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

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DrChinese

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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

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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

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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

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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.
 
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Demystifier

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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

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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

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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... :smile:
 

vanhees71

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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

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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

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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

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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.
 

atyy

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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

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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

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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

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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

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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

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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

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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

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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

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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.
 
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A. Neumaier

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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

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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

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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?
 

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