Entanglement swapping and Bohmian mechanics

[DrChinese]

I have a suggested Bohmian model of the DCES experiment. ...

I am studying, will respond tomorrow. Thanks!

-DrC

 

[PeterDonis]

I have a suggested Bohmian model of the DCES experiment.

I will note, btw, that although I've pitched this in terms of Bohmian mechanics, what I posted in that post only involves the wave function (which is why I said it was only a sketch, at least as far as the Bohmian model is concerned), so it should be applicable to any QM interpretation--indeed, it could be viewed as simply part of the basic math of QM. I am simply pointing out an aspect--including whatever quantum process determines whether or not a swap can occur, given that the experimenter has chosen to attempt a swap; what I called P in the wave function--that, AFAIK, has not been pointed out in the literature, but which seems to me to be relevant.

The specifically Bohmian part would be filling in the details about what happens with the particle positions and velocities as a result of the properties of the wave function (aka pilot wave) that I described.

 

[PeterDonis]

even if the experimenter decides this, he can't guarantee that a swap actually happens.

To follow up more on this: the Ma paper [1] says (bottom of p. 4):

Victor's detector coincidences with one horizontal and one vertical photon in spatial modes b’’ and c’’ indicate the states $\ket{HV}_{23}$ and $\ket{VH}_{23}$, which are always discarded because they are separable states independent of Victor’s choice and measurement.

It's unfortunate, at least for our discussion here , that they discard these runs, since that means they don't show data on what the photon 1 and 4 polarizations were. But since it is stated that these are "no swap" runs (photons 2 and 3 are in "separable states"), that means photon 2 remains entangled with 1, and photon 3 remains entangled with 4. Which means that the polarization states of 2 and 3, respectively, should be opposite to those that are measured on 1 and 4 (or were already measured, in the case where photons 1 and 4 are measured prior to the swap). In other words, $\ket{HV}_{23}$ and $\ket{VH}_{23}$, respectively, should correspond to $\ket{VH}_{14}$ and $\ket{HV}_{14}$. And note that those are exactly the photon 1 and 4 states (the antiparallel ones) that should prevent a swap from being able to happen, per my previous post.

And, as the quote above states, on these runs the photon 2 and 3 states that come out "are separable states independent of Victor's choice and measurement" (emphasis mine). In other words, even if the QRNG, which is what sets up the machinery to swap or not swap, sets things up to swap, we still get no swap on these runs.

What this suggests to me is that I might not have needed to add anything extra to the wave function (what I called "P" in my previous post) to enforce the "no swap if photons 1 and 4 were measured antiparallel" condition: if photons 1 and 4 were already measured, and their polarizations were antiparallel, that means photons 2 and 3 must also now have antiparallel polarizations, and that alone might be sufficient to prevent a swap regardless of Victor's settings. In other words, if photons 2 and 3 come into the swap/no swap machinery in a state that has zero amplitude for their polarizations being parallel, then they just go through the machinery without changing their polarization at all, regardless of how the machinery is set up (i.e., whether Victor has chosen swap or no swap).

[1] https://arxiv.org/pdf/1203.4834

 

[Demystifier]

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

[1] From Ma et al:

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

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

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

[2] They agree with this part.

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

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

Thanks.-DrC

It seems to me that you fail to grasp the notion of a Hilbert space at a sufficiently abstract level. To understand it properly you have to think like a theorist, not like an experimentalist. The Hilbert space is a large class of all possible states, irrespective of whether those states are realized in an actual experiment or not. My notes below should be understood with such abstract glasses put on.

First, a decomposition of a Hilbert space into a product can be understood purely mathematically, not depending on any physical event.

Second, the bolded part of your text is wrong. If a Hilbert space is decomposed as a product, it doesn't imply that a particular state on this space must be a product. The Hilbert space decomposed as a product is not merely a space of product states, it contains non-product states as well. You can think of a product decomposition as something akin to the choice of basis, in which every member of the basis is a product state. But general states are superpositions of these basis states, so general states are not product states, even though the basis states are.

 

[Demystifier]

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

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

Even if I accept that there is free will, not merely in some weak/illusional/emergent/compatibilist sense, but in the strong fundamental sense which makes the deterministic laws of physics wrong at some level, I still don't think that free decisions of the experimenter can influence the past in the causal sense, so I would still use the word "implies". Moreover, I think that free will of human experimenters is a red herring. You can consider a situation in which the decision to perform a measurement was not made by a human, but by an outcome of some auxiliary random process. In that case the question of causation does not depend on the existence of human free will.

 

[Demystifier]

Since @Sambuco talks about electron-positron annihilation, I would like to draw his attention to the Hardy setup
https://en.wikipedia.org/wiki/Hardy's_paradox#Setup_description_and_the_results
where an entangled electron-positron pair is created by postselection in which the annihilated pair is removed from the ensemble. Without postselection there is no entanglement, but after the postselection, that is, after the decision to ignore the cases when the pair annihilates and turns into photons, the resulting state ends up in a very interesting entangled state, called Hardy state, that Hardy used to show quantum nonlocality without using any inequalities.

 

[Roberto Pavani]

Standard entanglement-swapping setup: two Bell pairs
$|\Phi^+\rangle_{AB} \otimes |\Phi^+\rangle_{CD}$

BSM on $BC$, measurements on $A$ and $D$.

Without conditioning on the BSM outcome, the state of $AD$ is maximally mixed:

$\rho_{AD} = \tfrac{1}{4} \, \mathbb{I}$

and all correlators vanish: $E(a_i, d_j) = 0$ for all $i,j$.

The BSM on $BC$ partitions the runs into four mutually exclusive sub-ensembles. Each projects $AD$ onto a Bell state:

$|\Phi^+\rangle_{BC} ;\Rightarrow; \rho_{AD} = |\Phi^+\rangle\langle\Phi^+|_{AD}$
$|\Psi^+\rangle_{BC} ;\Rightarrow; \rho_{AD} = |\Psi^+\rangle\langle\Psi^+|_{AD}$
$|\Psi^-\rangle_{BC} ;\Rightarrow; \rho_{AD} = |\Psi^-\rangle\langle\Psi^-|_{AD}$
$|\Phi^-\rangle_{BC} ;\Rightarrow; \rho_{AD} = |\Phi^-\rangle\langle\Phi^-|_{AD}$

Every Bell state violates CHSH maximally ($|S|=2\sqrt{2}$), but each with a different sign pattern in the correlators.

Writing

$E_{ij} = \epsilon_{ij}/\sqrt{2}$ with $\epsilon_{ij}\in \{+1,-1\}$,

the four patterns are:

$|\Phi^+\rangle_{AD}:\quad (+,+,+,-) = (\epsilon_{00},\epsilon_{01},\epsilon_{10},\epsilon_{11})$
$|\Psi^+\rangle_{AD}:\quad (+,+,-,+)$
$|\Psi^-\rangle_{AD}:\quad (-,-,+,-)$
$|\Phi^-\rangle_{AD}:\quad (-,-,-,+)$

The general CHSH parameter is

$S_{\boldsymbol{\sigma}} = \sigma_1 E_{00} + \sigma_2 E_{01} + \sigma_3 E_{10} + \sigma_4 E_{11} = \frac{1}{\sqrt{2}}\sum_{k}\sigma_k \,\epsilon_k$

which reaches $2\sqrt{2}$ when $\boldsymbol{\sigma}=\boldsymbol{\epsilon}$ (all signs match) and $\approx 0$ otherwise.

BSM outcome​	Bell state on $AD$​	$\boldsymbol{\sigma}$ for $S=2\sqrt{2}$​
$|\Phi^+\rangle_{BC}$​	$|\Phi^+\rangle_{AD}$​	$(+1,+1,+1,-1)$​
$|\Psi^+\rangle_{BC}$​	$|\Psi^+\rangle_{AD}$​	$(+1,+1,-1,+1)$​
$|\Psi^-\rangle_{BC}$​	$|\Psi^-\rangle_{AD}$​	$(-1,-1,+1,-1)$​
$|\Phi^-\rangle_{BC}$​	$|\Phi^-\rangle_{AD}$​	$(-1,-1,-1,+1)$ ​

Each $\boldsymbol{\sigma}$ gives $2\sqrt{2}$ on its own sub-ensemble and $\approx 0$ on the other three.

 

[Morbert]

It's unfortunate, at least for our discussion here , that they discard these runs, since that means they don't show data on what the photon 1 and 4 polarizations were. But since it is stated that these are "no swap" runs (photons 2 and 3 are in "separable states"), that means photon 2 remains entangled with 1, and photon 3 remains entangled with 4.

Two points here:
i) A run where no swap is attempted will leave 2 entangled with 1 and 3 entangled with 4. But a run where a swap is attempted and fails will break entanglement between 1 and 2 (and between 3 and 4) and leave them classically correlated. This is because a failed measurement is not a complete failure. It's one that projects onto the $\Psi^\pm$ subspace.

ii) For Ma's experiment specifically, note that that since 1 and 4 have already been measured, entanglement is already broken.

What this suggests to me is that I might not have needed to add anything extra to the wave function (what I called "P" in my previous post) to enforce the "no swap if photons 1 and 4 were measured antiparallel" condition: if photons 1 and 4 were already measured, and their polarizations were antiparallel, that means photons 2 and 3 must also now have antiparallel polarizations, and that alone might be sufficient to prevent a swap regardless of Victor's settings.

Yes, under the scenario above, a Bell measurement must fail if it is attempted. I.e. Victor must obtain $\frac{1}{2}(\ket{\Psi^+}\bra{\Psi^+}_{23} + \ket{\Psi^-}\bra{\Psi^-}_{23})$ as the measurement result if he attempts a Bell measurement.

In other words, if photons 2 and 3 come into the swap/no swap machinery in a state that has zero amplitude for their polarizations being parallel, then they just go through the machinery without changing their polarization at all, regardless of how the machinery is set up (i.e., whether Victor has chosen swap or no swap).

If Victor attempts a swap, the polarizations of 2 and 3 will change as per the evolution in the "Bipartite state analyzer for Bell-state measurement" section of Ma's paper. If the detectors record same polarization and different spatial modes, or different polarization and same spatial modes, the Bell measurement was successful. But since, in this scenario, it must fail, the detectors will record the same spatial modes and the same polarization.

 

[DrChinese]

It seems to me that you fail to grasp the notion of a Hilbert space at a sufficiently abstract level. To understand it properly you have to think like a theorist, not like an experimentalist. The Hilbert space is a large class of all possible states, irrespective of whether those states are realized in an actual experiment or not. My notes below should be understood with such abstract glasses put on.

First, a decomposition of a Hilbert space into a product can be understood purely mathematically, not depending on any physical event.

Second, the bolded part of your text is wrong. If a Hilbert space is decomposed as a product, it doesn't imply that a particular state on this space must be a product. The Hilbert space decomposed as a product is not merely a space of product states, it contains non-product states as well. You can think of a product decomposition as something akin to the choice of basis, in which every member of the basis is a product state. But general states are superpositions of these basis states, so general states are not product states, even though the basis states are.

You are correct about my (lack of) understanding of Hilbert spaces. So I gladly defer to you.

However, your explanation doesn't seem to really lead to any particular way to look at a Hilbert space that relates to swaps. There's still 2 Bell states before the swap, and there's 2 completely different Bell states after. Presumably the Hilbert space changed too.

 

[DrChinese]

1. Even if I accept that there is free will, not merely in some weak/illusional/emergent/compatibilist sense, but in the strong fundamental sense which makes the deterministic laws of physics wrong at some level, I still don't think that free decisions of the experimenter can influence the past in the causal sense, so I would still use the word "implies".

2. Moreover, I think that free will of human experimenters is a red herring. You can consider a situation in which the decision to perform a measurement was not made by a human, but by an outcome of some auxiliary random process. In that case the question of causation does not depend on the existence of human free will.

1. I think "implies" is fair.

2. I would certainly include any automated "auxiliary random process" as being an expression of the free will of the experimenter. That's why I use that phrase, because the experimenter selects that random/pseudo-random algorithm.

 

[DrChinese]

1. For Ma's experiment specifically, note that that since 1 and 4 have already been measured, entanglement is already broken.

Yes, under the scenario above, a Bell measurement must fail if it is attempted. I.e. Victor must obtain $\frac{1}{2}(\ket{\Psi^+}\bra{\Psi^+}_{23} + \ket{\Psi^-}\bra{\Psi^-}_{23})$ as the measurement result if he attempts a Bell measurement.If Victor attempts a swap, the polarizations of 2 and 3 will change as per the evolution in the "Bipartite state analyzer for Bell-state measurement" section of Ma's paper. If the detectors record same polarization and different spatial modes, or different polarization and same spatial modes, the Bell measurement was successful. But since, in this scenario, it must fail, the detectors will record the same spatial modes and the same polarization.

1. That's a specific question I am asking, for which I am unsure your statement is correct. I would say entanglement is not yet completely broken for 2 and 3.

2. My reasoning for 1. is as follows, considering 1 and 4 have been measured, but not yet 2 and 3:

a) A Bell test can be performed on 2 which will of course confirm entanglement (as initially prepared) with 1 (say on the H/V basis). And similar with the 1/4 pair.

b) Alternately: A Bell state measurement (BSM) can be performed on 2 and 3 (say on the 1/0 basis, but not on the H/V basis), generating a swap.

So the future of 2 and 3 can either involve a measurement of one of two mutually exclusive options. I would say 2's (or 3's) entanglement is not completely broken. Were entanglement completely broken, you'd have to say 2 and 3 are now in a pure polarization state on the same basis 1 and 4 were measured. And I believe that's not possible, because: Only photons in a superposition on all polarization bases can participate in a BSM swap. Important note: That statement does not have explicit reference that I can locate, it is my deduction from reading experimental descriptions. I would greatly like to locate an explicit quote saying this, and I am hoping someone knows a specific source that would confirm (or deny) this.

So this is really one of the questions I am trying to solve here.

 

[PeterDonis]

Two points here:
i) A run where no swap is attempted will leave 2 entangled with 1 and 3 entangled with 4. But a run where a swap is attempted and fails will break entanglement between 1 and 2 (and between 3 and 4) and leave them classically correlated. This is because a failed measurement is not a complete failure. It's one that projects onto the $\Psi^\pm$ subspace.

Ah, I see. So my buckets 0 and 1 will not show the same statistics.

ii) For Ma's experiment specifically, note that that since 1 and 4 have already been measured, entanglement is already broken.

For an interpretation with actual collapse, yes. But we are using the Bohmian interpretation here, which only has "effective collapse". For that interpretation, you have to be more careful about statements like this: yes, we know from the photon 1 and 4 measurements which branch of the wave function is effective as far as the further dynamics of 1 and 4 are concerned, but that does not rule out 1 and 4 ending up showing entangled statistics if an entanglement swap is done.

Indeed, even in an interpretation with actual collapse (as long as it's a valid interpretation of QM, not an alternate theory like the GRW stochastic collapse model), saying that "entanglement is broken" does not preclude photons 1 and 4 showing entangled statistics in the appropriate subensembles if a swap is done. Such an interpretation would presumably account for this very differently from an interpretation in which there was no actual collapse, but if it's a valid interpretation of QM, it has to account for the fact that QM predicts, and experiments confirm, that such a thing can happen.

If Victor attempts a swap, the polarizations of 2 and 3 will change as per the evolution in the "Bipartite state analyzer for Bell-state measurement" section of Ma's paper.

I'll take a look; it's quite possible I didn't fully take into account all the aspects of that evolution. This would affect my bucket 1.

 

[PeterDonis]

Presumably the Hilbert space changed too.

The Hilbert space doesn't change just because you make a measurement or perform a unitary operation. The full Hilbert space describing the possible states of all four photons is the same throughout the entire experiment. @Demystifier has pointed out that there are different ways of describing this Hilbert space in terms of tensor products of Hilbert spaces of subsystems, and it might be that one such description makes things easier to analyze before the swap, while a different one makes things easier to analyze after the swap. But that doesn't change the Hilbert space itself. It just changes which description in terms of subsystems you choose to use. There is no change to the physics, just a change in which mathematical tool makes things easier to analyze.

 

[DrChinese]

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

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

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

So at the swap stage, we have three possibilities:

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

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

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

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

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

Bucket 0: No swap attempted.

Bucket 1: Swap attempted but no swap happened.

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

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

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

A few minor quibbles, not sure this changes anything important you are saying.

a) The experiment filters/reports 4 fold coincidences within the relevant time window. All of those (ideally) produce a swap. A truly failed swap is detection of less than 4 clicks in different detectors. That failure can occur when 2 photons appear "simultaneously" in a single detector, but only 1 click is registered - or if a click occurs in a detector outside the stated time window, or if a photon is lost in transit and never causes a click anywhere.
b) The measurement basis for the BSM on 2/3 must be on a different basis than on 1/4. Otherwise, as you pointed out, the result can be construed as being Separable rather than Entangled (or alternately you could refer to it as Indeterminate). In the Ma experiment, the Figure 3 graph of results shows this. There is no difference in correlation statistics when all 4 photons are measured on the same H/V basis.
c) In these experiments: The "no swap" condition is executed by causing distinguishability between the 2 and 3 photons. There is still 4 fold coincidence.

 

[DrChinese]

The Hilbert space doesn't change just because you make a measurement or perform a unitary operation. The full Hilbert space describing the possible states of all four photons is the same throughout the entire experiment. @Demystifier has pointed out that there are different ways of describing this Hilbert space in terms of tensor products of Hilbert spaces of subsystems, and it might be that one such description makes things easier to analyze before the swap, while a different one makes things easier to analyze after the swap. But that doesn't change the Hilbert space itself. It just changes which description in terms of subsystems you choose to use. There is no change to the physics, just a change in which mathematical tool makes things easier to analyze.

Understood. I thought @Demystifier was saying that description explains the change in physics. My misunderstanding.

 

[PeterDonis]

a) The experiment filters/reports 4 fold coincidences within the relevant time window. All of those (ideally) produce a swap.

Yes, as @Morbert pointed out, what I was calling my "bucket 1" is not really "swap attempted but no swap", but "swap that the experimental setup can't distinguish". If we had a setup that could distinguish all four possible Bell states for photons 2 and 3, then there would be no "no swap" runs when Victor selected a swap--i..e, no bucket 1 in my terminology (at least not in the ideal case--but see below)--but there would be four "swap" buckets (2A, 2B, 2C, and 2D in my terminology) instead of two (2A and 2B). And then the Bohmian account would be that, if photons 1 and 4 are measured first, the only possible swaps on each run will be the ones that are consistent with the 1 and 4 measurement results for that run (if they are parallel, only swaps into the $\Phi^\pm$ states can occur, and if they are antiparallel, only swaps into the $\Psi^\pm$ states can occur).

A truly failed swap is detection of less than 4 clicks in different detectors. That failure can occur when 2 photons appear "simultaneously" in a single detector, but only 1 click is registered - or if a click occurs in a detector outside the stated time window, or if a photon is lost in transit and never causes a click anywhere.

Yes, I glossed over all of these possibilities, but if they occur on runs where Victor attempted a swap, they would fall into a different "bucket 1" than what I was describing above. In other words, we would have bucket 1A, for "Victor attempted a swap, and a swap occurred, but it was into a Bell state that the experimental setup can't distinguish" (and this bucket would disappear in a setup that could distinguish all four Bell states), and bucket 1B, for "Victor attempted a swap but one of the things described in the quote above happened, so no swap occurred" (and this bucket would still be present even in a setup that could distinguish all four Bell states).

b) The measurement basis for the BSM on 2/3 must be on a different basis than on 1/4.

Yes, I glossed over this as well, because the important point for the analysis I was giving was the 1/4 measurement basis, and what that implies for the state of photons 2 and 3 in that basis as they go into the BSM/SSM apparatus (i.e., is there a nonzero amplitude for them to be parallel in that basis). The actual photon 2 and 3 results being in a different basis doesn't impact that analysis; it just means we have to be aware of the actual photon 2 and 3 measurement basis when we do statistics on the results to assess whether they are consistent with an entanglement swap into the Bell state that was signalled.

Otherwise, as you pointed out, the result can be construed as being Separable rather than Entangled (or alternately you could refer to it as Indeterminate).

What specific statement in my post are you referring to here?

 

[DrChinese]

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

To be clear about the experimental setup:

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

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

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

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

So at the swap stage, we have three possibilities:

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

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

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

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

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

Bucket 0: No swap attempted.

Bucket 1: Swap attempted but no swap happened.

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

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

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

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

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

Here is how we account for that:

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

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

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

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

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

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

It would not be fair to say case (2) does not occur, it occurs exactly (in principle) as often as case (1). And it is identified as such equally as (1). And yes, as you say, it's random and out of control of the experimenter. But I would use the concept "no swap" as 4 fold coincidences (according to quantum theory, ideal case) always produce a swap. It's not like the typical PDC mechanism, where only occasionally are entangled pairs produced. 4 clicks (implying all 4 photons detected) inside the stated time window leads to a swap. (Note that of course, the defined time window is a function of the distances (time of travel) to the related detector.)

In Ma's experiment, only the (1) case is presented - so I see why it might look as you describe. In the Megidish et al experiment, all 16 4-fold cases (permutations) are reported. See Figure 3 for the full state tomography (QST). Even though these experiments document a different measurement order, they both make the same essential point: measurement ordering does not change either the predicted or actual results.

You might fairly say: when 1 and 4 are as per your (1), only 2 of 4 Bell states can result. Vice versa, when 1 and 4 are as per your (2), only 2 different of 4 Bell states can result. The catch being that only one of the two possible compatible states can be identified using current technology.

Note: Were future technology to arrive that allowed identification of all 4 Bell states: that would not be expected to change any conclusions derived from existing experiments (according to the predictions of QM).

 

[PeterDonis]

Only photons in a superposition on all polarization bases can participate in a BSM swap.

I think this is somewhat too strong. Consider this from a previous post of mine:

if photons 1 and 4 are measured first, the only possible swaps on each run will be the ones that are consistent with the 1 and 4 measurement results for that run (if they are parallel, only swaps into the $\Phi^\pm$ states can occur, and if they are antiparallel, only swaps into the $\Psi^\pm$ states can occur).

Generalizing on this: for a swap into the $\Phi^\pm$ states to be possible, there must be a nonzero amplitude in the wave function for the polarizations of photons 2 and 3 to be parallel on any basis, and for a swap into the $\Psi^\pm$ states to be possible, there must be a nonzero amplitude in the wave function for the polarizations of photons 2 and 3 to be antiparallel on any basis.

So, for example, if photons 1 and 4 are measured and the result is $\ket{HH}_{14}$, that implies that photons 2 and 3 are in the state $\ket{VV}_{23}$. This is not a superposition on the H-V basis, but a swap into the $\Phi^\pm$ states is still possible. (This must be the case because it happens in the experiment!) That swap is still possible because the state $\ket{VV}_{23}$ gives a nonzero amplitude for parallel polarizations on any basis; there is no basis for which the polarizations must be antiparallel (although in any other basis than the H-V basis, there is a nonzero amplitude for antiparallel polarization).

But a swap into the $\Psi^\pm$ states is not possible for this case, because the state $\ket{VV}_{23}$ does have zero amplitude for antiparallel polarization in the H-V basis. So it is not the case that photons 2 and 3 are in a state with a nonzero amplitude for antiparallel polarization on any basis.

 

[DrChinese]

What specific statement in my post are you referring to here?

Hmmm, don't see that explicit or implied statement on re-reading your post. But whatever I saw (or hallucinated lol), I was agreeing.

When you measure all 4 photons on the same H/V basis, you cannot prove a swap occurred. That's because a "semi-classical" explanation is indistinguishable from an entanglement swap.

 

[PeterDonis]

It would not be fair to say case (2) does not occur,

I agree that, given the various complications that have been discussed in subsequent posts (some of which you allude to), saying "a swap will never happen" in case (2) is an oversimplification. A better phrasing would be that in Ma's particular experimental setup, the possible swaps in case (2) (into the $\Psi^\pm$ Bell states) can't be distinguished, so the statistics for that bucket of runs look like those of a separable state.

 

[DrChinese]

1. I think this is somewhat too strong. ... if photons 1 and 4 are measured and the result is $\ket{HH}_{14}$, that implies that photons 2 and 3 are in the state $\ket{VV}_{23}$. This is not a superposition on the H-V basis, but a swap into the $\Phi^\pm$ states is still possible.

2. (This must be the case because it happens in the experiment!)

1. Again, this is precisely an issue I am attempting to gain information about. Yes, I am familiar with this being a common description. So I consider what you are saying as "generally accepted" as stated, in theory.

2. And yet, the following statement I consider to be true and "generally accepted" with equal experimental support: If photons 2 and 3 are in the state $\ket{VV}_{23}$, they cannot lead to a swap.

Stated differently: It would not occur in the experiment if you fed $\ket{VV}_{23}$ photons into the swapping mechanism.

For if they could: Type I entanglement would not require 2 crystals to be useful in swapping experiments. And in Type II entanglement, it would not be necessary to overlap output cones - which significantly reduces the output of entangled pairs.

---

And again, I don't have explicit reference for my stance in 2. I am looking for one!

But there is no question that PDC crystals of Types I and II ONLY produce polarization unentangled pairs $\ket{VV}_{12 or 34}$ natively. Special procedures are required to convert them to entangled pairs, that can be seen in any paper on PDC entanglement (and I can cite plenty of those). Why execute those procedures, which significantly reduce output intensity, if such pairs can participate in a swap without them?

 

[DrChinese]

I agree that, given the various complications that have been discussed in subsequent posts (some of which you allude to), saying "a swap will never happen" in case (2) is an oversimplification. A better phrasing would be that in Ma's particular experimental setup, the possible swaps in case (2) (into the $\Psi^\pm$ Bell states) can't be distinguished, so the statistics for that bucket of runs look like those of a separable state.

They could have chosen to work with (2), as Megidish did, but there was some technical reason they chose not to. I don't know that reason. Megidish reports all 16 permutations of 4-fold outcomes, i.e. both your (1) and (2).

 

[PeterDonis]

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

But, as I think has already been noted, this implies backwards in time causation. Which means...

I believe the following description is fair under pretty much any viewpoint

...this can't possibly be true, since backwards in time causation is not allowed or believed to be possible under "pretty much any viewpoint". Yes, it does appear to be the viewpoint of the experimentalists who wrote the paper ("quantum steering into the past" at least appears to be a valid way to see things to them); but that's way, way, far from "pretty much any viewpoint" agreeing.

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

As a description of what the authors of the paper think, I agree.

However, I do not think that is a valid reason to consider what they claim as actually being independent of any QM interpretation, or saying that someone is being unscientific or ignoring objective facts by pointing out, as I do here (and as I have elsewhere), that their claims are interpretation dependent. Certainly I do not think that backwards in time causation, even as a "well, that's a valid viewpoint" kind of thing, is just basic QM independent of any interpretation. It is a specific interpretation, whether the authors of the paper say so or not.

Note that this is not a criticism of the authors of the paper. They aren't experts in QM interpretations. They're experimentalists, who are describing what they did and what they think it means, in the terms that seem suitable to them. That's to be expected. But it does not, IMO, justify simply adopting their apparent belief that everything they say is "objectively true". I would say they simply have not considered--because it's not their field--that their belief is based on implicitly adopting a particular interpretation of QM.

 

[PeterDonis]

Megidish reports all 16 permutations of 4-fold outcomes,

Yes, and something more like my post #76, where I described what my account would look like for a setup in which all four Bell states could be distinguished, would apply to that experiment.

 

[PeterDonis]

If photons 2 and 3 are in the state $\ket{VV}_{23}$, they cannot lead to a swap.

This might be one of those cases where we have to be very careful about distinguishing two different cases, which intuitively "look the same" but actually aren't:

(1) We say that photons 2 and 3 are in the state $\ket{VV}_{23}$ for the purposes of doing an analysis, because we prepared them to be in the singlet state with photons 1 and 4, respectively, and then measured photons 1 and 4 and got the result $\ket{HH}_{14}$. But we haven't actually done any operation yet on photons 2 and 3.

(2) We know that photons 2 and 3 are in the state $\ket{VV}_{23}$ because we explicitly did an operation on them that put them in that state--for example, we prepared them in the singlet state with photons 1 and 4, respectively, and then we passed each of them (photons 2 and 3) through a vertical polarizer.

I vaguely remember something like the latter case being mentioned in a previous thread some time ago on this topic. Do we have experiments that show that, in the latter case, no swap can occur?

 

[DrChinese]

Going back to the issue of

...this can't possibly be true, since backwards in time causation is not allowed or believed to be possible under "pretty much any viewpoint". Yes, it does appear to be the viewpoint of the experimentalists who wrote the paper ("quantum steering into the past" at least appears to be a valid way to see things to them); but that's way, way, far from "pretty much any viewpoint" agreeing.


As a description of what the authors of the paper think, I agree.

Not saying most viewpoints agree these experiments convince anyone there is quantum steering to the past, just simply that it does look like that at first glance. Most people would then reject that assessment, since most people believe a future action cannot change the past. I think that leads us to pretty close to the same assessment of the situation, but maybe not. I certainly agree that few interpretations are compatible with any concept of "retro-something"

We can certainly agree these authors are not portraying an expertise in Interpretations. And as best I can tell, their interpretation-related comments in other places do not particularly support a belief in retro-something. Or Bohmian-something either, for that matter.

 

[DrChinese]

This might be one of those cases where we have to be very careful about distinguishing two different cases, which intuitively "look the same" but actually aren't:

(1) We say that photons 2 and 3 are in the state $\ket{VV}_{23}$ for the purposes of doing an analysis, because we prepared them to be in the singlet state with photons 1 and 4, respectively, and then measured photons 1 and 4 and got the result $\ket{HH}_{14}$. But we haven't actually done any operation yet on photons 2 and 3.

(2) We know that photons 2 and 3 are in the state $\ket{VV}_{23}$ because we explicitly did an operation on them that put them in that state--for example, we prepared them in the singlet state with photons 1 and 4, respectively, and then we passed each of them (photons 2 and 3) through a vertical polarizer.

(3) Do we have experiments that show that, in the latter case, no swap can occur?

Agree with (1) for standard quantum mechanics. Norsen explicitly says otherwise.

Agree with (2) for basic analysis purposes, yes. Of course, their polarization before measurement does have that counterfactual element. That's the rub here!

I have yet to find explicit support for (3). But not for lack of trying!

 

[PeterDonis]

Norsen explicitly says otherwise.

Can you be more specific?

 

[PeterDonis]

as per the evolution in the "Bipartite state analyzer for Bell-state measurement" section of Ma's paper.

Unfortunately, after looking at this, it doesn't tell us what we need to know. This evolution starts with the two distinguishable Bell states, $\Phi^\pm$, and explains how they end up with the particular detection signatures shown. But what we need is an evolution that starts with the state of photons 2 and 3 going into the whole BSM/SSM setup, and ending with the Bell states for the case where a swap is done--and then we need that evolution to tell us what happens to the states of photons 2 and 3 when a swap is not done. (The evolution shown in the Ma paper happens after this.)

For example, if photons 2 and 3 are in the state $\ket{VV}_{23}$ going into the BSM/SSM setup, what are the amplitudes for the various possible states that can come out, if the setup has "swap" (BSM) or "no swap" (SSM) selected? I don't see anything in the Ma paper that gives this.

More generally, what we need is a description of the unitary operator that is realized by the BSM/SSM, taking photon 2 and 3 states in the input arms into either Bell states or separable states in the output arms (before the bipartite state analyzer kicks in).

 

[PeterDonis]

This evolution starts with the two distinguishable Bell states,

Indeed, looking at this and the diagram of the experimental setup (Figure 2 in the paper) makes me even more confused, because the evolution starts before BS1, i.e., the evolution assumes that photons 2 and 3 are in one of the Bell states before BS1--but then what puts them into those states? The only things in the experiment before BS1 are the preparations of the initial entangled pairs (1 and 2, 3 and 4), the time delay, and a half wave and quarter wave plate, which as far as I can tell are the same for both photons, so they should not change the phase relationship between them. What in all this can put photons 2 and 3 into a Bell state?