# Why does it take 5-10 steps to obtain reasonable orbitals during initialization in VASP?

I am trying to figure out the initialization process of VASP, and I found an INCAR tag NELMDL, on the wiki page of NELMDL, it is said that VASP requires 5-10 steps to obtain the reasonable orbitals. This seems strange, when setting ISTART=0 and INIWAV=1, the initial charge density is calculated by taking the superposition of atomic charge densities. So I could get reasonable charge density directly, and use it to calculate the reasonable orbitals by just one step.
My question is: Why does it take 5-10 steps to obtain reasonable orbitals?

Default: NELMDL
= -5 if ISTART=0, INIWAV=1, and IALGO=8
= -12 if ISTART=0, INIWAV=1, and IALGO=48
Description: NELMDL specifies the number of non-selfconsistent steps at the beginning.
If the orbitals are initialized using a random number generator (the default in VASP), the initial orbitals are usually unreasonable and the iterative matrix diagonalization will require 5-10 steps to obtain reasonable orbitals. The charge density corresponding to the initial orbitals is also, at best, erratic. It is hence advisable to perform a few electronic steps while keeping the initial Hamiltonian fixed. This initial Hamiltonian is usually determined from a superposition of atomic charge densities.

Here is the Schematic representation of the self-consistent loop for the solution of Kohn–Sham equations. I think it would only take just once loop to get reasonable orbitals. • Could you please expand on how you plan to perform "get reasonable charge density directly, and use it to calculate the reasonable orbitals by just one step" . Dec 15, 2021 at 12:50
• @IanBush, I just add a picture to show my idea, I think if I set the charge density as the superposition of all atoms' charge density. I could just do this calculation loop once and obtain reasonable orbitals.
– Jack
Dec 15, 2021 at 13:06
• And how are you doing the kinetic energy term given just the charge density? Dec 15, 2021 at 14:01
• p.s. I'm not a plane wave guy so I'm guessing, hence comments rather than an answer, but from the little I know I don't see how you can evaluate the KE term in your Hamiltonian given only the electron density Dec 15, 2021 at 14:06
• I guess the "5-10 steps" refer to iterations within the iterative diagonalization method, not the SCF iterations. In other words, the "Solve KS equation" box is itself an iterative procedure, and that procedure has to be iterated 5-10 steps Dec 15, 2021 at 14:57

addendum: you've added the flowchart in the question. The steps that the manual talks about happen in the "solve KS equation" phase. Once you have updated values for the occupied orbitals $$\psi_i^\sigma$$, you calculate the new electron density and the new potential, and solve again the KS equation with the new potential. Once the orbitals don't change any more, they have become self-consistent.