A Is Entanglement Swapping Driven by Post-Selection or Bell State Measurement?

[PeterDonis]

the photons arriving at a beam splitter are either both reflected or both transmitted in either scenario.

Yes, but if there are two beam splitters, the photons can't be put into a Bell state by being both reflected or both transmitted. This seems to me to be an obvious fact about the two different experimental setups. If it creates some sort of problem, I'm not sure what.

I don't think it is really interpretation dependent at this point when describing a delayed choice entanglement swap: it's physical and contradicts any causal explanation (where cause precedes effect).

I think one has to be extremely careful to distinguish what we actually can verify in experiments from what we can't.

We can verify in experiments that the initial processes at A and B each produce a pair of particles (one electron and one photon in the OP experiment, two photons in the post #12 "delayed choice" experiment) that, if we just measure them without doing anything else to them, are entangled within each pair, but the pairs are not entangled with each other.

We can verify in experiments that, if we do not do anything to the particles in each pair that are headed for C--i.e., we do not do a BSM or anything else, we just let those photons fly off into the environment and never be heard from again--then measurements of the left-over particles at A and B (the electrons in the OP experiment, or the photons in the post #12 experiment) will show them to be not entangled. We can similarly verify in experiments that if we do separate measurements on the two particles headed for C--such as separate beam splitters for each instead of them both passing through the same one--then, again, measurements of the left-over particles at A and B will show them to be not entangled.

We can verify in experiments that if the two photons at C both pass through a single beam splitter, then we will obtain at C one of two possible results ("event ready" or not) which allow us to separate the results at A and B into two subsets, one of which (the "event ready" one) shows statistics consistent with entanglement of A and B, and the other of which shows statistics consistent with A and B not being entangled.

We can record the times of the A, B, and C measurements, and we can verify that the above statistical results work out the same regardless of the spacetime relationships of those measurements.

We cannot verify in experiments "when the entanglement swap occurs".

We cannot verify in experiments "what the quantum state is" in between the preparations and measurements in the above scenarios.

I'll leave it to you to respond as to whether you think what I've said above is consistent with the statement of yours that I quoted.

 

[DrChinese]

1. Yes, but if there are two beam splitters, the photons can't be put into a Bell state by being both reflected or both transmitted. This seems to me to be an obvious fact about the two different experimental setups. If it creates some sort of problem, I'm not sure what.

2. I think one has to be extremely careful to distinguish what we actually can verify in experiments from what we can't.

We can verify in experiments that the initial processes at A and B each produce a pair of particles (one electron and one photon in the OP experiment, two photons in the post #12 "delayed choice" experiment) that, if we just measure them without doing anything else to them, are entangled within each pair, but the pairs are not entangled with each other.

We can verify in experiments that, if we do not do anything to the particles in each pair that are headed for C--i.e., we do not do a BSM or anything else, we just let those photons fly off into the environment and never be heard from again--then measurements of the left-over particles at A and B (the electrons in the OP experiment, or the photons in the post #12 experiment) will show them to be not entangled. We can similarly verify in experiments that if we do separate measurements on the two particles headed for C--such as separate beam splitters for each instead of them both passing through the same one--then, again, measurements of the left-over particles at A and B will show them to be not entangled.

We can verify in experiments that if the two photons at C both pass through a single beam splitter, then we will obtain at C one of two possible results ("event ready" or not) which allow us to separate the results at A and B into two subsets, one of which (the "event ready" one) shows statistics consistent with entanglement of A and B, and the other of which shows statistics consistent with A and B not being entangled.

3. We can record the times of the A, B, and C measurements, and we can verify that the above statistical results work out the same regardless of the spacetime relationships of those measurements.

We cannot verify in experiments "when the entanglement swap occurs".

We cannot verify in experiments "what the quantum state is" in between the preparations and measurements in the above scenarios.

I'll leave it to you to respond as to whether you think what I've said above is consistent with the statement of yours that I quoted.

1. I know this, and everyone agrees it is true (including of course me). But why does distinguishability matter if you AREN'T changing what happened in the past, you are merely selecting the 245 A/B pairs that already demonstrate entanglement? If one argues (quoting @vanhees71 but he is not the only person saying this):

...the point is to use entanglement swapping to select (or post-select, which doesn't really matter, if QT is correct, and there's no reason to doubt it, including the result of this experiment!) entangled electron pairs. Without this selection the electron pairs are not entangled at all!"

Entanglement swapping is not a "selection"! It's an action, an operation, and this experiment (and others like it) demonstrate so clearly. He even says that without this action (misleadingly labeled "selection"), the entanglement swap fails. 2. I agree with all of this and have never implied otherwise.3. I agree with all of this and have never implied otherwise.

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Who/what I am attacking is very simple: those who claim a delayed entanglement swap merely reveals which As and Bs are entangled. Were that true - and it's not - then there should be no need to do anything more on the photons arriving at C other than to test whether they are both transmitted (or both reflected) and whether they arrive near simultaneously. Why would you need to do anything else (like require them to overlap so they become indistinguishable) ? Again, *I* know that failing to have them overlap (so they are indistinguishable) means there is no swap. But for those that believe that a delayed swap can't affect the past (which it can), they must explain their reasoning here as to why the overlap is necessary - when the other requirements are otherwise met.

The photons from A and B arrive at C. There are 2 event qualifying possibilities (assuming near-simultaneous arrival times):

i) A is transmitted and B is transmitted, causing the "left" detector and the "right" detector to both click.
ii) A is reflected and B is reflected, causing the "left" detector and the "right" detector to both click.
If both detectors click, then you won't know whether it was case i) or case ii). We know this is a requirement (indistinguishability) per quantum theory, I am not questioning this point in any way. This scenario occurs once in 54 minutes on the average.

But why would indistinguishability (overlap) be a requirement if the action at C was not physically a part/cause of the A/B entanglement? So what if you learned whether it was scenario i) or scenario ii)? Would that identify different A/B pairs than the 245 that qualified? After all, you would be arguing that what occurred in the past was already fixed and in no way dependent on what you chose to do (or not do) at a later time. That's what I am asking - a defense of the contrary view.

Because I don't see that anything I am discussing is in fact interpretation dependent. It's the same facts in all interpretations, all interpretations require indistinguishability (again no one questions this). And yet indistinguishability can't really matter unless the entangled A/B pairs depend on that to become entangled, which requires an action in the future to affect the past. That must therefore be an element of any interpretation. Interpretations may describe it somewhat differently, but the essentials must be the same.

 

[vanhees71]

1. I know this, and everyone agrees it is true (including of course me). But why does distinguishability matter if you AREN'T changing what happened in the past, you are merely selecting the 245 A/B pairs that already demonstrate entanglement? If one argues (quoting @vanhees71 but he is not the only person saying this):

...the point is to use entanglement swapping to select (or post-select, which doesn't really matter, if QT is correct, and there's no reason to doubt it, including the result of this experiment!) entangled electron pairs. Without this selection the electron pairs are not entangled at all!"

Entanglement swapping is not a "selection"! It's an action, an operation, and this experiment (and others like it) demonstrate so clearly. He even says that without this action (misleadingly labeled "selection"), the entanglement swap fails.

The "entanglement-swapping protocol" is a typical selection process using projective measurement. You start with two entangled uncorrelated pairs. Then you use one piece of one pair and one piece of the other and perform a (local) measurement on these two pieces allowing to find these two pieces in the possible entangled states. Projecting out only one of these entangled state, i.e., working further with the two other pieces according to the local measurement, selects an entangled state of these two other pieces. The amazing thing is that these two now entangled pieces, which were before entirely uncorrelated, need not have to be in any causal contact to get entangled.

It's also irrelevant, whether you do the projective measurement with one of the pair before or after the measurement on the other (distant) pair or even when the two measurments are space-like separated.

My interpretation, arguing with microcausality of relativistic QFT is, that all this is just due to the entanglement of the initial pairs, i.e., the preparation of the objects you measure, before any of these measurements. There is no mutual causal action due to the measurement processes themselves.

2. I agree with all of this and have never implied otherwise.3. I agree with all of this and have never implied otherwise.

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Who/what I am attacking is very simple: those who claim a delayed entanglement swap merely reveals which As and Bs are entangled. Were that true - and it's not - then there should be no need to do anything more on the photons arriving at C other than to test whether they are both transmitted (or both reflected) and whether they arrive near simultaneously. Why would you need to do anything else (like require them to overlap so they become indistinguishable) ? Again, *I* know that failing to have them overlap (so they are indistinguishable) means there is no swap. But for those that believe that a delayed swap can't affect the past (which it can), they must explain their reasoning here as to why the overlap is necessary - when the other requirements are otherwise met.

The photons from A and B arrive at C. There are 2 event qualifying possibilities (assuming near-simultaneous arrival times):

i) A is transmitted and B is transmitted, causing the "left" detector and the "right" detector to both click.
ii) A is reflected and B is reflected, causing the "left" detector and the "right" detector to both click.
If both detectors click, then you won't know whether it was case i) or case ii). We know this is a requirement (indistinguishability) per quantum theory, I am not questioning this point in any way. This scenario occurs once in 54 minutes on the average.

But why would indistinguishability (overlap) be a requirement if the action at C was not physically a part/cause of the A/B entanglement? So what if you learned whether it was scenario i) or scenario ii)? Would that identify different A/B pairs than the 245 that qualified? After all, you would be arguing that what occurred in the past was already fixed and in no way dependent on what you chose to do (or not do) at a later time. That's what I am asking - a defense of the contrary view.

Because I don't see that anything I am discussing is in fact interpretation dependent. It's the same facts in all interpretations, all interpretations require indistinguishability (again no one questions this). And yet indistinguishability can't really matter unless the entangled A/B pairs depend on that to become entangled, which requires an action in the future to affect the past. That must therefore be an element of any interpretation. Interpretations may describe it somewhat differently, but the essentials must be the same.

 

[DrChinese]

1. The "entanglement-swapping protocol" is a typical selection process using projective measurement. You start with two entangled uncorrelated pairs. Then you use one piece of one pair and one piece of the other and perform a (local) measurement on these two pieces allowing to find these two pieces in the possible entangled states. Projecting out only one of these entangled state, i.e., working further with the two other pieces according to the local measurement, selects an entangled state of these two other pieces. The amazing thing is that these two now entangled pieces, which were before entirely uncorrelated, need not have to be in any causal contact to get entangled.

It's also irrelevant, whether you do the projective measurement with one of the pair before or after the measurement on the other (distant) pair or even when the two measurments are space-like separated.

2. My interpretation, arguing with microcausality of relativistic QFT is, that all this is just due to the entanglement of the initial pairs, i.e., the preparation of the objects you measure, before any of these measurements. There is no mutual causal action due to the measurement processes themselves.

1. Agreed, except for the 2 words in bold. These are loaded with meaning that is subject to debate.

2. The entanglement of A/B is caused by the later BSM at C. It cannot be otherwise. You are essentially saying the photons arrive at C with their quantum properties predetermined AND local (unable to be affected by anything outside a light cone). This violates everything we know about quantum mechanics post-Bell, and does not fit in with any interpretation I am aware of... other than yours.

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I was hoping you would address my explicitly constructed example (OP) with a spirited defense of your ideas. There were 245 events in which the criteria were met for a swap. In your view, when the 2 photons arrive at C, they already possessed the following attributes (in addition to arriving near simultaneously and like polarized):

i) A is transmitted and B is transmitted at a BS, causing the "left" detector and the "right" detector to both click; XOR...
ii) A is reflected and B is reflected at a BS, causing the "left" detector and the "right" detector to both click.

They must fulfill one of the above, regardless of whether they were indistinguishable, because that was precisely what was tested in the actual experiment. The requirement that they be indistinguishable is an ADDITIONAL requirement. So presumably, all 245 events would have also been identified in the less restrictive criteria when the additional requirement is waived. That is basic logic within the locally causal world you advocate.

Of course, if you have a less restrictive set of criteria, you might also pick up additional events over and above the 245 that were actual swaps. The additional number of events, X, might include events in which the A/B Bell test did not include entanglement. And for your idea to make sense, X would need to be large enough that the CHSH calculation on the (245+X) events would fall below 2. That because the X events are not entangled and are presumably uncorrelated.

I am not certain, but I think the CHSH S value for random polarized unentangled pairs is 0. Can anyone confirm? If so, I think you can see the problem here. When you average in the X events with an S of 0 with 245 events with an S of 2.42 (per the actual experiment) you still get a value above 0. But the CHSH calculation on the (245+X) events will be above 0. I say that if this experiment were to be performed, there will be no swaps at all. So you would see S=0 or whatever the expectation would be for uncorrelated A/B.

I say that if you are correct, waiving the indistinguishability requirement for a BSM swap will identify 245+X qualifying pairs in a 220 hour run, yielding an S>0 (due to the 245 entangled pairs in the mix). If I am correct, I say the S value will instead be near 0 (indicating no correlation, since there is no swap and therefore 0 entangled pairs). Not sure how many events to expect but I might guess around 245.

This experiment is physically realizable with no new technology required. It would be helpful if you addressed the substance of my argument.

 

[PeterDonis]

why does distinguishability matter if you AREN'T changing what happened in the past, you are merely selecting the 245 A/B pairs that already demonstrate entanglement?

I should have added this to the list of things we can't verify experimentally. We can't even verify that these are the only two possibilities.

If one argues (quoting @vanhees71 but he is not the only person saying this):

...the point is to use entanglement swapping to select (or post-select, which doesn't really matter, if QT is correct, and there's no reason to doubt it, including the result of this experiment!) entangled electron pairs. Without this selection the electron pairs are not entangled at all!"

Entanglement swapping is not a "selection"! It's an action

What you quote here seems to me to be saying that "entanglement swapping" is both a selection and an action. (Shimmer is a floor wax and a dessert topping...) It is a selection because the results of the C measurement are used to select the subsets of the A/B results. It is an action because, per the quote, "Without this selection the electron pairs are not entangled at all!" In other words, without the C measurement the A/B results will always show no entanglement. I don't think this contradicts any of the things I said, which you said you agreed with.

those who claim a delayed entanglement swap merely reveals which As and Bs are entangled.

Per the above, I don't think @vanhees71 is claiming this. If it were true, the A/B results would show entanglement even if the C measurement were never made. But of course they won't. And I don't think @vanhees71 was claimng that they would.

I don't see that anything I am discussing is in fact interpretation dependent

None of the things that I said can be verified by experiments are interpretation dependent (they can't be if they can be verified by experiments).

The things that I said cannot be verified by experiments are interpretation dependent.

 

[PeterDonis]

My interpretation, arguing with microcausality of relativistic QFT is, that all this is just due to the entanglement of the initial pairs, i.e., the preparation of the objects you measure, before any of these measurements.

I don't see how that can be true, since, if you don't make the C measurement, then the A and B particles are not entangled (and the measurements of them will show this). So the C measurement has a physical effect: it changes the statistics of the A and B measurement results from those that "the entanglement of the initial pairs" would produce.

As I have already pointed out, the effect of the C measurement on the statistics of the A and B measurement results is perfectly consistent with the QFT commutation relations (I use that term instead of "microcausality" because of the objections raised earlier to the latter term), because all of the measurements involved (at A, B, and C) commute.

 

[vanhees71]

I don't see how that can be true, since, if you don't make the C measurement, then the A and B particles are not entangled (and the measurements of them will show this). So the C measurement has a physical effect: it changes the statistics of the A and B measurement results from those that "the entanglement of the initial pairs" would produce.

Indeed, even if you make the C measurement and not select or post-select the outcomes of the measurements on the A and B particles, nothing changes, i.e., you simply see uncorrelated particles, but if you choose subensembles for the outcomes of measurements on the A and B particles, depending on the outcome of the C-measurement, you find that the A and B particles are entangled in each of these subsensembles. That's the astonishing point of "entanglement swapping" and something genuinely quantum.

As I have already pointed out, the effect of the C measurement on the statistics of the A and B measurement results is perfectly consistent with the QFT commutation relations (I use that term instead of "microcausality" because of the objections raised earlier to the latter term), because all of the measurements involved (at A, B, and C) commute.

Of course, everyting is consistent with the QFT commutation relations, which are guaranteed indeed by the microcausality constraints on local observables, and I don't understand, why there should be objections to this very fundamental principle of the theory.

So indeed everything is consistent with the assumption that there is not a causal influence of the C measurement on the outcome of the measurements at A and B but it's a selection, making use of the correlations already present in the initial state, where two entangled particle pairs were prepared, with the two pairs themselves independent from each other but with the subensembles being entangled. As I said, this is a generic quantum feature and cannot be explained with local realistic hidden-variable theories, and this is demonstrated once more by showing the predicted violation of Bell's inequality in accordance with Q(F)T and contradicting local realistic hidden-variable theories.

All the astonishing features of entanglement have been demonstrated today: "teleportation", "entanglement swapping", "violation of Bell's inequality", etc. Also three-particle (e.g., GHZ experiment) and higher entanglement has been realized very successfully. All this is of course in accordance with microcausal relativistic QFT which is used to describe the photons often used to realize these experiments.

 

[DrChinese]

Indeed, even if you make the C measurement and not select or post-select the outcomes of the measurements on the A and B particles, nothing changes, i.e., you simply see uncorrelated particles, but if you choose subensembles for the outcomes of measurements on the A and B particles, depending on the outcome of the C-measurement, you find that the A and B particles are entangled in each of these subsensembles. That's the astonishing point of "entanglement swapping" and something genuinely quantum.

Of course, everyting is consistent with the QFT commutation relations, which are guaranteed indeed by the microcausality constraints on local observables, and I don't understand, why there should be objections to this very fundamental principle of the theory.

So indeed everything is consistent with the assumption that there is not a causal influence of the C measurement on the outcome of the measurements at A and B but it's a selection, making use of the correlations already present in the initial state, where two entangled particle pairs were prepared, with the two pairs themselves independent from each other but with the subensembles being entangled. As I said, this is a generic quantum feature and cannot be explained with local realistic hidden-variable theories, and this is demonstrated once more by showing the predicted violation of Bell's inequality in accordance with Q(F)T and contradicting local realistic hidden-variable theories.

[Glad to see this re-opened... ]

You want it both ways. You say the entanglement exists between sub-ensembles upon preparation, and you say the swap operation is an essential operation which respects forward-in-time causality.

There is no pre-existing A/B entanglement waiting to be "uncovered" or "identified" or whatever as a sub-ensemble. Violation of a Bell inequality proves the A/B bond is strictly dependent on a later action to come into existence.

As I have shown in the OP: You can identify your sub-ensembles by an alternate scheme whereby the indistinguishability requirement is NOT met (but all other requirements are - what I call the Alt-BSM case). There will be no entanglement (according to the predictions of QM). What you assert is incorrect, and it would be helpful if you would address that point directly.

In your view of : The same 245 entangled pairs should be identified in the Alt-BSM experimental version, plus an additional X unentangled pairs (since the selection criteria is less restrictive). But QM does not predict that. There will be no entangled pairs, and the CHSH will reflect that.

 

[vanhees71]

[Glad to see this re-opened... ]

You want it both ways. You say the entanglement exists between sub-ensembles upon preparation, and you say the swap operation is an essential operation which respects forward-in-time causality.

Yes.

There is no pre-existing A/B entanglement waiting to be "uncovered" or "identified" or whatever as a sub-ensemble. Violation of a Bell inequality proves the A/B bond is strictly dependent on a later action to come into existence.

There is no pre-existing A/B entanglement in the full ensemble, independent of what's done at C, but there is A/B entanglement in each subensemble selected based on Bell-test measurements at C. That's the only interpretation I can imagine which does not contradict the microcausality constraint of relativistic QFT, which excludes causal influences between space-like separated events (the C measurement in relation to the A and the B measurements).

As I have shown in the OP: You can identify your sub-ensembles by an alternate scheme whereby the indistinguishability requirement is NOT met (but all other requirements are - what I call the Alt-BSM case). There will be no entanglement (according to the predictions of QM). What you assert is incorrect, and it would be helpful if you would address that point directly.

What do you mean by "indistinguishability requirement"? In any case you need to project the two photons at C to a Bell state, i.e., a superposition, which cannot be expressed as a product state. The photons need not be indistinguishable for that. E.g., you can use the singlet-polarization state, where one photon has the complementary polarization state than the other and they are thus not indistinguishable.

In your view of : The same 245 entangled pairs should be identified in the Alt-BSM experimental version, plus an additional X unentangled pairs (since the selection criteria is less restrictive). But QM does not predict that. There will be no entangled pairs, and the CHSH will reflect that.

What are you referring to here?

 

[grzegorzsz830402]

From 245 pairs that met requirements (coincidence window) 80% showed entanglement.
(Original experimental version).

So, is your predictions is saying that: higher number out of those pairs will show entanglement (example 90% out of 245) in Alt-BSM experimental version?

 

[grzegorzsz830402]

"indistinguishability requirement"

Example:

Alt-BSM location C

Detector 1
Spin UP

Detector 2
Spin Down

Alt-BSM location (AB)

Detector 3
Spin UP

Detector 4
Spin Down

Alt-BSM location C

Detector 1
Spin Down

Detector 2
Spin Up

Alt-BSM location (AB)

Detector 3
Spin Down

Detector 4
Spin Up

Yet, even if it would be 100% detectors correlation.

Then, you could possibly conclude, that whatever reaches D1 and D4 came from one source, and whatever reaches D2 and D3, came from another source.

I see some possibilities where one could disagree (and unfortunately logical ones), yet I will go a little bit further to test something else.

For "indistinguishability requirement" not to be met, you would have to be able to, identify source exactly.
(Or I am possibly wrong about that)

Example:

Source for:
D1 and D4

Location A
"Nitrogen-vacancy center diamond"

Source for:
D2 and D3

Location B
"Nitrogen-vacancy center diamond"And I don't think, I could do that in Alt-BSM experimental version.

 

[DrChinese]

What do you mean by "indistinguishability requirement"? In any case you need to project the two photons at C to a Bell state, i.e., a superposition, which cannot be expressed as a product state. The photons need not be indistinguishable for that. E.g., you can use the singlet-polarization state, where one photon has the complementary polarization state than the other and they are thus not indistinguishable.

From the referenced paper:

"The two photons are then sent to location C, where they are overlapped on a beam-splitter and subsequently detected. If the photons are indistinguishable in all degrees of freedom, the observation of one early and one late photon in different output ports projects the spins at A and B into the maximally entangled state |ψ −| ..."

Hopefully that is not ambiguous. It is standard for entanglement swaps using a BSM to require that the source of each detector click (at the BSM device) be unknown, even in principle. (That is, for example, why the polarizations are made to match.)

In my Alt-BSM version, the photons are distinguishable in one degree of freedom. But they have all the other attributes needed for a BSM. Since you say the BSM merely identifies a sub-ensemble of pairs that will demonstrate entanglement (245 in the experiment), loosening the identification requirements should identify that same sub-ensemble of 245 - as well as include an unknown number (X) of additional pairs. Those X additional pairs might demonstrate entanglement or not, but presumably not; because the experimenters needed the photons to be indistinguishable...

Or perhaps the BSM is a demonstration that quantum nonlocality does not obey Einsteinian causality, because here we have a future action (the BSM at C) affecting the past (Bell test results at A and B).

 

[PeterDonis]

In my Alt-BSM version, the photons are distinguishable in one degree of freedom. But they have all the other attributes needed for a BSM.

I'm not sure I would put it this way. "All the other attributes needed for a BSM", to me, includes both photons going through the same beam splitter. In your Alt-BSM version, they don't. I don't think @vanhees71 is claiming that you can do an entanglement swap if the two photons at "C" don't go through the same beam splitter. I think he's only claiming that you can still do an entanglement swap even if the two photons at C, both going through the same beam splitter in the same narrow time window, don't come into the beam splitter with the same values for all observables, such as polarization.

Whether the latter claim is actually correct is a different question. The key indistinguishability for the measurement at C is that it must not be possible to tell which photon goes to which output port--i.e., that we can't say "the photon in output port #1 came from A, and the photon in output port #2 came from B" (or vice versa). This will be the case if the two photons have the same values for all observables (e.g., polarization) when they come in, or, I think, if we don't measure at the output ports any observables in which they differ (e.g., if they differ in polarization when they come in, we can't measure polarization at the output ports). The latter seems to be the kind of case @vanhees71 is describing. But I haven't looked in detail at the math to see if it would actually work.

 

[DrChinese]

I'm not sure I would put it this way. "All the other attributes needed for a BSM", to me, includes both photons going through the same beam splitter. In your Alt-BSM version, they don't. I don't think @vanhees71 is claiming that you can do an entanglement swap if the two photons at "C" don't go through the same beam splitter. I think he's only claiming that you can still do an entanglement swap even if the two photons at C, both going through the same beam splitter in the same narrow time window, don't come into the beam splitter with the same values for all observables, such as polarization.

OK, no problem, then they go through the same beam splitter and go to the same detectors. (Again, I am not saying there will be entanglement if they are distinguishable.) All you need to do is add about 500 meters of fiber to one side so that one is delayed far past the point where they needed to be able to interact (or whatever it is they do when they are indistinguishable). Interacting shouldn't matter if we are simply revealing pre-existing attributes, right?

Photons coming to location C from A and B:
1. Same polarization: indistinguishable, check.
2. Same detectors click: indistinguishable, check.
3. Same beam splitter: indistinguishable, check.
4. Both reflect or both transmit: indistinguishable, check.
5. Both same wavelength: indistinguishable, check.
6. Same time narrow time window (assuming they were allowed to interact, otherwise adjusted for path length distance): indistinguishable, check.
7. Allowed to cross paths (and/or interact) within the narrow time frame: a) indistinguishable for actual BSM, b) distinguishable for the Alt-BSM version. Alt-BSM allows source to be determined because of the delay due to added fiber length (or you can distinguish them any other way you might choose, the result is the same).

The point is the same: all needed criteria met EXCEPT #7 (which allows the source to be determined). Classical logic (if that could be applied, which it can't in QM) dictates that if we are merely identifying the subset of events which meets certain criteria, then eliminating one criterion will identify the same 245 events plus X additional events, where X>=0.

With classical Einsteinian causality, no action (or lack thereof) can lead to Bell entanglement in the past. Obviously, quantum mechanics does NOT respect classical Einsteinian causality, as we know the indistinguishability requirement *must* be present for entanglement swapping. And none of the authors of any of these papers (for god's sake how many do I need to quote) say otherwise. I say that QFT is no more respecting of local causality than QM, regardless of how it is "constructed".

There has been a sea change, and the idea that Einsteinian causality can still be supported in quantum theory has flown the coop. We really knew that from Bell 1964 and Aspect 1981, but we have moved far past that in the past 10-20 years. Today's interpretations must account fully and explicitly for delayed (to the future) entanglement swapping. Or they must be left behind.

 

[PeterDonis]

Alt-BSM allows source to be determined because of the delay due to added fiber length.

Is the delay imposed on one photon before it passes through the same beam splitter as the other, or after?

If it's before, then of course the "narrow time window" requirement is no longer met, and I expect that @vanhees71 would agree that there would now be no entanglement swapping.

If it's after, then the "narrow time window" requirement is still met, it's just that one photon gets delayed before it is measured--but as long as that photon is kept isolated in the fiber or whatever it gets delayed in, the delay won't affect the results and entanglement swapping will still take place (for the runs where an "event ready" result is obtained). And I expect that @vanhees71 would agree with that as well.

Why do I think @vanhees71 would agree with the above? Because, as far as I can see, that's what the math of standard QM predicts, and he believes that whatever the math of standard QM predicts is what experiments will show.

 

[DrChinese]

Is the delay imposed on one photon before it passes through the same beam splitter as the other, or after?

If it's before, then of course the "narrow time window" requirement is no longer met, and I expect that @vanhees71 would agree that there would now be no entanglement swapping.

If it's after, then the "narrow time window" requirement is still met, it's just that one photon gets delayed before it is measured--but as long as that photon is kept isolated in the fiber or whatever it gets delayed in, the delay won't affect the results and entanglement swapping will still take place (for the runs where an "event ready" result is obtained). And I expect that @vanhees71 would agree with that as well.

Why do I think @vanhees71 would agree with the above? Because, as far as I can see, that's what the math of standard QM predicts, and he believes that whatever the math of standard QM predicts is what experiments will show.

Well, if they don't interact - what difference would it make if you argue that you are merely revealing pre-existing attributes? Look at the checklist again. You should walk away realizing that IF there is no interaction (because they are distinguishable) THEN in fact the past has changed and there is no entanglement swap.

Of course, QM says that the interaction is necessary for the 245! But again, if you drop that requirement - and STILL insist the attributes were pre-existing - then you will identify 245+X events. This is basic logic, except that it requires an invalid assumption that @vanhees71 is clinging to: that the critical attributes are all pre-existing. Obviously, this assumption must be dropped.

 

[PeterDonis]

if you make the C measurement and not select or post-select the outcomes of the measurements on the A and B particles, nothing changes, i.e., you simply see uncorrelated particles

No, you don't. You see a mixture of the "entangled" and "not entangled" statistics. "Uncorrelated particles" would be all "not entangled" statistics--i.e., what you see if you don't make the C measurement. The C measurement changes the observed statistics whether you post-select or not; post-selection just helps you to understand why the overall statistics changed: because a subset of the runs now have the particles measured at A and B entangled.

 

[PeterDonis]

an invalid assumption that @vanhees71 is clinging to: that the critical attributes are all pre-existing. Obviously, this assumption must be dropped.

I think what I pointed out in post #79 just now might be relevant to this. The only way the attributes could all be "pre-existing" is if the overall statistics are the same regardless of whether the C measurement is made or not. But they aren't; they can't be, because no pairs of particles measured at A and B are entangled if the C measurement is not made, but a subset of them are if the C measurement is made.

 

[akvadrako]

No, you don't. You see a mixture of the "entangled" and "not entangled" statistics. "Uncorrelated particles" would be all "not entangled" statistics--i.e., what you see if you don't make the C measurement. The C measurement changes the observed statistics whether you post-select or not; post-selection just helps you to understand why the overall statistics changed: because a subset of the runs now have the particles measured at A and B entangled.

Are you saying it changes the overall statistics? That cannot be true since it would allow signaling.

In the delayed choice version, the measurement results are already determined. Nothing about the statistics can change.

There is also no reason to think they do. It's always possible to bin measurement results into 4 buckets that have the same statistics as if they were entangled. Without QM you need the results to do that, so it can only be post-selection.

 

[vanhees71]

No, you don't. You see a mixture of the "entangled" and "not entangled" statistics. "Uncorrelated particles" would be all "not entangled" statistics--i.e., what you see if you don't make the C measurement. The C measurement changes the observed statistics whether you post-select or not; post-selection just helps you to understand why the overall statistics changed: because a subset of the runs now have the particles measured at A and B entangled.

This I don't understand. Can you provide the corresponding calculation that the statistics of measurements on the photons 1 and 4 change for the full enemble only by measuring photons 2 and 3? I don't see, how this can be.

The state of photons 1 and 4 is given by partially tracing the state over photons 2 and 3, and this leads to an non-entangled state of photons 1 and 4. The state is $|\Psi \rangle \langle \Psi$
$$|\Psi \rangle = |\psi_{12} \rangle \otimes |\psi_{34} \rangle$$
with $|\psi_{12} \rangle$ a Bell state for photon pair 1+2 and $|\psi_{34}$ a Bell state for photon pair 3+4 (or any other pure state for either pair for the matter of this exercise). Then the reduced state for photons 1 and 4 is
$$\hat{\rho}^{(14)} = \mathrm{Tr}_{23} \hat{\rho}.$$
Let $|\alpha \rangle$ be a complete one-photon basis. Then
$$\rho^{(14)}_{\alpha \delta,\alpha' \delta'}=\rho^{(1)}_{\alpha \alpha'} \rho^{(4)}_{\delta \delta'}$$
with
$$\rho_{\alpha \alpha'}^{(1)} = \sum_{\beta} \langle \alpha \beta|\psi_{12} \rangle \langle \psi_{12}|\alpha' \beta \rangle, \quad \rho_{\delta \delta'}^{(4)} = \sum_{\gamma} \langle \gamma \delta |\psi_{34} \rangle \langle \psi_{34}|\gamma \delta' \rangle,$$
i.e.,
$$\hat{\rho}^{(14)}=\hat{\rho}^{(1)} \otimes \hat{\rho}^{(2)}, \qquad (*)$$
i.e., the state of the pair 1+4 is factorizing, i.e., it's not entangled.

If you project however to any of the 4 possible Bell states of photons 2+3, you get an entangled state for the photons 1+4. The full ensemble is still described by the product state (*).

 

[vanhees71]

To be able to "put them in buckets according to the $P_i$", however, you have to perform the corresponding measurement on BC. If you ignore BC completely you just look at the full ensemble.

 

[vanhees71]

But I don't think that this is what's understood by "entanglement swapping", which has the aim to prepare entangled states of two particles (here photons 1+4) which were never "in causal contact" with each other, and this can only be achieved by making a measurement which allows to project the pair 2+3 into entangled states. Then also pair 1+4 are entangled in each of the corresponding subensembles. The math is pretty simple, as detailed in

https://doi.org/10.1103/PhysRevLett.80.3891

 

[DrChinese]

1. Are you saying it changes the overall statistics? That cannot be true since it would allow signaling.

2. In the delayed choice version, the measurement results are already determined. Nothing about the statistics can change.

3. The issue is very simple and @vanhees71 is completely right that entanglement swapping is perfectly consistent with locality.
...

4. (One misconception of @DrChinese is to confuse the full ensemble $\rho_{ABCD,\text{after}}$ with the subensemble ##\frac{P_i\rho_{ABCD,\text{before}}P_i}{\mathrm{Tr}(P_i\rho_{ABCD,\text{before}}P_i)}#

5. I just wanted to demonstrate that the full ensemble contains 4 entangled subensembles even if no measurement is performed at BC. @DrChinese and @PeterDonis denied this earlier.

1. No it never allows signaling. There is a random element introduced, and you need to know all the results from locations A, B *and* C to decode anything. Not much in the way of FTL possibilities there.2. That's completely wrong, and you (@akvadrako) are not addressing my example. If the "measurement results are already determined" (your words), then dropping a single criteria for indistinguishability (Alt-BSM scenario) will detect all 245 entangled A/B events (found in the actual experiment, CHSH/S=2.42) plus X additional that won't be entangled (S<=2, presumably S=0). QM actually predicts there will be the same 245 but they will not be entangled (S<=2) in the distinguishable case.

Indistinguishability is an absolute requirement, so my Alt version will demonstrate that the results are not predetermined. Your view is practically ignoring everything Bell has taught us anyway. It's a Bell test! The results cannot be predetermined unless there are FTL influences!! Remember the part about ruling out local realism? 3. Again, I guess we have forgotten the whole deal about local realism - what you describe as occurring here - being ruled out by thousands of experiments. Quantum nonlocality - whatever you think that is - has been demonstrated and must be accounted for in any and every interpretation. Even @vanhees71 acknowledges that a biphoton (entangled system of 2 photons) has spatial extent and cannot be considered "local" in any respect.4. Using the 1/2/3/4 notation you (@Nullstein) are using:

Before the swap, biphoton (1 & 2) and biphoton (3 & 4) are separately in maximally entangled states; together they are in a Product state (1 & 2) * (3 & 4). There are no 1 & 4 pairs in that group that are entangled, period. And obviously so. No particle can be maximally entangled with more than one other particle: monogamy of entanglement. If 1 is entangled with 2, it cannot be entangled with 4 also.

The only way to get 1 & 4 to be entangled is by a swap operation. That can be done in the future, defying Einsteinian causality.5. Hopefully point 4 explains why you are mistaken, there are never more than 2 entangled sub-ensembles. Or maybe you have a suitable quote otherwise?

 

[PeterDonis]

Are you saying it changes the overall statistics?

Yes.

That cannot be true since it would allow signaling.

No, it doesn't, because you need to have all the data in order to do the statistics that show the entanglement, and you don't have that until all of the measurement events are in your past light cone. There's no way to signal FTL. This is a feature of any set of measurements on entangled (or possibly entangled) particles.

 

[PeterDonis]

Can you provide the corresponding calculation that the statistics of measurements on the photons 1 and 4 change for the full enemble only by measuring photons 2 and 3? I don't see, how this can be.

It's simple:

If you don't do the BSM on photons 2 and 3, the statistics of all measurements on photons 1 and 4 are "not entangled".

If you do the BSM on photons 2 and 3, the statistics of all measurements on photons 1 and 4 are a mixture of "not entangled" (for the runs where the BSM did not give an "event ready" signal) and "entangled" (for the runs where the BSM did give an "event ready" signal).

These are different overall statistics.

 

[PeterDonis]

The state of photons 1 and 4 is given by partially tracing the state over photons 2 and 3

Partially tracing the overall state of all four photons, yes. But this overall state is different if you don't do the BSM at all (which is the state you are evaluating here), vs. if you do the BSM but don't look at its result (because you now have a mixture of the two possible results of the BSM).

 

[vanhees71]

Partially tracing the overall state of all four photons, yes. But this overall state is different if you don't do the BSM at all (which is the state you are evaluating here), vs. if you do the BSM but don't look at its result (because you now have a mixture of the two possible results of the BSM).

If you mean by "do the BSM" to include also the projection to the corresponding subensemble, we agree. If you do the measurement without selecting the subensemble, the full ensemble is still the same. That's why you get entangled 1&4-pairs only for a part of the prepared four-photon states in the entanglement-swapping protocol.

 

[PeterDonis]

If you do the measurement without selecting the subensemble, the full ensemble is still the same.

Your basis for the latter statement appears to be that there are four entangled subensembles, one for each of the four Bell states, and the correlations between them cancel out if all four are combined. But in the experiment as described in the paper, there is only one Bell state of photons 1 and 4 that is produced by the BSM (on the runs where the BSM gives an "event ready" signal). See p. 3 of the paper referenced in the OP, the second paragraph in the right column.

 

[DrChinese]

If you mean by "do the BSM" to include also the projection to the corresponding subensemble, we agree. If you do the measurement without selecting the subensemble, the full ensemble is still the same. That's why you get entangled 1&4-pairs only for a part of the prepared four-photon states in the entanglement-swapping protocol.

There is no "prepared four-photon state". There are 2 biphotons, and like any 2 independently prepared quantum objects, their overall state is a Product state.

 

[PeterDonis]

in the experiment as described in the paper, there is only one Bell state of photons 1 and 4 that is produced by the BSM (on the runs where the BSM gives an "event ready" signal). See p. 3 of the paper referenced in the OP, the second paragraph in the right column.

In an effort to dig further into this (since the discussion in the paper referenced in the OP is very sketchy), I have looked up two further references, from those given in the paper.

Further reference on the "event-ready" Bell setup (ref 16 in the paper) is here:

https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.71.4287

Further reference on the "Barrett-Kok scheme" for entanglement swapping (ref 26 in the paper):

https://arxiv.org/abs/quant-ph/0408040

Further comment after I've taken some time to digest these.