Translational and Rotational Invariance

One of the problems that arises when integration is performed numerically with a finite grid is that translational and rotational invariance can be lost. Luckily, using the grid points as defined above ensures translational invariance. The grid points are linked to atomic centers and therefore translate with nuclear movement. However a rotation of the molecule leaving the grid axes unrotated will cause a slight change in energy. This effect is even more pronounced with derivative calculations, in particular the calculation of harmonic frequencies [144].

An early attempt to solve rotational invariance was the use of randomly rotated angular grids, with the hope of averaging out any error [145]. A more rigorous solution was given by Johnson et al. [146] where the grid points are defined by
$$\begin{align}{\bf r}_{i} = {\bf R}_{A} + {\bf O s}_{i}, \end{align}$$
with ${\bf R}_{A}$ the atomic position, ${\bf s}_{i}$ the quadrature grid points, and ${\bf O}$ is the matrix formed from the eigenvectors, ${\bf M}$, of the charge moment tensor
$$\begin{align}{\bf M} = \sum_{A} Z_{A}[\vert{\bf R}_{A}-{\bf T}\vert^{2}{\bf I} - ({\bf R}_{A}-{\bf T})({\bf R}_{A}-{\bf T})^{T}] \end{align}$$
with Z_A the charge at nucleus A and
$$\begin{align}{\bf T} = \frac{\sum_{A}Z_{A}{\bf R}_{A}}{\sum_{A}Z_{A}}. \end{align}$$

With this new definition the grid points are defined in terms of the molecule, thus removing the problem of rotational invariance. This is only a problem when finite grids are used. If the grid is made exhaustively large the problem begins to disappear, though this will usually be computationally infeasible.