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I'm trying to derive the expression for the density of states for a hole, i.e: $$D_h(\epsilon)$$

To be able to this I need to make some assumptions, first of all, all the holes states can be expressed as $$N(K)=\frac{V_{sphere}}{V_k}$$ where $V_{sphere}$ simply is the fermi-sphere in the reciprocal space and $V_k$ is the volume of each state.

I'm from here on assumin that $V_{sphere}=\frac{4\pi k^3}{3}$ and $V_k = \frac {2 \pi}{L}$, this means that: $$N(K)=\frac {k^3 V}{3 \pi ^2}$$ where I've taken into account the degeneracy of the electrons (which basically mean I have to do the same for a hole). So far so good. Now letting the energy of a hole in the valence band being described as $\epsilon = E_v - \frac{\hbar^2 k^2}{2m_h}$ where $E_v$ is the energy at the upper part of the valence band.

Solving that one for $k^2$ will give me:

$$k^2 = (\frac{2m_h}{\hbar^2}(E_v - \epsilon))$$

Now subsituting this into my expression for N(K) gives me: $$N(\epsilon) = \frac{V}{3\pi^2}(\frac{2m_h}{\hbar^2})^{3/2}(E_v - \epsilon)^{3/2}$$

Using the identity $D(\epsilon) = \frac {dN}{d\epsilon} = -\frac{V}{2\pi^2}(\frac{2m_h}{\hbar^2})^{3/2}(E_v - \epsilon)^{1/2}$

This shouldn't be negative, but instead positive, where in my assumptions have I done wrong? Thankful for any help!

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Mathematically, there is nothing wrong with the derivation; it's just the consequence of your definition of energy, namely the number of states increases as the energy decreases (i.e. lowers in energy), therefore it results in a negative derivative (gradient).

The problem with your derivation is that the energy formula is not for holes but rather for electrons. In other words, you're trying to derive the holes DOS using an "electron perspective". According to the Dirac sea model, a hole is created by kicking out an electron from occupied valence states, since electrons in valence states have negative energy relative to the Fermi level ($E_F$), a hole created by removing an electron will have a positive energy. So in this case the energy for holes will have the same form as for electrons in the conduction bands; i.e.

$$\epsilon = \frac{\hbar^2 k^2}{2m_h}- E_F$$

If you re-derive your DOS With this formula, you will get a positive DOS which means that the higher the energy levels that holes go the more states are available.

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  • $\begingroup$ Ahh, i see what you mean, thanks a lot for clarifying this, its been bothering me for a while, wish I could give you reputation for this. I'll at least accept the answer. $\endgroup$ – John Skeet May 9 '18 at 11:08

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