Interpreting photon correlations from independent sources

[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.

 

[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.