Beating the Clauser-Horne-Shimony-Holt and the Svetlichny games with optimal states

We study the relation between the maximal violation of Svetlichny's inequality and the mixedness of quantum states and obtain the optimal state (i.e., maximally nonlocal mixed states, or MNMS, for each value of linear entropy) to beat the Clauser-Horne-Shimony-Holt and the Svetlichny games. For the two-qubit and three-qubit MNMS, we showed that these states are also the most tolerant state against white noise, and thus serve as valuable quantum resources for such games. In particular, the quantum prediction of the MNMS decreases as the linear entropy increases, and then ceases to be nonlocal when the linear entropy reaches the critical points $2/3$ and $9/14$ for the two- and three-qubit cases, respectively. The MNMS are related to classical errors in experimental preparation of maximally entangled states.

Figure 1

CHSH nonlocality versus linear entropy for two-qubit states. The red solid curve ${\mathcal{S}}_2\left({\rho }_2^{\mathrm{MNMS}}\right)=\sqrt{2-3{\mathcal{E}}_L/2}$ is for the MNMS, which maximizes ${\mathcal{S}}_2\left({\rho }_2^{\mathrm{MNMS}}\right)$ for each value of ${\mathcal{E}}_L$ (${\mathcal{E}}_L\le 2/3$ so that the CHSH inequality is violated). Pink and blue areas indicate, respectively, arbitrary two-qubit nonlocal and local states in the ${\mathcal{E}}_L\text{−}{\mathcal{S}}_2$ plane. The green dotted curve ${\mathcal{S}}_2=\sqrt{3-3{\mathcal{E}}_L}$ [corresponding to the state (11) with $\vec{a}=\vec{b}=0$ and $c_3=0]$ denotes the maximal ${\mathcal{S}}_2$ for each value of ${\mathcal{E}}_L>2/3$. The blue solid curve ${\mathcal{S}}_2=\sqrt{1-3{\mathcal{E}}_L/2}$ [corresponding to the states $\rho =p|00\rangle \langle 00|+(1-p)|11\rangle \langle 11|]$ denotes the minimal ${\mathcal{S}}_2$ for each fixed value of ${\mathcal{E}}_L\le 2/3$. For comparison, we also plot the MEMS denoted by the blue dashed curve ${\mathcal{S}}_2\left({\rho }_2^{\mathrm{MEMS}}\right)=(\sqrt{2}+\sqrt{2-3{\mathcal{E}}_L})/2$ for ${\mathcal{E}}_L\in [0,16/27]$ and ${\mathcal{S}}_2\left({\rho }_2^{\mathrm{MEMS}}\right)=\sqrt{25-27{\mathcal{E}}_L-min\{1,3(8-9{\mathcal{E}}_L)/2\}}/3$ for ${\mathcal{E}}_L\in (16/27,8/9]$.

Figure 2

The quantum winning probability in the CHSH game versus concurrence $\gamma$ for the MNMS (red solid curve), MEMS (blue dashed curve), and arbitrary states (pink region) that violate the CHSH inequality. Here the concurrence depicts degree of entanglement of two qubits. If the quantum state shared by Alice and Bob is the MNMS, the quantum winning probability equals ${\mathrm{Pr}}_2^{\mathrm{MNMS}}=(2+\sqrt{1+{\gamma }^2})/4$, while for the MEMS the winning probability is found to be ${\mathrm{Pr}}_2^{\mathrm{MEMS}}=(6+\sqrt{1+18{\gamma }^2-min\{1,9{\gamma }^2\}})/12$ for $\gamma \in [0,2/3]$ and ${\mathrm{Pr}}_2^{\mathrm{MEMS}}=(2+\sqrt{2}\gamma )/4$ for $\gamma \in (2/3,1]$. It is clearly shown that ${\mathrm{Pr}}_2^{\mathrm{MNMS}}$ surpasses ${\mathrm{Pr}}_2^{\mathrm{MEMS}}$ except $\gamma =1$, at which the MNMS and MEMS are the Bell states resulting in the largest probability $\frac{1}{4}(2+\sqrt{2})$.

Figure 3

The genuine multipartite nonlocality versus the linear entropy for three-qubit states. Point $A$ corresponds to the GHZ state $\left(\right|000\rangle +|111\rangle )/\sqrt{2},B$ to the MNMS (27) at ${\mathcal{E}}_L=9/14,C$ to $cos\pi /8|000\rangle +sin\pi /8|111\rangle$, and $D$ to the MNMS (27) at ${\mathcal{E}}_L=6/7$. The red curves $AB$ and $BD$ correspond to the maximal violation of SI with state (27) for each ${\mathcal{E}}_L$ [see also Eqs. (32) and (33)]. The solid line crossing points $B,C$, and $D$ is the classical bound. The blue points represent a great number of randomly chosen states.