Polygamy Inequalities for Qubit Systems

TL;DR

This paper explores the polygamy inequalities of entanglement in multi-qubit systems, extending previous results to broader parameter ranges for concurrence and entanglement of formation.

Contribution

It generalizes existing polygamy relations for entanglement measures in multi-qubit states to larger parameter regions.

Findings

Extended polygamy inequalities for concurrence and entanglement of formation.

Broadened the parameter region for entanglement polygamy relations.

Provided theoretical bounds for multi-qubit entanglement measures.

Abstract

Entanglement polygamy, like entanglement monogamy, is a fundamental property of multipartite quantum states. We investigate the polygamy relations related to the concurrence $C$ and the entanglement of formation $E$ for general $n$-qubit states. We extend the results in [Phys. Rev. A 90, 024304 (2014)] from the parameter region $\alpha\leq0$ to $\alpha\leq\alpha_0$, where $0<\alpha_0\leq2$ for $C$, and $0<\alpha_0\leq\sqrt{2}$ for $E$.

Equations

$${\epsilon }_{A|BC}\ge {\epsilon }_{AB}+{\epsilon }_{AC}.$$

$${\epsilon }_{A|BC}\le {\epsilon }_{AB}+{\epsilon }_{AC}.$$

$$C(|\psi {⟩}_{AB})=\sqrt{2[1-Tr({\rho }_A^2)]},$$

$$C({\rho }_{AB})=\underset{\{p_i,|{\psi }_i⟩\}}{\min }\sum _i{p}_{i}C(|{\psi }_i⟩),$$

$$C_a(|\psi {⟩}_{ABC})\equiv C_a({\rho }_{AB})=\underset{\{p_i,|{\psi }_i⟩\}}{\max }\sum _i{p}_{i}C(|{\psi }_i⟩),$$

$$C_{A|B_1{B}_2...B_{n-1}}^{\alpha }\ge C_{A{B}_1}^{\alpha }+...+C_{A{B}_{n-1}}^{\alpha },$$

$$C_{A|B_1...B_{n-1}}^{\alpha }<C_{A{B}_1}^{\alpha }+...+C_{A{B}_{n-1}}^{\alpha },$$

$$C^2(|\phi {⟩}_{A|B_1...B_{n-1}})\le \sum _{i=1}^{n-1}{C}_a^2({\rho }_{A{B}_i}).$$

$$C^{\alpha }({\rho }_{A|BC})\le C^{\alpha }({\rho }_{AB})+C^{\alpha }({\rho }_{AC}).$$

$$f(\alpha )=2{\left(\frac{2t}{3}-\sqrt{\frac{3-2t-t^2}{3}}\right)}^{\alpha }.$$

$$C^{\alpha }({\rho }_{A|B_1{B}_2...B_{n-1}})$$

$$C^{\beta }({\rho }_{A|B_1{B}_2...B_{n-1}})$$

$$C^{{\alpha }_1}({\rho }_{A|B_1{B}_2...B_{n-1}})$$

$$E({\rho }_{AB})=\underset{\{p_i,|{\psi }_i⟩\}}{\min }\sum _i{p}_{i}E(|{\psi }_i⟩),$$

$$E_a({\rho }_{AB})=\underset{\{p_i,|{\psi }_i⟩\}}{\max }\sum _i{p}_{i}E(|{\psi }_i⟩),$$

$$E^{\alpha }({\rho }_{A|B_1{B}_2...B_{n-1}})\ge \sum _{i=1}^{n-1}{E}^{\alpha }({\rho }_{A{B}_i})$$

$$E_a({\rho }_{A|B_1{B}_2...B_{n-1}})\le \sum _{i=1}^{n-1}{E}_a({\rho }_{A{B}_i})$$

$$[E_a({\rho }_{A|B_1{B}_2...B_{n-1}}){]}^{\beta }\le \sum _{i=1}^{n-1}{\beta }^i[E_a({\rho }_{A{B}_i}){]}^{\beta },$$

$$E_a({\rho }_{A{B}_i})\le \sum _{j=i+1}^{n-1}{E}_a({\rho }_{A{B}_j}).$$

$$E^{\alpha }({\rho }_{A|B_1{B}_2...B_{n-1}})$$

$$E^{\beta }({\rho }_{A|B_1{B}_2...B_{n-1}})$$

$$E_a^{\beta }(|\phi {⟩}_{A|B_1{B}_2...B_{n-1}})$$

$$E^{{\alpha }_1}({\rho }_{A|B_1{B}_2...B_{n-1}})$$

Full text

I Introduction

Quantum entanglement F ; K ; H ; J ; C lies at the heart of quantum information processing and quantum computation ma . The quantification of quantum entanglement has drawn much attention in the last decade. An fundamental difference between quantum entanglement and classical correlations is that a quantum system entangled with one of other systems limits its entanglement with the remaining systems. The monogamy and polygamy relations give rise to the structures of entanglement distribution in multipartite systems. They are also essential features allowing for security in quantum key distribution k3 .

For a tripartite system $A$, $B$ and $C$, the monogamy of an entanglement measure $\varepsilon$ implies that 022309 , the entanglement between $A$ and $BC$ satisfies

[TABLE]

Such monogamy relations are not always satisfied by any entanglement measures. Dually the polygamy inequality in literature is expressed as pol :

[TABLE]

It has been shown that the squared concurrence $C^{2}$ PRA80044301 ; C2 and the squared entanglement of formation $E^{2}$ PRLB ; PRA61052306 do satisfy such monogamy relations (1). In Ref. zhuxuena it has been shown that general monogamy inequalities are satisfied by the $\alpha$ $(\alpha\geq 2)$th power of concurrence $C^{\alpha}$ and the $\alpha$ $(\alpha\geq\sqrt{2})$th power of entanglement of formation $E^{\alpha}$ for $n-$qubit mixed states. If $C(\rho_{AB_{i}})\not=0$, $i=1,...,n-1$, $C^{\alpha}$ satisfies (2) for $\alpha\leq 0$. In Ref. jin tighter monogamy inequalities for concurrence, entanglement of formation have been given.

Ref. PRA97012334 shown that the $\alpha$th power of the square of convex-roof extended negativity (SCREN) provides a class of monogamy inequalities of multiqubit entanglement in a tight way for $\alpha\geq 1$, and further shown that the $\alpha$th power of SCREN also provides a class of tight polygamy inequalities for $0\leq\alpha\leq 1.$ By using the $\alpha$th power of entanglement of assistance for $0\leq\alpha\leq 1$, and the Hamming weight of the binary vector related with the distribution of subsystems, Ref. PRA97042332 established a class of weighted polygamy inequalities of multiparty entanglement in arbitrary dimensional quantum systems.

However, the polygamy properties of the $\alpha$th $(0<\alpha<2)$ power of concurrence and the $\alpha$th $(0<\alpha<\sqrt{2})$ power of entanglement of formation are still unknown. In this paper, we study the polygamy inequalities of $C^{\alpha}$ for $\alpha\in(0,2)$ and $E^{\alpha}$ for $\alpha\in(0,\sqrt{2})$.

II Polygamy relations for concurrence

For a bipartite pure state $|\psi\rangle_{AB}$, the concurrence is given by s7 ; s8 ; af ,

[TABLE]

where $\rho_{A}$ is reduced density matrix obtained by tracing over the subsystem $B$, $\rho_{A}=Tr_{B}(|\psi\rangle_{AB}\langle\psi|)$. The concurrence is extended to mixed states $\rho=\sum_{i}p_{i}|\psi_{i}\rangle\langle\psi_{i}|$, $p_{i}\geq 0$, $\sum_{i}p_{i}=1$, by the convex roof construction,

[TABLE]

where the minimum takes over all possible pure state decompositions of $\rho_{AB}$.

For a tripartite state $|\psi\rangle_{ABC}$, the concurrence of assistance (CoA) is defined by ca

[TABLE]

for all possible ensemble realizations of $\rho_{AB}=Tr_{C}(|\psi\rangle_{ABC}\langle\psi|)=\sum_{i}p_{i}|\psi_{i}\rangle_{AB}\langle\psi_{i}|$. When $\rho_{AB}=|\psi\rangle_{AB}\langle\psi|$ is a pure state, then one has $C(|\psi\rangle_{AB})=C_{a}(\rho_{AB})$.

For $n-$qubit quantum states, the concurrence satisfies zhuxuena

[TABLE]

for $\alpha\geq 2$, where $C_{A|B_{1}B_{2}...B_{n-1}}$ is the concurrence of $\rho$ under bipartite partition $A|B_{1}B_{2}...B_{n-1}$, and $C_{AB_{i}}$, $i=1,2...,n-1$, is the concurrence of the mixed states $\rho_{AB_{i}}=Tr_{B_{1}B_{2}...B_{i-1}B_{i+1}...B_{n-1}}(\rho)$. For $C_{AB_{i}}\not=0$, $i=1,...,n-1$, the concurrence satisfies

[TABLE]

for $\alpha\leq 0$. Further, in Ref. jin monogamy inequalities tighter than (3) are derived for the $\alpha$th $(\alpha\geq 2)$ power of concurrence.

Dual to the CKW inequality, the polygamy monogamy relation based on the concurrence of assistance for the $n-$qubit pure states $|\varphi\rangle_{A|B_{1}...B_{n-1}}$ was proved in du :

[TABLE]

Theorem 1

For any $2\otimes 2\otimes 2^{n-2}$ tripartite mixed state $\rho_{ABC}$, if $C(\rho_{AB})C(\rho_{AC})\not=0$, there exists a real number $\alpha_{0}\in(0,2]$, for any $\alpha\in[0,\alpha_{0}]$, we have

[TABLE]

[Proof] For arbitrary $2\otimes 2\otimes 2^{n-2}$ tripartite state $\rho_{ABC}$, if $C(\rho_{AB})C(\rho_{AC})\not=0$, denote $f(\alpha)=C^{\alpha}(\rho_{AB})+C^{\alpha}(\rho_{AC})$. Obviously, $f(0)=2$ and $f(2)=C^{2}_{AB}+C^{2}_{AC}\leq C^{2}(\rho_{A|BC})\leq 1$. Since the continuity of the $f(\alpha)$, there exists a real number $\alpha_{0}\in(0,2]$ such that $f(\alpha_{0})=1$. Together with the monotonicity of $f(\alpha)$, we have $f(\alpha)\geq 1$ for $\alpha\in[0,\alpha_{0}]$, i.e $C^{\alpha}_{AB}+C^{\alpha}_{AC}\geq 1\geq C^{\alpha}(\rho_{A|BC})$ for $\alpha\in[0,\alpha_{0}]$. $\Box$

Theorem 1 shows the polygamy of inequality (2) for arbitrary $2\otimes 2\otimes 2^{n-2}$ tripartite state $\rho_{ABC}$ in case of $C_{AB}C_{AC}\not=0$. Specifically, for $\alpha\in(\alpha_{0},2]$, from the proof of Theorem 1, we have $C^{\alpha}_{AB}+C^{\alpha}_{AC}\leq 1$. If $C_{AB}C_{AC}=0$, obviously we have $C^{\alpha}(\rho_{A|BC})\geq\max\{C^{\alpha}_{AB},C^{\alpha}_{AC}\}$ for any $\alpha\in[0,+\infty)$.

Example 1. Let us consider the three-qubit state, $\rho_{ABC}=\frac{1-t}{8}I_{8}+t|\psi\rangle_{ABC}\langle\psi|$ with $t\geq 0.783612$, where $|\psi\rangle_{ABC}=\frac{1}{\sqrt{3}}(|100\rangle+|010\rangle+|001\rangle),$ and $I_{8}$ is the $8\times 8$ identity matrix. We have $C(|\psi\rangle_{A|BC})=\frac{2\sqrt{2}}{3}$ and $C(\rho_{AB})=C(\rho_{AC})=\frac{2t}{3}-\sqrt{\frac{3-2t-t^{2}}{3}}$. Therefore,

[TABLE]

We have $f\left([\log_{2}(\frac{3}{2t-\sqrt{9-6t-3t^{2}}})]^{-1}\right)=1$, i.e., $\alpha_{0}=[\log_{2}(\frac{3}{2t-\sqrt{9-6t-3t^{2}}})]^{-1}$, see Fig. 1.

It is clear that $C^{\alpha}(\rho_{A|BC})\leq C^{\alpha}(\rho_{AB})+C^{\alpha}(\rho_{AC})$ for $\alpha\in[0,\alpha_{0}].$ In particular, take $t=1$. Then $\alpha_{0}\approx 1.70951$, see Fig. 2.

Generalizing the result of Theorem 1 , we have the following theorem for multipartite qubit systems.

Theorem 2

For any $n$-qubit quantum state $\rho$, if there are at least two substates $\rho_{AB_{i_{1}}}$ and $\rho_{AB_{i_{2}}}$ such that $C(\rho_{AB_{i_{1}}})C(\rho_{AB_{i_{2}}})\not=0$, $i_{1}\not=i_{2}$ and $i_{1},i_{2}\in\{1,...,n-1\}$, there must be a real number $\alpha_{0}\in(0,2]$ such that

[TABLE]

for $0\leq\alpha\leq\alpha_{0}$.

[Proof] For convenience, we denote $f(\alpha)=\sum_{i=1}^{n-1}C^{\alpha}(\rho_{AB_{i}})$ with $\alpha\geq 0$. For any $2\otimes 2\otimes 2\otimes...\otimes 2$ quantum states $\rho_{AB_{1}...B_{n-1}},$ $f(2)=\sum_{i=1}^{n-1}C^{2}(\rho_{AB_{i}})\leq C^{2}(\rho_{A|B_{1}...B_{n-1}})\leq 1$. Since $C(\rho_{AB_{i_{1}}})C(\rho_{AB_{i_{2}}})\not=0$, we have $f(0)\geq 2$. Taking into account that $f(\alpha)$ is continuous, we have that there must be a real number $\alpha_{0}\in(0,2]$ such that $f(\alpha_{0})=1$. As $f(\alpha)$ is monotonically decreasing, we have $f(\alpha)\geq 1$ for $\alpha\in[0,\alpha_{0}]$. $\Box$

From Theorem 2, inequalities (3) and (4), we have the following result for $n-$qubit quantum states $\rho_{AB_{1}...B_{n-1}}$:

(1) If there is only one substate $\rho_{AB_{i_{0}}}$, $i_{0}\in\{1,...,n-1\}$, is entangled, then $C^{\alpha}(\rho_{AB_{1}...B_{n-1}})\geq C^{\alpha}(\rho_{AB_{i_{0}}})$ for any $\alpha\geq 0$; and $C^{\alpha}(\rho_{AB_{1}...B_{n-1}})\leq C^{\alpha}(\rho_{AB_{i_{0}}})$ for any $\alpha\leq 0$;

(2) If there are at least two entangled substates, then there must be $\alpha_{0}\in(0,2]$, such that $C^{\alpha}(\rho_{AB_{1}...B_{n-1}})\geq\sum_{i=1}^{n-1}C^{\alpha}(\rho_{AB_{i}})$ for any $\alpha\geq 2$; and $C^{\alpha}(\rho_{AB_{1}...B_{n-1}})\leq\sum_{i=1}^{n-1}C^{\alpha}(\rho_{AB_{i}})$ for any $0\leq\alpha\leq\alpha_{0}$.

(3) If all the substates $\rho_{AB_{i}}$, $i=1,...,n-1$, are entangled, then there must be $\alpha_{0}\in(0,2]$, such that $C^{\alpha}(\rho_{AB_{1}...B_{n-1}})\geq\sum_{i=1}^{n-1}C^{\alpha}(\rho_{AB_{i}})$ for any $\alpha\geq 2$; and $C^{\alpha}(\rho_{AB_{1}...B_{n-1}})\leq\sum_{i=1}^{n-1}C^{\alpha}(\rho_{AB_{i}})$ for any $\alpha\leq\alpha_{0}$.

Example 2: We consider the $4$-qubit generalized $W$-class state, $|\psi\rangle=a|0000\rangle+b_{1}|1000\rangle+b_{2}|0100\rangle+b_{3}|0010\rangle+b_{4}|0001\rangle,$ with $a=b_{2}=\frac{1}{\sqrt{10}}$, $b_{1}=\frac{1}{\sqrt{15}}$, $b_{3}=\sqrt{\frac{2}{15}}$, $b_{4}=\sqrt{\frac{3}{5}}$. One has $C(\rho_{AB_{i}})=2|b_{1}||b_{i+1}|$, $i=1,2,3$, and $C(|\psi\rangle)=2|b_{1}|\sqrt{1-|b_{1}|^{2}}=\frac{2\sqrt{14}}{15}$. Denote $f(\alpha)=\sum_{i=1}^{4}C^{\alpha}(\rho_{AB_{i}})=(\frac{\sqrt{6}}{15})^{\alpha}+(\frac{2\sqrt{2}}{15})^{\alpha}+(\frac{2}{5})^{\alpha}$. We have $f(0.783586)=1$, i.e., $C^{\alpha}(|\psi\rangle)\leq C^{\alpha}(\rho_{AB_{1}})+C^{\alpha}(\rho_{AB_{2}})+C^{\alpha}(\rho_{AB_{3}})$ for $\alpha\in[0,0.783586]$, see Fig. 3.

From Theorem 2 and that $C(\rho_{AB})\leq C_{a}(\rho_{AB})$ for any quantum states, we have the following corollaries:

Corollary 1

For any $n$-qubit quantum state $\rho$, if there are at least two substates such that $C(\rho_{AB_{i_{1}}})C(\rho_{AB_{i_{2}}})\not=0$ for $i_{1}\not=i_{2}$ and $i_{1},i_{2}\in\{1,...,n-1\}$, there must be a real number $\beta_{0}\in(0,2]$,

[TABLE]

where $0\leq\beta\leq\beta_{0}$ and $\beta_{0}$ is a real number which satisfies $\sum_{i=1}^{n-1}C^{\beta_{0}}(\rho_{AB_{i}})=1.$

Denote $g(\alpha)=\sum_{i=1}^{n-1}C^{\alpha}(\rho_{AB_{i}})-C^{\alpha}(\rho_{A|B_{1}...B_{n-1}})$. For $n-$qubit states $\rho_{AB_{1}...B_{n-1}}$ with at least two entangled pairs of qubits, from Theorem 2, we have $g(\alpha_{0})\geq 0$ and $g(2)\leq 0$. There must be a real number $\alpha_{1}$ such that $g(\alpha_{1})=0$. Similar to Theorem 1 and 2, we have the following corollary.

Corollary 2

For any $n$-qubit quantum state $\rho_{A|B_{1}...B_{n-1}}$, if $C(\rho_{AB_{i_{1}}})C(\rho_{AB_{i_{2}}})\not=0$ for $i_{1}\not=i_{2}$ and $i_{1},i_{2}\in\{1,...,n-1\}$, there must be a real number $\alpha_{1}\in[\alpha_{0},2]$, such that

[TABLE]

where $\alpha_{0}$ is a real number which satisfies $\sum_{i=1}^{n-1}C^{\alpha_{0}}(\rho_{AB_{i}})=1$ for $0<\alpha_{0}\leq 2.$

III Polygamy inequalities for EoF

The entanglement of formation (EoF) C. H. Bennett ; D. P. DiVincenzo is a well-defined and important measure of quantum entanglement for bipartite systems. Let $H_{A}$ and $H_{B}$ be $m$- and $n$-dimensional $(m\leq n)$ vector spaces, respectively. The EoF of a pure state $|\psi\rangle\in H_{A}\otimes H_{B}$ is defined by $E(|\psi\rangle)=S(\rho_{A})$, where $\rho_{A}=Tr_{B}(|\psi\rangle\langle\psi|)$ and $S(\rho)=-Tr(\rho\log_{2}\rho)$. For a bipartite mixed state $\rho_{AB}\in H_{A}\otimes H_{B}$, the entanglement of formation is given by

[TABLE]

with the infimum taking over all possible decompositions of $\rho_{AB}$ in a mixture of pure states $\rho_{AB}=\sum_{i}p_{i}|\psi_{i}\rangle\langle\psi_{i}|$, where $p_{i}\geq 0$ and $\sum_{i}p_{i}=1$. $E_{a}(\rho_{AB})$ is the entanglement of assistance (EOA) of $\rho_{AB}$ defined as

[TABLE]

maximizing over all possible ensemble realizations of $\rho_{AB}=\sum_{i}p_{i}|\psi_{i}\rangle_{AB}\langle\psi_{i}|$.

It has been shown that the entanglement of formation does not satisfy monogamy inequality such as $E_{AB}+E_{AC}\leq E_{A|BC}$ PRA61052306 . In zhuxuena the authors showed that

[TABLE]

for $\alpha\geq\sqrt{2}$.

A general polygamy inequality of multipartite quantum entanglement was established as

[TABLE]

for any multipartite quantum state $\rho_{AB_{1}...B_{n-1}}$ of arbitrary dimension PRA85062302 . For any multipartite quantum state $\rho_{AB_{1}B_{2}...B_{n-1}}$, one has for any $\beta\in[0,1]$ PRA97042332 :

[TABLE]

conditioned that

[TABLE]

In fact, by using applying the approach for Theorems 1 and 2, we can prove the following results generally for EoF:

Theorem 3

For any $n$-qubit quantum state $\rho_{AB_{1}B_{2}...B_{n-1}}$, if there are at least two substates such that $C(\rho_{AB_{i_{1}}})C(\rho_{AB_{i_{2}}})\not=0$ for $i_{1}\not=i_{2}$ and $i_{1},i_{2}\in\{1,...,n-1\}$, there must be a real number $\alpha_{0}\in(0,\sqrt{2}]$, such that

[TABLE]

where $0\leq\alpha\leq\alpha_{0}$.

[Proof] For convenience, we denote $f(\alpha)=\sum_{i=1}^{n-1}E^{\alpha}(\rho_{AB_{i}})$ with $\alpha\geq 0$. Then $f(\sqrt{2})=\sum_{i=1}^{n-1}E^{\sqrt{2}}(\rho_{AB_{i}})\leq E^{\sqrt{2}}(\rho_{A|B_{1}...B_{n-1}})\leq 1$. Since $C(\rho_{AB_{i_{1}}})C(\rho_{AB_{i_{2}}})\not=0$, we have $E(\rho_{AB_{i_{1}}})E(\rho_{AB_{i_{2}}})\not=0$, i.e., $f(0)\geq 2$. As $f(\alpha)$ is continuous, there must be a real number $\alpha_{0}\in(0,\sqrt{2}]$ so that $f(\alpha_{0})=1$. Since $f(\alpha)$ is monotonically decreasing, we have $f(\alpha)\geq 1$ for $\alpha\in[0,\alpha_{0}]$. $\Box$

From Theorem 3, inequalities (10) and (11), we have the following results for $n$-qubit quantum states $\rho_{AB_{1}...B_{n-1}}$:

(1) If there is only one substate $\rho_{AB_{i_{0}}}$, $i_{0}\in\{1,...,n-1\}$, is entangled, then $E^{\alpha}(\rho_{A|B_{1}...B_{n-1}})\geq E^{\alpha}(\rho_{AB_{i_{0}}})$ for any $\alpha\geq 0$; and $E^{\alpha}(\rho_{A|B_{1}...B_{n-1}})\leq E^{\alpha}(\rho_{AB_{i_{0}}})$ for any $\alpha<0$.

(2) If at least two of the substates $\rho_{AB_{i}}$, $i=1,...,n-1$, are entangled, then there must be $\alpha_{0}\in(0,\sqrt{2}]$ so that $E^{\alpha}(\rho_{A|B_{1}...B_{n-1}})\geq\sum_{i=1}^{n-1}E^{\alpha}(\rho_{AB_{i}})$ for any $\alpha\geq\sqrt{2}$; and $E^{\alpha}(\rho_{A|B_{1}...B_{n-1}})\leq\sum_{i=1}^{n-1}E^{\alpha}(\rho_{AB_{i}})$ for any $0\leq\alpha\leq\alpha_{0}$.

Example 3. Consider the pure state in Example 1, $|\psi\rangle_{ABC}=\frac{1}{\sqrt{3}}(|100\rangle+|010\rangle+|001\rangle)$. We have $E_{A|BC}=0.918296$, $E_{AB}=E_{AC}=0.550048$. Let $f(\alpha)=E^{\alpha}_{AB}+E^{\alpha}_{AC}$. Then $f(1.15959)=1$. It is easily verified that $E^{\alpha}(|\psi\rangle_{A|BC})\leq E^{\alpha}(\rho_{AB})+E^{\alpha}(\rho_{AC})$ for $\alpha\leq 1.15959$, see Fig. 3.

From Theorem 3 and that $E(\rho_{AB})\leq E_{a}(\rho_{AB})$ for any quantum states, we have the follow result:

Corollary 3

For any $n$-qubit quantum state $\rho_{AB_{1}B_{2}...B_{n-1}}$, if there are at least two substates such that $C(\rho_{AB_{i_{1}}})C(\rho_{AB_{i_{2}}})\not=0$ for $i_{1}\not=i_{2}$, $i_{1},i_{2}\in\{1,...,n-1\}$, there must be a real number $\beta_{0}\in(0,\sqrt{2}]$, so that

[TABLE]

for $0\leq\beta\leq\beta_{0}$, where $\beta_{0}$ is a real number which satisfies $\sum_{i=1}^{n-1}E^{\beta_{0}}(\rho_{AB_{i}})=1.$

Since $E(|\varphi\rangle_{A|B_{1}...B_{n-1}})=E_{a}(|\varphi\rangle_{A|B_{1}...B_{n-1}})$ for any pure state $|\varphi\rangle_{AB_{1}...B_{n-1}}$, for pure states (14) becomes

[TABLE]

for $0\leq\beta\leq\beta_{0}$, where $\beta_{0}$ is a real number which satisfies $\sum_{i=1}^{n-1}E^{\beta_{0}}(\rho_{AB_{i}})=1.$ If $\beta_{0}>1$, we have

[TABLE]

for $\beta\in[1,\beta_{0}]$. (15) also gives the polygamy inequalities when $1<\beta\leq 1.15959$, see the example 3, while (12) fails in this case.

Similarly, for any $n$-qubit quantum state $\rho_{AB_{1}B_{2}...B_{n-1}}$, if there are at least two substates such that $C(\rho_{AB_{i_{1}}})C(\rho_{AB_{i_{2}}})\not=0$ for $i_{1}\not=i_{2}$, $i_{1},i_{2}\in\{1,...,n-1\}$, there must be a real number $\alpha_{1}\in[\alpha_{0},\sqrt{2}]$, so that

[TABLE]

For Example 3, Fig. 4 shows that the solid line and dashed line have only one intersection at $\alpha_{1}=1.35244$. The relations between $E^{\alpha}(|\psi\rangle)$ and $E^{\alpha}(\rho_{AB})+E^{\alpha}(\rho_{AC})$ fall into two classes: $E^{\alpha}(|\psi\rangle_{A|BC})\leq E^{\alpha}(\rho_{AB})+E^{\alpha}(\rho_{AC})$ for $0\leq\alpha\leq 1.35244$, and $E^{\alpha}(|\psi\rangle)\geq E^{\alpha}(\rho_{AB})+E^{\alpha}(\rho_{AC})$ for $\alpha\geq 1.35244$.

IV Conclusion

Like entanglement monogamy, entanglement polygamy is a fundamental property of multipartite quantum states. It characterizes the entanglement distribution in multipartite quantum systems. We have investigated the polygamy relations related to the concurrence $C$ and the entanglement of formation $E$ for general $n$-qubit states. We have extended the results (4) and (11) in Ref. zhuxuena from $\alpha\leq 0$ to $\alpha\leq\alpha_{0}$, where $0<\alpha_{0}\leq 2$ for $C$, and $0<\alpha_{0}\leq\sqrt{2}$ for $E$. When $\alpha_{0}>2$ ($\alpha_{0}>\sqrt{2}$), the polygamy relation of concurrence $C$ ($E$) can not be obtained. It remains an open question if for this case, like Example 3, there is only one intersection $\alpha_{1}$.

Acknowledgments    This work is supported by NSFC under numbers 11675113, 11605083, and the NSF of Beijing under Grant No. KZ201810028042.