The Bell inequality and CHSH

The last section ended on a dangling question: could the Bell state’s perfect correlations just mean the two particles copied the same answer sheet before parting? This section is the intellectual climax of the chapter — we will watch this seemingly untestable philosophical question be turned by John Bell, in 1964, into a piece of arithmetic, and then voted down by experiment, ballot by ballot, over the following sixty years.

EPR: a thoroughly reasonable objection

In 1935, Einstein, Podolsky and Rosen stared at entangled states and made the following argument. Take the spin singlet (the Bell state $\ket{\Psi^-}$, total spin zero):

$$\ket{\Psi^-}=\frac{\ket{\uparrow\downarrow}-\ket{\downarrow\uparrow}}{\sqrt2}\tag{9.4.1}$$

Send the two particles to a far-apart Alice and Bob. Alice measures along $z$ and gets $\uparrow$; she immediately knows that Bob, measuring along $z$, must get $\downarrow$ — one hundred percent, no exceptions. EPR argue: Bob’s particle is worlds away from Alice, so her measurement cannot instantly change it (locality); and a quantity you can predict with certainty without touching the system must be a property it carried all along (realism). Conclusion: the outcomes were determined before any measurement, and quantum mechanics’ “probabilities” merely reflect hidden determining factors we have not seen — hidden variables.

The claim sounds untestable: the crib sheet is hidden in the particle’s pocket — who could prove it isn’t there? Bell’s genius was to discover: if the crib sheet exists, then no matter what is written on it, it must obey an inequality.

What a local hidden-variable model actually promises

Write the claim as mathematics. Each particle pair sets out carrying a hidden variable $\lambda$ (the crib sheet — arbitrary content, arbitrary distribution). Alice has two possible measurement directions $a, a'$, Bob has $b, b'$, and every outcome is recorded as $\pm1$. Local realism promises the existence of four predetermined functions:

$$A(a,\lambda),\ A(a',\lambda),\ B(b,\lambda),\ B(b',\lambda)\ \in\ \{\pm1\}\tag{9.4.2}$$

Two points are crucial: $A$ does not depend on what Bob chose (locality), and all four values exist simultaneously for each $\lambda$ (realism — even if Alice only measured $a$ this round, “what would I have got measuring $a'$” has a definite answer). The correlation function is just an average over crib sheets:

$$E(a,b)=\int \dd\lambda\,p(\lambda)\,A(a,\lambda)B(b,\lambda)\tag{9.4.3}$$

The CHSH inequality: ∣S∣ ≤ 2basic~6 minexpand

Step 1: for one fixed crib sheet $\lambda$, examine the combination

$$s(\lambda)=A(a,\lambda)\big[B(b,\lambda)-B(b',\lambda)\big] +A(a',\lambda)\big[B(b,\lambda)+B(b',\lambda)\big]\tag{9.4.4}$$

Step 2: enumerate. $B(b,\lambda)$ and $B(b',\lambda)$ are each $\pm1$, so there are only two cases:

• they are equal: the first bracket is 0, the second is $\pm2$, so $s=\pm2$;
• they are opposite: the second bracket is 0, the first is $\pm2$, so $s=\pm2$.

However the crib sheet is written, always $s(\lambda)=\pm2$.

Step 3: average. Define

$$S\equiv E(a,b)-E(a,b')+E(a',b)+E(a',b')=\int\dd\lambda\,p(\lambda)\,s(\lambda)\tag{9.4.5}$$

A weighted average of $\pm2$‘s can never leave $[-2,2]$:

$$\boxed{\,|S|\le 2\,}\tag{9.4.6}$$

The derivation used exactly two assumptions: the outcomes are predetermined (all four values exist at once) and do not depend on the other party’s choice. Not a single ingredient of quantum mechanics was used — this is the ceiling for every local hidden-variable theory.

What quantum mechanics says: first compute the correlation function

The singlet's correlation function E(a, b) = −cos(θa − θb)advanced~8 minexpand

Let Alice measure the spin along $\hat a$ and Bob along $\hat b$ (outcomes $\pm1$, i.e. measuring $\hat a\cdot\vec\sigma$). The quantum correlation function is

$$E(a,b)=\braket{\Psi^-}{(\hat a\cdot\vec\sigma)\otimes(\hat b\cdot\vec\sigma)\,|\Psi^-}\tag{9.4.7}$$

Step 1: exploit the singlet’s rotational symmetry. $\ket{\Psi^-}$ has total spin zero and is invariant under simultaneous rotation of both particles (chapter 5). So the answer can depend only on the angle between $\hat a$ and $\hat b$. Take both in the $x$–$z$ plane, angle $\theta_{ab}$.

Step 2: three special cases first. Take $\hat a=\hat b=\hat z$:

$$(\sigma_z\otimes\sigma_z)\ket{\Psi^-} =\frac{(+1)(-1)\ket{\uparrow\downarrow}-(-1)(+1)\ket{\downarrow\uparrow}}{\sqrt2} =-\ket{\Psi^-}\tag{9.4.8}$$

So $E=-1$: parallel measurements are perfectly anticorrelated (total spin zero, made visible). The same method with $\hat a=\hat z,\hat b=\hat x$ (using $\sigma_x$ to flip basis states) makes the cross terms cancel in pairs, giving $E=0$.

Step 3: general angle. Decompose $\hat b\cdot\vec\sigma=\cos\theta_{ab}\,\sigma_{\parallel}+\sin\theta_{ab}\,\sigma_{\perp}$ along the directions parallel and perpendicular to $\hat a$, and assemble linearly from the two special cases:

$$E(a,b)=-\cos\theta_{ab}\tag{9.4.9}$$

Check the endpoints: $\theta_{ab}=0$ gives $-1$ (perfect anticorrelation), $90^\circ$ gives $0$ (independence), $180^\circ$ gives $+1$ (perfect correlation). A smooth cosine — compare it with the closest thing a hidden-variable model can produce, a piecewise-linear ramp: the cosine is steep in the middle and flat at the ends, and that is precisely the geometric reason it can violate CHSH.

Now pick angles to maximise $|S|$. Take (all in one plane, in degrees)

$$a=0,\quad a'=90,\quad b=45,\quad b'=135\tag{9.4.10}$$

Substitute each pair into $E=-\cos\theta_{ab}$:

Pair	Angle	$E$
$a,b$	45°	$-\tfrac{\sqrt2}{2}$
$a,b'$	135°	$+\tfrac{\sqrt2}{2}$
$a',b$	45°	$-\tfrac{\sqrt2}{2}$
$a',b'$	45°	$-\tfrac{\sqrt2}{2}$

$$S=E(a,b)-E(a,b')+E(a',b)+E(a',b') =-\frac{\sqrt2}{2}\times4=-2\sqrt2\approx-2.83\tag{9.4.11}$$ $$|S|=2\sqrt2>2\tag{9.4.12}$$

Quantum mechanics crashes through the ceiling of every local hidden-variable theory. And $2\sqrt2$ is no lucky choice of angles: one can prove that quantum mechanics cannot produce a larger value with any state and any measurements (the Tsirelson bound) — $2\sqrt2$ is the limit of quantum correlations.

The experimental tribunal: closing the loopholes one by one

An inequality can be violated on paper, but it must also be violated in the laboratory. From Freedman–Clauser in 1972 to Aspect’s photon experiments in 1982, the measured $S$ has always sided with quantum mechanics. But careful sceptics pointed out two loopholes — assumptions of the derivation that an experiment does not automatically enforce:

1. The locality loophole: if a signal carrying Alice’s choice of measurement direction has time to reach Bob’s particle (at light speed), the assumption ”$B$ does not depend on $a$” is not enforced. The fix: put the stations far enough apart and choose the measurement directions with fast random number generators only while the particles are in flight, so that any “tip-off” is relativistically too late. Aspect first implemented fast switching in 1982; the 1998 Innsbruck experiment nailed it with strict spacelike separation.
2. The detection loophole: photon detectors are inefficient, and if only a small fraction of particles is recorded, the recorded sample may be biased — hidden variables could conspire to let only the “cooperative” particles be detected (one must assume “fair sampling” to conclude anything). The fix: push the total detection efficiency above roughly 83%, or switch to matter qubits with near-unit detection efficiency.

The trouble was that early experiments each closed only one: photon experiments had the distance but poor detection; ion experiments detected everything but sat too close together. In 2015, three experiments closed both loopholes simultaneously: Delft used diamond colour centres 1.3 km apart (detection efficiency near 100% — extremely low event rate, but every event counts), while Vienna and NIST used entangled photons with high-efficiency superconducting detectors (hundreds of millions of events, with statistical significance at the level of $p$-values below $10^{-30}$). All three clearly violated the Bell inequality. The 2022 Nobel Prize in Physics, awarded to Clauser, Aspect and Zeilinger, put the official seal on this experimental line.

To pile up CHSH data with your own hands and watch $S$ climb past 2, go to Lab module 08: entanglement and Bell non-locality.

∎Key formulas

• CHSH combination

  $$S=E(a,b)-E(a,b')+E(a',b)+E(a',b')$$

  Four correlations measured separately, assembled afterwards

• Classical ceiling

  $$|S|\le 2$$

  All local hidden-variable theories; uses only "predetermined + local"

• Singlet correlation

  $$E(a,b)=-\cos\theta_{ab}$$

  Fully anticorrelated when parallel, uncorrelated when perpendicular; cosine beats the ramp

• Tsirelson bound

  $$|S|_{\text{QM}}\le 2\sqrt2\approx 2.83$$

  The quantum-mechanical limit, reached at the optimal angles 0°, 90°, 45°, 135°

?Self-check4 questions

1. 1.

   Which assumptions does the derivation of the CHSH inequality use? (Select all that apply.)

   Select all that apply

   • In each round, all four outcomes A(a), A(a′), B(b), B(b′) simultaneously have definite values
   • Alice’s outcome does not depend on which direction Bob chooses to measure
   • The Born rule of quantum mechanics
   • The two particles are in a maximally entangled state

   Submit
2. 2.

   At the optimal angles, what ∣S∣ does quantum mechanics predict? (Give three decimal places.)

   1% relative tolerance

   Submit
3. 3.

   What is the "detection loophole"?

   • With low detection efficiency, the recorded subsample may be biased, requiring an extra fair-sampling assumption
   • Detector noise can push S above 2
   • The detectors are close enough to communicate with each other
   • A single detector cannot distinguish +1 from −1

   Submit
4. 4.

   Once experiment confirms ∣S∣ > 2, which conclusion holds?

   • At least one of "outcomes are predetermined" and "unaffected by distant choices" must fail
   • Alice can use entanglement to send Bob a message instantaneously
   • All hidden-variable theories are ruled out
   • Quantum mechanics is proven to be the uniquely correct theory

   Submit

What comes next

The Bell experiments tell us: the correlations inside entanglement are stronger than any classical mechanism, and yet cannot carry a single bit. So what is this “super-strong but incommunicable correlation” actually good for? The next section gives the first astonishing answer: transferring an unknown quantum state from Alice’s hands to Bob’s — moving no particle at all, just making one two-bit phone call.