## 9.5 Tsirelson’s inequality

An upper bound on quantum correlations.

One may ask if |\langle S\rangle|= 2\sqrt{2} is the maximal violation of the CHSH inequality, and the answer is “yes, it is”: quantum correlations always satisfy the bound |\langle S\rangle|\leqslant 2\sqrt{2}. This is because, no matter which state |\psi\rangle we pick, the expected value \langle S\rangle = \langle\psi|S|\psi\rangle cannot exceed the largest eigenvalue of S, and we can put an upper bound on the largest eigenvalues of S. To start with, taking the largest eigenvalue (in absolute value) of a Hermitian matrix M, which we denote by \|M\|, gives a matrix norm, i.e. it has the following properties: \begin{aligned} \|M\otimes N\| & = \|M\| \|N\| \\\|MN\| & \leqslant\|M\| \|N\| \\\|M+N\| & \leqslant\|M\| + \|N\| \end{aligned}

Given that \|A_k\|=\|B_k\|=1 (for k=1,2), it is easy to use these properties to show that \|S\|\leqslant 4, but this is a much weaker bound than we want. However, one can show160 that S^2 = 4(\mathbf{1}\otimes\mathbf{1}) + [A_1,A_2]\otimes[B_1,B_2].

Now, the norms of the commutators \|[A_1, A_2]\| and \|[B_1, B_2]\| are bounded by161 2, and \|S^2\|=\|S\|^2. All together, this gives \begin{aligned} \|S^2\| &\leqslant 8 \\\implies \|S\| &\leqslant 2\sqrt{2} \\\implies |\langle S\rangle| &\leqslant 2\sqrt{2} \end{aligned} This result is known as the Tsirelson inequality.

In classical probability theory, the (absolute value of the) average value of the CHSH quantity S = A_1(B_1 - B_2) + A_2(B_1 + B_2) is bounded by 2 (and this bound can be attained); in quantum theory, it is bounded by 2\sqrt{2} (and this bound can also be attained).

1. Exercise. Prove this!↩︎

2. Exercise. Prove this!↩︎