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The resolved lines arise from simultaneous changes of vibration and rotation in hydrogen cyanide. A vibrational fundamental connects many populated lower rotational levels to different upper rotational levels, giving the fine structure of a rovibrational spectrum.

The prominent central Q branch, together with flanking P branch and R branch lines, identifies a perpendicular band. For linear HCN, this is the doubly degenerate bending fundamental, not either stretching fundamental. Bending creates a transition-dipole component perpendicular to the molecular axis; its vibrational angular momentum permits $\Delta J=0$ as well as $\Delta J=\pm 1$.

The two perpendicular bending coordinates share one vibrational frequency. Their degeneracy is different from the series of resolved rotational transitions superimposed on that fundamental. The stretches have their electric dipole transition moment along the molecular axis and would give the usual parallel-band P/R pattern without this strong central Q feature.

The transition strength depends on $|R_{ji}{|}^2$, so a necessary condition is that at least one Cartesian electric dipole transition moment is nonzero. An integral can be nonzero only if its integrand contains the totally symmetric representation:The harmonic vibrational ground state is totally symmetric, and a one-quantum state of normal mode $k$ transforms as ${\Gamma }^{(k)}$. For the real irreducible representations here, the fundamental is therefore symmetry-allowed precisely whenEquivalently, expand the electric dipole moment near equilibrium:The fundamental has $⟨1_k|Q_k|0_k⟩\ne 0$, so vibrational infrared activity requires a nonzero dipole derivative of the matching symmetry. In $D_{6h}$, $z\sim A_{2u}$ and $(x,y)\sim E_{1u}$, giving $A_{2u}$ and $E_{1u}$ as the IR-active species. Symmetry permission does not guarantee a large band intensity.

The initial vibronic state is $A_{1g}$ and the final one is $E_{1u}$, so an in-plane dipole component satisfiesThe product contains the totally symmetric representation. The combined electronic–vibrational transition is therefore symmetry-allowed, with an in-plane electric dipole transition moment.

A nonzero intensity still needs vibronic coupling. In a coordinate-independent separation, the forbidden electronic matrix element would remain zero. Its first derivative with respect to the $E_{2g}$ coordinate can instead be nonzero, and the vibrational factor $⟨1|Q|0⟩$ supplies the required one-quantum transition. This is vibrational activation of an electronic transition and explains why the band can be weak.

A coordinate $Q_s$ of an $A_{1g}$ normal mode is unchanged by every molecular symmetry operation. Every harmonic-oscillator excitation $H_n(Q_s)e^{-Q_s^2/2}$ therefore also has symmetry $A_{1g}$, including odd values of $n$. Consequentlyfor any number of quanta in that totally symmetric coordinate. The same electric-dipole permission survives every $A_{1g}$ excitation.

This statement concerns symmetry. Actual progression intensities depend on vibrational inner products, equilibrium displacements and the coordinate dependence of the electric dipole transition moment; a permitted line need not have measurable intensity for every vibrational quantum number.

An otherwise forbidden electronic transition gains intensity when accompanying molecular vibration supplies a symmetry factor that permits a nonzero electric dipole transition moment. A symmetry-allowed total vibronic state specifies a possible transition, while the coupling determines its actual strength.

Coupling between electronic and nuclear vibrational motion. Dependence of an electronic electric dipole transition moment on a normal mode coordinate can give intensity to a transition whose purely electronic matrix element vanishes at the equilibrium geometry.