Abstract
AbstractThe molecular spin–orbit coupling operator is brought into a simplified form through a convenient choice of origin for the orbital angular momentum operator. The eigenvalue problem of the Hamiltonian that includes the spin–orbit (SOC) operator as a perturbation is solved by means of a linear variational procedure in the basis of the spin‐pure molecular eigenstates. Test calculations on benzophenone are presented and the results are compared to experiment. We discuss the minimal size of the spin‐pure variational basis needed to achieve stable results as well as the amount of single‐excitation configurational mixing needed to describe the spin‐pure molecular eigenstates.
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