Minimum-Uncertainty States

The Heisenberg uncertainty principle states that for any quantum state,

$\sigma_x \,\sigma_p \;\geq\; \frac{\hbar}{2}.$

Most states give a product strictly larger than $\hbar/2$ . A natural question is: which states, if any, actually reach the lower limit? These are called minimum-uncertainty states, and they have a precise mathematical form.

The saturation condition

Recall that Robertson's inequality is derived from the Cauchy–Schwarz inequality applied to two vectors $|f\rangle = (\hat{x} - \langle x\rangle)|\psi\rangle$ and $|g\rangle = (\hat{p} - \langle p\rangle)|\psi\rangle$ . The Cauchy–Schwarz bound becomes an equality if and only if $|f\rangle$ and $|g\rangle$ are proportional — that is, if there exists a complex constant $\lambda$ such that

$(\hat{p} - \langle p\rangle)|\psi\rangle = \lambda\,(\hat{x} - \langle x\rangle)|\psi\rangle.$

Furthermore, the step that extracts the imaginary part of $\langle f|g\rangle$ to obtain the right-hand side $\hbar/2$ requires $\lambda$ to be purely imaginary, say $\lambda = i\mu$ with $\mu$ real and positive. Writing $\langle x\rangle = x_0$ and $\langle p\rangle = p_0$ , the condition becomes a first-order differential equation in the position representation:

$-i\hbar\,\frac{d\psi}{dx} - p_0\,\psi = i\mu\,(x - x_0)\,\psi.$

Gaussian wave packets

The differential equation above is separable. Rearranging,

$\frac{d\psi}{\psi} = \left(\frac{ip_0}{\hbar} - \frac{\mu}{\hbar}(x - x_0)\right)dx.$

Integrating and exponentiating yields a Gaussian,

$\psi(x) = N\,\exp\!\left(\frac{ip_0 x}{\hbar} - \frac{\mu}{2\hbar}(x - x_0)^2\right),$

where $N$ is a normalization constant. The probability density $|\psi(x)|^2$ is a Gaussian centered at $x_0$ with variance

$\sigma_x^2 = \frac{\hbar}{2\mu}.$

The momentum-space wave function $\tilde{\psi}(p)$ is the Fourier transform of $\psi(x)$ , and because the Fourier transform of a Gaussian is again a Gaussian, one finds

$\sigma_p^2 = \frac{\mu\hbar}{2}.$

Multiplying,

$\sigma_x\,\sigma_p = \sqrt{\frac{\hbar}{2\mu}} \cdot \sqrt{\frac{\mu\hbar}{2}} = \frac{\hbar}{2}.$

So the Gaussian wave packet saturates the Heisenberg bound exactly for every value of $\mu > 0$ . The parameter $\mu$ controls the width: large $\mu$ gives a narrow, well-localized packet ( $\sigma_x$ small) with a broad momentum distribution ( $\sigma_p$ large), while small $\mu$ spreads the packet spatially but sharpens it in momentum space. In both cases the product stays fixed at $\hbar/2$ .

Coherent states of the harmonic oscillator

The most important physical realization of minimum-uncertainty states arises in the quantum harmonic oscillator. For an oscillator with angular frequency $\omega$ and mass $m$ , the natural length scale is $x_0 = \sqrt{\hbar/(m\omega)}$ and the natural momentum scale is $p_0 = \sqrt{m\omega\hbar}$ . The ground state $|0\rangle$ of the oscillator is a Gaussian with $\sigma_x = x_0/\sqrt{2}$ and $\sigma_p = p_0/\sqrt{2}$ , giving

$\sigma_x\,\sigma_p = \frac{x_0}{\sqrt{2}} \cdot \frac{p_0}{\sqrt{2}} = \frac{x_0 p_0}{2} = \frac{\hbar}{2}.$

The ground state is therefore a minimum-uncertainty state. More generally, the coherent states $|\alpha\rangle$ are displaced versions of the ground state: $|\alpha\rangle$ is an eigenstate of the lowering operator $\hat{a}|\alpha\rangle = \alpha|\alpha\rangle$ , where $\alpha$ is any complex number. Each coherent state is a minimum-uncertainty state centered at $\langle x\rangle = \sqrt{2\hbar/(m\omega)}\,\text{Re}(\alpha)$ and $\langle p\rangle = \sqrt{2m\omega\hbar}\,\text{Im}(\alpha)$ , with the same spread $\sigma_x \sigma_p = \hbar/2$ as the ground state.

Coherent states also evolve cleanly under the harmonic oscillator Hamiltonian: the center $(\langle x\rangle, \langle p\rangle)$ traces an ellipse in phase space at angular frequency $\omega$ , just like a classical oscillator, while the shape of the Gaussian envelope is preserved. This combination of minimum uncertainty and classical-like behavior makes coherent states the quantum states closest to classical point particles.

Which states do NOT saturate the bound?

Any state whose wave function is not Gaussian in position space will have $\sigma_x \sigma_p > \hbar/2$ . Specific examples illustrate how much larger the product can be:

Infinite square well ground state: $\psi(x) \propto \cos(\pi x/L)$ on $(-L/2, L/2)$ . One can show $\sigma_x^2 = L^2(1/12 - 1/(2\pi^2))$ and $\sigma_p^2 = (\pi\hbar/L)^2$ , giving $\sigma_x \sigma_p \approx 0.568\,\hbar > \hbar/2$ .
Energy eigenstates $|n\rangle$ of the harmonic oscillator for $n \geq 1$ : they satisfy $\sigma_x \sigma_p = (n + 1/2)\hbar$ , which grows with $n$ and equals $\hbar/2$ only for $n = 0$ .
Superpositions of two separated Gaussians: the bimodal position distribution inflates $\sigma_x$ without a corresponding reduction in $\sigma_p$ , so the product rises above $\hbar/2$ .

The conclusion is that minimum-uncertainty states are not generic — they are a special, measure-zero family within the space of all quantum states, uniquely characterized by their Gaussian profile.

Summary

A state saturates $\sigma_x \sigma_p = \hbar/2$ if and only if its wave function is a Gaussian in position space. Equivalently, it is an eigenstate of a particular combination of $\hat{x}$ and $\hat{p}$ . The coherent states of the harmonic oscillator are the most physically important examples: they are minimum-uncertainty, they evolve like classical oscillators, and they are central to quantum optics, laser physics, and the description of the electromagnetic field in quantum field theory.