[CPMD-list] doubt of titanium Ti pseudopotential

[Axel Kohlmeyer]

>>> "EM" == Eduardo Ariel Menendez P <Eduardo> writes:


EM> Axel,
EM> Thank you very much. Some doubts about your explanations.

hello eduardo,

let's see if i can resolve them.

EM> I have the following two doubts. The follwing input and output refers
EM> to a calculation of Ti_6Si_2C_4. I use a supplied pseudopotential that has
EM> the valence shell 3s2 3p6 4s2 3d2. This makes a total of 10 electrons per
>> 
>> i count 12 valence electrons per Ti here.

EM> I apologize. 3s2 is core.

>> 
>> your output describes this potential differently. you have 5 core and
>> 2 valence states. probably the 4s and 3d state. so the initial guess is
>> correct. from my (still somewhat limited knowledge) my educated guess to
>> use this pseudopotential would be LMAX=D LOCAL=S SKIP=P (if anyone knows
>> better, please tell me/us).

EM> What is the meaning of SKIP=P? Manual reads "In addition it is possible to
EM> assign the local potential to a further potential with  SKIP=l". Does it
EM> means that with "LMAX=D LOCAL=S SKIP=P", both S and P chnnels are treated
EM> as local (and different) potentials?

you are right, from the text in the manual SKIP makes no sense here. sorry.

EM> This is the cpmd report of the pseudopotential:
EM> *   ATOM       MASS   RAGGIO NLCC              PSEUDOPOTENTIAL *
EM> *     Ti    47.9000   1.2000  YES    KLEINMAN      S  NONLOCAL *
EM> *                                                  P     LOCAL *
EM> *                                                  D  NONLOCAL *
EM> *                                                  F  NONLOCAL *
EM> I should have 3 valence states. F is not occupied. Why does CPMD ignores the P states?

since it will assume, that you want the 4p state, which is empty.
please look into the atom section of the potential:
&ATOM  
 Z  =   22
 ZV =    4
 XC = 1312        .666667
 TYPE = NORMCONSERVING NUMERIC
&END   
so you have a valence charge of 4. i.e. you have only 4s and 3d in 
the valence. also the pseudopotential report says 5 core and 2 valence
states. if the 3p would have been included, there would be 
3 valence states. also if you look at the pseudo wavefunctions from
the &WAVEFUNCTION section, you can see that the p- and f- wavefunctions
are diverging.

EM> And this is on the INFO section of the pseudopotential
EM> |  Full Potential Total Energy  -851.722440                |
EM> |  Trouiller-Martins normconserving PP                     |
EM> |     n    l        rc       energy                        |
EM> |     4    S    2.1000      -.16905                        |
EM> |     3    P    2.4000      -.16401                        |
EM> |     3    D    1.5000      -.16401                        |
EM> |     4    F    1.7076      -.16401                        |

well, isn't it suspicious, that the 3P,3D and 4F states have
the exact same energy? i can only assume this is an 'artefact' of 
the atomic code used to create the pseudopotential. this is only 
the result from the pseudopotential eigenvalue test, anyways.

EM> |  Nonlinear core correction                               |
EM> |  Core charge cutoff radius :   1.60000                   |
EM> |  Number of Mesh Points :   690                           |
EM> |  Pseudoatom Total Energy   -3.483142                     |

EM> Ti. Considering 4 electrons per Si and C, I have a total of
EM> 6*10+4*2+4*4=84 electrons
EM> However, looking at the section OCCUPATION I count 48 states. Looking at
EM> the section of basis sets one can see
EM> GENERATE ATOMIC BASIS SET
EM> Ti        SLATER ORBITALS
EM> 3D        ALPHA=   3.8833      OCCUPATION= 2.00
EM> 4S        ALPHA=   0.8514      OCCUPATION= 2.00
EM> It seems that the 3p orbitals are ignored. The occupation of 48 states is
EM> consistent with this.

EM> I think that if the pseudopotencial do not contain 3p, CPMD should
EM> calculate the 3P levels, unless the pseudopotential info is wrong and 3P
EM> means 4P.

>> 
EM> My second dounbt is related to the fact that the Ti pseudopotential has
EM> a proyector for level 4F. The manual of CPMD states that LMAX can be
EM> S,P,or D. However, the CPMD accepts F without complain. The output seems
>> 
>> of course, since the pseudopotential contains data at those places.
>> whether this data is meaningful, cpmd does not check, but the energy
>> output shows very clearly, that it is not the case.

EM> I am not sure of that. The F levels are not occupied and and should not
EM> appear in the report for the occupied states. However, CPMD must read
EM> data from the pp file, which are organized in columns, and LMAX is the
EM> parameter thet determines how many columns are read. How do I know that
EM> CPMD is reading the data correctly?

the only definitive answer is in the source code.
let me quote from recpnew.F, line 204-207:

        DO J=1,MESHV
          READ(IUNIT,*) RR(J),(VR(J,ISP,IL),IL=1,LMAX(ISP))
        ENDDO

cpmd reads in lines and all you need is enough columns. if you
read less, it does not matter (according to the fortran standard).


best regards,
        axel.

EM> Regards,
EM> Eduardo

[...]

--

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Axel Kohlmeyer       e-mail: axel.kohlmeyer at theochem.ruhr-uni-bochum.de
Lehrstuhl fuer Theoretische Chemie          Phone: ++49 (0)234/32-26673
Ruhr-Universitaet Bochum - NC 03/53         Fax:   ++49 (0)234/32-14045
D-44780 Bochum  http://www.theochem.ruhr-uni-bochum.de/~axel.kohlmeyer/
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If you make something idiot-proof, the universe creates a better idiot.