# Choosing operators for maximal Bell-CHSH violation for an arbitrary state

If I have the density matrix $$\rho$$ for a bipartite system (each system is two-dimensional), what set of observables should I choose so that I obtain the maximum possible violation of the Bell-CHSH inequality?

Say $$\rho$$ can be written as: $$\rho = \frac{1}{4} \sum_{n,m = 0}^{3} M_{nm} (\sigma_n \otimes \sigma_m)$$

Where $$\sigma_i$$ belongs to the vector $$\vec \sigma = \{1_2, \sigma_1, \sigma_s, \sigma_3 \}$$, i.e. the $$2 \times 2$$ identity matrix followed by the Pauli matrices. Then, according to section IV.B of this review by Horodecki et al., the maximum possible violation is $$2 \sqrt{\lambda_1^2 + \lambda_2^2}$$, where $$\lambda_1, \lambda_2$$ are the two largest singular values of the matrix $$M_{ij}$$, where $${i, j = 1, 2, 3}$$ (i.e the lower right $$3 \times 3$$ block). But I have not been able to find an algorithm to write down what operators will actually achieve this maximum violation.

How do I find these operators? Or is there any resource which explains the procedure to find these optimal observables?

Cross-posted on physics.SE

• I don't have time to write an answer but I will say that the relevant paper here is Violating Bell inequality by mixed spin-12 states. Unfortunately the paper is behind a paywall and not on arxiv so you may not be able to access it in a straightforward manner. The paper is very short and the proof is constructive so you can reverse engineer the measurement angles from the singular vectors of the correlation matrix. Oct 20, 2022 at 13:27
• @Rammus thanks for the paper
– Bard
Oct 20, 2022 at 13:49

Find the operation that diagonalises the $$3\times 3$$ matrix $$M_{ij}$$ for $$i,j=1,2,3$$. If $$M=V\Lambda W^\dagger$$ ($$\Lambda$$ is diagonal), there is a correspondence between the real $$3\times 3$$ matrices $$V$$ and $$W$$ and local $$2\times 2$$ complex unitaries $$U_A, U_B$$ such that when you calculate $$(U_A\otimes U_B)\rho(U_A\otimes U_B)^\dagger,$$ the two-body terms (i.e. those with no $$\sigma_0$$ component) become $$\lambda_1\sigma_1\otimes\sigma_1+\lambda_2\sigma_2\otimes\sigma_2+\lambda_3\sigma_3\otimes\sigma_3.$$ Now, if you had something diagonal like that, you'd be able to use the standard version of CHSH, $$S=\sqrt{2}\left(\sigma_1\otimes\sigma_1+\sigma_2\otimes\sigma_2\right).$$ Obviously this is not quite what you want, but it's getting close! So, let's assume that Alice's two measurements are $$\sigma_1$$ and $$\sigma_2$$ while Bob's measurements are $$(\lambda_1\sigma_1+\lambda_2\sigma_2)/\sqrt{\lambda_1^2+\lambda_2^2}$$ and $$(\lambda_1\sigma_1-\lambda_2\sigma_2)/\sqrt{\lambda_1^2+\lambda_2^2}$$. In this case, $$S'=\frac{2}{\sqrt{\lambda_1^2+\lambda_2^2}}\left(\lambda_1\sigma_1\otimes\sigma_1+\lambda_2\sigma_2\otimes\sigma_2\right).$$ Now if we calculate $$\text{Tr}\left(S'(U_A\otimes U_B)\rho(U_A\otimes U_B)^\dagger\right)=2\sqrt{\lambda_1^2+\lambda_2^2}$$ as required. So, we see that what you actually want to implement, to measure on $$\rho$$, is $$(U_A\otimes U_B)^\dagger S'(U_A\otimes U_B).$$ Thus Alice, for example, should measure $$U_A^\dagger\sigma_1 U_A$$ and $$U_A^\dagger\sigma_2 U_A$$.