Selection rules in the harmonic oscillator

If I have a molecule with two different atoms and this molecule gets excited by EM radiation, we can only see transitions with $$\Delta n=\pm1$$ in the molecule spectrum.

I found this in my physics textbook and searched for it online but couldn't find that much useful information about it. Is there any derivation out there for this selection rule?

I tried to do it on my on because I found out that the derivation of this could be done by calculating the transition dipole moment between two states but quickly got stuck.

It would be great if someone could link me to some good resources for this or could give me an advice on how to do it on my on. Thanks you!

1 Answer

Strictly speaking, this is not correct - one could obtain more complex transitions, especially with strong electromagnetic fields. However, for many problems it is sufficient to limit oneself to the dipole approximation. The reasoning is approximately like this:

• Suppose that atom/molecule is subject to an electromagnetic radiation with field $$\mathbf{E}(\mathbf{x},t)$$.
• For simplicity we take the field to be a plane wave: $$\mathbf{E}(\mathbf{x},t)=\mathbf{E}_0e^{i\mathbf{k}\cdot\mathbf{x}-i\omega t} + c.c.$$
• The absorption probability can be calculated using the Fermi golden rule: $$w_{i\rightarrow f} = \frac{2\pi}{\hbar}|\langle i|H_{int}|f\rangle|^2\delta(E_i +\hbar\omega-E_f)$$
• The interaction Hamiltonian in dipole approximation is $$H_{int} = -\mathbf{d}\cdot\mathbf{E}(\mathbf{x},t)$$ Atom here is considered as a collection of charged particles, which can be characterized by its multipole expansion. In principle, higher order electrostatic moments also couple to the field, but they are usually much smaller than the dipole moment $$\mathbf{d}$$ and neglected.
• The wave length of light is much bigger than the size of an atom, so one can expand the electric field as $$\mathbf{E}(\mathbf{x},t)=\mathbf{E}_0e^{i\mathbf{k}\cdot\mathbf{x}-i\omega t} + c.c. \approx \mathbf{E}_0(1 + i\mathbf{k}\cdot\mathbf{x}) + c.c.$$
• Now the matrix element in the Fermi golden rule becomes $$\langle i|H_{int}|f\rangle = \langle i|-\mathbf{d}\cdot\mathbf{E}(\mathbf{x},t)|f\rangle = \langle i|-\mathbf{d}\cdot\left[ \mathbf{E}_0(1 + i\mathbf{k}\cdot\mathbf{x}) + c.c.\right]|f\rangle = -i\mathbf{d}\cdot\mathbf{E}_0\langle i|\mathbf{k}\cdot\mathbf{x})|f\rangle,$$ (remonder: $$\langle i|f\rangle = 0$$). Thus, the matrix element is proportional to the matrix element for the position operator. For a harmonic oscillator this operator has non-zero matrix elements only between the states with $$n$$ and $$n\pm 1$$, i.e., only such transitions can occur with non-zero probability.