 Solution

Assume equal excess electron and electron hole number densities $N_t$, with no additional doping imbalance. At zero temperature, both occupied carrier spheres have the same Fermi wavevector $k_F$ because their spin counting is the same:For $k<k_F$, the conduction band is occupied and the valence band has electron holes, so $f_c-f_v=1$. Beyond $k_F$ the occupations reverse and the medium absorbs. Thus the upper edge of the positive-optical gain interval is the supremumEquivalently $E_{\mathrm{Fc}}=E_g+{\hslash }^2{k}_F^2/(2m_c^{\ast })$ and $E_{\mathrm{Fv}}=-{\hslash }^2{k}_F^2/(2m_v^{\ast })$. With the supplied effective masses,Using the vacuum wavelength $\lambda =hc/(\hslash \omega )$ reverses the endpoints:These boundaries are idealized and open: the lower-energy edge has zero joint density of states, and the upper edge is the zero-temperature occupation discontinuity, not a finite-width gain peak. The calculation gives positive material optical gain, not the stronger requirement of exceeding the losses at laser threshold.