Can a hidden plan fake quantum entanglement? The CHSH test, computed (lab notebook, 2026-11-01)
The question
Two particles are measured far apart and their answers agree more often than chance. Einstein's way out: the pair left the source carrying a shared plan (a "hidden variable") that fixes every answer in advance. Bell (1964) showed that any such local plan is capped, and Clauser, Horne, Shimony and Holt (1969) turned the cap into a test anyone can score: S = E(a,b) − E(a,b′) + E(a′,b) + E(a′,b′), where E is the average product of the two ±1 answers at one pair of detector settings. Every local plan gives |S| ≤ 2. Quantum mechanics predicts up to 2√2 ≈ 2.83 (Tsirelson's bound).
What we computed (our own calculation, script chsh.py, standard library, fixed seed)
Settings (polarisation angles): Alice a = 0°, a′ = 45°; Bob b = 22.5°, b′ = 67.5°.
Each figure two ways: closed form, and a simulation of 200,000 pairs per setting.
| Model | E(0°,22.5°) | E(0°,67.5°) | E(45°,22.5°) | E(45°,67.5°) | S closed form | S simulated | ||
|---|---|---|---|---|---|---|---|---|
| Quantum, entangled photons: E = cos 2(a − b) | +0.7071 | −0.7071 | +0.7071 | +0.7071 | 2.8284 | 2.8297 | ||
| Best shared-plan model of this kind: each pair carries an angle λ, each side answers sign cos 2(x − λ); E = 1 − 4 | a − b | /π | +0.5000 | −0.5000 | +0.5000 | +0.5000 | 2.0000 | 2.0000 |
- A third check of the cap: there are only 16 deterministic local strategies (each side fixes its answer for each of its two settings). Enumerating all 16, the largest |S| is exactly 2. Any random local plan is a mixture of these 16, so it cannot do better than 2 either.
- The simulated quantum S (2.8297) differs from 2√2 by 0.0013, inside the simulation's sampling error at this size (one standard error is about 0.003).
Where the gap closes (the two loopholes, computed)
- Noise. If a fraction of pairs is just noise, quantum S shrinks to 2√2 × v (v = visibility). It beats 2 only when v > 1/√2 = 70.7%. Below that, a local plan explains the data.
- Missed detections. If detectors miss pairs and the experiment counts only the pairs it caught, a local plan can fake a high S by "choosing" when to be missed. For maximally entangled pairs, the test is only honest when detector efficiency exceeds 2/(1 + √2) = 82.8% (Garg and Mermin, 1987).
- Worked example: a lab with 75% detection efficiency and 90% visibility gets S = 2.83 × 0.90 = 2.55 among the counted pairs, which looks like a clear violation, but at 75% (below 82.8%) a local plan can still produce it. Suppose the same lab reaches 85%: then 2.55 can no longer be faked this way.
What a real experiment found
Hensen et al. (2015, Delft) closed both loopholes at once with electron spins 1.3 km apart and measured S = 2.42 ± 0.20, above 2 (arXiv 1508.05949). The 2022 Nobel Prize in Physics went to Aspect, Clauser and Zeilinger "for experiments with entangled photons, establishing the violation of Bell inequalities and pioneering quantum information science".
What this does and does not show
- It shows: no plan fixed in advance and carried by the particles can reproduce the quantum correlations at these settings.
- It does not show that information travels faster than light: each side's own answers are 50/50 whatever the other does, so no message can be sent this way.
- It leaves out "superdeterminism" (the settings themselves were fixed in advance with the particles); the test cannot exclude a universe that conspires that way, and experiments choose settings with random number generators or distant starlight to make it implausible.
Interactive inputs for the reader (a simulation box)
- Visibility v (0.5 to 1.0, start 0.9): quantum S = 2√2 × v; result: beats 2 or not.
- Detector efficiency η (0.6 to 1.0, start 0.75): honest only when η > 82.8%; result: "a local plan could fake this" or not.
- Bob's first angle b (0° to 45°, start 22.5°; a = 0°, a′ = 45°, b′ = b + 45°): quantum S = cos 2b − cos 2(b + 45°) + cos 2(45° − b)
- cos 2(b); it peaks at 2.83 when b = 22.5° and falls to 2.00 at b = 0°.
Guess first
Before the numbers: if a shared plan can match the quantum answers 85% of the time at each setting, can it match them at all four settings together? (No: the 16 strategies cap S at 2, which corresponds to agreeing 75% of the time on average across the four.)
Sources
- Our own calculation (2026-11-01): chsh.py, both models closed form and simulated, 200,000 pairs per setting; all 16 local strategies enumerated.
- Clauser, J. F., Horne, M. A., Shimony, A., Holt, R. A. (1969). Proposed experiment to test local hidden-variable theories. Physical Review Letters 23, 880. https://doi.org/10.1103/PhysRevLett.23.880
- Garg, A., Mermin, N. D. (1987). Detector inefficiencies in the Einstein-Podolsky-Rosen experiment. Physical Review D 35, 3831. https://doi.org/10.1103/PhysRevD.35.3831
- Hensen, B. et al. (2015). Experimental loophole-free violation of a Bell inequality using entangled electron spins separated by 1.3 km. https://arxiv.org/abs/1508.05949
- The Nobel Prize in Physics 2022, press release. https://www.nobelprize.org/prizes/physics/2022/press-release/
Not sure / left out
- Tsirelson's bound (2√2) is stated without its own source here; it follows from the quantum formula at the optimal angles, which the table computes.
- Eberhard (1993) lowers the efficiency threshold to 66.7% with non-maximally entangled states; not computed here.