Effective Density of States for 2D Effective Mass Tensor

I have derived an effective mass tensor for conduction and valence electrons in a tight binding model. Specifically, for electrons in the conduction band: $$\begin{bmatrix} \hbar^2\left( \frac{\partial^2 E_c}{\partial k_x^2} \right )^{-1} & 0\\ 0 & \hbar^2\left( \frac{\partial^2 E_c}{\partial k_y^2} \right )^{-1} \end{bmatrix}=\begin{bmatrix} \frac{\hbar^2}{2a^2\cos(k_x a)} & 0\\ 0 & \frac{\hbar^2}{2a^2\cos(k_y a)} \end{bmatrix}.$$ And for valence holes $$\begin{bmatrix} \hbar^2\left( \frac{\partial^2 E_v}{\partial k_x^2} \right )^{-1} & 0\\ 0 & \hbar^2\left( \frac{\partial^2 E_v}{\partial k_y^2} \right )^{-1} \end{bmatrix}=\begin{bmatrix} -\frac{\hbar^2}{a^2\cos(k_x a)} & 0\\ 0 & -\frac{\hbar^2}{a^2\cos(k_y a)} \end{bmatrix}.$$

Now I know that in one dimension, the effective DOS is given by \begin{align*} g_C(E)&=\frac{1}{2\pi^2}\left ( \frac{2m_e^*}{\hbar^2}\right )^{3/2}(E-E_c)^{1/2} \\ g_V(E)&=\frac{1}{2\pi^2}\left ( \frac{2m_h^*}{\hbar^2}\right )^{3/2}(E_V-E)^{1/2} \end{align*} for the conduction and valence bands respectively. How does this translate to two dimensions, especially considering that I now have an effective mass tensor?

1 Answer

First, the formula you give for the effective DOS in one dimension is not correct. It is the usual 3D-formula. Your effective mass tensors have the same reciprocal masses in x and y directions. Therefore you have isotropic effective masses m* in the conduction and valence bands. If your isotropic effective masses are constant you should in 2D get an energy independent 2D density of states per unit area D(E)=m*/(πℏ^2).