Quantization of Lattice Vibrations

Statement

Starting from the harmonic lattice Hamiltonian, we diagonalise it into independent normal modes labelled by wavevector \(\mathbf{k}\) and branch \(s\), promote the normal coordinates and momenta to operators obeying canonical commutation relations, and introduce ladder operators \(\hat a_{\mathbf{k}s},\hat a_{\mathbf{k}s}^{\dagger}\) that recast the Hamiltonian as \(\hat H=\sum_{\mathbf{k},s}\hbar\omega_{\mathbf{k}s}\bigl(\hat a_{\mathbf{k}s}^{\dagger}\hat a_{\mathbf{k}s}+\tfrac12\bigr)\). The quanta created by \(\hat a_{\mathbf{k}s}^{\dagger}\) are phonons: bosonic quasiparticles of energy \(\hbar\omega_{\mathbf{k}s}\) and crystal momentum \(\hbar\mathbf{k}\).

Why it matters

Every thermal, transport and optical property of an insulating crystal — heat capacity, thermal conductivity, thermal expansion, Raman and infrared spectra, the electron–phonon coupling behind conventional superconductivity — is built on this result. The phonon is the archetype of a quasiparticle: a collective excitation of \(\sim 10^{23}\) coupled atoms that nevertheless behaves like a single free bosonic particle with a definite energy and (quasi-)momentum.

Second quantization of the lattice also provides the cleanest concrete example of promoting a classical field of oscillators to a quantum field. The same machinery — normal modes, ladder operators, occupation-number states — carries over verbatim to photons in QED and to spin waves, giving the student a reusable template rather than a one-off trick.

Assumptions

Derivation

Result

Limiting cases

• Single atom, \(\mathbf{k}=0\), one branch: the sum collapses to \(\hat H=\hbar\omega(\hat a^{\dagger}\hat a+\tfrac12)\), recovering the ordinary quantum harmonic oscillator.
• Long-wavelength acoustic branch (\(\mathbf{k}\to 0\)): \(\omega_{\mathbf{k}s}\to c_s|\mathbf{k}|\), so phonons become sound quanta with linear dispersion — the quantized analogue of classical elastic waves.
• Optical branch at \(\mathbf{k}=0\): \(\omega\to\omega_{\mathrm{opt}}\neq 0\), an Einstein-like dispersionless mode; each such mode is an independent oscillator of fixed frequency.
• Classical / high-\(T\) limit (\(k_BT\gg\hbar\omega\)): \(\langle n_{\mathbf{k}s}\rangle\to k_BT/\hbar\omega\), each mode holds \(k_BT\) of energy and the Dulong–Petit heat capacity \(3Nk_B\) is recovered.
• \(T\to 0\): all \(n_{\mathbf{k}s}=0\); only the zero-point energy \(\tfrac12\sum\hbar\omega_{\mathbf{k}s}\) remains.

Breaks when

Failure modes

• Pairing \(\mathbf{k}\) with itself in the ladder operators. Writing \(\hat Q_{\mathbf{k}}\propto(\hat a_{\mathbf{k}}+\hat a_{\mathbf{k}}^{\dagger})\) instead of \((\hat a_{\mathbf{k}}+\hat a_{-\mathbf{k}}^{\dagger})\) violates the reality constraint \(\hat Q_{\mathbf{k}}^{\dagger}=\hat Q_{-\mathbf{k}}\) and gives a non-Hermitian \(\hat H\).
• Double-counting the Brillouin zone. Summing over both \(\mathbf{k}\) and \(-\mathbf{k}\) as independent while also treating \(\hat Q_{\mathbf{k}},\hat Q_{-\mathbf{k}}\) as independent double-counts the degrees of freedom; there are exactly \(3pN\) modes.
• Dropping the zero-point energy. Discarding the \(\tfrac12\) because "it is a constant" loses real physics — it drives the isotope effect on lattice constants, quantum crystals (solid He), and the Casimir-like contributions.
• Confusing crystal momentum with true momentum. \(\hbar\mathbf{k}\) is conserved only modulo a reciprocal-lattice vector \(\mathbf{G}\); forgetting umklapp (\(\mathbf{G}\neq0\)) processes wrongly predicts infinite thermal conductivity.
• Treating phonons as conserved particles. Phonon number is not conserved (no chemical potential); their equilibrium distribution is Planck/Bose–Einstein with \(\mu=0\), unlike a real boson gas.
• Applying single-oscillator \(\langle x^2\rangle\) to \(\hat Q_{\mathbf{k}}\) without the complex-mode care. Because \(\hat Q_{\mathbf{k}}\) is complex (\(\hat Q_{\mathbf{k}}^{\dagger}=\hat Q_{-\mathbf{k}}\)), the mean-square amplitude uses \(\langle \hat Q_{\mathbf{k}}\hat Q_{-\mathbf{k}}\rangle\), not \(\langle \hat Q_{\mathbf{k}}^{2}\rangle\).

Discussion

The deep content of this derivation is that a many-body problem — \(3pN\) coupled ions — becomes trivial once written in the right variables. Diagonalising the harmonic Hamiltonian in normal coordinates is a purely classical step; quantization enters only through the single canonical commutator of Step 5. Everything distinctively quantum (discrete energy quanta \(\hbar\omega\), zero-point motion, Bose statistics) then follows algebraically. This "diagonalise, then quantize" pattern is the essence of free quantum field theory: the phonon field is a lattice-regularised scalar field, and the continuum limit \(a\to0\) of the acoustic branch is a relativistic-like scalar field with sound speed playing the role of \(c\).

The phonon is a quasiparticle, not a particle: it exists only as a coherent motion of the whole crystal and carries crystal momentum \(\hbar\mathbf{k}\), conserved only modulo a reciprocal-lattice vector. This weaker conservation law is what allows umklapp scattering to degrade a heat current and give a finite thermal conductivity — a fact with no analogue for genuine particles in free space. The mode count is rigidly fixed: \(p\) atoms per primitive cell in three dimensions give \(3p\) branches, of which \(3\) are acoustic (\(\omega\to0\) as \(\mathbf{k}\to0\), from the three rigid translations) and \(3p-3\) optical.

Because phonon number is not conserved — creating a phonon costs energy but is otherwise unconstrained — the chemical potential is zero and the thermal occupation is the Planck distribution \(\langle n_{\mathbf{k}s}\rangle=[\exp(\hbar\omega_{\mathbf{k}s}/k_BT)-1]^{-1}\). Feeding this into \(\hat H\) reproduces the Debye and Einstein heat-capacity models as the two idealisations of the dispersion \(\omega_{\mathbf{k}s}\), and the low-temperature \(C_V\propto T^{3}\) law is a direct consequence of the linear acoustic dispersion combined with Bose statistics.

A subtler point is that the harmonic phonon basis is only the zeroth order of an interacting field theory. Anharmonic terms, reinstated perturbatively, are cubic and quartic in \((\hat a+\hat a^{\dagger})\); they generate three- and four-phonon vertices whose imaginary self-energy gives each phonon a finite linewidth \(\Gamma=1/\tau\) and whose real part shifts \(\omega_{\mathbf{k}s}\) with temperature. In this language thermal expansion is the leading anharmonic correction (the Grüneisen effect), and the phonon acquires the full apparatus of a dressed particle — propagator, self-energy, spectral function — exactly as an electron does in an interacting medium.

Common misconceptions. A phonon is not a vibrating atom localised in space; a single-phonon Fock state \(\hat a_{\mathbf{k}s}^{\dagger}|0\rangle\) is a delocalised standing/travelling wave over the entire crystal. Nor does "quantizing the lattice" require the atoms to be quantum point particles orbiting anything — it is the collective coordinate \(\hat Q_{\mathbf{k}s}\) that is quantized. And the zero-point energy is physical, not a bookkeeping constant: it shifts equilibrium volumes and is measurable through the isotope dependence of lattice constants.

Worked examples

Example 1 — Thermal phonon occupation of an acoustic mode.

Example 2 — Energy quantum and Einstein temperature of a longitudinal-optical (LO) phonon.

Problems

1. Show that requiring the displacement field \(\hat u_{l\alpha}\) to be Hermitian, together with \(\mathbf{e}^{s}(-\mathbf{k})=\mathbf{e}^{s}(\mathbf{k})^{*}\), forces \(\hat Q_{-\mathbf{k}s}=\hat Q_{\mathbf{k}s}^{\dagger}\).
   Solution

   Take the Hermitian conjugate of \(\hat u_{l\alpha}=\tfrac{1}{\sqrt{NM}}\sum_{\mathbf{k}s}e^{s}_{\alpha}(\mathbf{k})\hat Q_{\mathbf{k}s}e^{i\mathbf{k}\cdot\mathbf{R}_l}\): since \(\hat u^{\dagger}=\hat u\), \(\sum_{\mathbf{k}s}e^{s}_{\alpha}(\mathbf{k})^{*}\hat Q_{\mathbf{k}s}^{\dagger}e^{-i\mathbf{k}\cdot\mathbf{R}_l}=\sum_{\mathbf{k}s}e^{s}_{\alpha}(\mathbf{k})\hat Q_{\mathbf{k}s}e^{i\mathbf{k}\cdot\mathbf{R}_l}\). Relabel \(\mathbf{k}\to-\mathbf{k}\) on the left; the exponentials then match and, using \(e^{s}_{\alpha}(-\mathbf{k})^{*}=e^{s}_{\alpha}(\mathbf{k})\), comparing coefficients of the independent plane waves gives \(\hat Q_{-\mathbf{k}s}^{\dagger}=\hat Q_{\mathbf{k}s}\), i.e. \(\hat Q_{\mathbf{k}s}^{\dagger}=\hat Q_{-\mathbf{k}s}\).
2. A crystal has \(p\) atoms per primitive cell in three dimensions with \(N\) cells. How many phonon branches are there, how many are acoustic, and how many total modes (states) exist?
   Solution

   Each cell contributes \(3p\) vibrational degrees of freedom, so there are \(3p\) branches. Three of them are acoustic (\(\omega\to0\) as \(\mathbf{k}\to0\), one per rigid-body translation direction); the remaining \(3p-3\) are optical. With \(N\) allowed \(\mathbf{k}\)-points in the Brillouin zone, the total number of normal modes is \(3pN\), matching the \(3pN\) classical degrees of freedom. For a monatomic crystal (\(p=1\)): 3 branches, all acoustic, \(3N\) modes.
3. Compute the mean thermal energy (including zero-point) of a single mode with \(f=2.0\) THz at \(T=100\) K.
   Solution

   \(\hbar\omega=\hbar(2\pi f)=(1.055\times10^{-34})(2\pi)(2.0\times10^{12})=1.325\times10^{-21}\,\mathrm J=8.27\,\mathrm{meV}\). \(k_BT=(1.381\times10^{-23})(100)=1.381\times10^{-21}\,\mathrm J\); ratio \(\hbar\omega/k_BT=0.959\). \(\langle n\rangle=1/(e^{0.959}-1)=1/(2.610-1)=0.621\). \(\bar E=\hbar\omega(\langle n\rangle+\tfrac12)=1.325\times10^{-21}(1.121)=1.49\times10^{-21}\,\mathrm J=9.27\,\mathrm{meV}\).
4. Verify the bosonic algebra: given \([\hat Q_{\mathbf{k}s},\hat P_{\mathbf{k}'s'}]=i\hbar\delta_{\mathbf{k}\mathbf{k}'}\delta_{ss'}\) and the definitions of Step 6, show \([\hat a_{\mathbf{k}s},\hat a_{\mathbf{k}s}^{\dagger}]=1\).
   Solution

   Invert Step 6: \(\hat a_{\mathbf{k}s}=\sqrt{\tfrac{\omega}{2\hbar}}\hat Q_{\mathbf{k}s}+\tfrac{i}{\sqrt{2\hbar\omega}}\hat P_{-\mathbf{k}s}\) and \(\hat a_{\mathbf{k}s}^{\dagger}=\sqrt{\tfrac{\omega}{2\hbar}}\hat Q_{-\mathbf{k}s}-\tfrac{i}{\sqrt{2\hbar\omega}}\hat P_{\mathbf{k}s}\) (using \(\hat Q_{\mathbf{k}s}^{\dagger}=\hat Q_{-\mathbf{k}s}\)). Then \([\hat a_{\mathbf{k}s},\hat a_{\mathbf{k}s}^{\dagger}]=\tfrac{\omega}{2\hbar}[\hat Q_{\mathbf{k}s},\hat Q_{-\mathbf{k}s}]+\tfrac{1}{2\hbar\omega}[\hat P_{-\mathbf{k}s},\hat P_{\mathbf{k}s}]-\tfrac{i}{2\hbar}[\hat Q_{\mathbf{k}s},\hat P_{\mathbf{k}s}]+\tfrac{i}{2\hbar}[\hat P_{-\mathbf{k}s},\hat Q_{-\mathbf{k}s}]\). The two \([\hat Q,\hat Q]\) and \([\hat P,\hat P]\) terms vanish; the remaining two each give \(\tfrac{i}{2\hbar}(i\hbar)\times(\mp1)\)... carefully: \(-\tfrac{i}{2\hbar}(i\hbar)=\tfrac12\) and \(+\tfrac{i}{2\hbar}[\hat P_{-\mathbf{k}s},\hat Q_{-\mathbf{k}s}]=\tfrac{i}{2\hbar}(-i\hbar)=\tfrac12\). Sum \(=1\).
5. In the Debye model the low-temperature heat capacity is \(C_V=\tfrac{12\pi^{4}}{5}Nk_B(T/\Theta_D)^{3}\). For copper (\(\Theta_D=343\) K), estimate \(C_V\) per mole at \(T=10\) K and compare with the Dulong–Petit value \(3R\).
   Solution

   Per mole \(N k_B\to R=8.314\,\mathrm{J\,mol^{-1}K^{-1}}\). \((T/\Theta_D)^{3}=(10/343)^{3}=(0.02915)^{3}=2.48\times10^{-5}\). Prefactor \(\tfrac{12\pi^{4}}{5}=\tfrac{12(97.41)}{5}=233.8\). So \(C_V=233.8\times8.314\times2.48\times10^{-5}=0.0482\,\mathrm{J\,mol^{-1}K^{-1}}\). Dulong–Petit: \(3R=24.9\,\mathrm{J\,mol^{-1}K^{-1}}\). The ratio \(C_V/3R\approx1.9\times10^{-3}\): at 10 K only the lowest-energy long-wavelength acoustic phonons are thermally accessible (Bose freeze-out of modes with \(\hbar\omega\gg k_BT\)), so the lattice heat capacity is four orders of magnitude below the classical equipartition value.