The quantum harmonic oscillator has Hamiltonian $$H=\frac{p^2}{2m}+\frac12m\omega^2x^2.$$ Instead of solving the differential equation directly, define ladder operators $$a=\sqrt{\frac{m\omega}{2\hbar}},x+\frac{i}{\sqrt{2m\hbar\omega}},p,$$ $$a^\dagger=\sqrt{\frac{m\omega}{2\hbar}},x-\frac{i}{\sqrt{2m\hbar\omega}},p,$$ which satisfy $[a,a^\dagger]=1$.

The Hamiltonian becomes $$H=\hbar\omega\left(a^\dagger a+\frac12\right).$$ If $|n\rangle$ is an eigenstate of the number operator $N=a^\dagger a$ with eigenvalue $n$, then $$E_n=\hbar\omega\left(n+\frac12\right),\qquad n=0,1,2,\ldots$$ The ladder operators connect neighboring energy eigenstates: $$a^\dagger|n\rangle=\sqrt{n+1}|n+1\rangle,\qquad a|n\rangle=\sqrt n|n-1\rangle.$$

The lowest state still has energy $$E_0=\frac12\hbar\omega,$$ called zero-point energy. A state with both exactly zero position and zero momentum spread is impossible, so the oscillator cannot sit motionless at the classical minimum.

Near any stable potential minimum, a smooth potential is approximately quadratic. The quantum harmonic oscillator therefore appears far beyond literal springs: molecular vibrations, lattice phonons, electromagnetic field modes and many perturbative approximations reduce locally to this structure.