# [molpro-user] spin orbit coupling

Thomas Ritschel ritschel at uni-potsdam.de
Thu Dec 2 10:30:13 GMT 2010

```Hi Lydia,

as you noticed, the 3P state (9-fold degenerated) splits into
three states with J=2,1,0, where the degeneracy of each state is 2J+1.
Therefore you find 9 (5+3+1) energy values in the soc output.

Hope that helps,

Thomas

> *************************************************************************
>  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
>
> ***************************************************************************

```