@@ -566,7+566,7 @@ Writing down the derivative of the total energy $E$ with respect to the position
indeed reveals a contribution to the change in total energy due to the change of the wave functions $\Phi_j$.
However, provided that the $\Phi_j$ are eigenstates of $H$, it is easy to show that the last two terms cancel each other and in the special case of $H=T+V$ the force is given by
\begin{equation}
indeed reveals a contribution to the change in total energy due to the change of the wave functions $\Phi_j$.
However, provided that the $\Phi_j$ are eigenstates of $H$, it is easy to show that the last two terms cancel each other and in the special case of $H=T+V$ the force is given by
This is called the Hellmann-Feynman theorem \cite{feynman39}, which enables the calculation of forces, called the Hellmann-Feynman forces, acting on the nuclei for a given configuration, without the need for evaluating computationally costly energy maps.
\text{ .}
\end{equation}
This is called the Hellmann-Feynman theorem \cite{feynman39}, which enables the calculation of forces, called the Hellmann-Feynman forces, acting on the nuclei for a given configuration, without the need for evaluating computationally costly energy maps.