\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: $\vec{v}1=\vec{v}$
\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}
\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}
-The inner product on a vector space $V$ over $K$ is a map $(\cdot,\cdot):V\times V \rightarrow K$, which satisfies
+\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 $K=\mathbb{R}$)
(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.
-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}
-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 the dual vector or linear functional of dual space $V^{\dagger}$
+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}
-(\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}(\vec{u})\vec{v}
+\lambda(\vec{u}+\vec{v}) \mapsto
+\varphi_{\lambda(\vec{u}+\vec{v})}=
+\lambda\varphi_{\vec{u}}+\lambda\varphi_{\vec{v}}\\
\end{equation}
-or the conjugate transpose in matrix formalism
+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}
-(\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}\vec{v} \text{ .}
+\varphi_{\lambda\vec{v}}(\vec{u})=
+(\lambda\vec{v},\vec{u})=
+\lambda^*(\vec{v},\vec{u})=
+\lambda^*\varphi_{\vec{v}}(\vec{u})
\end{equation}
-In doing so, conjugacy is associated with duality.
+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}
-If $\vec{u}\in U$, $\vec{v}\in V$ are vectors within the respective vector spaces and $\vec{y}^{\dagger}\in V^{\dagger}$ 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 $\vec{y}^{\dagger}$ and $\vec{u}$,
+\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{v}\mapsto\vec{y}^{\dagger}(\vec{v})\vec{u}
+\vec{w}\mapsto\varphi_{\vec{v}}(\vec{w})\vec{u}
\text{ ,}
\end{equation}
-where $\vec{y}^{\dagger}(\vec{v})$ denotes the linear functional $\vec{y}^{\dagger}\in V^{\dagger}$ on $V$ when evaluated at $\vec{v}\in V$, a scalar that in turn is multiplied with $\vec{u}\in U$.
+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, with respect to a given basis ${\vec{e}_i}$ of $\vec{u}$ and ${\vec{e}'_i}$ of $\vec{v}$,
+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 outer product can be written as matrix $A$ as
+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\\
+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\\
+u_mv_1^* & u_mv_2^* & \cdots & u_mv_n^*\\
\end{array}
\right)
-\text{ .}
+\text{ ,}
\end{equation}
-\end{definition}
-\begin{remark}
-The matrix can be equivalently obtained by matrix multiplication:
+which can be equivalently obtained by the rulrs 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, $\vec{v}^{\dagger}$ represents the conjugate transpose of $\vec{v}$.
-By definition, and as can be easily seen in the matrix representation, the following identity holds:
+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}
projects / lectures / latex.git / blobdiff