Heisenberg uncertainty principle

Remarks

The statement is the Heisenberg inequality for the Fourier transform in physical units. With the transform $\hat\psi(\xi) = \int \psi(x) e^{-2\pi i x\xi}\, dx$, momentum acts as multiplication by $2\pi\hbar\xi$, so $\sigma_P^2 = (2\pi\hbar)^2 \int (\xi - b')^2 \abs{\hat\psi(\xi)}^2 \, d\xi$ with $b' = \langle P \rangle/(2\pi\hbar)$. The Fourier inequality gives $\sigma_X^2 \sigma_P^2 \ge (2\pi\hbar)^2/(16\pi^2) = \hbar^2/4$. For $\psi$ in the Schwartz space, it also follows from the canonical commutation relation and the Robertson inequality: $\sigma_X\sigma_P \ge \frac12 \abs{\ip{\psi}{i\hbar\psi}} = \frac{\hbar}{2}$.

Example: for $\psi(x) = \pi^{-1/4} e^{-x^2/2}$, $\abs{\psi}^2$ is normal with variance $\frac12$, so $\sigma_X = 1/\sqrt{2}$. Also $P\psi = i\hbar x\psi$, so $\sigma_P^2 = \hbar^2 \int x^2 \abs{\psi}^2 = \hbar^2/2$. Hence $\sigma_X\sigma_P = \hbar/2$, the case of equality.

The inequality in this form was proved by E. H. Kennard in 1927, after Heisenberg's heuristic argument of the same year.

Going deeper

For a student

The principle is a Fourier-analytic fact: momentum is position seen through the Fourier transform, and a function and its transform cannot both be concentrated. Gaussian wave packets, the coherent states, are the minimisers. The Robertson inequality extends it to any two observables, with the commutator in place of $i\hbar$. Other measures of spread give other versions, such as the entropic uncertainty principle, which is stronger.