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.

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. $\square$