Entanglement swapping and Bohmian mechanics

[PeterDonis]

It matters in the sense that you are the one quoting Riedmatten.

Because you gave his paper as a reference. So if you think the point is relevant, I'd like to know what you think his answer to your question would be.

 

[PeterDonis]

Your line of thinking implies they are.

I don't see how. Basically you're invoking monogamy of entanglement. Nobody in this discussion has said anything whatever to contradict monogamy of entanglement. So I don't understand where you're getting this from.

 

[DrChinese]

Not sure if it's been brought up in this thread, but section 3/4 of this paper seems to work out the details in a toy model (I think stern-gerlach) using a configuration space: https://arxiv.org/pdf/0905.4036

I'm still slowly reading through it, but maybe it will be of help to you.

That's not a bad paper, trying to present the Bohmian perspective.

While I'm not particularly trying to critique it, it does fall victim to a couple of obvious issues:

a) It attempts to present a forward in time model that wouldn't really apply in the Delayed Choice case. Note that all the references are 1998 and older, prior to discussions of Delayed Choice.

b) One issue that the Bohmian analysis presented here fails on: After the "Bell-o-meter" (we call it a BSM), the other particles (we call them 1 and 4) are entangled in all spin degrees of freedom. But the BSM only measures one degree of freedom. So all the discussion presented - which purports to explain the remote effect ("the measurement of the spin of 2 splits its wavepacket into two" and similar commentary) doesn't explain entanglement outside that one degree of freedom.

This experimentally point is one of the amazing elements of entanglement swapping. The BSM is done on 1 basis: the swap occurs on 3 mutually unbiased spin bases.

 

[Matterwave]

After the "Bell-o-meter" (we call it a BSM), the other particles (we call them 1 and 4) are entangled in all spin degrees of freedom. But the BSM only measures one degree of freedom. So all the discussion presented - which purports to explain the remote effect ("the measurement of the spin of 2 splits its wavepacket into two" and similar commentary) doesn't explain entanglement outside that one degree of freedom.

Could you explain, for my understanding, how your degree of freedom count goes? What are you counting as a degree of freedom and what do you mean when you say a BSM only measures one degree of freedom?

 

[DrChinese]

Yes. Where in any of the experiments we've been discussing are photons 5 and 6? Who cares about them?

Trying answering a direct question for once instead of turning it around.

a) |Ψ〉1234 = 12(|Ψ+〉14⨂|Ψ+〉23 − |Ψ−〉14⨂|Ψ−〉23 − |Φ+〉14⨂|Φ+〉23 + |Φ−〉14⨂|Φ−〉23)
b) |Ψ〉1256 = 12(|Ψ+〉16⨂|Ψ+〉25 − |Ψ−〉16⨂|Ψ−〉25 − |Φ+〉16⨂|Φ+〉25 + |Φ−〉16⨂|Φ−〉25)
c) |Ψ〉1278 = 12(|Ψ+〉18⨂|Ψ+〉27 − |Ψ−〉18⨂|Ψ−〉27 − |Φ+〉18⨂|Φ+〉27 + |Φ−〉18⨂|Φ−〉27)

Which of these is more true prior to a BSM? What possibly makes you think a) is more special than any of the others prior to a BSM? Obviously, it is the ones going through the BSM that matter. The other representations are completely meaningless except as a blackboard exercise.

 

[javisot]

I don't see how. Basically you're invoking monogamy of entanglement. Nobody in this discussion has said anything whatever to contradict monogamy of entanglement. So I don't understand where you're getting this from.

The core of the problem.

 

[PeterDonis]

Trying answering a direct question for once instead of turning it around.

Your b) and c) have nothing to do with the experiments we're discussing. What's the point of asking about them? Of course I've been saying all along that a) is the state prior to the BSM. b) and c) are irrelevant.

The other representations are completely meaningless except as a blackboard exercise.

If by "the other representations" you mean your b) and c), of course they are, yes. But your b) and c) are not algebraically equivalent to your a). They don't even include the same photons.

But Ma's (2) being "another representation" of Ma's (1) is not a "blackboard exercise". Ma's (2) is algebraically equivalent to Ma's (1). Both include the same four photons, and straightforward algebra that has already been shown in this thread transforms one into the other. And as I have already explained more than once, to me, and apparently to everyone in this thread except you, that means Ma's (1) and Ma's (2) represent the same physical state. And that would be the pre-swap state.

You already agree that Ma's (1) represents the pre-swap state. I think you agree that the swap changes the state (but it would be nice if you would confirm or correct that). That seems to leave only disagreement about Ma's (2)--I, and I think everyone else in this thread except you, think it represents the pre-swap state, based on the fact that it's algebraically equivalent to Ma's (1); you think Ma's (2) represents the post-swap state, based on what I'm not sure.

Your b) and c) have nothing to do with any of this. So I cannot understand why you even bring them up.

 

[PeterDonis]

a) |Ψ〉1234 = 12(|Ψ+〉14⨂|Ψ+〉23 − |Ψ−〉14⨂|Ψ−〉23 − |Φ+〉14⨂|Φ+〉23 + |Φ−〉14⨂|Φ−〉23)
b) |Ψ〉1256 = 12(|Ψ+〉16⨂|Ψ+〉25 − |Ψ−〉16⨂|Ψ−〉25 − |Φ+〉16⨂|Φ+〉25 + |Φ−〉16⨂|Φ−〉25)
c) |Ψ〉1278 = 12(|Ψ+〉18⨂|Ψ+〉27 − |Ψ−〉18⨂|Ψ−〉27 − |Φ+〉18⨂|Φ+〉27 + |Φ−〉18⨂|Φ−〉27)

As I commented just now, these don't even include the same photons. Perhaps it might help me (and probably others) to understand what point you are trying to make if you would write all three of these including all eight photons, so that we could see what you think they mean in those terms.

 

[DrChinese]

Could you explain, for my understanding, how your degree of freedom count goes? What are you counting as a degree of freedom and what do you mean when you say a BSM only measures one degree of freedom?

Sure.

i) For a Spin 1/2 particle, say an electron, we have: spin-x, spin-y, and spin z. All 3 are independent degrees of freedom. Hopefully no one will question this point.

ii) For a Spin 1 particle, say a photon, we have: H/V polarization, L/R polarization, and +/-* polarization. Like with electrons, all 3 are independent degrees of freedom. These 3 are all reported on in the Ma et al paper.

---

Note that in experimental papers, all lot of these details are left out or glossed over. During the Bell State Measurement on photons 2 & 3, only one degree of freedom is being observed and it is kept constant. For convenience, it's often referred to as H/V. But the measurements on the 1 & 4 photons can be performed on any polarization basis, and entanglement (perfect correlation, ideal case) will be registered.

It's a little difficult to see, but look at Fig. 1 in Ma, specifically the Alice and Bob columns. They show measurement outcomes in all 3 degrees of freedom H/V L/R and +/-, varied randomly (in concept). But Victor only measures on the H/V basis. (I am necessarily giving a short explanation, read the full paper for more details.)

Note that you could have Alice and Bob orient their polarizers at any same/identical angle/polarization - 23 degrees, 78 degrees, 12 degrees, 59 degrees, etc. - and the results show will show entangled statistics. In the same vein: you can orient the polarizers of Alice and Bob differently and violate a CHSH inequality. Saying this a different way: The choice of polarization basis for the BSM does not alter the statistical outcome (for the identified Bell state) for measurements on 1 & 4. They will still be entangled on all polarization bases.

*Sometimes referred to as 1/0, or Diagonal/Anti-diagonal (D/AD) or +/- 45 degrees.

 

[DrChinese]

As I commented just now, these don't even include the same photons. Perhaps it might help me (and probably others) to understand what point you are trying to make if you would write all three of these including all eight photons, so that we could see what you think they mean in those terms.

Don't need to. If a) is true, by your logic b) and c) are equally true. Before the BSM, nothing has happened! Sorry if you can't see the algebraic substitution clearly.

 

[DrChinese]

The core of the problem.

If you believe |Ψ〉1234 = |Ψ−〉12⨂|Ψ−〉34 leads to (2) when no BSM has yet occurred, then by extension:

|Ψ〉12345678 = |Ψ−〉12⨂|Ψ−〉34⨂|Ψ−〉56⨂|Ψ−〉78 leads to:

a) |Ψ〉1234 = 12(|Ψ+〉14⨂|Ψ+〉23 − |Ψ−〉14⨂|Ψ−〉23 − |Φ+〉14⨂|Φ+〉23 + |Φ−〉14⨂|Φ−〉23)
b) |Ψ〉1256 = 12(|Ψ+〉16⨂|Ψ+〉25 − |Ψ−〉16⨂|Ψ−〉25 − |Φ+〉16⨂|Φ+〉25 + |Φ−〉16⨂|Φ−〉25)
c) |Ψ〉1278 = 12(|Ψ+〉18⨂|Ψ+〉27 − |Ψ−〉18⨂|Ψ−〉27 − |Φ+〉18⨂|Φ+〉27 + |Φ−〉18⨂|Φ−〉27)
d) Etc.

Of course a) is incompatible with b) and/or c). Hopefully you can see that a) and b) and c) cannot simultaneously be true in any physical sense. And only one can be true depending on overlapping arrivals at a BS.

So: Note of these have any physical meaning whatsoever until 2 photons overlap physically in a Beam Splitter. They are just blackboard exercises otherwise, equally true by any standard. But sure, elevate a) if you like.

 

[PeterDonis]

If a) is true, by your logic b) and c) are equally true.

I have no idea what logic you are attributing to me, but it isn't mine.

Sorry if you can't see the algebraic substitution clearly.

I can't.

 

[PeterDonis]

|Ψ〉12345678 = |Ψ−〉12⨂|Ψ−〉34⨂|Ψ−〉56⨂|Ψ−〉78 leads to:

Here at least you wrote down a state with all eight photons.

a) |Ψ〉1234 = 12(|Ψ+〉14⨂|Ψ+〉23 − |Ψ−〉14⨂|Ψ−〉23 − |Φ+〉14⨂|Φ+〉23 + |Φ−〉14⨂|Φ−〉23)
b) |Ψ〉1256 = 12(|Ψ+〉16⨂|Ψ+〉25 − |Ψ−〉16⨂|Ψ−〉25 − |Φ+〉16⨂|Φ+〉25 + |Φ−〉16⨂|Φ−〉25)
c) |Ψ〉1278 = 12(|Ψ+〉18⨂|Ψ+〉27 − |Ψ−〉18⨂|Ψ−〉27 − |Φ+〉18⨂|Φ+〉27 + |Φ−〉18⨂|Φ−〉27)

Here, however, none of the states have eight photons in them, so I have no idea what algebra you think you're doing. Any correct algebra would have to take an 8-photon state to another 8-photon state. So I don't see why you think any of these are algebraically equivalent to your $\ket{\psi}_{12345678}$.

 

[PeterDonis]

|Ψ〉12345678 = |Ψ−〉12⨂|Ψ−〉34⨂|Ψ−〉56⨂|Ψ−〉78 leads to:

What experiment was this state prepared in? A hypothetical experiment you have in your head is fine; I'd just like to know what it is.

 

[DrChinese]

i) For a Spin 1/2 particle, say an electron, we have: spin-x, spin-y, and spin z. All 3 are independent degrees of freedom. Hopefully no one will question this point.

ii) For a Spin 1 particle, say a photon, we have: H/V polarization, L/R polarization, and +/-* polarization. Like with electrons, all 3 are independent degrees of freedom. These 3 are all reported on in the Ma et al paper.

In the Bohmian view, since spin is an observable but not an independent property of an electron:

The parallel to the Ma experiment would be having the "BSM" be on the spin-x basis; and then having the spin-x, spin-y, and spin z components all become independently entangled (noting they are orthogonal). So we have (presumably) the spin-x of particles 2 and 3 - dependent on their unknown/unknowable x positions - leading to entanglement on independent bases spin-y and spin-z for particle 1 and 4.

Going back to our earlier Norsen reference on the Pilot Wave perspective: This concept is completely absent. See Fig. 6 and related text.

 

[DrChinese]

What experiment was this state prepared in? A hypothetical experiment you have in your head is fine; I'd just like to know what it is.

Any 4 independent events can be modeled as any permutation of any 1, 2, 3 or 4 of them. I think you are smart enough to take it from there.

So I don't see why you think any of these are algebraically equivalent to your $\ket{\psi}_{12345678}$.

Who said they were? Just read what I wrote.

I do not plan to continue discussing something we've discussed ad nauseum. You are harassing me. I am happy to discuss anything with anyone else here or in any thread. But this is the end of the line vis a vis your semantic attacks on my posts.

-DrC

 

[PeterDonis]

I do not plan to continue discussing something we've discussed ad nauseum.

That's fine.

You are harassing me.-DrC

This is uncalled for. I thought you had a point to make that I had not grasped. If you want to drop it, that's fine with me.

 

[Matterwave]

It's a little difficult to see, but look at Fig. 1 in Ma, specifically the Alice and Bob columns. They show measurement outcomes in all 3 degrees of freedom H/V L/R and +/-, varied randomly (in concept).

I see. Ma calls these "mutually unbiased bases". Calling them degrees of freedom was confusing me a bit.

But Victor only measures on the H/V basis.

In figure 1, I indeed see +/-, H/V, and L/R in Alice and Bobs columns. But what do you make of the $\Phi^+$ and $\Phi^-$ that shows up under Victor's column? Isn't that a measurement in a different basis than just H/V?

 

[Roberto Pavani]

It is said that two particles (e.g. two photons) with the same quantum numbers are indistinguishable, particles carry no label.

So take two runs in Victor's Φ⁺ group with identical recorded outcomes: (A1, V1, B1) and (A33, V33, B33). Since the photons carry no label, A1 and A33 are indistinguishable. I can therefore swap A1 with A33, i.e. pair A1 with B33 and A33 with B1, and I still get the same, maximal CHSH violation.

But nobody would say A1 is entangled with B33, or A33 with B1: they belong to different runs and never interacted. Yet the violation is unchanged.

 

[Morbert]

a) It attempts to present a forward in time model that wouldn't really apply in the Delayed Choice case. Note that all the references are 1998 and older, prior to discussions of Delayed Choice.

Forward-in-time models readily apply in the Delayed Choice case.

 

[Morbert]

Re/ Bohmian mechanics and entanglement swapping: I'm finding it conceptually straightforward but computationally fiddly. Considering spin/electrons instead of polarization/photons since that's more readily NRQM-friendly: The state takes a form like $$|\Psi_0\rangle
=
|\Phi_{\mathrm{spatial}}\rangle
|\Psi^-\rangle_{12}
|\Psi^-\rangle_{34}
\otimes
\left(
\frac{1}{\sqrt{3}}
\sum_{a=x,y,z}
|a\rangle_{q_A}
\right)
\otimes
\left(
\frac{1}{\sqrt{3}}
\sum_{b=x,y,z}
|b\rangle_{q_B}
\right)
\otimes
|A_0\rangle
|B_0\rangle
|V_0\rangle
$$where $q_A$ and $q_B$ are QRNGs modeling Alice's and Bob's choice of measurement axes, and $|A_0\rangle, |B_0\rangle,|V_0\rangle$ are the apparatuses. The Bohmian configuration is $$
Q_0
=
\left(
\mathbf{Q}_1,
\mathbf{Q}_2,
\mathbf{Q}_3,
\mathbf{Q}_4,
Q_{q_A},
Q_{q_B},
Q_A,
Q_B,
Q_V
\right)$$The Bohmian guidance equation will then give us trajectories and hence the data read off from the apparatuses including all observed correlations reported in these experiments.

The work seems to be in actually constructing the toy configuration space to model the essentials.

I wrote some basic code to visualize the branching structure of this global wavefunction, as the decoherent branches mark alternative coarse-grained paths for the Bohmian configuration. Results aren't too surprising but people might still be interested. Some examples follow:

First, a case where Victor performs his attempt at a BSM first, followed by Alice's measurement, then Bob's. If Alice and Bob both choose the RL basis, the branching structure will look like

It's unsurprising that, for example, if Victor records $\Phi^+$ or $\Phi^-$ then Alice and Bob will record the same outcomes, and hence there are only two branches for each $\Phi$. If Victor's BSM doesn't succeed, then Alice and Bob's results can be the same or different, and so there are four branches.

If they instead pick mutually unbiased bases, then the branching structure will look like

Here the BSM does not introduce a correlation between Alice's and Bob's outcome, and hence more branches have positive weights.

The above figures are for the standard ES experiment. But we can also look at the branching pattern of Ma's DCES experiment, where Victor makes his measurement last. Say Alice and Bob measure in the HV basis. We get a branching structure

We can follow the branches to see if Alice and Bob both record the same outcome, Victor's BSM must yield either $\Phi^+$ or $\Phi^-$. If Alice and Bob both record opposite outcomes,Victor's BSM must fail.

We can also look at the branching pattern when Victor makes his measurement after Alice but Before Bob, akin to Megidish's ordering. Considering the same as the above: Alice and Bob both measure in the HV basis

Here there is no further branching after Victor's measurement. What's interesting to note is regardless of the measurement order/branching structure, the final branches, each carrying an output record combination, have the same weights. This is why the statistics are insensitive to the measurement order.

 

[DrChinese]

1. I see. Ma calls these "mutually unbiased bases". Calling them degrees of freedom was confusing me a bit.


2. In figure 1, I indeed see +/-, H/V, and L/R in Alice and Bobs columns. But what do you make of the $\Phi^+$ and $\Phi^-$ that shows up under Victor's column? Isn't that a measurement in a different basis than just H/V?

...

1. I also use the term "mutually unbiased bases". Here I used the other term, although they are not precisely the same.

2. Good question.

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. Here are the caveats:

i) In the Ma et al reference, Victor always chooses the H/V basis. And holding the basis for the swap fixed is common in these experiments, usually instead varying the choices of Alice and Bob..
ii) When Alice and Bob measure on the same basis as Victor, there is still the a swap occurring. However, this is a special case in which both the Entangled State and Separable State statistics are the same. See Fig. 3 in which the a) and b) graphs for H/V are about the same. Ma: "separable states can be maximally correlated (ideal correlation function 1) only in one basis, the others being 0".

 

[DrChinese]

It is said that two particles (e.g. two photons) with the same quantum numbers are indistinguishable, particles carry no label.

So take two runs in Victor's Φ⁺ group with identical recorded outcomes: (A1, V1, B1) and (A33, V33, B33). Since the photons carry no label, A1 and A33 are indistinguishable. I can therefore swap A1 with A33, i.e. pair A1 with B33 and A33 with B1, and I still get the same, maximal CHSH violation.

But nobody would say A1 is entangled with B33, or A33 with B1: they belong to different runs and never interacted. Yet the violation is unchanged.

FYI: As it happens, photons carrying time stamp information (usually arrival time) can be distinguished - even if otherwise indistinguishable. This particular point is explicitly used the the Megidish et al paper Fig. 3: "(c) when the projection fails due to temporal distinguishability."

 

[Morbert]

In figure 1, I indeed see +/-, H/V, and L/R in Alice and Bobs columns. But what do you make of the $\Phi^+$ and $\Phi^-$ that shows up under Victor's column? Isn't that a measurement in a different basis than just H/V?

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 $\{HH,VV,HV,VH\}$. The experimental protocol says to discard runs where Victor obtains $HV$ or $VH$ whether or not his apparatus is configured to BSM or SSM. Hence, figure 1 shows the retained BSM outcomes $\{\Phi^+, \Phi^-\}$ and retained SSM outcomes $\{HH, VV\}$.

 

[DrChinese]

I wrote some basic code to visualize the branching structure of this global wavefunction, as the decoherent branches mark alternative coarse-grained paths for the Bohmian configuration. Results aren't too surprising but people might still be interested. Some examples follow:

First, a case where Victor performs his attempt at a BSM first, followed by Alice's measurement, then Bob's. If Alice and Bob both choose the RL basis, the branching structure will look like

1. View attachment 373972

It's unsurprising that, for example, if Victor records $\Phi^+$ or $\Phi^-$ then Alice and Bob will record the same outcomes, and hence there are only two branches for each $\Phi$. If Victor's BSM doesn't succeed, then Alice and Bob's results can be the same or different, and so there are four branches.

If they instead pick mutually unbiased bases, then the branching structure will look like

2. View attachment 373973

Here the BSM does not introduce a correlation between Alice's and Bob's outcome, and hence more branches have positive weights.

The above figures are for the standard ES experiment. But we can also look at the branching pattern of Ma's DCES experiment, where Victor makes his measurement last. Say Alice and Bob measure in the HV basis. We get a branching structure

3. View attachment 373974

We can follow the branches to see if Alice and Bob both record the same outcome, Victor's BSM must yield either $\Phi^+$ or $\Phi^-$. If Alice and Bob both record opposite outcomes,Victor's BSM must fail.

We can also look at the branching pattern when Victor makes his measurement after Alice but Before Bob, akin to Megidish's ordering. Considering the same as the above: Alice and Bob both measure in the HV basis

4. View attachment 373975

Here there is no further branching after Victor's measurement. What's interesting to note is regardless of the measurement order/branching structure, the final branches, each carrying an output record combination, have the same weights. This is why the statistics are insensitive to the measurement order.

Wow Morbert, nice work. I need you to come over and do a few of these for me.

- First graph (Alice/Bob "same basis"): I agree.
- Second graph (Alice/Bob "mutually unbiased bases"): I agree. Alice and Bob never show any entanglement on mutually unbiased bases.
- Third graph and fourth graphs: These I do not agree with. Actually, I think these two should be similar to your first graph, just in a different time order. IMHO there should be 8 final permutations, just like your first graph.

a) When Victor is measured as $\Phi^-$: Alice and Bob will agree for the H/V and R/L bases, but will be opposite on +/-. So you will see photons 1 and 4 as |HH> or |VV>, but never as |HV> or |VH>. Or photons 1 and 4 as |RR> or |LL>, but never as |RL> or |LR>. These results are most clearly seen in Ma's fig. 3, which only shows $\Phi^-$. (Which is all they reported, and they note "the magnitude of all correlation functions equals 1 ideally".

b) $\Phi^+$: This is a little foggier. But you can see that your first graph shows Alice and Bob as different (I agree). But the last 2 graphs show Alice and Bob the same for the $\Phi^+$ (which I think is backwards).

On the other hand: I could be confused... it happens. Anyway, look over what I say and give me your thoughts. Again, nice stuff.

Side note: This is not a criticism your labeling, I am simply saying this to make sure no readers misunderstand. What you label "No BSM" I would call "No Identifiable BSM". In principle: if 4 photons arrive with the coincidence time window, there is always a swap for Alice and Bob. But 2 of the 4 Bell states yield similar signatures, and cannot be distinguished as whether a $\Phi^-$ or $\Phi^+$ case occurs. Ma simply drops reporting of these cases, and filters on cases with affirmative signatures. My point does not in any way affect your statistics on your nice graphs. It's a nice format.

 

[DrChinese]

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

Note that the outputs you call "HV,VH" can also be labeled "|Ψ−>, |Ψ+>" since these are the other 2 Bell states. These cannot, however, be optically distinguished at this time.

 

[Morbert]

Third graph and fourth graphs: These I do not agree with. Actually, these two should be similar to your first graph, just in a different time order. IMHO there should be 8 final permutations, like your first graph.

a) When Victor is measured as $\Phi^-$: Alice and Bob will agree for the H/V and R/L bases, but will be opposite on +/-. So you will see photons 1 and 4 as |HH> or |VV>, but never as |HV> or |VH>. Or photons 1 and 4 as |RR> or |LL>, but never as |RL> or |LR>. These results are most clearly seen in Ma's fig. 3, which only shows $\Phi^-$. (Which is all they reported, and they note "the magnitude of all correlation functions equals 1 ideally".

b) $\Phi^+$: This is a little foggier. But you can see that your first graph shows Alice and Bob as different (I agree). But the last 2 graphs show Alice and Bob the same for the $\Phi^+$ (which I think is backwards).

On the other: I could be confused. Anyway, look over what I say and give me your thoughts. Again, nice stuff.

Side note: This is not a criticism your labeling, I am simply saying this to make sure no readers misunderstand. What you label "No BSM" I would call "No Identifiable BSM". In principle: if 4 photons arrive with the coincidence time window, there is always a swap. But you may not always see those 4 because 2 show up at the same detector too close together - ergo the labeling either way. Ma simply drops reporting of these cases. My point does not in any way affect your statistics on your nice graphs. It's a nice format.

The third and fourth graph correspond to Alice and Bob both measuring in the HV basis. And you can see there is no branch on either graph where Victor obtains $\Phi^+$/$\Phi^-$ and Alice and Bob obtain $HV$/$VH$. There are plenty of other graphs that can be constructed. (All graphs can be combined into a single graph but that is much too busy). If you request a choice of Bases for Alice and Bob and an ordering, I can supply it.

Note that the outputs you call "HV,VH" can also be labeled "|Ψ−>, |Ψ+>" since these are the other 2 Bell states. These cannot, however, be optically distinguished at this time.

They can't be labelled $\ket{\Psi^-},\ket{\Psi^+}$ as $\ket{HV} = \frac{1}{\sqrt{2}}(\ket{\Psi^+} + \ket{\Psi^-})$ and $\ket{VH} = \frac{1}{\sqrt{2}}(\ket{\Psi^+} - \ket{\Psi^-})$.

 

[DrChinese]

The third and fourth graph correspond to Alice and Bob both measuring in the HV basis. And you can see there is no branch on either graph where Victor obtains $\Phi^+$/$\Phi^-$ and Alice and Bob obtain $HV$/$VH$. There are plenty of other graphs that can be constructed. (All graphs can be combined into a single graph but that is much too busy). If you request a choice of Bases for Alice and Bob and an ordering, I can supply it.

I understood Alice and Bob are measuring on the HV basis. If you look at Ma Fig. 3, a (which is $\Phi^-$) - there is correlation. You have that too, all good. But you also show correlation for $\Phi^+$ in your 3 & 4.

On your graph 1, you show anti-correlation. I realize your graph 1 is R/L for Alice and Bob, but both H/V and R/L show anti-correlation for $\Phi^+$. I believe your graph 1 is correct.

Graph 1 also shows 4 final records for no BSM, which is correct. But graph 4 shows only 2. (Honestly, that really doesn't make much difference either way for our discussions.

Keeping in mind: The statistical results for Alice and Bob do not really vary according to choice of basis for Victor as long is Victor sees $\Phi^-$.

 

[Morbert]

I understood Alice and Bob are measuring on the HV basis. If you look at Ma Fig. 3, a (which is $\Phi^-$) - there is correlation. You have that too, all good. But you also show correlation for $\Phi^+$ in your 3 & 4.

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

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

 

[Morbert]

Graph 1 also shows 4 final records for no BSM, which is correct. But graph 4 shows only 2. (Honestly, that really doesn't make much difference either way for our discussions.

This is because the "No BSM" in these graphs correspond to the case where Victor has his apparatus configured for a BSM, but obtains a separable state outcome regardless. In these cases, there is some information conveyed when Alice and Bob measure in the HV basis, such that if Victor's BSM fails, Alice and Bob will have opposite HV outcomes. Hence graph 4 has 2. Graph 1 has 4 because in that graph Alice and Bob measure in the RL basis.

This in contrast to Victor never performing a measurement at all, in which case Alice and Bob would indeed see no correlation regardless of their choice.

If wanted, I can also supply graphs where Victor performs an SSM instead of a BSM.