theorems and hyperrefs

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

diff --git a/physics_compact/math_app.tex b/physics_compact/math_app.tex
index d097e70..f1f0ae3 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}
 
-\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: $\vec{v}1=\vec{v}$
@@ -36,7 +36,7 @@ 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 +55,9 @@ 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 $K=\mathbb{R}$)
@@ -77,22 +73,32 @@ Taking the complex conjugate $(\cdot)^*$ is the map from $K\ni z=a+bi\mapsto a-b
 
 \begin{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
 \begin{equation}
-V\rightarrow V^{\dagger}:\vec{v}\mapsto \vec{v}^{\dagger}
+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{ .}
 \end{equation}
-given by 
+Since the inner product is linear in the first argument, the same is true for the defined mapping.
 \begin{equation}
-v^{\dagger}()
+\lambda(\vec{u}+\vec{v}) \mapsto
+\varphi_{\lambda(\vec{u}+\vec{v})}=
+\lambda\varphi_{\vec{u}}+\lambda\varphi_{\vec{v}}\\
 \end{equation}
-indicating structural identity (isomorphism) of $V$ and $V^{\dagger}$.
+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 $\vec{v}\in V$.
 
 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 $V^{\dagger}$
@@ -104,18 +110,18 @@ or the conjugate transpose in matrix formalism
 \begin{equation}
 (\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}\vec{v} \text{ .}
 \end{equation}
-In doing so, conjugacy is associated with duality.
+In doing so, the conjugate transpose is associated with the dual vector.
 \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 $\vec{u}\in U$, $\vec{v}\in V$ are vectors within the respective vector spaces and $\vec{\varphi}\in V^{\dagger}$  is a linear functional of the dual space $V^{\dagger}$ of $V$,
+the outer product $\vec{u}\otimes\vec{v}$ is defined as the tensor product of $\vec{\varphi}$ 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{v}\mapsto\vec{\varphi}(\vec{v})\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 $\vec{\varphi}(\vec{v})$ 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 $\vec{u}\in U$.
 
 In matrix formalism, with respect to a given basis ${\vec{e}_i}$ of $\vec{u}$ and ${\vec{e}'_i}$ of $\vec{v}$,
 if $\vec{u}=\sum_i^m \vec{e}_iu_i$ and $\vec{v}=\sum_i^n\vec{e}'_iv_i$,
@@ -123,10 +129,10 @@ the outer product can be written as matrix $A$ as
 \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{ .}