CHSH inequality

For local outcomes $A_0,A_1,B_0,B_1\in \{-1,1\}$, the Clauser–Horne–Shimony–Holt inequality isAn arbitrary no-signalling correlation can reach the algebraic maximum $4$, while quantum correlations are bounded by $2\sqrt{2}$.

For a short proof of the local bound, fix a deterministic local hidden-variable model assignment. The combination is $A_0(B_0+B_1)+A_1(B_0-B_1)$. Exactly one bracket vanishes and the other is $\pm 2$, so every assignment contributes $\pm 2$. Averaging with a single setting-independent probability distribution preserves the interval $[-2,2]$. Mixtures also cover stochastic local hidden-variable theories. Rejecting a different sample at each setting would invalidate precisely that common-average step.

The quantum bound follows just as economically. For Hermitian operators with $A_i^2=B_j^2=I$, the CHSH inequality operator $C=A_0\otimes (B_0+B_1)+A_1\otimes (B_0-B_1)$ obeys $C^2=4I-[A_0,A_1]\otimes [B_0,B_1]$. Each commutator has operator norm at most two, giving $\Vert C\Vert \le 2\sqrt{2}$. The Polarization CHSH count model attains this bound without allowing either station to control the distant marginal probability distribution.

For the Bell state $|{\varphi }_{+}⟩$, let both zero-input measurements use the computational basis, and let the one-input bases be rotated by $\theta$ for Alice and $-\theta$ for Bob in the real qubit plane. In the CHSH game, the four conditional winning probabilities are $1,{\cos }^2\theta ,{\cos }^2\theta ,{\sin }^2(2\theta )$. They follow by contraction of the Bell state with vectors from the real orthonormal bases used for the basic measurements. Their average is the displayed expression. On $0\le \theta \le \pi /2$, this beats $3/4$ exactly when $0<\theta <\pi /4$ and reaches $13/16$ at $\theta =\pi /6$, since $t-t^2=1/4-(t-1/2{)}^2$.

Two separated players receive uniformly random input bits $a,b$ and output bits $x,y$ without communicating after receiving their inputs. They win when $x\oplus y=ab$, with exclusive or $\oplus$. For deterministic local response functions, the four desired equalities have inconsistent parity, so at most three can hold. Shared randomness only averages such strategies, proving classical winning probability at most $3/4$. In terms of correlations $E_{ab}=\mathbb{E}[(-1{)}^{x+y}\mid a,b]$, the winning probability is $1/2+(E_{00}+E_{01}+E_{10}-E_{11})/8$, directly connecting the game to the CHSH inequality.

The Hermitian operator $\mathcal{B}=\sqrt{2}(X\otimes X-Z\otimes Z)$ has eigenvalues $2\sqrt{2},0,0,-2\sqrt{2}$ on ${\Psi }^{+},{\Phi }^{+},{\Psi }^{-},{\Phi }^{-}$, respectively. Thus any density matrix satisfying $\mathrm{Tr}(\rho \mathcal{B})=2\sqrt{2}$ must be $|{\Psi }^{+}⟩⟨{\Psi }^{+}|$: an average can equal the largest eigenvalue only when all weight lies in its one-dimensional eigenspace. The purity decouples a subsystem from its purification lemma then excludes correlations with an external environment. This argument assumes the stated measurement operators; CHSH self-testing handles unknown representations with additional qualifications.

Maximal quantum CHSH inequality violation certifies an embedded maximally entangled qubit pair and its ideal measurement action, up to local isometries and unused ancillary systems, under the separated-device quantum assumptions. It does not identify every microscopic component of an arbitrary device. Near-maximal observed correlations require robust error bounds rather than exact identification.^[ref]

Let $H,V$ be orthogonal linear polarizations, and prepare the Bell state $|{\Phi }^{+}⟩=(|HH⟩+|VV⟩)/\sqrt{2}$. The $+1$ analyzer vector is $|a⟩=\cos a|H⟩+\sin a|V⟩$; its orthogonal companion has outcome $-1$. Direct inner products with the Bell state give $P_{++}=P_{--}={\cos }^2(a-b)/2$ and $P_{+-}=P_{-+}={\sin }^2(a-b)/2$. Thus the expectation value of the outcome product is $E(a,b)=\cos 2(a-b)$. These are physical polarizer angles; their corresponding Bloch sphere angles are doubled.

For a stipulated isotropic admixture, use the density matrixThe maximally mixed state contributes zero correlation, so $P_{st}(a,b)=[1+stv\cos 2(a-b)]/4$ for $s,t\in \{-1,1\}$. Each local marginal is $1/2$, independently of the other setting, illustrating quantum no-signalling. In this model $v$ is the interferometric visibility of the coincidence fringe; other noise models need not have one common visibility for every basis.

Choose $a_0=0^{\circ }$, $a_1=45^{\circ }$, $b_0=22.5^{\circ }$ and $b_1=-22.5^{\circ }$. The four correlations in the displayed CHSH inequality are $v/\sqrt{2},v/\sqrt{2},v/\sqrt{2},-v/\sqrt{2}$. Consequently $S=2\sqrt{2}v$, exceeding the local bound exactly when $v>1/\sqrt{2}$. Perfect correlation at a single common setting would not yield this comparison between incompatible quantum measurements.

To connect probabilities with expected counts, stipulate $M$ emitted pairs per setting pair and independent detection efficiencies ${\eta }_A,{\eta }_B$, independent also of settings, outcomes and hidden variables. The expected joint counts are $n_{st}=M{\eta }_A{\eta }_B{P}_{st}$. Their normalized correlation isFor original teaching inputs $M=1000$, ${\eta }_A=0.8$, ${\eta }_B=0.75$ and $v=0.8$, the $(a_0,b_0)$ expected counts are $n_{++}=n_{--}\simeq 234.85$ and $n_{+-}=n_{-+}\simeq 65.15$. They total $600$ and give $E_{\det }\simeq 0.56569$, hence $S\simeq 2.26274$. Fractional expected counts are not measured data. Efficiency cancels here because the loss model was specified to be independent; this algebra supplies no justification for discarding failures in an arbitrary Bell test with a detection loophole.

For spin-projective measurements on a two-qubit state, let ${\lambda }_1\ge {\lambda }_2\ge {\lambda }_3$ be the eigenvalues of $T^{T}T$, where $T$ is its Pauli correlation tensor. Bob's sum and difference of unit axes are orthogonal and have lengths $2\cos \varphi$ and $2\sin \varphi$. Maximizing Alice's two axes aligns them with the corresponding images under $T$. Maximizing $\varphi$ then gives $2\sqrt{\Vert T\mathbf{u}{\Vert }^2+\Vert T\mathbf{v}{\Vert }^2}$ for orthonormal $\mathbf{u},\mathbf{v}$. The maximum of this quadratic-form sum is ${\lambda }_1+{\lambda }_2$, by diagonalizing $T^{T}T$ and selecting its top two eigenvectors. Thus the displayed expression is both an upper bound and attainable. The same maximum applies to the form with the sum of two absolute values, since local outcome signs can select their positive branches.

Every entangled pure state of two qubits violates a CHSH inequality for a suitable choice of local measurement axes. If its two Schmidt coefficients are $c_0,c_1>0$, the optimal construction already gives a value $2\sqrt{1+(2c_0{c}_1{)}^2}>2$.

For the Schmidt-basis Pauli correlation tensor, choose Alice's axes ${\mathbf{e}}_z,{\mathbf{e}}_x$ and Bob's $({\mathbf{e}}_z\pm s{\mathbf{e}}_x)/\sqrt{1+s^2}$. The CHSH inequality expression is $2\sqrt{1+s^2}$. It exceeds two exactly when both Schmidt coefficients are nonzero, reaching $2\sqrt{2}$ for maximal quantum entanglement. A product state gives $s=0$ and saturates rather than violates the inequality.

A no-signalling box is a conditional probability distribution $P(x,y|a,b)$ whose marginal distribution for either party is independent of the other party's input. It can therefore exhibit correlations stronger than quantum correlations without transmitting information superluminally.

A Popescu–Rohrlich box is a bipartite no-signalling box attaining the algebraic maximum of the CHSH inequality.

Take a separated Boolean decomposition $f(x,y)={⨁}_{\ell }{u}_{\ell }(x)v_{\ell }(y)$. Feed $(u_{\ell },v_{\ell })$ to one independent PR box, obtaining outputs with $a_{\ell }\oplus b_{\ell }=u_{\ell }{v}_{\ell }$. Alice forms $A={⨁}_{\ell }{a}_{\ell }$, Bob forms $B={⨁}_{\ell }{b}_{\ell }$, and their parities satisfy $A\oplus B=f$. Sending the single bit $A$ lets Bob compute the answer with certainty. The truth-table decomposition needs at most $2^{\min (m,n)}$ boxes for input lengths $m,n$. Each box's local output is uniform, so Bob's uncommunicated output string carries no information about Alice's input; the final parity bit is essential for a general function depending on that input.