Tsirelson’s Bound, Derived

Quantum mechanics violates the CHSH inequality, but it does not violate it arbitrarily. The quantum value is capped at $S \le 2\sqrt2$ — Tsirelson's bound. We derive it from a single operator identity, the cleanest route to a result that underpins all of device-independent quantum information.

The Bell operator

Encode Alice's two measurements as $\pm1$-valued observables $A_0, A_1$ and Bob's as $B_0, B_1$. Being $\pm1$-valued and Hermitian, each squares to the identity: $A_i^2 = B_j^2 = I$. The CHSH value is the expectation of the Bell operator

so that $S = \langle\psi|\mathcal B|\psi\rangle$. The largest $S$ over all states is the operator norm $\|\mathcal B\|$ (its top eigenvalue), so bounding $\|\mathcal B\|$ bounds $S$.

Squaring the operator

Because Alice's operators act on a different factor than Bob's, $[A_i, B_j]=0$. Expanding $\mathcal B^2$ and using $A_i^2 = B_j^2 = I$ and $(B_0+B_1)(B_0-B_1)=B_0^2 - B_1^2 + [B_1,B_0] = [B_1,B_0]$, all the square terms collect into $4\,I$ and the cross terms collapse to a single commutator product:

This is an exact operator identity, valid in any quantum theory with the tensor-product structure.

Bounding the norm

Taking norms and using the triangle inequality together with $\|XY\|\le\|X\|\,\|Y\|$,

For a $\pm1$ observable, $\|A_i\|=1$, so $\|[A_0,A_1]\| = \|A_0A_1 - A_1A_0\| \le 2\|A_0\|\|A_1\| = 2$, and likewise $\|[B_0,B_1]\|\le2$. Therefore

Saturation

The bound is tight. For single-qubit Pauli observables $A_i = \hat a_i\cdot\vec\sigma$, the commutator is $[A_0,A_1] = 2i(\hat a_0\times\hat a_1)\cdot\vec\sigma$, with norm $2|\hat a_0\times\hat a_1|=2$ exactly when $\hat a_0\perp\hat a_1$. Choosing Alice's directions orthogonal, Bob's directions orthogonal, and the maximally entangled state aligns everything so that $\mathcal B^2 = 8\,I$ and $S = 2\sqrt2$ — the angles of the previous lesson.

The takeaway

The single identity $\mathcal B^2 = 4I - [A_0,A_1]\otimes[B_0,B_1]$ yields $S\le 2\sqrt2$ in two lines. Quantum nonlocality is real but bounded, and the bound is saturated by exactly the orthogonal- measurement strategy on a maximally entangled pair.

Try it

Plug the optimal orthogonal observable choices into the norm inequality: each commutator norm is $2$, so the bound is $\sqrt{4 + 2\cdot2}$. Return that number — you should get $2\sqrt2 \approx 2.828$, whose square is exactly $8$.

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