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

  • 1
    $\begingroup$ 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. $\endgroup$
    – Rammus
    Oct 20, 2022 at 13:27
  • $\begingroup$ @Rammus thanks for the paper $\endgroup$
    – Bard
    Oct 20, 2022 at 13:49

1 Answer 1


If memory serves...

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 $U_A^\dagger\sigma_2 U_A$.


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