DFT++ forces

Tairan Wang

The forces calculated by the DFT++ program are in the relative coordinates.

Let $\vec{a_1}$, $\vec{a_2}$ and $\vec{a_3}$ be the basis vectors of the crystal. The coordinates of the basis (the ions) are given in this basis $\vec{\tau}$.

So that the position of ion i is
$$\begin{align}\vec{R^i} &= [ \vec{a_1} \vec{a_2} \vec{a_3} ] \vec{\tau^i} \notag \\ &= A \vec{\tau^i} \notag \end{align}$$
or

$$R^i_j = \sum_k A_{jk} \tau^i_k$$

The jth component of the force on the ith ion is given by the Feynman-Hellman formula:

$$F^i_j = \frac{\partial E}{\partial R^i_j}$$

What the program calculates, however, is
$$\begin{align}f^i_j &= \frac{\partial E}{\partial \tau^i_j} \notag \\ &= \sum_k... ... R^i_k}{\partial \tau^i_j} \notag \\ &= \sum_k F^i_k A_{kj} \notag \end{align}$$

Thus, to get the real force from what the program outputs,

$$F^i_k = \sum_j f^i_j (A^{-1})_{jk}$$

Now, to move the ion under the influence of this force, we would like to project it to the relative coordinate system:
$$\begin{align}m \frac{d^2 R^i_j}{dt^2} &= F^i_j \notag \\ m \sum_k A_{jk} \frac... ..._k \notag \\ &= \sum_{qk} A^{-1}_{jq}(A^T)^{-1}_{qk} f^i_k \notag \end{align}$$

That is to say,

$$m \frac{d^2 \vec{\tau^i}}{dt^2} = (A^TA)^{-1} \vec{f^i}$$