Von Neumann QM Rules Equivalent to Bohm?

[vanhees71]

Let's assume for simplicity that Alice and Bob are using the same filter orientation. That is agreed-upon ahead of time. Then after Alice measures her photon to have polarization H, Alice knows something definite about Bob's future measurement: that he will measure the polarization to be H.

Classically, if Alice learns something definite about a future measurement performed by Bob, she can assume that that means that that result was pre-determined. If Alice learns that Bob will find a right shoe when he opens the box, she assumes that it was a right shoe before he opened the box.

Sure, that's what makes QT different from classical physics, where there are no indetermined observables, because all observables of a system always have definite values, which we sometimes don't know and thus use probabilistic descriptions of classical statistical physics.

Within QT there are always some observables indetermined even if we know the complete (pure) state of the system. Usually, if you have a pure state of a composite system the parts of it are not in a pure state, as it's the case here for the polarization-entangled two-photon state.

In EPR, Alice learns that Bob will measure polarization H. But she can't assume that it had polarization H before he measured it.

That's a pretty stark difference between the two cases.

Yes, it is, but this doesn't imply necessarily a collapse, because there's nothing preventing me from taking the point of view that the corresponding correlation was inherent from the very beginning in the entangled two-photon state. Thus it's a prediction of the model that when Alice finds her photon to be V-polarized, Bob must find his H-polarized, given the engangled initial state, $|HV \rangle-|VH \rangle$.

You don't need an "if" in the EPR case, if Bob agrees ahead of time to use a pre-arranged filter orientation.

Yes, but the important point is that, if he uses and appropriate other filter orientation, you can violate Bell's inequality, showing that the polarization have not been predetermined within a local deterministic hidden-variable theory. This clearly proves that QT is really fundamentally different from classical physics, as you stressed yourself above!

I don't think it's a matter of assuming a mechanism. Collapse is just a description of the situation, it seems to me. Assuming once again that Bob has agreed ahead of time to use the same pre-arranged filter orientation as Alice, Alice knows before Bob does what his measurement will be. She learns a fact about Bob's photon + filter + detector remotely. Under the assumption that Alice and Bob are using the same orientation, and that Alice observes a horizontally-polarized photon before Bob does his measurement, let $X$ be the claim: "Bob will observe a horizontally-polarized photon".

It seems to me that there are only three possibilities:

1. X was a fact before Alice observed her photon.
2. X only became a fact after Alice observed her photon.
3. X is not really a fact at all (the MWI tactic of rejecting unique outcomes)

By "fact" I mean something that is objectively true, independent of any observer. If you assume definite outcomes, then it seems to me that "Bob will observe a horizontally-polarized photon" is a fact, in this sense.

There's a possibility missing, namely precisely the one I follow!

X was not a fact before Alice's measurement (Bob's photon is unpolarized), but it was a fact that you had an entangled photon pair, so that there is this 100% correlation between A's and B's measurements. There's no problem with that point of view precisely when you don't take the collapse as a physical process (I guess, you'd call this ontological) but just an update of Alice's knowledge about the system (I guess, that you'd call epistemical). Nothing changes for Bob before his measurement, because he has not gained any information yet. From this point of view, there's no contradiction that for Bob his photon's state is still the maximum-entropy mixture $1/2 \mathbb{1}$ while for Alice's it's in the pure state $|H \rangle \langle H|$. So after A's measurement and before B's measurement X became a fact for Alice but not for Bob. That's all.

 

[vanhees71]

I think nearly any non-Crackpot interpretation is "more natural" than many worlds. What does it help anyway? It introduces simply infinitely many parallel universes for any measurement act. But what's a measurement act different that the interaction of a system and a measurement apparatus. So you cannot even say how many parallel universes are created per second, because you cannot clearly define what a measurement process distinguishes from any other kind of interaction. I don't see any merit in many worlds as compared to the standard minimal interpretation.

 

[Demystifier]

I think nearly any non-Crackpot interpretation is "more natural" than many worlds. What does it help anyway? It introduces simply infinitely many parallel universes for any measurement act. But what's a measurement act different that the interaction of a system and a measurement apparatus. So you cannot even say how many parallel universes are created per second, because you cannot clearly define what a measurement process distinguishes from any other kind of interaction. I don't see any merit in many worlds as compared to the standard minimal interpretation.

You misunderstood MWI. It does not introduce infinitely many parallel universes for any measurement act. It does not make a difference between measurement and interaction.

Due to Schrodinger evolution, almost any interaction with a system with a large number of degrees of freedom creates branching of the wave function into a finite number of new branches. These branches are mathematical objects derived from Schrodinger evolution of the wave function. Their mathematical existence does not depend on the interpretation. All what MWI adds is the claim that these branches are not only abstract mathematical objects, but also actual real worlds.

These very same branches are also essential for understanding the von Neumann theory of quantum measurements (before the collapse), Bohmian mechanics (before you calculate the trajectories), and even the mainstream theory of decoherence (before you calculate the reduced density matrix). Different interpretations of QM attribute different physical meaning to these branches (MWI takes them more seriously than any other interpretation), but all interpretations deal with these branches one way or another.

The main advantage of MWI compared to other interpretations (including the standard minimal interpretation) is the claim that it is the only interpretation for which everything can be described by the Schrodinger equation. (For instance, in the standard minimal interpretation (SMI), the existence of individual results of measurements cannot be described by Schrodinger equation alone, in the sense that the events themselves are not wave functions in SMI.)

 

[stevendaryl]

Yes, it is, but this doesn't imply necessarily a collapse, because there's nothing preventing me from taking the point of view that the corresponding correlation was inherent from the very beginning in the entangled two-photon state.

Yes, the correlation was true before Alice did her measurement. But the fact that "Bob will measure the photon to have horizontal polarization" was not true before Alice performed her measurement. So a fact about Bob's measurement becomes true as a result of Alice's actions. I don't see how it could be otherwise:

Yes, but the important point is that, if he uses and appropriate other filter orientation, you can violate Bell's inequality, showing that the polarization have not been predetermined within a local deterministic hidden-variable theory.

Let's make sure we're on the same page about the case where Alice and Bob pre-arrange to use the same orientation. Alice measures her photon to be horizontally polarized. Then, under the assumption that Bob is using the same filter orientation, would you say that the statement "Bob will measure horizontal polarization" is true after Alice's measurement? Was it true before her measurement, or not? If you say yes, you're assuming hidden variables. If you say no, then it sure seems that you are saying that something about Bob's measurement was changed by Alice's measurement.

X was not a fact before Alice's measurement (Bob's photon is unpolarized), but it was a fact that you had an entangled photon pair, so that there is this 100% correlation between A's and B's measurements.

X in this case being "Bob will measure a horizontally polarized photon". So the question is: Is X true after Alice's measurement? X is a logical consequence of the two true statements: (1) Alice and Bob's results are 100% correlated, and (2) Alice measured horizontal polarization. Normally, if something follows from true statements, then it is itself true.

 

[TrickyDicky]

But do relativistic QFTs exist? Do we have a gauge invariant and Lorentz invariant regulator for the standard model and Einstein gravity?

We don't, I guess the question was rhetorical.

I think most people assume that relativistic QFTs are possible, even if we haven't figured out a consistent one.

It would be very strange (but certainly possible) if it turned out that the only way to make QFT consistent is to chuck out SR and assume that there is an absolute notion of time. My intuition is that the problems in understanding what's going on in EPR-type experiments is orthogonal to the problems of making a consistent QFT. But it would certainly be exciting to find out that they are connected.

If discarding SR was the only way of making QFT consistent(I doubt that'd come to be the case), why should assuming absolute time be the only alternative? I think the choice is broader.
Regarding a connection between measurement problem and not having a consistent QFT I do believe they are connected. This can be seen clearer if one uses Penrose's RTR description of the issue in terms of the U and R processes. The non-unitary R process slips in the step from the asymptotic perturbative series to a finite terms( a valid calculation) series, preventing from keeping Poincare invariance.

 

[stevendaryl]

I think nearly any non-Crackpot interpretation is "more natural" than many worlds. What does it help anyway

To me, the issue is just one of logical consistency. The two claims

1. Only one outcome occurs.
2. There are no nonlocal effects.

seem to contradict the predictions of QM in the EPR case. You disagree because somehow you don't think that Bob's result going from "50/50 chance of being H or V" to "100% chance of H" is a physical change. I don't see how it could be otherwise. Saying that Bob will definitely measure H sure seems to be an objective fact about Bob's situation. If it wasn't true before Alice's measurement, I don't see how you can say that Alice's measurement didn't change the facts about Bob's situation.

If the two claims above are inconsistent with the predictions of QM, then one or the other (or both) is false. MWI rejects 1. Bohm rejects 2.

 

[vanhees71]

Yes, the correlation was true before Alice did her measurement. But the fact that "Bob will measure the photon to have horizontal polarization" was not true before Alice performed her measurement. So a fact about Bob's measurement becomes true as a result of Alice's actions. I don't see how it could be otherwise:
Let's make sure we're on the same page about the case where Alice and Bob pre-arrange to use the same orientation. Alice measures her photon to be horizontally polarized. Then, under the assumption that Bob is using the same filter orientation, would you say that the statement "Bob will measure horizontal polarization" is true after Alice's measurement? Was it true before her measurement, or not? If you say yes, you're assuming hidden variables. If you say no, then it sure seems that you are saying that something about Bob's measurement was changed by Alice's measurement.

After Alice finds V for her photon, she knows that Bob finds H for his for sure, and there is no hidden variable. Why should there be one?

X in this case being "Bob will measure a horizontally polarized photon". So the question is: Is X true after Alice's measurement? X is a logical consequence of the two true statements: (1) Alice and Bob's results are 100% correlated, and (2) Alice measured horizontal polarization. Normally, if something follows from true statements, then it is itself true.

After A's measurement, finding V, for sure B will measure H. That's due to the entangled initial state. This follows directly from the usual assumptions of QT.

 

[vanhees71]

To me, the issue is just one of logical consistency. The two claims

1. Only one outcome occurs.
2. There are no nonlocal effects.

seem to contradict the predictions of QM in the EPR case. You disagree because somehow you don't think that Bob's result going from "50/50 chance of being H or V" to "100% chance of H" is a physical change. I don't see how it could be otherwise. Saying that Bob will definitely measure H sure seems to be an objective fact about Bob's situation. If it wasn't true before Alice's measurement, I don't see how you can say that Alice's measurement didn't change the facts about Bob's situation.

If the two claims above are inconsistent with the predictions of QM, then one or the other (or both) is false. MWI rejects 1. Bohm rejects 2.

I don't know what you mean by "effects" under 2.

According to standard QFT (here QED for our photon case) there are no non-local interactions but there can be "non-local" correlations as described by entangled states. It's important to distinguish this. Concerning the Bohmian point of view, I've to study it for relativistic QFT, before I can say anything about it.

 

[atyy]

@vanhees71, just a comment: I don't think your intepretation is the most minimal, by saying collapse is epistemic.

Usually in a truly minimal interpretation, we say we don't know whether collapse is physical or epistemic or some unknown mixture of both. For example, Cohen-Tannoudj, Diu and Laloe (at least in the English translation of their textbook) are very careful not to exclude either possibility.

 

[vanhees71]

Good hint to look at this book!

 

[stevendaryl]

After Alice finds V for her photon, she knows that Bob finds H for his for sure,

Well, we're mixing up two different types of EPR here. I'm assuming the case where the two photons have the same polarization, but it doesn't matter.

and there is no hidden variable. Why should there be one?

Once again, X is the statemement "Bob will measure his photon to have horizontal polarization". I think we agree that after Alice's measurement. statement X is true. Yes or no? Was it true before Alice's measurement? Yes or no?

After A's measurement, finding V, for sure B will measure H. That's due to the entangled initial state. This follows directly from the usual assumptions of QT.

It doesn't matter what it is due to. The issue is just the truth value of the statement X = "Bob will measure his photon to have horizontal polarization". Is it true after Alice's measurement? Was it true before Alice's measurement?

 

[stevendaryl]

I don't know what you mean by "effects" under 2.

If there is a fact about Bob's situation that was not true at one time and becomes true at another time, I consider that to be a change in Bob's situation. I would say that Alice's actions have an effect on Bob if they cause a change to Bob's situation.

I suppose the disagreement is whether "Bob will measure result H" is a fact about Bob's situation, or not.

 

[atyy]

I suppose the disagreement is whether "Bob will measure result H" is a fact about Bob's situation, or not.

Especially in the Bell tests, since Bob chooses his measurement randomly at the last moment.

 

[stevendaryl]

Especially in the Bell tests, since Bob chooses his measurement randomly at the last moment.

In the particular exchange with vanhees71, I was talking about the situation in which Alice and Bob agree ahead of time to use the same filter orientation. For Bell's purposes, he didn't want to do this, but if we do stipulate that Alice and Bob use the same filter setting, and that it's the type of twin-pair photon generation that produces two photons with the same polarization, then we can reason as follows: (For definiteness, let's assume that it's asymmetric; Alice is much closer to the source of the photons than Bob is, so she gets her photon before Bob gets his.)

1. Immediately before Alice measure her photon's polarization, the most complete statement that can be made about Bob's measurement result is that it has a 50/50 chance of H polarization or V polarization.
2. Immediately after Alice measures her photon's polarization, and finds that it has polarization H, then she can claim, with 100% certainty: "Bob will measure polarization H for his photon"

So the issue is, what is the nature of the claim/prediction X = "Bob will measure polarization H for his photon"? Is is a fact about Bob's situation (photon + filter + detector)?

To me, it seems like it's as good a candidate for being a fact as anything else in quantum mechanics.

 

[TrickyDicky]

A clarification about my #145, I'm using atyy's distinction regarding necessity of collapse between sequential measurements and measurements to which the deferred measurement principle can be applied when I say that R(collapse) conditions effective perturbative QFT S-matrix time-ordered product, that must include microcausality(operators commuting at spacelike separation) to ensure coordinate independence(being spacetime fields).

 

[morrobay]

After Alice finds V for her photon, she knows that Bob finds H for his for sure, and there is no hidden variable. Why should there be one?
After A's measurement, finding V, for sure B will measure H. That's due to the entangled initial state. This follows directly from the usual assumptions of QT.

In your next post # 148, you refer to " non - local " correlations as described by entangled states. Could you elaborate on this in the context of whether Bob (above) would have measured H before Alice measured V
I see two options : 1. They were created at source , Alice V and Bob H (predetermined )
2. A superposition Alice 50/50 V / H and when Alice finds V Bob finds H ( non-local interaction or non local
correlation ) ?

 

[vanhees71]

Both. 1. and 2. are obviously wrong! Again: The biphoton was created by a local interaction of a laser beam with the non-linear birefringent crystal by parametric down conversion (LOCALITY of interactions). The polarization state is after some standard-optics manipulations given by
$$|\Psi \rangle=\frac{1}{\sqrt{2}} (|HV \rangle - |VH \rangle).$$
One of the photons is registered at A's place and one at B's place. Both are using polarization filters in the same $x$-direction.

Now the states of the single photons are given by tracing out the other photon respectively, i.e., they are in mixed states:
$$\hat{\rho}_A=\mathrm{Tr}_{B} |\Psi \rangle \langle \Psi |=\frac{1}{2} \mathbb{1}.$$
Also Bob finds
$$\hat{\rho}_B=\mathrm{Tr}_{A} |\Psi \rangle \langle \Psi |=\frac{1}{2} \mathbb{1}.$$
Both have maximally unpolarized single photons! The polarization state of each of the single photons is maximally indetermined.

Now, if A measures $H$ for her photon, she updates her state to
$$|\Psi_{A|H} \rangle = \left (|H \rangle \langle H| \otimes \mathbb{1} \right )|\Psi \rangle=\frac{1}{\sqrt{2}} |HV \rangle.$$
This happens with the probability
$$P_A(H)=\|\Psi_{A|H} \|^2=\frac{1}{2}.$$
Bob still doesn't know about this, and he still uses $|\Psi \rangle$. So he only knows that he'll find with 50% probability H and with 50% probability V, while Alice knows that he must find V.

In this minimal interpretation, nothing happened to Bob's photon, and it cannot happen anything to it before Bob measures it, if the registration events for the photons at A's and B's place are space-like separated (according to Einstein causality). This is a consistent (in my opinion the only consistent) interpretation of the formalism given by relativistic local QFTs.

The entanglement enables correlations between far-distant registration events for the photons without predetermination of the measured observables, as is demonstrated clearly by this example. In this sense there are NON-LOCAL correlations in a theory where you have only LOCAL interactions, and the latter is the case by construction of QED as a local microcausal relativistic QFT.

 

[atyy]

Now, if A measures $H$ for her photon, she updates her state to
$$|\Psi_{A|H} \rangle = \left (|H \rangle \langle H| \otimes \mathbb{1} \right )|\Psi \rangle=\frac{1}{\sqrt{2}} |HV \rangle.$$
This happens with the probability
$$P_A(H)=\|\Psi_{A|H} \|^2=\frac{1}{2}.$$
Bob still doesn't know about this, and he still uses $|\Psi \rangle$. So he only knows that he'll find with 50% probability H and with 50% probability V, while Alice knows that he must find V.

Alice's knowledge is frame dependent.

 

[vanhees71]

If the measurement events are space-like separated you are right, but that doesn't matter. The minimal interpretation is consistent in any frame, because QED is Lorentz invariant. This is, BTW, a very good argument for the fact that there's no collapse necessary!

 

[TrickyDicky]

If the measurement events are space-like separated you are right, but that doesn't matter. The minimal interpretation is consistent in any frame, because QED is Lorentz invariant. This is, BTW, a very good argument for the fact that there's no collapse necessary!

For the calculation of a specific outcome of a measurement perturbative QED is not Lorentz invariant, both by the nonlinearity of the perturbative operations and per a theorem by Haag that I've seen you call "a quibble", so in strict terms it is the other way around, the presence of something like collapse is granted mathematically.

 

[vanhees71]

Well, mathematically QED is not proven to exist. Nevertheless the S-matrix elements as calculated in renormalized perturbation theory and which are compared to observations are Lorentz invariant (and gauge invariant). So the theory I'm talking about is this theory and not some fictitious "exact solution" of QED which might exist or not. It's of course a somewhant unsatisfying state that one of the most successful theories (nowadays one can say the entire standard model including QFD and QCD is the most successful theory ever) is shaky in its mathematical foundations.

 

[atyy]

If the measurement events are space-like separated you are right, but that doesn't matter. The minimal interpretation is consistent in any frame, because QED is Lorentz invariant. This is, BTW, a very good argument for the fact that there's no collapse necessary!

Well, the minimal interpretation is consistent, and has predictions that are Lorentz invariant. But it has collapse, which may or may not be real. What I don't think is right about what you write is that relativistic quantum field theory preserves Einstein causality. Einstein causality, as considered by EPR, is formalized by Bell as separability. Einstein causality is either empty if the wave function is not real, or it is violated if the wave function is real.

 

[TrickyDicky]

Well, mathematically QED is not proven to exist. Nevertheless the S-matrix elements as calculated in renormalized perturbation theory and which are compared to observations are Lorentz invariant (and gauge invariant). So the theory I'm talking about is this theory and not some fictitious "exact solution" of QED which might exist or not. It's of course a somewhant unsatisfying state that one of the most successful theories (nowadays one can say the entire standard model including QFD and QCD is the most successful theory ever) is shaky in its mathematical foundations.

We must be talking about different things. AFAIK there is a QED that is perfectly Lorentz invariant but divergent, and a QED that manages to obtainn finite and very accurate results out of the calculations that uses regularization (the S-matrix elements you mention), this operation is mathematically very bad behaved, and Lorentz invariance is not rigorously recovered. So I'm not sure what other case you are referring to.

 

[vanhees71]

Now I don't understand what you are talking about. The renormalized n-point functions and thus the S-matrix elements, cross sections, etc. calculated from them are manifestly covariant.

 

[vanhees71]

Well, the minimal interpretation is consistent, and has predictions that are Lorentz invariant. But it has collapse, which may or may not be real. What I don't think is right about what you write is that relativistic quantum field theory preserves Einstein causality. Einstein causality, as considered by EPR, is formalized by Bell as separability. Einstein causality is either empty if the wave function is not real, or it is violated if the wave function is real.

I think, we won't ever come to a conclusion about this, because we seem not to talk the same language. Again, I don't know, what you mean by "collapse" here. For me it's the update of the description of the state due to new information gained by a measurement, it's not a physical process, and I don't need to call it collapse, because nothing collapses here.

I also don't see, where Einstein causality is violated at any place here. The only thing what can obey or violate causality are observable facts, and nowhere in the above description is Einstein causality in this sense violated, and it's not empty. If you can realize a faster-than light signal propagation, then it's violated. So far no such thing has been observed (not even any clear case of falsification of the Standard Model has been found, although it's vigorously searched for one).

 

[atyy]

I think, we won't ever come to a conclusion about this, because we seem not to talk the same language. Again, I don't know, what you mean by "collapse" here. For me it's the update of the description of the state due to new information gained by a measurement, it's not a physical process, and I don't need to call it collapse, because nothing collapses here.

I also don't see, where Einstein causality is violated at any place here. The only thing what can obey or violate causality are observable facts, and nowhere in the above description is Einstein causality in this sense violated, and it's not empty. If you can realize a faster-than light signal propagation, then it's violated. So far no such thing has been observed (not even any clear case of falsification of the Standard Model has been found, although it's vigorously searched for one).

That's fine - it is very unusual to call "on faster than light propagation of classical information" "Einstein causality". I still think you are wrong, because you keep referring to "local interactions" - as in the interaction between the measurement apparatus and the photon. This is the usual meaning of Einstein causality - that interactions are local. It is not the same as "no faster than light transmission of classical information". So I don't think you are being consistent: what do you mean by "Einstein causality"?

(1) if Einstein causality is local interactions - no, quantum theory does have non-local interactions. If you have the apparatus interact with the photon physically, then there is physical collapse, and Einstein causality is violated.

(2) if Einstein causality is no faster than light transmission of classical information, then yes, quantum field theory preserves this.

But notions (1) and (2) are distinct.

 

[vanhees71]

Quantum field theory as we know it is by assumption only using local interactions and thus cannot have non-local ones as you claimed. A QFT is called local when it's action is defined by a Lagrange density expressed as polynomials of fields and its derivatives that transform in a local way under Poincare transformations, i.e., like the classical fields. Further it's assumed to be microcausal, i.e., local observables commute at space-like distances. This guarantees the linked-cluster principle and thus there's no faster-than light propagation of observable signals and thus also no faster-than-light transmission of information (see Weinberg, QT of Fields, Vol. I).

I don't know, what you want to emphasize by the term "classical information". How do you distinguish classical information from just information?

 

[atyy]

Quantum field theory as we know it is by assumption only using local interactions and thus cannot have non-local ones as you claimed. A QFT is called local when it's action is defined by a Lagrange density expressed as polynomials of fields and its derivatives that transform in a local way under Poincare transformations, i.e., like the classical fields. Further it's assumed to be microcausal, i.e., local observables commute at space-like distances. This guarantees the linked-cluster principle and thus there's no faster-than light propagation of observable signals and thus also no faster-than-light transmission of information (see Weinberg, QT of Fields, Vol. I).

Yes, but the interactions you talk about are not real, just like wave function collapse. The interactions are essentially the Hamiltonian, which is not real. However, you talk about the photon interacting with the measurement apparatus as if it is real. That is misleading. Only the events and the probabilities of the events are real.

I don't know, what you want to emphasize by the term "classical information". How do you distinguish classical information from just information?

That is something quantum theory assumes we know.

 

[vanhees71]

Well, the interactions between photons and the charged particles in the measurement apparatus are described by QED, by what else?

 

[atyy]

Well, the interactions between photons and the charged particles in the measurement apparatus are described by QED, by what else?

At some stage in the minimal interpretation, we cannot include the whole universe in the wave function, which means there is also no Hamiltonian for the universe. Let's acll whatever is outside the wave function the measurement apparatus. The wave function is not real, and since the Hamiltonian describes the wave function evolution, the Hamiltonian is also not real. So all the local interactions you describe are not real. In contrast, the measurement apparatus is real. So the Hamiltonian does not describe the measurement apparatus.

 

[vanhees71]

Of course, within the minimal statistical interpretation the "state of the whole universe" makes no sense. I'd also not dare to say that our humble models are "complete" in the sense of a theory of everything. Also you can state a lot about the "whole universe", which after all is unobservable.

Nevertheless the measurement apparati used by us are tiny compared to the "whole universe", and they are described well enough by classical physics, which is a coarse grained limit of QT. Of course, you can argue that "local" then means "macroscopically local", and that's indeed right :-).

 

[atyy]

Of course, within the minimal statistical interpretation the "state of the whole universe" makes no sense. I'd also not dare to say that our humble models are "complete" in the sense of a theory of everything. Also you can state a lot about the "whole universe", which after all is unobservable.

Nevertheless the measurement apparati used by us are tiny compared to the "whole universe", and they are described well enough by classical physics, which is a coarse grained limit of QT. Of course, you can argue that "local" then means "macroscopically local", and that's indeed right :-).

The measurement apparatus cannot be a "coarse grained" version of something unreal like the Hamiltonian or the wave function, unless coarse graining can produce reality from non-reality.

 

[atyy]

Of course, within the minimal statistical interpretation the "state of the whole universe" makes no sense. I'd also not dare to say that our humble models are "complete" in the sense of a theory of everything. Also you can state a lot about the "whole universe", which after all is unobservable.

Nevertheless the measurement apparati used by us are tiny compared to the "whole universe", and they are described well enough by classical physics, which is a coarse grained limit of QT. Of course, you can argue that "local" then means "macroscopically local", and that's indeed right :-).

I'm not sure this is right, but let me see if this argument can persuade you that what you say is at odds with the Bell theorem.

The locality of the interactions in the Lagrangian ultimately becomes the locality in the Hamiltonian. The Hamiltonian in the Heisenberg picture evolves according to classical local equations of motion and deterministically. The Bell theorem forbids local deterministic explanations of the nonlocal correlations. So the local deterministic Hamiltonian cannot explain the nonlocal quantum correlations - it needs other things like the quantum state, the nonlocal observables, and the Born rule - but by the time we calculate that, it is hardly clear that the calculation is local.

 

[TrickyDicky]

Now I don't understand what you are talking about. The renormalized n-point functions and thus the S-matrix elements, cross sections, etc. calculated from them are manifestly covariant.

But what I'm saying is that you are using the free fields in the interaction picture for that, and that is precisely what Haag's theorem proves not to exist as a mathematical construction. So it doesn't really matter what regularization procedure you use to perform the calculations , the symmetries have already been lost.

 

[Haelfix]

I'm not sure this is right, but let me see if this argument can persuade you that what you say is at odds with the Bell theorem.

The locality of the interactions in the Lagrangian ultimately becomes the locality in the Hamiltonian. The Hamiltonian in the Heisenberg picture evolves according to classical local equations of motion and deterministically. The Bell theorem forbids local deterministic explanations of the nonlocal correlations. So the local deterministic Hamiltonian cannot explain the nonlocal quantum correlations - it needs other things like the quantum state, the nonlocal observables, and the Born rule - but by the time we calculate that, it is hardly clear that the calculation is local.

I'm not sure what you think the problem is. Suppose I took a pair of socks, one red the other blue. Put one in a box, and send it to Alpha Centauri. If I waited some time and opened the box, I would instantly know the state of the other pair. This is completely classical, and shows that there is nothing wrong with nonlocal correlations, indeed almost every correlation is nonlocal. There would only be a problem if I could wiggle a sock on Alpha Centauri, such that it would instantenously cause a wiggle on the earth. Then I would have acausal observable physics.

The difference between quantum mechanics and this classical example lies encoded in the fact that when we do GHZ or EPR experiments, we can infer that the original state was NOT either red or blue, but rather something new eg a state that is in a weird mixture, like half red and half blue. That further the dynamics must allow interference of states and the noncommutativity of operators. However all these things stay manifestly local, in the sense that the lagrangian stays written in a certain form and that operators at spacelike separation don't commute.