Entanglement swapping and Bohmian mechanics

[DrChinese]

1. I've already said I'm taking no moderation actions in this thread. I was speaking as a participant. And the paper in question was the Ma paper, which you already agreed was difficult to follow. Also, there have been a number of posts about that paper since the one of mine that you quoted, which have helped me to understand (at least for some value of "understand" ) the approach they are taking. In particular see @Morbert #124 and my #139.

2. The post you quoted that from was based on math that @Morbert posted. It looked like basic QM math to me, applying the rule that when you know the results of a measurement, you update the effective wave function you're using to reflect that result. I gave a reference to Ballentine supporting that approach, since that was the textbook I had handiest. @Morbert used notation that's a little different from what I've seen in textbooks, but he explained it clearly in post #40.

1. Great, thanks all around for saying this.

2. Of course I saw @Morbert's earlier post, in which he made the same assertion without citation. And yes, I am sure if I read enough textbooks, I might see something similar to what you and Morbert are claiming. But nowhere is there any authoritative support, and I am asking for it from either you, Morbert, or anyone else.

After all, this is basically the point in question: |VV> is not an entangled state. The other side is, and I don't believe that equation applies in a swapping protocol. No swapping paper I have ever read presents this, or anything like it. Certainly all the references we've been discussing don't.

Ergo: Please humor me and provide a suitable specific reference (not just a link). How hard could that be if it's in every textbook? Please, no more posts avoiding/wiggling out of my citation+quote request; but if you choose not to provide it, that's perfectly OK with me.

-DrC

 

[Demystifier]

Just to be clear, and mine is a serious question precisely because I thought there was a correspondence between the Hilbert representation and the system's "state" representation: I said "There's still 2 Bell states before the swap, and there's 2 completely different Bell states after. Presumably the Hilbert space changed too." But you say there's no change.

Norsen says: "...the physical state of (in particular) the waveguiding the particle is different before, and after, the intermediate SGx device. But in the pilot-wave theory, this difference – the physical influence of the measurement on the properties of the system in question – is a natural and straightforward consequence of the usual – the universal, the exceptionless – dynamical laws".

So is Norsen agreeing with you? Because: I would think the the Hilbert representation is meaningless if a swap occurs, and nothing changed. I an definitely missing the significance of bringing the Hilbert representation into the discussion in the first place, when state representations seem to suffice - including for most of the referenced authors.

We cannot have this conversation if you haven't mastered the basis concepts involved and the difference between Hilbert space, Hilbert space decomposition into product, Hilbert representation (which even I don't know what is this supposed to mean), etc. Anyway, the state representation may be sufficient, but the discussion in terms of the "Hilbert space decomposition into a product" may further deepen understanding for those who understand what that means.

 

[PeterDonis]

After all, this is basically the point in question: |VV> is not an entangled state. The other side is

No, it's actually not, at least not by the mathematical definition. The state can be expressed as a product state of photons 2 and 3--that's just $\ket{VV}_{23}$! The fact that we can mathematically write it down in the Bell state basis does not make it entangled, at least by that mathematical definition.

(@Demystifier might object to this phrasing--he might say it's entangled in the Bell state basis but not in the H-V basis. I'll let him weigh in on that if he wants to. But even if we take that position, that state can be entangled in one basis and not in another, that still would make the equation I wrote valid, at least mathematically.)

The reason we want to write it down in the Bell state basis is simple: because that's the basis that the BSM is going to project it into. From the Ma et al paper (p. 3, last paragraph):

Victor may perform a Bell-state measurement which projects photons 2 and 3 either onto |Φ+〉23 or onto |Φ−〉23 .

My emphasis.

So if we want to predict probabilities for BSM measurement results on photons 2 and 3, given that photons 1 and 4 were already measured with polarization $H$, we need to write down the photon 2 and 3 part of the effective wave function that we get from those photon 1 and 4 measurement results, which is $\ket{VV}_{23}$, in the Bell state basis. AFAIK that is just standard QM mathematical procedure as found in most textbooks; I gave a reference to Ballentine where he describes how to compute conditional probabilities for measurement B given that you know the result of measurement A, and that's what the above is doing. The prediction we get is that, for this case, we have an equal probability for each of the two Bell states $\Phi^\pm$.

Now, as you can see by the fact that I said "mathematically" several times above, all this is just math. The question is whether it's the right math, i.e., whether the prediction I just made above is what actuallly happens in experiments. Unfortunately we haven't found that anyone has actually tested it. But it matches the math that I understand Ballentine to be describing in his textbook (and it also matches what I've read in other textbooks, but I don't have any others handy right now to reference).

 

[martinbn]

(@Demystifier might object to this phrasing-

No, he is definitely not going to object to this phrasing.

 

[PeterDonis]

Now, as you can see by the fact that I said "mathematically" several times above, all this is just math. The question is whether it's the right math, i.e., whether the prediction I just made above is what actuallly happens in experiments.

I should expand on this some. The question that the math I was talking about is relevant to is whether putting photons 2 and 3 through vertical polarizers makes a difference, and if so, what difference it makes.

The math I wrote down, where we use $\ket{VV}_{23}$ as the effective wave function for photons 2 and 3 and then expand it in the Bell state basis to predict the probabilities of BSM results, is, as I said, what I take to be the standard textbook procedure for computing conditional probabilities, as given for example in Ballentine. But conditional on what? Generally speaking, conditional on knowing that photons 2 and 3 both have $\ket{V}$ as the polarization part of their wave function. But there are actually two ways we could know that, at least as far as computing conditional probabilities goes:

(1) We could have measured photons 1 and 4 and gotten $H$ for both results. That lets us assign the effective wave function $\ket{VV}_{23}$ to photons 2 and 3 because they were prepared in singlet states with photons 1 and 4.

(2) We could have put photons 2 and 3 through vertically oriented polarizers and had them come out the other side (instead of being absorbed). Then we assign $\ket{VV}_{23}$ as their joint wave function as a straightforward application of the projection postulate.

Case (1) above is what I implicitly assumed in my previous post, when I talked about photons 1 and 4 already being measured, but not about vertical polarizers. But as far as the math in that post is concerned, it also applies to case (2) above, and that was the connection in which I originally posted about the same math earlier in this thread.

In other words, if we restrict attention to scenarios where, for whatever reason, we are able to assign $\ket{VV}_{23}$ as the effective wave function going into the swap, we predict that we will get a swap into $\Phi^\pm$ with equal probabilities for each. The exact reason why we can assign that effective wave function doesn't matter.

So what is the difference between cases (1) and (2) above? To see that, we have to consider other scenarios besides the ones mentioned there. For example, consider these two alternatives:

(1') We measure photons 1 and 4 before the swap, but nothing else.

(2') We measure photons 1 and 4 before the swap, and, after those measurements but still before the swap, we put photons 2 and 3 through vertical polarizers.

Now the cases will be different. I'll expand on the math in a future post if needed, but from what's already posted it should be clear what the predictions will be. (Note that here I'm assuming an experimental setup that can distinguish all four Bell states.)

For case (1'), we get swaps into all four Bell states with equal probability; but on each run, the Bell states that appear are consistent with the photon 1 and 4 measurement results that were already recorded before the swap. As I've stated before, that means that if photons 1 and 4 are measured to be parallel ($HH$ or $VV$), we only get swaps into $\Phi^\pm$, and if photons 1 and 4 are measured to be antiparallel ($HV$ or $VH$), we only get swaps into $\Psi^\pm$.

For case (2'), we have a different breakdown of results:

1/4 of the time, photons 1 and 4 are measured with results $HH$. In every one of these runs, photons 2 and 3 make it through the vertical polarizers, and everything is as described for cases (1) and (2) above.

1/4 of the time, photons 1 and 4 are measured with results $VV$. In every one of these runs, photons 2 and 3 are absorbed at the vertical polarizers, so they never even make it to the swap device, and nothing is recorded at the swap device outputs.

1/2 of the time, photons 1 and 4 are measured antiparallel. Exactly one of photons 2 and 3 will make it through the vertical polarizers in these runs (photon 2 1/4 of the time, photon 3 1/4 of the time); the other will be absorbed. No swap can happen on these runs because one photon doesn't make it to the swap device; the other photon does get recorded at the swap device outputs, and its statistics will show the appropriate entanglement with the photon (1 or 4) that it was originally prepared in the singlet state with.

 

[PeterDonis]

Please humor me and provide a suitable specific reference

If you mean a reference that specifically walks through the math @Morbert and I have been posting in the context of an entanglement swapping experiment, I doubt if there is one. We haven't found any experiment that puts photons 2 and 3 through vertical polarizers, which is the case we're concerned about, and even such an experimental paper probably would not walk through the math to the level of detail we have been trying to, since it would be written for other experts in the field, not as pedagogy. Similar remarks would apply to the question of how to describe DCES experiments in general; the experimental papers you've already referenced give only a very sketchy presentation of the underlying theory, which is, again to be expected since they're not intended as pedagogy.

As far as theoretical work, I don't know if the question has arisen for any theorist to the point that would make them do the kind of analysis we've been trying to do. But if someone knows of a paper that treats the general theory of entanglement swapping experiments, by all means let's look at it.

In short, I don't think we'll be able to find a reference that will lay out what you're asking for. If that's the case, our only option, if we're going to have a discussion at all, is to apply the general principles of QM as best we can. I've already given a reference to a textbook that lays out those principles in a way that I can apply them, and I'm doing that as best I can.

For how Bohmian mechanics views such things, maybe we need to take a look at @Demystifier's paper on the subject, since the Norsen paper seems to be raising issues.

 

[PeterDonis]

No swapping paper I have ever read presents this, or anything like it.

As I said, the papers' presentation of theory is quite sketchy. But it should be noted that @Morbert's #40 and the other math we've been posting based on that agrees with those papers, and whatever math they present, on the key experimental facts. It agrees with the Bell state correlations when the swaps take place, and it agrees that the experimental results are the same regardless of the order in which the photon 1 and 4 measurements are done relative to the swap and to each other. It's just a different way of writing down the math that, at least for @Morbert and me, helps in doing the analysis.

 

[PeterDonis]

if someone knows of a paper that treats the general theory of entanglement swapping experiments, by all means let's look at it.

I have looked at the 1999 paper by Peres [1] (reference 4 in the Ma et al paper), which AFAIK first proposed the DCES concept. While I would not call its treatment a "general theory" containing all the details we have been discussing, it does have some good information and comments that I would like to share.

Note: Peres describes things in terms of spin-1/2 particles. I am going to describe them in terms of photons. But as Peres explains, the math is the same either way, at least for these experiments (because we are only concerned with photon polarizations, and those can be described by the same QM states that we use for spin-1/2 particles).

First, here is equation (6) in the paper:

$$
\Psi^-_A \otimes \Psi^-_B = \left( \Psi^+_E \otimes \Psi^+ - \Psi^-_E \otimes \Psi^- - \Phi^+_E \otimes \Phi^+ + \Phi^-_E \otimes \Phi^-\right) / 2
$$

This is the joint state of the pair of singlets originally prepared by Alice and Bob; on the LHS we just see that written out compactly: $\Psi^-_A$ is Alice's singlet (what we have been calling photons 1 and 2) and $\Psi^-_B$ is Bob's singlet (what we have been calling photons 3 and 4).

Now, what about the RHS of this? What is it doing?

First, note that the RHS is nothing but a refactoring of the LHS; the equality is just a matter of algebra. Nothing physical is being claimed; it's just a mathematical tool to make it easier to predict the results of the experiment that is going to be done.

Second, the experimental procedure is going to be that two of the photons, one from Alice and one from Bob, will be sent to Eve (we have been calling these photons 2 and 3). The states with the $E$ subscript on the RHS refer to that pair of photons. The other two photons are kept by Alice and Bob (we have been calling these photons 1 and 4); the states with no subscript on the RHS refer to that pair of photons.

Third, the reason for rewriting the state this way is that Eve is going to do a Bell state measurement on her two photons, i.e., she is going to project them into the Bell state basis. That means that, in order to compute the probabilities for the different possible results of all the measurements, we need to write the state with Eve's two photons in that basis. That is what the RHS of the above does (and it is the same thing that @Morbert and I were doing in previous posts).

Peres then computes, in section 2 of his paper, the predictions for the results when Alice and Bob each make polarization measurements on the photons they kept, and Eve does her BSM and then measures the polarization of her two photons. He shows how the CHSH inequalities will be violated and how that demonstrates that yes, indeed, Alice's and Bob's photons show the appropriate statistics for the Bell states that correspond to the ones Eve's BSM operation projected her photons into. And of course the experimental papers we've been looking at match these predictions, to within the limits of the various experimental setups.

Then, in section 3, Peres discusses what all this means. Note in particular this (second paragraph):

How can the appearance of entanglement arise in these circumstances? The point is that it is meaningless to assert that two particles are entangled without specifying in which state they are entangled, just as it is meaningless to assert that a quantum system is in a pure state without specifying that state [9]. If this simple rule is forgotten, or if we attempt to attribute an objective meaning to the quantum state of a single system, curious paradoxes appear: quantum effects mimic not only instantaneous action-at-a-distance but also, as seen here, influence of future actions on past events, even after these events have been irrevocably recorded.

What is Peres saying here? One thing he certainly does not appear to be saying is that these experiments show actual "quantum steering into the past" (as the Ma et al paper put it). Indeed, he appears to be saying they do not show "instantaneous action at a distance" or "influence of future actions on past events". He appears to be saying that those are not actual things the experiments show: those are "curious paradoxes" that only appear if we try to adopt an unwarranted interpretation.

Moreover, he says that we should not be trying to "attribute an objective meaning to the quantum state of a single system". In other words, he's using a statistical or ensemble type of interpretation, similar to Ballentine!

He also says "the appearance of entanglement", and follows that up with "it is meaningless to assert that two particles are entangled without specifying in which state they are entangled". What does that mean? I think it means that he is not saying that Alice's and Bob's photons (our photons 1 and 4) actually are entangled. As he says in the paragraph previous to the one quoted above,

There can be no doubt that the particles that were independently produced and tested by Alice and Bob were uncorrelated and therefore unentangled. Each one of these particles may well have disappeared (e.g., been absorbed) before the next particle was produced, and before Eve performed her tests. Only the records kept by the three observers remain, to be examined objectively.

In other words, those photons never were in an entangled state. So we can't say they actually were entangled. We can only say the statistics of their measurements show the appearance of entanglement when we post-select appropriate subensembles according to Eve's results. That's what I think he's saying.

Peres also discusses the fact that Eve has free will:

Note in particular that even after Alice and Bob have recorded the results of all their measurements, Eve still has the freedom of deciding which experiment she will perform.

And he goes on to discuss how this, too, creates no problem or "paradox", nor does it require any of the "curious paradoxes" he referred to.

Nor is "knowledge of the future" required when Alice and Bob make their measurements, even if they make them before Eve does her stuff:

It is not even necessary for Alice and Bob to know which experiments Eve will do.

[1] https://arxiv.org/abs/quant-ph/9904042

 

[DrChinese]

First, here is equation (6) in the paper:

$$
\Psi^-_A \otimes \Psi^-_B = \left( \Psi^+_E \otimes \Psi^+ - \Psi^-_E \otimes \Psi^- - \Phi^+_E \otimes \Phi^+ + \Phi^-_E \otimes \Phi^-\right) / 2
$$

This is the joint state of the pair of singlets originally prepared by Alice and Bob; on the LHS we just see that written out compactly: $\Psi^-_A$ is Alice's singlet (what we have been calling photons 1 and 2) and $\Psi^-_B$ is Bob's singlet (what we have been calling photons 3 and 4).

Now, what about the RHS of this? What is it doing?

First, note that the RHS is nothing but a refactoring of the LHS; the equality is just a matter of algebra. Nothing physical is being claimed; it's just a mathematical tool to make it easier to predict the results of the experiment that is going to be done.


[1] https://arxiv.org/abs/quant-ph/9904042

You cannot be serious. The RHS is not an algebraic refactoring of the LHS, period.

The LHS is before the swap. The RHS (with E for Eve) is after the swap ("At a later time, Eve performs joint tests on her pairs of particles."). You're better than this - copying and pasting without bothering to understand? And of course I'm familiar with this paper. It's seminal.

The LHS features two particles (2, 3) that have never interacted. These two particles could conceivably be from any source entangled pairs in the universe anywhere. If you (or any reader) think Peres is implying that this can be algebraically re-written mathematically to be the RHS (one of 4 entangled Bell states) without first making them indistinguishable, well, I think you'll need to re-read a few papers. Here is the correct math, which I have presented countless times previously (quoted from Ma et al):

|Ψ〉1234 = |Ψ−〉12⨂|Ψ−〉34 [1]

"... if Victor [or his sister Eve] subjects his photons 2 and 3 to a Bell-state measurement, they become entangled."

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

[1] is NOT equal or equivalent to [2] in any way, shape or form - completely in contradiction to your presentation. And of course Ma has your Peres citation as a reference, their 4. It would be fair to say they understood Peres correctly.

---

But knowing you, you'll do anything and everything to avoid saying you made any kind of boo boo here. Let me know when you, as a poster/participant, are willing to acknowledge the kind of mistake any person can make.

And once you have done that, we can go back to discussing the the point in question circa post #151 and a few earlier ones: |VV> is not an entangled state. The other side (of the substitution you claim) is, and I don't believe that equation applies in a swapping protocol - and of course I mean the algebraic equivalence, absent the actual swap. No swapping paper I have ever read presents such equivalence, or anything like it. Certainly all the references we've been discussing don't, as I demonstrate above. Again, references are requested from you - which I know don't exist. Quit saying they do, if you can't produce specifics. Pure states of independently produced particles cannot be made to be equal to an entangled state of those two without some kind of action/interaction. Their state changes when that happens, there is no mathematical trick.

-DrC

 

[PeterDonis]

The RHS is not an algebraic refactoring of the LHS, period.

Sorry, but it is. Try it. It's actually easiest to do it right to left--start with the RHS, multiply out, cancel terms, and collect what remains. The only somewhat subtle point is that you need to switch the order of factors in each term, but that's not an issue: the order of the factors is just a mathematical convenience for organizing things and doesn't change the meaning of the state.

The LHS features two particles (2, 3) that have never interacted.

No, it features two singlets. Those are pairs of particles, not single particles. Equation (5) writes down what singlet states are. They are two-particle states; that's why two numbers appear in the kets.

The LHS is before the swap. The RHS (with E for Eve) is after the swap ("At a later time, Eve performs joint tests on her pairs of particles.").

You're misreading the paper. He specifically states that his equation (6), the equation I wrote down, is "the joint state of a pair of singlets". Which pair of singlets? The pair prepared by Alice and Bob.

His equation (6) is an equality. It's not a description of any kind of time evolution. It's just two different ways of writing the same state: "the state of a pair of singlets".

Eve's swap operation happens "at a later time". But equation (6) is not at that time. It's at the time after Alice and Bob have prepared their "pair of singlets", and before anything else happens. Note that the statement you quoted, "At a later time..." is later in the paper than equation (6); it's in Section 2, and equation (6) is in Section 1. Section 2 is describing what happens after equation (6). it is not describing a part of equation (6).

 

[PeterDonis]

Pure states of independently produced particles cannot be made to be equal to an entangled state of those two without some kind of action/interaction.

There is one: the BSM. That's why I emphasized that the BSM is a projector (and gave a specific quote from the Ma et al paper to that effect). The BSM takes any state whatever of photons 2 and 3 and projects into one of the four Bell states. That is the action/interaction that changes their state.

 

[PeterDonis]

|VV> is not an entangled state. The other side (of the substitution you claim) is

No, it's not. It's an algebraic refactoring of the LHS state to express it in the Bell state basis, but that does not make it an entangled state. It's a linear combination of entangled states that turns out to be separable.

 

[PeterDonis]

That is the action/interaction that changes their state.

To be more precise about this: before the BSM, in the case under discussion, photons 2 and 3 are in the state (I'll write it this time in more conventional ket notation instead of the notation @Morbert used and include the normalization factor for the kets):

$$
\ket{VV}_{23} = \frac{1}{\sqrt{2}} \left( \ket{\Phi^+} - \ket{\Phi^-} \right)
$$

This state predicts that, if photons 2 and 3 are measured in the Bell state basis, which is what the BSM will do, there is an equal probability for each of the $\Phi^\pm$ Bell states. But the overall state before the BSM is separable; as I said in post #162, it's a linear combination (a superposition) of Bell states that turns out to be separable.

After the BSM, the state of photons 2 and 3 is either $\Phi^+$ or $\Phi^-$, with equal probability. Those of course are entangled states, and the BSM acts to project photons 2 and 3 into one or the other.

 

[Sambuco]

But if someone knows of a paper that treats the general theory of entanglement swapping experiments, by all means let's look at it.

@PeterDonis you might find this paper interesting, as it addresses Ma's and Megidish's experiments from a (solely) forwards-in-time perspective, with and without collapse.

Lucas.

 

[PeterDonis]

@PeterDonis you might find this paper interesting, as it addresses Ma's and Megidish's experiments from a (solely) forwards-in-time perspective, with and without collapse.

We discussed this paper in an earlier thread. I don't think we reached any consensus about it.

 

[Sambuco]

We discussed this paper in an earlier thread. I don't think we reached any consensus about it.

I know, but I'm bringing it up again because we're having the same kind of discussion now, and I think the article is relevant because its treatment of the DCES experiments is quite straightforward. If you don't think it's helpful, that's fine.

Lucas.

 

[Sambuco]

Moreover, he says that we should not be trying to "attribute an objective meaning to the quantum state of a single system". In other words, he's using a statistical or ensemble type of interpretation, similar to Ballentine!

As a minor comment, the fragment you quote certainly seems to suggest an interpretation like Ballentine's, but I'm not entirely sure that's the case, given that Peres was always closer to Copenhagen/QBism/information-based interpretations. In fact, a few months later, he published an article with Chris Fuchs where they say:

"Here, it is essential to understand that the validity of the statistical nature of quantum theory is not restricted to situations where there are a large number of similar systems. Statistical predictions do apply to single events. When we are told that the probability of precipitation tomorrow is 35%, there is only one tomorrow. This tells us that it is advisable to carry an umbrella. Probability theory is simply the quantitative formulation of how to make rational decisions in the face of uncertainty."

I believe the gist of his interpretation lies in the fact that the quantum state does not represent something objective about the system. It's a subtle detail, but Ballentine's interpretation, which associates the quantum state with a statistical property of an (idealized) ensemble, is closer to a frequentist interpretation, while Peres interpretation seems to suggest a subjective/bayesian interpretation of the probabilities assigned to events.

Lucas.

 

[Demystifier]

For how Bohmian mechanics views such things, maybe we need to take a look at @Demystifier's paper on the subject, since the Norsen paper seems to be raising issues.

So far I haven't written any paper about entanglement swapping or anything like that, but after this discussion maybe I will. (In this case I think it would be appropriate to mention in Acknowledgments "members of PhysicsForums" without explicit names.)

In my paper linked in my signature I point out that, in BM, trajectories of measured particles are not really important to understand how BM explains the results of measurements. What is really important are trajectories of particles constituting the macroscopic measuring apparatuses. In this way I combine some aspects of Copenhagen and Bohmian philosophies into an interpretation that I call instrumental Bohmian mechanics.

 

[PeterDonis]

The RHS (with E for Eve) is after the swap

This can't possibly be right, because the RHS describes a superposition of all four Bell states of Eve's photons (each one coupled with a matching Bell state of Alice's and Bob's photons). That's the state before the swap.

The swap projects Eve's photons into just one of the four Bell states--which also projects Alice's and Bob's photons into the matching Bell state. I.e., after the swap, the state is just one of the terms on the RHS of Peres's equation (6). Not the superposition of all of them.

 

[Roberto Pavani]

It seems there is not a common agreement on what the + sign means in this equation, whether it represents a coherent superposition (where terms can interfere) or a grouping of mutually exclusive alternatives (where only one term is ever realised).

 

[Morbert]

Here is the correct math, which I have presented countless times previously (quoted from Ma et al):

|Ψ〉1234 = |Ψ−〉12⨂|Ψ−〉34 [1]

"... if Victor [or his sister Eve] subjects his photons 2 and 3 to a Bell-state measurement, they become entangled."

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

[1] is NOT equal or equivalent to [2] in any way, shape or form

What you mean to say is

$$\begin{aligned}
|\Psi^{-}\rangle\langle\Psi^{-}|_{12}
\otimes
|\Psi^{-}\rangle\langle\Psi^{-}|_{34}
\neq
\frac{1}{4}\Big(
&|\Psi^{+}\rangle\langle\Psi^{+}|_{14}
\otimes
|\Psi^{+}\rangle\langle\Psi^{+}|_{23}+
|\Psi^{-}\rangle\langle\Psi^{-}|_{14}
\otimes
|\Psi^{-}\rangle\langle\Psi^{-}|_{23}
\\
+{}&
|\Phi^{+}\rangle\langle\Phi^{+}|_{14}
\otimes
|\Phi^{+}\rangle\langle\Phi^{+}|_{23}+
|\Phi^{-}\rangle\langle\Phi^{-}|_{14}
\otimes
|\Phi^{-}\rangle\langle\Phi^{-}|_{23}
\Big)
\end{aligned}$$

The LHS is the pure state representing the initial preparation. The RHS is a post-measurement mixed state obtained by tracing over environmental degrees of freedom. It's the state in my post #40, but with standard notation so that there is no confusion.

The difference between a pure state expressed in a Bell basis (what you argued against) and a mixed state diagonal in a Bell basis is important.

 

[Morbert]

It seems there is not a common agreement on what the + sign means in this equation, whether it represents a coherent superposition (where terms can interfere) or a grouping of mutually exclusive alternatives (where only one term is ever realised).

If it was the latter there would be no minus signs.

 

[Lord Jestocost]

It's just a different way of writing down the math that, at least for @Morbert and me, helps in doing the analysis.

Regarding PHYSICS, what are now the consequences in doing the "analysis"?

 

[Roberto Pavani]

I was referring to the + as the linear combination operation itself, regardless of the signs of individual terms.
The question is whether the terms coexist (and can interfere) or represent mutually exclusive alternatives, which is essentially the distinction you made in your previous post between the pure state and the diagonal mixed state.

 

[PeterDonis]

The LHS is the pure state representing the initial preparation. The RHS is a post-measurement mixed state obtained by tracing over environmental degrees of freedom.

Yes, agreed (for your equation). And what I was saying in my earlier post is that Peres's equation (6) has your LHS on its LHS, and on the RHS it has an algebraic refactoring of the same pure state, in the Bell basis. I.e., not the same as your RHS.

What @DrChinese seems to me to be saying is that Peres's equation (6) is the same as yours--his LHS is your LHS (the pure state pre-measurement) and his RHS is the same as your RHS (the mixed state post-measurement but before we know which result was observed). But you wrote (correctly) that your LHS is not equal to your RHS, whereas Peres's equation (6) is an equality. That's why I don't think @DrChinese's reading of Peres's equation(6) is correct.

 

[PeterDonis]

If it was the latter there would be no minus signs.

Meaning, Peres's equation (6) has minus signs on the RHS, which is another reason to think that @DrChinese's reading can't be correct. Your #171 has no minus signs on the RHS.

 

[PeterDonis]

was referring to the + as the linear combination operation itself, regardless of the signs of individual terms.

This makes no sense. When you write a state in QM as a linear combination of other states, the sign is not meaningless; it is part of the phase relationship between the terms. More generally, given two states $\psi_1$ and $\psi_2$, you can write a linear combination of them as $a \psi_1 + b \psi_2$ with any pair of complex numbers $a$ and $b$ (with appropriate normalization), and in principle you can distinguish the different possibilities with appropriate operations.

 

[PeterDonis]

Regarding PHYSICS, what are now the consequences in doing the "analysis"?

They've been stated in a number of posts, but this thread is getting long, so perhaps a brief summary, at least of what I think we've shown, is warranted.

For the general, idealized case of DCSE, where all the apparatus works with 100% accuracy and all four Bell states can be distinguished at the BSM, on runs where the experimenter chooses to swap, the forward in time Bohmian analysis says that the effective collapse of the wave function based on the results of the measurements on photons 1 and 4 "steers" the particle trajectories of photons 2 and 3 through the BSM in such a way as to produce a swap that is consistent with the photon 1 and 4 results. Specifically, if the photon 1 and 4 results are parallel, the swap will be into $\Phi^\pm$, and if the photon 1 and 4 results are antiparallel, the swap will be into $\Psi^\pm$. This is what enforces the correlations that appear when the final results are sorted into subensembles according to the photon 2 and 3 BSM results.

For cases where only some Bell states can be distinguished (for example, the Ma et al experiment where only $\Phi^\pm$ can be), some of the runs will go into a separate bucket, where a swap was attempted but could not be distinguished in the results.

For cases where the experimenter chooses not to swap, the "steering" of photons 2 and 3 by the photon 1 and 4 measurement results produces separable photon 2 and 3 states at the final output; the change is due to the change in configuration of the swap/no swap apparatus. The effective wave function of photons 2 and 3 going in is the same (for a given set of photon 1 and 4 measurement results), but the interaction of photons 2 and 3 with the swap/no swap apparatus is different, so the final trajectories are different.

 

[PeterDonis]

The difference between a pure state expressed in a Bell basis (what you argued against) and a mixed state diagonal in a Bell basis is important.

I was groping towards this in post #169, but you expressed it better.

 

[PeterDonis]

the fragment you quote certainly seems to suggest an interpretation like Ballentine's, but I'm not entirely sure that's the case, given that Peres was always closer to Copenhagen/QBism/information-based interpretations.

Yes, this is a fair point, and you give good context on Peres's viewpoint.