I have a Molpro script which computes $\ce{N2^+}$ potential energies.
The significant part is the input for the wave function:
{multi;
occ,3,2,2,0,3,2,2,0;
closed,1,0,0,0,1,0,0,0;
wf,13,5,1;
state,3;
}
Why is the active space specified by OCC
flag so large? It specifies 14 molecular orbitals, which means that both nitrogen should habe 7 atomic orbitals. Which atomic orbitals are additionally included?
The ground states of $\ce{N2^+}$ molecule has this electron configuration:
Thich can be specified as occ,3,1,1,0,2,0,0,0;
.
What advantage does occ,3,2,2,0,3,2,2,0;
have in the script and how can I understand and visualize those additional orbitals?
Explanation of OCC card
The OCC
card in Molpro specifies symmetries of the orbitals in active space. In this case, they're specified in $D_\mathrm{}$ point group, according the following table:
$$D_\mathrm{2h}\\ \begin{array}{ccc} \hline \text{No.} & \text{Name} & \text{Function}\\ 1& \mathrm{A_g} & s \\ 2& \mathrm{B_{3u}} & x \\ 3& \mathrm{B_{2u}} & y \\ 4& \mathrm{B_{1g}} & xy \\ 5& \mathrm{B_{1u}} & z \\ 6& \mathrm{B_{2g}} & xz \\ 7& \mathrm{B_{3g}} & yz \\ 8& \mathrm{A_u} & xyz \\ \hline \end{array} $$