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Orbital energies

asked by Helene (2020/03/31 14:05)

Hello,

I am used to looking at the energies of one-particle functions in DFT calculations. These make sense to me.

When doing ligand-field calculations with Quanty the energy of the configurations are written in terms of orbital energies. However, these values are not consistent with my assumptions. Are there other contributions to these energies that would make them “compatible” with the DFT results? If there are how can these be printed in Quanty.

Thank you for your help.

Answers

, 2020/04/17 10:42, 2020/04/17 10:43

Dear Helene,

Quanty is a many body code. The energy it produces is the total energy of the system. You can compare this to the energy you get in a DFT code when you sum the energies of all occupied orbitals.

For the comparison it is furthermore important to realise that DFT within the Kohn-Sham formalism and one of the implemented functionals is a mean-field approximation. The full Coulomb interaction is replaced by a potential. In Quanty we solve the full many body problem. (Naturally at a large computational cost).

A few examples: 1) If you have an dxy and dz2 orbital occupied and move the dz2 orbital to the dxxy2 orbital you need to consider not only the orbital energy difference between the dz2 and dx2y2 orbital, but also the difference in Coulomb interaction between the dxy and dx2y2 compared to the Coulomb interaction between the dxy and dz2 orbital.

2) Eigen-states in Quanty are not necessarily states with integer occupation of the different one particle orbitals (Single Slater Determinants). Due to Coulomb interaction different Single Slater Determinants can mix.

Effect 1 can be captured by time dependent mean field theories Effect 2 is a true manifestation of many body physics and allows us to build Quantum computers that are exponentially more efficient than regular computers. It also makes the calculations of the properties of Solids extremely difficult. And leads to a complexity in our natural world that we observe and would otherwise not be able to generate from the Schrödinger equations.

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