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\begin{document}

\title{Approximate substitutions and the normal ordering problem}

\author{HAYAT CHEBALLAH$^{\dag}$,
G\'ERARD H E Duchamp$^{\dag}$ and\\ 
KAROL A Penson$^{\diamondsuit}$ }
%
\address{
$^\dag$ LIPN - UMR 7030\linebreak CNRS - Universit\'e Paris
13\linebreak F-93430 Villetaneuse, France\vspace{2mm} }
\address
{$^\diamondsuit$ Laboratoire de Physique Th\'eorique des Liquides,\\
Universit\'e Pierre et Marie Curie, CNRS UMR 7600\\
Tour 24 - 2e \'et., 4 pl. Jussieu, F 75252 Paris Cedex 05, France}

\eads{\linebreak\mailto{cheballah@lipn-univ.paris13.fr},
\mailto{ghed@lipn-univ.paris13.fr},
\mailto{penson@lptl.jussieu.fr} \linebreak}

\begin{abstract}
......
\end{abstract}


\section{Introduction}

Quantum theory of fields takes its conceptual framework from
statistical physic which models the interactions between
particles. As now, two types of particles are known;  Fermions and
 Bosons. The symbolic computation on these particles is endowed with a relation of
$q$-commutation  where $q$ can take the values
$-1$ (fermions), or $1$ (bosons).\\

many works concern function s on  boson particles which have the
particular property of symmetry when the the particles are
exchanged.\\ We use the second quantification to replace the
complex coefficients of Fourier's expansion of scalar field by two
abstract operators: $a$, the \textit{annihilation operator} and
$a^\dag$, the \textit{creation operator}.\\

The countable Hilbert spaces represent the arena where quantum
phenomena can be described. A common realization is the Fock space
which is generated by a hilbertian  (\textit{i.e} orthonormal)
basis of
 vectors (kets) $(|n\rangle)_{n\geq 0}$ where $n$ gives the exact
number of particles.\\ The usual Fock space is thus defined as an
Hilbert space obtained by completed direct sumation of tensor
products of Hilbert spaces associated to one particle each. It
then reads
\begin{equation}
F_\nu(H)=F_\nu(H)=\widehat{\bigoplus_{n=0}^{\infty}} S_\nu
H^{\otimes n}
\end{equation}
where $S_\nu$ is the operator which symmetrises and $H$ is an
Hilbert
space for one particle.\\

Now, we consider creation and annihilation operators with the
commutation relation
\begin{eqnarray}\label{[a,a+]=1}
% \nonumber to remove numbering (before each equation)
  [a,a^\dag] &=& 1
\end{eqnarray}
Recall that
\begin{itemize}
    \item Annihilation operator, in seconde quantification,
    represents an operator which changes each state of Fock space containing $N>1$ particles
for another containing $(N-1)$ particles.
    $$a_i|N_1,N_2,...,N_i,...\rangle=
\sqrt{N_i}|N_1,N_2,...,N_i-1,...\rangle$$
    \item The hermitian conjugate of the annihilation operator is the creation
operator which changes each state of Fock space containing $N$
particles to another containing $(N+1)$ particles.
    $$a_i^\dag|N_1,N_2,...,N_i,...\rangle=
\sqrt{N_i+1}|N_1,N_2,...,N_i+1,...\rangle$$
\end{itemize}

Starting from the vacuum $|0\rangle$, we can reach all the
normalized states in the Fock space. There, indeed, are given by:
\begin{eqnarray}
% \nonumber to remove numbering (before each equation)
  |N_1,N_2,...\rangle &=&
  \prod_k\frac{(a_k^\dag)^{N_k}}{\sqrt{N_k}}|0\rangle
\end{eqnarray}

\section{Normally ordered form}
The noncommutativity of annihilation and creation operators may
cause problems in defining an operator function in quantum
mechanics. To solve these problems, we have to find some normal
form which allows computing, reduction and equality decision. One
of the simplest and widely used solution is the \textit{normally
ordered form} of the boson operators in which all $a^\dag$ stand
on the left of the factors $a$.\\There are two well procedures
defined on the boson expressions yielding a normally ordered form.
Namely $\mathcal N$, the \textit{Normal ordering}
 and  $::$, the \textit{double dot operation}.
\subsection{Normal ordering}
By  normal ordering of a general expression $F(a^\dag,a)$, we mean
 $\mathcal{N}[F(a^\dag,a)]$ which is obtained by moving all the
 annihilation operators $a$ on the right using the commutation
 relation (\ref{[a,a+]=1}).
 \begin{exemple} Let  $w\in\{a,a^\dag\}^*$ a word given by
$w=aa^\dag aaa^\dag a$. The normal ordering of $w$ is
\begin{eqnarray*}
   aa^\dag aaa^\dag a &=&  (1+a^\dag a)a(1+a^\dag a)a=a^2+a^\dag a^3+aa^\dag a^2+a^\dag a^2 a^\dag a^2\\
   &=& a^2+a^\dag a^3+(1+a^\dag a)a^2+a^\dag a(1+a^\dag a)a^2\\
   &=& a^2+a^\dag a^3+a^2+a^\dag a^3+a^\dag a^3+a^\dag aa^\dag
   a^3=2 a^2+3a^\dag a^3+a^\dag(1+a^\dag a)a^3\\
   &=& \underbrace{2a^2+3a^\dag a^3+(a^\dag)^2a^4}
\end{eqnarray*}
 \end{exemple}
 \begin{remark}
Note that all coefficients of the normal ordering of a word (more
precisely the coefficients of the decomposition of a word
 on the basis $\Big((a^\dag)^j a^l\Big)$)
are positive integers. This suggests that these quantities count
things.
 \end{remark}
\subsection{Double dot operation}
The double dot operation $:F(a^\dag,a):$ is a similar procedure
using another commutation relation i. e. $[a,a+]=0$ instead of
$[a,a+]=1$, \textit{i.e} moving all annihilation operators $a$ to
the right as if they were commuting with
the creation operators $a^\dag$.\\
\begin{remark}The double dot operation $: :$ is a linear operator
$$\C<a,a^\dag> \stackrel{::}{\longrightarrow} HW(\C)$$
which can be directly defined by:
$$\quad\quad\quad w\in\{a,a^\dag\}\longmapsto :w:=a^{\dag|w|_{a^\dag}}a^{|w|_{a}}$$
\end{remark}
\begin{exemple} We take the same example above
 $w=aa^\dag aaa^\dag a$.\\
The double dot operation gives
\begin{eqnarray*}
:aa^\dag aaa^\dag a: &=& \overbrace{a^\dag a^\dag
aaaa}^{\mbox{$a^\dag$ in left, $a$ in right}}
\end{eqnarray*}
\end{exemple}

\section{Combinatorics of the normal ordering}
Although Bell and Stirling numbers have a purely combinatorial
origin, in this communication, we will rather consider them as
coefficients of normal ordering problem.\\
The general word $w\in\{a,a^\dag\}^*$ with letters in
$\{a,a^\dag\}$ (Boson string) can be described by two sequences of
positive integers $r=(r_1,r_2,\cdots ,r_M)$ and $s=(s_1,s_2,\cdots
,s_M)$, so that we define
\begin{eqnarray}\label{H_rs_def}
% \nonumber to remove numbering (before each equation)
   w_{r,s} &=&
   (a^\dag)^{r_M}a^{s_M}\cdots(a^\dag)^{r_2}a^{s_2}(a^\dag)^{r_1}a^{s_1}
\end{eqnarray}
and $\displaystyle{d=\sum_{m=1}^n(r_m-s_m)\quad ,n=1,\cdots M}$
represents the excess (i. e. the number of creations in excess w.
r. t. the number of annihilations). Then the normally ordered form
of $w^n$ is
given by\\
$\mathcal N(w^n)=$$\left\{%
\begin{array}{ll}
     \displaystyle{(a^\dag)^{nd}\sum_{k=0}^{\infty}
S_{r,s}(n,k)(a^\dag)^ka^k,} & \hbox{if $d>0$;} \\
     \displaystyle{\Big(\sum_{k=0}^{\infty}
S_{r,s}(n,k)(a^\dag)^ka^k\Big) (a^\dag)^{n|d|},} & \hbox{else.} \\
\end{array}%
\right.$


 \vspace{0.5cm} where the quantities $S_{r,s}(n,k)$ are generalizations of
standard Stirling
numbers.\\
The generalized Bell polynomials $B_{r,s}(n,x)$ and the
generalized Bell numbers $B_{r,s}(n)$ are respectively
\begin{eqnarray}
% \nonumber to remove numbering (before each equation)
  B_{r,s}(n,x) &=& \sum_{k=0}^\infty S_{r,s}(n,k) x^k \\
  B_{r,s}(n,1) &=& B_{r,s}(n)=\sum_{k=0}^\infty S_{r,s}(n,k)
\end{eqnarray}
\begin{exemple}$ $\\
\begin{description}
    \item  [$\bullet\quad w=a^\dag a$]$ $, we get the usual matrix of
\textit{Stirling numbers of second kind}\\

    $$
\left(%
\begin{array}{cccccccc}
  1 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots  \\
  0 & 1 & 0 & 0 & 0 & 0 & 0 & \cdots \\
  0 & 1 & 1 & 0 & 0 & 0 & 0 & \cdots \\
  0 & 1 & 3 & 1 & 0 & 0 & 0 & \cdots \\
  0 & 1 & 7 & 6 & 1 & 0 & 0 & \cdots \\
  0 & 1 & 15 & 25 & 10 & 1 & 0 & \cdots \\
  0 & 1 & 31 & 90 & 65 & 15 & 1 & \cdots \\
 \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & &\ddots \\
\end{array}%
\right)
$$

    \item[$\bullet\quad w=a^\dag aa^\dag$], we have\\
$$
\left(%
\begin{array}{cccccccc}
  1 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots  \\
  1 & 1 & 0 & 0 & 0 & 0 & 0 & \cdots \\
  2 & 4 & 1 & 0 & 0 & 0 & 0 & \cdots \\
  6 & 18 & 9 & 1 & 0 & 0 & 0 & \cdots \\
  24 & 96 & 72 & 16 & 1 & 0 & 0 & \cdots \\
  120 & 600 & 600 & 200 & 25 & 1 & 0 & \cdots \\
  720 & 4320 & 5400 & 2400 & 450 & 36 & 1 & \cdots \\
 \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & &\ddots \\
\end{array}%
\right)
$$


\vspace{1.0cm}

    \item [$\bullet\quad w=a^\dag aaa^\dag a^\dag$], one gets\\
    $$
\left(%
\begin{array}{cccccccccc}
  1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots  \\
  2 & 4 & 1 & 0 & 0 & 0 & 0 & 0 & 0 &  \cdots \\
  12 & 60 & 54 & 14 & 1 & 0 & 0 & 0 & 0 &  \cdots \\
  144 & 1296 & 2232 & 1296 & 306 & 30 & 1 & 0 & 0 &  \cdots \\
  2880 & 40320 & 109440 & 105120 & 45000 & 9504 & 1016 & 52 & 1 &  \cdots \\
 \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & &\ddots \\
\end{array}%
\right)
$$
\end{description}
\end{exemple}

\vspace{0.5cm}

\begin{remark}
In each case, the matrix $S_w$ is of staircase form and the
dimension of the step is the number of $a$'s in the word $w$. Thus
all the matrices are row-finite and unitriangular iff the number
of its annihilation operators $r=1$. More over, the first column
is $(1,0,\cdots ,0,\cdots)$ iff $w$ ends with $a$ (this means that
$\mathcal N(w^n)$ has no constant term for all $n>0$).\\

A word $w$ with only one annihilation operator being written We
write $w$ in the unique form $w=(a^\dag)^{r-p}a (a^\dag)^p$, one
has:
\begin{itemize}
    \item if $p=0$, $S_w$ is the matrix of a \textit{unipotent
    substitution}
    \item if $p>0$, $S_w$ is the matrix of a \textit{unipotent substitution with prefunction}
\end{itemize}
\end{remark}

\section{Approximate substitutions}
We define here the  transformation matrices (space, topology and
decomposition), then we concentrate on the \textit{Riordan
subgroup} (\textit{i.e} transformations which are substitutions
with prefactor functions).\\

Let $\C^\N$ be the vector space of all complex sequences, endowed
with the Frechet product topology. The algebra $\mathcal L(\C^\N)$
of all continuous operators $\C^{\N}\longrightarrow \C^{\N}$ is
the space of row-finite matrices with complex coefficients. For a
sequence $A=(a_n)_{n\geq 0}$, the transformed sequence $B=MA$ is
given by $B=(b_n)_{n\geq 0}$ with:
\begin{eqnarray}\label{suite-transform�e-bn}
% \nonumber to remove numbering (before each equation)
  b_n &=& \sum_{k\geq 0} M(n,k)a_k
\end{eqnarray}
We may associate a univariate series with a given sequence
$(a_n)_{n\in\N}$ using a sequence of denominators $d_n=n!$ as
follow
\begin{eqnarray}\label{suite_transform_serie_univari�e}
% \nonumber to remove numbering (before each equation)
   \sum_{n\geq 0}a_n\frac{z^n}{n!}
\end{eqnarray}
\subsection{Substitution with prefunction}
We now examine an important class of transformations of $\mathcal
T$ as well as their diagonals: the substitutions with prefunctions

We consider, for a generating function $f$ the transformation
\begin{eqnarray}\label{transformation_prefonction}
% \nonumber to remove numbering (before each equation)
  \mathcal T_{g,\phi}[f](x) &=& g(x)f\Big(\phi(x)\Big)
\end{eqnarray}
where $g$ and $\phi$ are generating functions and $\mathcal T$ is
a linear application.\\
The matrix of this transformation $M_{g,\phi}$ is given by the
transforms of the monomials $\frac{x^k}{k!}$ hence
\begin{eqnarray}\label{transformation_matrice}
% \nonumber to remove numbering (before each equation)
  \sum_{n\geq 0}M_{g,\phi}(n,k)\frac{x^n}{n!} =
  \mathcal T_{g,\phi}\Big[\frac{x^k}{k!}\Big]=g(x)\Big[\frac{x^k}{k!}\Big](\phi(x))
  =g(x)\frac{\phi(x)^k}{k!}
\end{eqnarray}
\begin{remark}
Eq. (\ref{transformation_matrice}) is called the \textit{Sheffer
condition} and, for EGF it amounts to stating that
\begin{eqnarray}
% \nonumber to remove numbering (before each equation)
  \sum_{n,k\geq 0}T(n,k)\frac{x^n}{n!}y^k &=& g(x)e^{y\phi (x)}
\end{eqnarray}
where $T\in\mathcal L(\C^\N)$ is a matrix with non-zero two first
columns $(T=M_{g,\phi})$.
\end{remark}
\subsection{Approximate substitutions}

We give in this section the manner of determining if a unipotente
matrix  is a matrix of substitution.
\begin{definition}
Let $\mathcal M_{[n+1\times n+1]}$ a given unipotente matrix.
$\mathcal M$ is said unipotente matrix of substitution if it
satisfies the following condition:
\begin{eqnarray}
% \nonumber to remove numbering (before each equation)
  c_k &=& c_0\frac{\Big(\frac{c_1}{c_0}\Big)^k}{k!}
\end{eqnarray}
\begin{figure}[htbp]
\includegraphics[width=12cm]{presentation_calck}
\end{figure}
Where $c_k$ represents the exponantial generating serie of the
$k^{th}$ column and $c_0$, $c_1$ are respectively the exponantial
generating series of the $1^{st}$ and the $2^{nd}$ column.
\end{definition}



We consider now the set of matrices with complex c{\oe}fficients
noted by $\C^{\N\times\N}$ and let $\C^{[0\cdots n]\times [0\cdots
n]}$ all matrices of size $(n+1)$ obtained by the truncation of
$\C^{\N\times\N}$ with $r_n$. Then
$$r_n:\C^{\N\times\N}\longrightarrow \C^{[0\cdots n]\times [0\cdots
n]}$$ $r_n$ is not a morphism\\We consider now $\mathcal T(\N,\C)$
the algebra of lower triangular matrices and $T([0\cdots n],\C)$
the matrices of size $(n+1)$ obtained by the truncation $\tau_n$.
Then
$$\tau_n:\mathcal T(\N ,\C)\longrightarrow T([0\cdots n],\C)$$
$\tau_n$ is a morphism.
$$
\xymatrix@C=50pt{
\C^{\N\times\N}\ar[r]^{r_n}& \C^{[0\cdots n]\times [0\cdots n]} \\
\mathcal T(\N ,\C)\ar[r]^{\tau_n}\ar@{^{(}->}[u]^{\mathcal J_1}
&T([0\cdots n],\C)\ar@{^{(}->}[u]^{\mathcal J_2} }
$$
Where $\mathcal J_1$ and $\mathcal J_2$ are two canonical
injections.
\begin{remark}
We can write
\begin{eqnarray}
% \nonumber to remove numbering (before each equation)
  \mathcal T(\N,\C) &\displaystyle{=\lim_{\longleftarrow}}& (T(\N,\C))
\end{eqnarray}
wich means that $(T(\N,\C))$ is the projective limit of $ \mathcal
T(\N,\C)$
\end{remark}
\subsection{Random generation}
we are interested in this section in the probabilistic study of
the matrices of substitutions. To be done, we randomly generate
unipotente matrices  and we observe the number of apparition of
matrices of substitutions.\\We start by giving some examples of
our experiment which are summarized in the table below:

\begin{figure}[htbp]
\includegraphics[width=14cm]{presentation_tableau}
\end{figure}

\begin{remark}
According to the results obtained we can say that the matrices of
substitutions are not very frequent. However in meeting certain
conditions such as size , the numbers of drawing and the range of
variables, we can obtain positif probabilities  so that these
matrices appear.\\ Let us note that more the size of the matrix is
small more one is likely to obtain a matrix of substitution with
one a reasonable numbers of drawing.\\
 We also notice that if
we vary the range of variables, and this in increasing way  and by
keeping unchanged the number of drawing and size, the probability
tends towards 0. We also notice that all the unipotente matrices
of size 3 are matrices of substitution. This is easy to see
because the exponential generating series of the 3$^{rd}$ column
 will always have the form $c_k=\frac{x^2}{2!}$.\\ Thus we can say
that the test really starts   from the matrices of  size higher or
equal to 4.
\end{remark}
\begin{resultat}
Let $r$ represent the range of variables and $n$ the size of
matrix.\\According to the results obtained, we can say that the
probability $p$ of apparition of the matrices of substitutions
depends on $r$ and $n$ and is raised by:
\begin{eqnarray}
% \nonumber to remove numbering (before each equation)
  p_{n} &\leq& \frac{r^{2n-3}}{r^{\frac{n(n-1)}{2}}}\sim \frac{r^{2n}}{r^{\frac{n^2}{2}}}
\end{eqnarray}
And we have

\begin{center}
\begin{tabular}{c}

  % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
  $p_n\longrightarrow 0$ \\
  $n\longrightarrow \infty$ \\
\end{tabular}\end{center}

\end{resultat}

\begin{conjecture}
\begin{eqnarray}
% \nonumber to remove numbering (before each equation)
  p_{n} &\sim& \frac{r^{2n-3}}{r^{\frac{n(n-1)}{2}}}
\end{eqnarray}
\end{conjecture}

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\end{document}
