X-Git-Url: https://hackdaworld.org/gitweb/?p=lectures%2Flatex.git;a=blobdiff_plain;f=physics_compact%2Fmath.tex;h=4c8e6f9946efda96dfd874b2a2f90e79a0438fa5;hp=0e49a4010aaf2afc5a105b209cbeadf0d4cb9fd4;hb=0929321d46ef53429ae8d78a0f08e452e96b3cd2;hpb=e2bc34a23f03db9417fa6d7a0a9b1d4a1538c0e6 diff --git a/physics_compact/math.tex b/physics_compact/math.tex index 0e49a40..4c8e6f9 100644 --- a/physics_compact/math.tex +++ b/physics_compact/math.tex @@ -1,6 +1,53 @@ \part{Mathematical foundations} +Reminder: Modern Quantum Chemistry \& Sakurai \& Group Theory \ldots + \chapter{Linear algebra} -Reminder: Modern Quantum Chemistry \& Sakurai \& Group Theory \ldots +\section{Vectors and bases} + +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^N \vec{e}_i a_i +\label{eq:vec_sum} +\end{equation} +The scalar product for an $N$-dimensional real 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 enables to define a norm +\begin{equation} +||\vec{a}||=\sqrt{(\vec{a},\vec{a})} +\end{equation} +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}_i)a_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\cdot \vec{e}_i=\vec{1} +\end{equation} +if the basis is complete. +Indeed, the very important identity representation by the outer product ($\cdot$) in the second part of \eqref{eq:complete} is known as the completeness relation or closure.