Can I perform entanglement swapping on an entangled pair with another 2 pairs simultaneously?

Question

Suppose that I have 3 entangled pairs (a1, a2), (b1, b2), (c1, c2). Can I perform entanglement swapping on (a2, b1) and (b2, c1) simultaneously to get a new entanglement pair (a1, c2)?

The question is derived from the paper, On the waiting time in quantum repeaters with probabilistic entanglement swapping.

source: On the waiting time in quantum repeaters with probabilistic entanglement swapping. E.Shchukin et al. arXiv:1710.06214

In the paper, the authors exhibit two equivalent ways to get an entanglement pair (a1,c2) described above. The first way is to perform entanglement swapping on (a2, b1) and then on (b2, c1) sequentially. The seconde way is to perform entaglement swapping in reverse order. Neither is simultaneous.

Thank you!

Answer 1

Yes, you can make the two measurements you describe to produce an entangled state consisting of (a1,c2); they don't have to be sequential unless you want information from one of the two measurements to inform some controllable feature about the other measurement.

Let's say you aren't worried about sending information between those two measurements. And let's say each of your entangled particles is just a single qubit (say, a spin-1/2 state.). Now you measure (a2,b1) in an entangled basis (yielding one of four outcomes, A={0,1,2,3}) and you also measure (b2,c1) in an entangled basis (yielding one of four outcomes, B={0,1,2,3}). According to quantum theory, the consequence of those two measurements is that you have created a single state consisting of the remaining two qubits (a1,c2), and that state can generally be entangled.

The question of which precise entangled state you end up with depends on the results of those joint measurements -- in general, the outcomes AB can take 16 different combinations, so in general you end up with one of 16 different states. You don't know which one you have until you learn the values of A and B.

The order in which you make the measurements (A,B) or (B,A) doesn't matter as far as the final entangled state you end up with, given the same two outcomes. Now, it does make a difference as to the intermediate steps in your traditional quantum calculation; depending on which one happens first, you get a different account of what is happening in the middle. But that's all in the calculational mathematics; you can't see any difference in the lab. So you might as well treat them as happening simultaneously.

Interestingly, there is another way to make the calculation, using path integrals, where it doesn't matter which order you make those two measurements. The intermediate steps of the path integral calculation look the same regardless of which order you are using (A before B, or B before A, or simultaneously). See this paper for details.

Finally, note that this precise scenario can then bring the particles (a1,c2) together and make a third joint measurement on these two particles. This is known as the interesting "triangle network" (you can also find details about this in the above paper).

Answer 2

The answer is YES if you drop the requirement for simultaneity, and NO otherwise.

First, the ordering of the swaps makes no difference to the quantum predictions. In fact, they can be done after the entangled pair a1, c2 is already detected.

Second: The swap cannot be said to occur at any specific time for several reasons. Is it done at the time the Bell State measurement is performed? Or is it done when the a1, c2 pair becomes entangled? You can see that these connected events occur at what would be described as different times.

If you were to focus only on the Bell state measurement: the photons being measured must be indistinguishable for a swamp to occur. But in actual experiments, they never arrive simultaneously. As a result, there is no specific time that you can say the swap has occurred. The arrival times can be fairly close to each other, but they are always random. And so the difference in the arrival times is going to be nonzero. With modern avalanche photon detectors, those arrival times are clearly never simultaneous, and there is no way to make them so.

And even if you waited for all the detections to occur simultaneously, as measured by the detector timestamps: you still could not say they were simultaneous except by assumption. That’s because an even better and more accurate detector would have a different timestamp. And there are relativistic considerations too, which would make it impossible to use the word “simultaneous” to describe the swaps.