Transport Properties

The Drude model treats conduction electrons as a classical gas scattering off ions with a mean Free time $\tau$.

Under an electric field $\mathbf{E}$The equation of motion:

$m_e\frac{d\mathbf{v}}{dt} = -e\mathbf{E} - \frac{m_e\mathbf{v}}{\tau}$

In steady state ($d\mathbf{v}/dt = 0$): $\mathbf{v}_d = -\frac{e\tau}{m_e}\mathbf{E}$.

The current density: $\mathbf{J} = -ne\mathbf{v}_d = \frac{ne^2\tau}{m_e}\mathbf{E}$.

The Drude conductivity:

$\sigma = \frac{ne^2\tau}{m_e}$

The mean free path: $\ell = v_F\tau$.

Successes: Explains Ohm”s law ($\mathbf{J} = \sigma\mathbf{E}$) and the Wiedemann—Franz law ($\kappa/\sigma = LT$ with $L = \pi^2 k_B^2/(3e^2)$).

Failures: Predicts the wrong temperature dependence ($\rho \propto T$But experiments show $\rho \propto T^5$ at low $T$ for pure metals). Predicts $\gamma_{\mathrm{electron} = \frac{3}{2}nk_B}$ But experiments give $\gamma_{\mathrm{electron} = \frac{\pi^2}{2}nk_B(T/T_F)}$ (much smaller).

The semiclassical distribution function $f(\mathbf{r}, \mathbf{k}, t)$ satisfies:

$\frac{\partial f}{\partial t} + \mathbf{v}_{\mathbf{k}} \cdot \nabla_{\mathbf{r}} f - \frac{e\mathbf{E}}{\hbar}\cdot\nabla_{\mathbf{k}} f = \left(\frac{\partial f}{\partial t}\right)_{\mathrm{coll}}$

In the relaxation time approximation:

$\left(\frac{\partial f}{\partial t}\right)_{\mathrm{coll} = -\frac{f - f_0}{\tau}}$

Where $f_0$ is the equilibrium distribution.

Solution for conductivity. In a uniform electric field with $f = f_0 + f_1$:

$f_1 = e\tau\mathbf{E}\cdot\mathbf{v}_{\mathbf{k}}\frac{\partial f_0}{\partial\varepsilon}$

The conductivity becomes:

$\sigma = \frac{e^2}{3}\int \tau(\varepsilon)\,v^2(\varepsilon)\,g(\varepsilon)\left(-\frac{\partial f_0}{\partial\varepsilon}\right) d\varepsilon$

At low $T$, $-\partial f_0/\partial\varepsilon \approx \delta(\varepsilon - \varepsilon_F)$So only states Near $E_F$ contribute to transport. This explains why impurity scattering dominates at low $T$ (even a small concentration of impurities affects states near $E_F$).

Matthiessen’s rule. When multiple scattering mechanisms act independently, the total resistivity Is approximately additive:

$\rho(T) = \rho_0 + \rho_{\mathrm{ph}(T)}$

Where $\rho_0$ is the residual resistivity (temperature-independent, from impurities and defects) And $\rho_{\mathrm{ph}(T)}$ is the phonon contribution (proportional to $T$ at high $T$ and to $T^5$ At low $T$ via the Bloch—Grüneisen formula). The resistance ratio $RRR = \rho(300\ \mathrm{K})/\rho_0$ Is a measure of sample purity.

Bloch—Grüneisen formula. For electron—phonon scattering in a free electron metal:

$\rho_{\mathrm{ph}(T) \propto \left(\frac{T}{\Theta_D}\right)^5 \int_0^{\Theta_D/T} \frac{x^5}{(e^x - 1)(1 - e^{-x})}\,dx}$

At high $T$ ($T \gt \Theta_D$): $\rho_{\mathrm{ph} \propto T}$ (linear, agreeing with the Drude model). At low $T$ ($T \ll \Theta_D$): $\rho_{\mathrm{ph} \propto T^5}$Consistent with experiment.

The thermal conductivity of electrons:

$\kappa_e = \frac{1}{3}c_e v_F \ell_e$

Where $c_e = \frac{\pi^2}{2}nk_B(T/T_F)$ is the electronic specific heat. The phonon contribution:

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

The total thermal conductivity: $\kappa = \kappa_e + \kappa_{\mathrm{ph}}$.

When a magnetic field $\mathbf{B} = B\hat{\mathbf{z}}$ is applied perpendicular to a current $\mathbf{J} = J_x\hat{\mathbf{x}}$A transverse electric field develops:

$E_y = R_H J_x B$

The Hall coefficient: $R_H = -1/(ne)$ for a single carrier type.

The Hall angle: $\theta_H = \arctan(E_y/E_x) = \omega_c\tau$ where $\omega_c = eB/m^*$ is the Cyclotron frequency.

Near a band extremum, the energy can be expanded:

$\varepsilon(\mathbf{k}) = \varepsilon_0 + \frac{\hbar^2}{2}\sum_{ij}(m^{-1})_{ij}k_i k_j$

The effective mass tensor $(m^{-1})_{ij} = \frac{1}{\hbar^2}\frac{\partial^2 \varepsilon}{\partial k_i \partial k_j}$ Determines the response to external fields. For isotropic bands, $m^* = \hbar^2/(d^2\varepsilon/dk^2)$.

A large effective mass means a flat band (small group velocity). A small effective mass means a Steep band (high mobility).

Problem. A copper wire has residual resistivity $\rho_0 = 2 \times 10^{-10}\ \Omega\cdot$m from impurity scattering. At 300 K, the phonon contribution is $\rho_{\mathrm{ph}} = 1.7 \times 10^{-8}\ \Omega\cdot$m. What is the total resistivity and the resistance ratio RRR?

Solution. By Matthiessen’s rule: