From: hackbard Date: Tue, 7 Feb 2012 22:01:36 +0000 (+0100) Subject: more math, started operators X-Git-Url: https://hackdaworld.org/cgi-bin/gitweb.cgi?a=commitdiff_plain;h=1b61eac526bb34de8c8060551f70c6e7cd437d89;p=lectures%2Flatex.git more math, started operators --- diff --git a/physics_compact/math.tex b/physics_compact/math.tex index 4c8e6f9..c182a96 100644 --- a/physics_compact/math.tex +++ b/physics_compact/math.tex @@ -6,11 +6,12 @@ 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} +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{ ,} @@ -51,3 +52,23 @@ Inserting the expression for the coefficients into \eqref{eq:vec_sum}, the vecto 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. +\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} +