I Confused by nonlocal models and relativity

Lord Jestocost

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One should always keep in mind that “the Bell theorem furnishes a criterion to experimentally exclude the hypothesis that quantum correlations are established at the source (that is, to rule out an alternative description of quantum phenomena based on local variables).”

Valerio Scarani in “Quantum Physics: A First Encounter: Interference, Entanglement, and Reality”
 
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Good day.
I have a question about using for nonlocal connection not individual entanglement particles but common beam.
The main approach is depicted at following figure:
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Can we transmitt an information by means of distruction two photons entanglement in beams?
If we destroy entanglement (measure the polarization) before beam reaching the two slit wall then we will detect two peaks under the slits. Otherwise we will see an interference picture.
 

vanhees71

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1. It makes it difficult for myself and readers to follow a paper behind a paywall, but I see that the paper I cited did not have the equation you presented. Here's an freely available similar one from Zeilinger et al which does:

Experimental Nonlocality Proof of Quantum Teleportation and Entanglement Swapping

2. Now, you failed to mention a key point of formula (3) in this reference: This does NOT describe the state prior to photons 2 & 3 arriving together at the beamsplitter. So your "conclusion" that "nothing changes" for 1 & 4 after selection is tainted by this omission. If entanglement swapping does not happen (for example 2 & 3 are not brought together), there is no entanglement between 1 & 4 (and that has nothing to do with post selection).

3. There is no microcausality constraint. And certainly none mentioned in any entanglement swapping paper. Again, you are not quoting the authors here, and this is one of the things I object to. Please don't put your words in their mouth. What the authors say in no way supports your position.

4. And now you are mangling my position as well: I don't assert there are causes which propagate FTL, although it is possible as Bohmian Mechanics is a viable interpretation. What I am saying is that the action narrative for entanglement swapping does not follow any clear causal description, precisely because the detection ordering is not relevant, and because the detectors are spacelike separated. I don't know any more than anyone else what is occurring under the quantum covers.

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I notice you keep ignoring key points I am making:
a) Entanglement monogamy prevents photon 1 from being entangled with both 2 and 4 as you claim. It's one or the other, and I will challenge you to provide a reference to the contrary. That alone makes your description not viable.
b) My counter-example F from post #57. As mentioned previously, this counter-example is not theoretical: it is done as part of the tuning process for every entanglement swapping experiment. It must be performed, as this is how the 2 & 3 streams are calibrated so the the swap is enabled. It starts off with a timing difference and no swaps. As calibration improves, swaps occur and that continues until an optimal performance level is achieved. They simply don't publish those results because it is a null result (because 1 & 4 won't be entangled because no swap occurred).
Ok, let's discuss this paper (I use the PRL version however, since I hate revtex without the tightenlines switch envoked, but I guess it's identical with the arXiv version).

ad 2) The for our discussion crucial point is that, as the authors correctly state, (3) IS the prepared state of the four photons BEFORE any projections are done, and this time the authors got the signs right too. It's a bit of work to write it out, but indeed (3) is exactly the same vector as (1), it's just decomposed with respect to different bases.

I don't claim that nothing happens, but I claim that the projection accoring to A's finding of the pair 1&2 (note that now the photons are labelled differently from the paper we discussed before; instead of 1234 they now count 0123, everything else is identical) in one of the 4 maximally entangled (Bell) states, you get subensembles of the correspondingly entangled photon pair 0&3, though these two photons have not been entangled before nor have ever directly interacted with each other. The Bell-state measurement by A is local and doesn't causally influence photons 0&3, but due to the entanglement of the pairs 0&1 and 2&3 in the originally prepared 4-photon state each partial ensemble of the photon pair 0&3 is entangled due to the selection depending on the measurement outcome of A. Of course, all that has to be done is indeed to provide a measurement protocol to Victor with all the coincidence measurements properly, and you can sort the corresponding events into the four classes each showing the entanglement in each of the corresponding Bell states of the pair 0&3.

ad 3) Microcausality is a theoretical constraint on relativistic QFTs, and all in the paper is consistent with standard QED. That's why there's no contradiction between the experimental findings and microcausality.

ad 4) Good. Then at least in this point we agree. There's no FTL propagation, and this newer paper even underlines this again with the variant, where the choice of the state of Bob's photon pair 0&3 through Alice's Bell measurement on photons 1&2 is done after Bob has measured his photons.

Then I don't understand what you are saying in point a). By the preparation 0&1 are entangled as well as 2&3 but not 0&2 and 1&3. The whole point of the paper is that through the (post-)selection via A's measurements the pair 0&3 are entangled in the same state at which Alice found her pair 1&2. They even proved entanglement of photons 0&3 in ##|\psi_{03}^- \rangle## for the partial ensemble, for which Alice found her photons 1&2 in the state ##|\psi_{12}^-\rangle##.

I think I answered also b) with this. Note that I refer now to the labelling of the photon states according to the new paper, discussed in this posting, while you of course referred to the before discussed older paper.

The important point of the minimal interpretation is that the quantum state describes statistical properties of ensembles and that of course in the full ensemble of four photons 0123 only 0&1 and 2&3 are entangled. For the subensemble, where Alice found photons 1&2 in the state ##|\psi_{12}^- \rangle## the photons 0&3 are in the maximally entangled Bell state ##|\psi_{03}^- \rangle## either. That's the whole point of this great experiment, and that's what's called entanglement swapping and is a strong form of teleportation either as the authors stress in their paper too.
 

Lord Jestocost

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The important point of the minimal interpretation is that the quantum state describes statistical properties of ensembles.........
Allow me a comment on the side, in Abner Shimony’s words (in “Symposia on the Foundations of Modern Physics 1992 - The Copenhagen Interpretation and Wolfgang Pauli” (edited by K. V. Laurikainen and C. Montonen))

There is, for example, Ballentine, whom I mentioned yesterday. He says: ‘I am not a hidden variable theorist, I am only saying that quantum mechanics applies not to individual systems but to ensembles.’ I didn't put this down separately because I simply do not understand that position. Once you say that the quantum state applies to ensembles and the ensembles are not necessarily homogeneous you cannot help asking what differentiates the members of the ensembles from each other. And whatever are the differentiating characteristics those are the hidden variables. So I fail to see how one can have Ballentine's interpretation consistently. That is, one can always stop talking and not answer questions, but that is not the way to have a coherent formulation of a point of view. But to carry out the coherent formulation of a point of view, as I think Einstein had in mind, you certainly have to supplement the quantum description with some hypothetical extra variables.
 

vanhees71

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The members of the ensemble in the described experiment are easily differentiated simply by the time of the registration of the four photons for each prepared four-photon state. It's not about stopping talking (we talk a lot in fact ;-)), but it's the interpretation which just boils it down to what's described and what's observed (at least in experiments like the one we discuss here). I don't understand what "hypothetical extra variables" one "certainly has to supplement the quantum description with". Maybe at Einstein's time you might have the idea that QT is incomplete because of the inseparability and Einstein's quibbles with this implication of the QT formalism, but today where we are 84 years further in investigating it, it seems as if there's nothing of this kind to add since Q(F)T describes all observations accurately, including the one we are discussing here and which is a great example for the fact that the inseparability/entanglement is what's the case in nature. Einstein's gut feeling about the quibbles were put into a clearly observable prediction of an entire class of local deterministic hidden-variable theories by Bell, and then starting with Aspect's experiment and with many more until today, including the one we discuss here, clearly show that Q(F)T is the correct description. You may like it or not, it's a fact of nature revealed by hard work of many physicists, theorticians and experimentalists alike, and one has to accept it as a fact.

Whether there is some more comprehensive theory than Q(F)T, I don't know. From the here discussed kinds of experiments, I don't see the need for any such thing, but who knows, what a hopefully some day found solution of the puzzle of quantum gravity may come up with?
 

DrChinese

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Good day.
I have a question about using for nonlocal connection not individual entanglement particles but common beam.
The main approach is depicted at following figure:
View attachment 246141

Can we transmitt an information by means of distruction two photons entanglement in beams?
If we destroy entanglement (measure the polarization) before beam reaching the two slit wall then we will detect two peaks under the slits. Otherwise we will see an interference picture.
You might be surprised to learn that entangled photons do NOT produce interference as one might otherwise expect. You must stop the entanglement first. Once you do that, the ability to carry out the rest of your idea is lost. See Figure 2 in this great summary paper by Zeilinger:

 

DrChinese

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2) The for our discussion crucial point is that, as the authors correctly state, (3) IS the prepared state of the four photons BEFORE any projections are done, and this time the authors got the signs right too. It's a bit of work to write it out, but indeed (3) is exactly the same vector as (1), it's just decomposed with respect to different bases. 1.
...

ad 3) Microcausality is a theoretical constraint on relativistic QFTs, and all in the paper is consistent with standard QED. ...

ad 4) Good. Then at least in this point we agree. There's no FTL propagation, and this newer paper even underlines this again with the variant, where the choice of the state of Bob's photon pair 0&3 through Alice's Bell measurement on photons 1&2 is done after Bob has measured his photons.

Then I don't understand what you are saying in point a). By the preparation 0&1 are entangled as well as 2&3 but not 0&2 and 1&3. The whole point of the paper is that through the (post-)selection via A's measurements the pair 0&3 are entangled in the same state at which Alice found her pair 1&2. They even proved entanglement of photons 0&3 in ##|\psi_{03}^- \rangle## for the partial ensemble, for which Alice found her photons 1&2 in the state ##|\psi_{12}^-\rangle##.
I notice you keep ignoring key points I am making:
a) Entanglement monogamy prevents photon 1 from being entangled with both 2 and 4 as you claim. It's one or the other, and I will challenge you to provide a reference to the contrary. That alone makes your description not viable.
b) My counter-example F from post #57. As mentioned previously, this counter-example is not theoretical: it is done as part of the tuning process for every entanglement swapping experiment. It must be performed, as this is how the 2 & 3 streams are calibrated so the the swap is enabled. It starts off with a timing difference and no swaps. As calibration improves, swaps occur and that continues until an optimal performance level is achieved. They simply don't publish those results because it is a null result (because 1 & 4 won't be entangled because no swap occurred).

2. This is incorrect, and does not express what you claim. It assumes a swap occurs (i.e. 2 of the photons are indistinguishable). If no swap occurs, then the 0 and 3 photons (numbered per this paper) are NOT entangled on any basis (and in this form, they are entangled on one basis). The authors say the following in their conclusion (and notice how I quote the authors to support my statement):

"Thus depending on Alice’s later measurement [or choice not to measure], 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."

3. You keep quoting yourself. Microcausality is not a restriction on QFT. There are considerations for special relativity, of course. But as I keep repeating, quantum mechanics post-Bell is NOT local causal, and this is beyond contention. There is nothing different in QFT, and I challenge you for the nth time to produce a suitable quote from someone else that supports your position regarding causality in QFT.

4. We don't agree on this! I say there are no FTL causes because there is no causality!! There is no causal direction for anything in these experiments. And if there are, then it is FTL (which is viable in some interpretations). Either way, we don't agree.
 

vanhees71

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I'm not ignoring the points you are making, but I just quote the results of the very paper (PRL 88, 017903, 2002) you discuss to contradict you. Maybe I do not understand what you want to say. I try one last time:

ad a) The full ensemble of photons is described by the state in Eq. (2) which is indeed exactly the same as Eq. (3) is prepared such (as can be read off immediately in the form of Eq. (2)) that the photon pair 0&1 and the photon pair 2&3 are entangled, while neither the pair 0&2 nor the pair 1&3 are entangled (that's the meaning of a product state vs. a superposition of product states), and the entanglement for the pairs 0&1 and 2&3 is maximal, i.e., you have a Bell state. Now Alice projects the pair of photons 1&2 to the state ##|\psi_{12}^{-} \rangle##. As you can immediately read off of the state written in the form of Eq. (3) the photons 0&3 of this particularly partial ensemble (on average 1/4 of all measured four-photon states) are entangled, i.e., described by the state ##|psi_{03}^- \rangle##. That's precisely what the authors have demonstrated, and that's entanglement swapping and a particular nice example for quantum teleportation. The authors demonstrate that the pair 0&3 of the partial ensemble is really entangled even by proving the violation of the Bell inequality. They also demonstrate that there's no causal effect of Alice's measurement process on the pair 0&3 since the pairs 0&3 can be post-selected through A's measurement, i.e., A can do her measurement after the pair 0&3 is measured with the same result. As the authors also emphasize, all that's needed is a complete measurment protocol on an event-by-event basis, i.e., only Victor can choose the partial ensemble based on A's mesurement.

ad b) now you jump again to the other paper, but there's nothing different (only the labelling of the photons with 1234 rather than 0123). Thus b) is answered with a), and it's not answered by me but by the authors and the outcome of their real-lab experiment!

2) what do you mean by "the swap occurs"? As already expolained under a): You deny the very result of the paper! Based on Alice measurement Victor is able to choose the partial ensemble, for which photons 0&3 are entangled, and that's what the authors have shown experimentally and that's the whole purpose of this experiment, and that's what's called entanglement swapping and teleportation. I don't understand why you deny the main intention and conclusion of the very paper you yourself have chosen to discuss here.

3) Come on! Read any decent textbook on standard QED. There microcauality is in the very construction of the theory: You start with field operators which transform locally under the Poincare group as their classical analogues, and then you write down a Poincare invariant action with a Lagrangian consisting of a sum over local field monomials and this guarantees by construction the microcausality condition, i.e., if you have a local observable ##\hat{O}(x)##, then it commutes with the Hamilton density ##\hat{\mathcal{H}}(y)## at space-like separation
$$[\hat{O}(x),\hat{\mathcal{H}}(y)]=0 \quad \text{if} \quad (x-y)^2=(x^0-y_0)^2-(\vec{x}-\vec{y})^2<0.$$
This ensures that there are no spooky actions at a distance and formally that the time ordering in the Dyson series of perturbation theory is frame-independent and thus leads to Poincare-invariant S-matrix elements. It's a very fundamental property of all successful relativistic QFTs (including the Standard Model) and it's taught almost in the first lecture of the introductory QFT lecture worldwide!

4) Ok, since you don't understand the meaning of fundamental mathematical facts of QFT, as the microcausality principle, it's cear that you are not able to understand my argument either. Everything of the experiment is in perfect agreement with the predictions of QED. The whole experimental setup is based on QED and confirms it. For sure there are no spooky actions at a distance, and everything is about correlations described by the state given in Eq. (2), which is the same as given in Eq. (3), of the paper.
 

DrChinese

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1. I'm not ignoring the points you are making, but I just quote the results of the very paper (PRL 88, 017903, 2002) you discuss to contradict you. Maybe I do not understand what you want to say. I try one last time:

2. ad a) The full ensemble of photons is described by the state in Eq. (2) which is indeed exactly the same as Eq. (3) is prepared such (as can be read off immediately in the form of Eq. (2)) that the photon pair 0&1 and the photon pair 2&3 are entangled, while neither the pair 0&2 nor the pair 1&3 are entangled (that's the meaning of a product state vs. a superposition of product states), and the entanglement for the pairs 0&1 and 2&3 is maximal, i.e., you have a Bell state.
1. Referencing an entire paper - one that doesn't say anything like what your position is - well, that's not responsive. Every entanglement swap paper says something like one of the two statements below:

a. Your position: post selection identifies a state which already existed and was simply revealed, nothing changed in the state for photons 0 & 3.
b. My position: the swap changes the states of the 2 separate systems (pair 0 & 1 and pair 2 & 3) such that photons 0 & 3 (previously uncorrelated) are entangled. That change to the state of 0 & 3 is nonlocal.

To help clarify which of above matches standard science:

"We confirm successful entanglement swapping by testing the entanglement of the previously uncorrelated photons [0 & 3]. Violation of a CHSH inequality is not only of fundamental interest because it rules out local-hidden variable theories. It also proves that the swapped states are strongly entangled and, as a result, distillable [40]. The specific state of photons 1 and 4 after entanglement swapping depends on the result of the BSM, which can either be ψ + or ψ −."

"Initially, the system is composed of two independent entangled states... Alice subjects photons 1 and 2 to a measurement in a Bell-state analyzer (BSA), and if she finds them in the state |Ψ−>12, then photons 0 and 3 measured by Bob, will be in the entangled state |Ψ−>03. If Alice observes any of the other Bell-states for photons 1 and 2, photons 0 and 3 will also be perfectly entangled correspondingly. We stress that photons 0 and 3 will be perfectly entangled for any result of the BSA... [after the swap].

None of the above, or anything else in these papers, remotely supports your characterization.


2. Your narrative omits the key point that photons 0 & 3 are not entangled before the swap. I agree with your statements that 0 & 1 and 2 & 3 are entangled before the swap, and that 0 & 2 and 1 & 3 are not entangled before the swap. This is the correct state BEFORE the swap:

a. Before swap:
|Ψtotal> =
|Ψ −>01 ⊗ |Ψ −>23 =
1/2 [ |H0V1> |H2V3> + |V0H1> |H2V3> + |H0V1> |V2H3> + |V0H1> |V2H3>
[none of which terms can be recombined into any state where photons 0 & 3 are in an entangled state]

Where we agree that the state after the swap is as you describe.

b. After swap:
|Ψtotal> = 1/2 [|Ψ+>03|Ψ+>12 − |Ψ−>i03|Ψ−>12 − |Φ+>03|Φ+>12 + |Φ−>03|Φ−>12].

You need to be able to make photons 1 and 2 indistinguishable before you can move to state b. If they are distinguishable, they remain in state a. That's the whole point, that the state of 0 & 3 change as a result of the swap. And the statistical results change too!

The change (resulting from the swap action) to photons 0 & 3 is quantum nonlocal, and is not subject to a microcausality constraint. First, there is the demonstrated quantum nonlocality ruining the locality principle here. And second, the ordering of events prevents a discrete "cause" and a discrete "effect" from being identified. Which then ruins any notion of causality. Local causality (which should already have been thrown out post-Bell) is easily disproven with this experiment, and again, that is the [paradoxical] point. Photons from independent sources can be entangled by a remote action, and that action can occur before or after the detection of those photons.

Of course, all of this is standard QM/QFT. Relativity does not factor in other than everyone agrees there is no possibility of sending a message FTL. However, the relevance to quantum computing is that you can clone a quantum state FTL. This is a demonstration of spooky action at a distance. (Which is exactly the opposite of your position: that it is forbidden by construction.)
 
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vanhees71

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I think it's impossible to discuss this with you. You are simply not reading carefully what I'm writing and sometimes claim the opposite what I said!

What you claim to be my statement a) I've never said. I say:

The total ensemble is described by the state ket given in Eq. (2) by the authors (of course, I don't contradict the authors, because I want to discuss their paper and not something else). The state ket in Eq. (3) is exactly the same (also what the authors say). The pairs 0&1 and 2&3 are entangled but not 1&2 and 0&3 (that's what (2) directly implies).

I cannot read your formlae. So here are the facts from the paper, written out in LaTeX again:
$$|\psi_-^{12} \rangle \otimes |\psi_{03}^- \rangle.$$
For the following we need this in the form written as Eq. (3) by the authors:
$$|\Psi \rangle=\frac{1}{2} (|\psi_{03}^+ \rangle \otimes |\psi_{12}^+ \rangle - |\psi_{03}^- \rangle \otimes |\psi_{12}^- \rangle -\frac{1}{2} (|\phi_{03}^- \rangle \otimes |\phi_{12}^+ \rangle + |\phi_{03}^- \rangle \otimes |\phi_{12}^- \rangle).$$
Let me stress again this is mathematically identical and thus both formulae describe precisely the same state the four photons are prepared in before any measurement by Alice and/or Bob is made and no choice of an subensemble is made either. That comes now:

With the so prepared four-photon state Alice performs a Bell measurement on Photons 1&2, i.e., she chooses a partial ensemble by filtering out all findings except when she finds photons 1&2 to be in the Bell state ##|\psi_{12}^{-} \rangle## which is chosen for technical reasons: It's the one which is most simple to detect.

This is explained in the paper too, to understand this one must remember that the notation of the states above is a bit sloppy to simplify the explanation; in fact to understand how Alice filters out the specific entangled state of the pair 1&2, one has to take into account here that the photons are indistinguishable bosons and that thus ##|\psi_{12}^{-}## is the state where the spatial (momentum) part of the photon states is anti-symmetric and thus also the polarization part must be antisymmetric, and thus it's the very one of the four possible Bell states state, where at the beam splitter you have coincident detections in the detectors D1 and D2 at Alice's site, as drawn in Fig. 2 of the paper). To be clear, the correct notation for the states are rather like this (using photon creation and annihilation operators and the vacuum state ##\Omega \rangle##):
$$|\psi_{12}^- \rangle = \frac{1}{2} (\hat{a}_{1H}^{\dagger} \hat{a}_{2V}^{\dagger} - \hat{a}_{1V}^{\dagger} \hat{a}_{2H}^{\dagger}.$$
Anyway, it's clear that whenever Alice has a coincident click at both detectors D1 and D2 her photons are in the state ##|\psi_{12}^- \rangle## and thus due to the state preparation of the four-photon state, the corresponding partial ensemble must be described by the corresponding component:
$$|\Psi_' \rangle=- |\psi_{03}^- \rangle \otimes |\psi_{12}^- \rangle.$$
This clearly tells you that for this partial ensemble the photon pair 03 is maximally entangled and described by the state ##|\psi_{03}^- \rangle##. Alice will have such coincident clicks of her detectors D1 and D2 with probability
$$|\langle \Psi'|\Psi \rangle|^2=1/4.$$
The entanglement of photons 0&3 for this partial ensemble is clearly demonstrated by this experiment. The experimentalists even verified the violation of Bell's inequality which is a clear indication for entanglement.

That it cannot be Alices measurement/manipulations of photons 1&2 that causes the pair 0&3 being entangled when the above described subensemble is chosen, is also proven in the experiment: When A's photons are delayed such that the mesurements on the pair 0&3 are already finished and fixed when A is doing here measurement, precisely the entanglement of the pair 0&3 is still observed for this partial ensemble. The interaction of each photon with optical devices is local (according to QFT in the specific sense implemented by microcausality). What's nonlocal in a specific quantum sense is the correlation of the entangled pairs (0&1 as well as 2&3 for the initial total ensemble at the beginning of the experiment and 1&2 and 0&3 in the selected partial ensemble at the end of the experiment) described by entanglement. Nothing is contradicting standard QFT, nothing is faster-than-light action at a distance, and nothing violates causality in any way. The teleportation or entanglement swapping also cannot achieve faster-than-light communication, because to choose the partial ensemble you have to communicate about the measurement outcome about Alice's photon pair 1&2 before Bob can know that his photons 0&3 are in fact entangled, i.e., in the state ##|\psi_{03}^- \rangle## after choosing to consider only the pairs of that specific partial ensemble, where Alice had coincident clicks of her detectors D1 and D2 and thus has ensured that her photon pair 1&2 is in the state ##|\psi_{12}^{-} \rangle##.

So indeed by (post-)selecting the states as described the two photons 0&3 are to be described by the entangled state ##|\psi_{03}^- \rangle## though they never have been interacted or where in any "causal contact" with each other.

As you also see, I always even stress that the photons 0&3 have not been entangled before selecting the partial ensemble. That's indeed the key point of the entire paper, successfully aiming at entanglement swapping/teleportation of entangled states.

Last but not least there's no way to clone a quantum state. That's also true in the experiment here: The subensemble choosen by A is described by ##|\Psi' \rangle##. This clearly shows that in this subensemble the pairs 0&1 and 2&3 are NOT entangled at all. So by the very measurement enable entanglement swapping you destroy the original entanglement of these pairs 0&1 and 2&3. The correct term is not cloning but teleportation, as it is written in the paper. That's no surprise since Zeilinger and his group was indeed the first who demonstrated teleportation (in a somewhat simpler experiment) in the 1990ies. I'm sure you find the papers cited in his APS centennial RMP contribution you quoted above.
 

DrChinese

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1. I think it's impossible to discuss this with you. ...

2. ... nothing is faster-than-light action at a distance, and nothing violates causality in any way. ...

3. Last but not least there's no way to clone a quantum state. ...
1. Yes, and we may as well stop now as we are talking past each other. You have yet to cite a statement matching any of your [wrong] statements other than quoting yourself:

- Swapping does not change the state of photons 0 & 3 from uncorrelated to entangled. [incorrect]
- QM/QFT is causal. [incorrect]
- The authors agree with you and not with me. [incorrect]
etc.

2. This is exactly opposite of every swapping paper. There is no causal direction, and the action is quantum nonlocal.

3. What is "cloned" is a superposition. Particle 3 will now contain more information about 0 (which it never interacted with) than any preparation could otherwise do (unless it's entangled).

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

I have already quoted chapter and verse over and over again from the cited papers, and explained that actual experiments verify everything I said (per my post #57, 6F example) that clearly proves that swapping changes the statistical outcomes. I have nothing left to add on the matter, and will not respond further. So I will simply say "cheers" and move on. :smile:
 
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1. No, absolutely not possible if 1 & 2 are maximally entangled (which they are when exiting the PDC crystal). You can entangle N number of quantum particles (no specific limit), but they will not be maximally entangled in that case. In fact, PDC occasionally produces 4 entangled photons - but again, no 2 are maximally entangled.
I think part of the confusion comes from the way the word entanglement is used. Saying 1&4 are entangled seems to imply that it's some objective property of qubits 1&4, but it's only a property of the observer's view of 1&4. Before measuring 2&3, he doesn't know how to see 1&4 as entangled, so he says "they are not entangled". After measuring he does, so he says they are. Then if he throws away that information he gained, they "become unentangled" again. Nothing about 1&4 is changing, just his perspective.

To empathise the point, If another observer ##(O_B)## is casually disconnected from the 2&3 measuring observer ##(O_A)##, then 1&4 will never be entangled according to ##O_B##. Nothing ##O_A## can do will change that.

So entanglement is really about "known" or "knowable" correlations and the monogamy principle says that it's impossible to simultaneously see 1&2 and 1&4 as fully entangled – no view is compatible with both.

2. Post Bell, local realism has been roundly excluded along with most variations of same (depending on your particular exact definition, of course: local causality, local determinism).
It's possible to show step by step how Bell inequalities can be violated with local operations if you invoke many worlds. His theorem assumes single outcomes – that's why it doesn't apply. The interesting question about this experiment is if the entanglement swapping (not the CHSH test) can be explained locally within a single world.
 
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2. ... nothing is faster-than-light action at a distance, and nothing violates causality in any way. ...
- QM/QFT is causal. [incorrect]


2. This is exactly opposite of every swapping paper. There is no causal direction, and the action is quantum nonlocal.
As interesting as this discussion is I think (like an earlier poster) that it would benefit from some terminology clarification as it seems to me that the terms "microcausality" and "local causality" are still being interpreted differently by vanhees71 and DrChinese and for different aspects in both QFT and the experiments .

In fact there is one important additional subtlety in QFT with respect to Bell's mathematization of locality and causality, that is given by the use of the indefinite signature to discern different types of intervals, while in Bell's setting this distinction about which type of intervals should the nonlocality be attributed to gets lost.
And so in QFT microcausality is as much a nonlocal condition(on spacelike-separated operators) as can be interpreted as a local causal condition for timelike ones(time-ordered products) without contradiction wheras in Bell's formulation not having this distinction,
 
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Continued) there is no way to have both quantum nonlocality and causal locality without contradiction, mathematically nonlocality must violate Bell's inequalities based on causal locality.
 

DrChinese

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1. Nothing about 1&4 is changing... the monogamy principle says that it's impossible to simultaneously see 1&2 and 1&4 as fully entangled – no view is compatible with both.

2. It's possible to show step by step how Bell inequalities can be violated with local operations if you invoke many worlds.
1. Logically, monogamy should convince you that a physical change to the state of 1 & 4 must occur along with the swap. The statistics change depending on whether the system of 1 & 2 is allowed to interact with the system of 3 & 4. If they interact, then 1 & 4 well definitely become entangled as one of the four possible Bell states.

2. I have no disagreement with MWI or any other generally accepted interpretation. (I realize that MWI claims to be local, but it would be difficult to refer to it as causal in any meaningful manner.)
 
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Microcausality is not a restriction on QFT.
I'm not sure what you mean by this. The fact that quantum field operators at spacelike separated events have to commute, which is what "microcausality" is being used to mean in this discussion, most certainly is a "restriction" on QFT, since QFT satisfies it.

The change (resulting from the swap action) to photons 0 & 3 is quantum nonlocal, and is not subject to a microcausality constraint.
Are the events in question spacelike separated? If so, they have to commute--what happens at them cannot depend on the order in which they happen. And that is a microcausality constraint--see above.

Relativity does not factor in other than everyone agrees there is no possibility of sending a message FTL.
Why is this true? It's because spacelike separated measurements commute. So microcausality is a factor.

QM/QFT is causal. [incorrect]
I think you need to clarify what you mean by "causal". See above.
 

DrChinese

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1. I think you need to clarify what you mean by "causal". See above.

2. I'm not sure what you mean by this. The fact that quantum field operators at spacelike separated events have to commute, which is what "microcausality" is being used to mean in this discussion, most certainly is a "restriction" on QFT, since QFT satisfies it.
1. vanhees71 earlier (#38*) specified that "microcausal" was a normal use of the word as it applies to commuting of spacelike operators, same as you. But he then switches usage of the word "causal" in a different context. If there is a causal constraint, where A causes B, then A must precede B. In an entanglement swapping experiment, there is no clear causal direction but: A still causes B. The swap (which creates a new entangled pair - let's call that A) can occur before *or* after the entangled pair is detected (let's call that B). Ergo, there is no causal direction, and there is no description of this that contains causality as the word is normally used. That being: Cause A must precede effect B. Causality fails (if locality holds).

2. Now, if you say that "microcausal" in QFT means spacelike operators commute: well that would also mean that the order of results does *not* matter. Certainly not an issue in a swapping experiment, where the order of events does not matter (the only requirement is that the swap requires all 4 particles to be previously created). But he did not restrict his usage in that manner. He extended it to mean: local microcausality. And he further extends the meaning to say there is only quantum local action occurring. That extension is not a deduction from microcausality, and I'd again challenge anyone to provide a quote that says: QFT requires both that a) all causes must precede the effect, and b) also denies the existence of quantum nonlocal action such as swapping entails.

With entanglement swapping: There is a physical state change (if the remote systems interact), it is not quantum local, and the order of events is not relevant. Virtually every entanglement swapping paper more or less says this exact thing somewhere in the text. So it is not me that is mangling the use of the word "causality". I think I have been pretty clear in every post to indicate that I am referring to a post-Bell world, in which local causality is rejected, and it is that I object to. One should not say QM/QFT is local causal, and I think my reasoning is the norm.

If everyone agrees with my characterization of entanglement swapping, but also agrees with PeterDonis' definition of "microcausality", then we're all good.


So if someone wants to say that QFT requires that there be commuting between spacelike separated operators, fine. To me, that means they are separable. How that possibly applies to an entanglement swap, I don't know, because commuting is never mentioned in these papers - and Entangled state statistics are well different than separable Product state statistics. Entangled particles do not commute when measured, as they are not separable (the HUP applies).

This thread is about whether relativity restricts quantum actions to c. Since a swap can take previously uncorrelated spacelike separated particles and change them into a state in which the spooky action at a distance of entanglement occurs, and measurements on them are both correlated and do not commute, I would say the answer is NO. I am not saying that relativity is not relevant to QM, of course it is**. But you can't frame that as vanhees71 is doing in a post Bell world.


* In post #33, in reference to causality, vanhees71 says: "No spacelike separated events can be causally connected in any way." This is the assertion I object to, and I think he has been steady in standing by that. In the cited papers: We have spacelike separated events that depend on the decision to bring 2 independent systems together so they interact. The resulting statistics change for the newly entangled pair 0 & 3 [1 & 4 in some papers], which are in fact spacelike separated when they cease from being monogamously entangled to their birth partners. If you don't like Zeilinger et al's characterization of things, please consider the groundbreaking experiment of Henson et al, which uses swapping to simultaneously address multiple loopholes. Here the swap precedes detection, but every element is rigorously spacelike separated from the other. They confirm spooky action at a distance a la Bell, and it is not simply revealing "pre-existing" attributes. The entanglement generation (0 and 3, locations A and B) is caused by action at C on photons 1 & 2. They say:
"We generate entanglement between the two distant spins by entanglement swapping in the Barrett-Kok scheme using a third location C (roughly midway between A and B, see Fig. 1e). First we entangle each spin with the emission time of a single photon (timebin encoding). The two photons are then sent to location C, where they are overlapped on a beam-splitter and subsequently detected. If the photons are indistinguishable in all degrees of freedom, the observation of one early and one late photon in different output ports projects the spins at A and B into the maximally entangled state |ψ −> ... These detections herald the successful preparation and play the role of the event-ready signal in Bell’s proposed setup."
Nothing here about relativity preventing a causal (physical) connection between spacelike separated events.

**Keep in mind I am not a Bohmian, not that it should matter to this discussion.
 
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if you say that "microcausal" in QFT means spacelike operators commute: well that would also mean that the order of results does *not* matter
Yes. And it appears that you agree with that.

There is a physical state change (if the remote systems interact), it is not quantum local, and the order of events is not relevant.
Does "not quantum local" mean "violates the Bell inequalities"?
 
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You might be surprised to learn that entangled photons do NOT produce interference as one might otherwise expect. You must stop the entanglement first. Once you do that, the ability to carry out the rest of your idea is lost. See Figure 2 in this great summary paper by Zeilinger:

Thank you a lot for a precise remark. Indeed we will not detect interference at receive side as for entangled particles common function is sum of multiplications thus ortogonal addend (corresponding to slit path) will not give interference picture.
But what will be if we modify experement in the following way?
If 1 is transmitted then before beam arrival transmiited side measure polarization under two projections that have angle θ1.
For 0 transmissions we use axis θ0 for detection.
In all cases before arriving to double slit wall all particles will have their own independed fuction (no entanglement). In one case the beam will consist of two groups with eigenvalues for axis θ1 and in anouther from groups with eigenvalues for axis θ0. Angles are related to slit polarizers that have angels 0 and 90°.
I think that these two different cases give different interferences pictures.
 
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Entangled particles do not commute when measured
They don't? The results depend on the order in which they are measured? Bear in mind we are talking about spacelike separated measurements.
 
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If so, they have to commute--what happens at them cannot depend on the order in which they happen. And that is a microcausality constraint--see above.
Does this mean that if you apply operators that don't commute ([X, Y] 0) to a spacelike separated event (in which the order of application of the operators X and Y no longer makes sense), the opération must commute (The same result is obtained regardless of the order in which the operators are applied)?

/Patrick
 
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operators that don't commute ([X, Y] 0)
Operators that don't commute when applied to the same individual particle. But here we are talking about two spacelike separated operators applied to two different particles. Not the same thing.
 
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Operators that don't commute when applied to the same individual particle. But here we are talking about two spacelike separated operators applied to two different particles. Not the same thing.
Ok. It is about tensor product of operators (in particular here two spacelike separated operators applied on entangled (which is considered inseparable) particles or not entangled (factorizable into a tensor product of the state of the two particles) ) ? Thus the tensor product of these operators must commute [X1 x I2, I1 x Y2] = 0?

Not being a specialist in the field of quantum physics, it is a matter of understanding.

/Patrick
 
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vanhees71

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1. Yes, and we may as well stop now as we are talking past each other. You have yet to cite a statement matching any of your [wrong] statements other than quoting yourself:

- Swapping does not change the state of photons 0 & 3 from uncorrelated to entangled. [incorrect]
- QM/QFT is causal. [incorrect]
- The authors agree with you and not with me. [incorrect]
etc.

2. This is exactly opposite of every swapping paper. There is no causal direction, and the action is quantum nonlocal.

3. What is "cloned" is a superposition. Particle 3 will now contain more information about 0 (which it never interacted with) than any preparation could otherwise do (unless it's entangled).

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

I have already quoted chapter and verse over and over again from the cited papers, and explained that actual experiments verify everything I said (per my post #57, 6F example) that clearly proves that swapping changes the statistical outcomes. I have nothing left to add on the matter, and will not respond further. So I will simply say "cheers" and move on. :smile:
Ad 1) You always claim I'm saying the opposite of what I said. I said by projection you change the state of photons 0&3 to be entangled. You claim the opposite. Why are you doing this. Is it, because I'm a bit more careful in formulating it? I say

The partial ensemble chosen by the measurement of Alice (in short the "projection") is described by an entangled state of photons 0&3.

So indeed the state is changed but, there's no action at a distance it's just the selection of a partial ensemble out of the full ensemble.

Ad 2) QM/QFT is causal by construction. Read a textbook, before you claim esoteric ideas!

Ad 3) There's nothing cloned. It's a mathematical theorem that cloning is impossible. Also this you can read in many QM textbooks (e.g., Ballentine).

What do you mean by "particle 3 will now contain more information about 0"? First of all it's a photon, not a particle. The information about a system is described by the state the system is prepared in.
 

vanhees71

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They don't? The results depend on the order in which they are measured? Bear in mind we are talking about spacelike separated measurements.
It's of course utter nonsense. The very experiment we discuss here proves him wrong. The temporal order of the measurements on the pairs 0&3 and 1&2 to project to the state ##|\psi_{03}^- \rangle \otimes |\psi_{12}^- \rangle## is irrelevant for this description. This was explicitly shown by this experiment. I don't know, why @DrChinese is claiming the opposite (sometimes he agrees).
 

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