Interpreting photon correlations from independent sources

[msumm21]

No, it isn't. In your temperature example, the correlations will never violate the Bell inequalities. In the QM example, they will.

This is a different point. You could also argue temperature is not the same thing as polarization of course.

I think you already agreed above that the aggregate of all BSM results is the same as the aggregate of all SSM results. Do you still believe that? If so, then you must agree that any subset with particular statistics (including Bell violations) in one set is also in the other set, right?

No, you can't when it's apples to apples. The data is selected from 2&3 results being VV or HH. You get completely different results when the swap switch is on.

Yes, but again you're post selecting a subset with a particular P23 result and ignoring VH/HV. If you could determine BEFORE the experiment that the result would be VV or HH then I'd agree with you. The point is that you can't, there are 4 different resulting states for each measurement type (in the Ma case I think they'd discarded half of those for technical reasons, so perhaps you're looking at something that was filtered to remove that 50% of the data). The aggregate of all BSM result includes all 4 Bell states and looks completely random AND the aggregate of all SSM results looks completely random. Only when you pick out subsets based on the BSM/SSM results do you see differences.

 

[msumm21]

1. This concept is directly contradicted by co-author Zeilinger in an earlier paper. "We confirm successful entanglement swapping by testing the entanglement of the previously uncorrelated photons 1 and 4."

This is not contradictory. Please here me out. Zeilinger is testing entanglement within the subsets we've discussed, where there's a specific entangled state, e.g. phi- (a quarter of the aggregate data). Each P14 in this subset is certainly in this entangled state and no one is questioning that as far as I can tell. Everyone agrees this subset can show Bell violations, ....

What I think thomsj4 is saying is the same as what I'm trying to say: the aggregate of all the data is NOT in that entangled state. Furthermore, the aggregate of all the data is not even different swap/no swap (including all 4 measurement results in both cases).

2. It should be obvious that when there is no swap, there is no relationship whatsoever between the pairs 1&2

Sounds like everyone agrees on this. The disagreement is apparently in the BSM case, the fact that aggregate of all results also shows no relationship either (only the post-selected subsets pick out correlations associated with specific Bell states).


I need to move on to other things, but I'd urge doubters to do the math thomsj4 showed or read carefully Ma or better yet Peres on "delayed choice entanglement swapping" -- might be clearer in the Peres paper.

 

[PeterDonis]

This is a different point.

No, it's not. Correlations that violate the Bell inequalities is an essential part of why we need QM to explain the entanglement swapping results--classical models won't work. The same is not true of temperature, so your analogy with temperature is fatally flawed.

I think you already agreed above that the aggregate of all BSM results is the same as the aggregate of all SSM results.

No, that's not what I have agreed to. What I have agreed to is a much more limited claim.

you must agree that any subset with particular statistics (including Bell violations) in one set is also in the other set

No. That fact that the statistics of the particular subsets that are picked out by the four possible combinations of 2&3 results are not the same in the two cases is exactly the point of the much more limited claim I made. Are you reading my posts? Because it sure doesn't seem like it.

 

[PeterDonis]

the aggregate of all the data is not even different swap/no swap

Yes, it is. I have already explained why this claim is wrong: particular subsets show Bell state correlations in one case (swap made) but not in the other case (no swap made). That also means that in the swap case, Bell inequality violating correlations are present that are not present in the no swap case.

I strongly suggest that you stop posting until you have read and considered the above enough to either agree with it (and therefore withdraw the claims you have made to the contrary in this thread), or take the time to find a reference that explains why it is wrong (which you won't be able to because it isn't). If you continue to make the claims you have been making, and continue to ignore the rebuttals you have received, you are likely to receive a warning.

 

[PeterDonis]

Distinguishability, however, destroys these perfect correlations

the distinguishability of $P2$ and $P3$ determine which subset of the ensemble of $P1$ and $P4$ measurements will exhibit these perfect correlations.

These two statements of yours contradict each other.

 

[PeterDonis]

I'd urge doubters to do the math thomsj4 showed

@DrChinese has already responded to that post.

 

[iste]

What I think thomsj4 is saying is the same as what I'm trying to say: the aggregate of all the data is NOT in that entangled state. Furthermore, the aggregate of all the data is not even different swap/no swap (including all 4 measurement results in both cases).

Yes, I believe the 5th equation in their thomsj4's post implies what you arr saying about aggregates.

 

[iste]

I have some additional thoughts on entanglement swapping, I think.

Entanglement swapping is like a special case of quantum teleportation, no? i.e:

https://scholar.google.co.uk/scholar?cluster=201348324163392972&hl=en&as_sdt=0,5&as_vis=1

"Quantum physics predicts [1] that once particles 1 and 2 are projected into |ψ−>12, particle 3 is instantaneously projected into the initial state of particle 1."

And then the authors give the explanation next:

"The reason for this is as follows. Because we observe particles 1 and 2 in the state |ψ−>12 we know that whatever the state of particle 1 is, particle 2 must be in the opposite state, that is, in the state orthogonal to the state of particle 1. But we had initially prepared particle 2 and 3 in the state |ψ−>23, which means that particle 2 is also orthogonal to particle 3. This is only possible if particle 3 is in the same state as particle 1 was initially."

It appears to me this is entirely epistemic in the sense that all that is happening is that the conjunction lf the Bell state measurement and previous entangled state is allowing you to make an inference about the state of particle 3 through particle 1. The entanglement is a regular entanglement only possible due to an initial local interaction it seems; the Bell state measurement is also surely locally mediated. As entanglement swapping is just a special case of quantum teleportation, it seems to me we don't prima facie need any additional kind of non-locality beyond the regular kind which is entirely responsible for the quantum teleportation.

Entanglement swapping is nothing more than a juxtaposition or melding of two instances of this basic quantum teleportation scenario. Quantum teleportation scenario with particles 1, 2 and 3 is combined with another quantum teleportation scenario with particles a, b, c to make an entanglemet swapping scenario of particles c, 1=b, 2=a, 3 which then corresponds to the pairs that have been conventionally referred to in the forum as 1&2 (c, 1 = b) and 3&4 (2 = a, 3).

If you add a particle 4 to the previous described quantum teleportation with particles 1, 2, 3 mentioned in the quotes earlier (4 will be entangled to 2) - and then follow the reasoning of the quotes - it seems now that particle 1 is telling you about the states of both particle 3 and particle 4 (in conventional terminology of the forum this is then particle 2 telling you about both 1 and 4). We can also equally view this from the perspective of quantum teleportation scenario with particles a, b, c but adding particle d; particle a is now telling you things about particles c and d (conventional terminology this would be particle 3 telling you about particles 1 and 4).

Obviously, particles 1=b and 2=a (or 2&3 conventionally) are subject to a BSM and so are non-separable. In conventional terminology, we then just have this Bell state telling you information about 3 and 4. Surely then, conditioning on the Bell state, particle 3 is telling you directly about particle 4 and vice versa. Obviously, to do this does still require the coupling of 2&3. Its then an interesting coincidence that Mjelva (2024) suggests maximal Bell violations for 1 & 4 can be derived under this kind of consideration without 1 & 4 being explicitly projected onto a non-separable state (section 4), the basis of their claim that post-selection may be sufficient for entanglement-swapping phenomena - albeit this is a philosopher publishing in a philosophy journal:

https://scholar.google.co.uk/scholar?cluster=10636160464314492908&hl=en&as_sdt=0,5&as_vis=1

So ot doesn't seem to me that we need anything additional to the regular entanglement of 1&2 and regular entanglement of 3&4 to explain why the Bell state measurement results in the non-separability of 1&4, which is then only for a 1/4 subset of the original state. Combining all the subsets then results in a mixed / separable state with no correlations, consistent with the initial state. Based on the implications of the earlier two quotes in this post, entanglement swapping is epistemic, extending from the epistemic nature of quantum teleportation (at least this seems a tenable interpretation of quantum teleportation to me, once you take into account the contextuality / non-locality of regular entanglement). We are just learning about correlations in particles 1&4 that are unveiled in this particular network of systems when you condition on the non-separable Bell state measurement that couples 2 & 3 together in different subsets of the initial state.

 

[javisot20]

It appears to me this is entirely epistemic in the sense that all that is happening is that the conjunction lf the Bell state measurement and previous entangled state is allowing you to make an inference about the state of particle 3 through particle 1. The entanglement is a regular entanglement only possible due to an initial local interaction it seems; the Bell state measurement is also surely locally mediated. As entanglement swapping is just a special case of quantum teleportation, it seems to me we don't prima facie need any additional kind of non-locality beyond the regular kind which is entirely responsible for the quantum teleportation.

https://arxiv.org/abs/quant-ph/0201134

 

[iste]

https://arxiv.org/abs/quant-ph/0201134

I suggest you read the post more carefully.

 

[javisot20]

I suggest you read the post more carefully.

As entanglement swapping is just a special case of quantum teleportation, it seems to me we don't prima facie need any additional kind of non-locality beyond the regular kind which is entirely responsible for the quantum teleportation.

I agree with the relationship between entanglement (swapping) and quantum teleportation protocols, but I did not know that there is a bad relationship between non-locality and quantum teleportation protocols, in this paper I see the opposite. (Possibly what you say is true, I'm going to read your post more carefully, thank you)

 

[iste]

I agree with the relationship between entanglement (swapping) and quantum teleportation protocols, but I did not know that there is a bad relationship between non-locality and quantum teleportation protocols, in this paper I see the opposite. (Possibly what you say is true, I'm going to read your post more carefully, thank you)

I'm not saying there is no entanglement in 1&4, I am suggesting that you don't need anything to explain entanglement in 1&4 beyond the entanglements in 1&2, 3&4 and the Bell state of 2&3. From this view, entanglement in 1&4 is an emergent feature of those mechanisms without requiring some kind of additional element to cause the entanglement at 1&4 beyond the local interactions that caused non-separability for each of 1&2, 2&3, 3&4. The consequence is 2&3 is giving you information about both 1 and 4. So given 2&3, 1 is always telling you about 4 and vice versa, which might then be the resultant 1&4 entanglement.

 

[DrChinese]

1. Entanglement swapping is like a special case of quantum teleportation, no? i.e: ...

2. the Bell state measurement is also surely locally mediated...

1. This is a wild goose chase. The subject has been laid out concisely, we're discussiong entanglement swapping from independent sources. Yes, you can say that quantum teleportation is a related form of entanglement swapping, and vice versa. But absolutely nothing is different operationally between the two. All of the same nonlocality is present, and there is no causal ordering requirement.

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!

2. "Locally mediated" is a meaningless term here. The BSM overlap of photons 2&3 in the beamsplitter would naturally always be "local", otherwise they don't overlap (by definition). But the BSM itself is not local in any fashion to the observation of either photon 1 or photon 4.

 

[iste]

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!

Its not a problem because I never disagreed with that in the first place. 2 & 3 are entangled, and 1 was entangled to 2.

2. "Locally mediated" is a meaningless term here. The BSM overlap of photons 2&3 in the beamsplitter would naturally always be "local", otherwise they don't overlap (by definition). But the BSM itself is not local in any fashion to the observation of either photon 1 or photon 4.

Well all I mean by locally mediated is that the BSM is a local interaction in the same way that the 1&2 and 3&4 entanglements were mediated by an initial local interaction.