A Interpreting photon correlations from independent sources

[iste]

1. No, 1 and 2 and 3 are never entangled.

I never said this.

2. No, the BSM remotely switches 1&2 distant entanglement to distant 1&4 entangled. If there is no swap, that does not happen. Quite denying the obvious. Please seek an explanation that follows your preferred interpretation and explains this demonstrated behavior.

Didn't say anything that contradicts the basic description of entanglement swapping; maybe read the posts more carefully.

 

[PeterDonis]

Didn't say anything that contradicts the basic description of entanglement swapping; maybe read the posts more carefully.

This is uncalled for. @DrChinese is making a valid point: the BSM interaction can only be viewed as "local" if we only look at photons 2 & 3, and ignore the fact that whatever happens or doesn't happen at the BSM also affects photons 1 & 4, which are not locally present at the BSM. In other words, the quantum system involved is all four photons, and that system is inherently not local--it is composed of spatially separated subsystems that are entangled. Waving your hands and saying "well, the interactions are all just local" does not address this point.

 

[lodbrok]

2. No, the BSM remotely switches 1&2 distant entanglement to distant 1&4 entangled. If there is no swap, that does not happen. Quite denying the obvious. Please seek an explanation that follows your preferred interpretation and explains this demonstrated behavior.

This is contradicted by the article: https://arxiv.org/pdf/quant-ph/0201134 (see page 3). If there is no BSM result, Victor can't demonstrate any correlation. If there are BSM measurements, but the results are thrown away and not given to Victor, he can't demonstrate any correlation.

If Alice observes any of the other Bell-states for photons 1 and 2, photons 0 and 3 will also be perfectly entangled correspondingly. We stress that photons 0 and 3 will be perfectly entangled for any result of the BSA, and therefore, it is not necessary to apply a unitary operation to the teleported photon 3 as in the standard teleportation protocol. But it is certainly necessary for Alice to communicate to Victor her Bell-state measurement result. This will enable him to sort Bob’s data into four subsets, each one representing the results for one of the four maximally entangled Bell-states

 

[PeterDonis]

If there are BSM measurements, but the results are thrown away and not given to Victor

This is irrelevant. In any other area of science, nobody would throw away relevant data from an experiment and then claim that some effect wasn't present because the data didn't show it. Nor is @DrChinese claiming that the presence of entanglement between photons 1 & 4 can somehow be detected without knowing the photon 2 & 3 measurement results.

 

[lodbrok]

This is irrelevant. In any other area of science, nobody would throw away relevant data from an experiment and then claim that some effect wasn't present because the data didn't show it. Nor is @DrChinese claiming that the presence of entanglement between photons 1 & 4 can somehow be detected without knowing the photon 2 & 3 measurement results.

In fact, he is doing something worse; he is making the extraordinary claim, utterly unsupported by the cited experiments, that simply doing the 2 & 3 measurements changes the 1 & 4 results. In any other area of science, you don't get to claim that an effect is present without experimental evidence showing it. In every entanglement-swapping experiment, the 2 & 3 results must be transmitted to a third party and used to filter the 1 & 4 data before any correlation can be demonstrated. This is well-understood and entirely uncontroversial. There is no need to resort to wildly speculative fantastical explanations about an experiment at 2 & 3 remotely changing the 1 & 4 results without evidence.

 

[iste]

Suppose you teleport the unknown state of photon 2 to distant previously uncorrelated photon 4. You still have the problem how to explain the perfect correlation with photon 1 on a mutually unbiased basis to photons 2 & 3. In order words: the very act of teleportation is automatically nonlocal!

Woops, I actually misread this; but, this phenomenon is what the post was addressing. You're teleporting from 2 to 4; you're teleporting from 1 to 3; ergo, there is a definite relationship between 2&3 and 1, and a definite relationship between 2&3 and 4. So you can infer there should be a definite relationship between 1 and 4. Different bases doesn't matter. We are dealing with three different entanglements connected to each other, in the kind of manner described in the second quote of post #43, with correlations in different bases. The measurements have to respect the correlations in these bases that define the entanglements - if you have states that are entangled with correlations in these bases, actual nor measurements counterfactual measurements are not going to be inconsistent with those states.

 

[PeterDonis]

In any other area of science, you don't get to claim that an effect is present without experimental evidence showing it.

The experimental evidence certainly shows that choosing to do a swap or not do a swap at 2&3 affects the 1&4 measurement results: the subsets of 1&4 results corresponding to the four possible combinations of 2&3 results (HH, HV, VH, VV) show the predicted Bell state correlations if the experimenter chooses to do a swap; they show no correlations if the experimenter chooses not to do a swap. So the experimental evidence is that something is going on when a swap takes place at 2&3 that affects 1&4.

 

[Morbert]

the subsets of 1&4 results corresponding to the four possible combinations of 2&3 results (HH, HV, VH, VV) show the predicted Bell state correlations if the experimenter chooses to do a swap; they show no correlations if the experimenter chooses not to do a swap.

How can we say the subset "HH" when no swap occurs is the same subset as "HH" when a swap occurs? In other words, if the experimenter does no swap and measures both 2&3 photon polarizations to be "H", how can they say "If I had done a swap instead, the polarization of these photons would still have been "H"?

 

[PeterDonis]

How can we say the subset "HH" when no swap occurs is the same subset as "HH" when a swap occurs?

I'm not saying any such thing. I am just describing a straightforward procedure for running an experiment and analyzing the data. The math of QM makes a definite prediction for what will be found when that is done, and actual experiments match the prediction.

It is true that the math of QM does not make any claim about why the prediction is what it is, or why experiments match the prediction. If you are saying that some particular interpretation of QM makes the question you are asking relevant, what is that interpretation? What reference can you give for it?

if the experimenter does no swap and measures both 2&3 photon polarizations to be "H", how can they say "If I had done a swap instead, the polarization of these photons would still have been "H"?

Obviously you can't. So what? In any other area of science, nobody would bother asking such a question. The fact that such counterfactual "matching" cannot be done is not limited to QM.

 

[Morbert]

I'm not saying any such thing. I am just describing a straightforward procedure for running an experiment and analyzing the data. The math of QM makes a definite prediction for what will be found when that is done, and actual experiments match the prediction.

It is true that the math of QM does not make any claim about why the prediction is what it is, or why experiments match the prediction. If you are saying that some particular interpretation of QM makes the question you are asking relevant, what is that interpretation? What reference can you give for it?

Obviously you can't. So what? In any other area of science, nobody would bother asking such a question. The fact that such counterfactual "matching" cannot be done is not limited to QM.

Then, since the subsets are different, we can't say the BSM induces Bell-inequality-violationing correlations in any particular subset. We can only say the BSM lets the experimenter identify the appropriate subsets.

 

[lodbrok]

The experimental evidence certainly shows that choosing to do a swap or not do a swap at 2&3 affects the 1&4 measurement results: the subsets of 1&4 results corresponding to the four possible combinations of 2&3 results (HH, HV, VH, VV) show the predicted Bell state correlations if the experimenter chooses to do a swap; they show no correlations if the experimenter chooses not to do a swap. So the experimental evidence is that something is going on when a swap takes place at 2&3 that affects 1&4.

This is wrong. Without the BSM, you don't have the four subsets to decide whether they show a correlation. As the paper states, the BSM data must be transmitted to segregate the 1 & 4 data into four subsets. That's how the correlated four subsets are obtained. Any other subsets you may want to generate from the 1 & 4 data not conditioned on the swap BSM data won't be the same partitioning and thus not the same subsets.

 

[PeterDonis]

Without the BSM, you don't have the four subsets to decide whether they show a correlation.

Yes, you do. Photons 2 & 3 are measured after the BSM always, whether a swap is done at the BSM or not. You can always use those measurement results to get the four subsets.

the BSM data must be transmitted to segregate the 1 & 4 data into four subsets.

The results of the measurements on photons 2 & 3 must be transmitted. Those results are always there, whether a BSM swap is done or not. See above.

 

[lodbrok]

Yes, you do. Photons 2 & 3 are measured after the BSM always, whether a swap is done at the BSM or not. You can always use those measurement results to get the four subsets.


The results of the measurements on photons 2 & 3 must be transmitted. Those results are always there, whether a BSM swap is done or not. See above.

Here is how the experiment is described on Page 3-4 of https://arxiv.org/pdf/1203.4834:

In our experiment, the primary events are the polarization measurements of photons 1 and 4 by Alice and Bob. They keep their data sets for future evaluation. Each of these data sets by itself and their correlations are completely random and show no structure whatsoever. The other two photons (photons 2 and 3) are delayed until after Alice and Bob’s measurements, and sent to Victor for measurement. His measurement then decides the context and determines the interpretation of Alice and Bob’s data. In our setup, using two-photon interference on a beam splitter combined with photon detections, Victor may perform a Bell-state measurement which projects photons 2 and 3 either onto |Φ+〉23 or onto |Φ−〉23 . This would swap entanglement to photons 1 and 4. Instead of a Bell-state measurement, Victor could also decide to measure the polarization of these photons individually and project photons 2 and 3 either onto |𝐻𝐻〉23 or onto |𝑉𝑉〉23, which would result in a well-defined polarization for photons 1 and 4. These two measurements are mutually exclusive (complementary in the Bohrian sense) in the same way as measuring particle or wave properties in an interference experiment. The choice between the two measurements is made by using a quantum random number generator (QRNG). The QRNG is based on the intrinsically random detection events of photons behind a balanced beam splitter6 (for details, see the Supplementary Information). According to Victor’s choice of measurement (i.e. entangled or separable state) and his results (i.e. |Φ+〉23,|Φ−〉23, or |𝐻𝐻〉23, |𝑉𝑉〉23), Alice and Bob can sort their already recorded data into 4 subsets. They can now verify that when Victor projected his photons onto an entangled state (|Φ+〉23 or |Φ−〉23), each of their joint subsets behaved as if it consisted of entangled pairs of distant photons. When Victor projected his photons on a separable state (|𝐻𝐻〉23 or |𝑉𝑉〉23), Alice and Bob’s joint subsets behave as if they consisted of separable pairs of photons.

So you are incorrect, the four subsets generated by sorting according to (|Φ+〉23,|Φ−〉23) [BSM] are not always there. You have either four subsets generated according to (|Φ+〉23,|Φ−〉23) or four subsets generated according to (|𝐻𝐻〉23 or |𝑉𝑉〉23) [SSM]. These are not the same subsets. Therefore, it is wrong to suggest that data obtained by sorting according to (|Φ+〉23,|Φ−〉23) [BSM] changes "when the swap is done". What the authors mean by "swapping" is precisely the measurement of (|Φ+〉23,|Φ−〉23) and its subsequent use to generate subsets that appear correlated.

 

[PeterDonis]

you are incorrect

No, I'm not. You are misunderstanding my point.

These are not the same subsets.

"The same subsets" makes no sense in any case. Obviously on any given experimental run, either a swap is done or it is not. You can't have both on the same run. So obviously, the set of runs where a swap is done, partitioned into subsets by the photon 2&3 results, is going to give "different" subsets than the set of runs where a swap is not done, partitioned into subsets by the photon 2&3 results. They're physically different runs of the experiment. That's going to be true regardless of what interpretation you adopt.

 

[DrChinese]

How can we say the subset "HH" when no swap occurs is the same subset as "HH" when a swap occurs? In other words, if the experimenter does no swap and measures both 2&3 photon polarizations to be "H", how can they say "If I had done a swap instead, the polarization of these photons would still have been "H"?

Fair question. Let's examine closely what you are saying and see if it holds water:

a) What is the experimentally controlled difference between HH/VV datasets from the swap group as compared to the no-swap group? The difference is whether the 2 and 3 photons are indistinguishable (swap) or distinguishable (no swap).

b) How is this accomplished? By allowing the 2 and 3 photons to physically overlap (swap) or not (no swap). There are 2 techniques for this described in the papers: i) change the beam splitter reflectivity from 50:50 (swap) to 0:100 (no swap); or ii) delay either the 2 or 3 photon (but not both) sufficiently so they don't have the opportunity for both to overlap and physically interact in the beam splitter. Either way, there will be no possibility of overlap or other physical interaction between them.

c) That is the ONLY difference! And guess what? To answer your question (your hypothesis) about the H/V polarization being somehow changed by the choice of whether to swap or not: There is no known mechanism or quantum mechanical theory that ties linear polarization to interaction in a beam splitter. Regardless of whether there is overlap or not in the beam splitter, standard QM says: The V/H polarizations of neither the 2 or 3 photon changes in a beam splitter regardless of mutual interaction.

You must conclude: For your speculative hypothesis to make sense: We would need a H polarization to change to a V polarization (and vice versa) for one* - but not both photons - when there IS overlap, but NOT when there is no overlap**. Otherwise the HH and VV pairs being used to compare datasets will be properly apples to apples, exactly as the papers present them. And your changes would need to happen precisely 50% of the time to make the statistics work out correctly. But: there is NO known phenomena around beam splitters that would support this speculation.

In other words: The V/H polarizations for the 2 & 3 photons do not vary depending on whether there is physical overlap (or not) in the beam splitter of the BSM. The purpose of your hypothesis is to explain that the swap dataset's VV/HH pairs are not comparable to the no swap dataset's, but that is manifestly false. If you have any suitable reference to dispute this assessment, please share with me.

Because otherwise: What we have is a physical difference occurring in the swap group versus the no swap group, while holding all other variables constant***. According to locality considerations: There should be no change in the 1 & 4 pairs (left vs right circular polarization) depending on a physical change to the 2 & 3 pairs IF the V/H detection mechanism is otherwise held constant. Which, as I have shown, it is constant. And yet the experimental results say otherwise. The act of overlapping the 2 & 3 photons (swap case) produces some physical change to the 4-fold detection statistics as compared to the no overlap (no swap) case. The 4-fold context is nonlocal (in both space and time) and does not respect Einsteinian locality. (As before, this is how things appear to the naked eye - and I am still seeking an explanation that might restore locality to the description.)


* Hopefully you see why there must be a change for one but not both. If both changed, it would return VV instead of HH (and vice versa) which is the same bucket as HH anyway. Note also that you must also require 50% frequency (or something like that) for these changes to create the apparent correlation where there was otherwise none.

** Your hypothesis is in fact experimentally testable for your "swap" case. Just send H polarized photons as 2 and 3 into the BS and see if any change to V when there is physical overlap.

*** As @PeterDonis has pointed out repeatedly: This is textbook experimental methodology in all areas of science. You track the independent control variable, in this case swap vs. no swap.

 

[DrChinese]

DrChinese said: "2. No, the BSM remotely switches 1&2 distant entanglement to distant 1&4 entangled. If there is no swap, that does not happen. "

This is contradicted by the article: https://arxiv.org/pdf/quant-ph/0201134 (see page 3). If there is no BSM result, Victor can't demonstrate any correlation. If there are BSM measurements, but the results are thrown away and not given to Victor, he can't demonstrate any correlation.

The reference follows what I have been citing, and does not contradict me in any way. Obviously, in this experiment as in any scientific experiment, the data must be brought together to be analyzed. I have no idea why anyone would hide behind this requirement as a way to "prove" there is locality. We already acknowledged that entanglement swapping does not lead to FTL signaling. What it *appears* to do is "cause" (but not in the sense of Einsteinian causality) a distant change in photons 1 & 4 - they become entangled and display entangled state statistics as evidence.

[Note that as in the other papers by this same team (Zeilinger), the criteria for comparison is a single specific Bell state. There is nothing sinister or unusual or whatever about this, it in no way changes the scientific results. As long as fidelity is good, those results stand. If the fidelity was poor, you wouldn't get results that would match the quantum mechanical prediction. Of course, these do.]

 

[Morbert]

@DrChinese According to Ma:

https://arxiv.org/pdf/1203.4834

The two complementary measurements are realized in the following ways: The Bell-state measurement (BSM) corresponds to turning on the switchable quarter-wave plates. In this case, the phase of the interferometer is π/2 and the interferometer acts as a 50/50 beam splitter. Therefore, the two photons interfere and are projected onto a Bell state by polarization-resolving single-photon detections. The separable-state measurement (SSM) corresponds to turning off the switchable quarter-wave plates. The phase of the interferometer is 0 and the interferometer acts as a 0/100 beam splitter, i.e. a fully reflective mirror. Therefore, the two photons do not interfere and are projected onto a separable state by polarization-resolving single-photon detections. When used for a BSM, our BiSA can project onto two of the four Bell states, namely onto |Φ + 〉 23 = (|𝐻𝐻〉 23 + |𝑉𝑉〉 23)/√2 (both detectors in b’’ firing or both detectors in c’’ firing) and |Φ − 〉 23 = (|𝐻𝐻〉 23 − |𝑉𝑉〉 23)/√2 (one photon in b’’ and one in c’’ with the same polarization). When an SSM is made, we look at coincidences between b’’ and c’’ with the same polarization (as in the BSM case for |Φ − 〉 23) and the resultant projected states are |𝐻𝐻〉 23 and |𝑉𝑉〉 23.

Considering the scenarios where one photon fires b'' and one fires c'' with the same polarization. If Victor decided to carry out an SSM, this projects the photons into the state |𝐻𝐻〉 23 or |𝑉𝑉〉 23. If victor decided to carry out a BSM, this projects the photons onto the state |Φ − 〉 23 = (|𝐻𝐻〉 23 − |𝑉𝑉〉 23)/√2.

This means that if an SSM was carried out, the scenario (b'' = V and c'' = V) and the scenario (b'' = H and c'' = H) will be assigned to different partitions (|𝑉𝑉〉 23 and |𝐻𝐻〉 23 respectively). However, if a BSM was carried out, the scenario (b'' = V and c'' = V) and the scenario (b'' = H and c'' = H) will be assigned to the same partition (|Φ − 〉 23).

How would you square this with your claim that the partitions are the same even if the data is altered?

[edit] - expanded

 

[javisot20]

(2&3) is a locally entangled pair, a single system, therefore it can be entangled first with 1, forming (1&2&3) and then entangled with 4 forming (1&2&3&4), or first with 4 forming (2&3&4) and then with 1.
We can also consider that 2&3 interact simultaneously with 1&4, which implies synchronization.

In this context 1&4 always interact locally through 2&3, the entanglement of 1&4 is as local as that of 2&3. "Swap yes" or "swap no" will affect the results.

This is the "local" version, right?

 

[DrChinese]

@DrChinese According to Ma: https://arxiv.org/pdf/1203.4834

Considering the scenarios where one photon fires b'' and one fires c'' with the same polarization. If Victor decided to carry out an SSM, this projects the photons into the state |𝐻𝐻〉 23 or |𝑉𝑉〉 23. If victor decided to carry out a BSM, this projects the photons onto the state |Φ − 〉 23 = (|𝐻𝐻〉 23 − |𝑉𝑉〉 23)/√2.

This means that if an SSM was carried out, the scenario (b'' = V and c'' = V) and the scenario (b'' = H and c'' = H) will be assigned to different partitions (|𝑉𝑉〉 23 and |𝐻𝐻〉 23 respectively). However, if a BSM was carried out, the scenario (b'' = V and c'' = V) and the scenario (b'' = H and c'' = H) will be assigned to the same partition (|Φ − 〉 23).

How would you square this with your claim that the partitions are the same even if the data is altered?

Of course I agree completely with Ma et al. As I answered this question in the other thread, so I answer here.


1. Your question regarding the 2 & 3 photons (b and c detectors in the Ma paper): Is HH (swap) = HH' (no swap) and VV (swap) = VV' (no swap) ? In other words, are they measuring the same/equivalent characteristics (apples to apples)? Do the results stay the same after the beam splitter regardless of whether a swap is performed? Can we be sure a swapped HV or VH doesn't get wrongly reported as an HH or VV as a result of an hypothetical overlap effect in the beam splitter?

Well of course the answer is YES, we can be sure. It has to be, this is physically dependent on the polarization characteristics of a beam splitter - which are essentially none for our purposes. No amount of interference or interaction between 2 photons overlapping in the beam splitter is going to change H to V (or vice versa). There is no evidence otherwise, and there is no theory to support that speculative idea. It is easily testable, although I have never seen such an experiment. (Of course, there are many experiments one might perform to confirm the predictions of QM that have never been performed.)


2. Importantly: Even if there was the completely invented effect you imagine, there would be severe constraints on how it works. The constraints, in fact, would be easily discernable. i) The effect could only change the 2 photon or the 3 photon, but not both. Hopefully you can see why. Further, that specific results would necessarily require that exactly 1 photon change 50% of the time. That's to match the predictions of QM - and takes a bit of calculation to see why. Obviously, such a large and specific effect would have been noticed in the past 20 years or so.


3. Next, you ask: Are the selected [4 fold] subsets (what you also call partitions) the same? Answer is of course, NO, as the authors say in your quote. That's because the swap induces a change in the 1 & 4 pairs - they are correlated in a specific Bell state, when they weren't before. If I execute a swap, 1 & 4 will be entangled. If I don't execute a swap, they will not be entangled. When 2 & 3 are swapped as |Φ−〉 (which is indicated by the signature HH or VV result) , 1 & 4 will therefore marry up to a different correlation average (1.0 in the ideal case) than when the SSM is executed (0.0 ideal case).


So you have it backwards: 1 & 4 appear to change as a result of the swap, while the 2 & 3 signatures of HH or VV remains the same regardless of swap/no swap. The 2&3 datasets (swap/no swap) themselves are comparable, but the 4-fold results (including 1 & 4) are different - that is what the experiment was designed to demonstrate.

The [2&3] "partitions" must be considered differently than the [2&3 and 1&4] "partitions". We can't make sense trying to ask your questions and jumping from discussing one to the other. The experiment is only about 4-fold results when a single variable is changed - as happens in normal scientific experiments. We should strictly discuss their results. Which are: Swap->correlation, no swap->no correlation. You should explain this. I would assert, to the naked eye, it looks like the swap causes the correlation. In any other experiment (outside of QM), that would be the singular conclusion. Again, maybe there is another explanation...

...But there is NO evidence that overlap in a beam splitter has any material effect on the 2 & 3 polarization. If you dispute this well accepted point, please provide a reference to the contrary. Repeat: beam splitters do NOT change H polarization to V (or vice versa). And it does not matter whether there are 2 photons overlapping or not. It would be a major optical discovery to learn otherwise.

Of course, there are plenty of things that happen in a beam splitter of great interest, and which have been studied in great depth. An example is this 2015 paper by Henault which summarizes some of that. For example, a beam splitter can modify phase of a photon. However, that phase shift is always by a fixed amount (π) and does not change H to V or V to H. They also discuss HOM interference (2 entangled photons in a beam splitter), which again does not change H to V or V to H. So... there's that.

To summarize: You are speculating about a previously undiscovered quantum mechanical effect inside a beam splitter - and one that can be easily proven false.* Even in the Interpretations subforum, such nonstandard ideas must be getting close to (or over) the line.


*I'll be more than happy to provide details on how to do this.

 

[DrChinese]

(2&3) is a locally entangled pair, a single system, therefore it can be entangled first with 1, forming (1&2&3) and then entangled with 4 forming (1&2&3&4), or first with 4 forming (2&3&4) and then with 1.

The entanglement you describe is strictly prohibited in orthodox quantum theory. There can be no entangled (1&2&3), for example. See the seminal work here or a short proof here. The subject is Monogamy of Entanglement.

Nothing like what describe occurs in entanglement swapping.

 

[javisot20]

The entanglement you describe is strictly prohibited in orthodox quantum theory. There can be no entangled (1&2&3), for example. See the seminal work here or a short proof here. The subject is Monogamy of Entanglement.

Nothing like what describe occurs in entanglement swapping.

What I have said is that 2&3, which is 1 system, is entangled with 1, or 4 (or 1&4). One system is entangled with another, not three or four entangled systems.

(I say this playing devil's advocate, trying to understand the point of view of someone who defends the locallity to understand the experimental results, influenced by this thread and the consistent history and locality thread, mainly)

 

[PeterDonis]

What I have said is that 2&3, which is 1 system, is entangled with 1, or 4 (or 1&4).

No, that's not correct.

Initially (before the BSM), photons 1&2 form an entangled system, and photons 3&4 form an entangled system. Those two systems are separable; they are not entangled at all.

If no swap occurs, the entanglements remain as stated above.

If a swap does occur, then after the swap, photons 2&3 form an entangled system, and photons 1&4 form an entangled system, and those two systems are separable; they are not entangled at all.

There is never a time when photons 2&3, as a system (meaning they are entangled with each other), are entangled with 1, or 4, or both.

 

[Morbert]

@DrChinese I believe our primary disagreement is over 1., so I will address this for now and we can proceed to your other points once it is resolved. I suspect the resolution of our primary disagreement would impact how other points are addressed anyway.

Of course I agree completely with Ma et al. As I answered this question in the other thread, so I answer here.

1. Your question regarding the 2 & 3 photons (b and c detectors in the Ma paper): Is HH (swap) = HH' (no swap) and VV (swap) = VV' (no swap) ? In other words, are they measuring the same/equivalent characteristics (apples to apples)? Do the results stay the same after the beam splitter regardless of whether a swap is performed? Can we be sure a swapped HV or VH doesn't get wrongly reported as an HH or VV as a result of an hypothetical overlap effect in the beam splitter?

Well of course the answer is YES, we can be sure. It has to be, this is physically dependent on the polarization characteristics of a beam splitter - which are essentially none for our purposes. No amount of interference or interaction between 2 photons overlapping in the beam splitter is going to change H to V (or vice versa). There is no evidence otherwise, and there is no theory to support that speculative idea. It is easily testable, although I have never seen such an experiment. (Of course, there are many experiments one might perform to confirm the predictions of QM that have never been performed.)

HV or VH are less helpful as they must always be discarded.

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.

So to more accurately reflect the subsets, and if I am reading the paper correctly, they are {HH, VV, discarded} for the SSM and {Φ+, Φ-, discarded} for the BSM.

If your answer is still yes for the HH and VV cases, then why doesn't this run afoul of complemenetarity?

These two measurements are mutually exclusive (complementary in the Bohrian sense) in the same way as measuring particle or wave properties in an interference experiment.

E.g. say I perform a BSM and the detectors fire indicating (b'', H) and (c'', H). I can therefore infer the BSM result Φ- as described by Ma. You seem to be implying I can also infer the SSM result HH, but how can this be? These are results of complementary measurements. How can I infer HH from a result

Φ− = (HH − VV)/√2

 

[javisot20]

There is never a time when photons 2&3, as a system (meaning they are entangled with each other), are entangled with 1, or 4, or both.

This is the point at which many conclude, "nonlocality is proven," right?

 

[DrChinese]

This is the point at which many conclude, "nonlocality is proven," right?

“Once you eliminate the impossible, whatever remains, no matter how improbable, must be the truth.” - Arthur Conan Doyle

Just saying… we need to identify assumptions and question them. Mother Nature seems to be a trickster.

 

[javisot20]

“Once you eliminate the impossible, whatever remains, no matter how improbable, must be the truth.” - Arthur Conan Doyle

Just saying… we need to identify assumptions and question them. Mother Nature seems to be a trickster.

Therefore, a person who wants to deny non-locality must deny the above, and affirm that 1 and 4 did interact locally with each other, using 2 and 3 as intermediaries, and we already know that 2 and 3 are intermediaries between 1 and 4. At this point others will say that the locality is proven.

Whether we affirm locality or non-locality, everyone agrees that the swap or non-swap of 2&3 will affect 1&4.

 

[DrChinese]

Φ@DrChinese I believe our primary disagreement is over 1., so I will address this for now and we can proceed to your other points once it is resolved. I suspect the resolution of our primary disagreement would impact how other points are addressed anyway.

1. HV or VH are less helpful as they must always be discarded. So to more accurately reflect the subsets, and if I am reading the paper correctly, they are {HH, VV, discarded} for the SSM and {Φ+, Φ-, discarded} for the BSM.

2. If your answer is still yes for the HH and VV cases, then why doesn't this run afoul of complemenetarity?E.g. say I perform a BSM and the detectors fire indicating (b'', H) and (c'', H). I can therefore infer the BSM result Φ- as described by Ma. You seem to be implying I can also infer the SSM result HH, but how can this be? These are results of complementary measurements. How can I infer HH from a result

Φ− = (HH − VV)/√2

1. Again, you have this backwards. In the experiment, all 4 fold coincidences with HH or VV for the b” and c” polarizers are reported. Your hypothesis boils down to saying some of the HV or VH get misreported as HH or VV. If that happened, it would be reported as you say - and would bias the results, possibly overstating the correlation.

Keep in mind that what they are reporting is HH/VV (the signature) but that is an SSM if no swap and Φ− if BSM. We agree on this point.

2. Again, the HH/VV signature is identical for the reported BSM or SSM. In case this is not clear: there is no way to know if there was a swap or not by looking to the b” and c” results. Either way, there will be one click at a b” detector and one click at a c” - and they will both be H or both V. To know if a swap occurred, you also need to know what the setting of the beam splitter is.

So I think we agree on this. You therefore also know that if the setting had been the complementary option, you would get the same outcomes for b” and c”. No difference, which answers your question/criticism. The b”/c” results are the same, but the Alice/Bob results appear to change.

 

[DrChinese]

Therefore, a person who wants to deny non-locality must deny the above, and affirm that 1 and 4 did interact locally with each other, using 2 and 3 as intermediaries, and we already know that 2 and 3 are intermediaries between 1 and 4. At this point others will say that the locality is proven.

Whether we affirm locality or non-locality, everyone agrees that the swap or non-swap of 2&3 will affect 1&4.

Of course, 1 and 4 are never in a common light cone. And 2 and 3 can be observed long after 1 and 4 are measured. I would claim all measurements are causally disconnected from each other.

 

[PeterDonis]

1 and 4 did interact locally with each other, using 2 and 3 as intermediaries

I don't know of any feasible meaning of the word "locally" that would support such a claim. Note that, as @DrChinese has pointed out, photons 2&3 can interact long after photons 1&4 are measured. And the photon 1&4 measurements could, in principle, be many light-years apart, and many light-years separated from the photon 2&3 interaction.

 

[Morbert]

1. Again, you have this backwards. In the experiment, all 4 fold coincidences with HH or VV for the b” and c” polarizers are reported. Your hypothesis boils down to saying some of the HV or VH get misreported as HH or VV. If that happened, it would be reported as you say - and would bias the results, possibly overstating the correlation.

Keep in mind that what they are reporting is HH/VV (the signature) but that is an SSM if no swap and Φ− if BSM. We agree on this point.

2. Again, the HH/VV signature is identical for the reported BSM or SSM. In case this is not clear: there is no way to know if there was a swap or not by looking to the b” and c” results. Either way, there will be one click at a b” detector and one click at a c” - and they will both be H or both V. To know if a swap occurred, you also need to know what the setting of the beam splitter is.

So I think we agree on this. You therefore also know that if the setting had been the complementary option, you would get the same outcomes for b” and c”. No difference, which answers your question/criticism. The b”/c” results are the same, but the Alice/Bob results appear to change.

This is the time-evolution relevant to the BSM result Φ−, taken from Ma et al

So you can see that the issue is not HV or VH getting misreported as HH or VV. It's that "VV" could be "misreported" as "HH" or vice versa. To put it more correctly, when a BSM is carried out, a signature like (H b", H c'') only lets you infer (HH - VV)/√2. It does not let you infer HH.

 

[DrChinese]

… To put it more correctly, when a BSM is carried out, a signature like (H b", H c'') only lets you infer (HH - VV)/√2. It does not let you infer HH.

I don’t follow your thinking, so maybe you can help me.

For the BSM cases:

The HH and VV signatures (photons 2 and 3) are reported together. So this point doesn’t affect the experimental results and conclusion in any way. Hopefully this is not in question.

Now if you ask: what does it tell us about the associated 1 and 4 photons: you would be correct to say that those results need not match the 2 and 3 signatures, when measured on the same H/V basis. Of course, they will be perfectly correlated on any same basis. Hopefully I agree with you here as well.

So… what are you trying to get across?

 

[Morbert]

I don’t follow your thinking, so maybe you can help me.

For the BSM cases:

The HH and VV signatures (photons 2 and 3) are reported together. So this point doesn’t affect the experimental results and conclusion in any way. Hopefully this is not in question.

Now if you ask: what does it tell us about the associated 1 and 4 photons: you would be correct to say that those results need not match the 2 and 3 signatures, when measured on the same H/V basis. Of course, they will be perfectly correlated on any same basis. Hopefully I agree with you here as well.

So… what are you trying to get across?

What it means is the subset of experimental runs marked by the BSM result Φ− would be distributed across the HH and VV subsets if an SSM had been done instead. Similarly, the experimental runs marked by the SSM result HH would be distributed across the Φ− and Φ+ subsets if a BSM had been done instead. Hence, we don't have to conclude the data from the 1 & 4 measurements are altered by the choice to do a BSM. What is altered is the arrangement of the experimental runs into different sets.

[edit] - Deleted paragraph with bad counterfactual

 

[DrChinese]

What it means is the subset of experimental runs marked by the BSM result Φ− would be distributed across the HH and VV subsets if an SSM had been done instead. Similarly, the experimental runs marked by the SSM result HH would be distributed across the Φ− and Φ+ subsets if a BSM had been done instead. Hence, we don't have to conclude the data from the 1 & 4 measurements are altered by the choice to do a BSM. What is altered is the arrangement of the experimental runs into different sets.

That is incorrect. HH+BSM and VV+BSM (swap) are in a combined bucket, marked figure 3a in the paper. HH+SSM and VV+SSM (no swap) are also in a bucket, marked figure 3b in the paper. Apples to apples. The 1&4 stats change accordingly.

(Keep in mind the decision to swap or not can be made before or after 1 & 4 are measured.)

You can compare any permutation of the same signature (swap vs no swap) but a larger group is preferable. For the experimental results to be useful, they only need to place results in the proper bucket. So despite reading your comment a number of times, I have no clue as to where you are going with this. It’s experimental science 101. Record 4 fold results that match a predetermined criteria with switch set to swap, compare to results with switch set to no swap.


And just to be clear:

The BSM result Φ− dataset consists of the b" and c" detectors clicking H & H, or clicking V & V, whenever the apparatus is set to "swap". 4 fold coincidences are placed in figure 3a.

The SSM resultsets HH and VV consists of the b" and c" detectors clicking H & H, or clicking V & V, whenever the apparatus is set to "no swap". 4 fold coincidences are placed in figure 3b.

 

[Morbert]

That is incorrect. HH+BSM and VV+BSM (swap) are in a combined bucket, marked figure 3a in the paper. HH+SSM and VV+SSM (no swap) are also in a bucket, marked figure 3b in the paper. Apples to apples. The 1&4 stats change accordingly.

(Keep in mind the decision to swap or not can be made before or after 1 & 4 are measured.)

You can compare any permutation of the same signature (swap vs no swap) but a larger group is preferable. For this experimental results to be useful, they only need to place results in the proper bucket. So I have no clue as to where you are going with this. It’s experimentally science 101.

How can they be apples to apples? The combination of all BSM results (HH, b'' c'') or (VV, b'' c'') will correspond to the partition Φ+. But the combination of all SSM results (HH, b'' c'') or (VV, b'' c'') will not. The combination of all SSM results (HH, b'' c'') or (VV, b'' c'') will include experimental runs that could have produced the signatures (HV, b"b") or (HV, c"c") if a BSM had been done instead.

I.e. The combination of all SSM results (HH, b'' c'') or (VV, b'' c'') would lump together runs that would have been sorted into Φ+ and Φ- if a BSM had been done instead.

 

[DrChinese]

How can they be apples to apples? The combination of all BSM results (HH, b'' c'') or (VV, b'' c'') will correspond to the partition Φ+. But the combination of all SSM results (HH, b'' c'') or (VV, b'' c'') will not. The combination of all SSM results (HH, b'' c'') or (VV, b'' c'') will include experimental runs that could have produced the signatures (HV, b"b") or (HV, c"c") if a BSM had been done instead.

I.e. The combination of all SSM results (HH, b'' c'') or (VV, b'' c'') would lump together runs that would have been sorted into Φ+ and Φ- if a BSM had been done instead.

Uhhhh, we want to report the BSM results separate from the SSM results. And I am confused about your reference to Φ+ above, did you mean Φ-?

Φ- is both HH and VV signature.

 

[Morbert]

Uhhhh, we want to report the BSM results separate from the SSM results. And I am confused about your reference to Φ+ above, did you mean Φ-?

Φ- is both HH and VV signature.

Yes, sorry, I miswrote Φ+ for Φ-

 

[DrChinese]

How can they be apples to apples? The combination of all BSM results (HH, b'' c'') or (VV, b'' c'') will correspond to the partition Φ-. But the combination of all SSM results (HH, b'' c'') or (VV, b'' c'') will not. The combination of all SSM results (HH, b'' c'') or (VV, b'' c'') will include experimental runs that could have produced the signatures (HV, b"b") or (HV, c"c") if a BSM had been done instead.

OK, here is an issue where we differ. "signatures (HV, b"b") or (HV, c"c")" are NOT included in the reported results. Why would they? They aren't Φ-.

Repeating from post #83:

The BSM result Φ− dataset consists of the b" and c" detectors clicking H & H, or clicking V & V, whenever the apparatus is set to "swap". 4 fold coincidences are placed in figure 3a.

The SSM resultsets HH and VV consists of the b" and c" detectors clicking H & H, or clicking V & V, whenever the apparatus is set to "no swap". 4 fold coincidences are placed in figure 3b.

 

[Morbert]

OK, here is an issue where we differ. "signatures (HV, b"b") or (HV, c"c")" are NOT included in the reported results. Why would they? They aren't Φ-.

If a BSM is carried out, we can indeed discard runs that aren't Φ- by discarding all runs that don't have one of the signatures (HH, b'' c'') or (VV, b'' c''). All runs with one of these signatures correspond to the subset Φ- when a BSM is carried out.

If an SSM is carried out, we lose this filtering ability because the signatures (HH, b'' c'') or (VV, b'' c'') no longer correspond to Φ- when an SSM is carried out.

I think it is the counterfactual indefiniteness of QM that is the key. To make an apples to apples comparison, Victor would have to be able to make statements like "I have carried out an SSM and observed one of the signatures (HH, b'' c'') or (VV, b'' c''). If I had instead carried out a BSM, the signature observed would still be (HH, b'' c'') or (VV, b'' c'')"

 

[DrChinese]

1. If a BSM is carried out, we can indeed discard runs that aren't Φ- by discarding all runs that don't have one of the signatures (HH, b'' c'') or (VV, b'' c''). All runs with one of these signatures correspond to the subset Φ- when a BSM is carried out.

2. If an SSM is carried out, we lose this filtering ability because the signatures (HH, b'' c'') or (VV, b'' c'') no longer correspond to Φ- when an SSM is carried out.

3. I think it is the counterfactual indefiniteness of QM that is the key. To make an apples to apples comparison, Victor would have to be able to make statements like "I have carried out an SSM and observed one of the signatures (HH, b'' c'') or (VV, b'' c''). If I had instead carried out a BSM, the signature observed would still be (HH, b'' c'') or (VV, b'' c'')"

1. Good.

2. I don't follow, the signatures are the same in both cases (for a BSM and an SSM). You cannot tell them apart! To discern them, you need to know the beam splitter setting. That is a different piece of equipment, completely independent of the detectors.

3. Counterfactual definiteness is not a factor - not in any way. That might be an issue if Bell's Inequality is involved - but this experiment does not rely on Bell in any way. We don't need to assume the signature would have been the same if we had chose swap vs no swap. All we need to know is that the signature is a fair representation of what the apparatus encountered. If you believe that there is no "nonlocal influence" (which is what you say doesn't exist), then that should be no problem for you.

The only issue is whether something HV or VH is reported as HH or VV in error (and vice versa). This does not require an assumption, it is a physical fact one way or the other. That is why I explained the whole subject about beam splitters. Once we decide that the apparatus works as designed, we are gold.

The experiment does make a couple of assumptions. i) That the RNG's selection (swap vs no swap) does not affect the outcomes of the detectors; and ii) That the RNG's selection was "free" in the sense that there was nothing conspiratorial (a la superdeterminism) at play. Of course, assumptions like these exist in all experiments in science. (As I am fond of saying: maybe a form of superdeterminism exists that misleads us into thinking the speed of light corresponds to our value for c. How would we ever know otherwise?)

 

[Morbert]

2. I don't follow, the signatures are the same in both cases (for a BSM and an SSM). You cannot tell them apart! To discern them, you need to know the beam splitter setting. That is a different piece of equipment, completely independent of the detectors.

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

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

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?

I.e. The choice whether or not to perform a BSM doesn't affect the already-carried-out measurements on 1 and 4. It instead affects the runs we choose to discard and the runs we choose to keep.

3. Counterfactual definiteness is not a factor - not in any way. That might be an issue if Bell's Inequality is involved - but this experiment does not rely on Bell in any way. We don't need to assume the signature would have been the same if we had chose swap vs no swap. All we need to know is that the signature is a fair representation of what the apparatus encountered. If you believe that there is no "nonlocal influence" (which is what you say doesn't exist), then that should be no problem for you.

Counterfacctual 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.

 

[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.