Exploring Entanglement Swapping: Post-Selection vs. BSM?

[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.

 

[PeterDonis]

"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 −| ..."

In the paper referenced in the OP, it is not made very clear what happens on those runs of the experiment where "the observation of one early and one late photon in different output ports" does not occur. In the two further references I gave in post #62, and in other entangement swapping experiments where two pairs of photons are used instead of two photon-electron pairs, such as this one, it seems clear that on runs where the BSM at C (or its equivalent) does not produce a photon in different output ports, this could be due to either of two things: (1) the photons didn't arrive at all within the required time window, and the BSM did nothing; or (2) the photons were projected into one of the other three possible Bell states (one of the "triplet" states as opposed to the "singlet" state), which would mean that the particles at A and B would also be projected into some other entangled state. In other words, on some fraction of runs where no "event ready" signal was observed, there would still be entanglement swapping--just entanglement swapping that the experimenters chose not to try to detect.

Does this seem correct?

 

[akvadrako]

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.

By overall statistics I meant of the full 1&4 ensemble. The thing one can do is use the results of the BSM operation on 2&3 to select a sub-ensemble of 1&4, then perform a Bell test on that sub-ensemble and see that it violates the Bell inequalities, indicating that sub-ensemble was entangled. Without selecting that sub-ensemble, no statistical changes can be detected due to the operation on 2&3.

But there exist sub-ensembles of 1&4 which violates Bell inequalities even if you throw away 2&3. You can see this by just doing the Bell test on 1&4 and using the results to select the sub-ensemble. This is always possible and can even be done to select super-quantum correlations, though it can't be used to create usable entanglement.

So how do you know that your operation on 2&3 is not just doing selection? That is strongly suggested in the post-selection version where the measurements on 1&4 have already been done.

 

[akvadrako]

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.

Yes, you need the results of C to select a sub-ensemble of AB to see any new statistics.

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.

My answer is if you don't do the operation at C correctly, you are not selecting the correct sub-ensemble of AB. The BSM is required to do that.

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?

As I'm sure you are aware, having been discussed on the forum many times, Bell's theorem doesn't require FTL influences because it makes additional assumptions. There are explicitly local formulations of QM, but this is quite off-topic.

 

[vanhees71]

There are no faster-than-light signals necessary to explain the entanglement swapping, which is evident, because the exepriment is well described by relativistic microcausal QFT, where faster-than-light signals are excluded by construction. It's also underlined by the fact that many Bell tests are performed where the detection events are spacelike separated, i.e., there is no temporal order of the projection to the selected Bell state of photon pair 2+3 and the measurements on photons 1+4, i.e., you cannot call the one the cause of the other.

The possibility to entangle photons 1+4 without getting the into any "causal contact" with each other through making local measurements on photons 2+3 is due to the entanglement of the pairs 1+2 and 3+4.

Also see the 1st reference in @PeterDonis posting #60.

 

[PeterDonis]

there exist sub-ensembles of 1&4 which violates Bell inequalities even if you throw away 2&3.

If you mean, not do the 2&3 operation at all, no, this is not correct.

You can see this by just doing the Bell test on 1&4 and using the results to select the sub-ensemble.

Please give a reference to an actual experiment that does this.

On its face, it seems obviously false. After the initial preparation in the experiment, without any operation done on 2&3 afterwards, 1&4 are not entangled. That means the results of measurements done on them under those conditions will show no correlations and will not violate the Bell inequalities. No amount of jiggering with subensembles can change that, because there are no subensembles. There is nothing that picks out any subset of the runs, because nothing is done to any of the particles other than the initial preparation and the 1&4 measurement.

Doing the 2&3 measurement changes the correlations between 1&4; you now have subensembles, picked out by the measurement results at 2&3, and you can do statistics on the subensembles to show that, at least, the one corresponding to the "event ready" result at 2&3 violates the Bell inequalities. But this change cannot be detected until you know the measurement results from 2&3 so you can pick out the subensembles. That is why no FTL signaling is possible. But "no FTL signaling" is not the same as "no change in the statistics of the 1&4 measurement results". The latter is what you are claiming, and it is wrong.

 

[DrChinese]

1. By overall statistics I meant of the full 1&4 ensemble.2. The thing one can do is use the results of the BSM operation on 2&3 to select a sub-ensemble of 1&4, then perform a Bell test on that sub-ensemble and see that it violates the Bell inequalities, indicating that sub-ensemble was entangled. Without selecting that sub-ensemble, no statistical changes can be detected due to the operation on 2&3.

3. But there exist sub-ensembles of 1&4 which violates Bell inequalities even if you throw away 2&3. You can see this by just doing the Bell test on 1&4 and using the results to select the sub-ensemble.

4. This is always possible and can even be done to select super-quantum correlations, though it can't be used to create usable entanglement.

5. So how do you know that your operation on 2&3 is not just doing selection? That is strongly suggested in the post-selection version where the measurements on 1&4 have already been done.

1. Just so you know: In the Hensen et al swapping experiment, there were 19,762,613 pairs in the full 1 & 4 dataset (labeled as A/B pairs in that paper). Only 245 qualified for the swap. So that's the kind of ratio we are looking at, there are only a few successes compared to the total number of attempts.

2. That's true, except it's more than just selecting. The selection casts the associated 1/4 pairs into an entangled state.

3. If you did a Bell test on the full dataset without performing a swap, you would see absolutely NO correlation whatsoever.

And in fact there are literally no subsets of that same full dataset that would show any Bell inequality is violated unless you hand select the dataset to have that attribute. I could demonstrate entanglement of random pairs of cars in London using that technique.

4. I am scared to ask but... what do "super-quantum correlations" have to do with this discussion? I am not sure they even exist.

5. My assertion is that you can attempt to select on all of the necessary requirements except one. Look at that new, presumably larger dataset, and see whether a Bell inequality is violated, or close to it. Presumably a portion would be the same group that was actually selected, since the criteria is LESS restrictive than before. But I say there will be no correlation because no swap was executed.

 

[martinbn]

If you mean, not do the 2&3 operation at all, no, this is not correct.Please give a reference to an actual experiment that does this.

On its face, it seems obviously false. After the initial preparation in the experiment, without any operation done on 2&3 afterwards, 1&4 are not entangled. That means the results of measurements done on them under those conditions will show no correlations and will not violate the Bell inequalities. No amount of jiggering with subensembles can change that, because there are no subensembles. There is nothing that picks out any subset of the runs, because nothing is done to any of the particles other than the initial preparation and the 1&4 measurement.

But you can simply do the measurments on 1&4 and based on the outcomes select a subset with desired corelations.

Doing the 2&3 measurement changes the correlations between 1&4; you now have subensembles, picked out by the measurement results at 2&3, and you can do statistics on the subensembles to show that, at least, the one corresponding to the "event ready" result at 2&3 violates the Bell inequalities. But this change cannot be detected until you know the measurement results from 2&3 so you can pick out the subensembles. That is why no FTL signaling is possible. But "no FTL signaling" is not the same as "no change in the statistics of the 1&4 measurement results". The latter is what you are claiming, and it is wrong.

The way you phrase it, it is not true. The statistics of 1&4 does not change because the density matrix does not change. This is exactly the same as in the usuall Alice and Bob scenario. The measurements of A do not change the statistics of B.

 

[martinbn]

2. That's true, except it's more than just selecting. The selection casts the associated 1/4 pairs into an entangled state.

This is an interpretation dependent statement. Just as in the usual Bell case whether one measurment cast the other subsystem into a certain state depends on the interpretation you choose.

 

[akvadrako]

Please give a reference to an actual experiment that does this.

On its face, it seems obviously false. After the initial preparation in the experiment, without any operation done on 2&3 afterwards, 1&4 are not entangled. That means the results of measurements done on them under those conditions will show no correlations and will not violate the Bell inequalities. No amount of jiggering with subensembles can change that, because there are no subensembles. There is nothing that picks out any subset of the runs, because nothing is done to any of the particles other than the initial preparation and the 1&4 measurement.

My definition of sub-ensemble is just a subset of a full ensemble, so there are always sub-ensembles, like taking every odd result. When one performs the delayed-choice version of the experiment, for the 1&4 ensembles, one is left with a list of measurement settings and results.

There are two ways to pick out sub-ensembles which violate Bell inequalities. The first is to simply look at the results and pick the ones which give you the correlations you are looking for. @Nullstein gave an example of this above and it's also spelled out clearly in this paper, which I brought up last time this was discussed:

Bell Inequality Violation and Relativity of Pre- and Postselection

The second way is doing the BSM operation. The BSM operation might even select the same sub-ensemble that was selected by just looking at the results.

Another way to think about this is: I know there exist local interpretations of QM, in the sense of no FTL causal effects. That precludes the BSM operation at 2&3 affecting 1&4 as the cause, as 1&4 results are already determined. Same for 1&4 operations affecting 2&3, if far apart. The only way it can work is post-selection.

So that's two examples of ways to create Bell-inequality violating sub-ensembles via only selection. Since it's possible in these cases, I suspect that other cases work this way too, like the pre-selection version of BSM swapping in other interpretations.

 

[akvadrako]

And in fact there are literally no subsets of that same full dataset that would show any Bell inequality is violated unless you hand select the dataset to have that attribute. I could demonstrate entanglement of random pairs of cars in London using that technique.

See my last post for how to do this. You can indeed demonstrate entanglement of some random properties of a sub-ensemble of random pairs of cars using this technique, but only after the fact by looking at the results.

4. I am scared to ask but... what do "super-quantum correlations" have to do with this discussion? I am not sure they even exist.

Using post-selection on results one can violate Bell inequalities by more than the Tsirelson bound, all the way up to the logical maximum. It's not really relevant, though I would indeed be interested to figure out a principle which restricts one to the quantum limit.