\part{Mathematical foundations}

Reminder: Modern Quantum Chemistry \& Sakurai \& Group Theory \ldots

\chapter{Linear algebra}

\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}
i.e., if the basis set is complete, any vector can be written as a linear combination of these basis vectors.
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\otimes \vec{e}_i=\vec{1}
\end{equation}
if the basis is complete.
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}

