CHSH game sharing more general states

Question

In the CHSH game, when the players share EPR states they can achieve best the winning rate of approximately 85% which can be proved through the Tsirelson's bound. I wonder what the best winning rate will be if the players share more general states. In a previous question Worst Bell inequality violation with non-maximally entangled state? it was mentioned that if the players share state $c_0|00\rangle+c_1|11\rangle$, the maximal violation will be $2\sqrt{1+4|c_0c_1|^2)}$. Are there any more results like sharing a mixed state?

Answer 1

For a general two-qubit state, including mixed ones, the maximal violation was found by the Horodeccy in 1995. If you write the state in the Block decomposition as $$\rho = \frac{1}{4} \sum_{i,j = 0}^{3} M_{ij} (\sigma_i \otimes \sigma_j)$$ where $\sigma_i$ are the Pauli matrices, then the maximal violation is $2 \sqrt{\lambda_1^2 + \lambda_2^2}$, where $\lambda_1, \lambda_2$ are the two largest singular values of the lower right $3 \times 3$ block of $M$.

For a pure two-qudit state of local dimension $d$ the maximal violation is not known, but there exists a (probably correct) conjecture by Gisin and Peres from 1992. First write the state in the Schmidt decomposition $$|\psi\rangle = \sum_{i=1}^d c_i |ii\rangle,$$ and sort the Schmidt coefficients $c_i$ in decreasing order. Then define $$K = 2\sum_{i=1}^{\lfloor d/2 \rfloor} c_{2i-1}c_{2i}.$$ For even $d$ the violation is given by $$2\sqrt{1+K^2},$$ and for odd $d$ by $$2\sqrt{(1-c_d^2)^2 + K^2} + 2c_d^2$$