From 7fb9069b2ce11bdd32cc2489e98edd632263e944 Mon Sep 17 00:00:00 2001
From: hackbard <hackbard@hackdaworld.org>
Date: Tue, 20 Sep 2011 11:03:19 +0200
Subject: [PATCH] force, to be on the save side

---
 posic/thesis/basics.tex | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/posic/thesis/basics.tex b/posic/thesis/basics.tex
index 391af40..5da2eaa 100644
--- a/posic/thesis/basics.tex
+++ b/posic/thesis/basics.tex
@@ -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}
-\vec{F}_i=-\sum_j \langle \Phi_j | \Phi_j\frac{\partial V}{\partial \vec{R}_i} \rangle
+\vec{F}_i=-\sum_j \langle \Phi_j | \frac{\partial V}{\partial \vec{R}_i} \Phi_j \rangle
 \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.
-- 
2.39.5