# Why are overtones forbidden within the harmonic approximation?

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 $\Delta \nu = \pm 1$?

Parity is not sufficient to calculate this integral. The eigenfunctions of harmonic oscillator are mutually orthogonal meaning that $\left\langle \psi^n \big| \psi^m\right\rangle=0$ for every $n\neq m$ even if both are even.
Now, the important part of harmonic wavefunctions is Hermite polynomials. They have a particular recursive property that $$xHe_n(x)=He_{n+1}(x)+nHe_{n-1}(x)$$ As you said, the dipole moment operator behaves as $x$ and, since polynomials are orthogonal, you get nonzero integrals only for $\Delta\nu=\pm1$.
In other words, the $\hat{x}$ contains a single raising and lowering operator, so facilitates only transitions with $\Delta\nu=\pm1$.