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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]

The inferred dipole moment is approximately $0.626\mathrm{D}$. With the static field parallel to the radiation electric field, the microwave selection rule is $\Delta M_J=0$. A fully resolved three-component first-order rotational Stark effect pattern naturally identifies the $J=1\to 2$, $|K|=1$ transition: the three initial $M_J=-1,0,1$ values give the three components. $K=0$ would have no first-order shift.

Subtract the lower-state field shift from the upper-state field shift, rather than equating a transition shift to one level's shift:For $J=1$ and $|K|=1$, neighbouring components are separated by $\mu E_{\mathrm{field}}/(3h)$. HenceThis assignment assumes the stated splitting resolves all allowed first-order components, rather than three accidentally unresolved groups of a higher-$J$ transition.

The quadratic rotational Stark effect is a small-field expansion, rather than a response law valid at every field. For a polar linear rigid rotor with rotational constant $B$ in energy units and permanent electric dipole moment $p$, a static field $\mathcal{E}$ gives the Hamiltonian operatorThe dimensionless ratio $p\mathcal{E}/B$ measures competition between rotation and orientation. Once it is not small, the field mixes many field-free angular momentum eigenstates, so diagonalizing the fixed-magnetic quantum number matrix replaces a truncated second-order correction. Increasing the basis cutoff tests numerical convergence.

For a differentiable nondegenerate eigenvalue $E_{\alpha }(\mathcal{E})$, the Hellmann–Feynman theorem gives the mean dipole component $-d{E}_{\alpha }/d\mathcal{E}=p⟨\cos \theta ⟩$. Its magnitude cannot exceed $p$. Continuing the weak-field linear induced-dipole formula indefinitely would violate this bound. Strongly oriented low-lying pendular states instead describe restricted angular motion about the field direction.^[ref]