S ( ρ || σ ) reduces to ln 2 − ln x , which is less than ln 2 if x = 1. Therefore σ on ( o (+) 1 , o (+) 2 , o ( − ) 3 ) is not CSS of | β$_{1}$ 〉〈 β$_{1}$ | . By same way one can show that σ on ( o ( − ) 1 , o (+) 2 , o (+) 3 ) or ( o ( − ) 1 , o ( − ) 2 , o ( − ) 3 ) is not CSS of | β$_{1}$ 〉〈 β$_{1}$ | , which completes the proof. Theorem 2. The CSS of the any Bell-diagonal state ρ corresponds to the crossing point between the nearest surface of L from ρ and the straight line ℓ , which connects ρ and the nearest vertex of T from ρ . Proof. If σ is CSS of ρ , the CSS of ˜ ρ = xρ + (1 − x ) σ is also σ [12]. Let ρ be ρ = | β$_{1}$ 〉〈 β$_{1}$ | . Then, theorem 1 implies that σ can be any point on the surface ( o (+) 1 , o ( − ) 2 , o (+) 3 ). Let ˜ ρ belong to the small tetrahedron ( v$_{1}$, o (+) 1 , o ( − ) 2 , o (+) 3 ). Note that ˜ ρ corresponds to a internally dividing point of the line segment ρσ . Since Eq.(1.6) implies that the set of the entangled states which have same CSS should be represented by the straight line, the only possible σ as CSS of ˜ ρ is a crossing point between a line ρ ˜ and the surface ( o (+) 1 , o ( − ) 2 , o (+) 3 ), which completes the proof for the Bell-diagonal states. By making use of the Theorem 2 one can always find the CSS σ if ρ is a Bell-diagonal state. Fig. 2 shows how to find the CSS for the Bell-diagonal state. First, extend the line segment between ρ and the point corresponding to the nearest vertex of T . Second, compute the coordinate of the crossing point between the line and the nearest surface of the octahedron L . Finally, find the CSS which corresponds to the crossing point. This complete the reverse process of Ref.[22]. When the Bloch vectors r and s are non-zero, the geometrical objects T and L should be deformed. In this section we will discuss how T and L are deformed. In order to perform the following analysis analytically we consider in this paper the case where r and s are parallel to each other. It is worthwhile noting that if r and s are x - or y -direction, one can make them to be z -directional via the appropriate local-unitary transformation. For example, if they are x -direction, ρ ′ = ( U ⊗ U ) ρ ( U ⊗ U ) † with III. GEOMETRICAL DEFORMATION OF T AND L U = 1 √$_{2}$   1 1 − 1 1   7