1 \chapter{Mathematical tools}
3 \section{Vector algebra}
5 \subsection{Vector space}
6 \label{math_app:vector_space}
8 \begin{definition}[Vector space]
9 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:
11 \item $\forall \vec{v} \, \exists 1$ with: $\vec{v}1=\vec{v}$
12 (identity element of scalar multiplication)
13 \item $\vec{v}(\lambda_1+\lambda_2)=\vec{v}\lambda_1+\vec{v}\lambda_2$
14 (distributivity of scalar multiplication)
15 \item $(\vec{v}_1+\vec{v}_2)\lambda=\vec{v}_1\lambda + \vec{v}_2\lambda$
16 (distributivity of scalar multiplication)
17 \item $(\vec{v}\lambda_1)\lambda_2=\vec{v}(\lambda_1\lambda_2)$
18 (compatibility of scalar multiplication with field multiplication)
20 The elements $\vec{v}\in V$ are called vectors.
24 Due to the additive abelian group, the following properties are additionally valid:
26 \item $\vec{u}+\vec{v}=\vec{v}+\vec{u}$ (commutativity of addition)
27 \item $\vec{u}+(\vec{v}+\vec{w})=(\vec{u}+\vec{v})+\vec{w}$
28 (associativity of addition)
29 \item $\forall \vec{v} \, \exists \vec{0}$ with:
30 $\vec{0}+\vec{v}=\vec{v}+\vec{0}=\vec{v}$
31 (identity elemnt of addition)
32 \item $\forall \vec{v} \, \exists -\vec{v}$ with: $\vec{v}+(-\vec{v})=0$
33 (inverse element of addition)
35 The addition of two vectors is called vector addition.
38 \subsection{Dual space}
40 \begin{definition}[Dual space]
41 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 $K$
43 \varphi:V\rightarrow K \text{ .}
45 These type of linear maps are termed linear functionals.
46 The dual space $V^{\dagger}$ becomes a vector space over $K$ itself by the following additional definitions
48 (\varphi+\psi)(\vec{v}) & = & \varphi(\vec{v})+\psi(\vec{v}) \\
49 (\lambda\varphi)(\vec{v}) & = & \lambda\varphi(\vec{v})
51 for all $\vec{v}\in V$, $\varphi,\psi\in V^{\dagger}$ and $\lambda\in K$.
53 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 $\vec{v}\in V$.
56 \subsection{Inner and outer product}
57 \label{math_app:product}
59 \begin{definition}[Inner product]
60 The inner product on a vector space $V$ over $K$ is a map
61 $(\cdot,\cdot):V\times V \rightarrow K$, which satisfies
63 \item $(\vec{u},\vec{v})=(\vec{v},\vec{u})^*$
64 (conjugate symmetry, symmetric for $K=\mathbb{R}$)
65 \item $(\lambda\vec{u},\vec{v})=\lambda(\vec{u},\vec{v})$ and
66 $(\vec{u}'+\vec{u}'',\vec{v})=(\vec{u}',\vec{v})+(\vec{u}'',\vec{v})$
67 (linearity in first argument)
68 \item $(\vec{u},\vec{u})\geq 0 \text{, } ``=" \Leftrightarrow \vec{u}=0$
71 for $\vec{u},\vec{v}\in V$ and $\lambda\in K$.
72 Taking the complex conjugate $(\cdot)^*$ is the map from $K\ni z=a+bi\mapsto a-bi=z^*\in K$.
76 Due to conjugate symmetry, linearity in the first argument results in conjugate linearity (also termed antilinearity) in the second argument.
78 (\vec{u},\lambda(\vec{v}'+\vec{v}''))=(\lambda(\vec{v}'+\vec{v}''),\vec{u})^*=
79 \lambda^*(\vec{v}',\vec{u})^*+\lambda^*(\vec{v}'',\vec{u})^*=
80 \lambda^*(\vec{u},\vec{v}')+\lambda^*(\vec{u},\vec{v}'')
82 This is called a sesquilinear form.
83 If $K=\mathbb{R}$, conjugate symmetry reduces to symmetry and the sesquilinear form gets a bilinear form.
85 The inner product $(\cdot,\cdot)$ provides a mapping
87 V\rightarrow V^{\dagger}:\vec{v}\mapsto \varphi_{\vec{v}}
91 \varphi_{\vec{v}}(\vec{u})=(\vec{v},\vec{u}) \text{ .}
93 Since the inner product is linear in the first argument, the same is true for the defined mapping.
95 \lambda(\vec{u}+\vec{v}) \mapsto
96 \varphi_{\lambda(\vec{u}+\vec{v})}=
97 \lambda\varphi_{\vec{u}}+\lambda\varphi_{\vec{v}}\\
99 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.
100 Since the dimension of $V$ and $V^{\dagger}$ is equal, it is additionally surjective.
101 Then, $V$ is isomorphic to $V^{\dagger}$.
102 Vector $\vec{v}^{\dagger}\equiv \varphi_{\vec{v}}\in V^{\dagger}$ is said to be the dual vector of $\vec{v}\in V$.
104 In physics and matrix algebra, the inner product is often defined with linearity in the second argument and conjugate linearity in the first argument.
105 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 $V^{\dagger}$
107 (\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}(\vec{u})\vec{v}
110 or the conjugate transpose in matrix formalism
112 (\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}\vec{v} \text{ .}
114 In doing so, the conjugate transpose is associated with the dual vector.
117 \begin{definition}[Outer product]
118 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 $V$,
119 the outer product $\vec{u}\otimes\vec{v}$ is defined as the tensor product of $\vec{\varphi}$ and $\vec{u}$,
120 which constitutes a map $A:V\rightarrow U$ by
122 \vec{v}\mapsto\vec{\varphi}(\vec{v})\vec{u}
125 where $\vec{\varphi}(\vec{v})$ denotes the linear functional $\vec{\varphi}\in V^{\dagger}$ on $V$ when evaluated at $\vec{v}\in V$, a scalar that in turn is multiplied with $\vec{u}\in U$.
127 In matrix formalism, with respect to a given basis ${\vec{e}_i}$ of $\vec{u}$ and ${\vec{e}'_i}$ of $\vec{v}$,
128 if $\vec{u}=\sum_i^m \vec{e}_iu_i$ and $\vec{v}=\sum_i^n\vec{e}'_iv_i$,
129 the outer product can be written as matrix $A$ as
131 \vec{u}\otimes\vec{v}=A=\left(
132 \begin{array}{c c c c}
133 u_1v_1^* & u_1v_2^* & \cdots & u_1v_n^*\\
134 u_2v_1^* & u_2v_2^* & \cdots & u_2v_n^*\\
135 \vdots & \vdots & \ddots & \vdots\\
136 u_mv_1^* & u_mv_2^* & \cdots & u_mv_n^*\\
143 The matrix can be equivalently obtained by matrix multiplication:
145 \vec{u}\otimes\vec{v}=\vec{u}\vec{v}^{\dagger} \text{ ,}
147 if $\vec{u}$ and $\vec{v}$ are represented as $m\times 1$ and $n\times 1$ column vectors, respectively.
148 Here, $\vec{v}^{\dagger}$ represents the conjugate transpose of $\vec{v}$.
149 By definition, and as can be easily seen in the matrix representation, the following identity holds:
151 (\vec{u}\otimes\vec{v})\vec{w}=\vec{u}(\vec{v},\vec{w})
155 \section{Spherical coordinates}
157 \section{Fourier integrals}