Electric polarizability (${\mathbf{d}}_{\mathrm{ind}}=\mathbf{\alpha }\mathbf{E}$)

Electric polarizability relates an induced electric dipole moment to an applied electric field. In a weak-field model with linearity, ${\mathbf{d}}_{\mathrm{ind}}=\mathbf{\alpha }\mathbf{E}$. A tensor response allows different electric field directions to induce electric dipole moments in different directions. A response that changes with molecular rotation enables rotational Raman spectroscopy.

A bound charge in a harmonic restoring potential has induced electric dipole moment $d=(q^2/k)E$. Completing the square proves this for every static field within the strictly harmonic model. First-order nondegenerate perturbation theory gives the same coefficient because the position operator connects the unperturbed ground state only to the first excited state. This exact response coefficient does not make its two-state wavefunction approximation exact.

The linear response relating an induced electric dipole moment to an applied electric field. In ordinary nonresonant Raman spectroscopy, changes in its symmetric components during a normal mode determine activity. These components transform like $x^2,y^2,z^2,xy,xz,yz$.

A slender high-permittivity particle responds mainly along its axis $\mathbf{u}$, giving $\mathbf{\alpha }\simeq {\alpha }_{\parallel }\mathbf{u}{\mathbf{u}}^{\mathsf{T}}$. Far-field projection into an observation plane leaves one real polarization direction. Every nonzero output is therefore linearly polarized in this leading approximation, even for circular or unpolarized input.

The dynamic electric polarizability relates the induced electric dipole moment at a driving frequency to the applied electric field at that frequency in linear response theory. A complex coefficient can encode both a dispersive phase and absorption. For a charged one-dimensional quantum harmonic oscillator with charge magnitude $e$, mass $m$, and natural frequency ${\omega }_0$, the undamped response isOne derivation uses the oscillator electric-dipole selection rule and first-order time-dependent perturbation theory. The same result follows from the classical Lorentz oscillator model: $m\ddot{z}+m{\omega }_0^{2}z=-eE$ and $d=-ez$ imply $\ddot{d}+{\omega }_0^{2}d=e^{2}E/m$. At exact resonance there is no bounded periodic response without damping; a finite switch-on produces secular growth rather than a finite steady-state value of the pole expression.

For a localized quantum harmonic oscillator along $z$, let the transverse magnetic vector potential be $A_z(t)=2A_0\cos (\theta -\omega t)$ for $t>0$, with zero potential before the switch. The Coulomb gauge form of minimal coupling gives $H_1=e{A}_z{p}_z/m$. The creation operator matrix element $⟨1|p_z|0⟩=i\sqrt{m\hslash {\omega }_0/2}$ means that only $|1⟩$ is reached at first order from the ground state. Its interaction picture amplitude isIn particular $F_0=t$, so resonant finite-time expressions are nonsingular. The induced electric dipole moment is $-2e\sqrt{\hslash /(2m{\omega }_0)}\mathrm{Re}(e^{-i{\omega }_{0}t}{b}_1)$. It contains a driven component with the dynamic electric polarizability and free-frequency transients required by the initial conditions. The discontinuity in $A_z$ gives an impulsive electric field and an initial mechanical velocity $e{A}_z(0^{+})/m$ even though the initial canonical momentum has zero mean.

For a nondegenerate parity-symmetric ground state in an external electric field, write the perturbation as $V=-E\hat{d}$. The first-order state correction givesThe positive coefficient $\alpha$ is the electric polarizability along the applied direction. Meanwhile second-order nondegenerate perturbation theory gives $\delta E_0^{(2)}=-\alpha E^2/2$. Thus the induced electric dipole moment satisfies $d=-2\delta E_0^{(2)}/E$, not $-\delta E_0^{(2)}/E$. The Hellmann–Feynman theorem independently gives $d=-\partial E_0/\partial E$.

A neutral isotropic atom with static electric polarizability $\alpha >0$ in the field of a distant point charge $Q$ has induced dipole $\mathbf{p}=\alpha \mathbf{E}$ and energy $U=-\alpha E^2/2$. Since $E=|Q|/(4\pi {ϵ}_0{r}^2)$,The negative radial force is attractive for either sign of $Q$. Reversing $Q$ reverses the induced dipole but not the energy or leading force. The factor $1/2$ accounts for the work of inducing the dipole; using the permanent-dipole energy $-\mathbf{p}\cdot \mathbf{E}$ without that work would double the answer. The approximation assumes weak fields, a slowly varying field across the atom and negligible electronic overlap.