I Confused by nonlocal models and relativity

[vanhees71]

Does this mean that if you apply operators that don't commute ([X, Y] ≠ 0) to a spacelike separated event (in which the order of application of the operators X and Y no longer makes sense), the opération must commute (The same result is obtained regardless of the order in which the operators are applied)?

/Patrick

By construction in the usual relativistic QFTs, local operators at space-like distances commute. This is called the microcausality principle. It's important particularly for the case that one of the operators is the Hamilton density, because for this case the microcausality principle guarantees that there are no faster-than-light causal effects, which would not make sense since the time-ordering of space-like separated events are (in SRT) frame dependent and thus they cannot be causally connected (i.e., one event cannot be the cause of the other). Onlye time-like or light-like separated events can be causally connected, and that's why for interacting QFTs one doesn't use tachyonic realizations of the proper orthochronous Poincare group.

 

[vanhees71]

Ok. It is about tensor product of operators (in particular here two spacelike separated operators applied on entangled (which is considered inseparable) particles or not entangled (factorizable into a tensor product of the state of the two particles) ) ? Thus the tensor product of these operators must commute?

Not being a specialist in the field of quantum physics, it is a matter of understanding.

/Patrick

We are talking about relativistic QFT. We don't talk about tensor products of operators but simply about products: A local observable operator is simply a polynomial built of field operators with arguments at one point in spacetime. E.g., for the em. field the energy density is given by
$$\hat{\mathcal{H}}(x)=\frac{1}{2} (\hat{\vec{E}}^2(x) + \hat{\vec{B}}^2(x)),$$
and due to the canonical (in this case bosonic) commutation relations imposed on the field operators (using the usual Lagrange formalism) this operator commutes with any other local operator $\hat{O}(y)$ of this kind at space-like separated distances:
$$[\hat{\mathcal{H}}(x),\hat{O}(y)]=0 \quad \text{for} \quad (x-y)^2=(x^0-y^0)^2-(\vec{x}-\vec{y})^2<0.$$

 

[akvadrako]

1. Logically, monogamy should convince you that a physical change to the state of 1 & 4 must occur along with the swap. The statistics change depending on whether the system of 1 & 2 is allowed to interact with the system of 3 & 4. If they interact, then 1 & 4 well definitely become entangled as one of the four possible Bell states.

2. I have no disagreement with MWI or any other generally accepted interpretation. (I realize that MWI claims to be local, but it would be difficult to refer to it as causal in any meaningful manner.)

What the locality of WMI should make clear is there is no possibility of changing the objective state of 1&4 via a distant swap at 2&3. Only something local can be affected. In this case, it's about selecting a different sub-ensemble of 1&4. In the more general case of measuring one entangled qubit, it's also a process akin to sub-ensemble selection.

 

[Tendex]

Ad 2) QM/QFT is causal by construction. Read a textbook, before you claim esoteric ideas!

This causality is not Bell's "local causality" that DrChinese seems to be referring to. The latter doesn't see the distinction between causal(non-spacelike) and acausal (spacelike intervals) that relativistic QFT sees and this causes confusion in these discussions.[/QUOTE]

 

[vanhees71]

Can you refer to a (scientific!) paper, where "Bell's local causality" is defined?

If you mean Bell's definition of "local realistic hidden-varialbe models", then it's indeed contradicting QFT, and that's the ingenious idea of Bell's! He formulated a class of alternative models based very general assumptions (which is summarized in the term "local realistic HV model", where "realistic" means "deterministic" to be clear since "realism" is a word burnt by philosophical gibberish with an unclear meaning) that contradict standard Q(F)T, and that was the big step forward in this apparent problems with QT: It enabled to objectively test which concept is right "local determinism" or "relativistic QFT". As is well known and also impressively demonstrated by the very paper we are discussing about here, QFT lead to the correct predictions, and the violation of Bells inequality, which must hold if local deterministic HV theores were correct, has been demonstrated with astonishing statistical significance, while the predictions by QFT were confirmed at the same significance.

 

[Tendex]

Can you refer to a (scientific!) paper, where "Bell's local causality" is defined?

If you mean Bell's definition of "local realistic hidden-varialbe models", then it's indeed contradicting QFT, and that's the ingenious idea of Bell's! He formulated a class of alternative models based very general assumptions (which is summarized in the term "local realistic HV model", where "realistic" means "deterministic" to be clear since "realism" is a word burnt by philosophical gibberish with an unclear meaning) that contradict standard Q(F)T, and that was the big step forward in this apparent problems with QT: It enabled to objectively test which concept is right "local determinism" or "relativistic QFT". As is well known and also impressively demonstrated by the very paper we are discussing about here, QFT lead to the correct predictions, and the violation of Bells inequality, which must hold if local deterministic HV theores were correct, has been demonstrated with astonishing statistical significance, while the predictions by QFT were confirmed at the same significance.

Yes. But apparently DrChinese understands that when you say causal or microcausality you mean " local deterministic" in Bell's sense.

 

[vanhees71]

Of course I don't mean "local deterministic". The very point of Bell's work is to have provided the clear distinction of "local realistic HV models" vs. "QT models", and QFT is the realization of QT in the (special-)relativistic realm. Thus QFT cannot be "local realistic" in Bell's sense.

Maybe it's my fault to assume that everybody in the year 2019 has realized that "local realistic models" are ruled out with overwhelming statistical significance, while QFT is confirmed with the same overwhelming statistical significance. This is obviously a naive assumption in a time where the believe in "fake news" and "alternative facts" is an all too common phenomenon :-((.

 

[DrChinese]

They don't? The results depend on the order in which they are measured? Bear in mind we are talking about spacelike separated measurements.

Well, you can't demonstrate* that the order of measurements changes anything. Here's what you have with generic A and B entangled AND spacelike separated:

i) You measure p on A and get value P. You now know remote p on B, and q is completely indeterminate.
ii) You measure q on A and get value Q. You now know remote q on B, and p is completely indeterminate.

Clearly, the above outcomes are completely different as quantum descriptions of A and B. And that is entirely because p and q don't commute, and the observer's choice steers the results. Bell of course implies that the decision of what to measure on A changes the quantum state for B, and vice versa.

The entanglement swapping (teleportation) set up is really the same situation, except that the final entangled pair (0 & 3) never interact as they do in a typical PDC setup.*Not everyone will entirely agree that this demonstrates a quantum nonlocal effect. Different interpretations account for this differently.

 

[DrChinese]

This causality is not Bell's "local causality" that DrChinese seems to be referring to. The latter doesn't see the distinction between causal(non-spacelike) and acausal (spacelike intervals) that relativistic QFT sees and this causes confusion in these discussions.

You are correct. The problem (of course as I see it ) is that the term "microcausal" has no place in a discussion of entanglement swapping. Nowhere in any paper on the subject will you see a comment to the effect of: "QFT is local causal by construction".

What vanhees71 is *really* trying to say is: "there is no spooky action at a distance*". You can try to define and refine the meaning of the following 4 terms. These mean different things to different people; and yes, I have been interchanging them somewhat loosely. However, Bell clearly demonstrates there is spooky stuff occurring, and it takes an interpretation to make sense of that.

Local realism (this is realism as described in EPR)
Local hidden variables
Local determinism
Local causality*Quantum nonlocality is the term that is best used to describe what we have in a post Bell world. "Spooky action at a distance" obscures the fact that the action can cross the time dimension as well.

 

[DrChinese]

Does "not quantum local" mean "violates the Bell inequalities"?

Yes, that's what proves it. How can anyone with a straight face say that the following are both true?

i) Bell inequalities are violated.
ii) Particle states evolve from past to future in a single world, independently of spacelike separated systems.

 

[vanhees71]

Oh my. Only because microcausality isn't mentioned in this paper, this doesn't mean it's irrelevant for the argument. Again: The experiment (and all other experiments with photons and whatever else made till today) is consistent with relativistic QFT and the Standard Model, and this has microcausality built in. This in turn by construction rules out actions at a distance and faster-than-light causal effects of space-like separated events (including "detector clicks") on each other. That's all I'm saying, and it's a simple conclusion from the math of QFT, which describes all these experiments properly.

 

[vanhees71]

Yes, that's what proves it. How can anyone with a straight face say that the following are both true:

i) Bell inequalities are violated.
ii) Particle states evolve from past to future in a single world, independently of spacelike separated systems.

i) has been demonstrated in the experiment, and the temporal order of "projection" of the pair 1&2 by Alice and measuring the pair 0&3 by Bob to demonstrate entanglement of these two photons is shown to be irrelevant either.

I don't know, what you want to say with ii). Particle states evolve according to a unitary transformation, dependent on the specific picture of time evolution chosen. Observable quantities like probabilities for measurements are independent on the choice of the picture of time evolution of course. I've no clue what this has to do with spacelike separated systems (which of course cannot be causally affected by each other due to the mircocausality property built as a fundamental assumption into the theory).

 

[vanhees71]

You are correct. The problem (of course as I see it ) is that the term "microcausal" has no place in a discussion of entanglement swapping. Nowhere in any paper on the subject will you see a comment to the effect of: "QFT is local causal by construction".

What vanhees71 is *really* trying to say is: "there is no spooky action at a distance*". You can try to define and refine the meaning of the following 4 terms. These mean different things to different people; and yes, I have been interchanging them somewhat loosely. However, Bell clearly demonstrates there is spooky stuff occurring, and it takes an interpretation to make sense of that.

Local realism (this is realism as described in EPR)
Local hidden variables
Local determinism
Local causality*Quantum nonlocality is the term that is best used to describe what we have in a post Bell world. "Spooky action at a distance" obscures the fact that the action can cross the time dimension as well.
[/QUOTE]
Of course, all these are unclear philosophical terms, which I don't discuss at all. I discuss about quantum theory as physical theory, and there you have a clear mathematical formalism with a clear (probabilistic) physical meaning.

Bell does not demonstrate that there is "spooky stuff occurring". He demonstrate that the class of theories, which he calls "local realistic hidden-variable theories" contradicts Q(F)T, and that makes this class of theories testable against QFT. We all know now that QFT is the correct description and "local realistic HV theories" are not. My conclusion is that I rather use QFT to understand quantum-optics experiments as the ones discussed here, and that's what I'm talking about.

It's of course confusing for any discussion if you discuss some different theory without explicitly stating it. Since Bell's inequality is violated in the experiment and thus this experiment proves realistic local HV theories wrong I could not guess that you discuss the experiment in view of outruled theories. It's Bell's merit to have brought the EPR gibberish to a scientific question to nature, and it is decided in favor of standard micorcaual (usually called local!) QFT and in disfavor of locate HV theories. I don't discuss about the latter.

 

[DrChinese]

What the locality of MWI should make clear is there is no possibility of changing the objective state of 1&4 via a distant swap at 2&3. Only something local can be affected. ... it's also a process akin to sub-ensemble selection.

That is what vanhees71 is also asserting. And I say these experiments flat out exclude that possibility. Here is what happens:

Photon pairs of 1 & 4 start out uncorrelated (in all sub-ensembles). They become correlated (entangled) IF AND ONLY IF systems 1 & 2 and 3 & 4 are allowed to interact. If they interact, in all cases they will be entangled. Therefore the remote state of the 1 & 4 pairs physically changed (from uncorrelated to correlated).

This physical change is demonstrated by an abrupt change in the statistics for 4 fold coincidences. That change occurs exactly as the 2 & 3 arrival times in the remote beamsplitter become coincident so the 2 & 3 arrival times do not distinguish each other. That is the only change necessary to change the 1 & 4 stats from random to perfectly correlated (the 2 & 3 outcomes remaining constant as part of the 4 fold coincidences). You are selecting the same 4 fold groups in all scenarios, using the exact same criteria. The only variable is the difference in arrival times of 2 and 3, which certainly shouldn't matter to distant 1 & 4 according to those who think this is only about selection.

There is no swapping paper that will say otherwise to the above, other that to acknowledge there are different viable interpretations (and I do not dispute those). Every paper refers to the swap as an actual action that depends on the observer bringing remote systems into contact at a single point in spacetime. The decision to perform the swap physically changes the outcomes for 1 & 4, and is variously referred to by "project", "cast", "swap" and not by terms such as "reveal".

And in no swapping paper is what is occurring said to be restricted by locality, relativity, etc. So again, the reference to the construction of QFT to swapping papers is inappropriate. The experiment is objective; interpretations of QFT are subjective and must bow to what the experiment says.

 

[Lord Jestocost]

Yes. But apparently DrChinese understands that when you say causal or microcausality you mean " local deterministic" in Bell's sense.

I have the same impression. Maybe, all might boil down to the question: Does one assume that an observable has the same value just before the measurement as is obtained by the measurement or does one deny that an observable has any value at all before the measurement?

 

[DrChinese]

Maybe, all might boil down to the question: Does one assume that an observable has the same value just before the measurement as is obtained by the measurement or does one deny that an observable has any value at all before the measurement?

Ha ha that one's a bear. I get the impression everyone agrees that observables are indeterminate prior to measurement. But if so: how do entangled particles end up with perfect correlations AFTER the measurement if they are *not* part of a single physical system (and are therefore spacelike separated and fully independent) ? Lots of hand-waving* needed to explain that!*Just to be fair: a lot of hand-waving is also required to explain things even if you say they ARE part of a single physical system. For example: WHEN does the change occur?

 

[vanhees71]

That is what vanhees71 is also asserting. And I say these experiments flat out exclude that possibility. Here is what happens:

Photon pairs of 1 & 4 start out uncorrelated (in all sub-ensembles). They become correlated (entangled) IF AND ONLY IF systems 1 & 2 and 3 & 4 are allowed to interact. If they interact, in all cases they will be entangled. Therefore the remote state of the 1 & 4 pairs physically changed (from uncorrelated to correlated).

This physical change is demonstrated by an abrupt change in the statistics for 4 fold coincidences. That change occurs exactly as the 2 & 3 arrival times in the remote beamsplitter become coincident so the 2 & 3 arrival times do not distinguish each other. That is the only change necessary to change the 1 & 4 stats from random to perfectly correlated (the 2 & 3 outcomes remaining constant as part of the 4 fold coincidences). You are selecting the same 4 fold groups in all scenarios, using the exact same criteria. The only variable is the difference in arrival times of 2 and 3, which certainly shouldn't matter to distant 1 & 4 according to those who think this is only about selection.

There is no swapping paper that will say otherwise to the above, other that to acknowledge there are different viable interpretations (and I do not dispute those). Every paper refers to the swap as an actual action that depends on the observer bringing remote systems into contact at a single point in spacetime. The decision to perform the swap physically changes the outcomes for 1 & 4, and is variously referred to by "project", "cast", "swap" and not by terms such as "reveal".

And in no swapping paper is what is occurring said to be restricted by locality, relativity, etc. So again, the reference to the construction of QFT to swapping papers is inappropriate. The experiment is objective; interpretations of QFT are subjective and must bow to what the experiment says.

The misunderstanding is on your side!

What interacts in Alice's selective filter measurement are the photons 2&3 (note that @DrChinese now again flips the notation to the other older paper by Zeilinger et al he didn't want to discuss anymore, because it's not available as a preprint or a legal open source, but that other paper is just the very same experiment only that the four photons are now labeled with 1234 instead of 01234; just to avoid further confusion) with the beam splitter and the detectors. Since photons 1&2 as well as 3&4 are entangled, these photons are parts of the partially inseparable four-photon state, and that leads to the entanglement swapping and teleportation through selection. It's not that there are spooky actions at a distance via Alice's measurements on her photons 2&3. In this sense everything is local, as it's described by relativistic QFT. What's "nonlocal" in the very specific sense of QFT are the correlations described by entanglement. Taken both properties of QFT/photons together you have a causal description without spooky actions at a distance. That delicate balance between the nonlocal aspects of entanglement (describing strong correlations between inseparable parts of a quantum system) and the local description of interactions (microcausality) makes relativistic QFT (or more precisely stated this class of relativistic QFTs, namely local QFTs underlying the Standard Model) consistent with both the causality structure of relativity (no faster-than-light signals leading to causal effects) and the inseparability of correlations between far-distant parts of a quantum system, as described by entangled photon states.

Nobody denies that there are interactions leading to the (post-)selection of specific sub-ensembles as done in the two papers, discussed here. Just to cite one interpretational sentence from the more recent PRL by Jennewein et al (PRL 88, 017903 (2002)), discussing the variant of the experiment, where Alice's filtering manipulations are done after the photons 0&3 are registered, i.e., the post-selection or delayed-choice variant of the experient, which in my opinion clearly shows the correctness of the above features of standard QFT rather than non-local actions at a distance of some alternative models, which @DrChinese seems to prefer (I still don't know what precisely his model for the findings is, because he doesn't give a clear formulation, which indeed can only be given in a mathematical way). For the following note that this paper labels the four photons as 0123:

===================Quote Jennewein et al ====================================
A seemingly paradoxical situation arises — as suggested
by Peres [4]— when Alice’s Bell-state analysis is delayed
long after Bob’s measurements. This seems paradoxical,
because Alice’s measurement projects photons 0 and 3 into
an entangled state after they have been measured. Nev-
ertheless, quantum mechanics predicts the same correla-
tions. Remarkably, Alice is even free to choose the kind
of measurement she wants to perform on photons 1 and 2.
Instead of a Bell-state measurement she could also mea-
sure the polarizations of these photons individually. Thus
depending on Alice’s later measurement, Bob’s earlier re-
sults indicate either that photons 0 and 3 were entangled
or photons 0 and 1 and photons 2 and 3. This means that
the physical interpretation of his results depends on Alice’s
later decision.

Such a delayed-choice experiment was performed by
including two 10 m optical fiber delays for both outputs
of the BSA. In this case photons 1 and 2 hit the de-
tectors delayed by about 50 ns. As shown in Fig. 3, the
observed fidelity of the entanglement of photon 0 and pho-
ton 3 matches the fidelity in the nondelayed case within
experimental errors. Therefore, this result indicates that
the time ordering of the detection events has no influence
on the results and strengthens the argument of Peres [4]:
This paradox does not arise if the correctness of quantum
mechanics is firmly believed.

========================= end of quote ================

The cited reference by Peres is:

A. Peres, J. Mod. Opt. 47, 139 (2000).

 

[vanhees71]

Ha ha that one's a bear. I get the impression everyone agrees that observables are indeterminate prior to measurement. But if so: how do entangled particles end up with perfect correlations AFTER the measurement if they are *not* part of a single physical system (and are therefore spacelike separated and fully independent) ? Lots of hand-waving* needed to explain that!*Just to be fair: a lot of hand-waving is also required to explain things even if you say they ARE part of a single physical system. For example: WHEN does the change occur?

The funny thing with QT is that it is a formalism that shows that both is the case at the same time, and that's so unusual for our "common sense" that one gets into this useless debates all the time. The only way to understand QT is through the math and then carefully thinking about the meaning of the math in specific cases like the experiments we discuss here:

QT says both is the case: The polarization states of a single photons in an entangled photon pair, say in the polarization-singlet state $|\psi_{01}^{-} \rangle$ of the first pair in the experiment, are maximally indetermined, i.e., described by the mixed state $\hat{\rho}_1=\hat{\rho}_{2}=\hat{1}/2$. At the same time there's 100% correlation for the measurement outcomes, when both measure in polarization in the same (arbitrary!) direction: If one photon is found to be H-polarized the other is necessarily V-polarized. Though there's maximal uncertainty (i.e., a maximal degree of indeterminism) of the single-photon polarization state there's 100% correlation for joint measurements of these polarizations. In this sense of correlations (i.e., statistical properties) such an inseparable/entangled pair of photons is indeed one physical system, and it cannot be understood by measurements of only one part of the system, but one must have a set of coincidence measurements on both photons, i.e, the entire system, to reveal the entanglement. Of course, it's only possible if you have ensembles of photons. A measurement on a single photon pair doesn't reveal anything!

 

[akvadrako]

That is what vanhees71 is also asserting. And I say these experiments flat out exclude that

Where do they exclude that entanglement can be created via post-selection?

(and I do not dispute those).

You seem to be disputing all dynamically local interpretations.

The decision to perform the swap physically changes the outcomes for 1 & 4

This brings me back to my first point. Those measurements could already have been made and visible on a paper in front of the observer making the decision. So if his decision could change those measurements, it would mean the contents of that paper change based on what he decides. Is that a possibility you are open to?

 

[DrChinese]

You can't have your cake and eat it too. If you accept that an entangled pair of photons is indeed one physical system, demonstrating quantum non-locality: then you must accept that the nature of a measurement of one member of the pair physically changes the state of the other. How that occurs, I have no clue, but it does. After any 1 of an infinite number of possible measurements by Alice, Bob will be in a corresponding state (exhibiting quantum non-locality). It is absurd to assert that extreme coincidence is something that is merely "revealed" and also say Alice and Bob were indeterminate prior to the measurement.

So if you embrace quantum non-locality, then call it for what it is: it's exactly Einstein's spooky action at a distance. And pick an interpretation that explains it best for you. But don't turn around and deny the effect because you think it is ruled out by theory. Because the only way anyone accepts quantum non-locality in the first place is due to experiment. Your theory (and your interpretation) must be constrained by experiment, not the other way around.

"Quantum teleportation is a three stage protocol that enables a sender, Alice, to transmit a quantum state to a receiver, Bob, without a direct quantum channel." Reid et al, 2008

A portion of the transmission is FTL. That portion cannot be decoded without an additional classical signal, but that doesn't change the fact that a portion of the signal arrived far in advance of the classical portion. (Please keep in mind that useful information cannot be transmitted FTL.)

The GHZ program allows the demonstration of quantum non-locality without statistical analysis - a single instance is adequate.

"Specifically, the data of Fig. 16.4 indicate that the state of, say, photon 2 was teleported to photon 4 with a fidelity of 0.89. This clearly outperforms our earlier work [14] in this field, and for the first time fully demonstrates the non-local feature of quantum teleportation." Pan et al, 2002

 

[PeterDonis]

Not being a specialist in the field of quantum physics, it is a matter of understanding.

Yes, and your understanding appears to be mistaken. Here is a simple example to illustrate the correct understanding:

Suppose Alice and Bob each have one of a pair of entangled spin-1/2 particles. They are about to measure them at spacelike separated events.

The operators (for example) "Measure z-spin on Alice's particle" and "Measure x-spin on Alice's particle" do not commute. Similarly, the operators "Measure z-spin on Bob's particle" and "Measure x-spin on Bob's particle" do not commute.

But the operators, for example, "Measure z-spin on Alice's particle" and "Measure x-spin on Bob's particle" do commute.

 

[PeterDonis]

i) You measure p on A and get value P. You now know remote p on B, and q is completely indeterminate.
ii) You measure q on A and get value Q. You now know remote q on B, and p is completely indeterminate.

These descriptions assume that there is an invariant ordering to the measurements. But if the measurements are spacelike separated, there isn't. And if there isn't, you simply can't help yourself to a description that assumes that there is. So I do not accept either of these descriptions. I want to see a description that does not assume that the measurements occur in a particular order; only such a description can be consistent with relativity, since in relativity the ordering of spacelike separated events is not invariant.

For example:

i-a) You measure p on A and get value P. You measure p on B and get value not-P. This will always happen when those two measurements are combined.
i-b) You measure p on A and get value P. You measure q on B and get value Q. Then you repeat that pair of measurements many times, and the measurements on A and B show zero correlation.

ii-a) You measure q on A and get value Q. You measure q on B and get value not-Q. This will always happen when those two measurements are combined.
ii-b) You measure q on A and get value Q. You measure p on B and get value P. Then you repeat that pair of measurements many times, and the measurements on A and B show zero correlation.

the above outcomes are completely different as quantum descriptions of A and B. And that is entirely because p and q don't commute, and the observer's choice steers the results

Both observers' choices. Not just one. It's not either observer's choice in isolation, but the combination of the two, that makes the difference--in one case, they both choose to measure in the same direction, in the other, they choose to measure in two orthogonal directions. Neither choice by itself "steers" the results; only the combination of the two does.

Not everyone will entirely agree that this demonstrates a quantum nonlocal effect. Different interpretations account for this differently.

I guess that depends on what "quantum nonlocal" is supposed to mean. That's why I asked earlier if by that term you meant "violates the Bell inequalities". Any interpretation has to agree that the results violate the Bell inequalities, because that's an experimental fact.

 

[microsansfil]

Yes, and your understanding appears to be mistaken. Here is a simple example to illustrate the correct understanding:

Suppose Alice and Bob each have one of a pair of entangled spin-1/2 particles. They are about to measure them at spacelike separated events.

The operators (for example) "Measure z-spin on Alice's particle" and "Measure x-spin on Alice's particle" do not commute. Similarly, the operators "Measure z-spin on Bob's particle" and "Measure x-spin on Bob's particle" do not commute.

But the operators, for example, "Measure z-spin on Alice's particle" and "Measure x-spin on Bob's particle" do commute.

Ok Thank

The operators commute even if it doesn't apply in the same Hilbert space?

if you formulate the problem in the tensor product space of the two particles ε_A ⊗ ε_B because the two particles are entangled.

Does the tensor product extension "Measure z-spin on Alice's particle" Z_A ⊗ I_B and the tensor product extension "Measure x-spin on Bob's particle" I_A ⊗ X_B also commute ?

/Patrick

 

[PeterDonis]

The operators commute even if it doesn't apply in the same Hilbert's space?

They are operators on the same Hilbert space. See below.

if you formulate the problem in the tensor product space of the two particles εA ⊗ εB because the two particles are entangled.

Yes, that's the appropriate Hilbert space. There is no "if"; only one Hilbert space is valid, and it's that one.

Does the tensor product extension "Measure z-spin on Alice's particle" ZA ⊗ IB and the tensor product extension "Measure x-spin on Bob's particle" IA ⊗ XB also commute ?

What do you mean "also"? These are exactly the operators that I already said commute.

The operators that don't commute are pairs like $Z_A \otimes I_B$ and $X_A \otimes I_B$, or $I_A \otimes Z_B$ and $I_A \otimes X_B$.

 

[DrChinese]

But the operators, for example, "Measure z-spin on Alice's particle" and "Measure x-spin on Bob's particle" do commute.

I say they don't commute (assuming they are entangled). Yours is actually a variation of the EPR argument. So here is a specific point of departure, and I believe we are speaking the same terminology. A Bell inequality is violated in this case, indicating that Alice and Bob are not independent (i.e. separable). If they commute, then there are Product State statistics. If they don't, you see Entangled State statistics.EDIT: I'm sure you know that conjugate observables of entangled particles do not commute, and in fact I think you have written on that previously.

 

[microsansfil]

They are operators on the same Hilbert space. See below.

My understand is that Alice particle "live" in the Hilbert Space ε_A and Bob particle "live" in the Hilbert Space ε_B . However, Alice and Bob particles "live" in the Hilbert Space ε_A ⊗ ε_B which is different from ε _A or ε_B.

This is not a question of physics, but of mathematics.

/Patrick

 

[DrChinese]

1. Both observers' choices. Not just one. It's not either observer's choice in isolation, but the combination of the two, that makes the difference--in one case, they both choose to measure in the same direction, in the other, they choose to measure in two orthogonal directions. Neither choice by itself "steers" the results; only the combination of the two does.

2. I guess that depends on what "quantum nonlocal" is supposed to mean. That's why I asked earlier if by that term you meant "violates the Bell inequalities". Any interpretation has to agree that the results violate the Bell inequalities, because that's an experimental fact.

1. Concur completely.

2. I said yes earlier, which is why I label anything that violates a Bell Inequality as quantum non-local. That's pretty standard too. And of course I agree that any viable interpretation must follow experiment. I don't really know what quantum non-locality is, but I know how to identify it when I see it.

 

[PeterDonis]

I say they don't commute (assuming they are entangled).

Then I disagree with you and I have no idea why you would say that, since it appears to contradict basic QM. The two operators are $Z_A \otimes I_B$ and $I_A \otimes X_B$, which obviously commute since $Z_A$ and $I_A$ commute and $I_B$ and $X_B$ commute.

A Bell inequality is violated in this case, indicating that Alice and Bob are not independent (i.e. separable). If they commute, then there are Product State statistics. If they don't, you see Entangled State statistics.

I don't think this is correct. The Bell inequality is violated because the joint probability function does not factorize, i.e., it violates Bell's locality assumption. But that is not the same as the two operators not commuting.

 

[PeterDonis]

My understand is that Alice particle "live" in the Hilbert Space εA and Bob particle "live" in the Hilbert Space εB . However, Alice and Bob particles "live" in the Hilbert Space εA ⊗ εB which is different from ε A or εB.

This makes no sense. First you say each particle lives in its own Hilbert space. Then you say the two particles live in a single Hilbert space. Those two statements contradict each other.

The correct statement is that, if you have two particles, the Hilbert space of the two-particle system is the tensor product one. If the two particles are not entangled, then the state of the two particles is separable--it can be expressed as a single product of an "A" state and a "B" state. But such a state is still a state on the tensor product Hilbert space.

 

[kurt101]

But the operators, for example, "Measure z-spin on Alice's particle" and "Measure x-spin on Bob's particle" do commute.

The order you make spacelike separated measurements on entangled particles does not affect the resulting probability. However for a given instance of entangle particles, measuring one particle first may have given you a different result than if you had measured it second. In other words the order of measurements may commute for the probability, but not for a single pair of measurements. I think we should be clear on this point, because otherwise you might incorrectly be mislead into thinking a non-local causal interpretation is not allowed because of this.

 

[PeterDonis]

for a given instance of entangle particles, measuring one particle first may have given you a different result than if you had measured it second.

First, this is not what the math of QM says; I already gave that in a previous post. The math is clear that the two operators commute.

Second, there is no invariant sense in which either particle is measured first if the measurements are spacelike separated. So I don't see how to even formulate what you are claiming consistently.

And third, we have no way of measuring counterfactuals, so even if there were an invariant sense in which one measurement was before the other, we have no way of measuring what would have happened if the order had been opposite. Our only basis for talking about counterfactuals at all is theory, and the theory--the math of QM--says, as noted above, that the operators commute, which means that the results must be independent of the order in which the measurements are done.

you might incorrectly be mislead into thinking a non-local causal interpretation is not allowed

What do you mean by a "non-causal interpretation"?

 

[microsansfil]

This makes no sense. First you say each particle lives in its own Hilbert space. Then you say the two particles live in a single Hilbert space. Those two statements contradict each other.

I agree it's misspoken. I just want to point out that ε_A ⊗ ε_B is different to ε_A and also different to ε_B. in ε_A ⊗ ε_B you can have entangled particles. i. e. which is not factorized as the tensor product, and thus can't be expressed as a single product of an "A" state and a "B" state.

For the Operators in ε_A ⊗ ε_B which are tensor product of an operator of ε_A and an operator of ε_B, then this operator commute :

However, just as with vectors, there exist operators in ε_A ⊗ ε_B which are not tensor products of an operator of ε_A and an operator of ε_B.

Thus what about the commutation of this kind of operator?

/Patrick

 

[Tendex]

The order you make spacelike separated measurements on entangled particles does not affect the resulting probability. However for a given instance of entangle particles, measuring one particle first may have given you a different result than if you had measured it second. In other words the order of measurements may commute for the probability, but not for a single pair of measurements. I think we should be clear on this point, because otherwise you might incorrectly be mislead into thinking a non-local causal interpretation is not allowed because of this.

If by "non-local causal" interpretation you mean for instance the Bohmian Interpretation this indeed is allowed as long as the interpretation can not be made relativistic in which case you have the spacelike versus non-spacelike interval distinction that doesn't exist in a purely non-relativistic interpretation. Without this distinction the QFT microcausality prescription is moot and you can have "spooky actions at a distance" in the way DrChinese is using them to contradict vanhees71. To do it one just has to ignore that relativistic QFT is considered more fundamental and general than non-relativistic quantum mechanics.

One can only avoid confusing debates on Bell's theorem when this point is not ignored.

 

[vanhees71]

Where do they exclude that entanglement can be created via post-selection?

To the contrary, the here discussed experiment confirms the possibility to create entanglement by post-selection (as also other experimental realizations of delayed-choice protocols clearly demonstrate).

 

[Tendex]

If you accept that an entangled pair of photons is indeed one physical system, demonstrating quantum non-locality: then you must accept that the nature of a measurement of one member of the pair physically changes the state of the other.

I say they don't commute (assuming they are entangled). Yours is actually a variation of the EPR argument. So here is a specific point of departure, and I believe we are speaking the same terminology. A Bell inequality is violated in this case, indicating that Alice and Bob are not independent (i.e. separable). If they commute, then there are Product State statistics. If they don't, you see Entangled State statistics.EDIT: I'm sure you know that conjugate observables of entangled particles do not commute, and in fact I think you have written on that previously.

As explained by PeterDonis it is only if applied to spacelike-separated operators that they commute. If you treat the particles nonrelativistically as one system there is no spacelike separation consideration possible and the operators don't commute, but then mathematically there is no "spooky action at a distance" to be puzzled about. If one insists in factoring in that distance there is no other accepted way than going to the wider scope relativistic theory that acknowledges a spacelike separation for which the operators must commute.

 

[kurt101]

First, this is not what the math of QM says; I already gave that in a previous post. The math is clear that the two operators commute.

The QM math does not describe what happens when you measure a single instance of entangled particles. The QM math speaks to the probability you will get when you measure many samples of entangled particles.

Second, there is no invariant sense in which either particle is measured first if the measurements are spacelike separated. So I don't see how to even formulate what you are claiming consistently.

If the universe can be simulated (in a theoretical sense) and has hidden variables that describe its state and you can pause the simulation then you can say what measurement happened first in the simulation.

And third, we have no way of measuring counterfactuals, so even if there were an invariant sense in which one measurement was before the other, we have no way of measuring what would have happened if the order had been opposite.

Agreed and that is the point I am making.

Our only basis for talking about counterfactuals at all is theory, and the theory--the math of QM--says, as noted above, that the operators commute, which means that the results must be independent of the order in which the measurements are done.

What we observe in the experiment is also a basis for talking about counterfactuals.

What do you mean by a "non-causal interpretation"?

In the general sense like John Bell said in his Bertlmann's socks paper "Thirdly, it may be that we have to admit that causal influences - do go faster than light.".

In a more specific way as I previously described in a previous post on this thread about the entanglement swapping paper that DrChinese mentioned. For example I said "1 is measured and non-locally gives the result of this measurement to 2".

 

[kurt101]

If by "non-local causal" interpretation you mean for instance the Bohmian Interpretation this indeed is allowed as long as the interpretation can not be made relativistic in which case you have the spacelike versus non-spacelike interval distinction that doesn't exist in a purely non-relativistic interpretation. Without this distinction the QFT microcausality prescription is moot and you can have "spooky actions at a distance" in the way DrChinese is using them to contradict vanhees71. To do it one just has to ignore that relativistic QFT is considered more fundamental and general than non-relativistic quantum mechanics.

I am not referring to any formal interpretation, but to how we interpret what is observed in the entanglement experiments. And my point is that nothing in this informal interpretation of these entanglement experiments contradicts relativistic QFT or special relativity.

 

[Tendex]

I am not referring to any formal interpretation, but to how we interpret what is observed in the entanglement experiments. And my point is that nothing in this informal interpretation of these entanglement experiments contradicts relativistic QFT or special relativity.

Sure, but it is important to realize that this informal interpretation doesn't contradict relativistic QFT just as long as it ignores its relativistic content and the current consensus about non-relativistic quantum mechanics being less fundamental than relativistic QFT.

 

[Lord Jestocost]

... where "Bell's local causality" is defined?

Caslav Brukner in “Elegance and Enigma: The Quantum Interviews” (ed. by Maximilian Schlosshauer), p. 166:

"Bell’s theorem is a no-go theorem that states that no “local causal” or “local realistic” theories can ever reproduce all of the predictions of quantum mechanics. The desire for a local causal theory is based on the following three assumptions:

(1) There exist “causes” that determine measurement outcomes, or probabilities of outcomes, for all possible experiments that could be performed on an individual system, no matter whether any experiment — and which experiment — is actually performed (and so, in this sense, would be “real”).

(2) The actually measured outcome (or the probability for the outcome), and equally those outcomes that could be potentially measured, can only be influenced by local causes (that is, other events in the backward light cone) and not by any event in spacelike separated regions (“locality”).

(3) The experimenter’s choice of the measurement setting is independent of the causes that determine the actually measured outcome (“freedom of choice”)."

 

[DrChinese]

Then I disagree with you and I have no idea why you would say that, since it appears to contradict basic QM. The two operators are $Z_A \otimes I_B$ and $I_A \otimes X_B$, which obviously commute since $Z_A$ and $I_A$ commute and $I_B$ and $X_B$ commute.

The question is whether entangled Alice(p) commutes (or not) with Bob(q). Or substitute any conjugate pair for p and q (such as spin at various angles). To be specific, let's discuss entangled spin-1 particle spins x and z (singlet). These are not separable, do NOT commute, and the uncertainty principle should be applied.

a. There is no question that if Alice(x) and Bob(x) are measured, they will be opposite. Similarly, if Alice(z) and Bob(z) are measured, they too will be opposite. The uncertainty principle is not a constraint.

b. In the EPR-B case, as summarized by Weinberg, Lectures on Quantum Mechanics, 12.1 Paradoxes of Entanglement: "... the observer could have measured the x-component of the spin of particle 1 instead of its z-component, and by the same reasoning, if a value h/2 or −h/2 were found for the x-component of the spin of particle 1 then also the x-component of the spin of particle 2 must have been −h/2 or h/2 all along. Likewise for the y-components. So according to this reasoning, all three components of the spin of particle 2 have definite values, which is impossible since these spin components do not commute..."

Your recent comment (I know, this is cheating!) on the above quote was: "Weinberg is not talking about successive measurements. He's talking about alternative possibilities for single measurements (one on each particle). He's simply pointing out that, since measuring the same spin component of both particles will always give opposite results, no matter which component is measured (x or y or z), any hidden variable model, i.e., any model that attributes the correlations between these measurements to pre-existing properties of the particles, would have to attribute definite spin components (+1 or -1) in all three directions (x and y and z) to each particle, i.e., the pre-measurement state of each particle would have to have definite values for spin-x, spin-y, and spin-z. But that is not possible because no quantum state can have definite values for multiple non-commuting operators. "

I think you agree with Weinberg that the Uncertainty Principle applies to non-commuting components of entangled pairs. And again, any analysis of EPR-B is going to say virtually the same as Weinberg. Weinberg further comments: "There is a troubling weirdness about quantum mechanics. Perhaps its weirdest feature is entanglement, the need to describe even systems that extend over macroscopic distances in ways that are inconsistent with classical ideas." . One of the classical ideas being local causality.

c. And in fact, Weinberg goes on to say as follows: "Of course, according to present ideas a measurement in one subsystem does change the state vector for a distant isolated subsystem - it just doesn't change the density matrix." Which is what I assert: A measurement on Alice's particle changes the physical state of Bob's remote entangled particle (what is observed). Although if you and vanhees71 are instead referring to the density matrix, you would be right about that. (Weinberg personally believes in the reality of the reduced density matrix rather than the state vector, I believe, but that does not seem a common interpretation.)

d. This from Wikipedia, which is not cited as a source but rather to indicate the generally accepted viewpoint and related derivation:

EPR Paradox
"...how does Bob's [entangled] positron know which way to point if Alice decides (based on information unavailable to Bob) to measure x (i.e., to be the opposite of Alice's electron's spin about the x-axis) and also how to point if Alice measures z, since it is only supposed to know one thing at a time? The Copenhagen interpretation* rules that the wave function "collapses" at the time of measurement, so there must be action at a distance..." Of course, the mathematical presentation shows that Alice's x and Bob's z are constrained by the Uncertainty Principle and clearly do not commute. That portion is identical to nearly any presentation of this problem.*Copenhagen simply being one viable interpretation, there are many others too; such would not change the sense of this passage.

 

[kurt101]

Sure, but it is important to realize that this informal interpretation doesn't contradict relativistic QFT just as long as it ignores its relativistic content and the current consensus about non-relativistic quantum mechanics being less fundamental than relativistic QFT.

How does this informal interpretation contradict QFT by ignoring its relativistic content?

What does non-relativistic QM vs relativistic QFT have to do with this interpretation?

And the interpretation that I am discussing is what I said to PeterDonis:
In the general sense like John Bell said in his Bertlmann's socks paper "Thirdly, it may be that we have to admit that causal influences - do go faster than light.".

In a more specific way as I previously described in a previous post on this thread about the entanglement swapping paper that DrChinese mentioned. For example I said "1 is measured and non-locally gives the result of this measurement to 2".

 

[DrChinese]

As explained by PeterDonis it is only if applied to spacelike-separated operators that they commute. If you treat the particles nonrelativistically as one system there is no spacelike separation consideration possible and the operators don't commute, but then mathematically there is no "spooky action at a distance" to be puzzled about. If one insists in factoring in that distance there is no other accepted way than going to the wider scope relativistic theory that acknowledges a spacelike separation for which the operators must commute.

I don't believe relativity has anything to do with it. There are no predictive differences for a local setup versus a non-local setup for anything related to entanglement (at least AFAIK).

Conjugate operators of distant entangled particles do not commute. If they did, the EPR Paradox would not have been solved. EPR's argument was precisely that you could learn non-commuting values on a particle by reference to the entangled partner. If the partner's conjugate property DID commute, you could do that. But you can't.

Further, Bell experiments would have different outcomes. After all, the experiments have been done with strict Einsteinian non-locality a number of times. The predictions for Bob depend on the choice of measurement on distant Alice. The match statistics are dependent solely on the relative angle theta between, in a manner inconsistent with separability (Bell 1964). Bell showed there was a different relationship on theta if there were separability (i.e. and therefore they commuted). This difference does not show up in cases where the same measurement is performed on both Alice and Bob. Which makes it easy to gloss over.

 

[microsansfil]

Conjugate operators of distant entangled particles do not commute.

The commuting property of operators is independent of the states of the systems to which these operators may apply, isn't it? whether or not they're entangled.

Mathematically the two operators $Z_A \otimes I_B$ and $I_A \otimes X_B$ commute, regardless of whether Alice and Bob's particles are entangled, as shown here

/Patrick

 

[DrChinese]

The commuting property of operators is independent of the states of the systems to which these operators may apply, isn't it? whether or not they're entangled.

... as shown here

/Patrick

I am not sure where your citation is from (the image presented on the link). Could you pass that along for context? I admit that without any mention of entanglement, it is difficult to see how this relates.

On the other hand: I did mention the source for my (directly opposing) quotes in post #140, but did not provide a link. It is:

Weinberg, Lectures on Quantum Mechanics

 

[microsansfil]

I am not sure where your citation is from (the image presented on the link).

Quantum Mechanics by Claude Cohen-Tannoudji

/Patrick

 

[DrChinese]

Quantum Mechanics by Claude Cohen-Tannoudji

/Patrick

Thanks. The author is of course top notch. However, I believe this was originally written in 1973. It lacks a section on entanglement, even in the newer editions. I don't think this is a good fit for our discussion. The formulae presented do not describe entangled systems. (Or if it does, I couldn't locate it.)

 

[microsansfil]

It lacks a section on entanglement, even in the newer editions.

It is in volume 3.

/Patrick

 

[PeterDonis]

I just want to point out that εA ⊗ εB is different to εA and also different to εB.

Yes, and εA ⊗ εB is the only one of them that we need to be discussing here.

in εA ⊗ εB you can have entangled particles. i. e. which is not factorized as the tensor product, and thus can't be expressed as a single product of an "A" state and a "B" state.

Yes, this is the correct definition of an entangled state, as I think I've already said.

there exist operators in εA ⊗ εB which are not tensor products of an operator of εA and an operator of εB.

Yes, but none of them are relevant to the measurements we've been talking about.

what about the commutation of this kind of operator?

It would depend on the specific operators. But, as above, none of them are relevant to what we've been discussing. You're welcome to start a separate thread if you want to discuss operators like this, though it would be a good idea to first find a specific experiment that involves them.

 

[PeterDonis]

As explained by PeterDonis it is only if applied to spacelike-separated operators that they commute. If you treat the particles nonrelativistically as one system there is no spacelike separation consideration possible and the operators don't commute

You are misinterpreting what I said, and your statements are incorrect.

The operators I wrote down commute regardless of whether events A and B are spacelike separated or not.

Considering two entangled particles nonrelativistically as one system certainly does not rule out spacelike separation. It just means you don't make any use of it in your analysis. And the operators I wrote down commute regardless of whether you do a relativistic or a non-relativistic analysis.

 

[PeterDonis]

The QM math does not describe what happens when you measure a single instance of entangled particles.

That depends on which interpretation you adopt.

If the universe can be simulated (in a theoretical sense) and has hidden variables that describe its state and you can pause the simulation then you can say what measurement happened first in the simulation.

This is way off topic for this thread, and indeed this forum.

What we observe in the experiment is also a basis for talking about counterfactuals.

I don't see how, since we only observe what actually happened, not what might have happened counterfactually.

In the general sense like John Bell said in his Bertlmann's socks paper "Thirdly, it may be that we have to admit that causal influences - do go faster than light.".

Then I think you need to pick a specific interpretation that is based on this premise if you want to discuss it. Bell, IIRC, had Bohmian mechanics in mind as falling into this category.