Assuming we have some harmonic oscillator \begin{equation} H = \omega_0 (a^\dagger a + \frac{1}{2}) = \frac{p^2}{2m} + k x^2 \end{equation} for which the excitations have even wavefunctions $\Psi_n(x)=\langle x\vert (a^\dagger)^n \vert 0 \rangle$. The momentum and position operators can be expressed as differences/sums of the creation and annihilation operators i.e. $x \propto (a^\dagger + a)$. Upon minimal coupling to light $p\rightarrow p - e A$, we can consider the scattering amplitudes for IR, Raman and Reyleigh scattering, which can be derived via perturbation theory (see e.g. Louisell: Quantum Statistical Properties of Radiation).
In particular the Raman amplitude for scattering from $\vert n_i\rangle$ to $\vert n_f \rangle$ will have a numerator \begin{equation} \sum_{I=0}^\infty\langle n_f \vert \mu \vert n_I \rangle \langle n_I \vert \mu \vert n_i \rangle \end{equation} where $\mu= ex$ is the dipole operator. But due to the linearity of $x$ in $a$ and $a^\dagger$, $n_I$ gets restricted to $n_I = n_i\pm 1$ and therefore also $n_f$ gets restricted, namely to $n_f = n_i$ or $n_f = n_i \pm 2$. But the case $n_f = n_i$ means that the scattering was elastic hence Reyleigh scattering not Raman. Whereas $n_f = n_i \pm 2$ seems like it would correspond to a second order IR absorbtion(+) or emission (-). The second point is not entirely clear, because a priori I could also have a Stokes(+) or Anti-Stokes(-) Raman process ending in these states. However, at least in my mind, a Raman process requires a virtual state that absorbs the entire incoming photon i.e. significantly (more than 1 level) above the initial state so it can decay to a higher/lower energy final state and emit a photon with lower/higher energy. In the HO case the overlap to virtual states with energy more than 1 level removed from the initial state vanishes.
Am I correct in my deduction that a single harmonic oscillator even if it was Raman active from a symmetry perspective cannot Raman-scatter?
