From: hackbard 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