Entanglement swapping and Bohmian mechanics

[DrChinese]

I have provided you with a mathematical proof that Ma's eq. (1) and (2) are the same. If you don't understand we can help you, but stop talking nonsense.

Lucas.

Always humorous when someone quotes me, and proceeds to not address a single thing quoted. Oh and instead of addressing my point, resorts to ad hominem attack. Clever.

If you or others think Ma's initial pre-swap state (1) can be algebraically re-written as the same state (2): I'm scared to ask what you think the post-swap state looks like. Does it look anything like... (2)?

 

[Morbert]

If you or others think Ma's initial pre-swap state (1) can be algebraically re-written as the same state (2)

Ma explicitly says (2) is a rewriting of (1) in the basis of Bell states of photons 2 and 3. Did he misspeak?

 

[DrChinese]

"Orthodox QM" itself is a term for which there might well be at least as many meanings as there are quantum physicists.

I'll briefly summarize two accounts other than Bohmian that seem to me to be relevant here.

...

Well done! Thanks, very helpful to me.

I may be wrong, but you seem to be a fan of Ballentine (nothing wrong with that if you happen to be). All I have linked from him directly on his interpretation (called statistical back then) is that old 1970 paper. Do you have a link (not to a textbook) to an updated interpretation paper from him? My folder on Ensemble is woefully small.

-DrC

 

[PeterDonis]

Initially: Photons 1 and 2 are maximally entangled in their prepared state $\ket{\phi^-}_{12}$, and 3 and 4 similarly $\ket{\phi^-}_{34}$.

I think you meant $\psi$, not $\phi$. The initially prepared states are singlets.

So photons 2 and 3 - which can be prepared from independent sources distant to each other - cannot have any relationship.

Meaning, they cannot be entangled. That's true. And they are not entangled in the state @Sambuco wrote down (Ma's equation 2, which is equivalent to his equation 1). That state is the sum of four terms, each of which has photons 2 and 3 in a Bell state, but when you add all four terms together, all the interference terms between photons 2 and 3 cancel out and their joint state is separable. In other words, you can form a linear combination of Bell states of photons 2 and 3 in which photons 2 and 3 are not entangled. That's what Ma's equation (2) is. (And also the RHS of Peres' equation 6.) It is not saying photons 2 and 3 are entangled. They're not. It's only writing the initially prepared state, in which photons 1 and 2 are in the singlet state and photons 3 and 4 are in the singlet state, in a basis of Bell states of photons 2 and 3, for convenience. That's all it's doing.

Please take whatever time you need to read and re-read the bolded part above. It is not any kind of esoteric physical claim. It is a simple consequence of basic QM math: that you can take linear combinations of states to form other states, and that the properties of the linear combination do not have to be the same as the properties of the individual terms in it. As an example in the opposite direction, so to speak, the Bell states themselves are linear combinations of separable states of two qubits which, unlike the individual terms in them, are entangled. It can go both ways

@Sambuco showed you the basic algebra that demonstrates the bolded claim I made above. Doing basic algebra like he did on QM states is, again, a simple consequence of basic QM math, indeed it's just a version of taking linear combinations of states and manipulating them. There is nothing unusual or esoteric about it. It's just a useful mathematical tool.

 

[DrChinese]

Ma explicitly says (2) is a rewriting of (1) in the basis of Bell states of photons 2 and 3. Did he misspeak?

This is almost like deja vu all over again. Admittedly we've been trading posts rapidly in the last while.

I answered your identical question already in this post #197:

Graduate Post in thread 'Entanglement swapping and Bohmian mechanics'

Yesterday, 7:37 PM

So you're saying that, when Ma says "This can be seen by rewriting Eq. (1) in the basis of Bell states of photons 2 and 3" yielding (2), he's wrong?

Ma's physics is right. Your reading is completely wrong and ignores obvious physics. There is no connection of any kind between photons 2 and 3 prior to a swap, and no algebraic presentation can change that physics. So no, there is no Bell state basis between photons 2 and 3 implied or hidden in Ma's (1). Again, Ma's full context:

Ma: "As schematically shown in Fig. 1, if Victor subjects his photons 2 and 3 to a Bell-state...

• DrChinese

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[PeterDonis]

you seem to be a fan of Ballentine

I think his interpretation is useful to know because of what it does not claim (which I briefly described in my post). His textbook has lots of other useful stuff in it as well, of course.

 

[Sambuco]

Always humorous when someone quotes me, and proceeds to not address a single thing quoted.

As I said before, my previous #198 is not a suggestion that Ma's eq. (1) and (2) are the same, but rather a mathematical proof. Seriously, if you have any questions, I'd be happy to help.

Oh and instead of addressing my point, resorts to ad hominem attack.

It is not my intention to confront on personal terms. In fact, I appreciate the vehement way you discuss all of these topics because it shows that you're really passionate about them (Me too! ).

Lucas.

 

[PeterDonis]

All I have linked from him directly on his interpretation (called statistical back then) is that old 1970 paper. Do you have a link (not to a textbook) to an updated interpretation paper from him?

I don't offhand, but I think he still has a website at the university where he was a professor, that has links to all his papers. The reference of his that I normally use is his textbook (first edition was 1998, second edition was 2014).

 

[Morbert]

This is almost like deja vu all over again. Admittedly we've been trading posts rapidly in the last while.

I answered your identical question already in this post #197:

Graduate Post in thread 'Entanglement swapping and Bohmian mechanics'

Yesterday, 7:37 PM

So you're saying that, when Ma says "This can be seen by rewriting Eq. (1) in the basis of Bell states of photons 2 and 3" yielding (2), he's wrong?

Ma's physics is right. Your reading is completely wrong and ignores obvious physics. There is no connection of any kind between photons 2 and 3 prior to a swap, and no algebraic presentation can change that physics. So no, there is no Bell state basis between photons 2 and 3 implied or hidden in Ma's (1). Again, Ma's full context:

Ma: "As schematically shown in Fig. 1, if Victor subjects his photons 2 and 3 to a Bell-state...

• DrChinese

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• 

I responded to that in post post #201

Graduate Post in thread 'Entanglement swapping and Bohmian mechanics'

Yesterday, 7:45 PM

Ma's physics is right. Your reading is completely wrong and ignores obvious physics. There is no connection of any kind between photons 2 and 3 prior to a swap, and no algebraic presentation can change that physics. So no, there is no Bell state basis between photons 2 and 3 implied or hidden in Ma's (1). Again, Ma's full context:

Ma: "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. This can be seen...

• Morbert

• 

I just want this point to be clear. You are saying that while Ma's physics is right, he misspoke when he said (2) is a rewriting of (1)?

 

[DrChinese]

I think you meant $\psi$, not $\phi$. The initially prepared states are singlets.

You are so right! Correcting now...

 

[PeterDonis]

If you or others think Ma's initial pre-swap state (1) can be algebraically re-written as the same state (2)

As @Sambuco said, there's no "think" about it. It's a mathematical proof. It's the same algebra that I had already done before I posted, way back however many posts ago, about "algebraic refactoring" in Peres's equation (6).

I'm scared to ask what you think the post-swap state looks like

@Morbert has written that down more than once in this thread--his post #40 was the first one.

 

[DrChinese]

It is not my intention to confront on personal terms. In fact, I appreciate the vehement way you discuss all of these topics because it shows that you're really passionate about them (Me too! ).

Lucas.

Thanks for saying this. Yes, I get a bit caught up in these discussions. I really don't mind the back and forth, somehow I always gain some insight.

I'd actually be comfortable if you curse or insult my statements: "DrC - that's as bone-headed a statement as I've ever read from you". I'll take that over calling me bone-headed. LOL.

And I appreciate the time everyone is putting into this discussion. Even if I think you are dead wrong.

-DrC

 

[DrChinese]

I responded to that in post post #201

Graduate Post in thread 'Entanglement swapping and Bohmian mechanics'

Yesterday, 7:45 PM

Ma's physics is right. Your reading is completely wrong and ignores obvious physics. There is no connection of any kind between photons 2 and 3 prior to a swap, and no algebraic presentation can change that physics. So no, there is no Bell state basis between photons 2 and 3 implied or hidden in Ma's (1). Again, Ma's full context:

Ma: "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. This can be seen...

• Morbert

• 

I just want this point to be clear. You are saying that while Ma's physics is right, he misspoke when he said (2) is a rewriting of (1)?

Hmmm, I'm confused. It is deva vu.

What part of my basic message "read Ma's whole paragraph rather than just a cherry-picked portion" was not clear. Ma says (in my words): (1) is pre-swap. (2) is post-swap. I don't get how you read it otherwise.

 

[PeterDonis]

(1) is pre-swap. (2) is post-swap. I don't get how you read it otherwise.

We read it otherwise because we see that (1) is algebraically equivalent to (2). They are the same state, just algebraically refactored. @Sambuco posted the algebra explicitly.

Put aside for a moment the whole pre-swap vs. post-swap thing and just look at the algebra he posted. As a matter of algebra, do you agree with what he posted or not? If you agree, good, then we can move to the next step. If you disagree, which step do you think he did wrong?

 

[javisot]

So no, there is no Bell state basis between photons 2 and 3 implied or hidden in Ma's (1).

So yes.

which step do you think he did wrong?

I think what Dr. Chinese wants to convey is that, even though (1) can be rewritten as (2)—and that stems from the lack of an inherent order in the mathematics of QM—we know that an inherent order does exist. As I understand it, Dr. Chinese favors the "forward-in-time" perspective.

 

[PeterDonis]

(1) can be rewritten as (2)

If @DrChinese will confirm that he agrees with this, as a matter of algebra, that would be helpful, because from his posts so far I do not know whether he does or not.

and that stems from the lack of an inherent order in the mathematics of QM

No. At least, not in the view that I, @Morbert, and @Sambuco are taking. The view we are taking is that "time order" or lack thereof in QM has nothing whatever to do with (1) and (2) being equivalent. As I said in post #225, (1) and (2) are the same state. There is no "time" involved at all. (2) is not a "time evolution" of (1). It is not what (1) turns into "after the swap". It is the same state, at the same time, as (1). Just rewritten algebraically in a different basis, the Bell state basis, for convenience.

we know that an inherent order does exist. As I understand it, Dr. Chinese favors the "forward-in-time" perspective.

I'm not sure that's true, but he can tell us that himself. But the Ma and Megidish papers, whose description of the physics he appears to agree with, certainly don't appear to be taking such a perspective. The Ma paper has no problem talking about "quantum steering into the past", and the Megidish paper has no problem writing down states in which photons that never coexisted are entangled. The accounts they are giving of their experiments don't look at all like "forward in time" accounts to me.

 

[PeterDonis]

There is no connection of any kind between photons 2 and 3 prior to a swap, and no algebraic presentation can change that physics.

This is true. I put that in bold to make sure it's clear. I agree with this statement.

So no, there is no Bell state basis between photons 2 and 3 implied or hidden in Ma's (1).

However, I do not agree with this. Here's why:

Go back to the last part of the first sentence quoted above, that I agree with: "no algebraic presentation can change that fact". Which means that, if we use the algebra @Sambuco posted to change the "algebraic presentation" of Ma's state (1), we are not changing anything physical. We are not somehow making photons 2 and 3 entangled. We are just algebraically rewriting the state.

How are we rewriting the state? In the Bell state basis. We can do this because any state can always be written in any basis. The Bell states are a valid basis (each one is normalized, and they are all orthogonal, and they span the state space--every state can be written as a linear combination of them). And doing that does not change the physical meaning of the state at all. The fact that the states in the Bell state basis are entangled does not mean that every state we write as a linear combination of them is entangled. (I made this point in post #214.) So if we rewrite a state like Ma's (1) in a basis of Bell states of photons 2 and 3, which is what Ma's (2) is, we are not saying photons 2 and 3 are entangled in Ma's (1). They're not. And they're not in Ma's (2) either.

This seems to be a central point of disagreement.

 

[DrChinese]

As I said before, my previous #198 is not a suggestion that Ma's eq. (1) and (2) are the same, but rather a mathematical proof. Seriously, if you have any questions, I'd be happy to help.

In the opinion of several of you, Ma's (1) is mathematically equivalent ("the same") to their (2).

I don't have a particular objection to the algebraic derivations presented at this point (although I may in the future) since they have no physical meaning. (It's possible some of the intermediate steps in the proof might have such meaning.) To arrive at (2), a swap must be executed. If someone thinks there is any physical relevance going from (1) to (2) absent a swap, I'd be curious to know what it is.

1. My proof: I am merely stating generally accepted science using referencing and quotes from well-known sources. That is sufficient here per our standards. If you think Ma's literal quotes don't support the physics I have stated, how about you semantically (or otherwise) attack the Megidish paper. Their (2) is Ma's (1), and their (3) is Ma's (2). The bolded Megidish portion matches Ma's properly read statements IMHO (after all, it is referenced by Megidish).

"In order to project the second photon of the first pair and the first photon of the second pair onto a Bell state, the former is delayed by τ in a delay line. The same delay is also applied to the second photon of the second pair and the resulting state can be reordered and written as (3). When the two photons of time τ (photons 2 and 3) are projected onto any Bell state, the first and last photons (1 and 4) collapse also into the same state and entanglement is swapped. The first and last photons, that did not share between them any correlations, become entangled."

---

2. Further: Please do the same expansion as you go from (1) to (2) for the following:

|Ψ〉123456 = |Ψ−〉12⨂|Ψ−〉34⨂|Ψ−〉56

Hopefully you will see that there is no meaning in terms of physics to the expansion you arrive at. There is no more any Bell state representation present in the expansion of the above between photons 2 and 3, or photons 2 and 5, or photons 3 and 5, than the man in the moon. You must execute a swap.

---

3. My points here, as all along: There is a pre-swap state (1), and a post-swap state (2), and they are different. There is no connection of any kind pre-swap between the initially prepared |Ψ−〉 systems. And without a swap, there is no identifiable experimental correlations to be reported upon completion. How can anyone object to these statements?

I am hoping we can settle the point because a question I think we are still working on is: What is the state of photon 2 when it entered the Projecting beam splitter (BS), after photon 1 has already been determined? It was originally part of a biphoton in state |Ψ−〉12. Does it enter the BS retaining elements of its initial entanglement? Or has it collapsed to a pure state (maybe |V> or |H> depending on the basis of photon 1's measurement? Because it isn't yet in a Bell state with photon 3.

 

[PeterDonis]

I don't have a particular objection to the algebraic derivations presented at this point (although I may in the future) since they have no physical meaning.

Here we disagree. They do have a physical meaning: they mean that (1) and (2), physically, are the same state, just written in different bases.

This is generally true in QM: if I have two expressions for states, and I can algebraically transform one into the other, then physically, they are the same state.

So if you agree that @Sambuco did the algebra correctly, then you should also agree that Ma's (1) and (2) are physically the same state.

To arrive at (2), a swap must be executed.

Here we disagree again. See above: (1) and (2) are physically the same state. Executing a swap changes the state, so (2) cannot be what you get if you take (1) and execute a swap. Whatever you get when you execute the swap cannot be algebraically equivalent to what you had before the swap.

@Morbert's #40 showed what you get if you take (1) or (2)--they're the same, but it's a lot easier to see if you use (2), that's why it's convenient to rewrite (1) as (2)--and execute a swap, but don't yet know which result you got. The state he wrote there is a mixture of the four different possibilities for the result of the swap, each with equal probability (1/4).

Once you know which result you got, of course, you just have one term of that mixture, which will be a product of a Bell state of photons 2 and 3 and the same Bell state of photons 1 and 4. (At least, that's the math--what different interpretations say the math means is something else, I've posted a few takes on that already.)

 

[DrChinese]

1. If @DrChinese will confirm that he agrees with this, as a matter of algebra, that would be helpful, because from his posts so far I do not know whether he does or not.

2. This is true. [DrC: no connection between photons 2 and 3 absent a swap.] I put that in bold to make sure it's clear. I agree with this statement.

3. However, I do not agree with this. [DrC: there is no Bell state basis between photons 2 and 3 implied or hidden in Ma's (1).]

1. Consider for now that I have no objection to the derivation. I might later, but that's a separate issue and relates to an earlier reference in this thread that I chose not to address here.

2. Great.

3. Well, now you are saying that as we add more and more photons |Ψ〉12345678... and so on to the equation, we'll find a large number of implied/hidden Bell state permutations between all of these. That's a gosh darn lot of hidden stuff! And what relevant physical meaning do you ascribe to all of those?

No surprise here: I would say none whatsoever.

 

[PeterDonis]

without a swap, there is no identifiable experimental correlations to be reported upon completion.

Yes, agreed. So let's test this: let's take Ma's (2), and assume, just for the sake of argument, that it's a pre-swap state, and compute the photon 2 and 3 correlations it predicts.

What do we mean when we say that it's a pre-swap state? We mean that we do not put photons 2 and 3 through the BSM. That means we do not have any subensembles in our data picked out by any BSM results. We just have one ensemble of data: we do a large number of runs where we have 4 photons in the state described by Ma's (2)--without inquiring as to how they got to us, assume they come through a hole in the wall from a hidden lab whose internals we can't inspect, but the lab certifies, with a seal of approval stamped by Ma and Zeilinger, that every set of 4 photons that comes out is described by Ma's (2)--and we do nothing to them except measure the polarizations of photons (2) and (3). Once we have a large enought dataset, we do the statistics on the photon 2 and 3 results and compute their correlation.

How do we predict mathematically what results we expect from this? Standard QM says that, for each distinct possibility we are concerned with, we sum the amplitudes for all the terms in the state that realize that possibility, and then square the final amplitude to get the probability. For Ma's (2), just looking at photons 2 and 3, this is straightforward, at least for what we need. The only distinct possibilities we need to consider are photons 2 and 3 being parallel or antiparallel, since those are the only ways for them to be correlated (positive correlation or negative correlation). We have two terms where photons 2 and 3 are antiparallel (the $\Psi^\pm$ terms), with equal magnitudes and opposite signs: so the net amplitude for photons 2 and 3 to be negatively correlated is zero. And we have two terms where photons 2 and 3 are parallel (the $\Phi^\pm$ terms), with equal magnitudes and opposite signs, so the net amplitude for photons 2 and 3 to be positively correlated is zero. So the overall result is that we compute zero correlation between photons 2 and 3 in the state described by Ma's (2).

(Edited to add: note that the minus signs I referred to above, that cancel out the amplitudes for positive and negative correlations, reflect the destructive interference that @Sambuco describes in #232.)

 

[Sambuco]

There is a pre-swap state (1), and a post-swap state (2), and they are different.

In order not to distract the discussion too much, I would like to focus on this statement because I think it is the core of the confusion.

Before the swap, the quantum state is (eq. (1) in Ma's paper):

$\ket{\psi}_{1234} = \ket{\psi^-}_{12} \otimes \ket{\psi^-}_{34}$

As I demonstrated, it can be written also as (eq. (2) in Ma's paper):

$\ket{\psi}_{1234} = \frac{1}{4} (\ket{\psi^+}_{14} \ket{\psi^+}_{23} - \ket{\psi^-}_{14} \ket{\psi^-}_{23} - \ket{\phi^+}_{14} \ket{\phi^+}_{23} + \ket{\phi^-}_{14} \ket{\phi^-}_{23})$

What I understand from what you say is that, when you see this last equation, it gives you the impression that photons 2 and 3 are entangled because the Bell states appear. To put it in a way that I think may help, although each of those terms in the sum represents a Bell state where photons 2 and 3 are entangled, the truth is that when you add the four terms, the entanglement between photons 2 and 3 that each term contributes destructively interferes, such that the entire state shows no correlation between photons 2 and 3. Just as you said, photons 3 and 4 are equivalent to any other pair of photons, like photons 5 and 6. Your intuition is correct!

To make an analogy, in the typical two-slit experiment, the wave function of the particle emerging from each of the slits gives rise to a more or less continuous detection pattern on the screen. However, when both wavefunctions are added, dark points appear where the particle cannot be detected, even though the terms of the sum, each taken individually, say that it could be detected at that point. The same happens with the eq. (2) in Ma's paper. That is to say, although terms appear that seem to suggest a possible entanglement between photons 2 and 3, the truth is that when they are all added, said entanglement is completely cancelled.

Now, how can we see the transition between the pre-swap state (eq. (2)) and a state after the swap? Well, as @PeterDonis mentioned, what the swap does is project the state into a particular Bell state. That is, the physical effect of the swap, seen from the quantum state, is to eliminate all terms from the eq. (2), except one, which corresponds to the result that Víctor obtains at the BSM. When that happens, only one of the terms that were in the eq. (2) remains, and now there is entanglement, because those other terms that gave rise to that destructive interference are no longer there.

Lucas.

 

[PeterDonis]

now you are saying that as we add more and more photons |Ψ〉12345678... and so on to the equation, we'll find a large number of implied/hidden Bell state permutations between all of these.

In the sense that we could rewrite them in the appropiate Bell state bases, yes. That would not physically change the states at all, as I've said many times now. It just rewrites them in a different basis.

And of course as you add more dimensions to your linear vector space, you're adding more possible permutations of ways to algebraically rewrite states. Why would you expect anything else?

That's a gosh darn lot of hidden stuff!

If you think that having an infinite number of possible bases for a linear vector space, and being able to rewrite any state in any basis, is "a gosh darn lot of hidden stuff", then I guess it would be, yes. But to me it seems like making an awful lot of heavy weather about something very simple. Linear algebra is not some kind of esoteric cult of hidden secrets.

And what relevant physical meaning do you ascribe to all of those?

I've already said that: rewriting a state in a different basis does not physically change the state at all. It's still physically the same state.

 

[PeterDonis]

the physical effect of the swap, seen from the quantum state, is to eliminate all terms from the eq. (2), except one, which corresponds to the result that Víctor obtains at the BSM.

Just one thing to add to this, based on my #229: if you want to model the intermediate stage where Victor has put photons 2 and 3 through the BSM, but has not yet read off the result, @Morbert's #40 showed what the mathematical description of that looks like. Once you know the result, as you say, all but one of the terms drop out.

 

[Roberto Pavani]

Linear algebra is not some kind of esoteric cult of hidden secrets.

Agree

 

[Demystifier]

I've found an intuitive pictorial way to answer one of the @DrChinese 's central questions: How to describe the timeline of the full entanglement swapping experiment, step by step, in BM? Here is how.

Initially at time $t_1$ the wave function is the product $|\psi_{AB}\rangle|\psi_{CD}\rangle$. In BM I represent it with
$$ t_1: \; A \mathrel{-} B \;\;\;\; C \mathrel{-} D$$
The link "$\mathrel{-}$" between two particles is not a "minus" sign. Instead, this link denotes the Bohmian instantaneous mutual influence between the particles. $A$ is linked with $B$, and $C$ is linked with $D$, but there is no link between $B$ and $C$. Note that the link does not have an arrow, because at the fundamental microscopic level there is no time arrow, i.e., there is no "cause" and "effect".

Next, at time $t_2$, we independently measure particles $A$ and $D$, so they interact with the macroscopic measuring apparatuses $M_A$ and $M_D$, respectively. In BM I represent this with
$$ t_2: \; M_A \mathrel{-} A \mathrel{-} B \;\;\;\; C \mathrel{-} D \mathrel{-} M_D$$
or equivalently
$$ t_2:\; (M_A,A)_2 \mathrel{-} B \;\;\;\; C \mathrel{-} (M_D,D)_2$$
The notation $(M_A,A)_2$ denotes that the system $M_A+A$ is viewed as one system at time $t_2$.

Finally, at time $t_3$, we measure the system $B+C$, so
$$ t_3:\; (M_A,A)_3 \mathrel{-} B \mathrel{-} M_{BC} \mathrel{-} C \mathrel{-} (M_D,D)_3$$
or equivalently
$$ t_3:\; (M_A,A)_3 \mathrel{-} (B,M_{BC},C)_3 \mathrel{-} (M_D,D)_3$$
This finishes the description at the fundamental microscopic level.

However, at the emergent macroscopic level things look slightly different. The measurement outcomes encoded in $M_A$ and $M_D$ are stored on a computer, so they don't change much during the time. Also, after the measurement, the macroscopic system $M_A+A$ is practically indistinguishable from $M_A$ alone, i.e., the state of the particle $A$ after the measurement is practically irrelevant. The particle $A$ can even be destroyed, it doesn't matter as long as we keep the measurement outcome encoded in $M_A$. Thus, for practical purposes, we can use the approximation $(M_A,A)_2 \simeq (M_A,A)_3$. Likewise, we have $(M_D,D)_2 \simeq (M_D,D)_3$. Hence, effectively, the microscopic state at $t_3$ implies an effective macroscopic description
$$ (M_A,A)_2 \mathrel{-} (B,M_{BC},C)_3 \mathrel{-} (M_D,D)_2$$
At the statistical level, this explains why the results of measurements in the past at time $t_2$ are mutually correlated when combined with measurement results in the future at time $t_3$: $(M_A,A)_2$ is correlated with $(B,M_{BC},C)_3$, and $(B,M_{BC},C)_3$ is correlated with $(M_D,D)_2$, so it is not surprising that $(M_A,A)_2$ can be correlated with $(M_D,D)_2$. Furthermore, at the macroscopic level we also have a time arrow, due to which the past causes the future, so the correlation can also be interpreted causally as
$$ (M_A,A)_2 \rightarrow (B,M_{BC},C)_3 \leftarrow (M_D,D)_2$$
Thus we see that, at the macroscopic level, we can say that the results of measurements in the past cause the result of measurement in the future.

Finally, in the instrumental version of BM (see the paper in my signature and references therein), the Bohmian trajectories of the measured particles can be completely eliminated from the description, leaving only particle trajectories constituting the measuring apparatuses. In this description the last causal diagram reduces to
$$ (M_A)_2 \rightarrow (M_{BC})_3 \leftarrow (M_D)_2$$

This finishes my description of the entanglement swapping experiment from the Bohmian point of view, as an intriguing interplay between the fundamental microscopic and emergent macroscopic descriptions.

 

[Morbert]

Hmmm, I'm confused. It is deva vu.

What part of my basic message "read Ma's whole paragraph rather than just a cherry-picked portion" was not clear. Ma says (in my words): (1) is pre-swap. (2) is post-swap. I don't get how you read it otherwise.

What Ma actually says is (2) is arrived at by "rewriting Eq. (1) in the basis of Bell states of photons 2 and 3" (Ma's exact words). What you did is rewrite this to say (2) is "the newly swapped state" (your words, not Ma's).

The options are clear:
i) Ma misspoke when he said (2) is a rewriting of (1), and what he meant to say was (2) is the state post-swap.

ii) Ma meant what he said but he was wrong. (2) is not in fact a rewriting of (1). (2) is instead the state post-swap.

iii) Ma was perfectly correct when he said (2) is a rewriting of (1), as evidenced by undergraduate-level quantum theory.

I and others take position iii). You are taking position i). It's a brave move to both use Ma's eminence as an excuse to not learn basic QM, and to critique Ma's presentation of basic QM at the same time.

 

[martinbn]

The mathematical issues are just highschool algebra. If you have two variables, say $x$ and $y$ and consider the product $xy$. If you change the variables to $x=u+v$ and $y=u-v$, you get

$xy=(u+v)(u-v)=uu-uv+vu-vv$.

A product an be written as a sum of products.

 

[Demystifier]

Just for fun, let me also depict the Hardy setup in the notation above. Initially we have a non-entangled electron-positron pair
$$t_0: A \;\;\;\; B$$
Then we allow them to interact and possibly annihilate, which can be viewed as a joint measurement
$$t_1: A - M_{AB} - B$$
This establishes the correlation between $A$ and $B$, conditioned on the measurement outcome $M_{AB}$. This is the fundamental microscopic description.

At the effective macroscopic level, we can consider a subensemble in which the outcome $M_{AB}$ is always the same, saying that the annihilation did not occur. Since $M_{AB}$ is always the same, it is no longer treated as a variable, so can be ignored. Thus the effective description reduces to a correlation between $A$ and $B$ described as
$$t_1: A - B$$
which is the entangled Hardy state.