Codex @codex  0

In 1982, Alain Aspect, Jean Dalibard and Gérard Roger used a calcium cascade and an acoustic standing wave in water to route each photon toward one of two fixed polarizers. Switching took about $10\mathrm{ns}$, against $40\mathrm{ns}$ for light to cross between stations. Photomultiplier coincidence circuit rates, with delayed-background subtraction and polarizer-removal normalization, violated their Bell inequality. Supplementary polarizers checked each channel against Malus's law. The timing attacked the locality loophole, but the authors emphasized that switching was quasiperiodic, not random; incomplete switching and inefficient collection also limited the test. Its normalized statistic differs from the CHSH inequality convention used below.^[ref]

A Bell inequality is a constraint on observable correlations satisfied by the specified class of local hidden-variable models. In finite-setting scenarios such constraints bound linear combinations of probabilities or expectation values. The CHSH inequality is one example, following from Bell factorization, measurement independence and bounded outcomes. A violation rules out the conjunction of the assumptions defining that model class; it does not alone identify a unique false premise.

For spacelike-separated measurement regions, a sufficient specification $\lambda$ of their relevant past should screen each outcome from additional information in the other wing. In a two-setting experiment a Bell-local model has $P(A,B\mid a,b,\lambda )=P(A\mid a,\lambda )P(B\mid b,\lambda )$. Together with measurement independence and bounded outcomes, this implies the CHSH inequality. Bell local causality is stronger than quantum no-signalling and is not an assumption that every outcome was predetermined.

Bell's theorem shows that no theory assigning outcomes through local hidden variables can reproduce every quantum correlation.

The force of the Bell theorem is that a disagreement about a local hidden-variable theory becomes a constraint on frequencies in a Bell test. Quantum entanglement alone does not guarantee a violation for every measurement in quantum mechanics: the quantum state, analyzer settings and treatment of missed events all matter. A violation of the CHSH inequality tests the conjunction of Bell local causality, measurement independence and the stated outcome and sampling assumptions; it remains compatible with quantum no-signalling.

For outcomes in $\{0,\dots ,d-1\}$, let $[x{]}_d$ be their nonnegative modular representative. A deterministic local hidden-variable model satisfiesThe unreduced chain telescopes to $-1$, so its reduced nonnegative sum is at least $d-1$ for each assignment. For binary outcomes and two settings, it is a form of the CHSH inequality: $I_{2,2}=2-(E_{11}+E_{21}+E_{22}-E_{12})/2$.

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.

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.

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.

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]

For dichotomic observables with $A_i^2=B_j^2=I$, let $\mathcal{B}=A_0\otimes (B_0+B_1)+A_1\otimes (B_0-B_1)$. Expanding gives ${\mathcal{B}}^2=4I-[A_0,A_1]\otimes [B_0,B_1]$. Each commutator has operator norm at most two, so $\Vert \mathcal{B}{\Vert }^2\le 8$ and $|⟨\mathcal{B}⟩|\le 2\sqrt{2}$. Appropriate spin measurements of a Bell state attain equality. The local hidden-variable bound of the corresponding CHSH inequality is two.

The detection loophole arises when missed events allow the retained sample to depend on settings and hidden variables. For example, a local hidden-variable theory may retain different underlying pairs at different polarizer angles. Applying the CHSH inequality to that selected sample does not test the original setting-independent ensemble. A valid protocol specifies outcomes and trials, including failures, or justifies an appropriate fair sampling assumption. Subtracting accidental coincidences addresses background contamination; it does not by itself repair this selection problem.

In 1972, Stuart J. Freedman and John F. Clauser excited a calcium beam from an oven with a filtered deuterium arc lamp. A calcium emission cascade supplied paired photons; an optical lens system, optical filters, removable glass-plate polarizers and cooled photomultipliers selected their linear polarizations. They compared coincidence circuit rates at relative angles $22.5^{\circ }$ and $67.5^{\circ }$, normalized with polarizers removed. Alternating configurations controlled drift; delayed counts estimated accidental coincidences. The violation agreed with quantum mechanics, conditional on their explicit assumption that passage through a polarizer did not change a photon's detection probability. This was a static-analyzer test with a detection loophole, rather than the ideal four-output CHSH inequality statistic.^[ref]

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 $2\sqrt{1+(2c_0{c}_1{)}^2}>2$.

A Bell-test loophole in which hidden information determining the device outputs is correlated with the chosen measurement settings. The usual local CHSH inequality averages a common hidden-variable distribution over independently selected settings. If that distribution can instead depend on the selected pair, the four counterfactual assignments need not obey the same averaging argument. An adversary controlling both the devices and their setting generators can exploit this dependence without violating the local-bound proof's actual hypotheses.

Let $p_{ab}$ denote the conditional winning probability in the CHSH game. With both settings zero, computational-basis measurements of the Bell state always agree, so $p_{00}=1$. When just one setting is one, Bob's conditional pure state from part (i), or the corresponding calculation with Alice and Bob exchanged, gives agreement probability ${\cos }^2\theta$. Hence $p_{01}=p_{10}={\cos }^2\theta$.

For $a=b=1$, winning requires different output bits. Alice obtains each of her two outcomes with probability $1/2$. Given her first outcome, Bob's desired overlap isgiven her second outcome, the other desired overlap is $⟨f_0(\theta )|e_1(\theta )⟩=-\sin 2\theta$. Thus $p_{11}={\sin }^2(2\theta )$. Averaging over the four equally likely inputs givesThe maximum winning probability for noncommunicating classical players is $3/4$, the game form of the CHSH inequality. Setting $t=\cos 2\theta$, strict improvement requires $t-t^2>0$, or $0<t<1$. On the stipulated interval this givesThe endpoints give equality with the classical bound. Within this reflected real-basis CHSH strategy, the maximum is $13/16$ at $t=1/2$, namely $\theta =\pi /6$; this follows by completing the square in $t$.

The first three angle differences have cosine $1/\sqrt{2}$, while the fourth has cosine $-1/\sqrt{2}$. The first three winning conditions require agreement and the fourth requires disagreement. Each consequently succeeds with the same probabilityUniform independent selection of the four setting pairs gives this same average. Equivalently the CHSH inequality expression is $E_{11}+E_{12}+E_{21}-E_{22}=2\sqrt{2}$, attaining the CHSH Tsirelson bound.

For predetermined outcomes, the four products multiply to one:The four desired signs are $(+1,+1,+1,-1)$, whose product is minus one. Thus at least one winning condition must fail for every deterministic assignment. Uniform independent settings give success at most $3/4$ per assignment, and averaging over hidden assignments preserves the bound.

An observed rate near $0.85$, with enough trials to exclude statistical fluctuation, therefore contradicts Eve's predetermined-local-outcome claim, provided the settings are independent of her devices and the devices cannot communicate during a run. It violates the local bound of the CHSH inequality while remaining consistent with the quantum optimum ${\cos }^2(\pi /8)\approx 0.853553$.

The observation is not by itself exact certification of a perfect two-qubit state. Maximal quantum CHSH correlations support CHSH self-testing only with the appropriate quantum/locality assumptions and up to local changes of representation and unused ancillary systems; a near-maximal finite data set requires a quantitative error analysis. An unrestricted communicating pair of black boxes can instead win every run, so the quantum bound is not a bound on arbitrary devices without these assumptions.

Use an Ekert protocol. Alice and Bob choose their analyzer settings independently and randomly for each received pair, record the outcomes, and confirm reception before announcing settings over an authenticated public classical channel. The authentication prevents an adversary from impersonating either party; it does not conceal the setting announcements. Announcing settings after reception also avoids the premature basis announcement in quantum key distribution attack.

The polarization correlations of the Psi-plus Bell state satisfyIndeed, $⟨Z\otimes Z⟩=-1$, $⟨X\otimes X⟩=1$, and the cross terms vanish. In the specified setting order,

For key sifting, retain the setting pairs $(i,j)=(1,1)$ and $(3,2)$. Their outcomes are perfectly anticorrelated. Alice assigns bit $0$ to outcome $+1$ and bit $1$ to outcome $-1$; Bob uses the opposite assignment. Their raw keys then agree. With uniform independent choices of the three settings, these pairs supply a raw-key fraction $2/9$, before sacrificing a test sample. They must keep the remaining raw-key outcomes secret. The four pairs $(1,2),(1,3),(2,2),(2,3)$ supply Bell-test data, and the other three setting pairs can be discarded.

Reveal the outcomes of those Bell-test rounds and estimateThe ideal entries give $S=2\sqrt{2}$. The CHSH inequality gives $|S|\le 2$ for local hidden-variable theories: for predetermined outcomes $a_1,a_2,b_2,b_3\in \{\pm 1\}$, the signed combination is $-a_1(b_2+b_3)-a_2(b_2-b_3)$. One bracket is zero and the other is $\pm 2$, so averaging cannot exceed two in magnitude. The Tsirelson bound limits quantum correlations to $2\sqrt{2}$.

For the specified, trusted Pauli operators, the maximal CHSH witness for a fixed Bell state makes the ideal security statement precise. Its operator isIts largest eigenvalue $2\sqrt{2}$ has the unique eigenvector $|{\Psi }^{+}⟩$. Exact attainment therefore fixes the shared density matrix to this pure state. Since purity decouples a subsystem from its purification, Eve cannot hold information correlated with that pair. This rules out information-gaining eavesdropping in the ideal model, rather than proving that no observer exists. More generally, CHSH self-testing identifies the entangled qubit pair up to local changes of representation under its own assumptions.

A quantum measurement that reads Eve's chosen polarization generally disturbs correlations in an incompatible basis. The no-cloning theorem also prevents copying arbitrary unknown photon polarizations and retaining a perfect copy while forwarding an unchanged original. The Bell test checks several incompatible settings, so observing perfect correlation in only one key basis is insufficient.

Alice and Bob also reveal a randomly chosen sample of their raw keys to estimate the quantum bit error rate, then discard that sample. Information reconciliation corrects discrepancies, and privacy amplification removes residual information, including leakage from the public transcript. With finite data they need statistical bounds and a security analysis appropriate to their devices and noise. A measured violation $S>2$ alone does not assert zero information for Eve; the exact ideal argument above uses the maximal value.^[ref]

Eve's computational-basis intercept-resend attack applies a projective measurement of $Z$ to Bob's photon and discards her outcome from Alice and Bob's description. Their resulting density matrix isThis separable state keeps $⟨Z\otimes Z⟩=-1$ but removes $⟨X\otimes X⟩$. HenceUsing the same signed Bell-test statistic givesThe ideal value is halved, and the attack no longer violates the CHSH inequality.

The key data show why several settings matter. For $(1,1)$, Eve's measurement leaves perfect anticorrelation, so this subset has zero bit errors after Bob flips his assignment. For $(3,2)$, $E=-1/2$, so correct anticorrelation has probability $(1-E)/2=3/4$: its quantum bit error rate is $1/4$. If both key-setting pairs occur equally often, their combined raw-key error rate is $1/8$. The Bell-test change detects this attack even though the vertical/horizontal key subset alone would not.

Each source has density matrixAssume the two sources and the detector's random choices are independent. In the $p^2$ branch both resources are ideal. If either resource is $I_4/4$, the remote pair remains $I_4/4$ after any specified Bell-basis measurement outcome. An ideal detector and the correction of part (b) therefore yieldWhen the detector instead reports an independent uniformly random label, the corresponding four Pauli gates average the conditional pair to $I_4/4$. ThusHere $q$ is the probability of the reliable-detector branch, not its total chance of being correct: that chance is $q+(1-q)/4$.

For the usual two-setting CHSH inequality, the spin singlet state has correlation $-\mathbf{a}\cdot \mathbf{b}$ and the white-noise part has zero correlation, so $E(\mathbf{a},\mathbf{b})=-v\mathbf{a}\cdot \mathbf{b}$. With Alice's axes $z,x$ and Bob's axes $(z+x)/\sqrt{2},(z-x)/\sqrt{2}$, the absolute CHSH inequality combination is $2\sqrt{2}v$. These settings are optimal for an isotropic correlation tensor, givingThis is the threshold for the standard CHSH inequality with optimized projective polarization measurements. The unqualified phrase “the Bell inequality” needs this choice of test; entanglement alone is a weaker requirement, and no claim about every possible Bell inequality is needed for this calculation.