The case of an orthogonal basis

In the special case of an orthogonal basis $\{ D \}$ there is no distinction between covariant and contravariant quantities with respect to the expansion coefficients of this basis, and as such we use Latin suffixes to denote them: $D_{i}(\mathbf{r})$. In this case, Eq. (24) becomes

$$c'^{\ }_{i \alpha} = c^{\ }_{i \alpha} - \lambda \sum_{j} x^{-1}_{ij} g_{j}^{\ \beta} S^{\ }_{\beta \alpha},$$ (30)

where we have defined $g_{j}^{\ \beta} \equiv \partial E / \partial c^{\ast}_{j\beta}$.

Let $\mathbf{F}$ be that unitary transformation which diagonalises the (Hermitian) matrix $\mathbf{x}$, i.e., $\mathbf{FxF^{\dagger}} = \mathbf{\tilde{x}}$, where $\mathbf{\tilde{x}}$ is a matrix with eigenvalues $\xi_{p}$ on its diagonal:

$$\tilde{x}_{pq} = \xi_{p}\delta_{pq} .$$ (31)

Denoting transformed variables by $\tilde{v}_{p} = \sum_{j} F_{pj} v_{j}$, we apply $\mathbf{F}$ to Eq. (30) to obtain

$$\tilde{c}'^{\ }_{p\alpha} = \tilde{c}^{\ }_{p\alpha} - \la... ...de{x}^{-1}_{pq} \tilde{g}_{q}^{\ \beta} S^{\ }_{\beta \alpha}.$$ (32)

From Eq. (31) we see that $\tilde{x}^{-1}_{pq} = \xi^{-1}_{p} \delta_{pq}$ is diagonal, hence Eq. (32) becomes

$$\tilde{c}'^{\ }_{p\alpha} = \tilde{c}^{\ }_{p\alpha} - \lamb... ...}{\xi^{\ }_{p}} \tilde{g}_{p}^{\ \beta} S^{\ }_{\beta \alpha}.$$ (33)

In other words, for the case of an orthogonal basis $\{ D \}$, the transformed gradient is preconditioned by premultiplying by a diagonal matrix of inverse eigenvalues $\xi^{-1}_{p}$. Post-multiplication by the overlap matrix $\mathbf{S}$ is still present in order to account for the nonorthogonality of the localized functions $\{ \phi \}$.