I Realistic interpretation of QM

[kurt101]

@DrChinese here is a paper that refutes your claim. https://link.springer.com/article/10.1007/s10701-021-00511-3

 

[Nullstein]

The experiment is performed on the entangled sub-ensemble, the other pairs are not of any interest. You never consider B & C pairs that do not arrive together within a suitably small time window. The other pairs appear equally often, separated into 4 combinations. For all intents and purposes, all of those are considered.

Sure, that makes them subensembles, which are entangled. It's just that this kind of entanglement is entirely non-mysterious, since it is a statistical effect.

Obviously, the problem with your idea is that you want it both ways... the A & D pairs are in product states to begin with, but they become entangled by your purported distillation process. Were that true, then you would be agreeing with me and be inconsistent.

First of all, it's not my idea. It's well understood statistics, I provided a reference with thousands of citations. Also, A and D don't become entangled. The full system is in a product state, but the subensembles are entangled. It's just that this entanglement is entirely non-mysterious. What's mysterious still, is that A&B are entangled and that C&D are entangled.

As I mention, your distillation process - "entanglement swapping" or "quantum teleportation" is what everyone else calls it - doesn't have enough outcome options to match up photons with leading to perfect correlations for ultimately entangled A & D pairs. There are essentially a large (or perhaps infinite) number of angle settings at which they would need to matched for your idea to work. Your idea being that A & D happen to have identical predetermined orientations (which are reflected in their respective twins, B & C).

I never claim that A & D have identical predetermined orientations, so that's a non-starter and shows that you haven't understood the argument.

So let's say we have A & B entangled, and C & D entangled. We ALREADY know from Bell that the A & B (or C & D) entanglement cannot itself be explained by any local hidden variables. So basically, your idea is already failed (as you are attempting to push a local realistic explanation for entanglement swapping in which Bell's Inequality is violated).

Again, you misrepresent "my idea". I don't want to explain anything by local hidden variables. I don't believe in hidden variables to begin with. I totally agree that the entanglement between A & B (and C & D respectively) cannot be explained by a local realistic model. The entanglement between A & B is mysterious. It's just that the entanglement of a subensemble of A & D is not mysterious given that the initial pairs are entangled for whatever mysterious reason.

Regardless, and still going down your path: We compare B & C, considering only those pairs in which B & C arrive together at a single polarizing beam splitter within a small time window; and are therefore indistinguishable. There can be only 4 outcomes corresponding to the four maximally entangled Bell states |Φ+>, |Ψ+>, |Ψ−>, and |Φ −>. If your idea is correct, there is no "process" occurring - since there is nothing happening when B & C interact other than to "reveal" some pre-existing correlation.

Not true. There is no "pre-existing correlation" in A & B to begin with, so propagating this correlation by entanglement swapping won't suddenly make it "pre-existing." So you have completely misunderstood the argument.

But what would that pre-existing correlation be IF there are only 4 possibilities?

There is no pre-existing correlation and I didn't claim there was one.

The ideas you push are wrong, and it is more fully debunked in a generally accepted reference I gave in post #36. In their words:

https://arxiv.org/abs/0911.1314

"Starting from two independent pairs of entangled particles, one can measure jointly one particle from each pair, so that the two other particles become entangled, even though they have no common past history. The resulting pair is a genuine entangled pair in every aspect, and can in particular violate Bell inequalities. Intuitively, it seems that such entanglement swapping experiments exhibit nonlocal effects even stronger than those of usual Bell tests."

I fully agree that the subensembles are fully entangled and violate Bell's inequality and that there is no local realistic explanation for this. I'm just saying that entanglement swapping does not add any additional mystery on top of the initial entanglement of the particle pairs A&B and C&D. The swapping "process" itself is just a statistical artifact.

This is the Interpretations subforum and the rules are more relaxed here; but generally accepted science is still generally accepted science. The facts are: entanglement swapping is a quantum process, and it itself violates local realism. I say it is a physical process, and if you follow Bohmian Mechanics or MWI you should agree with me. I don't see how it can be viewed as OTHER than a physical process, but I would acknowledge there are plenty of other interpretations that might see things differently. But it is not a local realistic phenomena.

I don't disagree with any accepted science. It's just that you don't understand the argument I'm making. It is in full agreement with generally accepted science.

 

[PeterDonis]

The full system is in a product state, but the subensembles are entangled.

This doesn't make sense. "Subensembles" don't even have states. In each individual run of the experiment, the full system, consisting of A, B, C, D, is entangled, because pairs of subsystems, A&B, C&D, are entangled.

 

[Nullstein]

This doesn't make sense. "Subensembles" don't even have states. In each individual run of the experiment, the full system, consisting of A, B, C, D, is entangled, because pairs of subsystems, A&B, C&D, are entangled.

Subensembles do have states in the sense that the statistics of the subensemble is perfectly well described by a density matrix. Of course there is entanglement between A&B and C&D, but what I meant to say was that the composite system A&B is not entangled with the composite system C&D and that the subsystem A&D is in a product state.

 

[DrChinese]

1. Sure, that makes them subensembles, which are entangled. It's just that this kind of entanglement is entirely non-mysterious, since it is a statistical effect.

2. First of all, it's not my idea. It's well understood statistics, I provided a reference with thousands of citations. Also, A and D don't become entangled. The full system is in a product state, but the subensembles are entangled. It's just that this entanglement is entirely non-mysterious. What's mysterious still, is that A&B are entangled and that C&D are entangled.

3. I never claim that A & D have identical predetermined orientations, so that's a non-starter and shows that you haven't understood the argument.

4, The entanglement between A & B is mysterious. It's just that the entanglement of a subensemble of A & D is not mysterious given that the initial pairs are entangled for whatever mysterious reason.

1. And how do those sub-ensembles become entangled into 1 of 4 Bell states? By the swapping process, of course! In principle, there are no sub-ensembles that do not cast entangled A & D into an entangled state. So the idea you have - that only the "correlated" sub-ensembles are being selected - is totally wrong. You completely skipped over my explanation about how A & D come to be entangled. Again, there are 4 Bell states that result from a Bell State Analyzer, and they are not entangled unless they are indistinguishable. This idea does not relate to anything you describe, and of course, it can't.

2. It IS your idea... try quoting a source. You place a URL here, and then you extract a relevant quote that directly refutes my argument or directly supports yours. It's easy for me to request, because this is a lop-sided exercise. And please, don't tell me to Google something. You have it, or you don't.

3. Well, you claimed that there is no interaction between B & C that affects A & D. So... how do A & D magically become perfectly correlated?

4. The entire point of entanglement swapping is that it is equally "mysterious". Gisin et al say: "entanglement swapping experiments exhibit nonlocal effects even stronger than those of usual Bell tests." Maybe you have Gisin (or Zeilinger, or Pan) saying something different?

 

[PeterDonis]

what I meant to say was that the composite system A&B is not entangled with the composite system C&D

Yes.

and that the subsystem A&D is in a product state.

No. Since the subsystems A&B are entangled and the subsystems C&D are entangled, you cannot factor out a product state for the subsystem A&D.

 

[Nullstein]

1. And how do those sub-ensembles become entangled into 1 of 4 Bell states? By the swapping process, of course!

It's not a physical process, it's an artificial selection. Nothing becomes entangled in a physical sense.

In principle, there are no sub-ensembles that do not cast entangled A & D into an entangled state. So the idea you have - that only the "correlated" sub-ensembles are being selected - is totally wrong.

That's not the idea that I have. You are just putting words in my mouth.

You completely skipped over my explanation about how A & D come to be entangled. Again, there are 4 Bell states that result from a Bell State Analyzer, and they are not entangled unless they are indistinguishable. This idea does not relate to anything you describe, and of course, it can't.

Your explanation is based on invalid premises as I have pointed out.

2. It IS your idea... try quoting a source. You place a URL here, and then you extract a relevant quote that directly refutes my argument or directly supports yours. It's easy for me to request, because this is a lop-sided exercise. And please, don't tell me to Google something. You have it, or you don't.

I gave you a reference, just scroll up.

3. Well, you claimed that there is no interaction between B & C that affects A & D. So... how do A & D magically become perfectly correlated?

They don't become perfectly correlated, they are completely uncorrelated. Only a subensemble is perfectly correlated much like a random ensemble of heads and tails becomes perfectly deterministic if you choose only the subensemble of events consisting of only heads.

4. The entire point of entanglement swapping is that it is equally "mysterious". Gisin et al say: "entanglement swapping experiments exhibit nonlocal effects even stronger than those of usual Bell tests." Maybe you have Gisin (or Zeilinger, or Pan) saying something different?

No, there is no "whole point of entanglement swapping" in the first place, it's just a phenomenon with many applications. Entanglement swapping is a classical statistical phenomenon that happens on top of a quantum mechanical description involving two pairs of entangled particles. I gave you a reference that explains it (with several thousand citations), you just need to read if you are willing to learn.

By the way, nonlocal in the foundations community is just a term that means "violates Bell's inequality." It doesn't entail anything about causality and some people only keep using it for historic reasons. The less interpretation-laden contemporary term would be inseparable.

 

[Nullstein]

No. Since the subsystems A&B are entangled and the subsystems C&D are entangled, you cannot factor out a product state for the subsystem A&D.

That's not correct. First of all, you don't factor anything out. You get the state of a subsystem by taking partial traces. If the full system is in the state $\rho_{AB}\otimes\rho_{CD}$, then the partial trace with respect to B and C will just be $\mathrm{Tr}_B(\rho_{AB})\otimes\mathrm{Tr}_C(\rho_{CD})$, which is a product state.

 

[DrChinese]

1. I gave you a reference, just scroll up.

2. No, there is no "whole point of entanglement swapping" in the first place, it's just a phenomenon with many applications. Entanglement swapping is a classical statistical phenomenon that happens on top of a quantum mechanical description involving two pairs of entangled particles. I gave you a reference that explains it (with several thousand citations), you just need to read if you are willing to learn.

3. By the way, nonlocal in the foundations community is just a term that means "violates Bell's inequality." It doesn't entail anything about causality and some people only keep using it for historic reasons. The less interpretation-laden contemporary term would be inseparable.

1. LOL. Not surprising you can't produce anything to counter:

"Starting from two independent pairs of entangled particles, one can measure jointly one particle from each pair, so that the two other particles become entangled, even though they have no common past history. The resulting pair is a genuine entangled pair in every aspect, and can in particular violate Bell inequalities. Intuitively, it seems that such entanglement swapping experiments exhibit nonlocal effects even stronger than those of usual Bell tests."-Gisin et al.

If they are entangled, they are entangled. It is not "revealing a statistical coincidence". Please, maybe you have something different from Zeilinger, Tittel, Pan, Kwiat, Zurek, Weihs, Branciard, or any of the other pioneers in this area.2. The whole point is the whole point. That's why it's a thing. Otherwise, it wouldn't be a thing.3. I agree with this. Quantum nonlocality is what Bell's Theorem demonstrates. It doesn't require a rejection of locality, it could also be explained by a rejection of causal direction, by MWI, and any of a number of viable interpretations. Of course, swapping is a demonstration of a much deeper level of quantum nonlocality... see 2 above.

 

[johnsankey]

All I know is that when I was trapping single ions of barium or strontium, and playing Maxwell's daemon with one electron at a time, I saw quanta that were as clear and definite as golf balls. And when I use observations as the basis for inference with entangled states, I see an equally clear result; cf. [Link to personal website redacted by the Mentors]
Never forget the gap between mathematics based upon consistency, and physics based on observables.

 

[Nullstein]

1. LOL. Not surprising you can't produce anything to counter:

"Starting from two independent pairs of entangled particles, one can measure jointly one particle from each pair, so that the two other particles become entangled, even though they have no common past history. The resulting pair is a genuine entangled pair in every aspect, and can in particular violate Bell inequalities. Intuitively, it seems that such entanglement swapping experiments exhibit nonlocal effects even stronger than those of usual Bell tests."-Gisin et al.

I don't have to counter that because I fully agree with it. You still don't seem to understand that. The subensemble paris are genuinely entangled pairs in every aspect. The fact that you don't even understand what my position is just shows that you are still very far away from being able to formulate a counterargument. First you need to put some effort into understanding the actual argument.

Just to explain it again in simpler terms: Both A&B, C&D and the subensembles of A&D are entangled. But the point is that while we have no good explanation for the entanglement of A&B and C&D, the entanglement of the subensembles of A&D can be understood using classical statistics as long as one doesn't ask how A&B and C&D came to be entangled in the first place. Entanglement swapping thus doesn't add to the mystery.

If they are entangled, they are entangled. It is not "revealing a statistical coincidence". Please, maybe you have something different from Zeilinger, Tittel, Pan, Kwiat, Zurek, Weihs, Branciard, or any of the other pioneers in this area.

I agree that I the subensembles are entangled, so I don't have to counter these people. None of what I'm saying is in contradiction to them.

2. The whole point is the whole point. That's why it's a thing. Otherwise, it wouldn't be a thing.

There is no point to a physical phenomenon. There is no point to entanglement swapping just like there is no point to gravity. Nature doesn't want to make a point.

 

[PeterDonis]

First of all, you don't factor anything out.

You do if you want to avoid throwing away information.

You get the state of a subsystem by taking partial traces.

No, you get something which in some respects you can call a "state" of a subsystem, but which throws away information about the entanglement. So of course it won't tell you about the entanglement information that you threw away in getting it. In an entangled system, no individual subsystem has a state in the strict sense; only the full joint system does. Taking partial traces can be useful for some computations, but that does not mean you can interpret it physically the way you are claiming, precisely because doing it throws away important physical information.

 

[Nullstein]

You do if you want to avoid throwing away information.

I don't want to avoid throwing away information. I want the state of the subsystem, which is the state that contains all information about the subsystem, but no information beyond that. Otherwise it would just be the state of a larger system and not the subsystem under consideration.

No, you get something which in some respects you can call a "state" of a subsystem, but which throws away information about the entanglement. So of course it won't tell you about the entanglement information that you threw away in getting it. In an entangled system, no individual subsystem has a state in the strict sense; only the full joint system does. Taking partial traces can be useful for some computations, but that does not mean you can interpret it physically the way you are claiming, precisely because doing it throws away important physical information.

Not true, there is a state of the full system, which contains all information about the subsystem and the rest of the system and additionally information about their entanglement. And there is the state of the subsystem, which contains all information about the subsystem and nothing beyond it. That's absolute standard terminology in quantum theory. See e.g. Breuer, Petruccione "The Theory of Open Quantum Systems":

 

[PeterDonis]

I don't want to avoid throwing away information. I want the state of the subsystem, which is the state that contains all information about the subsystem, but no information beyond that.

I understand what the "state" obtained by partial tracing is. I just don't think it justifies the claims you are making.

See e.g. Breuer, Petruccione "The Theory of Open Quantum Systems":

This reference supports what I just said above. It explicitly says that the partial trace is only relevant if all you are interested in is measurements restricted to the subsystem in question. But that is not the case in the experiments under discussion here.

 

[Nullstein]

I understand what the "state" obtained by partial tracing is. I just don't think it justifies the claims you are making.

What claims am I making? I said that the subsystem A&D is not entangled, which is absolutely true given standard terminology and there is good reason for this choice of terminology. If a subsystem is not entangled, this will be reflected in the data, i.e. there will be zero correlations in experiments pertaining only to the subsystem.

This reference supports what I just said above. It explicitly says that the partial trace is only relevant if all you are interested in is measurements restricted to the subsystem in question. But that is not the case in the experiments under discussion here.

If I'm talking about entanglement between A&D, I'm only interested in measurements restricted to this system. Particle A is not entangled with particle D. Sure, I have thrown away the information that particle A is entangled with particle B and that particle C is entangled with particle D, but that's not relevant for the statement that particle A is not entangled with particle D. All information about the entanglement between A and D is fully contained in the reduced density matrix of the subsystem A&D. The subsystem A&D is not entangled. One really shouldn't have to discuss these basics in an intermediate level thread.

Sure, one needs the state of the full system to discuss the experiment, but one just needs the state of the subsystem A&D to make the statement that A&D is not entangled. This is only an aspect of the discussion, so it's perfectly valid to talk about the subsystem regarding this specific aspect.

The fact that the subsystem A&D is not entangled is fully in accordance with the experimental facts. If you don't have information about B&C, you cannot perform the post-selection and you will get find completely uncorrelated measurement results. As I said multiple times, only the subensembles are entangled. In the full system, there are zero correlations between A and D. You will absolutely see this fact in the data. It's akin to rolling two dice. The full dataset will be perfectly uncorrelated, but if you post-select only those throws that, say, add up to 5, then this subensemble will be correlated. (And nothing is mysterious about that.)

 

[Nullstein]

Here's a different way to see that A cannot possibly be entangled with D: Since we know that A&B is in a maximally entangled state (the standard EPRB singlet state), then by the monogamy of entanglement, A cannot possibly be entangled with anything else and in particular not with D.

 

[Maarten Havinga]

1. LOL. Not surprising you can't produce anything to counter:

"Starting from two independent pairs of entangled particles, one can measure jointly one particle from each pair, so that the two other particles become entangled, even though they have no common past history. The resulting pair is a genuine entangled pair in every aspect, and can in particular violate Bell inequalities. Intuitively, it seems that such entanglement swapping experiments exhibit nonlocal effects even stronger than those of usual Bell tests."-Gisin et al.

It is clear to me that also academia have different opinions on whether they are truly entangled and how to interpret it. It's simply a difficult subject and all of us can't know everything. I'm sure I don't, after googling around and reading your comments.

I'm afraid this discussion is not going to be constructive unless we allow for others disagreeing with ourselves - just leaving it like that. In any case, the best that can be achieved is an interpretation that matches observations and makes everything sort of understandable from our own beliefs. It seems to me like everyone already has that! And it's not like we need to convert others to those beliefs, I'd say.

Regardless of what is the truth, I'm myself intrigued by Mark v Raamsdonk's view: that entanglement is more fundamental to the geometry of spacetime and general relativity is the emergent description in a natural situation. It's not my goal to have you agree with me.

 

[vanhees71]

This doesn't make sense. "Subensembles" don't even have states. In each individual run of the experiment, the full system, consisting of A, B, C, D, is entangled, because pairs of subsystems, A&B, C&D, are entangled.

Of course "subensembles" have states. In this case the subensembles are operationally defined by doing coincidence measurements on photons B and C. The statistical operator of the subensembles is then given by the corresponding projections of the statistical operator describing the full ensemble (which is of course a bit idealized assuming ideal detectors).

The important point is that you can select the subensembles even after all measurements are done if you have a complete measurement protocol involving the coincidence measurements on photons B&C as well as on photons A&D. The entanglement between photons A&D in each of the subensembles is due to the initial state, i.e., the entanglement between A&B as well as C&D, while the initial state describing the full ensemble is of the form $\hat{\rho}=\hat{\rho}_{AB} \otimes \hat{\rho}_{CD}$, i.e., for the full ensemble photons A and D are indeed not entangled, and it's indeed all about correlations and statistics. In the subensembles A&D are entangled through selection (or even post-selection) due to measurements on B&C, i.e., you are able to prepare entangled states of photons A&D without A&D having ever been in direct "causal contact", but also without violating anywhere locality (in the usual sense of relativistic local QFT, which is used to describe this experiment).

 

[vanhees71]

It's not a physical process, it's an artificial selection. Nothing becomes entangled in a physical sense.

But for the (post-)selected subensembles, based on coincidence measurements on photons B&C, the photons A&D are indeed entangled. That's just what's called "entanglement swapping".

That's not the idea that I have. You are just putting words in my mouth.Your explanation is based on invalid premises as I have pointed out.

I gave you a reference, just scroll up.

They don't become perfectly correlated, they are completely uncorrelated. Only a subensemble is perfectly correlated much like a random ensemble of heads and tails becomes perfectly deterministic if you choose only the subensemble of events consisting of only heads.

That's correct, but it wouldn't work for any "classical ensembles". Entanglement swapping in this experiment only works because photons A&B as well as photons C&D are entangled. It's not explainable in terms of a "local realistic theory" in Bell's sense, and indeed for each subensemble photons A&D are entangled and thus can be used to demonstrate the violation of Bell's inequality.

No, there is no "whole point of entanglement swapping" in the first place, it's just a phenomenon with many applications. Entanglement swapping is a classical statistical phenomenon that happens on top of a quantum mechanical description involving two pairs of entangled particles. I gave you a reference that explains it (with several thousand citations), you just need to read if you are willing to learn.

It's NOT a "classical statistical phenomenon" but a generic "quantum statistical phenomenon".

By the way, nonlocal in the foundations community is just a term that means "violates Bell's inequality." It doesn't entail anything about causality and some people only keep using it for historic reasons. The less interpretation-laden contemporary term would be inseparable.

Indeed, but it's hopeless, because in this forum people insist on using these confusing "double meaning" of the words "local" and "non-local". When I say local I always mean it, of course, in the well-defined mathematical sense of relativistic local (sic!) QFT: the Hamilton density is built from field operators that transform locally (sic) under proper orthochronous Poincare transformations, and local (sic) observable-operators commute at space-like separation of their arguments (microcausality), leading to the well-established class of relativistic QTs with a unitary, Poincare covariant S-matrix, obeying the cluster-decomposition principle but of course are able to describe all the experiments with entangled states but at the same time fulfill the said constraints of relativistic causality by realizing them in terms of a local (sic!) relativistic QFT.

 

[akvadrako]

The interesting thing is that entanglement between sub-ensembles can be created via post-selection, even in the past. We know it's real entanglement, since they can be used to violate Bell inequalities. Obviously it's just a statistical phenomenon, not time travel, strongly implying all entanglement is just a statistical phenomenon.

 

[vanhees71]

Of course, it's a "statistical phenomenon", but it's a generic "quantum phenomenon", i.e., it cannot be modeled with "local realistic HV theories", exactly because Bell's inequalities are violated. It's all about correlations, not about a violation of locality in the sense of local relativistic QFTs, which by construction cannot imply any "spooky actions at a distance".

 

[Nullstein]

But for the (post-)selected subensembles, based on coincidence measurements on photons B&C, the photons A&D are indeed entangled. That's just what's called "entanglement swapping".

I never objected to that. I said from the very beginning that the subensembles are entangled. My only point is that the swapping itself is just a well understood statistical phenomenon and it adds nothing to the mystery of entanglement. All the mystery is contained in the initial entanglement of the A&B and C&D pairs.

That's correct, but it wouldn't work for any "classical ensembles". Entanglement swapping in this experiment only works because photons A&B as well as photons C&D are entangled. It's not explainable in terms of a "local realistic theory" in Bell's sense, and indeed for each subensemble photons A&D are entangled and thus can be used to demonstrate the violation of Bell's inequality.

Again, I fully agree with that. One needs initial pairs of entangled particles to find entanglement in the post-selected ensembles, so this is a genuine quantum situation. I'm just saying that the swapping process alone is a fully understood classical statistical phenomenon. The A&D subensembles just inherit the original entanglement of A&B, C&D by this well understood process. Hence, entanglement swapping is not a more severe version of entanglement, as DrChinese wants to imply. Instead, it's fully understood classical statistics and adds nothing to the mystery. The only mystery in an entanglement swapping scenario is the initial entanglement of A&B and C&D, precisely, because we can not just explain it by invoking classical statistics.

It's NOT a "classical statistical phenomenon" but a generic "quantum statistical phenomenon".

The swapping itself is a classical statistical phenomenon. The resulting entangled subensembles of A&D particles is a quantum statistical phenomenon, but not because of the swapping. The quantum aspecct is fully contained in the initial entanglement of the initial pairs.

 

[Nullstein]

Of course, it's a "statistical phenomenon", but it's a generic "quantum phenomenon", i.e., it cannot be modeled with "local realistic HV theories", exactly because Bell's inequalities are violated. It's all about correlations, not about a violation of locality in the sense of local relativistic QFTs, which by construction cannot imply any "spooky actions at a distance".

I fully agree and I never objected to that. All I'm saying is that the swapping itself is classical. All of the quantum is contained in the entanglement of the initial pairs. The swapping adds nothing peculiar on top of the entanglement of the initial pairs. The A&D subensembles are fully entangled and violate Bell's inequality, but this is just the result of post-selection and it's entirely non-surprising to someone who knows classical statistics. All the mystery is in the entanglement of the initial pairs. (If they weren't entangled in the first place, one could not post-select an entangled subensemble.)

 

[vanhees71]

I never objected to that. I said from the very beginning that the subensembles are entangled. My only point is that the swapping itself is just a well understood statistical phenomenon and it adds nothing to the mystery of entanglement. All the mystery is contained in the initial entanglement of the A&B and C&D pairs.

Well, there is no mystery to begin with. Entanglement is a pretty straight-forward consequence of the mathematical basic structure of QT. There is nothing mystical in why you can prepare two photons in various ways in an entangled state. In the very beginning of these investigations it done with a atomic cascade (Alan Aspect) et al. Nowadays one rather uses parametric downconversion or quantum dots for higher efficiency. The entanglement of the two photons in polarization and/or momentum is not at all mysterious.

Again, I fully agree with that. One needs initial pairs of entangled particles to find entanglement in the post-selected ensembles, so this is a genuine quantum situation. I'm just saying that the swapping process alone is a fully understood classical statistical phenomenon. The A&D subensembles just inherit the original entanglement of A&B, C&D by this well understood process. Hence, entanglement swapping is not a more severe version of entanglement, as DrChinese wants to imply. Instead, it's fully understood classical statistics and adds nothing to the mystery. The only mystery in an entanglement swapping scenario is the initial entanglement of A&B and C&D, precisely, because we can not just explain it by invoking classical statistics.

It's NOT classical, it's QUANTUM. Otherwise we seem to fully agree.

The swapping itself is a classical statistical phenomenon. The resulting entangled subensembles of A&D particles is a quantum statistical phenomenon, but not because of the swapping. The quantum aspecct is fully contained in the initial entanglement of the initial pairs.

Sure, all these quantum correlations are due to the preparation of the two photon pairs, and indeed there's no need to assume any "spooky actions at a distance" to understand these very correlations within a local relativistic QFT, which excludes any "spooky actions at a distance" by construction.

 

[CoolMint]

But those quantum correlations arise due to inseparability. Which is quite the mystery given that everything 'classical' appears completely separable.

 

[Nullstein]

Well, there is no mystery to begin with. Entanglement is a pretty straight-forward consequence of the mathematical basic structure of QT. There is nothing mystical in why you can prepare two photons in various ways in an entangled state. In the very beginning of these investigations it done with a atomic cascade (Alan Aspect) et al. Nowadays one rather uses parametric downconversion or quantum dots for higher efficiency. The entanglement of the two photons in polarization and/or momentum is not at all mysterious.

I agree that it is fully described my quantum mechanics, but for me and many people, the entanglement of the initial pairs is still mysterious, because the classical notion of causality seems to be inadequate in quantum mechanics and we don't have a well motivated quantum mechanical replacement. The entanglement can't be caused by any classical mechanism unless one accepts superdeterministic explanations (that's the content of Bell's theorem), so how did it came to be then if it wasn't caused? Personally, I do feel that a generalized notion of causality that is compatible with quantum mechanics will eventually be discovered, but until then, it remains a mystery, at least for me.

It's NOT classical, it's QUANTUM. Otherwise we seem to fully agree.

What is quantum about the mere act of selecting a subensemble out of an already recorded list of measurement results?

Sure, all these quantum correlations are due to the preparation of the two photon pairs, and indeed there's no need to assume any "spooky actions at a distance" to understand these very correlations within a local relativistic QFT, which excludes any "spooky actions at a distance" by construction.

I agree with that.

 

[vanhees71]

But those quantum correlations arise due to inseparability. Which is quite the mystery given that everything 'classical' appears completely separable.

Exactly, quantum correlations are inseparable, but there's no need for "spooky actions at a distance" to describe them. Einstein himself did not like the EPR paper, because the inseparability was his real concern, and it's Bell's merit to have found a way to empirically test whether nature is describable with a local, deterministic hidden-variable theory or whether the inseparability of QT is correct, and as is very well demonstrated in all experiments, the latter is the case.

 

[DrChinese]

1. I don't have to counter that because I fully agree with it. You still don't seem to understand that. The subensemble paris are genuinely entangled pairs in every aspect.

2, Just to explain it again in simpler terms: Both A&B, C&D and the subensembles of A&D are entangled. But the point is that while we have no good explanation for the entanglement of A&B and C&D, the entanglement of the subensembles of A&D can be understood using classical statistics as long as one doesn't ask how A&B and C&D came to be entangled in the first place. Entanglement swapping thus doesn't add to the mystery.

1. Great... except that in post #76, you say:

"Here's a different way to see that A cannot possibly be entangled with D: Since we know that A&B is in a maximally entangled state (the standard EPRB singlet state), then by the monogamy of entanglement, A cannot possibly be entangled with anything else and in particular not with D."

Your above statement of course is completely wrong. The reason it is called "entanglement swapping" is because A's monogamous entanglement is swapped from B to D.2. There are no "statistical sub-ensembles" of pairs A & B and C & D that have any properties that will re-produce the quantum mechanical results (at least not without knowing how A & D are to be measured first).
I have presented an example to explain this, and I have presented a paper by a top team which provides the formal theoretical no-go argument. Please, feel free to provide a counter-example. So far, you have failed to provide a single quote by someone in the field with suitable credentials. If you were representing the mainstream, you'd be able to reel that off with ease. In the hundreds of papers I have read on teleportation, none of them say anything OTHER than the Bell State Analyzer is responsible for the process whereby the A & D photons become entangled. (I didn't use the word "causes" in that sentence, for the reason that temporal order of the process can be ambiguous.)

"This procedure is also known as ”Entanglement Swapping” because one starts with two pairs of entangled photons A–B and C–D, subjects photons B and C to a Bell-state measurement by which photons A and D also become entangled."-Zeilinger et al

Try to explain what (heretofor unknown) properties an entangled PDC pair (A & B) must have such that it can be "selected" (by a measurement on B & C) into some subset so that photon A is now matched to photon D; yielding the usual quantum relationship. You will quickly see this is not possible, there are no subsets with these attributes. It requires the creation of a direct relationship between A & D to yield the statistical results, even though A & D have never existed within a common spacetime region.

Ask yourself: why exactly does the Bell State Measurement need to be done such that the B & C photons are indistinguishable? That seems an odd requirement, if all we are doing is selecting a subset.

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In all fairness, I have clearly demonstrated that Kurt101's premise below is false, and that entanglement of A & D (which are not local to each other) occurs by executing a process on distant B & C. For this thread, that should be enough. I will start a separate thread if necessary to demonstrate why there are no subsets of B & C that condition an entangled relationship between A & D (without of course there being changes to A & D on bringing bringing B & C together).

"3. Entanglement is a non-local behavior between particles that is created through local preparation."

 

[vanhees71]

I agree that it is fully described my quantum mechanics, but for me and many people, the entanglement of the initial pairs is still mysterious, because the classical notion of causality seems to be inadequate in quantum mechanics and we don't have a well motivated quantum mechanical replacement. The entanglement can't be caused by any classical mechanism unless one accepts superdeterministic explanations (that's the content of Bell's theorem), so how did it came to be then if it wasn't caused? Personally, I do feel that a generalized notion of causality that is compatible with quantum mechanics will eventually be discovered, but until then, it remains a mystery, at least for me.

Why should, in your opinion, nature behave classically? It's the other way round: Classical behavior is an effective, approximate description of properties of macroscopic systems. On a more fundamental level, it's however a many-body quantum system.

Also quantum theory is completely causal, i.e., given the state at an initial time and the Hamiltonian of the system the state is known at any later time. It's, however, indeterministic, because there's no state, where all observables take determined values, and thus the knowledge of the state only implies the probabilities for the outcomes of measurements on any observable of this system, i.e., nature is inherently probabilistic rather than deterministic. Imho there's no way out of this result of physical research over the centuries, and as soon as you accept this, there's no more mystery in entanglement. To the contrary, QT delivers an amazingly accurate description of these phenomena.

What is quantum about the mere act of selecting a subensemble out of an already recorded list of measurement results?

Well, this selection is of course entirely "classical". What's quantum is the entanglement of photons A&D in these subensembles. This cannot be explained in any classical way but only by the entanglement of photons A&B and C&D in the initial state.

 

[vanhees71]

1. Great... except that in post #76, you say:

"Here's a different way to see that A cannot possibly be entangled with D: Since we know that A&B is in a maximally entangled state (the standard EPRB singlet state), then by the monogamy of entanglement, A cannot possibly be entangled with anything else and in particular not with D."

Your above statement of course is completely wrong. The reason it is called "entanglement swapping" is because A's monogamous entanglement is swapped from B to D.2. There are no "statistical sub-ensembles" of pairs A & B and C & D that have any properties that will re-produce the quantum mechanical results (at least not without knowing how A & D are to be measured first).
I have presented an example to explain this, and I have presented a paper by a top team which provides the formal theoretical no-go argument. Please, feel free to provide a counter-example. So far, you have failed to provide a single quote by someone in the field with suitable credentials. If you were representing the mainstream, you'd be able to reel that off with ease. In the hundreds of papers I have read on teleportation, none of them say anything OTHER than the Bell State Analyzer is responsible for the process whereby the A & D photons become entangled. (I didn't use the word "causes" in that sentence, for the reason that temporal order of the process can be ambiguous.)

That's of course true. The point here is to use measurements on photons B&C (uncorrelated in the initial state!) to select (or even post-select) subensembles where A&D are entangled (but uncorrelated in the initial state, and this is also not changed by the measurements on B&C, i.e., the total ensemble of A&D is just described by the product state of two completely unpolarized photons). It's the statistics of different ensembles: For the full ensemble A&D are just uncorrelated photons. For each of the four subensembles, chosen by projection measurements on B&C, A&D are entangled, i.e., in the corresponding Bell state, and thus (in a sense maximally) correlated.

"This procedure is also known as ”Entanglement Swapping” because one starts with two pairs of entangled photons A–B and C–D, subjects photons B and C to a Bell-state measurement by which photons A and D also become entangled."-Zeilinger et al

Try to explain what (heretofor unknown) properties an entangled PDC pair (A & B) must have such that it can be "selected" (by a measurement on B & C) into some subset so that photon A is now matched to photon D; yielding the usual quantum relationship. You will quickly see this is not possible, there are no subsets with these attributes. It requires the creation of a direct relationship between A & D to yield the statistical results, even though A & D have never existed within a common spacetime region.

Indeed, the only properties the four photons must have is that A&B as well as C&D are prepared as entangled photon pairs. The initial state, describing the four photons is $\hat{\rho}=\hat{\rho}_{AB} \otimes \hat{\rho}_{CD}$, where $\hat{\rho}_{AB}=|\Psi_{AB} \rangle \langle |\Psi_{AB} \rangle$ with $\Psi_{AB}$ being a Bell state (and analogously of the pair C&D). Within Q(F)T there are no other properties (aka "hidden variables"), and there are also no other properties needed to describe the outcome of the experiment.

Ask yourself: why exactly does the Bell State Measurement need to be done such that the B & C photons are indistinguishable? That seems an odd requirement, if all we are doing is selecting a subset.

That's indeed not and odd requirement, but the prerequisite to really realize the "entanglement-swapping protocol".