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,
--
Aldo Ugolotti, Ph.D.
Post-doc fellow
Materials Science Dept. U5,
Università degli Studi di Milano-Bicocca
via Cozzi 55,
20125 Milano (MI)
Italy
e-mail: a.ugolo...@campus.unimib.it
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