A Hidden Assumptions in Bell's Theorem?

[vanhees71]

Because that's how the term is used in the literature in this area.Perhaps if one has never encountered the terminology before; but you don't have that excuse. We have had this discussion in multiple threads and you are perfectly well aware of what "non-locality" means in the context of discussions of Bell's Theorem and the Bell inequalities. If you don't like the terminology, then convince the community of physicists who are working in this area to change their usage. Continuing to belabor it in PF threads is pointless.

But I'm allowed to use the standard terminology of my scientific community too! That's why I always add my definition of "locality", namely the assumption of the microcausality constraint for operators that represent local observables as used in standard relativistic QFT since 1926!

 

[kurt101]

No! By construction there cannot be any causal influence between space-like separated events. In other words there's no faster-than-light signalling possible by definition within local relativistic QFT. It's called the microcausality principle and together with the assumption of the existence of a stable ground state predicts the CPT theorem, the spin-statistics theorem and makes the S-matrix unitary.

Thanks! That makes it clear what the answer is and it is like a thought, but I hear a lot of statements that seemingly contradict this, but I believe your answer is the correct one.

These questions are not well posed in relativistic QFT because the idea of "change" doesn't make sense in a relativistic theory the way you are using it. The relativistic model includes all of spacetime; it doesn't "change", it just is. Only one thing can happen at each event in spacetime, so the idea of something "changing" doesn't even make sense.

And thanks for the counter point.

What you can ask about in relativistic QFT is whether operators representing measurements made at different spacetime events commute. In a relativistic QFT, operators representing measurements at spacelike separated events will always commute--meaning their results cannot depend on the order in which they are made. Operators representing measurements at lightlike or timelike separated events do not have to commute in relativistic QFT, but they can; and it turns out that, in the particular experiments under discussion here, they do: the measurements commute regardless of the spacetime relationship between the events at which they take place.

And the fact that they can commute regardless of spacetime relationship in this case is the heart of this discussion. I don't think anyone denies they commute in this case. I think if we could develop the math that adequately simulates the relativistic QFT math of this experiment, we would find in the case where BSM test is done after the 1 & 4 measurement, that there would be no mathematical contribution between 1 & 4, but they will still commute out of principle. This is the only explanation that makes sense to me and keeps with the microcausality principle that @vanhees71 discusses.

 

[vanhees71]

I don't think there's any counter point in the, admittedly somewhat enigmatic, statement by @PeterDonis and mine. It's just, how an "event" is described with the local observables in relativistic QFT, i.e., by the spacetime arguments of the corresponding operators that describe these local observables.

I can only recommend again to read Coleman's great book, quoted above. There's also a version for free on the arXiv:

https://arxiv.org/abs/1110.5013

 

[DrChinese]

I don't understand this question. If the measurement events (i.e., the various registrations of photons in coincidence measurements) are space-like separated within relativistic microcausal QFT, they cannot causally influence each other. The explanation for the correlations described by entanglement is simply that they are simply properties described by the entangled state, which has been prepared in the very beginning by creating the corresponding entangled photon pairs (or more than 2 photons in other experiments).

Thus initially, we have in the Entanglement Swapping scenario: a) the maximally entangled [1 & 2] pairs - which cannot be entangled with anything else; and b) the distant independently created maximally entangled [3 & 4] pairs - which also cannot be entangled with anything else. Hopefully you* and I both agree on your explanation up to this point.

But then: The a) and b) pairs contain c) some subset in which the [1 & 4] pairs are also maximally entangled; and d) there's a subset in which the [2 & 3] pairs are maximally entangled (in the same Bell State as the matching [1 & 4] pairs per c). So the revelation of specific [2 & 3] entangled pairs as being in an entangled state (such as Psi-) via the BSM will identify a now distant matching [1 & 4] subset that will violate a Bell inequality. At no time, in your view, does the successful BSM on the [2 & 3] pairs CHANGE anything in the distant identified [1 & 4] subset.

I have already explained multiple times that your idea is falsified by Monogamy of Entanglement (reference and proof). [1] can never be maximally entangled with [2] and also be maximally entangled with [4] at the same time. So the initial preparation does not and cannot include any [1 & 4] pairs prepared in an initially entangled state. That must occur at a later time. Or maybe you deny Monogamy of Entanglement is a part of QFT?

I say the successful BSM "quantum causes" an entanglement swap. This is an objective fact, because only upon successful BSM do [1 & 4] Bell State correlations appear. Einsteinian causality is not respected.* And I can't believe that here I am replying to you in this thread, after promising I wouldn't anymore because you won't supply any quotes supporting your position. Oh well...

 

[mattt]

I have had this question about relativistic QFT for a long time now and I still don't have a clear answer, but I think it should be easy for someone who understands relativistic QFT math well to answer. The question is if there is a change outside of the light cone of the experiment being considered. How long does it take before that change makes a contribution (mathematically) to the experiment? Does it instantly make a mathematical contribution or does its contribution have to respect the speed of light?

To me it sounds like a simple question. One should be able to come up with a simple case and understand when the contribution of the change outside of the light cone of the experiment being considered makes any kind of contribution. Again: is the contribution instant or does it respect the speed of light? And I am looking for the relativistic QFT answer here.

I am thinking how to put it into words. There isn't a "change that is propagating..." the way you think it is.

It's more like, no matter how many versions of the whole universe (the whole space-time and all the mathematical operators on it, satisfying QM or QFT axioms if you wish) or parts of it that you consider (varying initial/boundary conditions), some sets of events must satisfy certain correlations (some statistical constraints) that for us humans are highly surprising.

And it is surprising for us because we were not used to it.

For those kinds of sets of events, the only mathematical way to express the statistical distribution of measurement results that satisfy those statistical constraints, is by taking into account things like "orientation of devices" at places and times (spacetime points) that can be space-like, time-like or whatever.

And you may say: "so what? Newtonian Gravitation also share that feature of depending on the whole distribution of masses"

Yes, but the validity domain of Newtonian Gravitation is reduced ( it fails in the relativistic regime) and the gravitational interaction in Newtonian Gravitation goes to zero with distance.

The way de Bohmian particles accomplish the satisfaction of those statistical constraints is very different, their velocities change in a way that doesn't go to zero with distance.

We are not used to anything like that, and some are still waiting for a more satisfying picture/ontology.

 

[vanhees71]

Thus initially, we have in the Entanglement Swapping scenario: a) the maximally entangled [1 & 2] pairs - which cannot be entangled with anything else; and b) the distant independently created maximally entangled [3 & 4] pairs - which also cannot be entangled with anything else. Hopefully you* and I both agree on your explanation up to this point.

That's indeed how the state is prepared. So I agree.

But then: The a) and b) pairs contain c) some subset in which the [1 & 4] pairs are also maximally entangled; and d) there's a subset in which the [2 & 3] pairs are maximally entangled (in the same Bell State as the matching [1 & 4] pairs per c). So the revelation of specific [2 & 3] entangled pairs as being in an entangled state (such as Psi-) via the BSM will identify a now distant matching [1 & 4] subset that will violate a Bell inequality. At no time, in your view, does the successful BSM on the [2 & 3] pairs CHANGE anything in the distant identified [1 & 4] subset.

The four-photon state before any Bell measurements is (by "preparation")
$$\hat{\rho}=\hat{\rho}_{12} \otimes \hat{\rho}_{34}.$$
Here $\hat{\rho}_{12}$ and $\hat{\rho}_{34}$ are, e.g., the pure states $|\Psi_{12}^{(-)} \rangle \langle \Psi_{12}^{(-)}|$ and $|\Psi_{34}^{(-)} \rangle \langle \Psi_{34}^{(-)}|$ where $\Psi_{jk}^{(-)}$ is the polarization-singlet state of the photon pair [j&k].

According to QT you can, however, make a measurement projecting [2&3] to the polarization-singlet state such that for the so created (!!!) subensemble of four-photon states now [2&3] is for sure in this state, but now also, for this subensemble also [1&4] are necessarily in this Bell state. The cause of this is not a faster-than light influence of the manipulations/measurements on photons 2 & 3 on the photons 1 and 4 but it's implied by the properties of the preparation of the original four-photon state in the state $\hat{\rho}$. In this sense this subensemble is "contained" in the ensemble described by $\hat{\rho}$.

I have already explained multiple times that your idea is falsified by Monogamy of Entanglement (reference and proof). [1] can never be maximally entangled with [2] and also maximally entangled with [4]. So the initial preparation does not and cannot include any [1 & 4] pairs prepared in an initially entangled state. Or maybe you deny Monogamy of Entanglement is a part of QFT?

But for the subensemble, of course 1 and 2 are not entangled at all, because due to the "swapping", i.e., the projection of [2&3] to the Bell state, this ensemble is described by
$$\hat{\rho}'=\hat{\rho}_{23} \otimes \hat{\rho}_{14}.$$

I say the successful BSM "quantum causes" an entanglement swap. This is an objective fact, because only upon successful BSM do [1 & 4] correlations appear. Einsteinian causality is not respected.

That's where we disagree: Nowhere in the above description is anything which contradicts microcausal (i.e., local!) relativistic QFT, and this theory respects Einstein causality. It's simply selection of a subensemble by a local measurement on photons [2&3]. The entanglement of [1&4] after this selection is due to the correlation between the pairs [1&2] and [3&4] described by the original state $\hat{\rho}$ but not due any "FTL causal effect" of the local projection measurement on [2&3] on either photon 1 or 4. There are simply no such "FTL causal effects" within relativistic microcausal QFT.

* And I can't believe that here I am replying to you in this thread, after promising I wouldn't anymore because you won't supply any quotes supporting your position. Oh well...

I quote and quote and quote all the time standard textbooks on QFT and original papers on Bell measurements. You obviously don't like to take notice of them, because you seem to be unwilling to accept that there is no contradiction between Einstein causality, which by construction is respected by relativistic QFT, and the observed phenomena of entanglement. I've no clue, why it is so hard to accept that there's no esoterics in entanglement but just beautiful math in accordance with all observed facts!

 

[gentzen]

I quote and quote and quote all the time standard textbooks on QFT

I can only recommend again to read Coleman's great book, quoted above. There's also a version for free on the arXiv:

https://arxiv.org/abs/1110.5013

Well, how to say this in friendly words: "quoted above" might be true, but you routinely leave the task to guess the parts you are actually quoting to the reader.

For you, "to quote" seems to mean more that you could come up with an actual quote if pressed. But for most others, to quote means to select a specific piece of text from a longer text to provide a focus point. Perhaps you did it in this specific case, and I would find it if I searched above (and had some special skill to guess right at which post to look). But more likely, your "quote above" was just a general recommendation to read that entire book.

 

[PeterDonis]

only if they are spacelike separated you can be sure that the measurements are not causally connected with each other.

But by your criterion, if the measurements commute, they can't be causally connected, since that's what rules out causal connection in the spacelike separation case. Being lightlike or timelike separated does not guarantee causal connection, it just makes it possible. But that just means you need to use other criteria (like whether or not the measurements commute) to determine whether a causal connection (in the sense of ordinary classical causality) exists.

The explanation for the correlations described by entanglement is simply that they are simply properties described by the entangled state, which has been prepared in the very beginning

But the state that was prepared at the very beginning does not have any entanglement between photons 1 & 4; that's the point of @DrChinese's monogamy of entanglement argument. The state that is prepared at the very beginning has 1 & 2 maximally entangled, and 3 & 4 maximally entangled. That means there cannot, by construction, be any entanglement between 1 & 4 in that state. So wherever the observed entanglement between 1 & 4 at the end comes from, it can't be from the initial state that was prepared.

Neither you nor anyone else that I can see in this thread has even tried to address this point.

 

[PeterDonis]

I'm allowed to use the standard terminology of my scientific community too!

In a discussion primarily focused on QFT, sure.

But this discussion is primarily focused on Bell's Theorem. So the context is different. Scientific terminology is and always has been context dependent.

 

[PeterDonis]

I think if we could develop the math that adequately simulates the relativistic QFT math of this experiment, we would find in the case where BSM test is done after the 1 & 4 measurement, that there would be no mathematical contribution between 1 & 4, but they will still commute out of principle.

I don't know what you mean here. The math is simple: it's the same regardless of the spacetime relationship between the measurements. That means it's the same regardless of whether the BSM test is done before (in the past light cone), after (in the future light cone), or spacelike separated from the 1 & 4 measurements (which, to be clear, are two separate measurements that do not have to take place at the same event). So there is no room in the math, for the scenario under discussion, for any difference between those cases at all. Any purported "explanation" that does not recognize that can't be correct.

 

[kurt101]

I have already explained multiple times that your idea is falsified by Monogamy of Entanglement (reference and proof). [1] can never be maximally entangled with [2] and also be maximally entangled with [4] at the same time. So the initial preparation does not and cannot include any [1 & 4] pairs prepared in an initially entangled state. That must occur at a later time. Or maybe you deny Monogamy of Entanglement is a part of QFT?

I pointed out the flaw in your monogamy entanglement argument for the case where the BSM test is done after measuring 1 & 4. Any response to this?

I say the successful BSM "quantum causes" an entanglement swap. This is an objective fact, because only upon successful BSM do [1 & 4] Bell State correlations appear. Einsteinian causality is not respected.

I have asked you twice if you would provide a definition of Einsteinian causality. Third times a charm? Will you provide a definition?

* And I can't believe that here I am replying to you in this thread, after promising I wouldn't anymore because you won't supply any quotes supporting your position. Oh well...

Is that also why you are not responding to my counter arguments? And to be fair, I am not even certain we disagree until you provide me your definition of Einsteinian causality. I can only guess what you mean.

 

[PeterDonis]

I pointed out the flaw in your monogamy entanglement argument for the case where the BSM test is done after measuring 1 & 4.

Your point was that if the BSM test is done after the 1 & 4 measurements, the entanglement of 1 & 4 no longer exists when the BSM test is done. That's true (at least at an appropriate heuristic level--there are always caveats and pitfalls lurking beneath such statements), but it does not at all refute the monogamy of entanglement argument. The monogamy of entanglement argument simply points out that, since in the intially prepared state, 1 & 2 are maximally entangled and 3 & 4 are maximally entangled, it is impossible for there to be any entanglement between 1 & 4 in the initially prepared state, so any purported "explanation" that attributes the observed entanglement between 1 & 4 to an entanglement between them in the intially prepared state cannot be correct. Your point does not address that at all.

 

[PeterDonis]

In this sense this subensemble is "contained" in the ensemble described by $\rho$.

Please show mathematically where in the initial ensemble $\rho$ the entanglement between 1 & 4 is contained. And you will then need to explain how this can be consistent with monogamy of entanglement, since in the initial ensemble 1 & 2 are maximally entangled and 3 & 4 are maximally entangled, leaving no room for any entanglement at all between 1 & 4.

It is no answer to this to continue to shout "microcausal QFT". Nobody is claiming that microcausal QFT is wrong. But you don't get to wave your hands and make an argument that appears to contradict monogamy of entanglement either; microcausal QFT still has to be consistent with monogamy of entanglement.

 

[kurt101]

Your point was that if the BSM test is done after the 1 & 4 measurements, the entanglement of 1 & 4 no longer exists when the BSM test is done. That's true (at least at an appropriate heuristic level--there are always caveats and pitfalls lurking beneath such statements), but it does not at all refute the monogamy of entanglement argument. The monogamy of entanglement argument simply points out that, since in the intially prepared state, 1 & 2 are maximally entangled and 3 & 4 are maximally entangled, it is impossible for there to be any entanglement between 1 & 4 in the initially prepared state, so any purported "explanation" that attributes the observed entanglement between 1 & 4 to an entanglement between them in the intially prepared state cannot be correct. Your point does not address that at all.

But why is it impossible for 1 & 4 to have entanglement between them in the initial state of the experiment if photon 1 has no connection to 3 & 4 and if photon 4 has no connection with 1 & 2 at initialization time? Are you saying for any particular photon, that it can't have multiple photons existing in the universe that happen to be in the same state as it (or be in a maximally entangled state with it)? And if that is what you are saying, how would you ever be able to tell?

 

[DrChinese]

1. That's indeed how the state is prepared. So I agree.

2. The four-photon state before any Bell measurements is (by "preparation")
$$\hat{\rho}=\hat{\rho}_{12} \otimes \hat{\rho}_{34}.$$
According to QT you can, however, make a measurement projecting [2&3] to the polarization-singlet state such that for the so created (!!!) subensemble of four-photon states now [2&3] is for sure in this state, but now also, for this subensemble also [1&4] are necessarily in this Bell state.

3. But for the [NEW] subensemble, of course 1 and 2 are not entangled at all, because due to the "swapping", i.e., the projection of [2&3] to the Bell state, this [NEW] ensemble is described by
$$\hat{\rho}'=\hat{\rho}_{23} \otimes \hat{\rho}_{14}.$$

Subensemble Before (Initial Preparation): $$\hat{\rho}=\hat{\rho}_{12} \otimes \hat{\rho}_{34}.$$
Subensemble After: $$\hat{\rho}'=\hat{\rho}_{23} \otimes \hat{\rho}_{14}.$$
Note that what you refer to as a "subensemble" did, in fact, change, and we AGREE on this! The before and after states are objectively (and experimentally) different as long as you believe QM (or QFT, which predicts the same result).

Now, once you admit that the BSM was responsible for "quantum causing" this - or something was, it really doesn't matter - then we can both sleep soundly tonight.

The change from the rho to rho' state occurs a) without regard to where the [1] measurement, the [4] measurement, and/or the BSM occur in spacetime; and b) without regard to the order (or reference frame) of those measurements. Therefore, the change in state does not respect Einsteinian causality at any level. QED.

But why is it impossible for 1 & 4 to have entanglement between them in the initial state of the experiment if photon 1 has no connection to 3 & 4 and if photon 4 has no connection with 1 & 2 at initialization time? Are you saying for any particular photon, that it can't have multiple photons existing in the universe that happen to be in the same state as it (or be in a maximally entangled state with it)?

Yes, no other particle (or particles) in the entire universe can be entangled with [1] when it is maximally entangled to [2]. You apparently are not following my references which say exactly this. Here is a proof, search and you will see plenty that say the same.

You ask: "Why is that so?" Because QM says so, and *all* Interpretations MUST say so too if they are to be considered a viable interpretation. Otherwise they are a different theory. Note some interpretations really are a different theory, and that's OK too. However, if they are a different theory, they are potentially distinguishable by experiment.

Some interpretations, such as Bohmian Mechanics, walk on a fine line in this regard. But most interpretations claim some extra assumption that attempts to restore either realism or locality or causality in some Einsteinian sense. Those, in my opinion, should be ruled out simply because of what @vanhees71 says above:
$$\hat{\rho}_{12} \otimes \hat{\rho}_{34} <> \hat{\rho}_{23} \otimes \hat{\rho}_{14}.$$

You ask my definition of Einsteinian causality? I would think anyone could fill that in for themselves, but here are 3 central ideas (any or all of which are Einsteinian):

i) Causes must precede effects
ii) Usually locality, effects must be in future time cone
iii) Some folks include some form of "realism" too, but that is not necessary

 

[PeterDonis]

why is it impossible for 1 & 4 to have entanglement between them in the initial state

Because 1 is maximally entangled with 2 and 4 is maximally entangled with 3. That means any additional entanglements are impossible.

Are you saying for any particular photon, that it can't have multiple photons existing in the universe that happen to be in the same state as it (or be in a maximally entangled state with it)?

If two photons are maximally entangled with each other, that means that they can't be entangled with anything else. This is basic QM.

 

[Jarvis323]

In a discussion primarily focused on QFT, sure.

But this discussion is primarily focused on Bell's Theorem. So the context is different. Scientific terminology is and always has been context dependent.

The thread is supposed to be about whether there are hidden assumptions in Bell's theorem which leave the possibility open for a local hidden variable theory.

The claimed local hidden variable model from Hess Et al. is disputed by Zellinger Et al. on the grounds that they use their own disagreeable definition of locality.

In this context, if we just define non-locality as violating the Bell inequality, then we are done with the subject of the thread before we even start aren't we?

 

[PeterDonis]

if we just define non-locality as violating the Bell inequality, then we are done with the subject of the thread before we even start aren't we?

Well, you're the OP of the thread. If you think that's a satisfactory resolution and you would like me to spin off the rest of the discussion into a separate thread, let me know and I can do that.

 

[Jarvis323]

Well, you're the OP of the thread. If you think that's a satisfactory resolution and you would like me to spin off the rest of the discussion into a separate thread, let me know and I can do that.

No it is fine to continue the discussion. I am not sure there is much to say about the original topic beyond what has been said.

 

[PeterDonis]

No it is fine to continue the discussion.

Ok.

 

[gentzen]

Note that what you refer to as a "subensemble" did, in fact, change, and we AGREE on this! The before and after states are objectively (and experimentally) different as long as you believe QM (or QFT, which predicts the same result).

QM (or QFT) predicts statistics of measurement results. Its predictions are not concerend with objective states of individual pairs of photons.

So your statements about objectively different states are interpretation dependent, believe it or not.

Yes, no other particle (or particles) in the entire universe can be entangled with [1] when it is maximally entangled to [2].

Which is actually a great thing, if you want to analyse the statistics of ("poor man") quantum teleportation. So you know that the particle (or qubit) which you feed into some remote quantum experiment (or quantum circuit) is totally independent (even "quantum independent") of all the other particles (or qubits) in the remote quantum experiment (or quantum circuit). So the effect of the particle (or qubit) on the measured statistics is maximally reduced. You get a maximal mixing of the different possible statistics you would get if the state of the "to be" teleported particle (qubit) would have been fed instead into the remote experiment. And from that mixed statistics, quantum teleportation can indeed pick-out a subensemble (roughly a quarter of the total ensemble) with the correct statistics.

And the statistics of the remote experiment can include many classical measurement results for each single instance, so the fact that there are subensembles with the correct statistics is non-trivial.

 

[PeterDonis]

QM (or QFT) predicts statistics of measurement results. Its predictions are not concerend with objective states of individual pairs of photons.

This depends on which interpretation you are using. You appear to recognize that in your next paragraph.

 

[kurt101]

Yes, no other particle (or particles) in the entire universe can be entangled with [1] when it is maximally entangled to [2]. You apparently are not following my references which say exactly this. Here is a proof, search and you will see plenty that say the same.

But the title of the paper says "Monogamy of entanglement means that an entangled state cannot be shared with many parties." Does this mean other photons can't happen to have the same state as photon 1 while it is entangled with 2? Is it even possible to prove or disprove that other photons can't have the same state as photon 1 while it is entangled with 2?

When I look at the proof itself, it sounds like it only applies to entangled particles prepared in the same state. Am I reading the proof wrong? It is short, but still hard for me to understand. Here is the statement at the end:
"Thus we proved that for entangled state, there exists a finite number N such that no n-extension can be found for n > N. Physically, entanglement in a given state can not be sharable in arbitrarily many parties. For the system of many identical particles, the state ρA1A2,··· ,An has the symmetry of permutation. It holds for any state that entanglement between any pair particles tends to zero as n → ∞."

 

[PeterDonis]

Does this mean other photons can't happen to have the same state as photon 1 while it is entangled with 2?

What do you mean by "the same state"?

You have reached the point where you really need to stop waving your hands and start looking at the actual math.

 

[kurt101]

What do you mean by "the same state"?

You have reached the point where you really need to stop waving your hands and start looking at the actual math.

Not sure what you are referring to. @DrChinese has made the following claim and I am trying to understand it:
"[1] can never be maximally entangled with [2] and also be maximally entangled with [4] at the same time. So the initial preparation does not and cannot include any [1 & 4] pairs prepared in an initially entangled state. That must occur at a later time. Or maybe you deny Monogamy of Entanglement is a part of QFT?"

His proof does not make sense. He says 1 can't be entangled with 2 and 4 at the same time. That is fine, at no time do I see where that actually occurs in the case where the BSM test is done last. He also says 1 & 4 can't be prepared in an initial entangled state because of Monogamy of Entanglement.

I am asking why monogamy of entanglement prevents 1 & 4 being in such an initial state. I read the proof he cited and it is not clear to me the paper is discussing anything beyond particles prepared in the same state. Clearly 1 & 4 are not being prepared in any common state. Their states are random. We know they are random because we can look at there measurements and there is no correlation between them. Only when we add the BSM measurement information and select out the pairs of 1 & 4 do we get the correlation. So the "hand waiving" claim that 1 & 4 can't be in certain initial states does not hold up to reality.

 

[PeterDonis]

@DrChinese has made the following claim and I am trying to understand it

It's basic QM.

He says 1 can't be entangled with 2 and 4 at the same time. That is fine, at no time do I see where that actually occurs in the case where the BSM test is done last.

If you agree that 1 is never entangled with 2 and 4 at the same time, then you agree that in the initial state that is prepared, 1 is not entangled with 4--because it is entangled with 2.

And that, all by itself, is sufficient to show that, whatever accounts for 1 being entangled with 4 at the time the 1 & 4 measurements are made, cannot be a property of the initial state.

You are not addressing this argument at all.

I am asking why monogamy of entanglement prevents 1 & 4 being in such an initial state.

What initial state? Don't just wave your hands. Write down, explicitly, the state you are asking about. You will find that it is impossible to write down an initial state in which 1 is maximally entangled with 2--which it is by construction--and also has any entanglement at all with 4. It is not possible.

If you disagree, do it. Explicitly. With math.

 

[PeterDonis]

I read the proof he cited and it is not clear to me the paper is discussing anything beyond particles prepared in the same state.

Again, what do you mean by "in the same state"? Be very careful and very explicit in your answer. You can't just wave your hands with this stuff. You need to actually do the math.

 

[PeterDonis]

If you disagree, do it. Explicitly. With math.

You can't just wave your hands with this stuff. You need to actually do the math.

Note also, @kurt101, that this is an "A" level thread, meaning graduate level understanding of the thread topic is assumed. That means that you are already supposed to understand basic facts about QM like what entanglement means and what maximal entanglement means and what monogamy of entanglement says. The statements @DrChinese is making about all of these things are correct, and should already be understood to be correct by anyone with the requisite background knowledge to post in an "A" level thread on this topic.

 

[kurt101]

Ok, maybe I am over my head here as I am not really comfortable posting in a "A" level thread since I don't have the education. I thought it would be easy to show @DrChinese that there is a simple explanation for realistic causal interpretations, but I can't say I have convinced a single person after all my effort. So I intend to stop like you suggest after this post.

Bottom line, anybody can take two random set of vectors and then hand pick the vectors that show maximum entanglement. It may not be what happens in nature but to not acknowledge this is a possibility and assert that something like this can't happen is misleading everyone IMHO. And to be clear I am only referring to the case where BSM is done last. If one of @DrChinese proofs is actually valid, I take it back, but so far the first one had issues (as far as I know he is still waiting on his correspondence with the experimenters on it) and the monogamy argument does not appear to me to have a clear path of logic behind it. I think I understand monogamy as it applies to something like a qubit, but I don't understand monogamy as you and @DrChinese seem to be using it in this experiment and I can't find a reference to confirm how you are using it and why it might be that way.

 

[PeterDonis]

maybe I am over my head here as I am not really comfortable posting in a "A" level thread since I don't have the education.

You are welcome to start a separate thread at a lower level ("I" would be appropriate) to ask questions based on things in the paper @DrChinese referenced that you are having trouble understanding.

anybody can take two random set of vectors and then hand pick the vectors that show maximum entanglement.

No, this is not correct, and shows a fundamental misunderstanding of how the math of QM works.

Please start a separate thread if you want to delve further into this.