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.
23 Due to the additive abelian group, the following properties are additionally valid:
25 \item $\vec{u}+\vec{v}=\vec{v}+\vec{u}$ (commutativity of addition)
26 \item $\vec{u}+(\vec{v}+\vec{w})=(\vec{u}+\vec{v})+\vec{w}$
27 (associativity of addition)
28 \item $\forall \vec{v} \, \exists \vec{0}$ with:
29 $\vec{0}+\vec{v}=\vec{v}+\vec{0}=\vec{v}$
30 (identity elemnt of addition)
31 \item $\forall \vec{v} \, \exists -\vec{v}$ with: $\vec{v}+(-\vec{v})=0$
32 (inverse element of addition)
34 The addition of two vectors is called vector addition.
37 \subsection{Dual space}
39 \begin{definition}[Dual space]
40 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 $K$
42 \varphi:V\rightarrow K \text{ .}
44 These type of linear maps are termed linear functionals.
45 The dual space $V^{\dagger}$ becomes a vector space over $K$ itself by the following additional definitions
47 (\varphi+\psi)(\vec{v}) & = & \varphi(\vec{v})+\psi(\vec{v}) \\
48 (\lambda\varphi)(\vec{v}) & = & \lambda\varphi(\vec{v})
50 for all $\vec{v}\in V$, $\varphi,\psi\in V^{\dagger}$ and $\lambda\in K$.
52 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 $\vec{v}\in V$.
55 \subsection{Inner and outer product}
56 \label{math_app:product}
58 \begin{definition}[Inner product]
59 The inner product on a vector space $V$ over $K$ is a map
60 $(\cdot,\cdot):V\times V \rightarrow K$, which satisfies
62 \item $(\vec{u},\vec{v})=(\vec{v},\vec{u})^*$
63 (conjugate symmetry, symmetric for $K=\mathbb{R}$)
64 \item $(\lambda\vec{u},\vec{v})=\lambda(\vec{u},\vec{v})$ and
65 $(\vec{u}'+\vec{u}'',\vec{v})=(\vec{u}',\vec{v})+(\vec{u}'',\vec{v})$
66 (linearity in first argument)
67 \item $(\vec{u},\vec{u})\geq 0 \text{, } ``=" \Leftrightarrow \vec{u}=0$
70 for $\vec{u},\vec{v}\in V$ and $\lambda\in K$.
71 Taking the complex conjugate $(\cdot)^*$ is the map from $K\ni z=a+bi\mapsto a-bi=z^*\in K$.
75 Due to conjugate symmetry, linearity in the first argument results in conjugate linearity (also termed antilinearity) in the second argument.
77 (\vec{u},\lambda(\vec{v}'+\vec{v}''))=(\lambda(\vec{v}'+\vec{v}''),\vec{u})^*=
78 \lambda^*(\vec{v}',\vec{u})^*+\lambda^*(\vec{v}'',\vec{u})^*=
79 \lambda^*(\vec{u},\vec{v}')+\lambda^*(\vec{u},\vec{v}'')
81 This is called a sesquilinear form.
82 If $K=\mathbb{R}$, conjugate symmetry reduces to symmetry and the sesquilinear form gets a bilinear form.
84 The inner product $(\cdot,\cdot)$ provides a mapping
86 V\rightarrow V^{\dagger}:\vec{v}\mapsto \varphi_{\vec{v}}
90 \varphi_{\vec{v}}(\vec{u})=(\vec{v},\vec{u}) \text{ .}
92 Since the inner product is linear in the first argument, the same is true for the defined mapping.
94 \lambda(\vec{u}+\vec{v}) \mapsto
95 \varphi_{\lambda(\vec{u}+\vec{v})}=
96 \lambda\varphi_{\vec{u}}+\lambda\varphi_{\vec{v}}\\
98 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.
99 Since the dimension of $V$ and $V^{\dagger}$ is equal, it is additionally surjective.
100 Then, $V$ is isomorphic to $V^{\dagger}$.
101 Vector $\vec{v}^{\dagger}\equiv \varphi_{\vec{v}}\in V^{\dagger}$ is said to be the dual vector of $\vec{v}\in V$.
103 In physics and matrix algebra, the inner product is often defined with linearity in the second argument and conjugate linearity in the first argument.
104 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 $V^{\dagger}$
106 (\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}(\vec{u})\vec{v}
109 or the conjugate transpose in matrix formalism
111 (\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}\vec{v} \text{ .}
113 In doing so, the conjugate transpose is associated with the dual vector.
116 \begin{definition}[Outer product]
117 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 $V$,
118 the outer product $\vec{u}\otimes\vec{v}$ is defined as the tensor product of $\vec{\varphi}$ and $\vec{u}$,
119 which constitutes a map $A:V\rightarrow U$ by
121 \vec{v}\mapsto\vec{\varphi}(\vec{v})\vec{u}
124 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 $\vec{u}\in U$.
126 In matrix formalism, with respect to a given basis ${\vec{e}_i}$ of $\vec{u}$ and ${\vec{e}'_i}$ of $\vec{v}$,
127 if $\vec{u}=\sum_i^m \vec{e}_iu_i$ and $\vec{v}=\sum_i^n\vec{e}'_iv_i$,
128 the outer product can be written as matrix $A$ as
130 \vec{u}\otimes\vec{v}=A=\left(
131 \begin{array}{c c c c}
132 u_1v_1^* & u_1v_2^* & \cdots & u_1v_n^*\\
133 u_2v_1^* & u_2v_2^* & \cdots & u_2v_n^*\\
134 \vdots & \vdots & \ddots & \vdots\\
135 u_mv_1^* & u_mv_2^* & \cdots & u_mv_n^*\\
142 The matrix can be equivalently obtained by matrix multiplication:
144 \vec{u}\otimes\vec{v}=\vec{u}\vec{v}^{\dagger} \text{ ,}
146 if $\vec{u}$ and $\vec{v}$ are represented as $m\times 1$ and $n\times 1$ column vectors, respectively.
147 Here, $\vec{v}^{\dagger}$ represents the conjugate transpose of $\vec{v}$.
148 By definition, and as can be easily seen in the matrix representation, the following identity holds:
150 (\vec{u}\otimes\vec{v})\vec{w}=\vec{u}(\vec{v},\vec{w})
154 \section{Spherical coordinates}
156 \section{Fourier integrals}