I Realistic interpretation of QM

[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.

 

[gentzen]

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

so far I agree

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?

I don't see why singling out a sublist should be a perfectly classical thing to do. And I see it even less when it comes to implementing a physical process that will do that singling out. (Also note that for me, postselection is a typically quantum thing. A typically classical thing would be statistics caused by an preexisting but unknown state.)

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.

Here I agree that the change in statistics caused by forming sublists can be analysed as a completely classical statistical situation. But I disagree that this would be the non-trivial part which needs a reference to back it up.

But overall, I agree that this constitutes a serious attempt "to bring forward a convincing argument".

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).

He explicitly wrote: "I think vanhees71 says it nicely and I agree with everything he says." So even if I would have made an explicit demand to clarify his stance, that demand would now be satisfied.

 

[Nullstein]

I don't see why singling out a sublist should be a perfectly classical thing to do.

So suppose you have a list of ~1000 measurement results, written on a piece of paper and your task is to strike through ~750 of them based on the rule that everything should be removed unless e.g. $B=1$ and $C=-1$, what is quantum about this task? You literally just remove lines from a macroscopic piece of paper by striking them through with a pencil.

And I see it even less when it comes to implementing a physical process that will do that singling out.

But it's not done by a physical process. It's done manually or by a classical computer (which strictly speaking can of course also be argued to be a physical process, albeit a very classical one and I don't think that is what you meant to say here?).

He explicitly wrote: "I think vanhees71 says it nicely and I agree with everything he says." So even if I would have made an explicit demand to clarify his stance, that demand would now be satisfied.

But in the very next post, he contradicted vanhees again, so it became unclear again, what his stance is, at least to me. He claimed that it is impossible to have it both ways: The subensembles are entangled and the full ensemble is non-entangled. Yet, vanhees said the exact opposite and it is indeed possible to have it both ways. So does he believe that the full ensemble of A&D is entangled or non-entangled? From the claim that it is impossible to have it both ways, taken together with his agreement with the statement that the subensembles are entangled, I would normally conclude that he believes that the full ensemble is entangled as well, which is in direct contradiction to vanhees (whom he claimed to agree with) and accepted mainstream science. If it is clear to you what his stance is, could you light me up?

 

[Nullstein]

Let me try to explain it in a different way:
You may be intruiged by the fact that the protocol includes a measurement at B&C. But really, you don't have to do this. Suppose you start with the two pairs A&B, C&D and don't perform a measurement at B&C. You do nothing there. You just measure A&D as before and compile a list of ~1000 results. Then you select ~250 of them just by chance, e.g. by coinflips. There is a tiny chance that you make a correct choice and this sublist of ~250 measurements result will be perfectly entangled, even though you have done precisely nothing to the photons at B&C. So it's really just the singling out of the sublist that introduces the entanglement in A&D and not the measurement at B&C. The operation at B&C does not in any way affect the state of A&D. It does not cause the entanglement of A&D.

The reason for performing the measurement at B&C is to consistently choose the right sublist and not having to rely on chance. There is no physical process occurring to A&D that would entangle them. We know this because $\rho_{AD,after} = \rho_{AD,before}$, so the "before"-state statistically already contains the entangled subensembles, even if we don't perform an operation at B&C.

The fact that $\rho_{AD,after} = \rho_{AD,before}$ proves that it is not the measurement at B&C that causes the entanglement of subensembles of A&D. But then why is it that a measurement at B&C is so helpful for chosing a correct sublist? That's because A is entangled with B and D is entangled with C. So the experimenter at B&C has access to the information which sublist will be the correct one, even though A&D may be far away. He just needs to devise a smart way to extract it and of course he is allowed to perform quantum operations in order to do that. But it is essential to understand that these operations do not affect A&D in any way and in particular don't cause any entanglement between A&D. That is excluded by the no-communication theorem.

 

[DrChinese]

1. And that is absolutely correct and also what vanhees said.

2. 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.

3. It would be nice if you could just complete the following two sentences:
a. The full ensemble of the A&D subsystem before measurement is __________. (entangled / not entangled)
b. 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.

1. vanhees71: "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. "

Nullstein: "The full ensemble of A&D is not entangled even after the measurement at B&C."2. I will leave your odd statement to speak for itself, just adding that I keep saying you want it both ways.3. I have stated this many a time above:
a. The full ensemble of the A&D subsystem before measurement is not entangled.
b. The SUB ensembles of the A&D subsystem after the PROJECTIVE measurement are entangled in a Bell state, of which there are 4 possibilities.

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I think one problem we are having in this discussion is that what you call a full ensemble is not the same thing is what everyone else calls it. When any Bell test is performed, there is a requirement that there be an element of indistinguishability. That is normally performed by requiring the arrival times of 2 photons to occur within a small time window. In our swapping case, that means the B & C photons arrive within that window. The "full" ensemble includes B & C pairs that arrive too far apart to be considered indistinguishable. (Note that it does not matter whether we actually know which is B and which is C.) The B & C pairs that arrive too far apart definitely do NOT lead to entangled A & D pairs (because B & C were distinguishable). The full A & D ensemble therefore consists of pairs that are definitely not entangled, as well as pairs that are definitely entangled (in principle, and not considering noise) depending on whether their B & C counterparts are distinguishable or not.

The next issue is the 4 Bell states that are possible outcomes when entanglement occurs. They are |Ψ+>, |Ψ−>, |Φ+>, and |Φ −>. Some entanglement swapping experiments identify all 4 of these sub-ensembles, and some identify only a single one. Regardless, you must know which type of entanglement is associated with each A & D pair. If you are ignoring one or more of the 4 Bell state ensembles, that is not a problem for the validity of the experiment. It just means a smaller sample size. For example, see https://arxiv.org/abs/quant-ph/0201134 and note that the quote below has been label-adjusted to match the example we are discussing

"We stress that photons A and D will be perfectly entangled for any result of the BSM [Bell-state measurement]... But it is certainly necessary ... to communicate to the Bell-state measurement result. This will enable ... sorting of data into four subsets, each one representing the results for one of the four maximally entangled Bell-states. ... In our experiment, we were restricted to only identifying the state |Ψ−> due to technical reasons. This reduction of the teleportation efficiency to 25% does not influence the fidelity."

Any sub-ensemble of A & D pairs that are identified as meeting a) the B & C indistinguishable requirement; and b) identified as being projected to the one or more of the desired Bell states; WILL be entangled.

Those sub-ensembles obviously cannot be entangled because they just happen to arrive at the same time - they are from independent source (in some of the experiments). If that was the trigger: then there would be no indistinguishability requirement and you would not need to allow the A & B biphoton system to interact with the C & D biphoton system. You would simply check the arrival times of the B & C photons, which can be done without allowing the 2 original biphoton systems to interact.

Again, the point of all this is that entanglement swapping creates new entangled A & D photon pairs, and the preparation process cannot be considered local since A & D are far apart when the swapping operation occurs as a result of a Bell State Measurement on B & C. Of course, another way of looking at it is to say that the A & B biphoton system (which has nonlocal extent) interacts locally with the C & D biphoton system (which also has nonlocal extent). That's a matter of interpretation.

 

[PeterDonis]

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