CHSH Inequality and the Tsirelson Bound

Derivation

CHSH Inequality and the Tsirelson Bound

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Statement

For four dichotomic observables \(A,A'\) (Alice) and \(B,B'\) (Bob), each taking values \(\pm 1\), define the CHSH combination \(S = \langle AB\rangle + \langle AB'\rangle + \langle A'B\rangle - \langle A'B'\rangle\). Any local hidden-variable theory obeys \(|S|\le 2\) (the CHSH inequality), whereas quantum mechanics permits \(|S|\) up to exactly \(2\sqrt{2}\) (the Tsirelson bound) and never larger.

Why it matters

The CHSH inequality turns a metaphysical question — whether nature is describable by pre-existing local values — into a number you can measure. A single experimental \(S>2\) falsifies the entire class of local hidden-variable models at once, and the loophole-free Bell tests of 2015 did exactly this. This is the sharpest empirical statement that quantum correlations are not classical.

The Tsirelson bound \(2\sqrt{2}\) is equally profound: quantum mechanics violates locality, but only by a bounded amount. It does not saturate the algebraic maximum of \(4\). That "quantum is nonlocal but not maximally so" is a structural fingerprint of Hilbert space, and it underlies device-independent cryptography and randomness certification.

Assumptions

Realism (definite values).If measurement outcomes are not predetermined functions of a hidden state \(\lambda\), the products \(A(\lambda)B(\lambda)\) below are undefined and the classical bound cannot be constructed.

Locality (no action at a distance).Alice's outcome must not depend on Bob's setting: \(A=A(a,\lambda)\), not \(A(a,b,\lambda)\). Drop this and \(S\) can reach the algebraic \(4\); the derivation collapses.

Outcomes are \(\pm 1\).The factorisation \(A(A'B\pm\ldots)\) relies on \(A^2=1\). Non-dichotomic observables need the CH form instead.

Measurement independence (free choice).The setting choices \(a,a',b,b'\) are statistically independent of \(\lambda\): \(\rho(\lambda)\) does not depend on which pair is measured. If \(\lambda\) correlates with the settings ("superdeterminism"), the whole inequality is void.

For Tsirelson: quantum kinematics.Observables are bounded self-adjoint operators with \(A^2=A'^2=B^2=B'^2=\mathbf 1\), Alice's operators commute with Bob's (tensor-product / commuting-operator structure), and states are density operators. Drop operator boundedness and \(2\sqrt2\) need not hold.

Derivation

\[ A,A',B,B' \in \{-1,+1\} \]

Each measurement yields one of two outcomes; encode them as \(\pm 1\). By realism these are definite functions of a shared state \(\lambda\). A

\[ s(\lambda) = A(B+B') + A'(B-B') \]

Form the CHSH quantity for a single \(\lambda\). Locality is what lets \(A\) be written without reference to Bob's setting, and vice versa. A

\[ B+B' \in \{-2,0,+2\},\qquad B-B' \in \{-2,0,+2\} \]

Since \(B,B'=\pm1\), exactly one of \((B+B')\) and \((B-B')\) is \(\pm 2\) and the other is \(0\). Both cannot be nonzero simultaneously. B

\[ s(\lambda) = \pm 2 A \;\text{ or }\; \pm 2 A' \;\Rightarrow\; |s(\lambda)| = 2 \]

One bracket vanishes; the surviving term is \((\pm1)(\pm2)=\pm2\). So for every hidden state, \(s(\lambda)=\pm2\) pointwise. B

\[ S = \int d\lambda\, \rho(\lambda)\, s(\lambda),\qquad \rho(\lambda)\ge 0,\ \int d\lambda\,\rho=1 \]

The measured correlator \(\langle AB\rangle=\int d\lambda\,\rho(\lambda)A(a,\lambda)B(b,\lambda)\); linearity of the average gives \(S\) as the mean of \(s(\lambda)\). Measurement independence lets one \(\rho\) serve all four settings. A

\[ |S| = \left|\int d\lambda\, \rho(\lambda)\, s(\lambda)\right| \le \int d\lambda\, \rho(\lambda)\, |s(\lambda)| = 2 \]

The average of a quantity bounded pointwise by \(2\) is itself bounded by \(2\) (triangle inequality plus normalisation of \(\rho\)). This is the CHSH inequality. A

\[ \hat A^2=\hat A'^2=\hat B^2=\hat B'^2=\mathbf 1,\qquad [\hat A,\hat B]=[\hat A,\hat B']=\dots=0 \]

Now pass to quantum mechanics: replace values by self-adjoint operators squaring to the identity (eigenvalues \(\pm1\)); Alice's and Bob's operators act on different tensor factors, hence commute. B

\[ \hat S = \hat A\otimes\hat B + \hat A\otimes\hat B' + \hat A'\otimes\hat B - \hat A'\otimes\hat B' \]

\[ \hat S^2 = 4\,\mathbf 1 - [\hat A,\hat A']\otimes[\hat B,\hat B'] \]

Expand \(\hat S^2\). All "diagonal" terms give \(\hat A^2\otimes\hat B^2=\mathbf1\) etc. (there are four, summing with signs to \(4\,\mathbf1\)); the cross terms, using \([\hat A,\hat B]=0\), collect into \(-[\hat A,\hat A']\otimes[\hat B,\hat B']\). C

\[ \big\|[\hat A,\hat A']\big\| \le \|\hat A\hat A'\|+\|\hat A'\hat A\| \le 2\|\hat A\|\,\|\hat A'\| = 2 \]

Norm of a sum and of a tensor product of the commutator term. Hence \(\|\hat S\|=\sqrt{\|\hat S^2\|}\le\sqrt 8=2\sqrt2\), because \(\hat S\) is self-adjoint. C

\[ |S| = |\langle\hat S\rangle| \le \|\hat S\| \le 2\sqrt 2 \]

Born-rule expectation of a self-adjoint operator is bounded by its largest eigenvalue in magnitude, i.e. its norm. This is the Tsirelson bound. B

\[ \langle A(a)B(b)\rangle_{\text{QM}} = -\cos\theta_{ab},\qquad (a,a',b,b')=(0^\circ,90^\circ,45^\circ,-45^\circ)\ \Rightarrow\ |S|=2\sqrt2 \]

For the singlet, spin measurements at relative angle \(\theta\) give \(-\cos\theta\). At these settings \(\langle AB\rangle=\langle AB'\rangle=\langle A'B\rangle=-\tfrac{1}{\sqrt2}\) and \(\langle A'B'\rangle=+\tfrac{1}{\sqrt2}\), so \(S=-\tfrac{3}{\sqrt2}-\tfrac{1}{\sqrt2}=-\tfrac{4}{\sqrt2}=-2\sqrt2\). The bound is attained, hence tight. B

Result

\[ \boxed{\,|S|\le 2 \ \text{(local realism)}\qquad\qquad |S|\le 2\sqrt2 \ \text{(quantum)}\,} \]

Reading. The same four-term correlator that any local-realistic world holds at or below \(2\) can, in quantum mechanics with an entangled state, climb to \(2\sqrt2\approx2.828\) — and no state or observables push it past that. The gap \([2,\,2\sqrt2]\) is the measurable quantum advantage; the ceiling \(2\sqrt2\) is a hard consequence of Hilbert-space structure, not of any specific state.

Units check. \(A,A',B,B'\) are dimensionless \(\pm1\) outcomes, so each correlator \(\langle\,\cdot\,\rangle\) and hence \(S\) is a pure number. Both bounds \(2\) and \(2\sqrt2\) are dimensionless, as required; \(\cos\theta\) is dimensionless with \(\theta\) in radians.

Limiting cases

Product (separable) state: \(\langle A\otimes B\rangle=\langle A\rangle\langle B\rangle\); the quantum \(S\) reduces to the classical form and \(|S|\le2\). Entanglement is necessary for violation.
Commuting local pairs \([\hat A,\hat A']=0\): the correction in \(\hat S^2\) vanishes, \(\|\hat S\|=2\); jointly measurable settings give no violation.
Maximally incompatible settings \(\theta=45^\circ\): commutators are largest, \(S\to2\sqrt2\); the bound is saturated by the singlet.
Algebraic (super-quantum) limit: a hypothetical PR box gives \(S=4\); forbidden by quantum mechanics but not by no-signalling alone.
Classical correlation \(\langle AB\rangle=-\cos\theta\) forced linear: if one wrongly used a triangle-wave correlation, \(S=2\) exactly at the classical corner.

Breaks when

Signalling / setting-dependence: if Alice's outcome depends on Bob's setting (locality violated), or the detector "knows" both settings, even the classical construction fails and \(S\) is unbounded up to \(4\). This is the physical content of the detection and locality loopholes.
Superdeterminism / measurement dependence: if \(\rho(\lambda)\) is correlated with the setting choices, the pointwise argument in Step 4 no longer averages to \(\le2\); no Bell inequality constrains such a theory.
Unbounded or non-\(\pm1\) observables: if \(\hat A^2\ne\mathbf1\), the identity \(\hat S^2=4\mathbf1-[\,,]\otimes[\,,]\) breaks and the \(2\sqrt2\) ceiling need not hold (though CHSH can be rescaled).
Post-quantum theories: generalised probabilistic theories (PR boxes) respect no-signalling yet reach \(S=4\); the derivation of \(2\sqrt2\) uses the operator identity specific to Hilbert space and fails there.

Failure modes

Sign placement error: putting the minus sign on the wrong term (e.g. \(-A'B\) instead of \(-A'B'\)) so the two brackets no longer factor as \(B\pm B'\); then \(s(\lambda)\) is not bounded by \(2\).
Confusing \(2\) with \(2\sqrt2\): claiming quantum mechanics reaches \(4\). The algebraic max is \(4\) but no quantum state attains it — Tsirelson is strictly below.
Adding correlators in quadrature: treating the four \(\langle\cdot\rangle\) as independent and bounding by \(4\times1\); ignores the anticorrelation forced by shared operators.
Forgetting \(\|\hat A\|=1\): using \(\hat A^2=\mathbf1\) but then bounding \(\|[\hat A,\hat A']\|\le\|\hat A\|^2\); the correct bound is \(2\|\hat A\|\|\hat A'\|\).
Using degrees in \(\cos\theta\): plugging \(45\) into a calculator in radian mode, getting a wrong correlator and a spurious \(S\).
Assuming entanglement suffices: a partially entangled state with bad settings gives \(S<2\); both the state and the angle choice matter.

Discussion

The two bounds sit on opposite sides of a conceptual divide. The classical \(|S|\le2\) is a statement about value assignments: it holds for any theory in which each particle carries pre-set answers to all questions, measured locally. The proof (Steps 1–6) never invokes probability beyond a normalised measure — it is essentially a combinatorial fact about \(\pm1\) numbers. That robustness is why a single violation is so damning: no local-realistic model of any complexity can escape it.

The Tsirelson bound is a statement about operators. The key identity \(\hat S^2=4\mathbf1-[\hat A,\hat A']\otimes[\hat B,\hat B']\) shows quantum mechanics gains its extra correlation precisely from the non-commutativity of incompatible local measurements — the very same incompatibility responsible for the uncertainty principle. Nonlocality and complementarity are two faces of one structure. And because the commutators are norm-bounded by \(2\), the advantage is capped: quantum weirdness is finite.

Why \(2\sqrt2\) and not \(4\)? No-signalling alone permits \(S=4\) (the Popescu–Rohrlich box), so the tighter quantum ceiling is not a relativity requirement — it is a fact about Hilbert space. Reconstructions of quantum theory (information causality, macroscopic locality, the exclusivity/CSW graph-theoretic bound) each single out \(2\sqrt2\) as the boundary of physically reasonable correlations, suggesting the number encodes something deep about why the world is quantum and not "super-quantum." Tsirelson's original proof recasts the operators as anticommuting generators of a Clifford algebra, whose spectral radius gives \(\sqrt2\) directly.

Common misconceptions. Violating CHSH does not permit faster-than-light signalling: the marginal statistics \(\langle A\rangle\) are independent of Bob's setting, so no message is carried. "Nonlocal" here means the joint correlations cannot be reproduced by any local hidden-variable model, not that a signal propagates. Also, entanglement is necessary but not sufficient — the correlator and the measurement angles must be chosen adversarially to reach \(2\sqrt2\).

Worked examples

Singlet state, optimal spin settings. \[ |\psi^-\rangle=\tfrac{1}{\sqrt2}\big(|{\uparrow\downarrow}\rangle-|{\downarrow\uparrow}\rangle\big),\quad \langle A(a)B(b)\rangle=-\,\hat a\!\cdot\!\hat b=-\cos\theta_{ab} \]

Standard singlet correlator (born-rule with spin projectors). Choose coplanar directions \(a=0^\circ,\ a'=90^\circ,\ b=45^\circ,\ b'=-45^\circ\). A

\[ \theta_{ab}=45^\circ,\ \theta_{ab'}=45^\circ,\ \theta_{a'b}=45^\circ,\ \theta_{a'b'}=135^\circ \]

Relative angles: \(a\text{-}b=0-45=-45\ (\to45^\circ)\), \(a\text{-}b'=0-(-45)=45^\circ\), \(a'\text{-}b=90-45=45^\circ\), \(a'\text{-}b'=90-(-45)=135^\circ\). B

\[ \langle AB\rangle=\langle AB'\rangle=\langle A'B\rangle=-\cos45^\circ=-\tfrac{1}{\sqrt2},\qquad \langle A'B'\rangle=-\cos135^\circ=+\tfrac{1}{\sqrt2} \]

Insert into \(-\cos\theta\). The subtracted term lands on the \(135^\circ\) correlator, which is the one of opposite sign — that is what makes all four contributions add constructively. B

\[ S=\langle AB\rangle+\langle AB'\rangle+\langle A'B\rangle-\langle A'B'\rangle = -\tfrac{1}{\sqrt2}-\tfrac{1}{\sqrt2}-\tfrac{1}{\sqrt2}-\tfrac{1}{\sqrt2} = -\tfrac{4}{\sqrt2} \]

Sum with the CHSH sign pattern (last term subtracted): four equal contributions of \(-\tfrac{1}{\sqrt2}\). B

\[ S = -\tfrac{4}{\sqrt2} = -2\sqrt2 \ \Rightarrow\ |S| = 2\sqrt2\approx 2.828 \]

Reading. The singlet at these angles saturates Tsirelson's bound: \(|S|=2\sqrt2>2\), a decisive violation of the classical ceiling (the sign of \(S\) is a convention-dependent label, only \(|S|\) is physical).

Units check. All correlators are pure numbers; \(2\sqrt2\) is dimensionless.

Werner state, visibility \(V\). \[ \rho = V\,|\psi^-\rangle\langle\psi^-| + (1-V)\tfrac{\mathbf1}{4},\qquad \langle A(a)B(b)\rangle = -V\cos\theta_{ab} \]

Mixing the singlet with white noise scales every correlator by the visibility \(V\in[0,1]\) (linearity of the trace). Take \(V=0.7\) and the same optimal angles. A

\[ S(V) = V\,\big(2\sqrt2\big) = 2\sqrt2\,V \]

Each of the four correlators carries the same factor \(V\); the optimal-angle combination gives \(2\sqrt2\) for \(V=1\), so it scales linearly. B

\[ S(0.7) = 2\sqrt2\times 0.7 = 1.980,\qquad \text{threshold: } S>2 \iff V>\tfrac{1}{\sqrt2}\approx0.707 \]

Compute numerically and compare to the classical bound. \(V=0.7<0.707\), so no violation. B

\[ \boxed{\,S(0.7)\approx 1.98 \le 2\,} \quad\text{— local-realistic model exists} \]

Reading. At \(70\%\) visibility the Werner state, though still entangled, is CHSH-local: noise has washed out the violation. The critical visibility is \(1/\sqrt2\).

Units check. \(V\) and \(S\) are dimensionless; \(2\sqrt2\times0.7=1.98\) is a pure number.

Problems

Show that for a product (unentangled) quantum state \(\rho_A\otimes\rho_B\), \(S\le2\).
Solution
For a product state \(\langle\hat A\otimes\hat B\rangle=\langle\hat A\rangle\langle\hat B\rangle\) with \(a\equiv\langle\hat A\rangle,\ a'\equiv\langle\hat A'\rangle,\ b\equiv\langle\hat B\rangle,\ b'\equiv\langle\hat B'\rangle\), all in \([-1,1]\). Then \(S=ab+ab'+a'b-a'b'=a(b+b')+a'(b-b')\). Maximising over \(a,a'\in[-1,1]\): \(|S|\le|b+b'|+|b-b'|\). For real \(b,b'\in[-1,1]\), \(|b+b'|+|b-b'|=2\max(|b|,|b'|)\le2\). Hence \(|S|\le2\): product states never violate CHSH.
The relative angles are fixed but the array is rotated rigidly by \(\phi\). Show \(S\) is unchanged.
Solution
Rotating all four directions by the same \(\phi\) sends each relative angle \(\theta_{ab}\to\theta_{ab}\) (differences of angles are invariant under a common shift). Since \(\langle AB\rangle=-\cos\theta_{ab}\) depends only on relative angles, every correlator is unchanged, so \(S\) is invariant. Physically, CHSH depends on measurement incompatibility, not on absolute lab orientation.
Find the settings that maximise \(|S|\) for the singlet, with coplanar directions parametrised by single angles.
Solution
With \(E(a,b)=-\cos(a-b)\), \(S=-[\cos(a-b)+\cos(a-b')+\cos(a'-b)-\cos(a'-b')]\). Stationarity requires each of the three "\(+\)" pairs to sit at the same relative angle and the subtracted pair at its supplement. Take \(a=0^\circ,\ a'=90^\circ,\ b=45^\circ,\ b'=-45^\circ\): the three \(+\) correlators are at \(45^\circ\) giving \(-\tfrac{1}{\sqrt2}\), and \(\langle A'B'\rangle\) is at \(135^\circ\) giving \(+\tfrac{1}{\sqrt2}\). Then \(S=-\tfrac{3}{\sqrt2}-\tfrac{1}{\sqrt2}=-2\sqrt2\), so \(|S|=2\sqrt2\). This is the optimum (unique up to global rotation and relabelling). Note the naive equally-spaced set \((0,45,90,135)\) puts the odd correlator in the wrong slot and yields \(S=0\).
A PR box outputs \(a\oplus b = x\cdot y\) (bits \(x,y\) settings, \(a,b\) outcomes). Compute its \(S\) in \(\pm1\) encoding.
Solution
The PR box perfectly correlates \(A B=+1\) for the three settings \((x,y)\ne(1,1)\) and anticorrelates \(AB=-1\) for \((1,1)\). Mapping to \(\pm1\), \(\langle AB\rangle=+1,\langle AB'\rangle=+1,\langle A'B\rangle=+1,\langle A'B'\rangle=-1\). Then \(S=1+1+1-(-1)=4\). This reaches the algebraic maximum, exceeding Tsirelson's \(2\sqrt2\), so PR boxes are super-quantum yet non-signalling.
An experiment measures \(S=2.4\pm0.1\). By how many standard deviations does it exceed the classical bound, and does it respect Tsirelson?
Solution
Excess over classical bound: \((2.4-2)/0.1=4.0\), so a \(4\sigma\) violation of local realism. Tsirelson's bound is \(2\sqrt2\approx2.828\); the measured \(2.4\) lies below it, consistent with quantum mechanics (the value \(2.4<2.828\) leaves headroom of about \(0.43\), i.e. \(4.3\sigma\) below the ceiling). The result both refutes local hidden variables and respects the quantum limit.