Robertson–Schrödinger uncertainty relation

Remarks

Put $A_0 = A - \langle A \rangle$, $B_0 = B - \langle B \rangle$ and $z = \ip{A_0\psi}{B_0\psi}$. The Cauchy–Schwarz inequality gives $\sigma_A^2\sigma_B^2 \ge \abs{z}^2 = (\operatorname{Re} z)^2 + (\operatorname{Im} z)^2$. Symmetry and the domain conditions give $z = \ip{\psi}{A_0B_0\psi}$ and $\bar z = \ip{\psi}{B_0A_0\psi}$. So $\operatorname{Re} z = \frac12 \langle \set{A, B} \rangle - \langle A \rangle\langle B \rangle$ and $\operatorname{Im} z = \frac{1}{2i} \langle [A, B] \rangle$. In particular $\langle \set{A, B} \rangle$ is real and $\langle [A, B] \rangle$ is purely imaginary.

Example: on $\C^2$ take the Pauli matrices $\sigma_x, \sigma_y, \sigma_z$, so $[\sigma_x, \sigma_y] = 2i\sigma_z$ and $\set{\sigma_x, \sigma_y} = 0$, and $\psi = (1, 0)$. Then $\langle \sigma_x \rangle = \langle \sigma_y \rangle = 0$, $\sigma_{\sigma_x} = \sigma_{\sigma_y} = 1$ and $\langle [\sigma_x, \sigma_y] \rangle = 2i$, so both sides equal $1$.

The domain conditions cannot be dropped. On $L^2(0, 2\pi)$, let $A$ be multiplication by $\theta$ and $B = -i\, d/d\theta$ with periodic boundary conditions, and $\psi = (2\pi)^{-1/2} e^{in\theta}$. Then $B\psi = n\psi$, so $\sigma_B = 0$, while formally $[A, B] = iI$ would force $\sigma_A\sigma_B \ge \frac12$. The inequality is saved because $A\psi = \theta\psi$ is not periodic, so $A\psi \notin D(B)$.

Going deeper

For a student

For position and momentum, $[X, P] = i\hbar$ on Schwartz functions, and the Robertson inequality gives Heisenberg's $\sigma_X\sigma_P \ge \hbar/2$. For spin components, $[\sigma_x, \sigma_y] = 2i\sigma_z$, so the bound depends on the state, and it vanishes when $\langle \sigma_z \rangle = 0$. The proof is only the Cauchy–Schwarz inequality, so the relation holds in any inner product space; the work in the unbounded case is in the domain conditions.