Given an arbitrary graph state $|G\rangle$ represented by the graph $G$, can one use the graphical structure to calculate the number of ebits (entanglement bits) present in the state?

If so, how?

  • $\begingroup$ What, precisely, do you mean by the number of ebits present in the state? For example, do you mean the minimum number of entangled pairs required to make the state (use local unitaries to convert to the LU-equivalent graph with the minimum number of edges, and it's the number of edges), or something about how many can be extracted under some constraints (single copy, asymptotic limit, local operations only, operations local under bipartitions of the graph,...)? $\endgroup$ – DaftWullie Sep 12 '18 at 8:53
  • $\begingroup$ The former. That was my intuition, but I couldn't find a reference to prove so. Do you know of one? $\endgroup$ – SLesslyTall Sep 12 '18 at 11:26
  • 2
    $\begingroup$ Now that I think about it, it's not so simple! $\endgroup$ – DaftWullie Sep 12 '18 at 13:15
  • $\begingroup$ Could you explain your thoughts on this? I think they would be constructive. $\endgroup$ – SLesslyTall Sep 12 '18 at 14:53

The number of Bell pairs required to construct a given graph state can easily be given an upper bound: $|V|-1$, where $V$ is the set of vertices. You do this simply by preparing the entire state at one site, and teleporting all the other qubits to the relevant party.

I wonder if this is actually all there is to it? If we assume that the entire graph is connected, then every individual qubit is (maximally) entangled with the rest of the graph, and that entanglement must be distributed somehow.

Lower bounds in the multipartite entanglement setting are quite difficult to prove. In a bipartite setting, you'd do it be showing how many Bell states you can extract from asymptotically many copies of the state of interest, but in the regime of multipartite entanglement, that rapidly leads you to MREGS (minimal reversible entanglement generating set), about which very little is known.

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