Expression for spontaneous emission in solids I have yet to find a reliable source for the expression of the rate of spontaneous emission in a solid. Can anyone confirm if the following is correct?
The basic ingredients of the calculation are the momentum matrix elements,
$$
p_{\textbf k}^{ij} = \langle\psi_{i\textbf k}|\hat{\textbf p}|\psi_{j\textbf k}\rangle,
$$
which can be obtained with some ab initio software (Abinit in my case). From these, I construct the position matrix elements,
$$
r_{\textbf k}^{ij} = \frac{i\hbar}{(\epsilon_j-\epsilon_i)m}p_{\textbf k}^{ij},
$$
and express the transition rate between the bands $j$ and $i$ at $\textbf k$ as
$$
W_{\textbf k}^{ij} = \frac{(\omega_j-\omega_i)^3 |qr_{\textbf k}^{ij}|^2}{3\pi\epsilon_0\hbar c^3}f_j(1-f_i),
$$
where $f$ are the occupation factors. I tried this for silicon, and got $W=1.7\times10^7$ s$^{-1}$ between the conduction and valence states (specifically, with $j=5$, $i=4$, $\textbf k = 0$), which sounds reasonable but I'm not sure.
 A: The spontaneous emission rate $[cm^{-3}s^{-1}J^{-1}]$ for photons of energy $\hbar\omega$ for parabolic semiconductors is, 
$$r_{sp}(\hbar\omega) =  A_{21}(\hbar\omega) g_{12}(\hbar\omega) (1 - f_1(E_l)) f_2(E_u) $$
where $A_{21}(\hbar\omega)$ [$s^{-1}$] is the inverse lifetime of the transition from energy state $E_u$  [$J$] in the upper band 2 to an energy state $E_l$ [$J$] in the lower band 1, $g_{12}(\hbar\omega)$ [$J^{-1}cm^{-3}$] is the joint optical density of states, $(1 - f_1(E_l))$ is the probability of the lower state being unoccupied and $f_2(E_u)$ is the probability of the upper state being occupied.
By equating the total rate (i.e. including simulated absorption and emission too) to the blackbody rate it can be shown that,
$$A_{21}(\hbar\omega) = \alpha_{0}(\hbar\omega) \frac{g_{\gamma}(\hbar\omega)}{g_{12}(\hbar\omega)}v_g $$
where $\alpha_{0}(\hbar\omega)$ is the absorption coefficient of the material, $g_{\gamma}(\hbar\omega)$ is the photon density of states and $v_g$ is the velocity of light in the material.
The joint optical density of states is related to the density of states and the effective masses,
$$g_{12}(\hbar\omega) = g_{2}(E_u)\frac{m_r}{m_2^{\star}}$$
where $g_{2}$ is the conduction band density of states, $m_r$ is the reduces effective mass and $m_2^{\star}$ is the conduction band effective mass.
With this approach you should be able to calculate an emission rate. I like this approach because it ties the emission rate to generally well known experimental parameters such as absorption coefficient and refractive index (strictly speaking that is one parameter I suppose). This is from some original research I did for this paper, http://dx.doi.org/10.1063/1.4916561.
