I Confused by nonlocal models and relativity

[vanhees71]

Bell's particular inequalities are, yes. But a lot of work in this area has been done since Bell, including finding cases where QM predicts results that are impossible according to local hidden variable models, so that the latter models can be ruled out with 100% certainty by observing such an "impossible" result, with no probabilities or statistics or ensembles required. For example, the GHZ experiment:

https://en.wikipedia.org/wiki/GHZ_experimentMore precisely, this:
Which, mathematically, refers to the fact that the reduced density matrix of 1&4 does not change. But the wave function does.

There's no wave function for photons!

What changes is the description of the sub-ensemble based on the selection depending on Alice's specific measurement of the pair 2&3.

Also in classical applications of statistics and probability theory a subensemble usually has some different probability distribution for some of its properties than the full ensemble. There's nothing mysterious in this.

 

[Tendex]

The 4-photon space is NOT this product space but it is the subspace spanned by all totally symmetrized (bosonic) product-basis states.

I don't see this, why can't the tensor product be used for the 4-photon system?

 

[vanhees71]

Because bosons are bosons. It's a well-established fact of nature.

 

[Tendex]

Because bosons are bosons. It's a well-established fact of nature.

I fail to see how the boson-fermion distinction is relevant to this discussion, we could be talking about electrons and spin instead of photons and polarization and the situation would be essentially the same.

 

[vanhees71]

Sure, but it's important to treat the photons as bosons (or electrons as fermions) particularly when it comes to entanglement.

 

[Tendex]

Sure, but it's important to treat the photons as bosons (or electrons as fermions) particularly when it comes to entanglement.

Ok, I agree it is important to always be precise in math descriptions, but in this particular experiment, does the Fock space instead of the tensor product description of the system add anything to the conceptual issue of swapping and entanglement we've been discussing with DrChinese? I just want to make sure I'm not missing something important.

 

[vanhees71]

In this case it doesn't make much of a difference.

 

[Tendex]

In this case it doesn't make much of a difference.

Thanks. I have seen such simplified descriptions in notes describing Zeilinger's experiment like in https://www.physik.hu-berlin.de/de/nano/lehre/copy_of_quantenoptik09/Chapter6 in 6.4.3

 

[Lord Jestocost]

...he is taking an ensemble interpretation viewpoint, in which QM does not describe individual photons or individual experimental runs, but only ensembles of them.

...I'll also repeat once more my suggestion to not use vague ordinary language but instead look at the math. The math is unambiguous, but there are many different ways of describing in ordinary language what the math is telling us...

Regarding the "Ensemble Interpretation", one can clearly speak out how "murky" this interpretation is. See comment https://www.physicsforums.com/threads/confused-by-nonlocal-models-and-relativity.973876/post-6202479 or listen to Maximilian Schlosshauer in “Decoherence, the measurement problem, and interpretations of quantum Mechanics”, Section B. 1. Superpositions and ensembles (https://arxiv.org/abs/quant-ph/0312059):

“Put differently, if an ensemble interpretation could be attached to a superposition, the latter would simply represent an ensemble of more fundamentally determined states, and based on the additional knowledge brought about by the results of measurements, we could simply choose a subensemble consisting of the definite pointer state obtained in the measurement. But then, since the time evolution has been strictly deterministic according to the Schrödinger equation, we could backtrack this subensemble in time and thus also specify the initial state more completely (“postselection”), and therefore this state necessarily could not be physically identical to the initially prepared state on the left-hand side of Eq. (2.1).“

 

[vanhees71]

I don't understand this criticism of the ensemble interpretation. Of course QED is T symmetric and thus all interactions of photons with charged matter are in principle reversible, but what has this to do with any specific interpretation?

 

[PeterDonis]

There's no wave function for photons!

Ok, then substitute "quantum state", which is more general and allows for the somewhat different mathematical objects that model photons in QFT.

If your criticism is that, strictly speaking, QFT has no invariant concept of "the quantum state of an extended system at an instant of time", that is true, but you can still pick a particular frame, such as the overall rest frame of the system + measurement apparatus in the experiment under discussion, and construct the quantum state of an extended system such as the multi-photon system in the experiment at an instant of time in that frame. That is implicitly what is being done when we talk about the "wave function" of the multi-photon system.

Also, as has been pointed out, you could do a similar experiment with electrons moving at non-relativistic speeds. Such an experiment could be modeled using only non-relativistic QM, in which the description in terms of wave functions is unproblematic and the issues you appear to be raising with photons not having wave functions do not arise.

 

[vanhees71]

The point is that photons have no position observable to begin with. All you can measure are probabilities to register a photon (with given properties depending on the measurement you do).

Surely you can do such experiments in principle also with electrons though I guess for technical reasons neutrons are better candidates. The formalism is nearly the same (up to electrons/neutrons being fermions rather than bosons).

 

[Mordred]

There was a recent experiment where superposition of photons were photographed just ran across it last night. Thought it interesting and applicable in this thread

https://www.sciencealert.com/scientists-just-unveiled-the-first-ever-photo-of-quantum-entanglement

 

[PeterDonis]

The point is that photons have no position observable to begin with.

Nor is one needed for the experiments under discussion. The states you wrote down earlier, where photons are distinguished by having momentum in different directions, with no position observable anywhere in sight, are quite sufficient to analyze the experiments.

 

[vanhees71]

Sure, that's what I'm saying!

 

[Elias1960]

And what could be the alternatives?

The straightforward alternative is simply a return to the Lorentz ether. Which is the original interpretation of special relativity, before Minkowski proposed his spacetime interpretation. Also known as the preferred frame hypothesis.

A straightforward candidate for that preferred system of coordinates exists - the CMBR frame.

Once you have a preferred frame, you can do standard QM as well as BM without any problems, and they will be equivalent in their physical predictions.

In the relativistic context, at least for bosons the field ontology is the most natural choice, see Bohm.D., Hiley, B.J., Kaloyerou, P.N. (1987). An ontological basis for the quantum theory, Phys. Reports 144(6), 321-375.

 

[PeterDonis]

The straightforward alternative is simply a return to the Lorentz ether. Which is the original interpretation of special relativity, before Minkowski proposed his spacetime interpretation.

Discussion of LET is off limits here.

In the relativistic context, at least for bosons the field ontology is the most natural choice, see Bohm.D., Hiley, B.J., Kaloyerou, P.N. (1987). An ontological basis for the quantum theory, Phys. Reports 144(6), 321-375.

As far as I can tell, this paper says nothing about any "preferred frame", and certainly does not hypothesize that the CMBR rest frame is such a thing.

 

[Elias1960]

Discussion of LET is off limits here.

As far as I can tell, this paper says nothing about any "preferred frame", and certainly does not hypothesize that the CMBR rest frame is such a thing.

That this has not been mentioned explicitly is not difficult to explain - the authors would clearly prefer to have a theory which is completely Lorentz-covariant but they are unable to present one. To present a version which requires a preferred frame is nothing scientists are proud of so that they would like to mention it, and they would even less like to refer to it using the e-word.

On the other hand, there is no necessity to mention this. The Bohmian field theory has been developed there for a relativistic scalar field theory. This has been done in a particular frame, without claiming or presenting anything which suggests that the construction could be made Lorentz-covariant. Look at the variables the wave functional \psi depends on, say, in the formula after (6) in part II:

\psi = R(\ldots \phi(x,t),\ldots,t)\exp[iS(\ldots \phi(x,t),\ldots,t)]

It explicitly contains a reference to time t, which is the time parameter of the Schroedinger theory, without any relation to any spacetime. So, Bohmian field theory as developed in the paper is a theory based on a field \phi(x), which defines the configuration, changing in an absolute time t.

Lorentz symmetry exists only for the Lagrange density and the resulting classical field equation.

Once there exist no Lorentz transformation rule even for the basic objects, in particular, for the wave functional, the theory can not even pretend to be more than a theory with a preferred frame, moreover, presented only in this particular preferred frame, without even caring about other frames.

The important point for this thread is that for relativistic bosonic field theories a Bohmian version exists, with the details worked out in the mentioned paper. This version is not fundamentally Lorentz covariant but uses a preferred frame.

There are those which consider this to be a decisive objection against it, and there are others who think there is no base to care, as long as all the observable effects are the same as predicted in standard QFT. It would be, indeed, off-topic to discuss here who is right.

But it is important to mention that relativistic field theory is something which not only potentially can be, but actually has been handled by a Bohmian version of field theory, even if it can be criticized for depending on a preferred frame.

 

[PeterDonis]

the authors would clearly prefer to have a theory which is completely Lorentz-covariant but they are unable to present one

relativistic field theory is something which not only potentially can be, but actually has been handled by a Bohmian version of field theory, even if it can be criticized for depending on a preferred frame

How is it relativistic if it's not Lorentz covariant?

 

[Elias1960]

How is it relativistic if it's not Lorentz covariant?

It is relativistic once the predictions for all observable effects are Lorentz covariant. The theory itself, on the fundamental level, does not have to be Lorentz covariant for this. Only observable effects matter, not?

Given that the question is not discussed in the paper, let's consider the straightforward way how to prove this, namely, by proving the equivalence with standard QFT. One uses a lattice regularization on a large cube with periodic boundary conditions in space (not time) to get rid of UV as well as IR infinities. The resulting theory is, then, finite-dimensional. The Lagrangian has then the standard form required for Bohmian theory (quadratic dependence on the momentum variables) as presented in the paper. Then one can use standard Bohmian theory to show the equivalence for anything observable of the Bohmian theory to the Schrödinger picture of QM and its equivalence to the Heisenberg picture too. So, we have equivalence with the Heisenberg picture of QM for a particular regularization. What remains to be proven (namely that the resulting theory is Lorentz covariant if we consider the continuous limit) is already part of standard QFT.

 

[PeterDonis]

It is relativistic once the predictions for all observable effects are Lorentz covariant

But are they? If the equations aren't Lorentz covariant, and the equations generate the predictions, how can the predictions be Lorentz covariant?

let's consider the straightforward way how to prove this

Has such a proof been given in any reference?

 

[Elias1960]

Has such a proof been given in any reference?

I'm not aware of such a proof being published, and I see no necessity, given that the part relevant to dBB theory is a trivial reference to the standard dBB equivalence results. What would be the non-trivial part of the proof? That a lattice regularization on a cube with periodic boundary conditions is finite-dimensional so that the standard theorems can be applied?

The third part, from the lattice regularization to the QFT itself, would be non-trivial, but this is already pure standard QFT.

But are they? If the equations aren't Lorentz covariant, and the equations generate the predictions, how can the predictions be Lorentz covariant?

The important non-covariant part, the trajectories, are not observable. If one removes important parts of reality from consideration, the symmetry will change.

 

[vanhees71]

But are they? If the equations aren't Lorentz covariant, and the equations generate the predictions, how can the predictions be Lorentz covariant?

There are plenty of examples for formulations of relativistic theories that are not manifestly Lorentz (or rather Poincare) covariant, while the results on observables perfectly are. An example are relativistic gauge theories (like QED or the Standard Model) in "non-covariant gauges" like the Coulomb gauge: The Feynman rules and the proper vertex functions are not manifestly covariant, the S-matrix elements referring to physical cross sections are. Sometimes non-covariant formulations have their merits!

 

[bhobba]

The straightforward alternative is simply a return to the Lorentz ether. Which is the original interpretation of special relativity, before Minkowski proposed his spacetime interpretation. Also known as the preferred frame hypothesis.

SR requires no interpretation. Its simply what the symmetries of nature require. That derivation just leaves the value of one constant to be determined - many many experiments show its the speed of light.

Thanks
Bill

 

[bhobba]

To present a version which requires a preferred frame is nothing scientists are proud of so that they would like to mention it, and they would even less like to refer to it using the e-word.

SR does not contradict an aether. Its simply superfluous. I have no issue with a preferred frame and its a perfectly legitimate conjecture. I am a little perplexed why anyone would worry about a theory that required such. Theoretically one would break the isotropy of an inertial frame but the CBMR already does - however it can be screened out. Any other preferred frame hopefully would be similar or basically undetectable - if not then we really do have a revolution on our hands similar to the discovery of broken parity symmetry. Maybe that's why some would dislike it.

Thanks
Bill

 

[Elias1960]

SR requires no interpretation. Its simply what the symmetries of nature require. That derivation just leaves the value of one constant to be determined - many many experiments show its the speed of light.

Symmetries it itself require nothing. And without interpretation, it is not even clear what is symmetric. SR certainly requires interpretation. Say, something non-symmetric may be in one interpretation only an irrelevant mathematical object, in another interpretation something really existing. Even if to mention other interpretations is forbidden here, the one interpretation which is allowed is nonetheless a nontrivial interpretation.

SR does not contradict an aether. Its simply superfluous. I have no issue with a preferred frame and its a perfectly legitimate conjecture. I am a little perplexed why anyone would worry about a theory that required such. Theoretically one would break the isotropy of an inertial frame but the CBMR already does - however it can be screened out. Any other preferred frame hopefully would be similar or basically undetectable - if not then we really do have a revolution on our hands similar to the discovery of broken parity symmetry. Maybe that's why some would dislike it.

But to mention the aether contradicts the rules of this forum, I have been told. So, there is something some people worry a lot. We are free to speculate about the reasons for ourselves, but cannot discuss it here.

 

[bhobba]

But to mention the aether contradicts the rules of this forum, I have been told. So, there is something some people worry a lot. We are free to speculate about the reasons for ourselves, but cannot discuss it here.

I don't know what's going on here. There is no rule about mentioning an aether. There is a rule about LET:
Generally, in the forums we do not allow the following:
Attempts to promote or resuscitate theories that have been discredited or superseded (e.g. Lorentz ether theory); this does not exclude discussion of those theories in a purely historical context

This is because modern SR has superseded LET and discussion about it will not lead anywhere except to why it was superseded, which is allowable in a historical context if that is what you are interested in. But it needs to remain in that context.

The aether is sometimes discussed when talking about the DBB interpretation because some think it implies an aether, and that is OK. But in order to avoid any confusion I personally call it preferred frame. In LET the aether, according to the theory, has physical effects, but could never actually be detected by experiment. That goes beyond preferred frame so what is meant could be a bit ambiguous.

Thanks
Bill.

 

[bhobba]

Symmetries it itself require nothing. And without interpretation, it is not even clear what is symmetric.

An inertial frame is, by definition, homogeneous in space and time and isotropic. That is a definition and in and of itself contains no physics. The physics lies in the many experiments that show, to a high degree of accuracy, such frames exist. The principle of relativity ie the laws of nature are the same in all inertial frames, or frames moving at constant velocity wrt to an inertial frame, are the same is also able to be experimentally checked and has been found correct.

That is all that goes into the derivation of SR. The value of the constant that appears in that derivation can be found in many ways, and that it is the speed of light is not the most obvious. But this is getting way beyond the scope of this thread. If you want to delve into the foundations of SR please start a thread in the relativity section.

Thanks
Bill

 

[Elias1960]

An inertial frame is, by definition, homogeneous in space and time and isotropic. That is a definition and in and of itself contains no physics. The physics lies in the many experiments that show, to a high degree of accuracy, such frames exist. The principle of relativity ie the laws of nature are the same in all inertial frames, or frames moving at constant velocity wrt to an inertial frame, are the same is also able to be experimentally checked and has been found correct.

Except that if gravity is considered, we have no such things as inertial frames and they survive only as local approximations. So, given the empirical evidence for GR, SR has already been found incorrect.

If you want to delve into the foundations of SR please start a thread in the relativity section.

To start a thread about the foundations of SR would be suicidal in a forum where even to mention one of the two established interpretations of SR is simply forbidden. One can hope for some possibility of discussing the foundations of quantum theory, but given that the most interesting (namely all realistic and causal) interpretations depend on the forbidden interpretation of relativity, this is nothing but a quite naive hope.

 

[bhobba]

To start a thread about the foundations of SR would be suicidal in a forum where even to mention one of the two established interpretations of SR is simply forbidden.

LET is an established modern interpretation? It has been well and truly superseded. It is on topic to discuss the history of why that is but these days to put it mildly it is very backwater. However, the reason, history wise, should be obvious - we have a justification in SR that requires no interpretation - it's simply a consequence of very well verified symmetries. These symmetries determine the geometry of space-time. I seem to recall the Enlargen Program was concerned about showing the strong connection between symmetries and geometry - the symmetries of SR implies Minkowski geometry. Its like asking what is the interpretation of Euclidean Geometry - there is no interpretation. Its just a consequence of Euclid's axioms that may be true or false. I am not talking about Hilbert's axioms, but the basic geometry you learned at school.

Thanks
Bill