# What is the point of introducing virtual orbitals in Hartree-Fock calculations?

Let us for simplicity discuss RHF formalism. For $2n$-electron system we have $n$ Hartree-Fock equations written for $n$ spatial orbitals $\{ \phi_{k} \}_{k=1}^{n}$ $$\newcommand{\mat}[1]{\boldsymbol{\mathbf{#1}}}$$ $$\hat{F}(1) \phi_{k}(1) = \varepsilon_{k} \phi_{k}(1) \, , \quad k = 1, 2, \dotsc, n \, .$$ Once we introduce finite basis $\{ \chi_{q} \}_{q=1}^{m}$ and express spatial orbitals as a linear combination of basis functions $\chi_{q}$ $$\phi_{k}(1) = \sum\limits_{q=1}^{m} c_{qk} \chi_{q}(1) \, , \quad k = 1, 2, \dotsc, n \, .$$ we end up with $n$ Roothaan–Hall equations $$\sum\limits_{q=1}^{m} F_{pq} c_{qk} = \varepsilon_{k} \sum\limits_{q=1}^{m} S_{pq} c_{qk} \, , \quad k = 1, 2, \dotsc, n \, ,$$ which can be rewritten in the following matrix form $$\mat{F} \mat{c}_{k} = \varepsilon_{k} \mat{S} \mat{c}_{k} \quad k = 1, 2, \dotsc, n \, .$$ The Fock matrix $\mat{F}$ and the overlap matrix $\mat{S}$ are both $m \times m$ square matrices, $\mat{c}_{k}$ is a column $m \times 1$ matrix, $\varepsilon_{k}$ is just a scalar value.

We can then collecl all $n$ $\mat{c}_{k}$ column $m \times 1$ matrices into one $m \times n$ matrix $\mat{C}$ and all $n$ values $\varepsilon_{k}$ into $n \times n$ square matrix $\mat{\varepsilon}$ $$\mat{F} \mat{C} = \mat{S} \mat{C} \mat{\varepsilon} \, .$$

In practice, however, we extend both $\mat{C}$ and $\mat{\varepsilon}$ to $m \times m$ matrices from $m \times n$ and $n \times n$ respectively, which results in having $m-n$ virtual (unoccupied) orbitals.

Taking into account that virtual orbitals are even more unphysical than their occupied counterparts the question is what is the point of such extension of $\mat{C}$ and $\mat{\varepsilon}$? Why do not we just leave them of $m \times n$ and $n \times n$ sizes respectively?

• Virtual orbitals are automatically 'given' by the solved eigenvalue equations. These orbitals are important for some post-HF methods such as CI and perturbation theory. – LordStryker Sep 11 '14 at 13:04
• Indeed, I'd just give that as the answer. If you think about a simple atomic orbital picture (i.e., a minimal basis) there's a conservation of orbitals when you compute the MOs. Since there are typically two electrons per occupied orbital, there will be unoccupied virtual orbitals. – Geoff Hutchison Sep 11 '14 at 22:35
• @GeoffHutchison That is absolutely true (even for larger basis sets). See also chemistry.stackexchange.com/a/15117/4945 – Martin - マーチン Sep 12 '14 at 13:25
• @GeoffHutchison, yes, the question is about why do we expand the matrices. I understand that we can do it, but not have to. At least for HF itself. – Wildcat Sep 12 '14 at 13:42
• The unitary transformation to form the canonical orbitals guarantees that you have a diagonal $\mathbb{F}$ matrix, and all of them are interdependent on their own solution (pseudo eigenwert), this means they are also dependent on the virtual orbitals. Hence for every function you plug into the formalism you get a molecular orbital. – Martin - マーチン Sep 12 '14 at 14:21

## 2 Answers

Virtual orbitals are automatically 'given' by the solved eigenvalue equations and are necessary for solving said equations. These orbitals are important for some post-HF methods such as CI and perturbation theory and coupled-cluster theory.

• On another note, the eigenvalue equations are of course no true eigenvalue equations, since they are dependent on their own solutions. – Martin - マーチン Sep 12 '14 at 14:23

We don't need virtual orbitals to solve the equation. For example Levine in chapter 14 solved the equation for the helium atom without recurring to Virtual orbitals. But using the Virtual orbitals the procedure is much easy.