A How do entanglement experiments benefit from QFT (over QM)?

[DrChinese]

TL;DR Summary

A number of posters have asserted that Quantum Field Theory (QFT) provides a better description of quantum entanglement than the non-relativistic Quantum Mechanics. Yet I don't see QFT references in experimental papers on entanglement. Why not?

I should first acknowledge 2 important points. I don't read papers on QFT, and therefore barely know how to spell it. And second, although I read many papers on entanglement (theory and experiment) I don't know if I have ever seen much reference to anything I might label QFT (that being something DIFFERENT than garden Quantum Mechanics). But I certainly don't know what I don't know, so perhaps I have overlooked the obvious for a long time. About all I understand is that in QFT, almost everything is entangled.

My question is this: what is an entanglement experiment that has been performed, that depends on QFT for a correct analysis - but that could NOT be analyzed suitably using QM? As QFT is relativistic, and QM is not, the first thing that comes to mind is that you need an entanglement experiment in which reference frames are critical to the outcome. I guess that might be relevant for entanglement where momentum is a factor, or perhaps energy. But I don't see how that would be a factor where spin entanglement is at play, or GHZ, quantum teleportation or the like. But again, I don't know what I don't know.

Does anyone have a reference handy that might enlighten me? @Cthugha got me started with this post, but that didn't have anything that helps with this particular question. Another way to phrase my question: when would we need the more complex QFT to get our answer to an experimental entanglement question, as opposed to the (presumably) simpler QM? (I'd like to limit this discussion to things like electrons and photons, and ignore discussions going into the strong or weak forces.)

 

[DarMM]

To my mind QFT doesn't make such a big deal to entanglement, except that it is ubiquitous as you said, but more the structure of the state space where it is hard to maintain the proper/improper distinction for mixed states and there is an absence of local (and possibly global) pure states. Entanglement in terms of its most crucial aspects seems the same to me, though perhaps others know more.

 

[zonde]

As I understand QFT can handle creation and annihilation of particles which NRQM can't do. If particles are conserved within the experiment QFT can not add much. And surely you can do Bell experiments without creating and annihilating particles. Even with photons, if you do not go deep into microscopic details of creation of entangled pair of photons and annihilation of photon in detector. You just replace the state with a more complicated field that has the same statistical properties.

 

[Demystifier]

I don't read papers on QFT

If you read papers on quantum optics (which I think you do), then you read papers on (a branch of) QFT. In fact, for a work in quantum fundations (such as Bell inequality violations) quantum optics is almost all one needs to know about QFT.

 

[Demystifier]

To understand how entanglement should be explained, QFT does not help. It can help to understand how entanglement should not be explained. But it is not so much QFT itself that helps (QFT is just QM applied to a very large, usually infinite, number of degrees of freedom), but relativity. Relativity (together with some additional assumptions) implies that information cannot travel faster than light, implying that entanglement cannot be explained by exchange of information (provided that relativity and those additional assumptions are true). The only role of QFT here is that we best understand how to combine QM and relativity when we use QM in the form of QFT. But it can also be misleading, because even if both QM and relativity are true (as they are in relativistic QFT, but note that not all QFT's are relativistic), it is not so obvious whether the other assumptions (that together with relativity prohibit superluminal exchange of information) are true as well.

 

[Cthugha]

Hm, I think the question should indeed be somewhat different. I agree that pretty much every quantum optics paper uses QFT. Any detailed description of SPDC will do so. Already the first one by Hong and Mandel in 1985 did this. However, people rarely use a relativistic formulation. I guess this is the major point here. Experiments in accelerated reference frames have been done (https://www.nature.com/articles/ncomms15304), but are in my opinion of limited usefulness for discussing basic matters.

I have a vague feeling that in a nutshell the discussion here will repeat this one:
https://www.physicsforums.com/threads/how-does-qft-handle-non-locality.849972/

So maybe, we can all save some time by starting from there. ;)

 

[DrChinese]

I have a vague feeling that in a nutshell the discussion here will repeat this one:
https://www.physicsforums.com/threads/how-does-qft-handle-non-locality.849972/

So maybe, we can all save some time by starting from there. ;)

And that thread led to yet another thread...
https://www.physicsforums.com/threads/cluster-decomposition-and-epr-correlations.409861/

 

[vanhees71]

Summary: A number of posters have asserted that Quantum Field Theory (QFT) provides a better description of quantum entanglement than the non-relativistic Quantum Mechanics. Yet I don't see QFT references in experimental papers on entanglement. Why not?

You mixing again up things.

There are two issues. First there's entanglement. That's common to all kinds of QT, relativistic and non-relativistic.

Relativistic QFT is simply a more comprehensive theory compared to non-relativistic QM, in the same sense as relativistic classical mechanics and field theory is more comprehensive than Newtonian classical mechanics. The reason why relativistic QT is formulated as a relativistic QFT is that it admits the ubiquitous case of particle-number changing processes when reactions exchange energies comparable or exceeding the masses of particles that can be produced in this reactions (according to the known conservation laws).

In non-relativistic QT for systems of fixed particle numbers QFT is still convenient for many-body systems of indistinguishable particles, because it takes care of the necessary symmetrization and antisymmetrization operations on many-body states due to Bose-Einstein or Fermi-Dirac statistics, respectively. Otherwise in this case 1st-quantized and 2nd-quantized (i.e., QFT) formulation of non-relativistic QM are completely equivalent.

Now we often discuss photons in the context of experiments with entangled states, simply because it's so easy to prepare entangled two- (or even many-)photon Fock space, particularly Bell states, with which Bell tests can be performed. Photons cannot be described in some non-relativistic approximation, and that's why all quantum optics is in fact an application of quantum electrodynamics (QED), i.e., a relativistic QFT. Of course some issues like the interaction of the em. field with lab equipment, including photodetectors, can be treated in the approximation, where the corresponding condensed matter (dielectrics, metals, semiconductors, or whatever equipment you have in an experiment) is described by non-relativistic many-body theory, and that usually simplifies the task. E.g., photodetection is often based on the photoelectric effect and since it's way easier to describe bound-state problems (like electrons in a semiconductor or a metal in this case of the photoeffect) such approximations are used, and they are well justified, because here the non-relativistic approximation for local (!) interaction processes is valid.

Then usually debates about instantaneous interactions, violating Einstein causalities, arise, which is natural in the context of entangled states with local experiments done at space-like separated space-time regions.

Of course, in the context of non-relativistic physics, and non-relativistic QM is no exception, you cannot expect the Einstein causality to hold. It doesn't hold in classical non-relativistic physics either, but you have absolute time and absolute space as postulated by Newton in the very dawn of modern physics. Thus there's no tension between instantaneous interactions and the causality structure of Newtonian spacetime and that's why you don't need to worry about it within a non-relativistic theory.

Now we know that nature is relativistic, and that's why it were a contradiction if there were instataneous interactions and thus faster-than-light signal propagation possible. To discuss whether QT obeys the causality constraints of relativity, you have to investigate the relativistic QT, and that's formulated in terms of relativistic QFT, and as discussed for a zillion of times, relativistic QFT by construction cannot violate Einstein causality, and it doesn't violate Einstein causality. It is also consistent with the finding that the strong correlations of far distant parts of quantum systems as described by entanglement. If this were not the case QED would have been ruled out for about 30 years when the first Bell tests have been successfully performed with the finding that QT (and also QED) make the correct predictions with an astonishing precision and significance, while the prediction of the Bell inequalitiy valid for local deterministic hidden-variable theories fails at the same level of accuracy and significance.

That's why we discuss physical systems which are utmost relativistic (photons) and fundamental questions about Einstein causality (which is specifically relativistic too and cannot be tested within non-relativistic approximations).

 

[vanhees71]

To my mind QFT doesn't make such a big deal to entanglement, except that it is ubiquitous as you said, but more the structure of the state space where it is hard to maintain the proper/improper distinction for mixed states and there is an absence of local (and possibly global) pure states. Entanglement in terms of its most crucial aspects seems the same to me, though perhaps others know more.

It doesn't make such a big deal to entanglement, because its efficient formalism takes care of it for you without any quibble. Whenever you write down and equation like
$$|\Psi \rangle=\hat{a}_1^{\dagger} \hat{a}_2^{\dagger} |\Omega \rangle,$$
you have written down the appropriate entangled to-body state, with the creation operators taking care of symmetrization (bosons) or antisymmetrization (fermions) due to the fundamental commutator relations,
$$[\hat{a}_1^{\dagger},\hat{a}_2^{\dagger}]_{\mp}=0.$$
For indinstinguishable particles it's pretty hard to have NO entanglement. The only immediate example for a two-particle state are two bosons in the same state, i.e.,
$$|\Psi \rangle=(\hat{a}_1^{\dagger})^2 |\Omega \rangle.$$
That's indeed a product state.

 

[vanhees71]

As I understand QFT can handle creation and annihilation of particles which NRQM can't do. If particles are conserved within the experiment QFT can not add much. And surely you can do Bell experiments without creating and annihilating particles. Even with photons, if you do not go deep into microscopic details of creation of entangled pair of photons and annihilation of photon in detector. You just replace the state with a more complicated field that has the same statistical properties.

Still, I'd advise to use QFT to describe the photons as any indistinguishable particles. To handle the necessary (anti-)symmetrization operations on product states can get cumbersome soon. In old-fashioned books you can read about this formalism, usually in connection with fermions, using Slater determinants to describe antisymmetrized product states.

 

[vanhees71]

Hm, I think the question should indeed be somewhat different. I agree that pretty much every quantum optics paper uses QFT. Any detailed description of SPDC will do so. Already the first one by Hong and Mandel in 1985 did this. However, people rarely use a relativistic formulation. I guess this is the major point here. Experiments in accelerated reference frames have been done (https://www.nature.com/articles/ncomms15304), but are in my opinion of limited usefulness for discussing basic matters.

I have a vague feeling that in a nutshell the discussion here will repeat this one:
https://www.physicsforums.com/threads/how-does-qft-handle-non-locality.849972/

So maybe, we can all save some time by starting from there. ;)

The photon part in quantum-optics books is of course completely relativistic. As I said already above, the part where you consider interactions with atoms, molecules, condensed matter (including describing lenses, mirrors, beam splitters, photodetectors, and also non-linear-optics processes like SPDC) you can use the non-relativistic approximation, which simplifies the task considerable.

One should, however note that even in atomic physics you cannot completely do without relativistic Q(F)T. For large enough $Z$ you need to do relativistic calculations. Otherwise you don't get the chemistry right, and you'd predict a wrong periodic table of elements. Only recently the proper use of relativistic atomic theory together with precision experiments have clarified the puzzle about the actinides! But that's another story.

Also the idea to test Q(F)T in non-inertial frames are very interesting. There are some interesting features like Unruh radiation to be seen (or maybe not seen ;-)) within special relativity. Also non-inertial reference frames in Minkowski space are some (modest) step towards general relativity.

 

[A. Neumaier]

And that thread led to yet another thread...

This is clearly an observable instance of retrocausality. 

 

[atyy]

QFT must be used, because it is the only way to combine relativity and QM. However, all we need is free QFT, which is rigorous and relativistic, and does predict violations of the Bell inequalities. Free QFT is so simple, one can simply use one's intuition from non-relativistic QM.

For other QFTs such as the standard model of particle physics, these are still not rigorously relativistic, so there is no advantage in rigour to using them, compared to just using the intuition from non-relativistic QM.

The Bell theorem itself does not assume QFT. However, if QFT violates the Bell inequalities, then the Bell theorem does apply to QFT, and says that QFT cannot be described by a local variable theory.

 

[A. Neumaier]

if QFT violates the Bell inequalities, then the Bell theorem does apply to QFT, and says that QFT cannot be described by a local variable theory.

... by a local, noncontextual hidden variable theory.

 

[DrChinese]

1. You're mixing again up things.

There are two issues. First there's entanglement. That's common to all kinds of QT, relativistic and non-relativistic.

Relativistic QFT is simply a more comprehensive theory compared to non-relativistic QM, in the same sense as relativistic classical mechanics and field theory is more comprehensive than Newtonian classical mechanics. The reason why relativistic QT is formulated as a relativistic QFT is that it admits the ubiquitous case of particle-number changing processes when reactions exchange energies comparable or exceeding the masses of particles that can be produced in this reactions (according to the known conservation laws).

In non-relativistic QT for systems of fixed particle numbers QFT is still convenient for many-body systems of indistinguishable particles, because it takes care of the necessary symmetrization and antisymmetrization operations on many-body states due to Bose-Einstein or Fermi-Dirac statistics, respectively. Otherwise in this case 1st-quantized and 2nd-quantized (i.e., QFT) formulation of non-relativistic QM are completely equivalent.

Now we often discuss photons in the context of experiments with entangled states, simply because it's so easy to prepare entangled two- (or even many-)photon Fock space, particularly Bell states, with which Bell tests can be performed. Photons cannot be described in some non-relativistic approximation, and that's why all quantum optics is in fact an application of quantum electrodynamics (QED), i.e., a relativistic QFT. Of course some issues like the interaction of the em. field with lab equipment, including photodetectors, can be treated in the approximation, where the corresponding condensed matter (dielectrics, metals, semiconductors, or whatever equipment you have in an experiment) is described by non-relativistic many-body theory, and that usually simplifies the task. E.g., photodetection is often based on the photoelectric effect and since it's way easier to describe bound-state problems (like electrons in a semiconductor or a metal in this case of the photoeffect) such approximations are used, and they are well justified, because here the non-relativistic approximation for local (!) interaction processes is valid.

Then usually debates about instantaneous interactions, violating Einstein causalities, arise, which is natural in the context of entangled states with local experiments done at space-like separated space-time regions.

Of course, in the context of non-relativistic physics, and non-relativistic QM is no exception, you cannot expect the Einstein causality to hold. It doesn't hold in classical non-relativistic physics either, but you have absolute time and absolute space as postulated by Newton in the very dawn of modern physics. Thus there's no tension between instantaneous interactions and the causality structure of Newtonian spacetime and that's why you don't need to worry about it within a non-relativistic theory.

2. Now we know that nature is relativistic, and that's why it were a contradiction if there were instantaneous interactions and thus faster-than-light signal propagation possible. To discuss whether QT obeys the causality constraints of relativity, you have to investigate the relativistic QT, and that's formulated in terms of relativistic QFT, and as discussed for a zillion of times, relativistic QFT by construction cannot violate Einstein causality, and it doesn't violate Einstein causality. It is also consistent with the finding that the strong correlations of far distant parts of quantum systems as described by entanglement. If this were not the case QED would have been ruled out for about 30 years when the first Bell tests have been successfully performed with the finding that QT (and also QED) make the correct predictions with an astonishing precision and significance, while the prediction of the Bell inequality valid for local deterministic hidden-variable theories fails at the same level of accuracy and significance.

That's why we discuss physical systems which are utmost relativistic (photons) and fundamental questions about Einstein causality (which is specifically relativistic too and cannot be tested within non-relativistic approximations).

Thanks for this, very helpful at a number of levels. Some I knew, some I did not. A couple of comments related to your sentences in bold.

1. I am specifically trying to understand how and why you are so focused on QFT as it relates to entanglement, when I don't think it is that critical (if relevant at all). Sure, a better theory is a better theory, and certainly advances are desired. But let's face it: entanglement scenarios (Bell tests for example) do not depend on time ordering or distance, so I don't see why a relativistic theory would be called for unless some additional benefit were derived. That doesn't seem to be the case, ergo my question.

Coming from a different angle: I would assume that a relativistic constraint added to QM would have difficulty explaining how signal locality is achieved, all the while allowing entangled quantum systems to exhibit quantum nonlocality. That seems to be an obvious problem with a theory purporting to respect c from its construction. You have made the case that QFT is consistent and does not have that problem, but I still wonder. I would guess the nonlocality of entanglement is not resolved in QFT; because I have said many times, we wouldn't need interpretations if it were. That would be big news indeed. So yes, I'd like to know if and how QFT explains the mechanism of entanglement better than QM.

(So I don't think I am mixing anything up.)2. And I think this is a significant point of departure between you and I. You are saying there isn't anything occurring FTL in entanglement experiments, because if it did, it would violate relativity - and more specifically relativistic QFT. While I see most entanglement experiments as a demonstration of quantum nonlocality.

I essentially deny that any classically local theory can explain this behavior, while you deny that the quantum nonlocal behavior occurs in the first place. Let me know if I am not representing your position fairly.Next question: Can you explain how perfect correlations occur in entanglement? (For sake of simplicity, can we assume that T1 < T2 < T3 in all reference frames? Let me know if this is not possible.)

a. We have spin entangled A and B, now distant from each other, at T1.
b. I presume you agree that at T1, neither has a well-defined spin.
c. Alice measures A at angle $\theta$ at time T2, giving A a well-defined spin.
d. Bob measures B at angle $\theta$ at time T3, giving B a well-defined spin if it didn't already have one as a result of c. Further, T3 is sufficiently near to time T2 that there is insufficient time for any classical signal to go from A to B.
e. How do Alice and Bob always have anti-correlated results, regardless of choice of $\theta$? One would assume that A and B need some kind of FTL signal, action, mutual rapport or something to accomplish this impressive feat. We know from Bell that it is not due to hidden variables.

Thanks, and this question is not intended to be confrontational. I'd really like to get a better understanding of what QFT says about this, and especially how it differs from QM (as you have said it matters).

 

[Demystifier]

However, all we need is free QFT

That's a good point, especially as a reply to @vanhees71 who argues that correlations are a result of non-local interactions. At the level of standard QM/QFT, one does not need interactions at all.

 

[A. Neumaier]

At the level of standard QM/QFT, one does not need interactions at all.

Well, one needs it in order to have measurement results at all. This requires interaction between system and detector. Only between preparation and measurement, free QFT suffices.

 

[Demystifier]

Well, one needs it in order to have measurement results at all. This requires interaction between system and detector.

Sure, but @vanhees71 rejects the idea that interactions during the measurement can be the cause of correlation (because otherwise it would imply that interactions are nonlocal, which he rejects).

The correlations are encoded in the non-product form of the state, and such a form may exist without interactions.

 

[A. Neumaier]

Sure, but @vanhees71 rejects the idea that interactions during the measurement can be the cause of correlation (because otherwise it would imply that interactions are nonlocal, which he rejects).

The correlations are encoded in the non-product form of the state, and such a form may exist without interactions.

But the correlations are produced by the preparation, which also involves interacting QED.
Free is only the dynamics of the prepared state until it reaches a detector.

 

[atyy]

But the correlations are produced by the preparation, which also involves interacting QED.
Free is only the dynamics of the prepared state until it reaches a detector.

And the detector too, and the observer :) Which means we have to include the observer in the wave function :) Which means MWI :)

 

[Demystifier]

But the correlations are produced by the preparation, which also involves interacting QED.
Free is only the dynamics of the prepared state until it reaches a detector.

More precisely, the non-product state is produced by the interactions. On the other hand, I would say that the Kochen-Specker theorem proves that the correlations themselves cannot be produced by preparation only, i.e. that the later measurement plays a role too. But that's of course a subtle point we had a lot of discussions before.

 

[A. Neumaier]

the product state is produced by the interactions.

No. The interactions (e.g., parametric down conversion) directly produce entangled states.

 

[A. Neumaier]

And the detector too, and the observer :)

No. The detector is treated by applying QED to a tiny subsystem - an electron interacting with the incident electromagnetic field - together with quasiclassical reasoning about the cumulative effect of a huge number of essentially independent electrons. This can be seen by looking at any textbook on quantum optics, e.g., Mandel & Wolf. The observer of the detector is nowhere needed.

Which means we have to include the observer in the wave function :) Which means MWI :)

It wouldn't mean MWI but only the dynamics of the universe. The MWI provides a very weird interpretation of the latter. My thermal interpretation provides a much more rational interpretation of the dynamics of the universe.

 

[DarMM]

And the detector too, and the observer :) Which means we have to include the observer in the wave function :) Which means MWI :)

There's @A. Neumaier's thermal interpretation where this can be done and acausal interpretations as well. So it wouldn't necessarily mean MWI.

Even in Copenhagen it can be done, observers can be included but I assume you mean some observer is left out.

 

[atyy]

No. The detector is treated by applying QED to a tiny subsystem - an electron interacting with the incident electromagnetic field - together with quasiclassical reasoning about the cumulative effect of a huge number of essentially independent electrons. This can be seen by looking at any textbook on quantum optics, e.g., Mandel & Wolf. The observer of the detector is nowhere needed.

Agreed. I was just joking. But if we include electrons, then is the theory still relativistic? Don't we run into the problem that there are still no 3+1D interacting relativistic QFTs?

 

[Demystifier]

No. The interactions (e.g., parametric down conversion) directly produce entangled states.

I meant to say non-product states, it was a typo.

 

[A. Neumaier]

But if we include electrons, then is the theory still relativistic? Don't we run into the problem that there are still no 3+1D interacting relativistic QFTs?

No.

As a fully relativistic but only approximate QFT, renormalized perturbative QED is perfectly valid and highly accurate (to 12 digits of accuracy). The approximate 2-point functions can be made fully local using Kallen-Lehmann based resummation (which also eliminates the Landau pole). No more is needed for the use in quantum optics.

The main open problem about QED (and other interacting 4D relativistic QFTs) is whether all uncharged n-point functions can be constructed in a way that the Wightman axioms hold. This would give locality for arbitrary n-point functions.

 

[A. Neumaier]

I meant to say non-product states, it was a typo.

But then your critique is empty. Entangled quantum states produce the standard correlations by Born's rule.

That they may violate Bell inequalities just means that these quantum correlations have no classical equivalent. The Kochen-Specker theorem also asserts only that certain quantum correlations cannot be reproduced with classical observables.

 

[Demystifier]

But then your critique is empty. Entangled quantum states produce the standard correlations by Born's rule.

That they violate Bell inequalities just means that the correlations have no classical equivalent. The Kochen-Specker theorem also asserts only that quantum correlations cannot be reproduced with classical observables.

Here the crucial word is "classical", because the validity of those statements depends on what exactly one means by "classical". My opinion (on which there is no universal agreement) is that the actual requirement of "classicality" needed for those statements to be true is in fact a very mild requirement. I think it is explained well in http://de.arxiv.org/abs/1501.04168

 

[A. Neumaier]

Here the crucial word is "classical", because the validity of those statements depends on what exactly one means by "classical". My opinion (on which there is no universal agreement) is that the actual requirement of "classicality" needed for those statements to be true is in fact a very mild requirement. I think it is explained well in http://de.arxiv.org/abs/1501.04168

These assumptions are perhaps desirable, but they are not necessary for quantum mechanics or quantum field theory, as the empirical contradiction with Bell's results show.

 

[atyy]

No.

As a fully relativistic but only approximate QFT, renormalized perturbative QED is perfectly valid and highly accurate (to 12 digits of accuracy). The approximate 2-point functions can be made fully local using Kallen-Lehmann based resummation (which also eliminates the Landau pole). No more is needed for the use in quantum optics.

The main open problem about QED (and other interacting 4D relativistic QFTs) is whether all uncharged n-point functions can be constructed in a way that the Wightman axioms hold. This would give locality for arbitrary n-point functions.

Approximate to what? Could they be approximate to 2-point functions of a non-relativistic theory?

 

[vanhees71]

Thanks for this, very helpful at a number of levels. Some I knew, some I did not. A couple of comments related to your sentences in bold.

1. I am specifically trying to understand how and why you are so focused on QFT as it relates to entanglement, when I don't think it is that critical (if relevant at all). Sure, a better theory is a better theory, and certainly advances are desired. But let's face it: entanglement scenarios (Bell tests for example) do not depend on time ordering or distance, so I don't see why a relativistic theory would be called for unless some additional benefit were derived. That doesn't seem to be the case, ergo my question.

Coming from a different angle: I would assume that a relativistic constraint added to QM would have difficulty explaining how signal locality is achieved, all the while allowing entangled quantum systems to exhibit quantum nonlocality. That seems to be an obvious problem with a theory purporting to respect c from its construction. You have made the case that QFT is consistent and does not have that problem, but I still wonder. I would guess the nonlocality of entanglement is not resolved in QFT; because I have said many times, we wouldn't need interpretations if it were. That would be big news indeed. So yes, I'd like to know if and how QFT explains the mechanism of entanglement better than QM.

(So I don't think I am mixing anything up.)2. And I think this is a significant point of departure between you and I. You are saying there isn't anything occurring FTL in entanglement experiments, because if it did, it would violate relativity - and more specifically relativistic QFT. While I see most entanglement experiments as a demonstration of quantum nonlocality.

I essentially deny that any classically local theory can explain this behavior, while you deny that the quantum nonlocal behavior occurs in the first place. Let me know if I am not representing your position fairly.Next question: Can you explain how perfect correlations occur in entanglement? (For sake of simplicity, can we assume that T1 < T2 < T3 in all reference frames? Let me know if this is not possible.)

a. We have spin entangled A and B, now distant from each other, at T1.
b. I presume you agree that at T1, neither has a well-defined spin.
c. Alice measures A at angle $\theta$ at time T2, giving A a well-defined spin.
d. Bob measures B at angle $\theta$ at time T3, giving B a well-defined spin if it didn't already have one as a result of c. Further, T3 is sufficiently near to time T2 that there is insufficient time for any classical signal to go from A to B.
e. How do Alice and Bob always have anti-correlated results, regardless of choice of $\theta$? One would assume that A and B need some kind of FTL signal, action, mutual rapport or something to accomplish this impressive feat. We know from Bell that it is not due to hidden variables.

Thanks, and this question is not intended to be confrontational. I'd really like to get a better understanding of what QFT says about this, and especially how it differs from QM (as you have said it matters).

I insist on the use of relativistic QFT when relativistic questions are asked, namely how to make the observed "non-locality" of the correlations described by entanglement consistent with the causality structure implied by Minkowski space, i.e., that there must not be faster-than-light causal effects, or in other words, that space-like separated measurement events cannot be causally connected. This is in fact satisfied by stanadard relativistic QFT, implmenting the microcausality feature, which implies Poincare invariance of the S-matrix and the cluster decomposition principle, i.e., precisely what's needed to respect the causal structure of spacetime. You cannot get this with some "addition to non-relativistic QM".

Ad 2) Of course, no classical (deterministic) theory can describe what entanglement within relativistic QFT describes. That's why we do QFT and not classical physics to describe the phenomenon. Why should we use a model that doesn't work?

The entanglement is due to the state preparation process in the very beginning of the experiment. The usual way you get it is through conservation laws from local interactions (in fact in QFT there are only local interactions by construction). E.g., in parametric downconversion you absorb a UV photon from a strong laser field in a birefringent chrystal, which in turn leads to the creation of two photons, obeying energy-momentum and angular-momentum conservation. Thus you get photon pairs that are both momentum and polarization entangled in one of the Bell states like, e.g., the polarization-singlet state,
$$|\Psi \rangle=\frac{1}{2} (\hat{a}^{\dagger} (\vec{p}_1 H) \hat{a}^{\dagger}(\vec{p_2},V) - \hat{a}^{\dagger} (\vec{p}_1 V) \hat{a}^{\dagger}(\vec{p_2},H))|\Omega \rangle,$$
where $|\Omega \rangle$ is the vacuum state of photons. Of course, this is idealized, because momentum eigenstates are no true states, und to have to smear somewhat to get properly normalized wave packets.

Concerning your example it's very simple. You have a spin-entangled state. Both A and B find completely unpolarized particles when measuring their spin. Due to the entanglement, when they compare their measurement protocol (taking carefully appropriate time stamps to know which of each spins are from one and the same entangled pair, they'll find 100% correlation, when measuring in the same direction $\theta$. This is due to the preparation at the very beginning, before any further manipulations where done. It doesn't matter in which temporal order A and B make their measurements. If the measurement events are space-like separated it's for sure that A's measurement cannot have in any causally influenced B's spin nor can B's measurment have in any way causally influenced A's measurement. That's ensured theoretically by the validity of the microcausality property and the only conclusion is that the correlation found is simply due to the preparation in the spin-entangled state in the very beginning.

 

[A. Neumaier]

Approximate to what? Could they be approximate to 2-point functions of a non-relativistic theory?

Approximate to the nonperturbative 2-point function of a local covariant QFT matching the perturbative QED expansion. Of course, nobody yet knows how to define the latter.

The renormalized perturbative gives only asymptotic series to the n-point functions, and partially summing these series (as is customary) produces not-quite local approximations to what are supposed to be the unknown true local n-point functions. My point was that on the 2-point function level it is known how to make the approximations truly local, whereas it is unknown how to do it in general.

 

[A. Neumaier]

This is in fact satisfied by standard relativistic QFT, implementing the microcausality feature, which implies Poincare invariance of the S-matrix and the cluster decomposition principle

Do you know where there is a proof for that for nonabelian gauge theories? Quarks surely do not satisfy the cluster decomposition principle.

 

[vanhees71]

Agreed. I was just joking. But if we include electrons, then is the theory still relativistic? Don't we run into the problem that there are still no 3+1D interacting relativistic QFTs?

QED is relativistic when everything is treated relativistically, including the electrons. Of course for the local interaction of the em. field with an atom within a solid (i.e., your detector) you don't need the full relativistic treatment. I guess, it's pretty difficult to describe the photoeffect within a fully relativistic model since you need the electron in a bound state being scattered into a continuum state by the interaction with the em. field, and bound states are hard to describe relativistically.

On the other hand, it can be done of course perturbatively too, e.g., for the hydrogen atom neglecting the motion of the proton. Then it boils down to use the Coulomb gauge and use an interaction picture, where $\hat{H}_0$ includes the Coulomb part of the interaction between the nucleus (treated as a fixed center) and the electron. Then you get nice approximate bound and scattering states for this Coulomb problem. The rest can then be done perturbatively in the usual way, leading to utmost precise predictions like the Lamb shift. See Weinberg QT of Fields Vol. I.

Of course you can also use perturbation theory to treat the photoeffect, i.e., the absorption of a photon by some bound state of the hydrogen atom leading to the emission of the photoelectron which can be used to register the photon.

 

[DrChinese]

Concerning your example it's very simple. You have a spin-entangled state. Both A and B find completely unpolarized particles when measuring their spin. Due to the entanglement, when they compare their measurement protocol (taking carefully appropriate time stamps to know which of each spins are from one and the same entangled pair, they'll find 100% correlation, when measuring in the same direction $\theta$. This is due to the preparation at the very beginning, before any further manipulations where done. It doesn't matter in which temporal order A and B make their measurements. If the measurement events are space-like separated it's for sure that A's measurement cannot have in any causally influenced B's spin nor can B's measurement have in any way causally influenced A's measurement. That's ensured theoretically by the validity of the microcausality property and the only conclusion is that the correlation found is simply due to the preparation in the spin-entangled state in the very beginning.

Here are points of departure:

1. If they evolved independently after state preparation, obviously they could be described by Product State statistics. This is the very definition of a Local Realistic explanation, and therefore flat out prohibited by Bell. Certainly you know all this, so why would you use this explanation?

2. Your explanation does not involve QFT. This same explanation was used in 1935 in EPR. The point of my question, phrased as it was, was to get an explanation of QFT's solution to the issue that works for a simple case (although obviously NOT using local hidden variables, which are excluded), and then you could walk me through how that is extended to a more complex case.

3. You assume: "If the measurement events are space-like separated it's for sure that A's measurement cannot have in any causally influenced B's spin nor can B's measurement have in any way causally influenced A's measurement." This argument is purely tautological, as this is the entire point in question.

In fact. that is almost verbatim what Bell started with and went on to disprove: "The vital assumption is that the result B for particle 2 does not depend on the setting a, of the magnet for particle 1, nor A on b." No theory, after Bell, and including QFT, can be local in the manner you describe (no FTL influences), with classic forward in time causality, without being contextual (i.e. observer/measurement dependent) in some non-classical manner. I am hoping you can explain how I am wrong about this point. I am definitely hoping to learn more from you, and if you have any specific quotes, that would be great too. Ah, and I just realized @bhobba shares your position (I'm sure others do too, but I never read them elsewhere).

 

[atyy]

QED is relativistic when everything is treated relativistically, including the electrons. Of course for the local interaction of the em. field with an atom within a solid (i.e., your detector) you don't need the full relativistic treatment. I guess, it's pretty difficult to describe the photoeffect within a fully relativistic model since you need the electron in a bound state being scattered into a continuum state by the interaction with the em. field, and bound states are hard to describe relativistically.

On the other hand, it can be done of course perturbatively too, e.g., for the hydrogen atom neglecting the motion of the proton. Then it boils down to use the Coulomb gauge and use an interaction picture, where $\hat{H}_0$ includes the Coulomb part of the interaction between the nucleus (treated as a fixed center) and the electron. Then you get nice approximate bound and scattering states for this Coulomb problem. The rest can then be done perturbatively in the usual way, leading to utmost precise predictions like the Lamb shift. See Weinberg QT of Fields Vol. I.

Of course you can also use perturbation theory to treat the photoeffect, i.e., the absorption of a photon by some bound state of the hydrogen atom leading to the emission of the photoelectron which can be used to register the photon.

There is have no theory at the moment, which is why most people take the effective field theory point of view of QED - including Weinberg.

 

[DarMM]

There is have no theory at the moment, which is why most people take the effective field theory point of view of QED - including Weinberg.

I would say it's the Landau pole that motivates that more so. There's also no theory for Yang Mills, but the view that it needs an effective field theory treatment isn't as common. QED is usually thought to be trivial due to this pole. Although in my opinion it's a weak argument in light of explicitly constructed models that have a perturbative Landau pole.

 

[atyy]

I would say it's the Landau pole that motivates that more so. There's also no theory for Yang Mills, but the view that it needs an effective field theory treatment isn't as common. QED is usually thought to be trivial due to this pole. Although in my opinion it's a weak argument in light of explicitly constructed models that have a perturbative Landau pole.

Yes, but I don't think considering interactions gives anything beyond the free theory. The free theory is relativistic and rigorous. The interacting theory considerations can either lead to a rigorous relativistic theory, or it could be consistent with a non-relativistic theory from the which the relativistic low energy theory is emergent. In the former case we would reach the same conclusions as for the free theory, in the latter case we would say relativity is not important for foundational considerations.

 

[Demystifier]

Yes, but I don't think considering interactions gives anything beyond the free theory. The free theory is relativistic and rigorous. The interacting theory considerations can either lead to a rigorous relativistic theory, or it could be consistent with a non-relativistic theory from the which the relativistic low energy theory is emergent. In the former case we would reach the same conclusions as for the free theory, in the latter case we would say relativity is not important for foundational considerations.

"I saw from this that to understand quantum field theories I would have to understand quantum field theories on a lattice."
- K. Wilson (Reviews of Modern Physics 55, 583 (1983))

 

[Lord Jestocost]

Both A and B find completely unpolarized particles when measuring their spin. Due to the entanglement, when they compare their measurement protocol (taking carefully appropriate time stamps to know which of each spins are from one and the same entangled pair, they'll find 100% correlation, when measuring in the same direction θ\theta. This is due to the preparation at the very beginning, before any further manipulations where done. [bold by LJ]

Maybe, you overlook something when pointing to the preparation.

J. Bell in “BERTLMANN’S SOCKS AND THE NATURE OF REALITY”:

“Let us summarize once again the logic that leads to the impasse. The EPRB correlations are such that the result of the experiment on one side immediately foretells that on the other, whenever the analyzers happen to be parallel. If we do not accept the intervention on one side as a causal influence on the other, we seem obliged to admit that the results on both sides are determined in advance anyway, independently of the intervention on the other side, by signals from the source and by the local magnet setting. But this has implications for non-parallel settings which conflict with those of quantum mechanics. So we cannot dismiss intervention on one side as a causal influence on the other.”

This is commented on https://www.mathpages.com/home/kmath731/kmath731.htm

“Here he is explaining why, if we rule out communication, the perfect anti-correlation at equal angles obliges us to admit that the results are determined in advance, by common cause, i.e., by extra variables. This is not an assumption, it is the only remaining causal option, deduced from the assumption of separability combined with the perfect anti-correlation at equal angles.”

 

[vanhees71]

Here are points of departure:

1. If they evolved independently after state preparation, obviously they could be described by Product State statistics. This is the very definition of a Local Realistic explanation, and therefore flat out prohibited by Bell. Certainly you know all this, so why would you use this explanation?

2. Your explanation does not involve QFT. This same explanation was used in 1935 in EPR. The point of my question, phrased as it was, was to get an explanation of QFT's solution to the issue that works for a simple case (although obviously NOT using local hidden variables, which are excluded), and then you could walk me through how that is extended to a more complex case.

3. You assume: "If the measurement events are space-like separated it's for sure that A's measurement cannot have in any causally influenced B's spin nor can B's measurement have in any way causally influenced A's measurement." This argument is purely tautological, as this is the entire point in question.

In fact. that is almost verbatim what Bell started with and went on to disprove: "The vital assumption is that the result B for particle 2 does not depend on the setting a, of the magnet for particle 1, nor A on b." No theory, after Bell, and including QFT, can be local in the manner you describe (no FTL influences), with classic forward in time causality, without being contextual (i.e. observer/measurement dependent) in some non-classical manner. I am hoping you can explain how I am wrong about this point. I am definitely hoping to learn more from you, and if you have any specific quotes, that would be great too. Ah, and I just realized @bhobba shares your position (I'm sure others do too, but I never read them elsewhere).

(1) The state and operators evolve according to the Hamiltonian of the system. If you have free photons the entangled state stays an entangled state and does not change into a product state. How you come to this idea from what I wrote, I don't know.

(2) My explanation does involve QFT. I wrote creation and annihilation operators for photons. The original EPR paper was not about relativistic QT. It was only among the first papers explicitly hinting at what was shortly thereafter called entanglement. The EPR paper is, by the way, much overrated. Einstein himself didn't like it much, because his main point has not been made clear. For him the main quibble was "inseparability" and not "non-locality".

(3) It's not tautological. It's of course the very assumption made to formulate the appropriate QFTs, i.e., those leading to Poincare invariant (co-variant) S-matrices fulfilling the cluster-decomposition principle.

Again: QFT is local in the interactions, but still enabling the inseparability of far-distant parts of quantum systems through entanglement. That's what's built in in the very beginning. Bell defines local deterministic hidden-variable models showing that these must obey the Bell inequalities, which are violated by all QTs (relativistic as well as non-relativistic). Relativistic QFT is local in the interactions but not deterministic, i.e., it violates the Bell inequalities in a way which is consistent with relativistic causality principles. All very accurate Bell tests confirm the predictions of Q(F)T, not the predictions of local deterministic hidden-variable theories. It's the great merit of Bell's idea to make in this way a before purely philosophical question scientific in the sense that the hypotheses (local deterministic HV theories versus relativistic microcausal QFT) can be objectively tested. The local deterministic HV theories are disproven by the corresponding experiments, while the predicts of relativistic microcausal QFT are confirmed. That's why the standard interpretation is that local deterministic HV theories are ruled out, and QFT is the correct description.

 

[vanhees71]

Maybe, you overlook something when pointing to the preparation.

J. Bell in “BERTLMANN’S SOCKS AND THE NATURE OF REALITY”:

“Let us summarize once again the logic that leads to the impasse. The EPRB correlations are such that the result of the experiment on one side immediately foretells that on the other, whenever the analyzers happen to be parallel. If we do not accept the intervention on one side as a causal influence on the other, we seem obliged to admit that the results on both sides are determined in advance anyway, independently of the intervention on the other side, by signals from the source and by the local magnet setting. But this has implications for non-parallel settings which conflict with those of quantum mechanics. So we cannot dismiss intervention on one side as a causal influence on the other.”

This is commented on https://www.mathpages.com/home/kmath731/kmath731.htm

“Here he is explaining why, if we rule out communication, the perfect anti-correlation at equal angles obliges us to admit that the results are determined in advance, by common cause, i.e., by extra variables. This is not an assumption, it is the only remaining causal option, deduced from the assumption of separability combined with the perfect anti-correlation at equal angles.”

The entangled state predicts the probabilities for the outcome of measurements. Each of the local observers can of course freely choose their local experimental setup, and the entangled state describes all correlations the preparation in this state implies. All these predictions are confirmed by very accurate experiments today, including the violation of Bell's inequality but confirming the prediction of this violation by QT precisely.

For me this implies clearly that all there is is the quantum state, and the prepation in the entangled state is the only cause for the correlations it describes. The single outcomes are maximally indetermined, but there's no other needs to describe the correlations accurately than the QT formalism itself. The conclusion that there must be hidden variables making a deterministic local world view consistent with these results is disproven by these experiments. There's only QT, and one must accept the irreducible randomness of nature or find a new non-local determinsitic theory which is conistent with both these findings and with Einstein causality, i.e., it must be at least as successful and consistent as relativistic QFT is. Obviously up to today nobody has found such a theory.

As long as this is the case we have to live with QFT, and indeed QFT is not contradicting any empirical findings. So there's no need for a new theory from this point of view. A real physical problem is the lack of understanding of gravity, and that may be even related to these questions of entanglement, correlations, locality and all that.

 

[DarMM]

Yes, but I don't think considering interactions gives anything beyond the free theory. The free theory is relativistic and rigorous. The interacting theory considerations can either lead to a rigorous relativistic theory, or it could be consistent with a non-relativistic theory from the which the relativistic low energy theory is emergent. In the former case we would reach the same conclusions as for the free theory, in the latter case we would say relativity is not important for foundational considerations.

I agree in the case of classifying mechanisms for entanglement. For other foundational issues I would say it does matter since it changes the structure of the state space quite a bit. Although I realize your remarks were only about entanglement mechanisms.

 

[bhobba]

I am definitely hoping to learn more from you, and if you have any specific quotes, that would be great too. Ah, and I just realized @bhobba shares your position (I'm sure others do too, but I never read them elsewhere).

I was unsure whether to reply here because both Vanhees and Dr Chinese are two of my favorite posters. I have posted my view on EPR many times. It is a minority view. It disputes nothing in Bells work who I consider a physicist on a par with the greats like Fermi, Feynman etc. It's just a different way of looking at it.

First, as can be seen in Charter 3 of Ballentine ordinary QM (ie Schrodinger's Equation etc) are derived from the assumption of the Galilean Transformation which automatically implies non-locality is possible. Of course that changes Bell in no way but does put non-locality in a different light - if there is non-locality its not really against QM like some seem to think. To further investigate the issue of locality in QM you really need QFT. But in QFT locality is replaced by the cluster decomposition property:
https://www.physicsforums.com/threads/cluster-decomposition-in-qft.547574/
In that principle to avoid possible problems IMHO its best to avoid correlated systems from discussions about locality in the first place. So my view is while Bell's Theorem is both interesting and true its not something people should worry about. You simply say including correlations in locality discussions makes QFT harder than it needs to be, and IMHO its already hard enough, then move on,

Just to be sure my position is understood - in discussions on locality you really need QFT. But including correlations in that, while allowable, complicates things. Bell would have known this, but chose to investigate including it. Others however did not seem to cotton onto one can take the position its not something you really need to worry about in issues of locality - and I personally do not.

Thanks
Bill

 

[DrChinese]

(1) The state and operators evolve according to the Hamiltonian of the system. If you have free photons the entangled state stays an entangled state and does not change into a product state. How you come to this idea from what I wrote, I don't know.

QFT is local in the interactions, but still enabling the inseparability of far-distant parts of quantum systems through entanglement.

I guess it's best to do this an idea at a time.

You said earlier on the one hand: that entangled pairs evolve such that neither it affected by a measurement of the other. To me, that says they evolve independently. Any pair of anything that evolve causally independently of each other would have Product State statistics, and could not have Entangled State statistics.

On the other hand: you just stated that the distant subsystems are inseparable (which of course I agree with). There is no useful meaning to your description of that single entangled system as "inseparable" unless a measurement on one component of the system affects another component of that system (regardless of actual mechanism). Of course, we call that effect "quantum nonlocality" and it cannot be limited to a forward looking light cone (as you imply).

So how are distant parts of an entangled system considered inseparable, while measurements on those parts are made locally without some kind of distant impact? Bell has something to say about that, and you seem to skip past that at every turn.

 

[DrChinese]

I was unsure whether to reply here because both Vanhees and DrChinese are two of my favorite posters.

Before I dig into the meat of your post, I'd like to address your kind comment.

I am not asking anyone to take sides with me against vanhees71, and I certainly respect almost everything vanhees71 adds to this forum. While his and my interactions of late seem acrimonious on some levels, I can assure you that I am truly interested in hearing the other side of opinions, viewpoints, perspectives, etc. I am perfectly happy to adjust my own perspective as I gain new knowledge. Please don't feel you need to sugarcoat anything on my behalf.

 

[bhobba]

So how are distant parts of an entangled system considered inseparable, while measurements on those parts are made locally without some kind of distant impact? Bell has something to say about that, and you seem to skip past that at every turn.

Its in the fact the observable of one part of an entangled system is a more complicated thing than a single non-entangled system.

Thanks
Bill

 

[A. Neumaier]

I guess it's best to do this an idea at a time.

You said earlier on the one hand: that entangled pairs evolve such that neither it affected by a measurement of the other. To me, that says they evolve independently. Any pair of anything that evolve causally independently of each other would have Product State statistics, and could not have Entangled State statistics.

On the other hand: you just stated that the distant subsystems are inseparable (which of course I agree with). There is no useful meaning to your description of that single entangled system as "inseparable" unless a measurement on one component of the system affects another component of that system (regardless of actual mechanism). Of course, we call that effect "quantum nonlocality" and it cannot be limited to a forward looking light cone (as you imply).

So how are distant parts of an entangled system considered inseparable, while measurements on those parts are made locally without some kind of distant impact? Bell has something to say about that, and you seem to skip past that at every turn.

Its inaccurate to talk about a pair of photons traveling. What travels is a single quantun system in an entangled 2-photon state. In QFT it is impossible to separate this system into two photons. This is called inseparability.

 

[bhobba]

Please don't feel you need to sugarcoat anything on my behalf.

I wasn't being 'kind' just telling the truth.

Many people here, including yourself and Vanhees know more physics than I do. I am sorry, when discussing physics with these people I am 'cautious' about what I say, which may come across as sugar-coating - its not something I can easily stop doing.

Thanks
Bill