Born rule

Remarks

Each $P^A(B)$ is an orthogonal projection, so $\operatorname{tr}(\rho P^A(B)) = \operatorname{tr}(\rho^{1/2} P^A(B) \rho^{1/2}) \ge 0$, and $\mu(\R) = \operatorname{tr}\rho = 1$. Countable additivity follows from that of $P^A$: write $\rho = \sum_k p_k \lvert e_k \rangle\langle e_k \rvert$, so $\mu(B) = \sum_k p_k \norm{P^A(B) e_k}^2$, a convex combination of probability measures. The support statement holds because $P^A(\R \setminus \sigma(A)) = 0$. If $\psi \in D(A)$, the mean of $\mu$ is the expectation value $\ip{\psi}{A\psi}$.

Example: on $\C^2$, let $\sigma_z = \mathrm{diag}(1, -1)$ with eigenvectors $e_1 = (1, 0)$, $e_2 = (0, 1)$, and let $\psi = (\cos\theta, e^{i\varphi}\sin\theta)$. The result $1$ has probability $\abs{\ip{e_1}{\psi}}^2 = \cos^2\theta$ and the result $-1$ has probability $\sin^2\theta$; the phase $\varphi$ does not enter.

The rule was proposed by Max Born in 1926. It is a postulate of the theory, not a theorem; within the theory, Gleason's theorem shows that the trace form is the only possible one in dimension at least $3$.

Going deeper

For a student

The rule combines two pieces of the formalism: the spectral measure of the observable, which splits the Hilbert space by measured value, and the state, which weighs the pieces. The mean of the resulting distribution is the expectation value $\operatorname{tr}(\rho A)$, and its variance is the quantity in the uncertainty relations. For a position measurement on $L^2(\R)$, the spectral projection of $B$ is multiplication by the indicator of $B$, and the rule gives $\int_B \abs{\psi(x)}^2 \, dx$, the usual reading of the wave function. Generalised measurements replace the projection-valued measure by a positive operator-valued measure, with the same trace formula.