Questions: Phonon Statistics and Dispersion Relations

Question 1

At temperatures well below the Debye temperature θ_D, only low-frequency acoustic phonons are excited. As temperature increases toward θ_D, what happens to the heat capacity C_V?

• A. C_V remains approximately constant because the total number of phonon modes is fixed by the crystal structure
• B. C_V decreases because higher-frequency optical phonons are suppressed by quantum statistics
• C. C_V rises toward the classical Dulong-Petit value 3Nk_B as more phonon modes become thermally accessible
• D. C_V jumps discontinuously when k_BT first exceeds ℏω_D

Answer: (C)

At low T, only acoustic phonons with ℏω ≲ k_BT are populated; C_V ∝ T³. As T increases, higher-frequency modes become accessible (their average occupation n_k = 1/[exp(ℏω_k/k_BT)−1] becomes non-negligible), and C_V rises continuously. In the high-temperature classical limit (T >> θ_D), all 3N modes are thermally activated and each contributes k_B of heat capacity, recovering the Dulong-Petit value C_V = 3Nk_B = 3R. The approach is continuous, not discontinuous.

Question 2

The chemical potential of phonons is zero because:

• A. Phonons have zero rest mass and therefore zero energy in the grand canonical ensemble
• B. Phonons are fermions and their Fermi energy happens to equal zero at all temperatures
• C. Phonon number is not conserved — phonons are freely created and destroyed — so there is no thermodynamic constraint that would fix μ ≠ 0
• D. The chemical potential of all bosons is identically zero by the rules of Bose-Einstein statistics

Answer: (C)

In the grand canonical ensemble, chemical potential μ arises as the Lagrange multiplier enforcing conservation of particle number. For phonons, there is no such conservation law: phonons are collective excitations of the lattice that are freely created and destroyed as atoms vibrate. With no constraint on total phonon number, the minimization of free energy sets μ = 0 automatically, simplifying the Bose-Einstein distribution to the Planck form n_k = 1/[exp(ℏω_k/k_BT)−1]. Option D is wrong: photons also have μ = 0 for the same reason (no conservation), but most bosons (e.g., ⁴He atoms) have nonzero μ.

Question 3

Optical phonons have higher energy than acoustic phonons at the same wavevector because optical phonons travel faster through the crystal lattice.

• T. True
• F. False

Answer: False

Optical phonons do not have higher energy because they travel faster — they are in fact nearly dispersionless (nearly flat ω vs. k), meaning they have nearly the same frequency across all wavevectors. Their high energy at k = 0 arises because they involve neighboring atoms vibrating against each other (out of phase), which requires more energy than the in-phase long-wavelength sound waves that acoustic branches describe. Acoustic branches start at ω = 0 (sound waves have zero frequency at zero wavevector); optical branches start at finite ω_optical even at k = 0.

Question 4

At sufficiently high temperatures (T >> θ_D), each phonon mode contributes approximately k_BT of thermal energy on average, recovering the classical Dulong-Petit law.

• T. True
• F. False

Answer: True

In the classical limit k_BT >> ℏω_k, the Planck distribution gives n_k ≈ k_BT/ℏω_k >> 1, and the mode energy ℏω_k · n_k ≈ k_BT. Summing over all 3N phonon modes gives total energy U ≈ 3Nk_BT, and heat capacity C_V = ∂U/∂T = 3Nk_B = 3R — the classical equipartition result, the Dulong-Petit law. This is the high-temperature limit of the quantum Bose-Einstein result and explains why classical thermodynamics works well for solids at room temperature when θ_D is low.

Question 5

Why does the phonon contribution to heat capacity scale as T³ at low temperatures, and what property of the dispersion relation is responsible?

Model answer: At low T, only modes with ℏω ≲ k_BT are significantly populated. Acoustic branches have linear dispersion (ω = v_s k), so the number of populated modes — those with k small enough that ℏv_s k ≲ k_BT — scales as k³ ~ T³. Integrating mode energies gives U ~ T⁴ and C_V = ∂U/∂T ~ T³.

The T³ law is a direct consequence of linear acoustic dispersion. Because ω ∝ k for long-wavelength acoustic phonons, the wavevectors excited at temperature T form a sphere of radius k ~ k_BT/ℏv_s, and the number of modes scales as the sphere's volume k³ ~ T³. Optical phonons contribute negligibly at low T because their energy ℏω_optical >> k_BT — they are frozen out. This T³ behavior is universal for systems with linearly dispersing bosonic excitations, including photons in a blackbody (where it yields the T⁴ Stefan-Boltzmann law for energy rather than T³).