Physics / Condensed Matter Solid State Physics I 100% Free Open Access
Chapter 5 • Theory & Derivations

Free Electron Theory of Metals & Fermi Surfaces

Comprehensive electronic transport in metals: classical Drude phenomenology, Sommerfeld quantum free Fermi gas, 1D/2D/3D density of states, Fermi-Dirac statistics, Fermi energy and velocity, linear electronic heat capacity, Pauli paramagnetism, Hall effect, Wiedemann-Franz law, and Matthiessen's rule.

§5.1 Drude Classical Phenomenology of Electrical & Thermal Conduction

1. Postulates of the Drude Model (1900)

Paul Drude treated valence electrons in a metal as an ideal classical gas of non-interacting point particles of mass $m$ and charge $-e$, moving through a static background of positive ion cores:

  1. Independent & Free Electron Approximation: Electron-electron and electron-ion interactions are neglected between collisions.
  2. Relaxation-Time Approximation: Collisions are instantaneous, random events occurring with an average probability per unit time $1/\tau$, where $\tau$ is the relaxation time (mean free time between collisions).
  3. Thermal Equilibrium: Electrons emerge from each collision in thermal equilibrium with the local lattice temperature, with zero average drift velocity.

2. Derivation of Ohm's Law and DC Electrical Conductivity

Under an applied macroscopic electric field $\vec{E}$, the equation of motion for the average drift velocity $\vec{v}_d$ of an electron is:

$$m \frac{d\vec{v}_d}{dt} = - e \vec{E} - \frac{m \vec{v}_d}{\tau}$$

In steady state ($d\vec{v}_d/dt = 0$):

$$\vec{v}_d = - \frac{e \tau}{m} \vec{E}$$

The macroscopic electric current density $\vec{J}$ carried by electron density $n$ is:

$$\vec{J} = - n e \vec{v}_d = \left( \frac{n e^2 \tau}{m} \right) \vec{E} = \sigma_0 \vec{E}$$

This derives microscopic Ohm's Law, with DC electrical conductivity $\sigma_0$ given by the Drude formula:

$$\sigma_0 = \frac{n e^2 \tau}{m} = n e \mu$$

where $\mu = e\tau/m$ is the electron drift mobility.

3. The Classical Wiedemann-Franz Law & Failure of Drude Model

Treating electrons as a classical Maxwell-Boltzmann gas with thermal conductivity $\kappa = \frac{1}{3} C_v v_{\text{th}} \ell$ and heat capacity $C_v = \frac{3}{2} n k_B$:

$$\frac{\kappa}{\sigma T} = \frac{3}{2} \left(\frac{k_B}{e}\right)^2 \approx 1.11 \times 10^{-8}\text{ W}\cdot\Omega\cdot\text{K}^{-2}$$

While this qualitatively explained the empirical Wiedemann-Franz law, Drude's classical model suffered from a catastrophic failure: it predicted that the electronic heat capacity should contribute $\frac{3}{2} R$ per mole, which was completely absent in room-temperature measurements.

§5.2 Sommerfeld Quantum Free Fermi Gas: Density of States in 1D, 2D & 3D

1. Schrödinger Equation in a 3D Potential Well

Arnold Sommerfeld (1928) resolved the Drude paradox by applying quantum mechanics and Fermi-Dirac statistics to the valence electrons. Consider $N$ free electrons confined within a box of volume $V = L_x L_y L_z$ with periodic Born-von Kármán boundary conditions:

$$- \frac{\hbar^2}{2m} \nabla^2 \psi(\vec{r}) = E \psi(\vec{r}) \implies \psi_{\vec{k}}(\vec{r}) = \frac{1}{\sqrt{V}} e^{i \vec{k}\cdot\vec{r}}$$

The energy eigenvalues are continuous parabolic free-particle dispersions:

$$E(\vec{k}) = \frac{\hbar^2 k^2}{2m} = \frac{\hbar^2}{2m}(k_x^2 + k_y^2 + k_z^2)$$

where allowed wavevectors form a uniform grid in reciprocal $\vec{k}$-space: $k_i = \frac{2\pi n_i}{L_i}$ ($n_i \in \mathbb{Z}$). Each allowed state occupies a volume in $k$-space of $\Delta k^3 = \frac{(2\pi)^3}{V}$.

2. Derivation of the Density of States $g(E)$ in 3D

Taking into account electron spin degeneracy ($g_s = 2$):

$$N(k) = 2 \times \frac{\frac{4}{3}\pi k^3}{(2\pi)^3 / V} = \frac{V}{3\pi^2} k^3$$

Substituting $k = \left(\frac{2mE}{\hbar^2}\right)^{1/2}$:

$$N(E) = \frac{V}{3\pi^2} \left( \frac{2mE}{\hbar^2} \right)^{3/2}$$

The Density of States (DOS) $g(E) = \frac{dN}{dE}$ per unit volume in three dimensions is:

$$g_{\text{3D}}(E) = \frac{1}{V} \frac{dN}{dE} = \frac{1}{2\pi^2} \left(\frac{2m}{\hbar^2}\right)^{3/2} E^{1/2} \propto \sqrt{E}$$

3. Dimensionality Comparison of Density of States

  • 1D (Quantum Wire): $g_{\text{1D}}(E) = \frac{\sqrt{2m}}{\pi \hbar} E^{-1/2} \propto E^{-1/2}$ (Van Hove singularity).
  • 2D (Quantum Well): $g_{\text{2D}}(E) = \frac{m}{\pi \hbar^2} = \text{constant}$ (energy-independent step functions).
  • 3D (Bulk Metal): $g_{\text{3D}}(E) = \frac{m}{\pi^2 \hbar^3} \sqrt{2mE} \propto E^{1/2}$.

§5.3 Fermi-Dirac Statistics, Fermi Energy & Quantum Degeneracy

1. The Fermi-Dirac Distribution Function

Because electrons are indistinguishable fermions obeying the Pauli exclusion principle, the probability of an electronic state of energy $E$ being occupied at absolute temperature $T$ is given by the Fermi-Dirac distribution:

$$f(E) = \frac{1}{e^{(E - \mu)/k_B T} + 1}$$

where $\mu(T)$ is the chemical potential. At $T = 0\text{ K}$, the chemical potential is defined as the Fermi Energy $E_F \equiv \mu(0)$. The distribution becomes an exact step function:

$$f(E) = \begin{cases} 1, & E < E_F \\ 0, & E > E_F \end{cases}$$

2. Calculation of the Fermi Energy $E_F$ and Fermi Surface

At $T = 0\text{ K}$, $N$ electrons fill all available states in $k$-space up to a maximum radius, the Fermi wavevector $k_F$, enclosing a sphere known as the Fermi Sphere:

$$n = \frac{N}{V} = \frac{1}{V} \int_0^{E_F} g(E) dE = \frac{k_F^3}{3\pi^2} \implies k_F = (3\pi^2 n)^{1/3}$$

The fundamental ground-state parameters of the free Fermi gas are:

  • Fermi Energy:
    $$E_F = \frac{\hbar^2 k_F^2}{2m} = \frac{\hbar^2}{2m} (3\pi^2 n)^{2/3}$$
  • Fermi Velocity:
    $$v_F = \frac{\hbar k_F}{m} = \frac{\hbar}{m} (3\pi^2 n)^{1/3}$$
  • Fermi Temperature:
    $$T_F = \frac{E_F}{k_B} \sim 10^4 - 10^5\text{ K}$$

Because room temperature ($T \approx 300\text{ K}$) satisfies $T \ll T_F$, valence electrons in typical metals form a highly degenerate Fermi gas.

§5.4 Electronic Heat Capacity & Pauli Paramagnetism

1. Quantum Suppression of Electronic Heat Capacity

When a metal is heated from $T = 0\text{ K}$ to temperature $T$, the Pauli exclusion principle dictates that electrons with energies deep below the Fermi surface ($E < E_F - k_BT$) cannot absorb thermal energy, because all states within $\sim k_BT$ above them are already occupied. Only electrons within a narrow thermal layer of width $\sim k_BT$ around $E_F$ can undergo transitions.

The fraction of thermally active electrons is approximately $\frac{k_B T}{E_F} = \frac{T}{T_F} \sim 10^{-2}$. Each of these excited electrons acquires thermal energy $\sim k_B T$, yielding an excess internal energy:

$$\Delta U_{el} \approx \left( N \frac{T}{T_F} \right) (k_B T) = N k_B \frac{T^2}{T_F}$$

Differentiating with respect to $T$ yields a heat capacity linear in temperature:

$$C_{el} = \frac{dU_{el}}{dT} \approx 2 N k_B \frac{T}{T_F} \propto T$$

2. Rigorous Sommerfeld Expansion for $C_{el}$

Performing a rigorous Sommerfeld expansion of the internal energy integral $U(T) = \int_0^\infty E g(E) f(E) dE$ leads to the exact theoretical formula for the electronic heat capacity:

$$C_{el} = \frac{\pi^2}{3} g(E_F) k_B^2 T = \frac{\pi^2}{2} N k_B \left(\frac{T}{T_F}\right) = \gamma T$$

where $\gamma = \frac{\pi^2}{3} g(E_F) k_B^2$ is the Sommerfeld coefficient.

At liquid helium temperatures ($T < 10\text{ K}$), the total heat capacity of a normal non-magnetic metal is the sum of electronic and phononic (Debye) contributions:

$$C_V = C_{el} + C_{\text{ph}} = \gamma T + A T^3 \implies \frac{C_V}{T} = \gamma + A T^2$$

Plotting $C_V/T$ against $T^2$ yields a straight line whose $y$-intercept is $\gamma$ and whose slope determines the Debye temperature $\Theta_D$.

3. Pauli Paramagnetism of Conduction Electrons

In an external magnetic field $B$, electron spin magnetic moments $\mu_B$ align parallel or antiparallel to the field, shifting the up- and down-spin sub-bands by $\mp \mu_B B$. At $T = 0\text{ K}$, electrons near $E_F$ flip their spins into the lower energy sub-band until the Fermi levels equalize:

$$\Delta N = \frac{1}{2} g(E_F) (\mu_B B) - \left(-\frac{1}{2} g(E_F) \mu_B B\right) = g(E_F) \mu_B B$$

The resulting net magnetic magnetization is $M = \Delta N \mu_B = g(E_F) \mu_B^2 B$. The Pauli paramagnetic susceptibility is:

$$\chi_{\text{Pauli}} = \frac{\mu_0 M}{B} = \mu_0 \mu_B^2 g(E_F) = \frac{3 \mu_0 n \mu_B^2}{2 E_F}$$

Unlike classical Curie paramagnetism ($\chi \propto 1/T$), Pauli paramagnetism is completely temperature-independent, in perfect agreement with experimental data for alkali and noble metals.

§5.5 The Hall Effect, Wiedemann-Franz Law & Matthiessen's Rule

1. The Hall Effect & Hall Coefficient $R_H$

When an electric current density $J_x$ flows along the $x$-direction of a conductor immersed in a transverse magnetic field $B_z$ in the $z$-direction, the magnetic Lorentz force $\vec{F}_B = q (\vec{v}_d \times \vec{B})$ deflects charge carriers along the $y$-direction:

$$F_{B,y} = q (v_{d,x} B_z) = - e v_{d,x} B_z$$

Charge accumulates on the lateral boundaries, generating a transverse electrostatic Hall electric field $E_y$. In steady state, the transverse electrostatic force exactly balances the magnetic Lorentz force ($F_y = 0$):

$$q E_y + q v_{d,x} B_z = 0 \implies E_y = - v_{d,x} B_z$$

Since $J_x = n q v_{d,x} \implies v_{d,x} = \frac{J_x}{n q}$:

$$E_y = - \frac{J_x B_z}{n q} = R_H J_x B_z$$

where $R_H$ is the Hall Coefficient:

$$R_H = \frac{E_y}{J_x B_z} = \frac{1}{n q} = - \frac{1}{n e} \quad (\text{for electrons})$$

The Hall effect provides an unambiguous experimental measurement of both the sign of the charge carriers (negative for electrons, positive for holes) and the carrier concentration $n$.

2. Quantum Wiedemann-Franz Law & The Lorenz Number

Applying Sommerfeld's degenerate Fermi gas theory to electronic thermal conduction ($\kappa = \frac{1}{3} C_{el} v_F^2 \tau$ with $C_{el} = \frac{\pi^2}{3} g(E_F) k_B^2 T$) and electrical conductivity ($\sigma = \frac{n e^2 \tau}{m}$) yields the quantum Wiedemann-Franz law:

$$\frac{\kappa}{\sigma T} = \frac{\pi^2}{3} \left(\frac{k_B}{e}\right)^2 = L \approx 2.443 \times 10^{-8}\text{ W}\cdot\Omega\cdot\text{K}^{-2}$$

where $L = \frac{\pi^2}{3}(k_B/e)^2$ is the universal Lorenz number, completely independent of carrier density, mass, and material parameters.

3. Matthiessen's Rule for Electrical Resistivity

Electrons in a real metal are scattered by both static structural crystal defects/impurities and dynamic thermal lattice vibrations (phonons). By Matthiessen's Rule, independent scattering rates add linearly:

$$\frac{1}{\tau_{\text{tot}}} = \frac{1}{\tau_{\text{impurity}}} + \frac{1}{\tau_{\text{phonon}}(T)}$$

Consequently, the total electrical resistivity $\rho = m / (ne^2\tau)$ separates into temperature-independent and temperature-dependent terms:

$$\rho(T) = \rho_{\text{residual}} + \rho_{\text{ideal}}(T)$$

where $\rho_{\text{residual}}$ is determined by impurity concentration, and $\rho_{\text{ideal}}(T) \propto T^5$ at low temperatures (Bloch-Grüneisen law) and $\rho_{\text{ideal}}(T) \propto T$ at high temperatures ($T > \Theta_D$).

Solved Problem Example 5.1: Fermi Energy, Fermi Velocity and Fermi Temperature of Copper

Copper is a monovalent metal (one conduction electron per atom) with atomic mass $M = 63.546 \text{ g/mol}$, density $\rho = 8.96 \text{ g/cm}^3$, and electron mass $m = 9.109 \times 10^{-31} \text{ kg}$. (a) Calculate the conduction electron number density $n$. (b) Determine the Fermi wavevector $k_F$, the Fermi energy $E_F$ in $\text{eV}$, the Fermi velocity $v_F$, and the Fermi temperature $T_F$.

Step 1: Calculate the Electron Density n
$$n = \frac{\rho N_A}{M} = \frac{(8.96 \times 10^6\text{ g/m}^3)(6.022 \times 10^{23}\text{ mol}^{-1})}{63.546\text{ g/mol}} \approx 8.492 \times 10^{28}\text{ electrons/m}^3$$

Compute conduction electron density from mass density and molar mass.

Step 2: Calculate the Fermi Wavevector k_F
$$k_F = (3\pi^2 n)^{1/3} = [3\pi^2 (8.492 \times 10^{28})]^{1/3} = (2.5144 \times 10^{30})^{1/3} \approx 1.360 \times 10^{10}\text{ m}^{-1} = 1.360\text{ \AA}^{-1}$$

Evaluate the radius of the spherical Fermi surface in reciprocal space.

Step 3: Calculate the Fermi Energy E_F in eV
$$E_F = \frac{\hbar^2 k_F^2}{2m} = \frac{(1.0546 \times 10^{-34}\text{ J}\cdot\text{s})^2 (1.360 \times 10^{10}\text{ m}^{-1})^2}{2(9.109 \times 10^{-31}\text{ kg})} = \frac{2.056 \times 10^{-47}}{1.8218 \times 10^{-30}}\text{ J} \approx 1.1286 \times 10^{-18}\text{ J} \approx 7.04\text{ eV}$$

Convert Joules to electron-volts by dividing by 1.6022 x 10^-19 J/eV.

Step 4: Calculate the Fermi Velocity v_F
$$v_F = \frac{\hbar k_F}{m} = \frac{(1.0546 \times 10^{-34}\text{ J}\cdot\text{s})(1.360 \times 10^{10}\text{ m}^{-1})}{9.109 \times 10^{-31}\text{ kg}} \approx 1.574 \times 10^6\text{ m/s}$$

Compute the velocity of electrons at the Fermi surface.

Step 5: Calculate the Fermi Temperature T_F
$$T_F = \frac{E_F}{k_B} = \frac{1.1286 \times 10^{-18}\text{ J}}{1.3806 \times 10^{-23}\text{ J/K}} \approx 81750\text{ K} \approx 8.18 \times 10^4\text{ K}$$

Compute the degeneracy temperature.

Final Answer & Physical Insight

n = 8.49 \times 10^{28} \text{ m}^{-3}, \quad E_F = 7.04 \text{ eV}, \quad v_F = 1.57 \times 10^6 \text{ m/s}, \quad T_F = 8.18 \times 10^4 \text{ K}

Solved Problem Example 5.2: Sommerfeld Electronic Heat Capacity Coefficient for Silver

Silver has a Fermi energy of $E_F = 5.49 \text{ eV}$ and atomic molar mass $M = 107.87 \text{ g/mol}$. (a) Calculate the theoretical Sommerfeld coefficient $\gamma$ per mole of conduction electrons. (b) Calculate the molar electronic heat capacity $C_{el}$ at liquid helium temperature $T = 4.2 \text{ K}$, and compare it with the classical value $3R/2$.

Step 1: Calculate the Fermi Temperature of Silver
$$T_F = \frac{E_F}{k_B} = \frac{5.49\text{ eV} \times 1.6022 \times 10^{-19}\text{ J/eV}}{1.3806 \times 10^{-23}\text{ J/K}} = \frac{8.796 \times 10^{-19}\text{ J}}{1.3806 \times 10^{-23}\text{ J/K}} \approx 63710\text{ K}$$

Convert Fermi energy to Fermi temperature.

Step 2: Calculate the Sommerfeld Coefficient gamma
$$\gamma = \frac{\pi^2}{2} R \frac{1}{T_F} = \frac{\pi^2}{2} \times (8.314\text{ J/mol}\cdot\text{K}) \times \frac{1}{63710\text{ K}} \approx \frac{41.028}{63710}\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-2} \approx 6.44 \times 10^{-4}\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-2} = 0.644\text{ mJ}\cdot\text{mol}^{-1}\cdot\text{K}^{-2}$$

Evaluate the theoretical Sommerfeld constant per mole.

Step 3: Evaluate Electronic Heat Capacity at T = 4.2 K
$$C_{el}(4.2\text{ K}) = \gamma T = (6.44 \times 10^{-4}\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-2}) \times (4.2\text{ K}) \approx 2.705 \times 10^{-3}\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1} = 2.71\text{ mJ}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$$

Multiply gamma by T = 4.2 K.

Step 4: Compare with Classical Prediction
$$\frac{C_{el}(4.2\text{ K})}{C_{\text{classical}}} = \frac{2.705 \times 10^{-3}\text{ J/mol}\cdot\text{K}}{\frac{3}{2}(8.314\text{ J/mol}\cdot\text{K})} = \frac{2.705 \times 10^{-3}}{12.471} \approx 2.17 \times 10^{-4}$$

The quantum electronic heat capacity is suppressed by nearly four orders of magnitude compared to the classical prediction.

Final Answer & Physical Insight

\gamma = 0.644 \text{ mJ}\cdot\text{mol}^{-1}\cdot\text{K}^{-2}, \quad C_{el}(4.2\text{ K}) = 2.71 \text{ mJ}\cdot\text{mol}^{-1}\cdot\text{K}^{-1} \quad (0.0217\% \text{ of classical } 3R/2)

Solved Problem Example 5.3: Hall Coefficient, Carrier Density and Mobility in Sodium Metal

A rectangular ribbon of sodium metal of thickness $t = 0.10 \text{ mm}$ and width $w = 5.0 \text{ mm}$ carries a longitudinal current $I_x = 10.0 \text{ A}$ in a transverse magnetic field $B_z = 1.20 \text{ T}$. A transverse Hall voltage of $V_H = -2.94 \text{ }μ\text{V}$ is measured across its width. The electrical resistivity of sodium is $\rho = 4.75 \times 10^{-8} \text{ }Ω\cdot\text{m}$. (a) Calculate the Hall coefficient $R_H$. (b) Determine the conduction electron density $n$. (c) Calculate the electron drift mobility $\mu$ and relaxation time $\tau$.

Step 1: Calculate the Hall Coefficient R_H
$$V_H = \frac{R_H I_x B_z}{t} \implies R_H = \frac{V_H t}{I_x B_z} = \frac{(-2.94 \times 10^{-6}\text{ V})(1.0 \times 10^{-4}\text{ m})}{(10.0\text{ A})(1.20\text{ T})} = \frac{-2.94 \times 10^{-10}}{12.0} = - 2.45 \times 10^{-11}\text{ m}^3/\text{C}$$

Solve for the Hall coefficient using the measured Hall voltage, sample thickness, current, and magnetic field.

Step 2: Determine the Conduction Electron Density n
$$n = - \frac{1}{e R_H} = \frac{1}{(1.6022 \times 10^{-19}\text{ C})(2.45 \times 10^{-11}\text{ m}^3/\text{C})} = \frac{1}{3.9254 \times 10^{-30}} \approx 2.548 \times 10^{28}\text{ electrons/m}^3$$

Extract the carrier density from R_H = -1 / (n e).

Step 3: Calculate the Electron Drift Mobility mu
$$\sigma = \frac{1}{\rho} = \frac{1}{4.75 \times 10^{-8}\text{ }\Omega\cdot\text{m}} \approx 2.105 \times 10^7\text{ }\Omega^{-1}\cdot\text{m}^{-1} \implies \mu = \sigma |R_H| = (2.105 \times 10^7)(2.45 \times 10^{-11})\text{ m}^2/\text{V}\cdot\text{s} \approx 5.158 \times 10^{-4}\text{ m}^2/\text{V}\cdot\text{s}$$

Compute mobility using mu = sigma * |R_H|.

Step 4: Calculate the Relaxation Time tau
$$\tau = \frac{m \mu}{e} = \frac{(9.109 \times 10^{-31}\text{ kg})(5.158 \times 10^{-4}\text{ m}^2/\text{V}\cdot\text{s})}{1.6022 \times 10^{-19}\text{ C}} \approx 2.93 \times 10^{-14}\text{ s}$$

Evaluate the electron collision relaxation time.

Final Answer & Physical Insight

R_H = -2.45 \times 10^{-11} \text{ m}^3/\text{C}, \quad n = 2.55 \times 10^{28} \text{ m}^{-3}, \quad \mu = 5.16 \times 10^{-4} \text{ m}^2/\text{V}\cdot\text{s}, \quad \tau = 2.93 \times 10^{-14} \text{ s}

EXAM SUCCESS WORKSHOP

Solved University Examination Problems

Step-by-step mathematical solutions to classic university honors examination questions.