Effective theory of Rayleigh scattering
Rayleigh scattering is the scattering of low energy light off a neutral atom. It can explain the color of the sky.
Following the EFT recipe:
Energy scales: We have low energy photons with $E_\gamma$. Their energy is lower than the bohr levels in the atom $\Delta E$ and the inverse of the bohr radius $a^{-1}0$. Therefore the photon cant resolve the structure of the atom. To the photon the atom is just a neutral particle so it scatters elastically. We have $E\gamma \ll \Delta E \ll a^{-1}0 \ll M\text{atom}$.
Degrees of freedom: Low energy photons $A_\mu$, neutral atom described by field operator $\phi$.
Symmetries: Lorentz invariance at low energies only resolved as Galilean invariance (?), since $E_\gamma \ll M_\text{atom}$ non-relativistic is ok. Gauge invariance: $A_\mu \to A_\mu + \partial_\mu\Lambda$, $\phi$ invariant since the atom is neutral. Also $C,P,T$ Symmetries.
Lagrangian: Kinetic terms for neutral atom field $\phi$ and photons, plus an interaction term $\mathcal{L}\text{int}$: $$\mathcal{L} = \phi^\dagger \left(i\partial_t - \frac{1}{2M}\partial^2\right) \phi - \frac{1}{4}F{\mu\nu}F^{\mu\nu} + … + \mathcal{L}\text{int}$$ The interaction term is given by $$\mathcal{L}\text{int} = \phi^\dagger\phi(c_1 E^2 + c_2 B^2) + …$$ Terms of the form $E\cdot B$ are forbidden due to parity. Since $[\mathcal{L}]=4, [\phi] = \frac{3}{2}, [E]=[B]=2$ we have $[c_i] = -3$
Power counting: Photons scale with their energy $E_\gamma$. The parameters in the EFT Lagrangian depend on the energy scales of the full theory ($\Delta E, a^{-1}0$). Therefore the EFT expansion is in the fraction of scales $\frac{E\gamma}{\Delta E}, E_\gamma a_0$.
Accuracy: The expansion is done up to order $\frac{E_\gamma}{\Delta E}$, since $\Delta E < a^{-1}0$. Valid since $\frac{E\gamma}{\Delta E} \ll 1$ for visible light (not for e.g. ultraviolet).
Matching: Matching coefficent could be any term consisting of the scales that were integrated out ($\Delta E, a^{-1}_0$). We assume $c_i \propto a^{3}_0$ since size of atom controls if it can be resolved.
Since $E$ and $B$ depend on $E_\gamma$ we deduce $\mathcal{A} \propto a^3_0 E^2_\gamma + \mathcal{O}(\frac{E_\gamma}{\Delta E})$ so that for the cross section $\sigma \propto a^6_0 E^4_\gamma \propto a^6_0\omega^4$ which is the known Reayleigh scattering law. Since $\sigma \propto \omega^4$ blue light scatters more than red light.