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.