Entanglement swapping and Bohmian mechanics

[Matterwave]

The answer is: Victor's Bell State Measurement, leading to $\Phi^+$ and $\Phi^-$ as you say, can be done on any basis. If Victor were to choose H/V, L/R, or +/-, all of those can cast Alice and Bob's photons into either $\Phi^+$ or $\Phi^-$ - regardless of what basis Alice and Bob choose.

When Victor's apparatus is set to perform a BSM, the basis is $\{\Phi^+,\Phi^-,HV,VH\}$. When Victor's apparatus is set to perform an SSM, the basis is $\{HH,VV,HV,VH\}$.

Understood. Up to here, I feel like you guys are telling me the same thing. Is there any disagreement on this?

I think my confusion was simply that I regarded the "H/V basis" in @DrChinese post #309 wording to mean Victor only ever measured H or V (I guess that's the SSM), and not that he measures in a basis where the possible outcomes are $\{\Phi^+,\Phi^-,HV,VH\}$ (the BSM). Thanks both for clarifying for me. :)

Here are the caveats:

i) In the Ma et al reference, Victor always chooses the H/V basis.

When you say "H/V basis" here do you mean the basis $\{\Phi^+,\Phi^-,HV,VH\}$?

ii) When Alice and Bob measure on the same basis as Victor, there is still the a swap occurring.

This is where I get a bit confused on the terminology "basis". Alice and Bob each just get 1 particle right? So they can't use the "same (4-state) basis" as Victor who has 2 particles? So here you mean whether Alice and Bob measure literally H/V or L/R or +/-?

 

[Morbert]

This is where I get a bit confused on the terminology "basis". Alice and Bob each just get 1 particle right? So they can't use the "same (4-state) basis" as Victor who has 2 particles? So here you mean whether Alice and Bob measure literally H/V or L/R or +/-?

We can model Alice's and Bob's measurement in the HV bases as a single measurement on the 1&4 subsystem which has the basis $\{HH, HV, VH, HH\}$. In which case the basis will align with Victor's SSM basis. But you are right: If Victor configures his apparatus for an imperfect BSM, the basis will be different from the Alice-Bob basis.

 

[Matterwave]

We can model Alice's and Bob's measurement in the HV bases as a single measurement on the 1&4 subsystem which has the basis $\{HH, HV, VH, HH\}$. In which case the basis will align with Victor's SSM basis.

Got it! Thanks for clarifying!

 

[DrChinese]

1. What's wrong with correlation if Alice and Bob both measure in the HV basis and Victor obtains $\Phi^+$?

2. PS Just in case: Here "No BSM" means no BSM when Victor has his apparatus set to attempt a BSM.

It may be helpful to simply jump to the bottom for the summary.

1. $\Phi^+$ generally yields opposite Alice/Bob correlation to $\Phi^-$, right? Actually this is a point is not entirely clear in Ma. I have always been under the impression that $\Phi^+$ and $\Phi^-$ yield opposite correlated Alice/Bob results on all bases. After, they have opposite signs in Ma's (2) and everywhere else.

Ma et al: You can see that per their Fig 1 (which you used): $\Phi^+$ and $\Phi^-$ yield opposite results when Alice and Bob are measured on the +/- or L/R bases, but yield the same results on the H/V basis. It looks like you kept this in mind. I don't know if Ma did this intentionally, or in error, as it doesn't see to quite fit with other sources. The specific entry I am questioning in Fig 1 is the last |Φ+〉 in the list, with results VV. I think that should be marked differently, but I acknowledge my uncertainty.

Megidish et al: Matching what I have always understood: "When the polarizations of the middle photons are correlated (hh or vv) they are projected onto a |Φ+〉 state. When they are anti-correlated (hv or vh) they are projected onto a |Φ-〉 state."

Also, keep in mind that for this experiment, they held Victor's measurement basis constant at H/V (Fig. 3). And of course, returning to their (2), we have the 4 equally likely Bell states we agree on (bolded representing ones being reported on):

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

I re-examined your 4 charts. I now realize I misinterpreted/misread some of your labeling. So I am now asking for confirmation on a couple of the charts.

Chart 1. I accept your presentation as correct and mine as "suspect" for now. As mentioned, I will continue looking for something that puts this question to bed.
Chart 2. I still agree with this, but it does not relate to the Ma experiment. (Since they don't ever measure Alice and Bob on different bases from each other.)
Chart 3. & 4. These 2 are basically the same, but showing a different timing order. I agree with this, conditioned on your treatment as being on the H/V basis for Alice and Bob. See also 2 below.


2. Since we're discussing the ideal case: all pairs 2 and 3 that overlap in the BS lead to a swap, or at least are indistinguishable from a swap. The true "failed BSM" cases are where there isn't enough overlap to satisfy the coincidence window condition. Note that those can't be the case, else your statistics for the "No BSM" case wouldn't work out. The reason that your statistics for the "No BSM" group are correct: Because only 2 of the 4 Bell states can be identified, the other 2 cases are grouped into your label. Since the output 4 Bell cases occur with equal probability (25%).

Put another way: Ma ignores all reporting on what you call the "No BSM" case, in either their Fig. 1 or Fig. 3. In the text, they make the comment: "Victor's detector coincidences with one horizontal and one vertical photon in spatial modes b’’ and c’’ indicate the states |𝐻𝐻〉23 and |𝑉𝑉〉23, which are always discarded because they are separable states independent of Victor’s choice and measurement." I might say instead that those cases always produce separable state statistics (i.e. no correlation) because it combines 2 unidentifiable Bell states that produce opposite results (netting to zero). After all, they also state: "|Φ+〉23 = (|𝐻𝐻〉23 + |𝑉𝑉〉23)/√2 (both detectors in b’’ firing or both detectors in c’’ firing)." So that's a case where you have both an |H> and a |V> result, but the Bell state is identifiable.

I would be happy if there is some other paper out there that clarifies the point. I've read quite a few*, and Ma is the only one that seems to disagree (if you call it that) with any others. See for example:

https://arxiv.org/pdf/0809.3991 Fig. 2, the HOM dip goes to zero. That dip does not discriminate between the 4 different Bell states, it combines all four and all four individually disappear. Two of those exactly cancel (destructive interference), and two do not trigger 4 fold coincidences. But all 4 states exist coming out of the BS, and two disappear due to strictly HOM quantum effects - so they can't be separable.

---

To make some sense out of the encyclopedia above :

1. I believe $\Phi^+$ and $\Phi^-$ yield opposite correlated Alice/Bob results on all bases.
2. Although what you call the "No BSM" case is not reported in Ma or elsewhere: I believe it is actually the indistinguishable $\Psi^+$ and $\Psi^-$ Bell state cases.

I'm not sure either of these points need to prevent us from moving forward. We completely agree on the reported results of Ma and Megidish, as far as I can tell.

 

[DrChinese]

1. Understood. Up to here, I feel like you guys are telling me the same thing. Is there any disagreement on this?

2. I think my confusion was simply that I regarded the "H/V basis" in @DrChinese post #309 wording to mean Victor only ever measured H or V (I guess that's the SSM), and not that he measures in a basis where the possible outcomes are $\{\Phi^+,\Phi^-,HV,VH\}$ (the BSM). Thanks both for clarifying for me. :)

3. When you say "H/V basis" here do you mean the basis $\{\Phi^+,\Phi^-,HV,VH\}$?

4. This is where I get a bit confused on the terminology "basis". Alice and Bob each just get 1 particle right? So they can't use the "same (4-state) basis" as Victor who has 2 particles? So here you mean whether Alice and Bob measure literally H/V or L/R or +/-?

More great questions. It's amazing how quickly a few words here or there can point in different directions without me realizing it.

1. A little disagreement. @Morbert uses the lingo "$\{\Phi^+,\Phi^-,HV,VH\}$" from Ma differently than I with "$\{\Phi^+,\Phi^-,\Psi^+,\Psi^-\}$.

a. Morbert: $\{\Phi^+,\Phi^-\}$ portion there is no disagreement about, we mean the same thing. I haven't found an exact analog to his $\{HV,VH\}$ portion, but perhaps Morbert is referring to Ma's text: "Victor's detector coincidences with one horizontal and one vertical photon in spatial modes b’’ and c’’ indicate the states |𝐻𝐻〉23 and |𝑉𝑉〉23 which are always discarded because they are separable states independent of Victor’s choice and measurement.". So we need Morbert to help us here. The statistics are the same for Morbert and I once we discard the last 2 terms from our sets. And note that the total of his last 2 terms and the total of my last 2 terms are exactly the same. But they are not exactly the same sets.
b. DrChinese: $\{\Phi^+,\Phi^-,\Psi^+,\Psi^-\}$ represent the 4 Bell states, any one of which can equally randomly emerge from the Beam Splitter portion of the BSM apparatus when there is suitable indistinguishable overlap. I don't believe there is anything controversial about this statement whatsoever.

2. Victor only ever measures on the H/V basis in Ma, regardless of BSM or SSM setup selection. In the BSM selected setup, Victor can identify 2 of 4 Bell states; those being $\{\Phi^+,\Phi^-\}$. The states $\{\Psi^+,\Psi^-\}$ cannot be identified during a BSM and those cases are discarded.

3. See 2., Victor only measures on the H/V basis for photons 2 and 3 used for the BSM or SSM. This is normal for swapping experiments, including Megidish.

4. Alice and Bob each get 1, yes. For Ma, the decision was made to always select the same basis for both Alice and Bob each time. Both H/V, or both L/R, or both +/-. They only report 4-fold coincidences in each trial: 2 results from Victor and 1 each from Alice and Bob.

 

[Morbert]

1. I believe $\Phi^+$ and $\Phi^-$ yield opposite correlated Alice/Bob results on all bases.
2. Although what you call the "No BSM" case is not reported in Ma or elsewhere: I believe it is actually the indistinguishable $\Psi^+$ and $\Psi^-$ Bell state cases.

I'm not sure either of these points need to prevent us from moving forward. We completely agree on the reported results of Ma and Megidish, as far as I can tell.

1. The correlations yielded can be inferred from the expansions. $\Phi^+$ expands as$$|\Phi^+\rangle=\frac{|HH\rangle+|VV\rangle}{\sqrt{2}}=\frac{|++\rangle+|--\rangle}{\sqrt{2}}=\frac{|RL\rangle+|LR\rangle}{\sqrt{2}}$$And so it will yield same correlation in the HV and +- bases, and opposite correlation in the RL basis. Similarly $$|\Phi^-\rangle
=
\frac{|HH\rangle-|VV\rangle}{\sqrt{2}}
=
\frac{|+-\rangle+|-+\rangle}{\sqrt{2}}
=
\frac{|RR\rangle+|LL\rangle}{\sqrt{2}}
$$so $\Phi^-$ will yield same correlation in the HV and RL bases and opposite correlation in the +- basis.

This is all consistent with Ma's Fig 1.

2. The label "No BSM" refers to Victor configuring his apparatus for a $\{\Phi^+,\Phi^-,HV,VH\}$ measurement and registering $HV$ or $VH$ which are not Bell states. (None of my graphs showed Victor performing an SSM though they are readily producible)

Victor's detector coincidences with one horizontal and one vertical photon in spatial modes b’’ and c’’ indicate the states |𝐻𝐻〉23 and |𝑉𝑉〉23 which are always discarded because they are separable states independent of Victor’s choice and measurement."

This looks like a typo. Ma mentions HV and VH, not HH and VV:

 

[DrChinese]

1. The correlations yielded can be inferred from the expansions. $\Phi^+$ expands as$$|\Phi^+\rangle=\frac{|HH\rangle+|VV\rangle}{\sqrt{2}}=\frac{|++\rangle+|--\rangle}{\sqrt{2}}=\frac{|RL\rangle+|LR\rangle}{\sqrt{2}}$$And so it will yield same correlation in the HV and +- bases, and opposite correlation in the RL basis. Similarly $$|\Phi^-\rangle
=
\frac{|HH\rangle-|VV\rangle}{\sqrt{2}}
=
\frac{|+-\rangle+|-+\rangle}{\sqrt{2}}
=
\frac{|RR\rangle+|LL\rangle}{\sqrt{2}}
$$so $\Phi^-$ will yield same correlation in the HV and RL bases and opposite correlation in the +- basis.

This is all consistent with Ma's Fig 1.

2. The label "No BSM" refers to Victor configuring his apparatus for a $\{\Phi^+,\Phi^-,HV,VH\}$ measurement and registering $HV$ or $VH$ which are not Bell states. (None of my graphs showed Victor performing an SSM though they are readily producible)

3. This looks like a typo. Ma mentions HV and VH, not HH and VV:
View attachment 374015

1. This looks pretty good, Morbert. And it is consistent with Ma Fig 1. So on the H/V basis there is same 1 and 4 correlations for both $\{\Phi^+,\Phi^-\}$, while for the other 2 bases they are opposite. For now, assume I agree with you. And thanks for taking the time to explain your thinking.

So I'm not saying that's wrong, but there is an oddity (to me ) following that thinking. $\{\Phi^+,\Phi^-\}$ then will yield same correlations for H/V for photons 1 and 4. OK. But for Victor, he gets both an H and a V for $\{\Phi^+\}$; and both H or both V for $\{\Phi^-\}$. That seems odd.

I also am interested in comparing to the Megidish experiment, but I will take that discussion elsewhere.


2. We'll continue disagreeing on this point. There is no such thing as configuring a BS to generate only 2 Bell states. All 4 always are equal random options, and 2 is the most identifiable at this time (although they are working on boosting it past 2). I don't think you will find a strong reference otherwise. Note in fact that for the $\{\Phi^+\}$ state, the signature is $HV$ or $VH$. And every reference shows something like Ma's (2) with all 4: 1/2(|Ψ+〉14⨂|Ψ+〉23 − |Ψ−〉14⨂|Ψ−〉23 − |Φ+〉14⨂|Φ+〉23 + |Φ−〉14⨂|Φ−〉23)

In fact, I think it is possible to experimentally prove all 4 states are present from the same BS source - although not all 4 at the same time. Some experiments configure their BSM for $\{\Phi^+,\Phi^-\}$, others (such as this) for $\{\Psi^+,\Psi^-\}$. You could switch from one configuration to the other, using a constant source. So all 4 must exist. I don't know myself what is done to change from one to the other.

And not that you are saying otherwise to this additional point: There are no entries in Ma's Fig 1 that relate to Victor seeing |HH> or |VV> (or |HV> or |VH> for that matter) while configured for a BSM. Entries so labeled are configured for identifiable Separable State Measurements. Look at caption for the column, and that becomes clear.


3. You're correct, it's a typo, thanks. Although I cut and pasted directly from the paper: |HV>23 and |VH>23 were instead incorrectly inserted as |HH>23 and |VV>23. I had to hand correct to what you correctly display. Never seen that before. Still doing it on my machine... which must contain an evil demon. (I assume yours didn't do that.)

 

[Morbert]

But for Victor, he gets both an H and a V for $\{\Phi^+\}$; and both H or both V for $\{\Phi^-\}$. That seems odd.

2. We'll continue disagreeing on this point. There is no such thing as configuring a BS to generate only 2 Bell states. All 4 always are equal random options, and 2 is the most identifiable at this time (although they are working on boosting it past 2). I don't think you will find a strong reference otherwise. Note in fact that for the $\{\Phi^+\}$ state, the signature is $HV$ or $VH$. And every reference shows something like Ma's (2) with all 4: 1/2(|Ψ+〉14⨂|Ψ+〉23 − |Ψ−〉14⨂|Ψ−〉23 − |Φ+〉14⨂|Φ+〉23 + |Φ−〉14⨂|Φ−〉23)

Victor's BiSA relates the input space to the output space via a time evolution like so: $$\begin{aligned}|\Phi^+\rangle_{bc}&\longrightarrow\frac{i}{\sqrt{2}}\left(|HV\rangle_{b''b''}-|HV\rangle_{c''c''}\right), \\[6pt]|\Phi^-\rangle_{bc}&\longrightarrow\frac{i}{\sqrt{2}}\left(|HH\rangle_{b''c''}-|VV\rangle_{b''c''}\right), \\[6pt]|HV\rangle_{bc}&\longrightarrow\frac{i}{\sqrt{2}}\left(|VV\rangle_{b''b''}-|HH\rangle_{c''c''}\right), \\[6pt]|VH\rangle_{bc}&\longrightarrow\frac{i}{\sqrt{2}}\left(|HH\rangle_{b''b''}-|VV\rangle_{c''c''}\right).\end{aligned}$$The first two are explicitly reported by Ma. The second two are the same time-evolution applied to the separable states $\ket{HV}_{bc}$ and $\ket{VH}_{bc}$. Hence the registering of signatures at b''c'' constitute a measurement in the basis $\{\Phi^+,\Phi^-,HV,VH\}$ at bc.

1. A little disagreement. @Morbert uses the lingo "$\{\Phi^+,\Phi^-,HV,VH\}$" from Ma differently than I with "$\{\Phi^+,\Phi^-,\Psi^+,\Psi^-\}$.

Victor's BiSA can't resolve $\Psi$ states. As can be seen by the time-evolutions$$\begin{aligned}|\Psi^+\rangle_{bc}&\longrightarrow\frac{i}{2}\left(|VV\rangle_{b''b''}+|HH\rangle_{b''b''}-|HH\rangle_{c''c''}-|VV\rangle_{c''c''}\right), \\[6pt]|\Psi^-\rangle_{bc}&\longrightarrow\frac{i}{2}\left(|VV\rangle_{b''b''}-|HH\rangle_{b''b''}+|VV\rangle_{c''c''}-|HH\rangle_{c''c''}\right).\end{aligned}$$Notice that a signature like $VV_{b''b''}$ (or any of the other three) does not uniquely select one $\Psi$ over the other, and so the apparatus cannot resolve measurements with a Bell basis spanning this subspace. It can however resolve measurements with a separable basis spanning this subspace. Hence the measurement basis I supplied.

 

[Matterwave]

This point is actually quite interesting to me.

From a theorerical perspective, is there a difference between Victor measuring on the $\{\Phi^+, \Phi^-, HV, VH\}$ basis and then discarding the $\{HV,\,VH\}$ results vs measuring in $\{\Phi^+, \Phi^-, \Psi^+,\Psi^-\}$ and then discarding the $\{\Psi^+,\Psi^-\}$ results?

It seems like @Morbert you are saying the two particle basis including $\{\Psi^+,\Psi^-\}$ can't be experimentally realized in Ma's set up, did I understand correctly?

 

[Morbert]

This point is actually quite interesting to me.

From a theorerical perspective, is there a difference between Victor measuring on the $\{\Phi^+, \Phi^-, HV, VH\}$ basis and then discarding the $\{HV,\,VH\}$ results vs measuring in $\{\Phi^+, \Phi^-, \Psi^+,\Psi^-\}$ and then discarding the $\{\Psi^+,\Psi^-\}$ results?

It seems like @Morbert you are saying the two particle basis including $\{\Psi^+,\Psi^-\}$ can't be experimentally realized in Ma's set up, did I understand correctly?

Yes, Ma's apparatus cannot perform a measurement in the basis $\{\Phi^+,\Phi^-,\Psi^+,\Psi^-\}$.

But since runs with measurement outcomes in the $\{\Psi^+,\Psi^-\}$/$\{HV,VH\}$ subspace are discarded in this protocol, it doesn't matter (though the protocol was of course motivated by this BSM limitation).

 

[DrChinese]

This point is actually quite interesting to me.

1. From a theorerical perspective, is there a difference between Victor measuring on the $\{\Phi^+, \Phi^-, HV, VH\}$ basis and then discarding the $\{HV,\,VH\}$ results vs measuring in $\{\Phi^+, \Phi^-, \Psi^+,\Psi^-\}$ and then discarding the $\{\Psi^+,\Psi^-\}$ results?

2. It seems like @Morbert you are saying the two particle basis including $\{\Psi^+,\Psi^-\}$ can't be experimentally realized in Ma's set up, did I understand correctly?

1. No difference at all.

2. That is true for Ma (2012), yes, but is changing. The current experimental record is at least 57.9%.

---

A BSM can normally be set to look for $\{\Phi^+,\Phi^-\}$ or for $\{\Psi^+,\Psi^-\}$. You currently can't differentiate all 4 with equal success. They are working on methods to increase the detection capability above the common 50% threshold. One paper has demonstrated a theoretical concept that raises it to 62.5%. At 100%, all 4 Bell states would be discernible. So then, there is nothing anyone would need to call $\{HV, VH\}$.

Bell-state measurement exceeding 50% success probability with linear optics (2022)
"As before in the standard scheme, we can still distinguish the states |Ψ±> with 100% probability. However, we get now in addition a subset of unique signatures for the states |Φ+> and |Φ−> in 25% of the cases (see Fig. 1 (d)). This means that the overall probability to correctly identify a Bell state for this scheme is p(c) = 62.5%."

Note that Morbert's method of defining the "unidentifiable" Bell states as $\{HV, VH\}$ is not "wrong". The sum of states he calls $\{HV, VH\}$ is exactly equal to the ones I call $\{\Psi^+,\Psi^-\}$ even though trial by trial, they would not be the same. Any one he identifies as $\{HV\}$ could be either $\{\Psi^+\}$ or $\{\Psi^-\}$. Similarly, any one I identify as $\{\Psi^+\}$ could be either $\{HV\}$ or $\{VH\}$.

For me, one method is more useful and matches theory "better". But obviously opinions vary. There is no question that Ma discards certain 4-fold coincidences, regardless of what you call them.

 

[DrChinese]

1. Let us illustrate this by a simple example. Suppose that, at a certain time $t$, $\psi(x_A,x_B)$ has a form
$$\psi(x_A,x_B)=\psi_A(x_A)\psi_B(x_B) \;\;\; {\rm for} \;\;\; x_B>0$$
$$\psi(x_A,x_B) \neq \psi_A(x_A)\psi_B(x_B) \;\;\; {\rm for} \;\;\; x_B\leq 0$$
This should be considered as one wave function, written separately for $x_B>0$ and $x_B\leq 0$. The full wave function, valid for all $x_A,x_B$, is not a product. It is an entangled wave function. Any yet, for $x_B>0$ it looks like a product wave function without entanglement. So if the B particle happens to have the Bohmian position $X_B>0$, then the rule says that the particle A has to move by velocity that does not depend on the position of B. But in order to obey this rule, the particle A has to know that $X_B>0$. So, the motion of A "independently" on B actually depends on B. This demonstrates that motion always depends on positions of all particles, even when this dependence is not manifest.

2. And now we can finally discuss the entanglement swapping. It involves 4 particles, A, B, C and D. Initially the wave function has the form $\psi_{AB}\psi_{CD}$, so A is entangled with B, and C is entangled with D, but there is no entanglement between A and D. Nevertheless, each particle knows positions of all other particles. In particular, A knows the position of D, and vice versa, but this knowledge does not have a direct manifestation. In other words, the motion of A depends on the position of D, and vice versa, but this dependence is not manifest. The entanglement swapping is a way to make this dependence manifest.

3. What seems to be confusing to DrChinese is how can suddenly the motion of A may start to depend on the position D, if there was no such dependence from the beginning? How can such a dependence be created without the interaction between A and D? And what exactly creates such a dependence?

4. The answer is that the dependence was there from the start, it was never really created. It is only that certain changes in the system changed the rules of the game (encoded in the change of the pilot wave), so that, at the time when B and C interacted, the dependence of A on D became manifest.

Yikes! I'm reaching back to post #1 in this thread!!

1. To be clear: The Bohmian spin observable - let's just call it spin - is some function of the particle's position and the positions/velocities/other properties of other particles. But for the experimental entangled spin rules to work, there must be something preferred about the relationship of A and B... correct? After A is measured to be "up", entangled B is never going to be measured as anything other than "down", correct? Assuming, of course, B later gets measured on that basis.

2. Switching to our swapping example: I admit I don't follow your reasoning. A and B are entangled initially, OK, but I don't see how anything about D relates to A. You refer to it as "not manifest", OK, that fits with 1. above. And you say A and D are not entangled. OK. But now D has been measured as well as A, and there is still nothing connecting them. Let's say A and D are measured on the +/- basis. We now should know that neither of the A and D outcomes are dependent in any way on the other, any more than B and C are mutually dependent.

3. Keep in mind that at this point, there is nothing in the future of B and C to indicate they will ever interact. In the related experiments, that decision (interact or not) is made by computer (or theoretically could be made by a person's free will). If you disagree, let's discuss.

4. I'll table this until I better understand the other points.

-DrC

 

[Demystifier]

I agree with your points above. But it will be much more illuminating if you jump to my post https://www.physicsforums.com/threa...bohmian-mechanics.1086070/page-8#post-7320980
The most interesting thing happens at $t_3$, when A and D become linked (entangled) through the link with $M_{BC}$.

 

[DrChinese]

1. ...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 $t_2$.

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.

Following your advice from the post above #343: I'm/we're covering a bit much here, so I am breaking my responses into 2 posts.

1. So, let's talk a bit more about t2. As you say later, A and D can be destroyed - their measured value is recorded. Now please understand: I agree that your notation is functionally correct here, and have no issue with it at all.

In the Bohmian (BM) perspective, there is no further ability for anything anywhere to change the recorded values of A & D. And I think we've agreed that the spin observable of B and C on the +/- basis (same basis A & D were measured) is essentially set in stone. All of this matches perfectly so far with experiment as to usual PDC entanglement.

But there is a point here which BM makes a statement, but that oQM does not. According to BM, particles B & C have firm definite values for H/V polarization at all times - which is presumably an unbiased basis relative to +/-. When A & D were measured, did B & C take on specific values related to those measurements? Are the H/V values for B & C likewise set in stone as their +/- values are? Or are they free to change, perhaps influenced by particles/systems elsewhere (following the BM pilot wave concepts)?


2. Here I see things a bit differently than you present them. A & D are measured on the +/- basis. B & C are to be measured on the H/V basis. This is a significant difference! In oQM, there is no correlation at all between the future outcome of a H/V measurement and the same photon's known +/- polarization, right? I don't think BM posits otherwise.

My point being: the relationship you are drawing $t_3:\; (M_A,A)_3 \mathrel{-} (B,M_{BC},C)_3 \mathrel{-} (M_D,D)_3$ does not account for this difference in basis. And I think it should. We are getting ready to measure B & C and the only thing we are looking for now is which Bell state is to be revealed. Although there are 4 Bell states, basically what we want to know is:

How does it happen that the interaction between B & C can only lead to a Bell State in which A & D are suitably correlated (either |++> or |-- >) or anti-correlated (either |+- > or |-+>). This of course when B & C are not being even measured on the +/- basis.

We know that there must be physical overlap between B & C, that has been demonstrated experimentally (Megidish et al). So from the forward in time only (FITO) view provided by BM, something must be causing the result of the interaction to somehow lead to the proper Bell state. That process being going through a suitably oriented beam splitter and follow-on polarizing beam splitters, on a different basis than which A & D were measured. I'm trying to understand how any of this is consistent with a) having nonlocal influences from other systems outside the lab, and b) having definitely values for H/V polarization at all times.

Of course, I'm not sure oQM answers these questions either!

-DrC

 

[DrChinese]

1. ...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 $t_2$.

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.

3. 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.

I couldn't figure out where to stick this...

2.5 When I try to imagine a word that sums up what polarization state B (or C) is in during the time between t2 and t3 (maybe t2.5?), I'm not sure what it is. It can't exactly be in state |+> or |- >, because a photon in that exact state cannot participate in a swap. But it is not exactly entangled either - or is it? Because I don't think the Bohmian perspective has a reasonable description here, since it is asserting Bohmian particles are in exact states - albeit unknown - at all times.


3. I'll come back to this later, I want to see where we're at for 1. and 2 (and 2.5).