From: hackbard <hackbard@hackdaworld.org>
Date: Mon, 6 Feb 2012 21:32:17 +0000 (+0100)
Subject: more math intro
X-Git-Url: https://hackdaworld.org/cgi-bin/gitweb.cgi?a=commitdiff_plain;h=3f852281225ee807881897349cf4ca6fba124a48;p=lectures%2Flatex.git

more math intro
---

diff --git a/physics_compact/math.tex b/physics_compact/math.tex
index 5f437fa..ddaebc6 100644
--- a/physics_compact/math.tex
+++ b/physics_compact/math.tex
@@ -8,19 +8,45 @@ Reminder: Modern Quantum Chemistry \& Sakurai \& Group Theory \ldots
 
 A vector $\vec{a}$ of an $N$-dimensional vector space (see \ref{math_app:vector_space} for mathematical details) is represented by its components $a_i$ with respect to a set of $N$ basis vectors ${\vec{e}_i}$.
 \begin{equation}
-\vec{a}=\sum_i \vec{e}_i a_i
+\vec{a}=\sum_i^N \vec{e}_i a_i
 \label{eq:vec_sum}
 \end{equation}
 The scalar product for an $N$-dimensional vector space is defined as
 \begin{equation}
 (\vec{a},\vec{b})=\sum_i^N a_i b_i \text{ ,}
+\label{eq:vec_sp}
 \end{equation}
-which introduces a norm
+which enables to define a norm
 \begin{equation}
 ||\vec{a}||=\sqrt{(\vec{a},\vec{a})}
 \end{equation}
-that correpsonds to the length of vector \vec{a}.
-Evaluating the scalar product $(\vec{a},\vec{b})$ by the sum representation of \eqref{eq:vec_sum} \ldots
+that just corresponds to the length of vector \vec{a}.
+Evaluating the scalar product $(\vec{a},\vec{b})$ by the sum representation of \eqref{eq:vec_sum} leads to
 \begin{equation}
+(\vec{a},\vec{b})=(\sum_i\vec{e}_ia_i,\sum_j\vec{e}_jb_j)=
+\sum_i\sum_j(\vec{e}_i,\vec{e}_j)a_ib_j \text { ,}
 \end{equation}
+which is equal to \eqref{eq:vec_sp} only if
+\begin{equation}
+(\vec{e}_i,\vec{e}_j)=
+\delta_{ij}  = \left\{ \begin{array}{lll}
+0 & {\rm for} ~i \neq j \\
+1 & {\rm for} ~i = j   \end{array} \right.
+\text{ (Kronecker delta symbol),}
+\end{equation}
+i.e.\  the basis vectors are mutually perpendicular (orthogonal) and  have unit length (normalized).
+Such a basis set is called orthonormal.
+The component of a vector can be obtained by taking the scalar product with the respective basis vector.
+\begin{equation}
+\vec{e}_j\vec{a}=\vec{e}_j \sum_i \vec{e}_ia_i=\sum_i \vec{e}_j\vec{e}_ia_i=
+\sum_i\delta_{ij}a_i=a_j
+\end{equation}
+Inserting the expression for the coefficients into \eqref{eq:vec_sum}, the vector can be written as
+\begin{equation}
+\label{eq:complete}
+\vec{a}=\sum_i \vec{e}_i (\vec{e}_i\vec{a}) \Leftrightarrow \sum_i\vec{e}_i\vec{e}_i=\vec{1}
+\end{equation}
+if the basis is complete.
+Thus, the very important second part of \eqref{eq:complete} is known as the completeness relation or closure.
 
+Todo: outer product ... + explicitly mark scalar product