\chapter{Mathematical tools}

\section{Vector algebra}

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

\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: $\vec{v}1=\vec{v}$
      (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}[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{ .}
\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}[Inner product]
The inner product on a vector space $V$ over $K$ is a map
$(\cdot,\cdot):V\times V \rightarrow K$, which satisfies
\begin{itemize}
\item $(\vec{u},\vec{v})=(\vec{v},\vec{u})^*$
      (conjugate symmetry, symmetric for $K=\mathbb{R}$)
\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
\begin{equation}
z=a+bi\mapsto z^*=a-bi \text{, } z,z^*\in K \text{.}
\end{equation}
\end{definition}

\begin{remark}
\label{math_app:ip_remark}
Due to conjugate symmetry, linearity in the first argument results in conjugate linearity (also termed antilinearity) in the second argument.
\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}
This is called a sesquilinear form.
If $K=\mathbb{R}$, conjugate symmetry reduces to symmetry and the sesquilinear form gets a bilinear form.

Furtermore, 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{ .}
\label{eq:ip_mapping}
\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}
If the inner product is nondegenerate, i.e.\  $\forall\vec{u}\, (\vec{v},\vec{u})=0 \Leftrightarrow \vec{v}=0$, as it applies for the scalar product for instance, the mapping is injective.
Since the dimension of $V$ and $V^{\dagger}$ is equal, it is additionally surjective.
Then, $V$ is isomorphic to $V^{\dagger}$.
Vector $\vec{v}^{\dagger}\equiv \varphi_{\vec{v}}\in V^{\dagger}$ is said to be the dual vector of $\vec{v}\in V$.
The dual pairing $[\vec{v}^{\dagger},\vec{u}]=[\varphi_{\vec{v}},\vec{u}]=\varphi_{\vec{v}}(\vec{u})$ is associated with the inner product $(\vec{v},\vec{u})$.

Now, in physics and matrix algebra, the inner product is often defined with linearity in the second argument and conjugate linearity in the first argument.
In this case, the antilinearity property is assigned to the element $\varphi_{\vec{v}}=\vec{v}^{\dagger}$ of dual space
\begin{equation}
\varphi_{\lambda\vec{v}}(\vec{u})=
(\lambda\vec{v},\vec{u})=
\lambda^*(\vec{v},\vec{u})=
\lambda^*\varphi_{\vec{v}}(\vec{u})
\end{equation}
and $V$ is found to be isomorphic to the conjugate complex of its dual space.
Then, the inner product $(\vec{v},\vec{u})$ is associated with the dual pairing of element $\vec{u}$ of the vector space and $\vec{v}^{\dagger}$ of its conjugate complex dual space
\begin{equation}
(\vec{v},\vec{u})\rightarrow 
[\varphi_{\vec{v}},\vec{u}]=
[\vec{v}^{\dagger},\vec{u}]
\text{ .}
\end{equation}

The standard sesquilinear form $\langle\cdot,\cdot\rangle$, also called Hermitian form, on $\mathbb{C}^n$ and linearity in the second argument, is given by
\begin{equation}
\langle\vec{v},\vec{u}\rangle=\sum_i^nv_i^*u_i
\text{ .}
\end{equation}
In this case, in matrix formalism, the inner product is reformulated
\begin{equation}
(\vec{v},\vec{u}) \rightarrow \vec{v}^{\dagger}\vec{u}
\text{ ,}
\end{equation}
where the dual vector is associated with the conjugate transpose $\vec{v}^{\dagger}$ of the corresponding vector $\vec{v}$
and the usual rules of matrix multiplication.
\end{remark}

\begin{definition}[Outer product]
If $\vec{u}\in U$, $\vec{v},\vec{w}\in V$ are vectors within the respective vector spaces and $\varphi_{\vec{v}}\in V^{\dagger}$ is a linear functional of the dual space $V^{\dagger}$ of $V$ (determined in some way by $\vec{v}$),
the outer product $\vec{u}\otimes\vec{v}$ is defined as the tensor product of $\varphi_{\vec{v}}$ and $\vec{u}$,
which constitutes a map $A:V\rightarrow U$ by
\begin{equation}
\vec{w}\mapsto\varphi_{\vec{v}}(\vec{w})\vec{u}
\text{ ,}
\end{equation}
where $\varphi_{\vec{v}}(\vec{w})$ denotes the linear functional $\varphi_{\vec{v}}\in V^{\dagger}$ on $V$ when evaluated at some $\vec{w}\in V$, a scalar that in turn is multiplied with $\vec{u}\in U$.
\end{definition}

\begin{remark}

In matrix formalism, if $\varphi_{\vec{v}}$ is defined as in \eqref{eq:ip_mapping} and
if $\vec{u}=\sum_i^m \vec{e}_iu_i$ and $\vec{v}=\sum_i^n\vec{e}'_iv_i$,
the standard form of the outer product can be written as the matrix
\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}
which can be equivalently obtained by the rules of 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, again, $\vec{v}^{\dagger}$ represents the conjugate transpose of $\vec{v}$.
By definition, and as can be easily seen in matrix representation, the identity
\begin{equation}
(\vec{u}\otimes\vec{v})\vec{w}=\vec{u}(\vec{v},\vec{w})
\end{equation}
holds.
\end{remark}

\section{Spherical coordinates}

Cartesian coordinates $\vec{r}(x,y,z)$ are related to spherical coordinates $\vec{r}(r,\theta,\phi)$ by
\begin{eqnarray}
x&=&r\sin\theta\cos\phi\textrm{ ,}\\
y&=&r\sin\theta\sin\phi\textrm{ ,}\\
z&=&r\cos\theta\textrm{ .}
\end{eqnarray}
Infinitesimal translations $dq_i$ and $dq'_i$ of the two coordinate systems are related by the partial derivatives.
\begin{equation}
dq_i=\sum_j \frac{\partial q_i}{\partial q'_j}dq'_j
\end{equation}
\begin{definition}[Jacobi matrix]
The matrix J with components
\begin{equation}
J_{ij}=\frac{\partial q_i}{\partial q'_j}
\end{equation}
is called the Jacobi matrix.
\end{definition}

For cartesian and spherical coordinates the relation of the translations are presented in detail
\begin{eqnarray}
dx&=&\frac{\partial x}{\partial r}dr +
     \frac{\partial x}{\partial \theta}d\theta +
     \frac{\partial x}{\partial \phi}d\phi\\
dy&=&\frac{\partial y}{\partial r}dr +
     \frac{\partial y}{\partial \theta}d\theta +
     \frac{\partial y}{\partial \phi}d\phi\\
dz&=&\frac{\partial z}{\partial r}dr +
     \frac{\partial z}{\partial \theta}d\theta +
     \frac{\partial z}{\partial \phi}d\phi\\
\end{eqnarray}
and the vector consisting of all or using the Jacobi matrix


\begin{equation}
   =\sin\theta\cos\phi dr + \\
\end{equation}

To obtain infinitesimal 
\begin{definition}[Jacobi matrix]

\end{definition}

\section{Fourier integrals}

