Concurrence and three-tangle of the graph

Abstract

In this article, we study the entanglement properties of two-qubit quantum states based on concurrence using the graph-theoretic approach. Entanglement properties of a density operator are obtained from the combinatorial Laplacian matrix which is constructed for a given graph. In the study of entanglement, we found that measure of entanglement is either $\frac{1}{ |{E}| }$ or zero for simple graphs. We further propose a simple method to evaluate the three-tangle and analyze inequivalent classes belonging to three-qubit pure states using graph-theoretic perspective. Our results allow a clear distinction between three-qubit separable states, genuinely entangled Greenberger–Horne–Zeilinger and W states, purely based on graphical interpretations.

Appendix

• 1. $det( L^A) = 0 \Rightarrow C(\rho ) = 0.$

Proof

Laplacian matrix of four vertices graph is given as

$$\begin{aligned} L = \left[ \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} a &{} b &{} c &{} d \\ b &{} e &{} f &{} g \\ c &{} f &{} h &{} i \\ d &{} g &{} i &{} j \end{array}\right] . \end{aligned}$$

The reduced form of Laplacian matrix is defined by

$$\begin{aligned} L^A = \left[ \begin{array}{rr} a+e &{} c+g \\ c+g &{} h+j \end{array}\right] . \end{aligned}$$

Since, $M = \rho \times p\times {\rho }^* \times p$ and $p = \sigma _y \otimes \sigma _y$.

$$\begin{aligned} M= & {} \frac{1}{a+e+h+j}\left[ \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} a &{} b &{} c &{} d \\ b &{} e &{} f &{} g \\ c &{} f &{} h &{} i \\ d &{} g &{} i &{} j \end{array}\right] \left[ \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 0 &{} 0 &{} 0 &{} -1 \\ 0 &{} 0 &{} 1 &{} 0\\ 0 &{} 1 &{} 0 &{} 0\\ -1 &{} 0 &{} 0 &{} 0 \end{array}\right] \\&\frac{1}{a+e+h+j} \left[ \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} a &{} b &{} c &{} d \\ b &{} e &{} f &{} g \\ c &{} f &{} h &{} i \\ d &{} g &{} i &{} j \end{array}\right] \left[ \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 0 &{} 0 &{} 0 &{} -1 \\ 0 &{} 0 &{} 1 &{} 0\\ 0 &{} 1 &{} 0 &{} 0\\ -1 &{} 0 &{} 0 &{} 0 \end{array}\right] .\\ M= & {} \frac{1}{({a+e+h+j)}^2} \\&\left[ \begin{array}{rrrrrrrrrr} d^2-cg+aj-bi &{} &{} &{} cf-cd-ai+bh &{} &{} &{} bf-ag-bd+ce &{} &{} &{} 2ad-2bc\\ dg+bj-fg-ei &{} &{} &{} f^2-cg-bi+eh &{} &{} &{} 2ef-2bg &{} &{} &{} bd+ag-bf-ce\\ cj+di-fi-gh &{} &{} &{} 2fh-2ci &{} &{} &{} f^2-cg-bi+eh &{} &{} &{} cd-cf+ai-bh \\ 2dj-2gi &{} &{} &{} fi-di-cj+gh &{} &{} &{} fg-bj-dg+ei &{} &{} &{} d^2-cg+aj-bi \end{array}\right] . \end{aligned}$$

For a Laplacian matrix

• (a) $\sum _{j=1}^{n} a_{ij} = 0 , ~~~ \forall ~i$ and $\sum _{i=1}^{n} a_{ij} = 0 , ~~~ \forall ~j$;

• (b) All non-diagonal elements are either negative or zero.

$$\begin{aligned} a+b+c+d= & {} 0. \end{aligned}$$

(25)

$$\begin{aligned} b+e+f+g= & {} 0. \end{aligned}$$

(26)

$$\begin{aligned} c+f+h+i= & {} 0. \end{aligned}$$

(27)

$$\begin{aligned} d+g+i+j= & {} 0. \end{aligned}$$

(28)

If $det(L^A)=0$, then

$$\begin{aligned} (a+e)(h+j)=(c+g)^2. \end{aligned}$$

(29)

$\square$

From Eqs. (25), (26), (27), (28), and (29), we get

$$\begin{aligned} \left[ \frac{2b+f+d}{c+g} +1 \right] \left[ \frac{2i+f+d}{c+g} +1 \right] = 1. \end{aligned}$$

$\Rightarrow b=d=f=i =0$.

Using all these conditions, we can prove that all the elements of matrix M are zero. $\Rightarrow C(\rho ) = 0.$

• 2. $det( L^A) \ne 0 \Rightarrow C(\rho ) = {\left\{ \begin{array}{ll} 0 &{}\quad {D(G) = D(G')},\\ \frac{1}{|{E}|} &{}\quad {D(G) \ne D(G')} \end{array}\right. }$.

Proof

This observation can be proved using Theorem 6 and Theorem 7. $\square$