after ice SO stuff

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

diff --git a/physics_compact/math.tex b/physics_compact/math.tex
index ddaebc6..bd2a888 100644 (file)
--- a/physics_compact/math.tex
+++ b/physics_compact/math.tex
@@ -1,26 +1,27 @@
 \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 ${\vec{e}_i}$.
+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 vector space is defined as
+i.e., if the basis set is complete, any vector can be written as a linear combination of these basis vectors.
+The scalar product in an $N$-dimensional Euclidean 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
+which satisfies the properties of an inner product (see \ref{math_app:product}) and 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}.
+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)=
@@ -38,15 +39,36 @@ i.e.\  the basis vectors are mutually perpendicular (orthogonal) and  have unit
 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=
+(\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\vec{e}_i=\vec{1}
+\vec{a}=\sum_i \vec{e}_i (\vec{e}_i,\vec{a}) \Leftrightarrow
+\sum_i\vec{e}_i\otimes \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.
+Indeed, the very important identity representation by the outer product ($\otimes$, see \ref{math_app:product}) in the second part of \eqref{eq:complete} is known as the completeness relation or closure.
+
+\section{Operators, matrices and determinants}
+
+An operator $O$ acts on a vector resulting in another vector
+\begin{equation}
+O\vec{a}=\vec{b} \text{ ,}
+\end{equation}
+which is linear if
+\begin{equation}
+O(\lambda\vec{a}+\mu\vec{b})=\lambda O\vec{a} + \mu O\vec{b} \text{ .}
+\end{equation}
+Thus, for a linear operator, it is sufficient to describe the effect on the complete set of basis vectors, which enables to describe the effect of the operator on any vector.
+Since the result of an operator acting on a basis vector is a vector itself, it can be expressed by a linear combination of the basis vectors
+\begin{equation}
+O\vec{e}_i=\vec{e}_jO_{ji}
+\text{ ,}
+\end{equation}
+with $O_{ji}$ determining the components of the new vector $O\vec{e}_i$ along $\vec{e}_j$.
+
+\section{Dirac notation}
 
-Todo: outer product ... + explicitly mark scalar product