[CPMD-list] free isolated atom energiesB
Oleg Yazyev
oleg.yazyev at epfl.ch
Fri Sep 17 11:44:48 CEST 2004
Dear Eduardo A. Menendez Proupin,
Once I also tried to dig such problems. Of course, I agree that the
only shperical atoms should be compared. Say, noble gases or nitrogen
atom in quartet spin state.
However, making comparison on helium atom I also noticed the there is
still quite large difference. Even if box size and pw cutoff seem to
be converged. Then I found that changing the integrator (KB to GH with different number
of integration points) gives rather big differences at this
conditions. I used norm-conserving pseudos in numerical form.
At that time I just concluded that one should not expect the same
accuracy from pw pp calculations as from numerical atomic ones...
However, this discussion is very interesting for me now.
Probably, someone can gives few valuable comments on this accuracy
issues.
Oleg
APS> 1) Like you noticed, the occupation numbers 2, 2, 1, 1 do not make really
APS> sense; well, this is leads to a non-spherical atom (p_x, p_y and p_z
APS> are occupied differently), which is used in the pseudo potential
APS> program. You should use an equal occupation of the p levels if you want
APS> to compare with Paolo Giannozzi's program (hmm, keyword 'OCCUPATION'?)
EAMP> During last days I have been involved in generation and testing of
EAMP> pseudopotentials for Cd and Te. After having performed the tests
EAMP> available in the pseudopp generation package
EAMP> (Giannozzi's) and convert the
EAMP> pseudopotentials to CPMD format, I find that the energies of the isolated
EAMP> atoms do not coincide precisely. The greatest difference betwwen CPMD and
EAMP> the atomic
EAMP> package is for the Te 5p level, where I found a difference of nearly
EAMP> 0.7eV, while 0.01 difference for Te 5s. I have tried increasing the
EAMP> cutoff and the lattice constant without improvement.
EAMP> Hence, I decided to
EAMP> use a database pseupotencial with the most common element on Earth: O.
EAMP> Even so I have two problems
EAMP> 1) Using a supplied MT pseudopotential (O_MT_LDA.psp)
EAMP> I have found .
EAMP> EIGENVALUES(EV) AND OCCUPATION:
EAMP> 1 -23.9467297NC 2.000 2 -8.7260317NC 2.000
EAMP> 3 -8.7260316NC 2.000
EAMP> CHEMICAL POTENTIAL = -8.7260316211 EV
EAMP> The energies are not precise, compared wih available references:
EAMP> http://physics.nist.gov/PhysRefData/DFTdata/Tables/08O.html
EAMP> 2s -0.871362 Ha= -23.70104 eV
EAMP> 2p -0.338381 Ha = -9.20 eV
EAMP> And the output of the atomic pp generation package og Giannozzi
EAMP> n l nl e(Ry) (eV)
EAMP> 1 0 1S( 2.00) -37.5178 -510.4594
EAMP> 2 0 2S( 2.00) -1.7424 -23.7069
EAMP> 2 1 2P( 4.00) -0.6766 -9.2055
EAMP> E(2P)-E(2S)=14.50eV
EAMP> CPMD: E(2P)-E(2S)=15.22eV.
EAMP> Morevoer, if I calculate Kohn-Sham empty state energies I find
EAMP> EIGENVALUES(EV) AND OCCUPATION:
EAMP> 1 -23.9412023 2.000 2 -10.3153247 2.000
EAMP> 3 -8.7206696 1.000 4 -8.7206695 1.000
EAMP> 5 -0.5820744 0.000
EAMP> CHEMICAL POTENTIAL = -8.7206708302 EV
EAMP> I would expect that the energies of the states 2,3 and 4
EAMP> be equal, corresponding to the
EAMP> three 2p levels. However, the state 2 has a differente. It is interesting that the
EAMP> occupation factors of 3 and 4 are 1, even when I am using
EAMP> LDA and not LSD. I guess it may be
EAMP> associated to the quasi-degeneracy.
EAMP> Here is the input file of the last calculation
EAMP> &CPMD
EAMP> RESTART DENSITY
EAMP> KOHN-SHAM ENERGIES
EAMP> 2
EAMP> MAXSTEP
EAMP> 150
EAMP> CENTER MOLECULE ON
EAMP> &END
EAMP> &DFT
EAMP> NEWCODE
EAMP> FUNCTIONAL LDA
EAMP> LDA CORRELATION PZ
EAMP> SLATER
EAMP> 0.666667
EAMP> GC-CUTOFF
EAMP> 0.1E-07
EAMP> &END
EAMP> &SYSTEM
EAMP> SYMMETRY
EAMP> 0
EAMP> ANGSTROM
EAMP> CELL
EAMP> 8.0 1.0 1.0 0 0 0
EAMP> CUTOFF SPHERICAL
EAMP> 95.0
EAMP> &END
EAMP> &ATOMS
EAMP> *O_MT_LDA.psp KLEINMAN-BYLANDER
EAMP> LMAX=D LOC=D #Also tried with LOC=P and LOC=S
EAMP> 1
EAMP> 0.000000 0.000000 0.000000
EAMP> &END
EAMP> 2) SECOND PROBLEM
EAMP> The second test I did was using ultrasoft pp.
EAMP> For example,
EAMP> performing optimization of the wavefunction for one O in
EAMP> a box of 8 Angstroms, and then calculating the Khon-Sham
EAMP> energies I found striking diferences:
EAMP> With this entry
EAMP> &CPMD
EAMP> RESTART DENSITY
EAMP> KOHN-SHAM ENERGIES
EAMP> 0
EAMP> output:
EAMP> EIGENVALUES(EV) AND OCCUPATION:
EAMP> 1 -11.2781946 2.000 2 0.7058206 2.000
EAMP> 3 0.7065389NC 2.000
EAMP> CHEMICAL POTENTIAL = 0.7065371502 EV
EAMP> *********************************
EAMP> With the same RESTART file and this input
EAMP> &CPMD
EAMP> RESTART DENSITY
EAMP> KOHN-SHAM ENERGIES
EAMP> 2
EAMP> y get this outpput:
EAMP> EIGENVALUES(EV) AND OCCUPATION:
EAMP> 1 -11.2781946 2.000 2 -0.3815193 2.000
EAMP> 3 -0.2300328 2.000 4 0.7065208 0.000
EAMP> 5 0.7065390 0.000
EAMP> CHEMICAL POTENTIAL = -0.2300326449 EV
EAMP> First problem: The energies do not seem to be correct.
EAMP> Now the "P energy is terribly wrong
EAMP> http://physics.nist.gov/PhysRefData/DFTdata/Tables/08O.html
EAMP> 2s -0.871362 Ha= -23.70104 eV
EAMP> 2p -0.338381 Ha = -9.20 eV
EAMP> Also, the states 2,3,and 4 have all diferent energies.
--
Best regards,
Oleg
______________________________________________________
Oleg Yazyev
Institute of Molecular and Biological Chemistry
Swiss Federal Institute of Technology in Lausanne (EPFL)
EPFL - BCH
CH-1015 Lausanne (Switzerland)
Tel.: +41 21 693 9881
E-mail: oleg.yazyev at epfl.ch
WWW: http://icmbcu001.epfl.ch/yazyev/index.html
_______________________________________________________
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