# 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 $U_{dd}$ 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 $2*\Delta+U_{dd}$

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

, 2020/02/03 18:50, 2020/02/03 18:54

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.

You could leave a comment if you were logged in.