Monogamy Inequality in Terms of Entanglement Measures Based on Distance for Pure Multiqubit States

Abstract

Using very general arguments, we prove that any entanglement measures based on distance must be maximal on pure states. Furthermore, we show that Bures measure of entanglement and geometric measure of entanglement satisfy the monogamy inequality on all pure multiqubit states. Finally, using the power of Bures measure of entanglement and geometric measure of entanglement, we present a class of tight monogamy relations for pure states of multiqubit systems.

Appendix: Proof of the Lemma

It is evident that the following inequalities

$$ \begin{aligned} \sqrt{1-x^{2}}+\sqrt{1-y^{2}}\geq 1+\sqrt{1-x^{2}-y^{2}} \end{aligned} $$

(A.1)

and

$$ \begin{aligned} \sqrt{(1-x^{2})(1-y^{2})}\geq \sqrt{1-x^{2}-y^{2}} \end{aligned} $$

(A.2)

hold on the domain D = {(x,y)|0 ≤ x,y,x² + y² ≤ 1}. By using the above inequalities, we can obtain the following conclusion.

Lemma

For η ≥ 1, we have

$$ \begin{aligned} B{^{\eta}}(\sqrt{x^{2}+y^{2}})\geq B{^{\eta}}(x)+B{^{\eta}}(y) \end{aligned} $$

(A.3)

and

$$ \begin{aligned} G{^{\eta}}(\sqrt{x^{2}+y^{2}})\geq G{^{\eta}}(x)+G{^{\eta}}(y) \end{aligned} $$

(A.4)

on the domain D = {(x,y)|0 ≤ x,y,x² + y² ≤ 1}.

Proof

Using the inequality (A.1), it is easy to verify that □

$$ \begin{aligned} \frac{1-\sqrt{1-x^{2}-y^{2}}}{2}\geq\frac{1-\sqrt{1-x^{2}}}{2}+\frac{1-\sqrt{1-y^{2}}}{2}. \end{aligned} $$

(A.5)

Then we get

$$ \begin{aligned} G(\sqrt{x^{2}+y^{2}})\geq G(x)+G(y). \end{aligned} $$

(A.6)

When η ≥ 1, by the inequality (A.6), one derives

$$ \begin{aligned} \left[\frac{1-\sqrt{1-x^{2}-y^{2}}}{2}\right]^{\eta} & \geq \left[\frac{1-\sqrt{1-x^{2}}}{2}+\frac{1-\sqrt{1-y^{2}}}{2}\right]^{\eta}\\ & \geq \left[\frac{1-\sqrt{1-x^{2}}}{2}\right]^{\eta}+\left[\frac{1-\sqrt{1-y^{2}}}{2}\right]^{\eta}, \end{aligned} $$

(A.7)

where in the second inequality we have used the property (1 + x)^η ≥ 1 + x^η for 0 ≤ x ≤ 1 and η ≥ 1. It implies that the inequality (A.4) holds, and completes the proof of the inequality (A.4).

Using the inequalities (A.1) and (A.2), we have

$$ \begin{aligned} (1+\sqrt{1-x^{2}})(1+\sqrt{1-y^{2}})\geq 2+2\sqrt{1-x^{2}-y^{2}}. \end{aligned} $$

(A.8)

Inequalities (A.5) and (A.8) together yield

$$ \begin{aligned} 1+\frac{\sqrt{1-x^{2}}}{2}+\frac{\sqrt{1-y^{2}}}{2}+\sqrt{(1+\sqrt{1-x^{2}})(1+\sqrt{1-y^{2}})}\geq 1+\frac{1+\sqrt{1-x^{2}-y^{2}}}{2}+\sqrt{2+2\sqrt{1-x^{2}-y^{2}}}. \end{aligned} $$

(A.9)

We can write (A.9) as

$$ \begin{aligned} \sqrt{\frac{1+\sqrt{1-x^{2}}}{2}}+ \sqrt{\frac{1+\sqrt{1-y^{2}}}{2}}\geq 1+ \sqrt{\frac{1+\sqrt{1-x^{2}-y^{2}}}{2}}. \end{aligned} $$

(A.10)

It follows that

$$ \begin{aligned} 2-2\sqrt{\frac{1+\sqrt{1-x^{2}-y^{2}}}{2}}\geq 2-2\sqrt{\frac{1+\sqrt{1-x^{2}}}{2}}+2-2\sqrt{\frac{1+\sqrt{1-y^{2}}}{2}}, \end{aligned} $$

(A.11)

that is

$$ \begin{aligned} B(\sqrt{x^{2}+y^{2}})\geq B(x)+B(y). \end{aligned} $$

(A.12)

When η ≥ 1, the proof of inequality (A.3) is similar to the proof of the inequality (A.4). It is now obvious that the lemma holds.