CHSH game sharing a general state

Question

For the CHSH game if the players share a general two-qubit state $\rho = \frac{1}{4} \sum_{i,j = 0}^{3} M_{ij} (\sigma_i \otimes \sigma_j)$, 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 $M$.

My question is what the maximal violation will be if the shared state $\rho$ which is of dimension $2^n$. If $\rho$ is $n$ copies of EPR state, the maximal violation is $2\sqrt{2}$ by Tsireson's bound. For a general state large dimension state $\rho$ does the violation will be easy to calculate?

Answer 1

Having the players share additional EPR pairs will not increase the probability of winning the game. This is because the Tsirelson bound of $2\sqrt{2}$ is the optimal value for the CHSH game amongst all quantum strategies, even those employing an arbitrary number of EPR pairs. Moreover, the self-testing properties of the CHSH game tell you that any quantum strategy (again, even one using any number of qubits) that achieves the Tsirelson bound is up to local isometries equivalent to strategy employing a single EPR pair tensored with an auxiliary "junk" register.