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I'm performing complete active space calculations for the prediction of X-ray absorption L edge spectra. This involves an excitation from 2p orbitals to the 3d orbitals of the metal. For the electronic configuration of 2p5 3d5 I can calculate the total number of microstates from "n choose k" to be 1512 however the total number of pure spin states is 600. Could someone please explain how to obtain this number?

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  • $\begingroup$ What do you mean by "microstates"? From $2K$ spin-orbitals with $N$ electrons, you get $\begin{pmatrix}2K\\N\end{pmatrix}$ Slater determinants but not all of them are eigenfunctions of the total electron spin operator. You can take appropriate linear combinations of determinants that form spin-adapted configurations or configuration state functions (CSFs). For configurations with a large number of unpaired electrons and a low total spin, the number of CSFs is significantly smaller than the number of determinants. $\endgroup$ – Antonio de Oliveira-Filho Sep 25 '18 at 19:54

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