In our posts on absorption in bulk we studied the situation where the semiconductor was excited by light with energy ($\hbar \omega$ per photon) larger than the energy gap, a situation in which unbound electron-hole pairs are created. The calculations were actually done assuming that the Coulomb interaction plays a minor role, and thus it was neglected.
A quite different situation arises when the light field has slightly lower energy than the bandgap. We call this below-bandgap excitation, and here the Coulomb interaction plays a crucial role and has to be taken into account in calculations. In the simplest interpretation, the mutual Coulomb attraction of electron and hole causes the particles to form a bound state: the exciton.
\begin{eqnarray}
\left(i\hbar\frac{d}{dt}-\Delta\right) \rho_{v\mathbf k',c\mathbf k}(t)
&&
+ \sum_{i\neq 0} V_{\mathbf i} \rho_{v\mathbf k'- \mathbf i,c\mathbf k- \mathbf i}(t)
\\
&&=
\langle c\mathbf k | [-(Q/m) \mathbf A(\mathbf r, t) \cdot \mathbf p] | v\mathbf k' \rangle
~~~~~~~~~~ (1)
\end{eqnarray}
where $\{Q,m,\mathbf p\}$ are the electron charge, mass and momentum, $\Delta$ is the energy difference between states in the valence and conduction bands, the second term in the left is the Coulomb potential, and $A(\mathbf r, t)$ is the vector potential.
In the vertical transition (VT) approximation, the matrix element of the light-matter interaction only exists for $\mathbf k'=\mathbf k$; then it makes sense to study the whole Eq. (1) only for $\mathbf k'=\mathbf k$. However, we want to take into account the momentum transfered by the light; we conclude in analogy to the VT case, that Eq. (1) should be studied only for quasi-momenta obeying $\mathbf k'=\mathbf k - \mathbf q_0$. For a plane wave $A(\mathbf r, t) = A_0(t) e^{i q_z z}+c.c.$ propagating in $z$ we have $\mathbf q_0 = q_z \hat{z}$ [3].
Performing a Fourier transform of Eq. (1) in the variable $\mathbf k$ one arrives at
\begin{eqnarray}
&& \left[
\hbar \omega -
\left(E_g + \frac{\hbar^2\,\mathbf q_0^2}{2|m_v^*|}\right) +
i\, \frac{\hbar^2\,\mathbf q_0}{|m_v^*|} \cdot \nabla +
\frac{\hbar^2}{2\, \mu} \nabla^2 +
V(\mathbf{r})
\right] \rho_{\mathbf q_0 } (\omega, \mathbf{r})
\nonumber \\
&& \hspace{30mm}
= L^3
\, \xi(\omega)\,
\delta(\mathbf{r})
\hspace{30mm} (2)
\,,
\end{eqnarray}
where $\xi(\omega) \propto A_0$, and the RHS was calculated for the specific case of plane waves.
In the VT limit $\mathbf q_0 = 0$, so the LHS (the homogeneous equation) is the Wannier equation, and its solutions $\{\psi_\nu(r)\}$ are the wave function for the relative motion of excitons.
When the momentum of light is not disregarded, we have two new terms:
T1: $\frac{\hbar^2\,\mathbf q_0^2}{2|m_v^*|}$ ,
T2: $i\, \frac{\hbar^2\,\mathbf q_0}{|m_v^*|} \cdot \nabla$ .
Term (T1) is just a renormalization of the energy, while term (T2) requires a little more work in order to solve Eq. (2). Using typical values of semiconductors (e.g. GaAs), one can see that (T2) is actually larger than (T1). This implies that, if we want to retain term (T1) we also have to keep term (T2).
To solve Eq. (2) we may either use perturbation theory on (T2) -since it is small-, or complete squares and apply a unitary transformation. Let us apply the former: consider first the homogeneous part of Eq. (2) -LHS- without (T2), which yields the common solution of excitons $\{\psi_\nu(\mathbf r)\}$ but with a renormalized energy. Now use (T2) as a perturbation. Without further calculations, we know that the wave function will be constructed as a superposition of exciton functions weighed by coefficients containing the perturbation term. We conclude that the new state compatible with the optical excitation that preserves the momentum is a superposition of exciton states. In other words, the relative-motion wave function is modified compared to the normal exciton state.
Using the unitary transformation [3] and going back to k-space:
\begin{eqnarray}\label{Eq_Coherence_TS_final}
\rho_{v \mathbf{k} - \mathbf q_0 ,c \mathbf{k}}(\omega)
&=& L^3\, \xi(\omega)\,
\times \nonumber \\
&& \hspace{-20mm}
\sum_\nu \,
\frac{\psi^*_\nu(\mathbf{r}=0)}
{\hbar\,\omega - E_g - \frac{\hbar^2 \mathbf q_0^2}{2 M} - E_\nu}
\psi_\nu \left(\mathbf{k} - \frac{\mu}{|m_v^*|} \mathbf q_0 \right)
\hspace{20mm} (3)
\,.
\end{eqnarray}
We see a new term in the denominator -coming from (T1)-, that signals the center-of-mass motion of the exciton (or exciton-like state).
Equation (3) is the building-block of several quantities describing both the electrons' kinetics/dynamics and the effect that electrons have on the EM-field. The polarization and electric currents deduced from Eq. (3) are given in Ref [3].
To conclude, one could say that the transfer of linear momentum from a plane wave to electrons causes the electron-hole bound states to: 1) gain center-of-mass motion, and 2) have a modified relative-motion wave function compared to that of excitons.
References
[1] Hartmut Haug, Stephan W. Koch, Quantum Theory of the Optical and Electronic Properties of Semiconductors (World Scientific Publishing Co. 1990).
[2] I use Cartesian coordinates since I want to compare the results with the common concept of excitons in bulk. Cylindrical coordinates were used in Absorption in Bulk. Part 2.
[3] G. F. Quinteiro, Below-bandgap excitation of bulk semiconductors by twisted light, EPL 91 27002 (2010).
0 comments:
Post a Comment