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Quantum phenomena in bipartite pure state systems like teleportation are pretty well understood. What I'm interested in is the following situation:

Alice, Bob and Charlie hold some general tripartite pure state $\rho_{ABC}$. As usual, Alice and Bob are trying to perform some task, and are allowed only LOCC. Charlie's subsystem, however, is completely "inaccessible", by which I mean that Charlie can't do anything at all- no local operations and no classical communication with either of the other two. His role is solely to hold onto his part of the system.

So my question is, is there any situation where Alice and Bob, with full knowledge of the tripartite state, would be able to perform some task 'better' (e.g. with a greater probability of success) with this setup than they would if they just traced out Charlie's system and were left with $\rho_{AB}$? If the answer is yes, then I'd be interested to know of any general results in this area.

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No, according to basic principles of quantum mechanics. Alice and Bob do not have access to any information about operations that Charlie might perform on his state. Therefore the outcome of any experiment they perform will be indistinguishable from the outcome that would be obtained if they shared the bipartite state $\mathrm{Tr}_C \rho_{ABC}$ between just the two of them.


The most general operation Alice & Bob can perform can be written

$$O(\rho) = \sum\limits_i (M_i\otimes 1_C)\rho (M_i^{\dagger}\otimes 1_C), $$ where $M_i$ acts on the Hilbert spaces of $A$ & $B$, satisfying $\sum_i M_i^{\dagger}M_i = 1_{AB}$, and $1$ means the identity operation. This includes any sequence of joint or local unitary operations, measurements etc... you name it. This specific tensor product structure comes about due to restriction to no LOCC involving Charlie. (If LOCC with Charlie is permitted, they can do operations using Kraus operators of the form $M_i \otimes N_i$, where $N_i$ acts non-trivially on $C$.) In order to perform a "task" using the state $\rho$ they must ultimately perform a measurement to obtain an outcome. The probability of obtaining outcome $i$ from the protocol is $$ p_i = \mathrm{Tr} \left[(M_i^{\dagger}\otimes 1_C)(M_i\otimes 1_C)\rho \right] = \mathrm{Tr}_{AB} \left[M_i^{\dagger}M_i \mathrm{Tr}_C(\rho) \right].$$ The probabilities of the possible outcomes of any allowed operation + measurement protocol Alice & Bob can do on the global state $\rho$ are the same as those obtained from the local state $\mathrm{Tr}_C\rho$. Therefore the two states are indistinguishable given the restrictions stated in the OP's question.

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  • $\begingroup$ Which basic principle do you mean? Maybe I'm wrongly guessing what you mean, but remember that Alice and Bob are allowed to know the full initial tripartite state, and are also allowed to assume that Charlie won't do anything with his subsystem. They're also not trying to determine anything that he might do. So if what you're talking about is regarding signalling, for example, I don't think it's directly relevant. $\endgroup$ – Ben Aaronson Oct 7 '13 at 20:40
  • $\begingroup$ Oh, hmm maybe I understand what you mean: if, e.g., Charlie did something to his system Alice and Bob shouldn't be able to detect it, which they'd be able to if him destroying his entanglement degraded their performance. So it'd only be possible if Charlie performing a measurement on his system increased the bipartite A-B entanglement by a corresponding amount. $\endgroup$ – Ben Aaronson Oct 7 '13 at 20:53
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    $\begingroup$ @Stereotomy See my edit containing more detail. The basic principles are just the most general form of quantum operations, combined with restrictions associated with locality in the context quantum information theory. One does not need to appeal to no-signalling , since quantum information theory has no-signalling built in, so to speak. However, of course, the above result does relate to no-signalling, in the sense that you could probably construct a protocol that allows superluminal signalling if A&B could actually do anything useful with "inaccessible entanglement", as you have defined it. $\endgroup$ – Mark Mitchison Oct 8 '13 at 9:01

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