Selection Rules from Matrix Elements

Every transition probability and rate we derived is proportional to $|V_{fi}|^2$ , the squared matrix element of the perturbation. When this matrix element vanishes by symmetry, the transition is forbidden at this order no matter how strong the drive. These vanishing conditions are the selection rules, and they explain the structured, gappy spectra of atoms.

The matrix element is the gatekeeper

For a perturbation $\hat H' = \hat V f(t)$ , the transition $|i\rangle\to|f\rangle$ has rate

W_{i\to f} \;\propto\; |V_{fi}|^2 = |\langle f|\hat V|i\rangle|^2 .

If $\langle f|\hat V|i\rangle = 0$ , then $W=0$ . Selection rules are precisely the statements of when this inner product is forced to vanish — and the cleanest way to see it is through the symmetry of the integrand.

The electric-dipole interaction

The dominant atom–light coupling is the electric dipole interaction. For a field along $\hat z$ ,

\hat H'(t) = -q\,\hat{\mathbf r}\cdot\boldsymbol{\mathcal E}(t) \;\sim\; \hat z ,

so the relevant operator is the position component, e.g. $\hat z = r\cos\theta$ . The matrix element between hydrogenic states $|n\,\ell\,m\rangle$ is

\langle n'\ell'm'|\,\hat z\,|n\ell m\rangle = \int \psi^*_{n'\ell'm'}\,(r\cos\theta)\,\psi_{n\ell m}\,d^3r .

The angular part decides whether this is zero.

Parity selection rule

The position operator $\hat{\mathbf r}$ is odd under spatial inversion $\mathbf r \to -\mathbf r$ . Atomic eigenstates have definite parity $(-1)^\ell$ . The integral of an odd operator between two states of the same parity vanishes because the integrand is odd over all space. Therefore the transition requires the parity to change:

\text{parity}(f) = -\,\text{parity}(i)\quad\Longrightarrow\quad \Delta\ell \ \text{must be odd}.

Angular-momentum selection rules

A sharper statement comes from the angular integrals (the Wigner–Eckart theorem, or directly from properties of spherical harmonics). For electric-dipole transitions the surviving cases are

\boxed{\;\Delta\ell = \pm 1, \qquad \Delta m = 0,\ \pm 1\;}

The $\Delta\ell=\pm 1$ rule reflects that the dipole operator carries one unit of angular momentum ( $\hat z$ and $\hat x\pm i\hat y$ transform like $\ell=1$ spherical harmonics), so absorbing or emitting a dipole photon changes $\ell$ by exactly one. The $\Delta m$ rule depends on the polarization: $\Delta m=0$ for linear ( $\hat z$ ) polarization and $\Delta m=\pm 1$ for circular polarization. Note $\ell=0\to\ell=0$ is strictly forbidden, consistent with parity.

Worked logic: why $1s \to 2s$ is forbidden

The hydrogen ground state $1s$ has $\ell=0$ and the $2s$ state also has $\ell=0$ . The dipole rule needs $\Delta\ell=\pm1$ , so $\langle 2s|\hat z|1s\rangle = 0$ : the $1s\to 2s$ transition is dipole-forbidden. It can still proceed via slower higher-order processes (e.g. two-photon emission), which is exactly why the metastable $2s$ state is so long-lived compared with $2p$ (which decays to $1s$ in nanoseconds because $2p\to 1s$ obeys $\Delta\ell=-1$ ).

The general recipe

To find any selection rule:

Identify the operator $\hat V$ in the perturbation and how it transforms under the system's symmetries (parity, rotations, spin).
Ask which quantum numbers must change so that $\langle f|\hat V|i\rangle$ is not forced to vanish by symmetry.
Anything else is forbidden at this order.

This is the same reasoning that gives spin selection rules ( $\Delta s = 0$ for spin-independent dipole operators) and the rules governing molecular vibrational and rotational spectra.