Why everyone cares about the oscillator
Near any stable equilibrium, the potential is approximately parabolic — Taylor-expand and the leading term is V\approx\tfrac12 V''x^2. So every small oscillation in physics is, to first approximation, a harmonic oscillator: molecular vibrations, the normal modes of a crystal (its phonons), even the modes of the electromagnetic field. Solve this one system and you have solved a huge slice of physics. You already met its classical version as simple harmonic motion.
The quantum harmonic oscillator Hamiltonian: kinetic energy plus a parabolic well of angular frequency \omega.
We could insert this \hat H into the time-independent Schrödinger equation and grind through Hermite's differential equation, whose solutions are the Hermite polynomials times a Gaussian. Instead we will extract every energy level with algebra alone — Dirac's ladder-operator trick, one of the most elegant arguments in physics.
Ladder operators
Define two non-Hermitian ladder operators, \hat a (lowering) and \hat a^{\dagger} (raising), as complex combinations of \hat x and \hat p. They very nearly factor the Hamiltonian into a product \hat a^{\dagger}\hat a — the only obstruction being the commutator, which leaves a constant remainder.
The lowering and raising operators as complex combinations of position and momentum.
Their commutator follows in one line from [\hat x,\hat p]=i\hbar — this single relation drives everything.
The Hamiltonian in terms of the number operator \hat N=\hat a^{\dagger}\hat a; the whole problem reduces to finding the eigenvalues of \hat N.
Climbing the ladder
The spectrum now falls out from pure operator algebra — not a single integral. Follow the five steps and watch the energy levels appear.
- Assume a rung. Let |n\rangle satisfy \hat N|n\rangle=n|n\rangle with \hat N=\hat a^{\dagger}\hat a; then \hat H|n\rangle=\hbar\omega(n+\tfrac12)|n\rangle, so eigenstates of \hat N are eigenstates of energy.
- Ladder commutators. From [\hat a,\hat a^{\dagger}]=1 one gets [\hat N,\hat a]=-\hat a and [\hat N,\hat a^{\dagger}]=+\hat a^{\dagger}. Hence \hat N(\hat a|n\rangle)=(n-1)(\hat a|n\rangle): \hat a steps down one rung and \hat a^{\dagger} steps up one — the origin of the name.
- A floor must exist. Norms cannot be negative: \langle n|\hat a^{\dagger}\hat a|n\rangle=n\ge 0. So the ladder cannot descend forever; there must be a lowest state |0\rangle with \hat a|0\rangle=0, forcing n=0 at the bottom.
- Build the whole tower. Apply \hat a^{\dagger} repeatedly to |0\rangle to generate n=0,1,2,3,\dots; the eigenvalues of \hat N are exactly the non-negative integers, no others.
- Read off the spectrum. E_n=\hbar\omega(n+\tfrac12): infinitely many levels, all evenly spaced by one quantum \hbar\omega — the energy of a single photon or phonon of frequency \omega.
The oscillator spectrum: unlike the box's n^2 ladder, these levels are perfectly evenly spaced.
The normalized ladder actions; iterating \hat a^{\dagger} builds |n\rangle=\dfrac{(\hat a^{\dagger})^{n}}{\sqrt{n!}}\,|0\rangle.
Zero-point energy, the states, and where this leads
The ground state does not have zero energy: E_0=\tfrac12\hbar\omega, the zero-point energy. This is demanded by the uncertainty principle — pinning the particle motionless at the bottom of the well would require both x and p sharp at zero, which \Delta x\,\Delta p\ge\hbar/2 forbids. The oscillator settles for the compromise that minimizes energy while respecting \Delta x\,\Delta p\approx\hbar/2.
Where does this lead? The very same algebra reappears across physics. Perturbation theory handles small anharmonic corrections to the oscillator; the angular-momentum operators are diagonalized by an identical raising/lowering trick; and in quantum field theory each mode of a field is an oscillator, with \hat a^{\dagger} and \hat a creating and destroying particles — the foundation of quantum electrodynamics. From here, Quantum Mechanics II applies this machinery to the hydrogen atom, spin and scattering.