 Solution

An electric dipole transition has amplitude proportional to $⟨f|\mathbf{d}\cdot \mathbf{ϵ}|i⟩$. The electric dipole operator is odd under parity, so the initial and final states must have opposite parity. As a rank-one spherical tensor it also requires $\Delta J=0,\pm 1$, excluding $J=0\leftrightarrow 0$, and $\Delta m_J=0,\pm 1$. These follow from angular-momentum addition and the Wigner-Eckart theorem; vanishing Clebsch-Gordan coefficients can impose further restrictions. Relative to the quantization axis, $\pi$ polarization drives $\Delta m=0$, while ${\sigma }^{\pm }$ drive $\Delta m=\pm 1$.

For a single active electron, parity and the orbital triangle rule combine to give $\Delta l=\pm 1$. In LS coupling, the spin-independent electric dipole operator gives $\Delta S=0$ and $\Delta L=0,\pm 1$, excluding $L=0\leftrightarrow 0$; total parity must still change. Spin-orbit mixing can weakly relax the approximate spin rule. For resolved hyperfine states the corresponding triangle rule is $\Delta F=0,\pm 1$, excluding $F=0\leftrightarrow 0$, with $\Delta m_F=0,\pm 1$. These labels require the appropriate coupling regime. There is no universal E1 rule demanding a particular change in principal quantum number. An E1-forbidden line may occur through another multipole or a multiphoton process.