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\title{\LARGE On the use of rational expressions in Combinatorial Physics: an example.}

\author{
G\'erard H. E. \sc Duchamp and Christophe \sc Tollu.\rm
\thanks{LIPN - UMR 7030
CNRS - Universit\'e Paris 13
F-93430 Villetaneuse, France}}
%            \null\ et 
%            Christophe \sc Reutenauer\rm\thanks{Math\'ematiques, 
%            Universit\'e du Qu\'ebec \`a Montr\'eal C.P. 8888, Succ. 
%            Centre-Ville Montr\'eal Qc Canada H3C 3P8. L'auteur a 
%            b\'en\'efici\'e 
%            d`une subvention CRSNG (Canada) 
%            et de l'hospitalit\'e de l'Universit\'e Louis Pasteur (Strasbourg).
%            }}

\date{}
%
\begin{document}

\maketitle

\begin{abstract} 
In a joint work with J. Katriel, one of us gave the solution of
the problem of matrix coefficients in the Fock space of carriers between
two levels. We give here a complete proof of these continued fractions-type
formulas and possible extensions of the method. 
\footnote[0]{Version compil\'ee le \isodayandtime}
\end{abstract}

\section{Introduction}

\section{Brief review of some known formulas}\label{known_formulas}

In \cite{D20} was considered, as Fock space, a general vector space $V$ over a field $k$ with basis 
$|e_n\rangle \;\; n= 0,\, 1,\, \cdots$, equipped with its natural grading
\begin{equation}
V=\oplus_{n\in {\bf Z}}V_n\ {\rm with}\ V_n:=ke_n;
\ V_{-n-1}:=(0)\ {\rm for}\ n\geq 0
\end{equation}
and scalar product defined by $\scal{e_n}{e_m}=\delta_{n,m}$.
and $f,\ g$ two linear operators in $V$ of degrees $-1,\ +1$, 
respectively. Generically, they read
\begin{equation}
\ {\rm for}\ n\geq 0;\ f|e_0\rangle :=
0;\ f|e_{k+1}\rangle =\alpha_{k+1}|e_k\rangle;
\ g|e_k\rangle :=\beta_{k+1}|e_{k+1}\rangle 	
\end{equation}
We consider the words in $f,\ g$:
\begin{equation}
w(f,g):=f^{p_1}g^{q_1}f^{p_2}g^{q_2}\cdots f^{p_n}g^{q_n}	
\end{equation}
the degree (excess) of which is $\sum_{k=1}^n (q_k-p_k)$ (this is, for algebraists, the degree of the graded operator $f^{p_1}g^{q_1}f^{p_2}g^{q_2}\cdots f^{p_n}g^{q_n}$). 
This provides a representation $\mu$ of a two letters free monoid $\{b_-,b_+\}^*$ on $V$ by 
$\mu(b_-)=f;\ \mu(b_+)=g$, which is graded for the weight on $\{b_-,b_+\}^*$. In order to keep the reading of a word 
from left to right, one performs the the action on the right. Thus $V$ becomes a $\{b_-,b_+\}^*$ right module by 
\begin{equation}
e_0.b_-=0\ ;\ e_{n+1}.b_-=\al_{n+1}e_n\ ;\ e_{n}.b_+=\be_{n+1}e_{n+1}
\end{equation}
one is interested by the matrix elements
\begin{equation}
	\scal{e_n.\{b_-,b_+\}^i}{e_m}=\om_{n\ra m}^{(i)}\ .
\end{equation}
The following propsition gives the support i. e. the set of words $w$ of length $i$ such that $\scal{e_n.w}{e_m}\not=0$.

\begin{proposition} Define $W_{n\ra m}^{(i)}$ as follows 
\begin{equation}
	W_{n\ra m}^{(i)}=\{w\in \{b_-,b_+\}^i\ |\ (\pi_e(w)=m-n)\ \mathrm{and}\ (w=uv\Longrightarrow \pi_e(w)\geq -n)\}
\end{equation}
then \\
i) If the all the weights $\al_n;\ n\geq 1$, $\be_n, n\geq 0$ are not zero $W_{n\ra m}^{(i)}$ is exactly the set of words of length $i$ such that $\scal{e_n.w}{e_m}\not=0$.\\
ii) In all cases the latter is a subset of $W_{n\ra m}^{(i)}$ i. e. 
\begin{equation}
\{w\in \{b_-,b_+\}^i\ |\ \scal{e_n.w}{e_m}\not=0\}\subset W_{n\ra m}^{(i)} 	
\end{equation}
\end{proposition}
As a result, 


\begin{equation}
\mu(W_{m-n}^{(i)})\ket{n}=\omega_{n\ra m}^{(i)}\ket{m}\ ;\ 
T_{n\rightarrow n+k}:=\sum_{i\geq 0}t^i\omega_{n\rightarrow n+k}^{(i)} \ .
\end{equation}
One can expand $T_{n\rightarrow n+k}$ as a product of continued fractions. Let 
\begin{eqnarray}\label{eq9}
                         F^+_n= {1\over\displaystyle 1-
{\strut t^2\alpha_{n+1}\beta_{n+1}\over\displaystyle 1-
{\strut t^2\alpha_{n+2}\beta_{n+2}\over\displaystyle 1-
{\strut t^2\alpha_{n+3}\beta_{n+3}\over\displaystyle 1-\cdots
}}}}={1\over 1-E^+_n}\cr\cr\cr
                         F^-_n= {1\over\displaystyle 1-
{\strut t^2\alpha_{n  }\beta_{n  }\over\displaystyle 1-
{\strut t^2\alpha_{n-1}\beta_{n-1}\over\displaystyle 1-
{\strut t^2\alpha_{n-2}\beta_{n-2}\over\displaystyle 1-\cdots
}}}}={1\over 1-E^-_n}
\end{eqnarray}
and 
\begin{equation}\label{eq10}
F_n= {1\over 1-E^+_n-E^-_n}\ .	
\end{equation}
Then, if $k\geq 0$, we have
\begin{equation}\label{eq11} 
T_{n\rightarrow n+k}=t^kF_{n+k}\prod_{i=0}^{k-1}F^-_{n+i}
=t^kF_n\prod_{i=1}^kF^+_{n+i}
\end{equation}
And, if $k\leq 0$
\begin{equation}\label{eq12}
T_{n\rightarrow n+k}=t^{-k}F_{n+k}\prod_{i=0}^{-k-1}F^+_{n-i}
=t^{-k}F_n\prod_{i=1}^{-k}F^-_{n-i}
\end{equation}

\section*{References}
\begin{thebibliography}{ABC}
\bibitem{D20} \textsc{Katriel J., Duchamp G.}, \textit{Ordering relations for q-boson operators, continued fractions techniques, 
and the q-CBH enigma}. Journal of Physics A {\bf 28} 7209-7225 (1995).
%
\end{thebibliography}

\end{document}