# Derivation of Optical Absorption Coefficient in Semiconductors

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.

• Thank you very much! – Harry Spratt Apr 5 at 18:20
• 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 – Harry Spratt Apr 8 at 11:49
• Don’t see any problem with signs: holes increase in energy as they go into deeper valence energy levels, but you are right about $k^2$. – boyfarrell Apr 8 at 14:36