Question 705069

SCQHARD

The band dispersion of electrons in a two-dimensional square lattice (lattice constant $a$ ) is given by, $E(k_x, k_y) = -2(t_x \cos k_x a + t_y \cos k_y a)$ where $t_x, t_y > 0$ . The effective mass tensor $m^* = \begin{pmatrix} m_{xx} & m_{xy} \\ m_{yx} & m_{yy} \end{pmatrix}$ of electrons at $\vec{k} = (\frac{\pi}{a}, \frac{\pi}{a})$ is

(A)

$\begin{pmatrix} 0 & \frac{h^2}{2a^2\sqrt{t_xt_y}} \\ \frac{h^2}{2a^2\sqrt{t_xt_y}} & 0 \end{pmatrix}$

(B)

$\begin{pmatrix} \frac{h^2}{2a^2t_x} & 0 \\ 0 & \frac{h^2}{2a^2t_y} \end{pmatrix}$

(C)

$\begin{pmatrix} -\frac{h^2}{2a^2t_x} & 0 \\ 0 & -\frac{h^2}{2a^2t_y} \end{pmatrix}$

(D)

$\begin{pmatrix} 0 & -\frac{h^2}{2a^2(t_x+t_y)} \\ \frac{h^2}{2a^2(t_x+t_y)} & 0 \end{pmatrix}$

Detailed Solution

The effective mass tensor is given by the inverse of the second derivative of the energy dispersion: $(m^*)_{ij}^{-1} = \frac{1}{\hbar^2} \frac{\partial^2 E}{\partial k_i \partial k_j}$ . Calculating the second partial derivatives at $\vec{k} = (\frac{\pi}{a}, \frac{\pi}{a})$ : $\frac{\partial^2 E}{\partial k_x^2} = 2t_x a^2 \cos(k_x a)$ . At $k_x = \frac{\pi}{a}$ , $\cos(\pi) = -1$ , so $\frac{\partial^2 E}{\partial k_x^2} = -2t_x a^2$ . Similarly, $\frac{\partial^2 E}{\partial k_y^2} = -2t_y a^2$ . The mixed derivatives $\frac{\partial^2 E}{\partial k_x \partial k_y} = 0$ . Thus, $m^*_{xx} = \hbar^2 / (-2t_x a^2) = -\frac{\hbar^2}{2a^2 t_x}$ and $m^*_{yy} = -\frac{\hbar^2}{2a^2 t_y}$ . With $\hbar = h/2\pi$ (noting standard definition conventions in condensed matter textbooks), the result corresponds to option 3.

Question 705069

Detailed Solution

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