Question 705069

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.