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-\chapter{Quantenmechanik}
+\part{Quantum mechanics}
+
+\chapter{Fundamental concepts}
+
+\section{Variational method}
+\label{sec:var_meth}
+
+The variational method constitutes a promising approach to estimate the ground-state energy $E_0$ of a system for which exact solutions are unknown.
+
+\begin{theorem}[Variational method]
+Considering a {\em trial ket} $|\tilde 0\rangle$, which tries to imitate the true ground-state ket $|0\rangle$, it can be shown that
+\begin{equation}
+\tilde E\equiv\frac{\langle \tilde 0|H|\tilde 0\rangle}{\langle \tilde 0|\tilde 0\rangle}
+\ge E_0 \textrm{ ,}
+\end{equation}
+i.e.\ an upper bound to the ground-state energy can be obtained by considering various kinds of $|\tilde 0\rangle$.
+\end{theorem}
+
+\begin{proof}
+The trial function $|\tilde 0\rangle$ is expanded by the exact energy eigenkets $|k\rangle$ with
+\begin{equation}
+H|k\rangle = E_k|k\rangle\text{ ,}
+\qquad E_0\leq E_1\leq\ldots\leq E_k\ldots \text{ ,}
+\qquad \langle k|k'\rangle=\delta_{k k'} \text{ ,}
+\label{sec:vm_d}
+\end{equation}
+which are unknown but, still, form a complete and orthonormal basis set, to read
+\begin{equation}
+|\tilde 0\rangle = \vec{1} |\tilde 0\rangle
+                 = \sum_{k=0}^{\infty} |k\rangle\langle k|\tilde 0\rangle
+\text{ .}
+\end{equation}
+Since $\langle k|k'\rangle=\delta_{k k'}$, $H|k\rangle = E_k|k\rangle$ and $E_k\geq E_0$ (see \eqref{sec:vm_d})
+\begin{equation}
+\tilde E=
+\frac{\sum_{k,k'}\langle \tilde 0|k\rangle\langle k|H|k'\rangle\langle k'|\tilde 0\rangle}
+     {\sum_{k,k'}\langle \tilde 0|k\rangle\langle k|k'\rangle\langle k'|\tilde 0\rangle}=
+\frac{\sum_k \left| \langle k|\tilde 0\rangle  \right|^2 E_k}
+     {\sum_k \left| \langle k|\tilde 0\rangle \right|^2} \geq
+\frac{\sum_k \left| \langle k|\tilde 0\rangle  \right|^2 E_0}
+     {\sum_k \left| \langle k|\tilde 0\rangle \right|^2}=E_0
+\text{ ,}
+\label{sec:vm_f}
+\end{equation}
+which proofs the variational theorem.
+\end{proof}
+
+Moreover, equality in \eqref{sec:vm_f} is only achieved if $|\tilde 0\rangle$ coincides exactly with $|0\rangle$, i.e.\ if the coefficients $\langle k|\tilde 0\rangle$ all vanish for $k\neq 0$.
+
+\chapter{Quantum dynamics}
+
+\chapter{Relativistic quantum mechanics}
+
+285 Schulten
+
+
+