A Interpreting photon correlations from independent sources

[DrChinese]

1. I don't know if we disagree here or are just misreading each other. On page 14 of Ma et al, the time-evolution through the set up is shown for when swap = on. You can see that the evolution relates the detected signatures at the end of the run, in spatial modes b'' and c'' to the state that was in the incident spatial modes b and c. This is also made clear by Fig. 2 where we can see the ordering of b c b' c' b'' c''. We see from these evolution rules that Victor uses the signatures in b" and c" to resolve a BSM in b and c. Similarly, when the quarter wave plate is off, the signatures in b'' and c'' resolve an SSM in b and c.

2. I don't understand [the quoted paragraph in bold]. ... If photons 1 and 2 are measured in different, unbiased bases, no correlation will be observed.

3. Yes, if a swap occurs, then photons 2 & 3 are projected onto Φ-, as are photons 1 & 4. If no swap occurs, no such projection occurs. We both agree with this.

4. Where we diverge is the interpretation of the projection. You interpret it as a very literal ontic event. But there are alternatives. E.g. The Copenhagen interpretation frames the projection as Victor updating the information he has about the outcomes of measurements on 1 and 4. The statistical ensemble interpretation frames the projection as the identification of a subensemble that will exhibit correlations between measurements on 1 and 4. When no swap is carried out, no such information can be learned about 1&4, and no such subensemble can be identified. At most, if an SSM is carried out, then a subensemble with correlations like Fig 3b is identified.

1. Ma, explaining how the EOMs I mentioned control the BSM vs SSM choice: "With opposite voltages on EOM1 and EOM2, we realize a π/2 phase change of the MZI (corresponding to Bell-state measurement) when EOM1 is applied with +EV and EOM2 with –EV, and no phase change(corresponding to separable state measurement) when EOM1 is applied with -EV and EOM2 with +EV."

However, this difference in how we understand turning swapping on and off has no real significance to our debate. How they do it isn't as important as we agree they are doing it as far away from Alice and Bob as needs be. This particular incarnation of Entanglement Swapping highlights delayed choice (failure of Einsteinian causality), but could just as easily highlight distance apart.


2. You: "If photons 1 and 2 are measured in different, unbiased bases, no correlation will be observed."

Me: "...the circular polarization of photon 1 is in a separable Product state with the vertical polarization of photon 2 after initial creation via PDC. ... There is no possibility of any statistical relationship in such a product state, [and they will be uncorrelated.]"

We're saying the exact same thing. It is a fact that photons 1 and 2 are initially entangled as Ψ-. However, photon 1 and photon 2 are entangled only on identical polarization bases. Bases L/R and H/V are different. Consequently, we must agree: there can be no correlation between Photon 1's L/R and Photon 2's H/V. They are in a Product state.


3. Yay!


4. We disagree on the interpretation of the projection (ontic vs epistemic), that too. You mention:

-The Copenhagen interpretation frames the projection as Victor updating the information he has about the outcomes of measurements on 1 and 4.
-The statistical ensemble interpretation frames the projection as the identification of a subensemble that will exhibit correlations between measurements on 1 and 4. When no swap is carried out, no such information can be learned about 1&4, and no such subensemble can be identified. At most, if an SSM is carried out, then a subensemble with correlations like Fig 3b is identified.

"Updating the information" is scientifically meaningless. Every experiment requires bringing information together. Hardly a description invoking locality.

"Identification of a subensemble" is precisely what I have said is impossible, and should be dropped as an argument to explain entanglement swapping using locality. We agreed that there is no statistical correlation between the L/R polarization of photon 1 with photon 2's H/V polarization; and the L/R polarization of photon 4 with photon 3's H/V polarization. Consequently, and there can't really be any question about this: you cannot learn anything about photon 1 and photon 2's L/R polarization by measuring the H/V polarizations of photon 2 and photon 3.

---

But where we really disagree is twofold:

A) A physical change to a setting *here* leads to a change in observed statistics *there*, holding all parameters constant other than the swap/no swap mechanism. Now I know you claim it is a different selection criteria being applied in the Ma experiment (you're wrong though). But the Megadish experiment reports the same result as Ma, but uses a different swap/no swap mechanism that is NOT subject to your critique. (There is no changing of a beam splitter, no change of wave plates, no optical change at all.)

B) It is canonically impossible for there to exist "hidden correlations" that are merely identified via BSMs. We should agree on this at this point - see point 2 above.

 

[DrChinese]

As a quick follow-up to my post #148 just now, note also that there is no interpretation that I'm aware of which just has a straightforward, intuitively comfortable story to tell about what is happening in these experiments.

This is a great point, there really aren't any good descriptions - and I certainly don't have one. So I'm not sure we get much farther towards understanding this vexing mystery even by acknowledging: There is nonlocality that essentially contradicts Einsteinian notions of locality and causality.

But maybe that's a start, I dunno.

 

[PeterDonis]

"Identification of a subensemble" is precisely what I have said is impossible

Not if it just means "picking out the swap and no-swap subsets of the data and doing statistics on each one separately". That's not impossible at all: it's what you do to verify that the swap has an effect.

should be dropped as an argument to explain entanglement swapping using locality.

Statistical interpretations, at least the ones I'm aware of, do not use "identification of a subensemble" to do this. That goes for both "weak" and "strong" ones. "Identification of a subensemble" just means what I said above, and only does what I said above.

I'm not sure exactly what interpretation @Morbert is using. Clarifying that might help.

 

[Lord Jestocost]

... And the Bohr/Copenhagen approach of saying, well, there just isn't anything more we can discover about that, isn't very satisfying either. Which is not to say that things can't possibly end up that way--but if they do, it will be disappointing.

Disappointing for whom? From the viewpoint of an experimental physicist, it’s fascinating that the “Bohr/Copenhagen approach” and its strict minimalist program can “model” experiments like delayed-choice entanglement swapping experiments.

 

[javisot20]

It would be strange if one of the two could prove that he is right with arbitrary precision, we would think that they were not doing QM.

 

[PeterDonis]

Disappointing for whom?

For people who aren't satisfied with just being able to construct models that make accurate predictions--who also want some kind of satisfying story to tell about "what's really happening". Which seems to be a lot of people.

 

[Morbert]

4. We disagree on the interpretation of the projection (ontic vs epistemic), that too.

"Updating the information" is scientifically meaningless. Every experiment requires bringing information together. Hardly a description invoking locality.

"Identification of a subensemble" is precisely what I have said is impossible, and should be dropped as an argument to explain entanglement swapping using locality. We agreed that there is no statistical correlation between the L/R polarization of photon 1 with photon 2's H/V polarization; and the L/R polarization of photon 4 with photon 3's H/V polarization. Consequently, and there can't really be any question about this: you cannot learn anything about photon 1 and photon 2's L/R polarization by measuring the H/V polarizations of photon 2 and photon 3.

I'm not sure exactly what interpretation @Morbert is using. Clarifying that might help.

The relevant statistical interpretation would be the kind professed by Allahverdyan, Balian, and Nieuwenhuizen. See here, here, and the physicsforum thread discussing it here.

The key feature is the splitting of the ensemble into subensembles upon the conclusion of (an infinite number of) experimental runs, where each subensemble is associated with relevant macroscopic outcomes, and not pre-existing microscopic properties. If pre-existing microscopic properties were naively used to predefine subensembles, this would indeed be impossible for reasons @DrChinese mentioned above.

To make it less conceptual and more concrete, instead of subensemble, I could say the identification of a subsample of experimental runs characterised by a 4-fold coincidence condition of 2&3 when the quarter-wave plate is on, and correlations between outcomes of measurements on 1&4 in all three mutually unbiased bases.

But where we really disagree is twofold:

A) A physical change to a setting *here* leads to a change in observed statistics *there*, holding all parameters constant other than the swap/no swap mechanism. Now I know you claim it is a different selection criteria being applied in the Ma experiment (you're wrong though). But the Megadish experiment reports the same result as Ma, but uses a different swap/no swap mechanism that is NOT subject to your critique. (There is no changing of a beam splitter, no change of wave plates, no optical change at all.)

B) It is canonically impossible for there to exist "hidden correlations" that are merely identified via BSMs. We should agree on this at this point - see point 2 above.

A)Whichever way the BSM is performed, what is relevant is the BSM is complementary to the SSM. We cannot simultaneously know the polarization of photons 1&4 (SSM) and their correlations in all three bases (BSM).

B) We don't disagree here. There are no hidden correlations identified via the BSMs. Instead, what the BSMs identify are the sets with expected macroscopic outcomes of tests on 1&4.

 

[PeterDonis]

The relevant statistical interpretation would be the kind professed by Allahverdyan, Balian, and Nieuwenhuizen. See here, here, and the physicsforum thread discussing it here.

Thanks for the references. This is helpful.

 

[PeterDonis]

We cannot simultaneously know the polarization of photons 1&4 (SSM) and their correlations in all three bases (BSM).

We can't simultaneously know their correlations in all three bases whether a swap is performed or not. Polarization measurements in different directions don't commute.

 

[Morbert]

We can't simultaneously know their correlations in all three bases whether a swap is performed or not. Polarization measurements in different directions don't commute.

To better phrase it: What I mean is know that they will be correlated regardless of which of the three bases they are measured in (and both measured in the same basis), as exemplified by Fig 3a in the Ma paper.

 

[DrChinese]

The relevant statistical interpretation would be the kind professed by Allahverdyan, Balian, and Nieuwenhuizen. See here, here, ...

To make it less conceptual and more concrete, instead of subensemble, I could say the identification of a subsample of experimental runs characterised by a 4-fold coincidence condition of 2&3 when the quarter-wave plate is on, and correlations between outcomes of measurements on 1&4 in all three mutually unbiased bases.

OK, there is a notable problem with this reference. Although it does technically address a "statistical" interpretation, it does not mention Entanglement Swapping in either reference. So your attempt to tie it to our discussion is just something you are pulling out of your hat. Nothing from them on the subject at all.

In fact: The first paper (2024) mentions "Entanglement" 1 time, and the second (2015) mentions it... none. Please, be serious. This is precisely the issue I am attempting to address. Specifically: Papers being written and addressing experiments performed last century. Versus papers with up-to-date theory and experiment.

 

[Morbert]

OK, there is a notable problem with this reference. Although it does technically address a "statistical" interpretation, it does not mention Entanglement Swapping in either reference. So your attempt to tie it to our discussion is just something you are pulling out of your hat. Nothing from them on the subject at all.

In fact: The first paper (2024) mentions "Entanglement" 1 time, and the second (2015) mentions it... none. Please, be serious. This is precisely the issue I am attempting to address. Specifically: Papers being written and addressing experiments performed last century.

Yes, the papers discuss the conventional EPRB experiment. The relevant excerpt on page 17 of the 2015 paper

Similar statements are made on page 275 in the 2024 paper.

In the conventional EPRB experiment, we have a pair in a Bell state, and a spin measurement on one particle projects the other onto a spin eigenstate. This is interpreted as the local selection of a subensembled tagged by an outcome of a spin measurement.

In Ma's experiment, we have 1&2 and 3&4 prepared in Bell states, and a BSM on 2&3 projects 1&4 onto a BSM eigenstate. This is interpreted as the local selection of a subensemble tagged by an outcome of a BSM measurement.

Entanglement swapping experiments, while interesting in their own right, don't pose an additional challenge to these minimalist interpretations. Issues you presumably have with such interpretations are just as present in EPRB experiments as they are in entanglement swapping experiments.

 

[Fra]

Sorry for beeing slow on the ball

Assuming you follow the above examples:
i) We know entanglement swaps (via a BSM on P2 & P3) can be executed after P1 and P4 cease to exist. (This is the delayed choice scenario.)
ii) We know entanglement swaps can be performed such that P1 & P4 have never interacted, and have never interacted with any common 3rd system in the past. (This is the strict locality test.)
iii) We know that the entanglement swap fails if there is distinguishability of the P2 & P3 photons. (As demonstrated in various experimental realizations such as THIS).
iv) We know Einsteinian causality should not allow a later choice (to execute a swap or not) to create correlations after the fact.

Something's gotta give! What gives? That's my question.

Why something has to go?

I think at (iv) is where the confusion lies: I would say what happens is that a "later choice" - adds information - that makes it possible to identify the pairs of measurements that are correlated. This is IMO, not a "causation".

I rather think that what is still missing is an understanding the difference between a "correlation" that are hidden to the whole environment except to the involved parts, and a "correlation" that is merely hidden to the "experimenter".

5. There is correlation when swap execution is chosen, no correlation when it is not. If there was always correlation waiting to become "apparent", then there should be no difference in the results for swap versus no-swap.

The scenario of swap has more information injected than the no-swap scenario, so what would it even mean to say that there would be no difference as they are totally different. Looking at ALL the pairs, the percentaged of "random correlations" is the same, where there is a swap or not, right?

The nice part about entanglement swapping is IMO rather that it allows us to "engineer" correlated remote pairs. This is cool and worthy attention, but does not resolve the underlying mystery of quantum entanglement.

But there is.

But only after select another subensemble with ther added information yes; so we are not comparing the same data.

But supposed the experimenters getting the swap data, make sure it's not transmitted at all.
Then, is there a difference?

/Fredrik

 

[DrChinese]

1. Yes, the papers discuss the conventional EPRB experiment. ...

2. In Ma's experiment, we have 1&2 and 3&4 prepared in Bell states, and a BSM on 2&3 projects 1&4 onto a BSM eigenstate. This is interpreted as the local selection of a subensemble tagged by an outcome of a BSM measurement.

3. Entanglement swapping experiments, while interesting in their own right, don't pose an additional challenge to these minimalist interpretations. Issues you presumably have with such interpretations are just as present in EPRB experiments as they are in entanglement swapping experiments.

1. These papers don't really meet the criteria for a suitable reference. One was placed in a French journal, the other in Annals of Physics (low impact score). If their argument makes sense, just make it yourself. And again, the EPR-B argument basically is now 60 years old. Certainly there is something better to quote.


2. I will present a counter-argument to this "interpretation" (which bears no resemblance to swapping) in my next post. Note once again, it is impossible for there to exist such "subensembles" within a larger ensemble that starts as a Ψ-⊗Ψ- Product state.


3. Hand-waving, there is not even an attempt to map their argument to the swapping.

 

[DrChinese]

In Ma's experiment, we have 1&2 and 3&4 prepared in Bell states, and a BSM on 2&3 projects 1&4 onto a BSM eigenstate. This is interpreted as the local selection of a subensemble tagged by an outcome of a BSM measurement.

In this post, I will demonstrate why "something" changes when a swap occurs. It is your assertion that a swap merely provides additional information - and no change to the overall results - allowing a "subensemble" to be selected from a larger sample. OK, let's assume that is true and see where it takes us. We will assume the ideal case, and look at the Quantum Mechanical predictions.


1. We agree that the 4 photon state in initially the Product of 2 entangled states (1&2 and 3&4). That is: |Ψ->12 ⊗ |Ψ->34, same as (1) in the Ma paper. If we denote this on the H/V basis, we get: (|𝐻〉1|𝑉〉2 − |𝑉〉1|𝐻〉2)/√2) ⊗ (|𝐻〉3|𝑉〉4 − |𝑉〉3|𝐻〉4)/√2). That produces the following outcomes per the usual QM, and no others:

a1) |𝐻〉1|𝑉〉2|V〉3|H〉4
b1) |𝐻〉1|𝑉〉2|𝐻〉3|𝑉〉4
c1) |V〉1|H〉2|V〉3|H〉4
d1) |V〉1|H〉2|𝐻〉3|𝑉〉4

Any disagreement so far?


2. If, as you claim, a BSM identifies a subset of these permutations that demonstrates Entangled State statistics, then at first glance: a1) and d1) would be compatible with Φ- Entanglement. That's because photons So that would work.

In reality, an Entanglement Swap can be from our initial state |Ψ->12 ⊗ |Ψ->34 to the final state |Φ->14 ⊗ |Φ->23 (or any of the other 3 Bell states). Agreed?


3. But here is where everything falls apart: This last state |Φ->14 ⊗ |Φ->23 produces the following possible outcomes:

a2) |𝐻〉1|𝑉〉2|V〉3|H〉4
b2) |𝐻〉1|H〉2|𝐻〉3|H〉4
c2) |V〉1|V〉2|V〉3|V〉4
d2) |V〉1|H〉2|𝐻〉3|𝑉〉4

Note that a2) and d2) are the same as a1) and d1), so that is good. But, we have 2 outcome permutations that do NOT match any of the 4 possible outcomes from our initial Product state. The b2) and c2) permutations cannot appear as experimental outcomes IF they subensemble is selected from a1-a4.

However, in actual experiments: The permutations b2) and c2) DO appear, and they appear randomly and as often as the a2) and d2) permutations. That's because the entanglement between photons 1 & 2 no longer exists after a swap (and same for photons 3 & 4).


4. We now have our contradiction: Assuming your premise regarding subensembles, listed are the actual experimental outcomes observed - but some are ones predicted to occur with zero frequency when entanglement is swapped. (Of course they don't appear when there is no swap.) Therefore, the premise must be rejected.

We are basically forced to see that something physically changes after a swap, as outcomes appear that were not part of the initial permutations.

 

[Morbert]

In this post, I will demonstrate why "something" changes when a swap occurs. It is your assertion that a swap merely provides additional information - and no change to the overall results - allowing a "subensemble" to be selected from a larger sample. OK, let's assume that is true and see where it takes us. We will assume the ideal case, and look at the Quantum Mechanical predictions.


1. We agree that the 4 photon state in initially the Product of 2 entangled states (1&2 and 3&4). That is: |Ψ->12 ⊗ |Ψ->34, same as (1) in the Ma paper. If we denote this on the H/V basis, we get: (|𝐻〉1|𝑉〉2 − |𝑉〉1|𝐻〉2)/√2) ⊗ (|𝐻〉3|𝑉〉4 − |𝑉〉3|𝐻〉4)/√2). That produces the following outcomes per the usual QM, and no others:

a1) |𝐻〉1|𝑉〉2|V〉3|H〉4
b1) |𝐻〉1|𝑉〉2|𝐻〉3|𝑉〉4
c1) |V〉1|H〉2|V〉3|H〉4
d1) |V〉1|H〉2|𝐻〉3|𝑉〉4

Any disagreement so far?

There might already be a disagreement. Outcomes need a measurement context. So those are the following outcomes per the usual QM, and no other if all four photons are measured in the H/V basis. If photons 1&4 are measured in the H/V basis and photons 2&3 are measured in the (partial) Bell basis, the possible outcomes are

1) |H〉1|Φ+〉23|H〉4
2) |V〉1|Φ+〉23|V〉4

3) |H〉1|Φ-〉23|H〉4
4) |V〉1|Φ-〉23|V〉4

5) |H〉1|H〉2|V〉3|V〉4

6) |V〉1|𝑉〉2|𝐻〉3|H〉4

Victor would then (via local action) instruct Alice and Bob to only keep outcomes 3) and 4)

So

2. If, as you claim, a BSM identifies a subset of these permutations that demonstrates Entangled State statistics

I don't claim the BSM identifies a subset of those permutations, because they are specific to a measurement context where a BSM is not carried out.

[edit] - Deleted +/- basis part of my response because of a potential pitfall

 

[DrChinese]

1. There might already be a disagreement. Outcomes need a measurement context. So those are the following outcomes per the usual QM, and no other if all four photons are measured in the H/V basis. ...

2. So I don't claim the BSM identifies a subset of those permutations, because they are specific to a measurement context where a BSM is not carried out.

1. Umm, that's exactly what I said. I specified the measurement context: H/V for all 4 photons.

2. The H/V measurement context where a BSM is NOT carried out produces these 4 fold outcomes and no others, per the usual QM (I'm sure you have done the math to verify):

a1) |𝐻〉1|𝑉〉2|V〉3|H〉4
b1) |𝐻〉1|𝑉〉2|𝐻〉3|𝑉〉4
c1) |V〉1|H〉2|V〉3|H〉4
d1) |V〉1|H〉2|𝐻〉3|𝑉〉4

When a BSM is carried out, you have those same permutations to select from for a subset. That's the definition of a subset. OTOH: I say that if a BSM occurs, there appear outcome permutations that are incompatible with the above. Something changes remotely, and this change does not respect Einsteinian locality.

The BSM result that projects photons 2&3 onto the state |Φ-〉 in the spatial modes b and c will identify a subset of 1&4 measurements that exhibit correlations in all three mutually unbiased bases, without the need to insist on nonlocal influence. This is the entanglement swap.

A local interpretation of events is possible if the SSM and BSM choices amount to different selection criteria, as we can say the BSM or SSM themselves are not responsible for any of the correlations seen in 1&4. Instead, the identification of alternative subsets is what is responsible.

This is interpreted as the local selection of a sub ensemble tagged by an outcome of a spin measurement.

[And there are plenty of additional statements of the like. There's only one way to interpret your comments.]

Hmmm, methinks that is what you asserted and have been arguing for all along. It's YOUR explanation of why nothing changes with a BSM. Because you say a BSM gives us "additional" information about certain 4 fold outcomes that presents a subset [of all results] with entangled statistics. But half of those outcomes don't appear in the full dataset (and please, we are talking about the H/V basis in this case). So how can it be a subset?

So either something DID change physically and remotely, as I say (and you almost seem ready to admit) - or nothing changes, and the contradiction with experiment remains. Which is it?

 

[DrChinese]

I would say what happens is that a "later choice" - adds information - that makes it possible to identify the pairs of measurements that are correlated. This is IMO, not a "causation".

The scenario of swap has more information injected than the no-swap scenario... Looking at ALL the pairs, the percentage of "random correlations" is the same, where there is a swap or not, right?

But only after select another subensemble with their added information yes; so we are not comparing the same data.

/Fredrik

Like Morbert and many others: You too are asserting that there is a larger dataset from which a subset is identified by obtaining "more information". If I have a larger dataset U with the members {1, 4, 5, 7, 11, 17}, then the set {4, 6, 8, 17} is not a subset of U. Agreed?

Well, in my post #165 I show that the set of predicted* outcomes E{a2, b2, c2, d2} is not a subset of the only possible outcomes U={a1, b1, c1, d1} because b2 and c2 are not present in U - even though a2 and d2 are.


*Also experimentally observed.

 

[Morbert]

1. Umm, that's exactly what I said. I specified the measurement context: H/V for all 4 photons. exactly as you say.

2. The H/V measurement context where a BSM is NOT carried out produces these 4 fold outcomes and no others, per the usual QM (I'm sure you have done the math to verify):

a1) |𝐻〉1|𝑉〉2|V〉3|H〉4
b1) |𝐻〉1|𝑉〉2|𝐻〉3|𝑉〉4
c1) |V〉1|H〉2|V〉3|H〉4
d1) |V〉1|H〉2|𝐻〉3|𝑉〉4

When a BSM is carried out, you have those same permutations to select from for a subset. That's the definition of a subset. OTOH: I say that if a BSM occurs, there appear outcome permutations that are incompatible with the above. Something changes remotely, and this change does not respect Einsteinian locality.

You don't have the same permutations. You have the ones I outlined:

1) |H〉1|Φ+〉23|H〉4
2) |V〉1|Φ+〉23|V〉4
3) |H〉1|Φ-〉23|H〉4
4) |V〉1|Φ-〉23|V〉4
5) |H〉1|H〉2|V〉3|V〉4
6) |V〉1|𝑉〉2|𝐻〉3|H〉4

The set has 6 possible outcomes, and Victor selects the subset 3) and 4), the subset associated with |Φ-〉23.

So either something DID change physically and remotely, as I say (and you almost seem ready to admit) - or nothing changes, and the contradiction with experiment remains. Which is it?

Different measurement contexts giving rise to different, incompatible sets of possible outcomes implies nonlocal influence only if parameters are added to quantum mechanics to determine the results of individual measurements, without changing the statistical predictions. This is what Asher Peres means when he says "It follows that any classical imitation of quantum mechanics is necessarily nonlocal. However Bell’s theorem does not imply the existence of any nonlocality in quantum theory itself."

 

[DrChinese]

You don't have the same permutations. You have the ones I outlined:

1) |H〉1|Φ+〉23|H〉4
2) |V〉1|Φ+〉23|V〉4
3) |H〉1|Φ-〉23|H〉4
4) |V〉1|Φ-〉23|V〉4
5) |H〉1|H〉2|V〉3|V〉4
6) |V〉1|𝑉〉2|𝐻〉3|H〉4

The set has 6 possible outcomes, and Victor selects the subset 3) and 4), the subset associated with |Φ-〉23.

1. OK, let's agree that the 4 photon state is initially the Product of 2 entangled states (1&2 and 3&4). That is: |Ψ->12 ⊗ |Ψ->34, same as (1) in the Ma paper. If we denote this on the H/V basis, we get: (|𝐻〉1|𝑉〉2 − |𝑉〉1|𝐻〉2)/√2) ⊗ (|𝐻〉3|𝑉〉4 − |𝑉〉3|𝐻〉4)/√2).

This cannot possibly be in question. The INITIAL STATE, BEFORE A BSM.


2. Let's agree that the initial state (|𝐻〉1|𝑉〉2 − |𝑉〉1|𝐻〉2)/√2) ⊗ (|𝐻〉3|𝑉〉4 − |𝑉〉3|𝐻〉4)/√2) expands to the following 4 permutations.

a1) |𝐻〉1|𝑉〉2|V〉3|H〉4
b1) |𝐻〉1|𝑉〉2|𝐻〉3|𝑉〉4
c1) |V〉1|H〉2|V〉3|H〉4
d1) |V〉1|H〉2|𝐻〉3|𝑉〉4

This cannot possibly be in question. The ONLY POSSIBLE OUTCOMES, BEFORE A BSM. And this is true regardless of whether a BSM is performed or not, and regardless of the resulting Bell state if one is performed.


3. Note that this is the universe you assert we have to start with regardless of whether a BSM is performed. Again, this are the ONLY permutations of the initial state. Since you say nothing physically changes with a BSM: ergo only the same 4 possible outcomes (or less) must be present after a BSM when selecting a subensemble.

Of course, they aren't (as I have shown by reference to the predictions of QM). The difference is due to a remote physical change from the a1-d1 permutations to the a2-d2 permutations (2 of those 4 being present in both sets, but 2 are not shared).

---

Please don't veer off from the BSM case we are discussing, which is |Φ-〉 and not |Φ+〉. (You have also inadvertently reversed the initial state for your 5) and 6).)

 

[DrChinese]

1. Different measurement contexts giving rise to different, incompatible sets of possible outcomes...

2. ...implies nonlocal influence only if parameters are added to quantum mechanics to determine the results of individual measurements, without changing the statistical predictions. This is what Asher Peres means when he says "It follows that any classical imitation of quantum mechanics is necessarily nonlocal.

3. However Bell’s theorem does not imply the existence of any nonlocality in quantum theory itself."

1. The measurement context for my proof is H/V. There is nothing to be "different" or "incompatible" as you imply - other than the predictions under your assumption versus predictions of QM.

2. You have slightly mixed up the analogy. The word "only" is not present in the original, and should not be included. There is no requirement that a nonlocal theory must include such parameters.

3. And for the Nth time: Bell's Theorem is not related to Ma's experiment in any way.

 

[DrChinese]

And just to be clear about my proof in post #165: It's conclusion is merely a restatement of what it means to swap entanglement in the |Φ-〉 Bell State Measurement.

Before: 1&2 entangled in polarization state |Ψ->, 3&4 also entangled in the state |Ψ->.
After: 2&3 entangled in polarization state |Φ-〉, 1&4 also entangled in the state |Φ-〉. But 1&2 are no longer entangled on any polarization basis, and 3&4 are also no longer entangled on any basis.

The "Before vs. After" constitutes an objective change of state, and is experimentally verifiable (besides being a prediction of QM). If an outcome is recorded that is not one of the initial 4-fold outcome permutations, then a swap cannot be labeled "the selection of a subset of the initial state" as an explanation of observed correlations.

 

[Morbert]

1. OK, let's agree that the 4 photon state is initially the Product of 2 entangled states (1&2 and 3&4). That is: |Ψ->12 ⊗ |Ψ->34, same as (1) in the Ma paper. If we denote this on the H/V basis, we get: (|𝐻〉1|𝑉〉2 − |𝑉〉1|𝐻〉2)/√2) ⊗ (|𝐻〉3|𝑉〉4 − |𝑉〉3|𝐻〉4)/√2).

This cannot possibly be in question. The INITIAL STATE, BEFORE A BSM.


2. Let's agree that the initial state (|𝐻〉1|𝑉〉2 − |𝑉〉1|𝐻〉2)/√2) ⊗ (|𝐻〉3|𝑉〉4 − |𝑉〉3|𝐻〉4)/√2) expands to the following 4 permutations.

a1) |𝐻〉1|𝑉〉2|V〉3|H〉4
b1) |𝐻〉1|𝑉〉2|𝐻〉3|𝑉〉4
c1) |V〉1|H〉2|V〉3|H〉4
d1) |V〉1|H〉2|𝐻〉3|𝑉〉4

This cannot possibly be in question. The ONLY POSSIBLE OUTCOMES, BEFORE A BSM. And this is true regardless of whether a BSM is performed or not, and regardless of the resulting Bell state if one is performed.

If by "only possible outcomes before a BSM" you mean these are the four possible outcomes of Alice's and Bob's measurements of photons 1&4 in the H/V basis. Sure.

So going back to your original proof.

2. If, as you claim, a BSM identifies a subset of these permutations that demonstrates Entangled State statistics, then at first glance: a1) and d1) would be compatible with Φ- Entanglement. That's because photons So that would work.

Physically, if Victor performs a BSM and selects all runs that yield the outcome Φ-, he will be selecting some runs from a1) and some runs from d1), but not all runs from a1) or from d1). We can split a1) and d1) to reflect this, so that when Victor attempts a BSM, the possible outcomes are

a1.1) |𝐻〉1|Φ-〉23|H〉4
a1.2) |𝐻〉1|Φ+〉23|H〉4
b1) |𝐻〉1|𝑉〉2|𝐻〉3|𝑉〉4
c1) |V〉1|H〉2|V〉3|H〉4
d1.1) |V〉1|Φ-〉23|𝑉〉4
d1.2) |V〉1|Φ+〉23|𝑉〉4

These are the six outcomes I presented in my previous post. So by selecting all runs that yield |Φ-〉23, victor is selecting the subset a1.1) and d1.1)

In reality, an Entanglement Swap can be from our initial state |Ψ->12 ⊗ |Ψ->34 to the final state |Φ->14 ⊗ |Φ->23 (or any of the other 3 Bell states). Agreed?

3. But here is where everything falls apart: This last state |Φ->14 ⊗ |Φ->23 produces the following possible outcomes:

a2) |𝐻〉1|𝑉〉2|V〉3|H〉4
b2) |𝐻〉1|H〉2|𝐻〉3|H〉4
c2) |V〉1|V〉2|V〉3|V〉4
d2) |V〉1|H〉2|𝐻〉3|𝑉〉4

Note that a2) and d2) are the same as a1) and d1), so that is good. But, we have 2 outcome permutations that do NOT match any of the 4 possible outcomes from our initial Product state. The b2) and c2) permutations cannot appear as experimental outcomes IF they subensemble is selected from a1-a4.

However, in actual experiments: The permutations b2) and c2) DO appear, and they appear randomly and as often as the a2) and d2) permutations. That's because the entanglement between photons 1 & 2 no longer exists after a swap (and same for photons 3 & 4).

This is where you lose me. Separable states like |H〉2|𝐻〉3 cannot be outcomes of a successful BSM because a BSM and SSM are complementary. Modelling Victor's choice with a QRNG like (|on〉 + |off〉)/√2, that determines whether the quarter-wave plate is on or off, these are the 10 possible outcomes

a1.1) |on〉|𝐻〉1|Φ-〉23|H〉4
a1.2) |on〉|𝐻〉1|Φ+〉23|H〉4
b1) |on〉|𝐻〉1|𝑉〉2|𝐻〉3|𝑉〉4
c1) |on〉|V〉1|H〉2|V〉3|H〉4
d1.1) |on〉|V〉1|Φ-〉23|𝑉〉4
d1.2) |on〉|V〉1|Φ+〉23|𝑉〉4
e1) |off〉|𝐻〉1|𝑉〉2|V〉3|H〉4
f1) |off〉|𝐻〉1|H〉2|𝐻〉3|H〉4
g1) |off〉|V〉1|V〉2|V〉3|V〉4
h1) |off〉|V〉1|H〉2|𝐻〉3|𝑉〉4

Victor gathers two subsets: The first subset is a1.1) and d1.1), the second subset is e1) and h1). The first subset is a swap, the second subset is not.

Please don't veer off from the BSM case we are discussing, which is |Φ-〉 and not |Φ+〉. (You have also inadvertently reversed the initial state for your 5) and 6).)

Victor cannot guarantee a BSM is successful. Even with the quarter-wave plate on, a SSM might still be performed. Hence 5) and 6), or as they are now labeled: b1) and c1)

[edit] - Renamed QRNG options for clarity

 

[DrChinese]

1. If by "only possible outcomes before a BSM" you mean these are the four possible outcomes of Alice's and Bob's measurements of photons 1&4 in the H/V basis. Sure.

So going back to your original proof.

2. Physically, if Victor performs a BSM and selects all runs that yield the outcome Φ-, he will be selecting some runs from a1) and some runs from d1), but not all runs from a1) or from d1).

3. We can split a1) and d1) to reflect this, so that when Victor attempts a BSM, the possible outcomes are...

4. This is where you lose me. Separable states like |H〉2|𝐻〉3 cannot be outcomes of a successful BSM because a BSM and SSM are complementary. ...


5. Victor cannot guarantee a BSM is successful.

1. and 2. Whew, good. We agree to start with:

a1) |𝐻〉1|𝑉〉2|V〉3|H〉4 [an SSM outcome that could lead to |Φ-〉 if a BSM is performed]
b1) |𝐻〉1|𝑉〉2|𝐻〉3|𝑉〉4 [an SSM outcome that could lead to |Φ+〉 if a BSM is performed]
c1) |V〉1|H〉2|V〉3|H〉4 [an SSM outcome that could lead to |Φ+〉 if a BSM is performed]
d1) |V〉1|H〉2|𝐻〉3|𝑉〉4 [an SSM outcome that could lead to |Φ-〉 if a BSM is performed]

Surely you can see that the a1 and d1 permutations are incompatible with a |Φ+〉 result. There can be no such component of either of the a1 or d1 permutations, since we need the 1 & 4 results to be the same with |Φ-〉 (which is an HH or VV result) if a BSM is performed. However, for purposes of your logic, my distinction doesn't really matter. But forget completely your proposed a1.2 and d1.2 since they can't occur. (Of course, we are talking about the ideal case.)


3. You are correct that there is a "splitting" of a1 into two parts though. But you are incorrect as to what they are, so I eliminated them. When a BSM occurs, there is indistinguishability of the 2 and 3 photons in the detectors. A |Φ-〉 result has the two V detectors firing if a2 occurs (i.e. a BSM). There are 2 ways that can happen and produce an a2 signature:

a2.1) |𝐻〉1|𝑉〉2b"|V〉3c"|H〉4
a2.2) |𝐻〉1|𝑉〉2c"|V〉3b"|H〉4

Similarly for d2:

d2.1) |V〉1|H〉2b"|H〉3c"|V〉4
d2.2) |V〉1|H〉2c"|H〉3b"|V〉4


4. Note that either way: The a2) outcome is still |𝐻〉1|𝑉〉2|V〉3|H〉4. And the d2) outcome is likewise |V〉1|H〉2|𝐻〉3|𝑉〉4. But we do not know which of a2.1 or a2.2 occurred (and similar for d2.1/d2.2). So you are also correct in one sense that "Separable states like |H〉2|𝐻〉3 cannot be outcomes of a successful BSM". I guess we could more correctly describe the a2) state as |𝐻〉1|VV〉23|H〉4. The label really doesn't matter, we agree on the concept.

But regardless, it is derived from |𝐻〉1|𝑉〉2|V〉3|H〉4 and according to your hypothesis, that is the permutation that the a2.1 and a2.2 must belong to. Your hypothesis being that nothing changes physically with a BSM vs an SSM, it's just a different subset.

So... what's missing here? The experiment returns 4 BSM outcomes that are not present in the above lists! They are:

|H〉1|H〉2b"|H〉3c"|H〉4 and |H〉1|H〉2c"|H〉3b"|H〉4 (which are indistinguishable)
|V〉1|V〉2b"|V〉3c"|V〉4 and |V〉1|V〉2c"|V〉3b"|V〉4(which are indistinguishable)

We actually see these outcomes as often as those previously predicted, and these are in fact predicted by QM. But they aren't predicted at all, if we follow your premise. Because now we have HHHH and VVVV outcomes, which don't/can't match any of the initial permutations.

Oops, something physically changed after a BSM. And notice that we don't even care about selection criteria, subsets/subensembles, incompatible bases, Bell, or any of the other things you have brought into this discussion. We simply have impossible outcomes from the initial 4-photon state IF nothing changes with a BSM/swap.


5. Disagree. In the ideal case, Victor chooses SSM (no swap) or BSM (swap) with certainty.

What Victor cannot do: i) steer which Bell state results from a swap; ii) determine which of 4 Bell states occurred for all swaps using current technology.

 

[Morbert]

1. and 2. Whew, good. We agree to start with:

a1) |𝐻〉1|𝑉〉2|V〉3|H〉4 [an SSM outcome that could lead to |Φ-〉 if a BSM is performed]
b1) |𝐻〉1|𝑉〉2|𝐻〉3|𝑉〉4 [an SSM outcome that could lead to |Φ+〉 if a BSM is performed]
c1) |V〉1|H〉2|V〉3|H〉4 [an SSM outcome that could lead to |Φ+〉 if a BSM is performed]
d1) |V〉1|H〉2|𝐻〉3|𝑉〉4 [an SSM outcome that could lead to |Φ-〉 if a BSM is performed]

Surely you can see that the a1 and d1 permutations are incompatible with a |Φ+〉 result. There can be no such component of either of the a1 or d1 permutations, since we need the 1 & 4 results to be the same with |Φ-〉 (which is an HH or VV result) if a BSM is performed. However, for purposes of your logic, my distinction doesn't really matter. But forget completely your proposed a1.2 and d1.2 since they can't occur. (Of course, we are talking about the ideal case.)

3. You are correct that there is a "splitting" of a1 into two parts though. But you are incorrect as to what they are, so I eliminated them. When a BSM occurs, there is indistinguishability of the 2 and 3 photons in the detectors. A |Φ-〉 result has the two V detectors firing if a2 occurs (i.e. a BSM). There are 2 ways that can happen and produce an a2 signature:

a2.1) |𝐻〉1|𝑉〉2b"|V〉3c"|H〉4
a2.2) |𝐻〉1|𝑉〉2c"|V〉3b"|H〉4

Similarly for d2:

d2.1) |V〉1|H〉2b"|H〉3c"|V〉4
d2.2) |V〉1|H〉2c"|H〉3b"|V〉4

I have been discussing Ma's setup specifically, and I don't see how any of your 2. and 3. follows. A |Φ-〉 result can have two H detectors firing even if Alice and Bob both got the outcome H. Similarly, it can have two V detectors firing even if d2 occurs. You are assuming that the photon polarization in the b" and c" modes must be the same as in the b and c modes even when the quarter-wave plate is on. This is not the case. When the quarter-wave plate is on, two detectors firing in the b" and c" modes with the same polarization (say, both V) only lets us infer |Φ-〉23, i.e. that the run belongs to either a1) or d1). Similarly, Ma says

Victor's detector coincidences with one horizontal and one vertical photon in spatial modes b’’ and c’’ indicate the states |𝐻𝑉〉 23 and |𝑉𝐻〉 23, which are always discarded because they are separable states independent of Victor’s choice and measurement

Hence my splitting.

This is an important disagreement over a technical aspect of Ma's apparatus. It is pointless to address 4. Until this is resolved.

 

[DrChinese]

1. I have been discussing Ma's setup specifically, and I don't see how any of your 2. and 3. follows. A |Φ-〉 result can have two H detectors firing even if Alice and Bob both got the outcome H. Similarly, it can have two V detectors firing even if d2 occurs.

2. You are assuming that the photon polarization in the b" and c" modes must be the same as in the b and c modes even when the quarter-wave plate is on. This is not the case. When the quarter-wave plate is on, two detectors firing in the b" and c" modes with the same polarization (say, both V) only lets us infer |Φ-〉23, i.e. that the run belongs to either a1) or d1).

3. Similarly, Ma says Hence my splitting.

4. This is an important disagreement over a technical aspect of Ma's apparatus. It is pointless to address 4. Until this is resolved.

1. Yes, of course a swap leading to a "|Φ-〉 result can have two H detectors firing even if Alice and Bob both got the outcome H". You'd get HHHH, exactly as you say. And yes, that same swap result |Φ-〉 can even feature VVVV. I'm not saying anything otherwise. We completely agree, this is in line with theory and experiment.

But... that still requires that the initial state had 2 V's and 2 H's. That's true of each and every initial state when we start out with |Ψ->12 ⊗ |Ψ->34 (as presented in Ma's (1), and you have already agreed to). No exceptions. So you now must agree that the BSM physically changes an initial state like HVHV (a viable initial state) to VVVV (a viable observation after a swap) occasionally. That's a change you are denying can occur - because that would indicate the very "nonlocal influence" that you argue against.


2. No, you are assuming that. I only assumed that in post #165 to demonstrate that it leads to this precise contradiction.


3. You have again starting talking about something other than |Φ-〉. The |Φ-〉 Bell state has VV or HH in separate sides (i.e. b" and c') while your out of context quote has it as VH or HV in b" and c". These have no connection to our discussion, which we decided to limit to |Φ-〉 to keep it simple - and to map it to the Ma paper, which only considers that swapped Bell state.


4. A BSM changes the initial state from 1&2 entanglement to 1&4 entanglement, which means that photon 2 need no longer maintain its prior relationship to photon 1. Accordingly, it can - and must change - because it is now entangled with photon 3 in the same Bell state as 1 & 4 are in.

There is no disagreement on any technical point. You are both agreeing (see 1.) and denying the same relevant issues. You cannot have an outcome of HHHH or VVVV when you start with two pairs in the |Ψ-> state WITHOUT A PHYSICAL CHANGE DUE TO A SWAP. There is no subset of outcomes leading to that permutation otherwise.

 

[Morbert]

1. Yes, of course a swap leading to a "|Φ-〉 result can have two H detectors firing even if Alice and Bob both got the outcome H". You'd get HHHH, exactly as you say. And yes, that same swap result |Φ-〉 can even feature VVVV. I'm not saying anything otherwise. We completely agree, this is in line with theory and experiment.

But... that still requires that the initial state had 2 V's and 2 H's. That's true of each and every initial state when we start out with |Ψ->12 ⊗ |Ψ->34 (as presented in Ma's (1), and you have already agreed to). No exceptions. So you now must agree that the BSM physically changes an initial state like HVHV (a viable initial state) to VVVV (a viable observation after a swap) occasionally. That's a change you are denying can occur - because that would indicate the very "nonlocal influence" that you argue against.

The state is first altered by the local unitary dynamics of Victor's apparatus. After the unitary evolution, and after detector responses are registered, the run is kept if a suitable outcome is registered, and hence entanglement swapping occurs. All of this can be described with local actions.

2. No, you are assuming that. I only assumed that in post #165 to demonstrate that it leads to this precise contradiction.

This discussion is becoming unfocused and is starting to repeat itself. You keep accusing me of making assumptions I do not make. So all I can do is leave an account of Ma's experiment motivated by the Allahverdyan et al papers I linked to earlier. This will be my last word unless a truly substantive critique is given.

---

Measurements on a quantum system are modeled as a joint process between the measured system and the measuring apparatus. In this delayed entanglement swapping experiment, we have a 4-photon system and three measurement apparatuses belonging to Alice, Bob, and Victor that each carry out a destructive measurement. The initial state describing the 4-photon system $\hat{r}(0) = \ket{\psi^-}_{12}\ket{\psi^-}_{34}\bra{\psi^-}_{12}\bra{\psi^-}_{34}$ along with Alice's Bob's and Victor's apparatuses is
$$\hat{\mathcal{D}}(0) = \hat{\mathcal{A}}(0)\otimes\hat{\mathcal{B}}(0)\otimes\hat{\mathcal{V}}(0)\otimes\hat{r}(0)$$We evolve the state via the appropriate dynamics to some time $t_f$ after all measurements, and we decompose the state to a sum of pointer states registering the measurement outcomes of Alice, Bob, and Victor.
$$\hat{\mathcal{D}}(t_f) = \sum p_i \hat{\mathcal{A}}_i\otimes\hat{\mathcal{B}}_i\otimes\hat{\mathcal{V}}_i$$As this is a destructive measurement, these pointer states are for the full Hilbert space. If Alice and Bob both measure polarization in the H/V basis, and Victor uses a QRNG to decide whether the quarter-wave plate is on or off, we can fully model the runs with 10 joint pointer states representing the 10 possible outcomes
\begin{eqnarray}
\hat{\mathcal{A}}_H&\otimes&\hat{\mathcal{B}}_H&\otimes&\hat{\mathcal{V}}_{\mathrm{on},\phi^+}\\
\hat{\mathcal{A}}_V&\otimes&\hat{\mathcal{B}}_V&\otimes&\hat{\mathcal{V}}_{\mathrm{on},\phi^+}\\
\hat{\mathcal{A}}_H&\otimes&\hat{\mathcal{B}}_H&\otimes&\hat{\mathcal{V}}_{\mathrm{on},\phi^-}\\
\hat{\mathcal{A}}_V&\otimes&\hat{\mathcal{B}}_V&\otimes&\hat{\mathcal{V}}_{\mathrm{on},\phi^-}\\
\hat{\mathcal{A}}_V&\otimes&\hat{\mathcal{B}}_H&\otimes&\hat{\mathcal{V}}_{\mathrm{on},HV}\\
\hat{\mathcal{A}}_H&\otimes&\hat{\mathcal{B}}_V&\otimes&\hat{\mathcal{V}}_{\mathrm{on},VH}\\
\hat{\mathcal{A}}_V&\otimes&\hat{\mathcal{B}}_V&\otimes&\hat{\mathcal{V}}_{\mathrm{off},HH}\\
\hat{\mathcal{A}}_H&\otimes&\hat{\mathcal{B}}_H&\otimes&\hat{\mathcal{V}}_{\mathrm{off},VV}\\
\hat{\mathcal{A}}_V&\otimes&\hat{\mathcal{B}}_H&\otimes&\hat{\mathcal{V}}_{\mathrm{off},HV}\\
\hat{\mathcal{A}}_H&\otimes&\hat{\mathcal{B}}_V&\otimes&\hat{\mathcal{V}}_{\mathrm{off},VH}\\
\end{eqnarray}
All experimental runs will register one of these outcomes. Victor tags all runs that register outcomes (3) and (4), and places them in one set. He also tags all runs that register outcomes (7) and (8) and places them in another set. Both sets exhibit correlation between Alice's and Bob's pointer states, even though only the first set constitutes entanglement swapping.

If Alice and Bob instead measure polarization in the +/- basis, then we can fully partition the runs into 16 possible outcomes
\begin{eqnarray}
\hat{\mathcal{A}}_+&\otimes&\hat{\mathcal{B}}_+&\otimes&\hat{\mathcal{V}}_{\mathrm{on},\phi^+}\\
\hat{\mathcal{A}}_-&\otimes&\hat{\mathcal{B}}_-&\otimes&\hat{\mathcal{V}}_{\mathrm{on},\phi^+}\\
\hat{\mathcal{A}}_+&\otimes&\hat{\mathcal{B}}_-&\otimes&\hat{\mathcal{V}}_{\mathrm{on},\phi^-}\\
\hat{\mathcal{A}}_-&\otimes&\hat{\mathcal{B}}_+&\otimes&\hat{\mathcal{V}}_{\mathrm{on},\phi^-}\\
\hat{\mathcal{A}}_+&\otimes&\hat{\mathcal{B}}_+&\otimes&\hat{\mathcal{V}}_{\mathrm{on},HV\lor VH}\\
\hat{\mathcal{A}}_-&\otimes&\hat{\mathcal{B}}_-&\otimes&\hat{\mathcal{V}}_{\mathrm{on},HV\lor VH}\\
\hat{\mathcal{A}}_+&\otimes&\hat{\mathcal{B}}_-&\otimes&\hat{\mathcal{V}}_{\mathrm{on},HV\lor VH}\\
\hat{\mathcal{A}}_-&\otimes&\hat{\mathcal{B}}_+&\otimes&\hat{\mathcal{V}}_{\mathrm{on},HV\lor VH}\\
\hat{\mathcal{A}}_+&\otimes&\hat{\mathcal{B}}_+&\otimes&\hat{\mathcal{V}}_{\mathrm{off},HH\lor VV}\\
\hat{\mathcal{A}}_-&\otimes&\hat{\mathcal{B}}_-&\otimes&\hat{\mathcal{V}}_{\mathrm{off},HH\lor VV}\\
\hat{\mathcal{A}}_+&\otimes&\hat{\mathcal{B}}_-&\otimes&\hat{\mathcal{V}}_{\mathrm{off},HH\lor VV}\\
\hat{\mathcal{A}}_-&\otimes&\hat{\mathcal{B}}_+&\otimes&\hat{\mathcal{V}}_{\mathrm{off},HH\lor VV}\\
\hat{\mathcal{A}}_+&\otimes&\hat{\mathcal{B}}_+&\otimes&\hat{\mathcal{V}}_{\mathrm{off},HV\lor VH}\\
\hat{\mathcal{A}}_-&\otimes&\hat{\mathcal{B}}_-&\otimes&\hat{\mathcal{V}}_{\mathrm{off},HV\lor VH}\\
\hat{\mathcal{A}}_+&\otimes&\hat{\mathcal{B}}_-&\otimes&\hat{\mathcal{V}}_{\mathrm{off},HV\lor VH}\\
\hat{\mathcal{A}}_-&\otimes&\hat{\mathcal{B}}_+&\otimes&\hat{\mathcal{V}}_{\mathrm{off},HV\lor VH}\\
\end{eqnarray}
Victor then tags all runs that register outcomes (13) and (14) and places them in one set, and tags all runs that register outcomes (19)-(22) and places them in a second set. This time the first set will exhibit anticorrelation between Alice's and Bob's measurement, while the second set will not exhibit any correlation.

This entanglement swapping is accounted for by local action: Victor tagging a subset of all experiment runs conditions on a successful BSM and a particular outcome, and discarding the rest.

 

[PeterDonis]

The state is first altered by the local unitary dynamics of Victor's apparatus.

But the state itself is inherently nonlocal, because it includes photons 1 & 4, which are spatially separated from Victor's apparatus. (Indeed, they could have already been measured before Victor does whatever operation he does with his apparatus).

All of this can be described with local actions.

Only if you adopt a definition of "local action" that includes actions that affect a state that is inherently nonlocal for the reason given above.

Note that this is not something that can be resolved by argument; it's a matter of personal opinion whether you think that such a definition of "local action" is acceptable. You think it is; @DrChinese thinks it isn't. I personally don't like it much either. But there's no way to resolve this disagreement, so continuing to discuss it is pointless. The best we can do is to define where the disagreement lies.

This discussion is becoming unfocused and is starting to repeat itself.

Yes, and I think that's because we are reaching the point where the only remaining disagreement is about matters of opinion regarding interpretations, which can't be resolved here.

 

[PeterDonis]

There is no disagreement on any technical point.

This does seem to be the case. And it means that the discussion has run its course, since disagreements over interpretation cannot be resolved here. They can only be described. It looks as though this disagreement has been sufficiently described, so I am closing this thread. Thanks to all who participated.