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The negativity of a quantum state, given by

$N(\rho)=\frac{||\rho^{\Gamma_A}||-1}{2}$

where $\rho^{\Gamma_A}$ is the partial transpose with respect to subsystem $A$ of the density matrix of a quantum state and $||.||$ is the trace norm. The measures the degree of entanglement between quantum states. I can plot this for a given quantum state but would like to check my answer is correct. I can't find anywhere what the maximum and minimum values of the negativity can be, and which refers to maximum entanglement. Does anyone have this information?

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Negativity is an LOCC monotone. Thus, it must be maximal on the maximally entangled state. It should be a straightforward exercise to compute this value.

On the other hand, negativity is zero on separable states. For the same reason, this is the lowest value. (More directly, the partial transpose is trace preserving, so the eigenvalues of $\rho^{\Gamma_A}$ add up to $1$. Thus, the sum of their absolute values must be $\ge 1$.)

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  • $\begingroup$ So a Bell state will have a negativity of 1? $\endgroup$ – JJH May 18 '18 at 8:52
  • $\begingroup$ @JJH I just got 1/2, which I found a bit puzzling ... but it seems to be correct. $\endgroup$ – Norbert Schuch May 18 '18 at 16:39
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    $\begingroup$ @FriendlyLagrangian Welcome to the party! Where in my answer do you find anything which is in contradiction to what you say? $\endgroup$ – Norbert Schuch Mar 24 at 10:46
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    $\begingroup$ Sure. That's for a 2-qubit state. $\endgroup$ – Norbert Schuch Mar 24 at 10:50
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    $\begingroup$ Negativity measures entanglement of bipartite states. A bipartite state is maximally entangled if all Schmidt coefficients are the same and it has maximal Schmidt rank. $\endgroup$ – Norbert Schuch Mar 24 at 12:34

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