A Interpreting photon correlations from independent sources

[DrChinese]

The signatures are the same, but our inference changes depending on what measurement is carried out.

1. When a BSM is carried out, then

(HH b"c") implies |Φ-⟩
(VV b"c") implies |Φ-⟩

2. but when an SSM is carried out

(HH b"c") implies |HH⟩ = (|Φ+⟩ + |Φ-⟩)/√2
(VV b"c") implies |VV⟩ = (|Φ+⟩ - |Φ-⟩)/√2

You can see that in one case we are discarding all runs that aren't |Φ-⟩, but in the other case we are instead discarding all runs that aren't (|Φ+⟩ + |Φ-⟩)/√2 or (|Φ+⟩ - |Φ-⟩)/√2. See how |Φ+⟩ gets mixed in to the SSM runs but not the BSM runs?


3. Counterfactual definiteness is unfortunately always a concern whenever we are considering scenarios where a choice has to be made between complementary measurements. In this case, if we perform an SSM and a run is kept, we cannot say that if we had instead done a BSM the run would still have been kept.

Yikes, interference in Product State statistics? In polarization outcomes on distinguishable photons in a beam splitter?


1. This is correct.


2. This is absolutely not correct. An HH (or VV) outcome from measurements on 2 photons that are distinguishable and have never interacted are NEVER a sum (or difference) of 2 entangled states. In fact, your assertion implies that all photons everywhere are in one or the other of a maximally entangled state - a result that violates MoE.

And besides, makes no sense at all. Again, these photons have not interacted so interference is obviously not a factor. And as I have already documented, a beam splitter will not change a V to H or vice versa anyway - entangled or not.

And because these photons are distinguishable in the SSM mode: They are in a Product state, so you cannot use terms like "(|Φ+⟩ + |Φ-⟩)/√2" to describe them. This is basic, nothing really to debate here.

The statement that would be correct is as follows: (HH b"c") implies we have an equal superposition of circular polarization Product states with |LL>, |LR>, |RL>, |RR> outcomes. Etc.

Hopefully, you will rethink this and retract without argument.


3. As I said, Counterfactual Definiteness (CD) is not an assumption in this experiment - any more than it is in any scientific experiment anywhere at any time. This is hand-waving at its worst. I can't stop you from believing it yourself, but I will challenge you to back it up for experiments of this type.

We are physically testing an A/B setup with a single independent variable. Flip a switch, and the results change. If I do this experiment with a classical setup, there is no scientific difference versus doing it with a quantum setup. I set my criteria for what I am testing, and record the results. With the switch is on, the results are different than when the switch if off. CD does not matter for scientific experiments of this type.

 

[Morbert]

2. This is absolutely not correct. An HH (or VV) outcome from measurements on 2 photons that are distinguishable and have never interacted are NEVER a sum (or difference) of 2 entangled states. In fact, your assertion implies that all photons everywhere are in one or the other of a maximally entangled state - a result that violates MoE.

And besides, makes no sense at all. Again, these photons have not interacted so interference is obviously not a factor. And as I have already documented, a beam splitter will not change a V to H or vice versa anyway - entangled or not.

And because these photons are distinguishable in the SSM mode: They are in a Product state, so you cannot use terms like "(|Φ+⟩ + |Φ-⟩)/√2" to describe them. This is basic, nothing really to debate here.

The statement that would be correct is as follows: (HH b"c") implies we have an equal superposition of circular polarization Product states with |LL>, |LR>, |RL>, |RR> outcomes. Etc.

Hopefully, you will rethink this and retract without argument.

We can expand the separable state |HH⟩ in whichever basis we like. E.g. In equation 2, Ma writes the initial state in a 23 Bell basis. It's a simple homework exercise to show that if |Φ+⟩ = (|HH⟩ + |VV⟩)/√2 and |Φ-⟩ = (|HH⟩ - |VV⟩)/√2 then by rearranging we have |HH⟩ = (|Φ+⟩ + |Φ-⟩)/√2.

The mistake you are making is interpreting this expansion as a claim that there is some secret entanglement between the photons. This is not true. The state is separable whether it is written as |HH⟩ or as (|Φ+⟩ + |Φ-⟩)/√2.

Instead, what this expansion shows is that the separable state |HH⟩ is not an eigenstate of the BSM, and |Φ-⟩ is not an eigenstate of the SSM. So if a run with an SSM produces the result |HH⟩, we cannot say this run would have produced the result |Φ-⟩ if a BSM had been done instead, and hence we cannot say the run would have been kept, as the run might have instead yielded |Φ+⟩ and been discarded.

3. As I said, Counterfactual Definiteness (CD) is not an assumption in this experiment - any more than it is in any scientific experiment anywhere at any time. This is hand-waving at its worst. I can't stop you from believing it yourself, but I will challenge you to back it up for experiments of this type.

We are physically testing an A/B setup with a single independent variable. Flip a switch, and the results change. If I do this experiment with a classical setup, there is no scientific difference versus doing it with a quantum setup. I set my criteria for what I am testing, and record the results. With the switch is on, the results are different than when the switch if off. CD does not matter for scientific experiments of this type.

The counterfactual indefiniteness of QM not handwaving. It's precise and well documented, and places a bound on what we can say about measurements that are not performed. If we perform an SSM and get the result |HH⟩, we cannot conclude the same run would have produced a |Φ-⟩ result if a BSM had been done instead. The conclusion of nonlocal influence assumes we can. It assumes the same runs would have been kept, and instead the 1&4 results would be influenced.

 

[javisot20]

2 photons that are distinguishable and have never interacted are NEVER a sum (or difference) of 2 entangled states. In fact, your assertion implies that all photons everywhere are in one or the other of a maximally entangled state - a result that violates MoE.

Does this imply that all the photons that have never interacted in any sense have exactly the same zero correlation?

In that group of "photons that never interacted in any sense", aren't there 2 that show minimal correlation in such a way that we can translate the notions of one to the other?

 

[PeterDonis]

We can expand the separable state |HH⟩ in whichever basis we like.

Sure, but that doesn't change the fact that it's a separable state and is not entangled. Nor does it change the physics of what happens when a pair of photons goes through a beam splitter. Your argument appears to hinge on denying those two obvious facts.

 

[Morbert]

Sure, but that doesn't change the fact that it's a separable state and is not entangled. Nor does it change the physics of what happens when a pair of photons goes through a beam splitter. Your argument appears to hinge on denying those two obvious facts.

My argument is compatible with these facts.

The mistake you are making is interpreting this expansion as a claim that there is some secret entanglement between the photons. This is not true. The state is separable whether it is written as |HH⟩ or as (|Φ+⟩ + |Φ-⟩)/√2.

Instead, what this expansion shows is that the separable state |HH⟩ is not an eigenstate of the BSM, and |Φ-⟩ is not an eigenstate of the SSM. So if a run with an SSM produces the result |HH⟩, we cannot say this run would have produced the result |Φ-⟩ if a BSM had been done instead, and hence we cannot say the run would have been kept, as the run might have instead yielded |Φ+⟩ and been discarded.

 

[DrChinese]

1. It's a simple homework exercise to show that:

If |Φ+⟩ = (|HH⟩ + |VV⟩)/√2 and |Φ-⟩ = (|HH⟩ - |VV⟩)/√2
Then ... |HH⟩ = (|Φ+⟩ + |Φ-⟩)/√2 and .

2. The counterfactual indefiniteness of QM not handwaving. It's precise and well documented, and places a bound on what we can say about measurements that are not performed.

3. If we perform an SSM and get the result |HH⟩, we cannot conclude the same run would have produced a |Φ-⟩ result if a BSM had been done instead.

1. That's a completely different IF/THEN statement than saying:

If |HH⟩ = (|Φ+⟩ + |Φ-⟩)/√2 and |VV⟩ = (|Φ+⟩ + |Φ-⟩)/√2
Then |Φ+⟩ = (|HH⟩ + |VV⟩)/√2 and |Φ-⟩ = (|HH⟩ - |VV⟩)/√2

Because the first statement is simply not true. An |HH> result is a Product state outcome. It is one of 4 possible outcomes when there is no swap, and does not indicate that it is the sum of 2 Entangled states.

Please read my next post for additional discussion of your idea.


2. Great. Where is such documentation? Because apparently a Nobel winner for his work in entanglement documented 2 other assumptions in the Ma paper, but completely missed yours.


3. This is not a statement being asserted. It doesn't need to be. All that needs to happen is that the stats change. They do.

 

[DrChinese]

@Morbert

I have been trying to understand where you are getting your ideas about the |HH⟩ = (|Φ+⟩ + |Φ-⟩)/√2 from. I think I can identify now and describe our point of departure.

Let's consider the Ma experiment specifically, and not the Megadish experiment for the time being. They both demonstrate essentially the same idea, and come to the exact same conclusion. And both of them agree with the predictions of QM. But there is a nuance of difference between their implementations that may be leading you astray. (Of course I say it's you, not me. )

In the Ma experiment, the "switch" being flipped (swap vs no swap) is a beam splitter (BS) which alternates between normal BS operation and full reflective operation (like a mirror). Normal BS being 50:50, the other being 0:100 (transmit:reflect).

Suppose we run 100 iterations and record 4 fold coincidences with the swap switch ON. In the ideal case, we'd expect 25 |Φ-⟩ outcomes (signatures being |HH⟩ and |VV⟩ for photons 2 and 3). After all, there are 4 Bell states and they occur randomly and equally often. I think you agree with this.

Suppose we run 100 iterations and record 4 fold coincidences with the swap switch OFF. In the ideal case, we'd expect 25 |HH⟩ outcomes (photons 2 and 3). After all, there are 4 Product state permutations of H and V (2 * 2) and they occur randomly and equally often. Similarly, there would be 25 |VV⟩ outcomes too. I think you agree with this.

Now, for the Ma sorting purposes, they would combine the |HH⟩ and |VV⟩ outcomes together for their reporting. That would be 50 results for swap=OFF, compared to only 25 results for swap=ON. Whoa, what gives? Those bad experimenters are trying to trick us somehow! Clearly, they are combining result subsets for the swap=OFF that have the effect of canceling out (and therefore hiding) correlations that would blow the entire experiment.

Of course I am kidding with that last bit. Obviously, no one pulling anything over on anybody. This situation is strictly an artifact of the specific method by which the swap/no swap switch operates in the Ma experiment. As long as we use the same criteria for comparison (3a vs 3b), all is good and the scientific method is preserved. And in fact, the Ma paper points this exact situation out explicitly:

Ma et al, page 5, text below (3): "Note that the reason why we use one specific entangled state [|Φ-⟩] but both separable states [|HH⟩ and |VV⟩] to compute the correlation function is that the measurement solely depends on the settings of the EOMs in the BiSA."

---

And if this somehow still bothers you, consider this point. The Megadish paper uses a different technique to turn the swap switch ON/OFF. Their technique restores the balance so that the same number of 4 fold coincidences would appear in the swap=ON group as is included in the swap=OFF group. The reason: Megadish has the 2 and 3 photons going through the exact same beam splitter regardless of the swap setting. The swap=ON setting has them physically overlapping in the beam splitter. In the swap=OFF setting: "we introduced a sufficient temporal delay between the two projected photons ... and the first and last photons [1 & 4] do not become quantum entangled but classically correlated [i.e. no swap]".

As in our earlier example, we'd see 25 out of 100 (swap=ON) just as before, the signature being in terms of the Ma experiment: (HH b"c") or (VV b"c"). But for swap=OFF, we'd now see 25 out of 100 (instead of 50 of 100 earlier), the signature also being: (HH b"c") or (VV b"c"). Where did the other 25 go? Those are discarded, because you end up with permutations like (HV b"b") or (VH b"b") or (HV c"c") or (VH c"c"). Those 25 must be discarded because they don't match the criteria.

Saying it a different way: The Ma and Megadish experiments use a different technique to achieve identical results. All good science, all properly documented and executed.

 

[PeterDonis]

My argument is compatible with these facts.

I don't see how. Your argument relies on interpreting the HH and VV states as though they were, physically, linear combinations of Bell states and therefore somehow contained some sort of entanglement. That's simply wrong. HH and VV are separable states. That's the physical fact. Your argument does not appear to me to be treating them that way.

 

[DrChinese]

All: I think this deep dive into the statistics missed numerous important points that are demonstrated in the Ma and Megadish papers. These specific points are best seen by a closer analysis of the Ma figure 3a/3b graph. It tells us a lot more than may be obvious.


1. Let's remember a couple of key points about ANY pair of entangled photons. There is absolutely NO correlation of any kind between the outcome of an H/V measurement on one (Alice, 1) versus L/R or +/- measurements on the other (Bob, 4). These are mutually unbiased bases, and any correlation would violate the Uncertainty principle. Because there is no such correlation, it is impossible to select* any subset of entangled 1 & 2 pairs and 3 & 4 pairs that would produce correlation between 1 & 4 - unless you select on the SAME basis.

In other words: if you test 2 & 3 on the H/V basis, you can never find a subset in which the 1 & 4 correlate on the L/R basis UNLESS a swap is executed. So all this talk about whether such and such is included or excluded is completely irrelevant. You can't find such a subset because it won't exist unless a swap is physically execute. And the Ma graph shows exactly this: correlation when there is a swap (3a, middle and right), no correlation without a swap (3b, middle and right).

The concept that there is hidden Entangled State statistics between un-swapped 1&2 pairs and 3&4 pairs is impossible. That's why the 3b middle/right graphs show no correlation. As predicted by QM.


2. And what about the equal correlation shown on the 3a and 3b left graph? What does this tell us?

This too is exactly as predicted, because the bases for all 4 photons being measured is the same. This shows that the correlation does exist when it should - which is merely coincidental. You would see the same correlation with any 2 entangled pair sets (without a swap) anywhere and anytime, regardless of experimental setting.

---

There is no selection method for cherry picking* 1 & 4 results that will show the correlations exactly per the Ma paper unless a swap is physically** executed. That swap can be performed at any time relative to the 1 & 4 measurements, and at any place relative to those measurements, regardless of the normal constraints of Einsteinian causality and locality.


*And of course I mean by some specific criteria, and not by hand. But I really didn't need to say this, did I?

**Not virtual. Physical overlap is a requirement, and it must be done so the detections are indistinguishable between 2 & 3.

 

[Morbert]

I don't see how.

See my quote in post #95.

What my argument relies on is the |HH⟩ and |VV⟩ not being eigenstates of the BSM, as seen by the Bell basis expansion, and hence the selection of runs conditioned on the coincidence of same polarization and different spatial modes amounts to a different selection procedure depending on whether a BSM or an SSM is carried out.

 

[PeterDonis]

What my argument relies on is the |HH⟩ and |VV⟩ not being eigenstates of the BSM

That claim is incorrect. "The BSM", as far as an actual measurement is concerned, is not what you appear to think it is.

There is a beam splitter that photons 2 & 3 both pass through, at which a swap might or might not occur, depending on the experimenter's choice. But no measurement takes place at that beam splitter; the beam splitter just realizes one of two unitary transformations on the two-photon input state, either the identity (no swap) or the swap operation that puts photons 2 & 3 into one of four possible Bell states. The two-photon output state of the beam splitter is the result of that unitary operation.

Then, after that beam splitter, there are two output channels, in each of which a polarization measurement is done in the H-V basis. That is the same whether a swap takes place at the beam splitter or not. So $\ket{HH}$ and $\ket{VV}$ are eigenstates of the measurement that is done on photons 2 & 3, whether a swap takes place at the beam splitter or not.

 

[Morbert]

So $\ket{HH}$ and $\ket{VV}$ are eigenstates of the measurement that is done on photons 2 & 3, whether a swap takes place at the beam splitter or not.

The signatures of interest correspond to $\ket{HH}_{b''c''}$ and $\ket{VV}_{b''c''}$ whether or not a swap takes place, but Ma is referring to $\ket{HH}_{bc}$ and $\ket{VV}_{bc}$ as two basis vectors of the SSM. Note the different spatial modes. This is what makes the BSM complementary to the SSM, as $\ket{\Phi^-}_{bc} = \frac{1}{\sqrt{2}}(\ket{HH}_{bc} - \ket{VV}_{bc})$

If we detect a coincidence in different spatial modes b’’ and c’’ but with same polarization in the |𝐻〉/|𝑉〉
basis, photons 2 and 3 are projected onto |Φ − 〉 in spatial modes b and c (up to a global phase)

 

[Morbert]

@DrChinese As the holidays are now over for me I will have to reduce my posting frequency but over the next couple days I will read your post and formulate a reply. I suspect this statement by Ma will be relevant but I will know for sure in my response.

When Victor performs a BSM, photons 1 and 4 are only entangled if there exists the information necessary for Victor to specify into which subensembles the data are to be sorted. In our case the subensembles correspond to |Φ − 〉 23 or |Φ + 〉 23. Without the ability for this specification, he would have to assign a mixture of these two Bell states to his output state which is separable, and thus he could not correctly sort Alice's and Bob's data into subensembles.

 

[DrChinese]

What my argument relies on is the |HH⟩ and |VV⟩ not being eigenstates of the BSM, as seen by the Bell basis expansion, and hence the selection of runs conditioned on the coincidence of same polarization and different spatial modes amounts to a different selection procedure depending on whether a BSM or an SSM is carried out.

This entire paragraph is mixing phrases and saying nothing. Really: spatial modes, eigenstates? I don’t mean to be flippant: but you are missing the forest for the trees.

The ONlY difference between the swap and no swap is physical overlap in a beam splitter. You need to explain how something you deny the existence of (remote swaps) magically allows the selection of correlated 1 and 4 pairs on the L/R basis.

Because there is no way to do that otherwise using SSM. Please: explain how the BSM “cheating“ occurs when an L/R basis outcome is canonically unrelated to an H/V measurement.

And for the Nth time: Where is the slightest theoretical support for what you assert in opposition to peer reviewed published papers by top researchers?

 

[javisot20]

I ask to understand better, if "swap yes" or "swap no" affects the results, doesn't that mean that the swap is also a form of measurement that breaks the entanglement between (1&4)?

Measuring and defining a state breaks the entanglement between 2 particles. In the case of 4 particles (pairwise entanglement in principle) and swap, if there is swap an apparent entanglement is also broken, or am I misunderstanding.

 

[DrChinese]

@DrChinese As the holidays are now over for me I will have to reduce my posting frequency but over the next couple days I will read your post and formulate a reply. I suspect this statement by Ma will be relevant but I will know for sure in my response.

No problem on timing, take all the time you need

But please don’t over focus on the Ma quote. Don’t forget that the Megadish paper demonstrates the same results. But it’s mechanism for stopping a swap is such that the issues you have been raising are not a factor.

 

[DrChinese]

I ask to understand better, if "swap yes" or "swap no" affects the results, doesn't that mean that the swap is also a form of measurement that breaks the entanglement between (1&4)?

The swap doesn’t break the entanglement, it creates the entanglement between 1 & 4. The swamp breaks the entanglement between 1 & 2 though (and between 3 & 4)

 

[javisot20]

The swamp breaks the entanglement between 1 & 2 though (and between 3 & 4)

It breaks them to entangled 1&4 and 2&3 (not mutually), right?

 

[gentzen]

All: I think this deep dive into the statistics missed numerous important points ...

4). These are mutually unbiased bases, and any correlation would violate the Uncertainty principle. Because there is no such correlation, it is impossible to select* any subset of entangled 1 & 2 pairs and 3 & 4 pairs that would produce correlation between 1 & 4 - unless you select on the SAME basis.

*And of course I mean by some specific criteria, and not by hand. But I really didn't need to say this, did I?

I a post started with "All:"? Of course you need to say this! I would even go further: since the meaning of "by hand" is not obvious, you should specify a spacetime point (relative to each individual run) where the selection of the subsets is possible if a swap was performed, but impossible if not.
Alternatively, you could specify explicitly which information is allowed to be used for the selection of the subsets.

 

[DrChinese]

I a post started with "All:"? Of course you need to say this! I would even go further: since the meaning of "by hand" is not obvious, you should specify a spacetime point (relative to each individual run) where the selection of the subsets is possible if a swap was performed, but impossible if not.
Alternatively, you could specify explicitly which information is allowed to be used for the selection of the subsets.

What I am saying is simple: there is no such criteria. Suppose you execute/record a 4 hour run of 4 fold coincidences without executing a swap and look for a pattern in the data that indicates correlations. You won’t find one. It’s canonically impossible.

Do the same with physical overlap of the 2 & 3 photons, and voila: a pattern jumps out that did not appear previously. This is predicted by QM.

If there was correlation hidden in the data: why is it predicted not to exist when there is no swap? And why does it appear as predicted when there is a swap?

“Something” changed! How is this not obvious? We already agreed that to see the pattern when there is a swap, look for the HH/VV signatures. So if the swap does not change anything, where does this “impossible” pattern come from?

 

[DrChinese]

It breaks them to entangled 1&4 and 2&3 (not mutually), right?

Yes, the new pairs are in the same Bell state but there is no other relationship remaining.

 

[gentzen]

What I am saying is simple: there is no such criteria. Suppose you execute/record a 4 hour run of 4 fold coincidences without executing a swap and look for a pattern in the data that indicates correlations. You won’t find one. It’s canonically impossible.

It is only impossible from your point of view, because you interpret "by hand" in a specific way. But this is a losing battle for anyone trying to argue against you. Whichever counterexample to your claim they would present, you would just say that it doesn't count.

I am sorry, but I will not play this game with you. Feel free to give whoever tries to play this game with you his medicine:

And for the Nth time: Where is the slightest theoretical support for what you assert in opposition to peer reviewed published papers by top researchers?

I don't care who is right or wrong here. But I am out. I have said my thing now, will unwatch this thread, and ignore any reactions in this thread.

 

[DrChinese]

1. It is only impossible from your point of view, because you interpret "by hand" in a specific way.

I am sorry, but I will not play this game with you.

2. I don't care who is right or wrong here. But I am out. I have said my thing now, will unwatch this thread, and ignore any reactions in this thread.

1. I’m only asking for someone - who thinks there are correlations hidden in data when there are no swaps - to present ANY hidden pattern. Easy to make such claim, but again this is canonically impossible per QM.

And I'm not sure how "by hand" can be interpreted other than one way.


2. As always, you are the best judge of how to allocate your time.

 

[lodbrok]

1. I’m only asking for someone - who thinks there are correlations hidden in data when there are no swaps - to present ANY hidden pattern. Easy to make such claim, but again this is canonically impossible per QM.

What do you mean by "data"? Please be precise. Are you referring to

1. the complete data from 1 & 4 pairs without any selection based on the 2 & 3 measurement?
2. Just the subset of 1 & 4 pairs for which a BSM measurement was performed irrespective of the BSM outcome?
3. Just the subset of 1 & 4 pairs for which an SSM measurement was performed regardless of the SSM outcome?
4. a subset of 1 & 4 data for which a particular BSM outcome was observed (i.e., selected based on |Φ+〉23, or |Φ−〉23)
5. a subset of 1 & 4 data for which a specific SSM outcome was observed (i.e., selected based on |𝐻𝐻〉23 or |𝑉𝑉〉23)

Secondly, what "data" do you claim "changed" as a result of the "swap"? Please answer based on the same categories above.

 

[PeterDonis]

what "data" do you claim "changed" as a result of the "swap"?

This is an ill-formed question. Each set of data is what it is; it can't be "changed". When we test a scientific theory, we don't ask "how does the data change when we do X?"

What we do ask is, what does the theory predict the data will look like when we do X, and when we do not do X? If the theory predicts that the data will look different when we do X vs. when we do not do X, then we say the theory predicts that doing X affects the data--or, in more colloquial language, that doing X "does something" to whatever thing we are measuring to obtain the data. That is how we test scientific theories.

In this case, the theory we are testing is QM, or more specifically QM's predictions about the correlations between measurements on entangled systems and about what the swap operation does when it is performed. The theory predicts that the data will look different when we do the swap operation vs. when we don't. The theory even tells us specifically how the data will look different: when we pick out subsets of the data according to the four possible photon 2 & 3 measurement results (HH, HV, VH, VV), the photon 1 & 4 measurement results in each subset will show Bell state correlations if we do the swap, but will show no correlations if we don't. Experiment bears out these predictions.

What more do you want?

As for your 1. through 5., as far as I can tell, none of them are picking out the correct subsets of data that the QM predictions I described above apply to. So they are all irrelevant.

 

[PeterDonis]

What you are missing is that the same BSM signature appears in the reported cases - the only difference being whether the swap is executed or not. Forget the subsets other than what they report on (the other 3 Bell states), they only report on the single Bell state |Φ->. For that state, the 2 & 3 signature is HH or VV. That same signature is reported on for both entangled (3a) and separable (3b) cases.

Just to re-emphasize this point, and maybe help to focus the discussion: in my previous posts, I've been talking about an idealized experiment where it's somehow possible to set things up so that each of the four possible 2 & 3 signatures (HH, HV, VH, VV) corresponds to one of the four Bell states. But AFAIK no actual experiment has actually done that. What actual experiments have done is what is described in the quote above, so let me restate the point I've made in multiple earlier posts (and which @DrChinese makes in the quote above) as clearly as I can for that specific scenario:

We do a bunch of runs, and for each run, we measure the polarization of the photons in the two output channels of the "BSM" beam splitter. In some runs, we don't get a signal at all in one of those output channels (because both photons, 2 & 3, went into the other one); we discard those runs. For runs where we do get a signal in both output channels, we pick out only those runs where the polarization measurement outcomes in both channels were the same: HH or VV.

Then we separate those runs into those where a swap was done by the experimenter, and those where no swap was done. QM predicts that, for the runs where a swap was done, the photon 1 & 4 measurements will show correlations indicating the Bell state $\ket{\Phi -}$; and QM predicts that, for the runs where no swap was done, the photon 1 & 4 measurements will show no correlation.

 

[DrChinese]

A. What do you mean by "data"? Please be precise. Are you referring to

1. the complete data from 1 & 4 pairs without any selection based on the 2 & 3 measurement?
2. Just the subset of 1 & 4 pairs for which a BSM measurement was performed irrespective of the BSM outcome?
3. Just the subset of 1 & 4 pairs for which an SSM measurement was performed regardless of the SSM outcome?
4. a subset of 1 & 4 data for which a particular BSM outcome was observed (i.e., selected based on |Φ+〉23, or |Φ−〉23)
5. a subset of 1 & 4 data for which a specific SSM outcome was observed (i.e., selected based on |𝐻𝐻〉23 or |𝑉𝑉〉23)

B. Secondly, what "data" do you claim "changed" as a result of the "swap"? Please answer based on the same categories above.

A. If you perform a continuous series of swap=OFF runs, say 1000 4 fold coincidences, the full data will include timing information (used to determine the 4 fold coincidences) along with information regarding the 4 photons' measurements. The 1 & 4 photons will be measured on L/R basis, and the 2 & 3 photons will be measured on the H/V basis.

I say the correlation between the 1 & 4 photons will be close to 0.0 for any criteria you might specify to filter those down to less than 1000, other than looking at the 1 & 4 photons' values themselves. You can look at the 2 & 3 outcomes, the time of day, the geographical location, or whatever else you think might help you establish a significant correlation.

On the other hand: I claim it is canonically impossible for there to exist any such criteria. Because LL and RR outcomes for 1 & 4 photons bear no known mathematical relationship to H/V outcomes for 2 & 3 photons. I hope there is no question about this statement, it's nothing more than a version of the uncertainty relations.

On the other hand: if you obtain a similar dataset of 1000 4 fold coincidences with swap-ON, you can easily filter down to a subset with high correlation. Simply look at the data points with the 2 and 3 photons measured as HH or VV and you will see that the LL and RR outcomes (photons 1 & 4) show high correlations - approaching 1.0 in the ideal case. That's because after a swap resulting in the Φ- Bell state, there is a mathematical relationship between the L/R polarization of the 1 & 4 photons. So all you do is look for that signature in the 2 & 3 data.

So how is it possible to do this "trick" if something wasn't different about the two full datasets? After all, it has been claimed that since there is no "nonlocal influence" (or whatever you choose to label it): There is nothing material that changed.


B. I can't say for a fact what specific things "change", if anything at all. I am not specifying that there is a counterfactual outcome. I just know that the population shows no correlation in one case (no swap), and strong correlation in the other (swap). And the only difference is physical overlap in a beam splitter, resulting in Entangled State statistics between photons that have never interacted nor have existed in a common light cone.

You tell me why that happens, that's what I am asking. Because I don't think there are any explanations that DON'T involve some kind of "nonlocal influence". But if you agree that the swap=OFF data cannot contain any markers that might allow us to filter down to a correlated subset; and you agree that the swap=ON dataset does: there really isn't much left as an option, is there?

 

[DrChinese]

Just to re-emphasize this point, and maybe help to focus the discussion: in my previous posts, I've been talking about an idealized experiment where it's somehow possible to set things up so that each of the four possible 2 & 3 signatures (HH, HV, VH, VV) corresponds to one of the four Bell states. But AFAIK no actual experiment has actually done that. ...

First, I think your entire post is concise.

Second, there is hope that one day there will be sufficient improvements to technology so as to allow all 4 Bell states to be identified in a single experiment. I am not aware of specific theoretical issues to prohibit this (but I am not an expert).

But that wouldn't really change our conclusion (whatever that is). Un-swapped 1 & 4 pairs are not entangled in any way, and show no mathematical relationship that can be determined from examining 2 & 3 pairs. Swapped pairs do show such a relationship when their Bell state is known. If that can be determined for even 1 Bell state, an experiment demonstrating such is valid and important.

 

[javisot20]

The entanglement between 1&4 was not "there waiting" in the initial state. Rather, it is created by the measurement of 2&3, which redistributes the correlations of the system according to quantum rules. The non-locality of 1&4 emerges because the measurement in 2&3 reconfigures the global state of the system. Quantum correlations are not limited to locally measured particles, but affect the entire system.

If the swap of 2&3 affects 1&4, aren't we violating MoE? (I understand that 2&3 is not maximally entangled with 1&4, but swapping or not swapping 2&3 affects 1&4, is that allowed without violating MoE?)

 

[DrChinese]

1. The entanglement between 1&4 was not "there waiting" in the initial state. Rather, it is created by the measurement of 2&3, which redistributes the correlations of the system according to quantum rules. The non-locality of 1&4 emerges because the measurement in 2&3 reconfigures the global state of the system. Quantum correlations are not limited to locally measured particles, but affect the entire system.

2. If the swap of 2&3 affects 1&4, aren't we violating MoE? (I understand that 2&3 is not maximally entangled with 1&4, but swapping or not swapping 2&3 affects 1&4, is that allowed without violating MoE?)

1. Yes.

2. Photon 1 (or any photon) can only be maximally entangled (the type we are working with here) to one other quantum system (particle in our case) at a time*. That's the MoE. So when the swap occurs, 1 is no longer entangled with 2 - and is now entangled with 4.


*There aren't words or terms to properly describe "when" the swap itself occurs. About all you can say is: Before it was X, later it was Y.