12
$\begingroup$

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

enter image description here

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 $\mat{\varepsilon}$ \begin{equation} \mat{F} \mat{C} = \mat{S} \mat{C} \mat{\varepsilon} \, . \end{equation}

enter image description here

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?

$\endgroup$
7
  • 5
    $\begingroup$ 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. $\endgroup$ Sep 11, 2014 at 13:04
  • $\begingroup$ 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. $\endgroup$ Sep 11, 2014 at 22:35
  • $\begingroup$ @GeoffHutchison That is absolutely true (even for larger basis sets). See also chemistry.stackexchange.com/a/15117/4945 $\endgroup$ Sep 12, 2014 at 13:25
  • $\begingroup$ @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. $\endgroup$
    – Wildcat
    Sep 12, 2014 at 13:42
  • 5
    $\begingroup$ 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. $\endgroup$ Sep 12, 2014 at 14:21

1 Answer 1

8
$\begingroup$

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.

$\endgroup$
1
  • 4
    $\begingroup$ On another note, the eigenvalue equations are of course no true eigenvalue equations, since they are dependent on their own solutions. $\endgroup$ Sep 12, 2014 at 14:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.