\chapter{Mathematical tools}

\section{Vector algebra}

\subsection{Vector space}
\label{math_app:vector_space}

\begin{definition}
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: $\vec{v}1=\vec{v}$
      (identity element of scalar multiplication)
\item $\vec{v}(\lambda_1+\lambda_2)=\vec{v}\lambda_1+\vec{v}\lambda_2$
      (distributivity of scalar multiplication)
\item $(\vec{v}_1+\vec{v}_2)\lambda=\vec{v}_1\lambda + \vec{v}_2\lambda$
      (distributivity of scalar multiplication)
\item $(\vec{v}\lambda_1)\lambda_2=\vec{v}(\lambda_1\lambda_2)$
      (compatibility of scalar multiplication with field multiplication)
\end{itemize}
The elements $\vec{v}\in V$ are called vectors.
\end{definition}
\begin{remark}
Due to the additive abelian group, the following properties are additionally valid:
\begin{itemize}
\item $\vec{u}+\vec{v}=\vec{v}+\vec{u}$ (commutativity of addition)
\item $\vec{u}+(\vec{v}+\vec{w})=(\vec{u}+\vec{v})+\vec{w}$
      (associativity of addition)
\item $\forall \vec{v} \, \exists \vec{0}$ with:
      $\vec{0}+\vec{v}=\vec{v}+\vec{0}=\vec{v}$
      (identity elemnt of addition)
\item $\forall \vec{v} \, \exists -\vec{v}$ with: $\vec{v}+(-\vec{v})=0$
      (inverse element of addition)
\end{itemize}
The addition of two vectors is called vector addition.
\end{remark}

\subsection{Dual space}

\begin{definition}
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{ .}
\end{equation}
These type of linear maps are termed linear functionals.
The dual space $V^{\dagger}$ becomes a vector space over $K$ itself by the following additional definitions
\begin{eqnarray}
(\varphi+\psi)(\vec{v}) & = & \varphi(\vec{v})+\psi(\vec{v}) \\
(\lambda\varphi)(\vec{v}) & = & \lambda\varphi(\vec{v})
\end{eqnarray}
for all $\vec{v}\in V$, $\varphi,\psi\in V^{\dagger}$ and $\lambda\in K$.

The map $V^{\dagger}\times V \rightarrow K: [\varphi,\vec{v}]=\varphi(\vec{v})$ is termed dual pairing of a functional $\varphi\in V^{\dagger}$ and an elemnt $\vec{v}\in V$.
\end{definition}

\subsection{Inner and outer product}
\label{math_app:product}

\begin{definition}
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
\begin{itemize}
\item $(\vec{u},\vec{v})=(\vec{v},\vec{u})^*$
      (conjugate symmetry, symmetric for $K=\mathbb{R}$)
\item $(\lambda\vec{u},\vec{v})=\lambda(\vec{u},\vec{v})$ and
      $(\vec{u}'+\vec{u}'',\vec{v})=(\vec{u}',\vec{v})+(\vec{u}'',\vec{v})$
      (linearity in first argument)
\item $(\vec{u},\vec{u})\geq 0 \text{, } ``=" \Leftrightarrow \vec{u}=0$
      (positive definite)
\end{itemize}
for $\vec{u},\vec{v}\in V$ and $\lambda\in K$.
Taking the complex conjugate $(\cdot)^*$ is the map from $K\ni z=a+bi\mapsto a-bi=z^*\in K$.
\end{definition}

\begin{remark}
Due to conjugate symmetry, linearity in the first argument results in conjugate linearity (also termed antilinearity) in the second argument.
This is called a sesquilinear form.
\begin{equation}
(\vec{u},\lambda(\vec{v}'+\vec{v}''))=(\lambda(\vec{v}'+\vec{v}''),\vec{u})^*=
\lambda^*(\vec{v}',\vec{u})^*+\lambda^*(\vec{v}'',\vec{u})^*=
\lambda^*(\vec{u},\vec{v}')+\lambda^*(\vec{u},\vec{v}'')
\end{equation}

The inner product $(\cdot,\cdot)$ provides a mapping
\begin{equation}
V\rightarrow V^{\dagger}:\vec{v}\mapsto \varphi_{\vec{v}}
\quad
\text{ defined by }
\quad
\varphi_{\vec{v}}(\vec{u})=(\vec{v},\vec{u}) \text{ .}
\end{equation}
Since the inner product is linear in the first argument, the same is true for the defined mapping.
\begin{equation}
\lambda(\vec{u}+\vec{v}) \mapsto
\varphi_{\lambda(\vec{u}+\vec{v})}=
\lambda\varphi_{\vec{u}}+\lambda\varphi_{\vec{v}}\\
\end{equation}
The kernel is $\vec{v}=0$, structural identity (isomorphism) of $V$ and $V^{\dagger}$ is .

In physics and matrix algebra, the inner product is often defined with linearity in the second argument and conjugate linearity in the first argument.
This allows to express the inner product $(\vec{u},\vec{v})$ as a product of vector $\vec{v}$ with a dual vector or linear functional of dual space $V^{\dagger}$
\begin{equation}
(\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}(\vec{u})\vec{v}
\text{ CHECK ! ! !}
\end{equation}
or the conjugate transpose in matrix formalism
\begin{equation}
(\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}\vec{v} \text{ .}
\end{equation}
In doing so, the conjugate transpose is associated with the dual vector.
\end{remark}

\begin{definition}
If $\vec{u}\in U$, $\vec{v}\in V$ are vectors within the respective vector spaces and $\vec{\varphi}^{\dagger}\in V^{\dagger}$  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 $\vec{\varphi}^{\dagger}$ and $\vec{u}$,
which constitutes a map $A:V\rightarrow U$ by
\begin{equation}
\vec{v}\mapsto\vec{\varphi}^{\dagger}(\vec{v})\vec{u}
\text{ ,}
\end{equation}
where $\vec{\varphi}^{\dagger}(\vec{v})$ denotes the linear functional $\vec{\varphi}^{\dagger}\in V^{\dagger}$ on $V$ when evaluated at $\vec{v}\in V$, a scalar that in turn is multiplied with $\vec{u}\in U$.

In matrix formalism, with respect to a given basis ${\vec{e}_i}$ of $\vec{u}$ and ${\vec{e}'_i}$ of $\vec{v}$,
if $\vec{u}=\sum_i^m \vec{e}_iu_i$ and $\vec{v}=\sum_i^n\vec{e}'_iv_i$,
the outer product can be written as matrix $A$ as
\begin{equation}
\vec{u}\otimes\vec{v}=A=\left(
\begin{array}{c c c c}
u_1v_1 & u_1v_2 & \cdots & u_1v_n\\
u_2v_1 & u_2v_2 & \cdots & u_2v_n\\
\vdots & \vdots & \ddots & \vdots\\
u_mv_1 & u_mv_2 & \cdots & u_mv_n\\
\end{array}
\right)
\text{ .}
\end{equation}
\end{definition}
\begin{remark}
The matrix can be equivalently obtained by matrix multiplication:
\begin{equation}
\vec{u}\otimes\vec{v}=\vec{u}\vec{v}^{\dagger} \text{ ,}
\end{equation}
if $\vec{u}$ and $\vec{v}$ are represented as $m\times 1$ and $n\times 1$ column vectors, respectively.
Here, $\vec{v}^{\dagger}$ represents the conjugate transpose of $\vec{v}$.
By definition, and as can be easily seen in the matrix representation, the following identity holds:
\begin{equation}
(\vec{u}\otimes\vec{v})\vec{w}=\vec{u}(\vec{v},\vec{w})
\end{equation}
\end{remark}

\section{Spherical coordinates}

\section{Fourier integrals}

