Solutions for Drude model exercises¶

The Hall voltage is measured across the sample width. It is given by

\[ V_H = V(W)-V(0) = -\int_{0}^{W} E_ydy = -E_y W \]

From the steady-state solution of the Drude equation of motion, we obtain $E_y = \rho_{yx}j_x = \frac{-B}{ne}j_x $. Using $j_x = I_x / W$ and $V_H=-E_y W$, we obtain $R_{yx} = V_H/I_x = \frac{B}{ne}$. so it does not depend on the sample geometry.
If the hall resistance and the magnetic field are known, we extract the charge density using $R_{yx} = \frac{B}{ne}$. Because $V_H = \frac{B}{ne}I_x$, a stronger field yields a larger Hall voltage, making it easier to measure. Likewise, a lower charge density yields a larger Hall voltage for a given bias current, making it easier to measure.
The longitudinal resistance is

\[ R_{xx} = \rho_{xx}\frac{L}{W} \]

with $\rho_{xx} = \frac{m_e}{ne^2\tau}$ the longitudinal Drude resistivity. Therefore, knowing the electron density $n$, we can extract the scattering time ($\tau$). We observe that $R_{xx}$ depends on the sample geometry ($L$ and $W$), whereas the Hall resistance $R_{yx}$ does not.

We first find the electron density using $n_e = \frac{ZnN_A}{W}$, where $Z=1$ is the number of free electrons per copper atom, n is the mass density of copper, $N_A$ is Avogadro's constant, and W is the atomic weight of copper. Then, we find the scattering time $\tau = 2.57 \cdot 10^{-14}$ s from the longitudinal Drude resistivity $\rho = \frac{m_e}{n_e e^2\tau}$.
Using $\lambda = \langle v \rangle\tau$, we get $\lambda =3$ nm.
The scattering time $\tau \propto \frac{1}{\sqrt{T}}$, such that we expect $\rho \propto \sqrt{T}$.
For most metals, the measured resistivity scales as $\rho \propto T$, in contrast with the $\rho \propto \sqrt{T}$ predicted by the Drude model. The linear scaling can be understood by assuming that the scattering is caused by phonons, as their number scales linearly with T at high temperature (recall that the high-temperature limit of the Bose-Einstein distribution functions is $n_B = \frac{kT}{\hbar\omega}$ leading to $\rho \propto T$). The inability to explain this linear dependence is a failure of the Drude model.

$\rho_{xx}$ is independent of B and $\rho_{xy} \propto B$
The conductivities are

\[ \sigma_{xx} = \frac{\rho_{xx}}{\rho_{xx}^2 + \rho_{xy}^2} = \frac{mne^2\tau}{m^2+e^2\tau^2 B^2} \]

\[ \sigma_{xy} = \frac{-\rho_{yx}}{\rho_{xx}^2 + \rho_{xy}^2} = \frac{Bne^3 \tau^2}{m^2+e^2 \tau^2 B^2} \]
These equations describe Lorentzian-like functions.
The Hall coefficient is $R_H = -\frac{1}{ne}$. The sign of the Hall coefficient depends on sign of the charge carriers (analyzed further in the next exercise).

The current is the sum of the currents carried by the positive and negative charge carriers:

\[ \mathbf{J} = -n_e e \mathbf{ v_e} + n_h e \mathbf{ v_h} \]
Write down the Drude equation of motion for the positive and negative charge carriers separately and use $ \mu_e = \frac{e\tau_e}{m_e}$ and $ \mu_h = \frac{e\tau_h}{m_h}$
From the assumption, we can ignore the $(\mathbf{ v_i} \times \mathbf{B})_x$ terms. Then combining question 1 and 2 gives the result.
The net current in the y-direction should be zero.
The Hall coefficient is

$$ R_H = \frac{n_h\mu_h^2 - n_e\mu_e^2}{e(n_h\mu_h + n_e \mu_e)^2} $$ We observe that the sign of $R_H$ is determined by the density and the mobilty of the positive and negative charge carriers. Therefore, a measurement of the Hall voltage reveals the dominant charge carrier.