Maximally entangled mixed states – Open Quantum Problems

Cite this problem as Problem 5.

Problem

Among all density operators of two qubits with the same spectrum one may look for those maximizing some measure of entanglement. It turns out [1] that for the entanglement of formation, the relative entropy of entanglement and the negativity one gets the same maximally entangled states.

Is this true for arbitrary entanglement monotones?

Obvious variants of this problem are for higher dimensional systems and weaker constraints on the spectrum, e. g., largest eigenvalue or entropy.

Background

(Refer to definitions of the measures of entanglement and `entanglement monotone’.)

Solution

The question is answered in the negative by de Vicente [2], who proves that some of the optimal states in [1] cannot be transformed via local operations and classical communication (LOCC) into all isospectral states. Let $\rho,\sigma$ be thus two-qubit isospectral states, with $\rho$ maximizing the entanglement of formation over all two-qubit states with the same spectrum, and such that $\rho$ cannot be transformed into $\sigma$ through LOCC. For any two states $\alpha_1,\alpha_2$ , define $\mbox{Prob}(\alpha_1\to\alpha_2)$ as the maximum probability of converting $\alpha_1$ into $\alpha_2$ through LOCC. Then, $E(\bullet):=\mbox{Prob}(\bullet\to\sigma)$ is an entanglement measure by [3]. Moreover, by construction, it satisfies the properties $E(\sigma)=1$ , $E(\rho)<1$ .

References

[1] F. Verstraete, K. Audenaert, and B. De Moor, “Maximally entangled mixed states of two qubits”, quant-ph/0011110 (2000).

[2] J. de Vicente, “Maximally Entangled Mixed States for a Fixed Spectrum Do Not Always Exist”, Phys. Rev. Lett. 133, 050202 (2024).

[3] G. Vidal, “Entanglement monotones”, J. Mod. Opt. 47, 355 (2000).