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Module 08

Entanglement and Bell non-locality

Refute, with your own data, the assumption that outcomes were already decided before the measurement.

What you will see

  • Each side alone is completely random, yet the two are strictly correlated
  • Rotate the measurement angles and the correlation follows a cosine, not the kinked line local realism allows
  • The CHSH value accumulates live; the moment it passes 2, the classical explanation is out

Assumed background

  • Single qubits and the Bloch sphere (module 05)
  • Measurement and projection (module 07)
  • Conditional and joint probability

Everything odd so far could still, at a stretch, be explained away by “the particle carries some property we have not identified”.

This module ends that option. The power of Bell’s theorem is that it does not ask you to believe quantum mechanics. It needs only two assumptions that look self-evident — locality (a measurement here does not affect the outcome there) and realism (measurement reads off a property that already exists) — and proves that a certain measurable quantity must then be at most 2.

Experiment gives 22≈2.832\sqrt2 \approx 2.83.

Entanglement and Bell: killing the assumption that outcomes were fixed in advance

Each pair flies to the two ends and each side records ±1. Either side alone is pure randomness; together they are strictly correlated. Once the CHSH value S climbs past 2, local realism is out.

Loading 3D scene…

Model

Sampling by quantum mechanics: P(B|A) = (1 − A·â·b̂)/2

Measurement angles
0°
90°
45°
135°

The defaults are the angles that maximise S (0°, 90°, 45°, 135°). Spin correlations go as −cos θ, so the optimal spacing is 45°.

Run
Correlation E(θ)
0°90°180°

Blue is the quantum prediction −cos θ; the dashed orange line is the best a local hidden-variable model can do; the green dots are the measured values for the four current settings. The two agree at 0°, 90° and 180° and differ most near 45° — which is exactly where the four CHSH angles sit.

CHSH value |S|0.000

Classical bound 2 · quantum bound 2√2 ≈ 2.828 · predicted 2.828

Samples 0Alice, fraction of +1: —Bob, fraction of +1: —
S=E(a,b)−E(a,b′)+E(a′,b)+E(a′,b′)S = E(a,b) - E(a,b') + E(a',b) + E(a',b')
  • outcome +1
  • outcome −1
  • Measurement axis

What to look for

  • Look at the two spheres first: both reduced states sit at the centre (maximally mixed), so either side on its own is pure noise carrying no information. All of it lives in the correlation.
  • Let a few hundred pairs go through at the default angles: |S| settles near 2.8, comfortably past the classical bound of 2.
  • Switch to the local hidden-variable model: at the same angles |S| sticks below 2 and refuses to climb. That is not a badly tuned model — no local hidden-variable model can get past it, and that is Bell's theorem.
  • Press "All parallel" and S drops to zero. With the wrong angles the experiment tests nothing at all — which is why the angles in a Bell test have to be chosen with care.
  • Compare the two curves: quantum mechanics gives a cosine, local hidden variables at best a straight-line kink, and they differ most near 45° — which is exactly why the four angles are spaced 45° apart.

✕ "Entanglement transmits information faster than light"

Alice cannot choose whether she gets +1 or −1, and Bob's side is always 50/50 on its own. Seeing the correlation requires bringing both records together over a classical channel. The no-signalling theorem guarantees this.

✕ "Alice's measurement changes Bob's particle"

Change inertial frame and the order of the two measurements reverses. "Which caused which" has no invariant meaning between space-like separated events. The safe statement is only this: the joint distribution does not factorise into two independent local ones.

✕ "Entanglement is just a very strong correlation"

Classical correlations can be perfect too — post one glove to each of two cities and looking at one tells you the other. The difference is that classical correlation strength versus angle is a kinked line while the quantum one is a cosine, and that difference is measurable.

Think it through

  1. In the glove story, measuring only "left or right" at both ends also gives 100% correlation. Why must you measure along at least two different directions to tell gloves from entanglement?
  2. The quantum bound is 2√2 (the Tsirelson bound) while the algebraic maximum is 4. Is there a theory that reaches 4 yet still forbids faster-than-light signalling? (Keyword: PR box.)
  3. If only 50% of pairs are detected and the losses are correlated with the hidden variable, can a local model fake |S| > 2? That is the detection loophole, closed for good only in 2015.

An entangled state: definite as a whole, random in its parts

Take the singlet (total spin zero):

∣Ψ−⟩=12(∣↑⟩A∣↓⟩B−∣↓⟩A∣↑⟩B).\ket{\Psi^-} = \frac{1}{\sqrt2}\left(\ket{\uparrow}_A\ket{\downarrow}_B - \ket{\downarrow}_A\ket{\uparrow}_B\right).

It cannot be written as ∣ϕ⟩A⊗∣χ⟩B\ket{\phi}_A\otimes\ket{\chi}_B, which is the definition of entanglement.

Consequently:

  • Alice looking at her particle alone gets ↑ and ↓ equally often, completely at random, whatever axis she picks;
  • yet whenever the two use the same axis, the results are always opposite — perfectly correlated.

The two spheres in the scene are the reduced states of A and B. Both sit at the centre (maximally mixed), which says that locally there is no information at all. All of it is in the correlation.

CHSH: turning philosophy into a number

Alice picks one of two angles a,a′a, a'; Bob picks one of b,b′b, b'; each records ±1\pm1. Define the correlation E(a,b)=⟨AB⟩E(a,b) = \langle A B\rangle and then

S=E(a,b)−E(a,b′)+E(a′,b)+E(a′,b′).S = E(a,b) - E(a,b') + E(a',b) + E(a',b').

What local hidden variables predict: if each particle carries predetermined values A(λ),B(λ)A(\lambda), B(\lambda), then for every λ\lambda

A(a)B(b)−A(a)B(b′)+A(a′)B(b)+A(a′)B(b′)=A(a)[B(b)−B(b′)]+A(a′)[B(b)+B(b′)]=±2,A(a)B(b) - A(a)B(b') + A(a')B(b) + A(a')B(b') = A(a)[B(b)-B(b')] + A(a')[B(b)+B(b')] = \pm 2,

because one bracket is 0 and the other is ±2\pm2. Averaging gives ∣S∣≤2|S| \le 2.

What quantum mechanics predicts: for the singlet, E(a,b)=−cos⁡θabE(a,b) = -\cos\theta_{ab}. Taking a=0°a=0°, a′=90°a'=90°, b=45°b=45°, b′=135°b'=135°, the four angles are 45°,135°,45°,45°45°, 135°, 45°, 45°, so

∣S∣=∣−22−22−22−22∣=22≈2.828.|S| = \left|{-\tfrac{\sqrt2}{2}} - \tfrac{\sqrt2}{2} - \tfrac{\sqrt2}{2} - \tfrac{\sqrt2}{2}\right| = 2\sqrt2 \approx 2.828 .

You can drag all four angles in the scene and watch SS respond. The defaults are the optimal set above.

Where the cosine comes fromadvanced

For the singlet, the joint expectation along a^\hat a and b^\hat b is

E(a^,b^)=⟨Ψ−∣(σ⃗⋅a^)⊗(σ⃗⋅b^)∣Ψ−⟩=− a^⋅b^=−cos⁡θab.E(\hat a,\hat b) = \bra{\Psi^-}(\vec\sigma\cdot\hat a)\otimes(\vec\sigma\cdot\hat b)\ket{\Psi^-} = -\,\hat a\cdot\hat b = -\cos\theta_{ab}.

A pretty property of the singlet is that it is anti-correlated along every direction (it is rotationally invariant), so only the relative angle matters.

The crucial difference is cosine versus kinked line: the best a local hidden-variable model can manage is piecewise linear, while quantum mechanics gives a cosine. They agree at 0°0°, 90°90° and 180°180° and differ most around 45°45° — exactly where the four CHSH angles are placed.

The loopholes, and how they were closed

Early Bell experiments left two classical escape routes:

  • Detection loophole: only a fraction of pairs are detected, and if the losses correlate with the hidden variable a classical model can still fit the data. High-efficiency ion and superconducting-qubit experiments closed it from 2013.
  • Locality loophole: if the two measurement choices have time to communicate, locality was never really tested. Aspect used fast switching in 1982; Weihs used space-like separation in 1998.

In 2015 three experiments (Delft, NIST, Vienna) closed both at once. The 2022 Nobel Prize went to Aspect, Clauser and Zeilinger.

The only remaining gap is superdeterminism — the assumption that even the experimenters’ choice of angles was fixed in advance. It cannot be falsified and it cannot be used to predict anything.

Key formulas

  • Singlet state

    ∣Ψ−⟩=12(∣↑↓⟩−∣↓↑⟩)\ket{\Psi^-}=\tfrac{1}{\sqrt2}(\ket{\uparrow\downarrow}-\ket{\downarrow\uparrow})

    Not a product ⇒ entangled

  • Quantum correlation

    E(a^,b^)=−a^⋅b^E(\hat a,\hat b) = -\hat a\cdot\hat b

    Depends only on the angle

  • Classical CHSH bound

    ∣S∣≤2|S| \le 2

    Forced by local realism

  • Tsirelson bound

    ∣S∣≤22|S| \le 2\sqrt2

    Quantum mechanics' own ceiling

Think it through

  1. Set all four angles equal. What is SS? Why does this choice test nothing non-classical?
  2. Quantum mechanics reaches 222\sqrt2, while the algebraic maximum is 4. Is there a theory that is more non-local than quantum mechanics yet still forbids signalling? (Keyword: PR box. Why nature stopped at 222\sqrt2 has no accepted answer.)
  3. If Alice measures first and Bob second, it is tempting to say “Alice’s measurement collapsed Bob’s particle”. Change inertial frame and the order reverses. What is wrong with the question?

Go deeper · matching textbook sections

The 3D scenes build the picture; the full derivations and exercises live in the textbook.

Having finished this module