It follows thatρ≥0 iff (x,y,z) lie in the intersection of the 4 half spaces:
x±y∓z≥−1
√
3
, −x±y±z≥−1
√
3
(2.4)
This defines a regular tetrahedron with vertices
1
√
3(1, 1 ,−1),
1
√
3
(1,− 1 ,1)
1
√
3
(− 1 , 1 ,1),
1
√
3
(− 1 ,− 1 ,−1) (2.5)
The vertices of the tetrahedron lie at the corners of the cube in Fig. 6, at unit
distance from the origin. It follows that the vertices of the tetrahedron represent
pure states. As the section represents states with maximally mixed subsystems,
the four pure states are maximally entangled: They are the 4 Bell states
The pairwise averages of the four corners of the tetrahedron give the
( 4
2)
= 6
vertices of the octahedron in Fig. (6). By Eq. (7.10) below, these averages represent
separable states. It follows that the octahedron represents separable states.
Figure 6: A 3D cross section in the space of states of 2 qubits. The 4 Bell states
lie at the vertices of the tetrahedron of states. The octahedron shows the sepa-
rable states. The vertices of the cube represent the extreme points of the, trace
normalized, entanglement witnesses. The cube is inscribed in the unit sphere (not
shown).
The cube, being the dual of the octahedron, represents the trace-normalized
entanglement witnesses (see section 7.6). Ifsis a vector inside the octahedron and
wa vector inside the cube then
0 ≤Tr(Wρs) =1 + 3s·w
4