Optimizing entanglement in two-qubit systems

To construct states that satisfy the $\Delta=0$ requirement, let us first consider that only condition (3) is satisfied. In this case the density operators (1) require only 7 real parameters and are X–shaped,

Since coherence is a basis-dependent concept, states represented by non-diagonal density operators contain a certain amount of coherence with respect to the chosen basis. To be concrete, as density operators (1) are written in terms of the computational basis, their nonzero coherences $r_{kj}$, $k\neq j$, determine the possibility of quantum interference with respect to the distinguishable alternatives $|e_{k}\rangle$ and $|e_{j}\rangle$. In this way, for X–states (5), interference is only allowed between $|e_{1}\rangle=|00\rangle$ and $|e_{4}\rangle=|11\rangle$, as well as between $|e_{2}\rangle=|01\rangle$ and $|e_{3}\rangle=|10\rangle$. Any other interference between the elements of the basis is not allowed. In the most general case both interferences occur simultaneously, but entanglement arises when at least one of them is activated.

To analyze the properties of X–states as generally as possible, it is necessary to ensure that $\rho_{X}$ is in fact a density operator. By construction, $\rho_{X}$ is self-adjoint and normalized, so we just need to make sure that it is positive semidefinite.

After calculating the eigenvalues of $\rho_{X}$, see explicit expressions in Eq. (A-1) of Appendix A, we find that they are non-negative as long as the following conditions are met [37]:

That is, $\rho_{X}$ represents an admissible quantum state, pure or mixed, only if the coherence amplitudes $|r_{14}|$ and $|r_{23}|$ are upper bounded by the population of the corresponding states. Note that the phases of $r_{14}$ and $r_{23}$ play no role in characterizing $\rho_{X}$ as a positive semidefinite matrix. Nevertheless, they characterize quantum interference with respect to the corresponding distinguishable alternatives.

The positive partial transposition of the density operator (1) obeys the following transformation rules

Without loss of generality, we will use the transposition with respect to the second qubit.

The transposition of the X–state (5) corresponds to make $r_{14}\rightarrow r_{23}$. Eigenvalues (A-1) and eigenvectors (A-2) must be transformed accordingly to solve the eigenvalue equation $\rho_{X}^{T_{2}}|\widetilde{\epsilon}\,\rangle=\widetilde{\Lambda}|\widetilde{\epsilon}\,\rangle$. Then, it is found that $\rho_{X}^{T_{2}}$ is positive semidefinite whenever $|r_{23}|\leq\sqrt{r_{11}r_{44}}$ and $|r_{14}|\leq\sqrt{r_{22}r_{33}}$. Applying the Peres–Horodecki criterion [19, 20], we find that $\rho_{X}$ is entangled as long as one of the following inequalities hold [37]:

This result imposes a lower limit on the coherence amplitudes that is unexpectedly determined by the populations of the complementary states. That is, to find entanglement, $|r_{14}|$ must be lower bounded by the population of states $|e_{2}\rangle=|01\rangle$ and $|e_{3}\rangle=|10\rangle$. Similar conclusions hold for the coherence amplitude $|r_{23}|$ and populations $r_{11}$ and $r_{44}$.

Thus, once the populations of the X–states are known, the coherences determine both the eligibility of $\rho_{X}$ as a quantum state and the possibility of finding entanglement.

In what follows we assume that $x_{0}$ and $y_{0}$ are known in advance. The description in terms of the point $p=(x,y)\in\mathcal{S}$ corresponds to determining the possible entanglement with respect to the coherences $r_{14}$ and $r_{23}$.

The convex set $\mathcal{S}$ can be divided into regions (subsets) whose points define separable states and regions where entanglement may be found. We distinguish three main classes, described below.

$\bullet$ Region $\mathcal{M}_{0}$. No entanglement is allowed if $r_{11}r_{44}=r_{22}r_{33}\neq 0$. That is, none of inequalities (28)-(29) apply if $x_{0}=y_{0}\neq 0$. Instead we have $0\leq x,y\leq x_{0}$, so we arrive at a square $\square_{x_{0}}$ of sides $x_{0}$, with $x_{0\max}=1/4$, see Figure 4. These points produce separable mixed states $\rho_{X}$ that can be identified by their rank. According to the eigenvalues (A-1), the vertex $q_{\star}=(x_{0},x_{0})$ represents rank-2 states. In turn, $(x_{0},y)$ and $(x,y_{0})$, located respectively along the right-vertical and top sides of the square, define rank-3 states. Any other point of $\square_{x_{0}}$ gives rank-4 states.

As an example consider the states $\rho_{B}=\rho_{S}$ shown in Figure 3(a). Entanglement is lost precisely at $q=1/2$, where $x_{0}=y_{0}=1/4$ puts vertex $q_{\star}=(x_{0},x_{0})$ to the middle of the hypotenuse of $\mathcal{S}$. The same is true for the states $\rho_{W_{k}}$ exhibited in Figure 2, at $q=0$.

$\bullet$ Region $\mathcal{M}_{1}$. To satisfy inequalities (28) it is necessary to take the ordering $0\leq y_{0}<x_{0}$, see for example the shaded-red area in Figures 2 and 3. The entanglement set is a rectangle lying on the horizontal leg of $\mathcal{S}$, as shown in Figure 5(a). The square $\square_{y_{0}}$ of separable states is located just to the left of such rectangle.

Using the eigenvalues (A-1) we find that vertex $q_{0}=(x_{0},y_{0})$ gives rank-2 states. Points along the right-vertical side and the top of the rectangle, $(x_{0},y)$ and $(x,y_{0})$ respectively, yield rank-3 states. Any other point of this entanglement set provides rank-4 states.

A very important subset $\mathcal{M}_{x}\subset\mathcal{M}_{1}$, defined by the coherence amplitudes $y=0$ (regardless of $y_{0}$), provides points restricted to the horizontal leg of $\mathcal{S}$. Hereafter, the states defined by points $(x,0)\in\mathcal{M}_{x}$ will be denoted by $\rho_{\mathcal{M}_{x}}$. Since $y=0$, to find entanglement, our interest is addressed to the coherence amplitudes $x$ that satisfy $y_{0}<x\leq x_{0}$. As an example consider the Werner–states $\rho_{W_{1,2}}$ illustrated in Figure 2(a), as well as states $\rho_{S}$ shown in the second row of Figure 3.

In general, by extending the domain of $x$ to the maximum we optimize the entanglement in $\mathcal{M}_{x}$. For this, it is enough to set $y_{0}=0$. We find three different sets of states

Neither of the first two matrices in (30) satisfy the factorization of their elements as required in Proposition Q, so they represent mixed states of rank-3 or rank-2, see Eq. (A-6). In turn, if the elements of the third matrix are factorized we have $x=x_{0}=|\alpha_{1}\alpha_{4}|$, so it represents the pure (rank-1) state $|\psi_{14}\rangle$ defined in (A-3). In particular, if $x=x_{0\max}=1/2$ we recover the Bell-states $|\beta_{1}\rangle$ and $|\beta_{2}\rangle$. If their elements cannot be factorized, this matrix represents a mixed state of rank-2, see Figure 5(b).

$\bullet$ Region $\mathcal{M}_{2}$. Inequalities (29) are satisfied by imposing the ordering $0\leq x_{0}<y_{0}$, see for example the shaded-blue area in Figures 2 and 3. The entanglement subset is also a rectangle, but in this case the base is placed just above the separable-states square $\square_{x_{0}}$, see Figure 6(a). Points along the right-vertical side and the top of the rectangle, respectively $(x_{0},y)$ and $(x,y_{0})$, give rise to rank-3 states. In turn, vertex $q_{0}=(x_{0},y_{0})$ corresponds to rank-2 states. Any other point provides rank-4 states.

In this case the subset $\mathcal{M}_{y}\subset\mathcal{M}_{2}$, defined by the coherence amplitudes $x=0$, provides points restricted to the vertical leg of $\mathcal{S}$ and defines states $\rho_{\mathcal{M}_{y}}$. Our interest now turns to the region $x_{0}<y\leq y_{0}$. As an example consider the Werner–states $\rho_{W_{3,4}}$ exhibited in Figure 2(b) and the $y$-version of the states $\rho_{S}$ shown in the second row of Figure 3. The optimization is achieved by setting $x_{0}=0$, which results in the following states

The first two matrices do not satisfy Proposition Q, so they represent mixed states of rank-3 or rank-2, see Eq. (A-7). If the elements of the third matrix are factorized we have $y=y_{0}=|\alpha_{2}\alpha_{3}|$, so it represents the pure (rank-1) state $|\psi_{23}\rangle$ defined in (A-4). If $y=y_{0\max}=1/2$, we recover the Bell-states $|\beta_{3}\rangle$ and $|\beta_{4}\rangle$. If their elements cannot be factorized, this matrix is a mixed state of rank-2, see Figure 6(b).

The classification developed above is very close to the parametrization of two-qubit X–density matrices with a fixed rank reported in Section 2.1 of [38]. The latter, based on the coefficients that define the characteristic equation for the density operator, considers the Newton-Girard formulae [52] and a very concrete parametrization of the matrix-elements of $\rho_{X}$. Beyond using the computational basis, our treatment is free of parametrization for the matrix elements $r_{jk}$, so it aims to be more practical and, therefore, universal. Nevertheless, the parameters used in [38] are very valuable as they comprise the information needed to search for a unitary transformation that converts arbitrary two-qubit states into their X counterparts.

In the previous sections we have identified the regions of the convex set $\mathcal{S}$ whose points would imply entanglement in state $\rho_{X}$. We know, for example, that squares $\square_{x_{0}}$ and $\square_{y_{0}}$ must be discarded because all their points represent separable states. However, we have not quantified the amount of entanglement that distinguishes some points from others in any other region, nor have we analyzed which region offers better entanglement conditions. To address these points we proceed as follows.

We seek to ensure that entangled states “resemble” separable states as little as possible. Translated into the language of the convex set $\mathcal{S}$, this statement means that if $p$ encodes entanglement then it should be as far away as possible from $\square_{x_{0}}$ and $\square_{y_{0}}$. The further remote, the greater the entanglement. With this in mind, consider the metric $d(\vec{r}_{1},\vec{r}_{2})=\max\{|x_{2}-x_{1}|,|y_{2}-y_{1}|\}$, where $\vec{r}_{k}=(x_{k},y_{k})\in\mathbb{R}^{2}$, $k=1,2$. We now introduce a geometric measure of entanglement for two-qubit systems.

Definition. ($L$-measure of entanglement) Given the extreme points $x_{0}$ and $y_{0}$, let $\mathcal{S}$ be the convex set (27) and $\square_{z_{0}}\subset\mathcal{S}$ be the subset of points defining separable states. The $L$-measure of entanglement, applied to the point $p\in\mathcal{S}$, is the distance between $p$ and the closest point $q\in\square_{z_{0}}$ given by

with $\xi_{0}$ a normalization factor.

Let us apply the measure $L$ to the points $p\in\mathcal{S}$ that define the states classified in the previous section.

States $\rho_{\mathcal{M}_{1}}$ are such that $\inf d(q,p)$ is given by the points $q=(y_{0},y)$, see Figure 7(a). Therefore

Note that $x=x_{0}$ defines the furthest point from the zone of separable states, therefore $L_{\max}(\rho_{\mathcal{M}_{1}})=2(x_{0}-y_{0})$. That is, among these states, the highest value of the $L$-measure is provided by the rank-3 states associated with the points $(x_{0},y)$.

For states $\rho_{\mathcal{M}_{2}}$ we find that $q=(x,x_{0})$ defines $\inf d(q,p)$, see Figure 7(b). Then

In this case the highest value of the $L$-measure is provided by the rank-3 states associated with the points $(x,y_{0})$, so that $L_{\max}(\rho_{\mathcal{M}_{2}})=2(y_{0}-x_{0})$.

Combining the previous results we obtain the general rule

To find entanglement it is enough for one of the subtractions of (34) to be positive.

As for systems composed of subsystems with equal entropy, see Section 2.1.1, we have immediate results:

To illustrate the applicability of the above expressions let us consider the state (11)-(12). From (35) one has

Therefore, $\left.\rho_{B}\right|_{b_{3}=b_{4}=\kappa}$ is disentangled if $|b_{1}-b_{2}|\leq 2\kappa=1-(b_{1}+b_{2})$. For these parameters the interference between $|e_{1}\rangle=|00\rangle$ and $|e_{4}\rangle=|11\rangle$ is so small that the system ends up behaving like a statistical mixture (a classical state) of the base states. On the other hand, for $|b_{1}-b_{2}|>2\kappa$ we find entanglement with measure $\left.L(\rho_{B})\right|_{b_{3}=b_{4}=\kappa}=|b_{1}-b_{2}|-2\kappa$ . The extreme condition $\kappa=0$ shows that the statistical mixture of the Bell–states $|\beta_{1}\rangle$ and $|\beta_{2}\rangle$ is entangled, with a measure $L$ that is greater as the interference between $|00\rangle$ and $|11\rangle$ is more intense. In particular, setting $b_{1}=q$ and $b_{2}=1-q$ we have $\left.L(\rho_{B})\right|_{b_{3}=b_{4}=0}=|2q-1|$. So state $\left.\rho_{B}\right|_{b_{3}=b_{4}=0}$ is disentangled for $q=1/2$ ($b_{1}=b_{2}$), as it is shown in Figures 3(a) and 3(d).

In general, if $x$ and $y$ take the maximum value in their respective domains we have

That is, when $x_{0}$ and $y_{0}$ are nonzero and different from each other, the strongest entanglement is exhibited by rank-3 states $\rho_{X}$. On the other hand, if $x_{0}=0$ or $y_{0}=0$, the $L$-measure identifies rank-1 (pure) states as those with greater entanglement. In particular, if $x_{0}=x_{0\max}=1/2$ or $y_{0}=y_{0\max}=1/2$, such states are not only pure but coincide with one of the Bell–states.

The most notable of the previous results is that the mathematical expression of the geometric measure $L$ coincides precisely with the Hill-Wootters concurrence calculated for the $\rho_{X}$ states [35, 36, 37]. In fact, according with [45, 46], the concurrence for any state $\rho$ is given by $C(\rho)=\max\{0,\,\sqrt{\ell_{1}}-\sqrt{\ell_{2}}-\sqrt{\ell_{3}}-\sqrt{\ell_{4}}\}$. The $\ell_{k}$’s are the eigenvalues (in decreasing order) of the matrix $R_{\rho}=\rho\left(\sigma_{2}\otimes\sigma_{2}\right)\rho^{*}\left(\sigma_{2}\otimes\sigma_{2}\right)$, with $\sigma_{2}$ the well-known anti-diagonal Pauli-matrix.

Using state $\rho_{X}$ to construct $R_{\rho}$, a simple calculation produces the eigenvalues $(x_{0}\pm x)^{2}$ and $(y_{0}\pm y)^{2}$, where no order has yet been established. From (26) we know that the quantities in parentheses are not negative, so we have two different ways of sorting $\sqrt{\ell_{k}}$ in descending order, with $y_{0}+y$ or $x_{0}+x$ leading. A little algebra produces the result