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Rayleigh scattering is elastic scattering of electromagnetic radiation by particles much smaller than the incident wavelength. In the stationary particle approximation, the scattered electromagnetic radiation keeps the incident frequency: its direction changes, rather than its color. The incident electric field induces an oscillating electric dipole moment, which produces electric dipole radiation. This is a useful example of how a microscopic response can explain a familiar large-scale sight, why the sky is blue. ^[ref].

The smallness condition for Rayleigh scattering by a small sphere of radius $a$ is $ka\ll 1$, where the wavenumber is $k=2\pi /\lambda$ and $\lambda$ is the surrounding medium's wavelength. For ordinary materials with modest refractive index, the incident electric field then varies little across the particle, making the electric-dipole approximation for radiation useful. For very large refractive index, also check the internal size $|m|ka\ll 1$ before applying the simple electric dipole model. Once these conditions fail, use Mie scattering for a sphere instead of extrapolating the Rayleigh scattering wavelength dependence.

Historically, the 1871 paper ^[ref] connected the color and polarization of an electromagnetic wave of scattered skylight with particles smaller than the wavelength. Its dimensional analysis anticipates scattered light intensity proportional to particle volume squared and to ${\lambda }^{-4}$. The original mechanical picture is historical; the modern explanation uses induced electric dipole radiation.

For incident unpolarized light and the isotropic electric dipole response of Rayleigh scattering by a small sphere, let $\theta$ be the angle between the incident beam's propagation direction and the outgoing direction. The differential scattering cross-section and its relative angular pattern arewhere $d\Omega$ is an element of solid angle. Averaging over the two transverse directions of incident polarization of an electromagnetic wave gives this pattern. Forward scattering at $0^{\circ }$ and backward scattering at $180^{\circ }$ are equally strong: $F=1$. Sideways scattering at $90^{\circ }$ is weakest, but nonzero: $F=1/2$. ^[ref].

The normalization is consistent becauseThus integrating $d\sigma /d\Omega$ gives the total Rayleigh scattering cross-section $\sigma$. The plotted $F$ has maximum one; it is not itself normalized to unit solid angle integral.

For a single electric dipole driven by incident linear polarization, the angular distribution of electric dipole radiation instead has a ${\sin }^2\alpha$ factor, where $\alpha$ is measured from the electric dipole moment direction, parallel to the incident electric field. That direction is perpendicular to the incident beam. Confusing $\alpha$ with the beam angle $\theta$ would incorrectly predict zero forward scattering. Averaging the dipole patterns over incident unpolarized light yields the displayed $F(\theta )$.

An ideal electric dipole at the origin with smooth electric dipole moment $\mathbf{p}(t)$ has Lorenz gauge electric potential and magnetic vector potential, for $r>0$,For a short proof, place electric charges $sq$ at $s\mathbf{p}/(2q)$, $s=\pm 1$, and let $q\to \infty$ with $\mathbf{p}$ fixed. The corresponding retarded times are $t_0+s\mathbf{n}\cdot \mathbf{p}(t_0)/(2qc)+O(q^{-2})$. Each Liénard–Wiechert potential has denominator $r-s(\mathbf{n}\cdot \mathbf{p}+r\mathbf{n}\cdot {\mathbf{p}}^{\prime }/c)/(2q)+O(q^{-2})$. The opposite $q/r$ terms cancel, leaving the displayed electric potential. The products of signed electric charge and velocity add, giving the displayed magnetic vector potential. Bounded time derivatives on the relevant interval justify the Taylor expansions.

The even, spherically symmetric hydrogen atom ground state has no permanent electric dipole, so a uniform electric field gives no first-order energy shift. The electric-dipole selection rules for hydrogen allow odd-parity excited states to mix in at second order, yielding a negative shift quadratic in the field.

Here there is a nondegenerate energy eigenvalue, so second-order nondegenerate perturbation theory applies. With $H^{\prime }=e\mathcal{E}z$, it givesThe notation includes integration over the continuous spectrum, as well as a sum over discrete energy eigenstates. All denominators are negative because $1s$ is the ground state. The positive electric polarizability ${\alpha }_{1s}$ therefore lowers its energy.^[ref]

An independent way to see the coefficient avoids truncating that sum. Define $E_{\mathrm{h}}=e^2/(4\pi {ϵ}_0{a}_0)$ and $f=e{a}_0\mathcal{E}/E_{\mathrm{h}}$. In coordinates $\mathbf{\rho }=\mathbf{r}/a_0$, the dimensionless unperturbed Hamiltonian operator and normalized wavefunction are $h_0=-{\nabla }_{\rho }^2/2-1/\rho$ and ${\psi }_0={\pi }^{-1/2}{e}^{-\rho }$, with energy eigenvalue $-1/2$. Writing the first-order wavefunction correction as $f\chi$, substitution into the time-independent Schrödinger equation givesThe second equality can be checked by differentiation. Its parity makes $\chi$ orthogonal to ${\psi }_0$. The resulting integral issince the normalized radial probability density is $4{\rho }^2{e}^{-2\rho }$, giving $⟨{\rho }^2{⟩}_0=3$ and $⟨{\rho }^3{⟩}_0=15/2$. Converting back to SI units gives the stated electric polarizability. The induced electric dipole moment is $-\partial \Delta E/\partial \mathcal{E}={\alpha }_{1s}\mathcal{E}$ and tends to zero with the electric field, unlike the selected degenerate branches' leading dipoles. Both calculations describe weak-field level shifts; a spatially uniform static electric field also allows eventual quantum tunnelling and ionization, which these real energy corrections do not calculate.

Consider an original ideal quantum system with positive scales $ϵ,d$ andInterpret $D$ as one component of an electric dipole moment, so its coupling to the corresponding electric field is the electric-dipole interaction $-\mathcal{E}(t)D$. The gaps are $2ϵ,3ϵ,5ϵ$, but $D_{02}=0$: this interaction has no direct first-order $2\leftrightarrow 0$ coupling. The nonzero squared matrix elements are $|D_{01}{|}^2=d^2$ and $|D_{12}{|}^2=4d^2$. These are coupling weights; actual spectral line intensities also depend on populations and radiation factors. In particular, there is no promise that an experiment populates the upper state of every allowed transition.

A measurement of $D$ itself has possible outcomes $0,\sqrt{5}d,-\sqrt{5}d$, sinceThese are eigenvalues of $D$, not the energy eigenvalues $0,2ϵ,5ϵ$. Moreover, $(D^2{)}_{02}=D_{01}{D}_{12}=2d^2$ although $D_{02}=0$: two factors connect through $|1⟩$. This algebraic statement does not turn $D_{02}$ into a nonzero direct coupling. Higher-order probability amplitudes for transitions require the appropriate dynamical calculation.

For potential energy homogeneous of degree $-2$, Newton's second law gives $d^2{r}^2/d{t}^2=4E/m$. Indeed, Euler's identity implies $\mathbf{r}\cdot \mathbf{F}=2U$, so the second derivative equals $2v^2+4U/m=4(K+U)/m$. Thus a trajectory with zero total energy and initially zero radial velocity remains on a sphere of constant radius. A positive electric charge released from rest on the equator of a fixed ideal electric dipole has these properties; in its meridional plane it oscillates on the semicircle on the negative-pole side. A finite two-charge electric dipole does not have exactly homogeneous potential energy and perturbs this result.