# How do I have to interpret the input to the OCC flag in Molpro?

I'm trying to understand the occupied orbital specification in Molpro. They're specified by the OCC flag, while it's stated in the documentation, that

OCC, m1, m2, ..., mn;

where the mᵢ are the number of occupied orbitals (including core/frozen and closed).

Let's say I want to compute energies in $$\ce{N2}$$, which has $$D_\mathrm{2h}$$ symmetry point group, i.e. the following character table: $$D_\mathrm{2h}\\ \begin{array}{ccc} \hline \text{No.} & \text{Name} & \text{Function}\\ 1& \mathrm{A_g} & s \\ 2& \mathrm{B_{3u}} & x \\ 3& \mathrm{B_{2u}} & y \\ 4& \mathrm{B_{1g}} & xy \\ 5& \mathrm{B_{1u}} & z \\ 6& \mathrm{B_{2g}} & xz \\ 7& \mathrm{B_{3g}} & yz \\ 8& \mathrm{A_u} & xyz \\ \hline \end{array}$$

Then, what exactly does the OCC specification in the following code mean?

{hf-scf;
occ,3,1,1,0,2,0,0,0;
wf,14,1,0;
}


In my understanding, it specifies:

• 3 orbitals of $$\mathrm{A_g}$$ symmetry, i.e. $$s$$ orbitals
• 1 orbital of $$\mathrm{B_{3u}}$$ symmetry, i.e. $$p_x$$ orbital
• 1 orbital of $$\mathrm{B_{2u}}$$ symmetry, i.e. $$p_y$$ orbital
• 2 orbitals of $$\mathrm{B_{1u}}$$ symmetry, i.e. $$z$$ orbital

What I find strange with this interpretation is the choice of orbitals - why would I choose 3 $$s$$ orbitals, while only one $$p_x$$ and 2 $$p_z$$?

• They specify molecular orbitals, not atomic orbitals; you have to first build the linear combinations. 1s+1s has Ag symmetry, as does 2s+2s, as does 2p+2p (X/y/z depending how the coordinate system is chosen)... – Martin - マーチン Oct 23 '18 at 20:22
• @Martin-マーチン Thank you! Now that makes more sense... Would you mind to write down the answer with explanation of the whole N2 case? – Eenoku Oct 23 '18 at 20:36

You are correct in your interpretation of the OCC flag (that command corresponds to $$3$$ electrons in $$\mathrm{a_g}$$ orbitals, $$1$$ in $$\mathrm{b_{3u}}$$, $$1$$ in $$\mathrm{b_{2u}}$$, and $$2$$ in $$\mathrm{b_{1u}}$$). (Because these are orbitals, not overall wavefuction symmetries we use lowercase letters)

Your confusion seems to be arising from visualizing the orbitals which you are including. Remember that these representations refer to the symmetry of the molecular orbital, not atomic orbitals. $$\mathrm{a_g}$$ refers to the set of orbitals which have no phase change under all of the symmetry operations, so that corresponds to the $$\sigma(\ce{1s})$$, $$\sigma(\ce{2s})$$, and $$\sigma(\ce{p_z})$$, orbitals. Similarly, $$\mathrm{b_{1u}}$$ refers to the filled $$\sigma^*(\ce{1s})$$, $$\sigma^*(\ce{2s})$$. Finally the $$\mathrm{b_{3u}}$$, and $$\mathrm{b_{2u}}$$ orbitals refer to the filled $$\pi(\ce{p_x})$$ and $$\pi(\ce{p_y})$$ orbitals.

As we know, the occupation of $$\ce{N2}$$ is: $$\ce{\sigma(1s)^2 \sigma(2s)^2 \sigma^*(1s)^2 \sigma^*(2s)^2 \pi(p_x)^2 \pi(p_y)^2 \sigma(p_z)^2},$$ and using the $$D_\mathrm{2h}$$ point group we can rewrite this as: $$\mathrm{1a_g^2\ 2a_g^2\ 1b_{1u}^2 \ 2b_{1u}^2 \ 1b_{3u}^2\ 2b_{3u}^2\ 3a_g^2}.$$ Hopefully, this rewriting of the occupation pattern in the chosen symmetry makes the choice of orbitals clear.

To really understand which orbitals fall under which irreducible representation you need to visualize them (or draw them in the case of a dimer) and then pull out the character table for the point group.

You can find a $$D_\mathrm{2h}$$ character table on the homepage of Dr. Achim Gelessus (Jacobs University Bremen).

• Please do not combine MathJax with normal text, it'll break on many devices (especially with the responsive design. – Martin - マーチン Oct 25 '18 at 19:31
• Group theory tables are available as the linked meta post. Warning: Uses MathJax heavily, very slow site. – Martin - マーチン Oct 25 '18 at 19:33
• Just a typo - $\pi(p_y)$ orbital has a $B_{2u}$ not a $B_{3u}$ symmetry. – Eenoku Oct 26 '18 at 22:13