Dear QE users,
I am actually working on a system with C and N atoms. Checking the
convergence of the total energy for finding the optimal values for
the cutoffs (i.e. DE ~ 1mRy), I found that, despite in the atomic
case the suggested values (for example the wfc cutoff are ~ 46 Ry for
both) are good enough, in a sample of my system which is already
relaxed (and whose geometry is in good agreement with reported
results) the same convergence check determines a cutoff which is,
again for example for the wavefunction, 2 to 3 times larger.
As I tried to modify the pseudo to make it softer, I have also run
some transferability tests, which I am curious to hear your opinion
about. In particular, the tests were running fine for the testing
configurations with less electrons (e.g. 2s2 2p1 for C) but there
were problems with tests with more electrons (e.g. 2s2 2p3 for C). In
those cases the scf cycles did not converge at all, both at AE or PS
level.
I found the same result with the pseudo US,PAW in the pslibrary of
different versions, namely 1.0.0, 0.3.1 and 0.1. I also tried to
change the radii, the local potential (adding a 3D empty orbital),
the configuration (e.g, Ztot=5.5, Zval 1.5 for C) or the pseudization
recipe (TM/RRKJUS).
Hence, I got few questions:
i) is it really a transferability issue, or do I need "only" to get
those scf cycles to converge? how?
ii) if the pseudo is not good to represent electronic configurations
with more electrons, that would be a viable explanation as to why the
cutoffs for a sample systems are so much larger than the atomic cases?
Below I am reporting the output for the test of the configtf(2)='2s2
2p3'
Message from routine scf:
warning: convergence not achieved
--------------------------- All-electron run
----------------------------
C
scalar relativistic calculation
atomic number is 6.00
dft =SLA PW PBX PBC lsd =0 sic =0 latt =0 beta=0.20 tr2=1.0E-14
Exchange-correlation = SLA PW PBX PBC ( 1 4 3 4 0 0)
mesh =1073 r(mesh) = 100.30751 a.u. xmin = -7.00 dx = 0.01250
1 Ry = 13.60569193 eV, c = 137.03599966
n l nl e(Ry) e(Ha) e(eV)
1 0 1S 1( 2.00) -19.6664 -9.8332 -267.5745
2 0 2S 1( 2.00) -0.6297 -0.3148 -8.5669
2 1 2P 1( 3.00) -0.0290 -0.0145 -0.3951
final scf error: 2.4E-01 reached in 201 iterations
Etot = -78.638531 Ry, -39.319266 Ha, -1069.931632 eV
Ekin = 73.218424 Ry, 36.609212 Ha, 996.187324 eV
Encl = -182.081805 Ry, -91.040902 Ha, -2477.348944 eV
Eh = 40.989732 Ry, 20.494866 Ha, 557.693668 eV
Exc = -10.764883 Ry, -5.382441 Ha, -146.463680 eV
normalization and overlap integrals
s(1S/1S) = 1.000000 <r> = 0.2707 <r2> = 0.0993 r(max) =
0.1730
s(1S/2S) = -0.000112
s(2S/2S) = 1.000000 <r> = 1.6236 <r2> = 3.2283 r(max) =
1.2315
s(2P/2P) = 1.000000 <r> = 2.1244 <r2> = 6.4268 r(max) =
1.2470
------------------------ End of All-electron run
------------------------
Message from routine run_pseudo:
Warning: convergence not achieved
---------------------- Testing the pseudopotential
----------------------
C
scalar relativistic calculation
atomic number is 6.00 valence charge is 4.00
dft =SLA PW PBX PBC lsd =0 sic =0 latt =0 beta=0.20 tr2=1.0E-14
mesh =1073 r(mesh) = 100.30751 xmin = -7.00 dx = 0.01250
n l nl e AE (Ry) e PS (Ry) De AE-PS (Ry)
1 0 2S 1( 2.00) -0.62966 -0.17919 -0.45046 !
2 1 2P 1( 3.00) -0.02904 -0.00000 -0.02904 !
eps = 3.2E-04 iter =201
Etot = -78.638531 Ry, -39.319266 Ha, -1069.931632 eV
Etotps = -18.974270 Ry, -9.487135 Ha, -258.158068 eV
dEtot_ae = -3.108582 Ry
dEtot_ps = -1.208418 Ry, Delta E= -1.900164 Ry
Ekin = 10.222924 Ry, 5.111462 Ha, 139.089950 eV
Encl = -31.022876 Ry, -15.511438 Ha, -422.087699 eV
Ehrt = 12.620743 Ry, 6.310371 Ha, 171.713935 eV
Ecxc = -10.795060 Ry, -5.397530 Ha, -146.874254 eV
(Ecc = -0.958640 Ry, -0.479320 Ha, -13.042955 eV)
---------------------- End of pseudopotential test
----------------------
-------------- Test with a basis set of Bessel functions
----------
Box size (a.u.) : 30.0
Cutoff (Ry) : 30.0
N = 1 N = 2 N = 3
E(L=0) = -0.1788 Ry 0.1213 Ry 0.1854 Ry
E(L=1) = 0.1263 Ry 0.1949 Ry 0.2715 Ry
Cutoff (Ry) : 60.0
N = 1 N = 2 N = 3
E(L=0) = -0.1789 Ry 0.1213 Ry 0.1854 Ry
E(L=1) = 0.1263 Ry 0.1949 Ry 0.2715 Ry
Cutoff (Ry) : 90.0
N = 1 N = 2 N = 3
E(L=0) = -0.1789 Ry 0.1213 Ry 0.1854 Ry
E(L=1) = 0.1263 Ry 0.1949 Ry 0.2715 Ry
Cutoff (Ry) : 120.0
N = 1 N = 2 N = 3
E(L=0) = -0.1789 Ry 0.1213 Ry 0.1854 Ry
E(L=1) = 0.1263 Ry 0.1948 Ry 0.2715 Ry
-------------- End of Bessel function test
------------------------
Thank you in advance,
