In vibrational spectroscopy only transitions between neighboring vibrational states ($\Delta \nu = \pm 1$, $\nu$ being the vibrational quantum number) are allowed within the harmonic approximation. Where does this selection rule come from?
I have some ideas but haven't come to a satisfactory conclusion.
I guess it has something to do with the transition dipole moment $P$ having to be nonzero.
This is given by
\begin{equation}
P = \left\langle \psi^{\text{final}} \big| \hat{\mu} \big| \psi^{\text{initial}} \right\rangle
\end{equation}
where $\hat{\mu}$ is the dipole moment operator and $\psi^{\text{final}}$ and $\psi^{\text{initial}}$ are the wave functions of the final and the initial state, respectively. Since the wave functions of the harmonic oscillator are even functions for $\nu = 2n$ and odd functions for $\nu = 2n + 1$ (with $n = 0, 1, \dots$) and $\hat{\mu}$ transforms as a vector (and so can be considered an uneven function) the integral for $P$ vanishes if $\psi^{\text{final}}$ and $\psi^{\text{initial}}$ are both even (or uneven).
So, if my ideas about the problem are correct (please correct me if they aren't) I would expect a selection rule $\Delta \nu = \pm 1, \pm 3, \pm 5, \dots$. But why is it only $\Delta \nu = \pm 1$?
Best Answer
Just to add to gigacyan's answer, the harmonic oscillator Hamiltonian may be written in terms of raising and lowering operators: \begin{eqnarray} \hat{H}\psi&=&-\frac{1}{2m}\frac{\partial^2\psi}{\partial x^2}+\frac{1}{2}m\omega^2x^2\psi\nonumber\\ &=&(a^{\dagger}a+\frac{1}{2})\omega\psi \end{eqnarray} where \begin{equation} a=\sqrt{\frac{m\omega}{2}}\hat{x}+\frac{i\hat{p}}{\sqrt{2m\omega}} \end{equation} and \begin{equation} a^{\dagger}=\sqrt{\frac{m\omega}{2}}\hat{x}-\frac{i\hat{p}}{\sqrt{2m\omega}} \end{equation} so \begin{equation} \hat{x}=\frac{1}{\sqrt{m\omega}}(a+a^{\dagger}) \end{equation}
In other words, the $\hat{x}$ contains a single raising and lowering operator, so facilitates only transitions with $\Delta\nu=\pm1$.