The Ground-State Gaussian

The quantum harmonic oscillator has a remarkable lowest-energy solution: the ground state is a Gaussian — a bell-shaped function that is simultaneously an eigenstate of energy and the state of minimum uncertainty allowed by the Heisenberg principle.

The energy eigenvalue equation

The Hamiltonian for a particle of mass $m$ in a parabolic potential $V(x) = \tfrac{1}{2}m\omega^2 x^2$ is

\hat{H} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + \frac{1}{2}m\omega^2 x^2.

The time-independent Schrödinger equation $\hat{H}\psi = E\psi$ must be solved on the entire real line, and the solutions must be normalisable (square-integrable). These two requirements together force the energies to be discrete: $E_n = \hbar\omega(n + \tfrac{1}{2})$ for $n = 0, 1, 2, \ldots$ .

The characteristic length

Before writing the ground-state wavefunction it is helpful to introduce the oscillator length

x_0 = \sqrt{\frac{\hbar}{m\omega}}.

This is the only combination of $\hbar$ , $m$ , and $\omega$ with dimensions of length. It sets the spatial scale of every eigenstate; for an electron in a typical atomic potential $x_0$ is on the order of angstroms, while for a macroscopic oscillator it is astronomically small, which is why quantum effects are invisible at everyday scales.

The ground-state wavefunction

Setting $n = 0$ and solving the Schrödinger equation directly — or, more elegantly, requiring that $\hat{a}\psi_0 = 0$ where $\hat{a}$ is the lowering operator — gives a first-order differential equation whose only normalisable solution is a Gaussian:

\psi_0(x) = \left(\frac{1}{\pi^{1/2}\,x_0}\right)^{1/2} \exp\!\left(-\frac{x^2}{2x_0^2}\right).

The prefactor is fixed by normalisation: $\int_{-\infty}^{\infty}|\psi_0(x)|^2\,dx = 1$ . Using the standard Gaussian integral $\int_{-\infty}^{\infty} e^{-u^2}\,du = \sqrt{\pi}$ with the substitution $u = x/x_0$ , one verifies that the prefactor $(\pi^{1/2} x_0)^{-1/2}$ is exactly correct.

The ground-state energy follows from the zero-point result: $E_0 = \tfrac{1}{2}\hbar\omega$ .

What the Gaussian shape tells us

The probability density $|\psi_0|^2$ is also a Gaussian, centred at $x = 0$ with a half-width of roughly $x_0$ :

|\psi_0(x)|^2 = \frac{1}{\sqrt{\pi}\,x_0}\,\exp\!\left(-\frac{x^2}{x_0^2}\right).

This means the particle is most likely found near the equilibrium point, and the probability of finding it far from $x = 0$ falls off exponentially.

Position uncertainty. A short calculation gives $\langle x \rangle = 0$ and $\langle x^2 \rangle = x_0^2/2$ , so the position uncertainty is

\Delta x = \sqrt{\langle x^2\rangle - \langle x\rangle^2} = \frac{x_0}{\sqrt{2}}.

Momentum uncertainty. Similarly, $\Delta p = \hbar/(\sqrt{2}\,x_0)$ .

The uncertainty product. Multiplying:

\Delta x\,\Delta p = \frac{x_0}{\sqrt{2}} \cdot \frac{\hbar}{\sqrt{2}\,x_0} = \frac{\hbar}{2}.

This is exactly the lower bound set by the Heisenberg uncertainty principle, $\Delta x\,\Delta p \geq \hbar/2$ . The ground-state Gaussian is therefore a minimum-uncertainty state — no state can simultaneously be more localised in position and momentum. It saturates the uncertainty principle.

Comparison with the classical oscillator

A classical oscillator of the same energy $E_0 = \tfrac{1}{2}\hbar\omega$ would be confined to the region $|x| \leq x_0$ (the classical turning points, where kinetic energy reaches zero). The quantum ground state extends beyond those points: the Gaussian has non-zero probability for all $x$ , including the classically forbidden region. This quantum tunnelling into the forbidden region is a direct consequence of the non-zero zero-point energy and has measurable effects in solid-state physics and quantum chemistry.