r = − s = 0 . 3 (Fig. 7b), r = s = 0 . 5 (Fig. 7c) and r = − s = 0 . 5 (Fig. 7d). For convenience, we plot T and L r,s together in each figure. Each line in Fig. 7 represents a set of t , whose CSS has same τ . As Fig. 7 exhibits, not all lines do pass one of the vertices of T . This fact indicates that unfortunately the nice property (ii) is not maintained for the arbitrary states. The non-maintenance of the property (ii) can be proved on the analytical ground by making use of the simpler model. Let us consider σ$_{Z}$ in Eq.(4.8) and ρ$_{Z}$ ( x ) in Eq.(4.9). Then, the Bloch vectors r , s and the correlation vector t of ρ$_{Z}$ ( x ) are r = (0 , 0 , r ), s = (0 , 0 , s ) and t = ( t$_{1}$, t$_{2}$, t$_{3}$ ), where Of course, if we take x → 0 limit in Eq.(6.1), the corresponding quantities are the Bloch vectors and the correlation vector of σ$_{Z}$ . Now, let us consider another state π$_{Z}$ , which can be obtained from σ$_{Z}$ by changing Y → Y ′ and R$_{i}$ → R ′ $_{i}$( i = 1 , · · · , 4). In order to ensure that π$_{Z}$ is CSS we require Y ′ = √ R ′ $_{1}$R ′ 4 and R ′ $_{2}$R ′ 3 ≥ R ′ $_{1}$R ′ $_{4}$. Then, the set of the entangled states ξ$_{Z}$ ( x $^{′}$), whose CSS are π$_{Z}$ , can be obtained from ρ$_{Z}$ ( x ) by changing Y → Y $^{′}$, R$_{i}$ → R ′ $_{i}$( i = 1 , · · · , 4) and x → x $^{′}$. Thus, Bloch vectors r $^{′}$, s ′ and correlation vector t ′ of ξ$_{Z}$ ( x $^{′}$) are r ′ = (0 , 0 , r $^{′}$), s ′ = (0 , 0 , s $^{′}$) and t ′ = ( t ′ $_{1}$, t ′ $_{2}$, t ′ $_{3}$), where Then, it is straightforward to show that the condition t = t ′ imposes where ˜ r = R$_{1}$ − R$_{2}$ − R$_{3}$ + R$_{4}$ and ˜ r ′ = R ′ 1 − R ′ 2 − R ′ 3 + R ′ $_{4}$. Thus, one can compute the crossing point t = t ′ = ( µ$_{1}$, µ$_{2}$, µ$_{3}$ ), where µ$_{i}$ becomes r = ( R$_{1}$ + R$_{2}$ − R$_{3}$ − R$_{4}$ ) − x ( ¯ R$_{2}$ − ¯ R$_{3}$ ) (6.1) s = ( R$_{1}$ − R$_{2}$ + R$_{3}$ − R$_{4}$ ) + x ( ¯ R$_{2}$ − ¯ R$_{3}$ ) t$_{1}$ = t$_{2}$ = 2 Y − 2 x ¯ Y t$_{3}$ = ( R$_{1}$ − R$_{2}$ − R$_{3}$ + R$_{4}$ ) − 4 x ¯ R$_{1}$. r ′ = ( R ′ 1 + R ′ 2 − R ′ 3 − R ′ $_{4}$) − x $^{′}$( ¯ R ′ 2 − ¯ R ′ $_{3}$) (6.2) s ′ = ( R ′ 1 − R ′ 2 + R ′ 3 − R ′ $_{4}$) + x $^{′}$( ¯ R ′ 2 − ¯ R ′ $_{3}$) t ′ 1 = t ′ 2 = 2 Y ′ − 2 x ′ ¯ Y ′ t ′ 3 = ( R ′ 1 − R ′ 2 − R ′ 3 + R ′ $_{4}$) − 4 x ′ ¯ R ′ $_{1}$. x = ¯ Y $^{′}$(˜ r − ˜ r $^{′}$) − 4( Y − Y $^{′}$) ¯ R ′ 1 4( ¯ Y ′ ¯ R$_{1}$ − ¯ Y ¯ R ′ $_{1}$) x ′ = ¯ Y (˜ r − ˜ r $^{′}$) − 4( Y − Y $^{′}$) ¯ R$_{1}$ 4( ¯ Y ′ ¯ R$_{1}$ − ¯ Y ¯ R ′ $_{1}$) (6.3) µ$_{1}$ = µ$_{2}$ = 4( Y ¯ Y ′ ¯ R$_{1}$ − Y ′ ¯ Y ¯ R ′ $_{1}$) − ¯ Y ¯ Y $^{′}$(˜ r − ˜ r $^{′}$) 2( ¯ Y ′ ¯ R$_{1}$ − ¯ Y ¯ R ′ $_{1}$) (6.4) µ$_{3}$ = 4( Y − Y $^{′}$) ¯ R$_{1}$ ¯ R ′ 1 − (˜ r ¯ Y ¯ R ′ 1 − ˜ r ′ ¯ Y ′ ¯ R$_{1}$ ) ¯ Y ′ ¯ R$_{1}$ ¯ Y ¯ R ′ 1 . 19