Understanding of onsite energies of ed and eL in multiplet ligand field theory.
asked by Ruiwen Xie (2025/03/04 14:42)
Dear Prof. Haverkort and Quanty developer,
I have a question regarding the way of interpreting the defined ed and eL in ligand field theory.
We have ed = (10*Delta-nd*(19+nd)*Udd/2)/(10+nd) and eL = nd*((1+nd)*Udd/2-Delta)/(10+nd) defined. Usually, for example, in the case of Ni, ed is negative while eL is positive. I am aware of that the ed and eL are defined for the d shell with nd localized electrons, which will be different from those given by DFT or Hartree Fock. However, my question is, when we decrease the energy difference between ed and eL |ed-eL| by changing the value of U or Delta, can we understand the decrease of |ed-eL| from the perspective of bonding/anti-bonding states? That is, smaller |ed-eL| would drive the bonding states closer to the Fermi energy, thus leading to stronger hybridization activities between TM d and ligand p orbitals?
Thanks a lot for your time!
Answers
Dear Ruiwen,
the reason $e_d < 0$ is mostly due to the $n_d * (19 + n_d)*U_{dd} / (10+n_d)$ term. In DFT or Hartree-fock the orbital energies include the mean-field average of the Coulomb interaction. In Quanty we add the Coulomb interaction explicitly. i.e. it is not included in $e_d$. You can either work with $U$ and $\Delta$ as done by Sawatzky et al. or use $e_d$ and $e_L$ as obtained from DFT or Hartree-Fock, but then, when you add the Coulomb interaction you do need to subtract the mean-field approximation of the Coulomb interaction.