Geometric design of roads

Free carrier absorption occurs when a material absorbs a photon and a carrier is excited from a filled state to an unoccupied state (in the same band). This is different from interband absorption in semiconductors because the excited electron is a conduction electron (i.e. it can move freely). In interband absorption the electron in question would be raised from a valence (nonconducting) band to a conducting one.

It is well known that the optical transition of electrons and holes in the solid state is a useful clue to understand the physical properties of the material. However, the dynamics of the carrier is affected by other carriers, not only by the periodic lattice potential. Moreover, the thermal fluctuation of each electron should be taken into account. Therefore a statistical approach is needed. To predict the optical transition in an appropriate precession, one should choose an approximation, called assumption of quasi-thermal distributions, of the electrons in the conduction band and of the holes in the valence band. In this case, the diagonal components of the density matrix become negligible after introducing thermal distribution function,

${\rho }_{\lambda \lambda }^0=\frac{1}{e^{({\epsilon }_{\lambda ,k}-\mu )\beta }+1}=f_{\lambda ,k}$

This is the famous Fermi-Dirac distribution for the distribution of electron energies. Thus, summing over possible l and k yields the total number of carriers N.

$N_{\lambda }=\sum _{\lambda }{f}_{\lambda ,k}$

The optical susceptibility

Using the above distribution function, the time evolution of density matrix does not have to be solved and the complexity is simplified.

${\rho }_{cv}^{\mathrm{i}\mathrm{n}\mathrm{t}}(k,t)=\int \frac{d\omega }{2\pi }\frac{d_{cv}\epsilon (\omega )e^{i({\epsilon }_{c,k}-{\epsilon }_{v,k}-\omega )t}}{\hslash ({\epsilon }_{c,k}-{\epsilon }_{v,k}-\omega -i\gamma )}(f_{v,k}-f_{c,k})$

The optical polarization is,

$P(t)=tr[\rho (t)d]$

With this relation and after adjusting the Fourier transformation, the optical susceptibility is $\chi (\omega )=-\sum _k\frac{{\left|d_{cv}\right|}^{2_{}}}{L^3}(f_{v,k}-f_{c,k})\left(\frac{1}{\hslash ({\epsilon }_{v,k}-{\epsilon }_{c,k}+\omega +i\gamma )}-\frac{1}{\hslash ({\epsilon }_{c,k}-{\epsilon }_{v,k}+\omega +i\gamma )}\right)$

Absorption coefficient

The transition amplitude corresponds to the absorption of energy and the absorbed energy is proportional to the optical conductivity which is the imaginary part of the optical susceptibility after frequency is multiplied. Therefore, in order to obtain the absorption coefficient that is crucial quantity for investigation of electronic structure, we can use the optical susceptibility.

$\alpha (\omega )=\frac{4\pi \omega }{n_{b}c}{\chi }^{″}(\omega )$

$=\frac{4\pi \omega }{n_{b}c}\sum _k{\left|d_{cv}\right|}^2(f_{v,k}-f_{c,k})\delta (\hslash ({\epsilon }_{v,k}-{\epsilon }_{c,k}+\omega ))$

Considering the gap energy Eg, energy dispersion relation of free carrier proportional to the square of momentum and the relation of electron-hole distribution function, we can obtain the absorption coefficient with some kind of mathematical calculation. The final result is $\alpha (\omega )={\alpha }_0^d\frac{\hslash \omega }{E_0}{\left(\frac{\hslash \omega -E_g-E_0^{(d)}}{E_0}\right)}^{(d-2)/2}\sum _k\mathrm{\Theta }(\hslash \omega -E_g-E_0^{(d)})A(\omega )$

The application of this result to semiconductor is important to understand the optical measurement data and the electronic properties. Some example shows the negative absorption coefficient that is fundamental presentation of semiconductor laser.

References

1. H. Haug and S. W. Koch, "Quantum Theory of the Optical and Electronic Properties of Semiconductors ", World Scientific (1994). sec.5.4 a