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The logarithmic derivatives

In pairing trial-function the variation of a parameter involves all terms of the matrix and so using D.3 the logarithmic derivatives will be:
$\displaystyle \frac{\partial \ln det(A)}{\partial \beta} = \sum_{ij}\frac{\part... \beta} = \sum_{ij} \frac{\partial \Phi(r_i,r_j)}{\partial \beta} A_{ij}^{-1}$      

If only the k-th orbital depends by $ \beta$ we obtain:

$\displaystyle \frac{\partial \Phi(r_i,r_j)}{\partial \beta} = \frac{\partial \p...
...r_j) + \frac{\partial \phi_k(r_j)}{\partial \beta}\sum_m\lambda_{mk}\phi_m(r_i)$ (B.15)

If $ \beta$ is one of the $ \lambda $ parameter the derivative will be:

$\displaystyle \frac{\partial \Phi(r_i,r_j)}{\partial \lambda_{ab}} =\phi_a(r_i)\phi_b(r_j) + \phi_b(r_i)\phi_a(r_j)$ (B.16)

because $ \lambda $ matrix is symmetric.

Claudio Attaccalite 2005-11-07