hopefully finished respective math stuff, finally!

[lectures/latex.git] / physics_compact / math_app.tex

diff --git a/physics_compact/math_app.tex b/physics_compact/math_app.tex
index 687f928..f1ac777 100644 (file)
--- a/physics_compact/math_app.tex
+++ b/physics_compact/math_app.tex
@@ -5,7 +5,7 @@
 \subsection{Vector space}
 \label{math_app:vector_space}
 
 \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: $\vec{v}1=\vec{v}$
 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: $\vec{v}1=\vec{v}$


@@ -19,6 +19,7 @@ A vector space $V$ over a field $(K,+,\cdot)$ is an additive abelian group $(V,+
 \end{itemize}
 The elements $\vec{v}\in V$ are called vectors.
 \end{definition}
 \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}
 \begin{remark}
 Due to the additive abelian group, the following properties are additionally valid:
 \begin{itemize}


@@ -36,7 +37,7 @@ The addition of two vectors is called vector addition.
 
 \subsection{Dual space}
 
 
 \subsection{Dual space}
 

-\begin{definition}
+\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{ .}
 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{ .}


@@ -55,13 +56,9 @@ The map $V^{\dagger}\times V \rightarrow K: [\varphi,\vec{v}]=\varphi(\vec{v})$
 \subsection{Inner and outer product}
 \label{math_app:product}
 
 \subsection{Inner and outer product}
 \label{math_app:product}
 

-\begin{definition}
+\begin{definition}[Inner product]

 The inner product on a vector space $V$ over $K$ is a map
 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
+$(\cdot,\cdot):V\times V \rightarrow K$, which satisfies

 \begin{itemize}
 \item $(\vec{u},\vec{v})=(\vec{v},\vec{u})^*$
       (conjugate symmetry, symmetric for $K=\mathbb{R}$)
 \begin{itemize}
 \item $(\vec{u},\vec{v})=(\vec{v},\vec{u})^*$
       (conjugate symmetry, symmetric for $K=\mathbb{R}$)


@@ -72,25 +69,31 @@ which satisfies
       (positive definite)
 \end{itemize}
 for $\vec{u},\vec{v}\in V$ and $\lambda\in K$.
       (positive definite)
 \end{itemize}
 for $\vec{u},\vec{v}\in V$ and $\lambda\in K$.

-Taking the complex conjugate $(\cdot)^*$ is the map from $K\ni z=a+bi\mapsto a-bi=z^*\in K$.
+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}
 \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.
 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}
 \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.

 
 

-The inner product $(\cdot,\cdot)$ provides a mapping
+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{ .}
 \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}
 \end{equation}
 Since the inner product is linear in the first argument, the same is true for the defined mapping.
 \begin{equation}


@@ -98,57 +101,81 @@ Since the inner product is linear in the first argument, the same is true for th
 \varphi_{\lambda(\vec{u}+\vec{v})}=
 \lambda\varphi_{\vec{u}}+\lambda\varphi_{\vec{v}}\\
 \end{equation}
 \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 .
+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}

 
 

-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 $V^{\dagger}$
+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}
 \begin{equation}

-(\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}(\vec{u})\vec{v}
-\text{ CHECK ! ! !}
+\langle\vec{v},\vec{u}\rangle=\sum_i^nv_i^*u_i
+\text{ .}

 \end{equation}
 \end{equation}

-or the conjugate transpose in matrix formalism
+In this case, in matrix formalism, the inner product is reformulated

 \begin{equation}
 \begin{equation}

-(\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}\vec{v} \text{ .}
+(\vec{v},\vec{u}) \rightarrow \vec{v}^{\dagger}\vec{u}
+\text{ ,}

 \end{equation}
 \end{equation}

-In doing so, the conjugate transpose is associated with the dual vector.
+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}
 
 \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 $\vec{u}$,
+\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}
 which constitutes a map $A:V\rightarrow U$ by
 \begin{equation}

-\vec{v}\mapsto\vec{\varphi}^{\dagger}(\vec{v})\vec{u}
+\vec{w}\mapsto\varphi_{\vec{v}}(\vec{w})\vec{u}

 \text{ ,}
 \end{equation}
 \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 $\vec{u}\in U$.
+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}

 
 

-In matrix formalism, with respect to a given basis ${\vec{e}_i}$ of $\vec{u}$ and ${\vec{e}'_i}$ of $\vec{v}$,
+\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 $\vec{v}=\sum_i^n\vec{e}'_iv_i$,
 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}
 \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\\
 \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)
 \end{array}
 \right)

-\text{ .}
+\text{ ,}

 \end{equation}
 \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.
 \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}
 \begin{equation}
 (\vec{u}\otimes\vec{v})\vec{w}=\vec{u}(\vec{v},\vec{w})
 \end{equation}

+holds.

 \end{remark}
 
 \section{Spherical coordinates}
 \end{remark}
 
 \section{Spherical coordinates}