I Entanglement swapping, monogamy, and realism

[Nullstein]

They most certainly are. Basing the signal on what happens at the BSM is very different from basing it on something from the initial preparation, which is what you said before.

Fine, if you want to study the more complicated situation, it just means that you are even less allowed to conclude causal relationships from the collected data. It means you need to to perform even more checks of the causal structure and take even more possible causal relationships into account, before any conclusion can be made. If I don't want to discuss these complications, it just means I'm being generous to you.

Not quite. The point of the event ready signal in these experiments is to select those events in which photons 2 & 3 are projected into the particular Bell state that the BSM is set up to uniquely detect. That doesn't mean the other events aren't successful measurements; they're just measurements whose results can't be distinguished by the humans reading the output of the apparatus. But they still involve projecting photons 2 & 3 into a Bell state and everything associated with that (for example, that photons 1 & 4 are also projected into a Bell state).

Well, I agree, but this just means that you need to perform an even more complicated analysis of the experimental setting, before you can perform proper causal inference. So it just got more complicated for you to argue for your position. Me agreeing to discuss the idealized experiment rather than the real world experiment is a generosity, not an excuse.

What you need to do to infer a causal relationship now is to draw a causal diagram and take all the effects into account that I explained in post #150. You then calculate the corrected correlations that account for these effects and you'd have to show that there is still correlation after the effects have bene taken into account.

 

[PeterDonis]

The paper https://arxiv.org/abs/1606.04523 shows that this is also well understood in the case of entanglement swapping.

Where is entanglement swapping discussed in the paper? I see a "swap" operator defined, but I don't see how it relates to what the BSM does in the entanglement swapping experiment we have been discussing.

 

[PeterDonis]

if you want to study the more complicated situation

What more complicated situation? I wasn't comparing two situations, I was correcting your misstatement about the "event ready" signal in the one situation we have been discussing all along.

 

[Nullstein]

Where is entanglement swapping discussed in the paper? I see a "swap" operator defined, but I don't see how it relates to what the BSM does in the entanglement swapping experiment we have been discussing.

The swap gate is a mathematical abstraction of the physical situation in entanglement swapping. It abstractly models the transition that we previously denoted by $\rho_{1234,\text{before}}\rightarrow\rho_{1234,\text{after}}$ in terms of a 4 qubit system. The paper applies to all such swapping experiments, independent of details such as e.g. the temporal and spatial ordering. (As discussed earlier, the temporal and spatial ordering has no effect on the measured statistics, so it can be abstracted away.)

What more complicated situation? I wasn't comparing two situations, I was correcting your misstatement about the "event ready" signal in the one situation we have been discussing all along.

The more complicated situation is what you get if you want to include the events, that did not constitute a proper BSM projection, into the description. In that case, you would need to introduce another variables X = "Has a proper BSM projection occured?" and study the causal graph that includes this variable, taking into account the issues mentioned in #150.

 

[PeterDonis]

It abstractly models the transition that we previously denoted by $\rho_{1234,\text{before}}\rightarrow\rho_{1234,\text{after}}$ in terms of a 4 qubit system.

I only see two qubit systems discussed in the paper.

 

[vanhees71]

I think so too. The only possible difference is that Fuchs and Peres are also willing to talk about the likelihood of single events as well as frequencies of events in ensembles, which I am not sure vanhees is willing to do.

"When we are told that the probability of precipitation tomorrow is 35%, there is only one tomorrow." -- Fuchs and Peres

That's the problem I have with interpretations of probability, no matter whether they are predicted by QT or other means of calculating statistical predictions (e.g., by classical statistical physics), which abandon the "frequentist interpretation", like "QBists".

What does it mean that the probability for rain tomorrow is 35%? Of course there's only one tomorrow, and I can't prepare the wheather situation today many times and check the frequency that it rains or not tomorrow. For me the prediction of some weather model means that at weather condistions like today, experience and/or projections via model simulations, tells me that in 35% of all such cases it will rain. A good statistical analysis also gives a level of significance and an uncertainty of this number, which also is based on experience with such weather models, and then I have some objective estimate for the probability of rain or good weather to plan my barbecue party tomorrow. Of course also this interpretation is after all based on the "frequentist interptation" of the statistical meaning of probabilities. From a single event, you of course can not decide whether your prediction that there's 35% chance for rain tomorrow is a good estimate or not.

 

[mattt]

That's the problem I have with interpretations of probability, no matter whether they are predicted by QT or other means of calculating statistical predictions (e.g., by classical statistical physics), which abandon the "frequentist interpretation", like "QBists".

What does it mean that the probability for rain tomorrow is 35%? Of course there's only one tomorrow, and I can't prepare the wheather situation today many times and check the frequency that it rains or not tomorrow. For me the prediction of some weather model means that at weather condistions like today, experience and/or projections via model simulations, tells me that in 35% of all such cases it will rain. A good statistical analysis also gives a level of significance and an uncertainty of this number, which also is based on experience with such weather models, and then I have some objective estimate for the probability of rain or good weather to plan my barbecue party tomorrow. Of course also this interpretation is after all based on the "frequentist interptation" of the statistical meaning of probabilities. From a single event, you of course can not decide whether your prediction that there's 35% chance for rain tomorrow is a good estimate or not.

Yes, that's what it usually means. The criteria of what exactly "conditions like today" means (among several other things) is what defines the probabilistic method used in each case.

 

[vanhees71]

1&2 are in a Bell state as well as 3&4, i.e., each of these pairs is maximally entangled. There are no correlations between 2&3 or 1&4 in the initial state. I haven't claimed anything like that.

Now if you project to a Bell state of 2&3, in the so prepared subensemble 2&3 are fully entangled, because they are prepared to be in a Bell state. Due to the entanglement of 1&2 and 3&4 this implies that for this subensemble also 1&4 are in a Bell state.

It's a easy, although some elaborate calculation, as described by Jennewein et al in

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

It's more easily seen in the 2nd-quantization notation. The four Bell states of a photon pair with momentum labels $j$ and $k$ are created from the Vakuum by
$$\hat{\Psi}_{jk}^{\dagger \pm} \rangle=\frac{1}{\sqrt{2}}[\hat{a}^{\dagger}(\vec{p}_j,H) \hat{a}^{\dagger}(\vec{p}_k,V) \pm \hat{a}^{\dagger}(\vec{p}_j,V) \hat{a}^{\dagger}(\vec{p}_k,H)],$$
$$\hat{\Phi}_{jk}^{\dagger \pm} \rangle=\frac{1}{\sqrt{2}}[\hat{a}^{\dagger}(\vec{p}_j,H) \hat{a}^{\dagger}(\vec{p}_k,H) \pm \hat{a}^{\dagger}(\vec{p}_j,V) \hat{a}^{\dagger}(\vec{p}_k,V)].$$
Then the initial four-photon state can then be written in two forms,
$$|\Psi_{1234} \rangle = \hat{\Psi}_{12}^{\dagger -} \hat{\Psi}_{34}^{\dagger-} |\Omega \rangle,$$
but this can as well be written as
$$|\Psi_{1234} \rangle=\frac{1}{2} (\hat{\Psi}_{23}^{\dagger +} \hat{\Psi}_{14}^{\dagger +} - \hat{\Psi}_{23}^{\dagger -} \hat{\Psi}_{14}^{\dagger -}-\hat{\Phi}_{23}^{\dagger +} \hat{\Phi}_{14}^{\dagger +} + \hat{\Phi}_{23}^{\dagger -} \hat{\Phi}_{14}^{\dagger -})|\Omega \rangle.$$
The former notation shows that photon pairs (12) and (34) are each in the polarization-singlet Bell state but (14) and (23), i.e, are uncorrelated.

The latter notation shows that if you project pair (23) to either of the four Bell state the pair (14) must be found in the same Bell state. In Pan et al's work, which we discuss here, (23) has been projected to the polarization-singlet state, and it has been demonstrated that then also the pair (14) is then in the same polarization-singlet state. This happens with probability 1/4.

Quantum theory is inherently nonlocal, although signal locality is respected. You - as many - have formulated requirements for some definition of nonlocality that cannot be met. Specifically, you basically reject experimental nonlocality proofs which some central observer does not experience until all of the information arrives - which is limited by signal. Circular reasoning - you have proved what you assume.

I don't know, how you can insist on "nonlocality" without clearly telling what you mean by it. It's a mathematical fact that the standard relativistic QFTs (used to formulate the standard model of HEP, which includes also QED, which is the theory describing all quantum-optics experiments) is a local theory, i.e., due to the enforced microcausality conditions on local observables, there are no causal influences between space-like separated events possible, and that's what's usually understood as "nonlocality". This discussion goes again in circles :-(((.

What is clearly nonlocal, as we have discussed here: An experimentalist here can create - or not - a distant biphoton from components that have never existed in a common light cone. There really is no disputing this experimental fact as I have characterized it. If you choose to reject this because it does not demonstrate signal nonlocality, that's... circular.

An experimentalist can create such the biphoton (14) only by projection due to a local measurement on (23), and you have to do coincidence measurements on all four photons. I.e., you need a measurement protocol which enables you to select the subensemble for which (14) is entangled. This is only possible when you bring together the information contained in all measurement protocols in their future light cone. The temporal order of measurements to obtain these measurement protocols is irrelevant, which shows that there's no faster-than-light influence of the projection measurement on (23) on photons 1 and 4 necessary to explain the entanglement of the biphotons in the corresponding subensemble. As @PeterDonis has said above, you can even do the measurement on photons 1 and 4 before you perform the projection of (23) to a Bell state, i.e., by "postselection".

Logically, if there is something called "quantum nonlocality" (as I say there is) and it can be demonstrated (as it can be), but it does not feature FTL signaling: then it cannot be judged by a "macroscopic" experiment where component results must be brought together before you accept the nonlocality. Macroscopic Alice can write her results down and compare them with the measurement results of Macroscopic Bob and Victor anytime. Obviously, their combined story indicates that there is quantum nonlocality. You can't wave your hands and ignore the obvious.

That's a contradiction in it self. If there's no FTL "signalling", i.e., no causal connection between space-like separated events in your model, then your model is not "nonlocal", and standard relativistic QFT is indeed not "nonlocal", as we've discussed zillions of times now.

You can interpret any theory in any way you like. The only restriction is that your interpretation must not contradict the mathematical properties of the model you want to interpret (here it's relativistic QFT).

 

[vanhees71]

I believe their approach to probabilities is Bayesian, which treats probabilities as states of epistemic belief and so has no problem assigning probabilities to single events.

Indeed, but what's the meaning of probabilities for single events? See #156 for further details concerning my question.

 

[Fra]

That's the problem I have with interpretations of probability, no matter whether they are predicted by QT or other means of calculating statistical predictions (e.g., by classical statistical physics), which abandon the "frequentist interpretation", like "QBists".

What does it mean that the probability for rain tomorrow is 35%? Of course there's only one tomorrow

For me the important question is:
What difference does it make what the we(or an agent) THINK is going to happend tomorrow? This question is motivated by the fact that noone can ever know until its too late!

I would say the difference is that, that a rational agents actions now is guided by it's expectations of the future, it encodes now. Decisions are typically based on incomplete information, so it's essentially a game of expectations. That in a nutshell is my view. And I think essentiall the qbist view as well.
Here one could even argue that wether the expectations are right or wrong, are in fact completely irrelevant. A badly inform agent, will indeed be expected to make "irrational decisions" judged by someone that has different expectations - which should be in principle observable even! This is not a "problem", it's part of the game and learning process.

In qbism the probability is guiding, rather than descriptive. Obviously we can not describe the future, only the past. And there is no deductive inference of the future from the past (or present). And the best inference seem unavoidably relative.

/Fredrik

 

[Fra]

Indeed, but what's the meaning of probabilities for single events? See #156 for further details concerning my question.

I think to elaborate on this in detail. Ie. to "explain" how one can infer a guiding probability from "emprics" without huge amounts of repeats and preparatations, may require a new formalism that unavoidably becomes speculative and involving hypothesis that doesn't belong here.

This is one of the things I expect from a future theory. And while you usually don't associate to it, for me it naturally relates to QG, as the cosmological perspective vs subatomic perspective exactly illustrates the difference between statistics that we can understnad as ensembles, and generalized probability that we can understand without ensembles/prepartion paradigm, and that makes sense even when the ensemble production cant be realized.

/Fredrik

 

[Simple question]

That's a contradiction in it self.

No, that was sound reasoning.
A contradiction is to name "local" a principle like "space-like observable should commute".
The contradiction is in bold.

If there's no FTL "signalling", i.e., no causal connection between space-like separated events in your model,

That's not what FLT "signaling" means. "FLT signaling" means "FLT signaling": communication with no delay.

standard relativistic QFT is indeed not "nonlocal", as we've discussed zillions of times now.

Aside being based on your philosophical misconception of its premises, and AFAIK, standard relativistic QFT did not get rid of entanglement and inseparability. So spatially-extended (non-local) two-particle properties is the bread and butter of QM.

You can interpret any theory in any way you like.

No you cannot. You have to use logic and reason. Unless your prefer bad philosophy and running in circle.

The only restriction is that your interpretation must not contradict the mathematical properties of the model you want to interpret (here it's relativistic QFT).

Mathematical facts don't exist in Nature. Mathematical truth exist in heads. Facts are collected in the laboratory, where your claims of the purported "mathematical facts" have been proved wrong.

 

[DrChinese]

1. These events taken together, i.e. prior to post-selection, constitute the full ensemble.

The photons 1&4 will be uncorrelated in the full ensemble. The statement is true for the post-selected subensembles only.

If you don't look at a subensemble, there will not be entanglement in the data. The correlation will be zero. This follows strictly from the math.

1. No, the full ensemble is maximally entangled. You just don't know how. If you run an experiment, and can identify 1/2 the cases as to Bell state (which is feasible), that portion will violate the CHSH bound. The other half will not show any correlation, as you say. That does not mean they aren't entangled, QM predicts they are.

2. You are being imprecise on purpose here. Quantum non-locality means nothing more than the fact that Bell's inequality is violated.

3. What you want to imply is that there are non-local cause and effect relationships. The broad consensus among physicists is that no non-local cause and effect relationship can be inferred from the data. QM can be interpreted either way.

2. Quantum nonlocality is a lot more than violating a Bell Inequality! The predictions of QM embed the nonlocality. Specifically: cos^2(Alice-Bob) prediction is inherently nonlocal, as Alice and Bob and their measurement settings need not be local to each other. Note that there are no other variables in this equation.

And as we have demonstrated repeatedly in our extended discussions: the particles that Alice and Bob are observing need never have interacted or even overlapped in a common light cone. That's a lot of nonlocality! And how do quantum repeaters work in an extended quantum network? They are built on this nonlocality.3. I never said there was a cause-and-effect relationship in the Einsteinian sense, because there isn't. The action at a distance: a) does not change when the ordering changes (effect precedes cause); and b) and the effect cannot be said to occur at any specific time. You can make a statement about the final context, and say the swap created the entangled 1 & 4 pair. But that is mostly a convenience in describing the situation.

There are other ways to describe it, but since you always have the option to execute a swap or not: this is the most useful. The decision to execute the swap gives you something to point to as the "cause" even though it may be after the "effect". If you feel more comfortable saying the "spontaneous" creation of the (1 & 4) biphoton causes the BSM to succeed, then that is your choice and is just as correct in the final analysis. But again, I would not say this description is useful because the BSM is where the "on/off switch" is located.

 

[PeterDonis]

what's the meaning of probabilities for single events?

Read the Bayesian literature for a detailed answer to this question. The short answer is, it's a measure of a degree of belief. If you want to tie it to something concrete, think of betting odds: a 35% probability of rain tomorrow means that a person would pay 35 cents for a bet that pays 1 dollar if it rains tomorrow.

 

[DrChinese]

This just in! From an experiment described in a paper deposited 15 Jan 2023:

https://arxiv.org/abs/2301.06091

"Photon-pair sources based on spontaneous parametric down-conversion (SPDC) serve as such a resource of entanglement, as they can be matched in frequency and bandwidth to the ionic transitions [4–7]. When entanglement between distant nodes has been established, quantum teleportation [8] provides a suitable protocol, alternative to direct transmission, for quantum state transfer [9]. Application of teleportation has been shown between photons over large distances [10–15], between quantum nodes [16–20], and from a photon to a quantum node [21–24]. In this paper we show quantum teleportation of an atomic qubit, encoded in the Zeeman sub-levels of the 40Ca+ ion, to the polarisation qubit of a single 854 nm photon. The teleportation protocol employs heralded absorption of a single photon by the ion [4, 9, 25, 26], which functions as a Bell measurement. In contrast to photonic teleportation protocols, this Bell measurement can identify all four Bell states."

All 4! Where's the subensemble "out" now?

They ran 500 million swaps (teleportations) over a 50 hour period. They identified all 4 Bell states with approximately 81% fidelity.

Pure optical setups may be limited to discerning a maximum of 2 Bell state due to limitations inherent in Avalanche Photon Detectors (APD). That's because a single APD cannot distinguish 2 photons that arrive in close proximity from a single photon, which would be necessary to identify 2 of the 4 Bell states. They can't reset fast enough - at least not at this time.

But technology advances, and red herrings fall. There is nothing about the BSM (pre- or post-) selection process that is can be considered "cherry picking".

As I was saying in the entanglement swapping scenarios: All of the cases (in principle) where we have (1 & 4) matched clicks indicate entanglement swapping. This can be seen in this experiment, although it is expressed differently than in the swaps we have been discussing. Hopefully everyone can see clearly: there are no (1 & 2) pairs where there is "coincidental" or "accidental" entanglement between 1 & 4 waiting to be revealed. If there is no swap (teleportation), there is no entanglement to be seen. If and only if...

 

[akvadrako]

All 4! Where's the subensemble "out" now?

There are 4 subensembles. You still need to know which runs belongs to which to get entanglement.

This doesn't change anything about the discussion; as several people have pointed out, it would have been clearer to just consider the ideal experiment, where there is no need for an event ready signal.

 

[DrChinese]

1. There are 4 subensembles. You still need to know which runs belongs to which to get entanglement.

2. This doesn't change anything about the discussion; as several people have pointed out, it would have been clearer to just consider the ideal experiment, where there is no need for an event ready signal.

1. And they did. Case closed.

2. The original post-selection "out" was that only *some* of the pairs were being considered. Most of the optical versions could only identify 1 or 2 of the Bell states, and some folks saw that as explaining that only the "cherry-picked" sample would show entanglement - but the full data set would not.

Obviously: If you can identify the entanglement of the full universe, then the cherry-picking argument fails. It never made sense anyway, as *all* of the 1 & 4 pairs are entangled according to the predictions of QM (perfect case of course). So unless you have alternate and different predictions to make, I don't see the point of that line of reasoning.

I hope the debate about "subensembles" will shift to something more appropriate. Like the obviously nonlocal nature of entanglement swapping in which a CHSH inequality is violated by systems that have never been in a common light cone (and need not have ever even communicated with systems in a common light cone).

 

[mattt]

For each of the four (1,4) subensembles, (1,4) will violate a Bell inequality (after all, for each of those (1,4) subensembles, the state that describes its statistics is, respectively, a maximally entangled state).

But for the four subensembles taken together, its state (which describes its statistics) tells you that it will not violate a Bell inequality (the whole (1,4) ensemble).

 

[Morbert]

1. No, the full ensemble is maximally entangled. You just don't know how. If you run an experiment, and can identify 1/2 the cases as to Bell state (which is feasible), that portion will violate the CHSH bound. The other half will not show any correlation, as you say. That does not mean they aren't entangled, QM predicts they are.

This distinction you are implying (unentangled pairs as distinct from hidden entangled pairs) isn't inherent in the formalism. It's also a double-edged sword. The consistent histories interpretation lets us split the full ensemble into subensembles of entangled (14) pairs even if the BSM on (23) is never performed.

 

[akvadrako]

2. The original post-selection "out" was that only *some* of the pairs were being considered. Most of the optical versions could only identify 1 or 2 of the Bell states, and some folks saw that as explaining that only the "cherry-picked" sample would show entanglement - but the full data set would not.

You are missing the point. It was always an aspect of the idealized experiment that you select 4 sub-ensembles, each entangled. And nobody was arguing that it was only the non-ideal experiment that allowed an "out" due to low fidelity.

One still needs to "cherry-pick" those 4 sub-ensembles to show entanglement. If you perform the entanglement swapping but then throw away the information of which runs are in which sub-ensemble, you will never be able to show any entanglement.

Which I would say is the same as saying there is no entanglement, because entanglement is not an objective property of pairs of systems, but a statistical correlation. It's some subjective information. The monogomy of entanglement means you can never have information like pairs A&B are fully entangled while B&C are fully entangled.

 

[DrChinese]

For each of the four (1,4) subensembles, (1,4) will violate a Bell inequality (after all, for each of those (1,4) subensembles, the state that describes its statistics is, respectively, a maximally entangled state).

But for the four subensembles taken together, its state (which describes its statistics) tells you that it will not violate a Bell inequality (the whole (1,4) ensemble).

This is one of the strangest comments I have ever seen, completely meaningless and essentially contradictory.

We're scientists. You may as well be saying the results from looking at only the 1 stream are random. True, but of no value whatsoever. I guess that means Bell tests are evidence of nothing.

 

[DrChinese]

This distinction you are implying (unentangled pairs as distinct from hidden entangled pairs) isn't inherent in the formalism. It's also a double-edged sword. The consistent histories interpretation lets us split the full ensemble into subensembles of entangled (14) pairs even if the BSM on (23) is never performed.

When the BSM is turned on, all of the 1 & 4 pairs are entangled. When the BSM is not performed, none of the 1 & 4 pairs are entangled. How many ways can I say the same thing?

 

[PeterDonis]

The consistent histories interpretation lets us split the full ensemble into subensembles of entangled (14) pairs even if the BSM on (23) is never performed.

Please give a reference for this very surprising statement.

 

[PeterDonis]

One still needs to "cherry-pick" those 4 sub-ensembles to show entanglement.

No, you don't. The subensembles for photons 1 & 4 are defined by the results of the BSM on photons 2 & 3. "Cherry picking" would be using the results on photons 1 & 4 to pick "subensembles" for photons 1 & 4.

 

[DrChinese]

One still needs to "cherry-pick" those 4 sub-ensembles to show entanglement. If you perform the entanglement swapping but then throw away the information of which runs are in which sub-ensemble, you will never be able to show any entanglement.

Which I would say is the same as saying there is no entanglement, because entanglement is not an objective property of pairs of systems, but a statistical correlation. It's some subjective information. The monogomy of entanglement means you can never have information like pairs A&B are fully entangled while B&C are fully entangled.

Again, a strange comment.

In science, we look for patterns and pattern exceptions. Yes, if you throw half of your experimental data away, then probably you won't see much of a pattern. Who does that? And why would any scientist suggest that?

This is state of the art experimentation, and if you choose to ignore the evidence of entanglement: I really have nothing more to discuss with you. Not sure why you would participate if you think looking at ALL of the data is "cherry-picking". And why would you say there is NO entanglement when all 4 Bell states shows entanglement?

 

[PeterDonis]

entanglement is not an objective property of pairs of systems, but a statistical correlation.

This is not correct as a statement about the actual math. The actual math contains an objective definition of an entangled state, and that definition has nothing to do with statistical correlation: it's an easily testable objective property of the mathematical state. Whether you think that mathematical state represents particular two-qubit systems, or only ensembles of such systems, is a matter of interpretation.

 

[PeterDonis]

This distinction you are implying (unentangled pairs as distinct from hidden entangled pairs) isn't inherent in the formalism.

Yes, it is. See my post #176 just now.

 

[Morbert]

This distinction you are implying (unentangled pairs as distinct from hidden entangled pairs) isn't inherent in the formalism. It's also a double-edged sword. The consistent histories interpretation lets us split the full ensemble into subensembles of entangled (14) pairs even if the BSM on (23) is never performed.

When the BSM is turned on, all of the 1 & 4 pairs are entangled. When the BSM is not performed, none of the 1 & 4 pairs are entangled. How many ways can I say the same thing?

In the original paper, the four bell states are $\psi^+_{ij},\psi^-_{ij},\phi^+_{ij},\phi^-_{ij}$. I will relabel them as $\psi^1_{ij},\psi^2_{ij},\psi^3_{ij},\psi^4_{ij}$ respectively.

Including the collective degrees of freedom of the BSM measurement apparatus, the experiment is characterised by the preparation $$|\Psi\rangle= |\psi^2_{12},\psi^2_{34},M^0\rangle =\sum_{i=1}^4c_i|\psi^i_{14},\psi^i_{23},M^0\rangle$$ where $M^0$ denotes the ready state of the detector (The detector registers the BSM result $\psi^i_{23}$ with the state $M^i$) and $c_i$ is +1/2 or -1/2. The measurement begins at time $0$ and concludes at time $t$. The time evolution is $$U(t,0)|\Psi\rangle =\sum_{i=1}^4c_i|\psi^i_{14},M^i\rangle$$We construct four history operators of the form $$C_i = P_{M^i}(t)P_{\psi^i_{14}}(0)$$ where $P$ are projectors. These operators are complete $$\sum_i C_i|\Psi\rangle = |\Psi\rangle$$ and consistent $$\mathrm{tr}\left[C^\dagger_iC_j|\Psi\rangle\langle\Psi|\right] = 0, i\neq j$$ and so they partition the ensemble of experimental runs into into four subensembles, resolved by the measurement outcome, each with probability $$p_i = \mathrm{tr}\left[C^\dagger_iC_i|\Psi\rangle\langle\Psi|\right]$$Each subensemble has the (14) pair in a bell state. No major controversy yet, except perhaps the history operators projecting to the bell states for (14) at time $0$, which is not an issue for CH.

Now consider an alternative "experiment", where the measurement apparatus never interacts with the (23) pair. The new time evolution is $$U(t,0)|\Psi\rangle = \sum_{i=1}^4c_i|\psi^i_{14},\psi^i_{23},M^0\rangle$$We can similarly construct the complete and consistent history operators $$C_i = P_{M^0}(t)P_{\psi^i_{14}}(0)$$that partition the experimental runs into four subensembles, each with the (14) pair in an entangled bell state. But these subensembles aren't useful to the experimenter, because they aren't resolved by the singular outcome $M^0$.

Please give a reference for this very surprising statement.

I don't know of a paper on the consistent histories interpretation that deals explicitly with entanglement swapping. But I have applied the consistent histories account of measurement presented in Roland Omnes's "Understanding Quantum Mechanics" (chapter 21) and RB Griffiths's paper here (pdf)

 

[akvadrako]

This is not correct as a statement about the actual math. The actual math contains an objective definition of an entangled state, and that definition has nothing to do with statistical correlation: it's an easily testable objective property of the mathematical state. Whether you think that mathematical state represents particular two-qubit systems, or only ensembles of such systems, is a matter of interpretation.

I think your view is contradicted by this post-selection swapping experiment, most clearly if I assume there are no FTL or backwards in time influences.

Are you trying to say that I must accept FTL influences? Because if not, the idea that entanglement is statistical correlation instead of an objective property is a self-consistent position to hold.

Also, how can entanglement be objective when it's frame dependent? From https://www.nature.com/articles/s41467-018-08155-0:

From the examples considered it is clear that the notions of superposition and entanglement are reference-frame dependent.
...
From A’s point of view B and C are entangled
...
A and B are entangled in an EPR state from C’s point of view

 

[akvadrako]

No, you don't. The subensembles for photons 1 & 4 are defined by the results of the BSM on photons 2 & 3. "Cherry picking" would be using the results on photons 1 & 4 to pick "subensembles" for photons 1 & 4.

I was just trying to use Dr. Chinese's terminology. My point was you still need to pick sub-ensembles; that's what the BSM results let you do.

Without knowing the sub-ensembles there is no entanglement, at least no possibility to show entanglement.

Edit: I think I get it now – Dr. Chinese thought the "cherry-picking" method only worked when picking 1 or 2 of the subsets of runs (I won't use the word sub-ensemble for this). This is incorrect; it works for all 4 subsets.