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Chapter 7 • Theory & Derivations

Dielectric Properties, Plasmons & Optical Phenomena

Comprehensive macroscopic and microscopic electrodynamics of solids: macroscopic field vs local Lorentz field, polarizabilities, Clausius-Mossotti relation, AC complex permittivity, Debye dielectric relaxation, plasma oscillations, plasma frequency, Thomas-Fermi screening, ferroelectricity, piezoelectricity, and Kramers-Kronig optical relations.

§7.1 Macroscopic Electric Field & The Microscopic Local Lorentz Field

1. Macroscopic Polarization $\vec{P}$ and Electric Displacement $\vec{D}$

In a dielectric solid subjected to an external electrostatic field, bound charges undergo micro-displacements, creating a volume dipole moment density known as the polarization vector $\vec{P}$:

$$\vec{P} = \lim_{\Delta V \to 0} \frac{1}{\Delta V} \sum_{i \in \Delta V} \vec{p}_i$$

The macroscopic electric displacement $\vec{D}$ and macroscopic electric field $\vec{E}$ are related in linear, isotropic media by:

$$\vec{D} = \varepsilon_0 \vec{E} + \vec{P} = \varepsilon_0 (1 + \chi_e) \vec{E} = \varepsilon_0 \varepsilon_r \vec{E}$$

where $\chi_e$ is the electric susceptibility and $\varepsilon_r = 1 + \chi_e$ is the relative dielectric permittivity.

2. Derivation of the Local Lorentz Field $\vec{E}_{\text{loc}}$

The actual electric field acting on an individual atom inside a condensed solid—the local microscopic field $\vec{E}_{\text{loc}}$—is not equal to the macroscopic Maxwell average field $\vec{E}$, because the atom is surrounded by polarized atomic neighbors. Following H. A. Lorentz, we construct a spherical cavity of microscopic radius $R_{\text{cav}}$ centered at the target atom:

$$\vec{E}_{\text{loc}} = \vec{E}_0 + \vec{E}_1 + \vec{E}_2 + \vec{E}_3$$
  1. $\vec{E}_0$: External field produced by fixed external charges.
  2. $\vec{E}_1$: Depolarization field produced by the macroscopic polarization charges on the external outer boundaries of the dielectric specimen ($\vec{E}_0 + \vec{E}_1 = \vec{E}$, the macroscopic field).
  3. $\vec{E}_2$: Surface polarization charge field on the spherical cavity boundary. Integrating the bound surface charge density $\sigma_b = - \vec{P} \cdot \hat{n} = P \cos\theta$ over the spherical surface yields:
    $$\vec{E}_2 = \frac{1}{4\pi\varepsilon_0} \int_0^\pi \frac{(P \cos\theta)(\cos\theta)}{R_{\text{cav}}^2} (2\pi R_{\text{cav}}^2 \sin\theta d\theta) \hat{z} = \frac{\vec{P}}{3\varepsilon_0}$$
  4. $\vec{E}_3$: Microscopic field produced by individual dipoles situated inside the cavity. For sites possessing cubic point-group symmetry (or random isotropic liquids), $\vec{E}_3 \equiv 0$.

Summing these terms yields the celebrated Lorentz Local Field Relation:

$$\vec{E}_{\text{loc}} = \vec{E} + \frac{\vec{P}}{3\varepsilon_0}$$

§7.2 Atomic Polarizabilities & The Clausius-Mossotti Relation

1. Mechanisms of Microscopic Dielectric Polarization

The total microscopic electric dipole moment $\vec{p}$ induced in an individual atom or molecule is directly proportional to the local field: $\vec{p} = \alpha \vec{E}_{\text{loc}}$, where $\alpha$ is the total polarizability, composed of three fundamental mechanisms:

  1. Electronic Polarizability ($\alpha_e$): Displacement of the negative valence electron cloud relative to the positive nucleus ($\sim 10^{-40}\text{ C}\cdot\text{m}^2/\text{V}$). Resonant at optical frequencies ($\sim 10^{15}\text{ Hz}$).
  2. Ionic (Atomic) Polarizability ($\alpha_i$): Relative displacement of positive and negative ions in ionic crystals ($\sim 10^{-39}\text{ C}\cdot\text{m}^2/\text{V}$). Resonant at infrared phonon frequencies ($\sim 10^{13}\text{ Hz}$).
  3. Dipolar (Orientation) Polarizability ($\alpha_d$): Thermal realignment of permanent molecular electric dipoles $\vec{p}_0$ against thermal randomization. Described by the Langevin-Debye formula:
    $$\alpha_d = \frac{p_0^2}{3 k_B T}$$

    Active at radio and microwave frequencies ($\sim 10^9 - 10^{11}\text{ Hz}$), vanishing at higher frequencies.

The total polarizability is: $\alpha = \alpha_e + \alpha_i + \frac{p_0^2}{3 k_B T}$.

2. Derivation of the Clausius-Mossotti Relation

For a non-polar dielectric with number density $N$ atoms per unit volume, the macroscopic polarization is $\vec{P} = N \vec{p} = N \alpha \vec{E}_{\text{loc}}$. Substituting the Lorentz local field $\vec{E}_{\text{loc}} = \vec{E} + \frac{\vec{P}}{3\varepsilon_0}$:

$$\vec{P} = N \alpha \left( \vec{E} + \frac{\vec{P}}{3\varepsilon_0} \right) \implies \vec{P} \left( 1 - \frac{N\alpha}{3\varepsilon_0} \right) = N \alpha \vec{E}$$

Recalling the macroscopic definition $\vec{P} = \varepsilon_0 (\varepsilon_r - 1) \vec{E}$, we equate:

$$\varepsilon_0 (\varepsilon_r - 1) = \frac{N \alpha}{1 - \frac{N\alpha}{3\varepsilon_0}} \implies \frac{\varepsilon_r - 1}{\varepsilon_r + 2} = \frac{N \alpha}{3 \varepsilon_0}$$

This is the Clausius-Mossotti Relation. It links a purely macroscopic, experimentally measurable property—the relative dielectric constant $\varepsilon_r$—directly to the microscopic atomic polarizability $\alpha$ and atomic density $N$. At optical frequencies where $\varepsilon_r = n^2$ ($n$ is the refractive index), it is called the Lorentz-Lorenz equation.

§7.3 AC Dielectric Response, Debye Relaxation & Dielectric Loss

1. Complex Dielectric Function $\tilde{\varepsilon}(\omega)$

Under an alternating sinusoidal electric field $\vec{E}(t) = \vec{E}_0 e^{-i\omega t}$, dipolar and ionic reorientations exhibit a phase lag relative to the driving field due to internal friction and damping. The dielectric response becomes a complex function of frequency:

$$\tilde{\varepsilon}(\omega) = \varepsilon_1(\omega) - i \varepsilon_2(\omega)$$
  • Real Part $\varepsilon_1(\omega)$: Quantifies reversible electrostatic energy storage (capacitance).
  • Imaginary Part $\varepsilon_2(\omega)$: Quantifies irreversible energy dissipation and dielectric heating loss.

The Loss Tangent (or dissipation factor) is defined as:

$$\tan\delta = \frac{\varepsilon_2(\omega)}{\varepsilon_1(\omega)}$$

2. The Debye Relaxation Equations

Peter Debye modeled the time-dependent relaxation of orientation polarization following the removal of a field as an exponential decay: $\frac{d\vec{P}_d}{dt} = - \frac{\vec{P}_d}{\tau_D}$, where $\tau_D$ is the Debye relaxation time. Solving under harmonic driving yields:

$$\tilde{\varepsilon}(\omega) = \varepsilon_\infty + \frac{\varepsilon_s - \varepsilon_\infty}{1 - i \omega \tau_D}$$

Separating into real and imaginary components yields the Debye equations:

$$\varepsilon_1(\omega) = \varepsilon_\infty + \frac{\varepsilon_s - \varepsilon_\infty}{1 + \omega^2 \tau_D^2}, \quad \varepsilon_2(\omega) = \frac{(\varepsilon_s - \varepsilon_\infty) \omega \tau_D}{1 + \omega^2 \tau_D^2}$$

where $\varepsilon_s$ is the static low-frequency dielectric constant ($\omega \tau_D \ll 1$) and $\varepsilon_\infty$ is the high-frequency electronic limit ($\omega \tau_D \gg 1$). The dielectric absorption loss $\varepsilon_2(\omega)$ reaches a sharp maximum at the resonance condition $\omega = 1/\tau_D$.

§7.4 Plasmons, Plasma Frequency & Thomas-Fermi Screening

1. Plasma Oscillations of the Free Electron Gas

Consider a free electron gas of density $n$ in a metal. If the entire electron cloud is displaced collectively as a rigid slab by a distance $u$ along $x$ relative to the positive ion core background, a surface charge density $\sigma = \pm n e u$ accumulates on the opposing ends of the slab. This generates an internal restoring electric field:

$$E = \frac{\sigma}{\varepsilon_0} = \frac{n e u}{\varepsilon_0}$$

The classical equation of motion for each electron in the displaced cloud is:

$$m \frac{d^2 u}{dt^2} = - e E = - \frac{n e^2}{\varepsilon_0} u \implies \frac{d^2 u}{dt^2} + \left(\frac{n e^2}{\varepsilon_0 m}\right) u = 0$$

This is a simple harmonic oscillator. Collective longitudinal density oscillations of the electron gas occur at the Plasma Frequency $\omega_p$:

$$\omega_p = \sqrt{\frac{n e^2}{\varepsilon_0 m}}$$

A quantum of plasma oscillation is a quasiparticle called a plasmon, carrying energy $\hbar \omega_p \sim 5 - 20\text{ eV}$ in typical metals.

2. Dielectric Function of a Metal & Ultraviolet Transparency

Neglecting damping at optical frequencies ($\omega \tau \gg 1$), the Drude dielectric function of a metal reduces to:

$$\varepsilon(\omega) = 1 - \frac{\omega_p^2}{\omega^2}$$
  • For $\omega < \omega_p$: $\varepsilon(\omega) < 0$. The refractive index $n = \sqrt{\varepsilon}$ is purely imaginary ($n = i \kappa$), causing total reflection ($R \approx 100\%$). Metals act as mirrors for visible light.
  • For $\omega > \omega_p$: $\varepsilon(\omega) > 0$. The refractive index is real and positive. Electromagnetic waves propagate freely through the metal without reflection: the metal undergoes an ultraviolet transparency transition.

3. Thomas-Fermi Electrostatic Screening

When a static test charge $Q$ is embedded in a degenerate electron gas, conduction electrons rearrange to screen its Coulomb potential. In the Thomas-Fermi approximation, the bare Coulomb potential $V_0(r) = \frac{Q}{4\pi\varepsilon_0 r}$ is screened exponentially:

$$V(r) = \frac{Q}{4\pi\varepsilon_0 r} e^{- k_{TF} r} = \frac{Q}{4\pi\varepsilon_0 r} e^{- r / \lambda_{TF}}$$

where the Thomas-Fermi screening wavevector $k_{TF}$ and screening length $\lambda_{TF}$ are given by:

$$k_{TF}^2 = \frac{e^2}{\varepsilon_0} g(E_F) = \frac{3 n e^2}{2 \varepsilon_0 E_F} \implies \lambda_{TF} = \frac{1}{k_{TF}} = \sqrt{\frac{2\varepsilon_0 E_F}{3 n e^2}} \sim 0.5 - 1.0\text{ \AA}$$

In typical metals, Coulomb interactions are completely screened within an atomic radius, explaining why electron-electron repulsion can often be neglected.

§7.5 Ferroelectricity, Piezoelectricity & Kramers-Kronig Relations

1. Ferroelectricity & The Soft Phonon Mode

A ferroelectric crystal (e.g., Barium Titanate $\text{BaTiO}_3$, $\text{PbTiO}_3$) exhibits a spontaneous, reversible electric polarization $\vec{P}_s$ in the absence of an applied electric field below a critical Curie temperature $T_C$. Above $T_C$, the crystal undergoes a structural phase transition to a paraelectric phase governed by the Curie-Weiss law:

$$\varepsilon_r = \varepsilon_0 + \frac{C}{T - T_C} \quad (T > T_C)$$

In the Lyddane-Sachs-Teller (LST) dynamical theory, the divergence of $\varepsilon_r(0)$ as $T \to T_C$ is driven by the condensation of a transverse optical phonon frequency, the soft mode:

$$\omega_{TO}^2(T) = \gamma (T - T_C) \to 0 \quad \text{as } T \to T_C^+$$

2. Piezoelectricity & Pyroelectricity

  • Piezoelectric Effect: Induction of electric polarization $\vec{P}$ proportional to applied mechanical stress $\sigma$ ($P_i = d_{ijk} \sigma_{jk}$), and conversely mechanical strain upon application of an electric field. Occurs exclusively in non-centrosymmetric crystal classes ($20$ of the $32$ point groups). Canonical example: Quartz ($\text{SiO}_2$), PZT.
  • Pyroelectric Effect: Spontaneous polarization changes as a function of temperature: $\Delta P_i = p_i \Delta T$. All pyroelectric materials are piezoelectric, but not all piezoelectrics are pyroelectric.

3. The Kramers-Kronig Dispersion Relations

By the fundamental principle of causality (an electric displacement response $\vec{D}(t)$ cannot precede the applied electric field $\vec{E}(t)$), Cauchy's residue theorem applied in the complex frequency plane establishes the Kramers-Kronig relations connecting the real and imaginary parts of the dielectric function:

$$\varepsilon_1(\omega) - 1 = \frac{2}{\pi} \mathcal{P} \int_0^\infty \frac{\omega' \varepsilon_2(\omega')}{\omega'^2 - \omega^2} d\omega'$$
$$\varepsilon_2(\omega) = - \frac{2\omega}{\pi} \mathcal{P} \int_0^\infty \frac{\varepsilon_1(\omega') - 1}{\omega'^2 - \omega^2} d\omega'$$

where $\mathcal{P}$ denotes Cauchy's principal value. Measuring the optical absorption spectrum $\varepsilon_2(\omega)$ across all frequencies allows exact determination of the real dielectric permittivity $\varepsilon_1(\omega)$ and refractive index without adjustable parameters.

Solved Problem Example 7.1: Clausius-Mossotti Polarizability Calculation for Solid Germanium

Solid Germanium crystallizes in the diamond structure with lattice constant $a = 5.658 \text{ Å}$ and measured static dielectric constant $\varepsilon_r = 16.0$. (a) Calculate the atomic number density $N$ of Germanium in atoms per $\text{m}^3$. (b) Using the Clausius-Mossotti relation, determine the electronic polarizability $\alpha$ of a Germanium atom in SI units ($\text{C}\cdot\text{m}^2/\text{V}$) and in volume units ($\text{Å}^3$).

Step 1: Calculate the Atomic Number Density N
$$N = \frac{8}{a^3} = \frac{8}{(5.658 \times 10^{-10}\text{ m})^3} = \frac{8}{1.8113 \times 10^{-28}\text{ m}^3} \approx 4.417 \times 10^{28}\text{ atoms/m}^3$$

Diamond cubic conventional unit cell contains 8 atoms.

Step 2: Solve the Clausius-Mossotti Equation for Polarizability alpha
$$\frac{\varepsilon_r - 1}{\varepsilon_r + 2} = \frac{N \alpha}{3\varepsilon_0} \implies \alpha = \frac{3\varepsilon_0}{N} \left( \frac{\varepsilon_r - 1}{\varepsilon_r + 2} \right)$$

Isolate the atomic polarizability alpha.

Step 3: Evaluate the Dimensionless Dielectric Factor
$$\frac{\varepsilon_r - 1}{\varepsilon_r + 2} = \frac{16.0 - 1}{16.0 + 2} = \frac{15.0}{18.0} = \frac{5}{6} \approx 0.8333$$

Compute (epsilon_r - 1) / (epsilon_r + 2).

Step 4: Compute Polarizability in SI Units
$$\alpha = \frac{3 \times (8.8542 \times 10^{-12}\text{ F/m})}{4.417 \times 10^{28}\text{ m}^{-3}} \times \left(\frac{5}{6}\right) = \frac{2.6563 \times 10^{-11} \times 0.8333}{4.417 \times 10^{28}} \approx 5.011 \times 10^{-40}\text{ C}\cdot\text{m}^2/\text{V}$$

Evaluate the numerical value in SI units.

Step 5: Convert Polarizability to Polarizability Volume alpha'
$$\alpha' = \frac{\alpha}{4\pi\varepsilon_0} = \frac{5.011 \times 10^{-40}\text{ C}\cdot\text{m}^2/\text{V}}{4\pi \times (8.8542 \times 10^{-12}\text{ F/m})} \approx 4.504 \times 10^{-30}\text{ m}^3 = 4.504\text{ \AA}^3$$

Compute the polarizability volume alpha' = alpha / (4 pi epsilon_0).

Final Answer & Physical Insight

N = 4.42 \times 10^{28} \text{ m}^{-3}, \quad \alpha = 5.01 \times 10^{-40} \text{ C}\cdot\text{m}^2/\text{V}, \quad \alpha' = 4.50 \text{ Å}^3

Solved Problem Example 7.2: Plasma Frequency and Ultraviolet Transmission Edge of Aluminum

Aluminum is a trivalent metal ($Z_{\text{val}} = 3$) with atomic mass $M = 26.98 \text{ g/mol}$ and density $\rho = 2.70 \text{ g/cm}^3$. (a) Calculate the free electron density $n$. (b) Determine the plasma frequency $\omega_p$, the plasmon energy $\hbar \omega_p$ in $\text{eV}$, and the critical ultraviolet transparency threshold wavelength $\lambda_p$.

Step 1: Calculate the Conduction Electron Density n
$$n = Z_{\text{val}} \frac{\rho N_A}{M} = 3 \times \frac{(2.70 \times 10^6\text{ g/m}^3)(6.022 \times 10^{23}\text{ mol}^{-1})}{26.98\text{ g/mol}} = 3 \times (6.026 \times 10^{28}) \approx 1.808 \times 10^{29}\text{ electrons/m}^3$$

Each aluminum atom contributes 3 conduction electrons.

Step 2: Calculate the Angular Plasma Frequency omega_p
$$\omega_p = \sqrt{\frac{n e^2}{\varepsilon_0 m}} = \sqrt{\frac{(1.808 \times 10^{29})(1.6022 \times 10^{-19})^2}{(8.8542 \times 10^{-12})(9.109 \times 10^{-31})}} = \sqrt{\frac{4.641 \times 10^{-9}}{8.065 \times 10^{-42}}} = \sqrt{5.754 \times 10^{32}} \approx 2.399 \times 10^{16}\text{ rad/s}$$

Evaluate the plasma frequency formula.

Step 3: Calculate the Plasmon Energy in eV
$$E_p = \hbar \omega_p = \frac{(1.0546 \times 10^{-34}\text{ J}\cdot\text{s})(2.399 \times 10^{16}\text{ rad/s})}{1.6022 \times 10^{-19}\text{ J/eV}} \approx \frac{2.530 \times 10^{-18}\text{ J}}{1.6022 \times 10^{-19}\text{ J/eV}} \approx 15.79\text{ eV}$$

Convert plasmon energy to electron-volts.

Step 4: Calculate the Critical Transparency Wavelength lambda_p
$$\lambda_p = \frac{2\pi c}{\omega_p} = \frac{2\pi \times (2.9979 \times 10^8\text{ m/s})}{2.399 \times 10^{16}\text{ s}^{-1}} \approx 7.852 \times 10^{-8}\text{ m} = 78.5\text{ nm}$$

Light with wavelength shorter than 78.5 nm (deep vacuum ultraviolet) passes freely through aluminum.

Final Answer & Physical Insight

\omega_p = 2.40 \times 10^{16} \text{ rad/s}, \quad \hbar\omega_p = 15.79 \text{ eV}, \quad \lambda_p = 78.5 \text{ nm}

Solved Problem Example 7.3: Thomas-Fermi Screening Length for Conduction Electrons in Copper

Copper has a conduction electron density $n = 8.49 \times 10^{28} \text{ m}^{-3}$ and Fermi energy $E_F = 7.04 \text{ eV} = 1.128 \times 10^{-18} \text{ J}$. (a) Calculate the Thomas-Fermi screening wavevector $k_{\text{TF}}$. (b) Determine the Thomas-Fermi screening length $\lambda_{\text{TF}}$ in Angstroms, and compare it with the interatomic spacing $a = 3.615 \text{ Å}$.

Step 1: State the Thomas-Fermi Screening Formula
$$k_{\text{TF}} = \sqrt{\frac{3 n e^2}{2 \varepsilon_0 E_F}}$$

Use the degenerate electron gas screening relation.

Step 2: Substitute Physical Constants and Material Parameters
$$k_{\text{TF}}^2 = \frac{3 \times (8.49 \times 10^{28}\text{ m}^{-3}) \times (1.6022 \times 10^{-19}\text{ C})^2}{2 \times (8.8542 \times 10^{-12}\text{ F/m}) \times (1.128 \times 10^{-18}\text{ J})} = \frac{6.538 \times 10^{-9}}{1.9975 \times 10^{-29}} \approx 3.273 \times 10^{20}\text{ m}^{-2}$$

Compute the square of the screening wavevector.

Step 3: Calculate the Screening Wavevector k_TF
$$k_{\text{TF}} = \sqrt{3.273 \times 10^{20}\text{ m}^{-2}} \approx 1.809 \times 10^{10}\text{ m}^{-1} = 1.809\text{ \AA}^{-1}$$

Take the square root.

Step 4: Determine the Screening Length lambda_TF
$$\lambda_{\text{TF}} = \frac{1}{k_{\text{TF}}} = \frac{1}{1.809 \times 10^{10}\text{ m}^{-1}} \approx 5.528 \times 10^{-11}\text{ m} = 0.553\text{ \AA}$$

The screening distance is approximately half an Angstrom, much smaller than the 3.615 Angstrom lattice constant of copper.

Final Answer & Physical Insight

k_{\text{TF}} = 1.81 \times 10^{10} \text{ m}^{-1}, \quad \lambda_{\text{TF}} = 0.553 \text{ Å} \quad (\text{Screening occurs within } 15\% \text{ of the unit cell size})

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