In an aromatic molecule of \(D_{2 \mathrm{h}}\) symmetry the lowest triplet term was identified as \(^{3} \mathrm{B}_{1 \mathrm{u}} .\) What is the polarization of its phosphorescence? Hint. Decide which singlet terms can mix with \(^{3} \mathrm{B}_{1 \mathrm{u}}\) and assess the polarization of the light involved in the return of that state to the \(^{1} \mathrm{A}_{\mathrm{g}}\) ground state.

First, identify the transition rules between electronic states. According to these rules, singlet and triplet terms are not allowed to mix directly. However, spin-orbit coupling can make them interact indirectly. Since \(D_{2\mathrm{h}}\) is centrosymmetric, mixing can occur by spin-orbit coupling through Herzberg–Teller vibronic coupling. The singlet terms that can interact and mix with \(^{3}\mathrm{B}_{1\mathrm{u}}\) can be determined from the symmetry of the spin-orbit operator as well as the singlet states accessible from \(^{1}\mathrm{A}_\mathrm{g}\) and symmetry conservation rules. All candidate singlet states are \(^{1}\mathrm{B}_{1\mathrm{u}}, ^{1}\mathrm{B}_{2\mathrm{u}},\) and \(^{1}\mathrm{B}_{3\mathrm{g}}\).

The corresponding matrix elements of the transition dipole moments can now be determined, which relate to the polarization of the emitted light during the phosphorescence process. For allowed electric dipole transitions, the integral \(\langle \mathrm{B}_{1\mathrm{u}} | z | \mathrm{A}_{\mathrm{g}} \rangle\) is non-zero. Thus, it indicates that the emitted light in the return of the state \(^{3} \mathrm{B}_{1\mathrm{u}}\) to the \(^{1}\mathrm{A}_\mathrm{g}\) ground state is z-polarized.