How does QFT handle non-locality?

[atyy]

Well, we seem to have different language. Microcausality+locality of the interactions indeed excludes superluminal signalling. Together with the dynamics of QT that implies relativistic causality, or what else do you need to establish it?

In order to have relativistic causality, the Bell inequalities cannot be violated. So "no superluminal signalling" is a weaker constraint than "relativistic causality".

One way to see that although technically, the conditions on the quantum Hamiltonian seem to have the same conditions we impose on a classical relativistic theory, it is not the same because in QFT the Hamiltonian is not real. In the Heisenberg picture, the Hamiltonian governs the time evolution of all observables, including non-commuting observables. But non-commuting observables cannot have simultaneous reality. So in the Heisenberg picture, the Hamiltonian is not real. In the Schroedinger, picture the Hamiltonian governs the evolution of the wave function, which is also not real (or at least not necessarily real).

In general, in the minimal interpretation, QFT and QM are not theories of reality. This is why relativistic causality is empty in the minimal interpretation. If QFT and QM are taken to be theories of reality, then the Bell theorem forces (except for the usual exceptions like MWI) QFT and QM to violate relativistic causality.

Whatever language one uses, there is the idea that the constraints in order of strength from weak to strong are:
-no superluminal signalling
-quantum causality or correlations
-relativistic causality

 

[vanhees71]

This sentence is very problematic in minimal ensemble interpretation (MEI). Namely, this sentence sounds as if the "beam" is a physical object existing even without our observations. On the other hand, using only MEI, I think you cannot answer whether the beam physically exists without our observations. Thus, the language you use does not seem compatible with MEI. So either
i) you really use something more than MEI (even if you fail to recognize it), or
ii) within MEI you have to answer whether the beam exists without our observations, or
iii) stay agnostic about this question and adopt your language accordingly, to prevent false impression of believing in beams existing without our observations.

So what is your choice, i), ii), or iii)?

Of course I have in mind the most simple example for a von Neumann filter measurement like a Stern-Gerlach experiment. Say, I have silver atoms as in the original experiment and want to prepare a pure state with spin up. I send the silver atoms (originally in a thermal state from the oven) through an appropriate magnetic field. This leads to a state, where the position of the atoms is entangled with its spin-z component. In other words the many particles of the ensemble of silver atoms are sorted into two well-separated locations, and at each location they have an almost perfectly prepared spin-z up or down. Now I put something in the beam at the location where the particles have spin down. What's then left are particles with spin up, i.e., I have prepared a pure state with determined spin-z component out of a thermal ensemble (or any other initial state you have in the beginning). I don't see, where I need more than just the postulates of quantum theory to come to this setup of a von Neumann filter preparation.

So I think (ii) is the right answer. The quantum dynamics, including several conservation laws, ensure that the silver atoms are where they should be in the spin-position entangled state after they've run through the Stern-Gerlach apparatus, and I can use the remaining beam to check, whether it is really a pure spin-up state by measuring the spin-z component again using a second Stern-Gerlach apparatus. I know that the particles are in a pure spin-up state after the preparation procedure due to the very natural laws, I've established by observation before. Any experiment in physics rests on the physical laws we use to construct the apparatus to perform it, and the very purpose of experiments is to check whether this works really out. If you find a discrepancy between your expectations and the outcome of the experiment, you have to figure out whether it's due to inaccuracies in your experiment or whether you found a violation to the physical laws known so far. Then you have a discovery ;-)).

 

[vanhees71]

In order to have relativistic causality, the Bell inequalities cannot be violated. So "no superluminal signalling" is a weaker constraint than "relativistic causality".

I think I give up. Obviously I cannot make this very simple argument clear. Just once more very brief: Relativistic local microcausal QFT by construction obeys relativistic causality, including no superluminal signalling, and it precisely describes all violations of the Bell inequality so far. This is no contradiction to Bell's theoryem, because QFT is not a local deterministic HV model of the world but a special realization of QT tailored to be consistent with relativistic causality. So relativistic causality does not exclude the violation of Bell's inequality but only local deterministic theories do so, and the observed violations in my opinion rule out any local deterministic theory. It seems that even the most sceptical physicists nowadays believe that loophole-free Bell tests have been performed (at least according to several recent publications titled as loophole-free Bell tests, but it's a Nature paper ;-))).

 

[atyy]

I think I give up. Obviously I cannot make this very simple argument clear. Just once more very brief: Relativistic local microcausal QFT by construction obeys relativistic causality, including no superluminal signalling, and it precisely describes all violations of the Bell inequality so far. This is no contradiction to Bell's theoryem, because QFT is not a local deterministic HV model of the world but a special realization of QT tailored to be consistent with relativistic causality. So relativistic causality does not exclude the violation of Bell's inequality but only local deterministic theories do so, and the observed violations in my opinion rule out any local deterministic theory. It seems that even the most sceptical physicists nowadays believe that loophole-free Bell tests have been performed (at least according to several recent publications titled as loophole-free Bell tests, but it's a Nature paper ;-))).

Sure, but then you don't mean anything by relativistic causality. You just mean whatever QFT is, in a way which is not microcausality. At best your definition is tautological. What is worse, is that it is very misleading.

 

[vanhees71]

Why is it misleading? Indeed, it's a tautology (although it's not trivial to prove that it is one). Again, what do you need in addition to the impossibility of faster-than-light propagation and a causal dynamical law that describes the time evolution of observables to call a theory "relativistically causal"?

Also "reality" is a pretty empty idea. After all the discussions here in the forum and also reading some papers, I could not make out what's the clear definition of what makes a theory "realistic". For me QFT in the minimal interpretation is very "realistic", because it describes all known phenomena concerning the behavior of elementary particles.

 

[atyy]

Why is it misleading? Indeed, it's a tautology (although it's not trivial to prove that it is one). Again, what do you need in addition to the impossibility of faster-than-light propagation and a causal dynamical law that describes the time evolution of observables to call a theory "relativistically causal"?

Also "reality" is a pretty empty idea. After all the discussions here in the forum and also reading some papers, I could not make out what's the clear definition of what makes a theory "realistic". For me QFT in the minimal interpretation is very "realistic", because it describes all known phenomena concerning the behavior of elementary particles.

Well, there are two concepts - no superluminal signalling and no local hidden variables. You can pick whichever one you wish to be "relativistic causality". But the fact remains that the minimal interpretation has collapse, and that violates "no local hidden variables". Also, collapse does not violate "no superluminal signalling". So whichever interpretation you choose for "relativistic causality", you cannot reject collapse on the your grounds. If you reject collapse because it violates "no local hidden variables", that is wrong since there is no way to save "no local hidden variables". If you reject collapse because it violates "no superluminal signalling", that is wrong because collapse is consistent with "no superluminal signalling".

 

[vanhees71]

No you confusing me even more. Standard QFT has no superluminal signalling and no local hidden variables (and also no collapse). Collapse explicitly violates "no superluminal signalling", because proponents claim that A's measurement of the polarization of the photon at her place instantaneously changes the polarization of B's photon at a far distant place. Nothing observed and also QFT doesn't necessarily justify this claim. The only thing that changes by A's measurement is her knowledge about B's measurement's outcome because of the polarization entanglement of the two photons observed. There is no action at a distance according to standard QFT because the interaction of one of the photons with A's measurement apparatus is local and cannot affect anything spacelike separated from the local detection event. So how can collapse be consistent with "no superluminal signalling"?

 

[atyy]

No you confusing me even more. Standard QFT has no superluminal signalling and no local hidden variables (and also no collapse). Collapse explicitly violates "no superluminal signalling", because proponents claim that A's measurement of the polarization of the photon at her place instantaneously changes the polarization of B's photon at a far distant place. Nothing observed and also QFT doesn't necessarily justify this claim. The only thing that changes by A's measurement is her knowledge about B's measurement's outcome because of the polarization entanglement of the two photons observed. There is no action at a distance according to standard QFT because the interaction of one of the photons with A's measurement apparatus is local and cannot affect anything spacelike separated from the local detection event. So how can collapse be consistent with "no superluminal signalling"?

Standard QFT has collapse. It just means that after Alice measures |uu>+|dd> to get the u result, the state collapses to |uu>. Of course you don't have to ascribe reality to the collapse. But even if you do, it doesn't violate no superluminal signalling, because Bob cannot tell by measuring his spin whether Alice has measured yet.

 

[Demystifier]

Collapse explicitly violates "no superluminal signalling",

Argh! Your ability of fast forgeting is amazing. Just a few posts above I explained you that it is not so, and you said that this post was helpful to you and you liked it, but now you wrongly repeat again that collapse violates no superluminal signalling.

 

[Demystifier]

So I think (ii) is the right answer.

I guess it means that you think that the beam does exist even if we don't measure it. Am I right? But then
1) You are not adherent of MEI (even if you think you are), and
2) The beam itself is a hidden variable (because it exists without measurement), so the Bell theorem implies that entangled beams involve a kind non-locality that you can't accept. But of course, you can't understand it as long as you keep repeating wrong claims such as those that "collapse contradicts signal locality".

 

[Demystifier]

Standard QFT has ... no collapse.

Yes it has. For example, a few posts above atyy quoted the precise equation in Weinberg's QFT I describing the collapse. What standard QFT does not have is an answer to the question whether the collapse is a real physical process or only a mental tool for information update. Standard QFT is agnostic about that. But it is precisely this agnosticism (namely refusing to make clear statements about certain interesting questions) that creates a lot of confusion about foundational issues among physicists who read only standard QFT/QM.

So if you want to really understand non-locality of QFT, collapse, hidden variables, etc ... forget standard books such as Weinberg. Instead, take a look at a text that more seriously deals with such questions. For example, F. Laloe, "Do We Really Understand Quantum Mechanics?" would be a good choice. After a while, you should realize that QFT is more non-local than you currently think.

 

[vanhees71]

Argh! Your ability of fast forgeting is amazing. Just a few posts above I explained you that it is not so, and you said that this post was helpful to you and you liked it, but now you wrongly repeat again that collapse violates no superluminal signalling.

No, I didn't forget that posting, but there you didn't mention the collapse but defined signal locality as being fulfilled by QT (which I agree with), but if you put the collapse hypothesis (which for me is clearly an addition to minimally interpreted QT) you explicitly assume signal nonlocality, because it implies that a quantum state instantaneously collapses after a measurement (even if this measurement involves only local interactions of (parts of) the system with the measurement appartus). It is a (for me fictitious) process outside of the quantumtheoretical dynamics.

 

[vanhees71]

Standard QFT has collapse. It just means that after Alice measures |uu>+|dd> to get the u result, the state collapses to |uu>. Of course you don't have to ascribe reality to the collapse. But even if you do, it doesn't violate no superluminal signalling, because Bob cannot tell by measuring his spin whether Alice has measured yet.

It is an assumption that after alices measurement the state collapses to $|uu \rangle$. How do you know that from quantum theory, and how can that be even independent of how Alice has measured her photon?

 

[zonde]

Relativistic local microcausal QFT by construction obeys relativistic causality, including no superluminal signalling, and it precisely describes all violations of the Bell inequality so far.

What about local microcausality of Fock states (or rather superposition of Fock states)? Fock spaces are by construction nonlocal when they incorporate Hilbertspaces of distant particles. I have asked something similar before but somehow I have not received any answer. Is there some problem with my question?

 

[atyy]

It is an assumption that after alices measurement the state collapses to $|uu \rangle$. How do you know that from quantum theory, and how can that be even independent of how Alice has measured her photon?

I know that from quantum theory, because quantum theory has exactly the same structure as quantum mechanics, which has collapse (eg. Landau & Lifshitz, Cohen-Tannoudji, Diu & Laloe, Sakurai, Nielsen & Chuang). The collapse is dependent on the result Alice gets when she measures her photon.

 

[mfb]

Sorry atyy, but repeating that claim over and over again won't make it true.
Most textbooks use the Copenhagen interpretation. So what?
Most textbooks about classical mechanics discuss its application on inclined planes. Does this mean inclined planes are a crucial part of the theory of classical mechanics? Does classical mechanics break down if you don't talk about inclined planes?

The analogy is not perfect as you can use the equations of classical mechanics to describe inclined planes, but you cannot use the equations of quantum mechanics to describe collapses.

 

[atyy]

Sorry atyy, but repeating that claim over and over again won't make it true.
Most textbooks use the Copenhagen interpretation. So what?
Most textbooks about classical mechanics discuss its application on inclined planes. Does this mean inclined planes are a crucial part of the theory of classical mechanics? Does classical mechanics break down if you don't talk about inclined planes?

The analogy is not perfect as you can use the equations of classical mechanics to describe inclined planes, but you cannot use the equations of quantum mechanics to describe collapses.

The claim is true. Are you assuming MWI in order to avoid collapse?

 

[mfb]

I'm not assuming anything, I am just aware that there are multiple interpretations with different answers to questions of locality, determinism and so on. And there is no particular reason to prefer one over the other. Copenhagen is so widespread mainly for historic reasons.

 

[atyy]

I'm not assuming anything, I am just aware that there are multiple interpretations with different answers to questions of locality, determinism and so on. And there is no particular reason to prefer one over the other. Copenhagen is so widespread mainly for historic reasons.

But are there? As far as I know, Copenhagen (in one flavour or another) is the only consensus interpretation. All other interpretations have some problem - a technical problem, so it is not a matter of taste.

 

[vanhees71]

I'm not assuming anything, I am just aware that there are multiple interpretations with different answers to questions of locality, determinism and so on. And there is no particular reason to prefer one over the other. Copenhagen is so widespread mainly for historic reasons.

There is not one "Copenhagen interpretation". I think the minimal interpretation (which doesn't use a collapse or unobservable parallel universes but just uses the quantum formalism and the probabilistic interpretation of the states a la Born) is also a flavor of the Copenhagen interpretation, but that doesn't matter too much. I don't know any example of an experiment, for which you need to invoke a collapse assumption, and since the collapse assumption is at least very problematic in the context of the EPR problem, I simply don't use it.

I think the minimal interpretation (which I consider to be a flavor of the Copenhagen interpretation) is the only one which is really consensus among physicists. All other interpretations (including some flavors of the Copenhagen interpretations) add some additional assumption, which in my opinion is just unnecessary to use
quantum theory as a physical description of the (so far known part) world.

Under collapse, I unserstand the assumption that there is a split of phenomena in a "quantum" and a "classical" part, and neither describes the dynamics of the described systems completely and the "collapse" is a process which again is inconsistent with either dynamics, because both classical and quantum dynamics in the relativistic realm by construction does not involve instantaneous interactions or signal propagation, while the collapse assumption exactly claims this: the measurement of A's photon's polarization immediately changes B's measurement of his photon at a far-distant place. This assumption is of course unnecessary, because the outcome of B's measurement is not affected by the collapse. The probabilities for finding a certain polarization state at Bob's place are given as well by the initial entangled state, in which the biphoton has been prepared, including the non-classical correlations violating Bell's inequality. This is the minimal interpretation, and the collapse even unobservable. So why should I assume it to happen?

 

[ShayanJ]

This assumption is of course unnecessary, because the outcome of B's measurement is not affected by the collapse. The probabilities for finding a certain polarization state at Bob's place are given as well by the initial entangled state, in which the biphoton has been prepared, including the non-classical correlations violating Bell's inequality. This is the minimal interpretation, and the collapse even unobservable. So why should I assume it to happen?

Correct me if I'm wrong, but what I understand from this is that when we send two spins to Alice and Bob, Alice is left with a spin in a state described by the density matrix $\rho_A=\frac{1}{2}(|\downarrow\rangle\langle \downarrow |+|\uparrow\rangle\langle \uparrow |)$ regardless of the fact that Bob has made any measurement or not. When a system is in such a state, we know that there is no axis that when Alice measures her spin along that axis, she gets +1 with certainty. So if we do this experiment over and over again, she'll get 50-50 distribution of ups and downs for any axis she chooses. But if collapse is correct, after Bob has measured his spin, Alice's spin will end up in one of the states $|\uparrow \rangle$ or $|\downarrow \rangle$, which means if we do this experiment over and over again, Alice is able to find an axis that continues to give her the same result +1 every time she measures her spin. This seems to me an experimental way to settle the issue whether collapse is really there or not, or maybe I'm just misunderstanding something!(Or maybe its not that much easy to say whether there exists such an axis as described above or not!)

 

[mfb]

But are there? As far as I know, Copenhagen (in one flavour or another) is the only consensus interpretation.

There is certainly no consensus interpretation involving collapses.
All interpretations have some problems.

I think the minimal interpretation (which I consider to be a flavor of the Copenhagen interpretation) is the only one which is really consensus among physicists.

Okay, depends on the definition of "Copenhagen interpretation". The description of collapses and Copenhagen are often combined.

 

[Demystifier]

No, I didn't forget that posting, but there you didn't mention the collapse

You obviously did forget a lot about that posting, since I did mention the collapse, several times, in items 4. and 6.

Explaining quantum non-locality takes a several steps. To understand it, one has to be able to have all the steps in one's mind at once.

 

[atyy]

There is certainly no consensus interpretation involving collapses.
All interpretations have some problems.

There is a consensus interpretation, and it involves collapse. This is why the textbooks have collapse. Can you give a consensus source for any interpretation without a collapse?

 

[mfb]

Can you give a consensus source for any interpretation without a collapse?

Of course not, because there is no consensus.
If there would be, this discussion and hundreds of papers discussing different interpretations would not exist.

 

[atyy]

Of course not, because there is no consensus.
If there would be, this discussion and hundreds of papers discussing different interpretations would not exist.

But if there isn't, then unless one believes the textbooks are wrong, the textbook version of Copenhagen is the only consensus interpretation (it's not even an interpretation, it's simply QM).

 

[mfb]

But if there isn't, then unless one believes the textbooks are wrong, the textbook version of Copenhagen is the only consensus interpretation

Where do the textbooks claim that Copenhagen with collapses is a consensus interpretation?
If they would claim that, they would be wrong, but they don't. They just do not cover all interpretations, and they do not have to.

 

[atyy]

Where do the textbooks claim that Copenhagen with collapses is a consensus interpretation?
If they would claim that, they would be wrong, but they don't. They just do not cover all interpretations, and they do not have to.

But all other interpretations have problems, to the point where it is unclear if they even work as scientific theories. So Copenhagen simply has no viable competitors. Can you name any viable interpretations except Copenhagen?

 

[mfb]

But all other interpretations have problems, to the point where it is unclear if they even work as scientific theories.

Collapse has collapse as problem. "We let fields evolve with a unitary, local, deterministic evolution. Then (at some arbitrary, unmeasurable point in time with unclear definition) a magical fairy comes and changes the wavefunction in some ill-defined, nonlocal, nondeterministic way, to the point that we suddenly have elements that are not described with a wave function any more but have to be treated in a macroscopic way."

Can you name any viable interpretations except Copenhagen?

All major interpretations are viable, and Wikipedia has a list.

Take any survey about favorite interpretations of scientists: collapses find a sizeable number of votes, but not the absolute majority. And consensus would be far more than an absolute majority. Claiming consensus where there is none is just wrong.

 

[atyy]

Collapse has collapse as problem. "We let fields evolve with a unitary, local, deterministic evolution. Then (at some arbitrary, unmeasurable point in time with unclear definition) a magical fairy comes and changes the wavefunction in some ill-defined, nonlocal, nondeterministic way, to the point that we suddenly have elements that are not described with a wave function any more but have to be treated in a macroscopic way."

Yes, but that is not a problem since Copenhagen acknowledges that it needs magical fairies.

All major interpretations are viable, and Wikipedia has a list.

Take any survey about favorite interpretations of scientists: collapses find a sizeable number of votes, but not the absolute majority. And consensus would be far more than an absolute majority. Claiming consensus where there is none is just wrong.

That is not true. Whether Bohmian Mechanics, for example, can work for all relativistic quantum theories is still a matter of research. Similarly, major proponents of MWI acknowledge that it has problems. Copenhagen is consensus in the sense that if these other interpretations work, then they must derive Copenhagen.

 

[mfb]

Yes, but that is not a problem since Copenhagen acknowledges that it needs magical fairies.

Okay, if you acknowledge that you have a problem it is not a problem any more?.

Similarly, major proponents of MWI acknowledge that it has problems.

Wait, the acknowledgment trick is fine for collapses, but not for MWI?

Copenhagen is consensus in the sense that if these other interpretations work, then they must derive Copenhagen.

That is not what "consensus" means at all.

Sorry, this discussion is getting too ridiculous, I'm out.

 

[atyy]

Okay, if you acknowledge that you have a problem it is not a problem any more?.

Yes, because the problem is not a technical problem, ie. the theory makes sense if there are magical fairies. Also, the magical fairies have been observed.

Wait, the acknowledgment trick is fine for collapses, but not for MWI?

In MWI the problems are technical, it is not clear whether any magical fairies can save MWI.

That is not what "consensus" means at all.

Sorry, this discussion is getting too ridiculous, I'm out.

One again, I am only defending textbook QM. If you are right, then there is no QM at all.

 

[Demystifier]

There is a consensus interpretation, and it involves collapse. This is why the textbooks have collapse. Can you give a consensus source for any interpretation without a collapse?

There is, indeed, a consensus that collapse is a useful bookkeeping tool. However, there is no consensus whether the collapse is anything more than that.

 

[vanhees71]

Correct me if I'm wrong, but what I understand from this is that when we send two spins to Alice and Bob, Alice is left with a spin in a state described by the density matrix $\rho_A=\frac{1}{2}(|\downarrow\rangle\langle \downarrow |+|\uparrow\rangle\langle \uparrow |)$ regardless of the fact that Bob has made any measurement or not. When a system is in such a state, we know that there is no axis that when Alice measures her spin along that axis, she gets +1 with certainty. So if we do this experiment over and over again, she'll get 50-50 distribution of ups and downs for any axis she chooses. But if collapse is correct, after Bob has measured his spin, Alice's spin will end up in one of the states $|\uparrow \rangle$ or $|\downarrow \rangle$, which means if we do this experiment over and over again, Alice is able to find an axis that continues to give her the same result +1 every time she measures her spin. This seems to me an experimental way to settle the issue whether collapse is really there or not, or maybe I'm just misunderstanding something!(Or maybe its not that much easy to say whether there exists such an axis as described above or not!)

I'd formulate the first sentence slightly differently: We prepare a spin-entangled two-particle state, and A and B at (perhaps far) distant locations measure a spin component of the particles.

No matter what Alice measures, without communicating with Bob there is no way for her predicting which outcome her spin-component measurement will have, regardless of the orientation of her Stern-Gerlach apparatus. If both experimenters take an accurate record of the time of their spin measurements and if they orient their SG apparati in the same direction they will find a 100% correlation when comparing their measurement protocols. It doesn't matter in which temporal order they do their measurements (they could be even spacelike separated, i.e., not having any time order at all). For me that's a clear indication that the local measurements of the spins at A's and B's places do not affect each other but that the correlation of the outcome of spin-component measurements is inherent in the preparation of the two-particle state in the given entangled way, and it's not possible to empirically justify or disprove the claim that a state collapse has occured.

Of course, you have to perform the experiment very often, because the predictions of QT are probabilistic, and this you can test empirically only by preparing a lot of such particle pairs stochastically independently and perform the measurement on a sufficiently large ensemble to get the statistical significance you want (for discovery in the HEP community you must aim for at least $5 \sigma$ significance before you can cry "heureka").