General monogamy relation of multiqubit systems in terms of squared Rényi-$\alpha$ entanglement

We prove that the squared Rényi-$\alpha$ entanglement ($\mathrm{SR}\alpha \mathrm{E}$), which is the generalization of entanglement of formation, obeys a general monogamy inequality in an arbitrary $N$-qubit mixed state. Furthermore, for a class of Rényi-$\alpha$ entanglement, we prove that the monogamy relations of the $\mathrm{SR}\alpha \mathrm{E}$ have a hierarchical structure when the $N$-qubit system is divided into $k$ parties. As a by-product, the analytical relation between the Rényi-$\alpha$ entanglement and the squared concurrence is derived for bipartite $2\otimes d$ systems. Based on the monogamy properties of $\mathrm{SR}\alpha \mathrm{E}$, we can construct the corresponding multipartite entanglement indicators, which still work well even when the indicators based on the squared concurrence and EOF lose their efficacy. In addition, the monogamy property of the $\mu \mathrm{th}$ power of Rényi-$\alpha$ entanglement is analyzed.

Figure 1

The indicator ${\tau }_{\alpha \left(3\right)}^{\left(1\right)}$ for the superposition state $\left|\psi \left(p\right)\right.〉$ with $\alpha =0.83$ (green line), $\alpha =1$ (blue line), and $\alpha =1.1$ (red line). As an comparison, we also plot the three-tangle of $\left|\psi \left(p\right)\right.〉$ with a black line.

Figure 2

The indicator ${\tau }_{\alpha \left(3\right)}^{\left(1\right)}$ with the order $\alpha \in (0.83,3)$ detects the existence of the genuine three-qubit entanglement in the mixed state ${\rho }_{ABC}$ when the parameter $p\le p_0$.

Figure 3

The plot of the dependence of $x$ with $\alpha$ which satisfies the equation (a) $\frac{\partial h_{\alpha }}{\partial x}=0$ and (b) $\frac{\partial h_{\alpha }}{\partial \alpha }=0$ respectively.

Figure 4

The function $h_{\alpha }$ is plotted as a function of $x$ and $\alpha$ for $0\le x\le 1,\alpha \ge 1$, which is positive, and as a result, the $\mathrm{SR}\alpha \mathrm{E}$ is a convex function of SC.

Figure 5

The plot of the dependence of $x$ with $\alpha$ using the equation (a) $\frac{{\partial }^2{E}_{\alpha }^2}{\partial x^2}=0$, (b) $\frac{{\partial }^2{E}_{\alpha }}{\partial x^2}=0$.