Entanglement swapping and Bohmian mechanics

[Morbert]

As I have said plenty of times, (2) is mathematically derived from (1). There is no physical meaning to that derivation.

I'm not asking you for physical meaning. That comes after. First we need basic linear algebra.

Ma says something more specific than "(2) is mathematically derived from (1)". He says (2) is a rewriting of (1). Do you accept this?

 

[Sambuco]

Because if you execute a BSM on photons 2 and 3, you do get one of the Bell States of i).

Yeahhhhh!!!!!! That's the key! Before the BSM on photons 2 and 3, the state of the entire system is:

i) |Ψ〉1234 (before BSM) = |Ψ−〉12⨂|Ψ−〉34 = 1/2(|Ψ+〉14⨂|Ψ+〉23 − |Ψ−〉14⨂|Ψ−〉23 − |Φ+〉14⨂|Φ+〉23 + |Φ−〉14⨂|Φ−〉23)

which does not show any entanglement between photons from the pairs (1,2) and (3,4). After the BSM, only one of these terms remains. For example, if Victor measures |Φ+〉23, the state of the entire system after the BSM is:

|Ψ〉1234 (after BSM) = |Φ+〉14⨂|Φ+〉23

Lucas.

 

[Morbert]

@DrChinese I suspect what is happening is when you see me ask whether you accept Ma's words when he says (2) is a rewriting of (1), you are jumping ahead and interpreting this as me asking whether you accept Ma is saying photons 1 and 4 were secretly entangled all along.

I.e. You're misunderstanding the physical significance of a state expansion and thereby refusing to acknowledge state expansions.

 

[DrChinese]

@DrChinese I suspect what is happening is when you see me ask whether you accept Ma's words when he says (2) is a rewriting of (1), you are jumping ahead and interpreting this as me asking whether you accept Ma is saying photons 1 and 4 were secretly entangled all along.

Funny you say that.

In Ma's (1), photons 1 and 4 are not entangled. In Ma's (2), photons 1 and 4 are entangled in all 4 Bell State terms. So... how is (2) a rewriting of (1)? Because they make contradictory assertions, right? Are they "secretly entangled" or not?

 

[danieltanfh95]

In the original setup of (1,2) and (3,4) being entangled, we can break it down to several situations.

1. No photons are measured BEFORE the beam-splitter correlating (2,3), then the result is probabilistic.
2. (2,3) is measured BEFORE the beamsplitter, so they are detected and gone, no beam splitter stage.
3. (1,4) is measured BEFORE the beamsplitter, and now we have information about polarisation of (2,3), and (2,3) is no longer entangled to (1,4). Even if we already know the polarisation of (2,3) here, the beam-splitter "re-entangles" (how entanglement happens here physically is the unknown) the photons so the result is still probabilistic. This is also where the confusion of the delayed choice framing comes from, basically a trivial misunderstanding over when the measurement happens.

 

[Demystifier]

And it looks to me as if the system's particle positions are in fact a function of other particles' probability current and probability density. Aren't those in turn a function of particle positions? Or particle velocities? Or is the issue that particle position of some particle N is influenced by the Pilot Wave, and not the other way around? Hopefully you can translate what Norsen is saying, to the extent I have it confused.

The wave function is a function of positions. But it is not a function of the positions of Bohmian particles. The wave function is defined on any possible position in space and it does not depend on whether there is a Bohmian particle on that position or not.

 

[Demystifier]

In Ma's (1), photons 1 and 4 are not entangled. In Ma's (2), photons 1 and 4 are entangled in all 4 Bell State terms. So... how is (2) a rewriting of (1)? Because they make contradictory assertions, right? Are they "secretly entangled" or not?

This is somewhat like saying that 16 can't be rewritten as 1+3+5+7, because 16 is even while each of the 4 terms in 1+3+5+7 is odd.
Just because each term is entangled doesn't mean that the whole sum is entangled. If you don't understand it, then you don't understand entanglement.

Here is a simpler example. Start with
$$|\psi\rangle=|1\rangle |1\rangle$$
which is clearly not entangled. Then rewrite it as
$$|\psi\rangle=\frac{1}{2}\left( |1\rangle |1\rangle + |2\rangle |2\rangle \right) + \frac{1}{2}\left( |1\rangle |1\rangle - |2\rangle |2\rangle \right)$$
Finally introduce the notation
$$|\pm\rangle = \frac{1}{\sqrt{2}}\left( |1\rangle |1\rangle \pm |2\rangle |2\rangle \right)$$
to rewrite $|\psi\rangle$ as
$$|\psi\rangle=\frac{1}{\sqrt{2}} |+\rangle + \frac{1}{\sqrt{2}} |-\rangle$$
The states $|+\rangle$ and $|-\rangle$ are entangled, so we see that the non-entangled state is a sum of two entangled states.

 

[Demystifier]

As I have said plenty of times, (2) is mathematically derived from (1). There is no physical meaning to that derivation. If you have $100 and you add $1000 to that using an adding machine, you get $1100. But you still only have $100. Note that when I start with |Ψ−〉12⨂|Ψ−〉34, I can do all kinds of operations on that and still not end up with photons 1 and 4 in a Bell state. In other words, you still don't have $1100. You have to earn it somewhere first.

Here is a better analogy. Suppose that initially you have $100 in cash. Write this as $100=$1100-$1000. What could it possibly mean physically? One interpretation is that you borrowed additional $1000 from a friend, so now you have $1100 in cash, but also a debt $1000 that you owe the friend, so in total you still have only $100. But then your friend suddenly dies, so you don't longer owe him. After that, you have $1100. Your debth has been erased by your friend's death.

In QM, the borrowing by the friend is a metaphor for a unitary operation on the quantum state, e.g., the splitting of the wave function by the magnets of Stern-Gerlach apparatus. The erasure of the debt is a metaphor for the projection, that is, the effective wave function collapse due to measurement.

 

[Morbert]

In Ma's (1), photons 1 and 4 are not entangled. In Ma's (2), photons 1 and 4 are entangled in all 4 Bell State terms. So... how is (2) a rewriting of (1)? Because they make contradictory assertions, right? Are they "secretly entangled" or not?

In Ma's (1), photons 1 and 4 are not entangled. In Ma's (2), photons 1 and 4 are not entangled.

 

[Lord Jestocost]

Maybe, an in-depth understanding of the following paper might help to disentangle some mess in this thread: “Experimental Entanglement Swapping: Entangling Photons That Never Interacted” by J.-W. Pan, D. Bouwmeester, H. Weinfurter, and A. Zeilinger, Phys. Rev. Lett. 80, 3891 (1998).
https://doi.org/10.1103/PhysRevLett.80.3891

A remark by the authors:

"We might also remark that the present results, taken together with those of our recent verification of quantum teleportation [15], are easily understood in the framework of the Copenhagen interpretation of quantum mechanics [16]. They cause no conceptual problems if one accepts that information about quantum systems is a more basic feature than any possible “real” properties these systems might have [17].”

 

[martinbn]

Maybe, an in-depth understanding of the following paper might help to disentangle some mess in this thread: “Experimental Entanglement Swapping: Entangling Photons That Never Interacted” by J.-W. Pan, D. Bouwmeester, H. Weinfurter, and A. Zeilinger, Phys. Rev. Lett. 80, 3891 (1998).
https://doi.org/10.1103/PhysRevLett.80.3891

A remark by the authors:

"We might also remark that the present results, taken together with those of our recent verification of quantum teleportation [15], are easily understood in the framework of the Copenhagen interpretation of quantum mechanics [16]. They cause no conceptual problems if one accepts that information about quantum systems is a more basic feature than any possible “real” properties these systems might have [17].”

It seems that entanglment swapping was well understood already in the 1990's, both theoretically and experimentaly.

 

[Demystifier]

"We might also remark that the present results, taken together with those of our recent verification of quantum teleportation [15], are easily understood in the framework of the Copenhagen interpretation of quantum mechanics [16]. They cause no conceptual problems if one accepts that information about quantum systems is a more basic feature than any possible “real” properties these systems might have [17].”

This corresponds to the information interpretation of QM, typical of Zeilinger. A Bohmian realist, on the other hand, is not satisfied with the notion of information living in some Platonic world of ideas, but seeks a physical object carrying this information. And since the relevant information is available in the future, the carrier of information cannot be the photon in the past. That's why, for a Bohmian, it's perfectly natural to accept that information is carried by the computer memory and its environment. To invert the quoted sentence above, that causes no conceptual problems if one accepts that real properties of a system are a more basic feature than information about quantum system that might be available to a particular observer.

For a more general critique of the Zeilinger's informational point of view, as seen by Bohmian realists, see https://arxiv.org/abs/quant-ph/0604173 .

 

[Matterwave]

For a more general critique of the Zeilinger's informational point of view, as seen by Bohmian realists, see https://arxiv.org/abs/quant-ph/0604173 .

That direct call out lol...

 

[DrChinese]

View attachment 374595
That direct call out lol...

You're right, don't see that too often!

I'm not criticizing Daumer et al - it's a strong group of scientists - but there is a bit of humor to be found in the paper. Obviously, they push the Bohmian "deterministic" perspective. Exactly what I'm interested in, especially vis a vis the Zeilinger experiments. But they have the nerve to say:

i) "Thus, the experimental facts of quantum mechanics do not establish indeterminism." Actually, essentially every quantum experiment ever run confirms the root nature of randomness in QM.* No experiment has ever been run that correctly predicts the outcome of a 50:50 Beam splitter using Bohmian theory. (oQM says that there is no more complete specification of the system.)

ii) To be fair, MWI proponents miss the ball on the same point. In case no one noticed: Only a single outcome of every quantum experiment ever run has been observed.*

Neither i) or ii) make the concepts of BM or MWI worthless. But I would not say experimental results haven't yet done either of them a favor. Their stronger points are on the conceptual side. Experiments show no hints of underlying determinism, or of other worlds. With both MWI and BM, the hope is that a future theoretical breakthrough might lead to a confirming experiment. That would be great.

Please keep in mind that I don't pay much attention to Zeilinger's views on QM outside of his experimental papers themselves. He is very careful to limit the scope of his comments to what is generally accepted science in those.

*If someone said that c was actually 400,000 km/sec, even though all measurements say it's closer to 300,000 km/sec: you might call them stubborn.

 

[DrChinese]

Maybe, an in-depth understanding of the following paper might help to disentangle some mess in this thread: “Experimental Entanglement Swapping: Entangling Photons That Never Interacted” by J.-W. Pan, D. Bouwmeester, H. Weinfurter, and A. Zeilinger, Phys. Rev. Lett. 80, 3891 (1998).
https://doi.org/10.1103/PhysRevLett.80.3891

A remark by the authors:

"We might also remark that the present results, taken together with those of our recent verification of quantum teleportation [15], are easily understood in the framework of the Copenhagen interpretation of quantum mechanics [16]. They cause no conceptual problems if one accepts that information about quantum systems is a more basic feature than any possible “real” properties these systems might have [17].”

Great reference. Free version: HERE

 

[DrChinese]

In Ma's (1), photons 1 and 4 are not entangled. In Ma's (2), photons 1 and 4 are not entangled.

I just don't get it, of course 1 and 4 are entangled in (2). The authors say exactly the opposite of you, not sure what is ambiguous here:

"...if Victor subjects his photons 2 and 3 to a Bell-state measurement [BSM], they become entangled. Consequently photons 1 (Alice) and 4(Bob) also become entangled and entanglement swapping is achieved. This can be seen by rewriting Eq. (1) in the basis of Bell states of photons 2 and 3:|Ψ〉1234 = 12(|Ψ+〉14⨂|Ψ+〉23 − |Ψ−〉14⨂|Ψ−〉23 − |Φ+〉14⨂|Φ+〉23 + |Φ−〉14⨂|Φ−〉23)..."

(1) -> BSM -> (2)

---

On the other hand, we may agree on a portion of what we both are saying.

(1)=(2) and 1&4 are not entangled [what you say above] -> BSM -> 1&4 are entangled in one of the 4 Bell states of (2) [what I say above]

 

[DrChinese]

Here is a better analogy. Suppose that initially you have $100 in cash. Write this as $100=$1100-$1000. What could it possibly mean physically? One interpretation is that you borrowed additional $1000 from a friend, so now you have $1100 in cash, but also a debt $1000 that you owe the friend, so in total you still have only $100. But then your friend suddenly dies, so you don't longer owe him. After that, you have $1100. Your debt has been erased by your friend's death.

Great analogy!

$100 -> Friend dies -> $1100

(1) -> BSM -> (2)

 

[DrChinese]

Just because each term is entangled doesn't mean that the whole sum is entangled. If you don't understand it, then you don't understand entanglement.

Here is a simpler example. Start with
$$|\psi\rangle=|1\rangle |1\rangle$$
which is clearly not entangled. Then rewrite it as ...
$$|\psi\rangle=\frac{1}{\sqrt{2}} |+\rangle + \frac{1}{\sqrt{2}} |-\rangle$$
The states $|+\rangle$ and $|-\rangle$ are entangled, so we see that the non-entangled state is a sum of two entangled states.

You must already know that I assign no physical meaning to this.

 

[DrChinese]

For a more general critique of the Zeilinger's informational point of view, as seen by Bohmian realists, see https://arxiv.org/abs/quant-ph/0604173 .

Having nothing whatsoever to do directly with Physics:

The authors, in 2006, said: "At a time when the forces of obfuscation in America are engaged in a campaign against the theory of evolution..." It is 20 years later, and the anti-science parade in the US has advanced further than anyone might have imagined in their worst case scenario. Sadly so.

 

[DrChinese]

1. ...
2. ...
3. (1,4) is measured BEFORE the beamsplitter, and now we have information about polarisation of (2,3), and (2,3) is no longer entangled to (1,4). Even if we already know the polarisation of (2,3) here, the beam-splitter "re-entangles" (how entanglement happens here physically is the unknown) the photons so the result is still probabilistic. This is also where the confusion of the delayed choice framing comes from, basically a trivial misunderstanding over when the measurement happens.

Welcome to the discussion!

You may not be aware of a technical point here. The (2,3) pair polarization is measured (via the BSM, Bell State Measurement) on a mutually unbiased basis relative to the (1,4) pair. This means there is no polarization information at all gained from one pair that tells you anything about the other pair. The BSM result instead indicates the nature of the Bell entanglement of the (1,4) pair. That is true regardless of the order of measurements on any of the 4 photons.

 

[Morbert]

I just don't get it, of course 1 and 4 are entangled in (2). The authors say exactly the opposite of you, not sure what is ambiguous here:

"...if Victor subjects his photons 2 and 3 to a Bell-state measurement [BSM], they become entangled. Consequently photons 1 (Alice) and 4(Bob) also become entangled and entanglement swapping is achieved. This can be seen by rewriting Eq. (1) in the basis of Bell states of photons 2 and 3:|Ψ〉1234 = 12(|Ψ+〉14⨂|Ψ+〉23 − |Ψ−〉14⨂|Ψ−〉23 − |Φ+〉14⨂|Φ+〉23 + |Φ−〉14⨂|Φ−〉23)..."

(1) -> BSM -> (2)

Ma and I are aligned perfectly. Ma is saying that if we rewrite $\Psi$, (1), in the basis of Bell stats of photons 2 and 3, we can see that a future BSM measurement acting on $\Psi$ that collapses 23 to a Bell state will also collapse 14 onto a Bell state.

I.e. Ma is not saying is that (2) is post BSM. He is saying (2) reveals what the state can be post-BSM.

 

[gentzen]

You must already know that I assign no physical meaning to this.

The problem is that the equality sign is fundamental to how math works. And you mess with the meaning of the equality sign. This cannot be excused by reference to unassigned physical meaning.
This is way worse than reinterpreting the meaning of the +/- sign, which is sometimes done informally as a shortcut.

 

[Demystifier]

You must already know that I assign no physical meaning to this.

How about a one-particle superposition, such as
$$|+\rangle = \frac{1}{\sqrt{2}}|1\rangle + \frac{1}{\sqrt{2}}|2\rangle ?$$
Do you assign a physical meaning to that?

 

[martinbn]

How about a one-particle superposition, such as
$$|+\rangle = \frac{1}{\sqrt{2}}|1\rangle + \frac{1}{\sqrt{2}}|2\rangle ?$$
Do you assign a physical meaning to that?

I asked that in post 501, you can see his answer somehere there.

 

[Demystifier]

If you have a |V> photon stream (say from a laser) that you can only measure on the $\{|+\rangle, |-\rangle\}$ basis, yes, that is meaningful. I would not call them the same state though. Naturally, I can distinguish them experimentally using a suitably oriented PBS.

Similarly, I can distinguish the difference between Ma's (1) and (2) experimentally. As they do. Megidish as well. Disagree?

Well, all physicists call them the same state, including Ma and Megidish. I understand your intuition behind your language, but you should be aware that you use a language that differs from the language of physicists, which makes the discussions between you and others rather challenging.

 

[DrChinese]

Ma is saying that if we rewrite $\Psi$, (1), in the basis of Bell stats of photons 2 and 3, we can see that a future BSM measurement acting on $\Psi$ that collapses 23 to a Bell state will also collapse 14 onto a Bell state.

I.e. Ma is not saying is that (2) is post BSM. He is saying (2) reveals what the state can be post-BSM.

"We can see that a future BSM measurement acting on $\Psi$ that collapses 23 to a Bell state will also collapse 14 onto a Bell state." Agreed.

"He is saying (2) reveals what the state can be post-BSM." Agreed.

"I.e. Ma is not saying is that (2) is post BSM." If this is your point, fine, you win. Agreed.

In that case: Ma curiously fails to mention that post-BSM, the new state is also (2). So you are interpreting Ma as saying "if Victor subjects his photons 2 and 3 to a Bell-state measurement, they become entangled" to refer to a conditional future. That is a way to read that, and yes, it is a true statement. I read it as "after Victor subjects his photons 2 and 3 to a Bell-state measurement, they become entangled." That too is true. The nuance escapes me as to the relevance of the science we are discussing. But hey, that's just me.

Any chance we can discuss the Bohmian view of the BSM now? I am still interested in looking into that more deeply.

 

[DrChinese]

Well, all physicists call them the same state, including Ma and Megidish. I understand your intuition behind your language, but you should be aware that you use a language that differs from the language of physicists, which makes the discussions between you and others rather challenging.

No, I don't believe there are any physicists calling streams of state A and state B "the same state" when they can be distinguished experimentally. Ma is able to distinguish BSM - using (2) - from SSM which is based on (1). Not in an individual case (as that is varied randomly in Ma), but certainly, they can for the segregated streams. Megidish sets their experiment up to be a stream, and uses quantum state tomography to demonstrate the difference between the BSM and SSM streams. So my "intuition" happens to exactly match... experiment.

Note that I certainly agree there are spin states which cannot be distinguished experimentally. And there are spin states that cannot be distinguished in individual trials. As you should know by now, I often hang my hat on experimental results. If I say something that does not match experiment, please, help me out with a useful quote. I don't see that as the case here, since I am the one quoting experiment.

 

[martinbn]

Ma curiously fails to mention that post-BSM, the new state is also (2).

Because it is not (2).

 

[Morbert]

In that case: Ma curiously fails to mention that post-BSM, the new state is also (2).

Post-BSM*, the new state is one of the four terms in (2), but it is not (2), as (2) is a linear combination of them. I.e. It is $\ket{\psi^+}_{14}\ket{\psi^+}_{23}$ or $\ket{\psi^-}_{14}\ket{\psi^-}_{23}$ or $\ket{\phi^+}_{14}\ket{\phi^+}_{23}$ or $\ket{\phi^-}_{14}\ket{\phi^-}_{23}$. None of these are (2).

*The usual caveats of the meaning behind "post" here given the retrocausal character of the account.

 

[DrChinese]

1. *The usual caveats of the meaning behind "post" here given the retrocausal character of the account.

2. Post-BSM*, the new state is one of the four terms in (2), but it is not (2), as (2) is a linear combination of them. I.e. It is $\ket{\psi^+}_{14}\ket{\psi^+}_{23}$ or $\ket{\psi^-}_{14}\ket{\psi^-}_{23}$ or $\ket{\phi^+}_{14}\ket{\phi^+}_{23}$ or $\ket{\phi^-}_{14}\ket{\phi^-}_{23}$. None of these are (2).

1. Agreed. I'm not trying to distinguish the order as we discuss this point. That is a different point.

2. I might call (2) a superposition. Again, the authors' usage of language does not really specify your point of view - or mine exactly. However, at this point it's clearly semantics and not science. The fact is, no one can make a provable statement about when the specific Bell state (one of the four) is selected. Is it when the Beam Splitter is encountered and there is physical overlap? Or is it later, when photons 2 and 3 get tested for polarization and the Bell state can be identified? I don't believe oQM makes a statement about this. It is possible an Interpretation might.

Where we agree: What we have all been using as standard language here - so far - is the BSM as "one black box". Most of the time, the black box approach is fine. And if it is more comfortable to you to specify that the post-BSM state is not (2) but rather one of the 4 terms of (2); then that's fine, no I have no specific disagreement with this distinction. What you say is true. I think everyone understands that we end up with one randomly selected Bell state, and that at some point prior the state was (2); and at some point prior to that it was (1).

And yet again, I find myself asking about the relationship of these nuances to Bohmian Mechanics. I get the feeling that there is a relevant point lurking around that is going to appear soon.