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. enter image description here

  • $\begingroup$ 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" . $\endgroup$
    – Ian Bush
    Dec 15, 2021 at 12:50
  • $\begingroup$ @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. $\endgroup$
    – Jack
    Dec 15, 2021 at 13:06
  • $\begingroup$ And how are you doing the kinetic energy term given just the charge density? $\endgroup$
    – Ian Bush
    Dec 15, 2021 at 14:01
  • $\begingroup$ 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 $\endgroup$
    – Ian Bush
    Dec 15, 2021 at 14:06
  • 1
    $\begingroup$ 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 $\endgroup$
    – wzkchem5
    Dec 15, 2021 at 14:57

1 Answer 1


Most quantum chemistry codes that employ a linear combination of atomic orbitals approach employ dense matrix diagonalization, meaning that once you have the Fock matrix i.e. potential, you get the orbitals it generates in a single step.

However, VASP, like every other plane wave code, employs an iterative diagonalizer / relaxation scheme to update the orbitals. Thus, even if you have a good guess for the electron density <=> the potential the electrons move in, you still have to perform several iterations to determine the orbitals that are eigenstates in this potential.

However, since the potential from the superposition of atomic densities (SAD) guess differs from the potential generated by the true ground state electron density, it is often more efficient to omit full optimization of the initial guess orbitals, since the orbitals are wrong anyway. So, instead of converging the guess orbitals (which may require 20-30 iterations), 5-10 is sufficient to get the orbitals "close enough".

Some codes employ an alternative approach in which the guess orbitals are obtained from a small to minimal-basis LCAO calculation, which is then projected onto the plane wave basis. However, this requires more code than the initialization with random numbers + iterative diagonalization with the SAD potential.

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.

  • 1
    $\begingroup$ Every step in the diagonalization updates the orbitals, which describe the electrons. $\endgroup$ Dec 16, 2021 at 18:09
  • $\begingroup$ @Jack does this answer your question? $\endgroup$ Dec 24, 2021 at 20:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .