= Solution
The quoted expansion is <nondegenerate perturbation theory> for a known isolated eigenstate of a time-independent <Hamiltonian>. A useful small-mixing condition is $|\langle k|H_1|i\rangle|\ll|E_i-E_k|$ 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.<Quantum Mechanics by Richard Fitzpatrick>[Non-degenerate perturbation theory and its breakdown at coupled degeneracies]{ref}
For the <rotational Stark effect>, $H_1=-pE\cos\theta$, where $\theta$ is the \b[<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 $J\pm1$ with the same $m$ contribute in second order. Using their <energy> differences $2BJ$ and $-2B(J+1)$,
$$
\delta E_{Jm}^{(2)}=\frac{(pE)^2}{2B}\left[\frac{J^2-m^2}{J(2J-1)(2J+1)}-\frac{(J+1)^2-m^2}{(J+1)(2J+1)(2J+3)}\right].
$$
For $J\ge1$ this simplifies to the <quadratic Stark shift of a polar linear rotor>,
$$
\boxed{\delta E_{Jm}^{(2)}=\frac{(pE)^2[J(J+1)-3m^2]}{2BJ(J+1)(2J-1)(2J+3)}.}
$$
For $J=0$, only $J'=1,m=0$ contributes, with squared angular <matrix element> $1/3$ and denominator $-2B$:
$$
\boxed{\delta E_{00}^{(2)}=-\frac{(pE)^2}{6B}.}
$$
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 $m=\pm1$. 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>:
$$
\Delta\nu=\frac{3(pE)^2}{20Bh},\qquad\boxed{p=\frac hE\sqrt{\frac{20}{3}\frac Bh\Delta\nu}=2.39\times10^{-30}\,\mathrm{C\,m}\simeq0.717\,\mathrm D.}
$$
The $\mathrm D$ unit is the <debye>. Using external $h=6.62607015\times10^{-34}\,\mathrm{J\,s}$ and $1\,\mathrm D=3.33564\times10^{-30}\,\mathrm{C\,m}$ gives $pE/B\simeq0.0593$, 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 $\pm1$. The inferred value is a magnitude; these shifts do not determine the dipole's sign.<CODATA recommended values of the fundamental physical constants: 2022 by National Institute of Standards and Technology>[Exact SI Planck constant]{ref}<Triatomic Spectral Database by National Institute of Standards and Technology>[Section5: <debye> conversion]{ref}
\Image[/past-exam-of-the-advanced-quantum-physics-course-of-the-university-of-cambridge/2014/paper-1-aqp2014-rotor-stark-shifts.png]
{title=Opposite-parity mixing produces quadratic rotor shifts and a polarization-dependent rotational doublet}
{source=paper-1-aqp2014-rotor-stark-shifts.py}
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