Entanglement swapping and Bohmian mechanics

[Morbert]

I understand them just fine. Mjelva’s paper opened my eyes to some of the conflicts between experiment and various interpretations. Your arguments essentially follow the same line as his. Of course, he also makes several other critical mistakes, not directly present in your mathematical development.

And you think there are these conflicts because you don't understand basic operational quantum theory, which readily reproduces the statistics of these experiments without recourse to retrocausal entanglement swapping.

 

[martinbn]

1. Your understanding of Entanglement Swapping is incorrect. When Swapping is enabled via BSM on photons 2/3, each and every 1/4 pair becomes entangled in one of 4 Bell states. Using current technology: Only half (or so) can be identified as belonging to a specific subset, and those (and only those) demonstrate Entangled State correlations. Those subsets are identified using specific criteria chosen in advance.

I agree with this. But you don't understand that the full ensemble is not entangled and shows no non-classical correlations. Only when you look at the subensembles determined by those four Bell state outcomes you see non-c;assical correlations.

2. When Swapping is NOT enabled on photons 2/3, there is no entanglement of any 1/4 pairs. Using the same criteria as for identifying the Entangled subsets, there is no correlation at all.

No, there isn't entanglement of the 1/4 pairs, just like in the swap case. I am not sayning that there is. You still fail to grasp that you need to look at subsets.

3. Also when Swapping is NOT enabled on photons 2/3 : There is absolutely no objective criteria under which non-classical correlations will appear. Product State experiments produce Product State statistics. I.e. there are no identifiable subsets of the type you are attempting to describe.

There is no objective criteria, but such subsets exist.

You keep posting as if what you say is so obvious, it needs no support. If you are correct, a suitable quote should be easy to locate. OTOH: Megidish: "The first and last photons, that did not share between them any correlations, become entangled." See their Fig. 3. "In order to fully characterize the first and last photons’ state, a quantum state tomography (QST) procedure is required."

Yes, what I post is obviously correct. No, I cannot give you a quote from a paper that shows that you have a misunderstanding. There are no such papers. Any QM text book explains how QM works. Any paper on swapping assumes basic knowledge of QM, none will explain what you want me to quote.

 

[Sambuco]

When Swapping is enabled via BSM on photons 2/3, each and every 1/4 pair becomes entangled in one of 4 Bell states.

This is obviously true, but we must remember that there is no quantum state we can assign to a system unless we know how it was prepared. In the case of photons 1 and 4, if we consider their initial preparation at the start of the experiment, each pair is in a non-entangled (product) state, whereas if we take the swap performed on photons 2 and 3 as the "preparation", then each pair of photons 1 and 4 is in an entangled state corresponding to the Bell state measured on photons 2 and 3. Both statements are true simultaneously. The fact that we can treat the swap performed on photons 2 and 3 as a form of "future preparation" for photons 1 and 4, thereby assigning them an entangled state, does not in any way mean that the swap physically affects photons 1 and 4. As clearly stated in Ma's paper:

"If one viewed the quantum state as a real physical object, one could get the paradoxical situation that future actions seem to have an influence on past and already irrevocably recorded events. However, there is never a paradox if the quantum state is viewed as no more than a `catalogue of our knowledge'."

In light of this, it is obvious that what @martinbn says is correct. As also noted in Ma's paper, the initial quantum state can be rewritten (in the Bell state basis) as:

$\ket{\psi}_{1234} = \frac{1}{2} (\ket{\psi^+}_{14} \otimes \ket{\psi^+}_{23} - \ket{\psi^-}_{14} \otimes \ket{\psi^-}_{23} - \ket{\phi^+}_{14} \otimes \ket{\phi^+}_{23} + \ket{\phi^-}_{14} \otimes \ket{\phi^-}_{23})$

which means that the outcomes obtained by Alice and Bob on photons 1 and 4, respectively, can be separated into four sets exhibiting Bell's inequalities violations. Victor's decision to perform the swap on photons 2 and 3 physically changes the state of photons 2 and 3, allowing us to identify (postselect) the subsets of photons 1 and 4 that will show violation of Bell's inequalities. The authors are very clear about this point:

"What, however, is important is to relate the lists of Alice, Bob and Victor's measurement results. On the basis of Victor's measurement settings and results, Alice and Bob can group their earlier and locally totally random results into subsets that each have a different meaning and interpretation. This formation of subsets is independent of the temporal order of the measurements."

As @martinbn said, if Víctor does not perform the swap, the measurement results for photons 1 and 4 will continue to show subsets that violate Bell's inequalities (as concluded from the equation I wrote above), but there will be no "preparation" that allows them to be distinguished beforehand.

Lucas.

 

[DrChinese]

... you don't understand basic operational quantum theory, which readily reproduces the statistics of these experiments without recourse to retrocausal entanglement swapping.

Ahhh, not so fast. Basic operational quantum theory DOES reproduce the statistics of these experiments. It is completely silent regarding retrocausality, instantaneous action at a distance, etc.

 

[Morbert]

Ahhh, not so fast. Basic operational quantum theory DOES reproduce the statistics of these experiments. It is completely silent regarding retrocausality, instantaneous action at a distance, etc.

Yep. If you want to read the calculations in a retrocausal way, you can, but you don't have to.

 

[DrChinese]

1. This is obviously true, but we must remember that there is no quantum state we can assign to a system unless we know how it was prepared. In the case of photons 1 and 4, if we consider their initial preparation at the start of the experiment, each pair is in a non-entangled (product) state, ...


2. ... whereas if we take the swap performed on photons 2 and 3 as the "preparation", then each pair of photons 1 and 4 is in an entangled state corresponding to the Bell state measured on photons 2 and 3. Both statements are true simultaneously.


3. The fact that we can treat the swap performed on photons 2 and 3 as a form of "future preparation" for photons 1 and 4, thereby assigning them an entangled state, does not in any way mean that the swap physically affects photons 1 and 4. As clearly stated in Ma's paper:

"If one viewed the quantum state as a real physical object, one could get the paradoxical situation that future actions seem to have an influence on past and already irrevocably recorded events. However, there is never a paradox if the quantum state is viewed as no more than a `catalogue of our knowledge'."


4. In light of this, it is obvious that what @martinbn says is correct.


5. Victor's decision to perform the swap on photons 2 and 3 physically changes the state of photons 2 and 3, allowing us to identify (postselect) the subsets of photons 1 and 4 that will show violation of Bell's inequalities. The authors are very clear about this point:
"What, however, is important is to relate the lists of Alice, Bob and Victor's measurement results. On the basis of Victor's measurement settings and results, Alice and Bob can group their earlier and locally totally random results [of photons 1 and 4] into subsets that each have a different meaning and interpretation. This formation of subsets is independent of the temporal order of the measurements."


6. ...if Víctor does not perform the swap, the measurement results for photons 1 and 4 will continue to show subsets that violate Bell's inequalities (as concluded from the equation I wrote above)...


7. but there will be no "preparation" that allows them to be distinguished beforehand.

Lucas.

1. Fine.


2. If you consider the preparation to be after/during the overlap at the Beam Splitter: then yes, I agree.


3. There is no such thing as "future preparation" in this context. You are just making this up. There is an initial preparation (either 1. or 2. as we agreed). There is an independent variable - to swap or not. My entire point - which really doesn't matter to a discussion of Bohmian Mechanics - is that there is no initial connection between photons 2 & 3 and/or any other photon in the entire universe. From an earlier post of mine:

---

Now, a little thought will explain why I vehemently reject Morbert's expansion as meaningless - even if it appears to be sound from an algebraic point of view. There is initially no connection between the 1&2 pair and the 3&4 pair. So they cannot be in a combined Bell state. And in fact, they are no more in a Bell state with each other than any other entangled photon pairs in the entire universe. We could just as easily start with another pair, photons 5&6, prepared as the others. This experimental permutation is experimentally realizable. Then we get (all unquestionably and simultaneously true by your own logic per 1.):

|Ψ〉1234 = |Ψ−〉12⨂|Ψ−〉34
|Ψ〉1256 = |Ψ−〉12⨂|Ψ−〉56
|Ψ〉5634 = |Ψ−〉56⨂|Ψ−〉34

and end up with the following using Morbert's expansion on each of these:

i) |Ψ〉1234 = 1/2(|Ψ+〉14⨂|Ψ+〉23 − |Ψ−〉14⨂|Ψ−〉23 − |Φ+〉14⨂|Φ+〉23 + |Φ−〉14⨂|Φ−〉23)
ii) |Ψ〉1256 = 1/2(|Ψ+〉16⨂|Ψ+〉25 − |Ψ−〉16⨂|Ψ−〉25 − |Φ+〉16⨂|Φ+〉25 + |Φ−〉16⨂|Φ−〉25)
iii) |Ψ〉5634 = 1/2(|Ψ+〉54⨂|Ψ+〉63 − |Ψ−〉54⨂|Ψ−〉63 − |Φ+〉54⨂|Φ+〉63 + |Φ−〉54⨂|Φ−〉63)

None of these last 3 have any physical significance. And they cannot be physically simultaneously true. (Hopefully that is obvious by MoE and needs no further explanation.) You need to first overlap photons in a BSM apparatus to give it physical significance.

---

4. What @martinbn says that is 100% wrong: There are no defined subsets (4-fold) in the Product State (Separable) results that exhibit non-classical correlations. This is not really a serious point of debate. This was demonstrated experimentally in the references. If there were such a subset, it would be news to a lot of experimenters around the world. If you want to see a paper that attempts - unsuccessfully - to locate hidden correlations in Bell test data sets, try this one from Huw Price (2024).


5. This is, of course, precisely correct. It's everything I have been saying.


6. This doesn't even make sense as a statement... "show"? You are assuming that which cannot be proved. The only subsets that you can "show" from 2 fold coincidences of photons 1 and 4 are |HH>, |VV>, |HV>, |VH>. You can make of those 4 subsets whatever you like, but non-classical correlations is not one of those things. They do not represent any connection whatsoever between the initial pairs 1&2 and 3&4. I can take any 2 lists of random numbers and group them by odd and even, and I will get 4 subsets too. Similarly, they don't "show" us anything meaningful.

Importantly: Just as the BSM changes the state of photons 2 & 3 (agreeing with you on this point): We cannot actually make a convincing statement that it does not also change the state of photons 1 & 4. You can, of course, make any assumption you like either way. There's no experimental proof. But the fact is:

Initially: |Ψ〉14 = |Ψ〉1⨂|Ψ〉4
After a detection of photons 2 and 3 as |Ψ−〉23: |Ψ〉12 = |Ψ-〉14
To the untrained eye, one might conclude that photons 1 and 4 changed states due to the BSM. Note that I am not making an assertion about what "really" happens. I don't know.


7. That's true whether or not there is a BSM. Only 4-fold coincidences are reported as meaningful. And the results are conditioned on the experimenter's choice of BSM/SSM, and nothing else.


Can we please return to a discussion of Bohmian Mechanics...

 

[DrChinese]

Yep. If you want to read the calculations in a retrocausal way, you can, but you don't have to.

Woo hoo! We see totally eye to eye!!

 

[DrChinese]

1. There is no objective criteria, but such _________ exist.

2. Yes, what I post is obviously correct. No, I cannot give you a quote from a paper that shows that you have a misunderstanding. There are no such papers.

1. I replaced your word "subsets" with a blank for anyone to fill in something they believe in. Possible alternatives: tooth fairy, Santa Claus, leprechauns... you get the idea. Exactly as you say, and with which I agree: There is no objective criteria to support your statement.


2. And here we agree as well: "There are no such papers." Obviously correct.


Can we return to Bohmian Mechanics?

 

[Sambuco]

Can we please return to a discussion of Bohmian Mechanics...

I am addressing your points because I think they are indeed relevant to how Bohmian mechanics explains the DCES experiment. I will focus on the disagreements that I believe are the most important.

We could just as easily start with another pair, photons 5&6, prepared as the others. This experimental permutation is experimentally realizable. Then we get (all unquestionably and simultaneously true by your own logic per 1.):

|Ψ〉1234 = |Ψ−〉12⨂|Ψ−〉34
|Ψ〉1256 = |Ψ−〉12⨂|Ψ−〉56
|Ψ〉5634 = |Ψ−〉56⨂|Ψ−〉34

and end up with the following using Morbert's expansion on each of these:

i) |Ψ〉1234 = 1/2(|Ψ+〉14⨂|Ψ+〉23 − |Ψ−〉14⨂|Ψ−〉23 − |Φ+〉14⨂|Φ+〉23 + |Φ−〉14⨂|Φ−〉23)
ii) |Ψ〉1256 = 1/2(|Ψ+〉16⨂|Ψ+〉25 − |Ψ−〉16⨂|Ψ−〉25 − |Φ+〉16⨂|Φ+〉25 + |Φ−〉16⨂|Φ−〉25)
iii) |Ψ〉5634 = 1/2(|Ψ+〉54⨂|Ψ+〉63 − |Ψ−〉54⨂|Ψ−〉63 − |Φ+〉54⨂|Φ+〉63 + |Φ−〉54⨂|Φ−〉63)

None of these last 3 have any physical significance. And they cannot be physically simultaneously true. (Hopefully that is obvious by MoE and needs no further explanation.)

Your statement regarding the monogamy of entanglement is just wrong. Any of the three states you wrote is correct, not only from an algebraic point of view but also from a physical one. None of those states exhibits entanglement between the photons of pair (1, 2) and photons from other pairs. In the past, I have shared several references with you on how to handle entangled states, including textbooks such as Zweibach’s (specifically chapter 18) so, given that your claim regarding entanglement between photons in the pair (1, 2) and photons from other pairs is wrong and contradicts what is stated in textbooks, I ask that you provide a reference to support your assertion. Since you are so certain that what you say is true, it should be very easy for you to find a reference demonstrating that type of entanglement. Until you do so, your claim will remain incorrect because it contradicts what is written in the references I have previously provided.

Just as the BSM changes the state of photons 2 & 3 (agreeing with you on this point): We cannot actually make a convincing statement that it does not also change the state of photons 1 & 4.

This is not true! All the interpretations I am aware of (many-worlds, bohmian mechanics, QBism, RQM, Copenhagen and so on) agree that the swap involving photons 2 and 3 does not physically change the state of photons 1 and 4. Indeed, as said in Ma's paper:

"If one viewed the quantum state as a real physical object, one could get the paradoxical situation that future actions seem to have an influence on past and already irrevocably recorded events. However, there is never a paradox if the quantum state is viewed as no more than a `catalogue of our knowledge'."

This doesn't even make sense as a statement... "show"? You are assuming that which cannot be proved. The only subsets that you can "show" from 2 fold coincidences of photons 1 and 4 are |HH>, |VV>, |HV>, |VH>. You can make of those 4 subsets whatever you like, but non-classical correlations is not one of those things.

This is also wrong. As I have cited, Ma's paper states that the measurement results for photons 1 and 4 have been "irrevocably recorded". What distinguishes the swap case from the case where no swap takes place is that, only in the swap scenario, we possess the information needed to reorder the data into four sets that demonstrate a violation of Bell's inequalities. The swap does not magically generate those results. One more time: those results have been "already irrevocably recorded", just as Ma's paper states. There is no debate on this point unless you wish to dispute what Ma's paper says, in which case you would need to provide a reference to support your claim.

All of this is highly relevant to the Bohmian interpretation of the experiments, given that you seem to believe that after the swap, the physical state of the photon pair (1, 4) changes due to the swap, which is wrong, as said by the authors in Ma's paper. Indeed, as I have mentioned in previous posts, in Bohmian mechanics there is a single wave function guiding the particles, and it evolves only forward in time. Consequently, the entanglement between photons 1 and 4 is as experimentally observed (violation of Bell inequalities) real, yet not ontic, since that wave function is not the one guiding the particles from the source to the detectors.

Lucas.

 

[martinbn]

1. I replaced your word "subsets" with a blank for anyone to fill in something they believe in. Possible alternatives: tooth fairy, Santa Claus, leprechauns... you get the idea. Exactly as you say, and with which I agree: There is no objective criteria to support your statement.


2. And here we agree as well: "There are no such papers." Obviously correct.

Ok, this is a good place for me to stop.

 

[DrChinese]

1. Your statement regarding the monogamy of entanglement is just wrong. Any of the three states you wrote is correct, not only from an algebraic point of view but also from a physical one. None of those states exhibits entanglement between the photons of pair (1, 2) and photons from other pairs.

2. In the past, I have shared several references with you on how to handle entangled states, including textbooks such as Zweibach’s (specifically chapter 18)...

3. Since you are so certain that what you say is true, it should be very easy for you to find a reference demonstrating that type of entanglement. Until you do so, your claim will remain incorrect because it contradicts what is written in the references I have previously provided.

4. This is not true! All the interpretations I am aware of (many-worlds, bohmian mechanics, QBism, RQM, Copenhagen and so on) agree that the swap involving photons 2 and 3 does not physically change the state of photons 1 and 4.
Indeed, as said in Ma's paper:

"If one viewed the quantum state as a real physical object, one could get the paradoxical situation that future actions seem to have an influence on past and already irrevocably recorded events. However, there is never a paradox if the quantum state is viewed as no more than a `catalogue of our knowledge'."

5. What distinguishes the swap case from the case where no swap takes place is that, only in the swap scenario, we possess the information needed to reorder the data into four sets that demonstrate a violation of Bell's inequalities. The swap does not magically generate those results.

6. ...in Bohmian mechanics there is a single wave function guiding the particles, and it evolves only forward in time [FITO] ...

Lucas.

1. That's what I've been saying all along: "None of those states exhibits entanglement between the photons of pair (1, 2) and photons from other pairs." You must first perform a BSM, which can produce one of the entangled states below:

i) |Ψ〉1234 = 1/2(|Ψ+〉14⨂|Ψ+〉23 − |Ψ−〉14⨂|Ψ−〉23 − |Φ+〉14⨂|Φ+〉23 + |Φ−〉14⨂|Φ−〉23)
ii) |Ψ〉1256 = 1/2(|Ψ+〉16⨂|Ψ+〉25 − |Ψ−〉16⨂|Ψ−〉25 − |Φ+〉16⨂|Φ+〉25 + |Φ−〉16⨂|Φ−〉25)
iii) |Ψ〉5634 = 1/2(|Ψ+〉54⨂|Ψ+〉63 − |Ψ−〉54⨂|Ψ−〉63 − |Φ+〉54⨂|Φ+〉63 + |Φ−〉54⨂|Φ−〉63)


2. I shouldn't need to point this out to longtime respected posters such as yourself, but... You provide no quote. You provide no link to the source material. You reference a chapter in a textbook I don't own, and I doubt most readers own either.


3. Here's how a suitable quote supporting my 1. should work:

Ma et al: "In entanglement swapping, two pairs of entangled photons 1&2 and 3&4 are each produced in the antisymmetric polarization-entangled Bell singlet state such that the total four-photon state has the form|Ψ〉1234 = |Ψ−〉12⨂|Ψ−〉34, (1)where |Ψ−〉12 = (|𝐻〉1|𝑉〉2 − |𝑉〉1|𝐻〉2)/√2 and likewise for |Ψ−〉34. ... As schematically shown in Fig. 1, if Victor subjects his photons 2 and 3 to a Bell-state measurement, they become entangled. Consequently photons 1 (Alice) and 4(Bob) also become entangled* and entanglement swapping is achieved. ... Note that after the entanglement swapping, photons 1&2(and 3&4) are not entangled with each other anymore, which manifests the monogamy of entanglement."


4. Agreed, most interpretations say exactly as you state, including Bohmian Mechanics. And certainly, in the delayed choice DCES case, the idea of changing the past seems unthinkable. As you say, those results are "already irrevocably recorded" (and thanks for a good quote here!). What I'm saying is merely that there is no experimental evidence that is that 1 & 4 don't physically change. I'm not asserting anything more. Ma says: "Consequently photons 1 (Alice) and 4(Bob) also become entangled..." "Consequently" here means: "following and as a result of".

Even the quote you presented references this point: "there is never a paradox if the quantum state is viewed as [i.e. assumed to be] no more than a `catalogue of our knowledge'." In other words: As long as you assume 1 & 4 aren't changed by a BSM, then 1 & 4 aren't changed by a BSM. It's a reasonable assumption, but of course, circular. Assumptions don't prove anything by themselves. They never actually say that 1 & 4 aren't changed by the BSM. In fact, since you asked for a quote to support my comment, Ma also says: "This can also be viewed as 'quantum steering into the past'." Personally, I'm not pushing their view on that one way or the other.


5. Some of this is incorrect as you have written it. The information (4-fold coincidences) to sort the data into comparable subsets is present equally in the swap case (Ma, Fig. 3a.) and the no swap case (Ma, Fig. 3.b). What you say that is correct: Only the swap case shows non-classical correlations. The no swap case doesn't. Obviously, the swap does "magically generate those results". It's reported experimental fact, what's there to question here?


6. Finally, something related to the Bohmian interpretation! As I mentioned previously, I am willing to accept whatever elements [forward in time only, FITO] you say are part of BM for discussion purposes. I am asking for clarity regarding BM, as different authors on the subject sometimes make apparently differing statements.


*To be clear: "become entangled" means their state changed because of Victor's remote action. Before, 1 & 4 were not entangled with each other.

 

[Morbert]

Two steps forward three steps back.

1. That's what I've been saying all along: "None of those states exhibits entanglement between the photons of pair (1, 2) and photons from other pairs." You must first perform a BSM, which can produce one of the entangled states below:

i) |Ψ〉1234 = 1/2(|Ψ+〉14⨂|Ψ+〉23 − |Ψ−〉14⨂|Ψ−〉23 − |Φ+〉14⨂|Φ+〉23 + |Φ−〉14⨂|Φ−〉23)
ii) |Ψ〉1256 = 1/2(|Ψ+〉16⨂|Ψ+〉25 − |Ψ−〉16⨂|Ψ−〉25 − |Φ+〉16⨂|Φ+〉25 + |Φ−〉16⨂|Φ−〉25)
iii) |Ψ〉5634 = 1/2(|Ψ+〉54⨂|Ψ+〉63 − |Ψ−〉54⨂|Ψ−〉63 − |Φ+〉54⨂|Φ+〉63 + |Φ−〉54⨂|Φ−〉63)

None of those three states i) ii) iii) are produced by a BSM. And i) is the prepared state before a BSM. Ma explicitly says i) is the prepared state rewritten in a specific basis.

 

[Sambuco]

I will split the post into two parts so as not to conflate the points of disagreement with your final point regarding the Bohmian interpretation.

1. That's what I've been saying all along: "None of those states exhibits entanglement between the photons of pair (1, 2) and photons from other pairs." You must first perform a BSM, which can produce one of the entangled states below:

i) |Ψ〉1234 = 1/2(|Ψ+〉14⨂|Ψ+〉23 − |Ψ−〉14⨂|Ψ−〉23 − |Φ+〉14⨂|Φ+〉23 + |Φ−〉14⨂|Φ−〉23)
ii) |Ψ〉1256 = 1/2(|Ψ+〉16⨂|Ψ+〉25 − |Ψ−〉16⨂|Ψ−〉25 − |Φ+〉16⨂|Φ+〉25 + |Φ−〉16⨂|Φ−〉25)
iii) |Ψ〉5634 = 1/2(|Ψ+〉54⨂|Ψ+〉63 − |Ψ−〉54⨂|Ψ−〉63 − |Φ+〉54⨂|Φ+〉63 + |Φ−〉54⨂|Φ−〉63)

Those states do not exhibit entanglement between photons from different pairs, which can be demonstrated trivially:

$\ket{\psi}_{1234} = \frac{1}{4} (\ket{\psi^+}_{14} \ket{\psi^+}_{23} - \ket{\psi^-}_{14} \ket{\psi^-}_{23} - \ket{\phi^+}_{14} \ket{\phi^+}_{23} + \ket{\phi^-}_{14} \ket{\phi^-}_{23})$
$\ket{\psi}_{1234} = \frac{1}{4} [(\ket{\psi^+}_{14} + \ket{\psi^-}_{14}) (\ket{\psi^+}_{23} - \ket{\psi^-}_{23}) - (\ket{\phi^+}_{14} + \ket{\phi^-}_{14}) (\ket{\phi^+}_{23} - \ket{\phi^-}_{23}) - (\ket{\phi^+}_{14} - \ket{\phi^-}_{14}) (\ket{\phi^+}_{23} + \ket{\phi^-}_{23}) + (\ket{\psi^+}_{14} - \ket{\psi^-}_{14}) (\ket{\psi^+}_{23} + \ket{\psi^-}_{23})]$
$\ket{\psi}_{1234} = \frac{1}{2} (\ket{H}_1 \ket{V}_4 \ket{V}_2 \ket{H}_3 - \ket{H}_1 \ket{H}_4 \ket{V}_2 \ket{V}_3 - \ket{V}_1 \ket{V}_4 \ket{H}_2 \ket{H}_3 + \ket{V}_1 \ket{H}_4 \ket{H}_2 \ket{V}_3)$
$\ket{\psi}_{1234} = \frac{1}{2} (\ket{H}_1 \ket{V}_2 \ket{H}_3 \ket{V}_4 - \ket{H}_1 \ket{V}_2 \ket{V}_3 \ket{H}_4 - \ket{V}_1 \ket{H}_2 \ket{H}_3 \ket{V}_4 + \ket{V}_1 \ket{H}_2 \ket{V}_3 \ket{H}_4)$
$\ket{\psi}_{1234} = \frac{1}{\sqrt{2}} (\ket{H}_1 \ket{V}_2 - \ket{V}_1 \ket{H}_2) \otimes \frac{1}{\sqrt{2}}(\ket{H}_3 \ket{V}_4 - \ket{V}_3 \ket{H}_4)$
$\ket{\psi}_{1234} = \ket{\psi^-}_{12} \otimes \ket{\psi^-}_{34}$

As you can see, the state i) you wrote is separable, in the sense that photons 1 and 2 are not entangled with photons 3 and 4. The same applies to states ii) and iii). Do you have any question about this?

2. I shouldn't need to point this out to longtime respected posters such as yourself, but... You provide no quote. You provide no link to the source material. You reference a chapter in a textbook I don't own, and I doubt most readers own either.

Whenever necessary, I provided the relevant quotes, and in the case of Zweibach's book, I even transcribed excerpts from the text demonstrating that what I was saying was correct according to a textbook! Of course, you are not obliged to follow that specific book. As I mentioned in my previous post, you could cite any other paper/book that supports your claims, yet you fail to do so. You keep repeating that the states i), ii), and iii) exhibit entanglement that isn't actually there. I have just mathematically demonstrated that this is incorrect. If you believe there is an error in my proof, point it out.

3. Here's how a suitable quote supporting my 1. should work:

Ma et al: "In entanglement swapping, two pairs of entangled photons 1&2 and 3&4 are each produced in the antisymmetric polarization-entangled Bell singlet state such that the total four-photon state has the form|Ψ〉1234 = |Ψ−〉12⨂|Ψ−〉34, (1)where |Ψ−〉12 = (|𝐻〉1|𝑉〉2 − |𝑉〉1|𝐻〉2)/√2 and likewise for |Ψ−〉34. ... As schematically shown in Fig. 1, if Victor subjects his photons 2 and 3 to a Bell-state measurement, they become entangled. Consequently photons 1 (Alice) and 4(Bob) also become entangled* and entanglement swapping is achieved. ... Note that after the entanglement swapping, photons 1&2(and 3&4) are not entangled with each other anymore, which manifests the monogamy of entanglement."

*To be clear: "become entangled" means their state changed because of Victor's remote action. Before, 1 & 4 were not entangled with each other.

We all agree on that. The problem is that you believe the "state" of photons 1 and 4 that "becomes entangled" represents a kind of real/ontic physical property of those photons, whereas the authors of the paper are very clear on this point: "the quantum state is viewed as no more than a `catalogue of our knowledge'". In other words, when the state of photons 1 and 4 becomes entangled, the only thing that changes is the information the observer possesses regarding the measurements performed by Alice and Bob, whose results cannot change as a result of the swap, a point clearly stated in Ma's paper: "If one viewed the quantum state as a real physical object, one could get the paradoxical situation that future actions seem to have an influence on past and already irrevocably recorded events".

What I'm saying is merely that there is no experimental evidence that is that 1 & 4 don't physically change. I'm not asserting anything more. Ma says: "Consequently photons 1 (Alice) and 4(Bob) also become entangled..." "Consequently" here means: "following and as a result of".

The key point is that "entanglement" is a property of the quantum state describing the photons, and according to the authors, they are using the quantum state merely as a tool to obtain information about the measurements.

They never actually say that 1 & 4 aren't changed by the BSM. In fact, since you asked for a quote to support my comment, Ma also says: "This can also be viewed as 'quantum steering into the past'." Personally, I'm not pushing their view on that one way or the other.

They explicitly say that the measurement results for photons 1 and 4 do not change as a result of the swap (already irrevocably recorded), and furthermore, that the entanglement arising between photons 1 and 4 is merely a change in the information the observer possesses regarding the measurement results of those photons (catalogue of our knowledge).

5. Some of this is incorrect as you have written it. The information (4-fold coincidences) to sort the data into comparable subsets is present equally in the swap case (Ma, Fig. 3a.) and the no swap case (Ma, Fig. 3.b). What you say that is correct: Only the swap case shows non-classical correlations. The no swap case doesn't. Obviously, the swap does "magically generate those results". It's reported experimental fact, what's there to question here?

No, what I have written is what the authors state in the paper. As they show in Fig. 3, the results change depending on whether or not Victor performs the swap on photons 2 and 3. Since the outcomes of photons 1 and 4 have been "already irrevocably recorded", the only difference is how the results for photons 1 and 4 are post-selected based on the results for photons 2 and 3. This is the origin of the entanglement swapping!

Lucas.

 

[Sambuco]

6. Finally, something related to the Bohmian interpretation! As I mentioned previously, I am willing to accept whatever elements [forward in time only, FITO] you say are part of BM for discussion purposes. I am asking for clarity regarding BM, as different authors on the subject sometimes make apparently differing statements.

As I have mentioned in other posts, I believe a key point in understanding how Bohmian mechanics interprets this experiment is that the physically real quantum state that guides the particles is the one corresponding to the system's initial preparation. This is because, as you said, Bohmian mechanics considers only a forward-in-time quantum state. Consequently, the entanglement between photons 1 and 4 which is deduced by the Bell's inequality violations in each subset is not ontic. Is there any aspect of this that you find questionable?

Lucas.

 

[DrChinese]

None of those three states i) ii) iii) are produced by a BSM. And i) is the prepared state before a BSM. Ma explicitly says i) is the prepared state rewritten in a specific basis.

That's absolutely false. I can prepare all three of these states simultaneously:

|Ψ〉1234 = |Ψ−〉12⨂|Ψ−〉34
|Ψ〉1256 = |Ψ−〉12⨂|Ψ−〉56
|Ψ〉5634 = |Ψ−〉56⨂|Ψ−〉34

Which is also |Ψ〉123456 = |Ψ−〉12⨂|Ψ−〉34⨂|Ψ−〉56.

Then, later, I can select a swap to occur as i) or ii) or iii) as previously listed, each mutually exclusive. The other neglected 2 have no physical meaning, and never did of course. They were merely possibilities, dependent solely on a future decision and nothing else. Note that 6 photon (3 entangled pair) synchronization has recently (lol) been realized, see their (1)*.

Multistage Entanglement Swapping (Goebel et al, 2008)


* ... which is fairly close to what I wrote, if only I weren't so math challenged.

 

[DrChinese]

As I have mentioned in other posts, I believe a key point in understanding how Bohmian mechanics interprets this experiment is that the physically real quantum state that guides the particles is the one corresponding to the system's initial preparation. This is because, as you said, Bohmian mechanics considers only a forward-in-time quantum state. Consequently, the entanglement between photons 1 and 4 which is deduced by the Bell's inequality violations in each subset is not ontic. Is there any aspect of this that you find questionable?

Lucas.

Of course it is questionable. Anything not ruled out by experiment is questionable.

And what you vs. other folks are saying about Bohmian mechanics is inconsistent. Saying "the physically real quantum state that guides the particles is the one corresponding to the system's initial preparation" is fine, makes perfect sense to me. To me what that means is the initial state is: |Ψ〉1234 = |Ψ−〉12⨂|Ψ−〉34. But the Bohmian viewpoint is that some/many/all nonlocal particle positions contribute to the pilot wave subsequently, thereby guiding evolution into the future. That's quite different than saying it is just the initial state guiding things.

 

[DrChinese]

Those states do not exhibit entanglement between photons from different pairs...

As you can see, the state i) you wrote is separable, in the sense that photons 1 and 2 are not entangled with photons 3 and 4. The same applies to states ii) and iii). Do you have any question about this?

I agree! Photons 1 and 2 are not initially entangled with photons 3 and 4. This is what I have stated many times.

As to photons 5 and 6, see my post #555 in which I make clear my i) ii) and iii) are physically realizable, IFF a BSM is performed in the future. This scenario has already been performed experimentally (slightly differently but using my basic initial entanglement |Ψ〉123456 = |Ψ−〉12⨂|Ψ−〉34⨂|Ψ−〉56). No ticket, no laundry. There is no entanglement between 1 and 4 without a BSM, which can be performed at any time (before or after 1 and 4 are measured).

The same applies to states ii) and iii)*. Do you have any question about this?

*Again, this is not hypothetical. It has already occurred. See (1) from Multistage Entanglement Swapping (Goebel et al, 2008)

 

[Sambuco]

And what you vs. other folks are saying about Bohmian mechanics is inconsistent.

I don't think so, but it doesn't matter, let's see

Saying "the physically real quantum state that guides the particles is the one corresponding to the system's initial preparation" is fine, makes perfect sense to me. To me what that means is the initial state is: |Ψ〉1234 = |Ψ−〉12⨂|Ψ−〉34.

Exactly!

But the Bohmian viewpoint is that some/many/all nonlocal particle positions contribute to the pilot wave subsequently, thereby guiding evolution into the future. That's quite different than saying it is just the initial state guiding things.

I'm not entirely sure I understood what you said here. It seems to me that you are referring to the case where one has two entangled particles. Certainly, in that situation, the velocity of one particle at a given instant depends on the wave function (and its derivative) evaluated at the position of the other particle at that same instant. Did you mean something like that?

Lucas.

 

[Sambuco]

I agree! Photons 1 and 2 are not initially entangled with photons 3 and 4. This is what I have stated many times.

As to photons 5 and 6, see my post #555 in which I make clear my i) ii) and iii) are physically realizable, IFF a BSM is performed in the future. This scenario has already been performed experimentally (slightly differently but using my basic initial entanglement |Ψ〉123456 = |Ψ−〉12⨂|Ψ−〉34⨂|Ψ−〉56). No ticket, no laundry. There is no entanglement between 1 and 4 without a BSM, which can be performed at any time (before or after 1 and 4 are measured).

The same applies to states ii) and iii)*. Do you have any question about this?

*Again, this is not hypothetical. It has already occurred. See (1) from Multistage Entanglement Swapping (Goebel et al, 2008)

I am not sure whether we agree or not

For example, in Goebel's paper, they write the initial state (before the BSM) twice, first in eq. (1) and then again in eq. (2) using a Bell-state basis for photons 1 through 4. If we agree on this, there is nothing more to discuss.

Lucas.

 

[Demystifier]

But the Bohmian viewpoint is that some/many/all nonlocal particle positions contribute to the pilot wave

That's a nonsense. First, there is no such thing as nonlocal particle positions. All particle positions are local. Second, particle positions don't contribute to the pilot wave. The pilot wave does not depend on the actual Bohmian particle positions. Perhaps you meant something else, but it's hard to understand what you meant if you don't use a reasonably precise language.

 

[DrChinese]

1. That's a nonsense. First, there is no such thing as nonlocal particle positions. All particle positions are local.

2. Second, particle positions don't contribute to the pilot wave. The pilot wave does not depend on the actual Bohmian particle positions. Perhaps you meant something else, but it's hard to understand what you meant if you don't use a reasonably precise language.

1. Hmmm, I guess my understanding of local vs nonlocal is different than yours. By nonlocal, I mean particles sufficiently distant that an influence traveling at c would take some amount of time t>0 to propagate between them. That would be almost any 2 particles we might be discussing. One of the key points of BM is that there is nonlocal action at a distance. Particle N's evolution is influenced by other particles nonlocal to it.


2. I guess I was confused by Norsen on this point (DrC: "the Bohmian viewpoint is that some/many/all nonlocal particle positions contribute to the pilot wave"). But if I have phrased this poorly, perhaps you can help me clean it up.

Norsen: (1) .... one also has definite positions for each particle in the system. For example, for a system of N spinless nonrelativistic particles, the position X~n of the nth particle will evolve according to... (2)

They refer to this as the Pilot Wave or Guidance Equation. And it looks to me as if the system's particle positions are in fact a function of other particles' probability current and probability density. Aren't those in turn a function of particle positions? Or particle velocities? Or is the issue that particle position of some particle N is influenced by the Pilot Wave, and not the other way around? Hopefully you can translate what Norsen is saying, to the extent I have it confused.

 

[DrChinese]

I am not sure whether we agree or not

For example, in Goebel's paper, they write the initial state (before the BSM) twice, first in eq. (1) and then again in eq. (2) using a Bell-state basis for photons 1 through 4. If we agree on this, there is nothing more to discuss.

Lucas.

Their experiment is somewhat different than what I am discussing. Yes, their (2) is correct under the following condition: They choose to perform the BSM on pairs 1&2 and 3&4. They said they did just that: "As a first step we perform a joint BSM on photons 2 and 3, that is, photons 2 and 3 are projected onto one of the four Bell states. This measurement also projects photons 1 and 4 onto a Bell state, in a form depending on the result of the BSM of photons 2 and 3." Note that according to the authors, photons 1 and 4 were not previously in a Bell state relative to each other. Their state changed, and it changed remotely without any direct particle interaction.

Were the setup arranged differently, they could choose which 2 of the 3 initial pairs would be selected for a BSM. It should be obvious that all 3 of the below were not simultaneously true prior to a BSM:

i) |Ψ〉1234 = 1/2(|Ψ+〉14⨂|Ψ+〉23 − |Ψ−〉14⨂|Ψ−〉23 − |Φ+〉14⨂|Φ+〉23 + |Φ−〉14⨂|Φ−〉23)
ii) |Ψ〉1256 = 1/2(|Ψ+〉16⨂|Ψ+〉25 − |Ψ−〉16⨂|Ψ−〉25 − |Φ+〉16⨂|Φ+〉25 + |Φ−〉16⨂|Φ−〉25)
iii) |Ψ〉5634 = 1/2(|Ψ+〉54⨂|Ψ+〉63 − |Ψ−〉54⨂|Ψ−〉63 − |Φ+〉54⨂|Φ+〉63 + |Φ−〉54⨂|Φ−〉63)

On the other hand, all 3 of the below are simultaneously true in Goebel:

|Ψ〉1234 = |Ψ−〉12⨂|Ψ−〉34
|Ψ〉1256 = |Ψ−〉12⨂|Ψ−〉56
|Ψ〉5634 = |Ψ−〉56⨂|Ψ−〉34

You agree?

 

[Morbert]

That's absolutely false. I can prepare all three of these states simultaneously:

|Ψ〉1234 = |Ψ−〉12⨂|Ψ−〉34
|Ψ〉1256 = |Ψ−〉12⨂|Ψ−〉56
|Ψ〉5634 = |Ψ−〉56⨂|Ψ−〉34

Which is also |Ψ〉123456 = |Ψ−〉12⨂|Ψ−〉34⨂|Ψ−〉56.

Then, later, I can select a swap to occur as i) or ii) or iii) as previously listed, each mutually exclusive. The other neglected 2 have no physical meaning, and never did of course. They were merely possibilities, dependent solely on a future decision and nothing else. Note that 6 photon (3 entangled pair) synchronization has recently (lol) been realized, see their (1)*.

Multistage Entanglement Swapping (Goebel et al, 2008)


* ... which is fairly close to what I wrote, if only I weren't so math challenged.

You're jumping around from experiment to experiment. Do you or do you not now accept that Ma's (2) is the same as Ma's (1)? I.e. Do you now accept that Ma's (2) is the state of the prepared system, prior to any measurement?

 

[Sambuco]

Note that according to the authors, photons 1 and 4 were not previously in a Bell state relative to each other. Their state changed, and it changed remotely without any direct particle interaction.

Totally agree!

Were the setup arranged differently, they could choose which 2 of the 3 initial pairs would be selected for a BSM. It should be obvious that all 3 of the below were not simultaneously true prior to a BSM:

i) |Ψ〉1234 = 1/2(|Ψ+〉14⨂|Ψ+〉23 − |Ψ−〉14⨂|Ψ−〉23 − |Φ+〉14⨂|Φ+〉23 + |Φ−〉14⨂|Φ−〉23)
ii) |Ψ〉1256 = 1/2(|Ψ+〉16⨂|Ψ+〉25 − |Ψ−〉16⨂|Ψ−〉25 − |Φ+〉16⨂|Φ+〉25 + |Φ−〉16⨂|Φ−〉25)
iii) |Ψ〉5634 = 1/2(|Ψ+〉54⨂|Ψ+〉63 − |Ψ−〉54⨂|Ψ−〉63 − |Φ+〉54⨂|Φ+〉63 + |Φ−〉54⨂|Φ−〉63)

On the other hand, all 3 of the below are simultaneously true in Goebel:

|Ψ〉1234 = |Ψ−〉12⨂|Ψ−〉34
|Ψ〉1256 = |Ψ−〉12⨂|Ψ−〉56
|Ψ〉5634 = |Ψ−〉56⨂|Ψ−〉34

But they are the same states! To be more explicit:

i) |Ψ〉1234 = |Ψ−〉12⨂|Ψ−〉34 = 1/2(|Ψ+〉14⨂|Ψ+〉23 − |Ψ−〉14⨂|Ψ−〉23 − |Φ+〉14⨂|Φ+〉23 + |Φ−〉14⨂|Φ−〉23)
ii) |Ψ〉1256 = |Ψ−〉12⨂|Ψ−〉56 = 1/2(|Ψ+〉16⨂|Ψ+〉25 − |Ψ−〉16⨂|Ψ−〉25 − |Φ+〉16⨂|Φ+〉25 + |Φ−〉16⨂|Φ−〉25)
iii) |Ψ〉5634 = |Ψ−〉56⨂|Ψ−〉34 = 1/2(|Ψ+〉54⨂|Ψ+〉63 − |Ψ−〉54⨂|Ψ−〉63 − |Φ+〉54⨂|Φ+〉63 + |Φ−〉54⨂|Φ−〉63)

Lucas.

 

[DrChinese]

Do you now accept that Ma's (2) is the state of the prepared system, prior to any measurement?

Of course isn't. You must perform a BSM, as quoted authors clearly indicate:

Goebel: "Note that photon pairs 1-2,3-4 and 5-6 are entangled in an antisymmetric polarization state. The states of the three pairs are factorizable from each other, namely, there is no entanglement among photons from different pairs. As a first step we perform a joint BSM on photons 2 and 3, that is, photons 2 and 3 are projected onto one of the four Bell states. This [Bell State] measurement also projects photons 1 and 4 onto a Bell state, in a form depending on the result of the BSM of photons 2 and 3. Close inspection shows that for the initial state given in Eq. (1), the emerging state of photons 1 and 4 is identical to the one that photons 2 and 3 collapse into."

If we start with |Ψ〉123456 = |Ψ−〉12⨂|Ψ−〉34⨂|Ψ−〉56 - as in an actual experiment I referenced - these are simultaneously physical:

|Ψ〉1234 = |Ψ−〉12⨂|Ψ−〉34
|Ψ〉1256 = |Ψ−〉12⨂|Ψ−〉56
|Ψ〉5634 = |Ψ−〉56⨂|Ψ−〉34

Your nonphysical logic says all of the below should also be simultaneously true:

i) |Ψ〉1234 = 1/2(|Ψ+〉14⨂|Ψ+〉23 − |Ψ−〉14⨂|Ψ−〉23 − |Φ+〉14⨂|Φ+〉23 + |Φ−〉14⨂|Φ−〉23)
ii) |Ψ〉1256 = 1/2(|Ψ+〉16⨂|Ψ+〉25 − |Ψ−〉16⨂|Ψ−〉25 − |Φ+〉16⨂|Φ+〉25 + |Φ−〉16⨂|Φ−〉25)
iii) |Ψ〉5634 = 1/2(|Ψ+〉54⨂|Ψ+〉63 − |Ψ−〉54⨂|Ψ−〉63 − |Φ+〉54⨂|Φ+〉63 + |Φ−〉54⨂|Φ−〉63)

But they clearly have no physical meaning whatsoever, because they cannot be simultaneously true. I agree that Ma's (2) is derived from (1), sure. A physical BSM operation is required to obtain a swap to one of the 4 Bell states. No ticket, no laundry. And once again note that the quantum state of the 1 and 4 photons change as a result of the BSM on photons 2 and 3.

 

[Morbert]

Of course isn't. You must perform a BSM, as quoted authors clearly indicate:

You keep ducking what Ma explicitly says.

Ma: "This can be seen by rewriting Eq. (1) in the basis of Bell states of photons 2 and 3"

Do you or do you not accept that Ma says that rewriting equation (1) in the bases of Bell states of photons 2 and 3 yields equation (2)?

 

[DrChinese]

1. Totally agree!


2. But they are the same states! To be more explicit:

i) |Ψ〉1234 = |Ψ−〉12⨂|Ψ−〉34 = 1/2(|Ψ+〉14⨂|Ψ+〉23 − |Ψ−〉14⨂|Ψ−〉23 − |Φ+〉14⨂|Φ+〉23 + |Φ−〉14⨂|Φ−〉23)
ii) |Ψ〉1256 = |Ψ−〉12⨂|Ψ−〉56 = 1/2(|Ψ+〉16⨂|Ψ+〉25 − |Ψ−〉16⨂|Ψ−〉25 − |Φ+〉16⨂|Φ+〉25 + |Φ−〉16⨂|Φ−〉25)
iii) |Ψ〉5634 = |Ψ−〉56⨂|Ψ−〉34 = 1/2(|Ψ+〉54⨂|Ψ+〉63 − |Ψ−〉54⨂|Ψ−〉63 − |Φ+〉54⨂|Φ+〉63 + |Φ−〉54⨂|Φ−〉63)

Lucas.

1. Yay!


2. My point is that photons 1 and 2 cannot be entangled with photons 3 and 4 in any of the Bell states of i); and then also be entangled with photons 5 and 6 in any of the Bell states of ii). That would, of course, violate Monogamy of Entanglement on the face of it. We conclude that i) and ii) cannot be simultaneously true. There are other reasons too, but this should be enough to convince. It doesn't make sense to prepare an equation involving 2 systems that have never interacted and then showing the results of a potential interaction as being "true". They must interact first.

 

[Morbert]

2. My point is that photons 1 and 2 cannot be entangled with photons 3 and 4 in any of the Bell states of i)

Here you also jump away the sticking point, which just results in tripping.

Before you can talk about what is entangled with what, you have to first accept simple state expansions. You need to accept that |Ψ〉1234 = |Ψ−〉12⨂|Ψ−〉34 = 1/2(|Ψ+〉14⨂|Ψ+〉23 − |Ψ−〉14⨂|Ψ−〉23 − |Φ+〉14⨂|Φ+〉23 + |Φ−〉14⨂|Φ−〉23)

 

[DrChinese]

You keep ducking what Ma explicitly says.

Ma: "This can be seen by rewriting Eq. (1) in the basis of Bell states of photons 2 and 3"

Do you or do you not accept that Ma says that rewriting equation (1) in the bases of Bell states of photons 2 and 3 yields equation (2)?

Me ducking? I proved here that yours is a false equivalence. Address that. I'm not even sure what relevance this even has to Bohmian Mechanics.

As I have said plenty of times, (2) is mathematically derived from (1). There is no physical meaning to that derivation. If you have $100 and you add $1000 to that using an adding machine, you get $1100. But you still only have $100. Note that when I start with |Ψ−〉12⨂|Ψ−〉34, I can do all kinds of operations on that and still not end up with photons 1 and 4 in a Bell state. In other words, you still don't have $1100. You have to earn it somewhere first.

 

[DrChinese]

DrChinese said:
2. My point is that photons 1 and 2 cannot be entangled with photons 3 and 4 in any of the Bell states of i)

Why would you provide a partial quote which completely changes the meaning of what I actually wrote. Which was:

DrC: My point is that photons 1 and 2 cannot be entangled with photons 3 and 4 in any of the Bell states of i); and then also be entangled with photons 5 and 6 in any of the Bell states of ii). That would, of course, violate Monogamy of Entanglement on the face of it.

Because if you execute a BSM on photons 2 and 3, you do get one of the Bell States of i). Which is exactly the opposite of what your (surprisingly) out of context quote above implies. So, do you accept Monogamy of Entanglement? Maybe that's the issue.