I have been researching how to derive an expression for the absorption coefficient in semiconductors. I know the absorption coefficient can be expressed as such $$\alpha = A(hf-E_g)^{n}$$ with $n = \frac{1}{2}$ and $n = 2$ for direct band gap and indirect band gap respectively. I have seen a few explanations via use of effective mass and momentum to infer this, but they all seem to take big steps with no clear and logical explanation. I am stumped on how to derive this equation. Any help would be much appreciated.


Starting with parabolic bands.

The absorbed photon has energy $h\nu$ and generates an electronic and hole at energy levels $E_2$ and $E_1$ respectively. Energy and moment balance imply,

$$ h\nu = E_2 - E_1 = E_c(k) - E_v(k)$$

where $k$ is the momentum of the photo-generated electron and hole (it’s the same for both carriers), $m_c$ and $m_v$ are the conduction and valence band effective masses,

$$ E_c(k) = E_g + \frac{\hbar^2 k^2 }{2m_c} $$

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

Solving these for $k$,

$$ k^2 = \frac{2m_r}{\hbar^2}\left(h\nu - E_g\right) $$

the reduced effective mass is defined as,

$$ \frac{1}{m_r} = \frac{1}{m_c} + \frac{1}{m_v} $$

The parabolic bands define the density of states of conduction $\rho_c(E) \propto \left(E - E_g\right)^{1/2} $ and valence $\rho_v(E)$ bands, however, not all of these states can couple to a photon of energy $h\nu$, only states which conserve both energy and momentum.

We need to know the optical joint density of states $\rho(\nu)$ which determines the electronic states which are coupled by a photon of energy $h\nu$.

There are a number of ways for deriving this. The simplest is relating an infinitesimal change in conduction band density of states at the electron energy to a infinitesimal change in joint optical density of states at the photon energy,

$$ \rho_c(E_2) dE_2 = \rho(\nu) d\nu $$

$$ \rho(\nu) = \frac{dE_2}{d\nu} \rho_c(E) $$

Therefore you end up with the joint optical density of states being proportional to,

$$ \rho(\nu) \propto \left(h\nu - E_g\right)^{1/2} $$

The linear absorption coefficient $\alpha$ is going to be proportional to joint optical density of states, so

$$ \alpha = A \left(h\nu - E_g\right)^{1/2} $$

The derivation for indirect semiconductors is much the same but phonons must be included to conserve momentum. This accounts for different exponents.

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  • $\begingroup$ Thank you very much! $\endgroup$ – Harry Spratt Apr 5 at 18:20
  • $\begingroup$ Sorry, just to check. Shouldn't $E_v(k)$ be positive. Also is is meant to be $k^2$ as apposed to just $k$ when rearranging? Thanks in advance $\endgroup$ – Harry Spratt Apr 8 at 11:49
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    $\begingroup$ Don’t see any problem with signs: holes increase in energy as they go into deeper valence energy levels, but you are right about $k^2$. $\endgroup$ – boyfarrell Apr 8 at 14:36

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