Bell’s inequality · Quantum Mechanics

Intuition

Bell asked what any local theory with pre-set answers must obey. Suppose each pair carries instructions $\lambda$ that fix the $\pm1$ results for every setting, $A$ ’s depending only on $A$ ’s setting and $B$ ’s only on $B$ ’s. With two settings on each side, $\mathbf{a}$ , $\mathbf{a}'$ and $\mathbf{b}$ , $\mathbf{b}'$ , the combination $A(B+B')+A'(B-B')$ is always $\pm2$ : one of the brackets is 0 and the other $\pm2$ . Its average, the CHSH combination $S$ of four correlations, therefore lies between $-2$ and 2. Quantum mechanics breaks the bound: for the singlet at the right angles $|S|=2\sqrt{2}\approx2.83$ . Experiments since the 1970s, and loophole-free ones since 2015, side with quantum mechanics: no local theory of pre-set answers describes nature.

If twins answered every question from a shared list written before they parted, their agreement over four pairs of questions would be capped by simple counting. Entangled pairs beat the cap, so they cannot be twins with lists.

The standard settings: $\mathbf{a}\perp\mathbf{a}'$ for one side, $\mathbf{b}\perp\mathbf{b}'$ for the other, turned $^{}$ from them. Three of the four pairs are apart and are apart, which gives the singlet .

The CHSH inequality

For settings $\mathbf{a},\mathbf{a}'$ at $A$ and $\mathbf{b},\mathbf{b}'$ at :

\begin{aligned}S&=E(\mathbf{a},\mathbf{b})+E(\mathbf{a},\mathbf{b}')+E(\mathbf{a}',\mathbf{b})-E(\mathbf{a}',\mathbf{b}')\\ |S|&\le2\ \ \text{(local)},\qquad |S|\le2\sqrt2\ \ \text{(quantum)}\end{aligned}

Properties

Local pre-set answers: $A(\mathbf{a},\lambda),B(\mathbf{b},\lambda)=\pm1$ , with $\lambda$ drawn from a distribution that does not depend on the settings.
For the singlet, $E=-\mathbf{a}\cdot\mathbf{b}$ ; at the settings in the figure .

The CHSH inequality

\begin{aligned}&A,A',B,B'=\pm1\ \Longrightarrow\\ &\bigl|\langle AB\rangle+\langle AB'\rangle+\langle A'B\rangle-\langle A'B'\rangle\bigr|\le2\end{aligned}

For each value of the hidden variables the four answers are fixed numbers $\pm1$ . Then $B+B'$ and $B-B'$ cannot both be nonzero: one is 0 and the other . So the combination is for every , and its average, which is , cannot exceed 2 in size.

Proof steps

A(\mathbf{a},\lambda),\ A(\mathbf{a}',\lambda),\ B(\mathbf{b},\lambda),\ B(\mathbf{b}',\lambda)\in\{+1,-1\}

Pre-set answers for every setting, each depending only on its own side.

\begin{aligned}&B+B'=0\ \ \text{and}\ \ B-B'=\pm2\\ &\text{or}\ \ B+B'=\pm2\ \ \text{and}\ \ B-B'=0\end{aligned}

Two numbers $\pm1$ are either equal or opposite.

A(B+B')+A'(B-B')=\pm2

One bracket vanishes and the other is $\pm2$ , multiplied by $\pm1$ .

S=\int p(\lambda)\bigl[A(B+B')+A'(B-B')\bigr]\,d\lambda

Each correlation is the average of a product; locality lets them share one $\lambda$ .

|S|\le\int p(\lambda)\cdot2\,d\lambda=2

The average of numbers of size 2 is at most 2 in size.

Applications

Practice

The Quantum Value

With $E=-\cos\theta$ and three of the four angles at $45^\circ$ and the fourth at $135^\circ$ , the four correlations add up, in size, to .

S=-\cos45^{\circ}-\cos45^{\circ}-\cos45^{\circ}+\cos135^{\circ}

Try it

With $E=-\cos\theta$ and the four angles $45^{\circ},45^{\circ},45^{\circ}$ and $135^{\circ}$ (the last for $\mathbf{a}',\mathbf{b}'$ ), what is $|S|$ ? Give three decimal places.