University Physics IV · Frontiers of Modern Physics · 15.2

Entanglement, CHSH & Bell Tests

Bell tests are where a philosophical argument acquired error bars. This topic gives you the working parts: write the singlet, predict its correlations, assemble the four numbers that make S, and say precisely which assumption a measured 2.4 destroys — and which one it leaves completely intact.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Entanglement says a two-particle state can be completely specified while neither particle separately has a state at all: for the singlet (|↑↓⟩ − |↓↑⟩)/√2 no pair of one-particle states reproduces the joint predictions, and each particle alone is the maximally mixed I/2. The cost only shows up when distant measurements are compared. Quantum mechanics gives the correlation E(θ) = −cos θ between analysers whose axes differ by θ; any account in which each particle carries its answers with it and neither setting reaches the other station is confined to |S| ≤ 2 for the combination S = E(a, b) − E(a, b′) + E(a′, b) + E(a′, b′).

That bound follows from two lines of ±1 algebra rather than from any particular model, so a single number tests the entire class at once. The singlet reaches 2√2 ≈ 2.83 when neighbouring analysers differ by 45°, and experiments from Freedman and Clauser in 1972 onward have landed above 2. What that buys is a genuine deletion — locality and pre-existing values cannot both be kept — and what it does not buy is a channel: the marginal at each station stays exactly 50/50 whatever the far setting, so the correlation appears only when the two lists of outcomes are brought together at ordinary speeds.

Simple definition: Two particles are entangled when their joint state cannot be written as one state for each particle separately, so the pair carries definite properties that neither member on its own possesses.
Example: Line both analysers up on a singlet pair and the two spins disagree on every single trial; turn one analyser by 45° and they still disagree on 85.4% of trials — cos²(22.5°) — where two independent coins would give 50%.

The singlet state|Ψ⁻⟩ = (|↑↓⟩ − |↓↑⟩)/√2

No product form exists: αγ = βδ = 0 cannot hold while αδ and βγ are both non-zero.

Amplitudes are dimensionless; the first ket is particle A, the second B. Normalised: (1/√2)² + (1/√2)² = 1.

Singlet correlationE(a, b) = ⟨(σ⋅â)(σ⋅b̂)⟩ = −cos θ_ab

−1 at 0°, 0 at 90°, +1 at 180°. This one curve is the entire prediction a Bell test checks.

θ_ab is the angle between the two analyser axes; E is a pure number between −1 and +1.

Correlation from the countsE = (N₊₊ + N₋₋ − N₊₋ − N₋₊)/N, P(opposite) = cos²(θ/2)

Turns a raw four-cell coincidence table into the single number the inequality constrains.

N₊₊ and the rest are coincidence counts for one setting pair, N their total; counts are integers, E dimensionless.

CHSH combination and the local boundS = E(a, b) − E(a, b′) + E(a′, b) + E(a′, b′), |S| ≤ 2

It holds for every local hidden-variable theory at once, so exceeding 2 refutes the whole class rather than one model.

Two settings per side, four correlation runs. The bound needs outcomes ±1, locality, and freely chosen settings.

Tsirelson's bound|S|_QM ≤ 2√2 ≈ 2.828

Quantum mechanics breaks the local bound but stops well short of the algebraic maximum 4, which is itself a testable claim.

Reached by a maximally entangled pair with neighbouring analysers 45° apart — 22.5° of physical rotation for polarisation.

No-signallingP(A = +|a, b) = P(A = +|a, b′) = ½

The distant setting never shifts a local histogram, so no protocol can turn the correlation into a message.

Marginal probability at Alice's station, for either of Bob's settings; ρ_A = Tr_B ρ = I/2 in every case.

Why the singlet has no product form

Write the most general product of two spin-½ states: (α|↑⟩ + β|↓⟩) ⊗ (γ|↑⟩ + δ|↓⟩) = αγ|↑↑⟩ + αδ|↑↓⟩ + βγ|↓↑⟩ + βδ|↓↓⟩. Matching it to |Ψ⁻⟩ = (|↑↓⟩ − |↓↑⟩)/√2 demands αδ = 1/√2 and βγ = −1/√2, so all four coefficients must be non-zero — yet it also demands αγ = 0 and βδ = 0, which forces one of them to vanish. The two requirements cannot both hold, so no product state exists, and that impossibility is the definition of entangled. Now trace out one particle: ρ_A = I/2, a fair coin with no preferred axis, identical in every basis. The pair is completely specified while each half separately carries nothing at all. Everything the state knows is stored in the relation between the two halves, which is exactly why one detector can never reveal it and a comparison of two detectors can.

The correlation you can actually count

A run produces a table, not a wavefunction. Fix the two analyser directions, record ±1 at each station, and count the four coincidence types: E = (N₊₊ + N₋₋ − N₊₋ − N₋₊)/N. For the singlet the prediction is E(θ) = −cos θ, equivalently P(opposite) = cos²(θ/2) and P(same) = sin²(θ/2), with θ the angle between the axes. Three cases anchor the curve. Parallel analysers, θ = 0: opposite on every trial, E = −1 — the perfect anticorrelation angular-momentum conservation already demands. Perpendicular, θ = 90°: P(opposite) = cos²45° = 0.5, the outcomes look independent, E = 0. Antiparallel, θ = 180°: identical every time, E = +1. In between, at θ = 30°, E = −cos 30° = −0.866, so 93.3% of pairs disagree. It is in this middle range, not at the anchor points, that the quantum and local predictions come apart — which is why a Bell test never uses aligned analysers.

Where the bound of 2 comes from

Bell's argument assumes three things and nothing else. Each pair carries some state λ, drawn from a distribution that does not depend on which settings will later be chosen; the outcome A(a, λ) = ±1 at one station does not depend on the distant setting b, nor B(b, λ) on a; and the settings are chosen freely. For a single λ, look at the quantity A(a, λ)[B(b, λ) − B(b′, λ)] + A(a′, λ)[B(b, λ) + B(b′, λ)]. Since B(b, λ) and B(b′, λ) are each ±1, they are either equal or opposite: one bracket is 0 and the other is ±2, so the whole expression is exactly ±2 for every λ. Averaging over λ can only shrink a quantity that never exceeds 2 in magnitude, giving |S| ≤ 2. Notice what is absent — no Hilbert space, no dynamics, no assumption about mechanism. The inequality constrains a class of theories, so measuring |S| > 2 does not adjust a model, it evacuates the class.

Choosing the settings: 45° for spins, 22.5° for photons

The four correlations are not equally useful; the settings must be staggered. Take a = 0°, a′ = 90°, b = 45°, b′ = 135°. Three of the pairs are 45° apart and the fourth, a with b′, is 135° apart, so E(a, b) = E(a′, b) = E(a′, b′) = −cos 45° = −0.7071 while E(a, b′) = −cos 135° = +0.7071. The combination subtracts that one term, so instead of cancelling, all four contributions reinforce: S = −0.7071 − 0.7071 − 0.7071 − 0.7071 = −2.8284. That is 2√2, Tsirelson's bound — the largest |S| quantum mechanics permits for any state and any observables, and well short of the algebraic maximum 4. One conversion trips people up. Polarisation correlations run on twice the angle, E = −cos 2Δ, because a photon returns to itself under a 180° rotation while a spin-½ needs 720°. So for entangled photons the same optimum sits at analyser settings 0°, 22.5°, 45° and 67.5°: half the rotation, identical physics.

What a violation settles, and what it leaves alone

Settled: no theory in which outcomes are fixed in advance by variables carried locally can produce the data. That is the whole content, and it is a great deal — it removes a kind of explanation rather than promoting one. Untouched: signalling. Compute Alice's marginal from the joint probabilities at θ = 45°. P(+,+) = ½sin²22.5° = 0.0732 and P(+,−) = ½cos²22.5° = 0.4268, which sum to exactly ½ — and the same ½ appears for whatever setting Bob picks, because his choice cannot alter ρ_A = I/2. His analyser never moves her histogram, so no protocol turns the correlation into a message and relativity's prohibition survives untouched. Also untouched: what replaces locality. Some accounts drop pre-existing values, others drop locality; the experiment does not choose between them. What it does end is the idea that the pair was merely correlated at the source, the way two shoes packed in separate boxes are correlated.

Loopholes, and what 2015 changed

An inequality constrains the data you actually collect, so an experiment has to earn its assumptions. The detection loophole: with detection efficiency η, the undetected pairs could be the awkward ones, and unless η exceeds about 82.8% — that is 2(√2 − 1) for a maximally entangled pair under CHSH, or 2/3 using Eberhard's inequality with a deliberately non-maximal state — the conclusion rests on a fair-sampling assumption rather than on the counts. The locality loophole: each measurement, setting choice included, must finish before light could carry news of it to the other station, and across the 1.3 km separating the two labs in the Delft experiment that budget is 4.3 μs. And freedom of choice: the settings must not be correlated with λ, which is why fast physical random generators drive them. The 2015 runs at Delft, NIST and Vienna closed detection and locality in the same experiment for the first time; Delft reported S = 2.42 from 245 event-ready trials, p = 0.039.

Change one variable at a time

Make the relationship visible.

Interactive model

Analyser step θ between settings45.0 °

Source visibility V1.00

Hold V at 1 and sweep θ: |S| peaks at 2.83 exactly at θ = 45° and nowhere else. Then pull V down — the bar crosses the local bound near V = 0.71, so a source below 71% visibility cannot violate the inequality at any angle.

CORRELATION E AT θ-0.707

CHSH |S|2.828

MARGIN OVER LOCAL BOUND0.828

PAIRS DISAGREEING AT θ85.4 %

Live interpretationCORRELATION E AT θ: −0.707. CHSH |S|: 2.828. MARGIN OVER LOCAL BOUND: 0.828. PAIRS DISAGREEING AT θ: 85.4 %

Catch the common trap

Explain before calculating.

Misconception check

The commonest error is reading a violation as a faster-than-light influence you could use. Watch one station alone and nothing happens: whichever setting the distant analyser takes, the local outcomes stay 50/50, because the reduced state there is I/2 for every choice the far side makes. Alice cannot even tell whether Bob has measured yet. The correlation exists only in the comparison of two lists, and those lists must be carried to one place at ordinary speeds, so nothing outruns light. What the violation does kill is the conjunction of locality with pre-existing values — one of the two must go — and it is silent on which. It also does not establish that a particle has no properties until it is looked at; that is one interpretation among several, not a measured result.

A spin-singlet source is measured with analyser settings a = 0°, a′ = 90°, b = 45° and b′ = 135°, every correlation obeying E = −cos(angle difference). What does S = E(a, b) − E(a, b′) + E(a′, b) + E(a′, b′) come to, and what does it establish?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA spin-singlet source sends one particle to each of two Stern–Gerlach analysers whose axes differ by 60°. Find the correlation E, the fraction of pairs whose outcomes disagree, and the expected split of 400 pairs.

The singlet prediction is E(θ) = −cos θ. With θ = 60°, E = −cos 60° = −0.500.
E is defined as P(same) − P(opposite), and the two sum to 1, so P(opposite) = (1 − E)/2 = 1.500/2 = 0.750 and P(same) = 0.250. The same value comes from cos²(θ/2) = cos²30° = 0.750.
Out of 400 pairs, expect 0.750 × 400 = 300 disagreeing and 100 agreeing, with a binomial spread of √(400 × 0.750 × 0.250) = √75 = 8.7 pairs.
Check the ends: θ = 0 gives E = −1 and disagreement every time; θ = 90° gives E = 0 and a 50/50 split. 60° sits between, nearer the anticorrelated end.

AnswerE = −0.500; 75.0% of pairs disagree, so about 300 opposite and 100 same out of 400, with a statistical spread near ±9 pairs.

MediumThe same source is analysed at a = 0°, a′ = 90°, b = 45° and b′ = 135°. Compute S = E(a, b) − E(a, b′) + E(a′, b) + E(a′, b′), compare it with the local bound, and state the analyser angles that reproduce the same test with polarisation-entangled photons.

List the angle differences: a to b is 45°, a to b′ is 135°, a′ to b is 45°, a′ to b′ is 45°. Cosine is even, so the sign of a difference does not matter.
Apply E = −cos θ: E(a, b) = −0.7071, E(a, b′) = −cos 135° = +0.7071, E(a′, b) = −0.7071, E(a′, b′) = −0.7071.
Combine, remembering the second term is subtracted: S = −0.7071 − (+0.7071) + (−0.7071) + (−0.7071) = −2.8284.
So |S| = 2.828 = 2√2, exceeding the local bound of 2 by a factor of 1.414. It is also the ceiling: Tsirelson's bound forbids any quantum state from doing better.
Polarisation correlations go as −cos 2Δ, so every angle halves: photons reach the same optimum at analyser settings 0°, 22.5°, 45° and 67.5°.

AnswerS = −2√2 = −2.828, so |S| = 2.83 against a local bound of 2. The equivalent photon settings are 0°, 22.5°, 45° and 67.5°.

HardA real source emits Werner states, an imperfect mixture whose correlations are reduced to E(θ) = −V cos θ for a visibility V. (a) Find S at the optimal settings. (b) Find the smallest V that can violate the inequality. (c) A run gives V = 0.780 with each of the four measured correlations uncertain by ±0.012; report |S| and how far above 2 it lies.

The unpolarised part of the mixture gives all four outcome pairs equal weight, so it contributes zero correlation at every angle; only the singlet fraction survives and every E carries the factor V.
(a) At the optimum each of the four terms has magnitude V cos 45° = 0.7071V and the CHSH signs make them reinforce: S = −4 × 0.7071V = −2.828V.
(b) Violation needs 2.828V > 2, so V > 2/(2√2) = 1/√2 = 0.7071. Below 70.7% visibility no choice of angles can break the inequality.
(c) |S| = 2.8284 × 0.780 = 2.206.
The four correlations are measured independently and enter with coefficients ±1, so their uncertainties add in quadrature: u_S = √(4 × 0.012²) = 2 × 0.012 = 0.024.
The margin over the bound is 2.206 − 2 = 0.206, which is 0.206/0.024 = 8.6 standard deviations. Significance is not the end of the argument: with detection efficiency below 82.8%, fair sampling would still be carrying part of the claim.

AnswerS = −2√2 V, so violation needs V > 0.707. At V = 0.780, |S| = 2.206 ± 0.024, which is 8.6σ above the local bound of 2.