Atoms are never still
Even at absolute zero, quantum mechanics forbids the atoms of a crystal from sitting perfectly still — they retain zero-point motion. At any finite temperature they oscillate about their equilibrium sites. Provided the displacements are small, each atom feels a restoring force linear in its displacement, so the whole crystal is a giant coupled system of harmonic oscillators. Diagonalizing it gives the normal modes: collective vibrations in which every atom moves at one shared frequency. This is the harmonic (small-oscillation) approximation — honest to flag it, because the small anharmonic corrections are exactly what give thermal expansion and finite thermal conductivity.
The one-dimensional chain and its dispersion
The cleanest example: a chain of identical masses m separated by a, each joined to its neighbours by springs of stiffness K. Newton's second law for atom n couples it to atoms n\pm1. Because the chain is periodic, we try a travelling-wave solution u_n = A\,e^{i(qna-\omega t)}; substituting turns the infinite set of coupled equations into a single algebraic relation between frequency \omega and wavevector q — the dispersion relation.
Equation of motion for one atom in the monatomic chain (nearest-neighbour springs).
The acoustic dispersion relation: linear (sound-like) near q=0, flat at the zone boundary q=±π/a.
Read the physics off the curve. Near q=0 the frequency is linear, \omega \approx a\sqrt{K/m}\,|q|: these long-wavelength modes are ordinary sound waves and their slope is the speed of sound. At the Brillouin-zone boundary q=\pi/a the curve flattens, the group velocity d\omega/dq falls to zero, and the wave becomes a standing wave — the atoms are Bragg-reflected by the lattice, exactly as in Guide 1. Note that q and q+2\pi/a describe the identical motion, so all distinct modes fit inside one Brillouin zone.
Quantizing the modes: phonons
Each normal mode is an independent harmonic oscillator, and quantum mechanics quantizes its energy in units of \hbar\omega. One quantum of a lattice vibration is a phonon — the particle of sound, exactly as the photon is the particle of light. A mode of frequency \omega holding n phonons has energy (n+\tfrac12)\hbar\omega; the \tfrac12 is the zero-point energy that never vanishes. Phonons are bosons: any number can occupy the same mode, following Bose-Einstein statistics.
Energy of a normal mode holding n phonons; n is the phonon occupation number.
Heat capacity: where classical physics fails
The classical equipartition theorem predicts a temperature-independent molar heat capacity of $3R (the Dulong-Petit law) — and this is badly wrong at low temperature, where every measured solid's heat capacity falls smoothly to zero. The resolution is quantum: as k_BT$ drops below \hbar\omega, high-frequency modes cannot be excited at all — they are 'frozen out'. Debye modelled the solid as a gas of phonons with a linear dispersion up to a cutoff frequency, giving the famous low-temperature result.
The Debye T³ law: the phonon heat capacity of an insulator vanishes as T³ at low temperature.
Be honest about the Debye model: it is an approximation — a single linear branch, a sharp cutoff at the Debye temperature \Theta_D, no real dispersion. It nails the T^3 scaling and interpolates well to the classical $3R$ at high T, but a real phonon spectrum has acoustic and optical branches and complicated structure, measured directly today by inelastic neutron scattering. The lesson to carry forward: whenever a classical prediction is a constant and experiment shows something freezing out toward zero, suspect that quanta of energy \hbar\omega are the culprit. Electrons will tell the same story next.