\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}

