Since we know, that the lattice parameter is 3.66 Angstrom, here is the structure for testing the sufficiency of the k-points sampling of the IBZ. The geometry is in Cu_111.bas file:
5 29 0.00000 1.49419 10.00000 29 0.00000 0.00000 12.11310 29 1.29401 0.74709 14.22620 29 0.00000 1.49419 16.33931 29 0.00000 0.00000 18.45241
And the lattice vector for the Cu (111) surface is in Cu_111.lvs:
2.58801 0.00000 0.00000 1.29401 2.24128 0.00000 0.00000 0.00000 99.99999
The last 2 layers will be fixed during optimization, therefore FRAGMENTS file contains:
0 1 2 4 1 1 1 5 1 1 1
The amount of the k-points is not known, therefore one should check, whether the total energy and geometry changes with amount of the k-points. For this reason 3 files containing 8, 18 and 32 k-points: Cu_111.8.kpts contains 4×4 k-points reduced to 8 k-points by inversion:
8 -0.91042715 -0.52563387 0.00000000 0.12500000 -0.91042715 0.17521401 0.00000000 0.12500000 -0.91042715 0.87606192 0.00000000 0.12500000 -0.91042715 -1.22648176 0.00000000 0.12500000 -0.30347571 -0.87605917 0.00000000 0.12500000 -0.30347571 -0.17521130 0.00000000 0.12500000 -0.30347571 0.52563661 0.00000000 0.12500000 -0.30347571 1.22648454 0.00000000 0.12500000
Cu_111.18.kpts contains 6×6 k-points reduced to 18 k-points by inversion:
18 -1.01158571 -0.58403766 0.00000000 0.05555556 -1.01158571 -0.11680572 0.00000000 0.05555556 -1.01158571 0.35042620 0.00000000 0.05555556 -1.01158571 0.81765813 0.00000000 0.05555556 1.41621998 -0.11681113 0.00000000 0.05555556 -1.01158571 -1.05126962 0.00000000 0.05555556 -0.60695142 -0.81765449 0.00000000 0.05555556 -0.60695142 -0.35042259 0.00000000 0.05555556 -0.60695142 0.11680934 0.00000000 0.05555556 -0.60695142 0.58404124 0.00000000 0.05555556 -0.60695142 1.05127323 0.00000000 0.05555556 -0.60695142 -1.28488645 0.00000000 0.05555556 -0.20231713 -1.05127132 0.00000000 0.05555556 -0.20231713 -0.58403945 0.00000000 0.05555556 -0.20231713 -0.11680753 0.00000000 0.05555556 -0.20231713 0.35042441 0.00000000 0.05555556 -0.20231713 0.81765634 0.00000000 0.05555556 -0.20231713 1.28488827 0.00000000 0.05555556
Cu_111.32.kpts contains 8×8 k-points reduced to 32 k-points by inversion:
32 -1.06216502 -0.61323953 0.00000000 0.03125000 -1.06216502 -0.26281559 0.00000000 0.03125000 -1.06216502 0.08760835 0.00000000 0.03125000 -1.06216502 0.43803230 0.00000000 0.03125000 -1.06216502 0.78845626 0.00000000 0.03125000 1.36564067 -0.26282105 0.00000000 0.03125000 1.36564067 0.08760300 0.00000000 0.03125000 -1.06216502 -0.96366343 0.00000000 0.03125000 -0.75868928 -0.78845221 0.00000000 0.03125000 -0.75868928 -0.43802825 0.00000000 0.03125000 -0.75868928 -0.08760429 0.00000000 0.03125000 -0.75868928 0.26281965 0.00000000 0.03125000 -0.75868928 0.61324358 0.00000000 0.03125000 -0.75868928 0.96366751 0.00000000 0.03125000 -0.75868928 1.31409144 0.00000000 0.03125000 -0.75868928 -1.13887617 0.00000000 0.03125000 -0.45521358 -0.96366483 0.00000000 0.03125000 -0.45521358 -0.61324090 0.00000000 0.03125000 -0.45521358 -0.26281694 0.00000000 0.03125000 -0.45521358 0.08760700 0.00000000 0.03125000 -0.45521358 0.43803096 0.00000000 0.03125000 -0.45521358 0.78845489 0.00000000 0.03125000 -0.45521358 1.13887882 0.00000000 0.03125000 -0.45521358 -1.31408879 0.00000000 0.03125000 -0.15173785 -1.13887751 0.00000000 0.03125000 -0.15173785 -0.78845352 0.00000000 0.03125000 -0.15173785 -0.43802959 0.00000000 0.03125000 -0.15173785 -0.08760565 0.00000000 0.03125000 -0.15173785 0.26281831 0.00000000 0.03125000 -0.15173785 0.61324227 0.00000000 0.03125000 -0.15173785 0.96366620 0.00000000 0.03125000 -0.15173785 1.31409013 0.00000000 0.03125000
The fireball.in should look like this:
&OPTION basisfile = Cu_111.bas lvsfile = Cu_111.lvs kptpreference = Cu_111.XXX.kpts nstepf = 5000 iquench = -6 ! -1 = MDmin, -6 = FIRE optimization icluster = 0 iqout = 1 dt = 0.5 &END &OUTPUT iwrtxyz = 1 &END
Now the optimization should be performed with different k-point sampling. The total energy could be simply checked from the output file:
grep "ETOT" name_of_the_output_file | tail -1
If the total energy and the geometry doesn't differ too much (comparing two different k-point sampling) then the lower amount of the k-points is considered as the sufficient one.
