# Monogamy of entanglement for tripartite systems

Xue-Na Zhu1 Gui Bao1 Zhi-Xiang Jin2 Shao-Ming Fei3 1School of Mathematics and
Statistics Science, Ludong University, Yantai 264025, China
2School of Computer Science and Technology, Dongguan University of Technology,
Dongguan, 523808, China
3School of Mathematical Sciences, Capital Normal University, Beijing 100048,
China

###### Abstract

We study the monogamy of arbitrary quantum entanglement measures $E$ for
tripartite quantum systems. Both sufficient and necessary conditions for $E$
to be monogamous in terms of the $\alpha$th power of $E$ are explicitly
derived. It is shown that such monogamy of a entanglement measure $E$ only
depends on the boundedness of the solution set of certain equations. Moreover,
the monogamy conditions have been also obtained with respect to certain
subsets of quantum states for a given quantum correlation. Detailed examples
are given to illustrate our results.

##  I Introduction

Quantum entanglement Q1  among the subsystems of a multipartite system play
significant roles in many information processing tasks. A fundamental property
of quantum entanglement is the monogamy. The monogamy relations give rise to
the distributions of quantum entanglement in the multipartite setting.
Monogamy is also an essential feature allowing for security in quantum key
distribution. For a tripartite system $A$, $B$ and $C$ associated with finite
dimensional Hilbert spaces $H_{A}$, $H_{B}$ and $H_{C}$, respectively, the
monogamy of an quantum entanglement measure $E$ implies that the correlation
$E(\rho_{A|BC})$ between $A$ and $BC$ satisfies m1

$E(\rho_{A|BC})\geq E(\rho_{AB})+E(\rho_{AC})$ (1)

for any $\rho_{ABC}\in H_{A}\otimes H_{B}\otimes H_{C}$, where
$\rho_{AB}=Tr_{C}(\rho_{ABC})$ and $\rho_{AC}=Tr_{B}(\rho_{ABC})$ are the
corresponding reduced bipartite states. An important issue is to determine
whether a given entanglement measure $E$ is monogamous or not. Considerable
efforts have been devoted to this problem E1 ; E2 ; E3 ; E4 ; PRL.117.060501 ;
G1 ; G2 ; G3 ; JIn  for quantum correlations such as entanglement of formation
(EOF) E3 , concurrence C1 ; C2 ; C3 , negativity N1 ; N2 ; N3  and concurrence
of assistance zhu2 .

If $E$ does not satisfy the relation (1), it is still possible to find a
positive power $\alpha\in(0,+\infty)$ such that $E^{\alpha}$ satisfies the
relation (1) for $\alpha>0$ G1 ,

$E^{\alpha}(\rho_{A|BC})\geq E^{\alpha}(\rho_{AB})+E^{\alpha}(\rho_{AC}).$ (2)

The $\alpha$th $(\alpha\geq 2)$ power of concurrence and the $\alpha$th
$(\alpha\geq\sqrt{2})$ power entanglement of formation for $n$-qubit states
satisfy the relation (2) zhu1 .

In Ref. PRL.117.060501  the authors show that $E$ is monogamous if there
exists a nontrivial continuous function $f$ such that the following
generalized monogamy relation

$E(\rho_{A|BC})\geq f(E(\rho_{AB}),E(\rho_{AC})),$ (3)

is satisfied for states $\rho_{ABC}$. In Ref. JIn  the authors presented a new
parameterized monogamy relation of entanglement with equality,

$E(\rho_{A|BC})=\mu E(\rho_{AB})+E(\rho_{AC}),$ (4)

where $\mu>0$ and $E(\rho_{AB})\leq E(\rho_{AC})$. Monogamy relations
satisfied by the $\alpha$th power of entanglement measures are presented based
on Eq. (4) Ref. JIn .

In this paper, we investigate the monogamy properties of arbitrary quantum
entanglement measures for tripartite quantum systems. We derive explicitly
both the sufficient and necessary conditions for $E$ to be monogamous in terms
of the $\alpha$th power of $E$, with respect to either all the quantum states.
And, the monogamy conditions have been also obtained with respect to certain
subsets of quantum states for quantum correlations.

##  II  Monogamy relations for tripartite systems

In the following we say a quantum entanglement measure $E$ is
$\alpha$-monogamous if there exists a real number $\alpha>0$ such that $E$
satisfies (2), and $E$ is non-monogamous if there is at least one state
$\rho_{ABC}\in H_{A}\otimes H_{B}\otimes H_{C}$ such that (2) is not satisfied
for any given $\alpha$.

With respect to each quantum state $\rho_{ABC}\in H_{A}\otimes H_{B}\otimes
H_{C}$, we define a state dependent parameter $x_{\rho_{ABC}}$ such that the
following equation is satisfied,

$x_{\rho_{ABC}}\left(E^{y}(\rho_{A|BC})-\max\\{E^{y}(\rho_{AB}),E^{y}(\rho_{AC})\\}\right)=\min\\{E^{y}(\rho_{AB}),E^{y}(\rho_{AC})\\},$
(5)

where $y$ is a positive number. If Eq.(5) has non-zero solution
$x_{\rho_{ABC}}$ for each $\rho_{ABC}$ with $y=1$, then Eq. (5) becomes
Eq.(4), i.e., $x_{\rho_{ABC}}=\frac{1}{\mu}$. Hereafter, we take
$x_{\rho_{ABC}}=0$ for any $y$ when
$E(\rho_{A|BC})=\max\\{E(\rho_{AB}),E(\rho_{AC})$ and
$\min\\{E(\rho_{AB}),E(\rho_{AC})\\}=0$. For simplicity, we denote the set of
parameters $x_{\rho_{ABC}}$ by

$X_{y}=\left\\{x_{\rho_{ABC}}|x_{\rho_{ABC}}\left(E^{y}(\rho_{A|BC})-\max\\{E^{y}(\rho_{AB}),E^{y}(\rho_{AC})\\}\right)=\min\\{E^{y}(\rho_{AB}),E^{y}(\rho_{AC})\\},\rho_{ABC}\in
H_{A}\otimes H_{B}\otimes H_{C}\right\\}.$

###### Theorem 1

A quantum entanglement measure $E$ is $\alpha$-monogamous if and only if there
exists a real number $y_{0}>0$ such that $X_{y_{0}}$ is a bounded set.

[Proof] Since as a measure of quantum entanglement, $E$ does not increase
under local operations and classical communication, we have
$E(\rho_{A|BC})\geq\max\\{E(\rho_{AB}),E(\rho_{AC})\\}$. Without loss of
generality, we assume $E(\rho_{AB})\geq E(\rho_{AC})$ (The case for
$E_{AC}\geq E_{AB}$ is similarly treated).

$(\Rightarrow)$ If $E$ is $\alpha$-monogamous, then $E$ satisfies (2) for any
$\rho_{ABC}\in H_{A}\otimes H_{B}\otimes H_{C}$.

(Case 1) For a given state $\rho_{ABC}$, if $E(\rho_{A|BC})>E(\rho_{AB})$, we
take $y=\alpha$, then
$x_{\rho_{ABC}}=\frac{E^{\alpha}(\rho_{AC})}{E^{\alpha}(\rho_{A|BC})-E^{\alpha}(\rho_{AB})}.$
From (2) we have $x_{\rho_{ABC}}\leq 1.$

(Case 2) If $E(\rho_{A|BC})=E(\rho_{AB})$, since $E^{\alpha}_{A|BC}\geq
E^{\alpha}_{AB}+E^{\alpha}_{AC}$, we have $E_{AC}=0$. The solution of Eq. (5)
is $x_{\rho_{ABC}}=0$ for any $y$, especially for $y=\alpha$.

From Case 1 and Case 2 we have $x_{\rho_{ABC}}\in[0,1]$ for each
$x_{\rho_{ABC}}\in X_{\alpha}$. Therefore, $X_{\alpha}$ is a bounded set.

$(\Leftarrow)$ If $X_{y_{0}}$ is a bounded set, then there exists a number
$M>0$ such that $0\leq x_{\rho_{ABC}}\leq M$ for any $x_{\rho_{ABC}}\in
X_{y_{0}}$.

(Case 1) If $x_{\rho_{ABC}}>0$, we have $E(\rho_{A|BC})>E(\rho_{AB})\geq
E(\rho_{AC})>0$. Let
$g(\beta)=E^{\beta}(\rho_{A|BC})-E^{\beta}(\rho_{AB})-E^{\beta}(\rho_{AC})$.
We get

$\displaystyle g(\beta)$
$\displaystyle=E^{\beta}(\rho_{A|BC})-E^{\beta}(\rho_{AB})-E^{\beta}(\rho_{AC})$
$\displaystyle=\left(E^{y_{0}}(\rho_{AB})+\frac{1}{x_{\rho_{ABC}}}E^{y_{0}}(\rho_{AC})\right)^{\frac{\beta}{y_{0}}}-E^{\beta}(\rho_{AB})-E^{\beta}(\rho_{AC})$
$\displaystyle=E^{\beta}(\rho_{AB})\left(1+\frac{1}{x_{\rho_{ABC}}}\frac{E^{y_{0}}(\rho_{AC})}{E^{y_{0}}(\rho_{AB})}\right)^{\frac{\beta}{y_{0}}}-E^{\beta}(\rho_{AB})-E^{\beta}(\rho_{AC})$
$\displaystyle\geq
E^{\beta}(\rho_{AB})\left(1+\frac{\beta}{x_{\rho_{ABC}}y_{0}}\frac{E^{y_{0}}(\rho_{AC})}{E^{y_{0}}(\rho_{AB})}\right)-E^{\beta}(\rho_{AB})-E^{\beta}(\rho_{AC})$
$\displaystyle=E^{y_{0}}(\rho_{AC})\left(\frac{\beta}{x_{\rho_{ABC}}y_{0}}E^{\beta-
y_{0}}(\rho_{AB})-E^{\beta-y_{0}}(\rho_{AC})\right),$

where the second equality is due to (5) and the first inequality is due to
that $(1+t)^{n}\geq 1+nt$ for $n\geq 1$ and $t\geq 0$. Setting
$\beta=\max\\{My_{0},y_{0}\\}$, we obtain $g(\beta)\geq 0$ for the given state
$\rho_{ABC}$ since $0<x_{\rho_{ABC}}\leq M$ and $E(\rho_{AB})\geq
E(\rho_{AC})$.

(Case 2) If $x_{\rho_{ABC}}=0$ for the quantum state $\rho_{ABC}$, one has
$E(\rho_{AC})=0$. Then $E^{\alpha}(\rho_{A|BC})\geq
E^{\alpha}(\rho_{AB})+E^{\alpha}(\rho_{AC})$ for any $\alpha>0$.

In conclusion, if there exists a real number $y_{0}$ such that $X_{y_{0}}$ is
a bounded set, we take
$\alpha=\max_{x_{\rho_{ABC}}>0}\\{\beta\\}=\max\\{My_{0},y_{0}\\}$. Then
$g(\alpha)\geq 0$, i.e., $E^{\alpha}(\rho_{A|BC})\geq
E^{\alpha}(\rho_{AB})+E^{\alpha}(\rho_{AC})$ for each $\rho_{ABC}\in
H_{A}\otimes H_{B}\otimes H_{C}.$

In fact, from the proof of the Theorem 1, we have $g(\beta)\geq 0$ for all
$\beta\geq\alpha.$ Namely, if $E$ is $l$-monogamous, $E$ must be
$L$-monogamous with $L\geq l$. From Theorem 1 we can conclude that the
monogamy of $E$ is determined by $X_{y}$.

As applications let us consider the concurrence and three-qubit systems. For
any arbitrary bipartite pure state $|\psi\rangle_{AB}$, the concurrence is
given by $C(|\psi\rangle_{AB})=\sqrt{1-Tr(\rho^{2}_{A})}$, where
$\rho_{A}=Tr_{B}(|\psi\rangle_{AB}\langle\psi|)$. We first consider the
solution of Eq. (5) for three-qubit with $E=C$ and $y=2$. Any three-qubit
state $|\psi\rangle_{ABC}$ can be written in the generalized Schmidt
decomposition gx ,

$\begin{array}[]{rcl}|\psi\rangle_{ABC}&=&\lambda_{0}|000\rangle+\lambda_{1}e^{i\varphi}|100\rangle+\lambda_{2}|101\rangle+\lambda_{3}|110\rangle+\lambda_{4}|111\rangle,\end{array}$
(6)

where $\lambda_{i}\geq 0$, $i=0,...,4$, and $\sum_{i=0}^{4}\lambda_{i}^{2}=1$.
We have
$C_{A|BC}=2\lambda_{0}\sqrt{\lambda^{2}_{2}+\lambda^{2}_{3}+\lambda^{2}_{4}},$
$C_{AB}=2\lambda_{0}\lambda_{2}$ and $C_{AC}=2\lambda_{0}\lambda_{3}.$

Denote $\lambda=\min\\{\lambda_{2},\lambda_{3}\\}$. Eq. (5) becomes
$x_{|\psi\rangle_{ABC}}\lambda^{2}_{0}(\lambda^{2}+\lambda^{2}_{4})=\lambda^{2}_{0}\lambda^{2}$.
We obtain the following solutions,

$x_{|\psi\rangle_{ABC}}=\left\\{\begin{aligned}
&~{}~{}~{}0,~{}~{}~{}~{}~{}~{}~{}~{}~{}\lambda_{0}=0;\\\
&~{}~{}~{}0,~{}~{}~{}~{}~{}~{}~{}~{}~{}\lambda_{0}\not=0~{}~{}and~{}~{}\lambda=0;\\\
&\frac{\lambda^{2}}{\lambda^{2}+\lambda^{2}_{4}},~{}~{}~{}\lambda_{0}\not=0~{}~{}and~{}~{}\lambda\not=0.\end{aligned}\right.$
(7)

We have
$X_{2}=\\{x_{|\psi\rangle_{ABC}}|x_{|\psi\rangle_{ABC}}\lambda^{2}_{0}(\lambda^{2}+\lambda^{2}_{4})=\lambda^{2}_{0}\lambda^{2}\\}$
and $\max\\{x_{|\psi\rangle_{ABC}}\\}=1$. According the proof of Theorem 1, we
take $M=1$. Then $\alpha=\max\\{2M,2\\}=2$, i.e.,
$C^{2}(|\psi\rangle_{A|BC})\geq C^{2}(\rho_{AB})+C^{2}(\rho_{AC})$ for any
pure states $|\psi\rangle_{ABC}$.

Since the concurrence of any mixed state
$\rho_{AB}=\sum_{i}p_{i}|\psi\rangle_{i}\langle\psi|$ is given by the convex
roof extension,
$C(\rho_{AB})=\min_{\\{p_{i},|\psi\rangle_{i}\\}}\sum_{i}p_{i}C(|\psi\rangle_{i})$,
for three-qubit mixed states $\rho_{A|BC}$, we have also
$C^{2}(\rho_{A|BC})\geq C^{2}(\rho_{AB})+C^{2}(\rho_{AC})$.

In Theorem 1, we established a necessary and sufficient condition for the
monogamy of a quantum entanglement measure $E$. In the following, we consider
the generalized measures of quantum correlation beside entanglement measure
$E$, such as entanglement of assistancega , quantum discord85040102  and
quantum deficit85012103  and so on. Hereafter, we also say $Q$ is
$\alpha$-monogamous if there exists a real number $\alpha>0$ such that $Q$
satisfies (2) and $Q$ is non-monogamous if there is at least one state
$\rho_{ABC}\in H_{A}\otimes H_{B}\otimes H_{C}$ such that (2) is not satisfied
for any given $\alpha$.

Different from the concurrence, the concurrence of assistance $C_{a}$ fails to
be $\alpha$-monogamous in general. For a tripartite state
$|\psi\rangle_{ABC}$, the concurrence of assistance is defined by ca

$C_{a}(|\psi\rangle_{ABC})\equiv
C_{a}(\rho_{AB})=\max_{\\{p_{i},|\psi_{i}\rangle\\}}\sum_{i}p_{i}C(|\psi_{i}\rangle),$
(8)

where the maximum takes over all possible ensemble realizations of
$\rho_{AB}=Tr_{C}(|\psi\rangle_{ABC}\langle\psi|)=\sum_{i}p_{i}|\psi_{i}\rangle_{AB}\langle\psi_{i}|$.
Consider the $2\otimes 2\otimes 2$ pure state $|\psi\rangle_{ABC}$ given in
(6). One has
$C_{a}(\rho_{AB})=2\lambda_{0}\sqrt{\lambda^{2}_{3}+\lambda^{2}_{4}}$ ,
$C_{a}(\rho_{AC})=2\lambda_{0}\sqrt{\lambda^{2}_{2}+\lambda^{2}_{4}}$ and
$C_{a}(|\psi\rangle_{A|BC}\rangle)=2\lambda_{0}\sqrt{\lambda^{2}_{2}+\lambda^{2}_{3}+\lambda^{2}_{4}}$.
In particular, when $\lambda_{0}\not=0,$ $\lambda_{2}=0$ and
$\lambda_{4}\not=0$, i.e.,
$C_{a}(|\psi\rangle_{A|BC}\rangle)=C_{a}(\rho_{AB})>0$ and
$C_{a}(\rho_{AC})>0$, the Eq. (5) has no bounded solution for any $y$. The set
$\Gamma=\left\\{\beta\,(0<\beta<+\infty)|C^{\beta}_{a}(\rho_{A|BC})\geq
C_{a}^{\beta}(\rho_{AB})+C_{a}^{\beta}(\rho_{AC})~{}for~{}all~{}\rho\in
2\otimes 2\otimes 2\right\\}$ is just an empty set.

###### Theorem 2

For a given measure of quantum correlation $Q$ with
$Q(\rho_{A|BC})\geq\max\\{Q(\rho_{AB}),Q(\rho_{AC})\\}$, if there exists a
state $\rho_{ABC}\in H_{A}\otimes H_{B}\otimes H_{C}$ such that
$Q(\rho_{A|BC})=\max\\{Q(\rho_{AB}),Q(\rho_{AC})\\}$ and
$\min\\{Q(\rho_{AB}),Q(\rho_{AC})\\}>0$, then $Q$ is non-monogamous for the
quantum system $ABC$.

[Proof] If $\rho_{ABC}$ satisfies
$Q(\rho_{A|BC})=\max\\{Q(\rho_{AB}),Q(\rho_{AC})\\}$ and
$\min\\{Q(\rho_{AB}),Q(\rho_{AC})\\}>0$, then
$Q^{\alpha}(\rho_{A|BC})<Q^{\alpha}(\rho_{AB})+Q^{\alpha}(\rho_{AC})$ for any
$\alpha>0$.

Therefore, there is no positive number $\alpha$ such that (2) is satisfied.

Example 1 Let us consider the $2\otimes 2\otimes 3$ pure state

$|\psi\rangle_{ABC}=\frac{1}{\sqrt{3}}(|000\rangle+|111\rangle+|\varphi^{+}\rangle|2\rangle),$
(9)

where $|\varphi^{+}\rangle=\frac{1}{\sqrt{2}}(|01\rangle+(|10\rangle))$ is one
of the Bell states. One has zhu3 ,
$C_{a}(|\psi\rangle_{A|BC}\rangle)=C_{a}(\rho_{AB})=1$ and
$C_{a}(\rho_{AC})=\frac{2\sqrt{2}}{3}$. According to Theorem 2, $C_{a}$ is
non-monogamous for the $2\otimes 2\otimes 3$ quantum system.

The above results show that the concurrence of assistance $C_{a}$ is non-
monogamous in general for the $2\otimes 2\otimes n$ ($n\in\\{2,3\\}$) quantum
systems. Nevertheless, for particular states, $C_{a}$ may be
$\alpha$-monogamous. In zhu2  it is shown that $C_{a}$ satisfies the monogamy
relation, $C_{a}^{\beta}(|\psi\rangle_{ABC})\geq
C_{a}^{\beta}(\rho_{AB})+C_{a}^{\beta}(\rho_{AC})$, for the three-qubit
W-class states $|\psi\rangle_{ABC}$ with $\beta\geq 2$. Therefore, with
respect to a given measure of quantum correlation $Q$, there may be exist
$\alpha$, which $Q$ satisfied $Q^{\alpha}(\rho_{A|BC})\geq
Q^{\alpha}(\rho_{AB})+Q^{\alpha}(\rho_{AC})$ for specifically states
$\rho_{ABC}$. We denote

$X_{y,\Omega}=\left\\{x_{\rho_{ABC}}|x_{\rho_{ABC}}\left(Q^{y}(\rho_{A|BC})-\max\\{Q^{y}(\rho_{AB}),Q^{y}(\rho_{AC})\\}\right)=\min\\{Q^{y}(\rho_{AB}),Q^{y}(\rho_{AC})\\},where~{}\rho_{ABC}\in\Omega\right\\}.$

Together with the Theorem 1 and Theorem 2, we have the following result.

###### Corollary 1

Given a given measure of quantum correlation $Q$ and a nonempty set of quantum
states
$\Omega=\left\\{\rho_{ABC}|Q(\rho_{A|BC})\geq\max\\{Q(\rho_{AB}),Q(\rho_{AC})\\}\right\\}$.
$Q$ is $\alpha-$monogamous for $\rho_{ABC}\in\Omega$ if and only if there
exists a real number $y_{0}$ such that $X_{y_{0},\Omega}$ is a bounded set.

The concurrence of assistance $C_{a}$ is an entanglement monotone for
$2\otimes 2\otimes n(n\geq 2)$ pure states72042329 , i,e
$C_{a}(|\psi\rangle_{A|BC})\geq\max\\{C_{a}(\rho_{AB}),C_{a}(\rho_{AC})\\}$
for quantum state $|\psi\rangle_{ABC}\in 2\otimes 2\otimes n(n\geq 2)$. Then
we can study the monogamy of $C_{a}$ for $2\otimes 2\otimes n(n\geq 2)$ pure
states by the corollary 1.

Example 2 Consider the three-qubit W-class states $|\psi\rangle_{ABC}$ zhu2 ,

$|\psi\rangle_{ABC}=b_{0}|000\rangle+b_{1}|100\rangle+b_{2}|010\rangle+b_{3}|001\rangle,$
(10)

where $\sum_{i=0}^{3}|b_{i}|^{2}=1$. Denote
$\Omega=\left\\{\rho_{ABC}|\rho_{ABC}=|\psi\rangle_{ABC}\langle\psi|\right\\}$.
One has $C_{a}(|\psi\rangle_{ABC})=2|b_{1}|\sqrt{|b_{2}|^{2}+|b_{3}|^{2}}$,
$C_{a}(\rho_{AB})=2|b_{1}||b_{2}|$ and $C_{a}(\rho_{AC})=2|b_{1}||b_{3}|$.
From Theorem 1 and Corollary 1 we have

$x_{|\psi\rangle_{ABC}}=\left\\{\begin{aligned}
&~{}~{}~{}0,~{}~{}~{}~{}~{}~{}~{}~{}~{}b_{1}b_{2}b_{3}=0;\\\
&~{}~{}~{}1,~{}~{}~{}~{}~{}~{}~{}~{}~{}b_{1}b_{2}b_{3}\not=0,\end{aligned}\right.$

then $X_{2,\Omega_{0}}=\\{0,1\\}$. Hence $C_{a}$ is $2$-monogamous for
$\rho_{ABC}\in\Omega$.

Theorem 1 and Corollary 1 provide the monogamy of $Q$ based on value of
$x_{\rho_{ABC}}\in X_{y}$ for given $y$. In the following, we provide the $y$
independent monogamy with respect to a given quantum state $\rho_{ABC}$. For
convenience, we denote $\Lambda=\\{\rho_{ABC}|Q(\rho_{A|BC})>Q(\rho_{AB})\geq
Q(\rho_{AC})>0~{}or~{}Q(\rho_{A|BC})>Q(\rho_{AC})\geq Q(\rho_{AB})>0\\}$.

###### Theorem 3

For a given measure of quantum correlation $Q$ and a quantum state
$\rho_{ABC}\in\Lambda$, $Q$ is $\alpha$-monogamous according to (2) with
$\alpha=\log_{b}2$, where
$b=\min\left\\{\frac{Q(\rho_{A|BC})}{Q(\rho_{AC})},\frac{Q(\rho_{A|BC})}{Q(\rho_{AB})}\right\\}$.

[Proof] Without loss of generality, we assume $Q(\rho_{A|BC})>Q(\rho_{AB})\geq
Q(\rho_{AC})>0$. Set $t_{1}=\frac{Q(\rho_{AB})}{Q(\rho_{A|BC})}$,
$t_{2}=\frac{Q(\rho_{AC})}{Q(\rho_{A|BC})}$ and
$k=\frac{Q(\rho_{AB})}{Q(\rho_{AC})}$. It is obvious that $t_{1}\geq t_{2}$,
$k\geq 1$ and $1=t^{y}_{1}+\frac{1}{x}t^{y}_{2}$, where $x$ is the solution of
Eq. (5) with a given $y>0$ for $\rho_{ABC}$. Let $\alpha$ satisfy
$(1+\frac{1}{k^{y}x})^{\frac{\alpha}{y}}=2$, i.e.,
$\alpha=\log_{\frac{Q(\rho_{A|BC})}{Q(\rho_{AB})}}2$. Next, we prove that
$g(\alpha)=Q^{\alpha}(\rho_{A|BC})(1-t_{1}^{\alpha}-t_{2}^{\alpha})\geq 0$ by
proving $(1-t_{1}^{\alpha}-t_{2}^{\alpha})\geq 0$,

$\displaystyle 1-t_{1}^{\alpha}-t_{2}^{\alpha}$
$\displaystyle=(t^{y}_{1}+\frac{1}{x}t^{y}_{2})^{\frac{\alpha}{y}}-t_{1}^{\alpha}-t_{2}^{\alpha}$
$\displaystyle=(k^{y}+\frac{1}{x})^{\frac{\alpha}{y}}t_{2}^{\alpha}-k^{\alpha}t_{2}^{\alpha}-t_{2}^{\alpha}$
$\displaystyle=k^{\alpha}\left((1+\frac{1}{k^{y}x})^{\frac{\alpha}{y}}-1-k^{-\alpha}\right)t_{2}^{\alpha}$
$\displaystyle=(k^{\alpha}-1)t_{2}^{\alpha}\geq 0,$

where the first equation is due to $1=t^{y}_{1}+\frac{1}{x}t^{y}_{2}$ and the
inequality is due to $k\geq 1$ and $\alpha>0$. The case for
$Q(\rho_{A|BC})>Q(\rho_{AC})\geq Q(\rho_{AB})>0$ is similarly proved.
Therefore, $Q$ is $\log_{b}2$-monogamous for the states $\rho_{ABC}$.

The entanglement cost $E_{C}$ of a bipartite state $\rho_{AB}$ is equal to the
regularized entanglement of formation $E_{f}$ EOF , namely m1 ,

$E_{C}(\rho_{AB})=\lim_{n\to\infty}\frac{1}{n}E_{f}(\rho^{\otimes n}_{AB}),$
(11)

where the entanglement of formation $E_{f}$ is defined by
$E_{f}(\rho_{AB})=\min_{\\{p_{i},|\psi_{i}\rangle\\}}\sum_{i}p_{i}S(Tr_{B}[|\psi_{i}\rangle\langle\psi_{i}|]),$
with $S(\rho)$ is the von Neumann entropy of density operator $\rho$ and the
minimum is taken over all ensembles $\\{p_{i},|\psi_{i}\rangle\\}$ satisfying
$\sum_{i}p_{i}|\psi_{i}\rangle\langle\psi_{i}|=\rho_{AB}$. It is worthwhile
noting that the inequality (1) does not hold for the entanglement cost $E_{C}$
m1 .

To illustrate the Theorem 3 we consider the purification of the totally
antisymmetric two-qutrit state,

$|\psi\rangle_{ABC}=\frac{1}{\sqrt{6}}(|123\rangle-|132\rangle+|231\rangle-|213\rangle+|321\rangle-|321\rangle).$
(12)

One has $E_{C}(|\psi\rangle_{A|BC})=\log_{2}3$ and
$E_{C}(\rho_{AB})=E_{C}(\rho_{AB})=1$ m1 . According to Theorem 3, we have
$b=\min\left\\{\frac{E_{C}(|\psi\rangle_{A|BC})}{E_{C}(\rho_{AC})},\frac{E_{C}(|\psi\rangle_{A|BC})}{E_{C}(\rho_{AB})}\right\\}=\log_{2}3$.
Then $\alpha=\log_{(\log_{2}3)}2\approx 1.51.$ Therefore, $E_{C}$ is
$\alpha$-monogamous for $|\psi\rangle_{ABC}$ for $\alpha\geq 1.51$. Denote
$f(\alpha)=E^{\alpha}_{C}(|\psi\rangle_{A|BC})-E^{\alpha}_{C}(\rho_{AB})-E^{\alpha}_{C}(\rho_{AC})$.
We have $f(\alpha)\geq 0$ for $\alpha\geq 1.51$, see the Fig. 1.

Fig. 1: $f(\alpha)$ as a function of $\alpha$ ($\alpha\geq 1.51$) for
$|\psi\rangle_{ABC}$.

On the other hand, for the state $|\psi\rangle_{ABC}$ in (12), using the
Theorem 1 and Corollary 1, we have
$X_{y,|\psi\rangle_{ABC}}=\\{\frac{1}{(\log_{2}3)^{y}-1}\\}$ for given $y>0$.
Then $E_{C}$ is $\beta$-monogamous for $|\psi\rangle_{ABC}$ with
$\beta=\max\\{\frac{1}{(\log_{2}3)^{y}-1}y,y\\}$. Obviously, the optimal value
of $\beta$ is obtained at the intersection of
$z_{1}(y)=\frac{1}{(\log_{2}3)^{y}-1}y$ and $z_{2}(y)=y$ with $y>0$, i.e.,
$\min_{y>0}\\{\beta\\}=\log_{(\log_{2}3)}2$, see Fig. 2.

Fig. 2: The dished line is for $z_{1}(y)$ and the solid line is for
$z_{2}(y)$.

In the proof of the Theorem 3, if we take
$1<c\leq\min\left\\{\frac{Q(\rho_{A|BC})}{Q(\rho_{AC})},\frac{Q(\rho_{A|BC})}{Q(\rho_{AB})}\right\\}$,
then we have that $Q$ is $\log_{c}2$-monogamous for the given quantum state.
For example, let $c=\frac{3}{2}\in(1,\log_{(\log_{2}3)}2]$ for the state (12).
Then $E_{C}$ is $\alpha$-monogamous for $|\psi\rangle_{ABC}$ with
$\alpha=\log_{\frac{3}{2}}2\approx 1.71$. More generally, for some states
$\rho_{ABC}\in\Lambda$, from the proof of Theorem 3 if
$T=\left\\{\frac{Q(\rho_{A|BC})}{Q(\rho_{AC})},\frac{Q(\rho_{A|BC})}{Q(\rho_{AB})}\big{|}\rho_{ABC}\right\\}$
has lower bound $c$ which is strictly greater than $1$, $Q$ is
$\log_{c}2$-monogamous for those states $\rho_{ABC}$.

##  III Conclusions and Remarks

We have addressed in general the question of whether entanglement measures $E$
are monogamous in the sense of (2) introduced in Ref. G1  for tripartite
quantum systems. We have presented the equations (5) and shown that the
monogamy of a entanglement measure $E$ only depends on the boundedness of the
solutions for the equations. Both sufficient and necessary conditions for $E$
being monogamous have been explicitly derived. Meanwhile, the monogamy
conditions for a measure of quantum correlation $Q$ have been also obtained
with respect to certain subsets of quantum states. According to the duality,
our approach may be also applied to study the polygamy of entanglement
measures $E$.

Acknowledgments This work is supported by the National Natural Science
Foundation of China under grant Nos. 12075159 and 12171044, Beijing Natural
Science Foundation (Z190005), and the Academician Innovation Platform of
Hainan Province.

## References

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