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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 quoted expansion is nondegenerate perturbation theory for a known isolated eigenstate of a time-independent Hamiltonian. A useful small-mixing condition is $|⟨k|H_1|i⟩|\ll |E_i-E_k|$ for coupled states. First order is a diagonal expectation value; second order describes virtual mixing. A degenerate or nearly degenerate coupled subspace must first be treated by diagonalization there. Symmetry can eliminate terms, and convergence still needs checking. Ground-state second-order shifts are nonpositive because all denominators are negative.^[ref]

For the rotational Stark effect, $H_1=-pE\cos \theta$, where $\theta$ is the polar angle to the field. The source calls it azimuthal, which is inconsistent with this coupling. Rotor states have parity $(-1{)}^J$, while $\cos \theta$ is odd, so every first-order matrix element within one fixed-$J$ manifold vanishes. The $m$ degeneracy therefore causes no first-order splitting.

Only $J\pm 1$ with the same $m$ contribute in second order. Using their energy differences $2BJ$ and $-2B(J+1)$,For $J\ge 1$ this simplifies to the quadratic Stark shift of a polar linear rotor,For $J=0$, only $J^{\prime }=1,m=0$ contributes, with squared angular matrix element $1/3$ and denominator $-2B$:Here $B$ is the rotational energy constant, unlike the magnetic field in Question3.

The unperturbed line gives $B/h=6.08\mathrm{GHz}$. The $J=1$ shifts are $(pE{)}^2/(10B)$ for $m=0$ and $-(pE{)}^2/(20B)$ for $m=\pm 1$. Subtracting the common lower-state shift gives transition shifts $4(pE{)}^2/(15Bh)$ and $7(pE{)}^2/(60Bh)$. Their separation determines the dipole moment from a rotational Stark doublet:The $\mathrm{D}$ unit is the debye. Using external $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$ and $1\mathrm{D}=3.33564\times 10^{-30}\mathrm{C}\mathrm{m}$ gives $pE/B\simeq 0.0593$, consistent with weak-field perturbation theory. The two shifted frequencies are about $12.165707$ and $12.162497\mathrm{GHz}$; polarization selects $\Delta m=0$ or $\pm 1$. The inferred value is a magnitude; these shifts do not determine the dipole's sign.^[ref][ref]

In the totally symmetric $(r,s)$ basis, the force constant matrix reduces toThe unperturbed eigenvalues are distinct within this symmetry block, so nondegenerate perturbation theory applies even though the full unperturbed matrix has degeneracies in different symmetry sectors. The first-order shifts are both $\gamma$. At second order the off-diagonal matrix element gives $-{\gamma }^2/(2k)$ for the lower mode and $+{\gamma }^2/(2k)$ for the upper one. ThereforeIn frequency form,The corresponding eigenvectors are $r+(\gamma /2k)s+\cdots$ and $s-(\gamma /2k)r+\cdots$. As a check, the exact squared-frequency eigenvalues arewhose expansions reproduce the perturbation theory result. The newly lifted zero frequency scales as $\sqrt{\gamma }$, so expanding ${\omega }_{-}$ as an ordinary Taylor series in $\gamma$ would be inappropriate.