Entanglement between particles of no common history

1. Dec 26, 2008

colorSpace

I'd like to confirm my understanding from one year ago, that the following is possible:

1) Using a combination of entanglement processes (entangling an already entangled particle with another already entangled particle, such that the outer particles, the other particles of each pair, will be entangled as well), one can effectively entangle two particles which have no common history.

(Think of a "W" shape where the two sources of each of the two entangled pairs are at the bottom, and the four particles fly towards the top, with the two inner particles meeting in the middle.)

2) This is possible to do in a way that the correlation effect between indirectly entangled particles can be established before any signal can travel the whole distance (since the inner particles of each pair travel in opposite directions).

And that it is therefore not possible that any "hidden variable" connects the outer particles within the speed of light before the measurement is taken, so that any local explanation for the correlation would have to explain this by what happens ("pairing-up") when the researchers meet afterwards to examine the results for correlations.

In case I'm getting the question across, is my interpretation of the research done so far correct?

2. Dec 26, 2008

Marcaias

Alice------Bob--------Charlie
1 -------- 2, 3 --------- 4

Let Alice and Bob entangle an EPR pair $$|\phi^+\rangle_{12} = |00\rangle + |11\rangle$$, and Bob and Charlie do the same, $$|\phi^+\rangle_{34} = |00\rangle + |11\rangle$$. Alice keeps qubit 1, Bob keeps qubits 2 and 3, and Charlie keeps qubit 4.

If Bob performs a Bell measurement on his pair of qubits 2&3, he will measure one of the Bell basis states $$|\xi\rangle_{23} \in \{ |\phi^+\rangle, |\phi^-\rangle, |\psi^+\rangle, |\psi^-\rangle \}$$. However, if you perform some algebraic magic on the initial combined state $$|\phi^+\rangle_{12} |\phi^+\rangle_{34}$$, you will see that the 1&4 qubit state will have to collapse to $$|\xi\rangle$$ as well.

In other words, Alice and Charlie end up with an entangled EPR pair despite never having to have met.

3. Dec 27, 2008

msumm21

So if particles p1 and p2 are entangled, you can subsequently entangle p1 with some new particle p3 without destroying the entanglement with p2? Any references about this?

4. Dec 27, 2008

Marcaias

That isn't what I described, but that is also possible.

What I described:

1. Initially the 1,2 system and the 3,4 system are entangled.

2. The relay (Bob) can perform only local operations (i.e. he only fiddles with his qubits #2 and #3) so that afterwards the 2,3 system and the 1,4 system are entangled. (But the 1,2 and 3,4 systems are no longer entangled.)

In other words, you "swap" the entanglement from one system to another, the amazing part being that the 1,4 system becomes entangled despite Alice and Charles never meeting each other.

What you describe is possible as follows:

Alice and Bob share an EPR-pair of qubits (Alice has qubit 1, Bob has qubit 2.) They want to create a three-way entangled "trio" with Charles, but only Bob can meet with Charles. (Alice has faraway places to be and takes qubit 1 with her.) Nonetheless it is still possible for Bob to "extend" the entangled system to include a third qubit, using only local operations.

1. Alice and Bob begin by creating an EPR-pair $$|00\rangle + |11\rangle$$. Alice takes the first qubit, and Bob takes the second qubit.

2. Bob meets with Charles, who has a third qubit, initially in the basis state $$|0\rangle$$. So the combined system is in the state $$(|00\rangle + |11\rangle)|0\rangle = |000\rangle + |110\rangle$$.

3. Bob and Charles perform a Controlled NOT operation* on the 2,3 system (with qubit 2 as the control). The result is $$|000\rangle + |111\rangle$$, an entangled trio of qubits.

* http://en.wikipedia.org/wiki/Cnot

5. Dec 27, 2008

colorSpace

Thanks for the reply! After thinking some more about it, it seems the answer to 2) is more complicated because the kind of correlation depends on the results of the Bell measurement at the location in the middle.

6. Dec 27, 2008

DrChinese

Here are a couple to start:

http://arxiv.org/abs/quant-ph/0501078v1" [Broken]
http://prola.aps.org/abstract/PRL/v80/i18/p3891_1" [Broken]

Last edited by a moderator: May 3, 2017
7. Dec 27, 2008

Marcaias

If I understand what you mean by question 2, the answer is yes: you can use entanglement swapping to speed up transmitting entangled pairs over great distances, which is useful because transporting a qubit in practice is a very slow process.

Let's say we have only two characters, Alice and David, separated by a distance of 1 glop, who want to share an EPR pair:

Alice -------------------------------------- David

Here Alice has to create an EPR pair, and sends one of the qubits to David at a speed of 1 glop / second ≪ c. Upon receiving it, David sends Alice a classical signal saying "Got it", at the speed of light c, which takes negligible time compared to the transmission of the qubit.

Thus, sharing an EPR pair takes 1 second.

Instead, suppose we have a man in the middle, Bob:

Alice ---------------- Bob ---------------- David

Here, Alice and Bob create EPR pairs $$|\phi^+\rangle_{12} = |\phi^+\rangle_{34}$$ as I described in my previous post. Alice sends qubit 2 to Bob, and Bob simultaneously sends qubit 4 to David. As soon as Bob receives his qubit, he performs the procedure I described in my earlier post (again, we assume that such constant-time operations are negligible) and messages Alice and David to let them know he's done.

This time, each qubit being transmitted only has to cover half the distance, so the transmission takes ≈ 1/2 second.

We can add another relay, Charles:

Alice ------- Bob -------- Charles ------ David

In this case, Alice, Bob and Charles each create EPR pairs and send one of their qubits rightward. Then Bob and Charles, upon receiving a qubit, perform the procedure I described earlier, and each signal Alice and David that they are done.

This time, each qubit only has to travel a third of the distance, so the transmission takes ≈ 1/3 second.

...And so on and so on... allowing us to transmit at a speed arbitrarily close to c.

There's a small catch with the method I just described: the pair with which Alice and David end up is a random Bell state, so they don't know whether they have, say, |00> + |11> at the end or |01> + |10>, which is useless if they want to use the pair for something like quantum teleportation.

However, this is easy to fix, since any Bell state can be transformed into any other by means of a local operation, with some information from the relays.

A neat little corollary to this method is that it allows us to do quantum teleportation arbitrarily quickly, since all we need in order to perform teleportation is a classical channel and an entangled pair.

8. Dec 28, 2008

colorSpace

When you say transporting a qubit is a slow process, are you speaking of using the spin of an electron? If you use the polarization of a photon, isn't it the speed of light?

So the idea is that if you use multiple pairs with multiple Bell measurements, you can make the connection in a time that would be much lower than the time it would require a signal to travel the whole distance even with the speed of light.

I'm not sure, however, if fixing the states due to the varying results of the Bell measurements could be equally optimized, so that all measurements could be taken within a time that is shorter than it would take a signal to travel the whole distance. Furthermore, the question would be whether it could really be said that there would be a non-local effect between the outer particles only, or whether the argument that no hidden variable could travel across the whole distance, in that time, would be negated by the implications of the Bell measurements in between.

EDIT: Or, assuming the "hidden variable" signal starts in the center, the measurements should be completed in a time required by light to travel half the distance, but that isn't really relevant to the core of the question.

Last edited: Dec 28, 2008
9. Dec 28, 2008

Marcaias

It takes only a constant time longer to fix the states than to simply measure them. Upon performing a measurement, the relay tells the man to his left what to do to his qubit as part of his "I'm done" message. If the state the relay sees is the phi+, then he says "do nothing". If it's a phi-, he says "apply a Z gate". If it's a psi+ he says "apply an X gate". If it's a psi- he says "apply an XZ gate".

This is because $$(I\otimes I)|\phi^+\rangle = (Z\otimes I) |\phi^-\rangle = (X\otimes I) |\psi^+\rangle = (XZ\otimes I) |\psi^-\rangle = |\phi^+\rangle$$, so you can always end up with the "canonical" EPR pair using this protocol.

Well, the protocol I described doesn't directly have much to say about hidden variable. If what you're looking for is an argument that rules out locality in a hidden variable theory, read up on Bell's Inequality. This lecture from Scott Aaronson's QC course talks at length about what you can't have in a hidden variable theory, including locality.