Study of a monogamous entanglement measure for three-qubit quantum systems

Abstract

The entanglement quantification and classification of multipartite quantum states is an important research area in quantum information. In this paper, in terms of the reduced density matrices corresponding to all possible partitions of the entire system, a bounded entanglement measure is constructed for arbitrary-dimensional multipartite quantum states. In particular, for three-qubit quantum systems, we prove that our entanglement measure satisfies the relation of monogamy. Furthermore, we present a necessary condition for characterizing maximally entangled states using our entanglement measure.

1 Introduction

Entanglement, as a significant feature of quantum mechanics, plays a vital role in quantum information, such as quantum key distribution, quantum teleportation, quantum dense coding, quantum secret sharing, quantum secure direct communication, quantum simulation and quantum computation [1–14]. Mathematically, a pure state in a quantum system is called entangled if it cannot be factorized into the direct product of states on the subsystems; a mixed state is entangled if it cannot be written as a convex mixture of direct products of local states.

Quantifying entanglement has attracted much attention in recent years. For bipartite system, quantum entanglement measures have been given, such as the von Neumann entropy of entanglement [15], the entanglement of formation [16], concurrence [17] and the negativity [18]. In the case of multipartite states, given that there does not exist a single measure that can successfully account for all possible entanglement characteristics and applications, each measure usually performs better for a specific purpose and is always needed to choose the one that better fits our needs [19–26].

One of the most important properties of entanglement is monogamy [27–37], which quantifies the relation of entanglement between different parties in multipartite setting. Monogamy is also a fascinating characterization related to many areas of physics, such as quantum key distribution [38, 39], the foundations of quantum mechanics [40, 41], statistical mechanics [40], condensed matter physics [42–44] and even black-hole physics [45]. Let E be an entanglement measure for a tripartite system ${\mathcal {H}}_\mathrm{A} \otimes {\mathcal {H}}_\mathrm{B} \otimes {\mathcal {H}}_\mathrm{C}$. If the entanglement of the particles A and BC satisfies the inequality

$$\begin{aligned} E_{\mathrm{A}\mid \mathrm{BC}}\geqslant & {} E_\mathrm{AB} + E_\mathrm{AC}, \end{aligned}$$

we call the entanglement measure E satisfies the monogamous relation. In this paper, we will propose an entanglement measure which itself has the monogamous relation. Moreover, as an application, we use our measure to establish the relation between maximally entangled states and single-qubit reduced states. We give a necessary condition for characterizing maximally entangled states.

We consider throughout this paper an n-partite system ${\mathcal {H}}={\mathcal {H}}^{d_1}\otimes {\mathcal {H}}^{d_2}\otimes \cdots \otimes {\mathcal {H}}^{d_n}$, where the dimension of a local space ${\mathcal {H}}^{d_i}$ is $d_i$ with $i=1,2,\cdots ,n$. A partition ${\mathscr {A}}$ of the system ${\mathcal {H}}={\mathcal {H}}^{d_1}\otimes {\mathcal {H}}^{d_2}\otimes \cdots \otimes {\mathcal {H}}^{d_n}$ is called a k-partition $(2\leqslant k\leqslant n)$ if it contains k disjoint nonempty subsets $\hbox {A}_1, \hbox {A}_2, \ldots , \hbox {A}_k$ such that $\left\{ {\mathcal {H}}^{d_1}, {\mathcal {H}}^{d_2}, \ldots , {\mathcal {H}}^{d_n}\right\} =\hbox {A}_1\bigcup \hbox {A}_2\bigcup \cdots \bigcup \hbox {A}_k$. Denote by ${\mathscr {A}}_k=\hbox {A}_1|\hbox {A}_2|\cdots |\hbox {A}_k$ the k-partition of ${\mathcal {H}}$. Every partition ${\mathscr {A}}_k=\hbox {A}_1|\hbox {A}_2|\cdots |\hbox {A}_k$ corresponds to a family of subsystems $\hbox {A}_1, \hbox {A}_2, \ldots , \hbox {A}_k$.

Let $|\psi \rangle \in {\mathcal {H}}$ be a pure state. It is called k-separable if there is a k-partition ${\mathscr {A}}_k=\hbox {A}_1|\hbox {A}_2|\cdots |\hbox {A}_k$ of ${\mathcal {H}}$ such that

$$\begin{aligned} |\psi \rangle= & {} \left| \psi _1\right\rangle _{\hbox {A}_1}\otimes \left| \psi _2\right\rangle _{\hbox {A}_2}\otimes \cdots \otimes \left| \psi _k\right\rangle _{\hbox {A}_k}, \end{aligned}$$

where $\left| \psi _l\right\rangle _{\hbox {A}_l}$ is a pure state in the subsystem $\hbox {A}_l$ $(l=1, 2, \ldots , k)$. An n-partite mixed state $\rho$ is called k-separable if there exist k-separable pure states $\left| \psi _j\right\rangle$ with respect to different subsets of parties and $p_j > 0$ with $\sum _j p_j =1$, such that

$$\begin{aligned} \rho= & {} \sum _j p_j \left| \psi _j\right\rangle \left\langle \psi _j\right| . \end{aligned}$$

An n-partite state is called genuinely entangled if it is not 2-separable. It is called fully separable if and only if it is n-separable.

The coefficient matrices, which are constructed through arrangement of the coefficients of pure states in lexicographical order, have been used as important tools in the research into entanglement. The mathematical connection between entanglement classification and the coefficient matrices was established in [46, 47]. In this work, the coefficient matrices of the pure state $\left| \psi \right\rangle$ are written as $M_{{s_1}\ldots {s_l}}(\left| \psi \right\rangle )$ (see Sect. 6), where $1\leqslant l\leqslant n$ and $\left\{ s_1, s_2,\ldots , s_l\right\} \in \left\{ 1,2,\ldots ,n\right\}$.

The following theorem was proved in [48].

Theorem

Let $|\psi \rangle$, $|\phi \rangle$ be any two pure states in the n-partite system ${{\mathcal {H}}}={{\mathcal {H}}}^{d_1}\otimes {{\mathcal {H}}}^{d_2}\otimes \cdots \otimes {{\mathcal {H}}}^{d_n}$. If there exist complex square matrices $\hbox {A}_i$ ($1\leqslant i\leqslant n$) such that

$$\begin{aligned} |\psi \rangle =\hbox {A}_1\otimes \hbox {A}_2\otimes \cdots \otimes \hbox {A}_n|\phi \rangle , \end{aligned}$$

then, for any $1\leqslant l<n$,

$$\begin{aligned} M_{1\cdots l}(\left| \psi \right\rangle )=\hbox {A}_1\otimes \cdots \otimes \hbox {A}_lM_{1\cdots l}(\left| \phi \right\rangle )(\hbox {A}_{l+1}\otimes \cdots \otimes \hbox {A}_n)^T , \end{aligned}$$

where $(\hbox {A}_{l+1}\otimes \cdots \otimes \hbox {A}_n)^T$ is the transpose matrix of the matrix $\hbox {A}_{l+1}\otimes \cdots \otimes \hbox {A}_n$.

A simple and effective application of the coefficient matrices is to concretely represent the reduced density matrix that provides a way to associate a density matrix with each component system. For $i\in \{1,\ldots ,k\}$. Denote by $\rho _{\hbox {A}_i}$ the reduced density matrix of $|\psi \rangle \langle \psi |$ on subsystem $\hbox {A}_i$. Then $\rho _{\hbox {A}_i}$ $(1\leqslant i\leqslant k)$ has a factorization in terms of the corresponding coefficient matrix and its conjugate transpose [49],

$$\begin{aligned} \rho _{\hbox {A}_i}=M_{\hbox {A}_i}M^\dag _{\hbox {A}_i}. \end{aligned}$$

Naturally, for the reduced density matrices of two pure states, the following corollary can be reached.

Corollary

Let $|\psi \rangle$, $|\phi \rangle$ be any two pure states in the n-partite system ${{\mathcal {H}}}={{\mathcal {H}}}^{d_1}\otimes {{\mathcal {H}}}^{d_2}\otimes \cdots \otimes {{\mathcal {H}}}^{d_n}$. If there exist unitary matrices $U_i$ ($1\leqslant i\leqslant n$) such that

$$\begin{aligned} |\psi \rangle =U_1\otimes U_2\otimes \cdots \otimes U_n|\phi \rangle , \end{aligned}$$

then their corresponding reduced density matrices $\rho _{1\cdots l}(\left| \psi \right\rangle )$ and $\rho _{1\cdots l}(\left| \phi \right\rangle )$ $(1\leqslant l<n)$ satisfy the relation that

$$\begin{aligned} \rho _{1\cdots l}(\left| \psi \right\rangle )=U_1\otimes \cdots \otimes U_l \rho _{1\cdots l}(\left| \phi \right\rangle ) \left( U_1\otimes \cdots \otimes U_l\right) ^\dag . \end{aligned}$$

This paper is organized as follows. In Sect. 2, an entanglement monotone (denoted by ${\mathcal {E}}_k$) for n-qudit states is constructed. Furthermore, we transform our entanglement monotone to a three-qubit monogamous entanglement measure in Sect. 3. By our entanglement measure, we give in Sect. 4 a necessary condition for characterizing maximally entangled states. Section 5 contains a brief summary.

2 An entanglement monotone

Let $\left| \psi \right\rangle$ be an n-qudit pure state in the n-partite quantum system ${\mathcal {H}}={\mathcal {H}}^{d_1}\otimes {\mathcal {H}}^{d_2}\otimes \cdots \otimes {\mathcal {H}}^{d_n}$. For arbitrary but fixed $k(2\leqslant k\leqslant n)$, we define a map ${\mathcal {E}}_k$ as

$$\begin{aligned} {\mathcal {E}}_k\left( \left| \psi \right\rangle \right) =\min _{{\mathscr {A}}_k} \left[ \frac{\sum _{i=1}^k \left( tr\sqrt{\rho _{\hbox {A}_i}}\right) ^2}{k}\right] , \end{aligned}$$

(1)

where the minimum $\min \limits _{{\mathscr {A}}_k}$ is taken over all possible k-partitions ${\mathscr {A}}_k=\hbox {A}_1|\cdots |\hbox {A}_k$ of the system ${\mathcal {H}}$.

Theorem 1

The ${\mathcal {E}}_k\left( \left| \psi \right\rangle \right)$ defined in Eq. (1) is an entanglement monotone for any pure state $\left| \psi \right\rangle$.

Proof

We will prove that ${\mathcal {E}}_k$ does not increase, on average, under local operations and classical communication (LOCC).

Because any protocol consists of a series of local positive operator valued measures (POVMs) such that only one subsystem be operated and ${\mathcal {E}}_k$ keeps invariant under permutations of the particles, it suffices to consider a general local POVM in only one part of a partition. Without loss of generality, we may assume that the local POVM is performed on the part $\hbox {A}_1$.

We note that any local POVM can be written as a sequence of two-outcome POVMs in analogy to the method in Ref. [50]. Let $F_1$ and $F_2$ be two POVM elements operated in the part $\hbox {A}_1$ such that $F_1 + F_2 =\mathbf{1 }_{\hbox {A}_1}$. Then there exist matrices $P_j$ such that $F_j=P_j^\dag P_j$ ($j = 1, 2$). Choose a proper unitary matrix V and decompose $P_j=U_j D_j V\ (j=1,2)$. Here $U_j\ (j=1,2)$ are unitary matrices; $D_1$ and $D_2$ are both diagonal matrices with nonnegative real numbers $\mu _1,\mu _2,\ldots ,\mu _{n_{\hbox {A}_1}}$ and $\sqrt{1-\mu _1^2},\sqrt{1-\mu _2^2},\ldots ,\sqrt{1-\mu _{n_{\hbox {A}_1}}^2}$ on their respective diagonals, where $n_{\hbox {A}_1}$ stands for the dimension of the part $\hbox {A}_1$.

Let $M_{\hbox {A}_1}\left( |\psi \rangle \right)$ be the coefficient matrix of $|\psi \rangle$ corresponding to the part $\hbox {A}_1$, then by the singular value decomposition $M_{\hbox {A}_1}\left( |\psi \rangle \right) =S \varOmega T^\dag$, where S, T are unitary matrices and $\varOmega$ is a matrix with the diagonal entries $\left\{ \omega _1,\omega _2,\ldots ,\omega _{n_{\hbox {A}_1}}\right\}$.

Because any local unitary operations do not cause a change of the entanglement, some local unitary operation H preceding the POVM can be implemented in the initial state $|\psi \rangle$. We select $H=V^\dag S^\dag$ only for simplicity of proof.

Hence, after local actions H and POVM, the initial state $\left| \psi \right\rangle$ is transformed into new states

$$\begin{aligned} \left| \eta _j\right\rangle= & {} \frac{\left( P_j H\otimes \mathbf{1 }_{\overline{\hbox {A}_1}}\right) \left| \psi \right\rangle }{\sqrt{p_j}}\nonumber \\= & {} \frac{\left( U_j D_j V V^\dag S^\dag \otimes \mathbf{1 }_{\overline{\hbox {A}_1}}\right) \left| \psi \right\rangle }{\sqrt{p_j}}, \end{aligned}$$

(2)

where $j=1,2$; $\overline{\hbox {A}_1}=\hbox {A}_2 \otimes \cdots \otimes \hbox {A}_k$ is the complement of $\hbox {A}_1$; $p_j=\left\langle \theta _j|\theta _j\right\rangle$ with $|\theta _j\rangle = \left( U_j D_j S^\dag \otimes \mathbf{1 }_{\overline{\hbox {A}_1}}\right) \left| \psi \right\rangle$; $p_1 +p_2 =1$.

By the theorem in the introduction, the coefficient matrices of new states $|\eta _j\rangle \ (j=1,2)$ are

$$\begin{aligned} M_{\hbox {A}_1}\left( |\eta _j\rangle \right) =\frac{1}{\sqrt{p_j}}U_j D_j S^\dag M_{\hbox {A}_1}\left( |\psi \rangle \right) =\frac{1}{\sqrt{p_j}}U_j D_j \varOmega T^\dag . \end{aligned}$$

(3)

It follows that

$$\begin{aligned} tr\sqrt{\rho _{\hbox {A}_1}\left( |\eta _1\rangle \right) }= & {} tr\sqrt{M_{\hbox {A}_1}\left( |\eta _1\rangle \right) M_{\hbox {A}_1}\left( |\eta _1\rangle \right) ^\dag } \nonumber \\= & {} tr\sqrt{\frac{1}{p_1}\left( U_1 D_1 \varOmega T^\dag \right) \left( U_1 D_1 \varOmega T^\dag \right) ^\dag } \nonumber \\= & {} tr\sqrt{\frac{1}{p_1}\left( D_1 \varOmega \right) \left( D_1 \varOmega \right) ^\dag } \nonumber \\= & {} \sum _{m=1}^{n_{\hbox {A}_1}} \frac{\omega _m \mu _m}{\sqrt{p_1}}, \end{aligned}$$

(4)

and similarly that

$$\begin{aligned} tr\sqrt{\rho _{\hbox {A}_1}\left( |\eta _2\rangle \right) }=\sum _{m=1}^{n_{\hbox {A}_1}} \frac{\omega _m \sqrt{1-\mu _m^2}}{\sqrt{p_2}} . \end{aligned}$$

(5)

We denote by $\left\langle {\mathcal {E}}_k(|\psi \rangle )\right\rangle$ the average entanglement after LOCC , then $\left\langle {\mathcal {E}}_k(|\psi \rangle )\right\rangle =p_1{\mathcal {E}}_k\left( \left| \eta _1\right\rangle \right) +p_2{\mathcal {E}}_k\left( \left| \eta _2\right\rangle \right)$.

Note that $p_1 + p_2=1$ and $\rho _{\hbox {A}_i}\left( |\psi \rangle \right) =\rho _{\hbox {A}_i}\left( |\eta _1\rangle \right) =\rho _{\hbox {A}_i}\left( |\eta _2\rangle \right) , i=2,3,\ldots ,k,$ it follows that

$$\begin{aligned}&p_1{\mathcal {E}}_k\left( \left| \eta _1\right\rangle \right) +p_2{\mathcal {E}}_k\left( \left| \eta _2\right\rangle \right) \nonumber \\&\quad =p_1\left\{ \min _{{\mathscr {A}}_k} \left[ \frac{\sum _{i=1}^k \left( tr\sqrt{\rho _{\hbox {A}_i}\left( |\eta _1\rangle \right) }\right) ^2}{k}\right] \right\} \nonumber \\&\qquad +p_2\left\{ \min _{{\mathscr {A}}_k} \left[ \frac{\sum _{i=1}^k \left( tr\sqrt{\rho _{\hbox {A}_i}\left( |\eta _2\rangle \right) }\right) ^2}{k}\right] \right\} \nonumber \\&\quad \leqslant \min _{{\mathscr {A}}_k} \left[ \frac{p_1\sum _{i=1}^k \left( tr\sqrt{\rho _{\hbox {A}_i}\left( |\eta _1\rangle \right) }\right) ^2}{k} +\frac{p_2\sum _{i=1}^k \left( tr\sqrt{\rho _{\hbox {A}_i}\left( |\eta _2\rangle \right) }\right) ^2}{k}\right] \nonumber \\&\quad =\min _{{\mathscr {A}}_k} \left[ \frac{\left( \sum _{m=1}^{n_{\hbox {A}_1}} \omega _m \mu _m\right) ^2+ \left( \sum _{m=1}^{n_{\hbox {A}_1}} \omega _m \sqrt{1-\mu _m^2}\right) ^2 }{k}\right. \nonumber \\&\qquad \left. +\frac{\sum _{i=2}^k \left( tr\sqrt{\rho _{\hbox {A}_i}\left( |\psi \rangle \right) }\right) ^2}{k}\right] . \end{aligned}$$

(6)

Because

$$\begin{aligned}&\left( \sum _{m=1}^{n_{\hbox {A}_1}} \omega _m \mu _m\right) ^2+ \left( \sum _{m=1}^{n_{\hbox {A}_1}} \omega _m \sqrt{1-\mu _m^2}\right) ^2 \nonumber \\&\quad = \sum _{m=1}^{n_{\hbox {A}_1}} \omega _m^2 \mu _m^2 + \sum _{\begin{array}{c} 1\leqslant k, l\leqslant {n_{\hbox {A}_1}}\nonumber \\ k\ne l \end{array}} \omega _k\mu _k\omega _l\mu _l +\sum _{m=1}^{n_{\hbox {A}_1}} \omega _m^2\left( 1-\mu _m^2\right) \nonumber \\&\qquad +\sum _{\begin{array}{c} 1\leqslant k, l\leqslant {n_{\hbox {A}_1}}\\ k\ne l \end{array}} \omega _k\sqrt{1-\mu _k^2}\omega _l\sqrt{1-\mu _l^2}\nonumber \\&\quad = \sum _{m=1}^{n_{\hbox {A}_1}} {\omega _m}^2 + \sum _{\begin{array}{c} 1\leqslant k, l\leqslant {n_{\hbox {A}_1}}\\ k\ne l \end{array}}{\omega _k \omega _l\left( \mu _k \mu _l + \sqrt{1-\mu _k^2}\cdot \sqrt{1-\mu _l^2} \right) } \nonumber \\&\quad \leqslant \sum _{m=1}^{n_{\hbox {A}_1}} {\omega _m}^2 + \sum _{\begin{array}{c} 1\leqslant k, l\leqslant {n_{\hbox {A}_1}}\\ k\ne l \end{array}} \omega _k \omega _l\sqrt{{\mu _k}^2 + \left( \sqrt{1- \mu _k^2}\right) ^2}\cdot \sqrt{{\mu _l}^2 + \left( \sqrt{1- \mu _l^2}\right) ^2} \nonumber \\&\quad =\sum _{m=1}^{n_{\hbox {A}_1}} {\omega _m}^2 + \sum _{\begin{array}{c} 1\leqslant k, l\leqslant {n_{\hbox {A}_1}}\\ k\ne l \end{array}}\omega _k \omega _l= \left( \sum _{m=1}^{n_{\hbox {A}_1}}\omega _m\right) ^2= \left( tr\sqrt{\rho _{\hbox {A}_1}\left( |\psi \rangle \right) }\right) ^2, \end{aligned}$$

(7)

it is obtained that

$$\begin{aligned} \left\langle {\mathcal {E}}_k(|\psi \rangle )\right\rangle \leqslant \min _{{\mathscr {A}}_k} \left[ \frac{\sum _{i=1}^k \left( tr\sqrt{\rho _{\hbox {A}_i}\left( |\psi \rangle \right) }\right) ^2}{k}\right] = {\mathcal {E}}_k(|\psi \rangle ). \end{aligned}$$

(8)

So the average entanglement $\left\langle {\mathcal {E}}_k(|\psi \rangle )\right\rangle$ does not increase after LOCC, and then ${\mathcal {E}}_k(|\psi \rangle )$ is an entanglement monotone.

This completes the proof of Theorem 1. $\square$

We now turn to consider the mixed states. For an n-qudit mixed state $\rho$ in the n-partite quantum system ${\mathcal {H}}$, we define

$$\begin{aligned} {\mathcal {E}}_k\left( \rho \right) =\inf _{\{p_i,|\psi _i\rangle \}}\sum _i p_i {\mathcal {E}}_k\left( \left| \psi _i\right\rangle \right) , \end{aligned}$$

(9)

where the infimum is taken over all possible pure state decomposition $\rho =\sum _i p_i\left| \psi _i\right\rangle \left\langle \psi _i\right| .$ On the basis of Theorem 1, it can be straightforwardly verified that ${\mathcal {E}}(\rho )$ defined in Eq. (9) is an entanglement monotone for any n-qudit mixed state $\rho$.

The entanglement monotone ${\mathcal {E}}_k$ has an physical interpretation in terms of the fidelity, which is defined by $F(\rho ,\sigma )\equiv tr\sqrt{\rho ^{1/2}\sigma \rho ^{1/2}}$. When $\rho$ and $\sigma$ are both pure states, the square of fidelity is the transition probability from $\sigma$ to $\rho$ [51]. In the general case of mixed states, a simple operational interpretation of the fidelity is also provided in Ref. [52]. If we re-write ${\mathcal {E}}_k$ as

$$\begin{aligned} {\mathcal {E}}_k\left( \left| \psi \right\rangle \right)= & {} \min _{{\mathscr {A}}_k} \left[ \frac{\sum _{i=1}^k n_{\hbox {A}_i} \left( tr\sqrt{\left( \frac{1}{n_{\hbox {A}_i}}I_{n_{\hbox {A}_i}}\right) ^{1/2}\rho _{\hbox {A}_i}\left( \frac{1}{n_{\hbox {A}_i}}I_{n_{\hbox {A}_i}}\right) ^{1/2}}\right) ^2}{k}\right] \\= & {} \min _{{\mathscr {A}}_k} \left[ \frac{\sum _{i=1}^k n_{\hbox {A}_i}F\left( \frac{1}{n_{\hbox {A}_i}}I_{n_{\hbox {A}_i}},\rho _{\hbox {A}_i}\right) ^2}{k}\right] , \end{aligned}$$

where $F\left( \frac{1}{n_{\hbox {A}_i}}I_{n_{\hbox {A}_i}\times n_{\hbox {A}_i}},\rho _{\hbox {A}_i}\right) ^2$ is the square of fidelity for the reduced state $\rho _{\hbox {A}_i}$ and its system’s totally mixed state $\frac{1}{n_{\hbox {A}_i}}I_{n_{\hbox {A}_i}}$, and $I_{n_{\hbox {A}_i}}$ denotes an identity matrix on subsystem ${\mathcal {H}}_{\hbox {A}_i}$. Then the ${\mathcal {E}}_k$ can be explained as the minimum of all weighed averages of the square of fidelity, corresponding to all possible k-partitions $(2\leqslant k\leqslant n)$ of the system ${\mathcal {H}}$.

3 A monogamous entanglement measure

Monogamous relation is an important criteria for the judgment of good measures of multipartite entanglement, because the entanglement measures that satisfy this relation can show us that quantum entanglement, differing from classical correlation, is not shareable at liberty when distributed among three or more parties. For an n-qudit pure state $\left| \psi \right\rangle$ in the n-partite quantum system ${\mathcal {H}}={\mathcal {H}}^{d_1}\otimes {\mathcal {H}}^{d_2}\otimes \cdots \otimes {\mathcal {H}}^{d_n}$, let

$$\begin{aligned} {\mathcal {E}}^M\left( \left| \psi \right\rangle \right) =\min _{2\leqslant k\leqslant n}\min _{{\mathscr {A}}_k} \sqrt{{{\widetilde{d}}}^n} \left[ \frac{\sum _{i=1}^k \left( tr\sqrt{\rho _{\hbox {A}_i}}\right) ^2}{k}-1\right] \ , \end{aligned}$$

(10)

where the minimum $\min \limits _{{\mathscr {A}}_k}$ is taken over all possible k-partitions ${\mathscr {A}}_k=\hbox {A}_1|\cdots |\hbox {A}_k$ of the system ${\mathcal {H}}$ and ${\widetilde{d}}=\frac{\sum _{i=1}^n d_i}{n}$. It is clear to see that ${\mathcal {E}}^M\left( \left| \psi \right\rangle \right) \geqslant 0$ for any pure state $\left| \psi \right\rangle$ and ${\mathcal {E}}^M\left( \left| \psi \right\rangle \right) = 0$ if and only if $\left| \psi \right\rangle$ is separable. By corollary in the introduction, ${\mathcal {E}}^M\left( \left| \psi \right\rangle \right)$ keeps invariant under local unitary transformations. Then, by the proof of Theorem 1, we know that ${\mathcal {E}}^M$ is an entanglement monotone. Therefore ${\mathcal {E}}^M$ becomes an entanglement measure.

For an n-qudit mixed state $\rho$ in the n-partite quantum system ${\mathcal {H}}$, we define

$$\begin{aligned} {\mathcal {E}}^M\left( \rho \right) =\inf _{\{p_i,|\psi _i\rangle \}}\sum _i p_i {\mathcal {E}}^M\left( \left| \psi _i\right\rangle \right) , \end{aligned}$$

(11)

where the infimum is taken over all possible pure state decomposition $\rho =\sum _i p_i\left| \psi _i\right\rangle \left\langle \psi _i\right| .$ According to the analysis of the preceding context, we can draw a conclusion that ${\mathcal {E}}^M(\rho )$ defined in Eq. (11) is an entanglement measure for any n-qudit mixed state $\rho$. Obviously, ${\mathcal {E}}^M$ satisfies the subadditivity [53]. In addition, we can verify that ${\mathcal {E}}^M$ satisfies the convexity by its definition :

$$\begin{aligned} {\mathcal {E}}^M\left( \sum _i p_i \rho _i\right) \leqslant \sum _i p_i{\mathcal {E}}^M\left( \rho _i\right) . \end{aligned}$$

Next we prove that ${\mathcal {E}}^M$ is monogamous for three-qubit systems.

Theorem 2

For a three-qubit system ${\mathcal {H}}_A \otimes {\mathcal {H}}_B \otimes {\mathcal {H}}_C$, ${\mathcal {E}}^M$ satisfies a monogamy inequality

$$\begin{aligned} {\mathcal {E}}^M_\mathrm{AB}+{\mathcal {E}}^M_\mathrm{AC}\leqslant {\mathcal {E}}^M_{\mathrm{A}|\mathrm{BC}} , \end{aligned}$$

where ${\mathcal {E}}^M_\mathrm{AB}$, ${\mathcal {E}}^M_\mathrm{AC}$, and ${\mathcal {E}}^M_{\mathrm{A}|\mathrm{BC}}$ mean the entanglement of the respective parts of the system.

Proof

First, analogously to Eq. (7) in [54], we use the Schmidt decomposition for a general pure state $\left| \psi \right\rangle _\mathrm{ABC}$ in the system ${\mathcal {H}}_A \otimes {\mathcal {H}}_B \otimes {\mathcal {H}}_C$,

$$\begin{aligned} \left| \psi \right\rangle _\mathrm{ABC}=\sqrt{p}\left| \phi _0\right\rangle _\mathrm{AB}\left| 0\right\rangle _C +\sqrt{1-p}\left| \phi _1\right\rangle _\mathrm{AB}|1\rangle _\mathrm{C} , \end{aligned}$$

with $0\leqslant p \leqslant 1$, $\left| \phi _0\right\rangle _\mathrm{AB}$ and $\left| \phi _1\right\rangle _\mathrm{AB}$ being the orthonormal states of biqubit system ${\mathcal {H}}_A \otimes {\mathcal {H}}_B$ and $\left| 0\right\rangle _C$ and $\left| 1\right\rangle _C$ being the orthonormal basis of qubit system ${\mathcal {H}}_C$. When $p=0$ or $p=1$, $|\psi \rangle _\mathrm{ABC}$ is separable, the inequality holds clearly. Now we assume that $0<p<1$. According to the Schmidt number of $\left| \phi _0\right\rangle _\mathrm{AB}$ and $\left| \phi _1\right\rangle _\mathrm{AB}$, they can be categorized into three classes:

• 1. there is no Schmidt rank-2 state,

• 2. there is only one Schmidt rank-2 state,

• 3. there are two Schmidt rank-2 states.

Case 1 There is no Schmidt rank-2 state in $\left| \phi _0\right\rangle _\mathrm{AB}$ and $\left| \phi _1\right\rangle _\mathrm{AB}$.

With a proper basis $\left\{ |{\widetilde{0}}\rangle _A, |{\widetilde{1}}\rangle _A\right\}$, $\left\{ |{\widetilde{0}}\rangle _B, |{\widetilde{1}}\rangle _B\right\}$ and $\left\{ |0\rangle _C, |1\rangle _C \right\}$ of ${\mathcal {H}}_A$, ${\mathcal {H}}_B$ and ${\mathcal {H}}_C$, respectively; $|\psi \rangle _\mathrm{ABC}$ can be expressed as

$$\begin{aligned} |\psi \rangle _\mathrm{ABC}= & {} \sqrt{p}|{\widetilde{0}}\rangle _A |{\widetilde{0}}\rangle _B |0\rangle _C \nonumber \\&+\sqrt{1-p}|{\widetilde{1}}\rangle _A (\sqrt{a}|{\widetilde{0}}\rangle _B +\sqrt{1-a}|{\widetilde{1}}\rangle _B)|1\rangle _C, \end{aligned}$$

where $0< p < 1$ and $0\leqslant a \leqslant 1$. A direct calculation implies that

$$\begin{aligned} {\mathcal {E}}^M_{\mathrm{A}|\mathrm{BC}}= & {} {\mathcal {E}}^M \left[ \rho _A \left( |\psi \rangle _\mathrm{ABC}\right) \right] =4\sqrt{2p(1-p)},\nonumber \\ {\mathcal {E}}^M_\mathrm{AB}= & {} {\mathcal {E}}^M \left[ \rho _\mathrm{AB}\left( |\psi \rangle _\mathrm{ABC}\right) \right] =0. \end{aligned}$$

It remains to calculate ${\mathcal {E}}^M_\mathrm{AC}$.

For the sake of simplicity, we write

$$\begin{aligned} |{\widetilde{0}}\rangle _A |0\rangle _C= & {} |e_{00}\rangle , \nonumber \\ |{\widetilde{0}}\rangle _A |1\rangle _C= & {} |e_{01}\rangle , \nonumber \\ |{\widetilde{1}}\rangle _A |0\rangle _C= & {} |e_{10}\rangle , \nonumber \\ |{\widetilde{1}}\rangle _A |1\rangle _C= & {} |e_{11}\rangle . \end{aligned}$$

Then the reduced density matrix of the subsystem ${\mathcal {H}}_A \otimes {\mathcal {H}}_C$ can be represented as

$$\begin{aligned} \rho _\mathrm{AC}\left( |\psi \rangle _\mathrm{ABC}\right)= & {} p|e_{00}\rangle \langle e_{00}|+\sqrt{ap(1-p)}|e_{00}\rangle \langle e_{11}| \nonumber \\&+ \sqrt{ap(1-p)}|e_{11}\rangle \langle e_{00}| + (1-p)|e_{11}\rangle \langle e_{11}|. \end{aligned}$$

Consider a pure state decomposition of $\rho _\mathrm{AC}$

$$\begin{aligned} \rho _\mathrm{AC}\left( |\psi \rangle _\mathrm{ABC}\right) =r_1|\varphi _1\rangle \langle \varphi _1|+r_2|\varphi _2\rangle \langle \varphi _2|, \end{aligned}$$

where

$$\begin{aligned} r_1= & {} p+a(1-p),\nonumber \\ r_2= & {} (1-a)(1-p),\nonumber \\ |\varphi _1\rangle= & {} \frac{1}{\sqrt{p+a(1-p)}}\left( \sqrt{p}|e_{00}\rangle +\sqrt{a(1-p)}|e_{11}\rangle \right) , \nonumber \\ |\varphi _2\rangle= & {} |e_{11}\rangle . \end{aligned}$$

Then we have

$$\begin{aligned} {\mathcal {E}}^M_\mathrm{AC}= & {} {\mathcal {E}}^M [\rho _\mathrm{AC}\left( |\psi \rangle _\mathrm{ABC}\right) ]\nonumber \\\leqslant & {} r_1{\mathcal {E}}^M\left( |\varphi _1\rangle \right) +r_2{\mathcal {E}}^M\left( |\varphi _2\rangle \right) \nonumber \\= & {} r_1\left\{ 2\left[ \left( tr\sqrt{\rho _\mathrm{A} \left( |\varphi _1\rangle \right) }\right) ^2-1\right] \right\} \nonumber \\= & {} \left[ p+a(1-p)\right] \left\{ 2\left[ \left( \frac{\sqrt{p}+\sqrt{a(1-p)}}{\sqrt{p+a(1-p)}}\right) ^2-1\right] \right\} \nonumber \\= & {} 4\sqrt{ap(1-p)}\nonumber \\< & {} 4\sqrt{2p(1-p)}= {\mathcal {E}}^M_\mathrm{A|BC}. \end{aligned}$$

This leads us to the conclusion that ${\mathcal {E}}^M_\mathrm{AB}+{\mathcal {E}}^M_\mathrm{AC}< {\mathcal {E}}^M_\mathrm{A|BC}$.

Case 2 There is only one Schmidt rank-2 state in $\left| \phi _0\right\rangle _\mathrm{AB}$ and $\left| \phi _1\right\rangle _\mathrm{AB}$. Without lose of generality, we might as well assume that $\left| \phi _0\right\rangle _\mathrm{AB}$ is a Schmidt rank-2 state.

With the proper choice of basis sets, $|\psi \rangle _\mathrm{ABC}$ can be expressed as

$$\begin{aligned} |\psi \rangle _\mathrm{ABC}= & {} \sqrt{p}(\sqrt{b}|{\widetilde{0}}\rangle _A|{\widetilde{0}}\rangle _B+\sqrt{1-b}|{\widetilde{1}}\rangle _A|{\widetilde{1}}\rangle _B)|0\rangle _C \nonumber \\&+\sqrt{1-p}(\alpha _1|{\widetilde{0}}\rangle _A+\alpha _2|{\widetilde{1}}\rangle _A)(\beta _1|{\widetilde{0}}\rangle _B+\beta _2|{\widetilde{1}}\rangle _B)|1\rangle _C, \end{aligned}$$

where $0< b < 1$ and the complex numbers $\alpha _i(i=1,2)$ and $\beta _i(i=1,2)$ satisfy $\sum _{i=1}^2 \left| \alpha _i\right| ^2=1$ and $\sum _{i=1}^2\left| \beta _i\right| ^2=1$, respectively. Similarly to the discussion in Case 1, we get

$$\begin{aligned} {\mathcal {E}}^M_{\mathrm{A}|\mathrm{BC}}= & {} {\mathcal {E}}^M \left[ \rho _A\left( |\psi \rangle _\mathrm{ABC}\right) \right] \nonumber \\= & {} 4\sqrt{2\left[ b(1-b)p^2+p(1-b)(1-p)\left| \alpha _1\right| ^2+bp(1-p)\left| \alpha _2\right| ^2\right] },\nonumber \\ {\mathcal {E}}^M_\mathrm{AB}= & {} {\mathcal {E}}^M \left[ \rho _\mathrm{AB}\left( |\psi \rangle _\mathrm{ABC}\right) \right] \leqslant 4p\sqrt{b(1-b)},\nonumber \\ {\mathcal {E}}^M_\mathrm{AC}= & {} {\mathcal {E}}^M \left[ \rho _\mathrm{AC}\left( |\psi \rangle _\mathrm{ABC}\right) \right] \nonumber \\\leqslant & {} 4\left| \alpha _2\beta _1\right| \sqrt{pb(1-p)}+4\left| \alpha _1\beta _2\right| \sqrt{p(1-p)(1-b)}. \end{aligned}$$

It can be directly checked that ${\mathcal {E}}^M_\mathrm{AB}+{\mathcal {E}}^M_\mathrm{AC}\leqslant {\mathcal {E}}^M_{\mathrm{A}|\mathrm{BC}}$.

Case 3 Both $\left| \phi _0\right\rangle _\mathrm{AB}$ and $\left| \phi _1\right\rangle _\mathrm{AB}$ are Schmidt rank-2 states.

By choosing a proper basis, we can get the expression of $|\psi \rangle _\mathrm{ABC}$

$$\begin{aligned} |\psi \rangle _\mathrm{ABC}= & {} \sqrt{p}(\sqrt{c}|{\widetilde{0}}\rangle _A|{\widetilde{0}}\rangle _B+\sqrt{1-c}|{\widetilde{1}}\rangle _A|{\widetilde{1}}\rangle _B)|0\rangle _C \nonumber \\&+\,\sqrt{1-p}(a_1|{\widetilde{0}}\rangle _A|{\widetilde{0}}\rangle _B+a_2|{\widetilde{0}}\rangle _A|{\widetilde{1}}\rangle _B\nonumber \\&+\,a_3|{\widetilde{1}}\rangle _A|{\widetilde{0}}\rangle _B+a_4|{\widetilde{1}}\rangle _A|{\widetilde{1}}\rangle _B)|1\rangle _C, \end{aligned}$$

where $0< c < 1$ and $\sum _{i=1}^4 \left| a_i\right| ^2=1$. A similar discussion just as in Case 1 implies that

$$\begin{aligned} {\mathcal {E}}^M_{\mathrm{A}|\mathrm{BC}}= & {} {\mathcal {E}}^M \left[ \rho _A\left( |\psi \rangle _\mathrm{ABC}\right) \right] \nonumber \\= & {} 4\sqrt{2}\left\{ c(1-c)p^2+p(1-p)\left[ (1-c)(|a_1|^2+|a_2|^2)+c(|a_3|^2+|a_4|^2)\right] \right. \nonumber \\&\left. +\, (1-p)^2(|a_1|^2|a_3|^2+|a_1|^2|a_4|^2+|a_2|^2|a_3|^2+|a_2|^2|a_4|^2)\right\} ^{1/2},\nonumber \\ {\mathcal {E}}^M_\mathrm{AB}= & {} {\mathcal {E}}^M \left[ \rho _\mathrm{AB}\left( |\psi \rangle _\mathrm{ABC}\right) \right] \nonumber \\\leqslant & {} 4\left[ (1-p)\left| a_1a_4-a_2a_3\right| +p\sqrt{c(1-c)}\right] ,\nonumber \\ {\mathcal {E}}^M_\mathrm{AC}= & {} {\mathcal {E}}^M \left[ \rho _\mathrm{AC}\left( |\psi \rangle _\mathrm{ABC}\right) \right] \nonumber \\\leqslant & {} 4|a_2|\sqrt{p(1-p)(1-c)}+4|a_3|\sqrt{p(1-p)c} ; \end{aligned}$$

which entails that ${\mathcal {E}}^M_\mathrm{AB}+{\mathcal {E}}^M_\mathrm{AC}\leqslant {\mathcal {E}}^M_{\mathrm{A}|\mathrm{BC}}$. The proof of Theorem 2 is complete. $\square$

4 Application

In Ref. [55], the authors conjecture that for an n-qubit state maximally entangled with respect to their entanglement measure, all single-qubit reduced states are totally mixed. In this section, by our entanglement measure, we prove that all single-qubit reduced states of a maximally entangled state are totally mixed. In order to do this, we first gives the boundedness of the entanglement measure ${\mathcal {E}}^M$.

Theorem 3

Let $\left| \psi \right\rangle$ be an n-partite pure state in the system ${\mathcal {H}}^{d_1}\otimes {\mathcal {H}}^{d_2}\otimes \cdots \otimes {\mathcal {H}}^{d_n}$, then

$$\begin{aligned} 0\leqslant {\mathcal {E}}^M(\left| \psi \right\rangle )\leqslant ({\widetilde{d}}-1)\sqrt{{{\widetilde{d}}}^n}, \end{aligned}$$

with ${\widetilde{d}}=\frac{\sum _{i=1}^n d_i}{n}$.

For a rigorous proof of this theorem the reader can refer to Sect. 7.

It can be seen from the theorem above that ${\mathcal {E}}^M$ is a bounded entanglement measure for any given system. For a pure state $\left| \psi \right\rangle$, it is separable if and only if ${\mathcal {E}}^M\left( \left| \psi \right\rangle \right) = 0$; if it is a genuine entangled state, then ${\mathcal {E}}^M\left( \left| \psi \right\rangle \right) > 0$; if its entanglement degree reaches the upper bound of ${\mathcal {E}}^M$, i.e., ${\mathcal {E}}^M(\left| \psi \right\rangle )= ({\widetilde{d}}-1)\sqrt{{{\widetilde{d}}}^n}$, then we say that it is maximally entangled. Recall that a state is totally mixed if its density matrix is the scalar multiplication of an identity matrix.

Theorem 4

If an n-qubit pure state $\left| \psi \right\rangle$ is maximally entangled with respect to ${\mathcal {E}}^M$, then all single-qubit reduced states of $\left| \psi \right\rangle$ are totally mixed.

Proof

We assert that

$$\begin{aligned} \min _{2\leqslant k\leqslant n}\min _{{\mathscr {A}}_k} \left[ \frac{\sum _{i=1}^k \left( tr\sqrt{\rho _{\hbox {A}_i}}\right) ^2}{k}-1\right] =\min _{{\mathscr {A}}_2} \left[ \left( tr\sqrt{\rho _{\hbox {A}_1}}\right) ^2-1\right] , \end{aligned}$$

(12)

Otherwise, it can be assumed that the minimum on the right side of Eq. (12) is obtained at a certain $k'$-partition with $2< k'\leqslant n$, i.e.,

$$\begin{aligned} \min _{2\leqslant k\leqslant n}\min _{{\mathscr {A}}_k} \left[ \frac{\sum _{i=1}^k \left( tr\sqrt{\rho _{\hbox {A}_i}}\right) ^2}{k}-1\right] = \frac{\sum _{i=1}^{k'} \left( tr\sqrt{\rho _{\hbox {A}_i}}\right) ^2}{k'}-1. \end{aligned}$$

Without loss of generality, we may as well assume that

$$\begin{aligned} tr\sqrt{\rho _{\hbox {A}_1}}=\min \left\{ tr\sqrt{\rho _{\hbox {A}_i}}|i=1,2,\ldots ,k'\right\} . \end{aligned}$$

Then we have

$$\begin{aligned} \min _{2\leqslant k\leqslant n}\min _{{\mathscr {A}}_k} \left[ \frac{\sum _{i=1}^k \left( tr\sqrt{\rho _{\hbox {A}_i}}\right) ^2}{k}-1\right]\geqslant & {} \left[ \left( tr\sqrt{\rho _{\hbox {A}_1}}\right) ^2-1\right] \nonumber \\\geqslant & {} \min _{{\mathscr {A}}_2} \left[ \left( tr\sqrt{\rho _{\hbox {A}_1}}\right) ^2-1\right] \nonumber \\> & {} \left[ \frac{\sum _{i=1}^{k'} \left( tr\sqrt{\rho _{\hbox {A}_i}}\right) ^2}{k'}-1\right] \nonumber \\= & {} \min _{2\leqslant k\leqslant n}\min _{{\mathscr {A}}_k} \left[ \frac{\sum _{i=1}^k \left( tr\sqrt{\rho _{\hbox {A}_i}}\right) ^2}{k}-1\right] . \end{aligned}$$

This leads to a contradiction.

So ${\mathcal {E}}^M(\left| \psi \right\rangle )=\min _{{\mathscr {A}}_2} \sqrt{{{\widetilde{d}}}^n} \left[ \left( tr\sqrt{\rho _{\hbox {A}_1}}\right) ^2-1\right]$. For an n-qubit pure state $|\psi \rangle$, assume that it is maximally entangled. Then

$$\begin{aligned} {\mathcal {E}}^M(\left| \psi \right\rangle )=\sqrt{2^n} \end{aligned}$$

and

$$\begin{aligned} \min _{{\mathscr {A}}_2} \left[ \left( tr\sqrt{\rho _{\hbox {A}_1}}\right) ^2-1\right] =1. \end{aligned}$$

Assume that there is a single-qubit reduced state $\rho _A$, satisfying that $(tr\sqrt{\rho _A})^2-1>1$. We might as well assume that $\rho _A$ has eigenvalues $\lambda _1$ and $\lambda _2$; then,

$$\begin{aligned} (\sqrt{\lambda _1}+\sqrt{\lambda _2})^2-1>1 \end{aligned}$$

and

$$\begin{aligned} \lambda _1+\lambda _2=1. \end{aligned}$$

This implies that

$$\begin{aligned} \left( \lambda _1-\frac{1}{2}\right) ^2<0. \end{aligned}$$

But it is impossible. So for all single-qubit reduced state $\rho _A$, we have $(tr\sqrt{\rho _A})^2-1=1$ and hence $\rho _A=\frac{1}{2}I$, namely, $\rho _A$ is totally mixed. The proof is finished. $\square$

It should be pointed out that Theorem 4 is not sufficient for maximal entanglement. In fact, there exists a separable state whose single-qubit reduced states are all totally mixed. For example, the state

$$\begin{aligned} |\alpha \rangle =\left( \frac{1}{\sqrt{2}}|00\rangle +\frac{1}{\sqrt{2}}|11\rangle \right) \otimes \left( \frac{1}{\sqrt{2}}|00\rangle +\frac{1}{\sqrt{2}}|11\rangle \right) \end{aligned}$$

is separable. However, all single-qubit reduced states of $|\alpha \rangle$ are

$$\begin{aligned}\frac{1}{2}|00\rangle \langle 00|+\frac{1}{2}|11\rangle \langle 11|,\end{aligned}$$

which is a totally mixed state.

In Ref. [56], the authors conjecture that the following state in four-qubit system ${\mathcal {H}}_A \otimes {\mathcal {H}}_B \otimes {\mathcal {H}}_C\otimes {\mathcal {H}}_D$

$$\begin{aligned} |\beta \rangle =\frac{1}{\sqrt{6}}\left[ |0011\rangle +|1100\rangle +\omega (|1010\rangle +|0101\rangle )+\omega ^2(|1001\rangle +|0110\rangle )\right] \end{aligned}$$

is maximally entangled, where $\omega =e^{\frac{2\pi i}{3}}$. Under our entanglement measure ${\mathcal {E}}^M$, this conjecture is true. A straightforward calculation shows that

$$\begin{aligned} \rho _A(|\beta \rangle )= & {} \rho _B(|\beta \rangle )=\rho _C(|\beta \rangle )=\rho _D(|\beta \rangle )= \frac{1}{2}|00\rangle \langle 00|+\frac{1}{2}|11\rangle \langle 11|,\\ \rho _\mathrm{AB}(|\beta \rangle )= & {} \frac{1}{6}|00\rangle \langle 00|+\frac{1}{3}|01\rangle \langle 01|+\frac{1}{6}|01\rangle \langle 10|+\frac{1}{6}|10\rangle \langle 01|\nonumber \\&+\,\frac{1}{3}|10\rangle \langle 10|+\frac{1}{6}|11\rangle \langle 11| \end{aligned}$$

and

$$\begin{aligned} \rho _\mathrm{AC}(|\beta \rangle )= & {} \rho _\mathrm{AD}(|\beta \rangle )=\frac{1}{6}|00\rangle \langle 00|+\frac{1}{3}|01\rangle \langle 01|-\frac{1}{6}|01\rangle \langle 10|-\frac{1}{6}|10\rangle \langle 01|\\&+\,\frac{1}{3}|10\rangle \langle 10|+\frac{1}{6}|11\rangle \langle 11|. \end{aligned}$$

Then,

$$\begin{aligned} \left( tr\sqrt{\rho _\mathrm{A}(|\beta \rangle )}\right) ^2=\left( tr\sqrt{\rho _B(|\beta \rangle )}\right) ^2=\left( tr\sqrt{\rho _C(|\beta \rangle )}\right) ^2=2 \end{aligned}$$

and

$$\begin{aligned} \left( tr\sqrt{\rho _\mathrm{AB}(|\beta \rangle )}\right) ^2=\left( tr\sqrt{\rho _\mathrm{AC}(|\beta \rangle )}\right) ^2=\left( tr\sqrt{\rho _\mathrm{AD}(|\beta \rangle )}\right) ^2=2+\sqrt{3}. \end{aligned}$$

Hence,

$$\begin{aligned} {\mathcal {E}}^M(|\beta \rangle )=\min _{{\mathscr {A}}_2} 4 \left[ \left( tr\sqrt{\rho _{\hbox {A}_1}(|\beta \rangle )}\right) ^2-1\right] =4. \end{aligned}$$

This means that the entanglement degree of $|\beta \rangle$ reaches the upper bound of ${\mathcal {E}}^M$ in four-qubit system; namely, $|\beta \rangle$ is maximally entangled with respect to ${\mathcal {E}}^M$.

5 Conclusion

In this paper, we propose an entanglement measure ${\mathcal {E}}^M$ for arbitrary-dimensional multipartite quantum states, starting with the entanglement monotone ${\mathcal {E}}_k$. Our entanglement measure is equipped with useful properties for any states, including boundedness, convexity and subadditivity. It vanishes for and only for the separable states. Furthermore, it satisfies the monogamous relation for three-qubit quantum systems. We hope that this result can be generalized to entanglement monogamy of n-qubit quantum states. We also establish a connection between a maximally entangled state and its single-qubit reduced states. A necessary condition to characterize maximally entangled states is obtained as an application of measure ${\mathcal {E}}^M$.

Appendices

Appendix 1

Here we introduce the concept of the coefficient matrix. Every pure state $\left| \psi \right\rangle$ in system ${\mathcal {H}}^{d_1}\otimes {\mathcal {H}}^{d_2}\otimes \cdots \otimes {\mathcal {H}}^{d_n}$ can be represented as

$$\begin{aligned} \left| \psi \right\rangle =\sum _ {j=0}^{\prod _{k=1}^n d_k -1}\lambda _j\left| t_j\right\rangle , \end{aligned}$$

where, for $j=0,1,\ldots ,{\prod _{k=1}^n d_k -1}$, coefficients $\lambda _j$ are complex numbers satisfying

$$\begin{aligned} \sum _ {j=0}^{\prod _{k=1}^n d_k -1}\left| \lambda _j\right| ^2=1 \end{aligned}$$

and $\left| t_j\right\rangle$ are the basis states in ${\mathcal {H}}$.

We denote the n systems by numbers $1,2,\ldots ,n$, respectively. Let $q_i \ (i=1,2,\ldots n)$ be positive integers such that $0\leqslant q_i \leqslant d_i -1$, then the state $\left| \psi \right\rangle$ can be rewritten as

$$\begin{aligned} \left| \psi \right\rangle =\sum _ {q_1 =0}^{d_1-1} \sum _{q_2=0}^{d_2-1}\cdots \sum _{q_n=0}^{d_n-1} a_{q_1,q_2,\ldots ,q_n}\left| q_1 q_2\ldots q_n\right\rangle , \end{aligned}$$

which induces the following $\left( \prod _{i=1}^l d_i\right) \times \left( \prod _{i=l+1}^n d_i\right)$ coefficient matrices whose entries $a_{q_1 q_2\ldots q_n}$ are arranged according to the subscript $q_1 q_2\ldots q_n$ in lexicographical ascending order

$$\begin{aligned}&M_{1\cdots l,l+1\ldots n}(\left| \psi \right\rangle )\nonumber \\&\quad =\left( \begin{array}{ccc} a_{\underbrace{0\cdots 0}_l \underbrace{0\cdots 0}_{n-l}} &{} \cdots &{} a_{\underbrace{0\cdots 0}_l \underbrace{d_{l+1} -1\cdots d_n -1}_{n-l}} \\ a_{\underbrace{0\cdots 1}_l \underbrace{0\cdots 0}_{n-l}} &{} \cdots &{} a_{\underbrace{0\cdots 1}_l \underbrace{d_{l+1} -1\cdots {d_n -1}}_{n-l}}\\ \vdots &{} \ddots &{} \vdots \\ a_{\underbrace{d_1-1\cdots d_l-1}_l \underbrace{0\cdots 0}_{n-l}} &{} \cdots &{} a_{\underbrace{d_1 -1\cdots d_l -1}_l \underbrace{d_{l+1} -1\cdots d_{n} -1}_{n-l} }\\ \end{array}\right) \end{aligned}$$

We abbreviate the coefficient matrix $M_{1\cdots l,l+1\cdots n}(\left| \psi \right\rangle )$ as $M_{1\cdots l}(\left| \psi \right\rangle )$ by omitting the column subscripts $l+1\cdots n$. Each realignment of the n particles, described simply as ${s_1s_2\cdots s_ls_{l+1}\cdots s_n}$, a permutation of the set $\{1, 2, \ldots , n\}$, generates correspondently a $\left( \prod _{i=1}^{l} d_{s_i}\right) \times \left( \prod _{i=l+1}^n d_{s_i}\right)$ coefficient matrix where l is an arbitrary but fixed positive integer satisfying $1\leqslant l\leqslant n$,

$$\begin{aligned}&M_{{s_1}\cdots {s_l}}(\left| \psi \right\rangle )\\&\quad =\left( \begin{array}{ccc} a_{\underbrace{0\cdots 0}_l \underbrace{0\cdots 0}_{n-l}} &{} \cdots &{} a_{\underbrace{0\cdots 0}_l \underbrace{d_{s_{l+1}} -1\cdots d_{s_n} -1}_{n-l}} \\ a_{\underbrace{0\cdots 1}_l \underbrace{0\cdots 0}_{n-l}} &{} \cdots &{} a_{\underbrace{0\cdots 1}_l \underbrace{d_{s_{l+1}} -1\cdots {d_{s_n} -1}}_{n-l}}\\ \vdots &{} \ddots &{} \vdots \\ a_{\underbrace{d_{s_1} -1\cdots d_{s_l} -1}_l \underbrace{0\cdots 0}_{n-l}} &{} \cdots &{} a_{\underbrace{d_{s_1} -1\cdots d_{s_l} -1}_l \underbrace{d_{s_{l+1}} -1\cdots d_{s_n} -1}_{n-l} }\\ \end{array}\right) \end{aligned}$$

Appendix 2

This appendix is devoted to prove Theorem 3. In order to prove this theorem, we need the following lemma.

Lemma 1

Let $S=\left\{ d_1, d_2,\ldots , d_n\right\}$ be a set of n positive numbers with $d_i\geqslant 1 \ (i=1,2,\ldots n)$. Divide S into any $k\ (1\leqslant k\leqslant n)$ subsets $S_j=\left\{ d_1^j,d_2^j,\ldots , d_{n_j}^j\right\}$, where $1\leqslant j\leqslant k$ and $\sum _{j=1}^k n_j =n$. Then,

$$\begin{aligned} \frac{\displaystyle \sum \nolimits _{i=1}^n d_i}{n}\leqslant \frac{\displaystyle \sum \nolimits _{j=1}^k \left( \displaystyle \prod \nolimits _{m=1}^{n_j}d_{m}^j\right) }{k}. \end{aligned}$$

Proof

It is sufficient to verify that for any $k\ (1\leqslant k\leqslant n-1)$ subsets $S_j=\left\{ d_1^j,d_2^j,\ldots , d_{n_j}^j\right\}$ of the set S with $1\leqslant j\leqslant k$ and $\sum _{j=1}^k n_j =n$, there exists $k+1$ subsets $T_l=\left\{ c_1^l,c_2^l,\ldots ,c_{h_l}^l\right\}$ of the set S with $1\leqslant l\leqslant k+1$ and $\sum _{l=1}^{k+1} h_l =n$, such that

$$\begin{aligned} \frac{\displaystyle \sum \nolimits _{l=1}^{k+1} \left( \displaystyle \prod \nolimits _{m=1}^{h_l}c_{m}^l\right) }{k+1}\leqslant \frac{\displaystyle \sum \nolimits _{j=1}^k \left( \displaystyle \prod \nolimits _{m=1}^{n_j}d_{m}^j\right) }{k}. \end{aligned}$$

For k subsets $S_j=\left\{ d_1^j,d_2^j,\ldots , d_{n_j}^j\right\}$ with $1\leqslant j\leqslant k$ and $\sum _{j=1}^k n_j =n$, without loss of generality we assume $n_1 \geqslant 2$. Suppose that

$$\begin{aligned} T_1= & {} \left\{ c_1^1\right\} =\left\{ d_1^1\right\} ,\nonumber \\ T_2= & {} \left\{ c_1^2,c_2^2\ldots ,c_{h_2}^2\right\} =\left\{ d_2^1,d_3^1,\ldots ,d_{n_1}^1\right\} ,\nonumber \\ T_l= & {} \left\{ c_1^l,c_2^l,\ldots , c_{h_l}^l\right\} =S_{l-1}=\left\{ d_1^{l-1},d_2^{l-1},\ldots , d_{n_{l-1}}^{l-1}\right\} , \end{aligned}$$

with $3\leqslant l\leqslant k+1$.

It is apparent from the condition that

$$\begin{aligned}&d_1^1\geqslant 1,\nonumber \\&\prod _{m=2}^{n_1} d_m^1\geqslant 1,\nonumber \\&\sum _{j=2}^k \left( \displaystyle \prod _{m=1}^{n_j}d_{m}^j\right) \geqslant k-1 . \end{aligned}$$

A routine computation gives rise to

$$\begin{aligned}&(k+1)\sum _{j=1}^k \left( \displaystyle \prod _{m=1}^{n_j}d_{m}^j\right) -k\left[ d_1^1 + \displaystyle \prod _{m=2}^{n_1}d_{m}^1+\sum _{j=2}^k \left( \displaystyle \prod _{m=1}^{n_j}d_{m}^j\right) \right] \nonumber \\&\quad = \left( k d_1^1 +d_1^1\right) \left( \prod _{m=2}^{n_1} d_m^1 -1\right) -k\prod _{m=2}^{n_1} d_m^1+\sum _{j=2}^k \left( \displaystyle \prod _{m=1}^{n_j}d_{m}^j\right) +d_1^1 \nonumber \\&\quad \geqslant \left( k d_1^1 +d_1^1\right) \left( \prod _{m=2}^{n_1} d_m^1 -1\right) -k\prod _{m=2}^{n_1} d_m^1 +k-1+ d_1^1 \nonumber \\&\quad = \left( d_1^1 -1\right) +\left[ k\left( d_1^1 -1\right) +d_1^1\right] \left( \prod _{m=2}^{n_1} d_m^1 -1\right) \nonumber \\&\quad \geqslant 0 . \end{aligned}$$

Rearranging the preceding inequality leads to

$$\begin{aligned} \frac{d_1^1 + \displaystyle \prod \nolimits _{m=2}^{n_1}d_{m}^1 +\sum \nolimits _{j=2}^k \left( \displaystyle \prod \nolimits _{m=1}^{n_j}d_{m}^j\right) }{k+1}\leqslant \frac{\displaystyle \sum \nolimits _{j=1}^k \left( \displaystyle \prod \nolimits _{m=1}^{n_j}d_{m}^j\right) }{k} . \end{aligned}$$

Thus we arrive at the conclusion that

$$\begin{aligned} \frac{\displaystyle \sum \nolimits _{l=1}^{k+1} \left( \displaystyle \prod \nolimits _{m=1}^{h_l}c_{m}^l\right) }{k+1}\leqslant \frac{\displaystyle \sum \nolimits _{j=1}^k \left( \displaystyle \prod \nolimits _{m=1}^{n_j}d_{m}^j\right) }{k}. \end{aligned}$$

This completes the proof of Lemma 1. $\square$

Now we turn to prove Theorem 3.

Proof of Theorem 3

It can be immediately seen that ${\mathcal {E}}^M(\left| \psi \right\rangle )\geqslant 0$ for any pure state $\left| \psi \right\rangle$. It remains to show that the upper bound of ${\mathcal {E}}^M(\left| \psi \right\rangle )$ is $({\widetilde{d}}-1)\sqrt{{{\widetilde{d}}}^n}$. For any $n_i$-partite component system $A= {\mathcal {H}}^{d_1^i}\otimes {\mathcal {H}}^{d_2^i}\otimes \cdots \otimes {\mathcal {H}}^{d_{n_i}^i}$ ($d_1^i, d_2^i,\ldots , d_{n_i}^i \in \{d_1, d_2,\ldots , d_n\}$), let $\rho _\mathrm{A}$ has the eigenvalues $\lambda _1 , \lambda _2 , \ldots ,\lambda _{\varPi _{m=1}^{n_i} d_{m}^i} .$ Therefore,

$$\begin{aligned} \lambda _1 + \lambda _2 + \cdots +\lambda _{\varPi _{m=1}^{n_i} d_{m}^i} =1 \end{aligned}$$

and

$$\begin{aligned} tr\sqrt{\rho _\mathrm{A}}= & {} \sum _{j=1}^{\varPi _{m=1}^{n_i} d_{m}^i} \sqrt{\lambda _j} \leqslant \sqrt{\left( \displaystyle \prod _{m=1}^{n_i}d_{m}^i\right) \sum _{j=1}^{{\varPi _{m=1}^{n_i} d_{m}^i}} \lambda _j} =\sqrt{\displaystyle \prod _{m=1}^{n_i}d_{m}^i} . \end{aligned}$$

Consequently, we infer that

$$\begin{aligned} \min _{{\mathscr {A}}_k} \left[ \frac{\displaystyle \sum \nolimits _{i=1}^k \left( \hbox {tr}\sqrt{\rho _{\hbox {A}_i}}\right) ^2}{k}-1\right] \leqslant \min _{{\mathscr {A}}_k} \left[ \frac{\displaystyle \sum \nolimits _{i=1}^k \left( \displaystyle \prod \nolimits _{m=1}^{n_i}d_{m}^i\right) }{k}-1\right] . \end{aligned}$$

Meanwhile, Lemma 1 tells us that

$$\begin{aligned} \min _{2\leqslant k\leqslant n}\min _{{\mathscr {A}}_k} \left[ \frac{\displaystyle \sum \nolimits _{i=1}^k \left( \displaystyle \prod \nolimits _{m=1}^{n_i}d_{m}^i\right) }{k}-1\right] =\frac{\sum _{i=1}^n d_i}{n}-1={\widetilde{d}}-1. \end{aligned}$$

Hence,

$$\begin{aligned} \min _{2\leqslant k\leqslant n}\min _{{\mathscr {A}}_k} \sqrt{{{\widetilde{d}}}^n} \left[ \frac{\sum _{i=1}^k \left( tr\sqrt{\rho _{\hbox {A}_i}}\right) ^2}{k}-1\right]\leqslant & {} ({\widetilde{d}}-1)\sqrt{{{\widetilde{d}}}^n}, \end{aligned}$$

which means that ${\mathcal {E}}^M\left( \left| \psi \right\rangle \right) \leqslant ({\widetilde{d}}-1)\sqrt{{{\widetilde{d}}}^n}$. Thus Theorem 3 is completed. $\square$