# Optical transition rate for quantum wells

1. Aug 2, 2009

### Kcant

So, let's say that you wanted to find the optical transition rate between subbands for electrons in a quantum well. To a good approximation, you can say that the wavefunctions are of the form

$\Psi = \psi(\vec r) u_c(\vec r)$

where u_c is the Bloch function and the lowercase psi is a slowly-varying envelope function that obeys an effective-mass Hamiltonian equation:

$\left( -\frac{\hbar^2}{2 m^*}\nabla^2+ V_{slow} \right)\psi = E\psi$

Obviously, finding the transition rates boils down to finding the momentum matrix elements between different states. Often, people simplify this by using the dipole moment instead of the momentum, using the following commutation relation:

$\vec p = \frac{i m}{\hbar}[H_0, \vec r]$

where H_0 is the crystal Hamiltonian. Then, to find the full momentum matrix element $\langle\Psi_f|\vec p|\Psi_i\rangle$, one can pull out a factor of the $i m/\hbar (E_f-E_i)$ from the element, becoming $i m/\hbar (E_f-E_i)\langle\Psi_f|\vec r|\Psi_i\rangle$. This can be further evaluated by exploiting the slow-varying nature of the envelope function and the orthonormality of the Bloch function, leaving $i m/\hbar (E_f-E_i)\langle\psi_f|\vec r|\psi_i\rangle$.

Ok, that's fine. But what if you don't use the commutation relation right away, and instead simplify the momentum element first? In that case,
$\langle\Psi_f|\vec p|\Psi_i\rangle = \int d^3\vec r \psi_f^* u_c^* \cdot -i \hbar (u_c \nabla \psi_i + \psi_i \nabla u_c)$.

Again using the slowness property of the envelope function, the second term is proportional to $\langle u_c|\vec p| u_c\rangle$, which vanishes for III-V materials. All you're left with is a factor of $\langle\psi_f|\vec p|\psi_i\rangle$. But wait! The commutation relation used before is now

$\vec p = \frac{i m^*}{\hbar}[H_{eff}, \vec r]$

where $H_{eff}$ is the effective-mass Hamiltonian from above. In that case, the matrix element reduces to $i m^*/\hbar (E_f-E_i)\langle\psi_f|\vec r|\psi_i\rangle$. So, in one case, the effective mass comes into the picture, while in the other, it is the free mass that does! What am I missing?