Entanglement swapping and Bohmian mechanics

[PeterDonis]

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.

I think it illustrates that there is not a general consensus among physicists about how to interpret these experiments. I don't have any particular comments to make about the details of what it presents; they do look straightforward to me, yes, they just don't suggest to me anything that hasn't already been said in these threads plenty of times. But others might see something that prompts a comment.

 

[DrChinese]

1. 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.

2. 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.

1. PeterDonis: Projector required...

2. PeterDonis: Projector not required.

And the answer is: 1.

I will not comment further back to you on this. I have quoted the masters (including Peres), and I will simply echo those quotes on the subject.

-DrC

 

[DrChinese]

1. What you mean to say is

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

3. The LHS is the pure state representing the initial preparation.

4. 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.

5. 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.

1. Hmmm... I directly quoted Ma et al. Since I agree with those pioneering experts: yes I meant to say what I wrote.

2. I agree this is also true. That's exactly what I told PeterDonis, expressed mathematically.

3. It's the initial preparation...

4. And this is the post-swap certain result (1 of 4 possible Bell states, featuring 2 and 3 entangled). Obviously not the same as the pre-swap preparation.

5. Admittedly I have a minor (hopefully) quibble with your #40. There, you present the equation we have as the RHS here - we both agree on this point.

However, there you qualify that by an additional comment to the effect of "that's assuming 2 and 3 are measured first". In actual fact, that's also the exact same equation if instead 1 and 4 are measured first. We know that because the RHS was quoted verbatim from the Ma et al Delayed Choice paper, where 1 and 4 are measured first. (Order in swapping, of course, doesn't matter.)

 

[Morbert]

1. Hmmm... I directly quoted Ma et al. Since I agree with those pioneering experts: yes I meant to say what I wrote.

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 [My emphasis]

Show me where Ma says that. You'll find it difficult since Ma explicitly says [1] can be rewritten as [2].

You're presenting your basic algebra mistakes [edit] as Ma's. What's frustrating is these are precisely the same mistakes you were making in earlier threads. No progress has been made.

 

[DrChinese]

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.

I said I wasn't going to comment back further to you, and yet here I am...

It is right. Correcting your verbage, which I hope resolves anyone's confusion: After the swap, and immediately coming out of the beam splitter, the RHS is the correct representation (showing all 4 possible Bell states). In orthodox theory, the measurement* of the indistinguishable pair (2/3) collapses that to a single one of those 4 terms. Of course, different interpretations would have a variety of things to say here; so I leave it at that. Your wording might be accurate for one of those interpretations, I can't say.

This is certainly no different than what we have with the initial singlet pairs, in a $\Phi^{-}$ state. That might later collapse to an |V> or |H> state, but initially it could be either of 2 terms (something like |V>+|H> or whatever).

*There are subsequent polarizers involved in the BSM measurement process, so those should be considered too.

 

[PeterDonis]

You're presenting your basic algebra mistakes is Ma's.

I think you mean "as" Ma's?

I agree with your post.

 

[Morbert]

I think you mean "as" Ma's?

I agree with your post.

fixed

 

[DrChinese]

1. Show me where Ma says that.

2. You'll find it difficult since Ma explicitly says [1] can be rewritten as [2].

3. You're presenting your basic algebra mistakes as Ma's.

1. Are you looking at the paper? I copied and pasted directly from Ma's (1) and (2)! Not really sure how many times I need to repeat these quotes, but this is not gonna be another.

2. But you'll have to show me where Ma says their pre-swap (1) is the same as ("can be re-written as") post-swap (2). The swap changes the state, so they are definitely not equivalent.

3. I kindly suggest you reconsider before you say it's my basic algebra mistake. I'm directly quoting a Nobel level paper.

 

[PeterDonis]

After the swap, and immediately coming out of the beam splitter, the RHS is the correct representation (showing all 4 possible Bell states).

The RHS of @Morbert's equation in #171 is, yes. But not the RHS of Peres' equation (6) or Ma's equation (2).

You appear to believe that all three of those are the same. @Morbert and I both agree that they're not.

You appear to believe that Ma et al are agreeing with you. @Morbert believes, and I agree with him, that Ma is saying what he and I are saying, and not what you are saying.

I copied and pasted directly from Ma's (1) and (2)!

Sure. Nobody is saying otherwise. You correctly copied and pasted those two equations.

Now here's a copy and paste from the text of Ma's paper just before (2):

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

Emphasis mine. Ma is saying the same thing about his (2) and (1), as I said about the RHS and LHS of Peres's equation (6). Ma's (1) is the same as the LHS of Peres' equation (6). Ma's (2) is the same as the RHS of Peres' equation (6). The only difference is the labeling of the photons in the subscripts. Ma's (2) is a rewriting (Ma's word) of the same state as (1) in a different basis (Ma's word). So is the RHS of Peres's equation (6) vs. the LHS.

That's my reading, and that's @Morbert's reading. Your reading is different. As @Morbert said, no progress has been made. We're in the same place we were before. I don't see how we're going to get anywhere if we can't even agree on such a basic point to the whole discussion.

 

[Morbert]

1. Are you looking at the paper? I copied and pasted directly from Ma's (1) and (2)! Not really sure how many times I need to repeat these quotes, but this is not gonna be another.

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

Ma says (1) can be written in the basis of Bell states of photons 2 and 3. You are misreading this as Ma saying (2) is the quantum state post-measurement.

 

[DrChinese]

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).

I think that may be tripping up some folks. To a certain degree, that is one of the issues of this thread: Demystifier raised the issue - which I had challenged in several ways in posts - of the timing of when the superposition becomes one of the mutually exclusive alternatives.

In the Bohmian perspective, I believe this is instantaneous and nonlocal. In the orthodox view, the timing is more blurred. All we know there is that by the end of the complete measurement context, we record the results. Intermediate "states" are not fully evident.

 

[DrChinese]

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

Ma says (1) can be written in the basis of Bell states of photons 2 and 3. You are misreading this as Ma saying (2) is the quantum state post-measurement.

If you think (1) and (2) are merely the same state, written differently outside of and in the absence of an intermediate BSM: I can't really help you. Ma clearly says that (2) is dependent on Victor's execution of a swap. Ma's "What can be seen" is the new swapped state.

Thinking that (1) somehow can be algebraically re-arranged to be (2) is wrong. Consider this simple point about the actual PHYSICS being we are discussing. Photons 2 and 3 initially have no relationship whatsoever. Pairs (1&2) and (3&4) could exist anywhere in the universe, at any time in history! Ma's (1) would still be absolutely correct, but (2) would not be. And ditto for dual pairs (1&2) and some other (5&6), or (1&2) and some (7&8), etc. Obviously, none of these pairs share any Bell state. This is elementary physics of entanglement. State (1) is not state (2) under any meaningful viewpoint.

But hey, feel free to stick to your guns.

 

[Morbert]

If you think (1) and (2) are merely the same state, written differently outside of and in the absence of an intermediate BSM: I can't really help you. Ma clearly says that (2) is dependent on Victor's execution of a swap. Ma's "What can be seen" is the new swapped state.

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?

 

[javisot]

Photons 2 and 3 initially have no relationship whatsoever.

Why?

If 2 and 3 were maximally entangled, they couldn't be entangled with 1 and 4 respectively, right? Because that would violate MoE.

Can't we flip everything around and say that 1 and 4 were the ones maximally correlated to begin with? If there is no inherent order, both assumptions should lead to the same result.

 

[DrChinese]

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

Great point! Thank you...

I believe that Bohmian perspective (initial subject of this thread, I think ) says/implies a specific ordering of intermediate events in the full context chain: That being: i) Initial preparation, ii) detection of photons 1/4, iii) collapse of photons 2/3, iv) execution of swap/BSM. To me, that appears to be inconsistent with experiment - but I'm not really sure. I am also trying to contrast that (BM) with the orthodox QM view; which as we are seeing, takes on different forms according to whom is presenting it.

So one of the questions - in any interpretation of the swapping protocol - is:

If photons 1 and 4 have already been measured (as in the Ma Delayed Choice paper), then what state are the 2 and 3 photons in PRIOR to the swap (specifically prior to the projecting Beam Splitter BS).
a) Are they still in the "singlet" state from when they were originally created? or
b) Are they now in a collapsed state, consistent with the observed outcome of their entangled partner? or
c) In some viewpoints, this might be considered a meaningless question.

- Assuming for discussion purposes c) is not to be considered: My intuition tells me that the orthodox answer must be a). But I'm not sure I can formulate a precise explanation of why. I believe @PeterDonis agrees the input photon 2 to the BSM is in the singlet state, but he should probably weigh in on that himself.

- I believe the Bohmian view to be b), but I also know that @Demystifier doesn't see that to be quite the case.

- And I believe if b) were to be asserted, it would be inconsistent with the physics of Parametric Down Conversion entanglement. I believe @Demystifier is aware of the point I am making on that, although I definitely think he disagrees with my characterization.

I have used "I believe" a lot in this post. As I have said many times, I am asking for thoughts and input on this - all of which is relevant to this thread. What I do know is what I read and quote from generally accepted sources on swapping. Clearly, some of my questions are not well-covered in the literature. So unfortunately, we have been reduced to debating the precise meaning of phrases. But there is still plenty of interesting physics to encounter along the way.

 

[Morbert]

Why?

If 2 and 3 were maximally entangled, they couldn't be entangled with 1 and 4 respectively, right? Because that would violate MoE.

Can't we flip everything around and say that 1 and 4 were the ones maximally correlated to begin with? If there is no inherent order, both assumptions should lead to the same result.

Photons 2 and 3 are not initially entangled, as can be seen by tracing 1 and 4 out of the initial state (equation (1), or equation (2), which is a rewriting of (1)). @DrChinese is misreading (2) as a state where 2 and 3 and entangled. Instead it's the initial state, where 2 and 3 are not entangled.

 

[DrChinese]

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 measurement, they become entangled. Consequently photons 1 (Alice) and 4 (Bob) also become entangled and entanglement swapping is achieved. This can be seen by rewriting Eq. (1) in the basis of Bell states of photons 2 and 3..."

If Ma had instead written (not that it needs to be changed) the following for the above, we probably would not be having this discussion:

"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. The newly swapped state |Ψ〉1234, in the basis of Bell states of photons 2 and 3, would be..."

 

[Sambuco]

State (1) is not state (2) under any meaningful viewpoint.

They are the same state!

First, some basic definitions from Ma's paper:

$\ket{\psi^\pm}_{ij} = \frac{1}{\sqrt{2}}(\ket{H}_i \ket{V}_j \pm \ket{V}_i \ket{H}_j)$
$\ket{\phi^\pm}_{ij} = \frac{1}{\sqrt{2}}(\ket{H}_i \ket{H}_j \pm \ket{V}_i \ket{V}_j)$
$\ket{H}_i \ket{V}_j = \frac{1}{\sqrt{2}}(\ket{\psi^+}_{ij} + \ket{\psi^-}_{ij})$
$\ket{V}_i \ket{H}_j = \frac{1}{\sqrt{2}}(\ket{\psi^+}_{ij} - \ket{\psi^-}_{ij})$
$\ket{H}_i \ket{H}_j = \frac{1}{\sqrt{2}}(\ket{\phi^+}_{ij} + \ket{\phi^-}_{ij})$
$\ket{V}_i \ket{V}_j = \frac{1}{\sqrt{2}}(\ket{\phi^+}_{ij} - \ket{\phi^-}_{ij})$

I will start from Ma's eq. (1) and arrive at eq. (2):

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

QED.

Lucas.

 

[DrChinese]

Photons 2 and 3 are not initially entangled, as can be seen by tracing 1 and 4 out of the initial state (equation (1), or equation (2), which is a rewriting of (1)). @DrChinese is misreading (2) as a state where 2 and 3 and entangled. Instead it's the initial state, where 2 and 3 are not entangled.

How many times do I need to say:

[2] presents an entangled state of photons 2 and 3. Therefore it cannot be the initial state.

|Ψ〉1234 = 12(|Ψ+〉14⨂|Ψ+〉23 − |Ψ−〉14⨂|Ψ−〉23 − |Φ+〉14⨂|Φ+〉23 + |Φ−〉14⨂|Φ−〉23) [2]

 

[javisot]

Photons 2 and 3 are not initially entangled

We all agree on that.

My question is whether it's really important.

 

[Morbert]

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 by rewriting Eq. (1) in the basis of Bell states of photons 2 and 3..."

If Ma had instead written (not that it needs to be changed) the following for the above, we probably would not be having this discussion:

"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. The newly swapped state |Ψ〉1234, in the basis of Bell states of photons 2 and 3, would be..."

So you're saying Ma's physics is right and he just misspoke? When he said "This can be seen by rewriting Eq. (1) in the basis of Bell states of photons 2 and 3", he misspoke, as (2) is not in fact a rewriting of (1)?

 

[DrChinese]

They are the same state!

...

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

QED.

Lucas.

Q: Why is photon 2 presented in a Bell state with photon 3 (such as your $\ket{\psi^+}_{23}$ above), and not in that same identical state with photon 5? Or photon 6? Or any other photon in the entire observable universe?

A: It can't be properly so represented! Why? Because initially photon 2 has a maximally entangled relationship with photon 1, and cannot have - or be expressed as having - a relationship with any other particle while entangled with photon 1.

Again, this is a basic rule of entanglement. If you don't know it, I'll gladly give you a reference on Monogamy of Entanglement.

 

[PeterDonis]

Bohmian perspective (initial subject of this thread, I think ) says/implies a specific ordering of intermediate events in the full context chain: That being: i) Initial preparation, ii) detection of photons 1/4, iii) collapse of photons 2/3, iv) execution of swap/BSM.

As long as you agree that "collapse" means "effective collapse" as Bohmian mechanics defines that (we've discussed this point previously, just a reminder), yes, this is consistent with what I wrote in post #178.

To me, that appears to be inconsistent with experiment - but I'm not really sure.

It's not. Post #178 explains how the Bohmian perspective you describe accounts for the experimental results. Specifically:

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.

As I noted in that post, I'm discussing in the above quote the idealized case where the experimental setup allows all four Bell states to be distinguished.

 

[Morbert]

They are the same state!

First, some basic definitions from Ma's paper:

$\ket{\psi^\pm}_{ij} = \frac{1}{\sqrt{2}}(\ket{H}_i \ket{V}_j \pm \ket{V}_i \ket{H}_j)$
$\ket{\phi^\pm}_{ij} = \frac{1}{\sqrt{2}}(\ket{H}_i \ket{H}_j \pm \ket{V}_i \ket{V}_j)$
$\ket{H}_i \ket{V}_j = \frac{1}{\sqrt{2}}(\ket{\psi^+}_{ij} + \ket{\psi^-}_{ij})$
$\ket{V}_i \ket{H}_j = \frac{1}{\sqrt{2}}(\ket{\psi^+}_{ij} - \ket{\psi^-}_{ij})$
$\ket{H}_i \ket{H}_j = \frac{1}{\sqrt{2}}(\ket{\phi^+}_{ij} + \ket{\phi^-}_{ij})$
$\ket{V}_i \ket{V}_j = \frac{1}{\sqrt{2}}(\ket{\phi^+}_{ij} - \ket{\phi^-}_{ij})$

I will start from Ma's eq. (1) and arrive at eq. (2):

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

QED.

Lucas.

@DrChinese is unfortunately caught in a closed loop where he will deny this QM math based on his misreading of Ma's paper, and his misreading of Ma's paper is based on his misunderstanding of QM math.

 

[Sambuco]

Q: Why is photon 2 presented in a Bell state with photon 3 (such as your |ψ+⟩23 above), and not in that same identical state with photon 5? Or photon 6? Or any other photon in the entire observable universe?

A: It can't be properly so represented! Why? Because initially photon 2 has a maximally entangled relationship with photon 1, and cannot have - or be expressed as having - a relationship with any other particle while entangled with photon 1.

Again, this is a basic rule of entanglement. If you don't know it, I'll gladly give you a reference on Monogamy of Entanglement.

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.

 

[DrChinese]

Why? [do photons 2 and 3 initially have no relationship]

If 2 and 3 were maximally entangled, they couldn't be entangled with 1 and 4 respectively, right? Because that would violate MoE.

Can't we flip everything around and say that 1 and 4 were the ones maximally correlated to begin with? If there is no inherent order, both assumptions should lead to the same result.

Initially: Photons 1 and 2 are maximally entangled in their prepared state $\ket{\psi^-}_{12}$, and 3 and 4 similarly $\ket{\psi^-}_{34}$. So photons 2 and 3 - which can be prepared from independent sources distant to each other - cannot have any relationship. They must first be brought together physically for the BSM/swap.

 

[DrChinese]

1. As long as you agree that "collapse" means "effective collapse" as Bohmian mechanics defines that (we've discussed this point previously, just a reminder), yes, this is consistent with what I wrote in post #178.

2. As I noted in that post, I'm discussing in the above quote the idealized case where the experimental setup allows all four Bell states to be distinguished.

1. Agree that when a Bohmian says "collapse", they often mean "FAPP collapse" or your "effective collapse".

2. That's perfect.

 

[javisot]

Q: Why is photon 2 presented in a Bell state with photon 3 (such as your $\ket{\psi^+}_{23}$ above), and not in that same identical state with photon 5? Or photon 6? Or any other photon in the entire observable universe?

A: It can't be properly so represented! Why? Because initially photon 2 has a maximally entangled relationship with photon 1, and cannot have - or be expressed as having - a relationship with any other particle while entangled with photon 1.

Again, this is a basic rule of entanglement. If you don't know it, I'll gladly give you a reference on Monogamy of Entanglement.

I understood that you were referring to that with those words in bold.

We've said that there is no inherent order (this statement wouldn't be an interpretation but rather part of orthodox QM). Forward in time or backward in time would be interpretations that must explain the same thing. I do not see MoE being violated in any case, in this case of four photons that we are dealing with. (I don't see MoE being violated regardless of whether (1) and (2) are the same state or not)

 

[Morbert]

Q: Why is photon 2 presented in a Bell state with photon 3 (such as your $\ket{\psi^+}_{23}$ above), and not in that same identical state with photon 5? Or photon 6? Or any other photon in the entire observable universe?

Equation (2) does not present photon 2 and 3 in a Bell state. It instead shows the initial state expanded in a Bell basis. An analogous mistake would be

@Morbert : "I have `2 + 2` dollars."

@DrChinese : "No you don't! You have `4` dollars!"

 

[PeterDonis]

I am also trying to contrast that (BM) with the orthodox QM view; which as we are seeing, takes on different forms according to whom is presenting it.

"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.

Ma and Megidish

First, write down the "obvious" wave function that takes into account the entire experimental context, disregarding any issues about the timing of the various events or whether that wave function "really is" the state of anything at any particular time. In the Ma paper, that wave function is Equation (2); in the Megidish paper, it's Equation (3). Use that to predict the results by the simple procedure of squaring the coefficient in front of each term to get the probability for that result (which will be 1/4 for each), and then using the appropriate statistics for each Bell state. Compare with experiment. And of course they agree.

Second, adopt an interpretation of all of the above in which the fact that the same process works regardless of the timing of the measurements of photons 1 and 4 (they can both be done before the swap--Ma--or they can be done so that the photons never coexist--Megidish) means that the straightforward realist description of the "ordinary" case--where the swap is done before all four photons are measured, and the obvious interpretation that after the swap, photons 1 and 4 "really are" entangled in the appropriate Bell state seems unproblematic--also applies in the other cases, even though it means accepting, or at least seriously considering, things like backwards in time causation.

Statistical/Ensemble Interpretation (a la Ballentine)

First, carefully distinguish, in the experimental process, two parts: state preparation and measurement. In this case, the state preparation is that Alice prepares photons 1 and 2 in the singlet state, Bob prepares photons 3 and 4 in the singlet state, and then photons 2 and 3 are passed to Victor, who puts them through the BSM. The measurement is that all four photons have their polarizations measured.

Now write down the wave function that corresponds to the state preparation. This is, of course, the same one that Ma and Megidish wrote down. Note that, like them, we are not concerned with the timing of the events; specifically, we are not concerned with the fact that a portion of the measurement can take place before the state preparation is completed. As long as each part is well defined, and as long as mathematically, we do the entire state preparation before we apply any mathematical operations regarding measurement, we will make correct predictions. (This is equivalent to taking into account the entire experimental context.) Use the wave function to make predictions as in Ma and Megidish, above.

Second--well, there is no second in this interpretation. This interpretation makes no claims about what is "really happening" in any individual run of the experiment. The wave function describes the state preparation process, or equivalently an abstract ensemble of systems prepared according to that process; it does not describe individual quantum systems in individual experimental runs. The predicted statistics match the experimental statistics, and that's it.