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% Coupling coefficients for Lie algebra representations and 
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% J. Van der Jeugt
% J. Math. Phys. 38 (1997), 2727-2740.
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%TITLE
{\Large \bf 
Coupling coefficients for Lie algebra representations and 
addition formulas for special functions
}

\vspace{2cm}
%AUTHORS
J.\ Van der Jeugt\footnote{Senior Research Associate of N.F.W.O.
(National Fund for Scientific Research of Belgium); 
E-mail address: Joris.VanderJeugt@rug.ac.be}

\vspace{0.5cm}
%ADDRESS
Department of Applied Mathematics and Computer Science \\ 
University of Ghent, Krijgslaan 281-S9, B-9000 Gent, Belgium 
\end{center}

\vspace{2cm}

\noindent
{\bf Abstract}:  
Representations of the Lie algebra
$su(1,1)$ and of a generalization of the oscillator algebra, $b(1)$, are
considered. The paper then
introduces polynomials which are related by the coupling
(or Clebsch-Gordan) coefficients of the Lie algebra in question; by making
a proper choice, these polynomials themselves are related to known
special functions. The coupling of two or three representations of the Lie
algebra then leads to interesting addition formulas for these special
functions. The polynomials appearing here are generalized Laguerre
and Jacobi polynomials for the $su(1,1)$ case, and Hermite polynomials
for the $b(1)$ algebra.

\vspace{1cm}

\noindent
Running title~: Addition formulas\\
PACS~: 02.20b, 02.30g.

\vspace{1cm}

%\noindent
%Submitted to {\em J.\ Math.\ Phys.}, June 1996.

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\section{Introduction}

In a short paper, Granovskii and Zhedanov~\cite{GZ} 
considered a new method to
construct $3nj$-symbols and applied it to the case of $su(1,1)$. Their
method is potentially powerful, and can in principle be applied to other Lie
algebras. On the other hand, their treatment of the $su(1,1)$ case is
rather schematic. Moreover, the most interesting application is missing
in their work. This application comes as a polynomial identity
following from the tensor product of two representations.

Here, we extend the method of Granovskii and Zhedanov, and apply
it to two algebras~: $su(1,1)$ and the oscillator algebra $b(1)$. 
The tensor product of two $su(1,1)$ (positive discrete series)
representations is treated in detail. 
Applying our method to this case yields an interesting
addition formula for generalized Laguerre polynomials
$L^{(\alpha)}_n(x)$, which can be
written in the following form~:
\bea
&&\sum_{k=0}^{n+j} Q_j(k;a,b,j+n) L_k^{(a)}(x) L_{n+j-k}^{(b)}(y) \nn\\
&&\qquad = {(-1)^j n!j!\over (a+1)_j(n+j)!} (x+y)^j
L_n^{(a+b+2j+1)}(x+y) P_j^{(a,b)}({y-x\over y+x});
\label{intro1}
\eea
herein, $Q$ stands for a Hahn polynomial and $P$ for a Jacobi
polynomial~\cite{As,Er,Ab}. Applying the method to the coupling of three
representations yields a more involved addition formula, given in
(\ref{new2}). As a side result, this method easily provides
explicit expressions for the
$su(1,1)$ coupling and recoupling coefficients.

The second Lie algebra considered in this paper is (a generalization of)
the oscillator algebra $b(1)$. For this algebra, a class of
unitary representations is considered; the tensor product of two
representations of this class can be completely reduced to a direct sum
of representations of the same class. The algebra is sufficiently
simple to calculate the Clebsch-Gordan (or coupling) and recoupling
coefficients explicitly. These turn out to be related to a ${\ }_2F_1$
hypergeometric function, or to a Wigner $d$-function. In this case, our
method leads to the following addition formula for Hermite polynomials~:
\bea
&&H_j(ax+by) H_n(ay-bx)=\nn\\
&&\sum_{k=0}^{j+n} \left(j+n\atop k\right) a^{n+k}b^{j-k}{\ }_2F_1\left(
{-n,-k\atop-j-n};{1\over a^2}\right) H_k(x) H_{j+n-k}(y),
\label{intro2}
\eea
where $a^2+b^2=1$.

For convenience, we recall here the definition of a (generalized)
hypergeometric function,
\beq
{}_pF_q\left({a_1,\cdots,a_p \atop b_1,\cdots,b_q};z\right) =
\sum_{k=0}^\infty {(a_1)_k \cdots (a_p)_k \over (b_1)_k\cdots (b_q)_k}
{z^k\over k!},
\eeq
where $(a)_k=a(a+1)\cdots(a+k-1)$ (and $(a)_0=1$) is the Pochhammer
symbol. The hypergeometric functions appearing in this paper will
always be terminating (that is, one of the numerator parameters is a
negative integer), so that no questions of convergence arise.

Finally, the outlay of the paper is as follows. In sections~2--4, the
case $su(1,1)$ is treated. Section~2 introduces some notation of
Granovskii and Zhedanov~\cite{GZ}.
The coupling of two representations, and the calculation of
the coupling coefficients, is given in Section~3, together with the
proof of (\ref{intro1}). Section~4 treats the coupling of three
representations, the determination of the recoupling coefficient, and
an interesting relation involving Jacobi polynomials and a $_4F_3$
series. The oscillator algebra $b(1)$, together with a class of unitary
representations, is introduced in Section~5. The coupling coefficients
of $b(1)$ are determined in Section~6, where (\ref{intro2}) is also
deduced. In Section~7, the tensor product of three $b(1)$
representations is considered, leading to an identification of the
recoupling coefficient with the coupling coefficient. In Section~8 some
concluding remarks are made. 


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\section{The algebra $su(1,1)$ and a class of representations}

The Lie algebra $su(1,1)$ is generated by $J_0, J_\pm$ subject to
the relations
\beq
[J_0, J_\pm]=\pm J_\pm,\qquad [J_+,J_-]=-2 J_0,
\eeq
with the conditions $J_0^\dagger=J_0$ and $J_\pm^\dagger=J_\mp$. The
positive discrete representations are labeled by a positive real number
$k$, and the orthonormal states or basis vectors of such a
representation can be denoted by $|k,n\rangle$, with $n=0,1,2,\cdots$.
The explicit action of the generators in the representation $(k)$ is
given by~\cite{GZ,Ba}
\beq
 \begin{array}{l}
 J_0 |k,n\rangle = (n+k) |k,n\rangle,\\
 J_+ |k,n\rangle = \sqrt{(n+1)(2k+n)} |k,n+1\rangle,\\
 J_- |k,n\rangle = \sqrt{n(2k+n-1)} |k,n-1\rangle.
 \end{array}
\label{su11action}
\eeq
The eigenvalue of the Casimir operator, ${\cal C}=-J_+J_-+J_0^2-J_0$,
on such a representation is $k(k-1)$. 

Instead of considering the eigenstates of $J_0$ in this representation
$(k)$, one can choose another operator, e.g.~\cite{GZ}
\beq
X= 2J_0 -J_+-J_- ;
\label{defX}
\eeq
we shall comment on what happens if one chooses another linear
combination of the generators later in this paper. The (generalized)
eigenstates of 
$X$ in the representation $(k)$ are labeled by $|k,x\rangle$, with
\beq
X |k,x\rangle = x |k,x\rangle.
\label{actionX}
\eeq
The overlap $\langle k,x|k,n\rangle$ can in principle be computed. The
ratio of $\langle k,x|k,n\rangle$ and $\langle k,x|k,0\rangle$ is a
quantity depending upon $n$, $x$ and $k$, and we write
\beq
 \langle k,x|k,n\rangle=\langle k,x|k,0\rangle Q_n(x;k),
\label{defQ}
\eeq
since it turns out that $Q_n(x,k)$ is a polynomial of degree $n$ in
$x$. To see this, calculate $\langle k,x|X|k,n\rangle$ first by acting
on the left, and then by acting on the right with $X$ (\ref{defX}).
Using (\ref{su11action}) and (\ref{actionX}) one obtains
\beq
x Q_n(x;k)=2(n+k)Q_n(x;k)-\sqrt{(n+1)(2k+n)}Q_{n+1}(x;k)-
\sqrt{n(2k+n-1)}Q_{n-1}(x;k). 
\eeq
Redefining
\beq
 Q_n(x;k)=\sqrt{ (2k)_n \over n! } q_n(x;k),
\label{defq}
\eeq
the polynomials $q_n$ are determined by
\beq
 (2k+n)q_{n+1}(x;k)=(2n+2k-x)q_n(x;k)-n q_{n-1}(x;k),
\label{recurq}
\eeq
together with $q_0(x;k)=1$ and $q_{-1}(x;k)=0$. This implies the
following expression for $q_n$ in terms of a hypergeometric function or
a generalized Laguerre polynomial~\cite{As,Er}~:
\beq
q_n(x;k)={\ }_1F_1\left({-n\atop 2k};x\right)= {n!\over (2k)_n}
L^{(2k-1)}_n(x). 
\eeq

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\section{Tensor product or coupling of two $su(1,1)$ representations}

The tensor product of two unitary representations $(k_1)$ and $(k_2)$
is again unitary and thus completely reducible. The decomposition can be
obtained by counting the multiplicities of the $J_0$-weights in the
tensor product, and reads
\beq
 (k_1)\otimes (k_2) = \bigoplus_{j=0}^\infty (k_1+k_2+j).
\eeq
The coefficients relating the uncoupled states to the coupled states are
known as Clebsch-Gordan coefficients~:
\beq
|(k_1,k_2)k,n\rangle = \sum_{n_1,n_2} C(k_1,n_1,k_2,n_2;k,n) 
 |k_1,n_1\rangle |k_2,n_2\rangle;
\label{coupl}
\eeq
herein, $k$ is of the form $k_1+k_2+j$ ($j$ a nonnegative integer), and
the sum is over all integers $n_1, n_2$ ($\geq 0$) with $(k_1+n_1)+
(k_2+n_2)=(k+n)$. 
The action of $J_-$ yields the following recurrence
relation for the Clebsch-Gordan coefficients~:
\bea
&&\sqrt{n(2k+n-1)}C(k_1,n_1,k_2,n_2;k,n-1) =\nn\\
&&\qquad\qquad  \sqrt{(n_1+1)(2k_1+n_1)}C(k_1,n_1+1,k_2,n_2;k,n) \nn\\
&&\qquad\qquad + \sqrt{(n_2+1)(2k_2+n_2)}C(k_1,n_1,k_2,n_2+1;k,n).
\eea
Putting herein $n=0$ leads to the following relation~:
\bea
&&C(k_1,n_1,k_2,n_2;k_1+k_2+n_1+n_2,0) = \nn\\
&&(-1)^{n_1}\left[ { (n_1+n_2)!(2k_2+n_2)_{n_1} \over n_1!n_2!
(2k_1)_{n_1} }\right]^{1/2} C(k_1,0,k_2,n_1+n_2;k_1+k_2+n_1+n_2,0).\nn\\
\label{recu}
\eea
Using the orthogonality (unitarity) condition
\beq
\sum_{n_1,n_2} C(k_1,n_1,k_2,n_2;k,n)^2 = 1,
\eeq
in (\ref{recu}) determines
\beq
 C(k_1,0,k_2,n_2;k_1+k_2+n_2,0)^2=(2k_1)_{n_2}/(2k_1+2k_2+n_2-1)_{n_2}.
\eeq
Choosing the phase factor for this coefficient positive leads by means
of (\ref{recu}) to the
expression for the Clebsch-Gordan coefficients with $n=0$~:
\beq
C(k_1,n_1,k_2,n_2;k_1+k_2+n_1+n_2,0)=(-1)^{n_1}
\left[(n_1+n_2)!(2k_1+n_1)_{n_2}(2k_2+n_2)_{n_1}\over
n_1!n_2!(2k_1+2k_2+n_1+n_2-1)_{n_1+n_2}\right]^{1/2}.
\label{sf}
\eeq
Using this last expression and the recurrence relation obtained
from the $J_+$ action, i.e.
\bea
&&\sqrt{(n+1)(2k+n)}C(k_1,n_1,k_2,n_2;k,n+1) =\nn\\
&&\qquad\qquad  \sqrt{n_1(2k_1+n_1-1)}C(k_1,n_1-1,k_2,n_2;k,n) \nn\\
&&\qquad\qquad + \sqrt{n_2(2k_2+n_2-1)}C(k_1,n_1,k_2,n_2-1;k,n),
\eea
together with the boundary conditions
that $C$ must vanish when one of
the $n$-values is negative, 
yields expressions for the Clebsch-Gordan coefficients with $n>0$.
The expression is then (provided $n_1+n_2=j+n$) given by 
\bea
&&C(k_1,n_1,k_2,n_2;k_1+k_2+j,n)=\nn\\
&&\left[{(2k_1)_{n_1}(2k_2)_{n_2}(2k_1)_j \over
n!n_1!n_2!j!(2k_1+2k_2+2j)_n(2k_2)_j(2k_1+2k_2+j-1)_j} \right]^{1/2}\nn\\
&&\times (j+n)! {\ }_3F_2\left( {2k_1+2k_2+j-1, -n_1, -j
\atop 2k_1, -n-j};1\right),
\label{3f2}
\eea
in terms of a terminating generalized hypergeometric series. Actually,
we will not need (\ref{3f2}) in what follows, but only (\ref{sf}).
Moreover, (\ref{3f2}) will be deduced in an easy way from our method at
the end of this Section. 

The action of each operator $Y=J_i$ ($i=0,\pm$) on the tensor product of
two representations $(k_1)\otimes (k_2)$ is given by $Y\otimes 1 +
1\otimes Y$; thus this also holds for $X$. This implies that
\beq
X\left(|k_1,x\rangle \otimes|k_2,y\rangle\right)=X|k_1,x,k_2,y\rangle =
(x+y) |k_1,x,k_2,y\rangle .
\eeq
Then it follows from (\ref{defQ}) that
\beq
\langle k_1,x,k_2,y|(k_1,k_2)k,n\rangle=
\langle k_1,x,k_2,y|(k_1,k_2)k,0\rangle Q_n(x+y;k).
\label{kQ}
\eeq
Using the tensor product decomposition for $|(k_1,k_2)k,0\rangle$, one
finds 
\bea
&&\langle k_1,x,k_2,y|(k_1,k_2)k,0\rangle=
 \sum_{n_1,n_2} C(k_1,n_1,k_2,n_2;k,0) \langle k_1,x|k_1,n_1\rangle
 \langle k_2,y|k_2,n_2\rangle \nn\\
&&= \sum_{n_1,n_2} C(k_1,n_1,k_2,n_2;k,0) Q_{n_1}(x;k_1) Q_{n_2} (y;k_2)
\langle k_1,x|k_1,0\rangle \langle k_2,y|k_2,0\rangle ,
\eea
or, by introducing the function
\beq
S(x,y;k_1,k_2,k)=\sum_{n_1,n_2} C(k_1,n_1,k_2,n_2;k,0) Q_{n_1}(x;k_1)
Q_{n_2} (y;k_2),
\label{defS}
\eeq
one can write
\beq
\langle k_1,x,k_2,y|(k_1,k_2)k,0\rangle=
S(x,y;k_1,k_2,k)\langle k_1,x|k_1,0\rangle \langle k_2,y|k_2,0\rangle.
\label{k0}
\eeq
Similarly, using the tensor product decomposition for
$|(k_1,k_2)k,n\rangle$, one deduces
\bea
&&\langle k_1,x,k_2,y|(k_1,k_2)k,n\rangle=
 \sum_{n_1,n_2} C(k_1,n_1,k_2,n_2;k,n)\nn\\
&&\qquad\qquad\times Q_{n_1}(x;k_1) Q_{n_2} (y;k_2)
\langle k_1,x|k_1,0\rangle \langle k_2,y|k_2,0\rangle.
\label{kn}
\eea
Substituting (\ref{k0}) and (\ref{kn}) in (\ref{kQ}), there comes
\beq
S(x,y;k_1,k_2,k) Q_n(x+y;k) =
\sum_{n_1,n_2} C(k_1,n_1,k_2,n_2;k,n) Q_{n_1}(x;k_1) Q_{n_2} (y;k_2),
\label{SQ}
\eeq
or, using orthogonality,
\beq
\sum_{k,n}C(k_1,n_1,k_2,n_2;k,n) S(x,y;k_1,k_2,k) Q_n(x+y;k) =
  Q_{n_1}(x;k_1) Q_{n_2} (y;k_2).
\label{QQ}
\eeq

Equations (\ref{defS}), (\ref{SQ}) and (\ref{SS}) are the basic
formulas for our technique.
It is interesting to see the consequences of these equations. First of
all, one wishes to find a proper form for $S$. Using the series
expansion 
\beq
\sum_{r,s \atop r+s=j} (-1)^r \left(j \atop r\right)
  {\ }_1F_1\left({-r\atop a};x\right)
 {\ }_1F_1\left({-s\atop b};y\right) =
{x^j \over (a)_j} {\ }_2F_1\left({1-j-a,-j\atop b};{-y\over x}\right),
\eeq
and the special form (\ref{sf}) in (\ref{defS}), one arrives at
\bea
&&S(x,y;k_1,k_2,k_1+k_2+j)=\nn\\
&&\qquad\left( (2k_2)_j \over
(2k_1)_j(2k_1+2k_2+j-1)_j\; j! \right)^{1/2} 
x^j {\ }_2F_1\left({1-j-2k_1,-j\atop 2k_2};{-y\over x}\right).
\label{2f1}
\eea
Using first the series transformation
\beq
{\ }_2F_1\left({a,-n\atop c};z\right)={(a)_n\over (c)_n}(1-z)^n
{\ }_2F_1\left({c-a,-n\atop 1-a-n};{1\over 1-z}\right),
\eeq
and next the following relation in terms of a Jacobi
polynomial~\cite{As,Er} 
\beq
{\ }_2F_1\left({1+a+b+n,-n\atop a+1};t\right)= {n!\over (a+1)_n}
P^{(a,b)}_n(1-2t), 
\eeq
one finds~:
\bea
&&S(x,y;k_1,k_2,k_1+k_2+j)=(-1)^j
\left[j!/((2k_1)_j(2k_2)_j(2k_1+2k_2+j-1)_j)\right]^{1/2} \nn\\
&&\qquad\qquad\times (x+y)^j P_j^{(2k_1-1,2k_2-1)}({y-x\over y+x}).
\eea

There are now two applications. First, consider (\ref{QQ}) and
substitute $y=-x$. In the lhs of (\ref{QQ}) the ${\ }_2F_1$ coming from
(\ref{2f1}) can be summed using Vandermonde's theorem~\cite{Sl}, 
and it reduces to~: 
\bea
&&\sum_{j=0}^{n_1+n_2} \left[
 (2k_1+2k_2+j-1)_j(2k_1+2k_2+2j)_{n_1+n_2-j} \over
(2k_1)_j(2k_2)_j(n_1+n_2-j)!j!
\right]^{1/2}\nn\\ 
&&\qquad\qquad\times C(k_1,n_1,k_2,n_2;k_1+k_2+j,n_1+n_2-j) x^j.
\label{lhs1}
\eea
In the rhs of (\ref{QQ}) there comes, apart from two square root
factors related to (\ref{defq}),
\bea
&&{\ }_1F_1\left({-n_1\atop 2k_1};x\right)
{\ }_1F_1\left({-n_2\atop 2k_2};-x\right) =
 \sum_{p=0}^{n_1}\sum_{q=0}^{n_2} {(-n_1)_p\over (2k_1)_p}{x^p\over p!}
{(-n_2)_q\over (2k_2)_q}  {(-x)^q\over q!}\nn\\
&&\qquad = \sum_{j=0}^{n_1+n_2} \left( n_1+n_2 \atop j\right) {x^j
\over (2k_2)_j} {\ }_3F_2\left({2k_1+2k_2+j-1, -n_1, -j \atop 2k_1,
-n_1-n_2} ; 1\right).
\label{rhs1}
\eea
The last expression is found by changing the summation variables $p$
and $q$ to $p$ and $j=p+q$, and elementary manipulations. Taking into
account the relevant square root factors, one finds back the expression
(\ref{3f2}) for the Clebsch-Gordan coefficients by comparing
(\ref{lhs1}) and (\ref{rhs1}). Note that we have used only the simple
coefficients (\ref{sf}) in this derivation;
so from these coefficients only this technique provides a way to get
expressions for all Clebsch-Gordan coefficients.

As a second application, consider (\ref{SQ}). Replacing $S$ by the 
Jacobi polynomial, $Q$ by the generalized Laguerre polynomials, and
$(2k_1-1, 2k_2-1)$ by $(a,b)$, the following addition formula comes~:
\bea
&&\sum_{k=0}^{n+j} {\ }_3F_2\left({a+b+j+1,-k,-j\atop a+1,-n-j};1\right)
L_k^{(a)}(x) L_{n+j-k}^{(b)}(y) \nn\\
&&\qquad = {(-1)^j n!j!\over (a+1)_j(n+j)!} (x+y)^j
L_n^{(a+b+2j+1)}(x+y) P_j^{(a,b)}({y-x\over y+x}).
\label{new1}
\eea

For $j=0$, this reduces to the known addition formula for generalized
Laguerre polynomials~\cite[eq.~(22.12.6)]{Ab}; 
also the case $n=0$ reduces to a known relation.
But the general case, as given here in (\ref{new1}), seems to be new.

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\section{Coupling of three $su(1,1)$ representations}

The tensor product of three representations, $(k_1)$, $(k_2)$ and
$(k_3)$, can be formed in two ways~: $\left( (k_1)\otimes (k_2) \right)
\otimes (k_3)$ or $(k_1)\otimes \left( (k_2) \otimes (k_3)\right)$. The
coefficients relating the decomposition of the first to the
decomposition of the second are recoupling coefficients, proportional
to 6-$j$ coefficients. In an obvious notation, one writes
\beq
|((k_1,k_2)k_{12},k_3)k,n\rangle = \sum_{k_{23}} 
U(k_1,k_2,k_{12},k_3,k,k_{23}) |(k_1,(k_2,k_3)k_{23})k,n\rangle.
\label{defU}
\eeq
The coefficient $U$ is independent of $n$ (Wigner-Eckart theorem). 

Using (\ref{k0}) twice in the coupling $\left( (k_1)\otimes (k_2)
\right) \otimes (k_3)$ yields~:
\bea
&&\langle k_1,x,k_2,y,k_3,z|((k_1,k_2)k_{12},k_3)k,0\rangle = \nn\\
&&S(x+y,z;k_{12},k_3,k) \langle k_1,x,k_2,y|(k_1,k_2)k_{12},0\rangle 
\langle k_3,z|k_3,0\rangle\nn\\
&&=S(x,y;k_1,k_2,k_{12})S(x+y,z;k_{12},k_3,k)
\langle k_1,x|k_1,0\rangle\langle k_2,y|k_2,0\rangle
\langle k_3,z|k_3,0\rangle.
\label{12-3}
\eea
Using it in the coupling $(k_1)\otimes \left( (k_2) \otimes
(k_3)\right)$ yields~:
\bea
&&\langle k_1,x,k_2,y,k_3,z|(k_1,(k_2,k_3)k_{22})k,0\rangle = \nn\\
&&=S(y,z;k_2,k_3,k_{23})S(x,y+z;k_1,k_{23},k)
\langle k_1,x|k_1,0\rangle\langle k_2,y|k_2,0\rangle
\langle k_3,z|k_3,0\rangle.
\label{1-23}
\eea
Relating (\ref{12-3}) and (\ref{1-23}) by means of (\ref{defU}) we obtain
\bea
&&S(x,y;k_1,k_2,k_{12})S(x+y,z;k_{12},k_3,k)= \nn\\
&&\sum_{k_{23}=k_2+k_3}^{k-k_1} U(k_1,k_2,k_{12},k_3,k,k_{23}) 
S(y,z;k_2,k_3,k_{23})S(x,y+z;k_1,k_{23},k).
\label{SS}
\eea
Note that in this formula $k_{23}$ assumes real values; by the
summation symbol we mean that $k_{23}$ takes all values $k_2+k_3$,
$k_2+k_3+1$, $\cdots$, $k-k_1$ in steps of one unit.

Again, this relation can be applied in two ways. First, choosing $x=1$
and $y=-z$, the rhs of (\ref{SS}) reduces to
\beq
\sum_{k_{23}=k_2+k_3}^{k-k_1} f(k_1,k_2,k_3,k_{23},k)
U(k_1,k_2,k_{12},k_3,k,k_{23}) z^{k_{23}-k_2-k_3},
\eeq
with $f$ some numerical factor containing square roots and Pochhammer
symbols. 
On the other hand,
the lhs of (\ref{SS}) reduces -- apart from square root
factors -- to a product of two $_2F_1$ series; one with argument $z$
and one with argument $z/(z-1)$, and a factor $(1-z)^{k-k_{12}-k_3}$.
Applying 
\beq 
(1-z)^m{\ }_2F_1\left( {a,-m\atop c};{z\over z-1}\right)=
 {\ }_2F_1\left( {c-a,-m\atop c};z\right)
\eeq
to the second $_2F_1$ combined with the power of $(1-z)$, the lhs
essentially reduces to a product of two $_2F_1$ series with argument
$z$. Then one can use
\bea
&&{\ }_2F_1\left({ b+d-c+m+n,-m\atop b};z\right) {\ }_2F_1\left( {c,-n\atop
d};z\right)= 
\label{gf} \\
&& \sum_{p=0}^{m+n}(-1)^p {(c-m)_p\over(d)_p}\left(m+n\atop p\right)
{\ }_4F_3\left({b+d+p-1,c-d-m-n,-m,-p\atop b,c-m,-m-n};1\right) z^p. \nn
\eea
The technique to find (\ref{gf})
 is essentially the same as in (\ref{rhs1}). Comparing
equal powers of $z$ in (\ref{SS}), under the given specialization (and
taking into account the appropriate numerical factors),
yields a single sum expression for the $su(1,1)$ recoupling
coefficients,
\bea
&& U(k_1,k_2,k_{12},k_3,k,k_{23})=
\bigl[ (2k_2)_{k_{12}-k_1-k_2} (2k_3)_{k-k_{12}-k_3}
(2k_2)_{k_{23}-k_2-k_3} (2k_1)_{k-k_1-k_{23}} \nn\\
&&\times (k_1+k_{23}+k-1)_{k-k_1-k_{23}}\bigr]^{1/2} 
\bigl[ (2k_1)_{k_{12}-k_1-k_2} (k_1+k_2+k_{12}-1)_{k_{12}-k_1-k_2}
(2k_{12})_{k-k_{12}-k_3}  \nn\\
&&\times (k_{12}+k_3+k-1)_{k-k_{12}-k_3}
(2k_3)_{k_{23}-k_2-k_3} (2k_{23})_{k-k_1-k_{23}}
(k_2+k_3+k_{23}-1)_{k_{23}-k_2-k_3} \nn\\
&&\times (k_{12}-k_1-k_2)!(k-k_{12}-k_3)!
(k-k_1-k_{23})!(k_{23}-k_2-k_3)! \bigr]^{-1/2} \nn\\
&&\times (k+k_1+k_2+k_3-1)_{k_{23}-k_2-k_3} (k-k_1-k_2-k_3)! \nn\\
&&\times {\ }_4F_3\left({k_1+k_2+k_{12}-1, k_2+k_3+k_{23}-1,
k_1+k_2-k_{12}, k_2+k_3-k_{23} \atop 2k_2, k+k_1+k_2+k_3-1,k_1+k_2
+k_3-k};1 \right).
\eea

As a second application, one replaces all $S$-functions in (\ref{SS})
by Jacobi polynomials, and uses the just obtained ${\ }_4F_3$
expression for the recoupling coefficient. Simplifying the numerical
factors, and defining the polynomials $p$ related to the usual Jacobi
polynomials $P$ by means of
\beq
p^{(a,b)}_n(x,y)={(x+y)^n \over (b+1)_n} P_n^{(a,b)}({y-x\over y+x}),
\eeq
this leads to
\bea
&&\sum_{j=0}^{m+n} \left(m+n\atop m\right)
{(a+b+c+m+n+2)_j\over(b+c+j+1)_j} 
{\ }_4F_3\left({a+b+m+1,b+c+j+1,-m,-j\atop b+1,a+b+c+m+n+2,-m-n};1\right)\nn\\
&&\times p^{(b,c)}_j(y,z) p^{(a,b+c+2j+1)}_{m+n-j}(x,y+z)=
 p^{(a,b)}_m(x,y) p^{(a+b+2m+1,c)}_n(x+y,z).
\label{new2}
\eea

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\section{The algebra $b(1)$ and a class of representations}

The second Lie algebra considered in this paper is a generalization
of the oscillator algebra (or boson algebra) in one dimension. It has a
basis of four elements $b^\pm$, $H$, $N$ with relations
\beq
\begin{array}{l}
[b^-,b^+] = H,\qquad [N,b^\pm]=\pm b^\pm,\\
{}[H,x]=0 \hbox{ for } x=b^\pm, N,
\end{array}
\label{b1}
\eeq
and unitarity conditions $H^\dagger=H$, $N^\dagger=N$, $(b^\pm)^\dagger
= b^\mp$. For the usual oscillator algebra, one takes
$H=1$ and requires $N=b^+b^-$. Here, we do not assume these extra
relations. The reason is that with these extra relations the
representation theory is very poor (only the Fock representation),
whereas for the given algebra~(\ref{b1}) it is much richer. The algebra
is also the Lie algebra of a Lie group of third order real triangular
matrices~\cite[Section~5.1.1]{VK}.
The algebra $b(1)$ has a class of lowest weight
representations~\cite{Mi} labeled
by two positive numbers $h$ and $k$, and orthonormal basis vectors
$|h,k,n\rangle$, with $n=0,1,2,\cdots$. The action of the Lie algebra
elements on the states of the representation $(h,k)$ is given by~:
\beq
 \begin{array}{l}
 H |h,k,n\rangle = h |h,k,n\rangle,\\
 N |h,k,n\rangle = (k+n) |h,k,n\rangle,\\
b^- |h,k,n\rangle = \sqrt{hn} |h,k,n-1\rangle,\\
 b^+ |h,k,n\rangle = \sqrt{h(n+1)} |h,k,n+1\rangle.
 \end{array}
\label{b1action}
\eeq
It is easy to verify that it is a representation, and that it is
unitary and irreducible if $h>0$ and $k\geq 0$. 
One finds back the usual Fock space for $h=1$ and $k=0$.

One can now apply the same technique as for the algebra $su(1,1)$. We shall
introduce again an operator $X$, polynomials $Q_n$ and $S$, coupling and
recoupling coefficients $C$ and $U$, etc. For these quantities, the
same notation is used for the $b(1)$ case as for the $su(1,1)$ case, in
order not to overload notation by adding an extra index referring to
the algebra in question. This should cause no confusion~: sections~2--4
deal with the $su(1,1)$ case, and sections~5--7 with the $b(1)$ case.

Consider here the operator 
\beq
X=b^++b^-,
\label{defX2}
\eeq
and its eigenstates
\beq
X|h,k,x\rangle= x|h,k,x\rangle
\label{54}
\eeq
in the representation $(h,k)$. Writing the overlap 
$\langle h,k,x|h,k,n\rangle$ as
\beq
\langle h,k,x|h,k,n\rangle =\langle h,k,x|h,k,0\rangle Q_n(x;h)
\label{b1Q}
\eeq
the action of $X$ to the left and right in $\langle
h,k,x|X|h,k,n\rangle$ yields
\beq
x Q_n(x;h)=\sqrt{h(n+1)} Q_{n+1}(x;h)+\sqrt{hn} Q_{n-1}(x;h),
\eeq
with $Q_0(x;h)=1$. This quantity is independent of $k$, which is why we
have left out the symbol $k$ from the beginning in (\ref{b1Q}).
Redefining
\beq
Q_n(x;h)=q_n(x;h)/\sqrt{n!},
\label{defb1Q}
\eeq
the polynomials $q_n$ are now defined through the recursion relation
\beq
q_{n+1}(x;h)={x\over\sqrt{h}} q_n(x;h)-n q_{n-1}(x;h).
\eeq
Together with the initial conditions, this implies that 
\beq
q_n(x;h) = H_n(x/\sqrt{h}),
\eeq
with $H_n$ the {\em monic} Hermite polynomials~\cite{Ab}.
The present can also be understood in quantum mechanical harmonic
oscillator language. For $h=1$ and $k=0$, (\ref{b1action}) are the
usual harmonic oscillator states. Then, (\ref{54}) are the eigenstates
of the position operator; (\ref{b1Q}) expresses that the wave function
in the position representation is equal to the ground state wave
function times a Hermite polynomial.

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\section{Tensor product of two $b(1)$ representations}

The tensor product of two unitary representations $(h_1,k_1)$ and
$(h_2,k_2)$ is unitary and completely reducible. By counting the
multiplicities of the $N$-weights in the tensor product, one obtains
the decomposition of it~:
\beq
(h_1,k_1)\otimes (h_2,k_2) = \bigoplus_{j=0}^\infty
(h_1+h_2,k_1+k_2+j). 
\eeq
The coefficients relating the uncoupled states to the coupled states are
the Clebsch-Gordan coefficients of $b(1)$~:
\beq
|(h_1,k_1;h_2,k_2)h_1+h_2,k_1+k_2+j,n\rangle 
= \sum_{n_1,n_2} C(h_1,h_2,j;n_1,n_2,n) 
 |h_1,k_1,n_1\rangle |h_2,k_2,n_2\rangle;
\eeq
herein, $n_1+n_2=j+n$, and the labels $k_i$ have been omitted in $C$ as
we shall see that it is independent of them.
To find expressions for the coefficients $C$, one can use a recurrence
relation following from the action of $b^-$~:
\bea
&&\sqrt{(h_1+h_2)n}C(h_1,h_2,j;n_1,n_2,n-1)=
\sqrt{h_1(n_1+1)}C(h_1,h_2,j;n_1+1,n_2,n)+\nn\\
&&\qquad\qquad\qquad \sqrt{h_2(n_2+1)}C(h_1,h_2,j;n_1,n_2+1,n).
\eea
Putting $n=0$, this equation implies~:
\beq
C(h_1,h_2,j;n_1,n_2,0)=(-1)^{n_1}\left[{h_2^{n_1}\over h_1^{n_1}}
\left( j\atop n_1\right)\right]^{1/2} C(h_1,h_2,j;0,j,0).
\eeq
Using orthogonality of the Clebsch-Gordan coefficients the above
formula implies~:
\beq
 C(h_1,h_2,j;0,j,0)^2=h_1^j/(h_1+h_2)^j;
\eeq
choosing the phase factor as $(-1)^j$ then leads to an explicit
expression for the Clebsch-Gordan coefficients with $n=0$ (where, of
course, $n_1+n_2=j$)~:
\beq
C(h_1,h_2,j;n_1,n_2,0)=(-1)^{j-n_1}\left[
\left( j\atop n_1\right) {h_1^{j-n_1}h_2^{n_1}\over 
(h_1+h_2)^j}\right]^{1/2}.
\eeq
Using the recurrence relation obtained from the $b^+$ action, i.e.
\bea
&&\sqrt{(h_1+h_2)(n+1)}C(h_1,h_2,j;n_1,n_2,n+1)=
\sqrt{h_1n_1}C(h_1,h_2,j;n_1-1,n_2,n)+\nn\\
&&\qquad\qquad\qquad \sqrt{h_2n_2}C(h_1,h_2,j;n_1,n_2-1,n),
\eea
one can determine the Clebsch-Gordan coefficients for $n>0$.
We give here two expressions (compare~\cite[paragraph~4-17]{Mi} or
\cite[section~8.6.2]{VK})~: 
\bea
C(h_1,h_2,j;n_1,n_2,n)&=& \delta_{n_1+n_2,j+n} (-1)^{n_2}
\sqrt{n_1!n_2!j!n!} \sum_{k} {(-1)^k a^{n+n_2-2k}b^{n_1-n+2k} \over
k!(n_2-k)!(n-k)!(j-n_2+k)!} \nn\\
&& \label{C1} \\
&=& \delta_{n_1+n_2,j+n} 
(-1)^{n_2} {(n_1+n_2)! a^{n+n_2}b^{n_1-n}
\over \sqrt{n_1!n_2!j!n!}}{\ }_2F_1\left({-n,-n_2 \atop -j-n};{1\over
a^2} \right),
\label{C2}
\eea
where
\beq
a=\sqrt{h_1\over h_1+h_2}, \qquad b=\sqrt{h_2\over h_1+h_2},
\qquad\hbox{ and thus } a^2+b^2=1.
\label{abcond}
\eeq
From the first expression, one can see that these coefficients coincide
with the Wigner $d$-functions~\cite{Va}~:
\bea
&&C(h_1,h_2,j;n_1,n_2,n)=\delta_{n_1+n_2,j+n} d^J_{MM'}(\beta),\nn\\
&&\hbox{ with } J=(n_1+n_2)/2,\; M=(j-n)/2,\; M'=(n_1-n_2)/2, 
\hbox{ and } \tan{\beta\over 2}=\sqrt{h_1/h_2}. \nn\\
&& \label{d-fun}
\eea

Since for the present algebra $b(1)$ we have again the action $Y\otimes
1 + 1\otimes Y$ for every element of $b(1)$, the same analysis as in
Section~3 can be made. Introducing the function~:
\beq
S(x,y;h_1,h_2,j)=\sum_{n_1,n_2} C(h_1,h_2,j;n_1,n_2,0) Q_{n_1}(x;h_1)
Q_{n_2}(x;h_2) ,
\label{bS}
\eeq
there holds the equality
\beq
S(x,y;h_1,h_2,j) Q_n(x+y;h_1+h_2) =
\sum_{n_1,n_2} C(h_1,h_2,j;n_1,n_2,n) Q_{n_1}(x;h_1) Q_{n_2} (y;h_2).
\label{bSQ}
\eeq
First, let us find an appropriate form for $S$. Using (\ref{defb1Q})
and (\ref{C1}) in (\ref{bS}), with $a$ and $b$ as given in
(\ref{abcond}), one finds 
\bea
S(x,y;h_1,h_2,j)&=&{1\over\sqrt{j!}}\sum_{n_1=0}^j (-1)^{j-n_1} \left(
j \atop n_1\right) a^{j-n_1} b^{n_1} H_{n_1}({x\over\sqrt{h_1}})
H_{j-n_1}({y\over\sqrt{h_2}}) \nn\\
&=& {1\over\sqrt{j!}} H_j(b{x\over\sqrt{h_1}} - a{y\over\sqrt{h_2}}).
\label{bSform}
\eea
In the last step, we have used the following
identity~\cite[eq.~(10.13.40)]{Er} 
\beq
H_j(\alpha u +\beta v)=\sum_{k=0}^j \left(j\atop k\right) \alpha^k
\beta^{j-k} H_k(u) H_{j-k}(v),
\label{ident}
\eeq
where $u$ and $v$ are arbitrary and $\alpha$ and $\beta$ should satisfy
$\alpha^2+\beta^2=1$. 

Having obtained the expression (\ref{bSform}) for $S$, we now have all
ingredients to write (\ref{bSQ}) in an appropriate form. Putting
$u=x/\sqrt{h_1}$ and $v=y/\sqrt{h_2}$, (\ref{bSQ}) reduces to
\bea
&&H_j(au+bv) H_n(av-bu)=\nn\\
&&\sum_{k=0}^{j+n} \left(j+n\atop k\right) a^{n+k}b^{j-k}{\ }_2F_1\left(
{-n,-k\atop-j-n};{1\over a^2}\right) H_k(u) H_{j+n-k}(v),
\label{gros}
\eea
where, of course, $a$ and $b$ stem from (\ref{abcond}) and thus satisfy
$a^2+b^2=1$. The identity has been derived using the monic Hermite
polynomials; however, when rescaled to the usual Hermite polynomials
the same factors appear in left and right hand side, so the same
identity is valid also for the usual Hermite polynomials.

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\section{Tensor product of three $b(1)$ representations}

It is then natural to consider the tensor product of three $b(1)$
representations and investigate the consequences for a relation like
(\ref{SS}) applied to the present algebra. In the first coupling
scheme, $|h_1,k_1,n_1\rangle$ and $|h_2,k_2,n_2\rangle$ are first
coupled to $|(h_1,k_1;h_2,k_2)h_1+h_2,k_1+k_2+j_{12},n_{12}\rangle$, and
then with $|h_3,k_3,n_3\rangle$ to a state denoted by
\beq
|(h_1,h_2)h_3;j_{12};j,n\rangle =
|\bigl((h_1,k_1;h_2,k_2)h_1+h_2,k_1+k_2+j_{12};h_3,k_3\bigr)h_1+h_2+h_3,
k_1+k_2+k_3+j,n\rangle.
\label{b12-3}
\eeq
In the second coupling scheme $|h_2,k_2,n_2\rangle$ and
$|h_3,k_3,n_3\rangle$ are first 
coupled to $|(h_2,k_2;h_3,k_3)\newline h_2 + h_3,k_2+k_3+j_{23},n_{23}\rangle$, and
then with $|h_1,k_1,n_1\rangle$ to a state denoted by
\beq
|h_1(h_2,h_3);j_{23};j,n\rangle =
|\bigl(h_1,k_1;(h_2,k_2;h_3,k_3)h_2+h_3,k_2+k_3+j_{23}\bigr)h_1+h_2+h_3,
k_1+k_2+k_3+j,n\rangle.
\label{b1-23}
\eeq
Note that $j\geq j_{12}\geq 0$ and $j\geq j_{23}\geq 0$, all $j$-values
being integer. The coefficients relating the basis (\ref{b12-3}) to the
basis (\ref{b1-23}) are recoupling coefficients, and we write
\beq
|(h_1,h_2)h_3;j_{12};j,n\rangle = \sum_{j_{23}=0} ^j
U(h_1,h_2,h_3;j_{12},j_{23},j) |h_1(h_2,h_3);j_{23};j,n\rangle,
\eeq
where $U$ is independent of $n$ (Wigner-Eckart theorem) and also
independent of the $k$-representations labels (since the Clebsch-Gordan
coefficients themselves are independent of the $k$-values).  

Expanding the two coupling schemes explicitly, one can obtain an
expression for the recoupling coefficient $U$ as a (double) sum over
the product of four Clebsch-Gordan coefficients. After lengthy
calculations, we realized that this complicated expression can be
simplified to a single sum; moreover, it reduces to a similar
expression as the $b(1)$ Clebsch-Gordan coefficient itself, but with
different parameters. The final result is~:
\beq
U(h_1,h_2,h_3;j_{12},j_{23},j) = C(h_1h_3,h_1h_2+h_2^2+h_2h_3,j_{23};
j-j_{12}, j_{12}, j-j_{23}),
\eeq
with $C$ given in (\ref{C1})--(\ref{abcond}).
Thus this algebra has the remarkable property that its recoupling
coefficients are the same as its coupling coefficients (see also
\cite[Section~8.6.6]{VK}). 

After this, there remains little to be done. An identity similar to
(\ref{SS}) is also valid for $b(1)$, but since the $U$ coefficients
reduce to $C$ coefficients and the $S$-functions to Hermite
polynomials, the analogue of (\ref{SS}) simply reduces to (\ref{gros}).

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\section{Comments and conclusions}

We have considered two Lie algebras, $su(1,1)$ and $b(1)$, together
with a class of unitary representations that is closed under the tensor
product (i.e.\ the tensor product of two representations of this class
can be decomposed as a direct sum of representations of the same
class). The relation with special functions stems from the introduction
of an operator $X$, as in (\ref{defX}) or (\ref{defX2}). The choice of
$X$ may seem somewhat arbitrary. It is, however, restricted by the fact
that the action of $X$ on the tensor product
representation should be given by $X\otimes 1 + 1\otimes X$, and
moreover $X$ must be unitary. Thus, for $su(1,1)$ the most general
choice for $X$ in (\ref{defX}) is (up to an overall constant)
\beq
X=a J_0 -J_+-J_- .
\eeq
Here $a=2$ was considered, leading to Laguerre polynomials. Making this
choice $a=2$ leads to the recurrence relation~(\ref{recurq}), which
coincides with the recurrence relation for Laguerre polynomials. For
arbitrary $a$, it is also possible to identify the recurrence relation
with one of known orthogonal polynomials, although the polynomials are
more complicated. Further
investigation~\cite{KV} has shown that for $-2<a<2$, the corresponding
polynomials are Meixner-Pollaczek polynomials, and for $|a|>2$ Meixner
polynomials (the case $a=-2$ again being related to Laguerre). Using a
related but more sophisticated technique as in this paper, a relation
similar to~(\ref{intro1}) has been derived~\cite{KV} with the Laguerre
(resp.\ Jacobi) polynomials replaced by Meixner-Pollaczek (resp.\
continuous Hahn) polynomials. 

A similar remark can be made for the choice of $X$ for $b(1)$. A more
general expression would lead to a replacement of Hermite polynomials
by Charlier polynomials. 

The present technique works for quantum algebras as well. In that case,
the choice for $X$ is restricted by the fact that it should be unitary
(invariant under the $*$-operation), and that it should be ``twisted
primitive'' \cite{KV}.

The main results of the paper are mathematical, but there are a number
of relations with physics. For the Lie algebra $su(1,1)$ and its
positive discrete series representations, the physicist's approach was
described by Bacry~\cite{Ba}. A consequence of our results is that the
$su(1,1)$ operators $J_0$ and $J_\pm$ have a realization as
differential operators in a variable $x$, that the states $|k,n\rangle$
have a realization in terms of Laguerre polynomials $L^{(2k-1)}_n(x)$,
such that the action~(\ref{su11action}) holds. The action
$J_0|k,n\rangle = (n+k) |k,n\rangle$ coincides with a Schrodinger
equation, with discrete spectrum $k+n$ ($n=0,1,\ldots$). For an operator
$X$ of the form $a J_0 -J_+-J_-$ the Schrodinger equation can also be
considered and in general $X$ has a continuous spectrum~\cite{Ba}. 
The coupling and recoupling coefficients
of $su(1,1)$ also have physical importance, although these have been
derived already by other methods. 
For the oscillator algebra $b(1)$ the relation with physics is clear~:
it is just a central extension of the boson algebra (harmonic
oscillator algebra). As a consequence, the representations considered
here are generalizations of the Fock space for the harmonic oscillator.

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\section*{Acknowledgements}

We have pleasure in thanking Profs.\ C.C.\ Grosjean and K.\ Srinivasa Rao
for stimulating discussions. Dr.\ H.T.\ Koelink is acknowledged for a
useful communication.
This research was partly supported by the E.E.C. (contract No.
CI1*-CT92-0101).



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