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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.

Let $Q_k$ be a vibrational normal mode coordinate with symmetry ${\Gamma }_k$. The harmonic vibrational ground state is totally symmetric; the one-quantum state transforms as $Q_k$. Integration over a symmetry-invariant domain annihilates any integrand lacking the totally symmetric representation. Thus a fundamental requiresfor some Cartesian component $\alpha$. For the one-dimensional irreducible representations used here, this means ${\Gamma }_k=\Gamma (x)$, $\Gamma (y)$ or $\Gamma (z)$. Equivalently, expanding the electric dipole moment givesthe derivative can be nonzero only if $Q_k$ and ${\mu }_{\alpha }$ have compatible symmetry. This is the symmetry criterion for vibrational infrared activity; it permits a transition without guaranteeing a large intensity.