Photons from separated sources can be entangled - after they were detected

[DrChinese]

If I am not mistaken according to QM entanglement is swapped between two interacting photons so that joined measurement of B&C entangle C with A and B with D and not A with D.
So you can easily go astray if you assume physical versus statistical connection between entangled photons.

A & D definitely get entangled. And in fact you could continue/extend the chain by adding a 3rd entangled pair, E & F. Then you could get A & F entangled (by performing the BSM on D & E) even though B & E never meet. And in fact the person who chooses to perform that experiment - or not - (let's call her Dee) could make the decision to do so before or after Charlie. Such a setup - where there is extension of the quantum protocol - is called a quantunm repeater. There is substantial work on this precise scenario - using entanglement repeaters - already in the literature. So that is strong confirmation - to me at least - that I have interpreted my proposed setup correctly.

http://arxiv.org/abs/0912.3871"

 

[DrChinese]

If you subselect from both Charlie's possible choices you still don't have correlations at A&D. You have correlations only when you make BSM at Charlie and detect simultaneously photons at both detectors. Check the paper you linked (fig.3) - there is not correlations for 3 photon coincidences. Correlations appear only for 4 photon coincidences.

4 fold coincidences allow you to see the sets where A & D are entangled. Sometimes there are photons at A & D but not both B & C, so the Bell State Measurement does not occur and there is no entanglement.

But Charlie can see several different Bell states. According to which detectors go off, the Bell state can imply correlated or anti-correlated entanglement of A & D. That occurs randomly.

 

[DrChinese]

Note that in my analysis of your thought experiment I have not used the Bohmian interpretation at all.

And yet: you don't see anything weird about your prediction that if Charlie performs the BSM after A & D are detected, then there is no entanglement. Because that prediction seems weird to me. To me it implies (and would confirm) the absolute existence of a uniform time frame. Something which as far as I know has never been observed.

And what about the scenario in which the order of measurements is Alice, Charlie, Bob? Is there entanglement then? I say order does not matter to outcome, period. And I say that is the prediction of SQM too.

 

[Demystifier]

And yet: you don't see anything weird about your prediction that if Charlie performs the BSM after A & D are detected, then there is no entanglement. Because that prediction seems weird to me.

But not to me.

To me it implies (and would confirm) the absolute existence of a uniform time frame. Something which as far as I know has never been observed.

The point is that this "absolute" time frame is not really absolute, but depends on the environment. For example, one time frame is defined by cosmic background radiation, another time frame by motion of Earth, etc. The physically relevant frame is the one which corresponds to the environment that causes physical decoherence.

And what about the scenario in which the order of measurements is Alice, Charlie, Bob? Is there entanglement then? I say order does not matter to outcome, period. And I say that is the prediction of SQM too.

I simply disagree. I say the order does matter the outcome, whenever there is an irreversible process such as decoherence involved.

Anyway, you are an experimentalist. Can you perform an experiment to see who is right?

 

[Dmitry67]

If order is important, and parts of the experimental setup are put on spaceships, then you can experimentally detect freferred frame for that experiment. And as ships can fly in different directions, that frame has nothing to do with the 'environment' or 'lab'. It is something really mysterious (not for dBB may be :) )

 

[SpectraCat]

And yet: you don't see anything weird about your prediction that if Charlie performs the BSM after A & D are detected, then there is no entanglement. Because that prediction seems weird to me. To me it implies (and would confirm) the absolute existence of a uniform time frame. Something which as far as I know has never been observed.

I am also fairly convinced by DeMystifier's arguments here, for the following reason.

When Alice and Bob measure A & D, then that destroys the entanglement between A&B, and between C & D. Thus, when Charlie performs his operation, he is working with photons that have already undergone decoherence, so there is nowhere to transfer the entanglement to. Sure, B & C become entangled if Charlie makes a BSM, but that no longer has any ramifications for A & D, because they are no longer entangled with B & C, respectively, when Charlie makes the BSM (or not).

So, based on this, I think there will be no violation of a Bell's inequality for A & D based on Charlie's choice of whether or not to conduct a BSM on B & C. But I really want to see the results of the experiment!

 

[Dmitry67]

decoherence can be used as explanation only if lab is small so Bob, Alice and Charlie constantly decohere with each other. In case of perfectly isolated spaceships it can not serve as explanation.

 

[SpectraCat]

decoherence can be used as explanation only if lab is small so Bob, Alice and Charlie constantly decohere with each other. In case of perfectly isolated spaceships it can not serve as explanation.

I can't see how that matters ... as was pointed out to me in the https://www.physicsforums.com/showthread.php?t=374888", the entire apparatus has to be considered as a single unit. So there is no way for the spaceships to be "perfectly isolated", because they have to be connected by the photon paths. AFAICS the relative locations of Alice, Bob, Charlie and the emitter are irrelevant to the phenomenon of decoherence.

 

[DrChinese]

I am also fairly convinced by DeMystifier's arguments here, for the following reason.

When Alice and Bob measure A & D, then that destroys the entanglement between A&B, and between C & D. Thus, when Charlie performs his operation, he is working with photons that have already undergone decoherence, so there is nowhere to transfer the entanglement to. Sure, B & C become entangled if Charlie makes a BSM, but that no longer has any ramifications for A & D, because they are no longer entangled with B & C, respectively, when Charlie makes the BSM (or not).

So, based on this, I think there will be no violation of a Bell's inequality for A & D based on Charlie's choice of whether or not to conduct a BSM on B & C. But I really want to see the results of the experiment!

OOOOO, fun...

We have a great question here, and now we have folks on different sides of the predictions. I honestly haven't seen the answer, but now I will need to look even deeper just to see... Does anyone else want to weigh in on an answer?

 

[Dmitry67]

They can be isolated ENOUGH. In order for spacehips to decohere they must exchange several photons. While you can make a small detector looking in precisely calculated direction, ready to detect a single photon. Difficult, but possible. As you know, it is possible to decohere - at least for short time - machroscopic systems on the lab table - an experiment with superconductive ring in the entangled state.

 

[zonde]

4 fold coincidences allow you to see the sets where A & D are entangled. Sometimes there are photons at A & D but not both B & C, so the Bell State Measurement does not occur and there is no entanglement.

Sometimes photons B & C take different paths out of BS and then there can be simultaneous detection at both detectors. But sometimes B & C take the same path out of BS and then there can be only one detection in principle. So you clearly make unfair postselection.

But Charlie can see several different Bell states. According to which detectors go off, the Bell state can imply correlated or anti-correlated entanglement of A & D. That occurs randomly.

Please explain the thing about different detectors and different correlations. How do you mean that?

 

[Demystifier]

If order is important, and parts of the experimental setup are put on spaceships, then you can experimentally detect freferred frame for that experiment. And as ships can fly in different directions, that frame has nothing to do with the 'environment' or 'lab'. It is something really mysterious (not for dBB may be :) )

You cannot determine the whole global preferred frame everywhere, but only 3 small pieces (for 3 local spaceships) of that frame. These pieces are determined by local labs. (I don't need dBB for that.)

 

[Dmitry67]

Demystifier, if ORDER of events in 3 spacially-separated locations affects the result, then different observers would not agree on the results. Let me say it more careful: i don't detect the frame, I don't care about it, I just show that nature is not Lorentz-invriant if you're right.

I can agree that we can be overlooking something and there won't be entanglement in any case, no matter what the order is, but the opposite, that order affects the result is absolutely impossible

 

[DrChinese]

I have a reference on the SQM interpretation. Will post it in a bit... everyone still has a chance to weigh in! Does decoherence mean the order of measurements matters?

 

[SpectraCat]

And what about the scenario in which the order of measurements is Alice, Charlie, Bob? Is there entanglement then? I say order does not matter to outcome, period. And I say that is the prediction of SQM too.

Based on my earlier arguments, I would say that, once Alice has made her measurements, B represents a free, unentangled photon with a definite polarization state, which we know because it must be opposite to whatever Alice measured. So, B is indistinguishable from any other random photon with a well-defined polarization that would become entangled with C when Charlie makes his measurement.

So, I would predict the results of that experiment should be the same as in the 3-photon case where Charlie and Bob are the only two measurements, B is a "free" photon and C & D are an entangled pair. I don't actually know what those results are ... I have seen 3-photon entanglement before (c.f. the Steinberg group's paper in Nature 429, p.161 [2004]), but never in the 2+1 case, i.e. from an entangled pair with a "free" photon.

 

[DrChinese]

http://arxiv.org/abs/quant-ph/0201134"

[Note: My Charlie is here labeled as Alice; and my photons A/B/C/D are labeled as 0/1/2/3. So the question to be answered below that compares to my question is: Can Alice delay her BSM measurement and still end up with photons 0 and 3 entangled?]

"A seemingly paradoxical situation arises — as suggested by Peres [4] — when Alice’s Bellstate analysis is delayed long after Bob’s measurements. This seems paradoxical, because
Alice’s measurement projects photons 0 and 3 into an entangled state after they have been measured. Nevertheless, quantum mechanics predicts the same correlations. Remarkably, Alice is even free to choose the kind of measurement she wants to perform on photons 1 and 2. Instead of a Bell-state measurement [BSM] she could also measure the polarizations of these photons individually. Thus depending on Alice’s later measurement, Bob’s earlier results either indicate that photons 0 and 3 were entangled or photons 0 and 1 and photons 2 and 3. This means that the physical interpretation of his results depends on Alice’s later decision.

"Such a delayed-choice experiment was performed by including two 10 m optical fiber delays for both outputs of the BSA. In this case photons 1 and 2 hit the detectors delayed by about 50 ns. As shown in Fig. 3, the observed fidelity of the entanglement of photon 0 and photon 3 matches the fidelity in the non-delayed case within experimental errors. Therefore, this result indicate that the time ordering of the detection events has no influence on the results and strengthens the argument of A. Peres [4]: this paradox does not arise if the correctness of quantum mechanics is firmly believed."

--------------------

So the experiment has been performed, and the ordering is not important. I think some beers are due me. Demystifier, don't think I won't come over there to collect. SpectraCat, not sure where you are but I'm in Texas.

 

[SpectraCat]

http://arxiv.org/abs/quant-ph/0201134"

[Note: My Charlie is here labeled as Alice; and my photons A/B/C/D are labeled as 0/1/2/3. So the question to be answered below that compares to my question is: Can Alice delay her BSM measurement and still end up with photons 0 and 3 entangled?]

"A seemingly paradoxical situation arises — as suggested by Peres [4] — when Alice’s Bellstate analysis is delayed long after Bob’s measurements. This seems paradoxical, because
Alice’s measurement projects photons 0 and 3 into an entangled state after they have been measured. Nevertheless, quantum mechanics predicts the same correlations. Remarkably, Alice is even free to choose the kind of measurement she wants to perform on photons 1 and 2. Instead of a Bell-state measurement [BSM] she could also measure the polarizations of these photons individually. Thus depending on Alice’s later measurement, Bob’s earlier results either indicate that photons 0 and 3 were entangled or photons 0 and 1 and photons 2 and 3. This means that the physical interpretation of his results depends on Alice’s later decision.

"Such a delayed-choice experiment was performed by including two 10 m optical fiber delays for both outputs of the BSA. In this case photons 1 and 2 hit the detectors delayed by about 50 ns. As shown in Fig. 3, the observed fidelity of the entanglement of photon 0 and photon 3 matches the fidelity in the non-delayed case within experimental errors. Therefore, this result indicate that the time ordering of the detection events has no influence on the results and strengthens the argument of A. Peres [4]: this paradox does not arise if the correctness of quantum mechanics is firmly believed."

--------------------

So the experiment has been performed, and the ordering is not important. I think some beers are due me. Demystifier, don't think I won't come over there to collect. SpectraCat, not sure where you are but I'm in Texas.

Ok, so I looked at the published version of that paper (PRL 88 [2002], art. 017903-1), and I don't think their case corresponds to the case we are discussing. I also don't think that all of the statements you quoted from their paper are correct. Their "delayed-choice" measurement involved adding 10 m extensions to the OUTPUTS of the BSA device. That delayed the detection of photons 1 and 2, but that is not the crucial event. The crucial event was the interaction of photons 1 & 2 in the BSA, which is defines the quantum teleportation event that we are discussing in this thread.

Our discussion here is debating what would happen if the extensions were added to the INPUTS of the BSA device described in the paper. Since the interaction of the photons in the BSA device constitutes a measurement in SQM (which is why I prefer the designmation BSM for this), for the purposes of our discussion, it doesn't matter how long one delays detection of photons 1 and 2 after the BSM.

So, I think my beer money is safe for now.

EDIT: I am currently digging through the 75 or so papers that cite that one, to see if any of them mention our case explicitly ... but it might take a while.

 

[DrChinese]

Ok, so I looked at the published version of that paper (PRL 88 [2002], art. 017903-1), and I don't think their case corresponds to the case we are discussing. I also don't think that all of the statements you quoted from their paper are correct. Their "delayed-choice" measurement involved adding 10 m extensions to the OUTPUTS of the BSA device. That delayed the detection of photons 1 and 2, but that is not the crucial event. The crucial event was the interaction of photons 1 & 2 in the BSA, which is defines the quantum teleportation event that we are discussing in this thread.

Our discussion here is debating what would happen if the extensions were added to the INPUTS of the BSA device described in the paper. Since the interaction of the photons in the BSA device constitutes a measurement in SQM (which is why I prefer the designmation BSM for this), for the purposes of our discussion, it doesn't matter how long one delays detection of photons 1 and 2 after the BSM.

So, I think my beer money is safe for now.

OK, now you are doubling the bet! :grin:

You see, it makes NO difference where in the BSM the extra 10m extensions are added as long as the delay occurs. Now, why do I make this claim? Because we already know that once a beam is split, it can be recombined to restore the quantum state of the original beam. (That is not easy to do, but it is feasible.) So the measurement cannot be the point at which the beamsplitter is encountered.

Ultimately, the measurement occurs at the point at which results can no longer be erased and the process of information gain is not reversible (at the quantum level). That is so even though it was the beamsplitter that was the key element in the measurement process. That might not be true for an electron, but it is true for a photon - it must be detected/detectable.

So line up them beers, I'm coming to collect! (Although I don't know where I need to head yet...)

 

[Frame Dragger]

OK, now you are doubling the bet! :grin:

You see, it makes NO difference where in the BSM the extra 10m extensions are added as long as the delay occurs. Now, why do I make this claim? Because we already know that once a beam is split, it can be recombined to restore the quantum state of the original beam. (That is not easy to do, but it is feasible.) So the measurement cannot be the point at which the beamsplitter is encountered.

Ultimately, the measurement occurs at the point at which results can no longer be erased and the process of information gain is not reversible (at the quantum level). That is so even though it was the beamsplitter that was the key element in the measurement process. That might not be true for an electron, but it is true for a photon - it must be detected/detectable.

So line up them beers, I'm coming to collect! (Although I don't know where I need to head yet...)

Can you elaborate on the point of just how that quantum state could reasonably be restored? I'm not disagreeing... I just don't understand.

 

[Demystifier]

Now I have found THE RIGHT solution of the problem,
completely different from my previous one.
It is so simple and obvious that, I am convinced,
everybody will accept it.
The solution consists of several conceptual steps.

0. Forget everything that I said in my previous posts
of this thread!

1. The standard delayed-choice experiment involves
two entangled particles. It cannot be used for FTL
transfer of information because the interference
is encoded in the COINCIDENCES between the entangled particles.
To observe the coincidences, one needs a CLASSICAL COMMUNICATION
between entangled systems, and classical communication
cannot be FTL.

2. Point 1. above is a special case of the general
property of QM: Without classical communication,
entanglement cannot be used to transmit information.
NOT EVEN SLOWER THAN LIGHT FORWARD IN TIME.

3. As we all know from everyday life,
by classical communication, information can be
transfered ONLY FORWARD IN TIME. This is related
to the second law of thermodynamics.

Now let us apply these facts to two variants of the
DrChinese setting.

4. Assume that Charlie does his choice BEFORE Alice and
Bob make their measurements. Can Alice and Bob observe
any consequences of this choice? Yes, but only if
Charlie sends a classical information to Alice and Bob.

5. Now consider a different situation.
Now assume that Charlie does his choice AFTER Alice and
Bob make their measurements. Can Alice and Bob observe
any consequences of this choice? They could if
Charlie could send a classical information to Alice and Bob.
However, Charlie cannot send classical information to the
past. Therefore, Alice and Bob cannot observe
any consequences of the Charlie's choice.

Q.E.D.

 

[Frame Dragger]

Now I have found THE RIGHT solution of the problem,
completely different from my previous one.
It is so simple and obvious that, I am convinced,
everybody will accept it.
The solution consists of several conceptual steps.

0. Forget everything that I said in my previous posts
of this thread!

1. The standard delayed-choice experiment involves
two entangled particles. It cannot be used for FTL
transfer of information because the interference
is encoded in the COINCIDENCES between the entangled particles.
To observe the coincidences, one needs a CLASSICAL COMMUNICATION
between entangled systems, and classical communication
cannot be FTL.

2. Point 1. above is a special case of the general
property of QM: Without classical communication,
entanglement cannot be used to transmit information.
NOT EVEN SLOWER THAN LIGHT FORWARD IN TIME.

3. As we all know from everyday life,
by classical communication, information can be
transfered ONLY FORWARD IN TIME. This is related
to the second law of thermodynamics.

Now let us apply these facts to two variants of the
DrChinese setting.

4. Assume that Charlie does his choice BEFORE Alice and
Bob make their measurements. Can Alice and Bob observe
any consequences of this choice? Yes, but only if
Charlie sends a classical information to Alice and Bob.

5. Now consider a different situation.
Now assume that Charlie does his choice AFTER Alice and
Bob make their measurements. Can Alice and Bob observe
any consequences of this choice? They could if
Charlie could send a classical information to Alice and Bob.
However, Charlie cannot send classical information to the
past. Therefore, Alice and Bob cannot observe
any consequences of the Charlie's choice.

Q.E.D.

To me it seems that your assumption (one from dBB) lies in Point #3. Dr. Chinese is essentially making the case that such is not the case, or at least, that it is not relevant in a DCQE setting. Isn't this a re-expression of your original objections?

 

[DrChinese]

Can you elaborate on the point of just how that quantum state could reasonably be restored? I'm not disagreeing... I just don't understand.

Sure, here is a diagram which shows what I am referring to, and a reference to where it originated:

http://www.pas.rochester.edu/~AdvLab/Eberly_Bell_Inequalities_AJP.pdf

"We employ an arrangement of polarization analyzer loops to derive several simple Bell inequalities and then discuss the violation of one of them in light of quantum and classical interpretations of the data recorded."

 

[Frame Dragger]

Sure, here is a diagram which shows what I am referring to, and a reference to where it originated:

http://www.pas.rochester.edu/~AdvLab/Eberly_Bell_Inequalities_AJP.pdf

"We employ an arrangement of polarization analyzer loops to derive several simple Bell inequalities and then discuss the violation of one of them in light of quantum and classical interpretations of the data recorded."

Ok. That looks straightforward, if, as you say... challenging. Hmmmm... I wonder if you could get a grant for this? (EDIT: 'this' being the original A-D experiment, not the one in the article here)

 

[DrChinese]

To me it seems that your assumption (one from dBB) lies in Point #3. Dr. Chinese is essentially making the case that such is not the case, or at least, that it is not relevant in a DCQE setting. Isn't this a re-expression of your original objections?

Yes, here is where the language gets so tricky we run the risk of spinning in circles. So if I say something someone doesn't like, it may just be in the language.

The paradox occurs IF you assume the photons A & B were independent to begin with. If they weren't, then it is no weirder (really) that A & D are no longer independent photons after Charlie makes the BSM early. Now, if that isn't a true paradox (they never are), then it shouldn't be weird that Charlie's decision can be made after the fact. Because although A & D no longer exist, they weren't independent anyway. They are - in actuality - part of a chain that consists of A-B-C-D and that chain requires Alice, Bob AND Charlie. In my view, the causal chain is time symmetric and traces a zig-zag path in spacetime.

It is easier to see this when you add quantum repeaters to the chain; and realize that there is no theoretical limit to the zig-zags. So A & Z are entangled after pairs A & B, C & D, ... , Y & Z are entangled and then BSMs are performed on adjoining pair members B & C, D & E, ... X & Y. That should NOT be able to happen in any realistic scenario, or even in a non-local scenario, as all the pairs don't even need to exist at anyone point in time (the only requirement is that adjoining pairs have spacetime overlap).

[Side comment: I mean, where do the pilot waves go after the photons cease to exist? If they are out there, and influence things, why are they otherwise unobservable? In the BM perspective, those pilot waves determine the polarization of A & Z as well as assuring they are correlated when entangled.]

Here is gross speculation on my part: IF there is time symmetry, AND we need classical channels to make sense of quantum non-local signaling (which is the SIGNAL locality requirement), THEN the same thing is probably true of information moving from the Future to the Past. You still need classical channels to interpret it as such. I have no idea why we seem to only send information to the future, and not vice versa. Because it would sure help me if I knew who was going to win this week's sports games...

 

[SpectraCat]

OK, now you are doubling the bet! :grin:

You see, it makes NO difference where in the BSM the extra 10m extensions are added as long as the delay occurs. Now, why do I make this claim? Because we already know that once a beam is split, it can be recombined to restore the quantum state of the original beam. (That is not easy to do, but it is feasible.) So the measurement cannot be the point at which the beamsplitter is encountered.

Ultimately, the measurement occurs at the point at which results can no longer be erased and the process of information gain is not reversible (at the quantum level). That is so even though it was the beamsplitter that was the key element in the measurement process. That might not be true for an electron, but it is true for a photon - it must be detected/detectable.

So line up them beers, I'm coming to collect! (Although I don't know where I need to head yet...)

Hmmm not so fast. I don't necessarily agree that the entangled state after the BSM can be recombined to "restore the quantum state of the original beam", because photons B & C do not come from the same source. If it were possible to do such a recombination, then it would be possible to create a scenario where we could know with 100% certainty the result of a measurement on entangled pairs.

Consider the following. Take two un-entangled photons with known polarizations and combine them in a BSM to create an entangled pair. As I understand it, one of the properties of an entangled pair is that the results of a measurement on the pair are correlated, but random. So AFAICS, there should be no way to "reconstitute" the original quantum state of two unentangled photons with *known* polarizations .. and by "known" here I mean it would be possible to perfectly predict the results before making the measurement.

 

[DrChinese]

Hmmm not so fast. I don't necessarily agree that the entangled state after the BSM can be recombined to "restore the quantum state of the original beam", because photons B & C do not come from the same source. If it were possible to do such a recombination, then it would be possible to create a scenario where we could know with 100% certainty the result of a measurement on entangled pairs.

Consider the following. Take two un-entangled photons with known polarizations and combine them in a BSM to create an entangled pair. As I understand it, one of the properties of an entangled pair is that the results of a measurement on the pair are correlated, but random. So AFAICS, there should be no way to "reconstitute" the original quantum state of two unentangled photons with *known* polarizations .. and by "known" here I mean it would be possible to perfectly predict the results before making the measurement.

We are mixing beers and shots, since you don't get entanglement from the BSM if there was no entanglement to begin with. 2 knowns in gives 2 knowns out.

The question is: does it matter whether the 10m length is added before the beamsplitter or after? The experimental group added it after, and considered the matter closed (at least I didn't think they left any doubt on the matter). And I say that the beamsplitter itself creates beams which, in principle, can be recombined to restore any state which existed prior to the beamsplitter. And I did provide a reference on that as well, specifically referencing entangled systems. So I think the detector is the point where the measurement is said to occur for timing purposes.

Now: I am 2-0 ahead on references. So I think I get to collect my beers (there are 2 at stake now), unless you have a bit more than this. But it is a bit early to start my drinking yet in this time zone... and by the way, I charge interest. But I am not a snob, so anything you will drink with me should suffice.

 

[DrChinese]

0. Forget everything that I said in my previous posts
of this thread!

I will take this as a concession of 1 delicious beer. If you come out this way and don't call me, I will be mad!

 

[peteratcam]

I am trying to understand what everybody is discussing here but there are no equations in this thread at all, and lots of acronyms.
So please help me with this:
Can I think of photons as a two state system? (I prefer to think of spin-1/2s)
Can I think of polarising filters as stern-gerlach aparatuses?
What is a BSM?

I would like to follow through the experimental protocol on paper using schrodinger equation unitary evolution, but I need to know what the degrees of freedom are and what is being measured at each stage, thanks.

 

[Frame Dragger]

I am trying to understand what everybody is discussing here but there are no equations in this thread at all, and lots of acronyms.
So please help me with this:
Can I think of photons as a two state system? (I prefer to think of spin-1/2s)
Can I think of polarising filters as stern-gerlach aparatuses?
What is a BSM?

I would like to follow through the experimental protocol on paper using schrodinger equation unitary evolution, but I need to know what the degrees of freedom are and what is being measured at each stage, thanks.

I can help you with precisely one of those... and the easy one. Go figure. BSM = Bell State Measurement. (edit: to be clear, as in Bell Inequalities)

 

[SpectraCat]

We are mixing beers and shots, since you don't get entanglement from the BSM if there was no entanglement to begin with. 2 knowns in gives 2 knowns out.

Ok, let's consider your statement above (which was exactly *my* point by the way, although I may not have communicated it effectively yet).

I think we both agree that once one member of an entangled pair is measured, the state of the other member is also known, irrespective of distance in space and time between the measuring events. All that matters is that the pair is entangled when the first measurement is made.

Now consider your original example. In that case, at the beginning we have two independent entangled pairs of photons (A/B) and (C/D). Once Alice and Bob make their measurements on photons A and D, which in your variation happens before[/B} B and C enter Charlie's BSM, then B and C are no longer entangled with anything ... they are independent photons with known polarizations. So, by your argument above, not even B & C are entangled by the BSM, since Alice and Bob have already destroyed their respective states entangled states. So that means that A and D aren't entangled either.

So, as I said, the first reference you posted does not cover the situation we are concerned with here, because in all variations discussed in that paper, the entangled pairs (0/1) and (2/3) still exist when photons 1 and 2 enter the BSM.

The question is: does it matter whether the 10m length is added before the beamsplitter or after? The experimental group added it after, and considered the matter closed (at least I didn't think they left any doubt on the matter). And I say that the beamsplitter itself creates beams which, in principle, can be recombined to restore any state which existed prior to the beamsplitter. And I did provide a reference on that as well, specifically referencing entangled systems. So I think the detector is the point where the measurement is said to occur for timing purposes.

That second reference you posted again does not deal with the case relevant to our example. It deals with the case where a single beam corresponding to one member of an entangled pair is split in a birefringent crystal, and then recombined in another matched crystal. That is different from our case, where two photons from different entangled pairs are mixed in a beam-splitter to produce a quantum teleportation event. Perhaps one can extend the example you cited to our case, but it is not at all obvious and should be explicitly proven or experimentally demonstrated. I did a quick literature search, but could find no such examples.

Now: I am 2-0 ahead on references. So I think I get to collect my beers (there are 2 at stake now), unless you have a bit more than this. But it is a bit early to start my drinking yet in this time zone... and by the way, I charge interest. But I am not a snob, so anything you will drink with me should suffice.

Well, you have posted 2 more references that I have, but they haven't really been on target. I'd say we're still even. I am in PA by the way.