Position–Momentum Uncertainty

Why position and momentum cannot both be sharp

Measure a particle's position very precisely and you inevitably disturb its momentum — and vice versa. This is not a limitation of instruments; it is a fundamental feature of any wave-like description. A state with a sharply defined position is a narrow spike in space, which requires a broad spread of spatial frequencies to build, and spatial frequency is proportional to momentum. A state with a sharply defined momentum is a pure plane wave that extends over all space, so its position is completely indefinite.

Heisenberg's Robertson–Schrödinger inequality makes this precise. For position $x$ and momentum $p$ :

$\Delta x \cdot \Delta p \;\geq\; \frac{\hbar}{2}.$

Here $\Delta x = \sqrt{\langle x^2 \rangle - \langle x \rangle^2}$ is the standard deviation of position in the given state, and $\Delta p$ is the same for momentum. Neither uncertainty is zero unless the state is not normalizable, and they cannot both be simultaneously small.

The minimum product $\Delta x \cdot \Delta p = \hbar/2$ is achieved by Gaussian (coherent) states — wave packets of the form

$\psi(x) \propto \exp\!\left(-\frac{(x - x_0)^2}{4\sigma^2}\right) e^{i p_0 x/\hbar},$

which have $\Delta x = \sigma$ and $\Delta p = \hbar/(2\sigma)$ . Every other normalizable state has a larger product.

A numerical feel for the bound

In SI units $\hbar \approx 1.055 \times 10^{-34}$ J·s, which is tiny. For a macroscopic ball ( $\Delta x \sim 1$ mm) the lower bound on $\Delta p$ is $\sim 5 \times 10^{-32}$ kg·m/s — undetectably small. At atomic scales ( $\Delta x \sim 10^{-10}$ m, one ångström) the minimum momentum spread is $\sim 5 \times 10^{-25}$ kg·m/s, corresponding to electron speeds of order $10^{6}$ m/s (comparable to the Bohr velocity $\alpha c \approx 2 \times 10^{6}$ m/s) — physically significant and observable in atomic spectra and ionization energies.

Try it

This is a numerical exercise — your code should return a number.

Work in natural units where $\hbar = 1$ . A particle has position uncertainty $\Delta x = 2.0$ . Compute the minimum allowed momentum uncertainty $\Delta p_{\min}$ from the Heisenberg bound.

Run your code to see the quantum state.