bibkey: higuchisudbery2000entangledcouples authors: A. Higuchi; A. Sudbery year: 2000 title: “How entangled can two couples get?” doi: 10.1016/S0375-9601(00)00506-4 url: https://arxiv.org/abs/quant-ph/0005013 claim: “Four-qubit marginal entropy: the Higuchi–Sudbery candidate, its known restrictions, and the unrestricted maximum question.” strata_touched:

• D5/S3/Quantum/Entanglement/HiguchiSudbery/HiguchiSudberyEntropyMaximum license: citation-only triage: anchor

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DOI: https://doi.org/10.1016/S0375-9601(00)00506-4

Source: https://arxiv.org/abs/quant-ph/0005013

How entangled can two couples get?

The average two-qubit entropy

Section 3, p. 5:

Given that a four-qubit state cannot have maximal entropy of entanglement for every two-qubit subset, we now ask what is the greatest possible average for such entropies, i.e. we seek to maximise

$$⟨E_2⟩=\frac{1}{6}(E_{AB}+E_{AC}+E_{AD}+E_{BC}+E_{BD}+E_{CD})=\frac{1}{3}(E_{AB}+E_{AC}+E_{AD}).$$

The second equality holds because complementary pairs have equal entropy.

We have not been able to solve this problem analytically. We will adopt a heuristic approach, using an (indefensible) analogy with the two-qubit system to obtain a candidate maximally entangled state and then showing that $⟨E_2⟩$ is indeed stationary at this state and appears to be maximal.

The explicit candidate on p. 5 is

$$|M_4⟩=\frac{1}{\sqrt{6}}[|0011⟩+|1100⟩+\omega (|1010⟩+|0101⟩)+{\omega }^2(|1001⟩+|0110⟩)],\omega =e^{2\pi i/3}.$$

Section 3, p. 6:

Hence the entanglement entropies are

$$E_{AB}=E_{AC}=E_{AD}=1+\frac{1}{2}{\log }_{2}3.$$

Section 3, p. 7:

We have searched numerically for states which maximise $⟨E_2⟩$ starting from several arbitrarily chosen states. All the states we have obtained in this manner are locally equivalent to $|M_4⟩$ or $|{\overline{M}}_4⟩$.

The amplitude index is $8a+4b+2c+d$ for the ordered qubits $A,B,C,D$. The three literal flattenings retain $AB$, $AC$, and $AD$, respectively. The entropy is the repository’s von Neumann entropy in natural logarithms divided by $\log 2$.