FIG. 1: The total Bell-diagonal states belong to the tetrahedron ( v$_{1}$, v$_{2}$, v$_{3}$, v$_{4}$ ) and the set of the separable states belong to the octahedron, whose vertices are o ± 1 , o ± 2 and o ± 3 . As this figure shows, the planes ( o (+) 1 , o ( − ) 2 , o ( − ) 3 ), ( o (+) 1 , o (+) 2 , o (+) 3 ), ( o ( − ) 1 , o ( − ) 2 , o (+) 3 ) and ( o ( − ) 1 , o (+) 2 , o ( − ) 3 ) are contained in the planes ( v$_{1}$, v$_{3}$, v$_{4}$ ), ( v$_{1}$, v$_{2}$, v$_{3}$ ), ( v$_{1}$, v$_{2}$, v$_{4}$ ) and ( v$_{2}$, v$_{3}$, v$_{4}$ ), respectively. Therefore, all entangled Bell-diagonal mixtures belong to the small four tetrahedra ( v$_{1}$, o (+) 1 , o ( − ) 2 , o (+) 3 ), ( v$_{2}$, o ( − ) 1 , o (+) 2 , o (+) 3 ), ( v$_{3}$, o (+) 1 , o (+) 2 , o ( − ) 3 ) and ( v$_{4}$, o ( − ) 1 , o ( − ) 2 , o ( − ) 3 ). Now, we show how to perform the reverse process of Ref.[22] when ρ is an entangled Bell-diagonal state. This can be achieved by following two theorems. Theorem 1. Every Bell state has infinite CSS, which cover fully the nearest surface of the octahedron L . Proof. It is sufficient to prove this theorem when ρ = | β$_{1}$ 〉〈 β$_{1}$ | . When ρ = | β$_{i}$ 〉〈 β$_{i}$ | ( i = 2 , 3 , 4), one can prove the theorem similarly. Let σ be a following Bell-diagonal state: with p = ( x, y, z ). Then, it is easy to show that the spectral decomposition of σ is The nearest surface of L from ρ = | β$_{1}$ 〉〈 β$_{1}$ | is ( o (+) 1 , o ( − ) 2 , o (+) 3 ), whose surface equation is x − y + z = 1. If σ belongs to the surface ( o (+) 1 , o ( − ) 2 , o (+) 3 ), it is easy to show that S ( ρ || σ ) = ln 2, which exactly coincides with the REE of the Bell states[11]. Therefore, σ on the surface ( o (+) 1 , o ( − ) 2 , o (+) 3 ) is the CSS of | β$_{1}$ 〉〈 β$_{1}$ | . Now, let us consider the case where σ belongs to other surface. For example, let us assume that σ belongs to the surface ( o (+) 1 , o (+) 2 , o ( − ) 3 ), whose surface equation is x + y − z = 1. Then, σ = 1 4 [ I ⊗ I + 3 ∑ n =1 p$_{n}$σ$_{n}$ ⊗ σ$_{n}$ ] (2.6) σ = 1 + x − y + z 4 | β$_{1}$ 〉〈 β$_{1}$ | + 1 − x + y + z 4 | β$_{2}$ 〉〈 β$_{2}$ | (2.7) + 1 + x + y − z 4 | β$_{3}$ 〉〈 β$_{3}$ | + 1 − x − y − z 4 | β$_{4}$ 〉〈 β$_{4}$ | . ( v$_{1}$, v$_{3}$, v$_{4}$ ), ( v$_{1}$, v$_{2}$, v$_{3}$ ), ( v$_{1}$, v$_{2}$, v$_{4}$ ) and ( v$_{2}$, v$_{3}$, v$_{4}$ ) are x − y − z = 1, x + y + z = 1, − x − y + z = 1 and − x + y − z = 1, respectively. It is easy to show that these plane equations are the same planes with the planes ( o (+) 1 , o ( − ) 2 , o ( − ) 3 ), ( o (+) 1 , o (+) 2 , o (+) 3 ), ( o ( − ) 1 , o ( − ) 2 , o (+) 3 ) and ( o ( − ) 1 , o (+) 2 , o ( − ) 3 ), respectively. 6