At very low temperatures, the heat capacity of a crystalline solid follows a T³ law (Debye model) rather than the classical constant value (Dulong-Petit). Why?
AAt low temperatures, atoms freeze in place and stop vibrating, so heat capacity drops to zero
BOnly long-wavelength, low-energy acoustic phonons are thermally accessible at low T; the number of excited modes grows as T³ because the density of states is quadratic in frequency
COptical phonons dominate at low temperatures because they have lower energy than acoustic phonons
DThe T³ law reflects the three spatial dimensions, with each dimension contributing T independently
In the Debye model, acoustic phonon frequencies extend from 0 up to a cutoff (the Debye frequency). At low temperatures, kT is much smaller than most phonon energies, so only the lowest-frequency (longest-wavelength) acoustic modes are thermally excited. The density of states for acoustic phonons scales as ω², meaning the number of available modes between ω and ω+dω grows quadratically. Combining this with the Bose-Einstein thermal occupation (which cuts off exponentially at high ω) gives a heat capacity ∝ T³. As temperature rises, more modes become accessible until all 3N modes are fully excited — then heat capacity saturates at the classical Dulong-Petit value of 3k_B per atom.
Question 2 Multiple Choice
A phonon mode has n_k = 0. What does this mean physically?
AThe atoms in that mode are at rest — there is no vibration in the lattice
BThe mode is in its quantum ground state: the atoms still undergo zero-point motion, but no additional thermal quanta have been added
CThe crystal has zero temperature and no thermal energy
DThat particular wavevector k does not exist in the crystal's phonon spectrum
n_k = 0 means zero phonons have been excited in mode k — the mode is in its quantum mechanical ground state. But this is not the same as no vibration: the quantum harmonic oscillator has a nonzero ground state energy of ℏω_k/2 (zero-point energy), and atoms undergo zero-point motion even at absolute zero. This is a purely quantum effect with no classical analog. Phonons are the excitation quanta above this ground state; n_k = 1 means one quantum of energy ℏω_k has been added, and so on. The zero-point motion of all modes is the reason solids have a nonzero energy even at T = 0.
Question 3 True / False
Phonons are not conserved particles: the total number of phonons in a system at thermal equilibrium is not fixed and changes freely with temperature.
TTrue
FFalse
Answer: True
Unlike electrons, phonons have no conservation law for their total number. They can be created and annihilated freely — when a solid is heated, more phonons are excited (created); when cooled, phonons are absorbed (annihilated). At thermal equilibrium, the mean occupation of each mode is given by the Bose-Einstein distribution n̄_k = 1/(exp(ℏω_k/kT)−1), which increases smoothly with temperature. This is why phonons obey Bose-Einstein statistics with zero chemical potential μ = 0: there is no conservation law constraining their number. This distinguishes them from atoms or electrons, which are conserved and have μ ≠ 0.
Question 4 True / False
Acoustic phonons are Goldstone modes of the crystal because the crystal breaks continuous translational symmetry down to discrete lattice translations.
TTrue
FFalse
Answer: True
Goldstone's theorem states that spontaneously breaking a continuous symmetry produces gapless (massless) excitations. A liquid has continuous translational symmetry; when it freezes into a crystal, this symmetry is spontaneously broken — atoms occupy specific positions rather than any position equally. The surviving symmetry is discrete lattice translation. The Goldstone modes of this symmetry breaking are the acoustic phonons: long-wavelength sound waves whose frequency goes to zero as k → 0 (linear dispersion ω ≈ v_s|k|). This connection to Goldstone's theorem explains why acoustic phonons are always gapless in any crystal, regardless of material specifics.
Question 5 Short Answer
Why is it useful to describe lattice vibrations as independent phonons rather than tracking the positions of individual atoms?
Think about your answer, then reveal below.
Model answer: A crystal of N atoms has 3N coupled degrees of freedom — tracking all atomic positions requires solving 3N coupled differential equations, which is intractable for 10²³ atoms. By transforming to normal modes (collective oscillations of the entire lattice), the problem decomposes into 3N independent harmonic oscillators. Each oscillator can be quantized separately, yielding phonons. The key insight is that in the harmonic approximation, the modes do not interact with each other — each phonon mode is independent, and thermodynamics becomes a sum of independent quantum harmonic oscillators with a known solution.
This is the power of the quasiparticle concept throughout condensed matter physics: complex many-body problems become tractable when the right collective coordinates (normal modes) are identified. The same strategy works for magnons in ferromagnets, plasmons in metals, and Cooper pairs in superconductors. The quasiparticle description is not just a computational convenience — it is often physically exact in the low-excitation limit where the harmonic approximation holds.