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Re: basic question

Alex Granovsky
gran@classic.chem.msu.su


Dear Remi,

Standard U/RO/RHF or U/RO/KS codes assume that electron density is
totally symmetric in the pont group that is used for computations.
The lower is the symmetry of the a group the less restrictions
are enforced on the solution of HF/KS equations the lower is the symmetry of solution and lower is its energy.

To obtain spherically symmetric solution for your system,
you need to use state-averaged MCSCF with 9 electrons on 5
degenerated orbitals.


Hope this helps.


Kind regards,
Alex Granovsky



On Thu Mar 31 '16 2:41pm, Remi wrote
------------------------------------
>Given a basis set, need expansion coefficients of exact AOs of isolated open shell atoms and ions (atomic MOs).
>The fixed potential is spherical (around the unique nucleus), but depending on the symmetry fixed in $data, calculated occupied orbitals do not satisfy the 3D spanning by proper degeneracy: the only formally respected degeneracy is that of the group Irreps, The R3 group (with S, P, D,...degeneracies) is not available, but the spatial degeneracy can however be recovered by approximating R3 by polyhedral groups like Oh (but the results obtained in C1 symmetry are restored by adding NOSYM=1 in $contrl).
>The explanation is likely simple, but I'd like to know why.
>Moreover, not only the total energy of the atom/ion depends on the fixed symmetry, but also the lowest value may not correspond to symmetry closest to R3.

>Example with Cu2+:

>C1 ->TOTAL ENERGY =   -1639.4442846853
>from:
>$contrl scftyp=uhf DFTTYP=B3PW91 runtyp=energy nprint=7 icharg=2
> mult=2 $end
> $system mwords=128 timlim=1200 $end
> $BASIS gbasis=TZV ndfunc=1 npfunc=1 $END
> $SCF NCONV=5 DIRSCF=.T. $END
> $data
>Cu2+ B3PW91/TZV energy C1
>C1
>Cu     29      0.000     0.0000          0.0000
> $end

>D6h -> TOTAL ENERGY =   -1639.4118631252

>OH -> TOTAL ENERGY =   -1639.4013245610  (with 3D spanning by degenerated OAs)      

>Thanks for any comment.
>Remi


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