FIG. 6: Fig. 6 shows how to find the CSS for the generalized Horodecki states. Fig. 6a and Fig. 6b correspond to r = − s = 0 . 3 and r = − s = 0 . 5 respectively. In order to find CSS one makes a straight line ℓ first, which connects the nearest vertex of T and a point t$_{H}$ . Secondly, we compute the coordinate for the nearest crossing point between ℓ and L $_{r,s}$. Thirdly, we identify the crossing point with τ$_{H}$ . Keeping u$_{H}$ = r$_{H}$ and v$_{H}$ = s$_{H}$ , one can find the CSS of the Horodecki state. points P$_{1}$ and P$_{2}$ for the Horodecki states. This is why the CSS condition min( µ Γ $_{−}$, ν Γ $_{−}$) = 0 gives two different solutions. Using the nearest crossing point ( P$_{1}$ in Fig. 6) one can derive τ$_{H}$ straightforwardly. Finally using Eq.(5.4) with imposing u$_{H}$ = r$_{H}$ and v$_{H}$ = s$_{H}$ , one can derive π$_{H}$ , the CSS of ρ$_{H}$ . In the previous sections we have shown how to find the CSS for the Bell-diagonal, gener- alized VP, and generalized Horodecki states. In fact, it is possible to find the CSS because those states exhibit the following nice features. Let r , s and t be Bloch and correlation vec- tors of those states. Let u , v and τ be Bloch and correlation vectors for the corresponding CSS of those states. Then, the features are: (i) u = r and v = s . (ii) τ can be computed from the crossing point between the straight line ℓ and the surface for a set of the separable states L $_{r,s}$. However, such simple but nice features are not maintained for the general mixtures. For example, let us consider the comparatively simple model introduced in Eq.(4.8) and Eq.(4.9). It is straightforward to show that the first property, i.e. u = r and v = s , is not maintained unless R$_{2}$ = R$_{3}$ $^{6}$. In order to find the CSS for the arbitrary states, therefore, we have to find the explicit relations between ( r , s ) and ( u , v ). As far as we know, still this is an unsolved problem. In addition, one can show that the second property is not maintained too for the general 6 When R$_{2}$ = R$_{3}$ , one can show [ ρ Z ( x ) , σ Z ] = 0. Therefore, E R ( ρ Z ( x ) ⊗ ˜ ρ ) = E R ( ρ Z ( x )) + E R (˜ ρ ) for all two-qubit mixture ˜ ρ [14]. VI. DIFFICULTIES IN FINDING CSS FOR ARBITRARY STATES 17