1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.