Second derivatives

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

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

now using equations D.3 and D.5

$$\frac{1}{\left\vert A \right\vert} \sum_{n,k,j,m} \frac{\partial ... ... \frac{\partial a_{jm}}{\partial \beta} \frac{\partial a_{nk}}{\partial \gamma} = \sum_{n,k,j,m} \left ( A^{-1}_{kn}A^{-1}_{mj}-A^{-1}_{km}A^{-1}_{... ... \frac{\partial a_{nk}}{\partial \beta} \frac{\partial a_{jm}}{\partial \gamma}$$

$$\frac{1}{\left\vert A \right\vert} \sum_{nk}\frac{\partial \left\... ...vert}{\partial{a_{nk}}}\frac{\partial^2 a_{nk}}{\partial \beta \partial \gamma} = \sum_{nk}A_{nk}^{-1}\frac{\partial^2 a_{nk}}{\partial \beta \partial \gamma}$$ (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:

$$\frac{\partial^2 \Phi(r_i,r_j)}{\partial \beta \partial \gamma} = \lambda_{kl}\frac{\partial \phi_k(r_i)}{\partial \beta} \frac{\pa... ...rtial \phi_l(r_i)}{\partial \beta} \frac{\partial \phi_k(r_j)}{\partial \gamma}$$ (B.19)

$$= \lambda_{kl} \left ( \frac{\partial \phi_k(r_i)}{\partial \beta} ... ...hi_l(r_i)}{\partial \beta} \frac{\partial \phi_k(r_j)}{\partial \gamma} \right)$$ (B.20)

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

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

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

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