Lattice Vibrations and Phonons

Consider $N$ atoms of mass $m$ connected by springs of constant $K$ with equilibrium spacing $a$.

The equation of motion for the $n$-th atom:

$$m\ddot{u}_n = K(u_{n+1} - u_n) + K(u_{n-1} - u_n) = K(u_{n+1} + u_{n-1} - 2u_n)$$

Derivation of the dispersion relation. Assuming solutions $u_n = u_0\, e^{i(qna - \omega t)}$:

$$-m\omega^2 = K(e^{iqa} + e^{-iqa} - 2) = 2K(\cos qa - 1) = -4K\sin^2\left(\frac{qa}{2}\right)$$ $$\omega(q) = 2\sqrt{\frac{K}{m}}\left|\sin\left(\frac{qa}{2}\right)\right|$$

$\blacksquare$

Key features:

• The Brillouin zone is $-\pi/a \leq q \leq \pi/a$.
• Linear for small $q$: $\omega \approx v_s \lvert q\rvert$ where $v_s = a\sqrt{K/m}$ is the speed of sound.
• Group velocity: $v_g = d\omega/dq = a\sqrt{K/m}\cos(qa/2)$.
• Maximum frequency: $\omega_{\mathrm{max} = 2\sqrt{K/m}}$.
• Phase velocity: $v_p = \omega/q$ Which exceeds $v_s$ and diverges as $q \to 0$.

For a chain with alternating masses $m_1$ and $m_2$ (e.g., NaCl):

$$\omega^2 = K\left(\frac{1}{m_1} + \frac{1}{m_2}\right) \pm K\sqrt{\left(\frac{1}{m_1} + \frac{1}{m_2}\right)^2 - \frac{4\sin^2(qa/2)}{m_1 m_2}}$$

This gives two branches:

• Acoustic branch ($-$ sign): $\omega \to 0$ as $q \to 0$. Atoms in the unit cell move in phase.
• Optical branch ($+$ sign): $\omega \neq 0$ at $q = 0$. Atoms in the unit cell move out of phase. Can interact with light (hence the name).

At $q = 0$ The optical frequency is $\omega_0 = \sqrt{2K(1/m_1 + 1/m_2)}$ and the acoustic branch Has $\omega = v_s q$ with $v_s = a\sqrt{2K/(m_1 + m_2)}$.

Lattice vibrations are quantised. Each normal mode of wave vector $\mathbf{q}$ and branch $s$ has Energy:

$$E_{\mathbf{q}s} = \left(n_{\mathbf{q}s} + \frac{1}{2}\right)\hbar\omega_{\mathbf{q}s}$$

Where $n_{\mathbf{q}s}$ is the phonon occupation number. Phonons are bosons obeying Bose-Einstein Statistics:

$$\langle n_{\mathbf{q}s} \rangle = \frac{1}{e^{\beta\hbar\omega_{\mathbf{q}s}} - 1}$$

In three dimensions, there are 3 acoustic branches (1 longitudinal, 2 transverse) and $3p - 3$ Optical branches for a crystal with $p$ atoms per primitive cell.

The Debye model approximates the phonon spectrum as linear ($\omega = v_s q$) up to a cutoff frequency $\omega_D$ (the Debye frequency):

$$\omega_D = v_s\left(\frac{6\pi^2 N}{V}\right)^{1/3}$$

The Debye temperature: $\Theta_D = \hbar\omega_D / k_B$.

Derivation of the phonon density of states. The number of modes with wave vector $\lvert\mathbf{q}\rvert \leq q$ In 3D is $N(q) = 3 \cdot \frac{V}{(2\pi)^3} \cdot \frac{4\pi q^3}{3}$ (factor of 3 for polarisations). Differentiating: $g(q)\,dq = dN/dq\,dq = (Vq^2/\pi^2)\,dq$. Converting to frequency with $\omega = v_s q$:

$$g(\omega)\,d\omega = \frac{Vq^2}{\pi^2}\frac{dq}{d\omega}\,d\omega = \frac{V\omega^2}{\pi^2 v_s^3}\,d\omega$$

Since there are $3N$ total modes, the cutoff is determined by $\int_0^{\omega_D} g(\omega)\,d\omega = 3N$ Giving $g(\omega) = \frac{3V\omega^2}{2\pi^2 v_s^3}$ For $0 \leq \omega \leq \omega_D$. $\blacksquare$

Lattice heat capacity:

$$C_V = 9Nk_B\left(\frac{T}{\Theta_D}\right)^3 \int_0^{\Theta_D/T} \frac{x^4 e^x}{(e^x - 1)^2}\,dx$$

High-temperature limit ($T \gg \Theta_D$): $C_V = 3Nk_B$ (Dulong—Petit law).

Low-temperature limit ($T \ll \Theta_D$): $C_V = \frac{12\pi^4}{5}Nk_B\left(\frac{T}{\Theta_D}\right)^3$ (Debye $T^3$ law).

The Einstein model treats all atoms as independent quantum harmonic oscillators with the same frequency $\omega_E$:

$$C_V = 3Nk_B\left(\frac{\Theta_E}{T}\right)^2 \frac{e^{\Theta_E/T}}{(e^{\Theta_E/T} - 1)^2}$$

Where $\Theta_E = \hbar\omega_E/k_B$.

High-temperature limit ($T \gg \Theta_E$): expanding $e^{\Theta_E/T} \approx 1 + \Theta_E/T + \frac{1}{2}(\Theta_E/T)^2$ gives $C_V \to 3Nk_B$ (Dulong—Petit), matching Debye.

Low-temperature limit ($T \ll \Theta_E$): $C_V \approx 3Nk_B(\Theta_E/T)^2 e^{-\Theta_E/T}$ Which is exponentially suppressed. This disagrees with the Debye $T^3$ law (and experiment).

Phonons carry heat through the lattice. By the kinetic theory formula:

$$\kappa_{\mathrm{ph} = \frac{1}{3}C_V v_s \ell_{\mathrm{ph}}}$$

Where $\ell_{\mathrm{ph}}$ is the phonon mean free path.

Scattering mechanisms that limit $\ell_{\mathrm{ph}}$:

1. Phonon—phonon scattering: At high $T$, $\ell_{\mathrm{ph} \propto 1/T}$ (Umklapp processes dominate, where the total phonon momentum is not conserved). At low $T$ Only normal processes ($N$-processes, conserving momentum) contribute, and $\ell_{\mathrm{ph}}$ grows exponentially.
2. Boundary scattering: At very low $T$, $\ell_{\mathrm{ph}}$ is limited by the sample size $L$.
3. Defect scattering: Point defects, dislocations, and grain boundaries scatter phonons, reducing $\kappa_{\mathrm{ph}}$.

Temperature dependence:

• Low $T$ ($T \ll \Theta_D$): $\kappa_{\mathrm{ph} \propto T^3}$ (from $C_V \propto T^3$ With $\ell_{\mathrm{ph}}$ limited by boundaries).
• Intermediate $T$: $\kappa_{\mathrm{ph}}$ peaks.
• High $T$ ($T \gtrsim \Theta_D$): $\kappa_{\mathrm{ph} \propto 1/T}$ (from $\ell_{\mathrm{ph} \propto 1/T}$ and $C_V \approx \mathrm{const}$).

The Debye model captures the correct low-$T$ behaviour because long-wavelength acoustic phonons Dominate the specific heat at low temperatures. The Einstein model is more appropriate for Describing the optical branch contribution, which is nearly flat (dispersionless) and hence well Approximated by a single frequency.

For crystals with optical branches (e.g., NaCl, SiO$_2$), a combined model using Debye for Acoustic modes and Einstein for optical modes gives better agreement with experiment across all Temperatures.

Worked Example: Debye Temperature of Copper

Copper has molar mass $M = 63.55$ g/mol, density $\rho = 8.96\ \mathrm{g}/cm^3$ And measured Speed of sound $v_s = 3810$ m/s (average of longitudinal and transverse).

Number density: $n = \frac{\rho N_A}{M} = \frac{8.96 \times 6.022 \times 10^{23}}{63.55} = 8.49 \times 10^{28}\ \mathrm{m}^{-3}$.

$$\Theta_D = \frac{\hbar v_s}{k_B}(6\pi^2 n)^{1/3}$$ $$(6\pi^2 n)^{1/3} = (6\pi^2 \times 8.49 \times 10^{28})^{1/3} = (5.03 \times 10^{30})^{1/3} = 1.71 \times 10^{10}\ \mathrm{m}^{-1}$$ $$\Theta_D = \frac{1.055 \times 10^{-34} \times 3810}{1.381 \times 10^{-23}} \times 1.71 \times 10^{10} = 2.91 \times 10^{-8} \times 1.71 \times 10^{10} = 498\ \mathrm{K}$$

The accepted experimental value is $\Theta_D = 343$ K. The discrepancy arises because the Debye Model uses a single average sound velocity, while the real phonon spectrum is anisotropic.

Phonons are quantized vibrations of atoms in a crystal lattice, much like photons are quantized light waves. When atoms in a crystal are displaced from their equilibrium positions, the restoring forces create collective oscillations that propagate through the lattice as waves. The acoustic branch corresponds to atoms moving in phase, producing sound waves, while the optical branch has atoms in adjacent cells moving out of phase, which can interact with light. Phonons carry thermal energy through the lattice, and their density of states determines the specific heat at low temperatures. The Brillouin zone boundary acts as a mirror for phonon wavevectors, limiting the range of distinct vibration modes.

Mistake 1: Confusing phonons with photons. Phonons are quantised lattice vibrations, while photons are quantised electromagnetic waves. Phonons exist only in materials with a lattice structure, while photons exist in vacuum. Do not assume that phonons behave exactly like photons; they have different dispersion relations and interactions.

Mistake 2: Assuming that phonons are particles. Phonons are quasiparticles, not fundamental particles. They are emergent excitations of the lattice and do not exist outside the material. Do not think of phonons as real particles; they are convenient mathematical descriptions of collective lattice vibrations.

Mistake 3: Forgetting that acoustic phonons have linear dispersion at small $q$. Acoustic phonons have $\omega \approx v_s |q|$ for small $q$, where $v_s$ is the speed of sound. This linear dispersion is crucial for understanding thermal conductivity and specific heat. Do not assume that all phonons have linear dispersion; optical phonons have a finite frequency at $q = 0$.

Mistake 4: Confusing the Brillouin zone with the reciprocal lattice. The Brillouin zone is the Wigner-Seitz cell of the reciprocal lattice. It is the set of all $q$-vectors that are closer to the origin than to any other reciprocal lattice point. Do not confuse the Brillouin zone with the reciprocal lattice itself; they are related but distinct concepts.

Mistake 5: Assuming that phonons can have any frequency. Phonons are confined to the Brillouin zone and have frequencies bounded by the maximum phonon frequency. Do not assume that phonons can have arbitrarily high frequencies; the lattice structure imposes a cutoff.

• Crystal Structures: The lattice geometry determines the phonon dispersion relations and the number of phonon branches.
• Reciprocal Lattice: Phonon wavevectors are defined in the Brillouin zone of the reciprocal lattice.
• Diffraction: Inelastic neutron scattering measures phonon dispersion relations using the same diffraction framework.
• Semiconductors: Phonon scattering limits carrier mobility and affects thermal conductivity in semiconductor materials.
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