Can entanglement swapping be performed on already-entangled photons, and if so, can it preserve this entanglement over the swap?

Question

Consider 2 uncorrelated photon pairs (a1,a2), (b1,b2) such that (a1,a2) are entangled, and separately (b1,b2) are entangled. We wish to entangle-swap so as to end up with a new entanglement (a1,b1) by using the ancillary photon pair (a3,b3), such that (a1,a3) are entangled and (b1,b3) are entangled. This suggests that we do a multipartite (3-way in this case) entanglement preparation prior to the swap, such that (a1,a2,a3) are entangled, and separately (b1,b2,b3) are entangled.

The swap proceeds in the usual way by Bell state measurement on a3 and b3, and we end up having entangled (a1,b1).

The question is whether, after this operation, the entanglements (a1,a2) and (b1,b2) remain intact? If not, can we modify the procedure to guarantee this? If yes, will (a2,b2) now be entangled as a result of the swap? (they were uncorrelated before it). If not, how can we arrange that they are entangled, by only manipulating the other 4 photons?

Answer 1

You have a problem with the requirements described in your question.

You say you a want to prepare the state of (a1,a2,a3) to have 3-way entanglement. In that case, the state of (a1,a3) will be a mixed state, since the system of (a1,a3) is entangled with a2. But you want to use the pair (a1,a3) for an entanglement swapping protocol. This protocol requires (a1,a3) to be in a state which is maximally entangled (such as a Bell state). But a mixed state is never maximally entangled, and therefore you cannot use the usual entanglement swapping protocol in this case. Of course the same reasoning applies to (b1,b2,b3).

The deeper reason for the problem here is that entanglement tends to be "monogamous", in the sense that if a1 is entangled with a2, it puts limits on the entanglement it can have with a3.

If you still want to try and perform a Bell state measurement on a3 and b3, even without satisfying the requirements of the entanglement swapping protocol, in general you'll get some sort of 4-way entanglement in the state of (a1,a2,b1,b2). The specifics of the state you will get depend on the states you started with for (a1,a2,a3) and (b1,b2,b3).

As to whether the entanglement in the systems (a1,a2) and (b1,b2) will remain intact, they couldn't have been very strongly entangled in the first place (again because of entanglement monogamy), but they would tend to be even less entangled after the above measurement since now they are a part of a 4-way entangled state instead of a 3-way entangled state.