Entanglement swapping and Bohmian mechanics

[Demystifier]

But of course in BM, the issue is:

i) Whether the future swap - which the universe knows is going to occur? - causes the perfect A/D correlation, thereby creating the illusion of action to the past; or
ii) The outcome of the perfect A/D correlation causes the swap to succeed.

If I have to choose one of those, I choose i). You cannot actually observe the A/D correlation before you do the swap. The data from which the correlation is extracted was there before the swap, but you can see this correlation only after the swap. In this sense, the correlation happens after the swap, so the swap is the cause and the correlation is the effect.

For an analogy consider the following string of letters: tlreeaef
Initially, it looks like a random string of letters. But after someone tells you that you should only read the even letters, suddenly you see "leaf". Alternatively, after someone tells you that you should only read the odd letters, you see "tree". Was there "leaf" or "tree" there before someone told you to read only even or only odd letters? It can be interpreted in several ways, but the interpretation that the future caused a change in the past is probably not an interpretation that many people would find plausible.

 

[martinbn]

If I have to choose one of those, I choose i). You cannot actually observe the A/D correlation before you do the swap. The data from which the correlation is extracted was there before the swap, but you can see this correlation only after the swap. In this sense, the correlation happens after the swap, so the swap is the cause and the correlation is the effect.

For an analogy consider the following string of letters: tlreeaef
Initially, it looks like a random string of letters. But after someone tells you that you should only read the odd letters, suddenly you see "leaf". Alternatively, after someone tells you that you should only read the even letters, you see "tree". Was there "leaf" or "tree" there before someone told you to read only odd or only even letters? It can be interpreted in several ways, but the interpretation that the future caused a change in the past is probably not an interpretation that many people would find plausible.

From the data you cannot tell if anything was done on B/C first or on A/D. But your/BM explanations are different!

 

[Roberto Pavani]

A simple 5-qubit circuit (GHZ-like + ancilla) demonstrates @PeterDonis' point concretely: the full ensemble of q1,q2 measurements shows |S|≤2, but conditioning on the ancilla outcome produces |S|=2√2 on the sub-ensemble, without any direct physical interaction between q1 and q2 after the conditioning.
The 'correlation' is entirely a property of the sub-ensemble selection, not of a direct physical mechanism between the qubits.

Python code attached (requires qiskit and qiskit-aer).
(edited: didn'r realize that .py extension was not accepted, renamed as txt)

 

[Demystifier]

One more point, that points to the core of the confusion. To decide whether a certain state in the Hilbert space ${\cal H}$ is entangled or not, one first has to decompose ${\cal H}$ into a product of two or more subspaces. The notion of entanglement depends on this decomposition, so the same state can be both entangled with respect to one decomposition and not entangled with respect to another decomposition.

A nice example is a hydrogen atom in the ground state. It contains two particles, electron and proton. Is there an entanglement in the ground state? It depends on the decomposition of the Hilbert space. If you decompose it as ${\cal H}_{h}={\cal H}_{p}\otimes {\cal H}_{e}$ (where $h$, $p$ and $e$ stand for hydrogen, proton and electron, respectively), then the corresponding wave function of the ground state is entangled
$$\phi_h(x_p,x_e) \neq \phi_p(x_p)\phi_e(x_e)$$
The electron is entangled with the proton. However, one can also introduce new coordinates
$$x_e'=x_e-x_p$$
$$x_p'=\frac{m_ex_e+m_px_p}{m_e+m_p}$$
where $x_e'$ is relative position of the electron with respect to the proton, while $x_p'$ is the position of the center of mass. The usual textbook analysis of the hydrogen atom is actually done in these primed coordinates. It is well known from textbooks that the ground state is a product state in these coordinates
$$\phi_h(x_p,x_e) = \psi_h(x_p',x_e') = \psi_p(x_p')\psi_e(x_e') $$
which corresponds to a different decomposition of the the Hilbert space ${\cal H}_{h}={\cal H}'_{p}\otimes {\cal H}'_{e}$. There is no entanglement with respect to this decomposition. Intuitively one can think of $x_p'$ and $x_e'$ as positions of quasi-particles, called quasi-proton and quasi-electron respectively, so one can say that in the ground state there is no entanglement between quasi-proton and quasi-electron.

With this insight in mind, now it is much easier to understand the origin of entanglement swapping. Now the relevant Hilbert space is a space of 4 particles ${\cal H}={\cal H}_A\otimes{\cal H}_B\otimes{\cal H}_C\otimes{\cal H}_D$. This can also be written as a product of two Hilbert spaces, but such a decomposition is not unique. One possible decomposition is
$${\cal H}={\cal H}_{AB}\otimes{\cal H}_{CD} \;\;\; (1)$$
where ${\cal H}_{XY}\equiv {\cal H}_X\otimes {\cal H}_Y$, while another possible decomposition is
$${\cal H}={\cal H}_{AD}\otimes{\cal H}_{BC} \;\;\; (2)$$
Now the origin of entanglement swapping is easy to explain. Is the state
$$|\psi\rangle = |\psi\rangle_{AB}\otimes |\psi\rangle_{CD}$$
entangled or not? It is not entangled under the decomposition (1), but it is entangled under many other decompositions. In particular, it is entangled under decomposition (2), i.e.
$$|\psi\rangle \neq |\phi\rangle_{AD}\otimes |\phi\rangle_{BC}$$
It is precisely the entanglement under the decomposition (2) that makes the state in ${\cal H}_{AD}$ dependent on the result of measurement in ${\cal H}_{BC}$. Furthermore, it turns out that post-collapse states in ${\cal H}_{AD}$ are themselves entangled with respect to the natural decomposition ${\cal H}_{AD} = {\cal H}_A\otimes {\cal H}_D$, which is why A is entangled with D after the measurement in the system B+C. But here "after" should be understood in the logical sense, not in the temporal sense, which is why it is not important what precedes what in the temporal sense.

 

[Sambuco]

However, I don't think that leads to the prediction you're making, because after the electron and positron spins are both measured, they aren't entangled, since you're specifying that nothing else is done to them after the measurement, and if they aren't entangled, they obviously can't be in the singlet state.

Sure. In fact, this same reasoning can be applied to DCES experiments, which indicate that once photons A and D are measured, they cease to be entangled, making it impossible to assign them a Bell state. However, what this Bell state does is a postdiction of the results of the measurements of photons A and D when these are post-selected based on the measured results of B and C.

Analogously, what I am trying to do is make a postdiction based on a subset of final results, in this case, the detection of exactly two photons, that is, those runs in which process $e^- + e^+ → \gamma + \gamma$ occurred. If we limit ourselves to this subset, an observer who has not read the results of the electron and positron spin measurements for each pair can take the reverse process ($\gamma + \gamma → e^- + e^+$) and conclude that these measurements are compatible with the predictions of a singlet state.

After writing this, I notice post #33 by @Roberto Pavani who seems to raise a similar idea.

Lucas.

 

[Sambuco]

From the data you cannot tell if anything was done on B/C first or on A/D. But your/BM explanations are different!

This is a particular feature of Bohmian mechanics that arises because it presupposes a preferred direction for time evolution. In contrast, in $\psi$-epistemic interpretations, the same "explanation" applies regardless of the order in which the measurements are made.

Personally, I see this as a disadvantage of Bohmian mechanics, which does not incorporate this symmetry into its fundamental equations, at least in its usual presentation. I believe that, to do so, it would have to adopt a relationalism that contradicts the original idea of the interpretation, that is the existence of an objective (observer-independent) reality.

Lucas.

 

[PeterDonis]

In fact, this same reasoning can be applied to DCES experiments, which indicate that once photons A and D are measured, they cease to be entangled

But this reasoning doesn't work for the case @DrChinese is really concerned about, where A and D are measured before the swap operation is done. In that case, at least according to the straightforward realist interpretation he is using, A and D are not entangled before they are measured, so saying they cease to be entangled when they are measured makes no sense. On the interpretation he is using, the swap operation "causes" A and D to become entangled after they are measured.

But that operation depends on A being entangled with B, and C being entangled with D, before they are measured. That is what allows post-selection on the B and C outcomes to show subensembles in which A and D are entangled. There is no such thing present in the electron-positron case you describe.

what I am trying to do is make a postdiction based on a subset of final results

But you are assuming that the final results will contain a subensemble that has the property you describe. I am saying that assumption is wrong. Your reason for making that assumption is based on the inverse process, but that is not the reasoning that is being used in the DCES experiments; there is no "inverse process" considered anywhere. As I said above, the reason the subensemble selection works in the DCES experiments is that there are other entanglements involved, which are not present in the electron-positron case.

 

[Sambuco]

In that case, at least according to the straightforward realist interpretation he is using, A and D are not entangled before they are measured, so saying they cease to be entangled when they are measured makes no sense.

It's the same in the case I described. The entire set of electron-positron pairs shows no correlation whatsoever in their spins. The correlation only appears when a subset of runs is post-selected (the subset where exactly two photons were produced).

But that operation depends on A being entangled with B, and C being entangled with D, before they are measured (...) There is no such thing present in the electron-positron case you describe.

True, the two cases are not identical, that's just an analogy (see below).

but that is not the reasoning that is being used in the DCES experiments; there is no "inverse process" considered anywhere.

In the DCES experiment, to obtain the non-separable wave function that shows the entanglement in the A-D pair, the measurement postulate (state reduction) must be applied based on the result of the BSM measurement on the B-C pair. Since that measurement was performed later, the measurement postulate is applied to a future measurement to obtain the wave function that allows for the postdiction of the results of two measurements in the past (those of photons A and D).

To be clear, I fully agree that the cases are not the same. The case I described has nothing to do with entanglement swapping. I am simply trying to discuss that the post-selection of a subset of runs where something occurred that could only have happened if the original pairs met certain conditions provides information about them, and that this information can be represented by a non-separable state.

Lucas.

 

[PeterDonis]

The correlation only appears when a subset of runs is post-selected

Again, you are assuming this is the case. I am saying I think that assumption is not justified.

I fully agree that the cases are not the same.

But you are still claiming they are analogous enough to justify your claim about a subensemble of electron-positron pairs being present that appears entangled. That's the claim I'm arguing against. You aren't addressing my arguments along those lines at all.

a subset of runs where something occurred that could only have happened if the original pairs met certain conditions

I disagree with this. I explained why in post #17. You have not responded at all to the argument I made there. You just keep repeating the claim I disagree with.

 

[Morbert]

b) Photons 2 and 3 are remotely subjected to a joint Bell-state measurement (beam splitter, PBSs), and they become entangled in one of 4 Bell states.

|Ψ〉1234 = 1/2 (|Ψ+〉14⨂|Ψ+〉23 − |Ψ−〉14⨂|Ψ−〉23 − |Φ+〉14⨂|Φ+〉23 + |Φ−〉14⨂|Φ−〉23)

To explicitly model* the measurement of 2 and 3 first, we would include the apparatus degrees of freedom and trace over them post-measurement, which would yield$$\rho_{1234} = \frac{1}{4}(\left[\Psi^+\right]_{14}\left[\Psi^+\right]_{23}+\left[\Psi^-\right]_{14}\left[\Psi^-\right]_{23}+\left[\Phi^+\right]_{14}\left[\Phi^+\right]_{23}+\left[\Phi^-\right]_{14}\left[\Phi^-\right]_{23})$$where $\left[\psi\right ] = \ket{\psi}\bra{\psi}$. Each term corresponds to a possible result, and so a standard approach would have us collapse the state to one of these terms depending on the measurement outcome. Each of the possible states has 1 and 4 in an entangled Bell state. Hence "entanglement swapping".

But if 1 and 4 are measured first (say, in the H,V basis), we get** $$\begin{equation*}\begin{aligned}\rho_{1234} = \frac{1}{8}\Big( &\left[HH\right]_{14}\left[\Phi^+-\Phi^-\right]_{23} + \left[VV\right]_{14}\left[\Phi^++\Phi^-\right]_{23} \\ &+ \left[HV\right]_{14}\left[\Psi^+-\Psi^-\right]_{23} + \left[VH\right]_{14}\left[\Psi^++\Psi^-\right]_{23}\Big)\end{aligned}\end{equation*}$$Here, the state would similarly collapse to one of these terms depending on the outcome, but none of the terms have 1 and 4 in an entangled state. All experimental correlations are nevertheless recovered.

* The model is a simple one: No destruction upon measurement, and a measurement capable of resolving any of the 4 Bell states.

** Similar expressions can be constructed for R,L or +,- bases.

 

[Sambuco]

Again, you are assuming this is the case. I am saying I think that assumption is not justified.

I disagree with this. I explained why in post #17. You have not responded at all to the argument I made there. You just keep repeating the claim I disagree with.

I think I understand it better now. Thank you for your patience!

Let me explain further and tell me if I'm wrong. If many runs are performed, I can carefully post-select only those where the final product is such that, considering the reverse process, the electron-positron pairs will show non-classical correlations (for example, the singlet state I mentioned earlier). Do you agree with that?

I understand that, for some people, this type of post-selection would be just cherry-picking. In fact, that's precisely my point.

Lucas.

 

[Roberto Pavani]

After writing this, I notice post #33 by @Roberto Pavani who seems to raise a similar idea.

I've edited the post adding the missing attachment


here "stripped" condensed output:
=== Full ensemble (q0 not measured) ===

TEST E(q1,q2)
A0B0 0.008
A0B1 -0.007
A1B0 0.711
A1B1 -0.712

Total CHSH = 1.42 (≤ 2, no violation)


=== Post-selected sub-ensembles (q0 measured) ===

TEST q0=0 q0=1
A0B0 -0.713 0.708
A0B1 -0.720 0.702
A1B0 0.710 0.710
A1B1 -0.705 -0.706

Conditional CHSH:
q0=0 : -0.018 (≈ 0)
q0=1 : 2.826 (≈ 2√2 = 2.828)

Same data, same qubits, same measurements.
The only difference: which events you include in the analysis.

 

[Demystifier]

where $\left[\psi\right ] = \ket{\psi}\bra{\psi}$.

Just a minor notational nitpicking. Is this your own notation, or have you seen it somewhere else? Wouldn't it be more common to use the notation $\rho_{\psi} = \ket{\psi}\bra{\psi}$?
Actually I like your notation, but I just haven't seen it before.

 

[DrChinese]

But this reasoning doesn't work for the case @DrChinese is really concerned about, where A and D are measured before the swap operation is done. In that case, at least according to the straightforward realist interpretation he is using, A and D are not entangled before they are measured, so saying they cease to be entangled when they are measured makes no sense. On the interpretation he is using, the swap operation "causes" A and D to become entangled after they are measured.

Yes, you are correct. I would argue that the swap operation causes the A/D entanglement. I believe the following description is fair under pretty much any viewpoint (considering the catch I mention); you can read my entire summary below and agree with it, or not.

DrChinese summary: The free will of an experimenter, by choosing to execute (or not) the B/C swap after A and D are detected, causes the A/D entanglement. The catch here being whether the experimenter truly has free will, or simply thinks she has it (which would include Demystifier's concept of Bohmian predetermination). According to references supplied, there is no correlation between the A/D pair for the relevant subensembles unless that swap occurs. In the explicitly stated view of the experimentalists: "It is also evidence that the [A] and [D] photons did not somehow share any entanglement before the projection [swap] of the middle photons [B/C]". Or this team: "This can also be viewed as “quantum steering into the past".*

And although I wouldn't label my description as you have ("realist"), I wouldn't exactly deny it either.

* Neither of these papers express any view on Interpretations at all, so I would refer to their characterizations simply as something they consider objectively true - even if others here would not.

 

[Demystifier]

I would argue that the swap operation causes the A/D entanglement.

I would say that the swap operation implies the A/D entanglement. The difference is that "implication" is logical rather than physical, and does not assume any temporal order.

 

[DrChinese]

I understand that, for some people, this type of post-selection would be just cherry-picking. In fact, that's precisely my point.

I don't know of any experimental team, anywhere, anytime, in any area of science, that considers a preconceived blind selection criteria as being "cherry-picking". If I filter all B/C pairs in which meet the following criteria: "photons and [C] ([detected] in b’’ and c’’ in Fig. 2)"; then the results as reported are considered scientifically valid.

Of course, you are free to reject experiments, or interpret them differently.

 

[PeterDonis]

If many runs are performed, I can carefully post-select only those where the final product is such that, considering the reverse process, the electron-positron pairs will show non-classical correlations (for example, the singlet state I mentioned earlier). Do you agree with that?

No. Again, you are assuming that getting a pair of photons out implies that the electron-positron pair was in the singlet state going in. I don't think that assumption is correct. To restate what I said in post #17:

If it is true that, in order to get a pair of photons out, the electron-positron pair must be in the singlet state going in, then I think you will not get any pairs of photons out, because you have specified that the electron-positron pair is prepared in a non-entangled state, not the singlet state.

However, I also don't think that it's actually true that the electron-positron pair must be in the singlet state in order to get a pair of photons out. I think that it is possible for the electron-positron pair to be in a state where their spins are uncorrelated, which is what you specified, but their orbital angular momentum is nonzero (and you made no specification about orbital angular momentum in the initial state), and produce a pair of photons. So if you run the experiment and you do get pairs of photons out, I don't think you can conclude from that that, in the subensemble of runs where a pair of photons came out, the electron-positron pair must have been in the singlet state going in. I think that instead you would have to conclude that, since you know the electron-positron pair had uncorrelated spins going in (since you prepared them that way), their orbital angular momentum on the runs in that subensemble must have been nonzero.

This is the argument that you still have not responded to at all.

 

[PeterDonis]

I understand that, for some people, this type of post-selection would be just cherry-picking.

To be clear, I do not think the type of post-selection used in the DCSE experiment is cherry-picking. I agree with @DrChinese that it's a perfectly valid experimental procedure which is used all the time in science.

However, I also do not think that your electron-positron pair scenario is analogous to the DCSE experiment in any useful way, so I don't think whatever post-selection you are proposing in that scenario is relevant to this thread.

 

[Roberto Pavani]

I'd point out that the interesting step is not the decomposition of ρ₁₂₃₄ (which everyone agrees on), but the passage ρ₁₂₃₄ → E(a,b) → S. That's where the number 2√2 actually comes from.

 

[Morbert]

Just a minor notational nitpicking. Is this your own notation, or have you seen it somewhere else? Wouldn't it be more common to use the notation $\rho_{\psi} = \ket{\psi}\bra{\psi}$?
Actually I like your notation, but I just haven't seen it before.

It's common in consistent histories literature, where you are often chaining many projectors together. See e.g. equation (2) in https://journals.aps.org/pra/abstract/10.1103/PhysRevA.96.032110 (preprint: https://arxiv.org/pdf/1704.08725 ).

 

[Sambuco]

No. Again, you are assuming that getting a pair of photons out implies that the electron-positron pair was in the singlet state going in.

No, I'm not assuming that, but I clearly didn't manage to explain myself well in the previous messages. What I tried to say is that, once only those runs in which there are two photons are chosen, according to the results obtained from the measurements made on those two photons, a fraction of those runs can be post-selected, which will have the characteristic that, if the process had been time-reversed, the product would have been entangled electron-positron pairs.

Lucas.

 

[PeterDonis]

once only those runs in which there are two photons are chosen, according to the results obtained from the measurements made on those two photons, a fraction of those runs can be post-selected, which will have the characteristic that, if the process had been time-reversed, the product would have been entangled electron-positron pairs.

How are you going to post-select that subensemble of runs?

 

[Sambuco]

I don't know of any experimental team, anywhere, anytime, in any area of science, that considers a preconceived blind selection criteria as being "cherry-picking". If I filter all B/C pairs in which meet the following criteria: "photons and [C] ([detected] in b’’ and c’’ in Fig. 2)"; then the results as reported are considered scientifically valid.

Of course, you are free to reject experiments, or interpret them differently.

I'm on your side on this! In effect, what I am trying to argue is that the entanglement observed in DCES is as real as any other, but adding that this is nothing more than a representation of the information we have about the correlation between certain events.

I'll try to explain myself better. In my opinion, the case I described of post-selection in the $e^- + e^+ → \gamma + \gamma$ process is not, in essence, different from the DCES case. Now, I believe that many people could say, at the same time, that (i) in the case of the electron-positron pair, post-selection is simply cherry-picking, while (ii) in the case of DCES it is not cherry-picking. Since I don't see a fundamental difference between the two, my opinion is that both are cherry-picking or neither is.

Now, the final part of my argument has to do with the fact that no one would say that, in the case I described, the measurement of the two photons produces backwards-in-time action that changes the state of the electron-positron pairs. From this, and the similarity with the DCES case, I allow myself to conclude that there is no backward-in-time action on photons A-D when the swap occurs.

It remains true that swap is only possible if the experimenter decides to do so, as you have said many times in different threads. My argument is that this swap only allows me to acquire some information, it does not generate a physical action.

Lucas.

 

[javisot]

QM does not require that every instance of entanglement must have been created through an interaction between the two particles. It merely requires that, if they are entangled, their joint state is non-factorizable.

 

[Sambuco]

To be clear, I do not think the type of post-selection used in the DCSE experiment is cherry-picking. I agree with @DrChinese that it's a perfectly valid experimental procedure which is used all the time in science.

I totally agree (see my post #53).

However, I also do not think that your electron-positron pair scenario is analogous to the DCSE experiment in any useful way, so I don't think whatever post-selection you are proposing in that scenario is relevant to this thread.

Yes I know. I still think it's relevant. In any case, if you think it's better to discuss it in a separate thread, that's fine.

How are you going to post-select that subensemble of runs?

Well, well, maybe we need someone more trained in QED than me to answer this. It will take me more than a few minutes to look up what conditions a pair of photons must meet for the resulting electron-positron pair to have a particular entangled state. At the end of the day, I am not a QFT expert, but rather my work is in device physics.

Lucas.

 

[emillindberg]

I think neither option quite follows. The future swap does not need to cause the earlier A/D outcomes, and the earlier outcomes do not cause the later swap to succeed. The swap result sorts the already recorded A/D data into subensembles, which is where the Bell-state correlations show up.

The part I find interesting is whether saying A and D "were entangled" before the swap has any meaning without first specifying the interpretation and which wave function or conditional state is being discussed. That seems closer to the real source of disagreement here.

 

[Roberto Pavani]

Similar idea was illustrated by the circuit in my earlier post.

The point of the circuit is actually quite simple.

Start from the uniform superposition

$|H\rangle = \frac{1}{2} \left( |00\rangle+|01\rangle+|10\rangle+|11\rangle \right).$

The ancilla q0 acts as a selector. It does not enter the CHSH test directly; rather, it partitions the runs into two subsets according to the observed values of (q1,q2).

The selector is used to post-select the subensembles

$q_0=0 \Rightarrow \{|01\rangle,|10\rangle\}$

and

$q_0=1 \Rightarrow \{|00\rangle,|11\rangle\}.$

Defining the conditional ensembles

$\rho_0=\rho_{12|q_0=0},$

$\rho_1=\rho_{12|q_0=1},$

the pair density matrix may be written as

$\rho_{12} = \frac12\rho_0 + \frac12\rho_1.$

The CHSH analysis gives

$S(\rho_{12}) \approx 0.71,$

while

$S(\rho_0) \approx 0,$

$S_{\rm alt}(\rho_0) \approx 2\sqrt2,$

and

$S(\rho_1) \approx \sqrt2.$

Thus the selector q0 allows the dataset to be analysed as two conditional ensembles rather than as a single unconditional ensemble.

The interesting step is therefore not only the decomposition of the global state, but the passage

$\rho_{12} \longrightarrow \{\rho_0,\rho_1\} \longrightarrow \{S,S_{\rm alt}\}.$

The selector provides the information needed to identify which correlation pattern is present in a given run.

q3 and q4 were introduced as control qubits.

They are not used in the CHSH calculation. Their role is to provide an independent verification of the selector mechanism implemented through q0.

Because q3 and q4 are generated symmetrically by the same circuit structure, any interpretation assigning a special physical role to q0 would also need to explain why analogous considerations should not apply to q3 and q4.

(Code and output attached.)

 

[PeterDonis]

I have a suggested Bohmian model of the DCES experiment. I'll sketch it for now, and possibly follow up with more detail in later posts.

To be clear about the experimental setup:

At the start, we prepare photons A&B in the singlet state $\Psi^-$ (I'll use the standard notation for Bell states throughout), and photons C&D in the same state.

Next, we measure the polarization of photons A and D in the $H/V$ basis (I'll assume we're doing all measurements in that basis) and record the results for later analysis.

Next, the experimenter decides whether or not to attempt a swap. (I say "attempt" for reasons which will be clear in a moment). If the experimenter decides not to attempt a swap, no further operation is done on any of the photons.

If the experimenter does decide to attempt a swap, then there are two possibilities, because even if the experimenter decides this, he can't guarantee that a swap actually happens. We have to look for a "swap occurred" signature in the final result data to know whether the swap actually happened or not. (In the case above where the experimenter doesn't attempt a swap, we assume this signature always shows "no swap".) This is why I said "attempt" above.

So at the swap stage, we have three possibilities:

The experimenter did not attempt a swap (so we get a "no swap" signature in the final results).

The experimenter attempted a swap, but it didn't happen (so we again get a "no swap" signature in the final results--but we still record that the experimenter attempted the swap, so this case is distinguishable from the previous one).

The experimenter attempted a swap and it happened (so we get a "swap" signature in the final results).

Finally, after all that, the polarizations of photons B and C are measured, and results are recorded.

We then analyze the data by sorting it into buckets as follows:

Bucket 0: No swap attempted.

Bucket 1: Swap attempted but no swap happened.

Bucket 2A: Swap attempted and happened, photons B and C ended up in Bell state $\Phi^+$. This signals that photons A and D should also show the appropriate correlations for the same Bell state.

Bucket 2B: Swap attempted and happened, photons B and C ended up in Bell state $\Phi^-$.This signals that photons A and D should also show the appropriate correlations for that Bell state.

I'm assuming that those are the only two possible Bell states (or at least the only ones we can distinguish in the data). Depending on the specific setup, it's possible that there are some runs that don't fit into any of the above buckets; any such runs are discarded.

Now, as far as a Bohmian model is concerned, I assume that Buckets 0 and 1 present no issue: nothing happens at the swap stage so we just have two pairs of entangled photons as prepared at the start, and we already have a Bohmian model of how each pair's measurement results end up with the appropriate correlations (a paper describing such a model was referenced, IIRC, in a previous thread).

Further, I'll assume there is no issue with describing how, in Buckets 2A and 2B, the photon B and C results end up correlated appropriately, assuming that the swap operation happens as intended. In other words, the only thing we really need to account for from the Bohmian viewpoint is what happens at the swap stage, given that photons A and D were already measured, to produce the results we actually observe.

Here is how we account for that:

When photons A and D are measured, there are two relevant possibilities for their results:

(1) Their measured polarizations are parallel ($HH$ or $VV$).

(2) Their measured polarizations are not parallel ($HV$ or $VH$).

In case (1), the photon A and D results are consistent with either of the two possible Bell states that a swap could produce. Which means that in this case, once more, we already have a Bohmian account of what happens! The key point is that the photon A and D results are consistent with either of the two possible Bell states--which means that in this case, the swap operation doesn't need to know, so to speak, anything about the photon A and D results! Whichever Bell state it produces, the photon A and D results will be consistent with it, and that's all that matters.

In case (2), by contrast, we have photon A and D results that are not consistent with either of the two possible Bell states that the swap could produce. So what happens then? Simple: for this case, a swap will never happen. In other words, no results in this category will appear in Buckets 2A or 2B--every run where the photon A and D measured polarizations are not parallel will be in Buckets 0 or 1.

How does that happen? Observe that the wave function of the overall setup has to include whatever it is that, once the experimenter decides to attempt a swap, determines whether or not the swap actually happens. In other words, since there is some kind of quantum process going on here which the experimenter can't control, we have to include that process as part of our model, and that means including it in the wave function. Call that part of the wave function P, and give it two possible outcome states: PN for no swap, and PS for swap. Then it's simple: there is zero amplitude in the wave function for the photon A and D measured polarizations to not be parallel and the final state of P to be PS. Which means that combination of outcomes will never occur.

 

[DrChinese]

[1] With this insight in mind, now it is much easier to understand the origin of entanglement swapping. Now the relevant Hilbert space is a space of 4 particles ${\cal H}={\cal H}_A\otimes{\cal H}_B\otimes{\cal H}_C\otimes{\cal H}_D$. This can also be written as a product of two Hilbert spaces, but such a decomposition is not unique. One possible decomposition is
$${\cal H}={\cal H}_{AB}\otimes{\cal H}_{CD} \;\;\; (1)$$
where ${\cal H}_{XY}\equiv {\cal H}_X\otimes {\cal H}_Y$, while another possible decomposition is
$${\cal H}={\cal H}_{AD}\otimes{\cal H}_{BC} \;\;\; (2)$$
Now the origin of entanglement swapping is easy to explain. Is the state
$$|\psi\rangle = |\psi\rangle_{AB}\otimes |\psi\rangle_{CD}$$
entangled or not? It is not entangled under the decomposition (1), but it is entangled under many other decompositions. In particular, it is entangled under decomposition (2), i.e.
$$|\psi\rangle \neq |\phi\rangle_{AD}\otimes |\phi\rangle_{BC}$$
It is precisely the entanglement under the decomposition (2) that makes the state in ${\cal H}_{AD}$ dependent on the result of measurement in ${\cal H}_{BC}$. Furthermore, it turns out that post-collapse states in ${\cal H}_{AD}$ are themselves entangled with respect to the natural decomposition ${\cal H}_{AD} = {\cal H}_A\otimes {\cal H}_D$, which is why A is entangled with D after the measurement in the system B+C.

[2] But here "after" should be understood in the logical sense, not in the temporal sense, which is why it is not important what precedes what in the temporal sense.

I will point out that the descriptions you give seem to be contradicted in the various papers I reference. They use a slightly different notation, but seem otherwise equivalent.

[1] From Ma et al:

|Ψ〉1234 = |Ψ−〉12⨂|Ψ−〉34 corresponding to your (1), agreed?

There is no decomposition into something like your (2) unless the experimenter chooses to execute a swap. . And in fact there is particular difference in your (1) featuring photons 1 & 2 and any other pair of entangled photons 5/6, 7/8, etc. as |Ψ−〉12⨂|Ψ−〉56⨂|Ψ−〉78". After" (agreeing with your [2] statement about this word) the swap:

|Ψ〉1234 = |Φ−〉14⨂|Φ−〉23) [here selecting the Bell state being post-selected]

[2] They agree with this part.

What am I misunderstanding between the presentations of you and Ma? It does not seem your deconstructions represent 2 sides of a "Hilbert" coin. You specifically say "It is not entangled under the decomposition (1), but it is entangled under ... decomposition (2)" I would say neither is any more or less entangled than the other. Both (1) and (2) feature Product states of entangled pairs.

You also say "post-collapse states in ${\cal H}_{AD}$ are themselves entangled with respect to the natural decomposition ${\cal H}_{AD} = {\cal H}_A\otimes {\cal H}_D$". While I would put forth that post-collapse states in ${\cal H}_{AD}$ are themselves entangled as you say, but cannot be decomposed into ${\cal H}_{AD} = {\cal H}_A\otimes {\cal H}_D$" precisely because A & D are entangled. Isn't that pretty much a definition of an entangled state? Please correct me if I am saying this wrong.

Thanks.-DrC

 

[DrChinese]

I would say that the swap operation implies the A/D entanglement. The difference is that "implication" is logical rather than physical, and does not assume any temporal order.

I am agreeing that "causes" versus "implies" is a direct result of taking on the Bohmian interpretation as you do. Again, in my view (and I don't think this is so much an interpretation), the word "causes" assumes the experimenter has free will. I think it is fair to say: Free will is by and large a generally accepted belief of scientists - but certainly not all.

Presumably: In a clockwork universe, nothing can be pointed to as a "cause" of anything else.