Nonlocality and the No-Signalling Principle

Two experimenters, Alice and Bob, sit far enough apart that no light signal can cross between them during a single run. Each receives one half of an entangled pair, freely dials a measurement setting, and records an outcome. When they later compare their logs, the joint statistics $P(a,b\,|\,x,y)$ are more strongly correlated than any story built from local causes and shared randomness can reproduce — this is quantum nonlocality, and Bell's theorem turns it into a hard experimental fact rather than a philosophical stance. The result looks, at first sight, like a threat to relativity. If measuring Alice's particle "instantly" fixes Bob's, can she wiggle her setting to send superluminal Morse code? The answer is a flat no, and the reason is structural. Everything Bob can access is encoded in his reduced density operator $\rho_B=\mathrm{Tr}_A(\rho_{AB})$ , and no local operation Alice performs — a unitary, an unsent measurement, even a measurement she then forgets — changes $\rho_B$ at all. The correlations are real, but their pull is invisible in either party's own histogram; it surfaces only when the two records are laid side by side over a slow, ordinary, subluminal channel. The no-signalling principle promotes that guarantee to an axiom stated on the probability table alone, with no mention of Hilbert space: each party's marginal must be independent of the other's input. Remarkably, this axiom is looser than quantum theory. Popescu–Rohrlich boxes obey no-signalling yet push CHSH to its algebraic ceiling of $4$ , past the quantum limit of $2\sqrt2\approx2.828$ . Nonlocality, no-signalling, and quantum theory are three nested rungs, and why nature stops on the middle one is still an open question.

Key takeaways

Entangled measurements produce correlations no local hidden-variable model can reproduce; the CHSH inequality certifies this nonlocality experimentally.
Yet quantum theory forbids faster-than-light signalling: a party's own outcome statistics are completely independent of the distant party's setting choice.
No-signalling is a clean condition on the table $P(a,b\,|\,x,y)$ — the marginal $\sum_b P(a,b\,|\,x,y)$ must not depend on $y$ — and quantum mechanics obeys it because reduced density operators are inert under remote local operations.
No-signalling is strictly weaker than quantum theory: PR boxes reach the algebraic value $S=4$ while staying no-signalling, so causality alone does not explain Tsirelson's bound $2\sqrt2$ .
Key idea: nonlocality lives entirely in the joint correlation table while every single-party marginal is flat and blind to the distant setting, so no local model explains the joint data yet no information ever outruns light.

The core insight

Nonlocality resides entirely in the joint correlation table; the single-party marginals are flat and independent of the remote setting, so no local model can explain the joint statistics and yet no information flows faster than light.

Where this lives today

Nonlocality left the philosophy seminar and became an engineering resource. In 2015 three groups — at Delft, NIST, and Vienna — reported the first loophole-free Bell tests, closing the locality and detection loopholes at once; the Delft experiment used electron spins in diamond nitrogen-vacancy centres separated by $1.3$ km. That underwrites device-independent quantum key distribution, where two parties certify a key's secrecy from the observed $S$ alone, trusting neither their devices nor their supplier. The same logic powers certified randomness: a violation $|S|>2$ cannot be faked by any deterministic local strategy, so the outputs must contain genuine unpredictability, and services now seed public randomness beacons with Bell violations. On the foundational side, the puzzle Tsirelson opened is very much alive: the Navascués–Pironio–Acín hierarchy bounds the quantum set from outside with semidefinite programs, and principles like information causality recover $2\sqrt2$ from information-theoretic axioms — yet why nature stops at $2\sqrt2$ and not $4$ remains open. The 2022 Nobel Prize in Physics, to Aspect, Clauser, and Zeilinger, recognized the program that turned Bell's Sunday hobby into the backbone of quantum information science.

The physical picture

Imagine two vending machines with no wire between them, dropped at opposite ends of a long hall. Each has a two-way switch (the setting) and spits out a token stamped $+1$ or $-1$ (the outcome). Watched alone, each machine looks like a fair coin, no matter how you flip its switch. The strangeness appears only when the two token streams are laid side by side afterward — the tokens agree or disagree in a pattern depending jointly on both switches, in a way no rulebook stuffed into the machines could produce. The "spooky" part is entirely in the comparison, never in either local stream.

Nonlocality is a statement about that joint table, not about any signal. The quantum state never lets Alice's choice $x$ tilt Bob's marginal $P(b\,|\,y)$ — the content of no-signalling — because tracing out Alice's system commutes with anything she does to it. Treating correlation tables as points in a high-dimensional space, tables reachable by local strategies form a polytope, quantum-achievable tables a larger convex set, no-signalling tables a still larger polytope. The CHSH quantity $S$ is one direction, and its ceiling climbs $2 \to 2\sqrt2 \to 4$ as you move outward through the three nested sets.

Figure

The three nested sets of correlations. Local hidden-variable tables sit inside the quantum set, which sits inside the no-signalling set. The CHSH value S measures how far out you are: local caps at 2, quantum at 2√2 ≈ 2.828 (Tsirelson), and no-signalling at the algebraic maximum 4 (reached by PR boxes). Quantum theory threads a strict middle path.

A remote measurement cannot signal for the same reason the reduced state is well defined: tracing out one wing commutes with any local operation on it, so $\mathrm{Tr}_A\big[(U_A\otimes I)\,\rho_{AB}\,(U_A^\dagger\otimes I)\big]=\rho_B$ . Causality is not an add-on axiom in quantum theory; it is baked into the tensor-product structure.

The core equations

A Bell scenario is fully described by a probability table. Alice chooses input $x$ , Bob chooses $y$ , and they obtain outcomes $a,b$ , all in $\{0,1\}$ in the simplest case. Everything downstream — nonlocality, no-signalling, the CHSH bounds — is a property of this one object.

P(a,b\,|\,x,y)\quad\text{with}\quad a,b,x,y\in\{0,1\} \qquad (1)

where: $x,y$ — Alice's and Bob's setting choices (inputs); $a,b$ — their outcomes (outputs); $P$ — joint conditional probability of the outcomes given the settings. Reading it: sixteen numbers, normalized for each $(x,y)$ . No-signalling demands that neither party's marginal leak the other's input: summing out Bob's outcome must give a distribution for Alice that does not know $y$ , and vice versa. These two conditions are the entire principle.

\sum_{b} P(a,b\,|\,x,y) = P_A(a\,|\,x)\ \ \forall y,\qquad \sum_{a} P(a,b\,|\,x,y) = P_B(b\,|\,y)\ \ \forall x \qquad (2)

where: $P_A(a\,|\,x)$ — Alice's marginal, well defined because it is independent of $y$ ; $P_B(b\,|\,y)$ — Bob's marginal, independent of $x$ . Reading it: a table failing either equation would let one party read the other's dial from local statistics alone — an instantaneous telegraph. Quantum mechanics generates its tables through the Born rule on a shared bipartite state, each measurement described by a POVM (positive operators summing to the identity) — the bridge from Hilbert space to the abstract table.

P(a,b\,|\,x,y)=\mathrm{Tr}\!\big[(M^x_a\otimes N^y_b)\,\rho_{AB}\big] \qquad (3)

where: $\rho_{AB}$ — the shared bipartite state, possibly entangled; $M^x_a$ — Alice's POVM element for outcome $a$ under setting $x$ , with $\sum_a M^x_a=I$ ; $N^y_b$ — Bob's POVM element, with $\sum_b N^y_b=I$ . Reading it: the tensor factor makes the two measurements act on separate subsystems. The engine of no-signalling is the invariance of the reduced state: Bob's accessible description is the partial trace over Alice, and nothing Alice does locally can move it.

\rho_B=\mathrm{Tr}_A(\rho_{AB}),\qquad \mathrm{Tr}_A\!\big[(U_A\otimes I)\,\rho_{AB}\,(U_A^\dagger\otimes I)\big]=\rho_B \qquad (4)

where: $U_A$ — any local unitary Alice applies (extends to any local CPTP map or unread measurement); $\rho_B$ — Bob's reduced state, invariant under Alice's action. Reading it: cyclicity of the trace over Alice's factor kills $U_A^\dagger U_A=I$ , leaving $\rho_B$ untouched. The maximally entangled Bell pair is the workhorse example: its global state is pure and perfectly correlated, yet each half, viewed alone, is completely random.

|\Phi^+\rangle=\tfrac{1}{\sqrt2}\big(|00\rangle+|11\rangle\big),\qquad \rho_A=\rho_B=\tfrac12 I \qquad (5)

where: $|\Phi^+\rangle$ — a Bell state carrying one full bit ( $\ln 2$ nat) of entanglement entropy; $\tfrac12 I$ — the maximally mixed single-qubit state. Reading it: the flat reduced state means every local measurement Alice performs yields $50/50$ statistics — exactly why she cannot signal. The CHSH combination packages four correlators into a single number that distinguishes the three theories, each correlator being the expectation of a product of $\pm1$ outcomes.

S=\langle A_0B_0\rangle+\langle A_0B_1\rangle+\langle A_1B_0\rangle-\langle A_1B_1\rangle \qquad (6)

where: $\langle A_xB_y\rangle=\sum_{a,b}(-1)^{a\oplus b}P(a,b\,|\,x,y)$ — the correlator for settings $x,y$ in $\pm1$ encoding. Reading it: the lone minus sign makes the classical maximum $2$ , not $4$ ; no local strategy makes all four terms favourable at once. Finally, the Popescu–Rohrlich box is the extremal no-signalling table, defined by a single logical relation between outcomes and inputs.

P_{\mathrm{PR}}(a,b\,|\,x,y)=\begin{cases}\tfrac12 & a\oplus b=x\cdot y\\[2pt]0 & \text{otherwise}\end{cases} \qquad (7)

where: $a\oplus b$ — outcome XOR (addition mod 2); $x\cdot y$ — product of the binary inputs, the target correlation. Reading it: outcomes are perfectly correlated ( $a=b$ ) unless both inputs are $1$ , where they are perfectly anticorrelated — the pattern that maximizes CHSH.

Derivation ANo-signalling from the quantum formalism

The goal is to prove, from the Born rule alone, that Bob's outcome statistics cannot depend on anything Alice does at her distant station — not on her setting $x$ , not on the outcome $a$ she happens to get, not even on whether she measures at all. We track Bob's effective state through Alice's measurement, average over her unread result, and watch the $x$ -dependence vanish identically once POVM completeness is applied — so relativistic causality is a theorem of quantum mechanics, not a postulate bolted on by hand.

Step 1. Suppose Alice measures setting $x$ and privately obtains outcome $a$ , which occurs with probability $P_A(a\,|\,x)$ . Bob's conditional state updates by the (renormalized) partial trace, $\rho^{(a)}_B=\mathrm{Tr}_A\big[(M^x_a\otimes I)\rho_{AB}\big]/P_A(a\,|\,x)$ . Why: conditioning on new information about the shared system is exactly the map that produces Bob's updated description.

Step 2. Bob now measures $y$ ; by the Born rule applied to his conditional state, $P(b\,|\,y,a,x)=\mathrm{Tr}\big[N^y_b\,\rho^{(a)}_B\big]$ . Why: this is the only statistic Bob can generate, so if it turns out to be $x$ -independent after averaging, no protocol of his can detect Alice's choice.

Step 3. Alice does not transmit $a$ , so Bob is described by the ensemble average over her hidden result, $P_B(b\,|\,y;x)=\sum_a P_A(a\,|\,x)\,P(b\,|\,y,a,x)=\sum_a \mathrm{Tr}\big[N^y_b\,\mathrm{Tr}_A((M^x_a\otimes I)\rho_{AB})\big]$ . Why: the normalizations cancel, leaving a clean sum of unconditioned partial traces — the physically relevant marginal when no classical message is sent.

Step 4. Push the sum inside and use completeness $\sum_a M^x_a=I_A$ , giving $P_B(b\,|\,y;x)=\mathrm{Tr}\big[N^y_b\,\mathrm{Tr}_A(\rho_{AB})\big]=\mathrm{Tr}\big[N^y_b\,\rho_B\big]$ . Why: completeness is what erases every trace of $x$ — the same collapse would occur for any setting Alice could have chosen, or for no measurement at all.

Step 5. The final expression contains neither $x$ nor any reference to Alice's measurement, so $\partial_x P_B(b\,|\,y;x)=0$ : Bob's statistics are frozen against Alice's every action. Why: this is condition (2) rederived from the state, and it holds no matter how strongly the joint table violates a Bell inequality. *Indeed, direct partial trace on $|\Phi^+\rangle\langle\Phi^+|$ gives $\rho_B=\tfrac12 I$ for every Alice POVM.*

Go deeper · Derivation B · 1 minTsirelson's bound — why quantum theory stops at 2√2

Derivation BTsirelson's bound — why quantum theory stops at 2√2

The goal is to show that quantum mechanics cannot reach the algebraic CHSH maximum of $4$ , but is capped at $2\sqrt2$ , and to locate that gap in the noncommutativity of each party's two observables. We square the Bell operator, watch the diagonal terms assemble a clean $4I$ , find the remaining terms collapse into a single product of commutators, and bound its norm — tracing the gap between $2\sqrt2$ and $4$ to a purely algebraic fact about bounded $\pm1$ observables that no quantum state can beat.

Step 1. Take $A_0,A_1,B_0,B_1$ as $\pm1$ observables, so $A_x^2=B_y^2=I$ , with $[A_x,B_y]=0$ across the cut, and form $\mathcal{B}=A_0B_0+A_0B_1+A_1B_0-A_1B_1$ . Squaring, the four diagonal terms each give $A_x^2B_y^2=I$ , summing to $4I$ . Why: the $\pm1$ property is what makes every squared term the identity, isolating a constant piece that already accounts for the classical bound.

Step 2. The sixteen cross terms cancel in pairs except for those mixing $A_0A_1$ with $B_0B_1$ ; carefully collecting signs gives $\mathcal{B}^2=4I+[A_0,A_1][B_1,B_0]$ (equivalently $4I-[A_0,A_1][B_0,B_1]$ ). Why: because Alice's operators commute with Bob's, the only surviving structure is a product of a commutator on each side — this is the entire quantum correction.

Step 3. Bound each commutator: for unit-norm observables, $\|[A_0,A_1]\|\le\|A_0A_1\|+\|A_1A_0\|\le 2$ , and likewise $\|[B_0,B_1]\|\le 2$ . Why: the triangle inequality plus $\|A_x\|=\|B_y\|=1$ caps how far two $\pm1$ observables can fail to commute.

Step 4. Combine: $\|\mathcal{B}^2\|\le 4+\|[A_0,A_1]\|\,\|[B_0,B_1]\|\le 4+4=8$ , so $\|\mathcal{B}\|\le\sqrt8=2\sqrt2$ . Why: the norm of a self-adjoint operator is the square root of the norm of its square, converting the bound on $\mathcal{B}^2$ into the CHSH bound.

Step 5. Therefore $|\langle\mathcal{B}\rangle|\le 2\sqrt2\approx2.828$ for every quantum state and every choice of measurements, with equality reached by $|\Phi^+\rangle$ and directions separated by $45^\circ$ . Why: the algebraic ceiling $4$ would require a commutator norm above $2$ , impossible for bounded $\pm1$ observables — which is why PR boxes are superquantum. *Numerically $2\sqrt2=2.8284\ldots$ ; the largest eigenvalue of the explicit Bell operator with those angles is $2.8284$ , and $\langle\Phi^+|\mathcal{B}|\Phi^+\rangle=2.8284$ .*

Go deeper · Derivation C · 1 minThe PR box saturates the algebraic maximum yet stays no-signalling

Derivation CThe PR box saturates the algebraic maximum yet stays no-signalling

The goal is to exhibit a correlation table that is a perfectly legitimate no-signalling object — flat marginals, causality intact — yet violates CHSH all the way to $S=4$ , decisively past the quantum ceiling. This proves no-signalling alone is too weak to pick out quantum theory: some extra principle must stop nature at $2\sqrt2$ . We check the marginals to confirm causality, evaluate the four correlators, then sum them with the CHSH signs.

Step 1. For the PR box, fix $a$ and $x$ and sum over $b$ : exactly one value of $b$ satisfies $a\oplus b=x\cdot y$ , and it carries weight $\tfrac12$ , so $P_A(a\,|\,x)=\tfrac12$ independent of $y$ ; symmetrically $P_B(b\,|\,y)=\tfrac12$ . Why: flat, input-blind marginals are precisely the no-signalling condition (2), so the box respects relativistic causality despite its extreme correlations.

Step 2. In $\pm1$ encoding the defining relation gives $\langle A_xB_y\rangle=(-1)^{x\cdot y}$ : the outcomes agree ( $\langle A_xB_y\rangle=+1$ ) for the three input pairs with $x\cdot y=0$ , and disagree ( $\langle A_xB_y\rangle=-1$ ) only for $x=y=1$ . Why: the product $x\cdot y$ singles out the one setting combination where the box flips from perfect correlation to perfect anticorrelation.

Step 3. Assemble the CHSH sum: $S=\langle A_0B_0\rangle+\langle A_0B_1\rangle+\langle A_1B_0\rangle-\langle A_1B_1\rangle=1+1+1-(-1)=4$ . Why: the lone minus sign in CHSH lands exactly on the one anticorrelated term, turning it into a positive contribution, so all four terms add constructively to the algebraic maximum.

Step 4. Since $4>2\sqrt2$ , no quantum state can reproduce this table, yet it violates no causality constraint. Why: the moral — nonlocality plus no-signalling does not equal quantum theory; the quantum set is a strict, still mysterious, subset. *Direct enumeration of the four input pairs confirms $S=1+1+1+1=4$ and flat $\tfrac12$ marginals.*

Limiting cases

The single quantity $S$ stratifies the three nested theories. Each row is a facet bound on how nonlocal a table in that class can be.

Case	Result	Meaning
Local hidden variables	$\lvert S\rvert \le 2$	The Bell/CHSH bound: any locally causal model with shared randomness is confined here. It is a facet of the local polytope.
Quantum mechanics	$\lvert S\rvert \le 2\sqrt2$	Tsirelson's bound. Reached by a maximally entangled pair with measurement angles separated by $45^\circ$ ; genuinely nonlocal yet sub-algebraic.
No-signalling (general)	$\lvert S\rvert \le 4$	The algebraic maximum, saturated by PR boxes. Proof that no-signalling alone is too weak to recover quantum correlations.
Product (unentangled) state	$\lvert S\rvert \le 2$	A separable $\rho_{AB}$ can never beat the local bound; entanglement is necessary (not sufficient) for violation.

Common trapthe misconception that measurement sends a detectable signal

It is tempting to say that measuring Alice's qubit "sends a signal" that collapses Bob's qubit into a definite state he can detect, letting them communicate instantly. This is wrong. Bob's local density operator is $\tfrac12 I$ both before and after Alice acts, and stays $\tfrac12 I$ whatever setting she chooses — a flat coin either way. He learns nothing until Alice's outcomes reach him over an ordinary subluminal channel and the records are compared. The "collapse" of Bob's qubit is how someone who already knows Alice's result updates their description; it is not a disturbance Bob can measure. Nothing travels faster than light — the correlation was there all along, waiting to be revealed by comparison.

Worked examples

1 · The Bell state reaches Tsirelson's bound (fully worked)

Take $|\Phi^+\rangle=\tfrac{1}{\sqrt2}(|00\rangle+|11\rangle)$ . Alice measures spin at angles $A_0=0^\circ$ and $A_1=90^\circ$ in the x–z plane; Bob at $B_0=45^\circ$ and $B_1=-45^\circ$ . Compute the four correlators and the CHSH value $S$ .

Step 1. For $|\Phi^+\rangle$ , an observable at angle $\alpha$ for Alice and $\beta$ for Bob has correlator $\langle A_\alpha B_\beta\rangle=\cos(\alpha-\beta)$ . This follows because $\langle Z\otimes Z\rangle=\langle X\otimes X\rangle=+1$ and $\langle Z\otimes X\rangle=\langle X\otimes Z\rangle=0$ , so only $\cos\alpha\cos\beta+\sin\alpha\sin\beta$ survives.

Step 2. Evaluate each term. $\langle A_0B_0\rangle=\cos(0-45^\circ)=\cos45^\circ=\mathbf{0.7071}$ ; $\langle A_0B_1\rangle=\cos(0+45^\circ)=\mathbf{0.7071}$ ; $\langle A_1B_0\rangle=\cos(90^\circ-45^\circ)=\mathbf{0.7071}$ ; $\langle A_1B_1\rangle=\cos(90^\circ+45^\circ)=\cos135^\circ=\mathbf{-0.7071}$ .

Step 3. Assemble with the CHSH signs: $S=0.7071+0.7071+0.7071-(-0.7071)=4\times0.7071=\mathbf{2.8284}$ . The lone minus sign lands on the one negative correlator and flips it positive, so all four terms add. This equals $2\sqrt2$ exactly, saturating Tsirelson's bound — the maximum any quantum system can achieve.

Step 4. This beats the local-realist ceiling of $2$ by a factor $\sqrt2$ . The entangled pair gives $S=2\sqrt2=2.828>2$ , ruling out every local hidden-variable model.

2 · Alice cannot signal to Bob, whatever she measures

Alice and Bob share $|\Phi^+\rangle$ . In run type I Alice measures in the computational basis ( $Z$ ); in run type II in the Hadamard basis ( $X$ ). She keeps her result secret. Show Bob's statistics are identical, so he cannot tell which setting she used — no signal.

Step 3. Whatever Bob then measures, his outcome probabilities are $\mathrm{Tr}[N^y_b\cdot\tfrac12 I]=\tfrac12\mathrm{Tr}[N^y_b]$ , depending only on his own POVM, never on Alice's basis. Bob's every statistic is identical in run types I and II, so Alice's setting is undetectable — no information crosses the gap.

Step 4. This is the operational content of equation (4): the two ensembles Alice prepares, $\{|0\rangle,|1\rangle\}$ and $\{|+\rangle,|-\rangle\}$ , share the same average $\tfrac12 I$ , and Bob accesses only that average. Different decompositions of the same $\rho_B$ are physically indistinguishable — the mathematical heart of no-signalling.

Physics of the portrait

The whole subject grows from one short paper — the person who made nonlocality precise turned a decades-old philosophical stalemate into laboratory physics.

Origins · 1 minJohn Stewart Bell and how this was discovered

John Stewart Bell

1928–1990 · quantum nonlocality and the foundations of physics

John Stewart Bell was born in Belfast in 1928, took his degree at Queen's University, and spent his career as a theoretical particle physicist at CERN, where his day job was accelerator design and quantum field theory. Foundations of quantum mechanics were, by his own description, a hobby he pursued "on Sundays." In 1964, dissatisfied with the vague truce over the Einstein–Bohr debate, he asked a sharp question: could any theory in which distant outcomes are fixed by local variables — carried with each particle and unaffected by the far-away setting — reproduce quantum predictions? His answer, the celebrated Bell inequality, was no: local hidden-variable theories obey a bound that entangled states demonstrably break, converting a matter of taste into an experiment. Bell was scrupulous about what the result showed; he stressed that its nonlocality cannot be used to signal. The Clauser–Horne–Shimony–Holt (CHSH) form made the inequality practical in 1969; Aspect's experiments closed key loopholes in the early 1980s; and Boris Tsirelson showed in 1980 that quantum theory has its own ceiling, $2\sqrt2$ , below the algebraic maximum. Bell died suddenly in 1990, the year he was reportedly nominated for the Nobel Prize — awarded in 2022 to Aspect, Clauser, and Zeilinger for the experiments his theorem made possible.

History

1935

— Einstein, Podolsky and Rosen argue that quantum mechanics is incomplete

1964

— Bell proves no local hidden-variable theory can reproduce quantum correlations

1969

— Clauser, Horne, Shimony and Holt cast the inequality in testable CHSH form

1980

— Tsirelson bounds the quantum CHSH value at 2√2

1994

— Popescu and Rohrlich introduce no-signalling boxes that exceed the quantum bound

Going further

Texts. Nielsen and Chuang, Quantum Computation and Quantum Information, Section 2.6 (EPR and Bell) · Preskill, Lecture Notes on Quantum Computation, Chapter 4 (Quantum Entanglement) · Watrous, The Theory of Quantum Information, chapter on nonlocal games · Wilde, Quantum Information Theory, Chapters 4 to 5 · Popescu and Rohrlich, Foundations of Physics 24, 379 (1994) · Mermin, Reviews of Modern Physics 65, 803 (1993) · Brunner, Cavalcanti, Pironio, Scarani and Wehner, Reviews of Modern Physics 86, 419 (2014). The key checkpoints in this article are: (1) $\mathrm{Tr}_B|\Phi^+\rangle\langle\Phi^+|=\tfrac12 I$ ; (2) the no-signalling sum follows from POVM completeness $\sum_b N^y_b=I$ ; (3) the PR-box marginals are flat at $\tfrac12$ and $S=4$ ; (4) the $\mathcal{B}^2=4I+[A_0,A_1][B_1,B_0]$ expansion yields $\|\mathcal{B}\|\le 2\sqrt2=2.8284\ldots$ ; (5) the Bell-state CHSH value is $2\sqrt2$ with angles $0^\circ, 90^\circ, 45^\circ, -45^\circ$ . All anchor numbers ( $2$ , $2\sqrt2$ , $4$ , $\ln 2$ ) are standard values.

Prerequisites. Bell States and EPR Pairs · CHSH Inequality and Its Quantum Violation · Bell's Theorem and Local Hidden Variables · Separable versus Entangled States. Leads to. Tsirelson's Bound · PR Boxes and Super-Quantum Correlations · Entanglement Witnesses.

Ask the tutor: Ask me to walk through why the PR box beats Tsirelson's bound while still refusing to send a single bit faster than light.

If measuring Alice's qubit changes Bob's, why isn't that a signal?

Because "changes Bob's qubit" is a statement about descriptions, not about anything Bob can measure. Before Alice acts, Bob's reduced state is $\tfrac12 I$ ; after she acts — with any setting, whether or not she tells him — it is still $\tfrac12 I$ . Everything Bob can observe is computed from that operator, so his histograms are identical no matter what Alice does. The correlation is real, but it shows up only when the two logs are compared over a classical channel no faster than light. So there is genuine nonlocality in the joint statistics and, at the same time, zero communication capacity.

What exactly does the no-signalling principle say?

It is a condition on the probability table $P(a,b\,|\,x,y)$ alone, with no mention of quantum mechanics. It demands that Alice's marginal $\sum_b P(a,b\,|\,x,y)$ be the same for every choice of Bob's input $y$ , and symmetrically for Bob. Physically: neither party can read the other's dial from their own statistics. If either condition failed, one party could encode a bit in their setting and the other decode it instantly — a faster-than-light telegraph. Because the principle lives on the table, it applies to any theory, which makes it a candidate axiom for reconstructing quantum mechanics from information-theoretic ideas.

How can correlations be "nonlocal" if nothing travels between the parties?

"Nonlocal" here is a technical verdict about the joint table, not a claim that a hidden influence zips across the gap. It means no model in which each particle carries pre-set instructions — fixed locally and untouched by the distant setting — can reproduce the observed correlations. Bell's theorem proves such local models obey $|S|\le2$ , and experiments see $|S|$ up to $2\sqrt2$ , stronger than any local mechanism allows. But this excess correlation carries no marginal bias and cannot be steered, so it transmits nothing. Nonlocality is the failure of local realism to explain the data — not a usable action at a distance.

Why does quantum theory stop at 2√2 instead of reaching 4?

Structurally, because each party's two observables are bounded $\pm1$ operators, and the Bell operator squares to $\mathcal{B}^2=4I+[A_0,A_1][B_1,B_0]$ . Each commutator of unit-norm observables has norm at most $2$ , so $\|\mathcal{B}^2\|\le8$ and $\|\mathcal{B}\|\le2\sqrt2$ ; reaching $4$ would need a commutator norm above $2$ , which bounded observables cannot supply. That is Tsirelson's bound. Why the world respects it rather than allowing PR-box correlations up to $4$ is a genuine open question — information causality reproduces $2\sqrt2$ , but a single agreed reason remains elusive.

What is a PR box and why do people care about it?

A Popescu–Rohrlich box is a hypothetical device whose outputs satisfy $a\oplus b=x\cdot y$ with otherwise uniform outcomes. Its marginals are flat, so it obeys no-signalling perfectly, yet it achieves $S=4$ , well beyond any quantum state. It is a thought experiment, not a lab object, but a useful one: it proves no-signalling alone does not single out quantum correlations, so some further principle must. It also collapses communication complexity — if PR boxes existed, many distributed computations would become trivial — one of the arguments for why nature is quantum and not "super-quantum."

Does entanglement always produce a Bell violation?

No. Entanglement is necessary for violating a Bell inequality — a separable (unentangled) state can never exceed $|S|=2$ — but it is not sufficient. Some mixed entangled states admit a local hidden-variable model for all projective measurements and so violate no Bell inequality, even though they are provably entangled by other tests. Maximally entangled pure states like $|\Phi^+\rangle$ do violate CHSH, up to the full $2\sqrt2$ , with the right measurements. The clean logical chain is: Bell violation $\Rightarrow$ entanglement, but entanglement $\not\Rightarrow$ Bell violation. This gap is why entanglement witnesses and nonlocality are studied as related but distinct resources.