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

[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.

 

[kurt101]

The way to investigate this question is for you to read the literature and see if you can find papers supporting what you say. If you can, by all means post references to them here. Writing your own personal code is not doing that; as I have said, it is personal theory and is off limits here.

Here is a paper that refutes @DrChinese claims https://link.springer.com/article/10.1007/s10701-021-00511-3#Sec21

From the Conclusions section in this paper:
"Under conventional assumptions—i.e., excluding retrocausality—the sensitivity of such experiments to this Collider Loophole depends on the temporal relation between the entanglement-swapping measurement C and the measurements A and B. CL a threat if the C is in the future of A and B, but not if it is in the past."

 

[vanhees71]

What you personally would call it is immaterial. In the QM literature, the "nonlocal" effects which some have called "spooky action at a distance" and which others have called "acausal" (not just @DrChinese, the term appears in multiple papers in the literature, some of which have been referenced in other threads--for example see some of the posts by @RUTA), are the basis for the violation of the Bell inequalities by QM.

Remember that PF does not allow discussion of personal theories. Your claims about your code amount to a personal theory. That means they are off limits for discussion here. Do not post further about them or you will receive a warning.

Now we are back at this fruitless discussion about what "non-locality" means. As we realized some time ago, it's not a uniquely defined term. In the HEP community we call our relativistic QFTs "local", including that all interactions are "local". This is implemented as a clearly defined mathematical property of the field operators that represent local observables, i.e., the microcausality condition, and this condition excludes faster-than-light causal influences, i.e., in this sense standard relativistic local QFT has no "spooky actions at a distance". In this sense of the term entanglement doesn't describe any non-local interactions but just correlations between far-distant parts of quantum systems prepared in the corresponding entangled states. It's rather what Einstein called "inseparability" than "non-locality".

 

[PeterDonis]

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

That's not a valid way of selecting a subensemble. It would be like me claiming I had a biased coin by flipping it a thousand times and then selecting the "subensemble" of all the flips that came up heads.

The statistics of 1&4 does not change because the density matrix does not change.

Yes, vague ordinary language isn't really suitable here. The density matrix of 1 alone or of 4 alone does not change if the 2&3 BSM is done, but the correlations between 1 and 4 do change--there are subensembles selected by the 2&3 BSM results that violate the Bell inequalities, that aren't there if the 2&3 BSM is not done. Both of those things could be referred to by the word "statistics", so it's good to be clear about exactly what doesn't change and what does.

 

[PeterDonis]

This is an interpretation dependent statement.

No, it isn't. Mathematically, if you want to correctly predict the correlations between 1&4, you have to use the appropriate state based on the result of the 2&3 measurement. That's true no matter what interpretation you adopt.

Different interpretations will tell different stories about why you have to do that.

 

[PeterDonis]

My definition of sub-ensemble is just a subset of a full ensemble,

That's the wrong definition. See the first part of my post #73.

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

This paper does not mean what you think it means. It is making the same error that I described in the first part of post #73.

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

This makes no sense. You can't pick out "the same results" in two physically different experiments.

 

[PeterDonis]

Here is a paper that refutes @DrChinese claims https://link.springer.com/article/10.1007/s10701-021-00511-3#Sec21

From the Conclusions section in this paper:
"Under conventional assumptions—i.e., excluding retrocausality

This isn't a refutation of anything. It's a proposed claim about a new "loophole" in such experiments. Such claims are ubiquitous in the literature, but they always boil down to one thing: claiming that if an experiment is done that closes the claimed loophole, the predictions of quantum mechanics will be violated. In other words, such claims are made by people who don't want to accept that the QM predictions are correct.

So far, every time a more sophisticated experiment has been done that closes one of these loopholes, the predictions of QM have continued to be confirmed. So the track record of these loophole claims is one of complete failure so far. If you want to maintain the position that this particular loophole won't share this fate, that's up to you; but unless and until an actual experiment is done that (a) closes the loophole, and (b) shows a violation of QM's predictions, you cannot claim that anything is refuted.

 

[PeterDonis]

Now we are back at this fruitless discussion about what "non-locality" means.

"Nonlocality" is just a word. The people who use that word in ways you don't like make it perfectly clear what it refers to in the actual math. If you prefer to call that thing in the math "inseparability", that's fine as far as it goes, but unless and until you have convinced everyone else in the QM community that uses the word "nonlocality" for that thing, to use the word "inseparabilty" instead, it's pointless to complain about it.

 

[DrChinese]

Here is a paper that refutes @DrChinese claims https://link.springer.com/article/10.1007/s10701-021-00511-3#Sec21

From the Conclusions section in this paper:
"Under conventional assumptions—i.e., excluding retrocausality—the sensitivity of such experiments to this Collider Loophole depends on the temporal relation between the entanglement-swapping measurement C and the measurements A and B. CL a threat if the C is in the future of A and B, but not if it is in the past."

LOL. They start by assuming that which they seek to prove. You can't exclude retrocausality by assumption because that is part of the question they seek to answer. One of the authors, Huw Price, has been arguing against retrocausality for over 25 years. Now please keep in mind that I am agnostic on the issue of retrocausality itself. All I assert is that quantum nonlocality violates strict Einsteinian locality, by creating a context that spans spacetime. I think a better term than "retrocausal" is "acausal". (They sometimes use the word "noncausal".)

As to the meat of their argument, it makes absolutely no sense at all. There is no such thing as the "collider loophole" (CL) in quantum physics. And again, this is precisely the point of experiments such as Delft. The CL examples they provide do not involve questions of action at a distance (AAD/AAAD) or locality. Those are classical examples. You cannot get violation of a Bell inequality from a classical preparation. It's that simple.

Quantum theory predicts that our 1 & 4 photons are monogamously bound (entangled) to other photons at creation. Either the swap operation "causes" 1 & 4 to be monogamously entangled (ceasing prior monogamous entanglement), or quantum theory is flat out wrong. You tell me.

 

[PeterDonis]

Thread closed for moderation.

 

[PeterDonis]

After moderator review, the thread will remain closed. Thanks to all who participated.