How to Derive the Density of States in Anisotropic Conduction Bands?

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The discussion focuses on deriving the density of states for anisotropic conduction bands using the provided band dispersion equation. The density of states is expressed as g(ε) = (m^(3/2)/(π^2 h^2)) √(2|ε - ε_c|). Two methods are suggested for the derivation: one involves calculating the total magnitude of k, while the second method uses an integral over a constant energy surface. The challenge lies in evaluating the integral, particularly in spherical coordinates, due to the complexity of the gradient of the energy. The conversation emphasizes the need to express the gradient in terms of ε - ε_c for simplification.
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Homework Statement



have band dispersion

\epsilon = \epsilon_c + \frac{h^2 k_x^2}{2 m_x} + \frac{h^2 k_y^2}{2 m_y} + \frac{h^2 k_z^2}{2 m_z}

Show density of states is

g(\epsilon) = \frac{m^{3/2}}{\pi^2 h^2} \sqrt{2|\epsilon - \epsilon_c|}

Homework Equations



2 \frac{d\vec{k}}{(2\pi)^3} = g(\epsilon) d\epsilon

or

g(\epsilon) = \frac{1}{4 \pi^3} = \int \frac{|d\vec{S}|}{|(grad)_k \epsilon|}
over a constant energy surface where the gradient is always perpendicular to the surface.

The Attempt at a Solution



For the first method I don't know how to solve for total magnitude of k so I can't procede.

Second method I get what appears to be a "not nice" integral, where

|(grad)_k \epsilon| = \sqrt{ \frac{h^4 k_x^2}{m_x^2} + \frac{h^4 k_y^2}{m_y^2} + \frac{h^4 k_z^2}{m_z^2}}

I don't see how to do this nicely in spherical coords over the Fermi surface.
 
Last edited:
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You can rewrite |\nabla_k \epsilon| in terms of \epsilon-\epsilon_c. Then remember that the integral that you have to do is over a constant energy surface.
 

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