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.

  • $\begingroup$ Does you value of $W$ correspond to the blackbody emission rate? You can test this using Planck's Law. $\endgroup$ – boyfarrell Aug 31 '15 at 11:35
  • $\begingroup$ My system is not in thermodynamic equilibrium, so this is not blackbody radiation. Basically, the question is: if I put one electron from the valence band to the conduction band, what is the probability of that electron returning to the ground state with the emission of a photon? $\endgroup$ – Raul Laasner Aug 31 '15 at 11:40
  • $\begingroup$ OK, do you know the hole concentration of the valence band? Or do you assume a state is available? $\endgroup$ – boyfarrell Aug 31 '15 at 11:43
  • $\begingroup$ I assume one state is available. The inverse of the number that I reported should be measurable as the luminescence decay time, I think. At least to first approximation. $\endgroup$ – Raul Laasner Aug 31 '15 at 11:48

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.

  • $\begingroup$ Thanks, although with $g_{12}$ it seems that a sum over k-points has been taken. I need an expression between any two states with the same wavevector. $\endgroup$ – Raul Laasner Aug 31 '15 at 12:27
  • $\begingroup$ This is the calculation for direct semiconductors where the transitions are vertical. This means that the initial and final k vectors are the same. For direct semiconductors there is a one to one correspondence between photon energy and k, no summation needed. $\endgroup$ – boyfarrell Aug 31 '15 at 14:12
  • $\begingroup$ I do not quite understand the meaning of $g_{12}$ when considering only two states. For clarity, let's consider a specific situation, where I put one electron from the top of the valence band to the first unoccupied state at $\Gamma$ (I'm thinking about silicon). The question is what is the probability for this electron to spontaneously return to the ground state? Can your formulas be applied here? $\endgroup$ – Raul Laasner Aug 31 '15 at 14:24
  • $\begingroup$ No, only for direct semiconductors. Equations assume vertical transitions in the E-k diagram. $\endgroup$ – boyfarrell Aug 31 '15 at 14:56
  • $\begingroup$ To answer the second part of your comment. $g_{12}$ tells you the density of states that can participate in optical transitions by conserving momentum. The formula I showed above is derived assuming direct band gap materials. In these materials there is a one to one correspondence between photon energy and k-space location which conserves momentum. $\endgroup$ – boyfarrell Aug 31 '15 at 17:17

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