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Gradients and Laplacian

Using the formula D.3 and the fact the only a column or a row may depends from a given electronic coordinate we obtain:
$\displaystyle \nabla^2_i \ln \left\vert A \right\vert$ $\displaystyle =$ $\displaystyle \sum_{k} \nabla^2_i \Phi(r_i,r_k) A_{ik}^{-1}$  
$\displaystyle \vec{\nabla}_i \ln \left\vert A \right\vert$ $\displaystyle =$ $\displaystyle \sum_{k} \vec{\nabla}_i \Phi(r_i,r_k) A_{ik}^{-1}$ (B.12)

where:
$\displaystyle \nabla^2_i \Phi(r_i,r_k)$ $\displaystyle =$ $\displaystyle \sum_{l,m}{\lambda_{l,m} \nabla^2_i \phi_{l}(r_i) \phi_{m}(r_j)}$ (B.13)
$\displaystyle \vec{\nabla}_i \Phi(r_i,r_k)$ $\displaystyle =$ $\displaystyle \sum_{l,m}{\lambda_{l,m} \vec{\nabla}_i \phi_{l}(r_i) \phi_{m}(r_j)}$ (B.14)

for spin down we have to exchange $ i$ with $ k$ in all these equations.


Claudio Attaccalite 2005-11-07