Realistic interpretation of QM

[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.

--------------------------------
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".

 

[Nullstein]

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.

You fail to understand the difference between the full ensemble and the subensembles. The full ensemble is not entangled. The subensembles are entangled. Nothing is actively swapped, the subensemble arises purely due to selection. The local measurement at B&C has no influence on the state of the A&D subsystem. The no communication theorem clearly proves that $\mathrm{Tr}_{BC}(P\rho) = \mathrm{Tr}_{BC}(\rho)$ if $P$ only acts on the B&C subsystem.

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).

That's not correct. Since the state of the A&D subsystem is not changed by the local measurement at B&C, it must already contain all the entangled subensembles that are post-selected later. You may just not know how to select them without the data from B&C. The full, non-entangled state of A&D will yield the entangled subensembles upon conditioning on the data from B&C.

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.

There is nothing odd about that. It just happens to be the procedure that leads to the correct post-selection rule. That just follows from the math.

 

[Nullstein]

Why should, in your opinion, nature behave classically?

It shouldn't in my opinion. It's just that we don't have any accepted notion of causality in quantum theory yet, so people, who want to understand the causal mechanism behind all of this, can't be satisfied quite yet. Sure, we have a theory that describes everything precisely, but it doesn't come with a causal mechanism. The situation was better in the classical theory.

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.

I agree that QT delivers an accurate description of the phenomena, but as you said, you have to accept the formalism and not ask any deeper questions in order to be happy. Not everyone is satisfied with that.

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.

I agree. When I'm saying that entanglement swapping is understood classically, I'm referring to the fact that the act of post-selection and thereby introducing correlation into the subensembles is a purely classical process. Of course, there are still quantum phenomena involved.

 

[vanhees71]

You fail to understand the difference between the full ensemble and the subensembles. The full ensemble is not entangled. The subensembles are entangled. Nothing is actively swapped, the subensemble arises purely due to selection. The local measurement at B&C has no influence on the state of the A&D subsystem. The no communication theorem clearly proves that $\mathrm{Tr}_{BC}(P\rho) = \mathrm{Tr}_{BC}(\rho)$ if $P$ only acts on the B&C subsystem.

No, in this case in the subensembles A&D are indeed entangled, and the two traces are not the same. For the rather simple calculation (working with the corresponding state vectors), see, e.g.,

https://doi.org/10.1103/PhysRevLett.80.3891

That's not correct. Since the state of the A&D subsystem is not changed by the local measurement at B&C, it must already contain all the entangled subensembles that are post-selected later. You may just not know how to select them without the data from B&C. The full, non-entangled state of A&D will yield the entangled subensembles upon conditioning on the data from B&C.

Of course you cannot select them without the data from B&C. It's important to always look at real experiments and not discuss about measurements that haven't been really done or are even physically impossible.

There is nothing odd about that. It just happens to be the procedure that leads to the correct post-selection rule. That just follows from the math.

That's true, but you must do the correct math ;-).

 

[gentzen]

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).

Yes, please do. But not for "convincing" Nullstein. It would have been "his" task to bring forward a convincing argument, including references to back it up.

Are you saying that the difficult Bell-state measurements on B & C change the relationship between A & D in other ways than by partitioning it into 4 subensembles? I sort of get your argument that the mystery of the entanglement between A & B and C & D alone cannot be enough to explain the mystery of entanglement swapping, because if "the rest" could just be explained classically, then the required measurement on B & C should be less difficult than it actually is. But if you would claim that this difficult measurement on B & C could change the state of A & D by some instantaneous action at a distance, then it would be "your" task to bring forward a convincing argument, including references to back it up.

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.

Despite starting with "That's of course true," vanhess71 makes it pretty clear that what is entangled are the "selected" the subensembles of A & D, but not the total ensemble of A & D.

 

[Nullstein]

No, in this case in the subensembles A&D are indeed entangled, and the two traces are not the same. For the rather simple calculation (working with the corresponding state vectors), see, e.g.,

https://doi.org/10.1103/PhysRevLett.80.3891

The subensembles are of course entangled, as I said multiple times. I don't deny that. But the full ensemble has no entanglement (between A&D) even after the measurement at B&C. There is no contradiction. In that paper, they project onto a Bell state to go from the full ensemble to a subensemble. But if you just perform the measurement and still describe everything using the full ensemble (which you have to do if you're interested in the statistics of the unconditioned A&D subsystem after measurement at B&C), then $P(\rho)$ is really given by
$$P(\rho) = \sum_i (\mathbb 1_A\otimes P_{i,BC}\otimes\mathbb 1_D)\rho(\mathbb 1_A\otimes P_{i,BC}\otimes\mathbb 1_D)$$ where the $P_{i,BC}$ are the four projectors onto the Bell basis and then by the no communication theorem, you get
$$\rho_{AD,after} = \mathrm{Tr}_{BC}(P(\rho_{AB}\otimes\rho_{CD})) =\mathrm{Tr}_{BC}(\rho_{AB}\otimes\rho_{CD}) =\rho_{AD,before}$$
The full ensemble of the subsystem A&D is the same as before. If you take $P$ to be a projection onto only one of the Bell states, say $P_{3,BC}$, then of course you will find an entangled state, but you also aren't talking about the full ensemble anymore.

(Just in case anyone asks for a citation for this simple computation, here it is: https://arxiv.org/pdf/quant-ph/0212023.pdf, sec. II.E)

Of course you cannot select them without the data from B&C. It's important to always look at real experiments and not discuss about measurements that haven't been really done or are even physically impossible.

Well, I said from the beginning that you need the data from B&C for this post-selection, so I don't see what your point is here.

That's true, but you must do the correct math ;-).

I did the correct math. I'm just talking about the transition from the full ensemble to the full ensemble, while the paper talks about the transition from the full ensemble to a subensemble. I'm in full agreement with the paper. As I always said, the subensembles are entangled. But it is also true that the full ensemble of the A&D subsystem is still in a product state.

Yes, please do. But not for "convincing" Nullstein. It would have been "his" task to bring forward a convincing argument, including references to back it up.

Which claim do you want me to provide references for that I haven't already provided? If DrChinese claims that the full ensemble of the A&D subsystem is entangled (before or after the Bell state projection at B&C, it doesn't matter), then its him who is in contradiction with accepted mainstream science and he should provide a reference.

 

[Nullstein]

Let me explain this in more detail:
If we perform a measurement on a state $\rho$ in a basis $\left|\Psi_i\right>$, then the state of the subensemble corresponding to the result $\left|\Psi_{i_0}\right>$ (where $i_0$ is fixed) is given by $$\rho_{i_0}=\frac{P_{i_0}\rho P_{i_0}}{\mathrm{Tr}(P_{i_0}\rho P_{i_0})}$$ where $P_{i_0}=\left|\Psi_{i_0}\right>\left<\Psi_{i_0}\right|$. We end up in this subensemble with probability $p_{i_0} = \mathrm{Tr}(P_{i_0}\rho P_{i_0})$.

The full ensemble after measurement is the mixed state given by $$\rho^\prime = \sum_i p_i \rho_i = \mathrm{Tr}(P_i\rho P_i) \frac{P_i\rho P_i}{\mathrm{Tr}(P_i\rho P_i)} = \sum_i P_i\rho P_i$$
In the case of a composite system, the density matrix is defined on a product Hilbert space $\mathcal H=\mathcal H_X\otimes H_Y$ and if we perform only local measurements, then the $P_i$ will have the form $P_i = \tilde P_i\otimes\mathbb 1$, where the $\tilde P_i$ act only on $\mathcal H_X$. In this situation, the formula for $\rho^\prime$ becomes $$\rho^\prime = \sum_i (\tilde P_i\otimes\mathbb 1)\rho (\tilde P_i\otimes\mathbb 1)$$
The state of the subsystem Y (before and after) is given by $\rho_Y = \mathrm{Tr}_X\rho$ and $\rho_Y^\prime = \mathrm{Tr}_X\rho^\prime$ respectively.

By the no communication theorem it then follows that $$\rho_Y^\prime = \mathrm{Tr}_X\rho^\prime = \mathrm{Tr}_X\rho = \rho_Y$$
That means that the full ensemble of subsystem Y is not changed by the local operation at subsystem X. All of this is absolute mainstream physics.

In our case, system Y is given by A&D and system X is given by B&C. Hence, the full ensemble of A&D is not entangled even after the measurement at B&C.

 

[vanhees71]

Are you saying that the difficult Bell-state measurements on B & C change the relationship between A & D in other ways than by partitioning it into 4 subensembles? I sort of get your argument that the mystery of the entanglement between A & B and C & D alone cannot be enough to explain the mystery of entanglement swapping, because if "the rest" could just be explained classically, then the required measurement on B & C should be less difficult than it actually is. But if you would claim that this difficult measurement on B & C could change the state of A & D by some instantaneous action at a distance, then it would be "your" task to bring forward a convincing argument, including references to back it up.

In each of the four subensembles, selected by (projective) measurements on photons B&D, photons A&D are entangled. They are not entangled in the full ensemble. There's nothing else needed to ensure this than the preparation of the original photons in the initial state. There is nothing mysterious. To the contrary it's well understood by quite simple application of the rules of QT.

 

[gentzen]

Which claim do you want me to provide references for that I haven't already provided?

Your initial claim at the end of your first post in this thread:

Entanglement swapping indeed adds nothing new to the mystery of entanglement. Those willing to learn can read more about this e.g. in Pearl's famous book titled "Causality."

If DrChinese claims that the full ensemble of the A&D subsystem is entangled

Then you should first ask him to confirm that he really wants to claim that. And if he insists, then try at least to untangle that discussion from the discussion about your initial claim. And if he neither explicitly confirms, nor denies that he wants to claim that, then I would accept that too, and not put words into his mouth, or guess at his intentions.

 

[Nullstein]

Your initial claim at the end of your first post in this thread:

Well, the sentence you quoted already contains the reference explicitely, so I don't understand what you are asking for.

Then you should first ask him to confirm that he really wants to claim that. And if he insists, then try at least to untangle that discussion from the discussion about your initial claim. And if he neither explicitly confirms, nor denies that he wants to claim that, then I would accept that too, and not put words into his mouth, or guess at his intentions.

Point 1 in his post #88 makes it very clear that he either does claim exactly that or that he doesn't understand the difference between the full ensemble and the subensemble. In the first case, he is contradicting accepted mainstream science. In the second case, it shows that he hasn't understood the argument and is therefore unable to meaningfully argue against it in the first place.

I can't untangle these discussions, because it is essential to the argument that the measurement at B&C does not change the full ensemble at A&D. The whole point is that the entanglement in the subensembles can be understood as mere conditioning of the full subensemble of A&D on results from B&C. If the full subensemble of A&D was changed by the measurement, the argument would no longer go through.

Moreover, he is the one who permanently claims that I contradict mainstream science and puts words in my mouth in the whole discussion even though I said numerous times that I fully agree with the contents of the papers he cited. So you might direct this criticism to him as well.

 

[gentzen]

Well, the sentence you quoted already contains the reference explicitely, so I don't understand what you are asking for.

Honestly, I guess that entanglement swapping adds something to the mystery of entanglement. I have read your reference, and it didn't convince me otherwise. But as I said, it doesn't even mention entanglement, so this is no real surprise.

So if DrChinese wants to open a new thread where he discusses why entanglement swapping adds something, I welcome this, especially since I believe that it is possible to do. But if all he wants is to "convince you," then I would not welcome it.

Point 1 in his post #88 makes it very clear that he either does claim exactly that or that he doesn't understand the difference between the full ensemble and the subensemble.

Maybe. What about the difference between an individual pair, and an ensemble of pairs? If he just takes one individual pair from anyone of the subensembles, couldn't he rightfully claim that this "resulting pair is a genuine entangled pair in every aspect, and can in particular violate Bell inequalities"? Or maybe the language used by Gisin et al here is misleading, and only ensembles of pairs can violate Bell inequalities?

 

[Nullstein]

Honestly, I guess that entanglement swapping adds something to the mystery of entanglement. I have read your reference, and it didn't convince me otherwise. But as I said, it doesn't even mention entanglement, so this is no real surprise.

Do you agree that the statistics of the A&D system is fully described by its reduced density matrix? Do you agree that the local operations at B&C don't change the reduced density matrix of A&D and therefore don't change the statistics of A&D? If you answer yes to these questions, then the full ensemble of A&D (before or after, it doesn't matter) already contains all the subensembles and they only need to be singled out by post-selection. Do you agree with that as well? Do you agree that singling out a sublist of a full list of already performed measurements is a perfectly classical thing to do? In that case, we're now in a completely classical statistical situation and the only thing left to understand is that singling out such a sublist may introduce a bias (new correlations) in the statistics of the resulting subensemble. This is what is explained in the book I referenced (in particular, what's interesting here is the collider bias).

Maybe. What about the difference between an individual pair, and an ensemble of pairs? If he just takes one individual pair from anyone of the subensembles, couldn't he rightfully claim that this "resulting pair is a genuine entangled pair in every aspect, and can in particular violate Bell inequalities"? Or maybe the language used by Gisin et al here is misleading, and only ensembles of pairs can violate Bell inequalities?

Entanglement is always a statistical property of an ensemble, it only appears in the statistics of many measurements. In order to violate Bell's inequality, one needs to collect data of many pairs and compute the empirical correlations.

 

[DrChinese]

In each of the four subensembles, selected by (projective) measurements on photons B&C, photons A&D are entangled. They are not entangled in the full ensemble. There's nothing else needed to ensure this than the preparation of the original photons in the initial state. There is nothing mysterious. To the contrary it's well understood by quite simple application of the rules of QT.

@gentzen asked if I would clarify my comments about A & D; I think vanhees71 says it nicely and I agree with everything he says. gentzen also suggests I move a more detail discussion to a new thread, which I plan to do. (Although sometimes I get derailed before I get there...)

I want to repeat a point I made previously. vanhees71 said there was nothing unusual about requiring indistinguishability of B & C as a part of the swapping operation, and he's right. However, for anyone who thinks that the Bell State Measurement of B & C is merely a statistical categorizing of the type of entanglement being observed, it should be a shock.

Why? Because the exact same measurements can be performed on B & C WITHOUT them being indistinguishable. You simply have 2 sets of BSM apparati, one for B and one for C. You will get the same results as to which type of entanglement you *would* get, and you will know which detector was triggered by B and which by C. (I.e. there are 2 ways the detectors will trigger for each of the 4 Bell states.) However, if you know which is which, A & D are not entangled and are instead fully Product State. In other words: the indistinguishable requirement is a purely QUANTUM requirement, while a CLASSICAL statistical treatment would not care about that. There is no (local) realistic interpretation possible in which you get around this quantum treatment.

 

[Nullstein]

@gentzen asked if I would clarify my comments about A & D; I think vanhees71 says it nicely and I agree with everything he says. gentzen also suggests I move a more detail discussion to a new thread, which I plan to do. (Although sometimes I get derailed before I get there...)

In that case you also agree with everything I say, because vanhees said exactly the same as I did in the post above his. But then it is unclear what you wanted to say with point 1 in post #88, since that is in direct contradiction to what I and vanhees said.

I want to repeat a point I made previously. vanhees71 said there was nothing unusual about requiring indistinguishability of B & C as a part of the swapping operation, and he's right. However, for anyone who thinks that the Bell State Measurement of B & C is merely a statistical categorizing of the type of entanglement being observed, it should be a shock.

Why? Because the exact same measurements can be performed on B & C WITHOUT them being indistinguishable. You simply have 2 sets of BSM apparati, one for B and one for C. You will get the same results as to which type of entanglement you *would* get, and you will know which detector was triggered by B and which by C. (I.e. there are 2 ways the detectors will trigger for each of the 4 Bell states.) However, if you know which is which, A & D are not entangled and are instead fully Product State. In other words: the indistinguishable requirement is a purely QUANTUM requirement, while a CLASSICAL statistical treatment would not care about that. There is no (local) realistic interpretation possible in which you get around this quantum treatment.

There is nothing quantum about indistinguishability. Also in classical statistics, indistinguishability is important sometimes. In order to get the correct extensive behavior of entropy (Gibbs' paradox), one needs to "integrate out" the distinguishability for example. Classically, one can take any distribution and sum over all permutations (and then normalize) to forget the distinguishability. Nothing about indistinguishability of particles is inherent to quantum theory.

 

[DrChinese]

In that case you also agree with everything I say, because vanhees said exactly the same as I did in the post above his. theory.

Nullstein, in that post: "The full ensemble of A&D is not entangled even after the measurement at B&C."

You keep trying to have it both ways! What vanhees71 actually said (and I say): "In each of the four subensembles, selected by (projective) measurements on photons B&C, photons A&D are entangled."

And despite everything you wrongly state about indistinguishability not being a quantum thing, you won't find that appearing in a suitable published quote. Of which, you have failed to provide a single one contradicting me. And why is that?

"...making the two entangled photon pairs indistinguishable in time [is] a necessary criterion for interfering photons from independent down conversions." - Zeilinger et al

 

[Nullstein]

Nullstein, in that post: "The full ensemble of A&D is not entangled even after the measurement at B&C."

And that is absolutely correct and also what vanhees said.

You keep trying to have it both ways! What vanhees71 actually said (and I say): "In each of the four subensembles, selected by (projective) measurements on photons B&C, photons A&D are entangled."

That's not the only thing vanhees said. He also said that the full ensemble is not entangled. I as well said multiple times that the subensembles are entangled. Here's just one of many examples:

Just to explain it again in simpler terms: Both A&B, C&D and the subensembles of A&D are entangled.

In fact it is both true. The subensembles of A&D are entangled and the full ensemble of A&D is not entangled, even after measurement. I can definitely have it both ways, because both propositions are true. In fact, after this post, it still seems that you disagree that thee full ensemble of A&D after measurement is not entangled. Is that correct? In that case, you would be in disagreement with accepted mainstream science.

If the only thing you agree with in vanhees' post is that the subensembles are entangled, then you still haven't satisfied @gentzen's demand to clarify your stance on the entanglement of the full ensemble of the A&D system (before and after measurement).

It would be nice if you could just complete the following two sentences:
1. The full ensemble of the A&D subsystem before measurement is __________. (entangled / not entangled)
2. The full ensemble of the A&D subsystem after measurement is __________. (entangled / not entangled)
Can you do that for me and @gentzen? Because from your posts I really can't extract what your stance is.

And despite everything you wrongly state about indistinguishability mot being quantum, you won't find that appearing in a suitable published quote.

Sorry, but Gibbs' paradox and its resolution by indistinguishbility is really a super standard fact in classical statistical mechanics and part of every undergrad curriculum. You'll find it in literally every introduction on statistical mechanics.

Of which, you have failed to provide a single one contradicting me.

I did provide you with references, it's just that you don't read them and generally only accept those that don't contradict you.

"...making the two entangled photon pairs indistinguishable in time [is] a necessary criterion for interfering photons from independent down conversions." - Zeilinger et al

I agree with that of course, but that doesn't make entanglement swapping a quantum phenomenon and it doesn't make indistinguishability a quantum phenomenon.