In the framework of the first strategy, for a given two-qubit Bell diagonal state (
5) we first identify the measurements
M that minimize the geometric quantum correlations (
3) and then we use these optimal measurements to compute the geometric classical correlations (
4). In other words, for a given point
\((c_{1}, c_{2}, c_{3})\) of tetrahedron (
6) we first identify unit vectors
\((n_{1}, n_{2}, n_{3})\) that maximize the fidelity (
9) and then we use these optimal vectors to compute the fidelity (
12).
It can be shown that under this strategy
-
if
\(c_{i}^{2} = c_{j}^{2} = c_{k}^{2}\) where
\(i \ne j \ne k\), then all measurements
M are optimal, and geometric quantum and classical correlations are given by
$$\begin{aligned} Q_{B}(\rho )&= \sqrt{2 - 2\sqrt{w_{k}}}, \end{aligned}$$
(13a)
$$\begin{aligned} C_{B}(\rho )&= \sqrt{2 - \sqrt{2 + 2\sqrt{1 - c_{k}^{2}}}}, \end{aligned}$$
(13b)
-
if \(c_{i}^{2} = c_{j}^{2} < c_{k}^{2}\) where \(i \ne j \ne k\), then only measurements M with \(n_{i}^{2} = n_{j}^{2} = 0\) and \(n_{k}^{2} = 1\) are optimal, and geometric quantum and classical correlations are given by Eqs. (13),
-
if \(c_{i}^{2} < c_{j}^{2} = c_{k}^{2}\) where \(i \ne j \ne k\), then only measurements M with \(n_{i}^{2} = 0\) and \(n_{j}^{2} + n_{k}^{2} = 1\) are optimal, and geometric quantum and classical correlations are given by Eqs. (13),
-
if \(c_{i}^{2}< c_{j}^{2} < c_{k}^{2}\) where \(i \ne j \ne k\), then only measurements M with \(n_{i}^{2} = n_{j}^{2} = 0\) and \(n_{k}^{2} = 1\) are optimal, and geometric quantum and classical correlations are given by Eqs. (13),
where coefficients
\(w_{1}\),
\(w_{2}\) and
\(w_{3}\) have the following form
$$\begin{aligned} w_{1}&= \frac{1}{16}\left( \sqrt{(1 - c_{1})(1 - c_{1} - c_{2} - c_{3})} + \sqrt{(1 - c_{1})(1 - c_{1} + c_{2} + c_{3})}\right. \nonumber \\&\phantom {w_{1}} +\,\left. \sqrt{(1 + c_{1})(1 + c_{1} - c_{2} + c_{3})} + \sqrt{(1 + c_{1})(1 + c_{1} + c_{2} - c_{3})}\right) ^{2}, \end{aligned}$$
(14a)
$$\begin{aligned} w_{2}&= \frac{1}{16}\left( \sqrt{(1 - c_{2})(1 - c_{1} - c_{2} - c_{3})} + \sqrt{(1 + c_{2})(1 - c_{1} + c_{2} + c_{3})}\right. \nonumber \\&\phantom {w_{1}} +\,\left. \sqrt{(1 - c_{2})(1 + c_{1} - c_{2} + c_{3})} + \sqrt{(1 + c_{2})(1 + c_{1} + c_{2} - c_{3})}\right) ^{2}, \end{aligned}$$
(14b)
$$\begin{aligned} w_{3}&= \frac{1}{16}\left( \sqrt{(1 - c_{3})(1 - c_{1} - c_{2} - c_{3})} + \sqrt{(1 + c_{3})(1 - c_{1} + c_{2} + c_{3})}\right. \nonumber \\&\phantom {w_{1}} +\, \left. \sqrt{(1 + c_{3})(1 + c_{1} - c_{2} + c_{3})} + \sqrt{(1 - c_{3})(1 + c_{1} + c_{2} - c_{3})}\right) ^{2}. \end{aligned}$$
(14c)
The above results show that geometric classical and quantum correlations based on the Bures distance for two-qubit Bell diagonal states are uniquely determined under the first strategy, despite the fact that for a wide class of two-qubit Bell diagonal states we have more than one optimal measurement
M which means that for those states the classical-quantum state
\(M(\rho )\) cannot be uniquely determined. Therefore, in general non-uniqueness of
\(M(\rho )\) does not necessarily imply non-uniqueness of geometric classical correlations under the first strategy. In the next section, we show that non-uniqueness of
\(M(\rho )\) may, however, imply non-uniqueness of geometric classical correlations under this strategy.
Let us note here that the problem of identification of unit vectors
\((n_{1}, n_{2}, n_{3})\) maximizing the fidelity (
9) for a given point
\((c_{1}, c_{2}, c_{3})\) of tetrahedron (
6) is closely related to the problem of finding classical-quantum states
\(\chi _{\rho }\) that maximize the fidelity
\(F(\rho , \chi _{\rho })\). This problem has been studied in the literature [
21,
22] in the context of the Bures geometric quantum discord in which the Bures distance was applied as the distance measure between a given quantum state and the closest classical-quantum state. Interestingly, in general the Bures geometric quantum discord
\(D_{B}(\rho )\) [
23] is less than or equal to geometric quantum correlations based on the Bures distance under the first strategy
\(Q_{B}(\rho )\), i.e.,
\(D_{B}(\rho ) \le Q_{B}(\rho )\), since
\(M(\rho )\) is always a classical-quantum state for a bipartite state
\(\rho \) [
24]. Of course, the relation
\(D_{B}(\rho ) \le Q_{B}(\rho )\) holds in the case of two-qubit Bell diagonal states as one can verify taking into account our results regarding
\(Q_{B}(\rho )\) and those found in the literature regarding
\(D_{B}(\rho )\) [
23]. Moreover, one can show that this inequality becomes an equality if and only if
\(\rho = M(\rho )\) for optimal measurement
M or
\(\rho \) is a mixture of two Bell states. It is also worth noting that considering the Bures geometric quantum discord for two-qubit Bell diagonal states one cannot uniquely determine the closest classical-quantum state for a wide class of two-qubit Bell diagonal states [
22]. More precisely, it can be done uniquely if and only if a two-qubit Bell diagonal state is represented by the point
\((c_{1}, c_{2}, c_{3})\) being interior point of tetrahedron (
6) and the index
k such that
\(c_{k}^2 = \max (c_{1}^{2}, c_{2}^2, c_{3}^{2})\) is uniquely given. For comparison, in the case of geometric quantum correlations based on the Bures distance under the first strategy the classical-quantum state
\(M(\rho )\) can be uniquely determined if and only if a two-qubit Bell diagonal state is represented by the point
\((c_{1}, c_{2}, c_{3})\) of tetrahedron (
6) and the index
k such that
\(c_{k}^2 = \max (c_{1}^{2}, c_{2}^2, c_{3}^{2})\) is uniquely given, as it was shown above. Moreover, it is worth noting that an alternative approach to geometric classical correlations based on the Bures distance under the first strategy has been considered in the literature [
23]. The classical correlations measure based on the Bures distance introduced in [
23] can be computed analytically for two-qubit Bell diagonal states, like geometric classical correlations based on the Bures distance under the first strategy. However, it can be shown that they are not directly comparable measures of classical correlations, because unlike for
\(D_{B}(\rho )\) and
\(Q_{B}(\rho )\) for which the relation
\(D_{B}(\rho ) \le Q_{B}(\rho )\) holds, a similar relation, valid for all two-qubit Bell diagonal states, cannot be established between these two measures of classical correlations.