The Born Rule, Formally

The Born rule is the bridge from amplitudes to observable probabilities. Postulate 3 stated it in terms of projectors; here we work through the formalism carefully and then compute a probability by hand.

The rule, written with projectors

Measure an observable whose eigenvalue $a_k$ has projector $\hat{P}_k$ . On a normalized state $|\psi\rangle$ the probability of outcome $a_k$ is

p(a_k) = \langle\psi|\hat{P}_k|\psi\rangle.

For a non-degenerate outcome with normalized eigenvector $|a_k\rangle$ , the projector is $\hat{P}_k = |a_k\rangle\langle a_k|$ , and the formula collapses to the familiar amplitude-squared form:

p(a_k) = \langle\psi|a_k\rangle\langle a_k|\psi\rangle = |\langle a_k|\psi\rangle|^2.

The quantity $\langle a_k|\psi\rangle$ is the amplitude for the outcome; its squared modulus is the probability. This is the precise statement of "probability is amplitude squared."

Why $p(a_k) = \langle\psi|\hat P_k|\psi\rangle$ is a probability

A projector satisfies $\hat{P}_k^{\dagger} = \hat{P}_k$ and $\hat{P}_k^2 = \hat{P}_k$ . Therefore

p(a_k) = \langle\psi|\hat{P}_k|\psi\rangle = \langle\psi|\hat{P}_k^{\dagger}\hat{P}_k|\psi\rangle = \big\|\hat{P}_k|\psi\rangle\big\|^2 \geq 0,

so probabilities are non-negative, and (using completeness $\sum_k\hat{P}_k = \hat{I}$ ) they sum to one. The Born rule is consistent precisely because observables are Hermitian and their eigenprojectors are complete.

Measuring in a rotated basis

A subtle point: the same state gives different probabilities depending on the measurement basis. Take the state

|\psi\rangle = \cos\theta\,|0\rangle + \sin\theta\,|1\rangle.

Measured in the computational ( $Z$ ) basis, $p(0) = \cos^2\theta$ and $p(1) = \sin^2\theta$ . But suppose we measure in the $X$ basis $\{|+\rangle, |-\rangle\}$ , where

|+\rangle = \tfrac{1}{\sqrt2}(|0\rangle + |1\rangle), \qquad |-\rangle = \tfrac{1}{\sqrt2}(|0\rangle - |1\rangle).

The amplitude for the $+$ outcome is

\langle +|\psi\rangle = \tfrac{1}{\sqrt2}\big(\langle 0| + \langle 1|\big)\big(\cos\theta|0\rangle + \sin\theta|1\rangle\big) = \tfrac{1}{\sqrt2}\big(\cos\theta + \sin\theta\big),

so by the Born rule

p(+) = |\langle +|\psi\rangle|^2 = \tfrac12\big(\cos\theta + \sin\theta\big)^2.

For $\theta = \pi/6$ this is $\tfrac12\big(\tfrac{\sqrt3}{2} + \tfrac12\big)^2 \approx 0.9330$ — quite far from the $\cos^2(\pi/6) = 0.75$ you would get in the $Z$ basis. The probability is a joint property of the state and the measurement.

Try it

Compute the Born-rule probability $p(+)$ for the state $|\psi\rangle = \cos\theta\,|0\rangle + \sin\theta\,|1\rangle$ with $\theta = \pi/6$ , measured in the $\{|+\rangle, |-\rangle\}$ basis. Form the amplitude $\langle +|\psi\rangle$ , then return its squared modulus.

Run your code to see the quantum state.