Word Count : 442

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We are given: ### Part A: Energy Dispersion Relation $$ \mathcal{E}(\mathbf{k}) = \frac{\hbar^2}{2} \left[ \frac{k_x^2 + k_y^2}{m_1} + \frac{k_z^2}{m_2} \right] $$ And we're told: $$ m_1 = 3m_2 $$ --- ### 1. Inverse Mass Tensor Calculation The energy dispersion relation is quadratic in $\mathbf{k}$, so we can compute the inverse mass tensor $\left[ \frac{1}{m^*_{ij}} \right]$ using the second derivative of $\mathcal{E}(\mathbf{k})$: $$ \left[ \frac{1}{m^*_{ij}} \right] = \frac{1}{\hbar^2} \cdot \frac{\partial^2 \mathcal{E}}{\partial k_i \partial k_j} $$ Since $\mathcal{E}(\mathbf{k})$ is separable and diagonal in $k_x, k_y, k_z$, the mixed derivatives are zero (i.e., off-diagonal terms are zero). Let's compute the diagonal elements: * $\frac{\partial^2 \mathcal{E}}{\partial k_x^2} = \frac{\hbar^2}{2} \cdot \frac{2}{m_1} = \frac{\hbar^2}{m_1}$ * __________ _________ ______ ____ _____ __________ ____ _____ ____ ____.
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