locality, in fact that's the generally accepted definition... period.

Cheers,

-DrC

I said that the terms "local" or "non-local" are used with different meanings in different communities, and I told you what HEP physicist mean when they say "local relativistic QFT". This has to be distinguished from what the interpretation-of-quantum-theory community (a mixture of physicists and, unfortunately, also philosophers) understand under "local" and "non-local". It's not an opinion of either group but just using the same words for different meanings.

I prefer to call correlations correlations and use locality/non-locality only in the clear mathematical meaning of the HEP community. That's all.

All I can say is that one has to carefully figure out, what's meant by any specific writer when he says "local" or "non-local".

Whether you draw the logical conclusion about the causality when measuring far-distant entangled observables or not is of course a matter of opinion and interpretation. So I shouldn't have written this part. I'll erase the corresponding statements. It belongs to the interpretational section of these forums, I'm no longer interested to participate in.

 

[vanhees71]

Just one short statement: I still believe, of course, that the mathematical properties of local relativistic QFTs (local in the HEP sense) inevitably excludes any fast-than-like propagation of signals.

 

[Lord Jestocost]

Just one short statement: I still believe, of course, that the mathematical properties of local relativistic QFTs (local in the HEP sense) inevitably excludes any fast-than-like propagation of signals.

An "agnostic" would be more tentative. As Peter Mittelstaedt remarks in “Quantum Holism, Superluminality, and Einstein Causality”: “Finally, we analyze these arguments and show that the micro-causality condition of relativistic quantum field theory excludes entanglement induced superluminal signals but that this condition is justified by the exclusion of superluminal signals. Hence, we are confronted here with a vicious circle, and the question whether there are superluminal EPR-signals cannot be answered in this way.”

 

[vanhees71]

This is interpretation, i.e., not subject to this forum.

Physicswise the microcausality condition together with the locality of the unitary representations of the proper orthochronous Poincare group indeed guarantees (a) causality, (b) the unitarity and Poincare invariance of the S-matrix, and (c) the validity of the linked cluster principle. Thus the inevitable consequence indeed is that there are no faster-than light signals and no violation of causality within such a "local" relativistic QFT.

One should be aware, as Weinberg stresses in his book, that this is a sufficient but not necessary condition. So it might well be that there are causal "non-local" relativistic QFTs. I'm not aware, whether somebody has constructed such a thing and which empirical consequences such a modification would imply in comparison to the local relativistic QFT making up the Standard Model.

I don't know, where, from a physics point of view should there be a "vicious circle" in this (purely mathematical) arguments.

 

[PeterDonis]

This is interpretation, i.e., not subject to this forum.

What are you saying is interpretation? It looks to me like both the Mittelstaedt remarks and your response to them are, as you say, purely mathematical arguments involving the math of QFT, and don't involve any choice of QM interpretation.

 

[vanhees71]

I meant the statement that it's a vicious circle. From a mathematical point of view there's no circle. It's the statement of a sufficient condition fulfilling the physically plausible properties (a)-(c). There's no circular argument. You can of course argue that also this is interpretation .

 

[atyy]

An "agnostic" would be more tentative. As Peter Mittelstaedt remarks in “Quantum Holism, Superluminality, and Einstein Causality”: “Finally, we analyze these arguments and show that the micro-causality condition of relativistic quantum field theory excludes entanglement induced superluminal signals but that this condition is justified by the exclusion of superluminal signals. Hence, we are confronted here with a vicious circle, and the question whether there are superluminal EPR-signals cannot be answered in this way.”

The Mittelstaedt statement is erroneous. The microcausality condition of relativistic QFT includes entanglement induced superluminal signals. There is no vicious circle. Nonetheless, the entanglement induced superluminal signals do not permit superluminal signalling, where "signals" and "signalling" are used in two different senses.

 

[vanhees71]

For me it's very clear that the microcausality condition excludes superluminal signals. You have to distinguish between causal effects (interactions) and correlations described by the (entangled) quantum states. Your last sentence admits this, and it's just somewhat confusing to say there's superluminal signalling but no superluminal signal.

If you want to introduce some new notion, you have to clearly define it. So what do you understand under "superluminal signalling"? It can for sure not mean causal effects by interactions because of the microcausality feature. That's a mathematical fact and indeed no question of opinion or "interpretation".

 

[martinbn]

The Mittelstaedt statement is erroneous. The microcausality condition of relativistic QFT includes entanglement induced superluminal signals. There is no vicious circle. Nonetheless, the entanglement induced superluminal signals do not permit superluminal signalling, where "signals" and "signalling" are used in two different senses.

So what does "signal" mean?

 

[EPR]

So what does "signal" mean?

That the entangled partner particle collapses to a 100% correlated state. Signalling doesn't take place, is what he is saying. 'Signals' is also a misnomer as no signals are going from one entangled partner particle to the other.
Assuming classicality and realism before measurement leads to the well-known EPR problem.

 

[vanhees71]

But collapsing cannot "signal" anything (despite the fact that the collapse assumption is highly problematic and fortunately unnecessary for the application of quantum theory as far as any physics is concerned).

 

[Lord Jestocost]

I meant the statement that it's a vicious circle. From a mathematical point of view there's no circle. It's the statement of a sufficient condition fulfilling the physically plausible properties (a)-(c). There's no circular argument. You can of course argue that also this is interpretation

The point is: You can believe in your physical theory, that's up to you. However, as Kent A. Peacock puts it in his thesis “Peaceful Coexistence or Armed Truce? Quantum Nonlocality and the Spacetime View of the World” (1991):

“….. that the proof of a result on the basis of a theory which was ‘constructed to ensure’ that result is no proof at all. As I have noted, all the proof shows is that the construction was successful: it accomplishes what it was intended to accomplish.”

 

[vanhees71]

It's no proof of course. All that counts is whether it stands stringent empirical tests.

 

[Demystifier]

Your examples have integrals of $\pi(x)$, for which you haven't given an explicit formula. As for examples take the Laplace operator, it is written down explicitly as $\sum\frac{\partial^2}{\partial x_i^2}$.

In that sense, we can represent it by a functional derivative
$$\pi(x)=-i\hbar\frac{\delta}{\delta\phi(x)}$$

 

[martinbn]

In that sense, we can represent it by a functional derivative
$$\pi(x)=-i\hbar\frac{\delta}{\delta\phi(x)}$$

And what is $\phi$? And what is the integral equal to? I just want to see e very specific example.

 

[Demystifier]

And what is $\phi$? And what is the integral equal to? I just want to see e very specific example.

If you want a crash course in functional calculus for physicists, please start a separate thread!

 

[martinbn]

If you want a crash course in functional calculus for physicists, please start a separate thread!

No, I just want an operator.

 

[Demystifier]

No, I just want an operator.

Since you are never satisfied with my answers, I would suggest you to look at the literature. See e.g. https://www.amazon.com/dp/0201360799/?tag=pfamazon01-20

 

[martinbn]

Since you are never satisfied with my answers, I would suggest you to look at the literature. See e.g. https://www.amazon.com/dp/0201360799/?tag=pfamazon01-20

OK, where is the definition of non-local operator?

 

[Demystifier]

OK, where is the definition of non-local operator?

See e.g.
https://en.wikipedia.org/wiki/Nonlocal_operator
https://en.wikipedia.org/wiki/Nonlocal_Lagrangian

 

[martinbn]

See e.g.
https://en.wikipedia.org/wiki/Nonlocal_operator
https://en.wikipedia.org/wiki/Nonlocal_Lagrangian

Thanks!

Just point out that this is yet another sense in which local/nonlocal is used. For example all differential operators are local, in this way, so Newtonian gravity can be considered local, because the Laplace operator that appears in the Poisson equation is local.

 

[Demystifier]

Thanks!

Just point out that this is yet another sense in which local/nonlocal is used. For example all differential operators are local, in this way, so Newtonian gravity can be considered local, because the Laplace operator that appears in the Poisson equation is local.

Never thought that way, but yes, in that sense Newtonian gravity is local.

 

[vanhees71]

Thanks!

Just point out that this is yet another sense in which local/nonlocal is used. For example all differential operators are local, in this way, so Newtonian gravity can be considered local, because the Laplace operator that appears in the Poisson equation is local.

That's the usual "no-nonsense" definition of a non-local operator.

Newtonian gravity is of course local when interpreted as a (non-relativistic!) field theory but also instantaneous in its action between far-distant objects. That's no problem, because in Newtonian physics there's no "speed limit" for causal effects. It's nevertheless amazing that Newton already felt pretty uneasy with such an "action at a distance"! Nevertheless it fitted all the known facts about gravity, i.e., about the motion of celestial bodies that he didn't ponder this issue too much further.

 

[LarryS]

Thank you all for your replies. I originally expected a simple, one-sentence answer and instead got a very rich discussion.

 

[Lynch101]

The statement that "QFT is local" is also a very common statement, and, as has been posted previously in this thread, in order to reconcile this statement with the statement that Bell inequality violations mean "nonlocality", one has to recognize that the term "local" is being used in two different senses.

Sabine Hossenfelder addresses this point in her video:


Please feel free to delete this if it's deemed unhelpful - this is partly to check that my own understanding is correct (or at least more correct than it was), but I also hope it might be helpful to the OP (assuming I have understood correctly).

The term "non-local" is used in two different, but related, ways. The use of the two terms are related in that they both make reference to the correlations observed in quantum experiments, correlations that violate Bell's inequality.

Bell's inequality essentially represents the experimental predictions of a (any??) local hidden variables theory, following the assumptions of Einstein, Podolsky, and Rosen (EPR). If the Universe were local, in the sense that EPR assumed, then Bell's inequality would not be violated by quantum experiments.

The fact that the observed correlations, of entanglement experiments, do violate Bell's inequality tells us that the Universe is not EPR local.

The other way in which the term is used is as a proposed explanation for those observed correlations. While the [here] first use of the term refers to correlations that violate those predicted according to EPR locality, the [here] second use of the term refers to some [undefined/unexplained] FTL causal mechanism, where an action performed on one particle has an instantaneous effect on a spatially separated entangled particle.Is that in the right ball park? Hopefully the video will, at least, be helpful.

 

[DrChinese]

While the [here] first use of the term refers to correlations that violate those predicted according to EPR locality, the [here] second use of the term refers to some [undefined/unexplained] FTL causal mechanism, where an action performed on one particle has an instantaneous effect on a spatially separated entangled particle.

I'd say that's pretty fair. We know there is "quantum nonlocality" per Bell; but we do not know if there are any FTL causal mechanisms. Nature could be otherwise "local", and in fact appears to be so.

 

[vanhees71]

To be brief I'd state it simply as:

"Relativistic QFTs are local" means that the interactions are local, i.e., the Hamilton density commutes with any local observable when the spacetime arguments of the corresponding operators are space-like separated. So a more concise formulation is:

Locality in QFT means that there are field operators realizing a unitary representation of the proper orthochronous Poincare transformations such that these field operators transform locally as their classical analogues and that the Hamilton density commutes with all local operators representing observables at space-like separated space-time arguments.

The locality of the unitary transformation representing Poincare trafos means that, e.g., for a vector field
$$\hat{U}(\Lambda) \hat{A}^{\mu}(x) \hat{U}^{\dagger}(\Lambda)={\Lambda^{\mu}}_{\nu} \hat{A}^{\mu}(\Lambda^{-1}x), \quad \Lambda \in \text{SO(1,3)}^{\uparrow}.$$
These properties lead to (a) a unitary Poincare covariant S-matrix and (b) the corresponding transition-probality rates obey the linked cluster principle.

The second meaning of (non-)locality does not refer to causal interactions but to correlations, i.e., as any quantum theory also a "local relativistic QFT" admits the description of "non-local correlations", described by entanglement. That means that if you prepare a quantum system in an entangled state like a momentum-polarization entangled photon pair, prepared in the state
$$|\Psi \rangle=\frac{1}{2} [\hat{a}^{\dagger}(\vec{k}_1,h=1) \hat{a}^{\dagger}(\vec{k}_2,h=-1)-\hat{a}^{\dagger}(\vec{k}_1,h=-1) \hat{a}^{\dagger}(\vec{k}_2,h=1)]|\Omega \rangle,$$
you can register the two photons at very far-distant places A and B and you have a 100% correlation for the polarization states, i.e., if the observer at A finds his photon having $h=1$, then the observer at B finds his photon having $h=-1$ and vice versa, although both photons are completely unpolarized before the measurement. It doesn't matter who measures his photon first, the 100% correlation of the polarizations is observed although the polarizations before the measurement are completely indetermined.

This together with the fact that a local relativistic QFT cannot describe any faster-than-light signal propagation (due to the microcausality built in this kind of relativistic QFTs) one must conclude that the correlation is not caused by the local measurements on each photon at far distant places but it is due to the preparation in the entangled state.

I'd prefer to call the "non-locality of correlations" rather "inseparability", as Einstein formulated it. Then a lot of misunderstanding were avoided by using different words for the different two meanings of locality vs. non-locality.

 

[atyy]

To be brief I'd state it simply as:

"Relativistic QFTs are local" means that the interactions are local, i.e., the Hamilton density commutes with any local observable when the spacetime arguments of the corresponding operators are space-like separated. So a more concise formulation is:

Locality in QFT means that there are field operators realizing a unitary representation of the proper orthochronous Poincare transformations such that these field operators transform locally as their classical analogues and that the Hamilton density commutes with all local operators representing observables at space-like separated space-time arguments.

The locality of the unitary transformation representing Poincare trafos means that, e.g., for a vector field
$$\hat{U}(\Lambda) \hat{A}^{\mu}(x) \hat{U}^{\dagger}(\Lambda)={\Lambda^{\mu}}_{\nu} \hat{A}^{\mu}(\Lambda^{-1}x), \quad \Lambda \in \text{SO(1,3)}^{\uparrow}.$$
These properties lead to (a) a unitary Poincare covariant S-matrix and (b) the corresponding transition-probality rates obey the linked cluster principle.

The second meaning of (non-)locality does not refer to causal interactions but to correlations, i.e., as any quantum theory also a "local relativistic QFT" admits the description of "non-local correlations", described by entanglement. That means that if you prepare a quantum system in an entangled state like a momentum-polarization entangled photon pair, prepared in the state
$$|\Psi \rangle=\frac{1}{2} [\hat{a}^{\dagger}(\vec{k}_1,h=1) \hat{a}^{\dagger}(\vec{k}_2,h=-1)-\hat{a}^{\dagger}(\vec{k}_1,h=-1) \hat{a}^{\dagger}(\vec{k}_2,h=1)]|\Omega \rangle,$$
you can register the two photons at very far-distant places A and B and you have a 100% correlation for the polarization states, i.e., if the observer at A finds his photon having $h=1$, then the observer at B finds his photon having $h=-1$ and vice versa, although both photons are completely unpolarized before the measurement. It doesn't matter who measures his photon first, the 100% correlation of the polarizations is observed although the polarizations before the measurement are completely indetermined.

This together with the fact that a local relativistic QFT cannot describe any faster-than-light signal propagation (due to the microcausality built in this kind of relativistic QFTs) one must conclude that the correlation is not caused by the local measurements on each photon at far distant places but it is due to the preparation in the entangled state.

I'd prefer to call the "non-locality of correlations" rather "inseparability", as Einstein formulated it. Then a lot of misunderstanding were avoided by using different words for the different two meanings of locality vs. non-locality.

Just as there are different definition of "local", there are different definitions of "cause". In one definition relativistic causality alone does not imply local causality (see Fig. 5 of https://arxiv.org/abs/1503.06413).

 

[Demystifier]

Just as there are different definition of "local", there are different definitions of "cause". In one definition relativistic causality alone does not imply local causality (see Fig. 5 of https://arxiv.org/abs/1503.06413).

The paper is very deep, but not easy to read.

 

[vanhees71]

Fortunately I've some holidays and time to read ;-).

 

[mattt]

The paper is very deep, but not easy to read.

I like it so far...(I'm halfway through)

 

[bhobba]

So what does "signal" mean?

The ability to send information, in particular information that can be used to sync clocks. These days the modern view of SR is as a geometry implied by the symmetries of the POR with a constant that needs to be determined by experiment - it turns out to be the speed of light. It was not always presented in such an elegant and transparent way, but instead how Einstein did it initially using thought experiments about syncing clocks. His arguments break down if you can sync clocks with a signal faster than the speed of light. Of course the modern method breaks down as well if it can be done. In fact things get really bad because the constant c can be determined by means having nothing to do with speeds and sending signals. Yet the equations imply you can't do it ie have speeds faster than that c. The whole edifice of SR would not only be wrong but a logical mess. A good exercise in understanding SR is working it out and why it is of no concern if that speed can't be used to send information. If you are like me and would like to see some detail about it before embarking on the journey, you can see the paper I often reference:
http://physics.umd.edu/~yakovenk/teaching/Lorentz.pdf

Then read the account given in Rindler:
https://www.amazon.com/dp/0198539525/?tag=pfamazon01-20

Happy contemplating . Just kidding - it will require your attention but is certainly not what I would call a mind boggling issue.

Dr Chinese is correct in saying 'Virtually the entirety of the physics community calls the correlations evidence of "quantum non-locality". No one understands the mechanism enough to say with certainty anything is happening that is FTL; certainly I don't.' I am the 'odd man out', in that I believe there is no non-locality involved . We have had many long threads about it - but as became clear it really is a semantic issue, and semantics is one of the silliest things there is to argue about.

I used to argue about it a lot here and on other forums, but not so much now because I understand its semantic nature.

Thanks
Bill

 

[bhobba]

I like it so far...(I'm halfway through)

I have read it before - it just reinforces my current view - a lot of it is semantics. But it is good to know how it is viewed using one lot of semantics, and contrasting it to another.

Thanks
Bill

 

[Tendex]

I'm not sure if the previous discussion was under the assumption that QFT was a deterministic(hidden variables) theory or not. If the premise is that it's not deterministic then Bell's quantum non-locality doesn't follow necessarily since it includes determinism as premise to conclude "EPR locality" is impossible.

With respect to the semantics of "local", I think that the sense in which QFT is usually said to be local is part of the "EPR locality" in Bell's theorem. So I guess the above commenters don't consider QFT as a deterministic theory.

 

[vanhees71]

Standard local relativistic QFT is a QT and not a deterministic HV theory. Locality in relativistic QFT means that the Hamilton density is built by field operators and their derivatives at one spacetime point and that all local observables commute with it a space-like distances of the arguments (microcausality condition).

 

[Tendex]

Standard local relativistic QFT is a QT and not a deterministic HV theory.

Fine, I guessed right in your case then.

Of course, justifying that QFT is not deterministic as a mathematical theory is a tall order(doing it on the grounds that it includes probabilities as some people do doesn't seem right). But maybe for another thread.

Locality in relativistic QFT means that the Hamilton density is built by field operators and their derivatives at one spacetime point and that all local observables commute with it a space-like distances of the arguments (microcausality condition).

Yes, this is consistent with EPR locality's assumption of separability.

 

[vanhees71]

I don't understand the 1st part. As any QT also local relativistic QFT is not deterministic, because the state provides probabilities for the outcome of measurements not determined values of all observables of the quantum system. Why do you think that the probabilistic interpretation of the state is not right? There's not the slightest evidence for such an idea to hold true.