Sample directory: adf/SCF_Ti2O4/
One can run into SCF convergence problems when calculating certain types of systems. Some of the notorious examples are transition metal oxides and lantanide compounds. Below, three approaches to solving the SCF convergence problem are demonstrated.
NewDIIS keyword
The first approach is to try a new DIIS algorithm, which will probably become default in a future version. The new algorithm is switched on by using the keyword NewDIIS anywhere in the input file:
$ADFBIN/adf << eor Title Ti2O4 SCF aid test (NewDIIS) Atoms Ti 1.730 0.000 0.000 Ti -1.730 0.000 0.000 O 0.000 1.224 0.000 O 0.000 -1.224 0.000 O 3.850 0.000 0.000 O -3.850 0.000 0.000 End XC GGA Becke Perdew End Basis Type DZ Core Small End SCF Iterations 300 End NewDIIS End input eor
Multi-step smearing
Second approach is an extension to the so-called "electron smearing" method. In this method, the electrons are distributed among orbitals around Fermi-level using a pseudo-thermal distribution function. Although the result with fractional occupation number has no physical sense, the method can be used to achieve integer occupation numbers by reducing the smearing parameter step-wise. In the example above, replace the NewDIIS keyword with the following line of text:
Occupations Smear=0.2,0.1,0.07,0.05,0.03,0.02,0.01,0.007,0.005,0.001
A few notes:
Steepest descent method
The third example demonstrates the use of the Occupations Steep= option (see the User's Guide for details). There are two differences from the previous example shown below:
SCF Iterations 300 Mixing 0.05 DIIS N=0 End Occupations Steep=0.5,0.3
One difference is, obviously, in the Occupations keyword. The other difference is more subtle. For stable convergence, it is often essential to switch off DIIS and set the mixing parameter to a low value. Of course, it will make convergence quite (sometimes very) slow. Ultimately you should get either an aufbau configuration or a configuration with exactly degenerate HOMO. In this example, the result is an aufbau solution.
Both methods should, in principle, give the same result, which is the case in this example.




