Difficulties in analytic computation for relative entropy of entanglement

It is known that relative entropy of entanglement for an entangled state $\rho$ is defined via its closest separable (or positive partial transpose) state $\sigma$. Recently, it has been shown how to find $\rho$ provided that $\sigma$ is given in a two-qubit system. In this article we study the reverse process, that is, how to find $\sigma$ provided that $\rho$ is given. It is shown that if $\rho$ is of a Bell-diagonal, generalized Vedral-Plenio, or generalized Horodecki state, one can find $\sigma$ from a geometrical point of view. This is possible due to the following two facts: (i) the Bloch vectors of $\rho$ and $\sigma$ are identical to each other; (ii) the correlation vector of $\sigma$ can be computed from a crossing point between a minimal geometrical object, in which all separable states reside in the presence of Bloch vectors, and a straight line, which connects the point corresponding to the correlation vector of $\rho$ and the nearest vertex of the maximal tetrahedron, where all two-qubit states reside. It is shown, however, that these properties are not maintained for the arbitrary two-qubit states.

Figure 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 $o_1^{\pm }$, $o_2^{\pm }$, and $o_3^{\pm }$. 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 $(v_4,o_1^{(-)},o_2^{(-)},o_3^{(-)})$.

Figure 2

How to find the CSS for the Bell-diagonal state. First, extend the line segment between $\rho$ and the point corresponding to the nearest vertex of $\mathcal{T}$. Second, compute the coordinate of the crossing point between the line and the nearest surface of the octahedron $\mathcal{L}$. Finally, find the CSS of $\rho$ which corresponds to the crossing point.

Figure 3

The deformation of $\mathcal{T}$ is plotted when $r=s=0.3$ (a), $r=-s=0.3$ (b), $r=s=0.5$ (c), and $r=-s=0.5$ (d). For comparison, we plot $\mathcal{T}$ together. The appearance of nonzero Bloch vectors generally shrinks the tetrahedron. The shrinking rate becomes larger with increasing norm of the Bloch vectors.

Figure 4

The deformation of $\mathcal{L}$ is plotted when $r=s=0.3$ (a), $r=-s=0.3$ (b), $r=s=0.5$ (c), and $r=-s=0.5$ (d). For comparison, we plot $\mathcal{L}$ together. The appearance of nonzero Bloch vectors generally shrinks the octahedron. The shrinking rate becomes larger with increasing norm of the Bloch vectors.

Figure 5

How to find the CSS for the VP states. Panels (a) and (b) correspond to $r=s=0.3$ and $r=s=0.5$, respectively. To find a CSS, make a straight line $\ell$ first, which connects the nearest vertex of $\mathcal{T}$ and a point ${\mathit{t}}_{\mathrm{vp}}$. Second, compute the coordinate for the intersection point between $\ell$ and ${\mathcal{L}}_{r,s}$. Third, identify the crossing point with ${\mathit{\tau }}_{\mathrm{vp}}$. Keeping ${\mathit{u}}_{\mathrm{vp}}={\mathit{r}}_{\mathrm{vp}}$ and ${\mathit{v}}_{\mathrm{vp}}={\mathit{s}}_{\mathrm{vp}}$, one can find the CSS of the VP state.

Figure 6

How to find the CSS for the generalized Horodecki states. Panels (a) and (b) correspond to $r=-s=0.3$ and $r=-s=0.5$ respectively. In order to find the CSS, make a straight line $\ell$ first, which connects the nearest vertex of $\mathcal{T}$ and a point ${\mathit{t}}_H$. Second, compute the coordinate for the nearest crossing point between $\ell$ and ${\mathcal{L}}_{r,s}$. Third, identify the crossing point with ${\mathit{\tau }}_H$. Keeping ${\mathit{u}}_H={\mathit{r}}_H$ and ${\mathit{v}}_H={\mathit{s}}_H$, one can find the CSS of the Horodecki state.

Figure 7

Illustration of how property (ii) is not maintained for the arbitrary two-qubit states. The four panels correspond to $r=s=0.3$ (a), $r=-s=0.3$ (b), $r=s=0.5$ (c), and $r=-s=0.5$ (d). For convenience, we plot $\mathcal{T}$ and ${\mathcal{L}}_{r,s}$ together in each figure. Each line represents a set of $\mathit{t}$ whose CSS has the same $\mathit{\tau }$. Not all lines pass one of the vertices of $\mathcal{T}$. This fact indicates that property (ii) is not maintained for the general mixtures.