Editing Tsirelson's bound (section)

==Bound for the CHSH inequality==

The first Tsirelson bound was derived as an upper bound on the correlations measured in the [[CHSH inequality]]. It states that if we have four ([[Self-adjoint operator|Hermitian]]) dichotomic observables <math>A_0</math>, <math>A_1</math>, <math>B_0</math>, <math>B_1</math> (i.e., two observables for [[Alice and Bob|Alice]] and two for [[Alice and Bob|Bob]]) with outcomes <math>+1, -1</math> such that <math>[A_i, B_j] = 0</math> for all <math>i, j</math>, then

: <math> \langle A_0 B_0 \rangle + \langle A_0 B_1 \rangle + \langle A_1 B_0 \rangle - \langle A_1 B_1 \rangle \le 2\sqrt{2}.</math>

For comparison, in the classical case (or local realistic case) the upper bound is 2, whereas if any arbitrary assignment of <math>+1, -1</math> is allowed, it is 4. The Tsirelson bound is attained already if Alice and Bob each make measurements on a [[qubit]], the simplest non-trivial quantum system.

Several proofs of this bound exist, but perhaps the most enlightening one is based on the Khalfin–Tsirelson–Landau identity. If we define an observable

: <math> \mathcal{B} = A_0 B_0  +  A_0 B_1  +  A_1 B_0  -  A_1 B_1, </math>

and <math>A_i^2 = B_j^2 = \mathbb{I}</math>, i.e., if the observables' outcomes are <math>+1, -1</math>, then

: <math> \mathcal{B}^2 = 4 \mathbb{I} - [A_0, A_1] [B_0, B_1]. </math>

If <math>[A_0, A_1] = 0</math> or <math>[B_0, B_1] = 0</math>, which can be regarded as the classical case, it already follows that <math>\langle \mathcal{B} \rangle \le 2</math>. In the quantum case, we need only notice that <math>\big\|[A_0, A_1]\big\| \le 2 \|A_0\| \|A_1\| \le 2</math>, and the Tsirelson bound <math>\langle \mathcal{B} \rangle \le 2\sqrt{2}</math> follows.