Measurement Settings and Correlations

A correlation function condenses the joint statistics of a two-particle measurement into a single number between $-1$ and $+1$ . It is the object that Bell and CHSH inequalities constrain, so we need to know how to compute it from quantum mechanics.

Definition

Alice and Bob each record an outcome $\pm 1$ . For fixed settings $a$ and $b$ , the correlation is the expectation value of the product of their outcomes:

E(a, b) = \langle A\, B \rangle = \sum_{A, B \in \{\pm 1\}} A\,B\;P(A, B \mid a, b).

Because $A B = +1$ when the outcomes agree and $-1$ when they disagree,

E(a, b) = P(\text{same}) - P(\text{opposite}).

So $E = +1$ means perfect correlation, $E = -1$ perfect anticorrelation, and $E = 0$ no correlation.

The singlet's joint probabilities

For the singlet $|\Psi^-\rangle = \tfrac{1}{\sqrt2}(|\!\uparrow\downarrow\rangle - |\!\downarrow\uparrow\rangle)$ , suppose Alice measures along an axis and Bob along an axis making angle $\theta$ with hers. The Born rule gives

P(\text{opposite}) = \cos^2\!\Big(\frac{\theta}{2}\Big), \qquad P(\text{same}) = \sin^2\!\Big(\frac{\theta}{2}\Big).

At $\theta = 0$ (same axis), $P(\text{opposite}) = 1$ — the spins are always opposite, the hallmark of the singlet. At $\theta = \pi$ (opposite axes), $P(\text{same}) = 1$ .

The correlation

Combine the two pieces:

E(\theta) = \sin^2\!\Big(\frac{\theta}{2}\Big) - \cos^2\!\Big(\frac{\theta}{2}\Big) = -\cos\theta,

using the identity $\cos\theta = \cos^2(\theta/2) - \sin^2(\theta/2)$ . This is the singlet correlation we have been quoting. Equivalently it is the expectation value of the joint spin operator, $E(\hat a, \hat b) = \langle \Psi^- | (\hat a \cdot \boldsymbol\sigma) \otimes (\hat b \cdot \boldsymbol\sigma) | \Psi^- \rangle = -\,\hat a \cdot \hat b$ .

Worked example

Take $\theta = \pi/3 = 60^\circ$ . Then $\theta/2 = \pi/6$ , so $\cos^2(\pi/6) = 3/4$ and $\sin^2(\pi/6) = 1/4$ . The correlation is

E = \tfrac14 - \tfrac34 = -\tfrac12 = -\cos\big(\tfrac{\pi}{3}\big).

Try it

This is a numerical exercise — your code should return a number. For $\theta = \pi/3$ , build the two probabilities $P(\text{same}) = \sin^2(\theta/2)$ and $P(\text{opposite}) = \cos^2(\theta/2)$ , then return the correlation $E = P(\text{same}) - P(\text{opposite})$ . You should get $-\tfrac12$ .

Run your code to see the quantum state.