| Commit | Date | |
|---|---|---|
| 2017-06-27 11:49:04 | Tree | |
| 2017-06-26 14:52:36 | Tree | |
|
[r298]
by
paulatz
quick return of the ij term when the fermi shift is zero (i.e.e high symmetry crystal) |
2017-06-26 14:52:29 | Tree |
|
[r297]
by
paulatz
better code alignment --This line, and thos below, will be ignored-- M incdrhoscf2.f90 |
2017-06-26 14:51:41 | Tree |
|
[r296]
by
paulatz
Tentative try to get the same result with nbnd_occ = nbnd and without, also without breaking the code in the insulating case when nbnd>nelec/2 I'm relatively convinced that the correct nbnd_occ to use in solve linter is the one at ik_wfc (i.e. at k) and not at ik_prj (i.e. k-q in the simple case), but I'm not 100% sure. However, it does go much better like this Also, it is still not perfect, and there are more residualse in the FFT when using this with respect to using nbnd_occ=nbnd |
2017-06-26 14:51:12 | Tree |
| 2017-06-26 14:48:13 | Tree | |
| 2017-06-26 14:47:37 | Tree | |
|
[r293]
by
paulatz
Checkpoint to solve the nbnd_occ problem one ancd for all, |
2017-06-26 09:16:15 | Tree |
| 2017-06-24 12:20:10 | Tree | |
| 2017-06-22 09:37:46 | Tree |