Quadratic Stark shift of a polar linear rotor

The dipole coupling $-pE\cos \theta$ connects only $J\pm 1$ at fixed $m$. For $J\ge 1$, second order gives $(pE{)}^2[J(J+1)-3m^2]/[2BJ(J+1)(2J-1)(2J+3)]$; the ground-state result is separately $-(pE{)}^2/(6B)$. Here $B$ has energy units and $\theta$ is the polar angle to the field.

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]

A pendular state is a field-dressed rotational quantum state in which a molecule's orientation is concentrated near a preferred direction, resembling small angular librations of a pendulum. It arises when the orienting or aligning field interaction competes strongly with rotational kinetic energy.

A permanent dipole in a static electric field has an interaction proportional to $-\cos \theta$, which distinguishes opposite directions. An anisotropic induced-polarization interaction often depends on $-{\cos }^2\theta$, aligning an axis without necessarily orienting its two ends. The field strength and rotational constant determine whether a weak-field perturbation treatment is adequate.

For $J=0\to 1$, the $m=0$ and $|m|=1$ frequency shifts are $4(pE{)}^2/(15Bh)$ and $7(pE{)}^2/(60Bh)$. Their difference is $3(pE{)}^2/(20Bh)$, so $p=(h/E)\sqrt{(20/3)(B/h)\Delta \nu }$. The common lower-state shift cancels from the splitting but not from either absolute transition frequency.

Radiation polarized along the static field drives the zero-projection change, while transverse polarization supplies the two circular components with projection changes plus or minus one. The opposite signs of the excited magnetic quantum number remain degenerate in a pure static electric field; they need not be separate spectral lines.

Every state in a fixed-$J$ manifold has parity $(-1{)}^J$, while the dipole component is odd. Therefore its complete matrix within that degenerate manifold vanishes. One must check the whole degenerate block, not only one diagonal element, before concluding that first-order splitting is absent.