,2, Shao-Ming Fei1,3, Zhi-Xi Wang,1 Corresponding authors: ?E-mail:chenbin5134@163.com;wangzhx@cnu.edu.cn
Received:2018-11-12Online:2019-05-1
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Long-Mei Yang, Bin Chen, Shao-Ming Fei, Zhi-Xi Wang. Tighter Constraints of Multiqubit Entanglement. [J], 2019, 71(5): 545-554 doi:10.1088/0253-6102/71/5/545
1 Introduction
Quantum entanglement is an essential feature of quantum mechanics, which distinguishes the quantum from the classical world and plays a very important role in quantum information processing.[1-4] One singular property of quantum entanglement is that a quantum system entangled with one of the other subsystems limits its entanglement with the remaining ones, known as the monogamy of entanglement (MoE).[5-6] MoE plays a key role in many quantum information and communication processing tasks such as the security proof in quantum cryptographic scheme[7] and the security analysis of quantum key distribution.[8]For a tripartite quantum state $\rho_{ABC}$, MoE can be described as the following inequality
where $\rho_{AB}={\rm tr}_C(\rho_{ABC})$ and $\rho_{{AC}}={\rm tr}_B(\rho_{ABC})$ are reduced density matrices, and $\mathcal{E}$ is an entanglement measure.However, it has been shown that not all entanglement measures satisfy such monogamy relations.It has been shown that the squared concurrence $\mathcal{C}^2$,[9-10] the squared entanglement of formation (EoF) $E^2$[11] and the squared convex-roof extended negativity (CREN) $\mathcal{N}_c^2$[12-13] satisfy the monogamy relations for multiqubit states.
Another important concept is the assisted entanglement, which is a dual amount to bipartite entanglement measure.It has a dually monogamous property in multipartite quantum systems and gives rise to polygamy relations.For a tripartite state $\rho_{ABC}$, the usual polygamy relation is of the form,
where $\mathcal{E}^a$ is the corresponding entanglement measure of assistance associated to $\mathcal{E}$.Such polygamy inequality has been deeply investigated in recent years,and was generalized to multiqubit systems and classes of higher-dimensional quantum systems.[12,14-20]
Recently, generalized classes of monogamy inequalities related to the $\beta$-th power of entanglement measures were proposed.
In Refs.[21-22], the authors proved that the squared concurrence and CREN satisfy the monogamy inequalities in multiqubit systems for $\beta\geq2$.It has also been shown that the EoF satisfies monogamy relations when $\beta\geq\sqrt{2}$.[21-23] Besides, the Tsallis-$q$ entanglement and Rényi-$\alpha$ entanglement satisfy monogamy relations when $\beta\geq1$[14,22-24] for some cases.Moreover, the corresponding polygamy relations have also been established.[16-18,20,25-26]
In this paper, we investigate monogamy relations and polygamy relations in multiqubit systems.We provide tighter constraints of multiqubit entanglement than all the existing ones,thus give rise to finer characterizations of the entanglement distributions among the multiqubit systems.
2 Tighter Constraints Related to Concurrence
We first consider the monogamy inequalities and polygamy inequalities for concurrence.For a bipartite pure state $|\psi\rangle_{AB}$ in Hilbert space $H_A\otimes H_B$, the concurrence is defined as[27-28] $\mathcal{C}(|\psi\rangle_{AB})=\sqrt{2(1-{\rm tr}\rho_{A}^2)}$ with $\rho_{A}={\rm tr}_B|\psi\rangle_{AB}\langle\psi|$.The concurrence for a bipartite mixed state $\rho_{AB}$ is defined by the convex roof extension,$$\mathcal{C}(\rho_{AB})=\min\limits_{\{p_i,|\psi_i\rangle\}}\sum\limits_{i}p_i\mathcal{C}(|\psi_i\rangle),$$
where the minimum is taken over all possible decompositions of $\rho_{AB}=\sum_{i}p_i|\psi_i\rangle\langle\psi_i|$
with $\sum p_i=1$ and $p_i\geq0$.For an $N$-qubit state $\rho_{AB_1\cdots B_{N-1}}\in H_A\otimes H_{B_1}\otimes\cdots\otimes H_{B_{N-1}}$,the concurrence $\mathcal{C}(\rho_{A|B_1\cdots B_{N-1}})$ of the state $\rho_{AB_1\cdots B_{N-1}}$ under bipartite partition $A$ and $B_1\cdots B_{N-1}$ satisfies[21]
for $\beta\geq2$, where $\rho_{AB_j}$ denote two-qubit reduced density matrices of subsystems $AB_j$ for $j=1,2,\ldots,N-1$.
Later, the relation (3) is improved for the case $\beta\geq2$[23] as
conditioned that $\mathcal{C}(\rho_{AB_i})\geq\mathcal{C}(\rho_{A|B_{i+1}\cdots B_{N-1}})$ for $i=1,2,\ldots,m$,and $\mathcal{C}(\rho_{AB_j})\leq\mathcal{C}(\rho_{A|B_{j+1}\cdots B_{N-1}})$ for $j=m+1,\ldots,N-2$.
The relation (4) is further improved for $\beta\geq2$ as[22]
with the same conditions as in Eq.(4).
For a tripartite state $|\psi\rangle_{ABC}$, the concurrence of assistance (CoA) is defined by[29-30]
where the maximun is taken over all possible pure state decompositions of $\rho_{AB}$, and $\mathcal{C}(|\psi\rangle_{AB})=\mathcal{C}_a(|\psi\rangle_{AB})$.
The generalized polygamy relation based on the concurrence of assistance was established in Refs.[16-17]
These monogamy and polygamy relations for concurrence can be further tightened under some conditions.To this end, we first introduce the following lemma.
Lemma 1 Suppose that $k$ is a real number satisfying $0< k\leq1$, then for any $0\leq t\leq k$ and non-negative real numbers $m$, $n$, we have
for $m\geq1$, and
for $0\leq n\leq 1$.
We first consider the function $f(m,x)=(1+x)^m-x^m$ with $x\geq{1}/{k}$ and $m\geq1$. Then $f(m,x)$ is an increasing function of $x$, since
$$ \frac{\partial f(m,x)}{\partial x}=m[(1+x)^{m-1}-x^{m-1}]\geq 0. $$
Thus,
Set $x={1}/{t}$ in Eq.(10), we get the inequality (8).
Similar to the proof of inequality (8), we can obtain the inequality (9), since in this case $f(n,x)$ is a decreasing function of $x$ for $x\geq {1}/{k}$ and $0\leq n\leq1$. $\Box$
In the next, we denote $\mathcal{C}_{AB_i}=\mathcal{C}(\rho_{AB_i})$ the concurrence of $\rho_{AB_i}$ and $\mathcal{C}_{A|B_1\cdots B_{N-1}}=\mathcal{C}(\rho_{A|B_1\cdots B_{N-1}})$ for convenience.
Lemma 2 Suppose that $k$ is a real number satisfying $0< k\leq1$.Then for any $2\otimes2\otimes2^{n-2}$ mixed state $\rho\in H_A\otimes H_B\otimes H_C$,if $\mathcal{C}_{AC}^2\leq k\mathcal{C}_{AB}^2$, we have
for all $\beta\geq 2$.
Since $\mathcal{C}_{AC}^2\leq k\mathcal{C}_{AB}^2$ and $\mathcal{C}_{AB}>0$,we obtain
where the first inequality is due to the fact, $\mathcal{C}_{A|BC}^2\geq\mathcal{C}_{AB}^2+\mathcal{C}_{AC}^2$ for arbitrary $2\otimes2\otimes2^{n-2}$ tripartite state $\rho_{ABC}$[9,31] and the second is due to Lemma 1. We can also see that if $\mathcal{C}_{AB}=0$, then $\mathcal{C}_{AC}=0$, and the lower bound becomes trivially zero. $\Box$
For multiqubit systems, we have the following Theorems.
Theorem 1 Suppose $k$ is a real number satisfying $0<k\leq 1$. For an $N$-qubit mixed state $\rho_{AB_1\cdots B_{N-1}}$, if $k\mathcal{C}_{AB_i}^2\geq\mathcal{C}_{A|B_{i+1}\cdots B_{N-1}}^2$ for $i=1,2,\ldots,m$, and $\mathcal{C}_{AB_j}^2\leq k\mathcal{C}_{A|B_{j+1}\cdots B_{N-1}}^2$ for $j=m+1,\ldots,N-2$, $\forall 1\leq m\leq N-3$, $N\geq 4$, then we have
for all $\beta\geq2$.
From the inequality (11), we have
Since $\mathcal{C}_{AB_j}^2\leq k\mathcal{C}_{A|B_{j+1}\cdots B_{N-1}}^2$,for $j=m+1,\ldots,N-2$, we get
Combining Eqs.(14) and (15), we get the inequality (13).
$\Box$
If we replace the conditions $k\mathcal{C}_{AB_i}\geq\mathcal{C}_{A|B_{i+1}\cdots B_{N-1}}$ for $i=1,2,\ldots,m$,and $\mathcal{C}_{AB_j}^2\leq k\mathcal{C}_{A|B_{j+1}\cdots B_{N-1}}^2$ for $j=m+1,\ldots,N-2$, $\forall 1\leq m\leq N-3$, $N\geq 4$,in Theorem 1 by $k\mathcal{C}_{AB_i}^2\geq \mathcal{C}_{A|B_{i+1}\cdots B_{N-1}}^2$ for $i=1,2,\ldots,N-2$,then we have the following theorem.
Theorem 2 Suppose $k$ is a real number satisfying $0<k\leq 1$. For an $N$-qubit mixed state $\rho_{AB_1\cdots B_{N-1}}$,if $k\mathcal{C}_{AB_i}^2\geq\mathcal{C}_{A|B_{i+1}\cdots B_{N-1}}^2$ for all $i=1,2,\ldots,N-2$, then we have
for $\beta\geq2$.
It can be seen that the inequalities (13) and (16) are tighter than the ones given in Ref.[22],since
$$ \frac{(1+k)^{\beta/2}-1}{k^{\beta/2}}\geq2^{{\beta/2}}-1,$$
for $\beta\geq2$ and $0<k\leq1$.
The equality holds when $k=1$. Namely, the result (5) given in Ref.[22] are just special cases of ours for $k=1$.
As ${[(1+k)^{\beta/2}-1]}/{k^{{\beta}/{2}}}$ is a decreasing function with respect to $k$ for $0< k\leq 1$ and $\beta\geq2$,
we find that the smaller $k$ is, the tighter the inequalities (11), (13), and (16) are.
Consider the three-qubit state $|\psi\rangle_{ABC}$ in generalized Schmidt decomposition form,[32-33]
where $\lambda_i\geq0$, $i=1,2,\ldots,4$, and $\sum_{i=0}^4\lambda_i^2=1$.Then we get $\mathcal{C}_{A|BC}=2\lambda_0\sqrt{\lambda_2^2+\lambda_3^2+\lambda_4^2}$, $\mathcal{C}_{AB}=2\lambda_0\lambda_2$, and $\mathcal{C}_{AC}=2\lambda_0\lambda_3$.Set $\lambda_0=\lambda_3={1}/{2}$, $\lambda_2={\sqrt{2}}/{2}$, and $\lambda_1=\lambda_4=0$.We have $\mathcal{C}_{A|BC}={\sqrt{3}}/{2}$, $\mathcal{C}_{AB}={\sqrt{2}}/{2}$, and $\mathcal{C}_{AC}={1}/{2}$. Then
One can see that our result is better than the result (5) in Ref.[22] for $\beta\geq2$, hence better than Eqs.(3) and (4) given in Refs.[21, 23],see Fig.1.
New window|Download| PPT slideFig.1(Color online) The $y$ axis is the lower bound of the concurrence $\mathcal{C}^\beta_{A|BC}$. The red (green) line represents the lower bound from our result for $k=0.6$ ($k=0.8$), and the blue line represents the lower bound of Eq.(5) from Ref.[22].
We now discuss the polygamy relations for the CoA of $\mathcal{C}_a(|\psi\rangle_{A|B_1\cdots B_{N-1}})$ for $0\leq\beta\leq 2$.We have the following Theorem.
Theorem 3 Suppose $k$ is a real number satisfying $0<k\leq 1$.For an $N$-qubit pure state $|\psi\rangle_{AB_1\cdots B_{N-1}}$, if $k\mathcal{C}^2_{aA|B_i}\geq\mathcal{C}^2_{aA|B_{i+1}\cdots B_{N-1}}$ for $i=1,2,\ldots,m$, and $\mathcal{C}^2_{aA|B_j}\leq k\mathcal{C}^2_{aA|B_{j+1}\cdots B_{N-1}}$ for $j=m+1,\ldots,N-2$, $\forall 1\leq m\leq N-3$, $N\geq 4$, then we have
for all $0\leq\beta\leq2$.
Proof The proof is similar to the proof of Theorem 1 by using inequality (9).\Box$
Theorem 4 Suppose $k$ is a real number satisfying $0<k\leq 1$.For an $N$-qubit pure state $|\psi\rangle_{AB_1\cdots B_{N-1}}$,if $k\mathcal{C}^2_{aAB_i}\geq\mathcal{C}^2_{aA|B_{i+1}\cdots B_{N-1}}$ for all $i=1,2,\ldots,N-2$, then we have
for $0\leq\beta\leq2$.
The inequalities (18) and (19) are also upper bounds of $\mathcal{C}(|\psi\rangle_{A|B_1\cdots B_{N-1}})$for pure state $|\psi\rangle_{AB_1\cdots B_{N-1}}$ since $\mathcal{C}(|\psi\rangle_{A|B_1\cdots B_{N-1}})=\mathcal{C}_a(|\psi\rangle_{A|B_1\cdots B_{N-1}})$.
4 Tighter Constraints Relate to EoF
Let $H_A$ and $H_B$ be two Hilbert spaces with dimension $m$ and $n$ $(m\leq n)$, respectively.Then the entanglement of formation (EoF)[34-35] is defined as follows:for a pure state $|\psi\rangle_{AB}\in H_A\otimes H_B$, the EoF is given bywhere $\rho_{A}={\rm tr}_B(|\psi\rangle_{AB}\langle\psi|)$ and $\mathcal{S}(\rho)=-{\rm tr}(\rho\log_2\rho)$.For a bipartite mixed state $\rho_{AB}\in H_A\otimes H_B$, the EoF is given by
with the minimum taking over all possible pure state decomposition of $\rho_{AB}$.
In Ref.[36], Wootters showed that
$$ E(|\psi\rangle)=f(\mathcal{C}^2(|\psi\rangle)) $$
for $2\otimes m \ (m\geq2)$ pure state $|\psi\rangle$, and $E(\rho)=f(\mathcal{C}^2(\rho))$ for two-qubit mixed state $\rho$,
where $f(x)=H[({1+\sqrt{1-x}})/{2}]$ and $H(x)=-x\log_2x-(1-x)\log_2(1-x)$.$f(x)$ is a monotonically increasing function for $0\leq x\leq1$, and satisfies the following relations:
where $f^{\sqrt{2}}(x^2+y^2)=[f(x^2+y^2)]^{\sqrt{2}}$.
Although EoF does not satisfy the inequality $E_{AB}$ $+E_{AC}\leq E_{A|BC}$,[37] the authors in Ref.[38] showed that EoF is a monotonic function satisfying
$$ E^2(\rho_{A|B_1B_2\cdots B_{N-1}})\geq \sum_{i=1}^{N-1}E^2(\rho_{AB_i}). $$
For $N$-qubit systems, one has[21]
for $\beta\geq\sqrt{2}$, where $E_{A|B_1B_2\cdots B_{N-1}}$ is the EoF of $\rho$ under bipartite partition $A|B_1B_2\cdots B_{N-1}$,
and $E_{AB_i}$ is the EoF of the mixed state $\rho_{AB_i}={\rm tr}_{B_1\cdots B_{i-1},B_{i+1}\cdots B_{N-1}}(\rho)$
for $i=1,2,\ldots,N-1$.Recently, the authors in Ref.[23] proposed a monogamy relation that is tighter than the inequality (23),
if $\mathcal{C}_{AB_i}\geq\mathcal{C}_{A|B_{j+1}\cdots B_{N-1}}$ for $i=1,2,\ldots,m$, and $\mathcal{C}_{AB_j}\leq\mathcal{C}_{A|B_{j+1}\cdots B_{N-1}}$ for $j=m+1,\ldots,N-2$, $\forall 1\leq m\leq N-3$, $N\geq4$ for $\beta\geq\sqrt{2}$.The inequality (24) is also improved to
under the same conditions as that of inequality (24).
In fact, these inequalities can be further improved to even tighter monogamy relations.
Theorem 5 Suppose $k$ is a real number satisfying $0<k\leq 1$.For any $N$-qubit mixed state $\rho_{AB_1\cdots B_{N-1}}$,if $kE^{\sqrt{2}}_{AB_i}\geq E^{\sqrt{2}}_{A|B_{i+1}\cdots B_{N-1}}$ for $i=1,2,\ldots,m$, and $E^{\sqrt{2}}_{AB_j}\leq kE^{\sqrt{2}}_{A|B_{j+1}\cdots B_{N-1}}$ for $j=m+1,\ldots,N-2$, $\forall 1\leq m\leq N-3$, $N\geq4$,the entanglement of formation $E(\rho)$ satisfies
for $\beta\geq\sqrt{2}$, where $t={\beta}/{\sqrt{2}}$.
Proof
For $\beta\geq\sqrt{2}$ and $k f^{\sqrt{2}}(x^2)\geq f^{\sqrt{2}}(y^2)$, we find
where the first inequality is due to the inequality (22), and the second inequality can be obtained from inequality (8).
Let $\rho=\sum_ip_i|\psi_i\rangle\langle\psi_i|\in H_A\otimes H_{B_1}\otimes\cdots H_{B_{N-1}}$ be the optimal decomposition of
$E_{A|B_1B_2\cdots B_{N-1}}(\rho)$ for the $N$-qubit mixed state $\rho$. Then[22]
Thus,
where the first inequality holds due to Eq.(28), the second inequality is similar to the proof of Theorem 1 by using inequality (27),and the last equality holds since for any $2\otimes 2$ quantum state $\rho_{AB_i}$, $E(\rho_{AB_i})=f[\mathcal{C}^2(\rho_{AB_i})]$. $\Box$
Similar to the case of concurrence, we have also the following tighter monogamy relation for EoF.
Theorem 6 Suppose $k$ is a real number satisfying $0<k\leq 1$.
For an $N$-qubit mixed state $\rho_{AB_1\cdots B_{N-1}}$,if $kE^{\sqrt{2}}_{AB_i}\geq E^{\sqrt{2}}_{A|B_{i+1}\cdots B_{N-1}}$ for all $i=1,2,\ldots,N-2$,we have
for $\beta\geq\sqrt{2}$ and $t={\beta}/{\sqrt{2}}$.As ${[(1+k)^t-1}]/{k^t}\geq2^t-1$ for $t\geq1$ and $0< k\leq 1$,our new monogamy relations (26) and (30) are tighter than the ones given in Refs.[21-23].Also, for $0< k\leq 1$ and $\beta\geq2$, the smaller $k$ is, the tighter inequalities (26) and (30) are.
Example 2 Let us again consider the three-qubit state $|\psi\rangle_{ABC}$ defined in Eq.(17)
with $\lambda_0=\lambda_3={1}/{2}$, $\lambda_2={\sqrt{2}}/{2}$ and $\lambda_1=\lambda_4=0$.Then $E_{A|BC}=2-\log_{2}3\approx0.811\,278$,
$$ E_{AB}=-\frac{2+\sqrt{2}}{4}\log_2\frac{2+\sqrt{2}}{4} \\\ -\frac{2-\sqrt{2}}{4}\log_2\frac{2-\sqrt{2}}{4}\approx0.600\,876 \\\ E_{AB}= -\frac{2+\sqrt{3}}{4}\log_2\frac{2+\sqrt{3}}{4} \\\ -\frac{2-\sqrt{3}}{4}\log_2\frac{2-\sqrt{3}}{4}\approx0.354\,579. $$
Thus,
$$ E^\beta_{AB}+\big(2^{\beta/2}-1\big)E^\beta_{AC} \\\ \quad =(0.600876)^\beta+\big(2^{\beta/2}-1\big)0.354579^\beta, \\\ E^\beta_{AB}+\frac{1.5^{\beta/2}-1}{0.5^{\beta/2}}E^\beta_{AC} \\\ \quad = (0.600876)^\beta+\frac{1.5^{\beta/2}-1}{0.5^{\beta/2}}0.354579^\beta,\\\ E^\beta_{AB}+\frac{1.7^{\beta/2}-1}{0.7^{\beta/2}}E^\beta_{AC} \\\ \quad = (0.600876)^\beta+\frac{1.7^{\beta/2}-1}{0.7^{\beta/2}}0.354579^\beta,\\\ E^\beta_{AB}+\frac{1.9^{\beta/2}-1}{0.9^{\beta/2}}E^\beta_{AC} \\\ \quad = (0.600876)^\beta+\frac{1.9^{\beta/2}-1}{0.9^{\beta/2}}0.354579^\beta.$$
One can see that our result is better than the one in Ref.[22] for $\beta\geq\sqrt{2}$, hencee better than the ones in Refs.[21, 23], see Fig.2.
New window|Download| PPT slideFig.2(Color online) The $y$ axis is the lower bound of the EoF $E^\beta_{A|BC}$. The red (green resp. blue) line represents the lower bound from our result for $k=0.5$ ($k=0.7$ resp. $k=0.9$), and the yellow line represents the lower bound from the result in Ref.[22].
We can also provide tighter polygamy relations for the entanglement of assistance.The entanglement of assistance (EoA) of $\rho_{AB}$ is defined as[39]
with the maximization taking over all possible pure decompositions of $\rho_{AB}$.For any dimensional multipartite quantum state $\rho_{AB_1B_2\cdots B_{N-1}}$,a general polygamy inequality of multipartite quantum entanglement was established as,[18]
Using the same approach as for concurrence, we have the following Theorems.
Theorem 7 Suppose $k$ is a real number satisfying $0<k\leq 1$.For any $N$-qubit mixed state $\rho_{AB_1\cdots B_{N-1}}$,if $kE_{AB_i}^a\geq E_{A|B_{i+1}\cdots B_{N-1}}^a$ for $i=1,2,\ldots,m$, and $E_{AB_j}^a\leq kE_{A|B_{j+1}\cdots B_{N-1}}^a$ for $j=m+1,\ldots,N-2$, $\forall 1\leq m\leq N-3$, $N\geq4$, we have
for $0\leq\beta\leq1$.
Theorem 8 Suppose $k$ is a real number satisfying $0<k\leq 1$.For any $N$-qubit mixed state $\rho_{AB_1\cdots B_{N-1}}$,if $kE_{AB_i}^a\geq E_{A|B_{i+1}\cdots B_{N-1}}^a$ for all $i=1,2,\ldots,N-2$,we have
for $0\leq\beta\leq1$.
4 Tighter Constraints Related to Negativity
The negativity, a well-known quantifier of bipartite entanglement, is defined as $\mathcal{N}(\rho_{AB})=\big(\|\rho_{AB}^{T_A}\|-1\big)/2$,[40] where $\rho_{AB}^{T_A}$ is the partial transposed matrix of $\rho_{AB}$ with respect to the subsystem $A$, and $\|X\|$ denotes the trace norm of $X$, i.e., $\|X\|={\rm tr}\sqrt{XX^{?}}$.For convenient, we use the definition of negativity as $\|\rho_{AB}^{T_A}\|-1$.Particularly, for any bipartite pure state $|\psi\rangle_{AB}$,$\mathcal{N}(|\psi\rangle_{AB})=2\sum_{i<j}\sqrt{\lambda_i\lambda_j}=({\rm tr}\sqrt{\rho_{A}})^2-1$,where $\lambda_i s$ are the eigenvalues of the reduced density matrix $\rho_{A}={\rm tr}_B|\psi\rangle_{AB}\langle\psi|$.The convex-roof extended negativity (CREN) of a mixed state $\rho_{AB}$ is defined bywhere the minimum is taken over all possible pure state decomposition of $\rho_{AB}$.Thus $\mathcal{N}_c(\rho_{AB})=\mathcal{C}(\rho_{AB})$ for any two-qubit mixed state $\rho_{AB}$.The dual to the CREN of a mixed state $\rho_{AB}$ is defined as
with the maximum taking over all possible pure state decomposition of $\rho_{AB}$.Furthermore, $\mathcal{N}^a_c(\rho_{AB})=\mathcal{C}^a(\rho_{AB})$ for any two-qubit mixed state $\rho_{AB}$.[12] Similar to the concurrence and EoF, we have the following Theorems.
Theorem 9 Suppose $k$ is a real number satisfying $0<k\leq 1$.For any $N$-qubit mixed state $\rho_{AB_1\cdots B_{N-1}}$,if $k\mathcal{N}_{cAB_i}^2\geq\mathcal{N}_{cA|B_{i+1}\cdots B_{N-1}}^2$ for $i=1,2,\ldots,m$, and $\mathcal{N}_{cAB_j}^2\leq k\mathcal{N}_{cA|B_{j+1}\cdots B_{N-1}}^2$ for $j=m+1,\ldots,N-2$,$\forall 1\leq m\leq N-3$, $N\geq 4$, then we have
for all $\beta\geq2$.
Theorem 10 Suppose $k$ is a real number satisfying $0<k\leq 1$.Theorem 10 Suppose $k$ is a real number satisfying $0<k\leq 1$.
For any $N$-qubit mixed state $\rho_{AB_1\cdots B_{N-1}}$,if all $k\mathcal{N}_{cAB_i}^2\geq\mathcal{N}_{cA|B_{i+1}\cdots B_{N-1}}^2$ for all $i=1,2,\ldots,N-2$, then
for $\beta\geq2$.
Example 3 Consider the state in Example 1 with $\lambda_0=\lambda_3={1}/{2}$, $\lambda_2={\sqrt{2}}/{2}$, and $\lambda_1=\lambda_4=0$.
We have $\mathcal{N}_{cA|BC}={\sqrt{3}}/{2}$, $\mathcal{N}_{cAB}={\sqrt{2}}/{2}$, and $\mathcal{C}_{cAC}={1}/{2}$.Then
$$ \mathcal{N}_{cAB}^\beta+\big(2^{\beta/2}-1\big)\mathcal{N}_{cAC}^\beta=\Big(\frac{\sqrt{2}}{2}\Big)^\beta+\big(2^{\beta/2}-1\big)\Big(\frac{1}{2}\Big)^\beta,\\\ \mathcal{N}_{cAB}^\beta+\frac{(1+k)^{\beta/2}-1}{k^{{\beta}/{2}}}\mathcal{N}_{cAC}^\beta \\\ \quad =\Big(\frac{\sqrt{2}}{2}\Big)^\beta+ \frac{(1+k)^{\beta/2}-1}{k^{{\beta}/{2}}}\Big(\frac{1}{2}\Big)^\beta.$$
One can see that our result is better than the one in Ref.[22] for $\beta\geq2$, thus also better than the ones in Refs.[21, 23], see Fig.3.
New window|Download| PPT slideFig.3(Color online) The $y$ axis is the lower bound of the negativity $\mathcal{N}_c(|\psi\rangle_{A|BC})$, which are functions of $\beta$.
The red (green) line represents the lower bound from our result for $k=0.6$ ($k=0.8$), and the blue line represents the lower bound from the result in Ref.[22].
For the negativity of assistance $\mathcal{N}_c^a$, we have the following results.
Theorem 11 Suppose $k$ is a real number satisfying $0<k\leq 1$.For an $N$-qubit pure state $|\psi\rangle_{A|B_1\cdots B_{N-1}}$,if $k(\mathcal{N}^a_{cA|B_i})^2\geq(\mathcal{N}^a_{cA|B_{i+1}\cdots B_{N-1}})^2$ for $i=1,2,\ldots,m$, and $(\mathcal{N}^a_{cAB_j})^2\leq k(\mathcal{N}^a_{cA|B_{j+1}\cdots B_{N-1}})^2$ for $j=m+1,\ldots,N-2$, $\forall 1\leq m\leq N-3$, $N\geq 4$, then we have
for all $0\leq\beta\leq2$.
Theorem 12 Suppose $k$ is a real number satisfying $0<k\leq 1$. For any $N$-qubit mixed state $|\psi\rangle_{AB_1\cdots B_{N-1}}$,
if $k(\mathcal{N}^a_{cAB_i})^2\geq(\mathcal{N}^a_{cA|B_{i+1}\cdots B_{N-1}})^2$ for all $i=1,2,\ldots,N-2$, then
for $0\leq\beta\leq2$.
5 Tighter Monogamy Relations for Tsallis-$q$ Entanglement and Rényi-$\alpha$ Entanglement
In this section, we study the Tsallis-$q$ entanglement and Rényi-$\alpha$ entanglement,and establish the corresponding monogamy and polygamy relations for the two entanglement measures, respectively.5.1 Tighter Monogamy and Polygamy Relations for Tsallis-$q$ Entanglement}
The Tsallis-$q$ entanglement of a bipartite pure state $|\psi\rangle_{AB}$ is defined as[14]where $q>0$ and $q\neq1$.
For the case $q$ tends to 1, $T_q(\rho)$ is just the von Neumann entropy, $\lim\limits_{q\rightarrow 1}T_q(\rho)=-{\rm tr}\rho\log_2\rho=S(\rho)$.The Tsallis-$q$ entanglement of a bipartite mixed state $\rho_{AB}$ is given by
$$ T_q(\rho_{AB})=\min\limits_{\{p_i,|\psi_i\rangle\}}\sum_{i}p_iT_q(|\psi_i\rangle) $$
with the minimum taken over all possible pure state decompositions of $\rho_{AB}$.
For $({5-\sqrt{13}})/{2}\leq q\leq({5+\sqrt{13}})/{2}$, Yuan et al. proposed an analytic relationship
between the Tsallis-$q$ entanglement and concurrence,
where
with $0\leq x\leq1$.[41] It has also been proved that $T_q(|\psi\rangle)=g_q(\mathcal{C}^2(|\psi\rangle))$ if $|\psi\rangle$ is a $2\otimes m$ pure state,and $T_q(\rho)=g_q(\mathcal{C}^2(\rho))$ if $\rho$ is a two-qubit mixed state.Hence, Eq.(42) holds for any $q$ such that $g_q(x)$ in Eq.(43) is monotonically increasing and convex.Particularly, one has that
for $2\leq q\leq 3$.
In Ref.[14], Kim provided a monogamy relation for the Tsallis-$q$ entanglement,
where $i=1,2,\ldots,N-1$ and $2\leq q\leq 3$. Later, this relation was improved as follows:if $\mathcal{C}_{AB_i}\geq\mathcal{C}_{A|B_{i+1}\cdots B_{N-1}}$ for $i=1,2,\ldots,m$, and $\mathcal{C}_{AB_j}\leq\mathcal{C}_{A|B_{j+1}\cdots B_{N-1}}$ for $j=m+1,\ldots,N-2$, $\forall 1\leq m\leq N-3$, $N\geq4$, then
where $\beta\geq1$ and $T^\beta_{qA|B_1B_2\cdots B_{N-1}}$ quantifies the Tsallis-$q$ entanglement under partition $A|B_1B_2\cdots B_{N-1}$, and $T^\beta_{qA|B_i}$ quantifies that of the two-qubit subsystem $AB_i$ with $2\leq q\leq3$.Moreover, for $({5-\sqrt{13}})/{2}\leq q\leq({5+\sqrt{13}})/{2}$, one has
We now provide monogamy relations which are tighter than Eqs.(45) and (46).
Theorem 13 Suppose $k$ is a real number satisfying $0<k\leq 1$.For an arbitrary $N$-qubit mixed state $\rho_{AB_1B_2\cdots B_{N-1}}$, if $kT_{qAB_i}\geq T_{qA|B_{i+1}\cdots B_{N-1}}$ for $i=1,2,\ldots,m$, and $T_{qAB_j}\leq kT_{qA|B_{j+1}\cdots B_{N-1}}$ for $j=m+1,\ldots,N-2$, $\forall 1\leq m\leq N-3$, $N\geq4$,then we have
for $\beta\geq1$ and $2\leq q\leq3$.
Theorem 14 Suppose $k$ is a real number satisfying $0<k\leq 1$.For any $N$-qubit mixed state $\rho_{AB_1\cdots B_{N-1}}$,
if all $kT_{qAB_i}\geq T_{qA|B_{i+1}\cdots B_{N-1}}$ for $i=1,2,\ldots,N-2$, then we have
for $\beta\geq1$ and $2\leq q\leq3$.
Example 4 Consider the quantum state given in Example 1 with $\lambda_0=\lambda_3={1}/{2}$, $\lambda_2={\sqrt{2}}/{2}$, and $\lambda_1=\lambda_4=0$.For $q=2$, one has $T_{2A|BC}={3}/{8}$, $T_{2AB}={1}/{4}$, and $T_{2AC}={1}/{8}$.Then
$$ T_{2AB}^\beta+\big(2^{\beta}-1\big)T_{2AC}^\beta =\Big(\frac{1}{4}\Big)^\beta+(2^{\beta}-1)\Big(\frac{1}{8}\Big)^\beta,\\\ T_{2AB}^{\beta}+\frac{(1+k)^\beta-1}{k^{\beta}}T_{2AC}^{\beta} \\\ \quad =\Big(\frac{1}{4}\Big)^{\beta} + \frac{(1+k)^\beta-1}{k^{\beta}}\Big(\frac{1}{8}\Big)^\beta.$$
It can be seen that our result is better than the one in Ref.[22] for $\beta\geq1$, and also better than the ones given in Refs.[21, 23], see Fig.4.
New window|Download| PPT slideFig.4(Color online) The $y$ axis is the lower bound of the Tsallis-$q$ entanglement $T_q^\beta(|\psi\rangle_{A|BC})$. The red (green resp. yellow) line represents the lower bound from our result for $k=0.5$ ($k=0.7$ resp. $k=0.9$), and the blue line represents the lower bound from the result in Ref.[22].
As a dual quantity to Tsallis-$q$ entanglement, the Tsallis-$q$ entanglement of assistance (TEoA) is defined by Ref.[14],
$$ T_q^a(\rho_{AB})=\max\limits_{\{p_i,|\psi_i\rangle\}}\sum\limits_ip_iT_q(|\psi_i\rangle),$$
where the maximum is taken over all possible pure state decompositions of $\rho_{AB}$.If $1\leq q\leq 2$ or $3\leq q\leq4$, the function $g_q$ defined in Eq.(43) satisfies
which leads to the Tsallis polygamy inequality
for any multi-qubit state $\rho_{A|B_1B_2\cdots B_{N-1}}$.[25] Here we provide tighter polygamy relations related to Tsallis-$q$ entanglement.We have the following results.
Theorem 15 Suppose $k$ is a real number satisfying $0<k\leq 1$.For any $N$-qubit mixed state $\rho_{AB_1\cdots B_{N-1}}$,
if $kT^a_{qAB_i}\geq T^a_{qA|B_{i+1}\cdots B_{N-1}}$ for $i=1,2,\ldots,m$, and $T^a_{qAB_j}\leq kT^a_{qA|B_{j+1}\cdots B_{N-1}}$ for $j=m+1,\ldots,N-2$, $\forall 1\leq m\leq N-3$, $N\geq4$, then
for $0\leq\beta\leq1$ with $1\leq q\leq2$ or $3\leq q\leq4$.
Theorem 16 Suppose $k$ is a real number satisfying $0<k\leq 1$.For any $N$-qubit mixed state $\rho_{AB_1\cdots B_{N-1}}$, if $kT^a_{qAB_i}\geq T^a_{qA|B_{i+1}\cdots B_{N-1}}$ for all $i=1,2,\ldots,N-2$,we have
for $0\leq\beta\leq1$ with $1\leq q\leq2$ or $3\leq q\leq4$.
5.2 Tighter Monogamy and Polygamy Relations for Rényi-$\alpha$ Entanglement
For a bipartite pure state $|\psi\rangle_{AB}$, the Rényi-$\alpha$ entanglement is defined as[42] $E(|\psi\rangle_{AB})=S_\alpha(\rho_{A})$,where $S_\alpha(\rho)=[{1}/({1-\alpha})]\log_2{\rm tr}\rho^\alpha$ for any $\alpha>0$ and $\alpha\neq1$, and $\lim\limits_{\alpha\rightarrow 1}S_\alpha(\rho)=S(\rho)=-{\rm tr}\rho\log_2\rho$.For a bipartite mixed state $\rho_{AB}$, the Rényi-$\alpha$ entanglement is given bywhere the minimum is taken over all possible pure-state decompositions of $\rho_{AB}$.For each $\alpha>0$, one has
$$ E_\alpha(\rho_{AB})=f_\alpha(\mathcal{C}(\rho_{AB})),$$
where
$$ f_\alpha(x)=\frac{1}{1-\alpha}\log\Big[\Big(\frac{1-\sqrt{1-x^2}}{2}\Big)^2 +\Big(\frac{1+\sqrt{1-x^2}}{2}\Big)^2\Big] $$
is a monotonically increasing and convex function.[24] For $\alpha\geq2$ and any $n$-qubit state $\rho_{A|B_1B_2\cdots B_{N-1}}$, one has[14]
We propose the following two monogamy relations for the Rényi-$\alpha$ entanglement, which are tighter than the previous results.
Theorem 17 Suppose $k$ is a real number satisfying $0<k\leq 1$. For an arbitrary $N$-qubit mixed state $\rho_{AB_1B_2\cdots B_{N-1}}$, if $kE_{\alpha AB_i}\geq E_{\alpha A|B_{i+1}\cdots B_{N-1}}$ for $i=1,2,\ldots,m$, and $E_{\alpha AB_j}\leq kT_{\alpha A|B_{j+1}\cdots B_{N-1}}$ for $j=m+1,\ldots,N-2$, $\forall 1\leq m\leq N-3$, $N\geq4$,then
for $\beta\geq1$ and $\alpha\geq2$.
Theorem 18 Suppose $k$ is a real number satisfying $0<k\leq 1$.For an arbitrary $N$-qubit mixed state $\rho_{AB_1B_2\cdots B_{N-1}}$,if $kE_{\alpha AB_i}\geq E_{\alpha A|B_{i+1}\cdots B_{N-1}}$ for all $i=1,2,\ldots,N-2$, then
for $\beta\geq1$ and $\alpha\geq2$.
Example 5 Consider again the state given in Example 1 with $\lambda_0=\lambda_3={1}/{2}$, $\lambda_2={\sqrt{2}}/{2}$, and $\lambda_1=\lambda_4=0$. For $\alpha=2$, we find $E_{2A|BC}=\log_2({8}/{5})\approx678\,072$, $E_{2AB}=\log_2({8}/{7})\approx0.415\,037$, and $E_{2AC}=\log_2({4}/{3})\approx0.192\,645$.Then
$$ E_{2AB}^\alpha+E_{2AC}^\alpha=0.415037^\alpha+0.192645^\alpha, \\ E_{2AB}^{\alpha}+\frac{(1+k)^\alpha-1}{k^{\alpha}}E_{2AC}^{\alpha} \\ \quad =0.415037^{\alpha}+\frac{(1+k)^\alpha-1}{k^{\alpha}}0.192645^\alpha.$$
One can see that our result is better than the result in Ref.[14], and the smaller $k$ is, the tighter relation is, see Fig.5.
New window|Download| PPT slideFig.5(Color online) The $y$ axis is the lower bound of the Rényi entropy entanglement $E^\beta_{2}(|\psi\rangle_{A|BC})$. The purple line represents the value of $E^\beta_2(|\psi\rangle_{A|BC})$, the red (green resp. yellow) line represents the lower bound from our result for $k=0.5$ ($k=0.7$ resp. $k=0.9$), and the blue line represents the lower bound from the result (55) in Ref.[14].
The Rényi-$\alpha$ entanglement of assistance (REoA),a dual quantity to Rényi-$\alpha$ entanglement, is defined as
$$E_\alpha^a(\rho_{AB})=\max\limits_{\{p_i,|\psi_i\rangle\}}\sum\limits_{i}p_iE_\alpha(|\psi_i\rangle),$$
where the maximum is taken over all possible pure state decompositions of $\rho_{AB}$.For $\alpha\in[({\sqrt{7}-1})/{2},({\sqrt{13}-1})/{2}]$ and any $n$-qubit state $\rho_{AB_1B_2\cdots B_{N-1}}$,a polygamy relation of multi-partite quantum entanglement in terms of REoA has been given by:[20]
We improve this inequality to be a tighter ones under some netural conditions.
Theorem 19 Suppose $k$ is a real number satisfying $0<k\leq 1$.For an arbitrary $N$-qubit mixed state $\rho_{AB_1B_2\cdots B_{N-1}}$, if $kE^a_{\alpha AB_i}\geq E^a_{\alpha A|B_{i+1}\cdots B_{N-1}}$ for $i=1,2,\ldots,m$, and $E^a_{\alpha AB_j}\leq kE^a_{\alpha A|B_{j+1}\cdots B_{N-1}}$ for $j=m+1,\ldots,N-2$, $\forall 1\leq m\leq N-3$, $N\geq4$,then
for $0\leq\beta\leq1$ with $({\sqrt{7}-1})/{2}\leq\alpha\leq({\sqrt{13}-1})/{2}$.
Theorem 20 Suppose $k$ is a real number satisfying $0<k\leq 1$.For an arbitrary $N$-qubit mixed state $\rho_{AB_1B_2\cdots B_{N-1}}$,if $kE^a_{\alpha AB_i}\geq E^a_{\alpha A|B_{i+1}\cdots B_{N-1}}$ for all $i=1,2,\ldots,N-2$, then
for $0\leq\beta\leq1$, with $({\sqrt{7}-1})/{2}\leq\alpha\leq({\sqrt{13}-1})/{2}$.
6 Conclusion
Both entanglement monogamy and polygamy are fundamental properties of multipartite entangled states.We have presented monogamy relations related to the $\beta$th power of concurrence, entanglement of formation,negativity, Tsallis-$q$ and Rényi-$\alpha$ entanglement.We also provide polygamy relations related to these entanglement measures.All the relations we presented in this paper are tighter than the previous results.These tighter monogamy and polygamy inequalities can also provide finer characterizations of the entanglement distributions among the multiqubit systems.Our results provide a rich reference for future work on the study of multiparty quantum entanglement.And our approaches are also useful for further study on the monogamy and polygamy properties related to measures of other quantum correlations and quantum coherence.[43-44]Reference By original order
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