Tighter monogamy relations of entanglement measures based on fidelity

Quantum entanglement is an indispensable resource for quantum information processing [1] which distinguishes quantum mechanics from classical mechanics. In contrast to classical correlations, quantum entanglement has an interesting feature in that entanglement cannot be freely shared among systems. For instance, if two parties are maximally entangled in a multipartite system, then none of them could share entanglement with any part of the remaining system. We call this phenomenon quantum entanglement monogamy [2]. The monogamy relation of entanglement serves to characterize different kinds of entanglement distribution.

Entanglement monogamy was first characterized as an inequality in three-qubit systems by Coffman–Kundu–Wootters (CKW) [3], i.e. $E(\rho_{A|BC})\geqslant E(\rho_{AB})+E(\rho_{AC})$ where $E(\rho_{A|BC}) = C^2(\rho_{A|BC})$ represents the squashed concurrence of $\rho_{A|BC}$ under bipartition A and BC, ρ_AB and ρ_AC are the reduced density matrices of the tri-qubit state ρ_ABC respectively. The concurrence of a two-qubit mixed state ρ is given by the analytic formula $C(\rho) = \max\{\lambda_1-\lambda_2-\lambda_3-\lambda_4, 0\}$, where $\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4}$, are the square roots of nonnegative eigenvalues of the matrix $\rho(\sigma_y\otimes\sigma_y)\rho^*(\sigma_y\otimes\sigma_y)$ arranged in nonincreasing order, σ_y is the Pauli matrix, and $\rho^*$ denotes the complex conjugate of ρ [3]. Later, Osborne and Verstraete proved that the CKW inequality also holds in an n-qubit system [4]. Other types of monogamy relations for entanglement were also proposed, notably entanglement negativity [5–8], entanglement of formation [9, 10], Tsallis q-entropy [11–13], Rényi-α entanglement [14–16] and unified-(q, s) entanglement [17]. In recent years, monogamy inequalities of one class of entanglement measure based on fidelity, such as the Bures measure of entanglement [18, 19] and the geometric measure of entanglement [20], were discussed in [21, 22]. All these monogamy relations were basically presented for qubit quantum states, however in higher-dimensional systems, some quantum states were found violating the CKW inequality [23, 24]. The monogamy relation of an entanglement measure for qudits was conjectured that seems to have no obvious counterexamples [6]. In this paper, we generalize the monogamy inequality in qudit quantum systems by projecting higher-dimensional states to qubit substates.

This article is organized as follows. In section 2, we derive the tightened monogamy inequalities in an arbitrary tripartite mixed state based on the Bures measure of entanglement and the geometric measure of entanglement. Then the monogamy relation is generalized to multipartite quantum systems. Using detailed examples, our results are seen to be superior to the previously published results. In section 3, we derive the monogamy inequalities of concurrence in an arbitrary dimensional tripartite system by projecting high-dimensional states to $2\otimes 2\otimes 2$ substates and we generalize the results for multipartite quantum systems. Comments and conclusions are given in section 4.

Recall that the fidelity of separability is defined by [25]:

$$\begin{align} F_S(\rho_{A_1A_2}) = \max_{\sigma_{A_1A_2}\in S}F(\rho_{A_1A_2},\sigma_{A_1A_2}) \end{align} \tag{ 1 }$$

where S is the set of separable states, the maximum is taken over all separable states $\sigma_{A_1A_2}$ in S and $F(\rho_{A_1A_2}, \sigma_{A_1A_2}) = [tr((\sqrt{\rho_{A_1A_2}}\sigma_{A_1A_2}\sqrt{\rho_{A_1A_2}})^{\frac12})]^2$. Now we consider the entanglement measures based on fidelity for the Bures measure of entanglement and the geometric measure of entanglement, which are defined respectively by [18, 20]:

$$\begin{align} M_B(\rho_{A_1A_2}) & = \min_{\sigma_{A_1A_2}\in S}\left(2-2\sqrt{F(\rho_{A_1A_2},\sigma_{A_1A_2})}\right)\nonumber\\ & = 2-2\sqrt{F_S(\rho_{A_1A_2})}, \end{align} \tag{ 2 }$$

$$\begin{align} M_G(\rho_{A_1A_2})& = \min_{\sigma_{A_1A_2}\in S}(1-F(\rho_{A_1A_2},\sigma_{A_1A_2})) = 1-F_S(\rho_{A_1A_2}). \end{align} \tag{ 3 }$$

For an arbitrary two-qubit mixed state, the analytical expressions for the Bures measure of entanglement and the geometric measure of entanglement in terms of the concurrence are given as follows [25]:

$$\begin{align} M_B(\rho_{A_1A_2})& = f_B(C(\rho_{A_1A_2})) = 2-2\sqrt{\frac{1+\sqrt{1-C^2(\rho_{A_1A_2})}}{2}}, \end{align} \tag{ 4 }$$

$$\begin{align} M_G(\rho_{A_1A_2})& = f_G(C(\rho_{A_1A_2})) = \frac{1-\sqrt{1-C^2(\rho_{A_1A_2})}}{2}, \end{align} \tag{ 5 }$$

where $f_B(x) = 2-2\sqrt{\frac{1+\sqrt{1-x^2}}{2}}$ and $f_G(x) = \frac{1-\sqrt{1-x^2}}{2}$ are monotonically increasing functions in $0\leqslant x\leqslant1$. For $2\otimes d\, (d\geqslant2)$ mixed states $\rho_{A_1A_2}$, one has the relations $M_B(\rho_{A_1A_2})\geqslant f_B(C(\rho_{A_1A_2}))$ and $M_G(\rho_{A_1A_2})\geqslant f_G(C(\rho_{A_1A_2}))$ in general, and the equalities hold for the special cases of pure states [22]

Lemma 1. 

• (a)  
  If $t\geqslant k^{\,\omega}\geqslant k\geqslant1$, $\omega\geqslant1$, and $0\leqslant x\leqslant\frac{1}{2}$, we have

  $$\begin{align} (1+t)^x\geqslant\left(\frac{1}{2}\right)^x+\frac{(1+k^{\,\omega})^x-\left(\frac{1}{2}\right)^x}{k^{\omega x}}t^{\,x}. \end{align} \tag{ 6 }$$
• (b)  
  If $0\leqslant t\leqslant k^{\,\omega}\leqslant k\leqslant1$, $\omega\geqslant1$, and $x\geqslant 1$, we have

  $$\begin{align} (1+t)^x\geqslant\left(\frac{1}{2}\right)^x+\frac{(1+k^{\,\omega})^x-\left(\frac{1}{2}\right)^x}{k^{\omega x}}t^{\,x}. \end{align} \tag{ 7 }$$

proof We prove these two inequalities in a similar manner. Consider $h(x,y) = (1+\frac{1}{y})^{x-1}-\left(\frac{1}{2}\right)^x$ where $0\leqslant x\leqslant\frac{1}{2}$ and $0\lt y\leqslant\frac{1}{k^{\,\omega}}$ with real numbers $k\geqslant1$ and $\omega\geqslant1$. Then $\frac{\partial h}{\partial x} = (1+\frac{1}{y})^{x-1}ln(1+\frac{1}{y})-\left(\frac{1}{2}\right)^xln\frac{1}{2}\gt0$ as $1+\frac{1}{y}\geqslant2$. So $h(x,y)$ is an increasing function of x when y is fixed, i.e. $h(x,y)\leqslant h(\frac{1}{2},y) = (1+\frac{1}{y})^{-\frac{1}{2}}-\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{1}{2}}\leqslant0$ as $0\lt(1+\frac{1}{y})^{-1}\leqslant\frac{1}{2}$. Let $g(x,y) = (1+y)^x-(\frac{1}{2}y)^x$ with $0\leqslant x\leqslant\frac{1}{2}$ and $0\lt y\leqslant\frac{1}{k^{\,\omega}}$. We have $\frac{\partial g}{\partial y} = xy^{x-1}[(1+\frac{1}{y})^{x-1}-\left(\frac{1}{2}\right)^x]\leqslant0$ and then $g(x,y)$ is a decreasing function of y for fixed x. Thus, for $t\geqslant k^{\,\omega}$, one has $g(x,\frac{1}{t})\geqslant g(x, \frac1{k^{\,\omega}})$. Therefore $(1+t)^x\geqslant\left(\frac{1}{2}\right)^x+\frac{(1+k^{\,\omega})^x-\left(\frac{1}{2}\right)^x}{k^{\omega x}}t^{\,x}$.

Lemma 2. For $0\leqslant x,y, x^2+y^2\leqslant1$, $0\leqslant\alpha\leqslant\frac{\eta}{2}$, $\eta\geqslant1$, $\omega\geqslant1$, and $k\geqslant1$,

• (a)  
  if $f_B^{\,\eta}(y)\geqslant k^{\,\omega} f_B^{\,\eta}(x)$, we have

  $$\begin{align} f_B^{\,\,\alpha}\left(\sqrt{x^2+y^2}\right)\geqslant\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}f_B^{\,\,\alpha}(x)+\frac{(1+k^{\,\omega})^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}-\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}}{k^{\genfrac{}{}{.3pt}{3}{\omega\alpha}{\eta}}}f_B^{\,\,\alpha}(y). \end{align} \tag{ 8 }$$
• (b)  
  if $f_G^{\,\eta}(y)\geqslant k^{\,\omega} f_G^{\,\eta}(x)$, we have

  $$\begin{align} f_G^{\,\,\alpha}\left(\sqrt{x^2+y^2}\right)\geqslant\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}f_G^{\,\,\alpha}(x)+\frac{(1+k^{\,\omega})^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}-\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}}{k^{\genfrac{}{}{.3pt}{3}{\omega\alpha}{\eta}}}f_G^{\,\,\alpha}(y). \end{align} \tag{ 9 }$$

proof When $f_B^{\,\eta}(y)\geqslant k^{\,\omega} f_B^{\,\eta}(x)$, we have

$$\begin{align} & f_B^{\,\,\alpha}\left(\sqrt{x^2+y^2}\right)\nonumber\\ &\quad \geqslant(f_B^{\,\eta}(x)+f_B^{\,\eta}(y))^\frac\alpha \eta = f_B^{\,\,\alpha}(x)\left(1+\frac{f_B^{\,\eta}(y)}{f_B^{\,\eta}(x)}\right)^\frac\alpha \eta\nonumber\\ &\quad \geqslant f_B^{\,\,\alpha}(x)\left[\left(\frac{1}{2}\right)^\frac\alpha \eta+\frac{(1+k^{\,\omega})^\frac\alpha\eta-\left(\frac{1}{2}\right)^\frac\alpha \eta}{k^{\genfrac{}{}{.3pt}{3}{\omega\alpha} \eta}}\left(\frac{f_B^{\,\eta}(y)}{f_B^{\,\eta}(x)}\right)^\frac\alpha \eta\right]\nonumber\\ &\quad = \left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}f_B^{\,\,\alpha}(x)+\frac{(1+k^{\,\omega})^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}-\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}}{k^{\genfrac{}{}{.3pt}{3}{\omega\alpha}{\eta}}}f_B^{\,\,\alpha}(y), \end{align} \tag{ 10 }$$

where $0\leqslant\alpha\leqslant\frac{\eta}{2}$, $\eta\geqslant1$, $\omega\geqslant1$, and $k\geqslant1$. The first inequality is obtained by $f_B^{\,\eta}\left(\sqrt{x^2+y^2}\right)\geqslant f_B^{\,\eta}(x)+f_B^{\,\eta}(y)$ for $0\leqslant x,y, x^2+y^2\leqslant1$ and $\eta\geqslant1$ [21] and the second one is due to (6) of lemma 1.

Theorem 1. In tri-qubit quantum systems, assuming real numbers $k\geqslant1$, $\omega\geqslant1$, $0\leqslant\alpha\leqslant\frac{\eta}{2}$ and $\eta\geqslant1$, then one has that

• (a)  
  if $M_B^{\,\eta}(\rho_{A_1A_3})\geqslant k^{\,\omega} M_B^{\,\eta}(\rho_{A_1A_2})$, then the Bures measure of entanglement satisfies

  $$\begin{align} M_B^{\,\,\alpha}(\rho_{A_1|A_2A_3})& \geqslant\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}M_B^{\,\,\alpha}(\rho_{A_1A_2})\nonumber\\ &\quad +\frac{(1+k^{\,\omega})^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}-\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}}{k^{\genfrac{}{}{.3pt}{3}{\omega\alpha}{\eta}}}M_B^{\,\,\alpha}(\rho_{A_1A_3}). \end{align} \tag{ 11 }$$
• (b)  
  if $M_B^{\,\eta}(\rho_{A_1A_2})\geqslant k^{\,\omega} M_B^{\,\eta}(\rho_{A_1A_3})$, then the Bures measure of entanglement satisfies

  $$\begin{align} M_B^{\,\,\alpha}(\rho_{A_1|A_2A_3})& \geqslant\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}M_B^{\,\,\alpha}(\rho_{A_1A_3})\nonumber\\ &\quad +\frac{(1+k^{\,\omega})^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}-\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}}{k^{\genfrac{}{}{.3pt}{3}{\omega\alpha}{\eta}}}M_B^{\,\,\alpha}(\rho_{A_1A_2}). \end{align} \tag{ 12 }$$

proof For an arbitrary tri-qubit state ρ under bipartite partition $A_1|A_2A_3$, one has [26]:

$$\begin{align} C^2(\rho_{A_1|A_2A_3})\geqslant C^2(\rho_{A_1A_2})+C^2(\rho_{A_1A_3}). \end{align} \tag{ 13 }$$

Suppose $M_B^{\,\eta}(\rho_{A_1A_3})\geqslant k^{\,\omega} M_B^{\,\eta}(\rho_{A_1A_2})$, $k\geqslant1$ and $\omega\geqslant1$, then

$$\begin{align} & M_B^{\,\,\alpha}(\rho_{A_1|A_2A_3})\nonumber\\ &\quad \geqslant f_B^{\,\,\alpha}(C(\rho_{A_1|A_2A_3}))\geqslant f_B^{\,\,\alpha}\left(\sqrt{C^2(\rho_{A_1A_2})+C^2(\rho_{A_1A_3})}\right)\nonumber\\ &\quad \geqslant\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}f_B^{\,\,\alpha}(C(\rho_{A_1A_2}))+\frac{(1+k^{\,\omega})^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}-\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}}{k^{\genfrac{}{}{.3pt}{3}{\omega\alpha}{\eta}}}f_B^{\,\,\alpha}(C(\rho_{A_1A_3})) \nonumber\\ &\quad = \left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}M_B^{\,\,\alpha}(\rho_{A_1A_2})+\frac{(1+k^{\,\omega})^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}-\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}}{k^{\genfrac{}{}{.3pt}{3}{\omega\alpha}{\eta}}}M_B^{\,\,\alpha}(\rho_{A_1A_3}), \end{align} \tag{ 14 }$$

where $0\leqslant\alpha\leqslant\frac{\eta}{2}$, $\eta\geqslant1$, the first inequality is obtained by $M_B(\rho_{A_1|A_2A_3})\geqslant f_B(C(\rho_{A_1|A_2A_3}))$ [22], the second one is due to inequality (13) and the fact that $f_B(x)$ is a monotonically increasing function, and the last inequality is due to lemma 2. The equality holds since $M_B(\rho) = f_B(C(\rho))$ for two-qubit states [22]. A similar proof gives inequality (12) by using lemma 2.

Remark 1. We have derived the monogamy relations for the Bures measure of entanglement and also for the geometric measure of entanglement by the same argument.

In the following, let $M(\rho_{A_1A_i}) = M_{A_1A_i}$, $C(\rho_{A_1A_i}) = C_{A_1A_i}$, $M(\rho_{A_1|A_{j+1}\ldots A_n}) = M_{A_1|A_{j+1}\ldots A_n}$, $C(\rho_{A_1|A_{j+1}\ldots A_n}) = C_{A_1|A_{j+1}\ldots A_n}$ where $i = 2,\ldots,n-1$ and $j = 1,\ldots,n-1$ and simply note the Bures measure of entanglement (M_B ) or the geometric measure of entanglement (M_G ) by M. We now generalize the monogamy inequalities of the αth ($0\leqslant\alpha\leqslant\frac{\eta}{2}$, $\eta\geqslant1$) power of the Bures measure of entanglement for n-qubit quantum states ρ under bipartite partition $A_1|A_2\ldots A_n$.

Theorem 2. In multi-qubit quantum systems, assuming real numbers $k\geqslant1$, $0\leqslant\alpha\leqslant\frac{\eta}{2}$, $\eta\geqslant1$, $\omega\geqslant1$, and $\mu = \frac{(1+k^{\,\omega})^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}-\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}}{k^{\genfrac{}{}{.3pt}{3}{\omega\alpha}{\eta}}}$, we have that

• (a)  
  if $k^{\,\omega} M_{A_1A_i}^{\,\eta}\leqslant M_{A_1|A_{i+1}\ldots A_n}^{\,\eta}$ for $i = 2,\ldots,m$ and $M_{A_1A_j}^{\,\eta}\geqslant k^{\,\omega} M_{A_1|A_{j+1}\ldots A_n}^{\,\eta}$ for $j = m+1,\ldots,n-1$, $\forall 2\leqslant m\leqslant n-2$, $n\geqslant4$, then we have

  $$\begin{align} & M_{A_1|A_2A_3\cdots A_n}^{\,\,\alpha}\nonumber\\[6pt] &\quad \geqslant \left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}(M_{A_1A_2}^{\,\,\alpha}+\mu M_{A_1A_3}^{\,\,\alpha}+\cdots+\mu^{m-2}E_{A_1A_m}^{\,\,\alpha})\nonumber\\[6pt] &\qquad +\mu^m\left[M_{A_1A_{m+1}}^{\,\,\alpha}+\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}M_{A_1A_{m+2}}^{\,\,\alpha}+\cdots\right.\nonumber\\[6pt] &\qquad\qquad\quad\left.+\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{(n-m-2)\alpha}{\eta}}M_{A_1A_{n-1}}^{\,\,\alpha}\right]\nonumber\\[6pt] &\qquad +\mu^{m-1}\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{(n-m-1)\alpha}{\eta}}M_{A_1A_n}^{\,\,\alpha}. \end{align} \tag{ 15 }$$
• (b)  
  If $k^{\,\omega} M_{A_1A_i}^{\,\eta}\leqslant M_{A_1|A_{i+1}\ldots A_n}^{\,\eta}$ for $i = 2,\ldots,n-1$ and $n\geqslant3$, then we have that

  $$\begin{align} & M_{A_1|A_2A_3\cdots A_n}^{\,\,\alpha}\nonumber\\[6pt] &\quad \geqslant\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}\left(M_{A_1A_2}^{\,\,\alpha}+\mu M_{A_1A_3}^{\,\,\alpha}+\cdots+\mu^{n-3}M_{A_1A_{n-1}}^{\,\,\alpha}\right)\nonumber\\[6pt] &\qquad+\mu^{n-2}M_{A_1A_n}^{\,\,\alpha}. \end{align} \tag{ 16 }$$
• (c)  
  If $M_{A_1A_i}^{\,\eta}\geqslant k^{\,\omega} M_{A_1|A_{i+1}\ldots A_n}^{\,\eta}$ for $i = 2,\ldots,n-1$ and $n\geqslant3$, then we have that

  $$\begin{align} & M_{A_1|A_2A_3\cdots A_n}^{\,\,\alpha}\nonumber\\[6pt] &\quad \geqslant \mu\left(M_{A_1A_2}^{\,\,\alpha}+\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}M_{A_1A_3}^{\,\,\alpha}+\cdots+\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{(n-3)\alpha}{\eta}}M_{A_1A_{n-1}}^{\,\,\alpha}\right)\nonumber\\[6pt] &\qquad+\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{(n-2)\alpha}{\eta}}M_{A_1A_n}^{\,\,\alpha}. \end{align} \tag{ 17 }$$

proof For an n-qubit quantum state ρ under bipartite partition $A_1|A_2A_3\ldots A_n$, if $k^{\,\omega} M_{A_1A_i}^{\,\eta}\leqslant M_{A_1|A_{i+1}\ldots A_n}^{\,\eta}$ for $i = 2,\ldots,m$, we have

$$\begin{align} M_{A_1|A_2A_3\cdots A_n}^{\,\,\alpha}&\geqslant f^{\;\alpha}(C_{A_1|A_2A_3\cdots A_n})\nonumber\\ &\geqslant f^{\;\alpha}\left(\sqrt{C_{A_1A_2}^2+C_{A_1|A_3\cdots A_n}^2}\right) \nonumber\\ &\geqslant \left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}f^{\;\alpha}(C_{A_1A_2})+\mu f^{\;\alpha}(C_{A_1|A_3\cdots A_n}) \nonumber\\ &\geqslant \cdots \nonumber \\ &\geqslant \left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}(f^{\;\alpha}(C_{A_1A_2})+\mu f^{\;\alpha}(C_{A_1A_3})\nonumber\\ &\quad +\cdots+\mu^{m-2}f^{\;\alpha}(C_{A_1A_m}))\nonumber\\ &\quad +\mu^{m-1}f^{\;\alpha}(C_{A_1|A_{m+1}\cdots A_n})\nonumber\\ & = \left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}\left(M_{A_1A_2}^{\,\,\alpha} + \mu M_{A_1A_3}^{\,\,\alpha}+\cdots+\mu^{m-2}M_{A_1A_m}^{\,\,\alpha}\right)\nonumber\\ &\quad +\mu^{m-1}f^{\;\alpha}(C_{A_1|A_{m+1}\cdots A_n}), \end{align} \tag{ 18 }$$

where the first inequality follows from $M_{A_1|A_2A_3\ldots A_n}\geqslant f(C_{A_1|A_2A_3\ldots A_n})$ [22], the second one is due to $C^2_{A_1|A_2A_3}\geqslant C^2_{A_1A_2}+C^2_{A_1A_3}$ for $2\otimes2\otimes2^{n-2}$ tripartite state [26], and f(x) being a monotonically increasing function. Using lemma 2, we get the third inequality. Other inequalities are consequences of lemma 2 and the last equality holds due to $M(\rho) = f(C(\rho))$ for two-qubit states.

For $M_{A_1A_j}^{\,\eta}\geqslant k^{\,\omega} M_{A_1|A_{j+1}\ldots A_n}^{\,\eta}$ for $j = m+1,\ldots,n-1$, a similar argument gives the following inequality by using lemma 2:

$$\begin{align} f^{\;\alpha}(C_{A_1|A_{m+1}+\cdots+A_n})&\geqslant \mu f^{\;\alpha}(C_{A_1A_{m+1}})+\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}f^{\;\alpha}(C_{A_1|A_{m+2}\cdots A_n})\nonumber\\ &\geqslant \cdots \nonumber\\ &\geqslant \mu\left[M_{A_1A_{m+1}}^{\,\,\alpha}+\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}M_{A_1A_{m+2}}^{\,\,\alpha}+\cdots\right.\nonumber\\ & \qquad\left. +\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{(n-m-2)\alpha}{\eta}}M_{A_1A_{n-1}}^{\,\,\alpha}\right]\nonumber\\ & \quad +\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{(n-m-1)\alpha}{\eta}}M_{A_1A_n}^{\,\,\alpha}. \end{align} \tag{ 19 }$$

Combining (18) and (19), one obtains (15). If all $k^{\,\omega} M_{A_1A_i}^{\,\eta}\leqslant M_{A_1|A_{i+1}\ldots A_n}^{\,\eta}$ for $i = 2,\ldots,n-1$ or $M_{A_1A_i}^{\,\eta}\geqslant k^{\,\omega} M_{A_1|A_{i+1}\ldots A_n}^{\,\eta}$ for $i = 2,\ldots,n-1$, we have the inequalities (16) and (17).

Remark 2. We use the Bures measure of entanglement as an example to compare our result with those in [22, 27, 28]. In tripartite quantum systems, when $M_B^{\,\eta}(\rho_{A_1A_3})\geqslant k^{\,\omega} M_B^{\,\eta}(\rho_{A_1A_2})\geqslant k M_B^{\,\eta}(\rho_{A_1A_2})$ for $k\geqslant1$, $\omega\geqslant1$, $0\leqslant\alpha\leqslant\frac{\eta}{2}$ and $\eta\geqslant1$, theorem 1 says that the αth power of the Bures measure of entanglement satisfies $M_B^{\,\,\alpha}(\rho_{A_1|A_2A_3})\geqslant\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}M_B^{\,\,\alpha}(\rho_{A_1A_2})+\frac{(1+k^{\,\omega})^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}-\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}}{k^{\genfrac{}{}{.3pt}{3}{\omega\alpha}{\eta}}}M_B^{\,\,\alpha}(\rho_{A_1A_3})$ denoted as m. On the other hand, the lower bounds of $M_B^{\,\,\alpha}(\rho_{A_1|A_2A_3})$ are $M_B^{\,\,\alpha}(\rho_{A_1A_2})+M_B^{\,\,\alpha}(\rho_{A_1A_3})\doteqdot m_1$ by using the Bures measure and similar method in [27, 28], we also derive the lower bounds of $M^{\,\,\alpha}_B(\rho_{A_1|A_2A_3})$ are $\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}M_B^{\,\,\alpha}(\rho_{A_1A_2})+\frac{(1+k)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}-\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}}{k^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}}M_B^{\,\,\alpha}(\rho_{A_1A_3})\doteqdot m_2$ and $M_B^{\,\,\alpha}(\rho_{A_1A_2})+\frac{(1+k^{\,\omega})^{\genfrac{}{}{.3pt}{3}{\alpha}{\eta}}-1}{k^{\genfrac{}{}{.3pt}{3}{\omega\alpha}{\eta}}}M_B^{\,\,\alpha}(\rho_{A_1A_3})\doteqdot m_3$, respectively. Let $\Delta m = m-m_i$, $i = 1, 2, 3$, in the following examples we will see that $\Delta m\geqslant0$, so our results are tighter than those in [22, 27, 28] for $0\leqslant\alpha\leqslant\frac{\eta}{2}$ and $\eta\geqslant1$.

Example 1. Let us consider the 3-qubit quantum state of generalized Schmidt decomposition $|\varphi\rangle$,

$$\begin{align} |\varphi\rangle = \lambda_0|000\rangle+\lambda_1e^{i\theta}|100\rangle+\lambda_2|101\rangle+\lambda_3|110\rangle+\lambda_4|111\rangle, \end{align} \tag{ 20 }$$

where $0\leqslant\theta\leqslant\pi$, $\lambda_i\geqslant0$, $i = 0,\ldots,4$ and $\sum_{i = 0}^4\lambda_i^2 = 1$. One computes that, one has $C(|\varphi\rangle_{A_1|A_2A_3}) = 2\lambda_0\sqrt{\lambda_2^2+\lambda_3^2+\lambda_4^2}$, $C(|\varphi\rangle_{A_1A_2}) = 2\lambda_0\lambda_2$, and $C(|\varphi\rangle_{A_1A_3}) = 2\lambda_0\lambda_3$. Let $\lambda_0 = \lambda_3 = \frac{\sqrt{2}}{3}$, $\lambda_2 = \frac{\sqrt{5}}{3}$, $\lambda_1 = \lambda_4 = 0$, k = 2, and ω = 1.5, then we have $M_B(|\varphi\rangle_{A_1|A_2A_3})\approx0.23617$, $M_B(|\varphi\rangle_{A_1A_2})\approx0.14989$, $M_B(|\varphi\rangle_{A_1A_3})\approx0.05279$. Therefore, $M_B^x(|\varphi\rangle_{A_1|A_2A_3}) = 0.23617^x$. By theorem 1, the lower bound of $M_B^x(|\varphi\rangle_{A_1|A_2A_3})$ is $z_1 = \left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{x}{y}}0.05279^x+\frac{3.82842^{\genfrac{}{}{.3pt}{3}{x}{y}}-\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{x}{y}}}{2.82842^{\genfrac{}{}{.3pt}{3}{x}{y}}}0.14989^x$. By theorem 1 in [22], the lower bound of $M_B^x(|\varphi\rangle_{A_1|A_2A_3})$ is $z_2 = 0.05279^x+0.14989^x$. Assuming $0\leqslant x\leqslant\frac{1}{2}$ and $1\leqslant y\leqslant10$, figure 1 verifies that our result is tighter than that of [22].

Example 2. Consider the three-qubit generalized W-class state $\rho = |\psi\rangle_{A_1A_2A_3}\langle\psi|$,

$$\begin{align} |\psi\rangle_{A_1A_2A_3} = \frac{1}{\sqrt{6}}|100\rangle+\frac{1}{\sqrt{6}}|010\rangle+\frac{2}{\sqrt{6}}|001\rangle. \end{align} \tag{ 21 }$$

By the definition of concurrence, we have $C(\rho_{A_1|A_2A_3}) = \frac{\sqrt{5}}{3}$, $C(\rho_{A_1A_2}) = \frac{1}{3}$, and $C(\rho_{A_1A_3}) = \frac{2}{3}$. Thus $M_B(\rho_{A_1|A_2A_3}) = 2-2\sqrt\frac56\approx0.17426$, $M_B(\rho_{A_1A_2}) = 2-2\sqrt\frac{3+2\sqrt2}6\approx0.02880$, and $M_B(\rho_{A_1A_3}) = 2-2\sqrt\frac{3+2\sqrt5}6\approx0.13166$. Let k = 2, ω = 2, and η = 2, then by theorem 1, the lower bound of $M_B^x(\rho_{A_1|A_2A_3})$ is $y_1 = \left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{x}{2}}0.02880^x+\frac{(1+4)^{\genfrac{}{}{.3pt}{3}{x}{2}}-\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{x}{2}}}{4^{\genfrac{}{}{.3pt}{3}{x}{2}}}0.13166^x$. Using the method in [27, 28], we have that the lower bound of $M_B^x(\rho_{A_1|A_2A_3})$ is $y_2 = 0.02880^x+\frac{(1+4)^{\genfrac{}{}{.3pt}{3}{x}{2}}-1}{4^{\genfrac{}{}{.3pt}{3}{x}{2}}}0.13166^x$ and $y_3 = \left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{x}{2}}0.02880^x+\frac{(1+2)^{\genfrac{}{}{.3pt}{3}{x}{2}}-\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{x}{2}}}{2^{\genfrac{}{}{.3pt}{3}{x}{2}}}0.13166^x$ respectively. Figure 2 depicts the value of y for $0\leqslant x\leqslant\frac{1}{2}$, which shows that theorem 1 supplies a better estimation of the Bures measure of entanglement than those of [27, 28].

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In the above section, we have given the monogamy inequalities of the Bures measure of entanglement and the geometric measure of entanglement, both of which can be expressed as functions of concurrence for qubit quantum states. In the following, we will present the monogamy relations of concurrence for qudit quantum states and present a method for higher-dimensional monogamy relations. We first introduce the following definition.

Let $H_{A_1}$ and $H_{A_2}$ be $d_{A_1}$- and $d_{A_2}$-dimensional Hilbert spaces. The concurrence of a bipartite quantum pure state $|\varphi\rangle_{A_1A_2}\in H_{A_1}\otimes\ H_{A_2}$ is defined by [29],

$$\begin{align} C(|\varphi\rangle_{A_1A_2}) = \sqrt{2\left[1-tr\left(\rho_{A_1}^2\right)\right]}, \end{align} \tag{ 22 }$$

where $\rho_{A_1}$ is the reduced density matrix of $\rho = |\varphi\rangle_{A_1A_2}\langle\varphi|$, i.e. $\rho_{A_1} = tr_{A_2}(\rho)$. For a mixed bipartite quantum state $\rho_{A_1A_2} = \sum_ip_i|\varphi_i\rangle_{A_1A_2}\langle\varphi_i|\in H_{A_1}\otimes H_{A_2}$, the concurrence is given by the convex roof

$$\begin{align} C(\rho_{A_1A_2}) = \min_{\{p_i,|\varphi_i\rangle\}}\sum_ip_iC(|\varphi_i\rangle_{A_1A_2}), \end{align} \tag{ 23 }$$

where the minimum is taken over all possible convex partitions of $\rho_{A_1A_2}$ into pure state ensembles $\{p_i,|\varphi_i\rangle\}$, $0\leqslant p_i\leqslant1$ and $\sum_ip_i = 1$.

Now consider the concurrence for a tripartite quantum state under bipartite partition $A_1|A_2A_3$. For a pure tripartite quantum state $|\varphi\rangle_{A_1|A_2A_3}^{\,d_1\otimes d_2\otimes d_3}\in\mathcal{H}_{A_1}^{\,d_1}\otimes\mathcal{H}_{A_2}^{\,d_2}\otimes\mathcal{H}_{A_3}^{\,d_3}$, it has the form as follows:

$$\begin{align} |\varphi\rangle_{A_1|A_2A_3}^{\,d_1\otimes d_2\otimes d_3} = \sum_{a = 1}^{\,d_1}\sum_{b = 1}^{\,d_2}\sum_{c = 1}^{\,d_3}h_{a|bc}|abc\rangle, \end{align} \tag{ 24 }$$

where $h_{a|bc}\in \mathbb{C}$, $\sum_{abc}h_{a|bc}h_{a|bc}^\ast = 1$. From the definition of concurrence in (22), the concurrence of $|\varphi\rangle_{A_1|A_2A_3}^{\,d_1\otimes d_2\otimes d_3}$ is given by:

$$\begin{align} C^2(|\varphi\rangle_{A_1|A_2A_3}^{\,d_1\otimes d_2\otimes d_3}) = \sum_{a,e = 1}^{\,d_1}\sum_{b,\,f = 1}^{\,d_2}\sum_{c,\,j = 1}^{\,d_3}|h_{a|bc}h_{e|\,fj}-h_{e|bc}h_{a|\,fj}|^2. \end{align} \tag{ 25 }$$

Next we analyse a pure substate of $|\varphi\rangle_{A_1|A_2A_3}^{\,d_1\otimes d_2\otimes d_3}$, i.e. $\,|\varphi\rangle_{A_1|A_2A_3}^{2\otimes 2\otimes 2}\, = \,\sum_{a\in\{a_1,a_2\}}\sum_{b\in\{b_1,b_2\}}\sum_{c\in\{c_1,c_2\}} h_{a|bc}|abc\rangle$, where $a_1\neq a_2\in\{1,2, \ldots, d_1\}$, $b_1\neq b_2\in\{1,2, \ldots, d_2\}$ and $c_1\neq c_2\in\{1,2, \ldots, d_3\}$. There are $\binom{d_1}{2}\binom{d_2}{2}\binom{d_3}{2}$ different substates, where $\binom{d_i}{2}$, $i = 1, 2, 3$, is the binomial coefficient, and we simply use $|\varphi\rangle_{A_1|A_2A_3}^{2\otimes2\otimes2}$ to denote one of the substates. It follows from (25) that

$$\begin{align} C^2(|\varphi\rangle_{A_1|A_2A_3}^{\,d_1\otimes d_2\otimes d_3})\geqslant\sum\frac{1}{\left(d_1-1\right)\left(d_2-1\right)\left(d_3-1\right)}C^2\left(|\varphi\rangle_{A_1|A_2A_3}^{2\otimes2\otimes2}\right), \end{align} \tag{ 26 }$$

where $\sum$ stands for summing over all possible pure substates $|\varphi\rangle_{A_1|A_2A_3}^{2\otimes2\otimes2}$. For a mixed state $\rho_{A_1|A_2A_3}^{\,d_1\otimes d_2\otimes d_3}$, its substate $\rho_{A_1|A_2A_3}^{2\otimes2\otimes2}$ has the following form,

$$\begin{align} \rho_{A_1|A_2A_3}^{2\otimes2\otimes2} = \left[ \begin{array}{cccccccc} \rho_{a_1|b_1c_1}{,}_{\,a_1|b_1c_1}&\rho_{a_1|b_1c_1}{,}_{a_1|b_1c_2}&\cdots&\rho_{a_1|b_1c_1}{,}_{a_2|b_2c_1}&\rho_{a_1|b_1c_1}{,}_{a_2|b_2c_2} \\ \rho_{a_1|b_1c_2}{,}_{\,a_1|b_1c_1}&\rho_{a_1|b_1c_2}{,}_{a_1|b_1c_2}&\cdots&\rho_{a_1|b_1c_2}{,}_{a_2|b_2c_1}&\rho_{a_1|b_1c_2}{,}_{a_2|b_2c_2} \\ \rho_{a_1|b_2c_1}{,}_{\,a_1|b_1c_1}&\rho_{a_1|b_2c_1}{,}_{a_1|b_1c_2}&\cdots&\rho_{a_1|b_2c_1}{,}_{a_2|b_2c_1}&\rho_{a_1|b_2c_1}{,}_{a_2|b_2c_2} \\ \vdots&\vdots&\vdots&\vdots&\vdots \\ \rho_{a_2|b_1c_2}{,}_{\,a_1|b_1c_1}&\rho_{a_2|b_1c_2}{,}_{a_1|b_1c_2}&\cdots&\rho_{a_2|b_1c_2}{,}_{a_2|b_2c_1}&\rho_{a_2|b_1c_2}{,}_{a_2|b_2c_2} \\ \rho_{a_2|b_2c_1}{,}_{\,a_1|b_1c_1}&\rho_{a_2|b_2c_1}{,}_{a_1|b_1c_2}&\cdots&\rho_{a_2|b_2c_1}{,}_{a_2|b_2c_1}&\rho_{a_2|b_2c_1}{,}_{a_2|b_2c_2} \\ \rho_{a_2|b_2c_2}{,}_{\,a_1|b_1c_1}&\rho_{a_2|b_2c_2}{,}_{a_1|b_1c_2}&\cdots&\rho_{a_2|b_2c_2}{,}_{a_2|b_2c_1}&\rho_{a_2|b_2c_2}{,}_{a_2|b_2c_2} \\ \end{array} \right ], \end{align} \tag{ 27 }$$

which is an unnormalized tripartite mixed state.

Lemma 3. In tripartite quantum systems $\mathcal{H}^{\,d_1}_{A_1} \otimes\mathcal{H}^{\,d_2}_{A_2}\otimes\mathcal{H}^{\,d_3}_{A_3}$, the concurrence of mixed state $\rho_{A_1|A_2A_3}^{\,d_1\otimes d_2\otimes d_3}$ satisfies:

$$\begin{align} C^2\left(\rho_{A_1|A_2A_3}^{\,d_1\otimes d_2\otimes d_3}\right)\geqslant\sum\frac{1}{\left(d_1-1\right)\left(d_2-1\right)\left(d_3-1\right)}C^2\left(\rho_{A_1|A_2A_3}^{2\otimes2\otimes2}\right), \end{align} \tag{ 28 }$$

where $\,\sum$ stands for summing over all possible mixed substates $\rho_{A_1|A_2A_3}^{2\otimes2\otimes2}$.

proof For a mixed state $\rho_{A_1|A_2A_3}^{\,d_1\otimes d_2\otimes d_3} = \sum_i p_i|\varphi_i\rangle_{A_1|A_2A_3}^{\,d_1\otimes d_2\otimes d_3}\langle\varphi_i|$, we have

$$\begin{align} C\left(\rho_{A_1|A_2A_3}^{\,d_1\otimes d_2\otimes d_3}\right)& = min \sum_ip_iC\left(|\varphi_i\rangle_{A_1|A_2A_3}^{\,d_1\otimes d_2\otimes d_3}\right)\nonumber\\ & \;\; \geqslant min \frac{1}{\sqrt{(d_1-1)(d_2-1)(d_3-1)}}\nonumber\\[3pt] & \;\; \quad \times \sum_ip_i\left(\sum C^2\left(|\varphi_i\rangle_{A_1|A_2A_3}^{2\otimes2\otimes2}\right)\right)^\frac{1}{2}\nonumber\\[1pt] & \;\; \geqslant min \frac{1}{\sqrt{(d_1-1)(d_2-1)(d_3-1)}}\nonumber\\[3pt] & \;\; \quad \times \left[\sum\left(\sum_ip_iC\left(|\varphi_i\rangle_{A_1|A_2A_3}^{2\otimes2\otimes2}\right)\right)^2\right]^\frac{1}{2} \nonumber \\[1pt] & \;\; \geqslant \frac{1}{\sqrt{(d_1-1)(d_2-1)(d_3-1)}}\nonumber\\[3pt] & \;\; \quad \times \left[\sum\left(min \sum_ip_iC\left(|\varphi_i\rangle_{A_1|A_2A_3}^{2\otimes2\otimes2}\right)\right)^2\right]^\frac{1}{2} \nonumber\\[1pt] & \;\; = \frac{1}{\sqrt{(d_1-1)(d_2-1)(d_3-1)}}\nonumber\\[3pt] & \;\; \quad \times \left[\sum C^2\left(\rho_{A_1|A_2A_3}^{2\otimes2\otimes2}\right)\right]^\frac{1}{2}, \end{align} \tag{ 29 }$$

where we have used the Minkowski inequality $\sum_m(\sum_n a_{mn}^2)^\frac{1}{2}\geqslant(\sum_n(\sum_ma_{mn})^2)^\frac{1}{2}$ in the second inequality, the minimum is taken over all possible pure state decompositions of mixed state $\rho_{A_1|A_2A_3}^{\,d_1\otimes d_2\otimes d_3}$ in the first three minimizations, while the minimum in the last inequality is taken over all pure state decompositions of $\rho_{A_1|A_2A_3}^{2\otimes2\otimes2}$.

By using $(1+t)^x\geqslant 1+t^{\,x}$ for $x\geqslant1$ and $0\leqslant t\leqslant1$, we can easily obtain $(\sum_{i = 1}^nc_i)^x\geqslant\sum_{i = 1}^nc_i^x$ for nonnegative numbers c_i , $i = 1, \ldots, n$, and $x\geqslant1$. The monogamy inequalities of the αth ($\alpha\geqslant2$) power of the concurrence for n-qudit quantum states are given as follows.

Theorem 3. In tripartite quantum systems $\mathcal{H}^{\,d_1} \otimes\mathcal{H}^{\,d_2}\otimes\mathcal{H}^{\,d_3}$, assuming $0\leqslant k^{\,\omega}\leqslant k\leqslant1$, $\omega\geqslant1$, and $\alpha\geqslant2$,

• (a)  
  if $C^2\left(\rho_{A_1A_3}^{2\otimes2}\right)\leqslant k^{\,\omega} C^2\left(\rho_{A_1A_2}^{2\otimes2}\right)$, the concurrence satisfies

  $$\begin{align} C^{\,\,\alpha}\left(\rho_{A_1|A_2A_3}^{\,d_1\otimes d_2\otimes d_3}\right)&\geqslant \sum\left[\frac1{(d_1-1)(d_2-1)(d_3-1)}\right]^\frac\alpha2\nonumber\\[6pt] & \quad \times \left[\left(\frac12\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{2}}C^{\,\,\alpha}\left(\rho_{A_1A_2}^{2\otimes2}\right)\right.\nonumber\\[6pt] & \qquad \left. +\frac{(1+k^{\,\omega})^{\genfrac{}{}{.3pt}{3}{\alpha}{2}}-\left(\frac12\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{2}}}{k^{\genfrac{}{}{.3pt}{3}{\omega\alpha}{2}}}C^{\,\,\alpha}\left(\rho_{A_1A_3}^{2\otimes2}\right)\right]. \end{align} \tag{ 30 }$$
• (b)  
  If $C^2\left(\rho_{A_1A_2}^{2\otimes2}\right)\leqslant k^{\,\omega} C^2\left(\rho_{A_1A_3}^{2\otimes2}\right)$, the concurrence satisfies

  $$\begin{align} C^{\,\,\alpha}\left(\rho_{A_1|A_2A_3}^{\,d_1\otimes d_2\otimes d_3}\right)& \geqslant \sum \left[\frac1{(d_1-1)(d_2-1)(d_3-1)}\right]^\frac\alpha2\nonumber\\[6pt] & \quad \times \left[\left(\frac12\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{2}}C^{\,\,\alpha}\left(\rho_{A_1A_3}^{2\otimes2}\right)\right.\nonumber\\[6pt] & \qquad \left. + \frac{(1+k^{\,\omega})^{\genfrac{}{}{.3pt}{3}{\alpha}{2}}-\left(\frac12\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{2}}}{k^{\genfrac{}{}{.3pt}{3}{\omega\alpha}{2}}}C^{\,\,\alpha}\left(\rho_{A_1A_2}^{2\otimes2}\right)\right]. \end{align} \tag{ 31 }$$

proof For $\alpha\geqslant2$, we have

$$\begin{align} C^{\,\,\alpha}\left(\rho_{A_1|A_2A_3}^{\,d_1\otimes d_2\otimes d_3}\right)&\geqslant \Bigg[\sum\frac{1}{(d_1-1)(d_2-1)(d_3-1)}\nonumber\\[6pt] & \qquad \times C^2\left(\rho_{A_1|A_2A_3}^{2\otimes2\otimes2}\right)\Bigg]^\frac\alpha2\nonumber\\[6pt] & = \left[\frac1{(d_1-1)(d_2-1)(d_3-1)}\right]^\frac\alpha2\nonumber\\[6pt] & \quad \times \left(\sum C^2\left(\rho_{A_1|A_2A_3}^{2\otimes2\otimes2}\right)\right)^\frac\alpha2\nonumber\\[6pt] &\geqslant \sum\left[\frac1{(d_1-1)(d_2-1)(d_3-1)}\right]^\frac\alpha2\nonumber\\[6pt] & \quad \times \left(C^2\left(\rho_{A_1|A_2A_3}^{2\otimes2\otimes2}\right)\right)^\frac\alpha2, \end{align} \tag{ 32 }$$

where the first inequality follows from lemma 3, the second inequality is due to $(\sum_{i = 1}^nc_i)^x\geqslant\sum_{i = 1}^nc_i^x$ for nonnegative numbers c_i , $i = 1, \ldots, n$, and $x\geqslant1$, and $\sum$ stands for summing over all possible $2\otimes 2\otimes 2$ mixed substates of $\rho_{A_1|A_2A_3}^{\,d_1\otimes d_2\otimes d_3}$. Assuming $C^2(\rho_{A_1A_3}^{2\otimes2})\leqslant k^{\,\omega} C^2(\rho_{A_1A_2}^{2\otimes2})$, by using $C^2(\rho_{A_1|A_2A_3})\geqslant C^2(\rho_{A_1A_2})+C^2(\rho_{A_1A_3})$ [26] and inequality (7) of lemma 1, one has

$$\begin{align} C^{\,\,\alpha}\left(\rho_{A_1|A_2A_3}^{2\otimes2\otimes2}\right) &\geqslant \left(C^2(\rho_{A_1A_2}^{2\otimes2})+C^2\left(\rho_{A_1A_3}^{2\otimes2}\right)\right)^\frac\alpha2\nonumber\\[4pt] & = C^{\,\,\alpha}\left(\rho_{A_1A_2}^{2\otimes2}\right)\left(1+\frac{C^2\left(\rho_{A_1A_3}^{2\otimes2}\right)}{C^2\left(\rho_{A_1A_2}^{2\otimes2}\right)}\right)^\frac\alpha2\nonumber\\[4pt] &\geqslant C^{\,\,\alpha}\left(\rho_{A_1A_2}^{2\otimes2}\right)\left[\left(\frac12\right)^\frac\alpha2+\frac{\left(1+k^{\,\omega}\right)^\frac{\alpha}{2}-\left(\frac12\right)^\frac{\alpha}{2}}{k^{\genfrac{}{}{.3pt}{3}{\omega \alpha}{2}}}\right.\nonumber\\[4pt] & \qquad\qquad\qquad \times \left.\left(\frac{C^2\left(\rho_{A_1A_3}^{2\otimes2}\right)}{C^2\left(\rho_{A_1A_2}^{2\otimes2}\right)}\right)^\frac\alpha2\right]\nonumber\\[3pt] & = \left(\frac12\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{2}}C^{\,\,\alpha}(\rho_{A_1A_2}^{2\otimes2})+\frac{(1+k^{\,\omega})^{\genfrac{}{}{.3pt}{3}{\alpha}{2}}-\left(\frac12\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{2}}}{k^{\genfrac{}{}{.3pt}{3}{\omega\alpha}{2}}}\nonumber\\[3pt] & \quad \times C^{\,\,\alpha}(\rho_{A_1A_3}^{2\otimes2}), \end{align} \tag{ 33 }$$

where $0\leqslant k^{\,\omega}\leqslant k\leqslant1$, $\omega\geqslant1$, and $\alpha\geqslant2$. Combining (32) and (33), one gets (30). Using similar methods, we obtain the inequality (31).

For a pure n-partite quantum state $|\varphi\rangle^{\,d_1\otimes d_2\otimes \cdots \otimes d_n}\in H_1^{\,d_1}\otimes H_2^{\,d_2}\otimes \cdots\otimes H_n^{\,d_n}$ under bipartite partition $A_1|A_2\ldots A_n$ has the form

$$\begin{align} |\varphi\rangle_{A_1|A_2\cdots A_n}^{\,d_1\otimes d_2\otimes\cdots\otimes d_n} = \sum_{a_1 = 1}^{\,d_1}\sum_{a_2 = 1}^{\,d_2}\cdots\sum_{a_n = 1}^{\,d_n}h_{a_1|a_2\cdots a_n}|a_1a_2\cdots a_n\rangle, \end{align} \tag{ 34 }$$

where $h_{a_1|a_2\ldots a_n}\in \mathbb{C}$, $\sum_{a_1a_2\ldots a_n}h_{a_1|a_2\ldots a_n}h_{a_1|a_2\ldots a_n}^\ast = 1$. So we get

$$\begin{align} C^2\left(|\varphi\rangle^{d_1\otimes d_2\otimes\cdots \otimes d_n}_{A_1|A_2\cdots A_n}\right)& = \sum_{a_1,b_1 = 1}^{\,d_1}\sum_{a_2,b_2 = 1}^{\,d_2}\cdots\sum_{a_n,b_n = 1}^{\,d_n}\nonumber\\[6pt] & \quad \times |h_{a_1|a_2\cdots a_n}h_{b_1|b_2\cdots b_n}-h_{b_1|a_2\cdots a_n}h_{a_1|b_2\cdots b_n}|^2. \end{align} \tag{ 35 }$$

According to (35), in multipartite quantum systems $\mathcal{H}^{\,d_1} \otimes\mathcal{H}^{\,d_2}\otimes\cdots\otimes\mathcal{H}^{\,d_n}$, the concurrence of the mixed state $\rho^{d_1\otimes d_2\otimes\cdots\otimes d_n}_{A_1|A_2\cdots A_n}$ satisfies: $C^2(\rho_{A_1|A_2\ldots A_n}^{\,d_1\otimes d_2\otimes\cdots\otimes d_n})\geqslant\sum\frac{1}{(d_1-1)(d_2-1)\cdots(d_n-1)}C^2(\rho_{A_1|A_2\ldots A_n}^{2\otimes2\otimes\cdots\otimes2})$, where $\sum$ sums over all possible mixed substates $\rho_{A_1|A_2\ldots A_n}^{2\otimes 2\otimes\cdots\otimes 2}$. For an n-qubit quantum states ρ under bipartite partition $A_1|A_2A_3\ldots A_n$, the concurrence satisfies $C^2(\rho_{A_1|A_2A_3\ldots A_n})\geqslant C^2(\rho_{A_1A_2})+C^2(\rho_{A_1A_3})+\cdots+C^2(\rho_{A_1A_n})$ [4]. Then using a similar method to that in theorem 3, we can generalize our result to n-partite quantum systems and obtain the following theorem:

Theorem 4. In multipartite quantum systems $\mathcal{H}^{\,d_1}_{A_1} \otimes\mathcal{H}^{\,d_2}_{A_2}\otimes\cdots\otimes\mathcal{H}^{\,d_n}_{A_n}$, assuming real numbers $0\leqslant k^{\,\omega}\leqslant k\leqslant1$, $\omega\geqslant1$, and $\alpha\geqslant2$, where $\sum$ represents the sum of all possible mixed substates $\rho_{A_1|A_2\ldots A_n}^{2\otimes 2\otimes\cdots\otimes 2}$ and $\mu = \frac{(1+k^{\,\omega})^{\genfrac{}{}{.3pt}{3}{\alpha}{2}}-\left(\frac{1}{2}\right)^{\genfrac{}{}{.3pt}{3}{\alpha}{2}}}{k^{\genfrac{}{}{.3pt}{3}{\omega\alpha}{2}}}$, we have the following results:

• (a)  
  if $k^{\,\omega} C^2\left(\rho_{A_1A_i}^{2\otimes2}\right)\geqslant\sum_{j = i+1}^nC^2\left(\rho_{A_1A_j}^{2\otimes2}\right)$ for $i = 2,\ldots,n-1$ and $n\geqslant3$, then

  $$\begin{align}\begin{array}{*{20}{c}} {}&{{C^\alpha }\left( {\rho _{{A_1}|{A_2} \cdots {A_n}}^{{d_1} \otimes {d_2} \otimes \cdots \otimes {d_n}}} \right)}\\ {}&{\mathop \geqslant \sum \nolimits^ {{\left[ {\frac{1}{{({d_1} - 1)({d_2} - 1) \cdots ({d_n} - 1)}}} \right]}^{\frac{\alpha }{2}}}}\\ {}&{ \times \left[ {{{\left( {\frac{1}{2}} \right)}^{\frac{\alpha }{2}}}\left( {{C^\alpha }\left( {\rho _{{A_1}{A_2}}^{2 \otimes 2}} \right) + \mu {C^\alpha }\left( {\rho _{{A_1}{A_3}}^{2 \otimes 2}} \right) + \cdots } \right.} \right.}\\ {}&{\left. {\left. { + {\mu ^{n - 3}}{C^\alpha }\left( {\rho _{{A_1}{A_{n - 1}}}^{2 \otimes 2}} \right)} \right) + {\mu ^{n - 2}}{C^\alpha }\left( {\rho _{{A_1}{A_n}}^{2 \otimes 2}} \right)} \right].} \end{array} \end{align} \tag{ 36 }$$
• (b)  
  If $C^2(\rho_{A_1A_i}^{2\otimes2})\leqslant k^{\,\omega}\sum_{j = i+1}^nC^2(\rho_{A_1A_j}^{2\otimes2})$ for $i = 2,\ldots,n-1$ and $n\geqslant3$, then

  $$\begin{align}\begin{array}{*{20}{c}} {}&{{C^\alpha }\left( {\rho _{{A_1}|{A_2} \cdots {A_n}}^{{d_1} \otimes {d_2} \otimes \cdots \otimes {d_n}}} \right)}\\ {}&{\mathop \geqslant \sum \nolimits^ {{\left[ {\frac{1}{{({d_1} - 1)({d_2} - 1) \cdots ({d_n} - 1)}}} \right]}^{\frac{\alpha }{2}}}}\\ {}&{ \times \left[ {\mu ({C^\alpha }\left( {\rho _{{A_1}{A_2}}^{2 \otimes 2}} \right) + {{\left( {\frac{1}{2}} \right)}^{\frac{\alpha }{2}}}{C^\alpha }\left( {\rho _{{A_1}{A_3}}^{2 \otimes 2}} \right) + \cdots } \right.}\\ {}&{\left. { + {{\left( {\frac{1}{2}} \right)}^{\frac{{(n - 3)\alpha }}{2}}}{C^\alpha }\left( {\rho _{{A_1}{A_{n - 1}}}^{2 \otimes 2}} \right)) + {{\left( {\frac{1}{2}} \right)}^{\frac{{(n - 2)\alpha }}{2}}}{C^\alpha }(\rho _{{A_1}{A_n}}^{2 \otimes 2})} \right].} \end{array} \end{align} \tag{ 37 }$$
• (c)  
  If $k^{\,\omega} C^2(\rho_{A_1A_i}^{2\otimes2})\geqslant\sum_{j = i+1}^nC^2(\rho_{A_1A_j}^{2\otimes2})$ for $i = 2,\ldots,m$ and $C^2(\rho_{A_1A_i}^{2\otimes2})\leqslant k^{\,\omega}\sum_{j = i+1}^nC^2(\rho_{A_1A_j}^{2\otimes2})$ for $i = m+1,\ldots,n-1$, $\forall 2\leqslant m\leqslant n-2$, $n\geqslant4$, we have

  $$\begin{align}\begin{array}{l} {C^{{\kern 1pt} {\kern 1pt} \alpha }}\left( {\rho _{{A_1}|{A_2} \cdots {A_n}}^{{\kern 1pt} {d_1} \otimes {d_2} \otimes \cdots \otimes {d_n}}} \right)\\ \;\; \geqslant \sum {{{\left[ {\frac{1}{{({d_1} - 1)({d_2} - 1) \cdots ({d_n} - 1)}}} \right]}^{\frac{\alpha }{2}}}} \\ \quad \times \left[ {{{\left( {\frac{1}{2}} \right)}^{\frac{\alpha }{2}}}\left( {{C^{{\kern 1pt} {\kern 1pt} \alpha }}\left( {\rho _{{A_1}{A_2}}^{2 \otimes 2}} \right) + \mu {C^{{\kern 1pt} {\kern 1pt} \alpha }}\left( {\rho _{{A_1}{A_3}}^{2 \otimes 2}} \right)} \right.} \right.\\ \qquad \;\;\left. { + \cdots + {\mu ^{m - 2}}{C^{{\kern 1pt} {\kern 1pt} \alpha }}\left( {\rho _{{A_1}{A_m}}^{2 \otimes 2}} \right)} \right)\\ \qquad \;\; + {\mu ^m}\left( {{C^{{\kern 1pt} {\kern 1pt} \alpha }}\left( {\rho _{{A_1}{A_{m + 1}}}^{2 \otimes 2}} \right) + {{\left( {\frac{1}{2}} \right)}^{3\alpha 2}}{C^{{\kern 1pt} {\kern 1pt} \alpha }}\left( {\rho _{{A_1}{A_{m + 2}}}^{2 \otimes 2}} \right)} \right.\\ \qquad \;\left. { + \cdots + {{\left( {\frac{1}{2}} \right)}^{\frac{{3(n - m - 2)\alpha }}{2}}}{C^{{\kern 1pt} {\kern 1pt} \alpha }}\left( {\rho _{{A_1}{A_{n - 1}}}^{2 \otimes 2}} \right)} \right)\\ \qquad \;\;\left. { + {\mkern 1mu} {\mu ^{m - 1}}{{\left( {\frac{1}{2}} \right)}^{\frac{{3(n - m - 1)\alpha }}{2}}}{C^{{\kern 1pt} {\kern 1pt} \alpha }}\left( {\rho _{{A_1}{A_n}}^{2 \otimes 2}} \right)} \right]. \end{array} \end{align} \tag{ 38 }$$

Remark 3. We have presented the monogamy inequalities of the αth ($\alpha\geqslant2$) power of concurrence. By adopting the inequalities (6) of lemma 1 and the above method, we can also obtain similar results for $0\leqslant\alpha\leqslant2$ which cover all the real numbers. As remark 2 shows, we have found tighter results for $0\leqslant\alpha\leqslant2$.

In this article, we have obtained the monogamy inequalities of the Bures measure of entanglement and the geometric measure of entanglement for tripartite quantum states and n-qubit quantum states. Using examples, we have shown that our monogamy relations are tighter than the existing ones. Moreover, we have discussed the monogamy inequalities of concurrence in higher-dimensional quantum systems which give rise to finer characterizations of the entanglement shareability and distribution among qudit systems. Our results may help us understand better the monogamy nature of multipartite higher-dimensional quantum entanglement.

This work is supported in part by Simons Foundation under Grant No. 523868 and NSFC under Grant Nos. 12126351 and 12126314.