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Second derivatives

In the general case in which all elements of the matrix depend from the parameters $ \beta,\gamma$ the derivative is:

$\displaystyle \frac{1}{\left\vert A \right\vert}\frac{\partial^2 \left\vert A \...
...vert}{\partial{a_{nk}}}\frac{\partial^2 a_{nk}}{\partial \beta \partial \gamma}$ (B.17)

now using equations D.3 and D.5
$\displaystyle \frac{1}{\left\vert A \right\vert} \sum_{n,k,j,m} \frac{\partial ...
... \frac{\partial a_{jm}}{\partial \beta} \frac{\partial a_{nk}}{\partial \gamma}$ $\displaystyle =$ $\displaystyle \sum_{n,k,j,m} \left ( A^{-1}_{kn}A^{-1}_{mj}-A^{-1}_{km}A^{-1}_{...
... \frac{\partial a_{nk}}{\partial \beta} \frac{\partial a_{jm}}{\partial \gamma}$  
$\displaystyle \frac{1}{\left\vert A \right\vert} \sum_{nk}\frac{\partial \left\...
...vert}{\partial{a_{nk}}}\frac{\partial^2 a_{nk}}{\partial \beta \partial \gamma}$ $\displaystyle =$ $\displaystyle \sum_{nk}A_{nk}^{-1}\frac{\partial^2 a_{nk}}{\partial \beta \partial \gamma}$ (B.18)

where the derivatives $ \partial a_{nk}/ \partial \gamma$ , $ \partial a_{nk}/ \partial \beta$ are given by B.15, B.16, B.14. The second derivatives can be evaluate from B.8.
If $ \beta$ and $ \gamma$ are two $ \lambda_{lm}$ using B.16 the second derivative is obviously zero.
If $ \beta$ and $ \gamma$ are both orbital parameters, and for example the $ k-th$ orbital depends from $ \beta$ and $ l-th$ orbital from $ \gamma$ , using B.15 we have:
$\displaystyle \frac{\partial^2 \Phi(r_i,r_j)}{\partial \beta \partial \gamma}$ $\displaystyle =$ $\displaystyle \lambda_{kl}\frac{\partial \phi_k(r_i)}{\partial \beta} \frac{\pa...
...rtial \phi_l(r_i)}{\partial \beta} \frac{\partial \phi_k(r_j)}{\partial \gamma}$ (B.19)
  $\displaystyle =$ $\displaystyle \lambda_{kl} \left ( \frac{\partial \phi_k(r_i)}{\partial \beta} ...
...hi_l(r_i)}{\partial \beta} \frac{\partial \phi_k(r_j)}{\partial \gamma} \right)$ (B.20)

because $ \lambda_{kl}=\lambda_{lk}$ .
If $ \beta$ and $ \gamma$ are parameters of the same orbital we have:

$\displaystyle \frac{\partial^2 \Phi(r_i,r_j)}{\partial \beta^2} = \frac{\partia...
...ial^2 \phi_k(r_j)}{\partial \beta \partial \gamma}\sum_m\lambda_{mk}\phi_m(r_i)$ (B.21)

If $ \gamma$ is one of the $ \lambda $ parameter, using the fact the $ \lambda $ matrix is symmetric we obtain:

$\displaystyle \frac{\partial^2 \Phi(r_i,r_j)}{\partial \beta \partial \lambda_{...
...ial \beta} \phi_b(r_j) +\phi_b(r_i) \frac{\partial \phi_a(r_j)}{\partial \beta}$ (B.22)


next up previous contents
Next: Derivatives of the Local Up: Pairing determinant Previous: The logarithmic derivatives   Contents
Claudio Attaccalite 2005-11-07