[molpro-user] spin orbit coupling
lydia
theochemly at gmail.com
Thu Dec 2 03:09:33 GMT 2010
Hi every expert,
I used the part of example in page 291 of 2010 molpro manule to
calculate the spin orbit coupling (soc).
***,SO calculation for the S-atom
geometry={s}
basis={spd,s,vtz}
!use uncontracted basis
{rhf;}
!rhf for 3P state
{multi
!casscf
wf,16,4,2;wf,16,6,2;wf,16,7,2}
!3P states
{ci;wf,16,4,2;save,3040.1;noexc}
{ci;wf,16,6,2;save,3060.1;noexc}
{ci;wf,16,7,2;save,3070.1;noexc}
lsint
!compute so integrals
{ci;hlsmat,ls,3040.1,3060.1,3070.1;print,hls=2,vls=0}
*************************************************************************
I get the following output:
Nr Sym E E-E0 E-E0 E-E(1)
E-E(1) E-E(1)
(au) (au) (cm-1) (au)
(cm-1) (eV)
1 1 -397.50241574 -0.00083187 -182.57 0.00000000
0.00 0.0000
2 1 -397.50241574 -0.00083187 -182.57 0.00000000
0.00 0.0000
3 1 -397.49992013 0.00166374 365.15 0.00249561
547.72 0.0679
4 4 -397.50241574 -0.00083187 -182.57 0.00000000
0.00 0.0000
5 4 -397.50075200 0.00083187 182.57 0.00166374
365.15 0.0453
6 6 -397.50241574 -0.00083187 -182.57 0.00000000
0.00 0.0000
7 6 -397.50075200 0.00083187 182.57 0.00166374
365.15 0.0453
8 7 -397.50241574 -0.00083187 -182.57 0.00000000
0.00 0.0000
9 7 -397.50075200 0.00083187 182.57 0.00166374
365.15 0.0453
E0 = -397.50158387 is the energy of the lowest zeroth-order state
E1 = -397.50241574 is the energy of the lowest SO-state
****************************************************************************************************
According to the theory, the 3P state will split to three soc states
(J=2,1,0). I am a little confused with so many energy values.
Does any expert give me some ideas how to analysis the results? I will
be really appreciate.
all the best,
lydia
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