# Solution of the Roothaan Equations of H2 by Symmetry Arguments

FWIW my background is in physics and maths, but I am just starting a chemistry PhD (the last time I took a chemistry class was high school). I have only some background in representation theory, and only an abstract taste, as it was in a maths class (bonus: suggest a reasonable representation/group theory reference with applications to chemistry, preferably a concise treatment).

In Szabo and Ostlund's Modern Quantum Chemistry, section 3.5.2 they solve the $${\rm H}_2$$ molecule (in Hartree-Fock theory) using a minimal STO-3G basis, $$\phi_1$$ and $$\phi_2$$ (centred on each proton). I believe I understand how to set up the Roothaan equations, guessing the density matrix, and solving self-consistently. However, they go on to say "The canonical molecular orbitals will form a representation of the point group of the molecule." Um, okay. Why? And what is meant by canonical here? They go on to say they can be labelled by their symmetry $$\sigma_u, \sigma_g, \pi_u, \pi_g$$, etc. (I know these terms from wikipedia, and it is discussed briefly in another chapter) I don't fully understand why they (can) do this.

Now, I understand we have two molecular orbitals so it makes sense that they choose only the 2 lowest energies. However, from the set of options above, they state without proof that the $$\sigma_g$$ is the lowest energy, followed by $$\sigma_u$$ (I also don't really understand where the wave function comes from). How do we conclude this? Is it because the hydrogen atoms are 1.4 Bohr radii away, hence close and hence we expect bonding (not antibonding?), which I understand to be represented by $$\sigma_g$$? If this is correct, then it's still not rigorous so I would appreciate any more explanation.

• If I'm interpreting this correctly, the main thing confusing you is how the MOs are associated with the symmetry labels...? So you understand that you can solve the SCF equations to get a bonding MO $\propto \phi_1 + \phi_2$ and an antibonding MO $\propto \phi_1 - \phi_2$, but don't know why one is $\sigma_g$ and the other $\sigma_u$? And the second question is why the bonding MO is lower in energy than the antibonding MO? – orthocresol Jan 28 at 22:58
• These things are usually covered in undergrad, and there are quite a lot of books for this: most undergrad inorganic chemistry books will have a chapter on "molecular symmetry", and there are some specialised books on symmetry. But they probably do not have the rigour which you might prefer (though you may well be able to fill it in yourself). If you want rigour, my recommendation is probably Albright et al. "Orbital Interactions in Chemistry": wiley.com/en-gb/… if you can get your hands on it, Chpt 4 is very good. – orthocresol Jan 28 at 23:11
• Cotton's book on Group Theory would probably help with this. – MaxW Jan 28 at 23:11
• @orthocresol Yes, I am wondering about the two questions you list. However, my main question is actually how they write $MO\propto\phi_1\pm\phi_2$ without solving the Roothaan equations (I understand how to get this numerically though, and purportedly it will converge to their symmetry argument). Also, thanks for the reference! I figured it is taught in undergrad and that's why they skim over it, and I'll need to do some catch up. – tmph Jan 28 at 23:19
• I'd also recommend Levine's Quantum Chemistry text as a better place to start than Szabo (which is still a very good text, just doesn't include the introductory material that you appear to want). Read Albright for the conceptual understanding of MOs, then Levine for the deeper math understanding, then come back to Szabo – Andrew Jan 29 at 14:32

## 1 Answer

After speaking to a coworker and reading some more I think I have a better understanding.

The symmetry of the wave function needs to follow the symmetry of the molecule (unless we are studying a broken symmetry solution, e.g. charge transfer, but that doesn't apply in this case). We are using a minimal basis, and hence we have only two orbitals. From atomic theory, these are s functions. Hence, they cannot form $$\pi$$ orbitals, and we are left with $$\sigma$$ orbitals. Between $$\sigma_g$$ and $$\sigma_u$$, $$\sigma_g$$ has the lower energy because it has no nodes, whereas $$\sigma_u$$ contains a node. From QM we know that this generally implies greater energy, but there's a nice explanation on this here (at the end he also specifically discusses a homonuclear diatomic molecule).

The specifics of how to get the symmetry representation for the molecule and match those with wave functions is still not entirely clear to me, but for this I will read a book on group theory and its chemical applications. It is, however, clear that $$\phi_1+\phi_2$$ and $$\phi_1-\phi_2$$ respect $$\sigma_g$$ and $$\sigma_u$$ symmetry respectively.