Hellmann-Feynman forces

Within the finite temperature, LDA forces are defined as the derivative of the generalized free energy. This quantity can be evaluated easily. The functional $F$ depends on the wavefunctions $\phi$, the partial occupancies $f$, and the positions of the ions $R$. In this section, we will shortly discuss the variational properties of the free energy and we will explain why we calculate the forces as a derivative of the free energy. The formulas given are very symbolic and we do not take into account any constraints on the occupation numbers or the wavefunctions. We denote the whole set of wavefunctions as $\phi$ and the set of partial occupancies as $f$.

The electronic groundstate is determined by the variational property of the free energy i.e.

$0 = \delta F(\phi,f,R)$

for arbitrary variations of $\phi$ and $f$. We can rewrite the right hand side of this equation as

$\frac{\partial F}{\partial \phi} \delta \phi + \frac{\partial F}{\partial f} \delta f.$

For arbitrary variations this quantity is zero only if $\frac{\partial F}{\partial \phi}=0$ and $\frac{\partial F}{\partial f}=0$, leading to a system of equations which determines $\phi$ and $f$ at the electronic groundstate. We define the forces as derivatives of the free energy with respect to the ionic positions i.e.

$\mbox{force} = \frac{d F(\phi,f,R)} {d R} = \frac{\partial F}{\partial \phi} \frac{\partial \phi}{\partial R} + \frac{\partial F}{\partial f} \frac{\partial f}{\partial R} + \frac{\partial F}{\partial R} .$

At the groundstate the first two terms are zero and we can write

$\mbox{force}= \frac{d F(\phi,f,R)} {d R} = \frac{\partial F}{\partial R}$

i.e. we can keep $\phi$ and $f$ fixed at their respective groundstate values and we have to calculate the partial derivative of the free energy with respect to the ionic positions only. This is relatively easy task.

Previously we have mentioned that the only physical quantity is the energy for $\sigma \to 0$. It is in principle possible to evaluate the derivatives of $E(\sigma \to 0)$ with respect to the ionic coordinates but this is not easy and requires additional computer time.

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