The Quantum Harmonic Oscillator

Pick up almost any physical system that sits happily at rest — a mass on a spring, the two atoms of a carbon-monoxide molecule held together by their chemical bond, an atom locked into the lattice of a crystal, even a mode of the electromagnetic field in a box — and give it a small nudge. It oscillates. Nudge it a little harder and it still oscillates, at the same frequency, with a restoring force that grows in proportion to how far you pushed it. This is the simple harmonic oscillator, and it is, without exaggeration, the single most important system in all of physics.

The reason is a piece of mathematics so useful it borders on cheating: near a minimum, every smooth potential looks like a parabola. So the moment you ask "what does this system do when it is disturbed only slightly from equilibrium?", the answer — whatever the system — is governed by the harmonic oscillator. This page treats that oscillator with the rules of quantum mechanics, and the result is one of the most beautiful in the subject: the energy of a quantum oscillator is not free to take any value it likes. It comes in a ladder of equally spaced rungs, and — astonishingly — the oscillator can never come fully to rest.

The universal potential of small oscillations

A simple harmonic oscillator is defined by a restoring force proportional to displacement, F = -kx, which comes from the potential energy

V(x) = \tfrac{1}{2}kx^2 = \tfrac{1}{2}m\omega^2 x^2,

a parabola with its minimum at x=0. Here m is the mass, k the stiffness, and \omega = \sqrt{k/m} the classical angular frequency of oscillation. Why should this one shape matter so much? Take any smooth potential V(x) with a minimum at x_0 and Taylor-expand it there:

V(x) = V(x_0) + V'(x_0)(x-x_0) + \tfrac{1}{2}V''(x_0)(x-x_0)^2 + \cdots

At a minimum the slope V'(x_0) is zero, and the constant V(x_0) just shifts the zero of energy. What is left, to leading order, is \tfrac{1}{2}V''(x_0)(x-x_0)^2 — a parabola, with effective stiffness k = V''(x_0). That is why the harmonic oscillator is everywhere: every bond, every trapped atom, every field mode, jiggling gently about its equilibrium, is — to first approximation — a simple harmonic oscillator. Solve this one problem and you have solved the low-energy behaviour of half of physics.

Solving it: an evenly spaced ladder of energies

We drop the potential into the time-independent Schrödinger equation \hat{H}\psi = E\psi, with \hat H = -\dfrac{\hbar^2}{2m}\dfrac{d^2}{dx^2} + \tfrac{1}{2}m\omega^2 x^2:

-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + \tfrac{1}{2}m\omega^2 x^2\,\psi = E\,\psi.

Demanding that \psi stay normalisable (it must decay to zero as x\to\pm\infty, not blow up) selects a discrete set of allowed energies. The result is remarkably clean:

Look hard at the spacing. In the particle in a box the levels go like n^2 and fan apart as you climb; here the levels march up in a perfectly uniform staircase, each step a clean \hbar\omega higher than the last. That single fact — uniform spacing — is the fingerprint of the oscillator, and it is exactly what makes the vibrational spectrum of a molecule a series of evenly spaced lines.

The wavefunctions: a Gaussian dressed with wiggles

Each energy level E_n carries a wavefunction \psi_n(x). They all share a common shape — a bell-shaped Gaussian that pins the particle near the bottom of the well — multiplied by a polynomial that adds wiggles:

\psi_n(x) = N_n\,H_n\!\left(\sqrt{\tfrac{m\omega}{\hbar}}\,x\right)\,e^{-m\omega x^2 / 2\hbar},

where H_n is the Hermite polynomial of degree n and N_n a normalisation constant. The ground state \psi_0 \propto e^{-m\omega x^2/2\hbar} is a pure Gaussian with no nodes — the smoothest, most compact wave the well allows. Climb the ladder and H_n adds exactly n zero-crossings (nodes), so higher states oscillate more and spread wider, pushing out toward the steep walls where a more energetic classical particle would indeed roam.

The chart draws the parabolic well, the horizontal energy rung for the level you pick, and the wavefunction \psi_n riding on that rung. Slide through n = 0 to 4 and watch the nodes multiply and the wave widen. Notice one quietly radical detail: the wave does not stop dead at the points where the rung meets the parabola (the classical turning points, where a classical particle of that energy would run out of kinetic energy and reverse). It leaks a little way past them, into the region that is classically forbidden — the same quantum tunnelling that lets particles escape where Newton says they cannot.

Worked examples

Example 1 — the energy of a level. A quantum oscillator has angular frequency \omega. What is the energy of its n=3 state? Straight from the formula,

E_3 = \left(3 + \tfrac{1}{2}\right)\hbar\omega = \tfrac{7}{2}\hbar\omega = 3.5\,\hbar\omega.

Seven half-quanta of \hbar\omega. No jumping straight to 4\hbar\omega or any value in between — only the rungs.

Example 2 — the spacing is constant. How much energy does it take to bump the oscillator from level n=5 up to n=6? And from n=99 to n=100?

\Delta E = E_{n+1} - E_n = \left[(n+1)+\tfrac12\right]\hbar\omega - \left[n+\tfrac12\right]\hbar\omega = \hbar\omega.

Both cost exactly \hbar\omega. The n cancels — the staircase is perfectly uniform, which is precisely why a vibrating molecule absorbs and emits at one sharp frequency \omega rather than a smear.

Example 3 — zero-point energy of a real bond. A carbon-monoxide molecule vibrates with \omega \approx 4.0\times 10^{14}\ \text{rad/s}. Its lowest possible vibrational energy is not zero but

E_0 = \tfrac{1}{2}\hbar\omega \approx \tfrac{1}{2}(1.05\times 10^{-34})(4.0\times 10^{14})\ \text{J} \approx 2.1\times 10^{-20}\ \text{J} \approx 0.13\ \text{eV}.

Even cooled to absolute zero, the bond still trembles with this much energy — it can never be frozen still. That is the zero-point energy made concrete.

Example 4 — a ratio, cleanly. How many times larger is the energy of level n=4 than the ground state? Working in units where \hbar\omega = 1, so E_n = n + \tfrac12 and E_0 = \tfrac12,

\frac{E_n}{E_0} = \frac{n + \tfrac12}{\tfrac12} = 2n + 1 \quad\Longrightarrow\quad \frac{E_4}{E_0} = 2(4)+1 = 9.

Level 4 holds nine times the zero-point energy — a tidy odd-number rule, 2n+1, that falls straight out of the half-integer offset.

Real bonds are not perfect parabolas — stretch a molecule far enough and it breaks, so the true potential (a Morse potential, say) flattens out at large separation instead of rising forever. The parabola is just the bottom of that curve, the \tfrac12 V''(x_0)x^2 term of the Taylor series. Down near the minimum, where the low-lying states live, the approximation is excellent and the vibrational lines are beautifully evenly spaced. But climb high up the ladder, toward the energies where the real potential starts to sag away from the parabola, and the higher-order Taylor terms (the anharmonicity) bite: the rungs slowly bunch closer together as you approach the dissociation energy, and eventually the ladder ends — the molecule comes apart. Spectroscopists measure exactly this gentle crowding of the top rungs and read off the bond's breaking energy from it. The harmonic oscillator is the honest first word on any bound system, not always the last.

Two classic traps, one page. First, do not carry over the pattern from the particle in a box. There the energies go as E_n \propto n^2 and fan apart as you climb. For the harmonic oscillator they go as E_n = (n+\tfrac12)\hbar\omega — linear in n, so the rungs are evenly spaced, separated by a constant \hbar\omega. Different well, different ladder. Mixing them up is the single most common mistake on this topic.

Second, the lowest energy is not zero. It is tempting to set n=0 and expect E_0 = 0, a particle sitting motionless at the bottom of the well. But the half — the +\tfrac12 — refuses to go away: E_0 = \tfrac12\hbar\omega. A quantum oscillator at rest at the exact centre with exactly zero momentum would pin down both position and momentum precisely, which the uncertainty principle forbids. The oscillator is compelled to jitter. The zero-point energy is not an accounting convention — it is the least the universe will let the oscillator hold.

Everywhere the very small and the very cold meet. It is why liquid helium never freezes under its own vapour pressure, no matter how close to absolute zero you take it: the zero-point jiggling of the light helium atoms is energetic enough to keep them sloshing past one another as a liquid. It sets the true bottom of every molecule's vibrational energy, shifting reaction rates and the exact colours in a vibrational spectrum. Quantise the modes of the electromagnetic field — each mode is itself a harmonic oscillator, and a quantum of excitation is a photon — and the sum of all their \tfrac12\hbar\omegas is the vacuum energy whose tiny differences produce the measurable Casimir force pulling two mirrors together in the dark. The same \tfrac12\hbar\omega that seemed a mathematical curiosity in our formula is, quite literally, a force you can measure.