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Configuration energies in ligand field calculations
asked by Helene (2020/02/03 17:42)
Dear developers,
I have two simple questions (hopefully) about ligand field calculations with Quanty.
Firstly I don't understand what is the origin of the Udd in the second equation below (this is a comment section from the XAS L2,3 calculation with ligand field).
-- The L^10 d^n configuration has an energy 0 -- The L^9 d^n+1 configuration has an energy Delta -- The L^8 d^n+2 configuration has an energy 2*Delta+Udd
Secondly, in the final state of core photoemission calculations the energy of which electronic configuration is set to zero? Is it the p^5 d^n configuration or another? Why?
All the best, Hélène
Answers
Dear Hélène,
two important questions indeed.
1) We follow the configuration energy definitions as for example used in the Zaanen Sawatzky Allen paper on band-gaps (fig 1.). If we define the onsite energy of the d-orbitals as ed and the onsite energy of the ligand orbitals as eL and add a Coulomb interaction $U$ to the electrons of the d-shell we have the configuration dependent energies:
$d^n L^{10} \to n \, ed + 10\, eL + \frac{( n)(n-1)}{2}\, U $
$d^{n+1} L^9 \to (n+1)\, ed + 9\, eL + \frac{(n+1)(n )}{2}\, U $
$d^{n+2} L^8 \to (n+2)\, ed + 8\, eL + \frac{(n+2)(n+1)}{2}\, U $
If you work this out and set the energy difference between the $d^n L^{10}$ and $d^{n+1} L^9$ configuration to be $\Delta$ you will find that the energy difference between the $d^{n+2} L^8$ and $d^{n+1} L^9$ configuration is $\Delta + U$
2) We normally do not calculate the absolute energy of a core level excitation. It is very hard to get this to be better than 1% accurate. This might sound good, but for a core level at 1 keV this is still of by 10 eV. That means we can choose a zero of energy. Any value you pick is possible and equally good. I standardly pick the $p^5 d^n$ configuration to be at zero. Nothing special other than convention and could have / can be done differently.