# Interpretation of negative density of states and fourth type Van Hove singularity?

If I have the energy of a free electron gas in three dimensions centered at energy $$E_0$$, $$E_n(k)=E_0+\frac{\hbar^2}{2}\left(\frac{k_x^2}{m_x}+\frac{k_y^2}{m_y}+\frac{k_z^2}{m_z}\right),$$ I can prove that the density of states in the Fermi level is $$d_n(E_F)=\frac{3}{2}\frac{n}{E_F-E_0},$$ where $$n$$ is the number of electrons per unit volume. Now, if all the $$m_i$$ are negative, I would have $$E_F and a negative density of states? Is that right? I've read that is called a van Hove singularity of the fourth type. Can anyone help me with a physical interpretation of the phenomenon?

I think you have a sign error; your numerator should be $$(E_{0}-E_{F})$$. Be careful in the derivation to keep track of the sign of $$m$$ and treating the $$\delta$$-function function composition rules carefully.
Then, for $$E_{F} < E_{0}$$, the density of states is finite and positive. For $$E_{F} = E_{0}$$, the DOS is infinite because of the van Hove singularity. For $$E_{F} > E_{0}$$ the DOS is actually zero at the Fermi energy, because the Fermi energy lies above the band (which is sloping downwards because of the negative $$m$$'s), and the formula no longer applies.