[molpro-user] Spin orbit coupling calculations
Rajagopala Reddy seelam
srgreddyseelam at yahoo.co.in
Fri Oct 31 04:42:41 GMT 2008
Dear Molpro Users,
I want to calculate Spin Orbit Coupling constant between triplet sigma and singlet sigma of carbon monoxide dication.I am getting -0.55 cm- but the reported value is 120 cm-.The ground electronic state is triplet pi.Can you kindly tell me where i am doing wrong.I am giving the input for the calculation.How to analyse the output for the spin orbit calculations.
Thanks in advance
The input is as follows
***,co
print,basis,orbitals;
basis={spdf,c,vqz;spdf,o,vqz}
geomtyp=zmat
geometry={
bohr
C
O 1 r1}
end
r1=4
hf;wf,12,1,2;
!set,charge=2;
casscf;wf,12,1,0,state,2;
casscf;wf,12,4,2,state,2;
mrci;
MAXITER,126,126;
wf,12,1,0;
state,2;
save,4010.1;
mrci;
MAXITER,256,256;
wf,12,4,2;
state,2;
save,4042.1;
ci;hlsmat,ls,4010.1,4042.1;
put,molden,test5.molden;
The out put is as follows
Eigenvalues of the spin-orbit matrix
....................................
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 -111.66658284 -0.00000251 -0.55 0.00000000 0.00 0.0000
2 1 -111.56933678 0.09724355 21342.49 0.09724607 21343.04 2.6462
3 1 -111.52300718 0.14357315 31510.66 0.14357567 31511.22 3.9069
4 1 -111.51933886 0.14724147 32315.77 0.14724398 32316.32 4.0067
5 2 -111.66658033 0.00000000 0.00 0.00000251 0.55 0.0001
6 2 -111.51935394 0.14722639 32312.46 0.14722890 32313.01 4.0063
7 3 -111.66658033 0.00000000 0.00 0.00000251 0.55 0.0001
8 3 -111.51935394 0.14722639 32312.46 0.14722890 32313.01 4.0063
E0 = -111.66658033 is the energy of the lowest zeroth-order state
E1 = -111.66658284 is the energy of the lowest SO-state
Rajagopala Reddy.Seelam
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