DFTtoMLFT Energy definition for XPS calculation
asked by Hamza (2025/01/28 12:53)
Dear Quantum Developer,
Inspired by the XAS script for DFTtoMLFT, we aim to calculate the XPS.
Regarding the energy definitions, we encountered different scenarios. Let’s consider the following configurations:
1. \( p^6L^{18}d^n \): ground state 2. \( p^5L^{18}d^n \): excitation from the \( p \)-orbital to the vacuum 3. \( p^6L^{17}d^{n+1} \): charge transfer (CT) from the ligand to the transition metal (TM)
Based on these scenarios, I derived the following equations:
\[ 6x + 18y + ndz + \frac{nd(nd-1)U_{dd}}{2} + 6ndU_{pd} = 0 \] \[ 5x + 18y + ndz + \frac{nd(nd-1)U_{dd}}{2} + 5ndU_{pd} = 0 \] \[ 6x + 17y + (nd+1)z + \frac{nd(nd+1)U_{dd}}{2} + 6(nd+1)U_{pd} = \Delta \]
Here, \( ep_{\text{final}} \), \( eL_{\text{final}} \), and \( ed_{\text{final}} \) correspond to \( x \), \( y \), and \( z \), respectively.
Are these equations correct?
After performing the calculations and plotting the XPS and XAS spectra on the same figure, I noticed a difference in the energy interval between the two peaks. The XPS peaks appear closer to each other compared to the XAS peaks, even though I used the same parameters for both calculations.
Could this difference be due to the coefficients in front of \( U_{pd} \) in the above equations, or is there another explanation?
Best regards, Hamza
Answers
Dear Hamza,
I'm not 100% sure I understand your question correctly, but there is a difference between a charge neutral experiment (XAS) and an experiment whereby you remove one electron (core level XPS). If you look at |fig 1.1 on page 3 at this link you get an energy level diagram for the different experiments. The $d^n$ configuration is at zero, the $\underline{L}^1 d^{n+1}$ configuration at an energy $\Delta$. With a core hole the energies shift by $U_{cd}$ with a core hole and extra $d$ electron by $U_{dd}-U_{cd}$.
Does this help?
PS you can test your script by multiplying all hoppings by 0.001, setting $F_2$ and $F_4$ to zero and taking $\Delta=0$. Then the energies you get should be only determined by the $U$'s and $\Delta$.
Best wishes, Maurits