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% The $q$-boson operator algebra and $q$-Hermite polynomials
% J. Van der Jeugt
% Lett. Math. Phys. 24 (1992), 267-274.
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\begin{document}
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\begin{center}
{\Large \bf The $q$-boson operator algebra and $q$-Hermite polynomials}\\[2cm]
J.~Van der Jeugt\footnote{Research Associate N.F.W.O. (National Fund
for Scientific Research of Belgium)} \\[8mm]
Laboratorium voor Numerieke Wiskunde en Informatica,
Universiteit Gent,\\
Krijgslaan 281--S9, B9000 Gent, Belgium
\end{center}

\vskip 3cm
\begin{abstract}
The $q$-boson algebra is defined as an associative algebra with
generators and relations. Some examples are given, and then the
$q$-boson algebra is extended such that roots of the ``diagonal
generators'' are also defined. It is shown that a family of
transformations exist mapping one set of standard generators of
the $q$-boson algebra to another set of standard generators.
Using such a transformation, one obtains expressions for
$q$-bosons for which the $k$th $q$-boson state is expressed
in terms of a $q$-Hermite polynomial $p_k(x;q)$ which reduces to
the ordinary Hermite polynomial of degree $k$ when $q=1$.
\end{abstract}
\newpage
\section{Introduction and definition}
The $q$-generalisation of boson creation and annihilation operators was
introduced by Biedenharn~[1] and Macfarlane~[2], who showed that Schwinger's
boson realisation of $su(2)$ could be modified in order to obtain
a realisation of the quantum enveloping algebra $su_q(2)$. In this
Letter, the $q$-boson operator algebra is defined as an associative
algebra with generators and relations, some examples of the algebra
and of representations are given, and a transformation preserving
the $q$-boson structure is presented.
For a particular transformation one obtains an interesting $q$-boson
algebra in which the $q$-boson creation and annihilation operators
$\bd$ and $b$ are expressed in terms of a coordinate $x$ and the
$q$-derivative $D_{q,x}$. In this case, the $k$th excited $q$-boson
state $|k\rangle$ can be written by means of a $q$-polynomial $p_k(x;q)$.
These polynomials reduce to the ordinary Hermite polynomials when
$q=1$, and they are related to a variety of definitions for the
$q$-Hermite polynomials and also to the continuous $q$-ultraspherical
polynomials.

In papers on quantum groups and $q$-bosons, it is common to use
${q^n-q^{-n} \over q-q^{-1}}$ as the definition for a $q$-number.
In the literature on $q$-special functions~[3], however, it is
common to use the quantity ${1-q^{n} \over 1-q}$. Wishing to keep our
$q$-boson operators equivalent with those defined elsewhere~[1,2],
we have chosen for $[n]={q^n-q^{-n} \over q-q^{-1}}$. Surprisingly,
it turns out that the $q$-special functions related to these
$q$-boson operators, appearing in Section~4, are nevertheless the
classical ones~[3, and references therein] in which the quantity
${1-\q^{n} \over 1-\q}$ with $\q=q^{-2}$ plays a role.

Let $q$ be a fixed complex number, $q\not=\pm 1$.
\begin{defi}
The $q$-boson algebra $B_q$ is the associative algebra over $\C$ with
standard generators $a, \ad, t, t^{-1}$ and relations
 \beq
  \begin{array}{ll}
  \rm (a) & tt^{-1}=t^{-1}t=1 , \\
  \rm (b) & ta-q^{-1}at=0 , \\
  \rm (c) & t\ad - q \ad t =0 , \\
  \rm (d) & a\ad - q^{-1} \ad a = t , \\
  \rm (e) & a\ad - q \ad a = t^{-1} .
  \end{array}
 \label{def}
 \eeq
\end{defi}
Note that relations (d) and (e) are equivalent to
\beq
  \begin{array}{ll}
  \rm (d') & a\ad = (qt-q^{-1}t^{-1})/(q-q^{-1}) , \\
  \rm (e') & \ad a = (t-t^{-1})/(q-q^{-1}) .
  \end{array}
  \label{altdef}
\eeq
In Macfarlane~[2] and other papers~[4], $t$ is written as $q^N$ with $N$ a
``number operator''; then some of the relations in (\ref{def}) can
be appropriately rewritten in terms of $N$.

\begin{defi}
The Fock space $F_q$ is a $B_q$ module with basis vectors $v_n$
($n\in\Nat$), and action given by
 \beq
  \begin{array}{l}
  t^{\pm1} v_n = q^{\pm n} v_n ,\\
  \ad v_n = \sqrt{[n+1]} v_{n+1} , \\
  a v_n = \sqrt{[n]} v_{n-1} .
  \end{array}
 \eeq
\end{defi}
Herein, $[r]$ stands for the $q$-number $(q^r-q^{-r})/(q-q^{-1})$.
The Fock space $F_q$ becomes an inner product space by
\beq
\langle v_m , v_n \rangle = \de_{m,n} .
\eeq
In fact, it is a Hilbert space, and the representatives of $a$ and
$\ad$ are each others adjoint, whereas those of $t$ and $t^{-1}$ are
self-adjoint operators.

\begin{defi}
A lowest weight module $V_\la$ is a cyclic $B_q$ module generated
by a vector $v$ satisfying $av=0$ and $tv=q^\la v$.
\end{defi}

The vector $v$ is called the lowest weight vector, or the vacuum vector.
From the relations (\ref{def}) one can easily deduce that a basis for
$V_\la$ is given by $\{ (\ad)^n v | n\in\Nat\}$, and that
\beq
 \begin{array}{l}
 t (\ad)^n v = q^{\la+n} (\ad)^n v . \\
 a (\ad)^n v =  [n+\la] (\ad)^{n-1} v .
 \end{array}
\label{low}
\eeq
Thus if $q$ is not a root of unity (which shall be assumed to be the case
throughout this Letter), then the lowest weight modules $V_\la$ are
irreducible $B_q$ modules if $\la\not\in\{-1,-2,-3,\ldots\}$.

\setcounter{equation}{0}
\section{Examples and extensions}

The two-dimensional quantum plane $A_q$ is defined as follows~[5].
\begin{defi}
$A_q$ is the associative algebra over $\C$ with generators $x,y,x^{-1},y^{-1}$
and relations
 \beq
  \begin{array}{ll}
  \rm (a) & x x^{-1} = x^{-1}x = y y^{-1} = y^{-1} y = 1 , \\
  \rm (b) & yx = q xy .
  \end{array}
 \eeq
\end{defi}
From these relations, one easily obtains the other commutation relations
among the generators, such as $y^{-1}x=q^{-1}xy^{-1}$.

\begin{prop}
Let $x,x^{-1},y,y^{-1}$ be the generators of the quantum plane. Then
$\ad = x$, $a=x^{-1}(y-y^{-1})/(q-q^{-1})$, $t=y$ and $t^{-1}=y^{-1}$
satisfy the defining relations (\ref{def}) of the $q$-boson operator
algebra.
\end{prop}

It is a straightforward exercise to verify that with the given
expressions all relations are indeed satisfied.

Another example of the $q$-boson operator algebra, which is
well-known~[6], is the following. Let $N=x\partial_x$; $x$ and $N$ are
supposed to be acting on differentiable functions of $x$.
Then $x, x^{-1}, y=q^N, y^{-1}=q^{-N}$ satisfy the defining relations
of the quantum plane generators. Hence $\ad=x$,
$a=x^{-1}(q^N-q^{-N})/(q-q^{-1})=D_{q,x}$, $t=q^N$ and $t^{-1}=q^{-N}$
satisfy the defining relations of the $q$-boson algebra.
Herein, $D_{q,x}$ is the $q$-differential operator with respect to $x$.
Note that the Fock space is $\C[x]$ with basis $v_n=x^n/\sqrt{[n]!}$,
where $[n]!=[n][n-1]\cdots[1]$.

In the literature, $q$-boson operator algebras are usually defined
together with an action on representations in which $t$ and
$t^{-1}$ are diagonal operators (such as the Fock space, or
lowest weight modules). When acting on such representations,
operators such as $t^{1/2}$ or $(t-t^{-1})^{-1}$ are well-defined,
although strictly speaking they do not belong to the associative
algebra generated by $a,\ad,t, t^{-1}$. Therefore, it is sometimes
useful to work in an extended $q$-boson algebra where such
expressions are defined. For this purpose, let $\C(t)$ be the
field of rational expressions in $t$ (by definition, $t^{-1}=1/t$
and $tt^{-1}=t^{-1}t=1$ is automatically satisfied).
Let $\C_t$ be the algebraic extension of $\C(t)$ (generally speaking,
its elements are roots of rational expressions in $t$).

\begin{defi}
The extended $q$-boson operator algebra $B'_q$ is the associative
algebra over $\C$ generated by $a,\ad$ and the elements $p(t)$ of
$\C_t$ subject to the relations
relations
 \beq
  \begin{array}{ll}
  \rm (a) & p(t)p'(t)=p(t)\cdot p'(t)
    \quad\hbox{(the rhs denotes the product in $\C_t$),} \\
  \rm (b) & p(t)a-ap(q^{-1}t)=0 , \\
  \rm (c) & p(t)\ad - \ad p(qt) =0 , \qquad \forall p(t),p'(t)\in \C_t\\
  \rm (d) & a\ad - q^{-1} \ad a = t , \\
  \rm (e) & a\ad - q \ad a = t^{-1} .
  \end{array}
 \eeq
\end{defi}
The $q$-boson operator algebra $B_q$ is naturally embedded in $B'_q$,
and every $(t,t^{-1})$-diagonal $B_q$ module $V$ can also be regarded
as a $B'_q$ module; note however, that the action of elements of
$\C_t$ which lead to ``division by zero'' is not defined.

\setcounter{equation}{0}
\section{Transformations for $q$-bosons}

It is the purpose of this section to find certain transformations
of the standard generators $a,\ad,t,t^{-1}$ that satisfy again the
relations of a $q$-boson operator algebra. The transformations we
intend to study are ``linear'' in $a$ and $\ad$, in the following
sense~:
\beq
 \begin{array}{l}
 b = \al(t)a+\be(t)\ad , \\
 \bd= \ga(t)a + \de(t) \ad ,
 \end{array}
\label{defb}
\eeq
where $\al(t), \be(t), \ga(t)$ and $\de(t)$ are elements of $\C_t$
(roots of rational forms in $t$). With (\ref{defb}), we define
$s=b\bd-q^{-1}\bd b$ and $s^{-1}=b\bd - q \bd b$. The conditions to
be satisfied are then~: $ss^{-1}=1$, $s^{-1}s=1$, $sb=q^{-1}bs$ and
$s\bd=q\bd s$. A careful analysis of these conditions shows that
$\al, \be, \ga$ and $\de$ have to satisfy~:
\beq
 \begin{array}{l}
 \al(t)/\al(qt) = q^{-1} \ga(t)/\ga(qt) , \\
 \be(t)/\be(qt) = q^{-1} \de(t)/\de(qt) , \\
 \al(t)\de(t) = -1/(qt-q^{-1}t^{-1}) , \\
 \be(t)\ga(t) = 1/(qt-q^{-1}t^{-1}) .
 \end{array}
\eeq

This implies the following result~:
\begin{prop}
The most general transformation of the $q$-boson of the form (\ref{defb})
is given by
 \beq
  \begin{array}{l}
  \ds b=\ep t \ga(t) a + {\ep'\over t-t^{-1}}\ad {1\over\ga(t)} , \\[5mm]
  \ds \bd = \ep'\ga(t) a - {\ep\over t-t^{-1}}\ad {t^{-1}\over\ga(t)} ,
  \end{array}
 \label{sol}
 \eeq
where $\ep^2=\ep'^2=1$ and $\ga(t)\in\C_t$.
The elements $(b,\bd,s,s^{-1})$, where
$s=b\bd-q^{-1}\bd b$ and $s^{-1}=b\bd - q \bd b$, satisfy the defining
relations of the $q$-boson operator algebra (\ref{def} a--e).
\end{prop}

Note that (\ref{sol}) has no ``classical'' counterpart in the
limit $q\rightarrow 1$.

The realisation of (\ref{sol}) has some interesting consequences. For
instance, having one $q$-boson with generators $a,\ad,t,t^{-1}$ and a
corresponding Fock space, one can study the Fock space of the
transformed $q$-boson with generators $b,\bd,s,s^{-1}$. This second
Fock space is usually very different from the first one. For example,
let $(a,\ad,t,t^{-1}) = ( D_{q,x}, x, q^N, q^{-N})$. Choosing
$\ga(q^N)=q^{-N/2}$ and $\ep=\ep'=1$, one has
\beq
 \begin{array}{l}
 \ds b=q^{N/2}D_{q,x} + {1\over q^N-q^{-N}} x  q^{N/2} ,\\[5mm]
 \ds \bd=q^{-N/2}D_{q,x} - {1\over q^N-q^{-N}} x  q^{-N/2} ,
 \end{array}
\eeq
Let the vacuum state $v_0$ for the new Fock space have coefficients $a_k$
in the old Fock space, i.e.
\beq
 v_0=\sum_{k=0}^\infty a_k x^k .
\eeq
The condition $bv_0=0$ leads to $a_{2k+1}=0$ and $a_{2k} =
(-1)^k a_0 / (\sqrt{q}r)^k [2k]! $, with $r= q-q^{-1}$. Thus one
finds
\begin{eqnarray}
v_0 &=& a_0 \sum_{k=0}^\infty {(-x^2/\sqrt{q}r)^k \over [2k]!} \nn\\
 &=&a_0 \cos_q ({q^{-1/4}x \over \sqrt{q-q^{-1}}}) .
\end{eqnarray}
Herein, $\cos_q$ is the $q$-generalised cosine function, studied by
Jackson~[7,8].

\setcounter{equation}{0}
\section{$q$-Hermite polynomials as boson states}

Consider again the case $(a,\ad,t,t^{-1})=(D_{q,x}\equiv D,x,
q^{N},q^{-N})$ and the transformation~(\ref{sol}). Choosing
$\ep=-1$ and $\ep'=1$, and
\beq
\ga(q^N)=  {c q^{-N/2} \over q^{N+1}-q^{-N-1} } ,
\eeq
where $c$ is to be determined later, one finds the following expressions
for $b$ and $\bd$~:
\begin{eqnarray}
b&=& {x\over c} q^{N/2} - {q^{N/2}\over q^{N+1}-q^{-N-1}} c D , \label{b}\\
\bd&=& {x\over c} q^{-N/2} + {q^{-N/2}\over q^{N+1}-q^{-N-1}} c D .\label{bd}
\end{eqnarray}
The vacuum state $v_0(x)=\sum_k a_k x^k$ is found by requiring
$b v_0(x)=0$. This leads to $a_{2k+1}=0$ and $a_{2k}=a_0(r/c^2q)^j$,
with again $r=q-q^{-1}$.
Thus $v_0(x)=a_0 \sum_{k=0}^\infty (rx^2/qc^2)^j$, which can
formally be rewritten as
\beq
v_0(x) = {a_0 \over 1 - rx^2/qc^2 } .
\eeq
It can be verified that $s^{-1}v_0(x)=v_0(x)$, thus also $sv_0=v_0$.
One can then define $|k\rangle = (\bd)^k v_0(x)$. Using~(\ref{low}),
the following relations are obtained~:
\begin{eqnarray}
b|k\rangle & = &[k] |k-1\rangle ,\label{bk}\\
\bd|k\rangle &=&  |k+1\rangle ,\label{bdk}\\
s|k\rangle &=& q^k |k\rangle  .\label{sk}
\end{eqnarray}

The following proposition shows that the $q$-boson states $|k\rangle$
can be written in terms of a polynomial of degree $k$.

\begin{prop}
The states $|k\rangle$ have the form
 \beq
 |k\rangle = q^{-k(N+1)/2+k^2/4} p_k(x;q) v_0(x) ,
 \label{state}
 \eeq
where $p_k(x;q)$ is a polynomial of degree $k$ in $x$ with coefficients
in $\C[q]$.
\end{prop}

\noindent {\sl Proof.} The statement is true when $k=0$. Assume that
it is valid for all values up to $k$; we shall prove that is then also
valid for $k+1$. Using~(\ref{bk}) and~(\ref{b}),
one finds the following expression~:
\beq
{q^{N/2}\over q^{N+1}-q^{-N-1}}cD|k\rangle =
{x\over c}q^{N/2} |k\rangle - [k] |k-1\rangle .
\eeq
This can now be used in the calculation of $\bd|k\rangle$, and we
find~:
\begin{eqnarray}
|k+1\rangle & = & \bd |k\rangle  \nn \\
&=& {x\over c} q^{-N/2} |k\rangle +
 q^{-N} {q^{N/2}\over q^{N+1}-q^{-N-1}} cD|k\rangle  \nn\\
&=& ({x\over c}q^{-N/2}+q^{-N}{x\over c}q^{N/2}) |k\rangle
 -q^{-N}[k] |k-1\rangle .
\label{rhs}
\end{eqnarray}
Using the induction argument, i.e.~expression~(\ref{state}), for
$|k\rangle$ and $|k-1\rangle$, the right hand side of~(\ref{rhs})
reduces to~:
\beq
q^{-(k+1)(N+1)/2+(k+1)^2/4}
\left( {q^{-1/4}(1+q)\over c} x p_k(x;q) - q^{-k+1}[k]p_{k-1}(x;q)\right)
v_0(x).
\eeq
Thus $|k+1\rangle = q^{-(k+1)(N+1)/2+(k+1)^2/4} p_{k+1}(x;q) v_0(x)$, where
\beq
p_{k+1}(x;q) = {1+q \over cq^{1/4}} x p_k(x;q) -
 {1-q^{-2k}\over 1-q^{-2}} p_{k-1}(x;q) .
\label{recursion}
\eeq
From this last equation the statement follows. \mybox

Equation~(\ref{recursion}), together with $p_{-1}(x;q)=0$ and
$p_0(x;q)=1$, can be used as
a recursion relation to calculate the $p_k(x;q)$.
Choosing $c=q^{-1/4}(1+q)$, this relation becomes
$p_{k+1}=xp_k-kp_{k-1}$ in the limit
$q\rightarrow 1$,
which is the recursion relation defining Hermite polynomials.
So, in the present case, the $q$-boson states are written in
terms of $q$-generalised Hermite polynomials satisfying~(\ref{recursion}).

Hermite polynomials have been $q$-generalised in different ways, and
there exists a huge amount of literature on the subject of
$q$-special functions~[3]. The polynomials~(\ref{recursion}) can be
related to many of the known orthogonal $q$-polynomials.
Choosing for example $c=q^{-1/4}(1+q)(1-q^{-2})^{1/2}$, one obtains
\beq
p_{k+1}(x;q)=(1-q^{-2})^{-1/2} x p_k(x;q) -
 {1-q^{-2k} \over 1-q^{-2}} p_{k-1}(x;q) ,
\eeq
or, with $\q=q^{-2}$ and $P_k(x;\q)=p_k(x;q)$~:
\beq
P_{k+1}(x;\q)=(1-\q)^{-1/2} x P_k(x;\q) -
 {1-\q^k \over 1-\q} P_{k-1} (x;\q).
\eeq
From now onwards, we shall continue to work with $\q$; however, for
sake of notation we shall denote $\q$ by $q$ in what follows.
The polynomials $P_k(x;q)$ are related to the $q$-Hermite polynomials
$R_k(x;q)$ of Allaway~[8] by
\beq
R_k(x;q)=(1-q)^{k/2} P_k(x;q) .
\eeq
These polynomials satisfy the recursion relation
\beq
R_{k+1}(x;q)=xR_k(x;q)-(1-q^k)R_{k-1}(x;q),
\label{R}
\eeq
with $R_{-1}=0$ and $R_0=1$, and have been studied in detail~[8,9].
They form a special case of the polynomials $R_k(x;q;a,b,c)$ introduced
by Al-Salam and Chihara~[9], and studied in a rather general context
by Dehesa~[10]. The $R_k(x)\equiv R_k(x;q;a,b,c)$ are defined by
\beq
R_{k+1}(x)=(x-aq^k)R_k(x)-(c-bq^{k-1})(1-q^k)R_{k-1}(x) ,
\eeq
with $R_{-1}(x)=0$ and $R_0(x)=1$. This coincides with~(\ref{R})
for $(a,b,c)=(0,0,1)$. Note that the so-called discrete $q$-Hermite
polynomials $H_k(x;q)$, as defined in~[3, p.~193], also fall in
this class since $H_k(x;q)=R_k(x;q;0,-1,0)$.

There is also a relation with the continuous $q$-ultraspherical
polynomials~[3, p.~26], defined by
\beq
C_k(x;\be|q)=\sum_{j=0}^k {(\be;q)_j (\be;q)_{k-j} \over
 (q;q)_j (q;q)_{k-j} } e^{i(k-2j)\th} ,
\eeq
where $x=\cos\th$ and
\beq
(a;q)_n = \cases {1 , & if $n=0$;\cr
          (1-a)(1-aq)\cdots(1-aq^{n-1}),& if $n=1,2,\ldots$\cr}
\eeq
The little $c_k$,
\beq
c_k(x;\be|q)={(q;q)_k\over (\be^2;q)_k} C_k(x;\be|q) ,
\eeq
introduced in~[3, p.~188], satisfy
\beq
(1-\be^2q^k)c_{k+1}(x;\be|q)=(1-\be q^k)2xc_k(x;\be|q)
-(1-q^k)c_{k-1}(x;\be|q),
\eeq
with $c_{-1}=0$ and $c_0=1$. Thus, the polynomials appearing here can also
be identified as follows~:
\beq
P_k(x;q)=(1-q)^{-k/2} c_k(x/2;0|q).
\eeq

Finally, it should be noted that in a recent publication~[11], a different
kind of $q$-generalised Hermite polynomials $h_n(w;q)$ were also
related to the oscillator quantum group. These polynomials are
different from~(\ref{recursion}) (e.g.~they satisfy
$h_n(w;q=1)=(1+w)^n$); moreover, the $q$-boson realisation
given in~[11] is also quite different from~(\ref{b}--\ref{bd}).

\section*{Acknowledgement}

The author would like to thank the referee for pointing out
references on $q$-special functions.

\section*{References}
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Biedenharn LC, {\it J.\ Phys.\ A : Math.\ Gen.} {\bf 22}, L873 (1989)
\item
Macfarlane AJ, {\it J.\ Phys.\ A : Math.\ Gen.} {\bf 22}, 4581 (1989)
\item
Gasper G and Rahman M, {\it Basic Hypergeometric Series,}
Encyclopedia of Mathematics and its Applications, vol.~{\bf 35},
Cambridge University Press, Cambridge (1990)
\item
Kulish PP and Damashinsky EV, {\it J.\ Phys.\ A : Math.\ Gen.} {\bf 23},
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\item
Manun Yu I, {\it Quantum groups and non-commutative geometry,}
Centre de Recherches Math\'ematiques, Montr\'eal, 1988
\item
Hong Yan, {\it J.\ Phys.\ A : Math.\ Gen.} {\bf 23}, L1155 (1990)
\item
Exton H, {\it $q$-Hypergeometric functions and applications,}
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\item
Jackson FH, {\it Proc.\ Edin.\ Math.\ Soc.} {\bf 22}, 28 (1904)
\item
Allaway WR, {\it Canad.\ J.\ Math.} {\bf 32}, 686 (1980)
\item
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\end{enumerate}

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