X-Git-Url: https://hackdaworld.org/gitweb/?p=lectures%2Flatex.git;a=blobdiff_plain;f=physics_compact%2Fmath_app.tex;h=f1ac777ea2a51d8168b426991d5fca0a05b3e8c1;hp=687f928df9145c3377402563658d5001d1090de2;hb=bc36ee251b928962cf2431d6582e5a5dd9cca682;hpb=f74a53b2acb28533ef22dbf3adb0c97dbb5dd958

diff --git a/physics_compact/math_app.tex b/physics_compact/math_app.tex
index 687f928..f1ac777 100644
--- a/physics_compact/math_app.tex
+++ b/physics_compact/math_app.tex
@@ -5,7 +5,7 @@
 \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: $\vec{v}1=\vec{v}$
@@ -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}
+
 \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}
 
-\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{ .}
@@ -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}
 
-\begin{definition}
+\begin{definition}[Inner product]
 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 $K=\mathbb{R}$)
@@ -72,25 +69,31 @@ which satisfies
       (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}
+\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}
+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{ .}
+\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}
@@ -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}
-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}
-(\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}
-or the conjugate transpose in matrix formalism
+In this case, in matrix formalism, the inner product is reformulated
 \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}
-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}
 
-\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}
-\vec{v}\mapsto\vec{\varphi}^{\dagger}(\vec{v})\vec{u}
+\vec{w}\mapsto\varphi_{\vec{v}}(\vec{w})\vec{u}
 \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 $\vec{v}=\sum_i^n\vec{e}'_iv_i$,
-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}