The harmonic oscillator by ladder operators
28 min read · Tier toggle above rewrites this page — Tier 1 intuition, Tier 2 standard, Tier 3 full rigour.
You need the commutator [x̂, p̂] = iℏ and the oscillator Hamiltonian Ĥ = p̂²/2m + ½mω²x̂².
Why this matters
The harmonic oscillator is the one problem physics keeps coming back to — any smooth potential looks like a spring near its minimum, so almost everything that vibrates is an oscillator in disguise.
You could solve it the hard way: write down the Schrödinger equation, wrestle with Hermite polynomials, and check the boundary conditions kill off the divergent solutions. It works, but the integrals hide what is really going on. This lecture takes the other road. We build two operators that step the energy up and down a ladder, and the entire spectrum — every allowed energy, evenly spaced — falls out of their algebra without evaluating a single integral. Once you have seen it, you will recognise the same trick powering angular momentum, quantum fields, and the photon itself.
Reframing the problem
The move that makes everything work is to stop thinking about position and momentum separately and package them into a single non-Hermitian operator. Define the lowering operator â as a complex combination of x̂ and p̂, and its Hermitian conjugate ↠as the raising operator. The Hamiltonian, which looked like a sum of two squares, collapses into Ĥ = ℏω(â†â + ½). All the physics now lives in the number operator N̂ = â†â and in one commutation relation, [â, â†] = 1.
The derivation
D-010 · Oscillator spectrum by ladder operators
Starting from [â, â†] = 1, the ladder operators generate the complete set of energy eigenstates and pin the ground state from below.
Reading the result
Two facts do all the work. The rungs are evenly spaced — every step up costs exactly ℏω, which is why a hot oscillator absorbs and emits light in identical quanta and why Planck's guess about black-body radiation turned out to be exact. And the ground state is not at zero: even in its lowest state the oscillator carries ½ℏω of energy. That zero-point energy is forced by the uncertainty principle — pinning the particle to the bottom of the well would fix both position and momentum, which the algebra forbids.
The ladder argument shows the spectrum is a subset of {½ℏω + nℏω}, but it does not by itself prove the spectrum is non-degenerate or that the eigenstates it builds are complete. That the states |n⟩ = (â†)n|0⟩/√(n!) span the whole Hilbert space needs a separate argument — that â annihilating |0⟩ has a unique normalisable solution in the position representation, which is where an integral finally sneaks back in. C
Three quick tests
1 · Show â†â|n⟩ = n|n⟩ follows from [â, â†] = 1. 2 · Why can there be no state below |0⟩? 3 · What are the units of â, and why must it be dimensionless up to a factor of √(mω/ℏ)?
PU-202 problem set →
Coherent states, the thermal partition function of the oscillator, and the two-dimensional isotropic case with its accidental degeneracy.
Cites D-010 · Threads: waves · energy