[molpro-user] enforcing spherical symmetries of atomic states in MRCI
Gershom Martin by way of Jacky LIEVIN (jlievin@ulb.ac.be)
gershom at weizmann.ac.il
Sat Sep 12 08:54:20 BST 2009
Dear Jan,
The problem comes from the internal contraction.
The symmetry breaking comes from the number of reference states for generating contracted pairs. It is not the same in the different symmetries: nstatr=2 in symmetry 1 and nstatr= 1 in symmetries 4, 6 and 7. This means that additional variational flexibility is given to symmetry 1 MRCI calculation, with the observed lowering of the energy.
You can check that {ci;wf,16,1,0;state,1} gives the same energy than symmetry 4,6 and 7. This energy corresponds to the nstatr=1 solution, and I think that the best you can do is to use this energy for all components of the singlet delta state.
I have tried to calculate state 2.1 with nstatr=1 using: {ci;wf,16,1,0;state,2;refstate,1,2} but the energy still differs from the previous ones.
By the way the nstatr problem should be also considered when calculating the singlet S state. State 3.1 is actually calculated with nstatr=3, thus with additional contracted pairs coming from 1.1 and 2.1 reference states. The refstate option should give the right result in this case because of symmetry block diagonalization.
{ci;wf,16,1,0;state,3;refstate,1,3} gives an energy of -397.522043497 to be compared to -397.52270174 with {ci;wf,16,1,0;state,3}
Note that projection could also be used:
{ci;wf,16,1,0;state,2;project,3000.3,-1}
{ci;wf,16,1,0;state,1;project,3000.3}
giving an energy of -397.52203926
best wishes
Jacky
>Dear Molpro MRCI gurus:
>
>When doing atomic MRCI calculations, I am having a devil of a time
>getting, say, the five components of a 1D state to come out with the
>same energy.
>
>For example, in a slightly modified version of the spin-orbit example
>in the manual
>
>geometry={s}
>basis={spd,s,vtz} !use uncontracted
>basis
>
>{rhf;occ,3,2,2,,2;wf,16,4,2} !rhf for 3P state
>
>{multi !casscf
>wf,16,4,2;wf,16,6,2;wf,16,7,2;wf,16,1,0;state,3; !1D and 1S states
>wf,16,4,0;wf,16,6,0;wf,16,7,0} !3P states
>
>
>{ci;wf,16,1,0;save,4010.1;state,3} !mrci calculations
>for 1D, 1S states
>ed=energy(1) !save energy for 1D
>state in variable ed
>edqes=energy(2)
>es=energy(3) !save energy for 1S
>state in variable es
>{ci;wf,16,4,0;save,5004.1}
>edbis=energy
>{ci;wf,16,6,0;save,5006.1}
>edter=energy
>{ci;wf,16,7,0;save,5007.1}
>edqtr=energy
>edt2g=(edbis+edter+edqtr)/3.0
>edeg=(ed+edqesl)/2.0
>edavg=0.4*edeg+0.6*edt2g
>{ci;wf,16,4,2;save,4042.1} !mrci calculations
>for 3P states
>ep=energy !save energy for 3P
>state in variable ep
>{ci;wf,16,6,2;save,4062.1} !mrci calculations
>for 3P states
>{ci;wf,16,7,2;save,4072.1} !mrci calculations
>for 3P states
>
>lsint
>{ci;hlsmat,ls,3010.1,3040.1,3060.1,3070.1,3042.1,3062.1,3072.1}
>
>
>edbis,edter,and edqtr all will get the same value (edt2g) and ed and
>edqes will both be equal to a different value (edeg), and of course
>this will mess up the subsequent spin-orbit matrix calc.
>
>A workaround is of course to "retouch" the diagonal of the SO matrix,
>i.e., follow lsint with
>
>hlsdiag=[ed,ed,es,ed,ed,ed,ep,ep,ep] !set variable hlsdiag
>to mrci energies
>
>or
>
>hlsdiag=[edavg,edavg,es,edavg,edavg,edavg,ep,ep,ep]
>
>but I was wondering if there is no way to properly symmetrize the wave
>function to begin with?
>
>(I tried "ref", to no avail.)
>
>Many thanks in advance for any ideas!
>
>Jan Martin
>
>_______________________________________________
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>Molpro-user at molpro.net
>http://www.molpro.net/mailman/listinfo/molpro-user
>
>
Jacky Liévin
Service de Chimie Quantique et Photophysique
CPi 160/09
50 av F.D. Roosevelt
B-1050 Bruxelles
Belgium
Tel: +32-2-650 4089
Fax: +32-2-650 4232
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