Tsirelson’s Bound

Quantum mechanics violates the classical CHSH bound $|S| \le 2$ , reaching $2\sqrt2$ with the right settings. A natural question: could a cleverer state or measurement push $S$ even higher? Tsirelson's bound answers no — $2\sqrt2$ is the absolute quantum ceiling.

The statement

Tsirelson's bound (1980). For any quantum state and any choice of $\pm 1$ -valued local observables, the CHSH value satisfies
$|S| \;\le\; 2\sqrt2 \approx 2.828.$

Compare the three regimes:

Local realism: $|S| \le 2$ .
Quantum mechanics: $|S| \le 2\sqrt2$ .
No-signalling (most general): $|S| \le 4$ .

Quantum theory sits strictly between classical and the most extreme no-signalling correlations.

Why $2\sqrt2$ — the operator argument

Math you'll use here: the operator norm. For a Hermitian operator $A$ , its operator (spectral) norm $\|A\|$ is the largest $|\lambda|$ over its eigenvalues — the biggest stretch $A$ can apply to any unit vector. Three facts drive the argument below:

An expectation can never exceed the norm: $|\langle A\rangle| \le \|A\|$ , since $\langle A\rangle$ is an average of eigenvalues.
A $\pm1$ -valued observable is involutory ( $A^2 = \mathbb{1}$ ), so every eigenvalue is $\pm1$ and $\|A\| = 1$ .
The norm is submultiplicative, $\|AB\| \le \|A\|\,\|B\|$ , which (with the triangle inequality on $AB - BA$ ) gives $\|[A,A']\| \le 2\,\|A\|\,\|A'\|$ , and $\|A^2\| = \|A\|^2$ .

So once we bound $\|\mathcal{S}^2\|$ , the chain $|S| = |\langle\mathcal{S}\rangle| \le \|\mathcal{S}\| = \sqrt{\|\mathcal{S}^2\|}$ converts it into a bound on the CHSH value.

Represent Alice's two measurements by Hermitian operators $A, A'$ and Bob's by $B, B'$ , each with eigenvalues $\pm 1$ , so $A^2 = A'^2 = B^2 = B'^2 = \mathbb{1}$ . Alice's operators commute with Bob's (they act on different particles). The CHSH operator is

\mathcal{S} = A \otimes B + A \otimes B' + A' \otimes B - A' \otimes B'.

Square it. Using $A^2 = \mathbb{1}$ etc. and that $A$ -operators commute with $B$ -operators, the cross-terms collapse and one finds the remarkable identity

\mathcal{S}^2 = 4\,\mathbb{1} - [A, A'] \otimes [B, B'].

The operator norm obeys $\|\mathcal{S}^2\| \le 4 + \|[A,A']\|\,\|[B,B']\|$ . For $\pm 1$ observables each commutator norm is bounded by $\|[A,A']\| \le 2\|A\|\,\|A'\| = 2$ , so

\|\mathcal{S}^2\| \le 4 + 2 \cdot 2 = 8 \quad\Longrightarrow\quad \|\mathcal{S}\| \le \sqrt8 = 2\sqrt2.

Since $|S| = |\langle \mathcal{S}\rangle| \le \|\mathcal{S}\|$ , the CHSH value can never exceed $2\sqrt2$ .

The bound is tight

The previous lesson exhibited a singlet state and measurement angles achieving exactly $2\sqrt2$ . So the bound is not merely an upper limit — it is attained. Maximally entangled two-qubit states with $45^\circ$ -spaced settings saturate it.

The takeaway

The factor $\sqrt2$ separating classical from quantum is not arbitrary: it is the operator-norm signature of non-commuting $\pm 1$ observables. The same commutator $[A, A']$ that forbids simultaneously sharp values — the very feature EPR found objectionable — is exactly what powers the violation, and exactly what caps it at $2\sqrt2$ .