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Question: imagine that Alice and Bob share, say, one ebit $\lvert\Phi_+\rangle_\text{AB}$, and that they are only able to carry out local operations and classical communication. Is there an algorithm that allows them to "spread the entanglement" of their maximally entangled pair over two ancillary subsystems without entangling those two subsystems with each other? To be precise, if we suppose that Alice and Bob have the ancillary states $\lvert 00\rangle_\text{A1A2}$ and $\lvert 00\rangle_\text{B1B2}$ respectively, is there a protocol in which Alice and Bob spend their ebit, and the following hold?

  • System A1 is entangled with system B1.
  • System A2 is entangled with system B2.
  • The bipartite system A1B1 is not entangled with system A2B2.

Motivation: I've been playing with variations of the CHSH game. We may consider a variation of the CHSH game in which Alice and Bob must play the game twice, they win the overall game if and only if they win both CHSH subgames, but they only have one shared ebit to make use of in their two-subgame game.

An obvious strategy that would clearly outperform any classical strategy would be to spend the ebit on one of the two games, and play the other game classically. But I'm wondering if there's a way to "split" the entanglement of a Bell state over two subsystems in such a way that allows both subgames to be slightly improved by spending one of the two non-maximally entangled subsystems on each of them.

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  • $\begingroup$ Have you tried experimenting with Nielsen's majorization criteria? It tells you if you can deterministically convert from one state to another via LOCC. You just need to work with the Schmidt coefficients of the two states. Or if you want probabilistic conversion, Vidal gave a generalisation to Nielsen's result. $\endgroup$
    – DaftWullie
    Commented Mar 30, 2023 at 6:32
  • $\begingroup$ @DaftWullie I had not heard of that criterion, thanks for the reference! I’ll play around with it. $\endgroup$
    – Franklin Pezzuti Dyer
    Commented Mar 30, 2023 at 8:21

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