# [molpro-user] spin orbit coupling

Timothy Wright Tim.Wright at nottingham.ac.uk
Thu Dec 2 10:30:38 GMT 2010

```The five states with a zero relative energy are the five 3-P-2 components; the next three degenerate one are the three 3-P-1 ones, and the last state is the single 3-P-0 component. There are (2J+1) components for each J. The total number of components = 5+3+1=9. This is equal to (2L+1)*(2S+1) = 3*3 = 9 components of the non-SO state. So it all adds up.

Hope that helps.

Tim

-----Original Message-----
From: molpro-user-bounces at molpro.net [mailto:molpro-user-bounces at molpro.net] On Behalf Of lydia
Sent: 02 December 2010 03:10
To: molpro-user at molpro.net
Subject: [molpro-user] spin orbit coupling

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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