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I understand that pure rotational spectra only exist for molecules which have a permanent dipole moment. The common explanation is "so that they can interact with the E-field of the incoming photons", but I don't think it is clear why photons can only interact with dipoles.

I'm also aware that the answer probably has something to do with the matrix element $\langle\psi_i|x|\psi_j\rangle$ where $\psi_i$ and $\psi_j$ are the initial and final states of the molecule, and $x$ comes from the dipole moment operator. I'm just not sure why we need this matrix element and not something else. And as an extension, why does Raman spectroscopy depend on the polarisability of a molecule, whose operator behaves like $x^2$?

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    $\begingroup$ In semi-classical terms think of the the electromagnetic wave (ir /microwave , visible uv etc) as an oscillating electric field, this can interact with the oscillating electric field of a rotating molecule caused by its permanent dipole, hence dependence on $x$. When the energies are the same, i.e. in resonance, energy can be absorbed from the EM wave into the molecule and vice versa. Raman is scattering, the polarisability is a volume, $x^2$ is due to the cross section (area) seen by the radiation. $\endgroup$ – porphyrin Nov 16 '20 at 10:55
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    $\begingroup$ The matrix element for rotational spectra comes from en.wikipedia.org/wiki/Fermi%27s_golden_rule. The relevant Hamiltonian is proportional to $\hat{x}$, hence your expression. (Well, I think it is $\propto \hat{x}$ as long as you define your coordinate system carefully; but in free space, we have the freedom to do that.) Not sure about Raman, though. $\endgroup$ – orthocresol♦ Nov 16 '20 at 11:04

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