CHSH Inequality and Its Quantum Violation

CHSH long ago stopped being only a test of philosophy and became engineering. The 2015 loophole-free experiments — Hensen and colleagues with electron spins in diamond NV centres 1.3 km apart, and Giustina and Shalm's groups with entangled photons — reported significant $S>2$ while closing the locality and detection loopholes at once — a fifty-year culmination. From there the violation became a resource. In device-independent quantum key distribution the parties spend a fraction of their entangled rounds estimating $S$; a value above $2$ certifies that no eavesdropper holds a classical copy of the key, with security that does not depend on trusting the boxes. In 2018 Bierhorst and colleagues at NIST used a loophole-free Bell test to generate 1024 bits of certified randomness — unpredictability guaranteed by physics, not by any assumption about the device — and public randomness beacons build on the same idea. Self-testing pushes the logic to its limit: observing $S=2\sqrt2$ forces the system to be, up to local isometry, a maximally entangled Bell pair measured along anticommuting observables, certifying quantum hardware as a black box. And the gap between $2\sqrt2$ and $4$ remains an open frontier — the search for principles (information causality, macroscopic locality) that single out Tsirelson's bound as the true edge of nature stays active.

Picture Alice and Bob far apart, each holding a box with a two-position knob and a lamp that flashes red ($+1$) or green ($-1$). A source in the middle sends each one half of a shared system, shot after shot; on every shot each freely picks setting $0$ or $1$ and records the flash. After many rounds they compute, for each of the four setting-pairs, the average product of outcomes — the correlator $E(A_xB_y)$. If a script written before the shot (the hidden variable $\lambda$) already fixed what each lamp would do, the four correlators are chained: you cannot make three pairs agree strongly while the fourth disagrees strongly, and the arithmetic of $\pm1$ caps $S$ at $2$. The first figure is that chaining made visible — the four axes and the $45^\circ$ tilt that quantum mechanics uses to beat the chain.

Quantum mechanics breaks the chain because the outcomes are not written in advance. Choosing a direction is choosing a basis, and for an entangled pair the joint statistics depend only on the relative angle between the axes, through $\cos\theta$. Tilting the four axes in $45^\circ$ steps makes three correlators equally large with one sign and the fourth equally large with the opposite sign — exactly what the $(+,+,+,-)$ pattern rewards, and what no shared script can imitate. The magic factor $\sqrt2$ is pure geometry: spin correlators act like inner products of unit vectors, and the best arrangement of two pairs of axes gives $4\cos45^\circ=2\sqrt2$. The second figure places that number on the ladder of theories.

The CHSH quantity is a single linear combination of four measured correlators, and every claim in this article is a statement about its value. Alice and Bob each hold two dichotomic observables; the correlator $E(A_xB_y)=\langle A_xB_y\rangle$ is the average product of their $\pm1$ outcomes. The asymmetric minus sign on the last term is not cosmetic — it is what creates the tension the whole inequality lives on.

where: $A_x,B_y$ — Alice's and Bob's outcomes $\pm1$ for settings $x,y\in\{0,1\}$; $E(A_xB_y)=\langle A_xB_y\rangle$ — the correlator, in $[-1,1]$; $S$ — the CHSH parameter. Reading it: three correlators enter with a plus, one with a minus, so a strategy making the first three agree must fight to make the fourth disagree. The first hard bound is the classical one. Under local realism — outcomes predetermined by a shared variable $\lambda$, Alice's result independent of Bob's setting — the parameter is trapped. This is Bell's theorem in CHSH form, assuming nothing quantum.

where: $S$ — the CHSH parameter built from correlators that factorize through $\lambda$. Reading it: any world in which measurement results are fixed in advance and locally, however elaborately, cannot push $|S|$ past $2$. Beating $2$ therefore rules out that entire class of worlds. Quantum correlations follow from the state and observables. For the singlet Bell state the correlator collapses to a simple geometric object — minus the cosine of the angle between the two measurement directions.

where: $|\Psi^-\rangle=\tfrac1{\sqrt2}(|01\rangle-|10\rangle)$ — the singlet; $\hat a,\hat b$ — unit vectors for Alice's and Bob's axes; $\vec\sigma=(\sigma_x,\sigma_y,\sigma_z)$ — the Pauli vector. Reading it: the entire dependence is on the relative angle $\theta$, with $E=-\cos\theta$; aligned axes give perfect anti-correlation $-1$. Feeding equation (3) into equation (1) turns the CHSH parameter into a geometric sum over four dot products, making the optimization a problem in trigonometry rather than operator algebra.

where: $\hat a_x,\hat b_y$ — the four chosen measurement directions. Reading it: choose the four unit vectors to make three dot products equal with one sign and the fourth with the other, and $|S_{QM}|$ is maximized. The maximum quantum value is Tsirelson's bound. It holds for any state (any dimension) and any dichotomic observables — a ceiling as rigid as the classical one, but higher.

where: $S_{\max}$ — the maximum of $|S|$ over all quantum states and $\pm1$ observables. Reading it: quantum mechanics beats $2$ but stops at $2\sqrt2$, a factor $\sqrt2$ above the classical bound and strictly below the algebraic maximum $4$. Finally, CHSH has an operational face as a cooperative game. A referee sends Alice and Bob random input bits $x,y$; they must return bits whose XOR equals $x\wedge y$. Their optimal winning probability is a linear function of $S$.

where: $p_{\text{win}}$ — probability of satisfying $a\oplus b=x\cdot y$ under the optimal strategy. Reading it: $S=2$ gives the classical ceiling $p=0.75$; $S=2\sqrt2$ gives $p=\tfrac12+\tfrac{\sqrt2}{4}\approx0.854$; the $\approx10.4$-point gap is nonlocality made operational. The optimal singlet settings are $\hat a_0=\hat z,\ \hat a_1=\hat x,\ \hat b_0=\tfrac{\hat z+\hat x}{\sqrt2},\ \hat b_1=\tfrac{\hat z-\hat x}{\sqrt2}$.

The value of $S$ acts as a coordinate that locates a theory or a state on the nonlocality ladder.

1 · The singlet at optimal settings, end to end (fully worked)

A source emits singlets $|\Psi^-\rangle$. Alice measures along $\hat a_0=\hat z$ or $\hat a_1=\hat x$; Bob along $\hat b_0=(\hat z+\hat x)/\sqrt2$ or $\hat b_1=(\hat z-\hat x)/\sqrt2$. Find all four correlators, the CHSH value $S$, and the corresponding CHSH-game winning probability.

Step 1. Use $E(\hat a,\hat b)=-\hat a\cdot\hat b$. Compute the dot products: $\hat a_0\cdot\hat b_0=\hat z\cdot\tfrac{\hat z+\hat x}{\sqrt2}=\tfrac1{\sqrt2}$; $\hat a_0\cdot\hat b_1=\hat z\cdot\tfrac{\hat z-\hat x}{\sqrt2}=\tfrac1{\sqrt2}$; $\hat a_1\cdot\hat b_0=\hat x\cdot\tfrac{\hat z+\hat x}{\sqrt2}=\tfrac1{\sqrt2}$; $\hat a_1\cdot\hat b_1=\hat x\cdot\tfrac{\hat z-\hat x}{\sqrt2}=-\tfrac1{\sqrt2}$.

Step 2. So the correlators are $E(A_0B_0)=E(A_0B_1)=E(A_1B_0)=-\tfrac1{\sqrt2}=\mathbf{-0.707}$ and $E(A_1B_1)=+\tfrac1{\sqrt2}=\mathbf{+0.707}$.

Step 3. Assemble with signs $(+,+,+,-)$: $S=-\tfrac1{\sqrt2}-\tfrac1{\sqrt2}-\tfrac1{\sqrt2}-\big(+\tfrac1{\sqrt2}\big)=-\tfrac4{\sqrt2}=-2\sqrt2$. The magnitude is what matters: |S| = 2√2 ≈ 2.828, saturating Tsirelson's bound. (Relabelling either party's two settings flips the overall sign; the physical content is $|S|=2\sqrt2$.)

Step 4. Convert to the game value with $p_{\text{win}}=\tfrac12+\tfrac{|S|}{8}=\tfrac12+\tfrac{2\sqrt2}{8}=\tfrac12+\tfrac{\sqrt2}{4}$. Numerically $\tfrac{\sqrt2}{4}=0.3536$, so p_win ≈ 0.854, versus the classical ceiling of $0.75$ — a gain of about $10.4$ percentage points that no local strategy can match.

2 · How noisy can the state be? The Werner threshold (fully worked)

Real sources produce mixed states. Model the output as a Werner state $\rho_W=p\,|\Psi^-\rangle\langle\Psi^-|+\tfrac{1-p}{4}\mathbb 1$, a singlet with visibility $p$ diluted by white noise. For what visibilities does $\rho_W$ still violate CHSH, and what is the minimum needed?

Step 1. The maximally mixed part $\tfrac14\mathbb 1$ gives zero correlation for every setting, so only the singlet fraction contributes. Each optimal correlator scales by $p$: $E_{\rho_W}(\hat a,\hat b)=-p\,\hat a\cdot\hat b$.

Step 2. The optimal CHSH value therefore scales linearly with visibility: $S_{\max}(\rho_W)=2\sqrt2\,p$. This is the singlet result of Example 1 weighted by $p$.

Step 3. Violation requires $S_{\max}>2$, i.e. $2\sqrt2\,p>2$, giving $p>\tfrac1{\sqrt2}=\mathbf{0.7071}$. Below this visibility the state may still be entangled — Werner states are entangled for $p>\tfrac13$ — yet it admits a local hidden-variable model for these measurements and shows no CHSH violation.

Step 4. Interpretation: an experiment needs to beat roughly $71\%$ visibility just to see any violation, and to approach the full $2\sqrt2$ it must push $p$ toward $1$. This is why loophole-free tests are so demanding — every percent of noise eats directly into $S$.

3 · Reading a real experimental number

A loophole-free Bell test reports $S=2.42\pm0.20$ (comparable to Hensen et al. 2015 with nitrogen-vacancy centres). Is local realism excluded, and what game-win probability does this represent?

Step 1. Compare to the classical bound: the excess $S-2=0.42$ is $0.42/0.20\approx\mathbf{2.1}$ standard deviations above $2$ — a genuine violation at modest significance, which is why such experiments accumulate many rounds and quote a $p$-value (Hensen et al. reported $p\approx0.039$).

Step 2. Convert to the game value: $p_{\text{win}}=\tfrac12+\tfrac{2.42}{8}=\mathbf{0.8025}$, exceeding the classical maximum $0.75$, so no local-realist strategy reproduces the observed win rate.

Step 3. Note the room left to Tsirelson: $2.42$ is well below $2.828$, the shortfall reflecting finite detector efficiency, decoherence, and noise added to close loopholes. The point is not to reach $2\sqrt2$ but to clear $2$ with the loopholes shut.

The CHSH inequality carries four names, but one of them turned Bell's abstract theorem into laboratory reality and, half a century later, into a Nobel Prize. John Clauser saw that Bell's 1964 inequality could not be tested with real, imperfect detectors — and he and his co-authors rebuilt it into a form that could.

Texts. Nielsen & Chuang, Quantum Computation and Quantum Information, §2.6 (Bell inequalities and the EPR paradox) · Preskill, Lecture Notes on Quantum Computation, Ch. 4 · Wilde, Quantum Information Theory, chapter on nonlocality · Watrous, The Theory of Quantum Information · Mermin, Quantum Computer Science.

Primary sources. Clauser, Horne, Shimony, Holt, Phys. Rev. Lett. 23, 880 (1969) — the inequality · Freedman & Clauser, Phys. Rev. Lett. 28, 938 (1972) — first test · Cirel'son (Tsirelson), Lett. Math. Phys. 4, 93 (1980) — the quantum bound · Werner, Phys. Rev. A 40, 4277 (1989) — separable-state bound · Hensen et al., Nature 526, 682 (2015) — loophole-free test · Bierhorst et al., Nature 556, 223 (2018) — certified randomness. The key results in this article are: the LHV bound $|S|\le2$ via the pointwise $s(\lambda)=\pm2$ argument; the singlet correlator $E=-\hat a\cdot\hat b$ and the $45^\circ$-settings computation giving $|S|=2\sqrt2$; Tsirelson's bound via $\mathcal B^2=4\mathbb1-[A_0,A_1][B_0,B_1]$ and $\|\mathcal B\|\le2\sqrt2$; the game-value relation $p_{\text{win}}=\tfrac12+S/8$ with classical ceiling $0.75$ and quantum value $\approx0.854$; and the Werner threshold $p>1/\sqrt2$ from $S_{\max}=2\sqrt2\,p$. The anchor constants throughout are $2$, $2\sqrt2\approx2.828$, $4$, $0.75$, $\approx0.854$, and $1/\sqrt2\approx0.7071$.

Prerequisites. Bell States and EPR Pairs · Separable versus Entangled States · Bell's Theorem and Local Hidden Variables. Leads to. Tsirelson's Bound · PR Boxes and Super-Quantum Correlations · Nonlocality and the No-Signalling Principle · Entanglement Witnesses.

Ask the tutor: Ask me to walk through why the operator square $\mathcal B^2=4\mathbb1-[A_0,A_1][B_0,B_1]$ forces the ceiling at exactly $2\sqrt2$, or to run the singlet correlators for any measurement angles you choose.