Phonons, Density of States, and Crystal Momentum

The classical harmonic crystal decomposes into independent normal modes, each a plane wave labelled by a wavevector $k$ and a branch $s$ oscillating at frequency $ω_{s} (k)$ . Quantum mechanics replaces each mode's continuous amplitude with a ladder of discrete energy levels. A single quantum of one mode is a phonon: a particle- like excitation carrying energy $ℏ ω$ and crystal momentum $ℏ k$ . This lesson quantizes the modes, counts the phonons at temperature $T$ , defines their momentum bookkeeping, and constructs the density of states that converts a dispersion relation into a thermal or spectroscopic prediction.

Normal modes as independent oscillators

A normal-mode transformation writes the crystal Hamiltonian as a sum over independent oscillators. Each mode $(k, s)$ contributes a term of the form

H_{k, s} = \frac{1}{2} P_{k, s}^{*} P_{k, s} + \frac{1}{2} ω_{s}^{2} (k) Q_{k, s}^{*} Q_{k, s},

a harmonic oscillator in the normal coordinate $Q_{k, s}$ with frequency $ω_{s} (k)$ . The full Hamiltonian is $H = \sum_{k, s} H_{k, s}$ with no coupling between modes. Reducing the crystal to independent oscillators is what the harmonic approximation buys. Each oscillator quantizes in the usual way, with energy levels

E_{k, s} (n) = (n + \frac{1}{2}) ℏ ω_{s} (k), n = 0, 1, 2, \dots

The integer $n$ is the number of phonons in the mode. Adding one phonon raises the mode's energy by exactly $ℏ ω_{s} (k)$ ; the half-quantum $\frac{1}{2} ℏ ω$ is the zero-point energy, present even at $T = 0$ .

Because the modes are independent oscillators and any number of phonons can share a mode, the total lattice energy is a sum over modes of their occupation times their quantum:

U = k, s \sum (⟨ n_{k, s} ⟩ + \frac{1}{2}) ℏ ω_{s} (k) .

Everything thermal in the lattice reduces to knowing the mean occupation $⟨ n_{k, s} ⟩$ and the spectrum $ω_{s} (k)$ .

Phonon statistics

Phonon number is not conserved: raising the temperature simply excites each oscillator to a higher rung, creating phonons. A gas of particles whose number is unconstrained and which obey Bose statistics has chemical potential zero, so the mean occupation of a mode of frequency $ω$ is the Bose–Einstein (Planck) distribution

⟨ n (ω)⟩ = \frac{1}{e ^{ℏ ω / k_{B} T} - 1} .

Two limits organize its behavior. When $ℏ ω ≪ k_{B} T$ the mode is classically excited, $⟨ n ⟩ \approx k_{B} T /ℏ ω ≫ 1$ , and its mean energy $ℏ ω ⟨ n ⟩ \approx k_{B} T$ recovers equipartition. When $ℏ ω ≫ k_{B} T$ the occupation is exponentially small, $⟨ n ⟩ \approx e^{- ℏ ω / k_{B} T}$ : the mode is frozen out, its contribution to the energy suppressed. The crossover at $ℏ ω \sim k_{B} T$ is what makes heat capacity fall below the classical value at low temperature.

A single vibrational mode as a quantum oscillator. Its equally spaced levels are separated by hbar omega; the mean occupation is the Bose-Einstein number, large at high temperature (many phonons) and exponentially small at low temperature (frozen out).

Crystal momentum

A phonon carries $ℏ k$ , but $k$ is defined only within the first Brillouin zone, because shifting $k$ by a reciprocal-lattice vector $G$ leaves the displacement pattern unchanged. The quantity $ℏ k$ is therefore not true mechanical momentum — a single phonon actually carries no net mass transport — but a bookkeeping label called crystal momentum, conserved only modulo $ℏ G$ .

The two ways the ledger can balance are physically distinct. When the wavevectors on both sides already add up within the zone, $G = 0$ , and the process is a normal process; the total crystal momentum is literally unchanged. When the sum of incoming wavevectors lands outside the first zone and must be folded back by a nonzero $G$ , the process is an Umklapp process (from the German for flipping over). Umklapp processes reverse the direction of momentum flow and are the reason lattice heat conduction is finite, as the thermal-transport lesson develops.

Two three-phonon collisions in the Brillouin zone (dashed square). In the normal process the summed wavevector stays inside the zone. In the Umklapp process the raw sum leaves the zone and a reciprocal-lattice vector G folds the outgoing phonon back inside, reversing its direction.

The density of states

Thermal averages and spectra are sums over the discrete allowed $k$ . With $N$ cells the $k$ points are spaced by $(2 π)^{3} / V$ and become dense, so sums convert to integrals over the zone. It is efficient to first collect all modes of a given frequency into the density of states $g (ω)$ , defined so that $g (ω) d ω$ counts the number of modes with frequency between $ω$ and $ω + d ω$ :

g (ω) = s \sum \frac{V}{( 2 π ) ^{3}} BZ \int δ (ω - ω_{s} (k)) d^{3} k = s \sum \frac{V}{( 2 π ) ^{3}} S_{ω} \int \frac{d S}{∣ \nabla _{k} ω _{s} ( k )∣} .

The last form rewrites the delta function as a surface integral over the constant- frequency surface $S_{ω}$ , weighted by the inverse of the group velocity $∣ \nabla_{k} ω ∣$ . Any thermodynamic average of a mode quantity $f (ω)$ becomes a single integral $\int f (ω) g (ω) d ω$ .

The group-velocity weighting has a sharp consequence. Wherever the dispersion is flat — at band extrema and saddle points, where $\nabla_{k} ω = 0$ — the integrand diverges and $g (ω)$ develops a van Hove singularity. In three dimensions $g (ω)$ stays finite but its slope jumps (a square-root kink); in lower dimensions the singularity is stronger. Every maximum and saddle of the dispersion prints a kink onto the density of states.

A phonon branch (left) and the density of states it produces (right). Flat regions of the dispersion at the zone center and boundary pile up many modes at nearly one frequency, printing van Hove kinks on g(omega); the sharp peak sits at the zone-boundary maximum where the group velocity vanishes.

Measuring dispersion by neutron scattering

The dispersion $ω_{s} (k)$ is measured directly by scattering a probe that can exchange both energy and momentum with a single phonon. Thermal neutrons are ideal: their de Broglie wavelength matches interatomic spacings, so their momentum $ℏ Q$ is comparable to a reciprocal-lattice vector, and their kinetic energy ( $\sim$ tens of meV) is comparable to $ℏ ω$ for phonons. A neutron that creates one phonon obeys

ℏ k_{i} - ℏ k_{f} = ℏ k + ℏ G, \frac{ℏ ^{2} k _{i}^{2}}{2 m _{n}} - \frac{ℏ ^{2} k _{f}^{2}}{2 m _{n}} = ℏ ω_{s} (k),

energy conservation and crystal-momentum conservation together. The neutron loses energy $ℏ ω$ (phonon creation, Stokes) or gains it (phonon absorption, anti-Stokes). Fixing the incident beam and measuring the outgoing energy and angle determines both $k$ and $ω_{s} (k)$ for one point on one branch. Sweeping the geometry maps the entire dispersion.

Inelastic neutron scattering. An incident neutron of wavevector k_i scatters to k_f, creating one phonon of wavevector k. The momentum triangle closes up to a reciprocal-lattice vector G, and the neutron's energy loss equals hbar omega of the phonon.

X-rays and electrons scatter from phonons on the same conservation laws, but neutrons have two experimental advantages: their energies natively match phonon energies, so the small energy loss is easy to resolve, and they scatter from nuclei rather than the electron cloud, giving strong signal from light atoms that X-rays barely see. The dispersion curves so measured feed directly into the thermal properties through the density of states $g (ω)$ built in this lesson.