\subsection{Vector space}
\label{math_app:vector_space}
-\begin{definition}
+\begin{definition}[Vector space]
A vector space $V$ over a field $(K,+,\cdot)$ is an additive abelian group $(V,+)$ and an additionally defined scalar multiplication of $\vec{v}\in V$ by $\lambda\in K$, which fullfills:
\begin{itemize}
\item $\forall \vec{v} \, \exists 1$ with: $\vec{v}1=\vec{v}$
\subsection{Dual space}
-\begin{definition}
+\begin{definition}[Dual space]
The dual space $V^{\dagger}$ of vector space $V$ over field $K$ is defined as the set of all linear maps from the vector space $V$ into its field $K$
\begin{equation}
\varphi:V\rightarrow K \text{ .}
\subsection{Inner and outer product}
\label{math_app:product}
-\begin{definition}
+\begin{definition}[Inner product]
The inner product on a vector space $V$ over $K$ is a map
-\begin{equation}
-(\cdot,\cdot):V\times V \rightarrow K
-\text{ ,}
-\end{equation}
-which satisfies
+$(\cdot,\cdot):V\times V \rightarrow K$, which satisfies
\begin{itemize}
\item $(\vec{u},\vec{v})=(\vec{v},\vec{u})^*$
(conjugate symmetry, symmetric for $K=\mathbb{R}$)
In doing so, the conjugate transpose is associated with the dual vector.
\end{remark}
-\begin{definition}
+\begin{definition}[Outer product]
If $\vec{u}\in U$, $\vec{v}\in V$ are vectors within the respective vector spaces and $\vec{\varphi}\in V^{\dagger}$ is a linear functional of the dual space $V^{\dagger}$ of $V$,
the outer product $\vec{u}\otimes\vec{v}$ is defined as the tensor product of $\vec{\varphi}$ and $\vec{u}$,
which constitutes a map $A:V\rightarrow U$ by
% units
\usepackage{units}
+% theorem environment
+\usepackage{framed}
+\usepackage[framed,thref,amsmath,hyperref]{ntheorem}
+
+% hyperlinks
+\usepackage[colorlinks=true,linkcolor=black,citecolor=black]{hyperref}
+
% shortcuts
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\newrgbcolor{hb}{0.75 0.77 0.89}
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\newrgbcolor{lachs}{1.0 .93 .81}
+%
+\newrgbcolor{ctheorem}{0.9 0.9 0.95}
+\newrgbcolor{cdefinition}{0.95 0.9 0.9}
+\newrgbcolor{cremark}{0.95 0.95 0.9}
+\newrgbcolor{cproof}{0.95 0.92 0.95}
% roman numbers
\newcommand{\RM}[1]{\MakeUppercase{\romannumeral #1{}}}
% vectors are simply represented by bold font characters
\renewcommand{\vec}[1]{{\bf #1{}}}
-
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-\newtheorem{proposition}[theorem]{Proposition}
-\newtheorem{corollary}[theorem]{Corollary}
-
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-
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-\ifdim\lastskip<1.5em \hskip-\lastskip
-\hskip1.5em plus0em minus0.5em \fi \nobreak
-\vrule height0.75em width0.5em depth0.25em\fi}
+%
+% theorem environment
+%
+% common
+\theoremstyle{plain}
+\theoremseparator{:}
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+% theorem
+\theoremsymbol{\ensuremath{\diamondsuit}}
+\shadecolor{ctheorem}
+\newshadedtheorem{theorem}{Theorem}[section]
+% definition
+\theoremsymbol{\ensuremath{\clubsuit}}
+\shadecolor{cdefinition}
+\newshadedtheorem{definition}[theorem]{Definition}
+% remark
+\theoremsymbol{\ensuremath{\diamondsuit}}
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+\newshadedtheorem{remark}[theorem]{Remark}
+% proof
+\theoremstyle{nonumberplain}
+\theoremsymbol{\rule{1ex}{1ex}}
+\shadecolor{cproof}
+\newshadedtheorem{proof}{Proof}
% author & title
\author{Frank Zirkelbach}
\frontmatter{}
\include{title}
\include{contents}
-%\include{lists}
\mainmatter{}
\include{intro}
\include{math_app}
\backmatter{}
+\include{lists}
\include{literature}
\include{ack}
\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 $|\tilde 0\rangle$.
-To proof this, $|\tilde 0\rangle$ is expanded by the exact energy eigenkets $|k\rangle$ with
+\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{ ,}
\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}
\end{equation}
In 1964, Hohenberg and Kohn showed the opposite and far less obvious result \cite{hohenberg64}.
-{\begin{theorem}
+\begin{theorem}[Hohenberg / Kohn]
For a nondegenerate ground state, the ground-state charge density uniquely determines the external potential in which the electrons reside.
\end{theorem}
\int n(\vec{r}) \left( V_2(\vec{r})-V_1(\vec{r}) \right) d\vec{r}
}_{=0}
\end{equation}
-is revealed, which proofs the Hohenberg Kohn theorem. \qed
+is revealed, which proofs the Hohenberg Kohn theorem.% \qed
\end{proof}
\vspace{60pt}
{\Large
- Augsburg, \today
+ Augsburg\\
+ \today
}
\end{center}
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