Maximally discordant separable two-qubit $X$ states

Abstract

In a recent article, Gharibian (Phys Rev A 86:042106 2012) has conjectured that no two-qubit separable state of rank greater than two could be maximally non-classical (defined to be those which have normalized geometric discord 1/4) and asked for an analytic proof. In this work, we prove analytically that among the subclass of $X$ states, there is a unique (up to local unitary equivalence) maximal separable state of rank two. For the general case, we derive some necessary conditions.

Appendix: Proof of the optimization in Proposition 1

We first note that the constraint (13a) implies $abcd\ne 0$, $a\ne c$, $b\ne d$. Now, let us try to parameterize $(a,b,c,d)$ using the constraints. To absorb the first constraint, without loss of generality, we can take $a=bk,c=dk,k>0$. Then, the constraint (13b) becomes

$$\begin{aligned} b+d=\frac{1}{k+1}, \end{aligned}$$

(19)

and we are left with only the following constraint

$$\begin{aligned} (a-c)^2+(b-d)^2&= 8bc\nonumber \\ \Rightarrow (b-d)^2(k^2+1)&= 8bdk, \end{aligned}$$

(20)

where we have just substituted $a$ and $c$ in terms of $bk$ and $dk$, respectively, in the first equation. Squaring (19) and subtracting (20) from it yields

$$\begin{aligned} 4bd=\frac{k^2+1}{(k+1)^4} \end{aligned}$$

(21)

Noting that $ad=bdk$, we have to find the maximum of the function

$$\begin{aligned} f(k)=\frac{k(k^2+1)}{(k+1)^4} \end{aligned}$$

subject to $k>0$. The derivatives are very easy to calculate. Indeed, $f'(k)=0$ only at $k=1$ and $f'(1)=f''(1)=f'''(1)=0$, but $f''''(1)=-3/16<0$. Hence, the unique maximum occurs at $k=1$. From (19) and (21), this corresponds to the solution $a=b=(2\pm \sqrt{2})/8,c=d=1/(32a)$.