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\date{}

\begin{document}
\title{Combinatorial analysis of magnetic configurations}
\author{W. Florek and T. Lulek\\
 	 Division of Mathematical Physics\\
	 The Institute of Physics\\
	 A.Mickiewicz University\\
	 Matejki 48/49\\
	 60-769 Pozna\'n, Poland}

\maketitle
\begin{abstract}
Properties of polynomial coefficients are applied to investigation of finite
spin systems. Number of spin-configurations states with a given total
magnetization are calculated by recurrence formulas for polynomial
coefficients and for sum of polynomial coefficients.
\end{abstract}


\section[]{Introduction}
There is a big interest in application of the so-called `finite lattice
method' due to easy computer calculations or simulations. A dimension of
state space is finite and, theoretically, one can obtain exact results. It is
very important to analyze a state space before starting computer program in
order to simplify calculations by, for example, decomposition of the space
into subspaces with given properties.  A spin system with its total
magnetization as `a given property' is a very nice example of such procedure. 

The finite spin system can be described as follows \cite{ker} (cf.\ also 
\cite{bl,lf} in this volume and references quoted therein):
\begin{enumerate}
  \item There is a given set of nodes of crystal lattice
    \be
      X:=\un:=\{1,2,\ldots,n\}.
    \ee
  \item Each node carries spin \vs\ determined by a spin number $s\geq0$
    (integer or half-integer).
  \item Possible spin projections on a quantization axis form an
    ($2s+1$)-element set
    \be
      Y:=[-s,+s]:=\{-s,-s+1,\ldots,+s\}.
    \ee
  \item The {\it spin-configurations}\/ of the considered spin system are the
    mappings $f:X\rightarrow Y$, hence the set of spin-configurations is
    determined as
    \be
      Y^X:={\rm Map}(\un,[-s,+s]):=\{f:X\rightarrow Y\}
    \ee
    and consists of $|Y^X|=(2s+1)^n$ elements.
\end{enumerate}
One can used Dirac's notation and writes a spin-configuration as the so-called
ket 
\be
\ket{i_1,i_2,\ldots,i_n},\;\;\;i_j\in Y,\;\;j\in X,
\ee
where $i_j=f(j)$. The set $Y^X$ is an orthonormal basis of a space $L$ of all
quantum states of magnet over the field of complex numbers \mathc\
\cite{bl,lf}. For $s=1/2$ we will use a set $Y=\{-,+\}$ instead of
$\{-1/2,+1/2\}$ for the sake of simplicity. In this case the set of all
configurations coincides with the power set $2^X$ of $X$.

The most important operator acting in the space $L$ is the energy operator,
i.e.\ the Hamiltonian. It is well known, that two commuting operators have
the same set of eigenfunctions, so the states can be labelled simultaneously
by two (or more) eigenvalues. These eigenvalues are called `good quantum
numbers'. For almost all magnetic systems (Hamiltonians) a total
magnetization is a good quantum number. It is the eigenvalue of an operator 
\be
  S^z:=\sum_{j\in X}\;s_j^z,
\ee
where $s_j^z$ is a one-site operator and $s_j^z\ket{i_j}=i_j\ket{i_j}$ so
\be
  S^z\ket{i_1,i_2,\ldots,i_n}=
    \left(\sum_{j\in X}\;i_j\right)\ket{i_1,i_2,\ldots,i_n}=
    M\ket{i_1,i_2,\ldots,i_n}.
\label{mag}    
\ee
It is obvious that all possible values of the total magnetization form a set
\be
  Z:=[-ns,+ns]=\{-ns,-ns+1,\ldots,+ns\},\;\;|Z|=2ns+1.
\label{totmag}  
\ee

In this paper we present a method for calculating number of configurations
with a given magnetization $M$. The Hamiltonian of the considered system is
irrelevant and it will be omitted in discussion. It is easy to notice that
the operator $S^z$ commutes with any permutation $\sigma\in\perg$, so this
group is a symmetry group of our problem. 

\section[]{Polynomial coefficients}
\label{polcoe}
Let $\omn\subset\mathz^{\um}$ be a set of all $m$-tuples
$(k_1,k_2,\ldots,k_m)$ such that
\be
\mathz^{\um}\supset\omn:=
  \{\uK=(k_1,k_2,\ldots,k_m)\mid k_i\geq0,\;\sum_{i\in\um}k_i=n\}.
\ee
The $n$-th power of polynomial ${\cal P}_{(m)}=\sum_{i\in\um}x_i$,
$x_i\in\mathc$, can be written as 
\be
{\cal P}^n_{(m)} 
  =\sum_{\uK\in\omn}\;\polcnk\prod_{i\in\um}\;x_i^{k_i},
\ee
where $\polcnk$ is a polynomial coefficient
\be
\polcnk:=\frac{n!}{\prod_{i\in\um}\:k_i!}.
\ee
For $m=2$ one obtains the Newton binomial coefficient
\be
\polc{n}{(k,n-k)}=\bink= \frac{n!}{k!(n-k)!}
\ee
and the well-known formula
\be
(x_1+x_2)^n=\sum_{k\in[0,n]}\;\bink\,x_1^kx_2^{(n-k)}.
\ee                  

The properties of these coefficients are gathered in Table \ref{polprop} and
compared with the properties of the binomial coefficients. A mnemonic scheme
corresponds to the first recurrence procedure. In the case $m=2$ the
well-known Pascal triangle (drawn in a two-dimensional space) is obtained,
whereas in the general case a similar scheme should be drawn in a
$m$-dimensional space and we can call it a {\it ($m$-dimensional) Pascal
`simplex'}. Figure \ref{psim} presents an example for $m=3$ (`Pascal
tetrahedron'). This general scheme has the same properties as the ordinary
Pascal triangle (cf.\ Table \ref{polprop}):
\begin{enumerate}
  \item Subsequent `bases' of simplex, labelled by $n$, consists of
    $\frac{(n+m-1)!}{n!(m-1)!}$ entries.
  \item The sum of entries in the $n$-th row is equal to $m^n$.
  \item The first recurrence formula (with respect to $n$) is satisfied.
\end{enumerate}

\begin{table}[thb]
{\footnotesize
\caption{Properties of polynomial coefficients}
\begin{center}
\begin{tabular}{|c|cc|}\cline{2-3}
\multicolumn{1}{c|}{} & general case & $m=2$\\ \hline
\parbox{32.5ex}{Number of different coefficients } 
    & $\bis{n+m-1}{m-1}$ 
    & $n+1$\\ \hline
\parbox{32.5ex}{\center Sum rules} 
    & $\begin{array}{c}\\
        {\displaystyle \sum_{\uK\in\omn}}\;\polcnk=m^n\\ \\
        {\displaystyle \sum_{\uK\in\omn}}\exp \left[\frac{2\pi i}{m}
          {\displaystyle \sum_{i\in\um}}ik_i\right]\polcnk\!=\!0\\ 
       \end{array}$ 
    & $\begin{array}{c}\\
        {\displaystyle \sum_{k\in[0,n]}}\bisnk=2^n\\ \\
        {\displaystyle \sum_{k\in[0,n]}}(-1)^k\bisnk=0\\
       \end{array}$ \\ \hline
\parbox{32.5ex}{~\\Recurrences:\\ 1.\ with respect to power $n$}
    & $\begin{array}{c}
       \polcnk={\displaystyle \sum_{i\in\um}}\polc{n\!-\!1}{\uK^{(j)}},\\ 
          \uK^{(j)}=(k_1,\ldots,k_j\!-\!1,\ldots,k_m)\\
          \uK^{(j)}\subset\Omega(m,n\!-\!1)
       \end{array} $
    & $ \bisnk=\bis{n-1}{k-1}\!+\!\bis{n-1}{k} $\\ 
    &&\\
\parbox{32.5ex}{~\\ 2.\ with respect to number of variables $m^1$} 
    & $\begin{array}{rcl}
       \polcnk&=&\polc{n}{(k_1,n\!-\!k_1)}\\
              &\times&\polc{n\!-\!k_1}{\uK'_1},\\
         \uK'_1&=&(k_2,k_3,\ldots,k_m)\\
         \uK'_1&\subset&\Omega(m\!-\!1,n\!-\!k_1)
       \end{array} $
    & $ \bisnk=\bisnk\frac{\textstyle (n\!-\!k)!}{\textstyle (n\!-\!k)!}$\\ 
\hline
\end{tabular}
\end{center}
\label{polprop}
}
\end{table}

It is easy to notice that each $m'$-dimensional, $m'\leq m$, `wall' of the
simplex is also a Pascal simplex corresponding to $m'$-nomial coefficients
(see Figure\ \ref{psim}). It is in the accordance with the second recurrence
formula\footnote{This formula is trivial for $m=2$ and for $m\geq3$ any
element $k_i$, $i\in\um$, can be considered.}.


Let ${\bf A}$ be a linear function ${\bf A}:\omn\rightarrow\mathr$ given as 
\be
{\bf A}(\uK)=\sum_{i\in\um}\;\alpha_ik_i
\label{afun}
\ee
and determined by real coefficients $\alpha_i\in\mathr$ (so this function can
be identified with an $m$-tuple $(\alpha_1,\alpha_1,\ldots,\alpha_m)$). For
each
$a\in {\bf A}(\omn)$ one can find its counter-image
\be
{\bf A}^{-1}(a)=\{\uK\in\omn\mid {\bf A}(\uK)=a\}.
\ee
For given $m$, $n$, and ${\bf A}$ we introduce the function
$\clanm:\mathr\rightarrow\mathz_+$ 
\be
\clanm(a):=\sum_{\uK\in A^{-1}(a)}\polcnk.
\label{acfun}
\ee
    
\begin{figure}
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\caption[]{An example of the Pascal simplex, $m=3$, $n\leq4$. Dashed lines
connect points with the same $\sum_{i\in\um}ik_i$ (cf.\ Sec.\ \ref{class}
Eq.\ (\ref{sumM})) }
\label{psim}
}
\end{figure}   

The recurrence formulas for polynomial coefficients yield similar ones for the
function \clanm. That is,
\be
\clanm(a)=\sum_{i\in\um}\cla{n-1}{m}(a-\alpha_i)
\label{frec}
\ee
with the initial condition
\be
\cla{1}{m}(a)=\sum_{i\in\um}\delta(\alpha_i,a),
\ee
and the second one in the form
\be
\clanm(a)=\sum_{k\in\Delta}\bink\cla{n-k}{m-1}(a-\alpha_1k),
\label{srec}
\ee
where
\be
\Delta:=\{k\in[0,n]\mid (n-k)\alpha\leq(a-\alpha_1k)\leq(n-k)\beta\}
\label{delta}
\ee
with $\alpha$ and $\beta$ being $\min(\alpha_2,\alpha_3,\ldots,\alpha_m)$ and
$\max(\alpha_2,\alpha_3,\ldots,\alpha_m)$, respectively. Of course, the other
coefficient $\alpha_i$ can be taken into account (cf.\ Table \ref{polprop}).
The initial formula for $m=2$ has the following form
\be
\cla{n}{2}(a)=\sum_{(k,n-k)\in {\bf A}^{-1}(a)}\bink = \left\{\begin{array}{ll}
       2^n & \mbox{if}\;\;\alpha_1=\alpha_2,\\
       \bisnk\;\; & \mbox{for}\;\; \alpha_1\neq\alpha_2,
         \:k=\frac{a-\alpha_2n}{\alpha_1-\alpha_2}.
      \end{array}\right.
\ee

Let's consider, e.g., a function (\ref{afun}) determined by the $m$-tuple
${\bf N}:=(1,1,\ldots,1)$, i.e.\ $\alpha_i=1$, $i\in\um$. Then
\be
{\bf N}(\uK)=\sum_{i\in\um}1\cdot k_i=n,\;\;\mbox{\rm for each}\; \uK\in\omn,
\label{nfun}
\ee
so ${\bf N}(\omn)=\{n\}$.
Therefore, for the function (\ref{acfun}) one obtains
\be
{\cal N}^{n,m}(n)=\sum_{\uK\in\omn}\polcnk=m^n.
\ee
The first recurrence formula (\ref{frec}) yields
\be
{\cal N}^{n,m}(n)=\sum_{i\in\um}{\cal N}^{n-1,m}(n-1)
         =\sum_{i\in\um}m^{n-1}=m\cdot m^{n-1},
\ee
so it is, of course, the well-known recurrence formula for the power $m^n$.
From the second recurrence formula (\ref{srec}) one obtains (the set $\Delta$
contains all integers $0,1,\ldots,n$, see Eq.\ (\ref{delta}))
\begin{eqnarray}
{\cal N}^{n,m}(n)&=&\sum_{k\in[0,n]}\bink{\cal N}^{n-k,m-1}(n-k)
         =\sum_{k\in[0,n]}\bink(m-1)^{n-k} \nonumber \\
         &=&\sum_{k\in[0,n]}\bink(m-1)^{n-k}\cdot1^k
         =(1+(m-1))^n.
\end{eqnarray}
So, in this case, the recurrence formula is less useful, but evidently true.

\section[]{Orbits of symmetric group}
\label{orbits}
The symmetric group $\perg$ acts on the set $Y^X$ of all spin-configurations
in the following way \cite{bl,lf}
\be
P:\perg\times Y^X \rightarrow Y^X,\;\;P(\sigma)=\left(
      \begin{array}{c} 
      f\\ f\circ\sigma^{-1} 
      \end{array}
      \right), \;\;f\in Y^X,\;\; \sigma\in\perg.
\ee
This action determines orbits
\be
O(f)=\{f\circ\sigma^{-1}\mid \sigma\in\perg\}.
\ee
For each $\uK\in\omn$ there exists an orbit with the representative
\be
f^0:=\ket{i^0_1,i^0_2,\ldots,i^0_n}:=
|\underbrace{-s,\ldots,-s}_{k_1 \:\mbox{\scriptsize times}},
     \underbrace{-s+1,\ldots,-s+1}_{k_2\: \mbox{\scriptsize times}},\ldots,
     \underbrace{+s,\ldots,+s}_{k_m \:\mbox{\scriptsize times}}\:\rangle,
\label{orb0}     
\ee
where $m=|Y|=2s+1$. It is evident that orbits can be labelled by the $m$-tuples
$\uK\in\omn$, so they will be denoted hereafter as $O_{\uK}$. These orbits have
the following properties:
\begin{enumerate}
\item Cardinality of an orbit: $|O_{\uK}|=\polcnk$.
\item Stabilizer: $G_{\uK}\cong\perg/\bigotimes_{i\in\um}S_{\ul{k}_i}$
  (quotient group of the symmetric group $\perg$ and the Young subgroup
  $\bigotimes_{i\in\um}S_{\ul{k}_i}$). 
\item Number of different orbits: $\polc{n+2s}{(n,2s)}$ ($n+1$ for
  $s=1/2$, i.e.\ for $m=2$).
\item Number of all configurations:
  \be
    \sum_{\uK\in\omn}\polcnk=(2s+1)^n=\sum_{\uK\in\omn}|O_{\uK}|.
  \ee  
\end{enumerate}

\section[]{Classification of configurations}
\label{class}
Let's consider a function (\ref{afun}) determined by the $m$-tuple
${\bf M}:=(-s,-s+1,\ldots,+s)$, i.e.\ $\alpha_i=i-s-1$, $i\in\um$. Then
\begin{eqnarray}
{\bf M}(\uK)&=& \sum_{i\in\um}(i-s-1)k_i=-n(s+1)+\sum_{i\in\um}ik_i=
         \sum_{\mu\in Y}\:\mu k_{\mu+s+1}\nonumber\\
       &=&k_1(-s) + k_2(-s+1) + \ldots + k_{2s+1}s = M_{\uK},
\label{sumM}       
\end{eqnarray}
where $M_{\uK}$ is a magnetization of any configuration in the orbit
$O_{\uK}$ (see Eqs.\ (\ref{mag}) and (\ref{orb0})). Therefore, the image of
the set $\omn$ is ${\bf M}(\omn)=Z$ given by Eq.\ (\ref{totmag}) and each value 
of
the function (cf.\ Sec.\ \ref{orbits})
\be
\clmn(M) = \sum_{\uK\in {\bf M}^{-1}(M)}\polcnk
                 = \sum_{\uK\in {\bf M}^{-1}(M)}|O_{\uK}|
\ee                 
determines a number of spin-configurations with a given total magnetization
$M\in Z$ (for a given number $n$ of spins $s$). It is important to underline
that for all $n$ and $s$
\be
\clmn(M)=\clmn(-M),\;\;M\in Z.
\ee

The recurrence formulas described by Eqs.\ (\ref{frec},\ref{srec}) yield the 
procedures for the calculation of $\clmn(M)$. From the first formula
one can easy find 
\be
\clmn(M)=\sum_{i\in\um}\clm{n-1}{2s+1}(M-i+s+1)
   =\sum_{\mu\in Y}\clm{n-1}{2s+1}(M-\mu)
\ee
with the starting condition
\be
\clm{1}{2s+1}(M)=\left\{\begin{array}{ll}
                          1\;\;&\mbox{\rm for}\;\;M\in[-s,+s]=Y,\\
                          0    &    \mbox{\rm in other cases}.
                       \end{array} \right.
\ee
One can introduce in a formal way the function $\clm{0}{2s+1}$ as
\be
\clm{0}{2s+1}(M):=\delta_{M,0}.
\label{def0}
\ee
It is easy to notice that with this definition (\ref{def0}) a mnemonic scheme,
like the Pascal triangle, can be constructed for this recurrence procedure.
Examples of such scheme are
presented in Figure \ref{expt12} and \ref{expt3} for $s=1/2, 1$ and $s=3/2$,
respectively. This `extended' Pascal triangle has the following properties
(of course, for $s=1/2$, $m=2$ they coincide with the properties of the
Pascal triangle):
\begin{enumerate}
\item There are $|Z|=|[-ns,+ns]|=2ns+1$ nonzero entries in the $n$-th row.
\item Each number in the $n$-th row is a sum of $|Y|=|[-s,+s]|=2s+1$ numbers
  from the $(n-1)$-th row. These numbers are the `nearest-neighbor' of
  calculated entry (see Figures \ref{expt12} and \ref{expt3}).
\end{enumerate} 


From the second recurrence formula the following rule is obtained
\be
\clmn(M)=\sum_{k\in\Delta}\bink\clm{n-k}{2s}(M+sk),
\ee
where
\be
\Delta=\{k\in[0,n]\mid n(1-s)-M\leq k \leq (ns-M)/2s\}
\ee
with the initial condition
\be
\clm{n}{2}(M)=\bi{n}{n/2-M}=\polc{n}{(\frac{n}{2}-M,\frac{n}{2}+M)}.
\ee
This procedure can be called {\it the recurrence with respect to spin} (since
one can introduce a new spin $s'=s-1/2$, so $2s'+1=2s=m-1$),
whereas the previous is {\it the recurrence with respect to number of spins}.


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\caption[]{`Extended' Pascal triangles for a) $s=1/2$ and b) $s=1$, $n\leq4$.
Arrows correspond to the first recurrence procedure \label{expt12}}
\end{figure}

\begin{figure}[th]
{\footnotesize
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\caption[]{`Extended' Pascal triangles for $s=3/2$, $n\leq4$. Arrows
correspond to the first recurrence procedure \label{expt3}}
}
\end{figure}

\section[]{Examples}

\begin{table}[bht]
{\small 
\caption{Results for $n=3$ and $s=3/2$ ($m=4$) for non-negative $M$}
\label{n3}
\begin{center}
\begin{tabular}{||c|cccc|c||}\hline\hline
Orbit number & $\uK$ & Representative & $M_{\uK}$ & $|O_{\uK}|$ &
$\cla{3}{4}(M)$ \\ \hline \hline
 1 & (1,0,1,1) & $\ket{-3/2,+1/2,+3/2}$ & 1/2 &  6 &    \\ 
 2 & (0,2,0,1) & $\ket{-1/2,-1/2,+3/2}$ & 1/2 &  3 &    \\ 
 3 & (0,1,2,0) & $\ket{-1/2,+1/2,+1/2}$ & 1/2 &  3 & 12 \\ \hline
 4 & (1,0,0,2) & $\ket{-3/2,+3/2,+3/2}$ & 3/2 &  3 &    \\ 
 5 & (0,1,1,1) & $\ket{-1/2,+1/2,+3/2}$ & 3/2 &  6 &    \\ 
 6 & (0,0,3,0) & $\ket{+1/2,+1/2,+1/2}$ & 3/2 &  1 & 10 \\ \hline
 7 & (0,1,0,2) & $\ket{-1/2,+3/2,+3/2}$ & 5/2 &  3 &    \\ 
 8 & (0,0,2,1) & $\ket{+1/2,+1/2,+3/2}$ & 5/2 &  3 &  6 \\ \hline
 9 & (0,0,1,2) & $\ket{+1/2,+3/2,+3/2}$ & 7/2 &  3 &  3 \\ \hline
10 & (0,0,0,3) & $\ket{+3/2,+3/2,+3/2}$ & 9/2 &  1 &  1 \\ \hline\hline
\end{tabular}
\end{center}
}
\end{table}

\begin{table}[htb]
{\small
\caption{Results for $n=4$ and $s=1$ ($m=3$) for non-negative $M$}
\label{n4}
\begin{center}
\begin{tabular}{||c|cccc|c||}\hline\hline
Orbit number & $\uK$ & Representative & $M_{\uK}$ & $|O_{\uK}|$ &
$\cla{4}{3}(M)$ \\ \hline \hline
 1 & (2,0,2) & $\ket{-1,-1,+1,+1}$ & 0 &  6 &    \\ 
 2 & (1,2,1) & $\ket{-1, 0, 0,+1}$ & 0 & 12 &    \\ 
 3 & (0,4,0) & $\ket{ 0, 0, 0, 0}$ & 0 &  1 & 19 \\ \hline
 4 & (1,1,2) & $\ket{-1, 0,+1,+1}$ & 1 & 12 &    \\ 
 5 & (0,3,1) & $\ket{ 0, 0, 0,+1}$ & 1 &  4 & 16 \\ \hline
 6 & (1,0,3) & $\ket{-1,+1,+1,+1}$ & 2 &  4 &    \\ 
 7 & (0,2,2) & $\ket{ 0, 0,+1,+1}$ & 2 &  6 & 10 \\ \hline
 8 & (0,1,3) & $\ket{ 0,+1,+1,+1}$ & 3 &  4 &  4 \\ \hline
 9 & (0,0,4) & $\ket{+1,+1,+1,+1}$ & 4 &  1 &  1 \\ \hline \hline
\end{tabular}
\end{center}
}
\end{table}

{\bf A. $n=2$}\\
One can easy find that
$$
\clm{2}{2s+1}=\sum_{\mu\in Y}\clm{1}{2s+1}(M-m)
  =\sum_{\mu\in Y}\;\sum_{\mu'\in Y}\delta(\mu',M-\mu)=2s+1-|M|.
$$
\vspace{1.5em}

\noindent{\bf B. $s=1/2$}\\
For each $M\in[-n/2,+n/2]$ there exists only one orbit with the representative
$$|\underbrace{-,-,\ldots,-}_{n/2-M\; \mbox{\scriptsize times}},
     \underbrace{+,+,\ldots,+}_{n/2+M\; \mbox{\scriptsize times}}\:\rangle,$$
so number of configurations with a given magnetization equals the cardinality
of the orbit 
$$O_{(n/2-M,n/2+M)}=\bi{n}{n/2-M}.$$     
\vspace{1.5em}

\noindent{\bf C. $M=ns$}  \\
There exists only one solution $\uK^0=(0,0,\ldots,0,n)$ of the equation 
${\bf M}(\uK)=ns$. Therefore
$$
\clmn(ns)=\polc{n}{\uK^0}=1.
$$
Hence, there is only one configuration --- $\ket{s,s,\ldots,s}$ --- with the
maximal total magnetization $M(\uK^0)=ns$.
\vspace{1.5em}

\noindent{\bf D. $M=ns-1$}  \\
In this case there is also one solution of the form
$\uK^1=(0,0,\ldots,0,1,n-1)$. Then the representative of the orbit
$O_{\uK^1}$ is $\ket{s-1,s,\ldots,s}$ and the cardinality of the orbit is
$$
|O_{\uK^1}|=\polc{n}{(0,0,\ldots,1,n-1)}=\bi{n}{1}=n.
$$
Therefore the dimension of a subspace $L_{ns-1}\subset L$, consisting of
states with $S^z\ket{\psi}=(ns-1)\ket{\psi}$ is $n$. Moreover,
this orbit does not decompose into subsets with the restriction $\perg|T$,
where $T$ is the translation symmetry group of a crystal lattice. This orbit
is the regular one of the group $T$. It is very important, since the
Hamiltonian commutes with any translation, so the subspace $L_{ns-1}$ is
always an eigenspace of the Hamiltonian.
\vspace{1.5em}

\noindent{\bf E. $n=3$, $s=3/2$ and $n=4$, $s=1$}  \\
The results obtained for these parameters are presented in Table~\ref{n3} and
\ref{n4}, respectively. Please compare them with Figures~\ref{psim},
\ref{expt12}, and \ref{expt3}.

\section[]{Final remarks}

The properties of the polynomial coefficients and
their application to the finite spin system investigation,
presented in this paper,  do not exhaust riches of this
structure. It should be pointed at first, that the spin system
is, very fine but, only one example of possible physics
applications. This approach can be also useful for any problem,
in which one has $n$ elements and $m$ equivalence classes. E.g.,
$\polcnk$ determines a number of different states for $n$ bosons
and $m$ admissible energy levels \cite{parry}. The second
example is the Ising model and equivalent systems: lattice gas
and two-component alloy \cite{hu}.

On the other hand, the analysis presented in Secs.\ \ref{polcoe} and
\ref{class} does not use the generating-function method. It seems that this
method would enable to perform more clear presentation and more efficient
application. Moreover, the polynomial coefficients are closely connected with
partitions (and then with Young diagrams) --- each $m$-tuply $\uK$
corresponds to a partition of $n$ into no more than $m$ non-negative integers
$k_i$. In the similar way, the condition (\ref{sumM}) determines a partition
of $M_{\uK}+n(s+1)$ into $n$ non-negative terms no longer than 
$m=2s+1$. (The sum $\sum_{i\in\um}ik_i$ determines a type of 
element $\pi\in S_{n(s+1)+M_{\uK}}$ \cite{kerlec}.)

The recurrence procedures for number of configuration are called the
recurrence with respect to number of nodes $n$ and to spin $s$,
respectively. In the first one, we `cut off' the $n$-th node, whereas in the
next one nodes carrying the minimal spin projections ($-s$ in the first step)
are removed, so the number of nodes decreases, too. A.~Kerber has proposed
lately a procedure in which the smallest (or the largest) projection $-s$ (or
$+s$) are substituted by the next one, i.e.\ by $-s+1$ or $s-1$, respectively
\cite{kerp}. Therefore, in this method only $m$ decreases while $n$ is
constant. 

The results obtained for spin-systems and presented here are the very first
step in solving an eigenproblem for a given Hamiltonian. However, this step is
important, since one finds dimensions of subspaces $L_M\subset L$, $M\in Z$,
which are eigenspaces of almost all magnetic Hamiltonians. For example, the
ground-state of the Heisenberg antiferromagnet should have the total
magnetization equal to 0.\@ From Table \ref{n4} one obtains that for $n=4$ and
$s=1$  the dimension of the subspace $L_0$ is 19, while whole space $L$ has
the dimension 81.\@ The next step, which is necessary, is determination of
bases of the subspaces $L_M$, i.e.\ one has to find a representative of each
orbit $O_{\uK}$. It appears that the presented recurrence procedures enable
us to determine these representatives, too \cite{wsf}. One of us (WF) has
used them investigating the finite Heisenberg magnets and the problems can be
solved very quick and in a very efficient way (see e.g. \cite{cube,conf}). 

At the end, we would like to show some extensions of the polynomial
coefficients. They can be written as a function
$\gamma:\mathz^r\rightarrow \mathz$ in the form
\be
\gamma(k_1,k_2,\ldots,k_r)
=\frac{(\sum_{i\in\ul{r}}k_i)!}{\prod_{i\in\ul{r}}k_i!}. 
\ee
This function can be extended to the space $\mathr^r$ introducing the Euler
function $\Gamma$ and for $\bar{x}=(x_1,x_2,\ldots,x_r)\in\mathr^r$ one
obtains 
\be
\gamma(\bar{x})
  =\frac{\Gamma(1+\sum_{i\in\ul{r}}x_i)}{\prod_{i\in\ul{r}}\Gamma(1+x_i)}. 
\ee
This function satisfies the recurrence formulas (see Table \ref{polprop})  
\begin{eqnarray}
\gamma(\bar{x})&=&\sum_{i\in\ul{r}}\gamma(x_1,\ldots,x_i-1,\ldots,x_r),\\
\gamma(\bar{x})&
    =&\gamma(x_1,\sum_{i=2}^rx_i)\gamma(x_2,x_3,\ldots,x_r).
\end{eqnarray}
For $r=2$ the sum-rule is also satisfied. For $x+y=n$ we have 
\be
\int\!\!\int\gamma(x,y)\,dx\,dy=\int_{-\infty}^\infty \gamma(x,n-x)\,dx = 2^n,
\ee
where we applied the formula (see, e.g., \cite{prud})
\be
\int_{-\infty}^{\infty}(\Gamma(c+x)\Gamma(d-x))^{-1}dx=
  \frac{2^{c+d+2}}{\Gamma(c+d-1)}.
\ee  

One also can assume that the Pascal simplex is drawn in $m$-dimensional
Euclidean space $\mathe^m$. It is easy to notice that counter-images of the
function (\ref{nfun}) lie in a $(m-1)$-dimensional hypersurface perpendicular
to the vector $[1,1,\ldots,1]$, so calculation of the function $\cal N$ is
simply connected with a projection of the space $\mathe^m$ onto the line
determined by this vector. The same is true for the second function defined
by the $m$-tuple $(-s,-s+1,\ldots,+s)$. Therefore number of
spin-configuration with a given total magnetization $M$, for a given number
of spins $n$, corresponds to a projection onto two-dimensional space spanned
by the orthogonal vectors $\bf N$ and $\bf M$. The `extended' Pascal
triangle is simply the projection of the Pascal simplex (for points with
non-negative integer coordinates, dashed line in Figure \ref{psim} correspond
to this projection). This also yields that points $\uK,\uK'\in\mathe^m$ has
the same value of $n={\bf N}(\uK)={\bf N}(\uK')$ and $M={\bf M}(\uK)={\bf
M}(\uK')$ iff the vector $\uK-\uK'\in\mathr^m$ is orthogonal to both vectors
$\bf N$ and $\bf M$. For example, when $n=4$ and $s=3/2$ ($m=4$) we have four
orbits with $M=0$: (2,0,0,2), (1,1,1,1), (1,0,3,0), (0,3,0,1). This points
(in $\mathe^4$) determines six vectors in $\mathr^4$: [1,-1,-1,1],
[1,0,\hbox{-3},2], [2,-3,0,1], [0,1,-2,1], [1,-2,1,0], [1,-3,3,-1]. These vectors
are orthogonal to vectors [1,1,1,1] and [-3/2,-1/2,1/2,3/2] and th first and
the sixth can be treated as the orthogonal basis in two-dimensional space
complementary to the subspace containing extended Pascal triangle. 

\begin{thebibliography}{99}
\bibitem{ker} A. Kerber, B. Lulek, T. Lulek, {\it Group actions,
  configurations and finite states}, in: W. Florek, T. Lulek, and M. Mucha
  (Eds.), {\it Symmetry and Structural Properties of Condensed Matter}, World
  Sci., Singapore 1991, p. 3.
  
\bibitem{bl} B. Lulek, {\it Impurities in the Heisenberg magnet and the
  general recipe of Weyl}, {\it 26. S\'eminaire Lotharingien de Combinatoire,
Thurnau (Germany), March 3--6, 1991}.
  
\bibitem{lf} T. Lulek and W. Florek, {\it Gauge symmetries in the Heisenberg
  model of a magnet}, {\it 26. S\'eminaire Lotharingien de Combinatoire,
Thurnau (Germany), March 3--6, 1991}.  

\bibitem{parry} W.E. Parry, {\it The Many--Body Problem}, Clarendon Press,
Oxford, 1973.

\bibitem{hu} K. Huang, {\it Statistical Mechanic}, J. Wiley and Sons, Inc.,
New York 1963, \S~16.2.

\bibitem{kerlec} A. Kerber, {\it Representations of Permutation Groups I},
  Lect. Notes in Math., vol. 240, Springer-Verlag, Berlin 1971, Sec.\ 1.

\bibitem{kerp} A. Kerber, {\it unpublished}.

\bibitem{wsf} W. Florek, {\it Eigenvectors of the operator of the projection
of the total spin for a finite Heisenberg magnet}, Acta Magnetica {\bf II}
(1985) 43.

\bibitem{cube} W. Florek, {\it s=1/2 antiferromagnetic Heisenberg cube},
accepted for publication in Acta Magnetica.

\bibitem{conf} W. Florek, {\it Ground-state properties of s=1/2 finite
Heisenberg antiferromagnets}, in: {\it Proceedings of the $VI^{th}$ National
Conference on Physics of Magnetism, Pozna\'n (Poland), June 19-21, 1990},
Acta Magnetica {\it Supplement'90} (1990) 114.

\bibitem{prud} A.P. Prudnikov, Ju.A. Brytchkov, and O.I. Maritchev, {\it
  Integrals and Series. Special Functions}, Nauka, Moscow 1983, p. 57 (in
  Russian). 
\end{thebibliography}


\end{document}
