Entanglement swapping, monogamy, and realism

[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.

 

[vanhees71]

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.

Yes, and this makes also sense, but it's not convincing in the application of probability theory to physics. In physics we aim at theories/models that describe/predict the outcome of experiments and observations, i.e., if I have a theory like QT that predicts probabilities for the outcome of measurements, I need to repeat that experiment under the same conditions ("preparation") many times and use statistical analysis (which itself is of course also based on probability theory) to decide, whether the prediction compares well or does not with experiment. In this sense quantum states describe ensembles rather than single realizations of measurements.

 

[vanhees71]

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.

But entanglement describes correlations, i.e., to test the predictions of entanglement you need to measure the predicted correlations, e.g., by testing for the validity of Bell's inequalities vs. the predictions of QT, and indeed the outcome is objectively in favor of QT in all experiments done yet.

 

[gentzen]

Dreischner calls the account of Peres + Fuchs a minimal instrumentalist interpretation and cites Friebe (though I do not have access to the cited text). By minimal statistical interpretation, do you mean a minimal ensemble interpretation? I don't know if they are as different as you imply, but the minimal interpretation as described in this thread and others has an instrumentalist character (for example, the association of a quantum state with a preparation procedure). If there is some important distinction or misrepresentation then please be explicit.

The paper of Peres + Fuchs simply does not live up to its title. It is so often quoted only for two reasons: first of course its title, and second that it gives you a convenient straw man to attack, if you associate some interpretation you don't like (for example QBism) with that paper. (See for example the "discussion" of QBism in "Making Sense of Quantum Mechanics" by Jean Bricmont. He extensively quotes from Peres + Fuchs. But it should be obvious that Peres is not a QBist.)

I am an instrumentalist, and I want to distance myself explicitly here from that paper by Peres + Fuchs. If you want, attack my instrumentalist position, but not by claiming that this paper would have anything to do with it. I am fine if you associate me with positions explained by Roland Omnès.

 

[vanhees71]

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

In theoretical physics we have clear mathematical definitions, and a QFT is called local if the microcausality constraint on local observables hold and that the Hamilton density of the theory is a local observable. Microcausality means that two local observable-operators commute if their arguments are space-like separated. For a thorough discussion of all this, see Weinberg, Quantum Theory of Fields, Vol. 1.

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

Of course not. Any QT has entanglement and inseparability. If QFT would not enable entanglement and inseparability it would have been proven wrong by all the Bell tests we discuss here. That's not the case. Obviously you misunderstand the meaning of locality and microcausality.

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

No, but there are mathematical facts about the theories we use to describe nature, and local QFTs are local, and any interpretation must not contradict the mathematical properties of the theory that it claims to interpret. Philosophy, of course, often is vague, because it doesn't consider the mathematical properties of the theories carefully enough, and that's why it rather causes confusion than clarification of the natural sciences.

 

[Morbert]

The paper of Peres + Fuchs simply does not live up to its title. It is so often quoted only for two reasons: first of course its title, and second that it gives you a convenient straw man to attack, if you associate some interpretation you don't like (for example QBism) with that paper. (See for example the "discussion" of QBism in "Making Sense of Quantum Mechanics" by Jean Bricmont. He extensively quotes from Peres + Fuchs. But it should be obvious that Peres is not a QBist.)

I am an instrumentalist, and I want to distance myself explicitly here from that paper by Peres + Fuchs. If you want, attack my instrumentalist position, but not by claiming that this paper would have anything to do with it. I am fine if you associate me with positions explained by Roland Omnès.

What does the article get wrong re/instrumentalism beyond perhaps the conceit of the title?

 

[Fra]

This is a side note in the thread, but i couldn't resist commenting.

No, but there are mathematical facts about the theories we use to describe nature,

Yes, but they have a limited domain of corroboration. Which means that the "mathematical model" that is corroborated for subatomic systems can not be universally applied in a deductive sense. One can certainly TRY it, and see where it leads us, this is FINE and in many cases the method is extremely powerful, but it's technically a speculation or hypothesis only; it's not "mathematical truth" in any way.

and local QFTs are local, and any interpretation must not contradict the mathematical properties of the theory that it claims to interpret. Philosophy, of course, often is vague, because it doesn't consider the mathematical properties of the theories carefully enough, and that's why it rather causes confusion than clarification of the natural sciences.

There flip side of the coin is that extrapolating "mathematical models" beyond it's domain of corroboration, may cause just as much confusion!

IMO a potential example of this is extrapolating mathematical symmetry principles from the smallest things we can distinguish, to the larger scale. For example applying the paradigm of QFT to arbitrary macro scales and cosmology. It may very well be that the symmetries for small subsystems seem solid, simply because they change too slow to be nocticable.

Smolin frequently has in his books, quoted Roberto Unger who calles this the "poisoned gift of mathematics to physics". https://www.newscientist.com/article/mg19125701-100-do-the-laws-of-nature-last-forever/
Which is also in a nutsheel the core topic of this book as well; we may have reached the time in theory development (when seeking unification of all forces) where this method is reaching its limits.
https://www.amazon.com/dp/0544245598/?tag=pfamazon01-20

/Fredrik

 

[vanhees71]

This is a side note in the thread, but i couldn't resist commenting.

Yes, but they have a limited domain of corroboration. Which means that the "mathematical model" that is corroborated for subatomic systems can not be universally applied in a deductive sense. One can certainly TRY it, and see where it leads us, this is FINE and in many cases the method is extremely powerful, but it's technically a speculation or hypothesis only; it's not "mathematical truth" in any way.

I'm not sure what you mean. Of course, current QT has the limitation of not being able to consistently describe the gravitational interaction. Here we talk about QED, and we apply it to quantum-optical phenomena. There are no known limits of validity of QED in this domain yet.

There flip side of the coin is that extrapolating "mathematical models" beyond it's domain of corroboration, may cause just as much confusion!

What we discuss here is a paradigmatic example for the application of QED. Nowhere have we applied it to anything that's not well established. To the contrary, the very experiments we discuss here confirm its validity at an amazing level of accuracy and significance.

IMO a potential example of this is extrapolating mathematical symmetry principles from the smallest things we can distinguish, to the larger scale. For example applying the paradigm of QFT to arbitrary macro scales and cosmology. It may very well be that the symmetries for small subsystems seem solid, simply because they change too slow to be nocticable.

The application of QFT to macroscopic physics is also amazingly successful. Ask any solid-state physicist, how well he can describe all kinds of macroscopic matter with QFT. In cosmology it's successfully applied to the description of the evolution of the universe. Of course, the further you go towards the big bang the more uncertain it gets, but at least after a few microseconds after the big bang we know at least about the known Standard Model particles. What we don't know is the complete mechanism behind CP violation (why are we here and not being annihilated already at the level of particles with an equal amount of antiparticles) and also what Dark Matter is and how to explain the value of the cosmological constant/Dark Energy content of the Universe. That are all open questions, but nothing indicates yet that QFT fails to describe it.

Smolin frequently has in his books, quoted Roberto Unger who calles this the "poisoned gift of mathematics to physics". https://www.newscientist.com/article/mg19125701-100-do-the-laws-of-nature-last-forever/
Which is also in a nutsheel the core topic of this book as well; we may have reached the time in theory development (when seeking unification of all forces) where this method is reaching its limits.
https://www.amazon.com/dp/0544245598/?tag=pfamazon01-20

/Fredrik

Well, Smolin...

 

[gentzen]

What does the article get wrong re/instrumentalism beyond perhaps the conceit of the title?

The conceit of the title is part of the problem, because Peres intentionally played with the expectations of its readers:

Asher Peres was a master of creating controversy for the sake of making a point. For instance, in 1982 he was asked to make a nomination for the Nobel prize in physics. He nominated Israeli prime minister Menachem Begin! Asher reasoned that Begin’s decision to invade Lebanon proved him as qualified for a Nobel physics prize as he was for his earlier peace prize.
It certainly wasn’t of the same magnitude, but Asher intended to make trouble when we wrote our 2000 “opinion piece” for Physics Today. Previous to our writing, the magazine had published a series of articles whose essential point was that quantum mechanics was inconsistent—it tolerated the unacceptable “measurement problem,” and what else could that mean but inconsistency? Quantum theory would need a patch to stay afloat, the wisdom ran—be it decoherence, consistent histories, Bohmian trajectories, or a paste of Everettian worlds.
To take a stand against the milieu, Asher had the idea that we should title our article, “Quantum Theory Needs No ‘Interpretation’.” The point we wanted to make was that the structure of quantum theory pretty much carries its interpretation on its shirtsleeve—there is no choice really, at least not in broad outline. The title was a bit of a play on something Rudolf Peierls once said, and which Asher liked very much: “The Copenhagen interpretation is quantum mechanics!” Did that article create some controversy! Asher, in his mischievousness, certainly understood that few would read past the title, yet most would become incensed with what we said nonetheless. And I, in my naiveté, was surprised at how many times I had to explain, “Of course, the whole article is about an interpretation! Our interpretation!

I certainly didn't knew this, when I wrote:

The conversation started, when I quoted Heisenberg’s own explanation:

… what one may call metaphysical realism. The world, i.e., the extended things, ‘exist’. This is to be distinguished from practical realism, and the different forms of realism may be described as follows: We ‘objectivate’ a statement if we claim that its content does not depend on the conditions under which it can be verified. Practical realism assumes that there are statements that can be objectivated and that in fact the largest part of our experience in daily life consists of such statements. Dogmatic realism claims that there are no statements concerning the material world that cannot be objectivated. Practical realism has always been and will always be an essential part of natural science. Dogmatic realism, however, …

and BT replied

That’s all a bit technical for me! I can’t tell whether he is in favour of “explanations” in the simple(-minded?) Deutsch sense or not.

After setting the passage in context, I got very concrete:

But I disagree that the quoted passage is technical. If he adheres to this passage, then Heisenberg cannot claim that Schrödinger’s cat would be both alive and dead, or that the moon would not be there if nobody watches.
Others, like Christopher A. Fuchs and Asher Peres in Quantum Theory Needs No Interpretation”, are apparently less sure whether (neo-Copenhagen) quantum theory is so clear about that fact. Hence they try to weasel out by claiming: “If Erwin has performed no observation, then there is no reason he cannot reverse Cathy’s digestion and memories. Of course, for that he would need complete control of all the microscopic degrees of freedom of Cathy and her laboratory, but that is a practical problem, not a fundamental one.”
This is non-sense, because the description of the experiment given previously was complete enough to rule out any possibility for Erwin to reverse the situation. Note the relevance of “… a consistent interpretation of QM as applied to what we do in a physical laboratory and how practitioners experience QM in that context.” If Erwin had access to a time machine enabling him to realistically reverse the situation, then it might turn out that Cathy and Erwin indeed lived multiple times through both situations (and experienced real macroscopic superpositions), as depicted in movies like “Back to the Future”.

According to Jan Faye in the SEP article on the Copenhagen Interpretation, even Bohr does not disagree with that conclusion: “Thus, Schrödinger’s Cat did not pose any riddle to Bohr. The cat would be dead or alive long before we open the box to find out.” However, I doubt that Bohr really said this, or replied to the content of Schrödinger’s article in any other direct way. (I read that he complained to Schrödinger about that article for indirectly supporting Einstein in his crusade against QM.)

You see, back then I assumed that Christopher A. Fuchs and Asher Peres were really unsure whether everything could be reversed as long as Erwin did not perform an observation. How should I have known that they used irony here? Or maybe they didn't, and really meant what they wrote? If you want, I can try to lookup the points that Jean Brickmont fell into to take serious. (On the other hand, maybe he wanted to fall for that trap. I certainly didn't want. I wanted to defend Heisenberg. But I am fully aware that vanhees71 won't join me in defending Heisenberg...)

 

[Fra]

I'm not sure what you mean.

My comment was mainly an principal comment, and a note that just like philosophy can be confusing, the apparent decuctive precision of mathematics can misguide. I prefer a balance. My personal view is that even in the hierarchy of the the interacitons besides gravity, some subtle symptoms of this appears. After all, there is not even a GUT. Because even within the GUT, there are large energy gaps between QCD and Electroweak. But it's harder to argue how this relates to the issue.

/Fredrik

 

[vanhees71]

Why should there be a GUT?

 

[Nullstein]

I only see two qubit systems discussed in the paper.

There are 3 systems, B, C and D in the paper. It discusses a general setting. System B corresponds to the 2&3 system in our case and systems C and D correspond to systems 1 and 4 respectively.

 

[Nullstein]

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.

No, the full system is not entangled. I have provided a mathematical proof several times and you had the opportunity to point at the equality sign that you do not comply with. You didn't bother to do that. On the other hand, you never provided any calculation at all. System 1&4 is in the product state $\frac 1 4\mathbb 1\otimes\mathbb 1$ and this is a mathematical fact. It takes at most 15 minutes of linear algebra, even for a beginner, to verify this.

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.

A correlation depends on spacelike separated variables. So what? Correlation does not imply causal relationships. In order to infer causal relationships, you must take the issues in post #150 into account. But if you do this, i.e. if you pass to the full ensemble, the correlation vanishes.

3. I never said there was a cause-and-effect relationship in the Einsteinian sense, because there isn't.

No cause and effect relationship can be inferred at all. That's the point. The statistical analysis is very clear on that.

 

[DrChinese]

No, the full system is not entangled. I have provided a mathematical proof several times and you had the opportunity to point at the equality sign that you do not comply with. You didn't bother to do that. On the other hand, you never provided any calculation at all. System 1&4 is in the product state $\frac 1 4\mathbb 1\otimes\mathbb 1$ ...

I don't follow your notation or what you what you are trying to communicate; but initially 1 and 4 are uncorrelated and therefore can be considered in a Product State as you say. I might represent that as: $\hat{\rho}_{14}=\hat{\rho}_{1} \otimes \hat{\rho}_{4}.$ But 1 is monogamously entangled with 2, and 4 is monogamously entangled with 3, so I prefer the notation: $\hat{\rho}_{1234}=\hat{\rho}_{12} \otimes \hat{\rho}_{34}$ which is more descriptive. Typically each pair is going to be in either the $\psi+ or \phi+$ Bell state, depending on the Source (i.e. Type I or Type II PDC, or whatever).

IFF the BSM occurs, each and every 1 & 4 are entangled (according to the predictions of QM). Then is new description is: $\hat{\rho'}_{1234}=\hat{\rho}_{14} \otimes \hat{\rho}_{23}.$ That can be simplified to $\hat{\rho}_{14}$ which is will be one of 4 possible Bell states. That system of 2 photons is a biphoton (Fock state = 2) and is nonseparable, therefore entangled. You can refer to the distant BSM results to learn which Bell state. Different BSMs can discern different of the Bell states, either 1 or 2 or probably soon to be all 4.

I don't think anyone is any longer disputing what I say here. The only thing left to discuss is whether the distant BSM should be considered "action at a distance". Is there any other open question at this point?

 

[DrChinese]

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.

There is entanglement whether or not you know the Bell state. That is a specific prediction of QM, there is no questioning this. The issue you have is whether it can be "proven" to be entangled in the cases when the Bell state cannot be determined. You are free to accept the predictions of QM, or say QM is wrong, but we all know what happens to those who deny the predictions of QM. You will need a towel to remove the egg on your face.

But you might want to know that recent (Jan 2023) experiments demonstrate teleportation with identification of all 4 Bell states. That will probably end up in a version someday in which the kind of swapping we are discussing can be identified similarly. Don't bet against improving technology either!

 

[gentzen]

IFF the BSM occurs, each and every 1 & 4 are entangled (according to the predictions of QM). Then is new description is: $\hat{\rho'}_{1234}=\hat{\rho}_{14} \otimes \hat{\rho}_{23}.$ ...

I don't think anyone is any longer disputing what I say here.

But your formula is simply wrong, especially if we take into account what you wrote in the next post:

But you might want to know that recent (Jan 2023) experiments demonstrate teleportation with identification of all 4 Bell states.

So you admit that there are 4 different Bell states, but still insist to write $\hat{\rho}_{14}$ for all of them? If they are 4 different states (even so all are maximally entangled), then you should also use 4 different symbols for them, for example $\hat{\rho}_{14}^1$, $\hat{\rho}_{14}^2$, $\hat{\rho}_{14}^3$, and $\hat{\rho}_{14}^4$. And then, wouldn't $\hat{\rho'}_{1234}= \frac{1}{4} \sum_{i=1}^4 \hat{\rho}_{14}^i \otimes \hat{\rho}_{23}^i$ be a more appropriate formula for the state, as long as no conditioning with respect to the 4 different states is done?

 

[Nullstein]

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?

This is literally the exact thing we have been discussing all along. Each Bell states corresponds to one of the four subensembles we have been discussing the whole time. Nothing this paper changes anything about the discussing.

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".

Nobody said that post-selection was cherry picking. Post-selection generates subensembles by conditioning on a common effect. Thus the correlations in these subensembles is spurious as every statistician will tell you.

 

[Nullstein]

I don't follow your notation or what you what you are trying to communicate

This is the whole problem. You take a step back and try to understand what is being discussed and confirm the calculations on your own before replying.

but initially 1 and 4 are uncorrelated and therefore can be considered in a Product State as you say. I might represent that as: $\hat{\rho}_{14}=\hat{\rho}_{1} \otimes \hat{\rho}_{4}.$ But 1 is monogamously entangled with 2, and 4 is monogamously entangled with 3, so I prefer the notation: $\hat{\rho}_{1234}=\hat{\rho}_{12} \otimes \hat{\rho}_{34}$ which is more descriptive. Typically each pair is going to be in either the $\psi+ or \phi+$ Bell state, depending on the Source (i.e. Type I or Type II PDC, or whatever).

I'm talking about the state of the 1&4 subsystem, which is given by $\rho_{14}=\mathrm{tr}_{23}\rho_{1234} = \frac 1 4 \mathbb 1\otimes\mathbb 1$. The state of the 1&4 system it not entangled, neither before nor after the BSM. This is the math and its simple linear algebra. It would be best if you tried that calculation on your own, since you don't seem to believe it. Nobody disputes it other than you. It's literally an undergrad calculation.

IFF the BSM occurs, each and every 1 & 4 are entangled (according to the predictions of QM). Then is new description is: $\hat{\rho'}_{1234}=\hat{\rho}_{14} \otimes \hat{\rho}_{23}.$

This is true for a subensemble. The full ensemble still is in a state such that the trace over 2&3 produces the state $\frac 1 4\mathbb 1\otimes\mathbb 1$.

I don't think anyone is any longer disputing what I say here.

Yes, we are. You deny that the full ensemble of 1&4 after the BSM is given by $\frac 1 4\mathbb 1\otimes\mathbb 1$. Everyone accepts that other than you.

 

[Fra]

Why should there be a GUT?

I guess that could or even should be questioned from a minimalist instrumentalist perspective - we could await building the GUT until explicitly driven by experimental deviations. This is a reasonable point. The same goes for QG. Who needs it anyway?

I am however driven by conceptual understanding of how physical interactions work and the nature of physical law from the perspective of inference. I do not work in a lab, I am not at the frontier of pushing subatomic interactions at higher energies. I see my personal quest as making sense of and find a coherent and efficient want to understand the massive body of data mankind already have. So far it's too much of a patchwork of effective theories, too many parameters that are experimentally determined that most likely should be predictable from a deeper theory. All the masses of the elementary particles for example. Predicting just a few of the manually tuned parameters or explaining some of the symmetries in the standard model would be an achievement worth asking the question even if no other experimental contradictions appear. It's how I see it. Is there a lack of ideas without experimental input? I don't think so?

From my perspective, the quest for a GUT goes hand in hand with the quest for QG, it's one quest, not two. Many programs, except string theory, try to solve them separately. String theory seems to have gotten stuck though IMO. I think solving the two problems together will be easier, than solving them separately. The problem does get "bigger", but I think some of the keys to make it work lies in seeing the relation.

/Fredrik

 

[vanhees71]

I don't follow your notation or what you what you are trying to communicate; but initially 1 and 4 are uncorrelated and therefore can be considered in a Product State as you say. I might represent that as: $\hat{\rho}_{14}=\hat{\rho}_{1} \otimes \hat{\rho}_{4}.$ But 1 is monogamously entangled with 2, and 4 is monogamously entangled with 3, so I prefer the notation: $\hat{\rho}_{1234}=\hat{\rho}_{12} \otimes \hat{\rho}_{34}$ which is more descriptive. Typically each pair is going to be in either the $\psi+ or \phi+$ Bell state, depending on the Source (i.e. Type I or Type II PDC, or whatever).

That's all right. Just once more for clarity in the notation you seem to prefer to the QFT-notation with creation operators: The notation for the Bell states of photon pairs $(ij)$ (where (ij) labels, e.g., the momentum):
$$|\Psi_{ij}^{\pm} \rangle= \frac{1}{\sqrt{2}} (|\vec{p}_i,H;\vec{p}_j, V \rangle \pm (|\vec{p}_i,V;\vec{p}_j, H \rangle,$$
$$|\Phi_{ij}^{\pm} \rangle= \frac{1}{\sqrt{2}} (|\vec{p}_i,H;\vec{p}_j, H \rangle \pm (|\vec{p}_i,V;\vec{p}_j, V \rangle.$$
The initial state is $\hat{\rho}_{1234}=|\Psi \rangle \langle \Psi |$
$$|\Psi \rangle=|\Psi_{12}^{-} \rangle \otimes |\Psi_{34}^{-} \rangle.$$
It is a bit of work but easy to see that this is the same as
$$|\Psi \rangle=\frac{1}{2} (|\Psi_{23}^{+} \otimes |\Psi_{14}^{+} \rangle - |\Psi_{23}^{-} \otimes |\Psi_{14}^{-} \rangle -|\Phi_{23}^{+} \otimes |\Psi_{14}^{+} \rangle +|\Phi_{23}^{-} \otimes |\Psi_{14}^{-} \rangle).$$
That means that the subsystem consisting of photons 1 and 4 are in the reduced state
$$\hat{\rho}_{14} = \mathrm{Tr}_{23} \hat{\rho}_{1234}=\frac{1}{4} \hat{1}_{14},$$
i.e., the photons (14) are both unpolarized and uncorrelated.

IFF the BSM occurs, each and every 1 & 4 are entangled (according to the predictions of QM). Then is new description is: $\hat{\rho'}_{1234}=\hat{\rho}_{14} \otimes \hat{\rho}_{23}.$ That can be simplified to $\hat{\rho}_{14}$ which is will be one of 4 possible Bell states. That system of 2 photons is a biphoton (Fock state = 2) and is nonseparable, therefore entangled. You can refer to the distant BSM results to learn which Bell state. Different BSMs can discern different of the Bell states, either 1 or 2 or probably soon to be all 4.

Indeed if 2&3 are projected to the state $|\Psi_{23}^- \rangle$, which is what you call BSM (Bell-state measurment), then for the so selected subensemble 1&4 are also in the state $|\Psi_{14}^- \rangle$. In math it's completely clear that you do something to photons 2&3 but nothing (!!) to photons 1 and 4, i.e., when projecting 2&3 to $|\Psi_{23}^- \rangle$, then there's nothing interacting with photons 1 and 4:
$$\mathcal{N} |\Psi' \rangle = |\Psi_{23}^{-} \rangle \langle \Psi_{23}^{-} \rangle \otimes \hat{1}_{14} |\Psi \rangle=\frac{1}{2} |\Psi_{23}^{-} \rangle \otimes |\Psi_{14}^{-1} \rangle.$$
Thus the projection to this specific Bell state occurs with probability $(1/2)^2=1/4$, and the (renormalized) state of the subensemble is
$$|\Psi' \rangle=|\Psi_{23}^{-} \rangle \otimes |\Psi_{14}^{-} \rangle,$$
i.e., now (23) and (14) are in Bell states, but the pairs are completely uncorrelated.

The state $|\Psi_{23}^{-} \rangle$ is chosen by the experimenters, because with a polarizing beam splitter it's simply identified by the fact that only in this state both output channels register a photon (since photons are bosons and the two photons are in an odd polarization state, the momentum state must be also odd, i.e., the photons cannot exit the beam splitter in the same momentum state).

I think about these facts we agree, and I hope we also agree on the math.

I don't think anyone is any longer disputing what I say here. The only thing left to discuss is whether the distant BSM should be considered "action at a distance". Is there any other open question at this point?

If we agree on the math and the meaning of a projection measurement, then nothing has be done to photons 1 or 4 when projecting photons 2 and 3 to a Bell state, which is indeed a local measurement, because you have to bring these two photons together at the polarizing beam splitter and the detectors in the two exit channels of this splitter, i.e., this is something where the two photons interact with equipment in the pretty small space-time region.

It is also completely irrelevant, in which temporal order you do the BSM and the measurements on photons 1 and 4. You can also perform the measurements space-like separated, so that there cannot be any actions at a distance by the BSM measurement on photons 1 and 4, if you accept the microcausality principle.

Interestingly enough it has been explicitly tested that indeed Q(F)T prevails also when tested against a type of "multisimultaneity model" (using energy-time entangled photon pairs):

https://arxiv.org/abs/quant-ph/0110117
https://doi.org/10.1103/PhysRevLett.88.120404

 

[Nullstein]

Just to complete @vanhees71's correct calculation: In total we then get 4 such subensembles, depending on the measurement outcome:
$$\rho_1 = P_{|\Psi_{23}^{+} \rangle \otimes |\Psi_{14}^{+} \rangle}\quad\rho_2 = P_{|\Psi_{23}^{-} \rangle \otimes |\Psi_{14}^{-} \rangle}\quad\rho_3 = P_{|\Phi_{23}^{+} \rangle \otimes |\Phi_{14}^{+} \rangle}\quad\rho_4 = P_{|\Phi_{23}^{-} \rangle \otimes |\Phi_{14}^{-} \rangle}\quad$$
We can now calculate the full ensemble by summing over them with weights given by the probabilities $p_i = \frac 1 4$, according to which they are distributed. This gives us the state of the full ensemble after the BSM:
$$\rho_{1234,\text{after}}=\frac 1 4\left(\rho_1+\rho_2+\rho_3+\rho_4\right)$$
It's then easy to see that the state of the 1&4 subsystem after the BSM is given by:
$$\rho_{14,\text{after}}=\mathrm{tr}_{23}\rho_{1234,\text{after}}=\frac 1 4\mathbb 1\otimes\mathbb 1$$
This is just the completeness relation, because the partial trace gives us the 4 Bell states, which constitute an orthogonal basis.

 

[martinbn]

Just to point out the the wikipedia article (although not an acceptable source) on teleportation has a good subsection on swapping.

 

[DrChinese]

1. Just once more for clarity in the notation you seem to prefer to the QFT-notation with creation operators: ...

That means that the subsystem consisting of photons 1 and 4 are in the reduced state
$$\hat{\rho}_{14} = \mathrm{Tr}_{23} \hat{\rho}_{1234}=\frac{1}{4} \hat{1}_{14},$$
i.e., the photons (14) are both unpolarized and uncorrelated.

2. Indeed if 2&3 are projected to the state $|\Psi_{23}^- \rangle$, which is what you call BSM (Bell-state measurment), then for the so selected subensemble 1&4 are also in the state $|\Psi_{14}^- \rangle$. In math it's completely clear that you do something to photons 2&3 but nothing (!!) to photons 1 and 4, i.e., when projecting 2&3 to $|\Psi_{23}^- \rangle$, then there's nothing interacting with photons 1 and 4:
$$\mathcal{N} |\Psi' \rangle = |\Psi_{23}^{-} \rangle \langle \Psi_{23}^{-} \rangle \otimes \hat{1}_{14} |\Psi \rangle=\frac{1}{2} |\Psi_{23}^{-} \rangle \otimes |\Psi_{14}^{-1} \rangle.$$
Thus the projection to this specific Bell state occurs with probability $(1/2)^2=1/4$, and the (renormalized) state of the subensemble is
$$|\Psi' \rangle=|\Psi_{23}^{-} \rangle \otimes |\Psi_{14}^{-} \rangle,$$
i.e., now (23) and (14) are in Bell states, but the pairs are completely uncorrelated.

The state $|\Psi_{23}^{-} \rangle$ is chosen by the experimenters, because with a polarizing beam splitter it's simply identified by the fact that only in this state both output channels register a photon (since photons are bosons and the two photons are in an odd polarization state, the momentum state must be also odd, i.e., the photons cannot exit the beam splitter in the same momentum state).

I think about these facts we agree, and I hope we also agree on the math.

3. If we agree on the math and the meaning of a projection measurement, then nothing has be done to photons 1 or 4 when projecting photons 2 and 3 to a Bell state, which is indeed a local measurement, because you have to bring these two photons together at the polarizing beam splitter and the detectors in the two exit channels of this splitter, i.e., this is something where the two photons interact with equipment in the pretty small space-time region.

4. ...if you accept the microcausality principle.

1. We are agreed as to this: 1 and 4 are initially completely uncorrelated and unentangled. There is no subset in which they are initially entangled with each other. We know this because of... Monogamy of Entanglement.

2. We disagree here. If we happen to project 2 & 3 into the $\psi-$ state, then it is true that 1 & 4 are changed into the entangled $\psi-$ state as well. They were previously in a completely uncorrelated and unentangled state, as we just agreed. That's the math!

And in the experiment we are discussing, the team is also able to discern the $\psi+$ state as well as the $\psi-$ state. The $\phi+/-$ states cannot be distinguished with APDs because they cannot reset fast enough using current technology to be able to count 2 photons going to the same detector in such close succession.

3. I don't know why you call the BSM a "local" measurement. It's not local to 1 or 4, and they are the ones affected because they have been projected into a Bell state that there were NOT previously in.

My assertion, which is supported by the math (which we just agreed upon):
Initial state (all cases): $|\Psi \rangle=|\Psi_{12}^{-} \rangle \otimes |\Psi_{34}^{-} \rangle$ [depending on Source type]
a. Final state (occurs 25%): $|\Psi' \rangle=|\Psi_{14}^{-} \rangle \otimes |\Psi_{23}^{-} \rangle$
b. Final state (occurs 25%): $|\Psi' \rangle=|\Psi_{14}^{+} \rangle \otimes |\Psi_{23}^{+} \rangle$
c. Final state (occurs 25%): $|\Psi' \rangle=|\Phi_{14}^{-} \rangle \otimes |\Phi_{23}^{-} \rangle$
d. Final state (occurs 25%): $|\Psi' \rangle=|\Phi_{14}^{+} \rangle \otimes |\Phi_{23}^{+} \rangle$

4. I don't, at least not as you apply it. It is contradicted by experiment, that is why we are having this debate. An action here objectively changes a state there. As we agree on the math, this should be easy... if you just ignore your wrong assumptions. So far, you haven't been able to explain how remotely created photons can violate a CHSH inequality without appealing to the (wrong) assumption that they were somehow already in that state previously. None of the 1 & 4 pairs were entangled to begin with, and there was no subset in which they were. Any more than you could take any 2 streams of photons from different sources and try to claim they are entangled.

5. Yes, we are. You deny that the full ensemble of 1&4 after the BSM is given by 𝟙𝟙141⊗1. Everyone accepts that other than you.

5. And just to dismiss @Nullstein's ridiculous argument: In all cases, the BSM leaves 1 & 4 in one of the entangled states per a, b c, or d above. That is NOT the same state by any means as you give (a random outcome).

Or what, you won't accept and use the available data? I guess when you do a typical Bell test, you only look at Alice's results... and throw away Bob's?

Here's a shocker: if you look at ANY stream of entangled particles, which are an equal mixture of $\psi-$ and $\psi+$ pairs, they will statistically show no correlation IF you discard the available information as to their state. But guess what: They are still entangled! It doesn't matter whether you know it or not. I don't know what kind of scientist you are that you require information to be discarded when running an experiment. While you are busy discarding useful information, the remainder of the scientific world is working on identifying all 4 Bell states after a BSM. And I expect someone will figure that out "soon", since there have already been proofs of concept.

 

[Nullstein]

5. And just to dismiss @Nullstein's ridiculous argument: In all cases, the BSM leaves 1 & 4 in one of the entangled states per a, b c, or d above. That is NOT the same state by any means as you give (a random outcome).

It speaks for itself that you find standard science "ridiculous." There is no evidence that the BSM has any effect on 1&4. The correlation in the post-selected ensembles is spurious, as any statistician will tell you. The conditioning on a common effect (the BSM result) introduces non-causal correlation. This is not controversial at all. It's the bread and butter of any statistician and a perfectly well-understood phenomenon.

Or what, you won't accept and use the available data? I guess when you do a typical Bell test, you only look at Alice's results... and throw away Bob's?

Nobody throws data away. All information is contained in the full ensemble. But you just haven't understood the difference between correlation and causation. You cannot infer causality from the existence of correlations. It is perfectly well-understood under what circumstances this is allowed. If there is conditioning on a common effect, then one must sum over the conditional variable. Otherwise, the correlation is meaningless for causal inference.

Here's a shocker: if you look at ANY stream of entangled particles, which are an equal mixture of $\psi-$ and $\psi+$ pairs, they will statistically show no correlation IF you discard the available information as to their state. But guess what: They are still entangled! It doesn't matter whether you know it or not.

No, the full data set is not entangled. The full ensemble is product state, as we have established multiple times now. One does not discard information. It's a necessary step in causal inference to sum over the common effect variables. It's not optional. It's the only correct way to perform causal inference. Only the subensembles are entangled, but no causal relationships can be inferred from this.

I don't know what kind of scientist you are that you require information to be discarded when running an experiment.

The information is not discarded. But if you don't sum over the common effect variable when performing causal inference, you are a bad scientist, because you are not taking into account that correlation doesn't imply causation.

While you are busy discarding useful information, the remainder of the scientific world is working on identifying all 4 Bell states after a BSM. And I expect someone will figure that out "soon", since there have already been proofs of concept.

Since we are discussing the idealized setting without any experimental complications, we have already taken care of that. As good theoretical physicists, we have already done the calculations and are waiting for the experimentalists to confirm them.--
And just so there are no misunderstandings: The fact that there is no evidence for non-local causal relationships doesn't imply that there can not be a non-local causal model. It just means that we are not forced to adopt one. This is also what the recent Nobel laureate Anton Zeilinger has chosen to do. If you think that that there is evidence for non-local causal relationships, then you are in direct contradiction with a recent Nobel laureate.

 

[martinbn]

5. And just to dismiss @Nullstein's ridiculous argument: In all cases, the BSM leaves 1 & 4 in one of the entangled states per a, b c, or d above. That is NOT the same state by any means as you give (a random outcome).

This is interpretation dependent. It may be how it is in some, say collapse, interpretations.

The interpretation independent statment is that it is a mixture of those four. The density matrix of 14 is the same before and after the measurement of 23. It is the partial trace that has been written many times in these theards.

 

[Fra]

They are still entangled! It doesn't matter whether you know it or not.

I assume that by "you", you mean the observer.

Are you actually saying that two quantum states can be entangled without the observers knowledge?? What does that even mean?

/Fredrik

 

[DrChinese]

1. This is interpretation dependent. It may be how it is in some, say collapse, interpretations.

2. The interpretation independent statement is that it is a mixture of those four. The density matrix of 14 is the same before and after the measurement of 23. It is the partial trace that has been written many times in these threads.

1. If each and every final 1 & 4 pair is identified as to its Bell state, and a Bell test is performed, and a CHSH inequality is violated: they are entangled. There is no interpretation I know of that denies this.

2. Why on earth you think they are not entangled if you throw away the information on the Bell state, I don't know. That is anti-science. There is absolutely no requirement that all entangled pairs after a swap be a single Bell state to consider them entangled.

3. The full ensemble is product state, as we have established multiple times now.

From Nullstein post #40:
I agree that the subensembles are entangled [into 1 of the 4 Bell states].

We have established many times that the initial 1 & 4 state is uncorrelated and unentangled, and that there is no relationship between them - hidden or otherwise. From the paper: "A successful entanglement swapping procedure will result in photons 1 and 4 being entangled, although they never interacted with each other." From @vanhees71: "the [initial] photons (14) are both unpolarized and uncorrelated."

Yes, it is trivially true that if you look at the 1 & 4 pairs and don't know their Bell state, no pattern pops out. I guess we need to tell this team that they are wasting their time when they run the experiment and announce: "We confirm successful entanglement swapping by testing the entanglement of the previously uncorrelated photons 1 and 4. Violation of a CHSH inequality is not only of fundamental interest because it rules out local-hidden variable theories. It also proves that the swapped states are strongly entangled and, as a result, distillable."

All I can say is "wow". Hundreds of teams out there are working to create quantum networks using swapping from point to point to point. They all seem to think there is entanglement, and they all seem to think the 2 & 3 BSM had something to do with it.

4. So you admit that there are 4 different Bell states, but still insist to write ρ^14 for all of them? If they are 4 different states (even so all are maximally entangled), then you should also use 4 different symbols for them, for example ρ^141, ρ^142, ρ^143, and ρ^144.

5. And then, wouldn't ρ′^1234=∑iρ^14i⊗ρ^23i be a more appropriate formula for the state, as long as no conditioning with respect to the 4 different states is done?

4. I am not "admitting" anything, since I have been saying there are 4 Bell states all along. I can spell them out (as I already did in later post #202), if it makes a difference to your understanding. But I think you know the answer already and don't need this:

Initial state (all cases): |Ψ⟩=|Ψ12−⟩ ⊗ |Ψ34−⟩ [depending on Source type]
a. Final state (occurs about 25%): |Ψ′⟩ = |Ψ14−⟩ ⊗ |Ψ23−⟩ [detected in the experiment]
b. Final state (occurs about 25%): |Ψ′⟩ = |Ψ14+⟩ ⊗ |Ψ23+⟩ [detected in the experiment]
c. Final state (occurs about 25%): |Ψ′⟩ = |Φ14−⟩ ⊗ |Φ23−⟩
d. Final state (occurs about 25%): |Ψ′⟩ = |Φ14+⟩ ⊗ |Φ23+⟩

Yes, all the 1 & 4 pairs are still entangled... exactly as I said since we started. Which is exactly as QM predicts for this experiment. Only 2 of 4 states are identified in this experiment, but it is only a matter of time before a version is run in which all 4 Bell states are identified. (They have already done this for remote teleportation.)

5. And why would we want to fail to condition? That would lead us to a completely useless formula. The authors of the paper do the exact opposite, they USE the conditioning to demonstrate entanglement: "Expectation values for the CHSH inequality depending on the outcome, ψ− or ψ+, of the BSM. Each of these values is calculated from four measurements of four photon coincidences integrated over 15000 seconds."

The scientists in this room use all the information they have to make their conclusions. I would everyone would. And one of those conclusions is that the BSM "causes" the projection of the random Bell State outcome onto the 1 & 4 pairs that were previously unentangled. The issue with the word "causes" is that time ordering doesn't matter. @vanhees71 is technically correct when he cites the "microcausality" condition to say there are no nonlocal causes. That's merely because quantum causality operates much differently than Einsteinian causality, and therefore cannot be applied like Einsteinian causality.

Any way you look at it: a BSM here creates an entangled biphoton there. Pretty hard to wave hands and make that result go away. 1 & 4 pairs: Not at all entangled before the BSM (as we all agree), 100% entangled after the BSM (in 1 of 4 Bell States) - which QM predicts.

 

[Morbert]

Everyone agrees that there is entanglement in the subensembles marked by the outcomes of the BSM. Where people disagree is the implication of this. There are three broad positions.

i) Under all established interpretations, it implies FTL influences
ii) Under all established interpretations, it doesn't imply FTL influences
iii) Under some established interpretations, it implies FTL influences, and under other established interpretations, it doesn't.

At the very least, there have been well-established interpretations presented here, consistent with experiment, that don't imply FTL. So at the very least, the right position is either ii) or iii)

 

[mattt]

The state of a Quantum System is a mathematical object that gives you a set of relative frequencies of possible results, for each of a set of measuring procedures, that can only be tested on a huge set of identically prepared copies of the system.

Dr Chinese seems to have problems understanding that a photon-pair can belong to a set (of identically prepared copies) collectively described as a Product State, and at the same time this photon-pair may also belong to a different set (for example, any of the four subensembles) collectively described (each of the four subsets) as a Maximally Entangled State (which are four different states by the way), without necessarily anything "changing" for that photon-pair.

The monogamy of Entanglement apply when you consider a given ensemble (of identically prepared copies), but it ceases to apply if you change the ensemble for a subensemble (if you change the state, that now describes a different collection).

Only if you consider the "state" as a "property" of each pair (regardless of the collection, the ensemble to which it may be considered to belong) you'll see a "non-local causation" (a "change of a property") in this experiment.

 

[DrChinese]

1. Everyone agrees that there is entanglement in the subensembles marked by the outcomes of the BSM.

2. Where people disagree is the implication of this. There are three broad positions.

i) It implies FTL
ii) It doesn't imply FTL
iii) The FTL implication is interpretation dependent

3. At the very least, there have been well-established interpretations presented here, consistent with experiment, that don't imply FTL. So at the very least, the right position is either ii) or iii)

1. I hope we have reached this point.

2. I agree with these basic positions as being represented here.

3. I don't see the range of interpretations as you do. Some continue to assert pre-Bell locality in the face of Bell, without stating exactly what they are giving up in the way of realism. Stated a different way: they present no mechanism as to how locality can be preserved in experiments such as Entanglement Swapping. Certainly the math of QM predicts without regard to spacetime distance.

Most interpretations attempt to maintain Einsteinian causality in all quantum contexts, when the evidence goes against it. That isn't a winning position. I don't know how nature pulls off its tricks, but I know that it displays quantum non-local/non-causal effects.

 

[DrChinese]

Dr Chinese seems to have problems understanding that a photon-pair can belong to a set (of identically prepared copies) collectively described as a Product State, and at the same time this photon-pair may also belong to a different set (for example, any of the four subensembles) collectively described (each of the four subsets) as a Maximally Entangled State (which are four different states by the way), without necessarily anything "changing" for that photon-pair.

The monogamy of Entanglement apply when you consider a given ensemble (of identically prepared copies), but it ceases to apply if you change the ensemble for a subensemble (if you change the state, that now describes a different collection).

I have quoted extensively on Monogamy and referenced papers. Where are your references disputing these quotes? According to QM: MoE applies to any and every single entangled pair. There is no requirement that there be a large set.

Further: of course you can model a system where you discuss an entangled pair (say 1 & 4) and other particles (let's say 5 & 6, which are not entangled). We would have a Product state with: (1 & 4) and (5) and (6). Unless there is some purpose to such a description, it isn't useful. And certainly has no relevance to our discussion. In our discussion, MoE applies exactly as follows:

If (1 & 2) are maximally entangled in any Bell state, then (1 & 4) are not at all entangled.
If (1 & 4) are maximally entangled in any Bell state, then (1 & 2) are not at all entangled.
If (3 & 4) are maximally entangled in any Bell state, then (1 & 4) are not at all entangled.
If (1 & 4) are maximally entangled in any Bell state, then (3 & 4) are not at all entangled.

Any other application is wrong. This is what is useful to our discussion.